Design Manual to CoP2004 (Housing)

Manual for Design andDetailings of Reinforced Concrete to Code of Practice for Structural Use of Concrete 2004 Housing Department February 2007 Contents Page 1.0 Introduction 1 2.0 Some highlighted aspects in Basis of Design 3 3.0 Beams 6 4.0 Slabs 35 5.0 Columns 52 6.0 Column Beam Joints 69 7.0 Walls 77 8.0 Corbels 86 9.0 Transfer Structures 93 10.0 Footings 99 11.0 Pile Caps 107 12.0 General R.C. Detailings 118 Appendices Appendix A – Clause by Clause Comparison between “Code of Practice for Structural Use of Concrete 2004” and BS8110 Appendix B – Assessment of Building Accelerations Appendix C – Derivation of Basic Design Formulae of R.C. Beam sections against Flexure Appendix D – Underlying Theory and Design Principles for Plate Bending Element Appendix E – Extracts of Design Charts for Slabs Appendix F – Derivation of Design Formulae for Rectangular Columns to Rigorous Stress Strain Curve of Concrete Appendix G – Derivation of Design Formulae for Walls to Rigorous Stress Strain Curve of Concrete Appendix H – Extracts of HKIE paper for plate bending analysis and design Appendix I – Derivation of Formulae for Rigid Cap Analysis 1 1.0 Introduction 1.1 Promulgation of the Revised Code A revised concrete code titled “Code of Practice for Structural Use of Concrete 2004” was formally promulgated by the Buildings Department of Hong Kong in late 2004 which serves to supersede the former concrete code titled “The Structural Use of Concrete 1987”. The revised Code, referred to as “the Code” hereafter in this manual will become mandatory by 15 December 2006, after expiry of the grace period in which both the revised and old codes can be used. 1.2 Main features of the Code As in contrast with the former code which is based on “working stress” design concept, the drafting of the Code is largely based on BS8110 1997 adopting the limit state design approach. Nevertheless, the following features of the Code in relation to design as different from BS8110 are outlined : (a) Provisions of concrete strength up to grade 100 are included; (b) Stress strain relationship of concrete is different from that of BS8110 for various concrete grades; (c) Maximum design shear stress of concrete ( max v ) are raised; (d) Provisions of detailings to enhance ductility are added, with the requirements of design in beam-column joints (Sections 9.9 and 6.8); (e) Criteria for dynamic analysis for tall building under wind loads are added (Clause 7.3.2). As most of our colleagues are familiar with BS8110, a comparison table highlighting differences between BS8110 and the Code is enclosed in Appendix A which may be helpful to designer switching from BS8110 to the Code in the design practice. 1.3 Outline of this Manual This Practical Design Manual intends to outline practice of detailed design and detailings of reinforced concrete work to the Code. Detailings of individual types of members are included in the respective sections for the types, though the last section in the Manual includes certain aspects in detailings which are 2 common to all types of members. Design examples, charts are included, with derivations of approaches and formulae as necessary. Aspects on analysis are discussed selectively in this Manual. In addition, as the Department has decided to adopt Section 9.9 of the Code which is in relation to provisions for “ductility”, conflicts of this section with others in the Code are resolved with the more stringent ones highlighted as requirements in our structural design. As computer methods have been extensively used nowadays in analysis and design, the contents as related to the current popular analysis and design approaches by computer methods are also discussed. The background theory of the plate bending theory involving twisting moments, shear stresses, and design approach by the Wood Armer Equations which are extensively used by computer methods are also included for analysis of slabs, flexible pile caps and footings. To make distinctions between equations quoted from the Code and the equations derived in this Manual, the former will be prefixed by (Ceqn) and the latter by (Eqn). 1.4 Reference of this Manual This Manual has made reference to the following documents : (a) The Code itself (b) Concrete Code Handbook by HKIE (c) Hong Kong Building (Construction) Regulations (d) The Code of Practice for Dead and Imposed Loads for Buildings (Draft) (e) BS8110 Parts 1, 2 and 3 (f) Code of Practice for Precast Concrete Construction 2003 (g) Code of Practice for Fire Resisting Construction 1996 (h) Reinforced and Prestressed Concrete 3 rd edition – Kong & Evans (i) Reinforced Concrete Design to CP110 – Simply Explained A.H. Allen (j) Ove Arup and Partners. Notes on Structures: 17. April 1989 (k) ACI Code ACI318-05. 3 2.0 Some highlighted aspects in Basis of Design 2.1 Ultimate and Serviceability Limit states The ultimate and serviceability limit states used in the Code carry the usual meaning as in BS8110. However, the new Code has incorporated an extra serviceability requirement in checking human comfort by limiting acceleration due to wind load on high-rise buildings (in Clause 7.3.2). No method of analysis has been recommended in the Code though such accelerations can be estimated by the wind tunnel laboratory if wind tunnel tests are conducted. Nevertheless, a design example is enclosed in Appendix B, based on some recognized empirical formulae in the absence of wind tunnel tests. 2.2 Design Loads The Code has made reference to the “Code of Practice for Dead and Imposed Loads for Buildings” for determination of characteristic gravity loads for design. However, the Code has not yet been formally promulgated. At the meantime, the design loads should be therefore taken from HKB(C)R Clause 17. Nevertheless, the designer may need to check for the updated loads by fire engine for design of new buildings, as required by FSD. The Code has placed emphasize on design loads for robustness which are similar to the requirements in BS8110 Part 2. The requirements include design of the structure against a notional horizontal load equal to 1.5% of the characteristic dead weight at each floor level and vehicular impact loads (Clause 2.3.1.4). The small notional horizontal load can generally be covered by wind loads if wind loads are applied to the structure. Identification of key elements and designed for ultimate loads of 34 kPa, together with examination for progress collapse in accordance with Clause 2.2.2.3 can be exempted if the buildings are provided with ties determined by Clause 6.4.1. The usual reinforcement provisions as required by the Code can generally cover the ties provision. Wind loads for design should be taken from Code of Practice on Wind Effects in Hong Kong 2004. It should also be noted that there are differences between Table 2.1 of the 4 Code that of BS8110 Part 1 in some of the partial load factors f γ . The beneficial partial load factor for earth and water load is 1. However, lower values should be used if the earth and water load are known to be over-estimated. 2.3 Materials – concrete Table 3.2 has tabulated a set of Young’s Moduli of concrete up to grade 100. The values are generally smaller than that in BS8110 and also slightly different from the former 1987 Code. The stress strain curve of concrete as given in Figure 3.8 of the Code, whose initial tangent is determined by these Young’s Moduli values is therefore different from Figure 2.1 of BS8110 Part 1. Furthermore, in order to achieve smooth (tangential connection) between the parabolic portion and straight portion of the stress strain curve, the Concrete Code Handbook (Clause 3.1.4) has recommended to shift the 0 ε value to c m cu E f ) / ( 34 . 1 γ instead of staying at m cu f γ 4 10 4 . 2 − × which is the value in BS8110. The stress strain curve for grade 35 is plotted as an illustration. Stress Strain Profile for Grade 35 0 2 4 6 8 10 12 14 16 18 0 0.2 0.4 0.6 0.8 1 Distance Ratio from Neutral axis S t r e s s ( M P a ) Design formulae for beams and columns based on these stress strain curves by 0.3769 where ε 0 = 0.001319 Figure 2-1 – Stress strain profile of grades 35 5 BS8110, strictly speaking, become inapplicable. A full derivation of design formulae and charts for columns and walls are given in Sections 5 and 7, together with Appendix F and G of this Manual. Table 4.2 of the Code tabulated nominal covers to reinforcements under different exposure conditions. However, reference should also be made to the “Code of Practice for Fire Resisting Construction 1996”. 2.4 Definitions of structural elements The Code has included definitions of slab, beam, column and wall in accordance with dimensions in Clause 5.2.1.1, 5.4 and 5.5 which are repeated for ease of reference : (a) Slab : the minimum panel dimension ≥ 5 times its thickness; (b) Beam : for span ≥ 2 times the overall depth for simply supported span and ≥ 2.5 times the overall depth for continuous span, classified as shallow beam, otherwise : deep beam; (c) Column : vertical member with section depth not exceeding 4 times its width and height is at least 3 times its section depth; (d) Wall : vertical member with plan dimensions other than that of column. (e) Shear Wall : walls contributing to the lateral stability of the structure. (f) Transfer Structure : horizontal element which redistributes vertical loads where there is a discontinuity between the vertical structural elements above and below. This Manual is based on the above definitions in delineating structural members for discussion. 6 3.0 Beams 3.1 Analysis (Cl. 5.2.5.1 & 5.2.5.2) Normally continuous beams are analyzed as sub-frames by assuming no settlements at supports by walls, columns (or beams) and rotational stiffness by supports provided by walls or columns as L EI / 4 (far end of column / wall fixed) or L EI / 3 , (far end of column / wall pinned). In analysis as sub-frame, Clause 5.2.5.2 states that the following loading arrangements will be adequate for seeking for the design moments : 1.0G K 1.0G K 1.4G K +1.6Q K 1.4G K +1.6Q K 1.0G K 1.0G K Figure 3.2c – To search for maximum hogging moment at support adjacent to spans with 1.4G K +1.6Q K 1.4G K +1.6Q K 1.0G K 1.4G K +1.6Q K 1.0G K 1.4G K +1.6Q K 1.0G K Figure 3.2b – To search for maximum sagging moment in spans with 1.4G K +1.6Q K Figure 3.1 – continuous beam analyzed as sub-frame 1.4G K +1.6Q K 1.4G K +1.6Q K 1.4G K +1.6Q K 1.4G K +1.6Q K 1.4G K +1.6Q K 1.4G K +1.6Q K Figure 3.2a – To search for maximum support reactions 7 However, most of the commercial softwares can actually analyze individual load cases, each of which is having live load on a single span and the effects on itself and others are analyzed. The design value of shears and moments at any location will be the summation of the values of the same sign created by the individual cases. With wind loads, the load cases to be considered will be 1.2(G K +Q K +W K ) and 1.0G K +1.4W K on all spans. 3.2 Moment Redistribution (Cl. 5.2.9) Moment redistribution is allowed for concrete grade not exceeding 70 under conditions 1, 2 and 3 as stated in 5.2.9.1. Nevertheless, it should be noted that there would be further limitation of the neutral axis depth ratio d x / if moment redistribution is employed as required by (Ceqn 6.4) and (Ceqn 6.5) of the Code which is identical to the provisions in BS8110. The rationale of such further limitations is discussed in Concrete Code Handbook 6.1.2. 3.3 Highlighted aspects in Determination of Design Parameters of Shallow Beam (i) Effective span (Cl. 5.2.1.2(b)) For simply supported beam, continuous beam and cantilever, the effective span can be taken as the clear span plus the lesser of half effective depth and half support width except that on bearing where the centre of bearing should be used to assess effective span. (Fig. 5.3 of the Code); (ii) Effective flange width of T- and L-beams (Cl. 5.2.1.2(a)) Effective flange width of T- and L-beams are as illustrated in Figure 5.2. of the Code as Effective width (b eff ) = width of beam (b w ) + 0.2 times the centre to centre width to the next beam (0.2b i ) + 0.1 times the span of zero moment (0.1l pi ), with the sum of the latter two not exceeding 0.2 times the span of zero moment and l pi taken as 0.7 times the effective span of the beam. An example for illustration is as indicated : 8 The effective spans are 5 m and they are continuous beams. The effective width of the T-beam is, by (Ceqn 5.1) of the Code 3500 5000 7 . 0 = × = pi l ; 750 3500 1 . 0 2000 2 . 0 2 , 1 , = × + × = = eff eff b b As 700 3500 2 . 0 750 2 , 1 , = × > = = eff eff b b 700 2 , 1 , = = eff eff b b ; 1800 1400 400 2 700 400 = + = × + = eff b So the effective width of the T-beam is 1800 mm. (iii) Support Moment Reduction (Cl. 5.2.1.2) The Code allows design moment of beam (and slab) monolithic with its support providing rotational restraint to be that at support face if the support is rectangular and 0.2Ø if the support is circular with diameter Ø. But the reduction should not be less than 65% of the support moment. An example 3.1 for illustration is given below : In Figure 3.4, the bending moment at support face is 200 kNm which 2000 2000 2000 400 400 400 400 Figure 3.3 – Example illustrating effective flange determination 800 250 kNm at 0.2 Ø into the support face 350 kNm at support 0.2×800 200 kNm at support face centre line of beam column elements idealized as line elements in analysis Bending Moment Diagram Figure 3.4 – Reduced moment to Support Face for support providing rotational restraint 9 can be the design moment of the beam if the support face is rectangular. However, as it is smaller than 0.65×350 = 227.5 kNm. 227.5 kNm should be used for design. If the support is circular and the moment at 0.2Ø into the support and the bending moment at the section is 250 kNm, then 250 kNm will be the design moment as it is greater than 0.65×350 = 227.5 kNm. For beam (or slab) spanning continuous over a support considered not providing rotational restraint (e.g. wall support), the Code allows moment reduction by support shear times one eighth of the support width to the moment obtained by analysis. Figure 3.5 indicates a numerical Example 3.2. The design support moment at the support under consideration can be reduced to 230 8 8 . 0 200 250 = × − kNm. (iv) Slenderness Limit (Cl. 6.1.2.1) The provision is identical to BS8110 as 1. Simply supported or continuous beam : Clear distance between restraints not to exceed 60b c or 250b c 2 /d if less; and 2. Cantilever with lateral restraint only at support : Clear distance from cantilever to support not to exceed 25b c or 100b c 2 /d if less F Ed,sup = 200 kN 800 250 kNm Figure 3.5 – Reduction of support moment by support shear for support considered not providing rotational restraint 10 where b c is the breadth of the compression face of the beam and d is the effective depth. Usually the slenderness limits need be checked for inverted beams or bare beam (without slab). (v) Span effective depth ratio (Cl. 7.3.4.2) Table 7.3 under Cl. 7.3.4.2 tabulates basic span depth ratios for various types of beam / slab which are deemed-to-satisfy requirements against deflection. The table has provisions for spans and end spans which are not specified in BS8110. Nevertheless, calculation can be carried out to justify deflection limits not to exceed span / 250. Support condition Rectangular Beam Flanged Beam b w /b < 0.3 One or two-way spanning solid slab Cantilever 7 5.5 7 Simply supported 20 16 20 Continuous 26 21 26 End span 23 18.5 23 (2) Note : 1. The values given have been chosen to be generally conservative and calculation may frequently show shallower sections are possible; 2. The value of 23 is appropriate for two-way spanning slab if it is continuous over one long side; 3. For two-way spanning slabs the check should be carried out on the basis of the shorter span. Table 3.1 – effective span / depth ratio The basic span depth ratios can be modified due to provision of tensile and compressive steels as given in Tables 7.4 and 7.5 which are identical to BS8110. (vi) Minimum spacing between bars should not be less than bar diameter, aggregate size + 5mm or 20mm (Cl. 8.2). (vii) Maximum spacing between bars in tension near surface, by Cl. 9.2.1, should be such that the clear spacing between bar is limited by clear spacing ≤ ≤ y b f β 70000 300 mm where b β is the ratio of moment redistribution. Or alternatively, clear spacing ≤ ≤ y f 47000 300 mm. 11 So the simplest rule is 152 460 1 70000 70000 = × = y b f β mm when using high yield bars and under no moment redistribution. (viii) Concrete covers to reinforcements (Cl. 4.2.4 and Cl. 4.3) Cl. 4.2.4 of the Code indicates the nominal cover required in accordance with Exposure conditions. However, we can, as far as our building structures are concerned, roughly adopt condition 1 (Mild) for the structures in the interior of our buildings (except for bathrooms and kitchens which should be condition 2), and to adopt condition 2 for the external structures. Nevertheless, the “Code of Practice for Fire Resisting Construction 1996” should also be checked for different fire resistance periods (FRP). So, taking into account our current practice of using concrete not inferior than grade 30 and maximum aggregate sizes not exceeding 20 mm, we may generally adopt the provision in our DEDD Manual (DEDD-404 Table 2) with updating by the Code except for compartment of 4 hours FRP. The recommended covers are summarized in the following table : Description Nominal Cover (mm) Internal 30 External 40 Simply supported (4 hours FRP) 55 Continuous (4 hours FRP) 45 Table 3.2 – Nominal Cover of Beams 3.4 Sectional Design for Rectangular Beam against Bending 3.4.1 Design in accordance with the Rigorous Stress Strain curve of Concrete The stress strain block of concrete as indicated in Figure 3.8 is different from Figure 2.1 of BS8110. Furthermore, in order to achieve smooth connection between the parabolic and the straight line portions, the Concrete Code Handbook has recommended to shift the ε 0 to the right to a value of c m cu E f γ 34 . 1 . With the values of Young’s Moduli of concrete, c E , as indicated in Table 3.2 of the Code, the stress strain block of concrete for various grades can 12 be determined. The stress strain curve of grade 35 is drawn as shown Stress Strain Profile for Grade 35 0 2 4 6 8 10 12 14 16 18 0 0.2 0.4 0.6 0.8 1 Distance Ratio from Neutral axis S t r e s s ( M P a ) Based on this rigorous concrete stress strain block, design formulae for beam can be worked out as per the strain distribution profile of concrete and steel as indicated in Figure 3.7. neutral axis d’ d x 0035 . 0 = ult ε Stress Diagram Strain Diagram Figure 3.7 – Stress Strain diagram for Beam 0.3769 where ε 0 = 0.001319 Figure 3.6 – Stress strain block of grades 35 13 The solution for the neutral axis depth ratio d x for singly supported beam is the root of the following quadratic equation where 0035 . 0 = ult ε for concrete grades not exceeding 60 (Re Appendix C for detailed derivation) : 0 3 1 1 67 . 0 12 1 3 1 2 1 67 . 0 2 0 2 2 0 0 = − | | . | \ | − + | . | \ | | | . | \ | − + − bd M d x f d x f ult m cu ult ult m cu ε ε γ ε ε ε ε γ (Eqn 3-1) With neutral axis depth ratio determined, the steel ratio can be determined by (Eqn 3-2) d x f f bd A ult m cu y st | | . | \ | − = ε ε γ 0 3 1 1 67 . 0 87 . 0 1 (Eqn 3-2) As d x is limited to 0.5 for singly reinforcing sections for grades up to 40 under moment redistribution not greater than 10% (Clause 6.1.2.4 of the Code), by (Eqn 3-1), cu f bd M 2 will be limited to ' K values as in 154 . 0 ' = K for grade 30 152 . 0 ' = K for grade 35 151 . 0 ' = K for grade 40 which are all smaller than 0.156 under the simplified stress block. However, for grade 45 and 50 where neutral axis depth ratio is limited to 0.4 for singly reinforced sections under moment redistribution not greater than 10% (Clause 6.1.2.4 of the Code), again by (Eqn 3-1) cu f bd M 2 will be limited to 126 . 0 ' = K for grade 45 125 . 0 ' = K for grade 50 which are more significantly below 0.132 under the simplified stress block. It should be noted that the d x ratio will be further limited if moment redistribution exceeds 10% by (Ceqn 6.4) and (Ceqn 6.5) of the Code as ( ) 4 . 0 − ≤ b d x β for 40 ≤ cu f ; and ( ) 5 . 0 − ≤ b d x β for 70 40 ≤ < cu f where b β us the ratio of the moment after and before moment redistribution. 14 When cu f bd M 2 exceeds the limited value for single reinforcement, compression reinforcements at ' d from the surface of the compression side should be added. The compression reinforcements will take up the difference between the applied moment and 2 Kbd and the compression reinforcement ratio is | . | \ | − | | . | \ | − = d d f f K f bd M bd A y cu cu sc ' 1 87 . 0 ' 2 (Eqn 3-3) And the same amount of reinforcement will be added to the tensile reinforcement : | . | \ | − | | . | \ | − + | | . | \ | − = d d f f K f bd M f f bd A y cu cu ult m cu y st ' 1 87 . 0 ' 3 1 1 67 . 0 87 . 0 1 2 0 η ε ε γ (Eqn 3-4) where η is the limit of neutral axis depth ratio which is 0.5 for grade 40 and below and 0.4 for grades up to and including 70 for moment redistribution not exceeding 10%. It follows that more compressive reinforcements will be required for grade 45 than 40 due to the limitation of neutral axis depth ratio, as illustrated by the following Chart 3-1 in which compression reinforcement decreases from grade 30 to 40 for the same 2 bd M , but increases at grade 45 due to the change of the limit of neutral axis depth ratio from 0.5 to 0.4 with moment redistribution not exceeding 10%. The same phenomenon applies to tensile steel also. With moment redistribution exceeding 10%, the same trend will also take place. 15 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.1 0 2 4 6 8 10 12 14 0 0.5 1 1.5 2 2.5 3 3.5 4 Reinforcement ratios (%) M / b d 2 Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bd Grade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Furthermore, as similar to BS8110, there is an upper limit of “lever arm ratio” d z which is the depth of the centroid of the compressive force of concrete to the effective depth of the beam section of not exceeding 0.95. To comply with this requirement, the minimum steel percentages will be 0.33%, 0.39%, 0.44% and 0.5% for grades 30, 35, 40 and 45 respectively which are all in excess of the minimum of 0.13% in accordance with Table 9.1 of the Code. Design Charts for grades 30 to 45 comprising tensile steel and compression steel ratios bd A st and bd A sc are enclosed at the end of Appendix C. 3.4.2 Design in accordance with the Simplified Stress Block The design will be simpler and sometimes more economical if the simplified rectangular stress block as given by Figure 6.1 of the Code is adopted. The design formula becomes For singly reinforced sections where ' 2 K bd f M K cu ≤ = where 156 . 0 ' = K for grades 40 and below and 132 . 0 ' = K for grades over 40 and not Chart 3-1 – Reinforcement Ratios of Doubly Reinforced Beams for Grade 30 to 45 with Moment Redistribution limited to 10% or below 16 exceeding 70. 95 . 0 9 . 0 25 . 0 5 . 0 ≤ − + = K d z ; 45 . 0 1 9 . 0 25 . 0 5 . 0 45 . 0 1 1 | | . | \ | − − = | . | \ | − = K d z d x ; z f M A y st 87 . 0 = (Eqn 3-5) For doubly reinforced sections ' 2 K bd f M K cu > = , 9 . 0 ' 25 . 0 5 . 0 K d z − + = 45 . 0 1 1 | . | \ | − = d z d x ( ) ( ) ' 87 . 0 ' 2 d d z f bd f K K A y cu sc − − = sc y cu st A z f bd f K A + = 87 . 0 ' 2 (Eqn 3-6) 3.4.3 Worked Examples for Rectangular Beam with Moment Redistribution < 10% Worked Example 3.3 Section : 500 (h) × 400 (w), 35 = cu f MPa (i) 286 1 = M kNm; (ii) 486 2 = M kNm; 444 16 40 500 = − − = d 0013192 . 0 23700 5 . 1 35 34 . 1 34 . 1 0 = × × = = c m cu E f γ ε 3769 . 0 0 = ult ε ε 152 . 0 104 . 0 444 400 35 10 286 2 6 2 1 < = × × × = bd f M cu , so singly reinforced Solving the neutral axis depth ratio by (Eqn 3-1) d x 38 . 60 12 1 3 1 2 1 67 . 0 2 0 0 − = | | . | \ | − + − ult ult m cu f ε ε ε ε γ ; 669 . 13 3 1 1 67 . 0 0 = | | . | \ | − ult m cu f ε ε γ ; 627 . 3 444 400 10 286 2 6 2 − = × × = − bd M ( ) ( ) ( ) 5 . 0 307 . 0 38 . 60 2 627 . 3 38 . 60 4 699 . 13 699 . 13 2 ≤ = − × − × − × − + − = d x 0105 . 0 307 . 0 699 . 13 460 87 . 0 1 3 1 1 67 . 0 87 . 0 1 0 = × × × = | | . | \ | − = d x f f bd A ult m cu y st ε ε γ 1865 = ⇒ st A mm 2 Use 2T32 + I T25 (ii) 440 20 40 500 = − − = d 17 0013192 . 0 23700 5 . 1 35 34 . 1 34 . 1 0 = × × = = c m cu E f γ ε 3769 . 0 0 = ult ε ε 152 . 0 179 . 0 440 400 35 10 486 2 6 2 2 > = × × × = bd f M cu , so doubly reinforced 50 10 40 ' = + = d 114 . 0 440 50 ' = = d d By (Eqn 3-3) ( ) ( ) 267 . 0 114 . 0 1 460 87 . 0 35 152 . 0 179 . 0 ' 1 87 . 0 2 = − × × × − = | . | \ | − | | . | \ | − = d d f f K f bd M bd A y cu cu sc % 469 440 400 00267 . 0 = × × = sc A mm 2 Use 2T20 By (Eqn 3-4) | . | \ | − | | . | \ | − + | | . | \ | − = d d f f K f bd M f f bd A y cu cu ult m cu y st ' 1 87 . 0 3 1 1 67 . 0 87 . 0 1 2 0 η ε ε γ 978 . 1 00236 . 0 5 . 0 699 . 13 460 87 . 0 1 = + × × = bd A st % 3481 440 400 01978 . 0 = × × = st A mm 2 Use 3T40 (i) and (ii) of Worked Example 3.3 are re-done in accordance with Figure 6.1 of the Code (the simplified stress) block by (Eqn 3-5) and (Eqn 3-6) (i) 867 . 0 444 400 35 10 286 25 . 0 5 . 0 9 . 0 25 . 0 5 . 0 2 6 = × × × − + = − + = K d z ( ) 01045 . 0 867 . 0 460 87 . 0 444 400 10 286 / 87 . 0 2 6 2 = × × × × × = × = d z f bd M bd A y st 1856 = ⇒ st A mm 2 Use 2T32 + I T25 (ii) 156 . 0 179 . 0 440 400 35 10 486 2 6 2 > = × × × = = bd f M K cu , so doubly reinforcing section required, 775 . 0 5 . 0 9 . 0 5 . 0 1 = × × − = d z ( ) ( ) ( ) ( ) 399 50 440 460 87 . 0 440 400 35 156 . 0 179 . 0 ' 87 . 0 ' 2 2 = − × × × × × − = − − = d d f bd f K K A y cu sc mm 2 > 0.2% in accordance with Table 9.2.1 of the Code; 3498 399 440 775 . 0 460 87 . 0 440 400 35 156 . 0 87 . 0 ' 2 2 = + × × × × × × = + = sc y cu st A z f bd f K A mm 2 Use 3T40 Summary for comparison : 18 Singly Reinforced Doubly Reinforced st A (mm 2 ) sc A (mm 2 ) st A (mm 2 ) Total (mm 2 ) Based on Rigorous Stress-strain profile 1865 469 3481 3950 Based on Simplified stress strain profile 1856 399 3498 3897 Results by the two approaches are very close. The approach based on the simplified stress block are slightly more economical. 3.4.4 Worked Example 3.4 for Rectangular Beam with Moment Redistribution > 10% If (ii) of the Worked Example 3.3 has undergone a moment redistribution of 20% > 10%, i.e. 8 . 0 = b β , by (Ceqn 6.4) of the Code ( ) 4 . 0 4 . 0 8 . 0 4 . 0 = − ≤ ⇒ − ≤ d x d x b β , 82 . 0 5 . 0 9 . 0 4 . 0 1 = × × − = d z 132 . 0 9 . 0 25 . 0 5 . 0 82 . 0 = ⇒ − + = K K ( ) ( ) ( ) ( ) 764 50 440 460 87 . 0 440 400 35 132 . 0 176 . 0 ' 87 . 0 ' 2 2 = − × × × × × − = − − = d d f bd f K K A y cu sc mm 2 > 0.2 % as required by Table 9.2.1 of the Code; 3242 347 440 82 . 0 460 87 . 0 440 400 35 132 . 0 87 . 0 ' 2 2 = + × × × × × × = + = sc y cu st A z f bd f K A mm 2 So total amount of reinforcement is greater. 3.5 Sectional Design of Flanged Beam against Bending 3.5.1 Slab structure adjacent to the beam, if in flexural compression, can be used to act as part of compression zone of the beam, thus effectively widen the structural width of the beam. The use of flanged beam will be particularly useful in eliminating the use of compressive reinforcements, apart from reducing tensile steel due to increase of lever arm. The principle of sectional design of flanged beam follows that rectangular beam with an additional flange width of w f b b − as illustrated in Figure 3.8. 19 Design formulae based on the simplified stress block are derived in Appendix C which are summarized as follows : (i) Singly reinforcing section where 0.9 × neutral axis depth is outside flange depth by checking d d K d x f ≤ | | . | \ | − − = 9 . 0 25 . 0 5 . 0 2 9 . 0 where 2 d b f M K f cu = (Eqn 3-7) If so, carry out design as if it is a rectangular beam of width f b . (ii) Singly reinforcing section where 0.9 × neutral axis depth is outside flange depth and ' 2 1 1 1 67 . 0 ' 2 r f w f f m f w K d d b b d d K d b M + | | . | \ | − | | . | \ | − = ≤ γ where 156 . 0 ' = r K for 40 ≤ cu f and 132 . 0 ' = r K for 70 40 ≤ < cu f d x be solved by the quadratic equation : 0 402 . 0 1809 . 0 2 2 = − + − | . | \ | d b M M d x f d x f w f cu cu (Eqn 3-8) where | | . | \ | − | | . | \ | − = d d b b d d f d b M f w f f m cu w f 2 1 1 1 67 . 0 2 γ (Eqn 3-9) If d d d x f × ≤ 9 . 0 , treat as a rectangular beam of width f b . x 9 . 0 m cu f γ 67 . 0 x d f d f b w b Figure 3.8 – Analysis of a T or L beam section 20 If d d d x f × > 9 . 0 , + | | . | \ | − = d x d d b b f f d b A f w f m cu y w st 9 . 0 1 67 . 0 87 . 0 1 γ (Eqn 3-10) (iii) Doubly reinforcing section : By following the procedure in (ii), if d x obtained by (Eqn 3-8) exceeds 0.5, then double reinforcements will be required by | . | \ | − | | . | \ | − = d d f f K f d b M bd A y cu f cu w sc ' 1 87 . 0 ' 2 (Eqn 3-11) | . | \ | − | | . | \ | − + + | | . | \ | − = d d f f K f d b M d d b b f f bd A y cu f cu w f w f m cu y st ' 1 87 . 0 ' 45 . 0 1 67 . 0 87 . 0 1 2 γ (Eqn 3-12) 3.5.2 Worked Examples for Flanged Beam (i) Worked Example 3.5 : Singly reinforced section where d d d x f ≤ 9 . 0 Consider the previous example done for a rectangular beam 500 (h) × 400 (w), 35 = cu f MPa, under a moment 486 kNm, with a flanged section of width = 1800 mm and depth = 150 mm,: 400 = w b , 440 20 40 500 = − − = d , 1800 = w b 150 = f d First check if d d d x f ≤ 9 . 0 based on beam width of 1800 = w b 0398 . 0 440 1800 35 10 486 2 6 2 = × × × = = d b f M K f cu By (Eqn 3-7), 341 . 0 440 150 103 . 0 45 . 0 1 9 . 0 25 . 0 5 . 0 9 . 0 = = ≤ = | | . | \ | − − = d d K d x f So 95 . 0 954 . 0 45 . 0 1 > = − = d x d z ; 95 . 0 = ∴ d z Thus ( ) 00367 . 0 95 . 0 460 87 . 0 440 1800 10 486 / 87 . 0 2 6 2 = × × × × × = × = d z f bd M bd A y st 21 2905 = st A mm 2 . As in comparison with the previous example based on rectangular section, it can be seen that there is saving in tensile steel (2905 mm 2 vs 3498 mm 2 ) and the compression reinforcements are eliminated (ii) Worked Example 3.6 – Singly reinforced section where d d d x f > 9 . 0 Beam Section : 1000 (h) × 600 (w), flange width = 2000 mm, flange depth = 150 mm 35 = cu f MPa under a moment 4000 kNm 600 = w b , 890 60 50 1000 = − − = d , 2000 = w b 150 = f d 169 . 0 890 150 = = d d f ; 333 . 3 600 2000 = = w f b b First check if d d d x f ≤ 9 . 0 based on beam width of 2000 = w b 0721 . 0 890 2000 35 10 4000 2 6 2 = × × × = = d b f M K f cu By (Eqn 3-7) 169 . 0 890 150 176 . 0 9 . 0 25 . 0 5 . 0 2 9 . 0 = = > = | | . | \ | − − = d d K d x f So 0.9 × neutral axis extends below flange. 65 . 2675 2 1 1 1 67 . 0 2 = ⇒ | | . | \ | − | | . | \ | − = f f w f f m cu w f M d d b b d d f d b M γ kNm Solve d x by (Eqn 3-8) 0 402 . 0 1809 . 0 2 2 = − + − | . | \ | d b M M d x f d x f w f cu cu ( ) 0 890 600 10 65 . 2675 4000 35 402 . 0 35 1809 . 0 2 6 2 = × × − + × − | . | \ | × d x d x 2198 . 0 = d x By (Eqn 3-10) 0231 . 0 9 . 0 1 67 . 0 87 . 0 1 = + | | . | \ | − = d x d d b b f f d b A f w f m cu y w st γ 12330 = st A mm 2 , Use 10-Y40 in 2 layers (iii) Worked Example 3.7 – Doubly reinforced section Beam Section : 1000 (h) × 600 (w), flange width = 1250 mm, flange depth = 150 mm 35 = cu f MPa under a moment 4000 kNm 22 600 = w b , 890 60 50 1000 = − − = d , 1250 = w b 150 = f d 169 . 0 890 150 = = d d f ; 083 . 2 600 1250 = = w f b b First check if d d d x f ≤ 9 . 0 based on beam width of 1250 = w b 115 . 0 890 1250 35 10 4000 2 6 2 = × × × = = d b f M K f cu By (Eqn 3-7) 169 . 0 890 150 302 . 0 9 . 0 25 . 0 5 . 0 2 9 . 0 = = > = | | . | \ | − − = d d K d x f So 0.9×neutral axis extends below flange. 26 . 1242 2 1 1 1 67 . 0 2 = ⇒ | | . | \ | − | | . | \ | − = f f w f f m cu w f M d d b b d d f d b M γ kNm Solve d x by (Eqn 3-8) 0 402 . 0 1809 . 0 2 2 = − + − | . | \ | d b M M d x f d x f w f cu cu ( ) 0 890 600 10 26 . 242 82 . 1337 4000 35 402 . 0 35 1809 . 0 2 6 2 = × × − + × − | . | \ | × d x d x 5 . 0 547 . 0 > = d x . So double reinforcement is required. By (Eqn 3-11) 0038 . 0 ' 1 87 . 0 ' 2 = | . | \ | − | | . | \ | − = d d f f K f d b M bd A y cu f cu w sc Use 0.4% as per Table 9.2.1 of the Code 2040 = sc A mm 2 . Use 2T25 + 4T20 0247 . 0 9 . 0 1 67 . 0 87 . 0 1 = + | | . | \ | − = d x d d b b f f d b A f w f m cu y w st γ 13196 = st A mm 2 , Use 11-Y40 in 2 layers 3.6 Detailings of longitudinal steel for bending The followings should be observed in placing of longitudinal steel bars for bending which will directly affect design (Cl. 9.2.1 and 9.9.1 of the Code, the requirements arising from “ductility” are marked with “D”) (i) Minimum tensile steel percentage : Table 9.1 of the Code where 0.13% 23 for rectangular beam and other percentages for others; (ii) Minimum compressive steel percentage : Table 9.1 of the Code where 0.2% for rectangular beam and other percentages for others; (iii) Minimum total percentage : total reinforcements to be not less than 0.3% (D); (iv) Maximum steel percentage : 2.5% (D); (v) At any section of a beam within a critical zone (e.g. a potential plastic hinge zone) the compression reinforcement should not be less than one-half of the tension reinforcement in the same region (D); (vi) Minimum clear spacing of bars should be the minimum of bar diameter, 20 mm and aggregate size + 5 mm; (vii) At laps, the sum of reinforcement sizes in a particular layer should not exceed 40% of the beam width; (viii) Maximum clear spacing between adjacent bars near tension face of a beam ≤ 70000β b /f y ≤ 300 mm where β b is the ratio of moment redistribution or alternatively ≤ 47000/f s ≤ 300 mm where b prov s req s y s A A f f β 1 3 2 , , × = ; (ix) Requirements for containment of compression steel bars is identical to that of columns (Cl. 9.5.2.2 of the Code) : Every corner bar and each alternate bar (and bundle) in an outer layer should be supported by a link passing around the bar and having an included angle < 135 o . No bar within a compression zone be more than 150 mm from a restrained bar; (x) For beam with depths > 750 mm, provision of sides bars > y b f b s / where b s is the side bar spacing and b is the lesser of the beam breadth under consideration and 500 mm. In addition, 250 ≤ b s mm and be distributed over two-thirds of the beam’s overall depth measured from its tension face; (xi) When longitudinal beam bars are anchored in cores of exterior columns or beam studs, the anchorage for tension shall be deemed to commence at the lesser of 1/2 of the relevant depth of the column or 8 times the bar diameter as indicated in Figure 3.9. Where it can be shown that the critical section of the plastic hinge is at a distance at least the beam depth or 500 mm, whichever is the less from the column face, the anchorage length can be considered to 24 commence at the column face. (D) For calculation of anchorage length, bars must be assumed to be fully stressed as a ductility requirement. As such, the anchorage and lap lengths as indicated in Tables 8.4 and 8.5 should be increased by 15%. Alternatively, the anchorage length can be determined by bu y b f f l 4 φ ≥ which is a modification from (Ceqn 8.4) of the Code where cu bu f f β = and β is 0.5 for tension anchorage and 0.63 for compression anchorage for high yield bars in accordance with Table 8.3 of the Code. Lap lengths can be taken as identical to anchorage lengths (D); (xii) Notwithstanding the adequacy of the anchorage of a beam bar in a column core or a beam stud, no bar shall be terminated without a vertical 90 o standard book or equivalent anchorage device as near as practically possible to the far side of the column core, or the end of the beam stud where appropriate, and not closer than 3/4 of the relevant depth of the column to the face of entry. Top beam bars shall be bent down and bottom bars must be bent up. Requirements by (xi) and (xii) are illustrated by Figure 3.9. 3.7 Design against Shear > 3/4 D >the lesser of D/2 or 8ø D h anchorage commence at this point Anchored bar of dia. ø Figure 3.9 – Anchorage of bar in exterior column for ductility enhancement 25 3.7.1 Checking of Shear Stress and provision of shear reinforcements Checking of shear in beam is based on the averaged shear stress calculated from (Ceqn 6.19) of the Code d b V v v = where v is the average shear stress, V is the ultimate shear, d is the effective depth of the beam and v b is beam width. v b should be taken as the averaged width of the beam below flange in case of flanged beam) If v is less than the values of c v , termed “design concrete shear stress” in Table 6.3 of the Code which is determined by the formula m v s cu c d d b A f v γ 1 400 100 25 79 . 0 4 1 3 1 3 1 | . | \ | | | . | \ | | . | \ | = listed in Table 6.3 with the following limitations : (i) 25 . 1 = m γ ; (ii) d b A v s 100 should not be taken as greater than 3; (iii) 4 1 400 | . | \ | d should not be taken as less than 0.67 for member without shear reinforcements and should not be taken as less than 1 for members with links; then shear links of ( ) yv c v sv f v v b s A 87 . 0 − = should be provided (Table 6.2 of the Code), or alternatively ( )( ) b yv sb b s d d f A V ' cot sin cos 87 . 0 − + = β α α in (Ceqn 6.20 of the Code) can be used for design of bent-up bars. Maximum shear stress not to exceed cu tu f v 8 . 0 = or 7 MPa, whichever is the lesser by Cl. 6.1.2.5. 3.7.2 Minimum shear reinforcements If c v v 5 . 0 < , no shear reinforcement is required; If ( ) r c c v v v v + < < 5 . 0 , minimum shear links of yv r v sv f bv s A 87 . 0 = along the whole length of the beam be provided where 4 . 0 = r v for 40 ≤ cu f and 26 ( ) 3 / 2 40 / 4 . 0 cu f for 40 > cu f , but not greater than 80; 3.7.3 Enhanced shear strength close to support At sections of a beam at distance d a v 2 ≤ from a support, the shear strength can be increased by a factor v a d 2 , bounded by the absolute maximum of cu tu f v 8 . 0 = or 7 MPa. 3.7.4 Worked Examples for Shears (i) Worked Example 3.8 – shear design without shear enhancement in concrete Section : 400 = b mm; 644 16 40 700 = − − = d mm; 5 . 1 100 = bd A st ; 35 = cu f MPa; 700 = V kN; 81 . 0 1 400 100 25 79 . 0 4 1 3 1 3 1 = | . | \ | | | . | \ | | . | \ | = m v s cu c d d b A f v γ MPa; where 4 1 400 | . | \ | d taken as 1. 72 . 2 644 400 10 700 3 = × × = v MPa, ( ) ( ) 91 . 1 460 87 . 0 81 . 0 72 . 2 400 87 . 0 = × − = − = yv c v sv f v v b s A ; Use Y12 – 200 c/c s.s. d v a Figure 3.10 – Shear enhancement near support section under consideration, shear strength enhanced to c v v a d 2 27 (ii) Worked Example 3.9 – shear design with shear enhancement in concrete. At section of 0.75 m from a support, as a heavy point load is acting so that from the support face to the point load along the beam, the shear is more or less constant at 700 kN. Section : 400 = b mm; 644 16 40 700 = − − = d mm; 5 . 1 100 = bd A st ; 35 = cu f MPa; 700 = V kN; 81 . 0 1 400 100 25 79 . 0 4 1 3 1 3 1 = | . | \ | | | . | \ | | . | \ | = m v s cu c d d b A f v γ MPa; where 4 1 400 | . | \ | d taken as 1. Concrete shear strength enhanced to 39 . 1 81 . 0 750 644 2 2 = × × = c v v a d MPa < 7 and 7 . 4 35 8 . 0 8 . 0 = = cu f MPa 72 . 2 644 400 10 700 3 = × × = v MPa, ( ) ( ) 33 . 1 460 87 . 0 39 . 1 72 . 2 400 87 . 0 = × − = − = yv c v sv f v v b s A ; Use Y12 – 150 c/c s.s (iii) Worked Example 3.10 – inclusion of bent-up bars If half of the shear in Worked Example 3.8 is to be taken up by bent-up bars, i.e. ( ) 246 10 644 400 81 . 0 72 . 2 5 . 0 3 = × × × − × − kN to be taken up by bent-up bars. By (Ceqn 6.20) of the Code, ( )( ) b yv sb b s d d f A V ' cot sin cos 87 . 0 − + = β α α ( ) 413 588 60 cot 45 sin 45 cos 160 87 . 0 441 246000 0 0 0 = × + × × = ⇒ sb A mm 2 s b = 441 s b = 441 S t = 882 β = 60 o α = 45 o d – d’= 644 – 40 – 16 = 588 28 Use 6 nos. of T10 at spacing of s b = 440 mm as shown. 3.8 Placing of Shear reinforcements The followings (in Cl. 6.1.2.5(d), Cl. 9.2.2 and Cl. 9.9.1.2 of the Code) should be observed for the placing of shear reinforcements which may directly affect design : (i) The minimum provision of shear reinforcements (links or bent up bars) in beams should be given by ( ) yv v v sv f s b A 87 . 0 / 4 . 0 ≥ ; (ii) At least 50% of the necessary shear reinforcements be in form of links; (iii) The maximum spacing of links in the direction of span of the beam should be the least of (a) 0.75d; (b) the least lateral dimension of the beam (D); (c) 16 times the longitudinal bar diameter (D); (iv) At right angle of the span, the horizontal spacing of links should be such that no longitudinal bar should be more than 150 mm from a vertical leg; this spacing should not exceed d; (v) Links or ties shall be arranged so that every corner and alternate longitudinal bar that is required to function as compression reinforcement shall be restrained by a leg (vi) Links should be adequately anchored by means of 135 o or 180 o hooks in accordance with Cl. 8.5 of the Code. Anchorage by means of 90 o hook is not permitted (D); 3.9 Design against Torsion 3.9.1 By Cl. 6.3.1 of the Code, in normal slab-and-beam and framed construction, checking against torsion is usually not necessary. However, checking needs be carried out if the design relies entirely on the torsional resistance of a member such as 29 3.9.2 Calculation of torsional rigidity of a rectangular section for analysis (in grillage system) is by (Ceqn 6.64) of the Code max 3 min h h C β = where β is to be read from Table 6.16 of the Code. 3.9.3 Calculation of torsional shear stress Upon arrival of the torsion on the rectangular section, the torsional shear stress is calculated by (Ceqn 6.65) of the Code | . | \ | − = 3 2 min max 2 min h h h T v t and in case of a section such as T or L sections made of rectangles, the section should be broken up into two or more rectangles such that the max 2 min h h is maximized and the total Torsional moment T be apportioned to each rectangle in accordance with (Ceqn 6.66) of the Code as | | . | \ | × ∑ max 3 min max 3 min h h h h T . If the total shear stress exceeds cu f 067 . 0 (but not more than 0.6MPa), torsional reinforcements will be required. Furthermore, the torsional shear stress should be added to the shear stress induced by shear force to ensure that the absolute maximum cu tu f v 8 . 0 = or 7MPa is not exceeded, though for small section where 1 y (the larger centre-to-centre dimension of a Beam carrying unbalanced torsion induced by slab needs be checked Figure 3.11 – Illustration for necessity of checking against torsion 30 rectangular link) < 550mm, tu v will be decreased by a factor 550 / 1 y . Revision of section is required if the absolute maximum is exceeded. 3.9.4 Calculation of torsional reinforcements Torsional reinforcement in forms of close rectangular links and longitudinal bars are to be calculated by (Ceqn 6.67) and (Ceqn 6.68) of the Code as ( ) yv v sv f y x T s A 87 . 0 8 . 0 1 1 = (Ceqn 6.67) ( sv A is the area of the 2 legs of the link) ( ) y v yv sv s f s y x f A A 1 1 + = (Ceqn 6.68) It should be noted that there is no reduction to shear strength ( c v ) by concrete. The derivation of the design formula (Ceqn 6.67) of the Code for close rectangular links is under the assumption of a shear rupture length of stirrup width + stirrup depth 1 1 y x + as shown in Figure 3.12. A spiral torsional failure face is along the heavy dotted line. It is shown in Figure 3.12 that the torsional moment of resistance by the stirrups within the Regions X and Y are identical and is the total resistance is therefore v y sv s y x f A 1 1 87 . 0 . y sv f A 87 . 0 5 . 0 y sv f A 87 . 0 5 . 0 y sv f A 87 . 0 5 . 0 y sv f A 87 . 0 5 . 0 shear rupture spiral face 1 x Moment provided by this stirrup in region Y is 1 87 . 0 5 . 0 x f A y sv . Total nos. of stirrup within Y is v s y / 1 . So total moment is v y sv s y x f A / 87 . 0 5 . 0 1 1 1 y Region Y Region X 1 y 1 x Moment provided by this stirrup in region X is 1 87 . 0 5 . 0 y f A y sv . Total nos. of stirrup within X is v s x / 1 . So total moment is v y sv s x y f A / 87 . 0 5 . 0 1 1 45 o 45 o 1 y 1 x Figure 3.12 – Derivation of Design Formulae for Torsional Resistance 31 So 1 1 1 1 87 . 0 87 . 0 y x f T s A s y x f A T y v sv v y sv = ⇒ = . An additional factor of 0.8 is added and the equation becomes (Ceqn 6.67). The derivation of the longitudinal bars is based on the use of same quantity of longitudinal bars as that of stirrups. 3.9.5 Worked Example 3.10 – Design for T-beam against torsion A total torsion of 200 = T kNm on a T-section as shown with an average vertical shear stress on the web of 0.82 N/mm 2 Option A is made up of two rectangles of 400 550× and one rectangle of 1000 400 400 1500 1000 400 400 1500 Option A 1000 400 400 1500 T-Section Option B Figure 3.13 – Section of a T section resisting torsion 32 1400 400× . From Table 6.16 of the Code, β for the 400 550× rectangle and 1400 400× are respectively 0.185 and 0.2675 So the total stiffness of Option A is 9 3 3 10 992 . 36 400 1400 2675 . 0 400 550 185 . 0 2 × = × × + × × × mm 4 Option B is made up of one rectangle of 400 1500× and one rectangle of 1000 400× . From Table 6.16 of the Code, β for the 400 1500× rectangle and 1000 400× are respectively 0.27125 and 0.245 So the total stiffness of Option B is 9 3 3 10 72 . 41 400 1000 245 . 0 400 1500 27125 . 0 × = × × + × × mm 4 As Option B has a large torsional stiffness, it is adopted for design. The torsional moment is apportioned to the two rectangles of Option B as : For the 400 1500× rectangle 832 . 124 72 . 41 04 . 26 200 1 = × = T kNm; Torsional shear stress is 142 . 1 3 400 1500 400 10 832 . 124 2 3 2 2 6 min max 2 min 1 1 = | . | \ | − × × = | . | \ | − = h h h T v t N/mm 2 > 396 . 0 067 . 0 = cu f N/mm 2 So torsional shear reinforcement is required 308 2 6 2 40 400 1 = × − × − = x ; 1408 2 6 2 40 1500 1 = × − × − = y ( ) 899 . 0 460 87 . 0 1408 308 8 . 0 10 832 . 124 87 . 0 8 . 0 6 1 1 1 = × × × × × = = yv v sv f y x T s A Use T12 – 250 s.s. ( ) ( ) 1543 460 1408 308 460 899 . 0 1 1 = + × × = + = y v yv sv s f s y x f A A mm 2 Use 8T16 For the 400 1000× rectangle 168 . 75 72 . 41 68 . 15 200 2 = × = T kNm 084 . 1 3 400 1000 400 10 168 . 75 2 3 2 2 6 min max 2 min 2 1 = | . | \ | − × × = | . | \ | − = h h h T v t N/mm 2 The total shear stress is 904 . 1 82 . 0 084 . 1 = + N/mm 2 < 88 . 2 550 / 1 = y v tu MPa As 396 . 0 067 . 0 904 . 1 = > cu f N/mm 2 33 So torsional shear reinforcement is required 308 2 6 2 40 400 1 = × − × − = x mm; 908 2 6 2 40 1000 1 = × − × − = y mm ( ) 84 . 0 460 87 . 0 908 308 8 . 0 10 168 . 75 87 . 0 8 . 0 6 1 1 1 = × × × × × = = yv v sv f y x T s A mm Use T12 – 250 s.s. It should be noted that the torsional shear link should be closed links of shape as indicated in Figure 9.3 of the Code. ( ) ( ) 1021 460 908 308 460 84 . 0 1 1 = + × × = + = y v yv sv s f s y x f A A mm 2 . Use 6T16 It should be borne in mind that these torsional reinforcements are in addition to others required for flexure and shear etc. 3.10 Placing of Torsional reinforcements The followings (in Cl. 6.3.7, Cl. 6.3.8 and Cl. 9.2.3 of the Code) should be observed for the placing of shear reinforcements which may directly affect design : (i) The shear link should form the closed shape as in Figure 9.3 in Cl. 9.2.3 of the Code; (ii) The value v s for the closed link should not exceed the least of 1 x , 2 / 1 y or 200 mm as per Cl. 6.3.7 of the Code; (iii) In accordance with Cl. 9.2.3 of the Code, provision of the longitudinal torsion reinforcement should comply the followings : 1000 400 400 1500 Figure 3.14 – Arrangement of torsional reinforcements 34 (a) The bars distributed should be evenly round the inside perimeter of the links; (b) Clear distance of the bars not exceed 300 mm; (c) Additional longitudinal bars required at the level of the tension or compression reinforcements may be provided by using larger bars than those required for bending alone; (d) The longitudinal bars should extend a distance at least equal to the largest dimension of the section beyond where it theoretically cases to be required. 35 4.0 Slabs 4.1 Types of Slabs Slabs can be classified as “one way slab”, “two way slab”, “flat slab”, “ribbed slab”. 4.1.1 One way slab is defined by the Code as one subjected predominantly to u.d.l. either (i) it possesses two free and parallel edge; or (ii) it is the central part of a rectangular slab supported on four edges with a ratio of the longer to shorter span greater than 2. So, strictly speaking, it is only the central part of (ii) be treated as one way slabs. Nevertheless, as we usually design for a slab based on the bending moment at the central part of the slab which is the maximum and assign reinforcements to the whole slab without curtailment, treating the whole rectangular slab as two ways bending in design is usually adequate. 4.1.2 Two way slab is a rectangular one supported on four sides with length to breadth ration smaller than 2. 4.1.3 Flat slab is a slab supported on columns without beam. 4.1.4 Ribbed or Waffled Slab is a slab with topping or flange supported by closely spaced ribs. The Code allows idealization of the ribbed slab or waffled slab as a single slab without treatment as discretized ribs and flanges in analysis. If the stiffness of the ribbed or waffled slab is required for input, the bending stiffness in the X and Y directions can be easily found by summing the total bending stiffness of the composite ribs and flange structure per unit width. The twisting stiffness is difficult to assess. However, it should be acceptable to set the twisting stiffness to zero which will end up with pure bending in the X and Y directions as the slab, with its ribs running in the X and Y directions are clearly predominantly strong in bending in the X and Y directions. 4.2 Analysis of Slabs without the use of computer method 4.2.1 One way slab can be analyzed as if it is a beam, either continuous or single span. As we aim at simple analysis for the slab, we tend to treat it as a single element without the necessity to consider the many loading cases for continuous spans, Cl. 6.1.3.2(c) of the Code allows the design against moment 36 and shear arising from the single-load case of maximum design load on all spans provided that (i) the area of each bay (defined in Figure 6.5 of the Code) not to exceed 30 m 2 ; (ii) the ratio of the characteristic imposed load to characteristic dead load not to exceed 1.25; and (iii) the characteristic imposed load does not exceed 5 kN/m 2 excluding partitions. 4.2.2 Two way rectangular slab is usually carried out by treating it as if it is a single slab in the absence of computer method. Bending moment coefficients for calculation of bending moments are presented in Table 6.6 of the Code for different support restraint conditions. Nevertheless, simplified formulae for the bending coefficients in case of rectangular simply supported two way slab are available in the Code (Ceqn 6.26 and 6.27). 4.2.3 Flat slabs, if of regular arrangement, can be analyzed as frames in the transverse and longitudinal directions by such methods as moment distribution method as if they are separate frames. Analyzed moments and shears should be apportioned to the “column strip” and “Middle strip” as per Table 6.10. In addition, the bending moment and shear force coefficients for the one way slab can also be used as a simplified approach. 4.2.4 More bending moment and shear force coefficients of rectangular slabs with various different support and loading conditions can be found from other published handbooks, the most famous one being “Tables for the Analysis of Plates, Slabs and Diaphragms based on the Elastic Theory”. Extracts of commonly encountered slabs are enclosed in Appendix E. 4.3 Analysis of Slabs with the use of the computer method Analysis of slabs with the use of the computer method is mainly by the finite element method in which the slab is idealized as an assembly of discrete “plate bending elements” jointed at nodes. The support stiffnesses by the supporting walls and columns are derived as similar to that for beams as “sub-frames”. A complete set of results including bending moments, twisting moment, shear force per unit width (known as “stress” in finite element terminology) can be obtained after analysis for design purpose. The design against flexure is most commonly done by the Wood Armer Equations which calculate design moments in two directions (conveniently in two perpendicular directions) which are adequate to cater for the complete set of bending and twisting 37 moments. The design based on node forces should be avoided due to its inadequacy to cater for twisting effects which will result in under-design. A discussion of the plate bending theory and the design approach by the Wood Armer Equations is enclosed in Appendix D, together with the “stress approach” for checking and designing against shear in the same appendix. An example of the mathematical modeling of a floor slab by the software SAFE and results of subsequent analysis is illustrated in Figure 4.1. Figure 4.1 – Modeling of an irregular floor slab as 2-D mathematical model, subsequent analytical results of bending moments and twisting moment, design of reinforcements by the Wood Armer Equations. 38 The finite element mesh of the mathematical model is often very fine. So it is a practice of “lumping” the design reinforcements of a number of nodes over certain widths and evenly distributing the total reinforcements over the widths, as is done by the popular software “SAFE”. However, care must be taken of not taking widths too wide for “lumping” as local effects may not be well captured. 4.4 Detailing for Solid Slabs Generally considerations in relation to determination of “effective span”, “effective span depth ratio”, “moment redistribution”, “reduced design moment to support”, “maximum and minimum steel percentages”, “concrete covers” as discussed in 3.3 for design of beam are applicable to design of slab. Nevertheless, the detailing considerations for slabs are listed as follows (Re 9.3.1.1 of the Code) : (i) Minimum steel percentage : Main Reinforcing bars: 0.24% for 250 = y f MPa and 0.13% for 460 = y f MPa; Distribution bars in one way slab : 20% of the main reinforcements; (ii) Maximum reinforcements spacing : (a) In general areas without concentrated loads : the principal reinforcement, max. spacing 400 3 ≤ ≤ h mm; and the secondary reinforcement, max. spacing 450 5 . 3 ≤ ≤ h mm. (b) In areas with concentrated loads or areas of maximum moment: the principal reinforcement, max. spacing 250 2 ≤ ≤ h mm; and for the secondary reinforcement, max. spacing 400 3 ≤ ≤ h mm. (iii) In addition to (ii), crack control can be carried out by crack widths calculations, Nevertheless, no further check on crack control is required on bar spacing if either : (a) max. spacing 250 ≤ ≤ h mm (grade 250 steel); (b) max. spacing 200 ≤ ≤ h mm (grade 460 steel); or (c) the percentage of required tension reinforcement is less than 0.3%. (iv) Requirements pertaining to curtailment and anchoring of tension reinforcements should be similar to that of beams; (v) Reinforcements at end supports 39 (a) Half of the span reinforcements should be provided and well anchored on end supports in simply supported slabs and continuous slabs; (b) If support shear c v v 5 . 0 < , the following arrangement can be considered as effective anchorage. (vi) Minimum bottom reinforcements at internal supports : 40% of the calculated mid-span bottom reinforcements. (vii) Reinforcements at free edge should be as shown : (viii) Shear reinforcements not to be used in slabs < 200 mm. 4.5 Structural Design of Slabs The structural of slab against flexure is similar to that of beam. The determination of reinforcements should be in accordance with Section 3.4 listing the options of either following the rigorous or simplified “stress strain” > 1/3b and 30 mm if c v v 5 . 0 < b Figure 4.2 – Anchorage of end span bar for c v v 5 . 0 < h ≥2h Figure 4.3 – Free edge reinforcements for Slabs 40 relationship of concrete. Design against shear for slabs under line supports (e.g. one-way or two-way) is also similar to that of beam. However for a flat slab, the checking should be based on punching shear in accordance with empirical method of the Code or by based on shear stresses revealed by the finite element method. They are demonstrated in the Worked Examples in the following sub-Section 4.6 : 4.6 Worked Examples Worked Example 4.1 – One Way Slab A one-way continuous slab with the following design data : (i) Live Load = 4.0 kN/m 2 ; (ii) Finishes Load = 1 kN/m 2 ; (iii) Concrete grade 35; (i) Fire rating : 1 hour, mild exposure; (ii) Span : 4 m Sizing : Limiting Span depth ratio = 6 . 40 56 . 1 26 = × (by Table 7.3 and Table 7.4 of the Code, assuming modification by tensile reinforcement to be 1.56 as the slab should be lightly reinforced). So effective depth taken as 120 5 25 150 = − − = d as 4000/120 = 33.3 < 40.6 (assuming 10mm dia. bars under 25 mm concrete cover.) Loading : D.L. O.W. = ×24 15 . 0 3.6kN/m 2 Fin. 1.0 kN/m 2 Total 4.6 kN/m 2 L.L. 4.0 kN/m 2 The factored load on a span is ( ) 36 . 51 4 0 . 4 6 . 1 6 . 4 4 . 1 = × × + × = F kN/m Based on coefficients of shear and bending in accordance with Table 6.4 of the 4m 4m 4m 4m Figure 4.4 – Slab in Example 1 41 Code listed as follows : (a) End span support moment = 22 . 8 4 36 . 51 04 . 0 = × × kNm/m 0163 . 0 120 1000 35 10 22 . 8 2 6 2 = × × × = = bd f M K cu 95 . 0 982 . 0 9 . 0 25 . 0 5 . 0 > = − + = K d z % 15 . 0 95 . 0 460 87 . 0 120 1000 10 22 . 8 / 87 . 0 2 6 2 = × × × × × = × = d z f bd M bd A y st 180 120 1000 100 15 . 0 = × × ÷ = st A mm 2 Use T10 – 300 ( st A provided = 262mm 2 ) End span shear = 63 . 23 36 . 51 46 . 0 = × kN/m. 2 . 0 120 1000 23630 = × = v N/mm 2 < ( ) 5 . 0 25 / 35 45 . 0 3 / 1 = × N/mm 2 . No shear reinforcement required. End span span moment = 41 . 15 4 36 . 51 075 . 0 = × × kNm/m 0306 . 0 120 1000 35 10 41 . 15 2 6 2 = × × × = = bd f M K cu 95 . 0 965 . 0 9 . 0 25 . 0 5 . 0 > = − + = K d z % 28 . 0 95 . 0 460 87 . 0 120 1000 10 41 . 15 / 87 . 0 2 6 2 = × × × × × = × = d z f bd M bd A y st 338 120 1000 100 28 . 0 = × × ÷ = st A mm 2 Use T10 – 225 ( st A provided = 349mm 2 ) (b) Interior span support moment = 67 . 17 4 36 . 51 086 . 0 = × × kNm/m 0351 . 0 120 1000 35 10 67 . 17 2 6 2 = × × × = = bd f M K cu 95 . 0 96 . 0 9 . 0 25 . 0 5 . 0 > = − + = K d z S.F. 0.46F 0.6F 0.6F 0.5F 0.5F 0.6F 0.6F 0.4F B.M. -0.04Fl 0.075 Fl -0.086Fl 0.063Fl -0.086Fl 0.063Fl -0.086Fl 0.075 Fl -0.04Fl Figure 4.5 – Bending Moment and Shear Force coefficients for Continuous Slab 42 % 13 . 0 % 32 . 0 95 . 0 460 87 . 0 120 1000 10 67 . 17 / 87 . 0 2 6 2 > = × × × × × = × = d z f bd M bd A y st 387 120 1000 100 32 . 0 = × × ÷ = st A mm 2 Use T10 – 200 ( st A provided = 393mm 2 ) Maximum shear = 82 . 30 36 . 51 6 . 0 = × kN/m. 25 . 0 120 1000 30820 = × = v N/mm 2 < ( ) 5 . 0 25 / 35 45 . 0 3 / 1 = × N/mm 2 . No shear reinforcement required. Interior span moment = 94 . 12 4 36 . 51 063 . 0 = × × kNm/m Use T10 – 225 ( st A provided = 349mm 2 ) Worked Example 4.2 – Two Ways Slab (4 sides simply supported) A two-way continuous slab with the following design data : (i) Live Load = 4.0 kN/m 2 ; (ii) Finishes Load = 1 kN/m 2 ; (iii) Concrete grade 35; (ii) Fire rating : 1 hour, mild exposure; (iii) Span : Long way : 4 m, Short way, 3 m. Sizing : Limiting Span depth ratio = 2 . 31 56 . 1 20 = × (by Table 7.3 and Table 7.4 of the Code, assuming modification by tensile reinforcement to be 1.56 as the slab should be lightly reinforced). So effective depth taken as 120 5 25 150 = − − = d as 3000/120 = 25 < 31.2. So the structural depth is 150 mm (assuming 10mm dia. bars under 25 mm concrete cover.) Loading : D.L. O.W. = ×24 15 . 0 3.6kN/m 2 Fin. 1.0 kN/m 2 Total 4.6 kN/m 2 L.L. 4.0 kN/m 2 The factored load is ( ) 84 . 12 0 . 4 6 . 1 6 . 4 4 . 1 = × × + × = F kN/m (Ceqn 6.26) and (Ceqn 6.27) of the Code are used to calculate the bending moment coefficients along the short and long spans : ( ) ( ) | | 095 . 0 / 1 8 / 4 4 = + = x y x y sx l l l l α ; ( ) ( ) | | 053 . 0 / 1 8 / 4 2 = + = x y x y sy l l l l α 43 So the bending moment along the short span is 98 . 10 3 84 . 12 095 . 0 2 = × × = x M kNm/m 0218 . 0 120 1000 35 10 98 . 10 2 6 2 = × × × = = bd f M K cu 95 . 0 975 . 0 9 . 0 25 . 0 5 . 0 > = − + = K d z % 13 . 0 % 2 . 0 95 . 0 460 87 . 0 120 1000 10 98 . 10 / 87 . 0 2 6 2 > = × × × × × = × = d z f bd M bd A y st 240 120 1000 100 2 . 0 = × × ÷ = st A mm 2 Use T10 – 300 ( st A provided = 262mm 2 ) 13 . 6 3 84 . 12 053 . 0 2 = × × = y M kNm/m 0145 . 0 110 1000 35 10 13 . 6 2 6 2 = × × × = = bd f M K cu % 13 . 0 % 11 . 0 95 . 0 460 87 . 0 110 1000 10 13 . 6 / 87 . 0 2 6 2 < = × × × × × = × = d z f bd M bd A y st 143 110 1000 100 13 . 0 = × × ÷ = st A mm 2 Use T10 – 300 ( st A provided = 262mm 2 ) Worked Example 4.3 – Two Ways Slab (3 sides supported) A two-way continuous slab with the following design data : (i) Live Load = 4.0 kN/m 2 ; (ii) Finishes Load = 1 kN/m 2 ; (i) Concrete grade 35; (ii) Span : Long way : 5 m, Short way, 4 m (iii) Fire rating : 1 hour, mild exposure; continuous edge 5 m 4 m free edge simply supported edge Figure 4.6 – Plan of 3-sides supported slab for Example 4.3 44 Sizing : Based on the same limiting Span depth ratio for one way and two way slab which is 2 . 31 56 . 1 20 = × (by Table 7.3 and Table 7.4 of the Code, assuming modification by tensile reinforcement to be 1.56 as the slab should be lightly reinforced). So effective depth taken as 130 5 25 160 = − − = d as 4000/130 = 25 < 30.8. Thus the structural depth is 160 mm (assuming 10mm dia. bars under 25 mm concrete cover.) Loading : D.L. O.W. = ×24 16 . 0 3.84kN/m 2 Fin. 1.0 kN/m 2 Total 4.84 kN/m 2 L.L. 4.0 kN/m 2 The factored load is ( ) 18 . 13 0 . 4 6 . 1 84 . 4 4 . 1 = × + × = F kN/m From Table 1.38 of “Tables for the Analysis of Plates, Slabs and Diaphragms based on Elastic Theory” extracted in Appendix E where 8 . 0 5 / 4 = = γ , the sagging bending moment coefficient for short way span is maximum along the free edge which 0.1104 (linear interpolation between 75 . 0 = γ and 0 . 1 = γ 28 . 23 4 18 . 13 1104 . 0 2 = × × = x M kNm/m 0394 . 0 130 1000 35 10 28 . 23 2 6 2 = × × × = = bd f M K cu 95 . 0 954 . 0 9 . 0 25 . 0 5 . 0 > = − + = K d z % 13 . 0 % 362 . 0 95 . 0 460 87 . 0 130 1000 10 28 . 23 / 87 . 0 2 6 2 > = × × × × × = × = d z f bd M bd A y st 471 130 1000 100 362 . 0 = × × ÷ = st A mm 2 Use T10 – 150 ( st A provided = 523mm 2 ) At 2 m and 4 m from the free edge, the sagging moment reduces to 798 . 17 4 18 . 13 0844 . 0 2 = × × kNm/m and 75 . 8 4 18 . 13 0415 . 0 2 = × × kNm/m and st A required are reduces to 360 mm 2 and 177 mm 2 . Use T10 – 200 and T10 – 300 respectively. The maximum hogging moment (bending along long-way of the slab) is at mid-way along the supported edge of the short-way span 01 . 24 5 18 . 13 0729 . 0 2 = × × = y M kNm/m 45 0476 . 0 120 1000 35 10 01 . 24 2 6 2 = × × × = = bd f M K cu 944 . 0 9 . 0 25 . 0 5 . 0 = − + = K d z % 13 . 0 % 44 . 0 944 . 0 460 87 . 0 120 1000 10 01 . 24 / 87 . 0 2 6 2 > = × × × × × = × = d z f bd M bd A y st 528 120 1000 100 44 . 0 = × × ÷ = st A mm 2 Use T10 – 125 ( st A provided = 628mm 2 ) The maximum sagging moment along the long-way direction is at 2 m from the free edge which is 18 . 6 5 18 . 13 01876 . 0 2 = × × = y M kNm/m. The moment is small. Use T10 – 300 Finally the reinforcement arrangement on the slab is (Detailed curtailment omitted for clarity) Worked Example 4.4 – Flat Slab by Simplified Method (Cl. 6.1.5.2(c)) Flat slab arrangement on rectangular column grid of 7.5 m and 6 m with the following design data : T 1 0 – 1 2 5 T 1 1200 1200 800 800 T10 – 300 B1 T10 – 200 B1 T10 – 150B1 T 1 0 – 3 0 0 B 2 2000 2000 1000 T10 – 300 T1 Figure 4.7 – Reinforcement Details for Example 4.3 46 (i) Finish Load = 1.5 kPa (ii) Live Load = 5.0 kPa. (iii) Column size = 550 × 550 (iv) Column Drop size = 3000 × 3000 with d h = 200 mm (v) Fire rating : 1 hour, mild exposure; Conditions as stipulated in Cl. 6.1.3.2(c) are satisfied as per Cl. 6.1.5.2(g). Hence the simplified method is applicable. In the simplified method, the flat slab is effectively divided into “column strips” containing the columns and the strips of the slabs linking and the columns and the “middle strips” between the “column strips” which are designed as beams in flexural design with assumed apportionment of moments among the strips. However, for shear checking, punching shear along successive “critical” perimeters of column are carried out instead. Effective dimension of column head ( ) 40 2 max − + = h c h d l l (Ceqn 6.37) ( ) 870 40 200 2 550 = − + = mm Effective diameter of column head (Cl. 6.1.5.1(c)) 982 4 870 2 = × = π c h mm Effective span in the long side is 6846 3 982 2 7500 3 2 = × − = − c h l mm 1500 1500 1500 1500 7500 7500 6000 6000 6000 Figure 4.8 – Flat Slab Plan Layout for Example 4 47 (Ceqn 6.38) Sizing : Based on the same limiting Span depth ratio for one way and two way slab which is 30 15 . 1 26 = × (by Table 7.3 and Table 7.4 of the Code, assuming modification by tensile reinforcement to be 1.15), 228 30 6846 = = d . As cover = 25 mm, assuming T12 bars, structural depth is 259 2 12 25 228 = ÷ + + mm, say 275 mm. Loading : D.L. O.W. = ×24 275 . 0 6.6kN/m 2 Fin. 1.5 kN/m 2 Total 8.1 kN/m 2 L.L. 5.0 kN/m 2 The factored load is ( ) 34 . 19 0 . 5 6 . 1 1 . 8 4 . 1 = × + × = F kN/m Design against Flexure (Long Way) The bending moment and shear force coefficients in Table 6.4 will be used as per Cl. 6.1.5.2(g). Total load on one panel is 3 . 870 6 5 . 7 34 . 19 = × × = Fl kN. Hogging Moment : Total moment at outer support is 46 . 238 85 . 6 3 . 870 04 . 0 = × × kNm Total moment at first interior support is 69 . 512 85 . 6 3 . 870 086 . 0 = × × kNm Total moment at middle interior support is 58 . 375 85 . 6 3 . 870 063 . 0 = × × kNm These moments are to be apportioned in the column and mid strips in accordance to the percentages of 75% and 25% as per Table 6.10, i.e. Column Strip Mid Strip Total Mt Mt/width Total Mt Mt/width Outer Support 178.85 59.62 59.62 19.87 1 st interior support 384.52 128.17 128.17 42.72 Middle interior support 281.69 93.9 93.9 31.30 The reinforcements – top steel are worked out as follows, ( 444 6 25 475 = − − = d over column support and 244 6 25 275 = − − = d in other locations) Column Strip Mid Strip Area (mm 2 )/m steel Area (mm 2 )/m steel Outer Support 640 T10 – 100 319 T10 – 225 1 st interior support 760 T12 – 125 459 T10 – 150 Middle interior support 577 T12 – 150 316 T10 – 225 48 Sagging Moment : Total moment near middle of end span is 12 . 447 85 . 6 3 . 870 075 . 0 = × × kNm Total moment near middle of interior span 58 . 375 85 . 6 3 . 870 063 . 0 = × × kNm These moments are to be apportioned in the column and mid strips in accordance to the percentages of 55% and 45% as per Table 6.10, i.e. Column Strip Mid Strip Total Mt Mt/width Total Mt Mt/width Middle of end span 245.92 81.97 201.20 67.07 Middle of interior span 206.57 68.86 169.01 56.34 The reinforcements – bottom steel are worked out as follows : Column Strip Mid Strip Area (mm 2 )/m steel Area (mm 2 )/m steel Middle of end span 910 T12 – 100 744 T10 – 100 Middle of interior span 764 T10 – 100 625 T10 – 125 Design against Shear The maximum shear is at the 1 st interior support which is 01 . 1201 3 . 870 15 . 1 2 6 . 0 = × × × kN (Note : 0.6 is the shear force coefficient for first interior support from Table 6.4; 2 is to account for shear from both sides of the slab; 1.15 is the coefficient to account for moment transfer in Cl. 6.1.5.6(b) in the absence of calculation. Alternatively, (Ceqn 6.40) can be used to arrive at a more accurate coefficient which is ( ) | | . | \ | × + × × × + × × × = | | . | \ | + = 2 666 . 0 55 . 0 3 . 870 7 . 0 52 . 384 5 . 1 1 3 . 870 2 6 . 0 5 . 1 1 sp t t t eff x V M V V . 8 . 1301 = kN > 1201.01 kN. The coefficient of 0.7 applied to t M is per note 2 of Cl. 6.1.5.6(b)) Check on 1 st critical perimeter – d 5 . 1 from column face, i.e. 666 . 0 444 . 0 5 . 1 = × . So side length of the perimeter is ( ) 1882 2 666 550 = × + Length of perimeter is 7528 1882 4 = × Shear force to be checked can be the maximum shear 1201.01 kN after deduction of the loads within the critical perimeter which is 49 51 . 1132 882 . 1 34 . 19 01 . 1201 2 = × − kN Shear stress = 339 . 0 444 7528 1132510 = × N/mm 2 . < 38 . 0 = c v N/mm 2 in accordance with Table 6.3 ( ( ) 38 . 0 25 / 35 34 . 0 3 / 1 = × ). No shear reinforcement is required. No checking on further perimeter is required. Nevertheless, if the higher shear of 8 . 1301 = eff V kN is adopted which becomes 3 . 1233 882 . 1 34 . 19 8 . 1301 2 = × − kN, the shear stress will be 37 . 0 444 7528 1233300 = × N/mm 2 . < 0.38 N/mm 2 . No shear reinforcement is required. Example 4.5 – Design for shear reinforcement (Ceqn 6.44) and (Ceqn 6.45) of the Code gives formulae for reinforcement design for different ranges of values of v . For c v v 6 . 1 ≤ , ( ) ∑ − > y c sv f ud v v A 87 . 0 sinα For c c v v v 0 . 2 6 . 1 ≤ < , ( ) ∑ − > y c sv f ud v v A 87 . 0 7 . 0 5 sinα As a demonstration, if 5 . 0 = v N/mm 2 in the first exterior column head which is < 61 . 0 6 . 1 = c v N/mm 2 but > 38 . 0 = c v N/mm 2 in the Example 4.4. By (Ceqn 6.64) of the Code, ( ) ( ) 1002 460 87 . 0 444 7528 38 . 0 5 . 0 87 . 0 sin = × × × − = − > ∑ y c sv f ud v v A α mm 2 . If vertical links is chose as shear reinforcement, 1 sin 90 0 = ⇒ = α α . So the 1002 mm 2 should be distributed within the critical perimeter as shown in Figure 4.6. In distributing the shear links within the critical perimeter, there are recommendations in Cl. 6.1.5.7(f) of the Code that (i) at least two rows of links should be used; (ii) the first perimeter should be located at approximately 0.5d from the face of the loaded area (i.e. the column in this case). So the first row be determined at 200 mm from the column face with total row length 3800 4 950 = × . Using T10 – 500 spacing along the row, the total steel area will be ( ) 597 500 / 3800 4 / 10 2 = × π mm 2 > 40% of 1002 mm 2 . The second row be at further 300 mm away where row length is 50 6200 4 1550 = × . Again using T10 – 500 spacing along the row, the total steel area will be ( ) 974 500 / 6200 4 / 10 2 = × π mm 2 . Total steel area is 597 + 974 = 1570 mm 2 Design for Shear when ultimate shear stress exceeds 1.6v c It is stated in (Ceqn 6.43) in Cl. 6.1.5.7(e) that if c c v v 0 . 2 6 . 1 < < , ( ) ∑ − > y c sv f ud v v A 87 . 0 7 . 0 5 sinα which effectively reduces the full inclusion of c v for reduction to find the “residual shear to be taken up by steel” at c v v 6 . 1 = to zero inclusion at c v v 0 . 2 = . Example 4.6 – when 1.6v c < v < 2.0v c In the previous Example 4.5, if the shear stress 68 . 0 = v N/mm 2 which lies between 61 . 0 38 . 0 6 . 1 6 . 1 = × = c v N/mm 2 and 76 . 0 38 . 0 2 0 . 2 = × = c v N/mm 2 300 275 200 550 666 666 300 ≈ 0.75d 200 ≈ 0.5d 666 5 . 1 = d 1 st critical perimeter T10 – 300 link 1 st row , using T10 – 500 (total area = 597mm 2 ) >40% of 1002 mm 2 2 nd row , using T10 – 250 (total area = 973mm 2 ) Figure 4.9 – Shear links arrangement in Flat Slab for Example 4.5 51 ( ) ( ) 4009 460 87 . 0 444 7528 38 . 0 68 . 0 7 . 0 5 87 . 0 7 . 0 5 sin = × × × − × = − > ∑ y c sv f ud v v A α mm 2 . If arranged in two rows as in Figure 4.6, use T12 – 250 for both rows : the inner row gives ( ) 1719 250 / 3800 4 / 12 2 = × π mm 2 > 40% of 4009 mm 2 ; the outer row gives ( ) 2802 250 / 6200 4 / 12 2 = × π mm 2 . The total area is 1719 + 2802 = 4521mm 2 > 4009 mm 2 . Cl. 6.1.5.7(e) of the Code says, “When c v v 2 > and a reinforcing system is provided to increase the shear resistance, justification should be provided to demonstrate the validity of design.” If no sound justification, the structural sizes need be revised. 52 5.0 Columns 5.1 Slenderness of Columns Columns are classified as short and slender columns in accordance with their “slenderness”. Short columns are those with ratios h l ex / and b l ey / < 15 (braced) and 10 (unbraced) in accordance with Cl. 6.2.1.1(b) where ex l and ey l are the “effective lengths” of the column about the major and minor axes, b and h are the width and depth of the column. As defined in Cl. 6.2.1.1, a column may be considered braced in a given plane if lateral stability to the structure as a whole is provided by walls or bracing or buttressing designed to resist all lateral forces in that plane. It would otherwise be considered as unbraced. The effective length is given by (Ceqn 6.46) of the Code as 0 l l e ⋅ = β where 0 l is the clear height of the column between restraints and the value β is given by Tables 6.11 and 6.12 of the Code which measures the restraints against rotation and lateral movements at the ends of the column. Generally slenderness limits for column : 60 / 0 < b l as per Cl. 6.2.1.1(f). In addition, for cantilever column b h b l 60 100 2 0 ≤ = . Worked Example 5.1 : a braced column of clear height 8 0 = l m and sectional dimensions 400 = b mm, 550 = h mm with its lower end connected monolithically to a thick cap and the upper end connected monolithically to deep transfer beams in the plane perpendicular to the major direction but beam of size 300(W) by 400(D) in the other direction. By Tables 6.11 and 6.12 of the Code Lower end condition in both directions : 1 Upper end condition about the major axis : 1 Upper end condition about the minor axis : 2 For bending about the major axis : (1 – 1) 75 . 0 = x β , 6 8 75 . 0 = × = ex l 15 91 . 10 550 / < = ex l . ∴ a short column. For bending about the minor axis : (1 – 2) 8 . 0 = y β , 53 4 . 6 8 8 . 0 = × = ey l 15 16 400 / > = ey l ∴ a slender column. 60 16 400 / < = ey l , O.K. For a slender column, an additional “deflection induced moment” add M will be required to be incorporated in design, as in addition to the working moment. 5.2 Design Moments and Axial Loads on Columns 5.2.1 Determination of Design moments and Axial Loads by sub-frame Analysis Generally design moments, axial loads and shear forces on columns are that obtained from structural analysis. In the absence of rigorous analysis, (i) design axial load may be obtained by the simple tributary area method with beams considered to be simply supported on the column; and (ii) moment may be obtained by simplified sub-frame analysis as follows : Worked Example 5.2 (Re Column C1 in Plan shown in Figure 5.2) Design Data : Slab thickness : 150 mm Finish Load : 1.5 kN/m 2 Live Load : 5 kN/m 2 Beam size : 550(D) × 400(W) Upper Column height : 3 m Lower Column Height : 4 m 1.0G k 1.4G k +1.6Q 2 1 5 . 0 5 . 0 b b u L u es u K K K K K M M + + + = 2 1 5 . 0 5 . 0 b b u L L es L K K K K K M M + + + = L K u K 2 b K 1 b K b K L K u K 1.4G k +1.6Q b u L u e u K K K K M M 5 . 0 + + = b u L L e L K K K K M M 5 . 0 + + = Symbols: e M : Beam Fixed End Moment. u K : Upper Column Stiffness es M : Total out of balance Beam Fixed End Moment. L K : Upper Column Stiffness u M : Upper Column Design Moment 1 b K : Beam 1 Stiffness L M : Upper Column Design Moment 2 b K : Beam 2 Stiffness Figure 5.1 – Diagrammatic illustration of determination of column design moments by Simplified Sub-frame Analysis 54 Column size : 400(W) × 600(L) Column Load from floors above D.L. 443 kN L.L. 129 kN Design for Column C1 beneath the floor Check for slenderness : End conditions of the column about the major axis are 1 at the lower end and 2 at the upper end whilst the end conditions about the minor axis are 1 at the lower end and 2 at the upper end. The clear height between restraints of the lower level column is 3450 550 4000 = − . So the effective heights of the column about the major and minor axes are 59 . 2 45 . 3 75 . 0 = × m and 76 . 2 45 . 3 8 . 0 = × m. So the slender ratios about the major and minor axes are 15 3 . 4 600 2590 < = and 15 9 . 6 400 2760 < = . Thus the column is not slender is both directions. Loads : Slab: D.L. O.W. 6 . 3 24 15 . 0 = × kN/m 2 Fin. 1.5 kN/m 2 5.1 kN/m 2 L.L. 5.0 kN/m 2 Beam B1 D.L. O.W. 12 . 21 4 24 55 . 0 4 . 0 = × × × kN End shear of B1 on C1 is D.L. 56 . 10 2 12 . 21 = ÷ kN B4 B3 B2 B1 5m C1 4m 3m 3m Figure 5.2 – Plan for illustration for determination of design axial load and moment on column by the Simplified Sub-frame Method 55 Beam B3 D.L. O.W. ( ) 84 . 3 24 15 . 0 55 . 0 4 . 0 = × − × kN/m Slab 85 . 17 5 . 3 1 . 5 = × kN/m 21.69 kN/m L.L. Slab 5 . 17 5 . 3 0 . 5 = × kN/m End shear of B3 on C1 D.L. 23 . 54 2 5 69 . 21 = ÷ × kN L.L. 75 . 43 2 5 50 . 17 = ÷ × kN Beam B4 D.L. O.W. ( ) 84 . 3 24 15 . 0 55 . 0 4 . 0 = × − × kN/m Slab 3 . 15 3 1 . 5 = × kN/m 19.14 kN/m L.L. Slab 0 . 15 3 0 . 5 = × kN/m End shear of B4 on B2 D.L. 85 . 47 2 5 14 . 19 = ÷ × kN L.L. 50 . 37 2 5 00 . 15 = ÷ × kN Beam B2 D.L. O.W. 68 . 31 6 24 55 . 0 4 . 0 = × × × kN B4 47.85 kN 79.53 kN L.L. B4 37.50 kN End shear of B2 on C1 , D.L. 77 . 39 2 53 . 79 = ÷ kN L.L. 75 . 18 2 5 . 37 = ÷ kN Total D.L. on C1 O.W. 12 . 21 4 24 55 . 0 4 . 0 = × × × kN B1 + B2 +B3 56 . 104 23 . 54 77 . 39 56 . 10 = + + kN Floor above 343.00 kN Sum 568.68 kN Total L.L. on C1 B1 + B2 +B3 5 . 62 75 . 43 75 . 18 0 = + + kN Floor above 129.00 kN Sum 191.50 kN So the factored axial load on the lower column is 55 . 962 5 . 191 6 . 1 68 . 468 4 . 1 = × + × kN Factored fixed end moment bending about the major axis (by Beam B3 alone): ( ) 6 . 121 5 5 . 17 6 . 1 69 . 21 4 . 1 12 1 2 = × × + × × = e M kNm Factored fixed end moment bending about the minor axis by Beam B2: 56 24 . 95 6 5 . 37 8 1 6 . 1 6 85 . 47 8 1 68 . 31 12 1 4 . 1 2 = × | . | \ | × × + × | . | \ | × + × × = eb M kNm Factored fixed end moment bending about the minor axis by Beam B1: 04 . 7 4 12 . 21 12 1 0 . 1 2 = × | . | \ | × × = eb M kNm So the unbalanced fixed moment bending about the minor axis is 2 . 88 04 . 7 24 . 95 = − kNm The moment of inertia of the column section about the major and minor axes are 0072 . 0 12 6 . 0 4 . 0 3 = × = cx I m 4 , 0032 . 0 12 4 . 0 6 . 0 3 = × = cy I m 4 The stiffnesses of the upper and lower columns about the major axis are : E E L EI K u cx ux 0096 . 0 3 0072 . 0 4 4 = × = = E E L EI K L cx Lx 0072 . 0 4 0072 . 0 4 4 = × = = The stiffnesses of the upper and lower columns about the minor axis are : E E L EI K u cy uy 004267 . 0 3 0032 . 0 4 4 = × = = E E L EI K L cy Ly 0032 . 0 4 0032 . 0 4 4 = × = = The moment of inertia of the beams B1, B2 and B3 are 005546 . 0 12 55 . 0 4 . 0 3 = × m 4 The stiffness of the beams B1, B2 and B3 are respectively E E 005546 . 0 4 05546 . 0 4 = × ; E E 003697 . 0 6 05546 . 0 4 = × ; and E E 004437 . 0 5 05546 . 0 4 = × Distributed moment on the lower column about the major axis is E E E E K K K K M M b Lx ux Lx ex ux 004437 . 0 5 . 0 0096 . 0 0072 . 0 0072 . 0 6 . 121 5 . 0 3 × + + × = + + = = 46.03 kNm Distributed moment on the lower column about the minor axis is ( ) ( )E E E E K K K K K M M b b Ly uy Ly ey uy 003697 . 0 005546 . 0 5 . 0 004267 . 0 0032 . 0 0032 . 0 2 . 88 5 . 0 2 1 + × + + × = + + + = 57 = 23.35 kNm So the lower column should be checked for the factored axial load of 962.55kN, factored moment of 46.03 kNm about the major axis and factored moment of 23.35 kNm about the minor axis. 5.2.2 Minimum Eccentricity A column section should be designed for the minimum eccentricity equal to the lesser of 20 mm and 0.05 times the overall dimension of the column in the plane of bending under consideration. Taking the example in 5.2.1, the minimum eccentricity about the major axis is 30 mm as 20 30 600 05 . 0 > = × mm and that of the minor axis is 20 400 05 . 0 = × mm. So the minimum eccentric moment to be designed for about the major and minor axes are 88 . 28 03 . 0 55 . 962 = × kNm and 25 . 19 02 . 0 55 . 962 = × kNm. As they are both less than the design moment of 46.03 kNm and 23.35 kNm, they can be ignored. 5.2.3 Check for Slenderness In addition to the factored load and moment obtained in 5.2.1, it is required by Cl. 6.2.1.2(b) to design for an additional moment add M if the column is found to be slender by Cl. 6.2.1.1. The arrival of add M is an eccentric moment created by the working axial load N multiplied by a pre-determined lateral deflection u a in the column as indicated by the following equations of the Code. u add Na M = (Ceqn 6.52) Kh a a u β = (Ceqn 6.48) 2 2000 1 | . | \ | = b l e a β (Ceqn 6.51) 1 ≤ − − = bal uz uz N N N N K (conservatively taken as 1) (Ceqn 6.49) sc y nc cu uz A f A f N 87 . 0 45 . 0 + = (Ceqn 6.50) bd f N cu bal 25 . 0 = Final design moment t M will be the greatest of (1) 2 M , the greater initial end moment due to design ultimate load; 58 (2) add i M M + where 2 2 1 4 . 0 6 . 0 4 . 0 M M M M i ≥ + = (with 2 M as positive and 1 M negative.) (3) 2 / 1 add M M + in which 1 M is the smaller initial end moment due to design ultimate load. (4) min e N × (discussed in 5.2.2) where the relationship between 1 M , 2 M , add M and the arrival of the critical combination of design moments due to add M are illustrated in Figure 6.16 of the Code. In addition to the above, the followings should be noted in the application of t M as the enveloping moment of the 4 cases described in the previous paragraph : (i) In case of biaxial bending (moment significant in two directions), t M should be applied separately for the major and minor directions with b replacing h for bending in the minor direction; (ii) In case of uniaxial bending about the major axis where 20 / ≤ h l e and b h 3 < , t M should be applied only in the major axis; (iii) In case of uniaxial bending about the major axis only where either 20 / ≤ h l e or b h 3 < is not satisfied, the column should be designed as biaxially bent, with zero i M in the minor axis; (iv) In case of uniaxial bending about the minor axis, add M obviously be applied only in the minor axis only. Worked Example 5.3 : A slender braced column of grade 35, cross sections 400 = b , 500 = h 8 = = ey ex l l m, 1500 = N kN (i) About the major axis only, the greater and smaller bending moments due to ultimate load are 153 kNm and 96 kNm respectively. As 20 16 / ≤ = h l ex ; 1200 3 500 = < = b h So needs only checked for bending in the major axis. 96 1 = M kNm; 153 2 = M kNm 2 M 1 M Illustration of 1 M and 2 M in column due to ultimate load 59 Taken 1 = K in the initial trial. 2 . 0 400 8000 2000 1 2000 1 2 2 = | . | \ | = | . | \ | = b l e a β 1 . 0 5 . 0 1 2 . 0 = × × = = Kh a a u β 150 1 . 0 1500 = × = = u add Na M The design moment will be the greatest of : (1) 153 2 = M (2) ( ) 4 . 53 153 6 . 0 96 4 . 0 6 . 0 4 . 0 2 1 = × + − × = + = M M M i 2 . 73 4 . 0 2 = < M (3) 171 2 / 150 96 2 / 1 = + = + add M M (4) 5 . 37 025 . 0 1500 min = × = ×e N as 20 25 500 05 . 0 min > = × = e So the greatest design moment is case (3) 171 2 / 1 = + add M M Thus the section need only be checked for uniaxial bending with 1500 = N kN and 171 = x M kNm bending about the major axis. (ii) Biaxial Bending, if in addition to the moments bending about the major axis as in (i), there are also moments of 72 1 = M kNm; 127 2 = M kNm bending in the minor axis. By Cl. 6.1.2.3(f), add M about the major axis will be revised as follows : Bending about the major axis : 128 . 0 500 8000 2000 1 2000 1 2 2 = | . | \ | = | . | \ | = h l e a β 064 . 0 5 . 0 1 128 . 0 = × × = = Kh a a u β 96 064 . 0 1500 = × = = u add Na M kNm. Thus item (3) which is 144 2 / 96 96 2 / 1 = + = + add M M is reduced and case (1) where 153 2 = M controls. Bending about the minor axis : 72 1 = M kNm; 127 2 = M kNm Taken 1 = K in the initial trial. 2 . 0 400 8000 2000 1 2000 1 2 2 = | . | \ | = | . | \ | = b l e a β 60 1 . 0 5 . 0 1 2 . 0 = × × = = Kh a a u β 150 1 . 0 1500 = × = = u add Na M The design moment will be the greatest of : (1) 127 2 = M (2) ( ) 4 . 47 127 6 . 0 72 4 . 0 6 . 0 4 . 0 2 1 = × + − × = + = M M M i 8 . 50 4 . 0 2 = < M (3) 147 2 / 150 72 2 / 1 = + = + add M M (4) 30 02 . 0 1500 min = × = ×e N as 20 400 05 . 0 min = × = e So the greatest design moment is case (3) 147 2 / 1 = + add M M Thus the ultimate design moment about the major axis is 153 kNm and that about the minor axis is 147 kNm. 5.3 Sectional Design Generally the sectional design of column utilizes both the strengths of concrete and steel in the column section in accordance with stress strain relationship of concrete and steel in Figure 3.8 and 3.9 of the Code respectively. Alternatively, the simplified stress block for concrete in Figure 6.1 can also be used. 5.3.1 Design for Axial Load only (Ceqn 6.55) of the Code can be used which is y sc cu f A f N 75 . 0 4 . 0 + = . The equation is particularly useful for a column which cannot be subject to significant moments in such case as the column supporting a rigid structure or very deep beams. There is a reduction of approximately 10% in the axial load carrying capacity as compared with the normal value of y sc cu f A f 87 . 0 45 . 0 + which accounts for the eccentricity of h 05 . 0 . Furthermore, (Ceqn 6.56) reading y sc cu f A f N 67 . 0 35 . 0 + = which is applicable to columns supporting an approximately symmetrical arrangement of beams where (i) beams are designed for u.d.l.; and (ii) the beam spans do 61 not differ by more than 15% of the longer. The further reduction is to account for extra moment arising from asymmetrical loading (by live load generally). 5.3.2 Design for Axial Load and Biaxial Bending : The general section design of a column is account for the axial loads and biaxial bending moments acting on the section. Nevertheless, the Code has reduced biaxial bending into uniaxial bending in design. The procedure for determination of the design moment, either ' x M or ' y M bending about the major or minor axes is as follows : (i) Determine ' b and ' h as defined by the diagram. In case there are more than one row of bars, ' b and ' h can be measured to the centre of the group of bars. (ii) Compare ' h M x and ' b M y . If ' ' b M h M y x ≥ , use y x x M b h M M ' ' ' β + = If ' ' b M h M y x < , use x y y M h b M M ' ' ' β + = where β is to be determined from Table 6.14 of the Code under the pre-determined bh N ; (iii) The ' x M or ' y M will be used for design by treating the section as either (a) resisting axial load N and moment ' x M bending about major axis; or (b) resisting axial load N and moment ' y M bending about minor axis as appropriate. 5.3.3 Concrete Stress Strain Curve and Design Charts The stress strain curve for column section design is in accordance with Figure 3.8 of the Code. It should be noted that the “Concrete Code Handbook” has recommended to shift 0 ε to ( ) c m cu E f γ / 34 . 1 in para. 3.1.4. With the recommended shifting, the detailed design formulae and design charts have been formulated and enclosed in Appendix F. Apart from the derivation for the normal 4-bar column, the derivation in Appendix F has also included steel ' h h ' b b 62 reinforcements uniformly distributed along the side of the column idealized as continuum of reinforcements with symbol sh A . The new inclusion has allowed more accurate determination of load carrying capacity of column with many bars along the side as illustrated in Figure 5.3 which is particularly useful for columns of large cross sections. The user can still choose to lump the side reinforcements into the 4 corner bars, with correction to the effective depth as in conventional design by setting 0 = sh A in the derived formulae. Figure 5.4 shows the difference between the 2 idealization. It can be seen that the continuum idealization is more economical generally except at the peak moment portion where the 4 bar column idealization shows slight under-design. Comparison of Load Carrying Capacities of Rectangular Shear Walls with Uniform Vertical Reinforcements Idealized as 4 bar column with d/h = 0.75 and Continuum to the Structural Use of Concrete 2004 Concrete Grade 35 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel as 4 bar column 0.4% steel as continuum 1% steel as 4 bar column 1% steel as continuum 4% steel as 4 bar column 4% steel as continuum 8% steel as 4 bar column 8% steel as continuum idealized as continuum of steel “strip” with area equivalent to the row of bars Figure 5.3 – Idealization of steel reinforcements in large column Figure 5.4 – Comparison between Continuum and 4-bar column Idealization 63 Worked Example 5.4 : Consider a column of sectional size 400 = b mm, 600 = h mm under an axial load 4000 = N kN, 250 = x M kNm, 150 = y M kNm, Assume a 4-bar column, 540 20 40 600 ' = − − = h mm; 340 20 40 400 ' = − − = b mm; 476 . 0 600 400 35 4000000 = × × = bh f N cu ; 446 . 0 = β from Table 6.14; 441 . 0 ' 463 . 0 ' = > = b M h M y x ; 356 150 340 540 446 . 0 250 ' ' ' = × × + = + = ∴ y x x M b h M M β kNm 67 . 16 = bh N ; 47 . 2 600 400 10 356 2 6 2 = × × = bh M ; 9 . 0 600 540 = = h d Use Chart F-10 in Appendix F, 1.8% steel is approximated which amounts to 4320 018 . 0 600 400 = × × mm 2 , or 6-T32 (Steel provided is 4826mm 2 ) Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35, 4-bar column, d/h = 0.9 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel The section design is as shown in Figure 5.5, with two additional T20 bars to avoid wide bar spacing. 64 5.3.4 Alternatively, the design of reinforcements can be based on formulae derived in Appendix F. However, as the algebraic manipulations are complicated (may involve solution of 4 th polynomial equations) and cases are many, the approach may only be practical by computer methods. Nevertheless, spread sheets have been prepared and 2 samples are enclosed at the end of Appendix F. 5.3.5 Direct sectional analysis to Biaxial Bending without the necessity of converting the biaxial bending problem into a uniaxial bending problem. Though the Code has provision for converting the biaxial bending problem into a uniaxial bending problem by (i) searching for the controlling bending axis and (ii) aggravate the moments about the controlling bending axis as appropriate to account for effects of bending in the non-controlling axis, a designer can actually solve the biaxial bending problem by locating the orientation of the extent of the neutral axis (which generally does not align with the resultant moment except for circular section) by balancing axial load and the bending in two directions. Theoretically, by balancing axial load and the 2 bending moments, 3 equations can be obtained for solution of the neutral axis orientation, neutral axis depth and the required reinforcement. However, the solution process will be very tedious and not possible for irregular section. Practical approach by computer method may be based on trial and error method. If the reinforcements are pre-determined, various combinations of the neutral axis depth and orientation will be presumed for calculation of axial load and moment carrying capacities of the section. The section will be considered adequate if a combination of neutral axis depth and orientation produce axial load and moments of resistance in excess of the applied loads and moments. Such approach can be applied to section of irregular shape. T32 Figure 5.5 – Design Chart and Worked Re-bar details for Example 5.4 Extra T20 65 Figure 5.6 illustrates the method of solution. 5.4 Detailing Requirements for longitudinal bars in columns (generally by Cl. 9.5 and Cl. 9.9.2.1(a) of the Code, the ductility requirements are marked with “D”) (i) Minimum longitudinal steel based on gross area of a column is 0.8%; (ii) Maximum longitudinal steel based on gross area of a column is 4% except at lap which can be increased to 5.2% (D); (iii) Longitudinal bars should have a diameter > 12 mm; (iv) The minimum number of longitudinal bars should be four in rectangular columns and six in circular columns. In columns having a polygonal cross-section, at least one bar be placed at each corner; (v) In any row of bars, the smallest bar diameter used shall not be less than 2/3 of the largest bar diameter used (D); (vi) At laps, the sum of reinforcement sizes in a particular layer should not exceed 40% of the breadth at that section; (vii) Where the columns bars pass through the beams at column beam joints, the bar diameter should be limited to y cu f f h 8 . 0 2 . 3 25 . 1 × ≤ φ as per (Ceqn 9.7) of the Code where h is the beam depth. For grade 35 concrete and based on high yield bar, the limiting bar diameter is simply strain profile across section concrete stress profile Strain profile on section Neutral axis The total sectional resistance of the section under the stress strain profile resists the applied axial loads and moments Figure 5.6 – General Biaxial Bending on irregular section 66 h 046 . 0 ≤ φ , i.e. if beam depth is 600 mm, 6 . 27 ≤ φ implying maximum bar size is 25 mm. The factor 1.25 can be removed if the column is not intended to form a plastic hinge (D); (viii) Where column bars terminate in a beam-column joints or joints between columns and foundation members where a plastic hinge in the column could be formed, the anchorage of the column bars into the joint region should commence at 1/2 of the depth of the beam/foundation member or 8 times the bar diameter from the face at which the bars enter the beam or the foundation member (D) and the bend should be “inwards”; (ix) No column bar should be terminated in a joint area without a horizontal 90 o standard hook or an equivalent device as near as practically possible to the far side of the beam and not closer than 3/4 of the depth of the beam to the face of entry (D); (x) Minimum clear spacing of bars should be the minimum of bar diameter, 20 mm and aggregate size + 5 mm; 5.5 Detailing Requirements for transverse reinforcements in columns (generally by Cl. 9.5.2 and Cl. 9.9.2.1(a) of the Code, the ductility requirements are marked with “D”. Items (i) to (iv) below are requirements for columns not within “critical regions” as defined in item (v) : (i) Diameter of transverse reinforcements > the greater of 6 mm and 1/4 of longitudinal bar diameter; (ii) The spacing of transverse reinforcement shall not exceed 12 times the diameter of the smallest longitudinal bar; (iii) For rectangular or polygonal columns, every corner bar and each alternate bar (or bundle) shall be laterally supported by a link passing >0.75D Beam or foundation element >0.5D or 8ø D Anchorage should commence at this point Figure 5.7 – Column Bar anchorage requirements as per 5.4 (viii) and (ix) 67 around the bar and having an included angle < 135 o . No bar within a compression zone be further than 150 mm from a restrained bar. Links be adequately anchored by hooks through angles < 135 o ; (iv) For circular columns, loops or spiral reinforcement satisfying (i) to (iii) should be provided. Loops (circular links) should be anchored with a mechanical connection or a welded lap by terminating each end with a 135 o hook bent around a longitudinal bar after overlapping the other end of the loop by a minimum length. Spiral should be anchored either by welding to the previous turn or by terminating each end with a 135 o hook bent around a longitudinal bar and at not more than 25 mm from the previous turn. Loops and spirals should not be anchored by straight lapping, which causes spalling of the concrete cover; (v) Transverse reinforcements in “Critical regions” within columns as defined in Figure 5.8 (Re Cl. 9.9.2.2 of the Code) shall have additional requirements as for the maximum spacing of transverse reinforcements as : (a) 1/4 of the least lateral column dimension (D) in case of rectangular or polygonal column and 1/4 of the diameter in case of a circular column; (b) 6 times the diameter of the longitudinal bar to be restrained (D); (c) 200 mm. 68 b h m max xM max M Normal transverse reinforceme H, Critical regions with enhanced transverse reinforcements H, Critical regions with enhanced transverse reinforcements Height of “critical region” , h, depends on N/A g f cu ratio : (a) 0<N/A g f cu ≤ 0.1, x = 0.85 H > h m H > h or D (b) 0.1 <N/A g f cu ≤ 0.3, x = 0.75 H > h m H > 1.5h or 1.5D (c) 0.3 <N/A g f cu ≤ 0.6, x = 0.65 H > h m H > 2h or 2D D h Figure 5.8 – “Critical Regions (Potential Plastic Hinge Regions)” in Columns 69 6.0 Column-Beam Joints 6.1 General The design criteria of a column-beam joint comprise (i) performance not inferior than the adjoining members at serviceability limit state; and (ii) sufficient strength to resist the worst load combination at ultimate limit state. To be specific, the aim of design comprise minimization of (a) the risk of concrete cracking and (b) spalling near the beam-column interface, and (c) checking provisions against diagonal crushing or splitting of the joint and where necessary, providing vertical and horizontal shear links within the joint and confinement to the longitudinal reinforcements of the columns adjacent to the joint. 6.2 The phenomenon of “diagonal splitting” of joint Diagonal crushing or splitting of column-beam joints is resulted from “shears” and unbalancing moment acting on the joints as illustrated in Figure 6.1(a) and 6.1(b) indicating typical loading arrangement acting on the joint. Figure 6.1(a) shows a large horizontal shear from the right, reversing the hogging moment on the beam on the left side of the joint. BR C BL C BL T BR T V c1 V c2 Potential failure surface (tension) V b1 V b2 hogging moment in beam sagging moment in beam Figure 6.1(a) – Phenomenon of Diagonal Joint Splitting by moments of opposite signs on both sides of joint 70 The unbalanced forces due to unbalanced flexural stresses by the adjoining beams on both sides of the joint tend “tear” the joint off with a potential tension failure surface, producing “diagonal splitting”. Though generally there are shears in the column usually tend to act oppositely, the effects are usually less dominant than that of moment and therefore can be ignored. Reinforcements in form of links may therefore be necessary if the concrete alone is considered inadequate to resist the diagonal splitting. 6.3 Design procedures : (i) Work out the total nominal horizontal shear force across the joint jh V in X and Y directions generally. jh V should be worked out by considering equilibrium of the joint. Generally the load due to unbalanced forces by the beams on both sides of the joints need be considered. The horizontal shears by the columns can generally be ignored as they are comparatively small and counteract the effects by bending. By the load capacity concept, the reinforcing bars in the beam will be assumed to have steel stress equal to 125% yield strength of steel if such assumption will lead to the most adverse conditions. Figure 6.2(a) and 6.2(b) indicates how to determine jh V by this criterion. BL BR T T > BR C BL C BR T BL T V c1 V c2 Potential failure surface (tension) V b1 V b2 hogging moment in beam hogging moment in beam Figure 6.1(b) – Phenomenon of Diagonal Joint Splitting by moments of same sign on both sides of joint 71 (ii) With the jh V determined, the nominal shear stress is determined by (Ceqn 6.71) in the Code. c j jh jh h b V v = where c h is the overall depth of the column in the direction of shear c j b b = or c w j h b b 5 . 0 + = whichever is the smaller when w c b b ≥ w j b b = or c c j h b b 5 . 0 + = whichever is the smaller when w c b b ≥ where c b is the width of column and w b is the width of the beam. The vector sum of jh v for both the X and Y directions should be worked out and check that it does not exceed cu f 25 . 0 . cu jhy jhx f v v 25 . 0 2 2 ≤ + sL y BL A f T 25 . 1 = sagging moment in beam hogging moment in beam BR C BL BL T C = BL T BR T Potential failure surface (tension) sR y BR A f T 25 . 1 = ( ) sR sL y BR BL jh A A f T T V + = + = 25 . 1 L z hogging moment in beam M L hogging moment in beam BR C BL C BL T sR y BR A f T 25 . 1 = Potential failure surface (tension) BL BR T T > L y L BL z f M T 87 . 0 = z f M A f T T V y L sR y BR BL jh 87 . 0 25 . 1 + = + = Figure 6.2(a) – Calculation of jh V , opposite sign beam moments on both sides Figure 6.2(b) – Calculation of jh V , same sign beam moments on both sides 72 (iii) Horizontal reinforcements based on (Ceqn 6.72) reading | | . | \ | − = cu g j yh jh jh f A N C f V A * 5 . 0 87 . 0 * should be worked out in both directions and be provided in the joint as horizontal links. It may be more convenient to use close links which can serve as horizontal reinforcements in both directions and confinements as required by (v). If the numerical values arrived by (Ceqn 6.72) becomes negative, no horizontal shear reinforcements will be required; (iv) Similarly vertical reinforcements based on (Ceqn 6.73) reading ( ) yv j jh c b jh f N C V h h A 87 . 0 * * / 4 . 0 − = should be worked out in both directions and be provided in the joint as horizontal links. It may be more convenient to use close links which can serve as horizontal reinforcements in both directions. If the numerical values arrived by (Ceqn 6.73) becomes negative, no horizontal shear reinforcements will be required; (v) Notwithstanding the provisions arrived at in (iii) for the horizontal reinforcements, confinements in form of closed links within the joint should be provided as : (d) Not less than that in the column shaft as required by Cl. 9.5.2 of the Code, i.e. Section 5.5 (i) to (iv) of this Manual if the joint has a free face in one of its four faces; (e) Reduced by half to that provisions required in (a) if the joint is connected to beams in all its 4 faces; (f) Link spacing φ 10 ≤ (diameter of smallest column bar) and 200 mm. 6.4 Worked Example 6.1: Consider the column beam joints with columns and beams adjoining as indicated in Figure 6.3 in the X-direction and Y-directions. Concrete grade is 40. 73 (i) Check nominal shear stress : X-direction The moments on the left and right beams are of opposite sign. So Figure 6.2(a) is applicable. The top steel provided on the left beam is 3T32, with 2413 1 = st A mm 2 whilst the bottom steel provided on the right beam is 4T20 with 1257 1 = st A mm 2 . 6000 * = N kN 800 6000 * = N kN Beam moment 300 kNm (hogging) Beam moment 550 kNm (hogging) Column size 900 × 800 Beam size 700 × 500 (effective depth 630) Beam size 700 × 500 (effective depth 630) Column size 900 × 800 Figure 6.3 – Design Example for Column Beam Joint 900 X-direction 900 Beam moment 300 kNm (hogging) Beam moment 550 kNm (hogging) Beam size 700 × 500 (effective depth 630) Column size 900 × 800 800 Y-direction Beam size 700 × 500 (effective depth 630) 74 1445 2413 460 25 . 1 = × × = BR T kN; BR BR T C = 723 1257 460 25 . 1 = × × = BL T kN; BL BL T C = So the total shear is 2168 = + = + = BL BR BR BL jx C T C T V kN In the X-direction 800 = c h As 500 900 = > = w c b b ,the effective joint width is the smaller of 900 = c b and 1300 800 5 . 0 900 5 . 0 = × + = + c w h b , so 900 = j b So 01 . 3 800 900 2168000 = × = = c j jx jx h b V v MPa Y-direction The moments on the left and right beams are of equal sign. So Figure 6.2(b) is applicable. As the moment on the right beam is higher, the potential plastic hinge will be formed on the right beam. 1445 2413 460 25 . 1 = × × = BR T kN; BR BR T C = BL T is to be determined by conventional beam design method. 512 . 1 2 = bd M MPa, 948 . 0 = d z , 49 . 1254 = sL A mm 2 ; 05 . 502 87 . 0 = = sL y BL A f T kN So 943 502 1445 = − = jy V kN In the Y-direction 900 = c h , similarly, 900 = j b , L z BR C BL C BL T BL BR T T > BR C BL C BL T BR T 75 So 31 . 1 800 900 943000 = × = = c j jy jy h b V v MPa. Combining X and Y direction, the total shear stress is 10 25 . 0 28 . 3 31 . 1 01 . 3 2 2 2 2 = < = + = + = cu jy jx jh f v v v MPa. (ii) To calculate the horizontal joint reinforcement by (Ceqn 6.72 of the Code), reading | | . | \ | − = cu g j yh jh jh f A N C f V A * 5 . 0 87 . 0 * where jy jx jh j V V V C + = X-direction 697 . 0 943 2168 2168 = + = + = jy jx jx jx V V V C | . | \ | × × × − × = | | . | \ | − = 40 800 900 5000000 697 . 0 5 . 0 460 87 . 0 2168000 * 5 . 0 87 . 0 * cu g jx yh jhx jhx f A N C f V A 1922 = mm 2 Y-direction 303 . 0 943 2168 943 = + = + = jy jx jy jx V V V y C | . | \ | × × × − × = | | . | \ | − = 40 800 900 5000000 303 . 0 5 . 0 460 87 . 0 943000 * 5 . 0 87 . 0 * cu g jx yh jhx jhx f A N C f V A 1029 = mm 2 Use 5T16 close stirrups (Area provided = 2011 mm 2 ) which can adequately cover shear reinforcements in both directions (iii) To calculate the vertical joint reinforcement by (Ceqn 6.73 of the Code), reading ( ) yh j jh c b jv f N C V h h A 87 . 0 * * / 4 . 0 − = X-direction ( ) ( ) 460 87 . 0 6000 697 . 0 2168 900 / 700 4 . 0 87 . 0 * * / 4 . 0 × × − = − = yh j jh c b jvx f N C V h h A 8 . 8 − = . So no vertical shear reinforcement is required. Y-direction 76 ( ) ( ) 460 87 . 0 6000 303 . 0 943 800 / 700 4 . 0 87 . 0 * * / 4 . 0 × × − = − = yh j jh c b jvy f N C V h h A 8 . 3 − = . So no vertical shear reinforcement is required. (iv) The provision of outermost closed stirrups in the column shaft is 10 T at 200 which is in excess of the confinement requirements. So no additional confinement requirement. Closed links 5T16 Figure 6.4 – Details of Column Beam Joint Detail 77 7.0 Walls 7.1 Design Generally 7.1.1 Similar to column by design to resist axial loads and biaxial moments. 7.1.2 The design ultimate axial force may be calculated on the assumption that the beams and slabs transmitting force to it are simply supported. (Re Cl. 6.2.2.2(a) and Cl. 6.2.2.3(a) of the Code). 7.1.3 Minimum eccentricity for transverse moment design is the lesser of 20 mm or 20 / h , as similar to columns. 7.2 Categorization of Walls Walls can be categorized into (i) slender walls; (ii) stocky walls; (iii) reinforced concrete walls; and (iv) plain walls. 7.3 Slender Wall Section Design 7.3.1 Determination of effective height e l (of minor axis generally which controls) – (i) in case of monolithic construction, same as that for column; and (ii) in case of simply supported construction, same as that for plain wall. 7.3.2 Limits of slender ratio (Re Table 6.15 of the Code) – (i) 40 for braced wall with reinforcements < 1%; (ii) 45 for braced wall with reinforcements ≥ 1%; (iii) 30 for unbraced wall. 7.3.3 Other than 7.3.1 and 7.3.2, reinforced concrete design is similar to that of columns. 7.4 Stocky Wall 7.4.1 As similar to column, stocky walls are walls with slenderness ratio < 15 for braced walls and slenderness ratio < 10 for unbraced walls; 78 7.4.2 Stocky reinforced wall may be designed for axial load w n only by (Ceqn 6.59) of the Code provided that the walls support approximately symmetrical arrangement of slabs with uniformly distributed loads and the spans on either side do not differ by more than 15%; sc y c cu w A f A f n 67 . 0 35 . 0 + ≤ 7.4.3 Other than 7.4.2 and the design for deflection induced moment add M , design of stocky wall is similar to slender walls. 7.5 Reinforced Concrete Walls design is similar to that of columns with categorization into slender walls and stocky walls. 7.6 Plain Wall - Plain wall are walls the design of which is without consideration of the presence of the reinforcements. 7.6.1 Effective height of unbraced plain wall, where 0 l is the clear height of the wall between support, is determined by : (a) 0 5 . 1 l l e = when it is supporting a floor slab spanning at right angles to it; (b) 0 0 . 2 l l e = for other cases. Effective height ratio for braced plain wall is determined by (a) 0 75 . 0 l l e = when the two end supports restraint movements and rotations; (b) 0 0 . 2 l l e = when one end support restraint movements and rotations and the other is free; (c) 0 l l e = when the two end supports restraint movements only; (b) 0 5 . 2 l l e = when one end support restraint movements only and the other is free; 7.6.2 The load carrying capacity of plain wall will depend on concrete strength alone. For detailed design criteria including check for concentrated load, shear, load carrying capacities etc, refer to Cl. 6.2.2.3 of the Code. 7.7 Sectional Design The sectional design of wall section is similar to that of column by utilizing stress strain relationship of concrete and steel as indicated in Figure 3.8 and 3.9 of the Code. (Alternatively, the simplified stress block of concrete as 79 indicated in Figure 6.1 can also be used.) 0 ε should be shifted as discussed in Section 5.3.3. Conventionally, walls with uniformly distributed reinforcements along its length can be treated as if the steel bars on each side of the centroidal axis are lumped into two bars each carrying half of the steel areas as shown in Figure 7.1 and design is carried out as if it is a 4 bar column. Nevertheless, it is suggested in the Manual that the reinforcements can be idealized as a continuum (also as shown in Figure 7.1) which is considered as a more realistic idealization. Derivation of the formulae for the design with reinforcements idealized as continuum is contained in Appendix G, together with design charts also enclosed in the Appendix. 7.8 Worked Example 7.1 Wall Section : thickness : 200 mm, plan length : 2000 mm; Wall Height : 3.6 m, Concrete grade : 35 Connection conditions at both ends of the wall : connected monolithically with floor structures shallower than the wall thickness. Check for slenderness d = 0.75h Shear Wall Section Current idealization based on 4-bar column design chart Proposed idealization with reinforcing bars as continuum with areas equal to the bars Figure 7-1 – Idealization of Reinforcing bars in shear wall 80 Only necessary about the minor axis. End conditions are 2 for both ends, 85 . 0 = β (by Table 6.11 of the Code); 06 . 3 6 . 3 85 . 0 = × = e l m Taken 1 = K (Ceqn 6.49 of the Code) in the initial trial. 117 . 0 200 3060 2000 1 2000 1 2 2 = | . | \ | = | . | \ | = b l e a β 0234 . 0 2 . 0 1 117 . 0 = × × = = Kh a a u β (i) Axial Load : 7200 = N kN, 110 = x M kNm at top and 110kNm at bottom , 25 = y M kNm at top and 24 kNm Final design moment t M about the major and minor axes will be the greatest of : (1) 2 M , the greater initial end moment due to design ultimate load; (2) add i M M + where 2 2 1 4 . 0 6 . 0 4 . 0 M M M M i ≥ + = (with 2 M as positive and 1 M negative.) (3) 2 / 1 add M M + in which 1 M is the smaller initial end moment due to design ultimate load. (4) min e N × (discussed in 5.2.2) For bending about the major axis, 15 53 . 1 2000 / 3060 / < = = h l e , so 0 = add M , x M will be the greatest of (1) 110 2 = M ; (2) 106 0 110 6 . 0 100 4 . 0 = + × + × = + add i M M ; (3) 100 0 100 1 = + = + add M M and (4) 144 02 . 0 7200 min = × = ×e N . So 144 = x M kNm for design. For bending about the minor axis, 15 3 . 15 200 / 3060 / > = = b l e , 117 . 0 200 3060 2000 1 2000 1 2 2 = | . | \ | = | . | \ | = b l e a β 0234 . 0 2 . 0 1 117 . 0 = × × = = Kh a a u β 48 . 168 0234 . 0 7200 = × = = u add Na M kNm, y M will be the greatest of (1) 25 2 = M ; (2) 08 . 193 48 . 168 25 6 . 0 24 4 . 0 = + × + × = + add i M M ; (3) 48 . 192 48 . 168 24 1 = + = + add M M and (4) 144 02 . 0 7200 min = × = ×e N . So 08 . 193 = y M kNm for design. So the factored axial load and moments for design are 81 7200 = N kN; 144 = x M kNm; 08 . 193 = y M kNm 165 10 25 200 ' = − − = b ; 1500 75 . 0 2000 ' = × = h 514 . 0 2000 200 35 7200000 = × × = bh f N cu ; 403 . 0 = β from Table 6.14; 17 . 1 ' 096 . 0 ' = < = b M h M y x ; 46 . 199 144 1500 165 403 . 0 08 . 193 ' ' ' = × × + = + = ∴ x y y M h b M M β kNm 18 = bh N ; 4933 . 2 200 2000 10 46 . 199 2 6 2 = × × = hb M ; 825 . 0 200 165 = = b d Choosing the most appropriate chart (Chart F-7) in Appendix F which is Grade 35 with d/h = 0.8, and plot 18 = bh N and 494 . 2 2 = hb M on it, the estimated steel percentage is 2.5% which is 10000 2000 200 025 . 0 = × × mm 2 or 5000 mm 2 /m Use T20 – 125 (B.F.) Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35, 4-bar column, d/h = 0.8 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel (ii) Axial Load : 7200 = N kN, 1800 = x M kNm at top and 1200kNm at bottom , 25 = y M kNm at top and 24 kNm Final design moment t M about the major and minor axes will be the greatest of : (1) 2 M , the greater initial end moment due to design ultimate load; 82 (2) add i M M + where 2 2 1 4 . 0 6 . 0 4 . 0 M M M M i ≥ + = (with 2 M as positive and 1 M negative.) (3) 2 / 1 add M M + in which 1 M is the smaller initial end moment due to design ultimate load. (4) min e N × (discussed in 5.2.2) For bending about the major axis, 15 53 . 1 2000 / 3060 / < = = h l e , so 0 = add M , x M will be the greatest of (1) 1800 2 = M ; (2) 1560 0 1800 6 . 0 1200 4 . 0 = + × + × = + add i M M ; (3) 1200 0 1200 1 = + = + add M M and (4) 144 02 . 0 7200 min = × = ×e N . So 1800 = x M kNm for design. For bending about the minor axis, 15 3 . 15 200 / 3060 / > = = b l e , 117 . 0 200 3060 2000 1 2000 1 2 2 = | . | \ | = | . | \ | = b l e a β 0234 . 0 2 . 0 1 117 . 0 = × × = = Kh a a u β 48 . 168 0234 . 0 7200 = × = = u add Na M kNm, y M will be the greatest of (1) 25 2 = M ; (2) 08 . 193 48 . 168 25 6 . 0 24 4 . 0 = + × + × = + add i M M ; (3) 48 . 192 48 . 168 24 1 = + = + add M M and (4) 144 02 . 0 7200 min = × = ×e N . So 08 . 193 = y M kNm for design. So the factored axial load and moments for design are 7200 = N kN; 1800 = x M kNm; 08 . 193 = y M kNm 165 10 25 200 ' = − − = b ; 1500 75 . 0 2000 ' = × = h 514 . 0 2000 200 35 7200000 = × × = bh f N cu ; 403 . 0 = β from Table 6.14; 17 . 1 ' 2 . 1 ' = < = b M h M y x ; 6 . 2415 08 . 139 165 1500 403 . 0 1800 ' ' ' = × × + = + = ∴ y x x M b h M M β kNm 18 = bh N ; 0195 . 3 2000 200 10 6 . 2415 2 6 2 = × × = hb M ; Choosing the wall design chart for grade 35 for wall (Chart G-2) in Appendix G which is Grade 35 with d/h = 0.8, and plot 18 = bh N and 83 494 . 2 2 = hb M on it, the estimated steel percentage is 3% which is 12000 2000 200 03 . 0 = × × mm 2 or 6000 mm 2 /m Use T20 – 100 (B.F.) Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004, Concrete Grade 35 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Alternatively, use Column Design Chart F-6 in Appendix F by treating it as column of d/h = 0.75, the estimated steel percentage is 3.1% amounting to 12400 2000 200 031 . 0 = × × mm 2 or 6200 mm 2 /m which is slightly higher. Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35, 4-bar column, d/h = 0.75 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Design can be performed by calculations with the formulae derived in Appendices F and G. However, the calculations are too tedious and cases to try 84 are too many without the use of computer methods. Spread sheets have been devised to tackle the problem and a sample is enclosed in Appendix G. 7.9 Detailing Requirements There are no ductility requirements in Cl. 9.9 of the Code for walls. The detailing requirements are summarized from Cl. 9.6 of the Code : Vertical reinforcements for reinforced concrete walls (i) Minimum steel percentage : 0.4%. When this reinforcement controls the design, half of the steel area be on each side; (ii) Maximum steel percentage : 4%; (iii) Maximum distance between bars : the lesser of 3 times the wall thickness and 400 mm; Horizontal and transverse reinforcements for reinforced concrete walls (i) Minimum horizontal reinforcements : 0.25% for 460 = y f MPa and 0.3% for 250 = y f MPa if the required vertical reinforcement does not exceed 2%; (ii) If the required vertical reinforcement > 2%, links be provided as : (g) to anchor every vertical compression longitudinal bar; (h) no bar be at a distance further than 200 mm from a restrained bar at which a link passes round at included angle ≤ 90 o ; (i) minimum diameter : the greater of 6 mm and 1/4 of the largest compression bar; (j) maximum spacing : twice the wall thickness in both the horizontal and vertical directions. In addition, maximum spacing not to exceed 16 times the vertical bar diameter in the vertical direction; ≤ 200 ≤ 200 bars not required to take compression bars required to take compression Figure 7.2 – Anchorage by links on vertical reinforcements 85 Plain walls If provided, minimum reinforcements : 0.25% for 460 = y f MPa and 0.3% for 250 = y f MPa in both directions generally. 86 8.0 Corbels 8.1 General – A corbel is a short cantilever projection supporting a load-bearing member with dimensions as shown : 8.2 Basis of Design 8.2.1 According to Cl. 6.5.2 of the Code, the basis of design method of a corbel is that it behaves as a “Tie-and-strut” model as illustrated in Figure 8.2. The strut action (compressive) is carried out by concrete and the tensile force at top is carried by the top steel. Concrete ultimate strain 0035 . 0 = ult ε Concrete stress block at corbel support neutral axis v x β Balancing force polygon u V t F c F Tie action by reinforcing bar Strut action by concrete Applied Load u V d v a Figure 8.2 – Strut-and-Tie Action of a Corbel v x 9 . 0 Steel strain to be determined by linear extrapolation Top steel bar Applied Load h 5 . 0 ≥ d d a v < Figure 8.1 – Dimension requirement for a Corbel h 87 8.2.2 In addition to the strut-and tie model for the determination of the top steel bars, shear reinforcements should be provided in form of horizontal links in the upper two thirds of the effective depth of the corbel. The horizontal links should not be less than one half of the steel area of the top steel. 8.2.3 Bearing pressure from the bearing pad on the corbel should be checked and properly designed in accordance with “Code of Practice for Precast Concrete Construction 2003” Cl. 2.7.9. In short, the design ultimate bearing pressure to ultimate loads should not exceed (i) cu f 4 . 0 for dry bearing; (ii) cu f 6 . 0 for bedded bearing on concrete; (iii) cu f 8 . 0 for contact face of a steel bearing plate cast on the corbel with each of the bearing width and length not exceeding 40% of the width and length of the corbel. The net bearing width is obtained by stress bearing ultimate length bearing effective load ultimate × The Precast Concrete Code has specified that the effective bearing length be the least of (i) physical bearing length; (ii) one half of the physical bearing length plus 100 mm; (iii) 600 mm. 8.3 Design Formulae for the upper steel tie The capacity of concrete in providing lateral force as per Figure 8.2 is bx f x b f cu cu 405 . 0 9 . 0 45 . 0 = × × where b is the length of the corbel The force in the compressive strut is therefore β cos 405 . 0 bx f F cu c = By the force polygon, u cu u c V bx f V F = ⇒ = β β β cos sin 405 . 0 sin As v a x d 45 . 0 tan − = β ; ( ) 2 2 45 . 0 cos x d a a v v − + = β ( ) ( ) 2 2 45 . 0 45 . 0 sin x d a x d v − + − = β So ( ) ( ) ( ) ( ) 2 2 2 2 45 . 0 45 . 0 405 . 0 45 . 0 45 . 0 405 . 0 x d a x d bxa f V V x d a x d a bx f v v cu u u v v cu − + − = ⇒ = − + − 88 Expanding and grouping ( ) ( ) ( ) 0 45 . 0 9 . 0 18225 . 0 2025 . 0 2 2 2 = + + + − + d a V x ba f V d x ba f V v u v cu u v cu u Putting v cu u ba f V A 18225 . 0 2025 . 0 + = ; ( ) v cu u ba f V d B 45 . 0 9 . 0 + − = ( ) 2 2 d a V C v u + = A AC B B x 2 4 2 − − − = (Eqn 8-1) By the equilibrium of force, the top steel force is x d a V V T v u u 45 . 0 cot − = = β (Eqn 8-2) The strain at the steel level is, by extrapolation of the strain diagram in Figure 8.2 is 0035 . 0 × − = − = x x d x x d ult s ε ε (Eqn 8-3) 8.4 Design Procedure : (iii) Based on the design ultimate load and v a , estimate the size of the corbel and check that the estimated dimensions comply with Figure 8.1; (iv) Check bearing pressures; (v) Solve the neutral axis depth x by the equation (Eqn 3-1). (vi) By the assumption plane remains plane and that the linear strain at the base of the corbel is the ultimate strain of concrete 0035 . 0 = ult ε , work out the strain at the top steel level as s ε ; (vii) Obtain the steel stress as s s s E ε σ = where 6 10 205× = s E kPa. However, the stress should be limited to y f 87 . 0 at which 002 . 0 = s ε ; (viii) Obtain the force in the top steel bar T by (Eqn 8-2) (ix) Obtain the required steel area of the top steel bars st A by s st T A σ = (x) Check the shear stress by ( )bd V v u 3 / 2 = . If c v v > (after enhancement as applicable), provide shear reinforcements by ( ) y c v sv f v v b s A 87 . 0 − = over the upper d 3 2 where sv A is the cross sectional area of each link and v s is the link spacing. However, the total shear area provided which is 89 d s A v sv 3 2 × should not be less than half of the top steel area, i.e. sl v sv A d s A 2 1 3 2 ≥ × . 8.5 Detailing Requirements (i) Anchorage of the top reinforcing bar should (a) either be welded to a bar of equivalent strength or diameter; or (b) bent back to form a closed loop. In the former case, the bearing area of the load should stop short of the transverse bar by a distance equal to the cover of the tie reinforcement. In the latter case, the bearing area of the load should not project beyond the straight portion of the bars forming the tension reinforcements. (ii) Shear reinforcements be provided in the upper two thirds of the effective depth and total area not less than half of the top bars. Typical detailings are shown in Figure 8.3(a) and 8.3(b) Shear reinforcements d 3 2 c, cover to transverse bar d u V c > v a Figure 8.3(a) – Typical Detailing of a Corbel transverse bar welded to the main tension bar of equal diameter or strength Top main bar Additional bar for shear link anchorage 90 8.6 Worked Example 8.1 Design a corbel to support an ultimate load of 600 kN at a distance 200 mm from a wall support, i.e. 600 = u V kN, 200 = v a mm. The load is transmitted from a bearing pad of length 300 mm. Concrete grade is 40. (a) The dimensions of the corbel are detailed as shown which comply with the requirement of Cl. 6.5.1 of the Code with length of the corbel Shear reinforcements d 3 2 d u V 0 > v a Figure 8.3(b) – Typical Detailing of a Corbel Top main bar Net bearing width 500 = h 600 = u V h 5 . 0 250 ≥ 450 = d 450 200 = < = d a v Figure 8.4 – Worked Example. 8.1 91 300 = b mm. (b) Check bearing stress : Design ultimate bearing stress is 32 40 8 . 0 8 . 0 = × = cu f MPa Net bearing width is 5 . 62 32 300 10 600 3 = × × mm. So use net bearing width of bearing pad 70 mm. (c) With the following parameters, 600 = u V kN; 40 = cu f MPa; 300 = b mm; 200 = v a mm 450 = d mm substituted into (Eqn 8-1) ( ) ( ) ( ) 0 45 . 0 9 . 0 18225 . 0 2025 . 0 2 2 2 = + + + − + d a V x ba f V d x ba f V v u v cu u v cu u Solving 77 . 276 = x mm. (d) The strain at steel level 002 . 0 00219 . 0 0035 . 0 77 . 276 77 . 276 450 > = × − = − = ult s x x d ε ε (e) The stress in the top steel is y f 87 . 0 as 002 . 0 > s ε ; (f) The force in the top steel is 71 . 368 77 . 276 45 . 0 440 200 600 45 . 0 = × − × = − = x d a V T v u kN; (g) Steel area required is 32 . 921 460 87 . 0 368710 = × mm 2 , provide 3T20; (h) Check shear stress over the upper two thirds of d , i.e. 300 mm. ( ) 33 . 3 600 450 3 / 2 600000 = × MPa > 5 . 0 = c v MPa. So shear reinforcement ( ) ( ) 12 . 2 460 87 . 0 5 . 0 33 . 3 300 87 . 0 = × − = − = y c v sv f v v b s A mm So use T12 closed links – 100 ( v sv s A provided is 2.26) over the top 300 mm. Total steel area is 679 3 2 113 = × × mm 2 > 461 921 5 . 0 = × mm 2 . 92 T12 closed links 300 450 u V v a Figure 8.5 – Detailing of Worked Example 8.1 3T20 T20 anchor bar 93 9.0 Transfer Structures 9.1 According to Cl. 5.5 of the Code, transfer structures are horizontal elements which redistribute vertical loads where there is a discontinuity between the vertical structural elements above and below. 9.2 In the analysis of transfer structures, consideration should be given to the followings : (i) Construction and pouring sequence – the effects of construction sequence can be important in transfer structures due to the comparative large stiffness of the transfer structure and sequential built up of stiffness of structures above the transfer structure as illustrated in Figure 9.1; (ii) Temporary and permanent loading conditions – especially important when it is planned to cast the transfer structures in two shifts and use the lower to support the upper shift as temporary conditions, thus creating locked-in stresses; (iii) Varying axial shortening of elements supporting the transfer structures – which leads to redistribution of loads. The phenomenon is more serious as the transfer structure usually possesses large flexural stiffness in comparison with the supporting structural members, behaving somewhat between (a) flexible floor structures on hard columns; and (b) rigid structures (like rigid cap) on flexible columns; (iv) Local effects of shear walls on transfer structures – shear walls will stiffen up transfer structures considerably and the effects should be taken into account; (v) Deflection of the transfer structures – will lead to redistribution of loads of the superstructure. Care should be taken if the structural model above the transfer structure is analyzed separately with the assumption that the supports offered by the transfer structures are rigid. Re-examination of the load redistribution should be carried out if the deflections of the transfer structures are found to be significant; (vi) Lateral shear forces on the transfer structures – though the shear is lateral, it will nevertheless create out-of-plane loads in the transfer structures which need to be taken into account; (vii) Sidesway of the transfer structures under lateral loads and unbalanced gravity loads should also be taken into account. The effects should be considered if the transfer structure is analyzed as a 2-D model. 94 Stage (1) : Transfer Structure (T.S.) just hardened G/F Stage (2) : Wet concrete of G/F just poured G/F Stage (3) : G/F hardened and 1/F wet concrete just poured 1/F G/F Stage (4) : 1/F hardened and 2/F wet concrete just poured 2/F 1/F G/F Stage (5) : 2/F hardened and 3/F wet concrete just poured 3/F 2/F 1/F G/F Stage (6) and onwards Structure above transfer structure continues to be built. Final force induced on T.S. becomes {F n } + {F n-1 } + {F n-2 } + ........... + {F 2 } + {F 1 } Figure 9.1 – Diagrammatic illustration of the Effects of Construction Sequence of loads induced on transfer structure Stress/force in T.S. being {F 1 } due to own weight of T.S. and stiffness of the T.S. Stress/force in T.S. being {F 1 } + {F 2 }, {F 2 } being force induced in transfer structure due to weight of G/F structure and stiffness of the T.S. only Stress/force in T.S. being {F 1 } + {F 2 } + {F 3 }, {F 3 } being force induced in transfer structure due to weight of 1/F structure and stiffness of the T.S. + G/F Stress/force in T.S. being {F 1 } + {F 2 } + {F 3 } + {F 4 }, {F 4 } being force induced in T.S. due to weight of 2/F structure and stiffness of the T.S. + G/F + 1/F Stress/force in T.S. being {F 1 } + {F 2 } + {F 3 } + {F 4 } + {F 5 }, {F 5 } being force induced in T.S. due to weight of 1/F structure and and stiffness of the T.S. + G/F + 1/F + 2/F 95 9.3 Mathematical modeling of transfer structures as 2-D model (by SAFE) : The general comments in mathematical modeling of transfer structures as 2-D model to be analyzed by computer methods are listed : (i) The 2-D model can only examine effects due to out-of-plane loads, i.e. vertical loads and moments. Lateral loads have to be analyzed separately; (ii) It is a basic requirement that the transfer structure must be adequately stiff so that detrimental effects due to settlements of the columns and walls being supported on the transfer structure are tolerable. It is therefore necessary to check such settlements. Effects of construction sequence may be taken into account in checking; (iii) The vertical settlement support stiffness should take the length of the column/wall support down to level of adequate restraint against further settlement such as pile cap level. Reference can be made to Appendix H discussing the method of “Compounding” of vertical stiffness and the underlying assumption; (iv) Care should be taken in assigning support stiffness to the transfer structures. It should be noted that the conventional use of either L EI / 4 or L EI / 3 have taken the basic assumption of no lateral movements at the transfer structure level. Correction to allow for sidesway effects is necessary, especially under unbalanced applied moments such as wind moment. Fuller discussion and means to assess such effects are incorporated in Appendix H; (v) Walls which are constructed monolithically with the supporting transfer structures may help to stiffen up the transfer structures considerably. However, care should be taken to incorporate such stiffening effect in the mathematical modeling of the transfer structures which is usually done by adding a stiff beam in the mathematical model. It is not advisable to take the full height of the wall in the estimation of the stiffening effect if it is of many storeys as the stiffness can only be gradually built up in the storey by storey construction so that the full stiffness can only be effected in supporting the upper floors. Four or five storeys of walls may be used for multi-storey buildings. Furthermore, loads induced in these stiffening structures (the stiff beams) have to be properly catered for which should be resisted by the wall forming the stiff beams; 96 9.4 Modeling of the transfer structure as a 3-dimensional mathematical model can eliminate most of the shortcomings of 2-dimensional analysis discussed in section 9.3, including the effects of construction sequence if the software has provisions for such effects. However, as most of these softwares may not have the sub-routines for detailed design, the designer may need to “transport” the 3-D model into the 2-D model for detailed design. For such “transportation”, two approaches can be adopted : (i) Transport the structure with the calculated displacements by the 3-D software (after omission of the in-plane displacements) into the 2-D software for re-analysis and design. Only the displacements of the nodes with external loads (applied loads and reactions) should be transported. A 2-D structure will be re-formulated in the 2-D software for re-analysis by which the structure is re-analyzed by forced displacements (the transported displacements) with recovery of the external loads (out-of-plane components only) and subsequently recovery of the internal forces in the structure. Theoretically the results of the two models should be identical if the finite element meshing and the shape functions adopted in the 2 models are identical. However, as the finite element meshing of the 2-D model is usually finer than that of the 3-D one, there are differences incurred between the 2 models, as indicated by the differences in recovery of nodal forces in the 2-D model. The designer should check consistencies in reactions acting on the 2 models. If large differences occur, revealing lesser loads on the structure, the designer should review his approach; 2-D model (usually finer meshing) with nodal forces recovered by forced displacement analysis at nodes marked with External nodal force is {F 2D } ≠ {F 3D } after re-analysis External nodal force is {F 3D } 3-D model (usually coarser meshing) with displacements at nodes with external loads marked with Figure 9.2 – 3-D model to 2-D with transportation of nodal displacements 97 (ii) Transport the out-of-plane components of the external loads (applied loads and reactions) acting on the 3-D model to the 2-D model for further analysis. This type of transportation is simpler and more reliable as full recovery of loads acting on the structure is ensured. However, in the re-analysis of the 2-D structure, a fixed support has to be added on any point of the structure for analysis as without which the structure will be unstable. Nevertheless, no effects due to this support will be incurred by this support because the support reactions should be zero as the transported loads from the 3-D model are in equilibrium. 9.5 Structural Sectional Design and r.c. detailing The structural sectional design and r.c. detailing of a transfer structure member should be in accordance with the structural element it simulates, i.e. it should be designed and detailed as a beam if simulated as a beam and be designed and detailed as a plate structure if simulated as a plate structure. Though not so common in Hong Kong, if simulation as strut-and-tie model is employed, the sectional design and r.c. detailing should be accordingly be based on tie and strut forces so analyzed. The commonest structural simulation of a transfer plate structure is as an assembly of plate bending elements analyzed by the finite element method. As such, the analytical results comprising bending, twisting moments and out-of-plane shears should be designed for. Reference to Appendix D can be The out-of-plane components of all loads acting on the structure including reactions be transported 3-D model with external loads obtained by analysis 2-D model with out-of-plane components of external forces transported from 3-D model and re-analyzed with a fixed support at any point Figure 9.3 – 3-D model to 2-D with transportation of nodal displacements 98 made for the principles and design approach of the plate bending elements. 99 10.0 Footings 10.1 Analysis and Design of Footing based on the assumption of rigid footing Cl. 6.7.1 of the Code allows a footing be analyzed as a “rigid footing” provided it is of sufficient rigidity with uniform or linearly varying pressures beneath. As suggested by the Code, the critical section for design is at column or wall face as marked in Figure 10.1. As it is a usual practice of treating the rigid footing as a beam in the analysis of its internal forces, the Code sets the some rules in requiring concentration of reinforcing bars in areas with high concentration of stresses. Re Figure 10.2 : Footing under pure axial load creating uniform pressure beneath Footing under eccentric load creating linearly varying pressure beneath critical sections for design c c area with 2/3 of the required reinforcements area with 2/3 of the required reinforcements 1.5d 1.5d 1.5d 1.5d l c is the greater of l c1 and l c2 d is the effective depth l c2 2l c1 Plan Figure 10.1 – Assumed Reaction Pressure on Rigid Footing Figure 10.2 – Distribution of reinforcing bars when l c > (3c/4 + 9d/4) 100 Cl. 6.7.2.4 of the Code requires checking of shear be based on (i) section through the whole width of the footing (as a beam); and (ii) local punching shear check as if it is a flat slab. (Re Example 4.5 in Section 4). 10.2 Worked Example 10.1 Consider a raft footing under two column loads as shown in Figure 10.3. Design data are as follows : Column Loads (for each): Axial Load: D.L. 2400 kN L.L. 600 kN Moment D.L. 300kNm L.L. 80 kNm Overburden soil : 1.5 m deep Footing dimensions : plan dimensions as shown, structural depth 500 mm Concrete grade of footing : grade 35 (i) Loading Summary : D.L. Column: = ×2400 2 4800 kN; O.W. 108 24 5 . 0 0 . 2 5 . 4 = × × × kN Overburden Soil = × × × 20 5 . 1 2 5 . 4 270 kN Total 5178 kN Moment (bending upwards as shown in Figure 10.3) 600 300 2 = × kNm L.L. Column 1200 600 2 = × kN. Moment (bending upwards as shown in Figure 10.3) 160 80 2 = × kNm Factored load : Axial load 2 . 9169 1200 6 . 1 5178 4 . 1 = × + × kN Moment 1096 160 6 . 1 600 4 . 1 = × + × kNm (ii) The pressure beneath the footing is first worked out as : D.L. 300 kNm L.L. 80kNm for each column 1000 1000 400 1000 400 400 1000 2500 Plan Figure 10.3 – Footing layout for Worked Example 10.1 101 At the upper end : 13 . 1384 2 5 . 4 1096 6 2 5 . 4 2 . 9169 2 = × × + × kN/m 2 At the lower end : 47 . 653 2 5 . 4 1096 6 2 5 . 4 2 . 9169 2 = × × − × kN/m 2 At the critical section 87 . 1091 12 / 2 5 . 4 2 . 0 1096 2 5 . 4 2 . 9169 3 = × × + × kN/m 2 The pressures are indicated in Figure 10.3(a) (iii) At the critical section for design as marked in Figure 10.3(a), the total shear is due to the upward ground pressure minus the weight of the overburden soil which is 8 . 4348 5 . 4 8 . 0 5 . 1 20 5 . 4 8 . 0 2 87 . 1091 13 . 1384 = × × × − × × | . | \ | + kN The total bending moment is ( ) 5 . 4 3 2 8 . 0 2 87 . 1091 13 . 1384 5 . 4 2 8 . 0 5 . 1 20 87 . 1091 2 2 × × × | . | \ | − + × × × − 66 . 1809 = kNm (iv) Design for bending : Moment per m width is : 15 . 402 5 . 4 66 . 1809 = kNm/m; 5 . 412 5 . 12 50 500 = − − = d mm 363 . 2 5 . 412 1000 10 15 . 402 2 6 2 = × × = = bd M K , By the formulae in Section 3 for Rigorous Stress Approach, 655 . 0 0 = p %; 2702 = st A mm 2 /m As 125 . 1228 4 / 5 . 412 9 4 / 400 3 4 / 9 4 / 3 1250 = × + × = + > = d c l c , two thirds of the reinforcements have to be distributed within a zone of d 5 . 1 1091.87 kN/m 2 Section for critical design 653.47 kN/m 2 1384.13 kN/m 2 400 1000 400 400 1000 2000 Figure 10.3(a) – Bearing Pressure for Worked Example 10.1 102 from the centre of column, i.e. 2375 . 1 2 5 . 412 5 . 1 = × × , Total flexural reinforcements over the entire width is 12159 5 . 4 2702 = × mm 2 , 2/3 of which in 475 . 2 2 2375 . 1 = × m. So 3275 475 . 2 / 3 / 2 12159 = × mm 2 /m within the critical zone. So provide T25 – 150. Other than the critical zone, reinforcements per metre width is ( ) 2001 475 . 2 5 . 4 / 3 / 12159 = − mm 2 /m. Provide T25 – 225 (v) Design for Strip Shear : Shear per m width at the critical section is 4 . 966 5 . 4 8 . 4348 = kN Shear stress at the critical section is 343 . 2 5 . 412 1000 966400 = × = v N/mm 2 > 614 . 0 = c v N/mm 2 as per Table 6.3 of the Code. So shear reinforcement required is ( ) ( ) 32 . 4 460 87 . 0 614 . 0 343 . 2 1000 87 . 0 = × − = − = yv c v sv f v v b s A mm/m width Use T12 – 150 (B.Ws), v sv s A provided is 5.02 (vi) Check punching shear by first locating the first critical perimeter for punching shear checking. The first critical perimeter is at 1.5d from the column face. Factored load by the column is 4320 600 6 . 1 2400 4 . 1 = × + × kN Weight of overburden soil is ( ) 64 . 75 5 . 1 20 4 . 0 6375 . 1 2 2 = × × − kN 400+1.5d×2=1.637.5 Critical perimeter for punching shear checking 400+1.5d×2=1637.5 400 1000 1000 400 1000 400 Plan Figure 10.3(b) – checking punching shear for Worked Example 10.1 103 Weight of concrete is ( ) 26 . 30 5 . 0 24 4 . 0 6375 . 1 2 2 = × × − kN Upthrust by ground pressure is 82 . 2731 6375 . 1 2 5 . 4 2 . 9169 2 = × × kN Net load along the critical perimeter is 08 . 1694 82 . 2731 26 . 30 64 . 75 4320 = − + + kN Effective shear force for punching shear stress checking is, by (Ceqn 6.40) of the Code : 06 . 2196 6375 . 1 08 . 1694 548 5 . 1 1 08 . 1694 5 . 1 1 = | . | \ | × × + = | | . | \ | + = sp t t t eff x V M V V kN Punching shear stress is 81 . 0 5 . 412 4 5 . 1637 10 06 . 2196 3 = × × × N/mm 2 > 614 . 0 = c v N/mm 2 . So shear reinforcements in form of links required. As 98 . 0 6 . 1 81 . 0 = < c v N/mm 2 , full reduction by c v can be carried out. Total shear reinforcement is, by (Ceqn 6.44) of the Code, ( ) ( ) 1323 460 87 . 0 5 . 412 5 . 1637 4 614 . 0 81 . 0 87 . 0 = × × × × − = − ≥ yv c sv f ud v v A mm 2 Comparing with the shear reinforcements obtained in (v) by which the total reinforcements within the perimeter is ( ) 12670 4 . 0 6375 . 1 2 . 0 2 . 0 201 2 2 = − × × mm 2 . So shear reinforcements in (v) controls (vii) Checking of bending and shear in the direction parallel to the line joining the columns can be carried out similarly. However, it should be noted that there is a net “torsion acting on any section perpendicular to the line joining the two columns due to linearly varying ground pressure. To be on the conservative side, shear arising due to this torsion should be checked and designed accordingly as a beam as necessary. Nevertheless, one can raise a comment that the design has to some extent be duplicated as checking of bending has been carried out in the perpendicular direction. Furthermore, for full torsion to be developed for design in accordance with (Ceqn 6.65) to (Ceqn 6.68) of the Code, the “beam” should have a free length of beam stirrup width + depth to develop the torsion (as illustrated in Figure 3.12 in Section 3) which is generally not possible for footing of considerable width. As unlike shear where enhancement can be adopted with “shear span” less than d 2 or d 5 . 1 , 104 no similar strength enhancement is allowed in Code, though by the same phenomenon there should be some shear strength enhancement. So full design for bending in both ways together with torsion will likely result in over-design. (viii) The flexural and shear reinforcements provisions for the direction perpendicular to the line joining the columns is 10.3 Flexible Footing Analysis and Design As contrast to the footing analyzed under the rigid footing assumption, the analysis of footing under the assumption of its being a flexible structure will also take the stiffness of the structure into account by which the deformations of the structure itself will be analyzed. The deformations will affect the distribution of the internal forces of the structure and the reactions which are generally significantly different from that by rigid footing analysis. Though it is comparatively easy to model the cap structure, it is difficult to model the surface supports provided by the ground because : (i) the stiffness of the ground with respect to the hardness of the subgrade and geometry of the footing are difficult to assess; (ii) the supports are interacting with one another instead of being independent “Winkler springs” supports. However, we are currently lacking computer softwares to solve the problem. As the out-of-plane deformations and forces are most important in footing analysis and design, flexible footings are often modeled as plate bending Shear links T12 – 150 BWs in the whole footing T25 – 225(B1) T25 – 150(B1) 400 1237.5 1237.5 400 400 Plan Figure 10.3(c) – Reinforcement Details for Worked Example 10.1 (in the direction perpendicular to the line joining the two columns only) 105 elements analyzed by the finite element method discussed in 10.4 in more details. 10.4 Analysis and Design by Computer Method The followings are highlighted for design of footing modeled as 2-D model (idealized as assembly of plate bending elements) on continuous surface supports: (i) The analytical results comprise bending, twisting moments and out-of-plane shears for consideration in design; (ii) As local “stresses” within the footing are revealed in details, the rules governing distribution of reinforcements in footing analyzed as a beam need not be applied. The design at any location in the footing can be based the calculated stresses directly. However, if “peak stresses” (high stresses dropping off rapidly within short distance) occur at certain locations which are often results of finite element analysis at points with heavy loads or point supports, it would be reasonable to “spread” the stresses over certain width for design. Nevertheless, care must be taken of not taking widths too wide for “spreading” as local effects may not be well captured. (iii) The design against flexure should be done by the “Wood Armer peak stress width over which the peak stress is designed for Figure 10.4 – Spreading of peak stress over certain width for design 106 Equations” listed in Appendix D, together with discussion of its underlying principle. As the finite element mesh of the mathematical model is often very fine, it is a practice of “lumping” the design reinforcements of a number of nodes over certain widths and evenly distributing the total reinforcements over the widths, as is done by the popular software “SAFE”. Again, care must be taken of not taking widths too wide for “lumping” as local effects may not be well captured. The design of reinforcements by SAFE is illustrated on the right portion of Figure 10.3; (iv) The principle together with a worked example for design against shear is included in Appendix D, as illustrated in Figure D5a to D5c. It should be noted that as the finite element analysis give detailed distribution of shear stresses on the structure, it is not necessary to calculate shear stress as done for flat slab under empirical analysis in accordance with the Code. The checking of shear and design of shear reinforcements can be based directly on the shear stresses revealed by the finite element analysis. 107 11.0 Pile Caps 11.1 Rigid Cap analysis Cl. 6.7.3 of the Code allows a pile cap be analyzed and designed as a “rigid cap” by which the cap is considered as a perfectly rigid structure so that the supporting piles deform in a planar structure. As the deformations of the piles are governed, the reactions by the piles can be found. If it is assumed that the piles are identical, the reactions of the piles follow a linearly varying pattern. Appendix I contains derivation of formulae for solution of pile loads under rigid cap assumption. Upon solution of the pile loads, the internal forces of the pile cap structure can be obtained with the applied loads and reactions acting on it. The conventional assumption is to consider the cap as a beam structure spanning in two directions and perform analysis and design separately. It is also a requirement under certain circumstances that some net torsions acting on the cap structure (being idealized as a beam) need be checked. As the designer can only obtain a total moment and shear force in any section of full cap width, there may be under-design against heavy local effects in areas having heavy point loads or pile reactions. The Code (Cl. 6.7.3.3) therefore imposes a condition that shear enhancement due to closeness of applied load and support cannot be applied. Cl. 6.7.3.5 of the Code requires checking of torsion based on rigid body theory. pile loads, magnitude follows linear profile under assumption of equal pile stiffness Figure 11.1 – Pile load profile under rigid cap assumption 108 11.2 Worked Example 11.1 (Rigid Cap Design) The small cap as shown in Figure 11.2 is analyzed by the rigid cap assumption and will then undergo conventional designed as beams spanning in two directions. Design data : Pile cap plan dimensions : as shown Pile cap structural depth : 2 m Pile diameter : 2 m Column dimension : 2 m square Factored Load from the central column : 50000 = P kN 2000 = x M kNm (along X-axis) 1000 = y M kNm (along Y-axis) (i) Loads : from Columns : 50000 = P kN 2000 = x M kNm (along X-axis) 1000 = y M kNm (along Y-axis) O.W. of Cap 4752 24 2 9 11 = × × × kN Weight of overburden soil 2970 20 5 . 1 9 11 = × × × kN Factored load due to O.W. of Cap and soil is ( ) 10811 2970 4752 4 . 1 = + kN So total axial load is 60811 10811 50000 = + kN; 400 Y 3000 3000 1500 1500 1500 1500 4000 4000 X Figure 11.2 – Pile cap layout of Worked Example 11.1 P1 P2 P3 P4 P5 P6 critical sections for shear checking 109 (ii) Analysis of pile loads – assume all piles are identical (Reference to Appendix I for analysis formulae) x I of pile group = 54 3 6 2 = × y I of pile group = 32 0 2 4 2 2 = × + × Pile Loads on P1 : 69 . 9903 54 3 1000 32 4 2000 6 60811 = × + × − kN P2: 69 . 10153 54 3 1000 32 0 2000 6 60811 = × + × − kN P3: 69 . 10403 54 3 1000 32 4 2000 6 60811 = × + × + kN P4: 65 . 9866 54 3 1000 32 4 2000 6 60811 = × − × − kN P5: 65 . 10116 54 3 1000 32 0 2000 6 60811 = × − × − kN P6: 65 . 10366 54 3 1000 32 4 2000 6 60811 = × − × + kN (iii) Design for bending in the X-direction The most critical section is at the centre line of the cap Moment created by Piles P3 and P6 is ( ) 33 . 83081 4 65 . 10366 69 . 10403 = × + kNm Counter moment by O.W. of cap and soil is ( ) 75 . 10617 75 . 2 2 2970 4752 = × ÷ + kNm The net moment acting on the section is 58 . 76423 75 . 10617 33 . 83081 = − kNm 1865 60 75 2000 = − − = d ; 9000 = b 441 . 2 1865 9000 10 58 . 76423 2 6 2 = × × = = bd M K ; % 668 . 0 = p 112070 = st A mm 2 , requiring T40 – 200 (2 layers, B1 and B3) (iv) Design for shear in the X-direction By Cl. 6.7.3.2 of the Code, the critical section for shear checking is at 20% of the diameter of the pile inside the face of the pile as shown in Figure 11.2 Total shear at the critical section is : Upward shear by P3 and P6 is 33 . 20770 65 . 10366 69 . 10403 = + kN Downward shear by cap’s O.W. and soil is 110 ( ) 2 . 1474 11 1 . 2 2970 4752 = × + kN Net shear on the critical section is 13 . 19299 2 . 1474 33 . 20770 = − kN 17 . 1 1865 9000 10 13 . 19299 3 = × × = v N/mm 2 > 618 . 0 = c v N/mm 2 by Table 6.3 of the Code. No shear enhancement can be effected. Shear reinforcements in form of links per metre width is ( ) ( ) 379 . 1 460 87 . 0 618 . 0 17 . 1 1000 87 . 0 = × − = − = yv c v sv f v v b s A Use T12 links – 200 in X-direction and 400 in Y-direction by which v sv s A provided is 1.41. (v) Design for bending in the Y-direction The most critical section is at the centre line of the cap Moment created by Piles P1, P2 and P3 ( ) 17 . 91383 3 06 . 30461 3 69 . 10403 69 . 10153 69 . 9903 = × = × + + kNm Counter moment by O.W. of cap and soil is ( ) 25 . 8687 25 . 2 2 2970 4752 = × ÷ + kNm The net moment acting on the section is 92 . 82695 25 . 8687 17 . 91383 = − kNm 1825 40 60 75 2000 = − − − = d ; 11000 = b 257 . 2 1825 9000 10 92 . 82695 2 6 2 = × × = = bd M K ; % 613 . 0 = p 122422 = st A mm 2 , requiring T40 – 200 (2 layers, B2 and B4) (vi) Checking for shear in the Y-direction By Cl. 6.7.3.2 of the Code, the critical section for shear checking is at 20% of the diameter of the pile inside the face of the pile as shown in Figure 11.2 Total shear at the critical section is : Upward shear by P1, P2 and P3 is 06 . 30461 kN Downward shear by cap’s O.W. and soil is ( ) 8 . 1801 9 1 . 2 2970 4752 = × + kN 111 Net shear on the critical section is 26 . 28259 8 . 1801 06 . 30461 = − kN 417 . 1 1825 11000 10 26 . 28259 3 = × × = v N/mm 2 < 601 . 0 = c v N/mm 2 by Table 6.3 of the Code. No shear enhancement can be effected. Shear reinforcements in form of links per metre width is ( ) ( ) 038 . 2 460 87 . 0 601 . 0 417 . 1 1000 87 . 0 = × − = − = yv c v sv f v v b s A As v sv s A in Y-direction is greater than that in X-direction, so adopt this for shear reinforcement design Use T12 links – 200 BWs by which v sv s A provided is 2.82. (vii) Punching shear : As the critical perimeter for punching shear checking of the piles and column overlap with each other, punching shear cannot be effected. So shear checking as a beam in (iv) and (vi) is adequate. (viii) Checking for torsion : There are unbalanced torsions in any full width sections at X-Y directions due to differences in the pile reactions. However, as discussed in sub-section 10.2(vii) of this Manual for footing, it may not be necessary to design the torsion as for that for beams. Anyhow, the net torsion is this example is small, being 12 . 1611 3 04 . 537 = × kNm (537.04 is the difference in pile loads between P3 and P4), creating torsional shear stress in the order of 097 . 0 3 2000 9000 2000 10 12 . 1611 2 3 2 2 6 min max 2 min = | . | \ | − × × = | . | \ | − = h h h T v t N/mm 2 . So the torsional shear effects should be negligible. (ix) Reinforcement details are as shown in Figure 11.3 112 11.3 Strut-and-Tie Model Cl. 6.7.3.1 of the Code allows pile cap be designed by the truss analogy, or more commonly known as “Strut-and-Tie Model” (S&T Model) in which a concrete structure is divided into a series of struts and ties which are beam-like members along which the stress are anticipated to follow. In a S&T model, a strut is a compression member whose strength is provided by concrete compression and a tie is a tension member whose strength is provided by added reinforcements. In the analysis of a S&T model, the following basic requirements must be met (Re ACI Code 2002): (i) Equilibrium must be achieved; (ii) The strength of a strut or a tie member must exceed the stress induced on it; (iii) Strut members cannot cross each other while a tie member can cross another tie member; (iv) The smallest angle between a tie and a strut joined at a node should not exceed 25 o . The Code has specified the following requirements : (i) Truss be of triangular shape; (ii) Nodes be at centre of loads and reinforcements; (iii) For widely spaced piles (pile spacing exceeding 3 times the pile diameter), only the reinforcements within 1.5 times the pile diameter T 4 0 - 2 0 0 ( T 1 , B 1 , B 3 ) T40-200 (T2, B2,,B4) Y 3000 3000 1500 1500 1500 1500 4000 4000 X Figure 11.3 – Reinforcement Design of Worked Example 11.1 P1 P2 P3 P4 P5 P6 Shear links T12 – 200 BWs on the whole cap 113 from the centre of pile can be considered to constitute a tension member of the truss. 11.4 Worked Example 11.2 (Strut-and-Tie Model) Consider the pile cap supporting a column load of 6000kN supported by two piles with a column of size 1m by 1 m. The dimension of the cap is as shown in Figure 11.4, with the thickness of cap 1.5 m. (i) Determine the dimension of the strut-and-tie model Assume two layers of steel at the bottom of the cap, the centroid of both layers is at 135 20 40 75 = + + mm from the base of the cap. So the effective width of the tension tie is 270 2 135 = × mm. The tie and strut dimension is drawn in Figure 11.5. (ii) A simple force polygon is drawn and the compression in the strut can be simply worked out as 75 . 5028 6000 62 . 36 sin 2 0 = ⇒ = C C kN; 6000kN 1500 2500 1000 dia. 1000 dia. 3000 3000 Figure 11.4 – Pile Cap Layout of Worked Example 11.2 Plan Elevation 114 And the tension in the bottom tie is 87 . 4035 62 . 36 cos 0 = = C T kN. (iii) To provide the bottom tension of 4035.87 kN, the reinforcement steel required is 10085 460 87 . 0 10 87 . 4035 87 . 0 10 87 . 4035 3 3 = × × = × y f mm 2 . Use 9-T40. (iv) Check stresses in the struts : Bottom section of the strut, the strut width at bottom is 26 . 813 62 . 36 cos 270 62 . 36 sin 1000 0 0 = + mm As the bottom part is in tension, there is a reduction of compressive strength of concrete to 08 . 10 35 8 . 1 8 . 1 = = cu f MPa as suggested top strut width = 596.57mm 1000 Bottom strut width = 813.26mm bottom tie, strength be provided by steel concrete strut 36.62 o 36.62 o 2230 2230 3000 3000 4000 kN 4000 kN 6000 kN 6000kN 270 2500 1000 dia. 1000 dia. 3000 3000 Elevation Figure 11.4 – Analysis of strut and tie forces in Worked Example 11.2 115 by OAP, which is an implied value of the ultimate concrete shear strength of cu f 8 . 0 as stated in BS8110. As a conservative approach, assuming a circular section at the base of the strut since the pile is circular, the stress at the base of the strut is 69 . 9 4 / 813 10 75 . 5028 2 3 = × π MPa < 10.08MPa For the top section of the strut, the sectional width is 57 . 596 62 . 36 sin 500 2 0 = × mm As the sectional length of the column is 1 m, it is conservative to assume a sectional area of 1000mm × 596.57 mm. The compressive stress of the strut at top section is 439 . 8 57 . 59 1000 10 75 . 5028 3 = × × MPa < 75 . 15 45 . 0 = cu f MPa 11.5 Flexible Cap Analysis A pile cap can be analyzed by treating it as a flexible structure, i.e., as in contrast to the rigid cap assumption in which the cap is a perfectly rigid body undergoing rigid body movement only with no deformation upon the application of loads, the flexible pile cap structure will deform and the 5T40 B1 & 4T40 B2 1000 2500 1000 dia. 1000 dia. 3000 3000 Elevation T16 s.s. – 200 5T32 T1 Figure 11.5 – Reinforcement Details of Worked Example 11.2 116 deformations will affect the distribution of internal forces of the structure and the reactions. Analysis of the flexible cap structure will require input of the stiffness of the structure which is comparatively easy. However, as similar to that of footing, the stiffness of the pile cap itself which is mainly offered by the supporting pile is often difficult, especially for the friction pile which will interact significantly with each other through the embedding soil. Effects by soil restraints on the piles can be considered as less significant in end-bearing piles such large diameter bored piles. Similar to the flexible footing, as the out-of-plane loads and deformation are most important in pile cap structures, most of the flexible cap structures are modeled as plate structures and analyzed by the finite element method. 11.6 Analysis and Design by Computer Method Analysis and design by computer method for pile cap are similar to Section 10.3 for footing. Nevertheless, as analysis by computer methods can often account for load distribution within the pile cap structure, Cl. 6.7.3.3 of the Code has specified the followings which are particularly applicable for pile cap in respect of design : (i) shear enhancement may be applied to a width of φ 3 for circular pile, or pile width plus 2 × least dimension of pile. (ii) averaging of shear force shall not be based on a width > the effective depth on either side of the centre of a pile, or as limited by the actual dimension of the cap. v a v a B B B Area where shear enhancement may apply v a v a φ φ φ Figure 11.6 – Effective width for shear enhancement in pile cap around a pile 117 Figure 11.7 can be a guideline for “effective widths” adopted in averaging “peak stresses” as will often be encountered in finite element analysis for pile cap structure modeled as an assembly of plate bending elements under point loads and point supports, as in the same manner as that for footing as discussed in 10.4(ii) of this Manual. Shear force diagram Width over which the shear force can be averaged in the cap for design φ ≤ φ φ peak shear at pile centre Figure 11.7 – Width in cap over which shear force at pile can be averaged for Design 118 12.0 General Detailings 12.1 In this section, the provisions of detailing requirements are general ones applicable to all types of structural members. They are mainly taken from Section 8 of the Code. Requirements marked with (D) are ductility requirements. 12.2 Minimum spacing of reinforcements – clear distance is the greatest of (i) maximum bar diameter; (ii) maximum aggregate size (h agg ) + 5 mm; (iii) 20 mm. 12.3 Permissible bent radii of bars. The purpose of requiring minimum bend radii for bars are (i) avoid damage of bar; (ii) avoid overstress by bearing on concrete in the bend. Generally, the minimum bend radii are φ 3 for 20 ≤ φ mm and φ 4 for 20 > φ mm. If the above requirement is not fulfilled, bearing pressure inside the bend should be checked so that (Ceqn 8.1) listed as below is fulfilled | | . | \ | + ≤ b cu bt a f r F φ φ 2 1 2 where bt F is the tensile force in the bar; r the internal bend radius of the bar; φ is the bar diameter, b a is centre to centre distance between bars perpendicular to the plane of the bend and in case the bars are adjacent to the face of the member, + = φ b a cover. Take an example of a layer of T40 bars of centre to centre separation of 100 mm and internal bend radii of 120mm in grade 35 concrete. 503051 1257 460 87 . 0 = × × = bt F N 8 . 104 40 120 503051 = × = φ r F bt 89 . 38 100 40 2 1 35 2 2 1 2 = | . | \ | × + × = | | . | \ | + > b cu a f φ So the (Ceqn 8.1) is not fulfilled. 12.4 Anchorage of longitudinal reinforcements 119 (i) Anchorage is derived from ultimate anchorage bond stress with concrete assessed by the (Ceqn 8.3) of the Code. cu bu f f β = where for high yield bars 5 . 0 = β for tension and 65 . 0 = β for compression. For example, 96 . 2 35 5 . 0 = = bu f MPa for grade 35. For a bar of diameter φ , the total force up to y f 87 . 0 is | | . | \ | 4 87 . 0 2 π φ y f . The required bond length L will then be related by φ φ β φ πφ β π φ 34 8 . 33 4 87 . 0 4 87 . 0 2 ≈ = = ⇒ = | | . | \ | cu y cu y f f L L f f which agrees with Table 8.5 of the Code; (ii) In addition to (i), it has been stated in 9.9.1.1(c) of the Code which contains ductility requirements for beams that “For the calculation of anchorage length the bars must be assumed to be fully stressed.” The steel bars should therefore be carrying stress of y f instead of y f 87 . 0 in determination of anchorage (and bond) lengths, resulting in increase of bond and anchorage lengths for beam should be increased by % 15 1 87 . 0 / 1 = − . (D) (iii) Bearing stress on concrete by bend must be checked by (Ceqn 8.1) of the Code if anchorage of bar requires a length more than φ 4 beyond the bend Code (Re Figure 8.1 of the Code) (iv) With the minimum support width requirements as stated in Cl. 8.4.8 of the Code, bends of bars can start beyond the centre line of supports offered by beams, columns and walls. By the same clause the requirement can be considered as not confining to simply supported beam as stated in Cl. 9.2.1.7 of the Code; 12.5 Anchorage of links – Figure 8.2 of the Code displays bend of links of bend angles from 90 o to 180 o . However, it should be noted that the Code requires anchorage links in beams and columns to have bend angles not more than 135 o as ductility requirements; (D) 120 12.6 Laps – the followings should be noted (i) If sum of reinforcement sizes in a particular layer exceeds 40% of the breadth of the section at that level, the laps must be staggered; (ii) Details of requirements in bar lapping are indicated in Figure 8.4 of the Code reproduced in Figure 12.1 for ease of reference; (iii) Compression and secondary reinforcements can be lapped in one section. 12.7 Lap Length – the followings should be noted : (i) Absolute minimum lap length is the greater of φ 15 and 300 mm; (ii) Tension lap length should be at least equal to the design tension anchorage length and be based on the diameter of the smaller bar; (iii) Lap length be increased by a factor of 1.4 or 2.0 as indicated in Figure 8.5 of the Code. ≥20mm >4ø ≥0.3l 0 l 0 ≤50mm ≤4ø Figure 12.1 – Lapping arrangement for tension laps Appendix A Clause by Clause Comparison between “Code of Practice for Structural Use of Concrete 2004” and BS8110 Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-1 1.1 – Scope The clause has explicitly stated that the Code applies only to normal weight concrete, with the exclusion of (i) no fines, aerated, lightweight aggregate concrete etc; (ii) bridge and associated structures, precast concrete (under the separate code for precast concrete); and (iii) particular aspects of special types of structures such as membranes, shells. Pt. 1 1.1 – Scope The clause only explicitly excludes bridge structures and structural concrete of high alumina cement. The exclusion of CoPConc2004 should also be applied to BS8110. In addition, BS8110 does not apply to high strength concrete. 2.1.5 – Design working life The clause states that the Code assumes a design working life of 50 years. Where design working life differs from 50 years, the recommendations should be modified. – Nil No similar statement in BS8110. 2.2.3.3 – Response to wind loads The clause refers to clause 7.3.2 for the usual limits of H/500 to lateral deflection at top of the building and accelerations of 1-in-10 year return period of 10 minutes duration of 0.15m/sec 2 for residential and 0.25m/sec 2 for office. However, there is no requirement on the inter-storey drift, though the draft steel code has a requirement of storey height/400. Pt. 1 2.2.3.3 – Response to wind loads Reference to specialist literature is required. In addition Pt. 2 3.2.2.2 stipulates a limit on inter-storey drift of Storey height/500 for avoidance of damage to non-structural elements. CoPConc2004 is more specific. However, method for determination of the acceleration is not given in the Code and in the HKWC-2004. 2.3.2.1 – Loads for ultimate limit state Table 2.1 is generally reproduced from Table 2.1 of BS8110 except that the partial factor for load due to earth and water is 1.0 for the beneficial case. Pt. 1 2.4.3.1.2 – Partial factors for earth pressures It is stated in the clause that when applying the load factor, no distinction should be made between adverse and beneficial loads. 1.0 in CoPConc2004 may not be adequately conservative as there may even be over-estimation in the determination of the unfactored soil load. It is even a practice to set the load to zero in beneficial case. ICU has raised this comment during the comment stage when the draft Code has exactly the content of BS8110. 2.3.2.3 & 2.3.2.4 – The clauses explicitly state that these effects need only be considered when they are significant for – No similar clauses in BS8110 The clause in CoPConcrete2004 affirms engineers to ignore Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-2 Differential settlement of foundations, creep, shrinkage, temperature effects ULS. In most other cases they need not be considered provided ductility and rotational capacity of the structure sufficient. consideration of these effects in normal cases which are the usual practices. 2.4.3.2 – Values of γ m for ULS Table 2.2 gives γ m for ULS for concrete and re-bars. γ m for re-bars is 1.15, implying strength of re-bars for design remain as 0.87f y . Pt. 1 2.4.4.1 – Values of γ m for ULS Table 2.2 gives γ m for ULS for concrete and re-bars. γ m for re-bars is 1.05, implying strength of re-bars for design remain as 0.95f y . BD has been insisting on the use of 0.87f y even if BS8110 was used before the promulgation of the new concrete code 3.1.3 – Strength grades Table 3.1 states concrete strength grades from 20 MPa up to 100 MPa which is the range covered by the Code. – BS8110 has not explicitly stated the concrete grades covered by the BS, However, concrete grades covered by the design charts in Part 3 of the Code range from grade 25 to 50 whilst other provisions such as v c (Pt. 1 Table 3.8), lap lengths (Pt. 1 Table 3.27) are up to grade 40. The coverage of CoPConc2004 is wider. 3.1.4 – Deformat- ion of Concrete It is stated in the 1 st paragraph of the clause that for ULS, creep and shrinkage are minor, and no specific calculation are required. Pt. 1 2.4.3.3 – creep, shrinkage and temperature effects The clause states that “For the ULS, these effects will usually be minor and no specific calculations will be necessary. BS 8110 has included temperature effects be a minor one that can be ignored in calculation. 3.1.5 – Elastic deformat- ion Table 3.2 stipulates short term static Young’s Modulus of concrete of various grades based on the formula 3.46√f cu +3.21 in MPa derived from local research. Pt. 1 Figure 2.1 The determination of short term static Young’s Modulus of concrete is given by the slope gradient in the figure which is 5.5√(f cu /γ m ). Values in the CoPConc2004 should be used as it is based on local research and concrete E values are affected by constituents which are of local materials. Nevertheless, it should be noted that E values in the new Code are slightly Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-3 higher than the previous ones in “The Structural Use of Concrete – 1987” (Table 2.1). 3.1.7 & 3.1.8 – Creep and shrinkage Though it is stated in 3.1.4 that for ULS creep and shrinkage are minor and require no specific calculations, these clauses contain detailed formulae and charts for prediction of creep and shrinkage strain. The approach is identical to the “Structural Design Manual” issued by Highways Department and the charts are extracted from BS5400:Pt 4:1990. – No account has been given for creep and shrinkage, as stated in 2.4.3.3. As stated in the CoPConc2004 Cl. 2.3.2.4, effects need only be considered if they are significant. 3.1.9 – Thermal Expansion The linear coefficient of thermal expansion given in the Code for normal weight concrete is 10×10 -6 / o C whilst that stated in the “Structural Design Manual” issued by Highways Department is 9×10 -6 / o C in Clause 2.4.4. – No account has been given for temperature, as stated in 2.4.3.3. The linear coefficient of thermal expansion given by both CoPConc2004 is slightly higher than SDM and both are independent of concrete grades. 3.1.10 – Stress -strain relationship for design The short term design stress-strain curve of concrete (Fig. 3.8) follows closely the traditional one in BS8110, though there are differences in the values of the Young’s Moduli. Furthermore, the ultimate strain is limited to below 0.0035 for concrete grade exceeding C60. Nevertheless, the “plastic strain”, strain beyond which stress is constant remains identical as BS8110; Pt. 1 2.5.3 – Analysis of section for ULS Pt. 1 Fig. 2.1 shows stress-strain relation of normal weight concrete. (i) The Young’s Moduli of concrete stipulated in CoPConc2004 should be adopted as they are based on local data and cover up to grade 100 concrete, together with the decrease of ultimate strain for grade above C60, to account for the brittleness of high strength concrete; (ii) “Smooth” connection between the parabolic curve and the straight line portion of the stress-strain curve cannot be effected if ε 0 = 2.4×10 -4 √f cu /γ m is kept Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-4 and the Young’s moduli in Table 3.2 of CoPConc2004 are used. For smooth connection, ε 0 should be revised 2σ ult /E where σ ult =0.67f cu /γ m . 3.2.7 – Weldability (of re-bars) The clause states that re-bars can be welded provided the types of steel have the required welding properties given in acceptable standards and under Approval and inspection by competent person. Further provisions for welding are given in 10.4.6. Pt. 1 3.12.8.16 7.6 There are provisions for lapping re-bars by welding in Pt. 1 3.12.8.16, and general welding requirements in Pt. 1 7.6. Nevertheless, another BS7123 titled “Metal arc welding of steel for concrete reinforcement” requires the re-bars be in compliance with BS4449 or BS4482. Steel complying CS2 is likely weldable as CS2 does not differ significantly from BS4449. Section 4 – Durability and fire resistance The requirements are general. The followings are highlighted :- (i) In 4.1.1, it is stated that requirements are based on design working life of 50 years; (ii) In Table 4.1 under 4.2.3.2, exposure conditions 1 to 5 are classified with headings similar to that in Pt. 1 3.3.4 of BS8110. However, detailed descriptions are different except the last one “abrasive”; (iii) In 4.2.7.3, control of AAR has been incorporated from the previous PNAP 180; (iv) Concrete covers are given in Table 4.2 for various concrete grades and exposure conditions; (v) By 4.3, the user has to refer to the current fire code for additional requirements against fire resistance. Pt. 1 2.2.4 – Durability; Pt. 1 2.2.6 – Fire resistance; Pt. 1 3.1.5.2 – Design for Durability Pt. 1 3.3 – Concrete cover to rebars Pt. 2 Section 4 The requirements are general. However, the followings are highlighted : (i) 3.1.5.2 states that when cement content > 400 kg/m 3 and section thicker than 600 mm, measures for temperature control should be implemented; (ii) A full description for fire resistance control is given in Pt. 2 Section 4, outlining methods of determining fire resistance of structural elements and with reference to BS476 Pt. 8; The approaches of the two codes are quite different. However, CoPConc2004 is more related to local practice. 5.1.3.2 – Load cases and 3 simplified load cases for Dead + Live loads are recommended (with DL always present): (i) all spans loaded with LL; (ii) alternate spans loaded with LL; Pt. 1 3.2.1.2.2 2 simplified load cases are considered sufficient for design of spans and beams (with DL always present): (i) all spans loaded with LL; (ii) alternate spans CoPConc2004 more reasonable, though not truly adequate. Nevertheless, the current Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-5 combinations for beams and slabs (iii) adjacent spans loaded with LL. The 2 nd case is to seek for max. sagging moment and the 3 rd case is likely to seek for max. hogging moment. But true max hogging moment should also include alternate spans from the support being loaded. loaded with LL; softwares mostly can account for the load case to search for max. hogging moment at support. 5.2 – Analysis of Structure The clause contains definition of beam, slab (included ribbed, waffle, one-way or two-ways), column in accordance with their geometries. The following differences with BS8110 are highlighted : (i) Conditions in relations to rib spacings and flange depths for analysis of a ribbed or waffle slab as an integral structural unit are given in 5.2.1.1(d); (ii) By definition, the effective flange widths in T and L-beams included in 5.2.1.2 (a) are slightly greater than that in BS8110 by 0.1×clear rib spacing unless for narrow flange width controlled by rib spacing. It is also stated that the effective flange can be adopted in structural analysis; (iii) Clearer definition of effective spans of beams is also included in 5.2.1.2(a) with illustration by diagrams, together with reduction of span moments due to support width in 5.2.1.2(b). In principle, the effective span of a simply supported beam, continuous beam or cantilever is the clear span plus the minimum of half effective depth and half support width except that on bearing where the span should be up to the centre of the bearing. (iv) Furthermore reduction in support moments due to support width is allowed which is not mentioned in BS8110; (v) The definition of effective flange for T- and 3.2 – Analysis of Structures 3.4 – Beams The followings are highlighted : (i) Generally provisions are applicable to normal strength concrete; (ii) Consideration for high-rise buildings (second order effects) and structures such as shear walls, transfer structures are not given; (iii) The effective span of a simply supported beam is the clear span plus the lesser of half effective depth and half of support width whilst that of continuous beam and cantilever are span between supports (presumably mid-support width) except at end span in case of continuous beam and a cantilever forming the end of a continuous beam, the effective span is the clear span plus mid-support width should be used. The following remarks are made : (i) CoPConc2004 is more suitable for use in Hong Kong as it cover high strengths concrete, high rise buildings, shears, transfer structures. (ii) Extended use of effective flange in beam in structural analysis and moment reduced to support shear are explicitly stated in CoPConc2004. (iii) Method of analysis in both codes are old-fashioned ones that can be performed by hand calculations. Use of computer methods (extensively adopted currently in Hong Kong) are not mentioned. Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-6 L-beams are slightly different from BS8110 though the upper bound values are identical to that of BS8110. So more stringent. (vi) Moment redistribution is restricted to grade below C70 as per 5.2.9.1; (vii) Condition 2 in 5.2.9.1 relation to checking of neutral axis depths of beam in adopting moment re-distribution is different from Cl. 3.2.2.1 b) of BS8110. More stringent limit on neutral axis is added for concrete grade higher than C40 as per 6.1.2.4(b); (viii) Provision for second order effects with axial loads in 5.3 is added. The provisions are quite general except the statement “second order effects can be ignored if they are less than 10%”; (ix) Provisions for shears walls and transfer structures are added in 5.4 and 5.5 though the provisions are too general. 6.1 – Members in Flexure The followings differences with BS8110 are identified : (i) Ultimate strains are reduced below 0.0035 for concrete grade exceeding C60 as per Fig. 6.1; (ii) Neutral axis depths limited to less than 0.5d for concrete grade higher than C40 as per 6.1.2.4(b); (iii) The limitations on neutral axis depth ratio under redistribution of moments are further restricted for grade > 40, as different from BS8110 which makes not differences among concrete grades in Cl 3.4.4.4 which if redistribution of moment exceed 10%; (iv) Different design formulae are used for the higher grades (40<f cu ≤70; 70<f cu ≤100) concrete Pt. 1 3.4 – Beams In the design aspects, the Code is limited to grade 40. The followings are highlighted : (i) Provisions are made in CoPConc2004 for concrete grades higher than 40 including limitation of neutral axis depths etc. However, it is also noted that the ultimate stress in Fig. 6.1 remains unchanged for high concrete grade though the ultimate strain is reduced; (ii) Ultimate concrete shear strengths increased in CoPConc2004; Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-7 based on simplified stress blocks as per 6.1.2.4(c). Design charts similar to that BS8110 based on stress-strain relationship in figure 3.8 are not available; (v) Table 6.2 in relation to v c of concrete as related to tensile reinforcements is identical to Pt. 1 Table 3.8 of BS8110 except that applicability is extended to C80 whereas BS8110 limits to grade 40. The minimum amount of shear reinforcements required is reduced to that can provide shear resistance of 0.4(f cu /40) 2/3 MPa for grade over 40 whilst that for concrete at and below grade 40 remains to that can provide 0.4 MPa; (vi) Partial strength factor for steel remains as 0.87f y in accordance with BS8110:1985 instead of 0.95f y as in accordance with BS8110:1997, for both flexure and shear; (vii) Ultimate shear strength of concrete increased to 7.0 MPa as compared 5.0 MPa in BS8110. The other limitation of 0.8√f cu is identical to BS8110; (viii) By 6.1.3.5, the minimum shear reinforcement to cater for shear strength of 0.4 MPa is for concrete grade below C40 (requirement by BS8110). Above grade C40, the required shear strength is increased by factor (f cu /40) 2/3 as per Table 6.2; (ix) By 6.1.4.2, ribbed slabs (6.1.4.2) can be designed as two-ways spanning as similar to flat slab if they have equal structural properties in two mutually perpendicular directions. BS8110 does not explicitly require equal structural properties in mutually perpendicular directions in adopting design method as flat slab (BS8110, (iii) Due to the more stringent limitation on neutral axis depth for high grade concrete in CoPConc2004, different formulae have been devised for x > 0.5d for rectangular beams. However, similar formulae are not given for flanged beams; (iv) 6.1.4.2 of CoPConc2004 is similar to BS8110 Pt. 1 3.6.2. However, it is ICU’s comment to qualify that if ribbed slabs are to be designed as two-ways spanning as similar to a flat slab, it should be qualified that the slab has equal structural properties in two mutually perpendicular directions. Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-8 3.6.2). 6.2 – Members axially loaded with or without flexure The followings are highlighted: (i) The CoP contains no design charts for column design with moments. Strictly speaking, the design charts in BS8110 are not applicable (a) the Young’s Moduli of concrete are different; and (b) the ultimate strain for concrete grade > C60 are reduced; (ii) Due to the use of lower partial strength factor of steel (γ m = 1.15), the factors on steel in equations for strengths of column sections (6.55, 6.56) are lower than BS8110 (Eqn 28, 29); (iii) The provisions for more accurate assessment of effective column height in accordance with BS8110 Pt. 2 2.5 are not incorporated in the Code. Pt. 1 3.8 – Columns Pt 2 2.5 – Effective column height The followings are highlighted : (i) In the design aspects, the provisions are limited to grade 40; (ii) A more tedious approach for determination of effective height of column by consideration stiffness of the connecting beams and columns is outlined in Pt. 2 2.5. The followings are highlighted : (i) The limitation of neutral axis depth as for beam is clearly not applicable to columns. However, it should be noted that members with axial loads creating axial stress < 0.1f cu may be regarded as beam in design. Re 6.1.2.4 of CoPConc2004. 6.3 – Torsion and Combined Effects The followings are highlighted : Table 6.17 in relation to limitation of torsional shear stresses contains specific values for grades 25 to 80. For values below grade 40, they are identical to BS8110. Above grade 40, the Code provides values for v tmin and v t for different grades up to 80, beyond which the values remain constant whilst BS8110 set the values to that of grade 40 for grades above 40; Pt. 2 2.4 The followings are highlighted (i) Table 2.3 contains specific values for v tmin and v t up to grade 40. The values remain constant from grade 40 thereon; (ii) CoPConc2004 contains more specific values for ultimate concrete shear stress. 6.4 – Design for robustness The followings are highlighted : (i) In 6.4.2 in relation to design of “bridging elements” which is identical to BS8110 Pt.2 2.6.3, the words “where required in buildings of five or more storeys” have been deleted. So the Code is more stringent as consideration to loss of elements is required for all buildings; Pt 2 2.6.3 The followings are highlighted : (i) Pt.2 2.6.3 in relation to design of “bridging elements” applies to buildings of five or more storeys; CoPConcrete 2004 is more stringent. Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-9 6.6 – Staircase The followings are highlighted : (i) 6.6.1 is in relation to design of staircase. There is no stipulation that the staircase may include landing; Pt 1 3.10.1 The followings are highlighted : (i) A note in 3.10.1 in relation to design of staircase has stipulated that a staircase also include a section of landing spanning in the same direction and continuous with flight; It is more reasonable to assume staircase should include landing, as in CoPConc2004. 6.7 – Foundations The following differences with BS8110 are highlighted : (i) In 6.7.1.1, the assumptions of uniform reaction or linearly varying reaction of footing and pile cap are based on use of rigid footings or pile caps. So a pre-requisite for the use of these assumptions is stated at the end of the clause which reads “if a base or pile cap is considered be of sufficient rigidity.”; (ii) In 6.7.3.1, a statement has been inserted that a pile cap may be designed as rigid or flexible, depending on the structural configuration. No similar provision is found in BS8110; (iii) In 6.7.3.3, 2 nd dot, it is stated that “where the shear distribution across section has not been considered, shear enhancement shall not be applied.” No similar provision is found in BS8110; (iv) In 6.7.3.3 3 rd dot, shear enhancement in pile cap can only be applied where due consideration has been given to shear distribution across section. No similar provision is found in BS8110; (v) In 6.7.3.3 4 th dot, determination of effective width of section for resisting shear is included. No similar provision is found in BS8110; (vi) In 6.7.3.3 5 th dot, it is explicitly stated that no shear reinforcement is required if v<v c . No Pt 1 3.11 The followings are highlighted : (i) In Pt 1 3.11.2.1, the assumption of uniform or linearly varying assumption is without the pre-requisite that the footing or pile cap is sufficiently rigid. This is not good enough as significant errors may arise if the footing or pile cap is flexible; (ii) In Pt 1 3.11.4.1, there is no mention of pile cap rigidity which affects design; (iii) In Pt 1 3.11.4.3, there is no mention that shear enhancement shall not be applied to where shear distribution across section has not been considered; (iv) In Pt 1 3.11.4.4 b), there is no mention that shear enhancement in pile cap shall be applied to under the condition that shear distribution across section has not been duly considered; (v) There is no explicit stipulation on the limit of effective width for checking shear; (vi) No explicit statement that shear reinforcement is required if v<v c though it is a normal practice of not providing shear stirrups in pile caps; (vii) No explicit statement that torsion is required to be checked, if any. The followings are highlighted : (i) CoPConc2004 is generally more reasonable as it makes provision for the modern analysis by treating the pile cap as a flexible structure by computer methods; (ii) Apparently CoPConc2004 forces the designer to check torsion as if the cap or footing is a beam under the rigid cap (footing). This is not too sound as the formulae for beam are under the assumption that torsional cracks can be fully developed for a beam length of b+d. If such length is not available which is very common for cap or footing structures which are usually wide structures. There should be shear enhancement if the full length cannot mobilized, as similar to direct stress. However, no Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-10 similar stipulation is found in BS8110; (vii) In 6.7.3.4, the ultimate shear stress is increased to 7 MPa, as compared with BS8110; (viii) In 6.7.3.5, it is stated that torsion for a rigid pile cap should be checked based on rigid body theory and where required, torsional reinforcements be provided. study data is available for torsional shear enhancement. 6.8 – Beam Column Joints The Code contains detailed provisions for design of beam column joints. No similar provision found in BS8110. – No similar provision. Design checking on beam-column shall be done if CoPConc2004 is used. Section 7 – Serviceability Limit States The followings are highlighted : (i) This section contains provisions in BS8110 Pt. 2 Section 3 and the deem-to-satisfy requirements (for deflections) in BS8110 Pt. 1 Section 3; (ii) Limits of calculated crack widths are given in 7.2.1 for member types and exposure conditions. The limited values are mostly 0.3mm as compared with BS8110 except water retaining structures and pre-stressed concrete (0.2 mm); (iii) 7.2.4.1 is identical to BS8110 Pt. 2 3.8.4.1 except that the last paragraph and Table 3.2 of BS8110 Pt. 2 3.8.4.1 have been omitted. The omitted portion is in relation to estimated limiting temperatures changes to avoid cracking; (iv) Equation 7.3 in 7.2.4.2 is different from Equation 14 of BS8110 Pt. 2 3.8.4.2 where the difference in temperatures is divided into 2 parts and a factor of 0.8 is employed in the estimation of thermal strain. Also it is allowed in the clause to take reinforcements into Pt. 2 Section 3 The followings are highlighted : (i) General provisions for determination of deflections, cracks (including thermal cracking) are given; (ii) Some of the provisions may be applicable to UK only; (iii) Limit of calculated crack widths is given as 0.3mm in 3.2.4.1 as a guide. The following remarks are made : (i) The omission of the last paragraph and Table 3.2 of BS8110 Pt. 2 3.8.4.1 is likely because of the different climate and material properties in Hong Kong; (ii) Equation 7.3 in 7.2.4.2 appears to be a refined version of Equation 14 of BS8110 Pt. 2 3.8.4.2. The factor 0.8 accounts for discount of strain due to the long term strain. Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-11 consideration; (v) In 7.3, pre-camber is limited to span/250 to compensate excessive deflection. The limit is not given in BS8110 Pt. 2 3.2.1. Also, deflection limit after construction for avoidance of damage to structure is limited to span/500, whilst BS8110 Pt. 2 3.2.1.2 specifies span/500 or span/350 in accordance with brittleness of finishes for avoidance of damage to non-structural elements. In addition, 20 mm as an absolute value is also imposed in BS8110; (vi) In 7.3.2, limits (0.15 and 0.25 m/s 2 ) on accelerations (10 years return period on 10 min. duration) are given as avoidance of “excessive response” to wind loads whilst no numerical values are given in BS8110 Pt. 2 3.2.2.1. Furthermore, deflection limit due to wind load is given as H/500 whilst BS8110 Pt. 2 3.2.2.2 indicates limit of h/500 as inter-storey drift for avoidance of damage to non-structural members; (vii) In 7.3.3, excessive vibration should be avoided as similar in BS8110 Pt. 2 3.2.3. However, there is extra requirement in 7.3.3 that dynamic analysis be carried out in case structural frequency less than 6 Hz for structural floor and footbridge less than 5 Hz; (viii) Table 7.3 under 7.3.4.2 in relation to deem-to-satisfy requirement for basic span/depth ratios of beam and slab contains, requirements for “two-way slabs” and “end spans” are included, as in comparison with Table 3.9 of BS8110 Pt 1; (ix) Table 7.4 in relation to modification factor to effective span depth ratio by tensile Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-12 reinforcement is identical to Table 3.10 of BS8110 except that the row with service stress 307 (f y = 460) has replaced that of 333 (f y = 500); (x) 7.3.4.6 is identical to BS8110 Pt. 1 3.4.6.7 except that the last sentence in BS8110 is deleted; (xi) The provision of deflection calculation in 7.3.5 is identical to BS8110 Pt 2 Cl. 3.7; (xii) Equation 7.7 in 7.3.6 is not identical to equation 9 in BS8110 Pt. 2 in the derivation of shrinkage curvature; Section 8 – Reinf’t requirements The followings are highlighted : (i) In 8.1.1, it is declared that the rules given in the Section for re-bars detailings do not apply to seismic, machine vibration, fatigue etc and epoxy, zinc coated bars etc. No similar exclusion is found in BS8110; (ii) In 8.1.2, it is stated that bar scheduling should be in accordance with acceptable standards whilst BS8110 Pt. 1 3.12.4.2 requires standard be in accordance with BS4466; (iii) In 8.2, the minimum spacing of bars should be the lesser of bar diameter, h agg +5 mm and 20 mm. BS8110 Pt. 1 3.12.11.1 does not include 20 mm and bar diameter is only required when bar size > h agg +5 mm; (iv) In 8.2, it is stated that the distance between horizontal layer of bars should be sufficient without quantification whilst BS8110 Pt. 1 3.12.11.1 requires minimum be 2h agg /3; (v) 8.3 is essentially identical to BS8110 Pt. 1 3.12.8.22, 24, 25, except that a single Table 8.2 (in relation to minimum bend radii of re-bars) Pt. 1 Section 3 Pt. 1 4.10 in relation to anchorage of tendons in prestressed concrete The followings are highlighted : (i) The provisions are general; (ii) Consideration for ductility is not adequate. The followings are highlighted : (i) CoPConc2004 requires bar scheduling to acceptable standards. However, provisions in 8.3, 9.2.3 etc have requirements for bend of bars; (ii) ICU has commented that requirement in BS8110 Pt. 1 3.12.9.4(a) should extend to all support conditions. 8.4.8 of the Code seems to incorporate the comment; (iii) ICU has also suggested the use of torsional links similar to that in ACI code (135 o hood) which is of shape other than shape code 77 of BS4466. Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-13 to replace Table 3 of BS4466 to which BS8110 is making reference. As such, no distinction is made between mild steel bars and HY bars – both adopting minimum radii of HY bars. Provisions to the newer BS – BS8666 where the minimum bend radii are generally smaller is not adopted; (vi) 8.4 is essentially identical to BS8110 Pt. 1 3.12.8 except (a) It is mentioned in 8.4.1 that when mechanical device is used, their effectiveness has to be proven. No similar provision is found in BS8110; (b) Type 1 bars are not included in the Code; (c) 8.4.6 is added with Figure 8.1 for illustration of bend anchorage; (d) 8.4.8 in relation to minimum support widths requires any bend inside support be beyond the centre line of the support. This effectively extend the requirement of BS8110 Pt. 1 3.12.9.4(a) to support conditions other than simply support; (vii) 8.5 in relation to anchorage of links and shear reinforcements contains more stringent requirement for the length of the link beyond bends of bars than BS8110 Pt. 1 3.12.8.6. – (1) the greater of 4φ or 50 mm for 180 o bend in the Code but only 4φ in BS8110; (2) the greater of 10φ or 70 mm for 90 o bend in the Code but only 8φ in BS8110. Provisions for 135 o bend and welded bars are also added; (viii) 8.6 contains requirements for anchorage by welded bars which is not found in BS8110; (ix) Except for the requirement that the sum of re-bar sizes at lapping < 40% of the breadth of Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-14 the section, 8.7 contains more requirements in terms of “staggering laps”, minimum longitudinal and transverse distances between adjacent lapping of bars which are not found in BS8110 in case that sum of re-bar sizes at lapping > 40% of the breadth of the section in not using staggered laps. The requirements are also schematically indicated in Fig. 8.4; (x) 8.7.4.1 contains different and more detailed requirements for transverse reinforcements in lapped zone than BS8110 Pt. 2 3.12.8.12; (xi) 8.8 and 8.9 in relation to large diameter bars (>40φ) and bundle bars which are not found in BS8110; (xii) The provision in BS8110 Pt. 1 3.12.8.16 for butt joints of re-bars is not found in the Code. Section 9 – Detailing of Members and particular rules The followings are highlighted : (i) Table 9.1 under Cl. 9.2 tabulates minimum steel percentage equal to that of BS8110; (ii) In 9.2.1.4 in relation to maximum distance of bars in tension as similar to BS8110 Pt. 1 3.12.11.2, the stipulation in BS8110 that demonstration of crack width < 0.3 mm can be accepted is omitted; (iii) In 9.2.1.5, a requirement of 15% span moment be used to design beam support even under simply supported assumption is not found in BS8110. Furthermore, it is also stated in the clause that total tension re-bars of a flanged beam over intermediate supports can be spread over the effective width of the flange provided that half of the steel within the web width. There is also no such provision in BS8110; (iv) 9.2.1.8 requiring 30% of the calculated Pt. 1 Section 3 and 5.2.7 The followings are highlighted : (i) The analysis procedures are largely old-fashioned relying on old theories of Johansen, Hillerborg. Detailings to cater for behaviours not well understood or quantified are thus provided, though the determination of which are largely empirical or from past experiences; (ii) Though ductility is not a design aid explicitly stated in the BS, the BS does requires 135 o bend of links in anchoring compression bars in columns and beams (Pt. 1 3.12.7.2). The followings are highlighted : (i) The stress reduction in design of cantilevered projecting structures in PNAP173 is not incorporated is likely because the PNAP is based on working stress design method. So there should be some other approaches and this is not mentioned in the CoPConc2004; (ii) Ductility is more emphasized in CoPConc2004 9.9 which largely stem from seismic design. Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-15 mid-span re-bars be continuous over support appears to be adopted from Fig. 3.24 a) of BS8110. However, the circumstances by which the Figure is applicable as listed in 3.12.10.2 of the BS is not quoted; (v) 9.2.1.9 requires top steel of cantilever to extend beyond the point of contraflexure of the supporting span whilst Fig. 3.24 c) requires at least half of the top steel to extend beyond half span of the cantilever or 45φ; (vi) In 9.2.2, maximum spacing of bent-up bars is stipulated whilst no such requirement is found in BS8110; (vii) Torsional links has to be closed links (shape code 77 of BS4466) as required by BS8110 Pt. 2 2.4.8. However, 9.2.3 of the Code provides an alternative of using closed links of 135 o bend; (viii) In 9.3.1.1 (b) in relation to maximum spacing of re-bars in slab is more detailed than BS8110 Pt. 1 3.12.11.2.7 and appears more reasonable. The provisions are, in fact, more stringent : (a) for principal re-bars, 3h ≤ 400 mm whilst BS8110 is 3h ≤ 750 mm; (b) for secondary re-bars 3.5h ≤ 450 mm whilst no provision in BS8110; (c) more stringent requirements are added for slabs with concentrated loads or areas of maximum moments whilst no similar requirements are found in BS8110; (ix) The first para. in 9.3.1.3 requires half of the area of the calculated span re-bars be provided at the simply supported slabs and end support of continuous slabs. The requirement is identical to BS8110 Pt. 1 3.12.10.3.2. However, the provision in BS8110 is under the condition Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-16 listed in 3.12.10.3.1 that the slabs are designed predominantly to carry u.d.l. and in case of continuous slabs, approximately equal span. These conditions are not mentioned in the Code; (x) In 9.3.1.3, there is also a provision that if the ultimate shear stress < 0.5v c at support, straight length of bar beyond effective anchorage for 1/3 of support width or 30 mm (whichever is the greater) is considered effective anchorage. No similar provision is found in BS8110; (xi) 9.3.1.6 requiring closed loops of longitudinal rebars at free edge of slab is not found in BS8110; (xii) 9.3.2 is in relation to shear in slabs which should be identical to that for beams. However it is stated that shears should be avoided in slabs < 200 mm thick; (xiii) 9.4 in relation to cantilevered projecting structures has incorporated requirements for minimum thickness, minimum steel areas, maximum bar spacing, anchorage length from PNAP173. However, other requirements such as design with reduced stresses are not found; (xiv) 9.5.2.3 contains more stringent requirements of links in circular columns than that in BS8110 Pt. 1 3.12.7.3 as the termination of links should be at 135 o hook; (xv) There is an extra requirement in the maximum spacing of traverse reinforcement in wall in 9.6.3 which is 400 mm, as in comparison with BS8110 Pt. 1 3.12.7.4; (xvi) 9.6.5 in relation to reinforcement provisions to plain walls include BS8110 Pt. 1 3.9.4.19 to 22. However, 3.9.4.23 in relation to plain walls Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-17 with more than 1/10 in tension to resist flexure is not included. Anyhow, this is not important as the wall should not be regarded as plain wall; (xvii) In 9.7.1 and 9.7.2 in relation to pile caps and footings, it is stipulated that the minimum steel percentage should refer to Table 9.1 though Table 9.1 is under the heading “Beam”. So the minimum steel percentage of 0.13% for HY bars should be observed. There is no explicit provision in BS8110 for minimum steel in pile caps and footings; (xviii) 9.7.3 in relation to tie beams has included a requirement that the tie beam should be designed for a load of 10 kN/m if the action of compaction machinery can cause effects to the tie beams. This is not found in BS8110; (xix) 9.8 in relation to design of corbels carries the same content as 6.5.2.2 to 6.5.2.4 and is identical to BS8110 Pt. 1 5.2.7; (xx) 9.9 contains requirements more stringent than BS8110 in detailing with the aim to enhance ductility. The followings are highlighted : (a) 9.9.1.1(a) requires steel percentage in beam > 0.3% and percentage of tensile steel < 2.5%; (b) 9.9.1.1(c) requires anchorage of beam bar into exterior column to commence beyond centre line of column or 8φ instead of column face unless the moment plastic hinge can be formed at 500 mm or half beam depth from column face; (c) 9.9.1.1(d) imposes restriction in locations of laps and mechanical couplers – (i) not within column/beam joints, (ii) not within Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-18 1 m from potential plastic hinge locations; (iii) reversing stresses exceeding 0.6f y unless with specified confinement by links. In addition, bars be terminated by a 90 o bend or equivalent to the far face of column; (d) 9.9.1.2(e) requires distribution and curtailment of flexural re-bars be attained in critical sections (potential plastic hinge regions); (e) 9.9.1.2(a) states the link spacing in beam < the lesser of 16φ and beam width or depth and corner and alternate compression re-bars be anchored by links; (f) 9.9.1.2(b) states that links be adequately anchored by means of 135 o or 180 o hooks. Anchorage by 90 o hooks or welded cross bars not permitted; (g) 9.9.2.1(a) states min. (0.8%) and max. steel% (4% with increase to 5.2% at lap)in column; (h) 9.9.2.1(a) requires the smallest dia. of any bars in a row > 2/3 of the largest bar; (i) 9.9.2.1(a) limits max. dia. of column re-bar through beam by (eqn 9.7) dependent on beam depth, with increase by 25% if not forming plastic hinge; (j) 9.9.2.1(b) requires spacing of links to longitudinal bars not be spaced further than 1/4 of the adjacent column dimension or 200 mm; (k) 9.9.2.1(c) requires anchorage of column bar into exterior beam or foundation to commence beyond centre line of beam or foundation or 8φ instead of interface unless Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-19 the moment plastic hinge can be formed at 500 mm or half beam depth from column face; (l) 9.9.2.1(d) states restrictions in locations of laps (m) 9.9.2.2 describes the establishment of “critical regions” in columns where there are extra requirements on links – (i) link spacing in column < the lesser of 6φ and least 1/4 of column lateral dimension; (ii) each longitudinal bar be laterally supported by a link passing around the bar and having an included angle < 135 o . (Regions other than “critical regions” fallow 9.5.2) Section 10 The followings are highlighted : (i) 10.2 lists figures for construction tolerances whilst BS8110 refers most of the requirements to other BS; (ii) 10.3.4 in relation to sampling, testing and compliance criteria of concrete. They are extracted from HKB(C)R but with incorporation of 100 mm test cubes. Such provision is not found in BS8110; (iii) The sub-clause on “Concreting in cold weather” in BS8110 is not incorporated. 10.3.7 on “Concreting in hot weather” is modified from BS8110 Pt. 1 6.2.5 (reference to BS deleted); (iv) Table 10.4 is similar to BS8110 Pt. 1 Table 6.1. However the parameter t (temperature) is deleted and the categorization of cement is OPC and “others” instead of the few types of cement in BS8110; (v) 10.3.8.1 contains general requirements for Pt. 1 2.3, Section 6, Section 7, Section 8 The followings are highlighted : (i) Pt. 1 2.3 lists general requirements for inspection of construction; (ii) References to other BS are often stated in Section 6 and 7; (iii) Provisions of works in extreme temperatures are given which are deleted in CoPConc2004. The followings are highlighted : (i) The first part of Section 10 of CoPConc2004 mainly stems from HKB(C)R, CS1, CS2 whilst the second part incorporates workmanship requirements listed in BS8110 Pt. 1 Section 6; (ii) (iii) Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-20 “Formwork and falsework” similar (but not identical) to BS8110 Pt. 1 6.2.6.1; (vi) 10.3.8.2 lists criteria for striking of formwork identical to that in BS8110 Pt. 1 6.2.6.3.1. In addition, provisions for using longer or shorter striking time for PFA concrete and climbing formwork are included; (vii) Minimum striking time in 10.3.8.2 are in accordance with HKB(C)R Table 10 (with the addition of props to cantilever requiring 10 days striking time) instead of BS8110 Pt. 1 Table 6.2. Furthermore, BS8110 Pt. 1 Table 6.2 gives temperature dependent striking time whilst the striking times in CoPConc2004 are not temperature dependent; (viii) The contents of 10.3.9 in relation to surface finish are extracted from BS8110 Pt. 1 6.2.7. However, the general requirements are differently written and the “classes of finish” have been deleted; (ix) 10.3.10 and 10.3.11 in relation to joints are identical to BS8110 Pt. 1 6.2.9 and 6.2.10 though the wordings are different, except the last sentence of 6.2.9 last para. in relation to treating vertical joint as movement joint; (x) 10.4.1 contains general requirements on re-bars to standards CS2 and other acceptable standards whilst BS8110 Pt. 1 7.1 requires conformance to other BS; (xi) 10.4.2 in relation to cutting and bending of re-bars is identical to BS8110 Pt. 1 7.2 except (a) conformance is not restricted to BS but to acceptable standards; and (b) the requirement of pre-heating re-bars at temperatures below 5 o C is deleted; Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-21 (xii) 10.4.3 is effectively identical to BS8110 Pt. 1 7.3 except that the requirement for spacer blocks be of concrete of small aggregates of equal strength to the parental concrete is replaced by spacer blocks to acceptable standards; (xiii) 10.4.6 is effectively identical to BS8110 Pt. 1 7.6 except (a) conformance to BS changed to acceptable standards; (b) detailed descriptions of the types of welding omitted; and (c) requirement to avoid welding in re-bar bends omitted; (xiv) 10.5.1 is identical to BS8110 Pt. 1 8.1 except conformance to BS is changed to acceptable standards; (xv) 10.5.5.3 in relation to tensioning apparatus of prestressing tendons is effectively identical to BS8110 Pt. 1 8.7.3 except that CoPConc2004 has an additional requirements that apparatus be calibrated within 6 months; (xvi) 10.5.5.4 in relation to pre-tensioning of deflected tendons, compressive and tensile stresses should be ensured not to exceed permissible limits during transfer of prestressing force to the concrete with the release of holding-up and down forces. BS8110 Pt. 1 8.7.4.3 has omitted the compressive forces; (xvii) 10.5.5.5(b) requires anchorage of post tensioning wires to conform to acceptable standards whilst BS8110 Pt. 1 8.7.5.2 requires compliance with BS4447 (xviii) 10.5.5.5(d) in relation to tensioning procedures which is identical to BS8110 Pt. 1 8.7.5.4, the requirement of not carrying out Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-22 tensioning below 0 o C is omitted. Further, the paragraph in BS8110 stipulating that when full force cannot be developed in an element due to breakage, slip during the case when a large no. of tendons is being stressed is omitted in CoPConc2004; (xix) 10.5.7 contains detailed provisions for grouting of prestressed tendons whilst BS8110 Pt. 1 8.9 requires compliance to BS EN 445, 446. Section 11 – Quality Assurance and Control This section outlines measures and procedures in general quality assurance and control, with reference to local practice. The followings are highlighted : (i) Control are on design, construction and completed products; (ii) Control can be by independent organization; (iii) Concrete must be from supplier certified under the Quality Scheme for the Production and Supply of Concrete (QSPSC); (iv) Control on construction includes surveillance measures. – No similar provisions in BS8110. The control in CoPConc2004 are summaries of local good practice. Section 12 – Prestressed Concrete This section is basically identical to Section 4 of BS8110 Pt. 1. The followings are highlighted : (i) 12.1.5 in relation to durability and fire resistance makes reference to previous recommendations in Sections 4 and 10 whilst BS8110 makes reference also to Part 2 of BS8110; (ii) 12.2.3.1 in relation to redistribution of moments is restricted to concrete grade C70, as in consistency with reinforced concrete. BS8110 Pt. 1 4.2.3.1 does not have this limitation. But the BS covers grades up to 40; (iii) The first loading arrangement in 12.3.3 for Pt. 1 Section 4 The provisions are general. CoPConc2004 follows quite closely the provisions in BS8110 except for minor changes. Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Appendix A HK CoP Structural Use of Concrete 2004 BS8110:1997 Clause No. Contents Clause No. Contents Remark A-23 continuous beam is not found BS8110 Pt. 1 4.3.3. The loading arrangement is in consistency with 5.1.3.2 for reinforced concrete beams. Though not truly adequate (per similar argument as above), CoPConc2004 is more conclusive than BS8110; (iv) 12.3.8.2 gives ultimate concrete stress 7.0 MPa, as similar to r.c. works; (v) 12.8.2.2 in relation to 1000 h relaxation value which is identical to BS8110 Pt. 1 4.8.2.2, “UK” has been deleted in description of manufacturer’s appropriate certificate; (vi) 12.8.4 and 12.8.5 in relation to shrinkage and creep of concrete make reference to 3.1.8 and 3.1.7 whilst BS8110 Pt 1. 4.8.4 and 4.8.5 list UK data; (vii) 12.10 makes reference to 8.10.2.2 for transmission lengths in pre-stressed members which is titled “transfer of prestress” which is identical to BS8110 Pt. 1 4.10.1 except that the 2 nd paragraph of the BS in relation to the difficulty of determination of transmission length has been deleted; (viii) 12.12.3.1(a) is identical to BS8110 Pt. 1 4.12.3.1.1 except that not only protection against corrosion is added. In 12.12.3.1(c), reference for protection against fire is not identical to BS8110; Section 13 – Load Tests of Structures or parts of structures This section contains testing of structures during construction stage under circumstances such as sub-standard works are suspected and visible defects are identified. – No similar provisions in BS8110. Appendix B Assessment of Building Accelerations Appendix B B-1 Assessment of acceleration of Buildings Underlying principles : Two Approaches are outlined in this Appendix : (i) The first one is based on the assumption that the building will undergo simple harmonic motion under wind loads. Thus the equation of x x 2 ω − = & & where x& & is the acceleration and x is the displacement of the motion, ω is the circular frequency of the building where f π ω 2 = ( f is the natural frequency of the building) can be used. However, the acceleration that will be causing discomfort to the occupants is the “dynamic resonant component” of the motion. If the G factor which is equal to ζ SE g B g I G f v h 2 2 2 1 + + = in Appendix F of the Wind Code 2004 is used to arrive at a total displacement, it can be considered that the total displacement is made of up of three components : (a) the static part which is 1 in the equation; (b) the dynamic background component which is B g I v h 2 2 ; and (c) the dynamic resonant component which is ζ SE g I f h 2 2 . It is the last displacement that should be multiplied to 2 ω to obtain the acceleration by which the building is responding to wind loads. So it is only necessary to calculate the dynamic resonant component. ω of the building can either be obtained by detail dynamic analysis or by some empirical formula such as 460/h. (ii) The second approach is that listed in Australian Wind Code AS/NSZ 1170.2:2002 Appendix G2 as extracted at the end of this Appendix. The approach is also based on the same principle as (i) though the formula G2 listed in the Australian Code has incorporated empirical parameters 0 m and h for assessment of the dynamic properties of the building. The formula G2 shows clearly it utilizes the dynamic resonant component Furthermore, it should also be noted that ζ SE g R for assessment of acceleration. Nevertheless, there is a denominator of h v I g 2 1+ in the formula in the Appendix B B-2 Australian Code as different from Hong Kong Wind Code, the reason being that the Australian Code is based on θ , des V which is 3 second gust whilst Hong Kong Code is based on hourly mean wind speed. So this denominator can be omitted. In addition, it should be noted that the Concrete Code requires the wind load for assessment of acceleration to be 1-in-10 year return period of 10 minutes duration whilst the wind load arrived for structural design in the Hong Kong Wind Code is based on 1-in-50 year return period of hourly duration. For conversion, the formula listed in Appendix B can be used (as confirmed by some experts that the formula can be used for downward conversion from 1-in-50 year to 1-in-10 year return periods). The 10 minutes mean speed can also be taken as identical to that of hourly mean speed (also confirmed by some experts.) Worked Example Data : Building height 4 . 96 = h m; Building width 3 . 37 = b m Building natural frequency 384 . 0 = a n Hz Total displacement at the uppermost floor by the approach listed in Appendix G of the Hong Kong Wind Code is 74 mm Total dead load is 251555 kN and total live load is 48171 kN (A) To compute the dynamic resonant component of the total force Computation of the G factor, the dynamic background and dynamic resonant components in accordance with HKWC 2004: 4 . 96 = h m 3 . 37 = b m 384 . 0 = a n Hz 1047 . 0 90 4 . 96 1055 . 0 90 1055 . 0 11 . 0 11 . 0 = | . | \ | = | . | \ | = − − h I h 7 . 3 = v g 00 . 4 ) 3600 ln( 2 = = a f n g 05 . 1762 10 1000 25 . 0 = | . | \ | = h L h Appendix B B-3 73 . 0 64 36 1 1 2 2 = + + = h L b h B 6 . 49 = h V m/sec 64 . 13 = = h h a V L n N 1285 . 0 4 1 5 . 3 1 1 = + + = h a h a V b n V h n S ( ) 0816 . 0 2 47 . 0 6 / 5 2 = + = N N E 02 . 0 = ζ So 898 . 1 2 1 2 2 = + + = ζ SE g B g I G f v h The dynamic background component is 662 . 0 2 2 = = B g I G v h The dynamic resonant component is 607 . 0 2 2 = = ζ SE g I G f h t dynresonan (B) To scale the wind loads from 50 years return period 1 hour duration to 10 years return period 10 minutes duration The formula in Appendix B of the Wind Code can be used for scaling to wind loads of shorter return period, as confirmed by some experts that the formula could be used for period shorter than 50 years. Thus the factor in scaling down the wind load from 50 years to 10 years is 6714 . 0 50 ln 5 10 ln 5 2 = | . | \ | + + The wind pressures of 10 minutes duration and 1 hour duration can be taken as identical. Thus no adjustment needs be carried out. Appendix B B-4 Assessment of accelerations By the two approaches (I) By x x 2 ω − = & & , based on assumption of simple harmonic motion. Fundamental natural frequency = 0.384 Hz Circular frequency = 412 . 2 2 = = f π ω /sec Dynamic Magnification factor = 1.898 Total Displacement = 0.074 m. The dynamic resonant component is 024 . 0 898 . 1 607 . 0 074 . 0 = × = x m The acceleration is 138 . 0 2 = x ω m/sec 2 which is for 50 year return period and one hour mean wind speed. Scaling down to 10 year return period, the acceleration for 10 years return period and 10 minute duration as required by Clause 7.3.2 of the Concrete Code 2004 is 093 . 0 6714 . 0 138 . 0 = × m/sec 2 < 0.15 m/sec 2 . (II) By b M h m a ˆ 3 2 0 = in accordance with the Australian Code As total dead load is 251555 kN and total live load is 48171 kN The weight used is full dead load + 0.25 live load = 263598 kN = h 96.4 m The weight per m height is 2734 4 . 96 263598 = kN/m which is equal to 279022 8 . 9 2734000 = kg/m 279022 0 = ∴m kg/m The total bending moment about X axis is 562163 kNm for full static + dynamic load. For the dynamic portion 179786 898 . 1 607 . 0 562163 = × kNm Scaled to 10 minutes duration and 10 years return period 120708 6714 . 0 179786 ˆ = × = b M kNm Appendix B B-5 So the acceleration is 14 . 0 10 120708 4 . 96 279022 3 3 2 = × × × m/sec 2 < 0.15 m/sec 2 . Explanatory Note : The existence of the denominator h v I g 2 1+ in the formula G2 listed in the Australian Code is because θ , des V used in the wind pressure calculation in the same formula is based on 3 second gust. As our code is based on hourly mean speed, downscaling by the factor is not necessary. Appendix C Derivation of Basic Design Formulae of R.C. Beam sections against Flexure Appendix C C-1 Derivation of Basic Design Formulae of R.C. Beam sections against Flexure The stress strain relationship of a R.C. beam section is illustrated in Figure C-1. In Figure C-1 above, the symbols for the neutral axis depth, effective depth, cover to compressive reinforcements are x, d, and d’, as used in BS8110 and the Code. To derive the contribution of force and moment by the concrete stress block, assume the parabolic portion of the concrete stress block be represented by the equation ε ε σ B A + = 2 (where A and B are constants) (Eqn C-1) So B A d d + = ε ε σ 2 (Eqn C-2) As c c E B E d d = ⇒ = =0 ε ε σ where c E is the tangential Young’s Modulus of concrete listed in Table 3.2 of the Code. Also 0 0 0 2 2 0 2 0 0 ε ε ε ε σ ε ε c E B A B A d d − = − = ⇒ = + ⇒ = = (Eqn C-3) As m cu f γ σ 67 . 0 = when 0 ε ε = m c cu c m cu c c m cu E f E f E E f γ ε ε γ ε ε ε γ 34 . 1 2 67 . 0 2 67 . 0 0 0 0 0 2 0 = ⇒ = ⇒ − = − ∴ (Eqn (C-4) (accords with 3.14 of the Concrete Code Handbook) neutral axis d’ d x 0035 . 0 = ult ε Stress Diagram Strain Diagram Figure C-1 – Stress Strain diagram for Beam Appendix C C-2 0 2ε c E A − = where m c cu E f γ ε 34 . 1 0 = So the equation of the parabola is ε ε ε σ c c E E + − = 2 0 2 for 0 ε ε ≤ Consider the linear strain distribution At distance u from the neutral axis, x u ult ε ε = So stress at u from the neutral axis up to ult x ε ε 0 is u x E u x E x u E x u E E E ult c ult c ult c ult c c c ε ε ε ε ε ε ε ε ε σ + − = | . | \ | + | . | \ | − = + − = 2 2 0 2 2 0 2 0 2 2 2 (Eqn C-5) Based on (Eqn C-5), the stress strain profiles can be determined. A plot for grade 35 is included for illustration : Stress Strain Profile for Grade 35 0 2 4 6 8 10 12 14 16 18 0 0.2 0.4 0.6 0.8 1 Distance Ratio from Neutral axis S t r e s s ( M P a ) h x 0 ε ε = u ult x ε ε / 0 0035 . 0 = ult ε Figure C-2 – Strain diagram across concrete section 0.3769 where ε 0 = 0.001319 Figure C-3 – Stress strain profile of grades 35 Appendix C C-3 Sectional Design of rectangular Section to rigorous stress strain profile Making use of the properties of parabola in Figure C-4 offered by the parabolic section as 1 c F given by bx f f x b F ult m cu m cu ult c ε γ ε γ ε ε 3 34 . 1 67 . 0 3 2 0 0 1 = = (Eqn C-6) and the moment exerted by 1 c F about centre line of the whole section | | . | \ | − − = − | | . | \ | − − = ult c ult ult c c x h F x x h F M ε ε ε ε ε ε 0 1 0 0 1 1 8 5 1 2 8 3 1 2 (Eqn C-7) The force by the straight portion is | | . | \ | − = | | . | \ | − = ult m cu ult m cu c bx f b x x f F ε ε γ ε ε γ 0 0 2 1 67 . 0 67 . 0 (Eqn C-8) The moment offered by the constant part about the centre line of the whole section is | | . | \ | − − = 2 1 2 0 2 2 x h F M ult c c ε ε (Eqn C-9) The compressive force by concrete as stipulated in (Eqn C-6) and (Eqn C-8) is | | . | \ | − = | | . | \ | − + = + = ult m cu ult m cu ult m cu c c c bx f bx f bx f F F F ε ε γ ε ε γ ε γ ε 0 0 0 2 1 3 3 67 . 0 1 67 . 0 3 34 . 1 For singly reinforcing sections, moment by concrete about the level of the tensile steel is, by (Eqn 2-7) and (2-9) Area = ab 3 2 centre of mass a 8 3 b a Figure C-4 – Geometrical Properties of Parabola Appendix C C-4 | | . | \ | − − + | | . | \ | − − = + = 2 1 8 5 1 0 2 0 1 2 1 x d F x d F M M M ult c ult c c c ε ε ε ε | | . | \ | − − | | . | \ | − + | | . | \ | − − = 2 1 1 67 . 0 8 5 1 3 34 . 1 0 0 0 0 x d bx f x d bx f ult ult m cu ult ult m cu ε ε ε ε γ ε ε ε γ ε | | . | \ | − − | | . | \ | − + | | . | \ | − − = ⇒ d x d x f d x d x f bd M ult ult m cu ult ult m cu ε ε ε ε γ ε ε ε γ ε 0 0 0 0 2 1 2 1 1 1 67 . 0 8 5 1 1 3 34 . 1 ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | . | \ | − + − + | | . | \ | − = ⇒ d x d x f bd M ult ult ult m cu 2 0 0 0 2 12 1 3 1 2 1 3 1 1 67 . 0 ε ε ε ε ε ε γ 0 3 1 1 67 . 0 12 1 3 1 2 1 67 . 0 2 0 2 2 0 0 = − | | . | \ | − + | . | \ | | | . | \ | − + − ⇒ bd M d x f d x f ult m cu ult ult m cu ε ε γ ε ε ε ε γ (Eqn C-9) which is a quadratic equation in d x As d x is limited to 0.5 for singly reinforcing sections for grades up to 40 under moment distribution not greater than 10% (Clause 6.1.2.4 of the Code), by (Eqn C-9), cu f bd M 2 will be limited to K values as in 154 . 0 ' = K for grade 30 152 . 0 ' = K for grade 35 151 . 0 ' = K for grade 40 which are all smaller than 0.156 under the simplified stress block. However, for grades 45 and 50 and 70 where d x is limited to 0.4 for singly reinforcing sections under moment distribution not greater than 10% (Clause 6.1.2.4 of the Code), again by (Eqn 3-1) cu f bd M 2 will be limited to 126 . 0 ' = K for grade 45 125 . 0 ' = K for grade 50 1199 . 0 ' = K for grade 70 again less than more significantly less than 0.132 under the simplified stress block. Consider grade 100 where d x is limited to 0.33, 0958 . 0 ' = K , again more significantly less than that arrived by the simplified stressed block which is 0.113. The implication is that design based on the rigorous stress block approach requires compression steel at smaller moments than the simplified stress approach. Appendix C C-5 With the d x analyzed by (Eqn C-9), the forces in concrete d x f bd F bx f bx f F F F ult m cu c ult m cu ult m cu c c c | | . | \ | − = ⇒ | | . | \ | − + = + = ε ε γ ε ε γ ε γ ε 0 0 0 2 1 3 1 1 67 . 0 1 67 . 0 3 34 . 1 can be calculated which will be equal to the required force to be provided by steel, thus d x f f bd A d x f bd A f ult m cu y st ult m cu st y | | . | \ | − = ⇒ | | . | \ | − = ε ε γ ε ε γ 0 0 3 1 1 67 . 0 87 . 0 1 3 1 1 67 . 0 87 . 0 (Eqn C-10) When cu f bd M 2 exceeds the limited value for single reinforcement. Compression reinforcements at ' d from the surface of the compression side should be added. The compression reinforcements will take up the difference between the applied moment and 2 ' bd K | . | \ | − | | . | \ | − = ⇒ | | . | \ | − = | . | \ | − d d f f K f bd M bd A K f bd M d d bd A f y cu cu sc cu sc y ' 1 87 . 0 ' ' ' 1 87 . 0 2 2 (Eqn C-11) And the same amount of steel will be added to the tensile steel. | . | \ | − | | . | \ | − + | | . | \ | − = d d f f K f bd M f f bd A y cu cu ult m cu y st ' 1 87 . 0 ' 3 1 1 67 . 0 87 . 0 1 2 0 η ε ε γ (Eqn C-12) where η is the limit of d x ratio which is 0.5 for grade 40 and below and 0.4 for grades up to and including 70. Furthermore, there is a limitation of lever arm ratio not to exceed 0.95 which requires 95 . 0 3 1 1 67 . 0 3 1 1 67 . 0 12 1 3 1 2 1 67 . 0 0 0 2 2 0 0 ≤ | | . | \ | − | | . | \ | − + | . | \ | | | . | \ | − + − d x f d x f d x f ult n cu ult m cu ult ult m cu ε ε γ ε ε γ ε ε ε ε γ Appendix C C-6 | | . | \ | + − | | . | \ | − ≥ ⇒ 2 0 0 0 12 1 3 1 2 1 3 1 1 05 . 0 ult ult ult d x ε ε ε ε ε ε (Eqn C-13) Thus the lower limits for the neutral axis depth ratio are 0.112, 0.113, 0.114 and 0.115 for grades 30, 35, 40, 45 respectively. With such lower limits of neutral axis depth ratio, the tensile steel required will be 0.33%, 0.39%, 0.44% and 0.5% for grades 30, 35, 40 and 45 respectively which are in excess of the minimum of 0.13% in accordance with Table 9.1 of the Code. As illustration for comparison between the rigorous and simplified stress block approaches, plots of 2 bd M against steel percentages is plotted as Comparison of Reinforcement Ratios for Grade 35 according to the Rigorous and Simplified Stress Block 0 2 4 6 8 10 12 14 0 1 2 3 4 5 Steel Percentage (%) M / b d 2 Ast/bd - Rigorous Stress Approach Ast/bd - Simplified Approach Asc/bd - Rigorous Approach Asc/bd - Simplified Approach It can be seen that the differences are very small, maximum error is 1%. Appendix C C-7 Determination of reinforcements for Flanged Beam Section – T- or L-Sections For simplicity, only the simplified stress block in accordance with Figure 6.1 of the Code is adopted. The exercise is first carried out by treating the width of the beam as f b and analyze the beam as if it is a rectangular section. If the 90% of neutral axis depth is within the depth of the flange, i.e. d d d x f ≤ 9 . 0 , the reinforcement so arrived is adequate for the section. The requirement for d d d x f ≤ 9 . 0 is d d K d x f ≤ | | . | \ | − − = 9 . 0 25 . 0 5 . 0 2 9 . 0 (Eqn C-14) If, however, d d d x f > 9 . 0 , the section has to be reconsidered with reference to Figure C-5. For singly reinforced sections, taking moment about the level of the reinforcing steel, ( ) ( )( ) x d x b f d d d b b f M w m cu f f w f m cu 45 . 0 9 . 0 67 . 0 2 67 . 0 − + | | . | \ | − − = γ γ (Eqn C-15) | . | \ | − | . | \ | + | | . | \ | − | | . | \ | − = ⇒ d x d x f d d d d b b f d b M m cu f f w f m cu w 45 . 0 1 9 . 0 67 . 0 2 1 1 1 67 . 0 2 γ γ x 9 . 0 m cu f γ 67 . 0 x d f d f b w b Figure C-5 – Analysis of a T or L beam section Appendix C C-8 Putting | | . | \ | − | | . | \ | − = d d b b d d f d b M f w f f m cu w f 2 1 1 1 67 . 0 2 γ , the equation becomes 0 9 . 0 67 . 0 9 . 0 45 . 0 67 . 0 2 2 = − + × − | . | \ | × × ⇒ d b M M d x f d x f w f m cu m cu γ γ 0 402 . 0 1809 . 0 2 2 = − + − | . | \ | ⇒ d b M M d x f d x f w f cu cu (Eqn C-16) As the neutral axis depth ratio is limited to 0.5 or 0.4 depends on concrete grade in accordance with (Eqn 6.1) to (Eqn 6.3) of the Code, the maximum moment of resistance is, by (Eqn C-15). For grades 40 and below : | . | \ | − | . | \ | + | | . | \ | − | | . | \ | − = = d x d x f f d d b b d d f f d b f M K cu m cu f w f f cu m cu w cu f 45 . 0 1 9 . 0 67 . 0 2 1 1 1 67 . 0 ' 2 γ γ 156 . 0 2 1 1 1 67 . 0 + | | . | \ | − | | . | \ | − = d d b b d d f w f f m γ which is simply 2 d b f M w cu f plus ' K of the rectangular section Similarly for grade above 40 and up to 70 132 . 0 2 1 1 1 67 . 0 ' + | | . | \ | − | | . | \ | − = d d b b d d K f w f f m f γ For tensile steel ratio : ( ) ( ) x b f d b b f d b A f w m cu f w f m cu w st y 9 . 0 67 . 0 67 . 0 87 . 0 γ γ + − = + | | . | \ | − = ⇒ d x d d b b f f d b A f w f m cu y w st 9 . 0 1 67 . 0 87 . 0 1 γ (Eqn C-16) When cu w f d b M 2 exceeds the limited value for single reinforcement. Compression reinforcements at ' d from the surface of the compression side should be added. The compression reinforcements will take up the difference between the applied moment and 2 Kbd | . | \ | − | | . | \ | − = ⇒ | | . | \ | − = | . | \ | − d d f f K f d b M bd A K f bd M d d bd A f y cu f cu w sc cu sc y ' 1 87 . 0 ' ' 1 87 . 0 2 2 (Eqn C-17) Appendix C C-9 And the same amount of steel will be added to the tensile steel. | . | \ | − | | . | \ | − + + | | . | \ | − = d d f f K f d b M d d b b f f bd A y cu f cu w f w f m cu y st ' 1 87 . 0 ' 45 . 0 1 67 . 0 87 . 0 1 2 γ (Eqn C-18) Chart-C1 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.05 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) M / b d 2 Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bd Grade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Chart-C2 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) M / b d 2 Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bd Grade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Chart-C3 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) M / b d 2 Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bd Grade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Chart-C4 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) M / b d 2 Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bd Grade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Chart-C5 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.25 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) M / b d 2 Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bd Grade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Appendix D Underlying Theory and Design Principles for Plate Bending Element Appendix D D-1 Underlying Theory and Design Principles for Plate Bending Element By the finite element method, a plate bending structure is idealized as an assembly of discrete elements joined at nodes. Through the analysis, “node forces” at each node of an element, each of which comprises two bending moments and a shear force can be obtained, the summation of which will balance the applied load at the node. Figures D-1a and D-1b illustrates the phenomena. F Z at Node 2 F A , external load applied at the common node F 4 F 2 F 3 F 1 M X at Node 2 M X at Node 1 F Z at Node 1 F Z at Node 4 F Z at Node 3 M Y at Node 3 3 M Y at Node 2 M X at Node 2 Y Z X 2 1 Figure D-1a – Diagrammatic illustration of the Node Forces at the four Nodes of a Plate Bending Element 1234. Note : M X , M Y , F Z represents respectively the bending moments about X, Y axes and the force in the Z axis at the nodes of the plate bending element For equilibrium, F 1 , F 2 , F 3 and F 4 which are node vertical shear of 4 elements at the common node will sum up to balance the externally applied force F A such that F 1 + F 2 + F 3 + F 4 = F A . Balancing of moments is similar. Figure D-1b – Diagrammatic illustration of balancing of Node shear forces at a common node to 2 or more adjoining elements. The four elements joined at the common node are displaced diagrammatically for clarity. Appendix D D-2 The finite element method goes further to analyze the “stresses” within the discrete elements. It should be noted that “stresss” is a terminology of the finite element method which refer to bending moments, twisting moments and shear forces per unit width in plate bending element. They represent the actual internal forces within the plate structure in accordance with the plate bending theory. R.H. Woods (1968) has developed the famous Wood-Armer Equations to convert the bending moments and twisting moments (both are moments per unit width) at any point to “design moments” in two directions for structural design purpose. Outline of the plate bending theory Apart from bending moment in two mutually perpendicular directions as well known by engineers, a twisting moment can be proved to be in existence by the plate bending theory. The bending and twisting moments constitutes a “moment field” which represents the actual structural behaviour of a plate bending structure. The existence of the twisting moment and its nature are discussed in the followings. Consider a triangular element in a plate bending structure with two of its sides aligning with the global X and Y directions as shown in Figure D-2 where moments X M and Y M (both in kNm per m width) are acting respectively about X and Y. A moment B M will generally be acting on the hypotenuse making an angle of θ with the X-axis as shown to achieve equilibrium. However, as the resultant of X M and Y M does not necessarily align with B M , so there will generally be a moment acting in the perpendicular direction of B M to achieve equilibrium which is denoted as T M . The vector direction of T M is normal to the face of the hypotenuse. So instead of “bending” the element like X M , Y M and B M which produces flexural stresses, it “twists” the element and produce shear stress in the in-plane direction. The shear stress will follow a triangular pattern as shown in Figure D-2 for stress-strain compatibility. T M is therefore termed the “twisting moment”. Furthermore, in order to achieve rotational equilibrium about an axis out of plane, the shear stress will have to be “complementary”. As the hypotenuse can be in any directions of the plate structure, it follows that at any point in the plate bending structure, there will generally be two bending moments, say X M and Y M in two mutually perpendicular directions coupled with a complementary twisting moment XY M as indicated in Figure 11a. The phenomenon is in exact analogy to the in-plane stress problem where generally two direct stresses coupled with a shear stress exist and these components vary with directions. The equations relating B M , T M with X M , Y M , XY M and θ derived from equilibrium conditions are stated as follows: Appendix D D-3 ( ) ( ) ( ) θ θ θ θ 2 sin 2 sin 2 1 2 sin 2 cos 2 1 2 1 XY Y X T XY Y X Y X B M M M M M M M M M M − − = + − + + = (Eqn D-1) In addition, if θ is so varied that T M vanishes when φ θ = , then the element will be having pure bending in the direction. The moments will be termed the “principal moments” and denoted as 1 M , 2 M , again in exact analogy with the in-plane stress problem having principal stresses at orientations where shear stresses are zero. The angle φ can be worked out by ) ( 2 tan 2 1 1 Y X XY M M M − = − φ (Eqn D-2) Complementary shear stress pattern Y X θ Plate Structure M B M T M Y M X Figure D-2 – Derivation and nature of the “Twisting Moment” Appendix D D-4 Again, as similar to the in-plane stress problem, one may view that the plate bending structure is actually having principal moments “bending” in the principal directions which are free of “twisting”. Theoretically, it will be adequate if the designer designs for these principal moments in the principal directions which generally vary from point to point. However, practically this is not achievable for reinforced concrete structures as we cannot vary the directions of the reinforcing steels from point to point and from load case to load case. The “stress” approach for design against flexure would therefore involve formulae for providing reinforcing steels in two directions (mostly in orthogonal directions) adequate to resist the “moment field” comprising the bending moments and twisting moments. The most popular one is the “Wood Armer” Equations by Woods (1968), the derivation of which is based on the “normal yield criterion” which requires the provided reinforcing steels at any point to be adequate to resist the normal moment which is the bending moment B M in any directions as calculated from (Eqn D-1). The effects of the twisting moments have been taken into account in the formulae. The Wood Armer Equations are listed as follows. For bottom steel reinforcement provisions: XY X X M M M + = ∗ if 0 ≥ ∗ X M ; XY Y Y M M M + = ∗ if 0 ≥ ∗ Y M If 0 < ∗ X M , then 0 = ∗ X M and X XY Y Y M M M M 2 + = ∗ M Y X M 1 M 1 φ M Y M XY M XY M X M X M XY M XY M 2 M 2 Y Figure D-3a – General co-existence of bending moments and twisting moment in a plate bending structure Figure D-3b – Principal moment in a plate bending structure Appendix D D-5 If 0 < ∗ Y M , then 0 = ∗ Y M and Y XY X X M M M M 2 + = ∗ For top steel reinforcement provisions: XY X X M M M − = ∗ if 0 ≤ ∗ X M ’ XY Y Y M M M − = ∗ if 0 ≤ ∗ Y M If 0 > ∗ X M , then 0 = ∗ X M and X XY Y Y M M M M 2 − = ∗ If 0 < ∗ Y M , then 0 = ∗ Y M and Y XY X X M M M M 2 − = ∗ (Eqn D-3) The equations have been incorporated in the New Zealand Standard NZS 3101:Part 2:1995 as solution approach for a general moment field. The “stress” approach is therefore based on the actual structural behaviour of the plate bending structure which is considered as a direct and realistic approach. The approach is particularly suitable for structures analyzed by the finite element method which produces a complete set of components of the internal forces of the plate bending structures including twisting moments, max Q . Design has to cater for all these components to ensure structural adequacy. Design against shear As an alternative to checking or designing punching shear in slab in accordance with 6.1.5.7 of the Code by which the punching shear load created by column (or pile in pile cap) is effectively averaged over a perimeter, more accurate design or checking can be carried out which is based on finite element analysis by which an accurate shear stress distribution in the slab structure can be obtained. The finite element analysis outputs the “shear stresses” (shear force per unit width) in accordance with the general plate bending theory at the “X-face” and “Y-face” of an element which are respectively y M x M Q XY Y XZ ∂ ∂ + ∂ ∂ − = and y M y M Q XY X YZ ∂ ∂ − ∂ ∂ = , as diagrammatically illustrated in Figure D-4. It can be easily shown that the maximum shear after “compounding” these two components will occur in a plane at an orientation         = − YZ XZ Q Q 1 tan θ on plan and the value of the maximum shear is Appendix D D-6 2 2 max YZ XZ Q Q Q + = as per the illustration in the same Figure. Thus one can view that both XZ Q and YZ Q are components of the actual shears in a pre-set global axis system. The actual shear stress is max Q , the action of which tends to produce shear failure at the angle θ on plan as shown in Figure D-3. So the designer needs to check or design for max Q at the spot. There is no necessity to design for XZ Q and YZ Q separately. Following the usual practice of designing against shear in accordance with the Code, if the max Q does not exceed allowable shear strength of concrete based on v c (the max Q X θ θ Plan Out-of plane shear Q XZ , Q YZ and Q θ YZ Q Plate structure Z Y X XZ Q YZ Q θ Q XZ Q θ Derivation For vertical equilibrium, Q θ can be expressed as θ θ θ cos sin 1 YZ XZ Q Q Q + = × For θ Q to be maximum set YZ XZ YZ XZ Q Q Q Q d dQ = ⇒ = − ⇒ = θ θ θ θ θ tan 0 sin cos 0 2 2 2 2 2 2 max YZ XZ YZ YZ XZ XZ Q Q Q Q Q Q Q Q + + + = = ⇒ θ 2 2 YZ XZ Q Q + = A rectangular element extracted from a plate structure showing shear per unit width in the X and Y directions The triangular element formed from the rectangular element by cutting into half. By varying the angle θ, the maximum shear on the hypotenuse is obtained. Potential shear failure at the orientation ( ) YZ XZ Q Q / tan 1 − = θ max Q Figure D-4 – Diagrammatic illustration of shear “stresses” in the X and Y faces of an element in Plate bending structure, potential shear failure and Derivation of the magnitude and orientation of the design shear stress Appendix D D-7 design concrete shear stress with enhancement as appropriate) no shear reinforcements will be required. Otherwise, reinforcements will be required to cater for the difference. The “stress” approach for shear design based on max Q can best be carried out by graphical method, so as to avoid handling the large quantity of data obtainable in the finite element analysis. An illustration of the method for a raft footing is indicated in Figure D-5 as : (i) an enveloped shear stress (shear force per unit width) contour map of a structure due to applied loads is first plotted as shown in Figure D-5(a); (ii) the concrete shear strength contour of the structure which is a contour map indicating the shear strength of the concrete structure after enhancement of the design concrete shear stress (v c ) to closeness of supports in accordance with established code requirements (say BS8110) is plotted as shown in Figure D-5(b); (iii) locations where the stresses exceed the enhanced strengths be reinforced by shear links as appropriate in accordance with established code requirements as shown in Figure D-5(c). Figure D-5a – Stress contour of enveloped shear “stresses” of a raft footing due to applied load Appendix D D-8 Figure D-5b – Strength contour of the raft footing with enhancement Figure D-5c – Arrangement of shear reinforcements Appendix E Extracts of Design Charts for Slabs Appendix F Derivation of Design Formulae for Rectangular Columns to Rigorous Stress Strain Curve of Concrete Appendix F F-1 Derivation of Design Formulae for Rectangular Columns to Rigorous Stress Strain Curve of Concrete (I) Computing stress / force contribution of concrete stress block Assuming the parabolic portion of the concrete stress block as indicated in Fig. 3.8 of HKCoP2004 be represented by the equation ε ε σ B A + = 2 (where A and B are constants) (Eqn F-1) So B A d d + = ε ε σ 2 (Eqn F-2) As c c E B E d d = ⇒ = =0 ε ε σ where c E is the tangential Young’s Modulus of concrete listed in Table 3.2 of the Code. Also 0 0 0 2 2 0 2 0 0 ε ε ε ε σ ε ε c E B A B A d d − = − = ⇒ = + ⇒ = = (Eqn F-3) As m cu f γ σ 67 . 0 = when 0 ε ε = m c cu c m cu c c m cu E f E f E E f γ ε ε γ ε ε ε γ 34 . 1 2 67 . 0 2 67 . 0 0 0 0 0 2 0 = ⇒ = ⇒ − = − ∴ (Eqn (F-4) (accords with 3.14 of the Concrete Code Handbook) 0 2ε c E A − = where m c cu E f γ ε 34 . 1 0 = So the equation of the parabola is ε ε ε σ c c E E + − = 2 0 2 for 0 ε ε ≤ Consider the linear strain distribution h x 0 ε ε = u ult x ε ε / 0 0035 . 0 = ult ε Figure F-1 – Strain diagram across concrete section Appendix F F-2 At distance u from the neutral axis, x u ult ε ε = So stress at u from the neutral axis up to ult x ε ε 0 is u x E u x E x u E x u E E E ult c ult c ult c ult c c c ε ε ε ε ε ε ε ε ε σ + − = | . | \ | + | . | \ | − = + − = 2 2 0 2 2 0 2 0 2 2 2 (Eqn F-5) Based on (Eqn F-5), the stress strain profiles for grade 35 within the concrete compression section are plotted in Figure F2 for illustration. Stress Strain Profile for Grade 35 0 2 4 6 8 10 12 14 16 18 0 0.2 0.4 0.6 0.8 1 Distance Ratio from Neutral axis S t r e s s ( M P a ) By the properties of parabola as shown in Figure F-3, we can formulate total force offered by the parabolic section as 1 c F given by Area = ab 3 2 centre of mass a 8 3 b a Figure F-3 – Geometrical Properties of Parabola 0.3769 where ε 0 = 0.001319 Figure F-2 – Stress strain profile of grades 35 Appendix F F-3 bx f f x b F ult m cu m cu ult c ε γ ε γ ε ε 3 34 . 1 67 . 0 3 2 0 0 1 = = (Eqn F-6) and the moment exerted by 1 c F about centre line of the whole section | | . | \ | − − = − | | . | \ | − − = ult c ult ult c c x h F x x h F M ε ε ε ε ε ε 0 1 0 0 1 1 8 5 1 2 8 3 1 2 (Eqn F-7) The force by the straight portion is | | . | \ | − = | | . | \ | − = ult m cu ult m cu c bx f b x x f F ε ε γ ε ε γ 0 0 2 1 67 . 0 67 . 0 (Eqn F-8) The moment offered by the constant part about the centre line of the whole section is | | . | \ | − − = 2 1 2 0 2 2 x h F M ult c c ε ε (Eqn F-9) Thus if full section of concrete in compression exists in the column section h x f bh F bh F bh F ult m cu c c c | | . | \ | − = + = ε ε γ 0 2 1 3 1 1 67 . 0 (Eqn F-10) | | . | \ | − | . | \ | − | . | \ | = | | . | \ | − − = ult ult m cu ult ult m cu c h x h x f bh x h bx f bh M ε ε ε γ ε ε ε ε γ ε 0 0 2 0 0 2 1 8 5 1 2 1 3 34 . 1 1 8 5 1 2 3 34 . 1 | | . | \ | − − | | . | \ | − | . | \ | = | | . | \ | − − | | . | \ | − = h x h x f bh x h bx f bh M ult ult m cu ult ult m cu c ε ε ε ε γ ε ε ε ε γ 0 0 2 0 0 2 2 1 1 1 2 67 . 0 1 2 1 2 1 67 . 0 | | . | \ | − − | | . | \ | − | . | \ | + | | . | \ | − | . | \ | − | . | \ | = + = h x h x f h x h x f bh M M bh M ult ult m cu ult ult m cu c c c ε ε ε ε γ ε ε ε γ ε 0 0 0 0 2 2 1 2 1 1 1 2 67 . 0 8 5 1 2 1 3 34 . 1 ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | . | \ | − + − + − | . | \ | = h x h x f ult ult ult m cu 2 0 0 0 12 1 3 1 2 1 6 1 2 1 67 . 0 ε ε ε ε ε ε γ (Eqn F-11) (II) Derivation of Basic Design Formulae of R.C. column sections Cases 1 to 7 with different stress / strain profile of concrete and steel across the column section due to the differences in the neutral axis depth ratios, h x , are investigated. The section is reinforced by continuous reinforcements sh A along its length h idealized as continuum and reinforcements at its end faces sb A with cover ' d . Pursuant to the derivation of the stress strain relationship of concrete and steel, the Appendix F F-4 stress strain diagram of concrete and steel for Cases 1 to 7 are as follows, under the definition of symbols as : b : width of the column h : length of the column x : neutral axis depth of the column sb A : total steel area at the end faces of the column ' d : concrete cover to the centre of the end face steel sh A : total steel area along the length of the column Case 1 – where x/h ≤ 7/3(d’/h) Pursuant to the derivation of the stress strain relationship of concrete and steel, the stress strain diagram of concrete and steel for Case 1 is as indicated in Figure F-4.1 : Steel compressive force in the portion steel elastic zone by sb A is sb y sb y sc A f x d A f x d x F 5 . 0 87 . 0 ' 1 4 7 5 . 0 87 . 0 7 / 4 ' 1 × × | . | \ | − = × × | . | \ | − = (Eqn F-12) Steel compressive force in the portion steel plastic zone by sh A is | . | \ | × = | . | \ | × = h x A f x h A f F sh y sh y sc 7 3 87 . 0 7 3 87 . 0 2 (Eqn F-13) Steel compressive force in the portion steel elastic zone by sh A is | . | \ | × = × | . | \ | × = h x A f x h A f F sh y sh y sc 7 2 87 . 0 2 1 7 4 87 . 0 3 (Eqn F-14) Steel tensile force in the portion steel plastic zone by sb A is sb y st A f F 5 . 0 87 . 0 1 × = (Eqn F-15) Steel tensile force in the portion steel plastic zone by sh A is | . | \ | − × = | . | \ | − × = h x A f x h h A f F sh y sh y st 7 11 1 87 . 0 7 11 87 . 0 2 (Eqn F-16) Steel tensile force in the portion steel elastic zone by sh A is | . | \ | × = × | . | \ | × = h x A f x h A f F sh y sh y st 7 2 87 . 0 2 1 7 4 87 . 0 3 (Eqn F-17) To balance the external load u N bh F bh F bh F bh F bh F bh F bh F bh F bh N st st st sc sc sc c c u 3 2 1 3 2 1 2 1 − − − + + + + = bh F bh F bh F bh F bh F bh N st st sc sc c u 2 1 2 1 − − + + = as bh F sc3 and bh F st 3 are equal and opposite. By (Eqn 10) Appendix F F-5 | . | \ | × + × × | . | \ | − + | | . | \ | − = ⇒ h x bh A f bh A f x d h x bx f bh N sh y sb y ult m cu u 7 3 87 . 0 5 . 0 87 . 0 ' 1 4 7 3 1 1 67 . 0 0 ε ε γ | . | \ | − × − × − h x bh A f bh A f sh y sb y 7 11 1 87 . 0 5 . 0 87 . 0 bh A f x h h d bh A h x f h x f bh N sb y sh y ult m cu u 87 . 0 ' 8 7 8 3 1 2 87 . 0 3 1 1 67 . 0 0 | . | \ | − + | . | \ | − + | | . | \ | − = ⇒ ε ε γ (Eqn F-18) Re-arranging (Eqn F-18) F sc1 4x/7 3x/7 ε s = 0.002 F st2 F st3 F sc3 F sc2 F c2 F c1 4x/7 h–11x/7 4x/7 0.87f y 3x/7 4x/7 m cu f γ 67 . 0 ' d ' d h x 0 ε ε = u ult x ε ε / 0 0035 . 0 = ult ε Concrete stress Block Steel stress Block Strain diagram across whole section Figure F-4.1 – Concrete and steel stress strain relation for Case 1 F st1 Appendix F F-6 h x bh A bh A f bh N h x bh A f f sb sh y u sh y ult m cu | . | \ | − + − | . | \ | × + | | . | \ | − 8 3 87 . 0 87 . 0 2 3 1 1 67 . 0 2 0 ε ε γ 0 ' 87 . 0 8 7 = − h d bh A f sb y (Eqn F-19) (Eqn F-19) can be used for solve for h x To balance the external load u M 2 3 2 2 2 1 2 3 2 2 2 1 2 2 2 1 2 bh M bh M bh M bh M bh M bh M bh M bh M bh M st st st sc sc sc c c u + + + + + + + = (Eqn F-20) By (Eqn F-11) ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | . | \ | − + − + − | . | \ | = + = h x h x f bh M bh M bh M ult ult ult m cu c c c 2 0 0 0 2 2 2 1 2 12 1 3 1 2 1 6 1 2 1 67 . 0 ε ε ε ε ε ε γ | . | \ | − × × | . | \ | − = | . | \ | − × × | . | \ | − = h d bh A f x h h d d h bh A f x d bh M sb y sb y sc ' 2 1 5 . 0 87 . 0 ' 1 4 7 ' 2 5 . 0 87 . 0 ' 1 4 7 2 2 1 (Eqn F-21) | . | \ | − | . | \ | = | . | \ | − | . | \ | × = h x h x bh A f bh x h h x A f bh M sh y sh y sc 14 3 2 1 7 3 87 . 0 1 14 3 2 7 3 87 . 0 2 2 2 (Eqn F-22) | . | \ | − | . | \ | = | . | \ | − | . | \ | × = h x h x bh A f bh x h h x A f bh M sh y sh y sc 21 13 2 1 7 2 87 . 0 1 21 13 2 7 2 87 . 0 2 2 3 (Eqn F-23) | . | \ | − × = | . | \ | − × = h d bh A f d h bh A f bh M sb y sb y st ' 2 1 5 . 0 87 . 0 ' 2 5 . 0 87 . 0 2 2 1 (Eqn F-24) | . | \ | | . | \ | − × = | . | \ | | . | \ | − × = h x h x bh A f x h x bh A f bh M sh y sh y st 14 11 7 11 1 87 . 0 14 11 7 11 1 87 . 0 2 2 2 (Eqn F-25) | . | \ | − | . | \ | × = | . | \ | − | . | \ | × = 2 1 21 29 7 2 87 . 0 1 2 21 29 7 2 87 . 0 2 2 3 h x h x bh A f bh h x h x A f bh M sh y sh y st (Eqn F-26) | . | \ | − × + | . | \ | − × × | . | \ | − = + h d bh A f h d bh A f x h h d bh M M sb y sb y st sc ' 2 1 5 . 0 87 . 0 ' 2 1 5 . 0 87 . 0 ' 1 4 7 2 1 1 | . | \ | − | . | \ | − = x h h d h d bh A f sb y ' 8 7 8 11 ' 2 1 87 . 0 (Eqn F-27) | . | \ | | . | \ | − × + | . | \ | − | . | \ | = + h x h x bh A f h x h x bh A f bh M M sh y sh y st sc 14 11 7 11 1 87 . 0 14 3 2 1 7 3 87 . 0 2 2 2 | . | \ | − | . | \ | = 2 49 65 87 . 0 h x h x bh A f sh y (Eqn F-28) | . | \ | − | . | \ | × + | . | \ | − | . | \ | = + 2 1 21 29 7 2 87 . 0 21 13 2 1 7 2 87 . 0 2 3 3 h x h x bh A f h x h x bh A f bh M M sh y sh y st sc 2 147 32 87 . 0 | . | \ | = h x bh A f sh y (Eqn F-29) Appendix F F-7 Total | . | \ | − | . | \ | + | . | \ | − | . | \ | − = 2 2 147 163 87 . 0 ' 8 7 8 11 ' 2 1 87 . 0 h x h x bh A f x h h d h d bh A f bh M sh y sb y s ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | − | . | \ | + | . | \ | − | . | \ | − = ⇔ 2 2 147 163 ' 8 7 8 11 ' 2 1 87 . 0 h x h x bh A x h h d h d bh A f bh M sh sb y s (Eqn F-30) Substituting into (Eqn F-20) ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | . | \ | − + − + − | . | \ | = h x h x f bh M ult ult ult m cu u 2 0 0 0 2 12 1 3 1 2 1 6 1 2 1 67 . 0 ε ε ε ε ε ε γ ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | − | . | \ | + | . | \ | − | . | \ | − + 2 147 163 ' 8 7 8 11 ' 2 1 87 . 0 h x h x bh A x h h d h d bh A f sh sb y (Eqn F-31) Case 2 – where 7/3(d’/h) < x/h ≤ 7/11 (1 – d’/h) There are two sub-cases to be considered in Case 2, i.e. Case 2(a) – 14 3 ' ≥ h d and Case 2(b) – 14 3 ' < h d For Case 2(a), where 14 3 ' ≥ h d . However, 2 1 ' 3 7 ≥ h d and 2 1 ' 1 11 7 < | . | \ | − h d . So this case doesn’t exist. For Case 2(b), where 14 3 ' < h d both 1 sc A and 1 st A are in the plastic zone as shown in Figure 3.2. The various components of stresses in concrete and steel are identical to that of Case 1 except that of 1 sc A where sb y sc A f F 5 . 0 87 . 0 1 × = and | . | \ | − × = h d A f M sb y sc ' 2 1 5 . 0 87 . 0 1 It can be seen that 1 st F and 1 sc F are identical but opposite in direction, so cancel out. By formulation similar to the above, bh A h x f h x f bh N sh y ult m cu u | . | \ | − + | | . | \ | − = 1 2 87 . 0 3 1 1 67 . 0 0 ε ε γ (Eqn F-32) × + | | . | \ | − + = ⇒ bh A f f bh A f bh N h x sh y ult m cu sh y u 87 . 0 2 3 1 1 67 . 0 87 . 0 0 ε ε γ (Eqn F-33) Appendix F F-8 ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | . | \ | − + − + − | . | \ | = h x h x f bh M ult ult ult m cu u 2 0 0 0 2 12 1 3 1 2 1 6 1 2 1 67 . 0 ε ε ε ε ε ε γ ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | − | . | \ | + | . | \ | − + 2 147 163 ' 2 1 87 . 0 h x h x bh A h d bh A f sh sb y (Eqn F-34) Case 3 – where 7/11(d’/h) < x/h ≤ 7/11 for d’/h > 3/14 and 7/11(1 – d’/h) < x/h ≤ 7/11 for d’/h < 3/14 The concrete / steel stress / strain diagram is worked out as shown in Figure F-4.3. It F st2 F st3 F sc3 F sc2 F c2 F c1 4x/7 h–11x/7 4x/7 0.87f y 3x/7 4x/7 m cu f γ 67 . 0 ' d ' d h x 0 ε ε = u ult x ε ε / 0 0035 . 0 = ult ε Concrete stress Block Steel stress Block ε s = 0.002 Strain diagram across whole section Figure F-4.2 – Concrete and steel stress strain relation for Case 2(b) Appendix F F-9 should be noted that the components of stresses are identical to that of Case 2 except that by 1 st A which is ( ) ( ) sb y sb y sb y st A x d x h f A x d x h f A x d x h f F 8 7 1 ' 87 . 0 8 7 ' 87 . 0 5 . 0 7 / 4 ' 87 . 0 1 | . | \ | − − = − − = × − − = (Eqn F-35) By derivation similar to the above in balancing axial load, bh A h x f x d x h bh A f h x f bh N sh y sb y ult m cu u | . | \ | − + | . | \ | + − + | | . | \ | − = 1 2 87 . 0 ' 8 7 8 7 8 11 87 . 0 3 1 1 67 . 0 0 ε ε γ (Eqn F-36) F st2 F st3 F sc3 F sc2 F c2 F c1 4x/7 h–11x/7 4x/7 0.87f y 3x/7 4x/7 m cu f γ 67 . 0 ' d ' d h x 0 ε ε = u ult x ε ε / 0 0035 . 0 = ult ε Concrete stress Block Steel stress Block ε s = 0.002 Strain diagram across concrete section Figure F-4.3 – Concrete and steel stress strain relation for Case 3 Appendix F F-10 Re-arranging (Eqn F-36) | . | \ | − | . | \ | − + | . | \ | × + | | . | \ | − h x bh N bh A bh A f h x bh A f f u sh sb y sh y ult m cu 8 11 87 . 0 87 . 0 2 3 1 1 67 . 0 2 0 ε ε γ 0 1 ' 8 7 87 . 0 = | . | \ | − + h d bh A f sb y (Eqn F-37) By balancing moments : ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | . | \ | − + − + − | . | \ | = h x h x f bh M ult ult ult m cu u 2 0 0 0 2 12 1 3 1 2 1 6 1 2 1 67 . 0 ε ε ε ε ε ε γ ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | − | . | \ | + − | . | \ | | . | \ | − | . | \ | | . | \ | − + bh A h x h x bh A x h h d x h h d f sh sb y 2 147 163 8 3 ' 8 7 8 7 ' 2 1 87 . 0 (Eqn F-38) Case 4 – where x < h < 11x/7, i.e. 7/11 < x/h ≤ 1 The concrete / steel stress / strain diagram is worked out as shown in Figure F-4.4. It should be noted that the components of stresses are identical to that of Case 3 except that by 2 st A which vanishes. Hence by derivation similar to the above by balancing axial load, | . | \ | + − − + | . | \ | + − + | | . | \ | − = 4 7 8 7 56 9 87 . 0 ' 8 7 8 7 8 11 87 . 0 3 1 1 67 . 0 0 x h h x bh A f x d x h bh A f h x f bh N sh y sb y ult m cu u ε ε γ (Eqn F-39) Re-arranging | . | \ | − | . | \ | + + | . | \ | × − | | . | \ | − h x bh N bh A bh A f h x bh A f f u sh sb y sh y ult m cu 4 7 8 11 87 . 0 87 . 0 56 9 3 1 1 67 . 0 2 0 ε ε γ 0 1 ' 8 7 87 . 0 = − | . | \ | − + bh A h d bh A f sh sb y (Eqn F-40) which can be solved as a quadratic equation. By balancing moment : ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | | | . | \ | − | . | \ | + − | . | \ | − | . | \ | = h x h x h x h x f bh M ult ult ult m cu u 2 0 0 0 2 12 1 3 1 6 1 1 2 1 67 . 0 ε ε ε ε ε ε γ ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | + − + − | . | \ | | . | \ | − | . | \ | | . | \ | − + bh A h x h x x h bh A x h h d x h h d f sh sb y 2 392 9 112 9 48 7 8 3 ' 8 7 8 7 ' 2 1 87 . 0 (Eqn F-41) Appendix F F-11 Case 5 – where x>h>(1 – ε 0 /ε ult )x, i.e 1 < x/h ≤ 1/(1 – ε 0 /ε ult ) The concrete / steel stress / strain diagram is worked out as follows. The whole section is in compression so that the neutral axis depth ratio is greater than unity and hence becomes a hypothetical concept. The contribution by 1 c F becomes partial and need be calculated by integration to partial limits. The contribution by 1 st F become compression ' 1 sc F and 2 st F , 3 st F vanish. ( ) x x h − = 0035 . 0 1 ε F st3 F sc3 F sc2 F c2 F c1 4x/7 h – x 0.87f y 3x/7 4x/7 m cu f γ 67 . 0 ' d ' d h x 0 ε ε = u ult x ε ε / 0 0035 . 0 = ult ε Concrete stress Block Steel stress Block ε s = 0.002 Strain diagram across concrete section Figure F-4.4 – Concrete and steel stress strain relation for Case 4 Appendix F F-12 Concrete compressive stresses and forces bdu F ult x h x c ∫ − = ε ε σ / 1 0 where u x E u x E ult c ult c ε ε ε σ + − = 2 2 0 2 2 (Eqn F-42) du u x b E du u x b E bdu u x E u x E F ult ult ult x h x ult c x h x ult c x h x ult c ult c c ∫ ∫ ∫ − − − + − = + − = ε ε ε ε ε ε ε ε ε ε ε ε / / 2 2 0 2 / 2 2 0 2 1 0 0 0 2 2 | | . | \ | − + | | . | \ | − − | | . | \ | − = ⇒ 0 2 3 3 0 0 2 2 2 0 1 2 1 6 1 2 ε ε ε ε ε ε ε ε ε ε ult c ult c ult ult c ult ult c c E E h x E E bh F ( ) x x h − = 0035 . 0 1 ε F sc3 F sc2 F c2 F c1 4x/7 0.87f y 3x/7 4x/7 m cu f γ 67 . 0 ' d ' d h x 0 ε ε = u ult x ε ε / 0 0035 . 0 = ult ε Concrete stress Block Steel stress Block ε s = 0.002 Strain diagram across concrete section Figure F-4.5 – Concrete and steel stress strain relation for Case 5 Appendix F F-13 2 0 2 0 2 6 2 2 | . | \ | − | | . | \ | − + x h E x h E E ult c ult c ult c ε ε ε ε ε (Eqn F-43) h x f F ult m cu c | | . | \ | − = ε ε γ 0 2 1 67 . 0 (same as the previous cases) (Eqn F-44) | | . | \ | − + | | . | \ | − + | | . | \ | − − | | . | \ | − = + = 0 2 0 0 3 3 0 0 2 2 2 0 2 1 2 1 2 1 6 1 2 ε ε ε ε ε ε ε ε ε ε ε ε ε ult c ult c ult c ult ult c ult ult c c c c E E h x E E E bh F bh F bh F 2 0 2 0 2 6 2 2 | . | \ | − | | . | \ | − + x h E x h E E ult c ult c ult c ε ε ε ε ε (Eqn F-45) Steel compressive force in the portion steel plastic zone by sb A is bh A f bh F A f F sb y sc sb y sc 5 . 0 87 . 0 5 . 0 87 . 0 1 1 × = ⇔ × = (Eqn F-46) ( ) bh A h d x h f bh F A x d h x f F sb y sc sb y sc 8 7 ' 1 1 87 . 0 5 . 0 7 / 4 ' 87 . 0 ' 1 ' 1 | . | \ | − − = ⇒ × + − = (Eqn F-47) bh A h d x h f bh F bh F sb y sc sc | . | \ | − − = + ' 1 8 7 8 11 87 . 0 ' 1 1 (Eqn F-48) Steel compressive force in the portion steel plastic zone by sh A is bh A h x f bh F x h A f F sh y sc sh y sc | . | \ | = ⇔ | . | \ | × = 7 3 87 . 0 7 3 87 . 0 2 2 (Eqn F-49) Steel compressive force in the portion steel elastic zone by sh A is 2 1 7 3 7 / 4 1 87 . 0 7 3 7 / 4 87 . 0 3 × | . | \ | − × | . | \ | − − + | . | \ | − × − × = x h h A x h x f x h h A x h x f F sh y sh y sc bh A h x x h f bh F sh y sc | . | \ | − − = ⇒ 56 33 8 7 4 7 87 . 0 3 (Eqn F-50) bh A h x x h f bh A h x x h f bh A h x f bh F bh F sh y sh y sh y sc sc | . | \ | − − = | . | \ | − − + | . | \ | = + 56 9 8 7 4 7 87 . 0 56 33 8 7 4 7 87 . 0 7 3 87 . 0 3 2 (Eqn F-51) As bh F bh F bh F bh F bh F bh N sc sc sc sc c u 3 2 ' 1 1 + + + + = bh A h x x h f bh A h d x h f x h E x h E E E E h x E E E bh N sh y sb y ult c ult c ult c ult c ult c ult c ult ult c ult ult c u | . | \ | − − + | . | \ | − − + | . | \ | − | | . | \ | − + | | . | \ | − + | | . | \ | − + | | . | \ | − − | | . | \ | − = 56 9 8 7 4 7 87 . 0 ' 1 8 7 8 11 87 . 0 6 2 2 2 1 2 1 6 1 2 2 0 2 0 2 0 2 0 0 3 3 0 0 2 2 2 0 ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε h x bh A f E bh N sh y ult ult ult ult c u − | | . | \ | + − + − = ⇔ 56 9 87 . 0 6 1 2 1 2 1 6 1 0 0 2 2 0 ε ε ε ε ε ε ε Appendix F F-14 | | . | \ | + + − + bh A f bh A f E E sh y sb y ult c ult c 4 7 87 . 0 8 11 87 . 0 2 0 2 ε ε ε 2 0 2 0 2 6 8 7 87 . 0 1 ' 8 7 87 . 0 2 2 | . | \ | − − | . | \ | − + − + x h E x h bh A f h d bh A f E E ult c sh y sb y ult c ult c ε ε ε ε ε (EqnF-52) Re-arranging (Eqn F-52) 3 0 0 2 2 0 56 9 87 . 0 6 1 2 1 2 1 6 1 | . | \ | − | | . | \ | + − + − h x bh A f E sh y ult ult ult ult c ε ε ε ε ε ε ε 2 0 2 4 7 87 . 0 8 11 87 . 0 2 | . | \ | | | . | \ | − + + − + h x bh N bh A f bh A f E E u sh y sb y ult c ult c ε ε ε 0 6 8 7 87 . 0 1 ' 8 7 87 . 0 2 2 0 2 0 2 = − − | . | \ | − + − + ε ε ε ε ε ult c sh y sb y ult c ult c E h x bh A f h d bh A f E E (Eqn F-53) which is a cubic equation in h x that can be solved by formulae or by numerical method with pre-determined bh A sh , bh A sb and under applied u N . Summing the Moments as follows : Concrete compressive stresses and moments | . | \ | + − = ∫ − u x h bdu M ult x h x c 2 / 1 0 ε ε σ where u x E u x E ult c ult c ε ε ε σ + − = 2 2 0 2 2 du u x h b u x E u x E M ult x h x ult c ult c c | . | \ | + − + − = ∫ − 2 2 / 2 2 0 2 1 0 ε ε ε ε ε (Eqn F-54) Integrating and expanding | . | \ | − + − + | . | \ | | | . | \ | − + | . | \ | + − + | | . | \ | − | . | \ | − − + − + | . | \ | | | . | \ | − + − + | | . | \ | − | . | \ | − = 2 2 4 4 0 2 3 3 0 0 2 2 3 3 0 2 2 0 2 1 4 6 4 1 4 1 3 3 1 2 1 3 1 2 3 3 1 3 1 2 1 2 1 2 1 x h x h h x h x x h x h h x h x E x h h x h x x h h x h x E bh M ult ult ult c ult ult ult c c ε ε ε ε ε ε ε ε ε ε ε | | . | \ | − − | | . | \ | − | . | \ | = | | . | \ | − − | | . | \ | − = h x h x f bh x h bx f bh M ult ult m cu ult ult m cu c ε ε ε ε γ ε ε ε ε γ 0 0 2 0 0 2 2 1 1 1 2 67 . 0 1 2 1 2 1 67 . 0 | . | \ | − × = | . | \ | − × = h d bh A f d h bh A f bh M sb y sb y sc ' 2 1 5 . 0 87 . 0 ' 2 5 . 0 87 . 0 2 2 1 bh A h d x h h d x h f h d h bh A x d x h f bh M sb y sb y sc | . | \ | − | . | \ | + − − = | . | \ | − | . | \ | + − − = ' 2 1 ' 1 8 7 87 . 0 1 ' 2 8 7 ' 1 87 . 0 2 ' 1 bh A h d x h h d f sb y | . | \ | − | . | \ | − + − = ' 2 1 1 ' 1 8 7 87 . 0 Appendix F F-15 bh A h d x h h d f h d bh A f bh M bh M sb y sb y sc sc | . | \ | − | . | \ | − + − | . | \ | − × = + ' 2 1 1 ' 1 8 7 87 . 0 ' 2 1 5 . 0 87 . 0 2 ' 1 2 − | . | \ | − | . | \ | − × = + ⇒ 8 3 ' 1 8 7 ' 2 1 87 . 0 2 ' 1 2 x h h d h d bh A f bh M bh M sb y sc sc (Eqn F-55) bh A h x h x f bh A h x h x f bh M sh y sh y sc | . | \ | − = | . | \ | − | . | \ | = 2 2 2 98 9 14 3 87 . 0 14 3 2 1 7 3 87 . 0 | . | \ | − | . | \ | + − × × − | . | \ | − − × × − = 6 1 7 2 56 9 4 3 8 7 87 . 0 14 3 4 3 4 7 2 5 87 . 0 2 3 h x h x x h bh A f h x h x x h bh A f bh M sh y sh y sc | . | \ | + − = 2 392 45 112 33 48 7 87 . 0 h x h x x h bh A f sh y | . | \ | + − = + ∴ 2 2 3 2 2 392 9 112 9 48 7 87 . 0 h x h x x h bh A f bh M bh M sh y sc sc (Eqn F-56) Total Moment 2 3 2 2 2 ' 1 2 1 2 2 2 1 2 bh M bh M bh M bh M bh M bh M bh M sc sc sc sc c c + + + + + = bh A h x h x x h f x h h d h d bh A f h x h x f x h x h h x h x x h x h h x h x E x h h x h x x h h x h x E bh M sh y sb y ult ult m cu ult ult ult c ult ult ult c u | . | \ | + − + − | . | \ | − | . | \ | − × + | | . | \ | − − | | . | \ | − | . | \ | + | | . | \ | | . | \ | − + − + | . | \ | | | . | \ | − + | | . | \ | | . | \ | + − + | | . | \ | − | . | \ | − − | | . | \ | + − + | . | \ | | | . | \ | − + | | . | \ | − + | | . | \ | − | . | \ | − = 2 0 0 2 2 4 4 0 2 3 3 0 0 2 2 3 3 0 2 2 0 2 392 9 112 9 48 7 87 . 0 8 3 ' 1 8 7 ' 2 1 87 . 0 1 1 1 2 67 . 0 4 6 4 1 4 1 3 3 1 2 1 3 1 2 3 3 1 3 1 2 1 2 1 2 1 ε ε ε ε γ ε ε ε ε ε ε ε ε ε ε ε 2 0 0 2 2 0 3 3 0 2 24 1 6 1 4 1 6 1 24 1 | . | \ | | | . | \ | − + − + − = ⇒ h x E bh M ult ult ult ult ult c u ε ε ε ε ε ε ε ε ε 2 0 0 0 0 2 2 0 24 1 12 1 12 1 12 1 4 1 4 1 12 1 | . | \ | + | | . | \ | − + + − + − + x h E x h E h x E ult ult c ult ult c ult ult ult ult c ε ε ε ε ε ε ε ε ε ε ε ε ε bh A h x h x x h f x h h d h d bh A f sh y sb y | . | \ | + − + − | . | \ | − | . | \ | − × + 2 392 9 112 9 48 7 87 . 0 8 3 ' 1 8 7 ' 2 1 87 . 0 (Eqn F-57) Case 6 – where (1 – ε 0 /ε ult )x>h>3x/7, i.e. 1/(1 – ε 0 /ε ult ) < x/h ≤ 7/3 The concrete / steel stress / strain diagram is worked out as follows : Appendix F F-16 Comparing with Case 5, contribution by 1 sc F vanishes. So formulation as similar to above : bh A h x x h f bh A h d x h f f bh N sh y sb y m cu u | . | \ | − − + | . | \ | − − + = 56 9 8 7 4 7 87 . 0 ' 1 8 7 8 11 87 . 0 67 . 0 γ (Eqn F-58) Re-arranging (Eqn F-58) | . | \ | | . | \ | + − − + | . | \ | h x bh A bh A f f bh N h x bh A f sh sb y m cu u sh y 4 7 8 11 87 . 0 67 . 0 56 9 87 . 0 2 γ ε s = 0.002 F sc3 F sc2 F c2 4x/7 0.87f y 3x/7 4x/7 m cu f γ 67 . 0 ' d ' d h x 0035 . 0 = ult ε Concrete stress Block Steel stress Block Strain diagram across concrete section Figure F-4.6 – Concrete and steel stress strain relation for Case 6 Appendix F F-17 0 8 7 1 ' 8 7 87 . 0 = − | . | \ | − − bh A bh A h d f sh sb y (Eqn F-59) which is a quadratic equation in h x which can be solved In moment balancing moment 0 2 = bh M c , whilst that by steel is identical to Case 5 Total Moment 2 3 2 2 2 ' 1 2 1 2 bh M bh M bh M bh M bh M sc sc sc sc + + + = | . | \ | + − + − | . | \ | − | . | \ | − × = 2 392 9 112 9 48 7 87 . 0 8 3 ' 1 8 7 ' 2 1 87 . 0 h x h x x h bh A f x h h d h d bh A f sh y sb y (Eqn F-60) Case 7 – where x/h > 7/3 In this case, the concrete and steel in the entire column section are under ultimate stress. The axial load will be simply | . | \ | + + = bd A bd A f f bh N sh sb y m cu u 87 . 0 67 . 0 γ (Eqn F-61) and the moment is zero. 0 2 = bh M u (Eqn F-62) (III) Design formulae for 4-bar column sections for determination of reinforcement ratios It is the aim of the section of the Appendix to derive formulae for the determination of bh A sb against applied axial load and moment under a pre-determined sectional size. In the following derivations, bh A sh are set to zero. The process involves : (i) For the 7 cases discussed in the foregoing, eliminate bh A sb between equations h b Appendix F F-18 obtained from balancing bh N u and 2 bh M u by making bh A sb subject of formulae in the equation for balancing of bh N u substitute into the equation for balancing of 2 bh M u . The equation obtained in a polynomial in h x which can be solved by equations (if quadratic or cubic or even 4 th power) or by numerical methods. Solution in h x will be valid if the value arrived at agree with the pre-determined range of the respective case; (ii) Back substitute the accepted value of h x into the equation obtained by balancing bh N u to solve for bh A sb . Case 1 – where x/h ≤ 7/3(d’/h) By (Eqn F-18) with 0 = bh A sh 0 ' 87 . 0 8 7 8 3 87 . 0 3 1 1 67 . 0 2 0 = − − − | . | \ | | | . | \ | − h d bh A f h x bh A f bh N h x f sb y sb y u ult m cu ε ε γ h x h d h x bh N h x f bh A f u ult m cu sb y 8 3 ' 8 7 3 3 67 . 0 87 . 0 2 0 − − | . | \ | | | . | \ | − = ⇒ ε ε γ (Eqn F-63) Substituting into (Eqn F-34) again with 0 = bh A sh ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | | | . | \ | − − + | | . | \ | − | . | \ | = h x h x f bh M ult ult ult m cu u 2 0 0 0 2 12 1 2 1 3 1 6 1 2 1 67 . 0 ε ε ε ε ε ε γ | . | \ | − | . | \ | − − − | . | \ | | | . | \ | − + x h h d h d h x h d h x bh N h x f u ult m cu ' 8 7 8 11 ' 2 1 8 3 ' 8 7 3 1 1 67 . 0 2 0 ε ε γ 3 2 0 0 12 1 2 1 3 1 8 3 67 . 0 | . | \ | | | . | \ | − − ⇒ h x f ult ult m cu ε ε ε ε γ 2 2 0 0 0 ' 96 7 ' 4 3 24 7 ' 16 29 2 1 67 . 0 | . | \ | ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | . | \ | + − + + − + h x h d h d h d f ult ult ult m cu ε ε ε ε ε ε γ 0 ' 2 1 ' 8 7 3 1 1 ' 8 7 67 . 0 ' 2 1 8 11 2 2 0 2 = | . | \ | − − + ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ − | | . | \ | − | . | \ | − | . | \ | − + h d bh N bh M h d h x bh M h d f h d bh N u u u ult m cu u ε ε γ Appendix F F-19 The equation can be solved for h x and substituted into (Eqn F-18) | . | \ | − − | . | \ | | | . | \ | − = h x h d f h x bh N h x f bh A y u ult m cu sb 8 3 ' 8 7 87 . 0 3 3 67 . 0 2 0 ε ε γ Case 2 – where 7/3(d’/h) < x/h ≤ 7/11 – 7/11(d’/h) (applicable to d’/h > 3/14 only) By (Eqn F-33) with 0 = bh A sh , | | . | \ | − ÷ = ult m cu u f bh N h x ε ε γ 0 3 1 1 67 . 0 Substituting h x into (Eqn F-34) obtained and calculate bh A sb as | . | \ | − ÷ ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | | | . | \ | − | . | \ | + − | . | \ | − | . | \ | − = h d f h x h x h x h x f bh M bh A y ult ult ult m cu u sb ' 2 1 87 . 0 12 1 3 1 6 1 1 2 1 67 . 0 2 0 0 0 2 ε ε ε ε ε ε γ (Eqn F-64) Case 3 – where 7/11(d’/h) < x/h ≤ 7/11 for d’/h > 3/14 and 7/11(1 – d’/h) < x/h ≤ 7/11 for d’/h < 3/14 and Case 4 – where 7/11 < x/h ≤ 1 By (Eqn F-36) and setting 0 = bh A sh . | . | \ | + − + | | . | \ | − = x d x h bh A f h x f bh N sb y ult m cu u ' 8 7 8 7 8 11 87 . 0 3 1 1 67 . 0 0 ε ε γ | . | \ | + − | | . | \ | − − = x h h d x h f h x f bh N bh A y ult m cu u sb ' 8 7 8 7 8 11 87 . 0 3 1 1 67 . 0 0 ε ε γ Substituting into (Eqn F-38) and again setting 0 = bh A sh ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | | | . | \ | − | . | \ | + − | . | \ | − | . | \ | = h x h x h x h x f bh M ult ult ult m cu u 2 0 0 0 2 12 1 3 1 6 1 1 2 1 67 . 0 ε ε ε ε ε ε γ | . | \ | + − | | . | \ | − − − | . | \ | | . | \ | − | . | \ | | . | \ | − + x h h d x h f h x f bh N x h h d x h h d f y ult m cu u y ' 8 7 8 7 8 11 87 . 0 3 1 1 67 . 0 8 3 ' 8 7 8 7 ' 2 1 87 . 0 0 ε ε γ 3 2 0 0 12 1 2 1 3 1 8 11 67 . 0 | . | \ | | | . | \ | − − ⇔ h x f ult ult m cu ε ε ε ε γ Appendix F F-20 2 2 0 0 2 0 0 ' 96 7 ' 12 5 96 7 ' 16 13 12 7 16 21 67 . 0 | . | \ | | | . | \ | − + | | . | \ | + − − + h x h d h d h d f ult ult ult ult m cu ε ε ε ε ε ε ε ε γ | . | \ | ) ` ¹ ¹ ´ ¦ − | . | \ | − − | | . | \ | + − − | . | \ | − + h x bh M bh N h d h d h d h d f u u ult ult m cu 2 0 0 8 11 ' 2 1 8 3 ' 3 1 ' 3 1 1 1 ' 8 7 67 . 0 ε ε ε ε γ 0 1 ' 8 7 1 ' ' 2 1 8 7 2 = | . | \ | − − | . | \ | − | . | \ | − − h d bh M bh N h d h d u u (Eqn F-65) which is a cubic equation in | . | \ | h x Upon solving | . | \ | h x lying between | . | \ | − h d' 1 11 7 and 1, bh A sb can be obtained by | . | \ | + − | | . | \ | − − = x h h d x h f h x f bh N bh A y ult m cu u sb ' 8 7 8 7 8 11 87 . 0 3 1 1 67 . 0 0 ε ε γ (Eqn F-66) Case 5 – where 1 < x/h ≤ 1/(1 – ε 0 /ε ult ) Solving bd A sb by (Eqn F-52) | . | \ | − − ÷ | . | \ | + | | . | \ | − − | | . | \ | − − + | . | \ | − − ÷ | | . | \ | + + − − − = x h h d x h E x h E E E E x h h d h x bh N bh A f ult c ult c ult c ult c ult c ult ult ult u sb y 1 ' 8 7 8 11 6 2 2 2 1 ' 8 7 8 11 2 1 6 1 6 1 2 1 87 . 0 2 0 2 0 2 0 2 0 0 2 2 0 ε ε ε ε ε ε ε ε ε ε ε ε ε ε Substituting into (Eqn F-57) − | . | \ | − | . | \ | − × + | | . | \ | − − | | . | \ | − | . | \ | + | | . | \ | | . | \ | − + − + | . | \ | | | . | \ | − + | | . | \ | | . | \ | + − + | | . | \ | − | . | \ | − − ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ + − + | . | \ | | | . | \ | − + − + | | . | \ | − | . | \ | − = 8 3 ' 1 8 7 ' 2 1 87 . 0 1 1 1 2 67 . 0 4 6 4 1 4 1 3 3 1 2 1 3 1 2 3 3 1 3 1 2 1 2 1 2 1 0 0 2 2 4 4 0 2 3 3 0 0 2 2 3 3 0 2 2 0 2 x h h d h d bh A f h x h x f x h x h h x h x x h x h h x h x E x h h x h x x h h x h x E bh M sb y ult ult m cu ult ult ult c ult ult ult c ε ε ε ε γ ε ε ε ε ε ε ε ε ε ε ε to solve for h x . Back-substituting into the previous equation to solve for bd A sb by trial and error Appendix F F-21 Case 6 – where (1 – ε 0 /ε ult )x > h > 3x/7 i.e. 1/(1 – ε 0 /ε ult ) < x/h ≤ 7/3(1–d’/h) Referring to (Eqn F-58) and setting 0 = bd A sh bh A h d x h f f bh N sb y m cu u | . | \ | − − + = ' 1 8 7 8 11 87 . 0 67 . 0 γ | . | \ | − − ÷ | | . | \ | − = ⇒ h d x h f bh N bh A f m cu u sb y ' 1 8 7 8 11 67 . 0 87 . 0 γ Substituting into (Eqn F-60) and again setting 0 = bd A sh and re-arranging | . | \ | − | | . | \ | − + + | . | \ | − | | . | \ | − | . | \ | − = ⇒ 8 3 ' 2 1 67 . 0 8 11 ' 2 1 67 . 0 ' 1 8 7 2 2 h d f bh N bh M bh M h d f bh N h d h x m cu u m cu u γ γ With h x determined, calculate bh A sb by (Eqn F-58) bh A h d x h f f bh N sb y m cu u | . | \ | − − + = ' 1 8 7 8 11 87 . 0 67 . 0 γ | . | \ | − − − = ⇒ h d x h f f bh N bh A y m cu u sb ' 1 8 7 8 11 87 . 0 67 . 0 γ (Eqn F-67) Case 7 – where x/h > 7/3 y m cu u sb sb y m cu u f f bh N bh A bh A f f bh N 87 . 0 1 67 . 0 87 . 0 67 . 0 | | . | \ | − = ⇒ + = γ γ (Eqn F-68) 0 2 = bh M u Appendix F Chart F-1 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30, 4-bar column, d/h = 0.75 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-2 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30, 4-bar column, d/h = 0.8 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-3 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30, 4-bar column, d/h = 0.85 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-4 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30, 4-bar column, d/h = 0.9 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-5 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30, 4-bar column, d/h = 0.95 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-6 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35, 4-bar column, d/h = 0.75 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-7 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35, 4-bar column, d/h = 0.8 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-8 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35, 4-bar column, d/h = 0.85 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-9 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35, 4-bar column, d/h = 0.9 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-10 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35, 4-bar column, d/h = 0.95 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-11 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40, 4-bar column, d/h = 0.75 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-12 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40, 4-bar column, d/h = 0.8 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-13 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40, 4-bar column, d/h = 0.85 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-14 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40, 4-bar column, d/h = 0.9 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-15 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40, 4-bar column, d/h = 0.95 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-16 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45, 4-bar column, d/h = 0.75 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-17 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45, 4-bar column, d/h = 0.8 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-18 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45, 4-bar column, d/h = 0.85 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-19 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45, 4-bar column, d/h = 0.9 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-20 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45, 4-bar column, d/h = 0.95 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-21 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50, 4-bar column, d/h = 0.75 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-22 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50, 4-bar column, d/h = 0.8 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-23 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50, 4-bar column, d/h = 0.85 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-24 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50, 4-bar column, d/h = 0.9 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-25 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50, 4-bar column, d/h = 0.95 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-26 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55, 4-bar column, d/h = 0.75 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-27 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55, 4-bar column, d/h = 0.8 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-28 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55, 4-bar column, d/h = 0.85 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-29 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55, 4-bar column, d/h = 0.9 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-30 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55, 4-bar column, d/h = 0.95 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-31 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60, 4-bar column, d/h = 0.75 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-32 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60, 4-bar column, d/h = 0.8 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-33 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60, 4-bar column, d/h = 0.85 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-34 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60, 4-bar column, d/h = 0.9 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Appendix F Chart F-35 Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60, 4-bar column, d/h = 0.95 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 18 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Rectangular Column R.C. Design to Code of Practice for Structural Use of Concrete 2004 Project : Column Floor f cu = 35 N/mm 2 f y = 460 N/mm 2 E c = 23700 N/mm 2 b = 1500 h = 2000 b' = 1285.17 h' = 1684.31 cover= 50 Steel provided : 15 Y 40 (Along each long sides h, excluding corner bars) 12 Y 40 (Along each short sides b, excluding corner bars) 4 Y 40 (Corner bars) Total Steel Area = 72885 mm 2 Steel Percentage = 2.43 % Max. Ultimate Load = 76069 kN Area of Steel per mm length for the long sides bars (including corner bars) = 21.36 mm 2 /mm Area of Steel along long sides (excluding corner bars) = 37699 mm 2 Area of Steel per mm length for the short sides bars (including corner bars) = 23.46 mm 2 /mm Area of Steel along short sides (excluding corner bars) = 30159 mm 2 Basic Load Case 1 2 3 4 5 6 D.L. L.L. Wx Wy W45 W135 37872 1101 -3628.1 -2611.1 -5692.3 8209.2 -291.3 -37.11 470.81 -3700 -1750.3 4892.9 -31.33 16.09 5.17 2700 2764 -3520.2 P M x M y Load Comb 1 1.4D+1.6L 54782 -467.2 -18.118 Mx' = 476.55 Mux = 16452 Section OK Load Comb 2 1.2(D+L+Wx) 42413 170.88 -12.084 Mx' = 179.2 Mux = 23474 Section OK Load Comb 3 1.2(D+L-Wx) 51121 -959.06 -24.492 Mx' = 973.01 Mux = 18743 Section OK Load Comb 4 1.2(D+L+Wy) 43634 -4834.1 3221.7 Mx' = 6999.6 Mux = 22861 Section OK Load Comb 5 1.2(D+L-Wy) 49900 4045.9 -3258.3 My' = 4639 Muy = 14871 Section OK Load Comb 6 1.2(D+L+W45) 39936 -2494.5 3298.5 My' = 4352.2 Muy = 18945 Section OK Load Comb 7 1.2(D+L-W45) 53598 1706.3 -3335.1 My' = 3865.6 Muy = 13130 Section OK Load Comb 8 1.2(D+L+W135) 56618 5477.4 -4242.6 My' = 5801.2 Muy = 11590 Section OK Load Comb 9 1.2(D+L-W135) 36916 -6265.5 4206 Mx' = 9507.3 Mux = 26076 Section OK Load Comb 10 1.4(D+Wx) 47941 251.31 -36.624 Mx' = 273.77 Mux = 20575 Section OK Load Comb 11 1.4(D-Wx) 58099 -1067 -51.1 Mx' = 1090.8 Mux = 14185 Section OK Load Comb 12 1.4(D+Wy) 49365 -5587.8 3736.1 Mx' = 7805.2 Mux = 19771 Section OK Load Comb 13 1.4(D-Wy) 56676 4772.2 -3823.9 My' = 5179.4 Muy = 11560 Section OK Load Comb 14 1.4(D+W45) 45051 -2858.3 3825.7 My' = 4911.9 Muy = 16951 Section OK Load Comb 15 1.4(D-W45) 60989 2042.6 -3913.4 My' = 4416.8 Muy = 9164 Section OK Load Comb 16 1.4(D+W135) 64513 6442.2 -4972.2 My' = 6446.8 Muy = 7132 Section OK Load Comb 17 1.4(D-W135) 41527 -7257.9 4884.4 Mx' = 10685 Mux = 23910 Section OK Load Comb 18 1.0D+1.4Wx 32792 367.83 -24.092 Mx' = 387.89 Mux = 27896 Section OK Load Comb 19 1.0D-1.4Wx 42951 -950.43 -38.568 Mx' = 976.72 Mux = 23206 Section OK Load Comb 20 1.0D+1.4Wy 34216 -5471.3 3748.7 Mx' = 8512.2 Mux = 27278 Section OK Load Comb 21 1.0D-1.4Wy 41527 4888.7 -3811.3 My' = 5808.5 Muy = 18343 Section OK Load Comb 22 1.0D+1.4W45 29902 -2741.8 3838.3 My' = 5236.3 Muy = 22460 Section OK Load Comb 23 1.0D-1.4W45 45841 2159.2 -3900.9 My' = 4707.8 Muy = 16626 Section OK Load Comb 24 1.0D+1.4W135 49364 6558.7 -4959.6 Mx' = 9502.2 Mux = 19771 Section OK Load Comb 25 1.0D-1.4W135 26379 -7141.3 4897 Mx' = 11689 Mux = 29921 Section OK Moment M y (kNm) Load Case No. Load Case Axial Load P (kN) Moment M x (kNm) P versus Mx and My of the Column Section 0 10000 20000 30000 40000 50000 60000 70000 80000 0 5000 10000 15000 20000 25000 30000 35000 M (kNm) P ( k N ) P - Mx P - My Actual Loads Mx control Actual Loads My control Rectangular Column R.C. Design to Code of Practice for Structural Use of Concrete 2004 - 4 bar column Project : Column Mark Floor f cu = 35 N/mm 2 f y = 460 N/mm 2 E c = 23700 N/mm 2 b = 400 h = 500 b' = 325.00 h' = 425.00 cover= 55 bar size = 40 Basic Load Case 1 2 3 4 5 6 D.L. L.L. Wx Wy W45 W135 2304.7 582.1 -362.17 -545.1 56.92 82.09 29.13 32.11 47.1 -75.12 98.1 8.93 -31.33 16.09 2.15 44.2 76.99 35.21 N Mx My N/bh M/bh 2 d/h / d/b x/h / y/h Steel Steel area (kN) (kNm) (kNm) (N/mm 2 ) (N/mm 2 ) (%) (mm 2 ) Load Comb 1 1.4D+1.6L 4157.9 92.158 -18.118 Mx' = 99.437 20.79 0.9944 0.85 1.1601 1.9538 3907.6 Load Comb 2 1.2(D+L+Wx) 3029.6 130.01 -15.708 Mx' = 140.15 15.148 1.4015 0.85 0.9671 0.8 1600 Load Comb 3 1.2(D+L-Wx) 3898.8 16.968 -20.868 My' = 25.431 19.494 0.3179 0.8125 1.31 1.1989 2397.7 Load Comb 4 1.2(D+L+Wy) 2810 -16.656 34.752 My' = 41.482 14.05 0.5185 0.8125 1.0602 0.8 1600 Load Comb 5 1.2(D+L-Wy) 4118.3 163.63 -71.328 Mx' = 192.92 20.591 1.9292 0.85 1.0304 2.5083 5016.5 Load Comb 6 1.2(D+L+W45) 3532.5 191.21 74.1 Mx' = 231.37 17.662 2.3137 0.85 0.9381 2.0771 4154.3 Load Comb 7 1.2(D+L-W45) 3395.9 -44.232 -110.68 My' = 125.44 16.979 1.5679 0.8125 0.9731 1.4265 2852.9 Load Comb 8 1.2(D+L+W135) 3562.7 84.204 23.964 Mx' = 97.029 17.813 0.9703 0.85 1.0902 1.1552 2310.4 Load Comb 9 1.2(D+L-W135) 3365.7 62.772 -60.54 My' = 81.714 16.828 1.0214 0.8125 1.0402 0.9675 1935 Load Comb 10 1.4(D+Wx) 2719.5 106.72 -40.852 Mx' = 135.77 13.598 1.3577 0.85 0.9306 0.8 1600 Load Comb 11 1.4(D-Wx) 3733.6 -25.158 -46.872 My' = 54.182 18.668 0.6773 0.8125 1.1502 1.2111 2422.2 Load Comb 12 1.4(D+Wy) 2463.4 -64.386 18.018 Mx' = 78.233 12.317 0.7823 0.85 0.9637 0.8 1600 Load Comb 13 1.4(D-Wy) 3989.7 145.95 -105.74 Mx' = 192.42 19.949 1.9242 0.85 1.0205 2.3331 4666.2 Load Comb 14 1.4(D+W45) 3306.3 178.12 63.924 Mx' = 215.78 16.531 2.1578 0.85 0.9268 1.6862 3372.3 Load Comb 15 1.4(D-W45) 3146.9 -96.558 -151.65 My' = 186.76 15.734 2.3345 0.8125 0.8694 1.7268 3453.6 Load Comb 16 1.4(D+W135) 3341.5 53.284 5.432 Mx' = 56.444 16.708 0.5644 0.85 1.1402 0.8 1600 Load Comb 17 1.4(D-W135) 3111.7 28.28 -93.156 My' = 103.56 15.558 1.2945 0.8125 0.9764 0.8543 1708.6 Load Comb 18 1.0D+1.4Wx 1797.7 95.07 -28.32 Mx' = 121.06 8.9883 1.2106 0.85 0.7208 0.8 1600 Load Comb 19 1.0D-1.4Wx 2811.7 -36.81 -34.34 My' = 49.207 14.059 0.6151 0.8125 1.0401 0.8 1600 Load Comb 20 1.0D+1.4Wy 1541.6 -76.038 30.55 Mx' = 105.83 7.7078 1.0583 0.85 0.585 0.8 1600 Load Comb 21 1.0D-1.4Wy 3067.8 134.3 -93.21 Mx' = 193.77 15.339 1.9377 0.85 0.9181 1.2341 2468.2 Load Comb 22 1.0D+1.4W45 2384.4 166.47 76.456 Mx' = 226.58 11.922 2.2658 0.85 0.7798 0.8 1600 Load Comb 23 1.0D-1.4W45 2225 -108.21 -139.12 My' = 191.13 11.125 2.3891 0.8125 0.7229 0.8 1600 Load Comb 24 1.0D+1.4W135 2419.6 41.632 17.964 Mx' = 55.614 12.098 0.5561 0.85 0.9885 0.8 1600 Load Comb 25 1.0D-1.4W135 2189.8 16.628 -80.624 My' = 88.693 10.949 1.1087 0.8125 0.8755 0.8 1600 Steel required = 2.5083 5016.5 Moment M y (kNm) Load Case No. Load Case Axial Load P (kN) Moment M x (kNm) Mx My b h P versus Mx and My of the Column Section 0 1000 2000 3000 4000 5000 6000 0 100 200 300 400 500 600 M (kNm) P ( k N ) P - Mx P - My Actual Loads Mx control Actual Loads My control Appendix G Derivation of Design Formulae for Walls to Rigorous Stress Strain Curve of Concrete Appendix G G-1 Derivation of Design Formulae for Shear Walls to Rigorous Stress Strain Curve of Concrete As similar to the exercise in Appendix F for columns, the exercise in this Appendix is repeated for walls by which bh A sb are set to zero in the various cases 1 to 7, using the equations summarized in Appendix F. Cases 1 to 3 – where x/h ≤ 7/11 By (Eqn F-18) or (Eqn F-32) or (Eqn F-36) of Appendix F bh A h x f h x f bh N sh y ult m cu u | . | \ | − + | | . | \ | − = 1 2 87 . 0 3 3 67 . 0 0 ε ε γ | . | \ | − | | . | \ | − − = 1 2 87 . 0 3 3 67 . 0 0 h x f h x f bh N bh A y ult m cu u sh ε ε γ (Eqn G-1) Substituting into (Eqn F-31) or (Eqn F-34) or (Eqn F-38) and putting 0 = bh A sb | . | \ | − | . | \ | + ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | | | . | \ | − | . | \ | + − | . | \ | − | . | \ | = 2 2 0 0 0 2 147 163 87 . 0 12 1 3 1 6 1 1 2 1 67 . 0 h x h x bh A f h x h x h x h x f bh M sh y ult ult ult m cu u ε ε ε ε ε ε γ 2 2 0 0 3 2 0 0 147 163 12 1 3 1 2 1 67 . 0 6 1 441 131 147 16 67 . 0 | . | \ | ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ − | | . | \ | + − + | . | \ | ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | . | \ | − + ⇒ h x bh N f h x f u ult ult m cu ult ult m cu ε ε ε ε γ ε ε ε ε γ 0 6 1 2 1 67 . 0 2 2 0 2 = + | | . | \ | − − − + bh M h x f bh M bh N u ult m cu u u ε ε γ (Eqn G-2) Upon solving h x , back substituting into (Eqn G-1) to calculate bh A sh Case 4 – where 7/11 < x/h ≤ 1 h b Appendix G G-2 Referring to (Eqn F-39) of Appendix F and setting 0 = bh A sb bh N x h h x bh A f h x f u sh y ult m cu = | . | \ | + − − + | | . | \ | − 4 7 8 7 56 9 87 . 0 3 3 67 . 0 0 ε ε γ | . | \ | + − − | | . | \ | − − = ⇒ 4 7 8 7 56 9 87 . 0 3 3 67 . 0 0 x h h x f h x f bh N bh A y ult m cu u sh ε ε γ (Eqn G-3) Substituting into (Eqn F-41) with 0 = bh A sb ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | | | . | \ | − | . | \ | + − | . | \ | − | . | \ | = h x h x h x h x f bh M ult ult ult m cu u 2 0 0 0 2 12 1 3 1 6 1 1 2 1 67 . 0 ε ε ε ε ε ε γ bh A h x h x x h f sh y | . | \ | + − + 2 392 9 112 9 48 7 87 . 0 | . | \ | − − ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | . | \ | | | . | \ | − − + | | . | \ | − | . | \ | = | . | \ | − − ⇒ x h h x h x h x f x h h x bh M ult ult ult m cu u 8 7 56 9 4 7 12 1 2 1 3 1 6 1 2 1 67 . 0 8 7 56 9 4 7 2 0 0 0 2 ε ε ε ε ε ε γ | | . | \ | − − | . | \ | + − + h x f bh N h x h x x h ult m cu u ε ε γ 0 2 3 3 67 . 0 392 9 112 9 48 7 4 2 0 0 224 3 196 9 784 45 67 . 0 | . | \ | | | . | \ | − + − ⇒ h x f ult ult m cu ε ε ε ε γ 3 2 0 0 392 9 48 7 12 7 8 7 67 . 0 | . | \ | − | | . | \ | + − + h x bh N f u ult ult m cu ε ε ε ε γ 2 2 0 0 2 112 9 12 1 3 2 5 . 1 8 7 67 . 0 56 9 | . | \ | + | | . | \ | − + − + − + h x bh N f bh M u ult ult m cu u ε ε ε ε γ 0 48 7 8 7 3 12 7 3 67 . 0 4 7 2 0 2 = − − + − + + bh N bh M h x f bh M u u ult m cu u ε ε γ (Eqn G-4) Upon solving h x , back-substitute into (Eqn G-3) to solve for bh A sh Case 5 – where 1 < x/h ≤ 1/(1 – ε 0 /ε ult ) Referring to (Eqn F-52) and setting 0 = bd A sb | | . | \ | − + + + − − = 0 0 0 2 2 0 2 1 1 2 1 6 1 6 1 2 1 ε ε ε ε ε ε ε ε ε ε ult ult c ult ult ult ult c u E h x E bh N Appendix G G-3 bh A h x x h f x h E x h E sh y ult c ult ult c | . | \ | − − + | . | \ | − | | . | \ | − + 56 9 8 7 4 7 87 . 0 6 2 1 2 1 2 0 2 0 ε ε ε ε ε (Eqn G-5) Substituting for bh A f sh y 87 . 0 into (Eqn F-57), again setting 0 = bd A sb 2 0 0 2 2 0 3 3 0 2 24 1 6 1 4 1 6 1 24 1 | . | \ | | | . | \ | − + − + − = h x E bh M ult ult ult ult ult c ε ε ε ε ε ε ε ε ε 2 0 0 0 0 2 2 0 24 1 12 1 12 1 12 1 4 1 4 1 12 1 | . | \ | + | | . | \ | − + + − + − + x h E x h E h x E ult ult c ult ult c ult ult ult ult c ε ε ε ε ε ε ε ε ε ε ε ε ε bh A h x h x x h f sh y | . | \ | + − + 2 392 9 112 9 48 7 87 . 0 2 0 0 2 2 0 3 3 0 2 56 9 8 7 4 7 24 1 6 1 4 1 6 1 24 1 56 9 8 7 4 7 | . | \ | | . | \ | − − | | . | \ | − + − + − = | . | \ | − − ⇒ h x h x x h E h x x h bh M ult ult ult ult ult c ε ε ε ε ε ε ε ε ε x h h x x h E h x h x x h E ult ult c ult ult ult ult c | . | \ | − − | | . | \ | − + | . | \ | − − + − + − + 56 9 8 7 4 7 12 1 12 1 56 9 8 7 4 7 12 1 4 1 4 1 12 1 0 0 0 2 2 0 ε ε ε ε ε ε ε ε ε ε | . | \ | + − + | . | \ | | . | \ | − − + 2 2 0 392 9 112 9 48 7 56 9 8 7 4 7 24 1 h x h x x h bh N x h h x x h E u ult ult c ε ε ε | . | \ | + − + + − − − 2 0 0 2 2 0 392 9 112 9 48 7 2 1 6 1 6 1 2 1 h x h x x h h x E ult ult ult ult c ε ε ε ε ε ε ε | . | \ | + − | | . | \ | − − | . | \ | + − | | . | \ | − − 2 0 2 0 392 9 112 9 48 7 2 1 2 1 392 9 112 9 48 7 2 1 1 h x h x x h x h E h x h x x h E ult ult c ult ult c ε ε ε ε ε ε | . | \ | + − | . | \ | + 2 2 0 2 392 9 112 9 48 7 6 h x h x x h x h E ult c ε ε 3 0 0 2 2 0 3 3 0 3136 9 196 3 1568 45 392 9 448 3 | . | \ | | | . | \ | + − + − ⇒ h x E ult ult ult ult ult c ε ε ε ε ε ε ε ε ε 2 0 0 2 2 0 3 3 0 392 9 4704 289 294 79 16 7 24 7 96 7 | . | \ | + | | . | \ | − + − + − + h x bh N E u ult ult ult ult ult c ε ε ε ε ε ε ε ε ε | . | \ | − + | | . | \ | + − + − + h x bh N bh M E u ult ult ult ult ult c 112 9 56 9 9408 1229 588 289 32 21 24 7 193 7 2 0 0 2 2 0 3 3 0 ε ε ε ε ε ε ε ε ε x h bh N bh M E bh M E u ult c ult ult ult ult c | | . | \ | + + − + − | | . | \ | − + − + 48 7 8 7 1344 125 4 7 7056 281 21 5 24 7 72 7 2 0 2 2 0 0 2 2 0 ε ε ε ε ε ε ε ε ε 0 576 7 96 7 3 0 2 2 0 2 = | . | \ | − | . | \ | + x h E x h E ult c ult c ε ε ε ε (Eqn G-6) which is in fact an equation of 6 th power in h x . h x is to be solved by numerical Appendix G G-4 method. By back-substituting h x into (Eqn G-5) | | . | \ | − + + + − − = 0 0 0 2 2 0 2 1 1 2 1 6 1 6 1 2 1 ε ε ε ε ε ε ε ε ε ε ult ult c ult ult ult ult c u E h x E bh N bh A h x x h f x h E x h E sh y ult c ult ult c | . | \ | − − + | . | \ | − | | . | \ | − + 56 9 8 7 4 7 87 . 0 6 2 1 2 1 2 0 2 0 ε ε ε ε ε | . | \ | − − | . | \ | + | | . | \ | − − | | . | \ | − − + + − − − = h x x h f x h E x h E E h x E bh N bh A y ult c ult ult c ult ult c ult ult ult ult c u sh 56 9 8 7 4 7 87 . 0 6 2 1 2 1 2 1 1 2 1 6 1 6 1 2 1 2 0 2 0 0 0 0 2 2 0 ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε Case 6 – where 1/(1 – ε 0 /ε ult ) < x/h ≤ 7/3 Consider (Eqn F-58) and substituting 0 = bh A sb | . | \ | − − − = ⇒ | . | \ | − − + = h x x h f f bh N bh A bh A h x x h f f bh N y m cu u sh sh y m cu u 56 9 8 7 4 7 87 . 0 67 . 0 56 9 8 7 4 7 87 . 0 67 . 0 γ γ (Eqn G-7) Substituting into (Eqn F-60) | . | \ | + − = 2 2 392 9 112 9 48 7 87 . 0 h x h x x h bh A f bh M sh y u 2 2 3 67 . 0 112 9 56 9 67 . 0 392 9 | . | \ | | | . | \ | − − + | . | \ | | | . | \ | − ⇒ h x f bh N bh M h x f bh N m cu u u m cu u γ γ 0 8 7 67 . 0 48 7 4 7 2 2 = + | | . | \ | − + − bh M f bh N h x bh M u m cu u u γ (Eqn G-8) Solving the cubic equation for h x and substituting into (Eqn G-7) to solve for bh A sh Case 7 – where x/h > 7/3 By (Eqn F-61) and setting 0 = bh A sb y m cu u sh sh y m cu u f f bh N bh A bh A f f bh N 87 . 0 1 67 . 0 87 . 0 67 . 0 | | . | \ | − = ⇒ + = γ γ (Eqn G-9) 0 2 = bh M u Chart-G1 Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004, Concrete Grade 30 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Chart-G2 Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004, Concrete Grade 35 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Chart-G3 Design Chart of Rectangular Shear Wall with Uniform Vertical reinforcements to Code of Practice for Structural Use of Concrete 2004, Concrete Grade 40 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Chart-G4 Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004, Concrete Grade 45 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Chart-G5 Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004, Concrete Grade 50 0 5 10 15 20 25 30 35 40 45 50 55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Chart-G6 Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004, Concrete Grade 55 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Chart-G7 Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004, Concrete Grade 60 0 5 10 15 20 25 30 35 40 45 50 55 60 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 M/bh 2 N/mm 2 N / b h N / m m 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel Shear Wall R.C. Design to Code of Practice for Structural Use of Concrete 2004 - Uniform Reinforcements Project : Wall Mark Floor f cu = 35 N/mm 2 f y = 460 N/mm 2 E c = 23700 N/mm 2 b = 200 h = 2000 b' = 165.00 h' = 1500.00 cover= 25 bar size = 20 Basic Load Case My Mx 1 2 3 4 5 6 D.L. L.L. Wx Wy W45 W135 3304.7 1582.1 -362.17 -245.1 56.92 82.09 29.13 32.11 2047.1 -1275.1 1098.1 888.93 -31.33 16.09 2.15 44.2 206.5 35.21 N Mx My N/bh M/bh 2 d/h / d/b x/h / y/b Steel Steel area (kN) (kNm) (kNm) (N/mm 2 ) (N/mm 2 ) (%) (mm 2 ) Load Comb 1 1.4D+1.6L 7157.9 92.158 -18.118 My' = 22.238 17.895 0.278 0.825 1.28 0.7541 3016.2 Load Comb 2 1.2(D+L+Wx) 5429.6 2530 -15.708 Mx' = 2607.8 13.574 3.2597 - 0.7152 2.3059 9223.4 Load Comb 3 1.2(D+L-Wx) 6298.8 -2383 -20.868 Mx' = 2473.2 15.747 3.0915 - 0.7772 2.5637 10255 Load Comb 4 1.2(D+L+Wy) 5570 -1456.7 34.752 Mx' = 1624.9 13.925 2.0311 - 0.8365 1.0928 4371.1 Load Comb 5 1.2(D+L-Wy) 6158.3 1603.6 -71.328 Mx' = 1918.9 15.396 2.3986 - 0.8314 1.7861 7144.5 Load Comb 6 1.2(D+L+W45) 5932.5 1391.2 229.51 My' = 306.62 14.831 3.8328 0.825 0.7508 2.7599 11040 Load Comb 7 1.2(D+L-W45) 5795.9 -1244.2 -266.09 My' = 336.52 14.49 4.2065 0.825 0.7255 3.0079 12032 Load Comb 8 1.2(D+L+W135) 5962.7 1140.2 23.964 Mx' = 1249.5 14.907 1.5618 - 0.9185 0.9046 3618.4 Load Comb 9 1.2(D+L-W135) 5765.7 -993.23 -60.54 Mx' = 1277.8 14.414 1.5972 - 0.9032 0.8122 3248.8 Load Comb 10 1.4(D+Wx) 4119.5 2906.7 -40.852 Mx' = 3150.7 10.299 3.9384 - 0.6025 2.5143 10057 Load Comb 11 1.4(D-Wx) 5133.6 -2825.2 -46.872 Mx' = 3068 12.834 3.835 - 0.6673 2.7996 11199 Load Comb 12 1.4(D+Wy) 4283.4 -1744.4 18.018 Mx' = 1849.7 10.709 2.3121 - 0.6995 0.7417 2966.7 Load Comb 13 1.4(D-Wy) 4969.7 1826 -105.74 Mx' = 2387.4 12.424 2.9842 - 0.7031 1.7913 7165.1 Load Comb 14 1.4(D+W45) 4706.3 1578.1 245.24 My' = 350.54 11.766 4.3818 0.825 0.6532 2.626 10504 Load Comb 15 1.4(D-W45) 4546.9 -1496.6 -332.96 My' = 435.07 11.367 5.4384 0.825 0.62 3.4305 13722 Load Comb 16 1.4(D+W135) 4741.5 1285.3 5.432 Mx' = 1315.1 11.854 1.6439 - 0.8274 0.4 1600 Load Comb 17 1.4(D-W135) 4511.7 -1203.7 -93.156 Mx' = 1731.6 11.279 2.1645 - 0.7373 0.6743 2697.1 Load Comb 18 1.0D+1.4Wx 2797.7 2895.1 -28.32 Mx' = 3093.4 6.9942 3.8667 - 0.5051 2.1902 8760.9 Load Comb 19 1.0D-1.4Wx 3811.7 -2836.8 -34.34 Mx' = 3050.1 9.5293 3.8126 - 0.5843 2.2839 9135.8 Load Comb 20 1.0D+1.4Wy 2961.6 -1756 30.55 Mx' = 1966 7.4039 2.4576 - 0.5307 0.607 2428.1 Load Comb 21 1.0D-1.4Wy 3647.8 1814.3 -93.21 Mx' = 2405.2 9.1196 3.0065 - 0.5941 1.3272 5308.9 Load Comb 22 1.0D+1.4W45 3384.4 1566.5 257.77 My' = 381.82 8.461 4.7727 0.825 0.5596 2.3848 9539.2 Load Comb 23 1.0D-1.4W45 3225 -1508.2 -320.43 My' = 442.13 8.0625 5.5266 0.825 0.5456 2.9149 11660 Load Comb 24 1.0D+1.4W135 3419.6 1273.6 17.964 Mx' = 1390.7 8.5491 1.7384 - 0.6436 0.4 1600 Load Comb 25 1.0D-1.4W135 3189.8 -1215.4 -80.624 Mx' = 1755.3 7.9744 2.1941 - 0.5706 0.4 1600 Steel required = 3.4305 13722 Moment M y (kNm) Load Case No. Load Case Axial Load P (kN) Moment M x (kNm) b h Plot of P (kN) versus M (kNm) 0 2000 4000 6000 8000 10000 12000 14000 0 500 1000 1500 2000 2500 3000 3500 4000 4500 M (kNm) P ( k N ) P - Mx P - My Actual Loads Mx control Actual Loads My control Appendix H Extracts of HKIE paper for plate bending analysis and design Appendix H H-1 Some Problems in Analysis and Design of Thin/Thick Plate (Extracts) Y.M. Cheng 1 and C.W. Law 2 Department of Civil and Structural Engineering, Hong Kong Polytechnic University, Hong Kong 1 Housing Department, Hong Kong SAR Government 2 III. Simulation of Support Stiffness in Plate Bending Structure For support stiffness, we are referring to the force or moment required to produce unit vertical movement or unit rotation at the support which are denoted by Z K , X K θ , Y K θ for settlement stiffness along the Z direction, and rotational stiffnesses about X and Y directions. These stiffnesses are independent parameters which can interact only through the plate structure. Most softwares allow the user either to input numerical values or structural sizes and heights of the support (which are usually walls or columns) by which the softwares can calculate numerical values for the support stiffnesses as follows : (i) For the settlement stiffness Z K , the value is simply L AE where A is the cross sectional of the support which is either a column or a wall, E is the Young’s Modulus of the construction material and L is the free length of the column / wall. The L AE simply measures the ‘elastic shortening’ of the column / wall. Strictly speaking, the expression L AE is only correct if the column / wall is one storey high and restrained completely from settlement at the bottom. However, if the column / wall is more than one storey high, the settlement stiffness estimation can be very complicated. It will not even be a constant value. The settlement of the support is, in fact, ‘interacting’ with that of others through the structural frame linking them together by transferring the axial loads in the column / wall to others through shears in the linking beams. Nevertheless, if the linking beams (usually floor beams) in the structural frame are ‘flexible’, the transfer of loads from one column / wall through the linking beams to the rest of the frame will generally be negligible. By ignoring such transfer, the settlement Appendix H H-2 stiffness of a column / wall can be obtained by ‘compounding’ the settlement stiffness of the individual settlement stiffness at each floor as ∑ = + + + = i i i n n n Z E A L E A L E A L E A L E A L K 1 ...... 1 3 3 3 2 2 2 1 1 1 (ii) For the rotational stiffness, most of the existing softwares calculate the numerical values either by L EI 4 or L EI 3 , depending on whether the far end of the supporting column / wall is assumed fixed or pinned (where I is the second moment of area of the column / wall section). However, apart from the assumed fixity at the far end, the formulae L EI 4 or L EI 3 are also based on the assumption that both ends of the column / wall are restrained from lateral movement (sidesway). It is obvious that the assumption will not be valid if the out-of-plane load or the structural layout is unsymmetrical where the plate will have lateral movement. The errors may be significant if the structure is to simulate a transfer plate under wind load which is in the form of an out-of-plane moment tending to overturn the structure. Nevertheless, the errors can be minimized by finding the force that will be required to restrain the slab structure from sideswaying and applying a force of the same magnitude but of opposite direction to nullify this force. This magnitude of this restraining force or nullifying force is the sum of the total shears produced in the supporting walls / columns due to the moments induced on the walls / columns from the plate analysis. However, the analysis of finding the effects on the plate by the “nullifying force” has to be done on a plane frame or a space frame structure as the 2-D plate bending model cannot cater for lateral in-plane loads. This approach is adopted by some local engineers and the procedure for analysis is illustrated in Figure 13. IV. Effects of Lateral Shears If the plate structure is also subjected to an in-plane shear, it will sway under the shear force (typically wind load) unless it is effectively restrained from lateral movement. If there are moment connections between the plate and the supporting walls/columns/piles, there will be frame actions creating extra loads on the plate when it sways. As the 2-D plate model cannot cater for the in-plane load, such effects have to be assessed separately by plane frame or space frame mathematical Appendix H H-3 models. The effects due to lateral shears as obtained from such analysis should be superimposed onto that of the 2-D plate mathematical model for design. Out-of- plane support reactions as derived from the plane frame or space frame models due to the lateral shear can be added onto the 2-D plate model with all support stiffnesses removed but adding a single support with restraints in settlement and rotations at any point of the plate structure. The purpose of adding the single support is to prevent the stiffness matrix of the 2-D plate model from being singular. However, as the out-of-plane support reactions (on the actual supports) derived from the plane frame or space frame models should have a net effect equal to zero, the reactions on the added support should also be zero, thus creating no effects on the 2-D model. The 2-D model is then re-analyzed and effects can be conveniently superimposed onto the original one which has been used to solve for the out-of-plane loads. V. General discussions on design based on “Nodal Forces” and “Stresses” As pursuant to completion of analysis by the finite element method based on discretized elements joined at nodes, generally (i) the forces at the joining nodes of every element and; (ii) “stresses” within every element can be obtained. It should be perfect if the designer relies on “stresses” for structural design. However, he will soon find that the “stresses” at any nodes joined by more than one element (which should have unique values except under point loads or supports) are generally different when the results are generated from different adjoining elements. The differences can be accounted for as errors in the finite element analysis and it should diminish as the meshing of elements is getting fine. It is however well known that the average stresses from all the adjoining elements of a node are reasonably close to the exact values and can be based on for design. Design based on “stresses” is however not popular as many commercial programs generate the design forces from nodal forces. Appendix H H-4 A popular approach in design in Hong Kong is based on nodal forces of the elements. The nodal force at each node is to be resisted by the “tributary” structural width extending from the node to mid-way to the adjacent node of the same element. Such design approach possesses the beauty of ensuring structural adequacy (but not necessarily an accurate distribution of forces in the structure) as perfect equilibrium among all elements is achieved. However, one may view that the design forces so chosen are not representing the actual behaviour of the structure. For example, in a plate bending problem where there should be “stresses” in form of bending, twisting moments and out-of-plane shears per unit width at every point in the structure, the “nodal force” approach simply reduces the structure into an assembly of discretized elements joined at nodes with nodal out-of-plane shears and moments. Twisting moment does not appear as a component in the nodal force and the out-of-plane nodal shear cannot be resolved into its original components at the X and Y faces. The problems encountered in design based on nodal forces will be discussed in fuller details in Section VII. Actually, design based on “stresses” instead of nodal reactions is very popular in the past when finite element analysis is first introduced to Hong Kong for plate analysis. Some recent popular commercial programs adopt nodal reactions instead of “stresses” in the design and most of the local engineers just simply adopt this approach without greater thought. The problems that are mentioned are actually very common for recent design (though the error is not great). VI. Structural Design in Plate Bending Structure Before carrying out more detailed discussion in the structural design of a plate bending structure, it is worthwhile to conduct a brief review in the structural behaviour of a plate bending structure. The problem relating to bending and twisting moment is first discussed. Basically, one may view that at any point inside the plate bending structure, there exist bending moments in two mutually perpendicular directions and a twisting moment (all per unit widths) which are termed “stresses”. A perfect analogy can be drawn to an “in-plane” load problem where at any point there exist normal stresses in two mutually perpendicular Appendix H H-5 directions and a shear stress. Whilst the normal stresses and the shear stress at the point can be compounded to the “principal stresses” in the directions where the shear stress is zero, likewise the “principal moments” can be derived by compounding the bending and twisting moment in certain directions (known as the principal directions) free of twisting. One can then view that the plate is under “pure bending” in the principal directions without twisting. Of course the principal directions do not necessarily align with the artificial X or Y directions pre-determined by the designer. Theoretically the designer can perform his design in the principal directions as the twisting moment is not involved. However, this approach is impractical for the engineers to design as the reinforcements have to follow the principal directions which change from point to point and load case to load case and is therefore not adopted in practice. The Wood-Armer Equations developed by Wood [19] serve to calculate the optimum design moments at any point in two directions (not necessarily, but preferably mutually perpendicular) in a “moment field”. The “components” of the optimum moments in any orientation, which are derived by Johansen’s Yield Criterion [10] can adequately cover the bending moments normal and perpendicular to the plane of the orientation. During the process, the twisting moments have been catered for and the design is structurally adequate. On the contrary, design by nodal moments (current local practice) can never capture the effects of twisting correctly, as there is no twisting moment in the nodal forces. Furthermore, if the designer goes further to sum up the nodal forces across a number of adjacent nodes and design the forces against the section formed from the summed tributary length of the nodes (currently the most popular approach in Hong Kong as provided by some commercial softwares), he will face the problem of how to design adequately for the net torsion on the section due to the generally unbalanced effects of the out-of-plane nodal forces and nodal moments bending in the plane of the section. Theoretically this net torsion is not a real force in the structure. So it is neither correct to design the torsion for the section by treating the section as a beam section. Wood-Armer equations are recently introduced to Appendix H H-6 some commercial programs but nearly all the local designs are still based on the nodal moments in the bending reinforcement design. To provide a sound design approach against out-of-plane shears, it is required to calculate the design shear based on the components of “stresses” in relation to shear derived from structural analysis, as similar to that in establishing the Wood- Armer Equations. By the general plate bending theory, the shear force per unit width at the “X-face” and “Y-face” of an element are respectively y M x M Q XY Y XZ ∂ ∂ + ∂ ∂ − = and y M y M Q XY X YZ ∂ ∂ − ∂ ∂ = , as diagrammatically illustrated in Fig. 14(a). It can be easily shown that the maximum shear will occur in a plane at an orientation θ on plan such that         = − YZ XZ Q Q 1 tan θ and the value of the maximum shear is 2 2 max YZ XZ Q Q Q + = as per the illustration in Fig. 14(b). So the designer needs to check or design for the maximum shear of magnitude max Q at the point, as it is the shear “stress” under which the plate will most likely to fail with failure surface at the orientation θ. Following the usual practice of designing against shear, if the max Q does not exceed allowable shear stress of concrete v c , no shear reinforcements will be required. Otherwise, reinforcements will be required. The designer may initially wonder that he has to arrange the vertical shear links in the direction of the orientation θ as shown in Fig. 14(c) which is practically impossible to be achieved. However, as the required shear resistance depends only on the total cross sectional area of links per unit area, the orientation of the shear links arrangement on plan does not really matter so long as the same total cross sectional area is provided. Thus it will be at the designer’s discretion to arrange the shear links on plan in any convenient orientations, say in the global X- direction as shown in Fig. 14(d). Turning back to the current practice in design of shear by nodal forces, the designer will easily find that the vertical shear force in a node cannot be split into Appendix H H-7 components in the adjacent faces of the element. A practical way for design appears to be lumping a series of adjacent nodal shear forces along a cut-section and design the summed shear force against the section. However, problems again arise as to the determination of the orientation of the cut-section on plan. The designer will find that, in a slab simply supported on two opposite sides. The summed shears along any sections in the direction of the span of the slab are small and that in the direction perpendicular to the span are close to the actual shear values in the slab structure. So, apparently the designer has to identify the main span direction for his determination of the orientation of the cut-sections. And yet, it is difficult to determine appropriate guidelines for adequate design as, in the first place, the slab structure does not behave as discretized elements with nodal forces. Lateral force, S , to prevent sidesway 1 S 1 U M 2 S 2 U M 3 S 3 U M h 1 h 2 h 3 1 S 3 S 2 S Note : 1. If the wall / column is prismatic and the lower end is restrained from rotation, the moment at the lower end will be Ui Li M M 5 . 0 = (carry-over from the top); if the lower end is assumed pinned, the moment at it will be zero; 2. The shear on the wall / column will be i Li Ui i h M M S + = where Ui M is obtained from plate bending analysis and the total restraining shear is ∑ = i S S Figure 13. Diagrammatic illustration of the restraining shear or nullifying shear Appendix H H-8 Z Q XZ Y Q YZ X Figure 14(a). Diagrammatic illustration of shear “stresses” in the X and Y faces of an element Q θ Q XZ θ Q YZ Plan Figure 14(b). Derivation of the direction and magnitude of the maximum shear “stress” s v s v max Q Q XZ θ θ Q YZ Plan Figure 14(c). Diagrammatic illustration of arrangement of theoretical shear links on Plan against max Q s v max Q s v Q XZ Q YZ Plan Figure 14(d). Diagrammatic illustration of convenient arrangement of shear links on Plan that will produce the identical shear capacity against max Q For vertical equilibrium, Qθ can be expressed as θ θ θ cos sin 1 YZ XZ Q Q Q + = × For θ Q to be maximum set YZ XZ YZ XZ Q Q Q Q d dQ = ⇒ = − ⇒ = θ θ θ θ θ tan 0 sin cos 0 2 2 2 2 2 2 max YZ XZ YZ YZ XZ XZ Q Q Q Q Q Q Q Q + + + = = ⇒ θ 2 2 YZ XZ Q Q + = A sv total link area in one row A sv total link area in one row θ Appendix I Derivation of Formulae for Rigid Cap Analysis I-1 Derivation of Formulae for Rigid Cap Analysis Underlying Principles of the Rigid Cap Analysis The “Rigid Cap Analysis” method utilizes the assumption of “Rigid Cap” in the solution of pile loads beneath a single cap against out-of-plane loads, i.e. the cap is a perfectly rigid body which does not deform upon the application of loads. The cap itself may settle, translate or rotate, but as a rigid body. The deflections of a connecting pile will therefore be well defined by the movement of the cap and the location of the pile beneath the cap, taking into consideration of the connection conditions of the piles. Consider a Pile i situated from a point O on the pile cap as shown in Figure I-1 with settlement stiffness iZ K As the settlement of all piles beneath the Cap will lie in the same plane after the application of the out-of-plane load, the settlement of Pile i denoted by iZ ∆ can be defined by i i O y b x b b 2 1 + + which is the equation for a plane in ‘co-ordinate geometry’ where O b , 1 b and 2 b are constants. The upward reaction by Pile i is ) ( 2 1 i i O iZ y b x b b K + + Summing all pile loads : Balancing the applied load ( ) ∑ + + = i i O iZ y b x b b K P 2 1 i iZ i iZ iZ O y K b x K b K b P ∑ ∑ ∑ + + = ⇒ 2 1 Balancing the applied Moment ( ) i i i O iZ X y y b x b b K M 2 1 + + − = ∑ +ve X M and Y M Y O i y i x Pile i X Figure I-1 – Derivation of Pile Loads under Rigid Cap I-2 2 2 1 i iZ i i iZ i iZ O X y K b y x K b y K b M ∑ ∑ ∑ − − − = ⇒ Balancing the applied Moment ( ) i i i O iZ Y x y b x b b K M ∑ + + = 2 1 i i iZ i iZ i iZ O Y y x K b x K b x K b M ∑ ∑ ∑ + + = ⇒ 2 2 1 It is possible to choose the centre O such that 0 = = = ∑ ∑ ∑ i i iZ i iZ i iZ y x K y K x K . So the three equations become ∑ = iZ O K b P ⇒ ∑ = iZ O K P b 2 2 i iZ X y K b M ∑ − = ⇒ ∑ − = 2 2 i iZ X y K M b 2 1 i iZ Y x K b M ∑ = ⇒ ∑ = 2 1 i iZ Y x K M b The load on Pile i is then ) ( 2 1 i i O iZ y b x b b K P + + = ∑         − + = ∑ ∑ ∑ i i iZ X i i iZ Y iZ iZ y y K M x x K M K P K 2 2 i i iZ iZ X i i iZ iZ Y iZ iZ y y K K M x x K K M K PK ∑ ∑ ∑ − + = 2 2 To effect 0 = = = ∑ ∑ ∑ i i iZ i iZ i iZ y x K y K x K , the location of O and the orientation of the axes X-X and Y-Y must then be the “principal axes” of the pile group. Conventionally, designers may like to use moments along defined axes instead of moments about defined axes. If we rename the axes and U-U and V-V after translation and rotation of the axes X-X and Y-Y such that the condition 0 = = = ∑ ∑ ∑ i i iZ i iZ i iZ v u K v K u K can be satisfied, then the pile load become i i iZ iZ Y i i iZ iZ U iZ iZ iZ v v K K M u u K K M K PK P ∑ ∑ ∑ − + = 2 2 If all piles are identical, i.e. all iZ K are equal, then the formula is reduced i i V i i U iZ v v M u u M N P P ∑ ∑ − + = 2 2 where N is the number of piles. Or if we do not wish to rotate the axes to U and V , then only ∑ ∑ = = 0 i iZ i iZ y K x K and the moment balancing equations becomes I-3 2 2 1 i iZ i i iZ i iZ O X y K b y x K b y K b M ∑ ∑ ∑ − − − = 2 2 1 i iZ i i iZ X y K b y x K b M ∑ ∑ − − = ⇒ and i i iZ i iZ i iZ O Y y x K b x K b x K b M ∑ ∑ ∑ + + = 2 2 1 i i iZ i iZ Y y x K b x K b M ∑ ∑ + = ⇒ 2 2 1 Solving ∑ ∑ = ⇒ = iZ O iZ O K P b K b P ( ) ( ) ∑ ∑ ∑ ∑ ∑ + − + = 2 2 2 2 1 i iZ i iZ i i iZ i iZ Y i i iZ X y K x K y x K y K M y x K M b ( ) ( ) ∑ ∑ ∑ ∑ ∑ + − − − = 2 2 2 2 2 i iZ i iZ i i iZ i iZ X i i iZ Y y K x K y x K x K M y x K M b So the pile load becomes ( ) ( ) i iZ i iZ i iZ i i iZ i iZ Y i i iZ X iZ iZ iZ x K y K x K y x K y K M y x K M K PK P ∑ ∑ ∑ ∑ ∑ ∑ + − + + = 2 2 2 2 + ( ) ( ) i iZ i iZ i iZ i i iZ i iZ X i i iZ Y y K y K x K y x K x K M y x K M ∑ ∑ ∑ ∑ ∑ + − − − 2 2 2 2 . If all piles are identical, i.e. all K iZ are equal, then the formula is reduced ( ) ( ) ( ) ( ) i i i i i i X i i Y i i i i i i Y i i X iZ y y x y x x M y x M x y x y x y M y x M N P P ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ + − − − + + − + + = 2 2 2 2 2 2 2 2 For a symmetrical layout where 0 = ∑ i i y x , the equation is further reduced to i i X i i Y iZ y y M x x M N P P ∑ ∑ − + + = 2 2 Contents Page 1 3 6 35 52 69 77 86 93 99 107 118 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 Introduction Some highlighted aspects in Basis of Design Beams Slabs Columns Column Beam Joints Walls Corbels Transfer Structures Footings Pile Caps General R.C. Detailings Appendices Appendix A – Appendix B – Appendix C – Appendix D – Appendix E – Appendix F – Appendix G – Appendix H – Appendix I – Clause by Clause Comparison between “Code of Practice for Structural Use of Concrete 2004” and BS8110 Assessment of Building Accelerations Derivation of Basic Design Formulae of R.C. Beam sections against Flexure Underlying Theory and Design Principles for Plate Bending Element Extracts of Design Charts for Slabs Derivation of Design Formulae for Rectangular Columns to Rigorous Stress Strain Curve of Concrete Derivation of Design Formulae for Walls to Rigorous Stress Strain Curve of Concrete Extracts of HKIE paper for plate bending analysis and design Derivation of Formulae for Rigid Cap Analysis 1.0 1.1 Introduction Promulgation of the Revised Code A revised concrete code titled “Code of Practice for Structural Use of Concrete 2004” was formally promulgated by the Buildings Department of Hong Kong in late 2004 which serves to supersede the former concrete code titled “The Structural Use of Concrete 1987”. The revised Code, referred to as “the Code” hereafter in this manual will become mandatory by 15 December 2006, after expiry of the grace period in which both the revised and old codes can be used. 1.2 Main features of the Code As in contrast with the former code which is based on “working stress” design concept, the drafting of the Code is largely based on BS8110 1997 adopting the limit state design approach. Nevertheless, the following features of the Code in relation to design as different from BS8110 are outlined : (a) (b) (c) (d) (e) Provisions of concrete strength up to grade 100 are included; Stress strain relationship of concrete is different from that of BS8110 for various concrete grades; Maximum design shear stress of concrete ( v max ) are raised; Provisions of detailings to enhance ductility are added, with the requirements of design in beam-column joints (Sections 9.9 and 6.8); Criteria for dynamic analysis for tall building under wind loads are added (Clause 7.3.2). As most of our colleagues are familiar with BS8110, a comparison table highlighting differences between BS8110 and the Code is enclosed in Appendix A which may be helpful to designer switching from BS8110 to the Code in the design practice. 1.3 Outline of this Manual This Practical Design Manual intends to outline practice of detailed design and detailings of reinforced concrete work to the Code. Detailings of individual types of members are included in the respective sections for the types, though the last section in the Manual includes certain aspects in detailings which are 1 common to all types of members. Design examples, charts are included, with derivations of approaches and formulae as necessary. Aspects on analysis are discussed selectively in this Manual. In addition, as the Department has decided to adopt Section 9.9 of the Code which is in relation to provisions for “ductility”, conflicts of this section with others in the Code are resolved with the more stringent ones highlighted as requirements in our structural design. As computer methods have been extensively used nowadays in analysis and design, the contents as related to the current popular analysis and design approaches by computer methods are also discussed. The background theory of the plate bending theory involving twisting moments, shear stresses, and design approach by the Wood Armer Equations which are extensively used by computer methods are also included for analysis of slabs, flexible pile caps and footings. To make distinctions between equations quoted from the Code and the equations derived in this Manual, the former will be prefixed by (Ceqn) and the latter by (Eqn). 1.4 Reference of this Manual This Manual has made reference to the following documents : (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) The Code itself Concrete Code Handbook by HKIE Hong Kong Building (Construction) Regulations The Code of Practice for Dead and Imposed Loads for Buildings (Draft) BS8110 Parts 1, 2 and 3 Code of Practice for Precast Concrete Construction 2003 Code of Practice for Fire Resisting Construction 1996 Reinforced and Prestressed Concrete 3rd edition – Kong & Evans Reinforced Concrete Design to CP110 – Simply Explained A.H. Allen Ove Arup and Partners. Notes on Structures: 17. April 1989 ACI Code ACI318-05. 2 2.0 2.1 Some highlighted aspects in Basis of Design Ultimate and Serviceability Limit states The ultimate and serviceability limit states used in the Code carry the usual meaning as in BS8110. However, the new Code has incorporated an extra serviceability requirement in checking human comfort by limiting acceleration due to wind load on high-rise buildings (in Clause 7.3.2). No method of analysis has been recommended in the Code though such accelerations can be estimated by the wind tunnel laboratory if wind tunnel tests are conducted. Nevertheless, a design example is enclosed in Appendix B, based on some recognized empirical formulae in the absence of wind tunnel tests. 2.2 Design Loads The Code has made reference to the “Code of Practice for Dead and Imposed Loads for Buildings” for determination of characteristic gravity loads for design. However, the Code has not yet been formally promulgated. At the meantime, the design loads should be therefore taken from HKB(C)R Clause 17. Nevertheless, the designer may need to check for the updated loads by fire engine for design of new buildings, as required by FSD. The Code has placed emphasize on design loads for robustness which are similar to the requirements in BS8110 Part 2. The requirements include design of the structure against a notional horizontal load equal to 1.5% of the characteristic dead weight at each floor level and vehicular impact loads (Clause 2.3.1.4). The small notional horizontal load can generally be covered by wind loads if wind loads are applied to the structure. Identification of key elements and designed for ultimate loads of 34 kPa, together with examination for progress collapse in accordance with Clause 2.2.2.3 can be exempted if the buildings are provided with ties determined by Clause 6.4.1. The usual reinforcement provisions as required by the Code can generally cover the ties provision. Wind loads for design should be taken from Code of Practice on Wind Effects in Hong Kong 2004. It should also be noted that there are differences between Table 2.1 of the 3 8 1 Distance Ratio from Neutral axis 0. in order to achieve smooth (tangential connection) between the parabolic portion and straight portion of the stress strain curve. The beneficial partial load factor for earth and water load is 1. 2. Furthermore.6 0. The stress strain curve of concrete as given in Figure 3.4 × 10 − 4 Ec γm which is the value in BS8110.8 of the Code.3 Materials – concrete Table 3.3769 where ε0 = 0.001319 Figure 2-1 – Stress strain profile of grades 35 Design formulae for beams and columns based on these stress strain curves by 4 .Code that of BS8110 Part 1 in some of the partial load factors γ f .34( f cu / γ m ) f cu instead of staying at 2.4 0. Stress Strain Profile for Grade 35 18 16 14 Stress (MPa) 12 10 8 6 4 2 0 0 0. The stress strain curve for grade 35 is plotted as an illustration.2 has tabulated a set of Young’s Moduli of concrete up to grade 100.4) has recommended to shift the ε 0 value to 1.1.2 0.1 of BS8110 Part 1. The values are generally smaller than that in BS8110 and also slightly different from the former 1987 Code. lower values should be used if the earth and water load are known to be over-estimated. However. whose initial tangent is determined by these Young’s Moduli values is therefore different from Figure 2. the Concrete Code Handbook (Clause 3. (f) Transfer Structure : horizontal element which redistributes vertical loads where there is a discontinuity between the vertical structural elements above and below. This Manual is based on the above definitions in delineating structural members for discussion. together with Appendix F and G of this Manual.5 which are repeated for ease of reference : (a) Slab : the minimum panel dimension ≥ 5 times its thickness. 2. Table 4.2. classified as shallow beam. However.BS8110. column and wall in accordance with dimensions in Clause 5. A full derivation of design formulae and charts for columns and walls are given in Sections 5 and 7.1. otherwise : deep beam. (e) Shear Wall : walls contributing to the lateral stability of the structure. 5.2 of the Code tabulated nominal covers to reinforcements under different exposure conditions.5 times the overall depth for continuous span.4 and 5. 5 . reference should also be made to the “Code of Practice for Fire Resisting Construction 1996”. strictly speaking. (c) Column : vertical member with section depth not exceeding 4 times its width and height is at least 3 times its section depth. beam. (d) Wall : vertical member with plan dimensions other than that of column.1.4 Definitions of structural elements The Code has included definitions of slab. (b) Beam : for span ≥ 2 times the overall depth for simply supported span and ≥ 2. become inapplicable. 0GK 1.0GK 1. Figure 3.6QK 1.2a – To search for maximum support reactions 1.6QK 6 .4GK+1.2c – To search for maximum hogging moment at support adjacent to spans with 1.0GK 1.0GK 1.6QK 1.4GK+1.5.4GK+1.4GK+1.6QK 1.2) Normally continuous beams are analyzed as sub-frames by assuming no settlements at supports by walls.6QK 1.6QK 1.4GK+1. columns (or beams) and rotational stiffness by supports provided by walls or columns as 4 EI / L (far end of column / wall fixed) or 3EI / L .2.4GK+1.3. 5.0GK Figure 3.2.6QK Figure 3. (far end of column / wall pinned).1 – continuous beam analyzed as sub-frame In analysis as sub-frame.6QK 1.4GK+1.1 & 5.4GK+1.0GK 1.6QK 1. Clause 5.6QK 1.6QK 1.2 states that the following loading arrangements will be adequate for seeking for the design moments : 1.2b – To search for maximum sagging moment in spans with 1.4GK+1.4GK+1.0GK Figure 3.4GK+1.6QK 1.5.2.4GK+1.4GK+1.0 3.6QK 1.1 Beams Analysis (Cl.5. However. 3. The rationale of such further limitations is discussed in Concrete Code Handbook 6.1.2.1 times the span of zero moment (0.2. continuous beam and cantilever.2.2(b)) For simply supported beam. An example for illustration is as indicated : 7 .9) Moment redistribution is allowed for concrete grade not exceeding 70 under conditions 1. it should be noted that there would be further limitation of the neutral axis depth ratio x / d if moment redistribution is employed as required by (Ceqn 6.5) of the Code which is identical to the provisions in BS8110. The design value of shears and moments at any location will be the summation of the values of the same sign created by the individual cases.1.2bi) + 0.1. (ii) Effective flange width of T. 5.2(GK+QK+WK) and 1. 2 and 3 as stated in 5.1. with the sum of the latter two not exceeding 0.9.2.2 times the span of zero moment and lpi taken as 0. Nevertheless. most of the commercial softwares can actually analyze individual load cases.3 of the Code). With wind loads.2. 5.2 Moment Redistribution (Cl.0GK+1. 5. the load cases to be considered will be 1.1lpi). of the Code as Effective width (beff) = width of beam (bw) + 0.and L-beams are as illustrated in Figure 5.4WK on all spans. each of which is having live load on a single span and the effects on itself and others are analyzed.7 times the effective span of the beam.4) and (Ceqn 6.2(a)) Effective flange width of T.2.2 times the centre to centre width to the next beam (0.3 Highlighted aspects in Determination of Design Parameters of Shallow Beam (i) Effective span (Cl. 5. 3.and L-beams (Cl. the effective span can be taken as the clear span plus the lesser of half effective depth and half support width except that on bearing where the centre of bearing should be used to assess effective span. (Fig. 1 = beff .4 – Reduced moment to Support Face for support providing rotational restraint In Figure 3.2×800 800 Bending Moment Diagram Figure 3. 2 = 750 > 0.2Ø if the support is circular with diameter Ø.400 2000 400 2000 400 2000 400 Figure 3.1 for illustration is given below : 350 kNm at support 250 kNm at 0.1) of the Code l pi = 0. the bending moment at support face is 200 kNm which 8 . beff .1 × 3500 = 750 As beff .1.2 × 3500 = 700 beff .2. An example 3. But the reduction should not be less than 65% of the support moment. (iii) Support Moment Reduction (Cl.3 – Example illustrating effective flange determination The effective spans are 5 m and they are continuous beams.2 Ø into the support face 200 kNm at support face centre line of beam column elements idealized as line elements in analysis 0.1 = beff .7 × 5000 = 3500 . The effective width of the T-beam is.4. 5.2 × 2000 + 0.2) The Code allows design moment of beam (and slab) monolithic with its support providing rotational restraint to be that at support face if the support is rectangular and 0. 2 = 700 .1 = beff . beff = 400 + 700 × 2 = 400 + 1400 = 1800 So the effective width of the T-beam is 1800 mm. by (Ceqn 5. 2 = 0. However. If the support is circular and the moment at 0.1) The provision is identical to BS8110 as 1.5 indicates a numerical Example 3. 250 kNm FEd. then 250 kNm will be the design moment as it is greater than 0.65×350 = 227.can be the design moment of the beam if the support face is rectangular.5 kNm.1. Cantilever with lateral restraint only at support : Clear distance from cantilever to support not to exceed 25bc or 100bc2/d if less 9 . 8 (iv) Slenderness Limit (Cl. wall support). 227.8 = 230 kNm.2.5 kNm.5 – Reduction of support moment by support shear for support considered not providing rotational restraint The design support moment at the support under consideration can be reduced to 250 − 200 × 0.sup = 200 kN 800 Figure 3. Figure 3.g. For beam (or slab) spanning continuous over a support considered not providing rotational restraint (e. the Code allows moment reduction by support shear times one eighth of the support width to the moment obtained by analysis. as it is smaller than 0. and 2.5 kNm should be used for design.65×350 = 227.2.2Ø into the support and the bending moment at the section is 250 kNm. 6. Simply supported or continuous beam : Clear distance between restraints not to exceed 60bc or 250bc2/d if less. 3 under Cl.where bc is the breadth of the compression face of the beam and d is the effective depth. The table has provisions for spans and end spans which are not specified in BS8110.4 and 7. 2. 3.3 5. (vi) Minimum spacing between bars should not be less than bar diameter. Usually the slenderness limits need be checked for inverted beams or bare beam (without slab). aggregate size + 5mm or 20mm (Cl. For two-way spanning slabs the check should be carried out on the basis of the shorter span.1. should be such that the clear spacing between bar is limited by 70000β b clear spacing ≤ ≤ 300 mm where β b is the ratio of moment fy redistribution.5 16 21 18.5 One or two-way spanning solid slab 7 20 26 23(2) Support condition Cantilever Simply supported Continuous End span Rectangular Beam 7 20 26 23 Note : 1. Maximum spacing between bars in tension near surface. Flanged Beam bw/b < 0. (v) Span effective depth ratio (Cl.1 – effective span / depth ratio The basic span depth ratios can be modified due to provision of tensile and compressive steels as given in Tables 7. clear spacing ≤ 47000 ≤ 300 mm.3.2) Table 7.2).2 tabulates basic span depth ratios for various types of beam / slab which are deemed-to-satisfy requirements against deflection.3. 7. Or alternatively. calculation can be carried out to justify deflection limits not to exceed span / 250.4. The values given have been chosen to be generally conservative and calculation may frequently show shallower sections are possible. fy (vii) 10 . 9.5 which are identical to BS8110.4. 8. The value of 23 is appropriate for two-way spanning slab if it is continuous over one long side.2. Table 3. Nevertheless. 7. by Cl. 1 Design in accordance with the Rigorous Stress Strain curve of Concrete The stress strain block of concrete as indicated in Figure 3. 4.So the simplest rule is 70000β b 70000 × 1 = = 152 mm when using fy 460 high yield bars and under no moment redistribution.1 of BS8110.2 – Nominal Cover of Beams 3.4 of the Code indicates the nominal cover required in accordance with Exposure conditions. taking into account our current practice of using concrete not inferior than grade 30 and maximum aggregate sizes not exceeding 20 mm. the Concrete Code Handbook has recommended to shift the ε0 to the right to a value of 1. we may generally adopt the provision in our DEDD Manual (DEDD-404 Table 2) with updating by the Code except for compartment of 4 hours FRP.4 Sectional Design for Rectangular Beam against Bending Nominal Cover (mm) 30 40 55 45 3.2 of the Code. in order to achieve smooth connection between the parabolic and the straight line portions. we can. So. The recommended covers are summarized in the following table : Description Internal External Simply supported (4 hours FRP) Continuous (4 hours FRP) Table 3. Furthermore. However. (viii) Concrete covers to reinforcements (Cl. and to adopt condition 2 for the external structures.3) Cl. E c .4 and Cl.8 is different from Figure 2.4. the “Code of Practice for Fire Resisting Construction 1996” should also be checked for different fire resistance periods (FRP). as far as our building structures are concerned. With the values of Young’s Moduli of concrete.2.34 f cu . Nevertheless. as indicated in γ m Ec Table 3. 4.2. the stress strain block of concrete for various grades can 11 . 4. roughly adopt condition 1 (Mild) for the structures in the interior of our buildings (except for bathrooms and kitchens which should be condition 2). 6 0.be determined.7. ε ult = 0.001319 Figure 3.6 – Stress strain block of grades 35 Based on this rigorous concrete stress strain block. design formulae for beam can be worked out as per the strain distribution profile of concrete and steel as indicated in Figure 3.4 0.8 1 Distance Ratio from Neutral axis 0.3769 where ε0 = 0.7 – Stress Strain diagram for Beam 12 .2 0.0035 d’ x d neutral axis Stress Diagram Strain Diagram Figure 3. The stress strain curve of grade 35 is drawn as shown Stress Strain Profile for Grade 35 18 16 14 Stress (MPa) 12 10 8 6 4 2 0 0 0. 67 f cu  1 1 ε 0 1 ε  − + −  0 γ m  2 3 ε ult 12  ε ult      2  x  2 0.4 of the Code). again by (Eqn 3-1) bd 2 f cu to K ' = 0.1. bd 2 f cu K ' = 0.1. 13 . It should be noted that the x ratio will be further limited if moment d redistribution exceeds 10% by (Ceqn 6.5) of the Code as x ≤ (β b − 0.154 for grade 30 K ' = 0.87 f y γ m  1 ε0 1 −  3ε ult  x  d  (Eqn 3-2) As x is limited to 0.151 for grade 40 which are all smaller than 0.125 for grade 50 which are more significantly below 0.4 for singly reinforced sections under moment redistribution not greater than M will be limited 10% (Clause 6.67 f cu = bd 0.4) and (Ceqn 6.5 for singly reinforcing sections for grades up to 40 d under moment redistribution not greater than 10% (Clause 6. However.132 under the simplified stress block.2. and d x ≤ (β b − 0.2.5) for 40 < f cu ≤ 70 d where β b us the ratio of the moment after and before moment redistribution.4 of the Code).67 f cu   + d γm    1 ε0 1 −  3ε ult  x M  − 2 =0  d bd  (Eqn 3-1) With neutral axis depth ratio determined. the steel ratio can be determined by (Eqn 3-2) Ast 1 0.4) for f cu ≤ 40 .156 under the simplified stress block.152 for grade 35 K ' = 0.0035 for concrete The solution for the neutral axis depth ratio grades not exceeding 60 (Re Appendix C for detailed derivation) : 0. M will be limited to K ' values as in by (Eqn 3-1). for grade 45 and 50 where neutral axis depth ratio is limited to 0.126 for grade 45 K ' = 0.x for singly supported beam is d the root of the following quadratic equation where ε ult = 0. With moment redistribution exceeding 10%.87 f y 1 −   d (Eqn 3-3) And the same amount of reinforcement will be added to the tensile reinforcement :  M   2   bd f − K '  f cu  1 ε0  cu   1 −  3 ε η +   d' ult   0. The same phenomenon applies to tensile steel also. as illustrated by the following Chart 3-1 in which compression reinforcement decreases from grade 30 to 40 for the same M .67 f cu = bd 0. compression reinforcements at d ' from the surface of the compression side should be added.87 f y 1 −   d Ast 1 0. It follows that more compressive reinforcements will be required for grade 45 than 40 due to the limitation of neutral axis depth ratio. The compression reinforcements will take up the difference between the applied moment and Kbd 2 and the compression reinforcement ratio is  M   2 − K '  f cu  Asc  bd f cu   = d'  bd  0. 14 .4 for grades up to and including 70 for moment redistribution not exceeding 10%. but increases at grade 45 due to the change of the bd 2 limit of neutral axis depth ratio from 0.4 with moment redistribution not exceeding 10%. the same trend will also take place.87 f y γ m (Eqn 3-4) where η is the limit of neutral axis depth ratio which is 0.5 to 0.When M bd 2 f cu exceeds the limited value for single reinforcement.5 for grade 40 and below and 0. 5 4 Grade 30 Asc/bd Grade 40 Asc/bd Grade 35 Ast/bd Grade 45 Ast/bd Grade 35 Asc/bd Grade 45 Asc/bd M/bd 2 Chart 3-1 – Reinforcement Ratios of Doubly Reinforced Beams for Grade 30 to 45 with Moment Redistribution limited to 10% or below Furthermore.1 of the Code is adopted. 40 and 45 respectively which are all in excess of the minimum of 0. bd bd 3.44% and 0.4. as similar to BS8110. there is an upper limit of “lever arm ratio” z which is the depth of the centroid of the compressive force of concrete to d the effective depth of the beam section of not exceeding 0.1 Grade 30 Ast/bd Grade 40 Ast/bd 14 12 10 8 6 4 2 0 0 0.13% in accordance with Table 9. The design formula becomes For singly reinforced sections where K = M ≤ K ' where K ' = 0. 0.5% for grades 30.33%.5 2 Reinforcement ratios (%) 2. To comply with this requirement.1 of the Code.5 1 1.156 f cu bd 2 for grades 40 and below and K ' = 0. 35.132 for grades over 40 and not 15 . the minimum steel percentages will be 0. Design Charts for grades 30 to 45 comprising tensile steel and compression steel ratios Ast A and sc are enclosed at the end of Appendix C.5 3 3.Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0. 0.39%.2 Design in accordance with the Simplified Stress Block The design will be simpler and sometimes more economical if the simplified rectangular stress block as given by Figure 6.95. 307 ≤ 0. so singly reinforced f cu bd 2 35 × 400 × 444 2 x Solving the neutral axis depth ratio by (Eqn 3-1) d 2 0.627 γ m  3 ε ult  bd 400 × 444 2   M1 = 2 x − 13.25 − 0 .87 f y γ m ⇒ Ast = 1865 mm2  1 ε0  x 1 1 −  3 ε  d = 0.699 × 0.45  0.87 f y z (Eqn 3-5) For doubly reinforced sections K = z K' = 0.5 d 2 × (− 60. d  d  0.95 .34 f cu ε0 1.25 − = 1 −  .0013192 = 0.699 + 13.5 + 0.3 f cu = 35 MPa Section : 500 (h) × 400 (w).3 Worked Examples for Rectangular Beam with Moment Redistribution < 10% Worked Example 3.87 × 460 ×13.25 − ≤ 0.87 f y z (Eqn 3-6) 3.5 − 0.104 < 0. d 0 .5 × 23700 γ m Ec ε ult 286 × 10 6 = 0.67 f cu  1 ε 0  M 286 × 10 6 1 −  = 13.87 f y z (d − d ') K ' f cu bd 2 Ast = + Asc 0.9  x  z 1 K  1  =  0. (ii) d = 500 − 40 − 16 = 444 1.38) Ast 1 0. z K = 0. γ m  2 3 ε ult 12  ε ult       0.5 + 0.9 d M > K'.152 .38 .exceeding 70. (i) M 1 = 286 kNm.699 − 4 × (− 60.45   Ast = M 0.627 ) = = 0.0105  ult   Use 2T32 + I T25 (ii) d = 500 − 40 − 20 = 440 16 .669 .67 f cu = bd 0.45 Asc (K − K ') f cu bd 2 = 0.38) × (− 3.34 × 35 ε0 = = = 0.4. M 2 = 486 kNm. − 2 = = −3. f cu bd 2 x  z 1 = 1 −  d  d  0.67 f cu  1 1 ε 0 1 ε   − + −  0   = −60.9  0.3769 1.307 = 0. 156 × 35 × 400 × 440 2 Ast = + Asc = + 399 = 3498 mm2 0.34 f cu 1.87 f y 1 −   d Asc = 0.87 × 460 × 0.87 f y z 0.5 + 0.5 + 0.152 .114 d 440  M   2 − K  f cu  Asc  bd f cu (0.9 35 × 400 × 444 2 Ast M 286 × 10 6 = = = 0.3 are re-done in accordance with Figure 6.1 of the Code.978 % = bd 0.87 × 460 × 0.5 = 0.267 %   By (Eqn 3-3) = = bd 0.87 f y γ m   M  2 − K  f cu   bd f  1 ε0  cu  1 − η +   3ε   d' ult   0. K ' f cu bd 2 0.3769 ε ult 486 ×10 6 = 0.87 f y 1 −   d By (Eqn 3-4) Ast 1 13.87 × 460 × (440 − 50 ) 0.87 f y ( z / d ) 400 × 444 2 × 0.179 − 0.775 × 440 Use 3T40 Summary for comparison : 17 .152)× 35 = 0.0013192 1.179 − 0. so doubly reinforcing = f cu bd 2 35 × 400 × 440 2 z = 1 − 0.2.9 × 0. so doubly reinforced f cu bd 2 35 × 400 × 440 2 d ' 50 d ' = 40 + 10 = 50 = = 0.156)× 35 × 400 × 440 2 Asc = = = 399 mm2 > 0.25 − = 0.114)  d' 0.87 × 460 × (1 − 0.87 f y (d − d ') 0.34 × 35 = = 0.ε0 = 1.00267 × 400 × 440 = 469 mm2 Use 2T20 M2 = Ast 1 0.25 − d 0.67 f cu = bd 0.5 × 0.179 > 0.5 + 0.867 ⇒ Ast = 1856 mm2 Use 2T32 + I T25 (ii) K = 486 × 10 6 M = 0.156 .2% in accordance with Table 9. d (K − K ') f cu bd 2 (0.00236 = 1.179 > 0.775 section required.699 × 0.01978 × 400 × 440 = 3481 mm2 Use 3T40 (i) and (ii) of Worked Example 3.87 × 460 Ast = 0.01045 bd bd 2 × 0.5 × 23700 γ m Ec ε0 = 0.1 of the Code (the simplified stress) block by (Eqn 3-5) and (Eqn 3-6) (i) z K 286 × 10 6 = 0.867 = 0. The principle of sectional design of flanged beam follows that rectangular beam with an additional flange width of b f − b w as illustrated in Figure 3.5 = 0.1 of the Code. β b = 0. if in flexural compression.e.8 − 0.82 d d d K ⇒ K = 0.4 .5 Sectional Design of Flanged Beam against Bending 3.9 (K − K ') f cu bd 2 (0. i.Singly Reinforced Ast (mm2) Based on Rigorous Stress-strain profile Based on Simplified stress strain profile 1865 1856 Asc (mm2) 469 399 Doubly Reinforced Ast (mm2) 3481 3498 Total (mm2) 3950 3897 Results by the two approaches are very close.4 × 0.3 has undergone a moment redistribution of 20% > 10%.4) ⇒ ≤ 0. = 1 − 0.8.1 Slab structure adjacent to the beam. The use of flanged beam will be particularly useful in eliminating the use of compressive reinforcements.2. The approach based on the simplified stress block are slightly more economical.87 f y z 0.9 × 0.87 × 460 × 0.4 = 0.4 for Rectangular Beam with Moment Redistribution > 10% If (ii) of the Worked Example 3. 0.4. thus effectively widen the structural width of the beam.176 − 0. by (Ceqn 6. can be used to act as part of compression zone of the beam.4) of the Code x x z ≤ (β b − 0.25 − 3.5. 18 .2 % Asc = = 0.82 × 440 So total amount of reinforcement is greater.132 × 35 × 400 × 440 2 Ast = + 347 = 3242 mm2 + Asc = 0. apart from reducing tensile steel due to increase of lever arm.132)× 35 × 400 × 440 2 = 764 mm2 > 0.4 Worked Example 3.8 .5 + 0.82 = 0.132 0 .87 f y (d − d ') 0. 3.87 × 460 × (440 − 50) as required by Table 9. K ' f cu bd 2 0. 25 −  0.8 – Analysis of a T or L beam section Design formulae based on the simplified stress block are derived in Appendix C which are summarized as follows : (i) Singly reinforcing section where 0.67 f cu γm 0 .67 d f  b f  − 11 −  + Kr ' ≤ K f '= 2 b γ m d  w  2 d  bw d   where K r ' = 0.402 f cu + =0 (Eqn 3-8) d bw d 2 d  where Mf bw d 2 =  1 d f  0.67 f cu d f  b f   − 11 −  2 d  γm d  bw    for : (Eqn 3-9) If df x ≤ 0. (ii) Singly reinforcing section where 0.9  d d   where K = M (Eqn 3-7) f cu b f d 2 If so.156 for f cu ≤ 40 and K r ' = 0. treat as a rectangular beam of width b f .9 x x bw Figure 3. d d 19 .9 × neutral axis depth is outside flange depth and  1 d f  M 0.5 − 0. carry out design as if it is a rectangular beam of width b f .9 × .1809 f cu   − 0.bf df d 0.9 × neutral axis depth is outside flange depth by checking 0.9  x K  df ≤ = 2 0.132 40 < f cu ≤ 70 x be solved by the quadratic equation d 2 x M −M f x 0. 9 =  0.87 f y γ m d d  b f df x  − 1 b  d + 0.341  0.95 20 .5. d f 150 x  K  1  0. if 3.9 × .87 f y (z / d ) 1800 × 440 2 × 0.67 f cu  b f 1    = − 1 + 0.95 . d = 500 − 40 − 20 = 440 .87 f y γ m  b w  d'     0.87 × 460 × 0. under a moment 486 kNm.25 − = 0.87 f y 1 −   d (Eqn 3-12) By following the procedure in (ii).5. ∴ = 0.0398 f cu b f d 2 35 × 1800 × 440 2 By (Eqn 3-7).5 : Singly reinforced section where 0.9 d   w   (Eqn 3-10) (iii) Doubly reinforcing section : x obtained by (Eqn 3-8) d exceeds 0.9 x df ≤ d d Consider the previous example done for a rectangular beam 500 (h) × 400 (w).00367 bd bd × 0.45 = 0.5 − 0. f cu = 35 MPa.67 f cu = > 0.9 bw = 1800 d f = 150 section of width = 1800 mm and depth = 150 mm. First check if 0.2 Worked Examples for Flanged Beam (i) Worked Example 3.95 d d d Thus Ast M 486 × 10 6 = 2 = = 0. bw d 0.9  0.45 d 440 d   z z x So = 1 − 0. then double reinforcements will be required by  M   − K f '  f cu 2  Asc  bw d f cu   (Eqn 3-11) = bd  d'  0.: x df ≤ based on beam width of bw = 1800 d d M 486 × 10 6 K= = = 0.87 f y 1 −   d  M   − K f '  f cu    bw d 2 f cu df Ast 0. with a flanged bw = 400 .45 +   d bd 0.954 > 0.103 ≤ = = 0.If df Ast x 1 0. 890 bf bw = 2000 = 3. flange depth = 150 mm f cu = 35 MPa under a moment 4000 kNm 21 . bw = 2000 d f = 150 Worked Example 3.25 − = 0.169 .1809 × 35  − 0. d = 1000 − 50 − 60 = 890 .9 × neutral axis extends below flange.65 kNm d  bw γm bw d    x by (Eqn 3-8) Solve d 2 x M −M f x 0. flange width = 1250 mm.2198 d By (Eqn 3-10) df Ast 1 0.402 f cu + =0 d bw d 2 d  x (4000 − 2675.402 × 35 + =0 d 600 × 890 2 d  x = 0.67 f cu  b f x + 0.67 f cu d f  b f    = − 11 − 2  2 d  ⇒ M f = 2675.65)× 10 6 x 0.9 df d = 150 = 0. flange depth = 150 mm f cu = 35 MPa under a moment 4000 kNm bw = 600 .176 > f = 0.9  = 0. Mf  1 d f  0.6 – Singly reinforced section where 0.Ast = 2905 mm2.9 x df ≤ based on beam width of bw = 2000 d d M 4000 × 10 6 K= = = 0. it can be seen that there is saving in tensile steel (2905 mm2 vs 3498 mm2) and the compression reinforcements are eliminated (ii) x df > d d Beam Section : 1000 (h) × 600 (w). Use 10-Y40 in 2 layers (iii) Worked Example 3.169  0. As in comparison with the previous example based on rectangular section.0231 =  − 1  d d bw d 0.0721 f cu b f d 2 35 × 2000 × 890 2 By (Eqn 3-7) d  x K  150   = 0.87 f y γ m  bw    2 Ast = 12330 mm2.7 – Doubly reinforced section Beam Section : 1000 (h) × 600 (w).9  d d 890  So 0.333 600 First check if 0.5 − 0. flange width = 2000 mm.1809 f cu   − 0.9 = 2 0. 26 × 10 6 x 0.2.5 .1 and 9.26 kNm d  bw γm bw d 2    x by (Eqn 3-8) Solve d 2 x M −M f x 0. d = 1000 − 50 − 60 = 890 .1 of the Code.1809 f cu   − 0.25 − = 0.169 0.9  = 0.9  d d 890   So 0.87 f y 1 −   d the Code Asc = 2040 mm2.1 of bd  d' 0. 9.9 x df ≤ based on beam width of bw = 1250 d d M 4000 ×10 6 K= = = 0.1809 × 35  − 0.67 f cu d f  b f   = − 11 −  2 d  ⇒ M f = 1242.82)242. Use 11-Y40 in 2 layers 3.87 f y γ m  b f df x + 0.9.169 . d By (Eqn 3-11)  M     b d 2 f − K f '  f cu Asc  w cu  = = 0.6 Detailings of longitudinal steel for bending The followings should be observed in placing of longitudinal steel bars for bending which will directly affect design (Cl.9×neutral axis extends below flange.4% as per Table 9.1 of the Code where 0.0038 Use 0.402 × 35 + =0 d 600 × 890 2 d  x = 0.9 = 2 0.bw = 600 .67 f cu = bw d 0.302 > f = 0.5 − 0. Mf  1 d f  0. 890 bf bw = bw = 1250 d f = 150 1250 = 2. the requirements arising from “ductility” are marked with “D”) (i) Minimum tensile steel percentage : Table 9.547 > 0.13% 22 .0247  − 1  d  d  bw    2 Ast = 13196 mm . Use 2T25 + 4T20 Ast 1 0.402 f cu + =0 d bw d 2 d  2 x (4000 − 1337. So double reinforcement is required. df d = 150 = 0.115 f cu b f d 2 35 ×1250 × 890 2 By (Eqn 3-7) d  x K  150   = 0.083 600 First check if 0.2. 1 of the Code where 0. sb ≤ 250 mm and be distributed over two-thirds of the beam’s overall depth measured from its tension face. 9. the sum of reinforcement sizes in a particular layer should not exceed 40% of the beam width. (iii) Minimum total percentage : total reinforcements to be not less than 0. (vi) Minimum clear spacing of bars should be the minimum of bar diameter. (viii) Maximum clear spacing between adjacent bars near tension face of a beam ≤ 70000βb/fy ≤ 300 mm where βb is the ratio of moment redistribution or alternatively ≤ 47000/fs ≤ 300 mm where fs = (ix) 2 f y As . (v) At any section of a beam within a critical zone (e. the anchorage for tension shall be deemed to commence at the lesser of 1/2 of the relevant depth of the column or 8 times the bar diameter as indicated in Figure 3. a potential plastic hinge zone) the compression reinforcement should not be less than one-half of the tension reinforcement in the same region (D). Where it can be shown that the critical section of the plastic hinge is at a distance at least the beam depth or 500 mm. prov × 1 βb .g. For beam with depths > 750 mm. the anchorage length can be considered to 23 (xi) .for rectangular beam and other percentages for others.req 3 As . Requirements for containment of compression steel bars is identical to that of columns (Cl.2% for rectangular beam and other percentages for others.5.2 of the Code) : Every corner bar and each alternate bar (and bundle) in an outer layer should be supported by a link passing around the bar and having an included angle < 135o.9. (vii) At laps.5% (D). (iv) Maximum steel percentage : 2. provision of sides bars > (x) sb b / f y where sb is the side bar spacing and b is the lesser of the beam breadth under consideration and 500 mm. No bar within a compression zone be more than 150 mm from a restrained bar. 20 mm and aggregate size + 5 mm. (ii) Minimum compressive steel percentage : Table 9.3% (D). whichever is the less from the column face. In addition.2. When longitudinal beam bars are anchored in cores of exterior columns or beam studs. Notwithstanding the adequacy of the anchorage of a beam bar in a column core or a beam stud. Top beam bars shall be bent down and bottom bars must be bent up. f yφ Alternatively.5 for tension anchorage and 0.5 should be increased by 15%. or the end of the beam stud where appropriate. and not closer than 3/4 of the relevant depth of the column to the face of entry. D anchorage commence at this point >the lesser of D/2 or 8ø Anchored bar of dia.commence at the column face. As such.4 and 8. Requirements by (xi) and (xii) are illustrated by Figure 3. the anchorage and lap lengths as indicated in Tables 8.9 – Anchorage of bar in exterior column for ductility enhancement 3.7 Design against Shear 24 . bars must be assumed to be fully stressed as a ductility requirement.63 for (xii) compression anchorage for high yield bars in accordance with Table 8. Lap lengths can be taken as identical to anchorage lengths (D). no bar shall be terminated without a vertical 90o standard book or equivalent anchorage device as near as practically possible to the far side of the column core.9. the anchorage length can be determined by lb ≥ 4 f bu which is a modification from (Ceqn 8. ø h > 3/4 D Figure 3. (D) For calculation of anchorage length.3 of the Code.4) of the Code where f bu = β f cu and β is 0. (ii) bv d  400  4 (iii)   should not be taken as less than 0.2 Minimum shear reinforcements If v < 0.7.4 for f cu ≤ 40 and 25 .1.7.5v c < v < (v c + v r ) .1 Checking of Shear Stress and provision of shear reinforcements Checking of shear in beam is based on the averaged shear stress calculated from (Ceqn 6. 100 As should not be taken as greater than 3.87 f yv )(cos α + sin α cot β ) 6.67 for member without shear  d  reinforcements and should not be taken as less than 1 for members with links. d − d' in (Ceqn sb Maximum shear stress not to exceed v tu = 0.19) of the Code V v= bv d where v is the average shear stress.2.3 of the Code which is determined by the formula  f  3  100 As v c = 0. 3.25 . termed “design concrete shear stress” in Table 6.20 of the Code) can be used for design of bent-up bars.87 f yv 1 Code). whichever is the lesser by Cl. no shear reinforcement is required.2 of the sv 0.5. minimum shear links of Asv bv r = along the sv 0.5v c .8 f cu or 7 MPa. or alternatively Vb = Asb (0. V is the ultimate shear.79 cu     25   bv d 1 1 1  3  400  4 1     d  γ listed in Table 6. 6. d is the effective depth of the beam and bv is beam width.3. If 0. A b(v − v c ) then shear links of sv = should be provided (Table 6.87 f yv whole length of the beam be provided where v r = 0. bv should be taken as the averaged width of the beam below flange in case of flanged beam) If v is less than the values of v c .3 with the following  m  limitations : (i) γ m = 1. 7.87 f yv 0.10 – Shear enhancement near support 3. d av section under consideration. sv 0.81) = = = 1.3 Enhanced shear strength close to support At sections of a beam at distance a v ≤ 2d from a support. the shear strength 2d can be increased by a factor .  d  v= 700 × 10 3 = 2. Use Y12 – 200 c/c s. V = 700 kN. 1 1 100 Ast = 1. d = 700 − 40 − 16 = 644 mm. f cu = 35 MPa.72 MPa.87 × 460 26 . but not greater than 80. 3. 400 × 644 Asv b(v − vc ) 400(2.4( f cu / 40 ) 2/3 for f cu > 40 . bounded by the absolute maximum of av v tu = 0.81 MPa.s.91 .79  f cu  3  100 As   25   bv d  1  3  400  4 1     d  γ = 0.72 − 0.5 .0.8 – shear design without shear enhancement in concrete Section : b = 400 mm.  m   400  4 where   taken as 1. shear strength 2d enhanced to vc av Figure 3.8 f cu or 7 MPa.7.4 Worked Examples for Shears (i) Worked Example 3. bd 1  v c = 0. 5 .81) × 400 × 644 × 10 −3 = 246 kN to be taken up by bent-up bars.72 MPa.87 f yv 0.8 f cu = 0.5 × (2.8 is to be taken up by bent-up bars.20) of the Code.39 MPa av 750 700 × 10 3 = 2.s 0.72 − 1.79 cu     25   bv d 1 1  3  400  4 1     d  γ = 0. d = 700 − 40 − 16 = 644 mm.87 × 160(cos 45 + sin 45 cot 60 )× 588 27 .72 − 0. 400 × 644 Asv b(v − v c ) 400(2. Use Y12 – 150 c/c s. 100 Ast = 1 .10 – inclusion of bent-up bars If half of the shear in Worked Example 3.39) = = = 1. V = 700 kN. as a heavy point load is acting so that from the support face to the point load along the beam. bd f cu = 35 MPa. the shear is more or less constant at 700 kN.81 = 1. Vb = Asb (0.7 MPa v= 2d 2 × 644 vc = × 0.8 35 = 4. i. Section : b = 400 mm.  m  1 1  400  4 where   taken as 1.(ii) Worked Example 3.  f  3  100 As v c = 0.  d  Concrete shear strength enhanced to < 7 and 0.75 m from a support.87 f yv )(cos α + sin α cot β ) ⇒ Asb = d − d' sb 246000 × 441 = 413 mm2 0 0 0 0. sb = 441 sb = 441 d – d’= 644 – 40 – 16 = 588 β = 60o α = 45o St = 882 By (Ceqn 6. At section of 0.9 – shear design with shear enhancement in concrete.e. 0.33 .81 MPa.87 × 460 sv (iii) Worked Example 3. checking needs be carried out if the design relies entirely on the torsional resistance of a member such as 28 .3.2. 6. However.9 Design against Torsion 3.1.5(d). Cl. this spacing should not exceed d. Links or ties shall be arranged so that every corner and alternate longitudinal bar that is required to function as compression reinforcement shall be restrained by a leg Links should be adequately anchored by means of 135o or 180o hooks in accordance with Cl.1 of the Code. checking against torsion is usually not necessary.4bv sv / (0.9.2. the horizontal spacing of links should be such that no longitudinal bar should be more than 150 mm from a vertical leg. 9. The maximum spacing of links in the direction of span of the beam should be the least of (a) 0. 6. in normal slab-and-beam and framed construction. At right angle of the span. (iv) (v) (vi) 3.2 of the Code) should be observed for the placing of shear reinforcements which may directly affect design : (i) The minimum provision of shear reinforcements (links or bent up bars) in beams should be given by Asv ≥ 0. 3. 9.8 Placing of Shear reinforcements The followings (in Cl.2 and Cl. 8.5 of the Code.75d. Anchorage by means of 90o hook is not permitted (D). (ii) (iii) At least 50% of the necessary shear reinforcements be in form of links.87 f yv ) .Use 6 nos.1 By Cl.1. (b) the least lateral dimension of the beam (D).9. (c) 16 times the longitudinal bar diameter (D). of T10 at spacing of sb = 440 mm as shown. 64) of the Code C = βhmin hmax where β is to be read from Table 6. 3 3. the section should be broken up into two or more rectangles such that the hmin hmax is maximized and the total Torsional moment T be apportioned to each  hmin 3 hmax rectangle in accordance with (Ceqn 6.66) of the Code as T ×   ∑ h 3h min max   .11 – Illustration for necessity of checking against torsion 3.9. the torsional shear stress is calculated by (Ceqn 6.   2 If the total shear stress exceeds 0.Beam carrying unbalanced torsion induced by slab needs be checked Figure 3. though for small section where y1 (the larger centre-to-centre dimension of a 29 .3 Calculation of torsional shear stress Upon arrival of the torsion on the rectangular section. torsional reinforcements will be required.067 f cu (but not more than 0.8 f cu or 7MPa is not exceeded.2 Calculation of torsional rigidity of a rectangular section for analysis (in grillage system) is by (Ceqn 6. the torsional shear stress should be added to the shear stress induced by shear force to ensure that the absolute maximum v tu = 0.6MPa).16 of the Code. Furthermore.65) of the Code 2T vt = h  2 hmin  hmax − min  3   and in case of a section such as T or L sections made of rectangles.9. of stirrup within Y is y1 / sv .12 – Derivation of Design Formulae for Torsional Resistance 30 .5 Asv 0. So total moment is 0.87 f y 0.87 f y Moment provided by this stirrup in region X is 0.5 Asv 0.8 x1 y1 (0.87 f y x1 y1 identical and is the total resistance is therefore .4 Calculation of torsional reinforcements Torsional reinforcement in forms of close rectangular links and longitudinal bars are to be calculated by (Ceqn 6.67) sv 0.5 Asv 0.67) and (Ceqn 6. 3. Total nos.12.87 f y x1 .rectangular link) < 550mm.5 Asv 0.87 f y y1x1 / sv x1 45o 45o 0.67) of the Code for close rectangular links is under the assumption of a shear rupture length of stirrup width + stirrup depth x1 + y1 as shown in Figure 3. sv 0. of stirrup within X is x1 / sv . v tu will be decreased by a factor y1 / 550 .9.87 f y y1 shear rupture spiral face 0.5 Asv 0.5 Asv 0.87 f yv ) ( Asv is the area of the 2 legs of the link) As = Asv f yv ( x1 + y1 ) sv f y (Ceqn 6.68) It should be noted that there is no reduction to shear strength ( v c ) by concrete. It is shown in Figure 3.68) of the Code as Asv T = (Ceqn 6.5 Asv 0.12 that the torsional moment of resistance by the stirrups within the Regions X and Y are Asv 0.87 f y y1 . So total moment is 0. Revision of section is required if the absolute maximum is exceeded. Total nos. A spiral torsional failure face is along the heavy dotted line. The derivation of the design formula (Ceqn 6.5 Asv 0.87 f y Moment provided by this stirrup in region Y is 0.87 f y x1 y1 / sv y1 y1 Region Y x1 x1 Region X Figure 3. 10 – Design for T-beam against torsion A total torsion of T = 200 kNm on a T-section as shown with an average vertical shear stress on the web of 0.So T = Asv 0.13 – Section of a T section resisting torsion Option A is made up of two rectangles of 31 550 × 400 and one rectangle of .8 is sv 0.5 Worked Example 3.87 f y x1 y1 sv ⇒ Asv T = .9. An additional factor of 0.67). The derivation of the longitudinal bars is based on the use of same quantity of longitudinal bars as that of stirrups.87 f y x1 y1 added and the equation becomes (Ceqn 6. 3.82 N/mm2 1500 400 1000 400 T-Section 1500 1500 400 400 1000 1000 400 400 Option A Option B Figure 3. 04 = 124.067 f cu = 0.27125 and 0.8 × 308 ×1408 × 0.084 N/mm2 400   400 2 1000 −  3   As 1. From Table 6.185 × 550 × 400 3 + 0. As = Asv f yv ( x1 + y1 ) sv f y = 0.2675 So the total stiffness of Option A is 2 × 0. From Table 6.400 × 1400 .72 h  2 hmin  hmax − min  3   The total shear stress is 1. β for the 1500 × 400 rectangle and 400 × 1000 are respectively 0.396 N/mm2 32 .245 ×1000 × 400 3 = 41.2675 × 1400 × 400 3 = 36. 41.832 kNm.87 f yv ) 0.899 × 460 × (308 + 1408) = 1543 mm2 460 Use 8T16 For the 1000 × 400 rectangle T2 = 200 × v t1 = 2T2 15.832 ×10 6 = 1.72 ×10 9 mm4 As Option B has a large torsional stiffness.27125 ×1500 × 400 3 + 0.245 So the total stiffness of Option B is 0.142 N/mm2 v t1 = = 400  h   2 hmin  hmax − min  400 2 1500 −  3  3    > 0.899 sv 0. β for the 550 × 400 rectangle and 400 × 1400 are respectively 0. The torsional moment is apportioned to the two rectangles of Option B as : For the 1500 × 400 rectangle T1 = 200 × 26.396 N/mm2 So torsional shear reinforcement is required x1 = 400 − 40 × 2 − 6 × 2 = 308 .084 + 0.832 × 10 6 = = = 0.72 Torsional shear stress is 2T1 2 ×124.16 of the Code.s.88 MPa = 2 × 75.16 of the Code.185 and 0. it is adopted for design.992 × 10 9 mm4 Option B is made up of one rectangle of 1500 × 400 and one rectangle of 400 × 1000 .904 N/mm2 < v tu y1 / 550 = 2.68 = 75. y1 = 1500 − 40 × 2 − 6 × 2 = 1408 Asv T1 124.82 = 1.168 × 10 6 = 1.904 > 0.168 kNm 41.067 f cu = 0.87 × 460 Use T12 – 250 s.8 x1 y1 (0. y1 / 2 or 200 mm as per Cl.3. 6.3.3 of the Code. The value s v for the closed link should not exceed the least of x1 .7.168 ×10 6 = = = 0. 6. 9.2.8 × 308 × 908 × 0. 9. It should be noted that the torsional shear link should be closed links of shape as indicated in Figure 9.8 and Cl.2.3.8 x1 y1 (0.87 f yv ) 0.10 Placing of Torsional reinforcements The followings (in Cl. y1 = 1000 − 40 × 2 − 6 × 2 = 908 mm Asv T1 75. In accordance with Cl.87 × 460 Use T12 – 250 s.3 in Cl.14 – Arrangement of torsional reinforcements It should be borne in mind that these torsional reinforcements are in addition to others required for flexure and shear etc.s.3 of the Code) should be observed for the placing of shear reinforcements which may directly affect design : (i) (ii) (iii) The shear link should form the closed shape as in Figure 9.84 mm sv 0.3 of the Code. 9. provision of the longitudinal torsion reinforcement should comply the followings : 33 . Cl. 3.So torsional shear reinforcement is required x1 = 400 − 40 × 2 − 6 × 2 = 308 mm. 6.3 of the Code.7 of the Code.2.84 × 460 × (308 + 908) = 1021 mm2. Use 6T16 460 1500 400 1000 400 Figure 3. As = Asv f yv ( x1 + y1 ) sv f y = 0. The longitudinal bars should extend a distance at least equal to the largest dimension of the section beyond where it theoretically cases to be required. 34 . Additional longitudinal bars required at the level of the tension or compression reinforcements may be provided by using larger bars than those required for bending alone.(a) (b) (c) (d) The bars distributed should be evenly round the inside perimeter of the links. Clear distance of the bars not exceed 300 mm. 6. 4. Cl. it should be acceptable to set the twisting stiffness to zero which will end up with pure bending in the X and Y directions as the slab. Nevertheless. “flat slab”. the bending stiffness in the X and Y directions can be easily found by summing the total bending stiffness of the composite ribs and flange structure per unit width. 4.2 Two way slab is a rectangular one supported on four sides with length to breadth ration smaller than 2.1 One way slab is defined by the Code as one subjected predominantly to u.l. either (i) it possesses two free and parallel edge.2 Analysis of Slabs without the use of computer method 4. “ribbed slab”. with its ribs running in the X and Y directions are clearly predominantly strong in bending in the X and Y directions. as we usually design for a slab based on the bending moment at the central part of the slab which is the maximum and assign reinforcements to the whole slab without curtailment. As we aim at simple analysis for the slab. 4. The Code allows idealization of the ribbed slab or waffled slab as a single slab without treatment as discretized ribs and flanges in analysis. If the stiffness of the ribbed or waffled slab is required for input. either continuous or single span. The twisting stiffness is difficult to assess.1 Types of Slabs Slabs can be classified as “one way slab”.1.d. we tend to treat it as a single element without the necessity to consider the many loading cases for continuous spans.1.4 Ribbed or Waffled Slab is a slab with topping or flange supported by closely spaced ribs. 4.4.1. strictly speaking.2(c) of the Code allows the design against moment 35 . it is only the central part of (ii) be treated as one way slabs. So.1 One way slab can be analyzed as if it is a beam.3. or (ii) it is the central part of a rectangular slab supported on four edges with a ratio of the longer to shorter span greater than 2.1. “two way slab”.1.2. 4.0 Slabs 4. treating the whole rectangular slab as two ways bending in design is usually adequate. However.3 Flat slab is a slab supported on columns without beam. and (iii) the characteristic imposed load does not exceed 5 kN/m2 excluding partitions. Bending moment coefficients for calculation of bending moments are presented in Table 6. A complete set of results including bending moments. The design against flexure is most commonly done by the Wood Armer Equations which calculate design moments in two directions (conveniently in two perpendicular directions) which are adequate to cater for the complete set of bending and twisting 36 .3 Analysis of Slabs with the use of the computer method Analysis of slabs with the use of the computer method is mainly by the finite element method in which the slab is idealized as an assembly of discrete “plate bending elements” jointed at nodes.2 Two way rectangular slab is usually carried out by treating it as if it is a single slab in the absence of computer method. the bending moment and shear force coefficients for the one way slab can also be used as a simplified approach.and shear arising from the single-load case of maximum design load on all spans provided that (i) the area of each bay (defined in Figure 6.3 Flat slabs.10. Nevertheless. if of regular arrangement.25. The support stiffnesses by the supporting walls and columns are derived as similar to that for beams as “sub-frames”. can be analyzed as frames in the transverse and longitudinal directions by such methods as moment distribution method as if they are separate frames. (ii) the ratio of the characteristic imposed load to characteristic dead load not to exceed 1. twisting moment. 4.5 of the Code) not to exceed 30 m2.2. shear force per unit width (known as “stress” in finite element terminology) can be obtained after analysis for design purpose.2.2.27). 4. 4. Slabs and Diaphragms based on the Elastic Theory”. simplified formulae for the bending coefficients in case of rectangular simply supported two way slab are available in the Code (Ceqn 6. Analyzed moments and shears should be apportioned to the “column strip” and “Middle strip” as per Table 6. In addition. 4.6 of the Code for different support restraint conditions.4 More bending moment and shear force coefficients of rectangular slabs with various different support and loading conditions can be found from other published handbooks.26 and 6. the most famous one being “Tables for the Analysis of Plates. Extracts of commonly encountered slabs are enclosed in Appendix E. moments. A discussion of the plate bending theory and the design approach by the Wood Armer Equations is enclosed in Appendix D.1 – Modeling of an irregular floor slab as 2-D mathematical model. subsequent analytical results of bending moments and twisting moment. together with the “stress approach” for checking and designing against shear in the same appendix. design of reinforcements by the Wood Armer Equations.1. An example of the mathematical modeling of a floor slab by the software SAFE and results of subsequent analysis is illustrated in Figure 4. The design based on node forces should be avoided due to its inadequacy to cater for twisting effects which will result in under-design. Figure 4. 37 . (b) max. spacing ≤ 3. However. “effective span depth ratio”. Nevertheless. Nevertheless.4 Detailing for Solid Slabs Generally considerations in relation to determination of “effective span”. Maximum reinforcements spacing : (a) In general areas without concentrated loads : the principal reinforcement. spacing ≤ 3h ≤ 400 mm.The finite element mesh of the mathematical model is often very fine. Reinforcements at end supports 38 (c) (iv) (v) . So it is a practice of “lumping” the design reinforcements of a number of nodes over certain widths and evenly distributing the total reinforcements over the widths.13% for f y = 460 MPa.3%. 4.3 for design of beam are applicable to design of slab. the detailing considerations for slabs are listed as follows (Re 9. (b) In areas with concentrated loads or areas of maximum moment: the principal reinforcement. spacing ≤ h ≤ 250 mm (grade 250 steel). spacing ≤ h ≤ 200 mm (grade 460 steel). spacing ≤ 2h ≤ 250 mm. “reduced design moment to support”. or the percentage of required tension reinforcement is less than 0.1. Requirements pertaining to curtailment and anchoring of tension reinforcements should be similar to that of beams. Distribution bars in one way slab : 20% of the main reinforcements. and for the secondary reinforcement. as is done by the popular software “SAFE”. “concrete covers” as discussed in 3.3.24% for f y = 250 MPa and 0. max. crack control can be carried out by crack widths calculations. and the secondary reinforcement. no further check on crack control is required on bar spacing if either : (a) max. max. “maximum and minimum steel percentages”. “moment redistribution”. (ii) (iii) In addition to (ii).5h ≤ 450 mm. max. spacing ≤ 3h ≤ 400 mm. care must be taken of not taking widths too wide for “lumping” as local effects may not be well captured.1 of the Code) : (i) Minimum steel percentage : Main Reinforcing bars: 0. max. 5v c (vi) (vii) Minimum bottom reinforcements at internal supports : 40% of the calculated mid-span bottom reinforcements. The determination of reinforcements should be in accordance with Section 3.2 – Anchorage of end span bar for v < 0. If support shear v < 0.(a) (b) Half of the span reinforcements should be provided and well anchored on end supports in simply supported slabs and continuous slabs. the following arrangement can be considered as effective anchorage.5 Structural Design of Slabs The structural of slab against flexure is similar to that of beam. > 1/3b and 30 mm if v < 0.5vc . 4. Reinforcements at free edge should be as shown : h ≥2h Figure 4.3 – Free edge reinforcements for Slabs (viii) Shear reinforcements not to be used in slabs < 200 mm.5vc b Figure 4.4 listing the options of either following the rigorous or simplified “stress strain” 39 . Fin.) Loading : D. assuming modification by tensile reinforcement to be 1.6 + 1. (i) Fire rating : 1 hour. Design against shear for slabs under line supports (e.4 × 4. (ii) Span : 4 m 4m 4m 4m 4m Figure 4.0 kN/m2.3 and Table 7.56 = 40. (iii) Concrete grade 35.56 as the slab should be lightly reinforced).0 )× 4 = 51.W. However for a flat slab.relationship of concrete.6 (by Table 7. one-way or two-way) is also similar to that of beam.6 : 4.36 kN/m Based on coefficients of shear and bending in accordance with Table 6.g. So effective depth taken as d = 150 − 25 − 5 = 120 as 4000/120 = 33.L.15 × 24 = 3.0 kN/m2 The factored load on a span is F = (1. (ii) Finishes Load = 1 kN/m2. bars under 25 mm concrete cover. the checking should be based on punching shear in accordance with empirical method of the Code or by based on shear stresses revealed by the finite element method.0 kN/m2 4.1 – One Way Slab A one-way continuous slab with the following design data : (i) Live Load = 4.4 of the Code. They are demonstrated in the Worked Examples in the following sub-Section 4. mild exposure.L.6 × 4.4 – Slab in Example 1 Sizing : Limiting Span depth ratio = 26 × 1. 4. O.4 of the 40 .6 kN/m2 L.6 Worked Examples Worked Example 4. Total 0.6 (assuming 10mm dia.3 < 40.6kN/m2 1. 15% = 2 = bd bd × 0.87 f y z / d 1000 × 120 2 × 0.6F 0.6F 0.Code listed as follows : B. -0.87 × 460 × 0. End span span moment = 0.36 × 4 = 17.0163 K= f cu bd 2 35 ×1000 ×120 2 z K = 0.95 0.F.063Fl -0.04Fl 0.5 – Bending Moment and Shear Force coefficients for Continuous Slab (a) End span support moment = 0.5 + 0.063Fl -0.95 0 .25 − = 0.95 0 .95 Ast = 0.96 > 0.95 Ast = 0.25 − = 0.9 d 41 .0306 f cu bd 2 35 × 1000 ×120 2 z K = 0.5 + 0.075 Fl -0.41×10 6 M = = 0.36 = 23.25 − = 0.04 × 51.15 ÷ 100 × 1000 × 120 = 180 mm2 Use T10 – 300 ( Ast provided = 262mm2) End span shear = 0.67 ×10 6 M = = 0.28% = = bd bd 2 × 0.41 kNm/m K= 15.63 kN/m.46F -0.87 × 460 × 0.22 × 10 6 M = 0.5F 0.075 Fl S.41×10 6 M = 0. v= 23630 1/ 3 = 0.5 + 0.22 × 10 6 M = = 0.6F 0.2 N/mm2 < 0.5F 0.075 × 51.086Fl 0. 1000 ×120 No shear reinforcement required.36 × 4 = 15.36 × 4 = 8.6F 0.982 > 0.9 d Ast 15.4F Figure 4.22 kNm/m 8.M.67 kNm/m K= 17.5 N/mm2.04Fl 0.0351 f cu bd 2 35 ×1000 ×120 2 z K = 0.46 × 51.086Fl 0.45 × (35 / 25) = 0.28 ÷ 100 × 1000 ×120 = 338 mm2 Use T10 – 225 ( Ast provided = 349mm2) (b) Interior span support moment = 0.086Fl 0.086 × 51.87 f y z / d 1000 × 120 2 × 0. 0.965 > 0.9 d Ast 8. 27) of the Code are used to calculate the bending moment coefficients along the short and long spans : α sx = 8 1 + (l y / l x ) [ (l y /lx) 4 4 ] = 0.36 × 4 = 12. 1000 ×120 No shear reinforcement required.6kN/m2 1. Short way.2.94 kNm/m Use T10 – 225 ( Ast provided = 349mm2) Worked Example 4. 42 α sy = 8 1 + (l y / l x ) [ (l y /lx) 2 4 ] = 0.2 – Two Ways Slab (4 sides simply supported) A two-way continuous slab with the following design data : (i) Live Load = 4. 4.095 .6 kN/m2 L.56 = 31.5 N/mm2.Ast 17.32% > 0.32 ÷ 100 × 1000 ×120 = 387 mm2 Use T10 – 200 ( Ast provided = 393mm2) Maximum shear = 0.0 kN/m2 The factored load is F = (1.L. Sizing : Limiting Span depth ratio = 20 × 1.W.87 f y z / d 1000 × 120 2 × 0.0 )× = 12.87 × 460 × 0.L.6 × 4.6 × 51. (iii) Span : Long way : 4 m.4 of the Code.36 = 30. 3 m.67 × 10 6 M = = = 0. (iii) Concrete grade 35. Interior span moment = 0. Total 0.3 and Table 7. mild exposure.0 kN/m2 4.25 N/mm2 < 0.6 + 1. So effective depth taken as d = 150 − 25 − 5 = 120 as 3000/120 = 25 < 31. O.82 kN/m.13% bd bd 2 × 0.84 kN/m (Ceqn 6.95 Ast = 0. (ii) Fire rating : 1 hour.0 kN/m2.15 × 24 = 3.56 as the slab should be lightly reinforced).26) and (Ceqn 6.45 × (35 / 25) = 0.4 × 4.063 × 51. assuming modification by tensile reinforcement to be 1. bars under 25 mm concrete cover. So the structural depth is 150 mm (assuming 10mm dia. (ii) Finishes Load = 1 kN/m2. Fin. v= 30820 1/ 3 = 0.) Loading : D.2 (by Table 7.053 . 84 × 3 2 = 6. (i) Concrete grade 35.13 × 10 6 = = = 0.13% bd bd 2 × 0.84 × 3 2 = 10.11% < 0.9 d Ast 10.6 – Plan of 3-sides supported slab for Example 4.98 × 10 6 M = 2 = = 0.13% bd bd × 0. Short way.2% > 0.0218 f cu bd 2 35 ×1000 ×120 2 z K = 0.95 0.13 kNm/m Use T10 – 300 ( Ast provided = 262mm2) 6.3 43 .87 f y z / d 1000 ×120 2 × 0.975 > 0.0 kN/m2.98 ×10 6 M = = 0. (ii) Finishes Load = 1 kN/m2.2 ÷ 100 × 1000 ×120 = 240 mm2 M y = 0.0145 K= f cu bd 2 35 × 1000 × 110 2 Ast M 6.87 f y z / d 1000 ×110 2 × 0. mild exposure. 4 m (iii) Fire rating : 1 hour.So the bending moment along the short span is M x = 0.98 kNm/m K= 10. 5m free edge 4m simply supported edge continuous edge Figure 4.95 Ast = 0.3 – Two Ways Slab (3 sides supported) A two-way continuous slab with the following design data : (i) Live Load = 4.87 × 460 × 0.095 × 12.13 ÷ 100 × 1000 ×110 = 143 mm2 Use T10 – 300 ( Ast provided = 262mm2) Worked Example 4. (ii) Span : Long way : 5 m.13 × 10 6 M = = 0.5 + 0.25 − = 0.053 × 12.95 Ast = 0.87 × 460 × 0. 798 kNm/m and 0.28 kNm/m K= 23.8 .87 f y z / d 1000 ×130 2 × 0.87 × 460 × 0.75 and γ = 1.18 × 5 2 = 24.13% bd bd × 0. The maximum hogging moment (bending along long-way of the slab) is at mid-way along the supported edge of the short-way span M y = 0.0 kN/m2 The factored load is F = (1.4 of the Code.362 ÷ 100 × 1000 × 130 = 471 mm2 Use T10 – 150 ( Ast provided = 523mm2) At 2 m and 4 m from the free edge.28 × 10 6 M = = 0.362% > 0.0844 ×13.18 × 4 2 = 23. Thus the structural depth is 160 mm (assuming 10mm dia.01 kNm/m 44 . the sagging moment reduces to 0.84 kN/m2 L.1104 ×13.84 + 1.) Loading : D.0 kN/m2 4.W. 4.8. assuming modification by tensile reinforcement to be 1.56 as the slab should be lightly reinforced).95 Ast = 0.4 × 4. the sagging bending moment coefficient for short way span is maximum along the free edge which 0.954 > 0.5 + 0.18 kN/m From Table 1.16 × 24 = 3.18 × 4 2 = 8.95 d 0 .Sizing : Based on the same limiting Span depth ratio for one way and two way slab which is 20 × 1.18 × 4 2 = 17.0415 ×13. So effective depth taken as d = 160 − 25 − 5 = 130 as 4000/130 = 25 < 30.28 × 10 6 = 2 = = 0.38 of “Tables for the Analysis of Plates.6 × 4.75 kNm/m and Ast required are reduces to 360 mm2 and 177 mm2. Use T10 – 200 and T10 – 300 respectively. O.0 M x = 0.0729 × 13.0 ) = 13.L.0394 f cu bd 2 35 × 1000 × 130 2 z K = 0.56 = 31. Slabs and Diaphragms based on Elastic Theory” extracted in Appendix E where γ = 4 / 5 = 0. Fin.3 and Table 7.L.9 Ast M 23. Total 0.2 (by Table 7.25 − = 0. bars under 25 mm concrete cover.84kN/m2 1.1104 (linear interpolation between γ = 0. 87 f y z / d 1000 ×120 2 × 0.944 Ast = 0.01876 × 13.2(c)) Flat slab arrangement on rectangular column grid of 7.01×10 6 = 2 = = 0.01× 10 6 M = = 0.1.7 – Reinforcement Details for Example 4.3 Worked Example 4.4 – Flat Slab by Simplified Method (Cl.9 Ast M 24. Use T10 – 300 Finally the reinforcement arrangement on the slab is (Detailed curtailment omitted for clarity) T10 – 300 T1 1200 T10 – 300 B1 T10 – 125T1 800 T10 – 300B2 800 T10 – 200 B1 T10 – 150B1 1200 1000 2000 2000 Figure 4.18 × 5 2 = 6.87 × 460 × 0.0476 K= f cu bd 2 35 × 1000 × 120 2 z K = 0. The moment is small.13% bd bd × 0.5 m and 6 m with the following design data : 45 .24.5.44% > 0.944 d 0 .5 + 0.18 kNm/m.25 − = 0.44 ÷ 100 × 1000 ×120 = 528 mm2 Use T10 – 125 ( Ast provided = 628mm2) The maximum sagging moment along the long-way direction is at 2 m from the free edge which is M y = 0. 6. 8 – Flat Slab Plan Layout for Example 4 Conditions as stipulated in Cl.5. the flat slab is effectively divided into “column strips” containing the columns and the strips of the slabs linking and the columns and the “middle strips” between the “column strips” which are designed as beams in flexural design with assumed apportionment of moments among the strips. 6.1(c)) hc = 870 2 × 4 π = 982 mm 2hc 2 × 982 = 7500 − = 6846 mm 3 3 Effective span in the long side is l − 46 . Column size = 550 × 550 Column Drop size = 3000 × 3000 with dh = 200 mm Fire rating : 1 hour. punching shear along successive “critical” perimeters of column are carried out instead. mild exposure.2(c) are satisfied as per Cl. 6.37) Effective diameter of column head (Cl.1. However. Effective dimension of column head l h max = l c + 2(d h − 40 ) = 550 + 2(200 − 40 ) = 870 mm (Ceqn 6. In the simplified method.1.2(g). 6. Hence the simplified method is applicable.0 kPa.3.5 kPa Live Load = 5. for shear checking.(i) (ii) (iii) (iv) (v) Finish Load = 1. 7500 7500 1500 1500 1500 1500 6000 6000 6000 Figure 4.1.5. 5 × 6 = 870. Total 0.6 × 5.1 + 1.46 kNm Total moment at first interior support is 0.2(g).69 The reinforcements – top steel are worked out as follows.17 93. 5.5.086 × 870.9 178.85 = 238. O. ( d = 475 − 25 − 6 = 444 over column support and d = 275 − 25 − 6 = 244 in other locations) Column Strip Area (mm2)/m Outer Support 1st interior support Middle interior support 640 760 577 47 steel T10 – 100 T12 – 125 T12 – 150 Mid Strip Area (mm2)/m 319 459 316 steel T10 – 225 T10 – 150 T10 – 225 . assuming modification by tensile reinforcement to be 1.1.34 × 7. 6.9 Mt/width 19. d = As cover = 25 mm.3 × 6.62 128.3 × 6.38) Sizing : Based on the same limiting Span depth ratio for one way and two way slab which is 26 × 1.W.4 of the Code.58 kNm These moments are to be apportioned in the column and mid strips in accordance to the percentages of 75% and 25% as per Table 6. structural 6. Total load on one panel is Fl = 19.30 Mt/width 59.(Ceqn 6.6kN/m2 1. assuming T12 228 + 25 + 12 ÷ 2 = 259 mm.L. Fin.62 128.85 384.17 93.3 × 6. 30 depth is bars.85 = 512. Loading : D.69 kNm Total moment at middle interior support is 0.4 × 8.4 will be used as per Cl.34 kN/m Design against Flexure (Long Way) The bending moment and shear force coefficients in Table 6.10.0 ) = 19.87 42.72 31.e.0 kN/m2 The factored load is F = (1. say 275 mm. Column Strip Total Mt Outer Support 1 interior support Middle interior support st Mid Strip Total Mt 59.275 × 24 = 6846 = 228 .L.3 kN.3 and Table 7.15 = 30 (by Table 7.52 281. i.04 × 870.85 = 375.063 × 870. Hogging Moment : Total moment at outer support is 0.15).1 kN/m2 L.5 kN/m2 8. The coefficient of 0. 1.6(b) in the absence of calculation.3 × 6.6 × 2 × 1.01 kN. 1.6 is the shear force coefficient for first interior support from Table 6.92 206. Column Strip Total Mt Middle of end span Middle of interior span 245.12 kNm Total moment near middle of interior span 0. 2 is to account for shear from both sides of the slab.10.3 × 6.666 .7 applied to M t is per note 2 .01 kN after deduction of the loads within the critical perimeter which is 48   1.5 × 0.8 kN > 1201. i.07 56.3 × (0. So side length of the perimeter is (550 + 666 × 2 ) = 1882 Length of perimeter is 4 × 1882 = 7528 Shear force to be checked can be the maximum shear 1201.444 = 0.e.5d from column face.97 68.5 × 384. Alternatively.5M t which is Veff = Vt 1 +  Vt xsp  of Cl.57 Mt/width 81.3 × 1 +   870.1.063 × 870.1.      = 1301.52 × 0.85 = 447.34 The reinforcements – bottom steel are worked out as follows : Column Strip Area (mm2)/m Middle of end span Middle of interior span 910 764 steel T12 – 100 T10 – 100 Mid Strip Area (mm2)/m 744 625 steel T10 – 100 T10 – 125 Design against Shear The maximum shear is at the 1st interior support which is 0. i.3 = 1201.58 kNm These moments are to be apportioned in the column and mid strips in accordance to the percentages of 55% and 45% as per Table 6.Sagging Moment : Total moment near middle of end span is 0.15 is the coefficient to account for moment transfer in Cl.20 169.6(b)) Check on 1st critical perimeter – 1.55 + 0.40) can be used to arrive at a more accurate coefficient  1. (Ceqn 6.666 × 2)  .4. 6.85 = 375.01 kN (Note : 0. 6.6 × 2 × 870.5.15 × 870.e.075 × 870.5.7  = 0.86 Mid Strip Total Mt 201.01 Mt/width 67. 5 N/mm2 in the first exterior column head which is < 1.0vc .87 × 460 y is chose as shear reinforcement. (v − v )ud (0.01 − 19.3 kN.7v − vc )ud 0.8 kN is adopted which becomes 1301.4.34 × (35 / 25) 1/ 3 = 0. α = 90 0 ⇒ sin α = 1 . there are recommendations in Cl. the shear stress will be 1233300 = 0.339 N/mm2.37 N/mm2.7(f) of the Code that (i) at least two rows of links should be used.6vc < v ≤ 2. In distributing the shear links within the critical perimeter. The second row be at further 300 mm away where row length is 49 . 6.5 – Design for shear reinforcement (Ceqn 6.61 N/mm2 but > vc = 0. the column in this case).6vc = 0. No shear reinforcement is required.3 ( 0.38 N/mm2. if the higher shear of Veff = 1301.8 − 19.87 f y As a demonstration. Using T10 – 500 spacing along the row.44) and (Ceqn 6.34 × 1.38 ). ∑ Asv sin α > 0. So the 1002 mm2 should be distributed within the critical perimeter as shown in Figure 4.87c f = 0. ∑ Asv sin α > 0. Nevertheless. (ii) the first perimeter should be located at approximately 0. the total steel area will be (10 2 / 4 )π × 3800 / 500 = 597 mm2 > 40% of 1002 mm2.e.882 2 = 1132. < 0.6vc .38 N/mm2 in the Example 4.882 2 = 1233.45) of the Code gives formulae for reinforcement design for different ranges of values of v . By (Ceqn 6.38 N/mm2 in accordance 7528 × 444 with Table 6.1201.38)× 7528 × 444 = 1002 mm2 If vertical links . (v − v )ud For v ≤ 1.34 ×1.1. ∑A sv sin α > 5(0. No shear reinforcement is required.6. No checking on further perimeter is required. 7528 × 444 Example 4.5d from the face of the loaded area (i.51 kN 1132510 Shear stress = = 0.5. So the first row be determined at 200 mm from the column face with total row length 950 × 4 = 3800 .64) of the Code.87c f y For 1.5 − 0. if v = 0. < vc = 0. 1550 × 4 = 6200 . Again using T10 – 500 spacing along the row, the total steel area will be 10 2 / 4 π × 6200 / 500 = 974 mm2. ( ) Total steel area is 597 + 974 = 1570 mm2 666 1st row , using T10 – 500 (total area = 597mm2) >40% of 1002 mm2 550 2nd row , using T10 – 250 (total area = 973mm2) 666 275 200 ≈ 0.5d 300 ≈ 0.75d 200 300 1st critical perimeter T10 – 300 link Figure 4.9 – Shear links arrangement in Flat Slab for Example 4.5 1.5d = 666 Design for Shear when ultimate shear stress exceeds 1.6vc It is stated in (Ceqn 6.43) in Cl. 6.1.5.7(e) that if 1.6 c < v < 2.0vc , 5(0.7v − v )ud which effectively reduces the full inclusion of ∑ Asv sin α > 0.87 f c y vc for reduction to find the “residual shear to be taken up by steel” at v = 1.6vc to zero inclusion at v = 2.0vc . Example 4.6 – when 1.6vc < v < 2.0vc In the previous Example 4.5, if the shear stress v = 0.68 N/mm2 which lies between 1.6vc = 1.6 × 0.38 = 0.61 N/mm2 and 2.0vc = 2 × 0.38 = 0.76 N/mm2 50 ∑A sv sin α > 5(0.7v − vc )ud 5(0.7 × 0.68 − 0.38)× 7528 × 444 = = 4009 mm2. If 0.87 f y 0.87 × 460 arranged in two rows as in Figure 4.6, use T12 – 250 for both rows : the inner row gives (12 2 / 4)π × 3800 / 250 = 1719 mm2 > 40% of 4009 mm2; the outer row gives (12 2 / 4)π × 6200 / 250 = 2802 mm2 . The total area is 1719 + 2802 = 4521mm2 > 4009 mm2. Cl. 6.1.5.7(e) of the Code says, “When v > 2vc and a reinforcing system is provided to increase the shear resistance, justification should be provided to demonstrate the validity of design.” If no sound justification, the structural sizes need be revised. 51 5.0 Columns 5.1 Slenderness of Columns Columns are classified as short and slender columns in accordance with their “slenderness”. Short columns are those with ratios l ex / h and l ey / b < 15 (braced) and 10 (unbraced) in accordance with Cl. 6.2.1.1(b) where l ex and l ey are the “effective lengths” of the column about the major and minor axes, b and h are the width and depth of the column. As defined in Cl. 6.2.1.1, a column may be considered braced in a given plane if lateral stability to the structure as a whole is provided by walls or bracing or buttressing designed to resist all lateral forces in that plane. It would otherwise be considered as unbraced. The effective length is given by (Ceqn 6.46) of the Code as l e = β ⋅ l 0 where l 0 is the clear height of the column between restraints and the value β is given by Tables 6.11 and 6.12 of the Code which measures the restraints against rotation and lateral movements at the ends of the column. Generally slenderness limits for column : l 0 / b < 60 as per Cl. 6.2.1.1(f). In addition, for cantilever column l 0 = 100b 2 ≤ 60b . h Worked Example 5.1 : a braced column of clear height l 0 = 8 m and sectional dimensions b = 400 mm, h = 550 mm with its lower end connected monolithically to a thick cap and the upper end connected monolithically to deep transfer beams in the plane perpendicular to the major direction but beam of size 300(W) by 400(D) in the other direction. By Tables 6.11 and 6.12 of the Code Lower end condition in both directions : 1 Upper end condition about the major axis : 1 Upper end condition about the minor axis : 2 For bending about the major axis : (1 – 1) β x = 0.75 , l ex = 0.75 × 8 = 6 l ex / 550 = 10.91 < 15 . ∴ a short column. For bending about the minor axis : (1 – 2) β y = 0.8 , 52 l ey = 0.8 × 8 = 6.4 l ey / 400 = 16 > 15 ∴ a slender column. l ey / 400 = 16 < 60 , O.K. For a slender column, an additional “deflection induced moment” M add will be required to be incorporated in design, as in addition to the working moment. 5.2 Design Moments and Axial Loads on Columns 5.2.1 Determination of Design moments and Axial Loads by sub-frame Analysis Generally design moments, axial loads and shear forces on columns are that obtained from structural analysis. In the absence of rigorous analysis, (i) design axial load may be obtained by the simple tributary area method with beams considered to be simply supported on the column; and (ii) moment may be obtained by simplified sub-frame analysis as follows : Ku Ku 1.4Gk+1.6Q Kb 1.0Gk K b2 1.4Gk+1.6Q K b1 KL Mu = M e Ku K L + K u + 0.5K b MeKL ML = K L + K u + 0.5 K b KL Mu = ML = M es K u K L + K u + 0.5 K b1 + 0.5 K b 2 M es K L K L + K u + 0.5 K b1 + 0.5 K b 2 Symbols: M e : Beam Fixed End Moment. M es : Total out of balance Beam Fixed End Moment. K u : Upper Column Stiffness K L : Upper Column Stiffness K b1 : Beam 1 Stiffness K b 2 : Beam 2 Stiffness M u : Upper Column Design Moment M L : Upper Column Design Moment Figure 5.1 – Diagrammatic illustration of determination of column design moments by Simplified Sub-frame Analysis Worked Example 5.2 (Re Column C1 in Plan shown in Figure 5.2) Design Data : Slab thickness : 150 mm Finish Load : 1.5 kN/m2 Live Load : 5 kN/m2 Upper Column height : 3 m 53 Beam size : 550(D) × 400(W) Lower Column Height : 4 m Column size : 400(W) × 600(L) Column Load from floors above D.L. 443 kN L.L. 129 kN 5m B3 B4 C1 B1 4m 3m B2 3m Figure 5.2 – Plan for illustration for determination of design axial load and moment on column by the Simplified Sub-frame Method Design for Column C1 beneath the floor Check for slenderness : End conditions of the column about the major axis are 1 at the lower end and 2 at the upper end whilst the end conditions about the minor axis are 1 at the lower end and 2 at the upper end. The clear height between restraints of the lower level column is 4000 − 550 = 3450 . So the effective heights of the column about the major and minor axes are 0.75 × 3.45 = 2.59 m and 0.8 × 3.45 = 2.76 m. So the slender ratios about the major and minor axes are 2590 2760 = 4.3 < 15 and = 6.9 < 15 . Thus the 600 400 column is not slender is both directions. Loads : Slab: D.L. O.W. Fin. L.L. 0.15 × 24 = 3.6 kN/m2 1.5 kN/m2 5.1 kN/m2 5.0 kN/m2 Beam B1 D.L. O.W. 0.4 × 0.55 × 24 × 4 = 21.12 kN End shear of B1 on C1 is D.L. 21.12 ÷ 2 = 10.56 kN 54 O.55 − 0. 17.50 kN 0.69 + 1.0 kN/m D. Slab 0.15) × 24 = 3.55 × 24 × 4 = 21.50 × 5 ÷ 2 = 43.L.6 kNm 12 Factored fixed end moment bending about the minor axis by Beam B2: 55 .85 kN 79.5 kN/m D.56 + 39. B4 End shear of B2 on C1 .0 × 3 = 15. Slab End shear of B3 on C1 21.0 × 3.55 kN Factored fixed end moment bending about the major axis (by Beam B3 alone): Me = 1 × (1.00 kN 568.5 kN Total L. O.00 kN 191.L.69 kN/m 5.L. O. Slab L.L.75 = 62. 0.4 × 0.W. 19.68 kN 0 + 18.56 kN Floor above Sum 343.L.55 × 24 × 6 = 31.W.75 kN Total D.W.4 × 21. 47.84 kN/m 5.4 × 468.75 kN 0.53 kN 37.55 − 0. 79.L. 21.L. B4 L.1 × 3 = 15.50 kN So the factored axial load on the lower column is 1.85 kN/m L.L.69 × 5 ÷ 2 = 54.6 × 17.4 × 0.3 kN/m Beam B4 D.L.23 kN L.77 + 54.14 kN/m 5.5 = 17.85 kN L.68 kN Beam B2 D.4 × (0. 37.4 × (0.14 × 5 ÷ 2 = 47.53 ÷ 2 = 39. on C1 B1 + B2 +B3 Floor above Sum 129.68 + 1.L.15) × 24 = 3.1 × 3.23 = 104.L.00 × 5 ÷ 2 = 37.5) × 5 2 = 121.Beam B3 D. on C1 O.5 = 17.12 kN B1 + B2 +B3 10.75 + 43.6 × 191. 15.84 kN/m 5.L.77 kN L.5 ÷ 2 = 18.5 = 962.50 kN D. Slab End shear of B4 on B2 19.L.W.L. 4 × 0. B2 and B3 are respectively 4 E × 0.2 × 0.5 × 0.85  × 6 + 1.0 ×  × 21.0072 = = 0.6 × 0.24 − 7.03 kNm Distributed moment on the lower column about the minor axis is 4 E × 0.6 × 0.5(K b1 + K b 2 ) M ey K Ly = 88.003697 E .5 K b 3 0. 4 4 E × 0.12  × 4 = 7.004437 E 5 Distributed moment on the lower column about the major axis is M ex K Lx 121.1 1  1  M eb 2 = 1.0096 E Lu 3 K Lx = 4 EI cx 4 E × 0.004437 E = 46.0032 = = 0.0072 E M ux = = K ux + K Lx + 0.6 ×  × 37.0096 E + 0.0032 E + 0.2 kNm The moment of inertia of the column section about the major and minor axes 0.4 3 4 are I cx = = 0.68 + × 47.55 3 = 0.0072 E LL 4 The stiffnesses of the upper and lower columns about the minor axis are : 4 EI cy 4 E × 0.05546 = 0. B2 and B3 are 0.4 ×  × 31.003697 )E 56 . and 6 M uy = K uy + K Ly + 0.0072 K ux = = = 0.0032 E 0.004267 E + 0.004267 E K uy = 3 Lu 4 EI cy 4 E × 0.0072 m .005546 + 0.05546 = 0.5 × (0.6 3 0.04 kNm  12  So the unbalanced fixed moment bending about the minor axis is 95.0032 m4 12 12 The stiffnesses of the upper and lower columns about the major axis are : 4 EI cx 4 E × 0.04 = 88.005546 E .05546 = 0.5  × 6 = 95.005546 m4 12 The stiffness of the beams B1. I cy = = 0.0072 E + 0.0032 E K Ly = 4 LL The moment of inertia of the beams B1.24 kNm 8  12  8  Factored fixed end moment bending about the minor axis by Beam B1: 1  M eb 2 = 1.0032 = = 0.4 × 0. = 23.35 kNm So the lower column should be checked for the factored axial load of 962.55kN, factored moment of 46.03 kNm about the major axis and factored moment of 23.35 kNm about the minor axis. 5.2.2 Minimum Eccentricity A column section should be designed for the minimum eccentricity equal to the lesser of 20 mm and 0.05 times the overall dimension of the column in the plane of bending under consideration. Taking the example in 5.2.1, the minimum eccentricity about the major axis is 30 mm as 0.05 × 600 = 30 > 20 mm and that of the minor axis is 0.05 × 400 = 20 mm. So the minimum eccentric moment to be designed for about the major and minor axes are 962.55 × 0.03 = 28.88 kNm and 962.55 × 0.02 = 19.25 kNm. As they are both less than the design moment of 46.03 kNm and 23.35 kNm, they can be ignored. 5.2.3 Check for Slenderness In addition to the factored load and moment obtained in 5.2.1, it is required by Cl. 6.2.1.2(b) to design for an additional moment M add if the column is found to be slender by Cl. 6.2.1.1. The arrival of M add is an eccentric moment created by the working axial load N multiplied by a pre-determined lateral deflection au in the column as indicated by the following equations of the Code. M add = Nau au = β a Kh 1  le  βa =   2000  b  N uz − N K= ≤ 1 (conservatively taken as 1) N uz − N bal N uz = 0.45 f cu Anc + 0.87 f y Asc 2 (Ceqn 6.52) (Ceqn 6.48) (Ceqn 6.51) (Ceqn 6.49) (Ceqn 6.50) N bal = 0.25 f cu bd Final design moment M t will be the greatest of (1) M 2 , the greater initial end moment due to design ultimate load; 57 (2) (3) M i + M add where M i = 0.4M 1 + 0.6M 2 ≥ 0.4M 2 (with M2 as positive and M 1 negative.) M 1 + M add / 2 in which M 1 is the smaller initial end moment due to design ultimate load. N × emin (discussed in 5.2.2) (4) where the relationship between M 1 , M 2 , M add and the arrival of the critical combination of design moments due to M add are illustrated in Figure 6.16 of the Code. In addition to the above, the followings should be noted in the application of M t as the enveloping moment of the 4 cases described in the previous paragraph : (i) In case of biaxial bending (moment significant in two directions), M t should be applied separately for the major and minor directions with b replacing h for bending in the minor direction; In case of uniaxial bending about the major axis where l e / h ≤ 20 and h < 3b , M t should be applied only in the major axis; In case of uniaxial bending about the major axis only where either l e / h ≤ 20 or h < 3b is not satisfied, the column should be designed as biaxially bent, with zero M i in the minor axis; In case of uniaxial bending about the minor axis, M add obviously be applied only in the minor axis only. (ii) (iii) (iv) Worked Example 5.3 : A slender braced column of grade 35, cross sections b = 400 , h = 500 l ex = l ey = 8 m, N = 1500 kN Illustration of M1 and M 2 in column due to ultimate load M2 (i) About the major axis only, the greater and smaller bending moments due to ultimate load are 153 kNm and 96 kNm respectively. As l ex / h = 16 ≤ 20 ; h = 500 < 3b = 1200 So needs only checked for bending in the major axis. M 1 = 96 kNm; M 2 = 153 kNm 58 M1 Taken K = 1 in the initial trial. 1  le  1  8000  βa =   =   = 0 .2 2000  b  2000  400  2 2 au = β a Kh = 0.2 × 1 × 0.5 = 0.1 M add = Nau = 1500 × 0.1 = 150 The design moment will be the greatest of : M 2 = 153 (1) M i = 0.4M 1 + 0.6M 2 = 0.4 × (− 96) + 0.6 × 153 = 53.4 (2) (3) < 0.4 M 2 = 73.2 M 1 + M add / 2 = 96 + 150 / 2 = 171 N × emin = 1500 × 0.025 = 37.5 as emin = 0.05 × 500 = 25 > 20 (4) So the greatest design moment is case (3) M 1 + M add / 2 = 171 Thus the section need only be checked for uniaxial bending with N = 1500 kN and M x = 171 kNm bending about the major axis. (ii) Biaxial Bending, if in addition to the moments bending about the major axis as in (i), there are also moments of M 1 = 72 kNm; M 2 = 127 kNm bending in the minor axis. By Cl. 6.1.2.3(f), M add about the major axis will be revised as follows : Bending about the major axis : 1  le  1  8000  βa =   =   = 0.128 2000  h  2000  500  2 2 au = β a Kh = 0.128 × 1 × 0.5 = 0.064 M add = Nau = 1500 × 0.064 = 96 kNm. Thus item (3) which is M 1 + M add / 2 = 96 + 96 / 2 = 144 is reduced and case (1) where M 2 = 153 controls. Bending about the minor axis : M 1 = 72 kNm; M 2 = 127 kNm Taken K = 1 in the initial trial. 1  le  1  8000  βa =   =   = 0 .2 2000  b  2000  400  2 2 59 au = β a Kh = 0.2 × 1 × 0.5 = 0.1 M add = Nau = 1500 × 0.1 = 150 The design moment will be the greatest of : M 2 = 127 (1) M i = 0.4M 1 + 0.6M 2 = 0.4 × (− 72 ) + 0.6 × 127 = 47.4 (2) (3) < 0.4 M 2 = 50.8 M 1 + M add / 2 = 72 + 150 / 2 = 147 N × emin = 1500 × 0.02 = 30 as emin = 0.05 × 400 = 20 (4) So the greatest design moment is case (3) M 1 + M add / 2 = 147 Thus the ultimate design moment about the major axis is 153 kNm and that about the minor axis is 147 kNm. 5.3 Sectional Design Generally the sectional design of column utilizes both the strengths of concrete and steel in the column section in accordance with stress strain relationship of concrete and steel in Figure 3.8 and 3.9 of the Code respectively. Alternatively, the simplified stress block for concrete in Figure 6.1 can also be used. 5.3.1 Design for Axial Load only (Ceqn 6.55) of the Code can be used which is N = 0.4 f cu + 0.75 Asc f y . The equation is particularly useful for a column which cannot be subject to significant moments in such case as the column supporting a rigid structure or very deep beams. There is a reduction of approximately 10% in the axial load carrying capacity as compared with the normal value of 0.45 f cu + 0.87 Asc f y which accounts for the eccentricity of 0.05h . Furthermore, (Ceqn 6.56) reading N = 0.35 f cu + 0.67 Asc f y which is applicable to columns supporting an approximately symmetrical arrangement of beams where (i) beams are designed for u.d.l.; and (ii) the beam spans do 60 not differ by more than 15% of the longer. The further reduction is to account for extra moment arising from asymmetrical loading (by live load generally). 5.3.2 Design for Axial Load and Biaxial Bending : The general section design of a column is account for the axial loads and biaxial bending moments acting on the section. Nevertheless, the Code has reduced biaxial bending into uniaxial bending in design. The procedure for determination of the design moment, either M x ' or M y ' bending about the major or minor axes is as follows : b (i) Determine b' and h' as defined by the diagram. In case there are more than one row of bars, b' and h' can be measured to (ii) (iii) h' the centre of the group of bars. My Mx Compare and . h' b' Mx M y h' ≥ , use M x ' = M x + β M y If h' b' b' b' Mx M y b' < , use M y ' = M y + β M x If h' b' h' where β is to be determined from Table 6.14 of the Code under the N pre-determined ; bh The M x ' or M y ' will be used for design by treating the section as h either (a) resisting axial load N and moment M x ' bending about major axis; or (b) resisting axial load N and moment M y ' bending about minor axis as appropriate. 5.3.3 Concrete Stress Strain Curve and Design Charts The stress strain curve for column section design is in accordance with Figure 3.8 of the Code. It should be noted that the “Concrete Code Handbook” has 1.34( f cu / γ m ) recommended to shift ε 0 to in para. 3.1.4. With the Ec recommended shifting, the detailed design formulae and design charts have been formulated and enclosed in Appendix F. Apart from the derivation for the normal 4-bar column, the derivation in Appendix F has also included steel 61 75 and Continuum to the Structural Use of Concrete 2004 Concrete Grade 35 50 45 40 35 30 25 20 15 10 5 0 0 0.3 – Idealization of steel reinforcements in large column Figure 5.4 – Comparison between Continuum and 4-bar column Idealization 62 .5 3 3.reinforcements uniformly distributed along the side of the column idealized as continuum of reinforcements with symbol Ash .4 shows the difference between the 2 idealization.5 1 1.5 9 9.5 8 8.4% steel as continuum 1% steel as 4 bar column 1% steel as continuum 4% steel as 4 bar column 4% steel as continuum 8% steel as 4 bar column 8% steel as continuum N/bh N/mm 2 5. The user can still choose to lump the side reinforcements into the 4 corner bars.4% steel as 4 bar column 0. idealized as continuum of steel “strip” with area equivalent to the row of bars Figure 5. Comparison of Load Carrying Capacities of Rectangular Shear Walls with Uniform Vertical Reinforcements Idealized as 4 bar column with d/h = 0.5 4 4. with correction to the effective depth as in conventional design by setting Ash = 0 in the derived formulae.5 2 6 6.5 7 7.5 M/bh N/mm Figure 5.5 2 2. The new inclusion has allowed more accurate determination of load carrying capacity of column with many bars along the side as illustrated in Figure 5.3 which is particularly useful for columns of large cross sections. It can be seen that the continuum idealization is more economical generally except at the peak moment portion where the 4 bar column idealization shows slight under-design.5 5 2 0. 63 .67 .463 > = 0.5 11 11.5 13 13.5 7 7.5 14 14. 4-bar column.5 15 M/bh N/mm The section design is as shown in Figure 5. b' = 400 − 40 − 20 = 340 mm.14. d/h = 0. h' b' ∴M x '= M x + β β = 0.5 9 9.47 . h' = 600 − 40 − 20 = 540 mm.5 4 4.476 . f cu bh 35 × 400 × 600 My Mx = 0.9 h 600 Use Chart F-10 in Appendix F.5 12 12. 1.441 .5 10 10. with two additional T20 bars to avoid wide bar spacing. bh 2 400 × 600 2 N = 16.9 50 45 40 35 30 25 20 15 10 5 0 0 0.5 2 2. h = 600 mm under an axial load N = 4000 kN.5.5 2 0.5 1 1.018 = 4320 mm2. M y = 150 kNm.4 : Consider a column of sectional size b = 400 mm.Worked Example 5.8% steel is approximated which amounts to 400 × 600 × 0. or 6-T32 (Steel provided is 4826mm2) Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35.5 6 6. bh d 540 = = 0.446 from Table 6. 540 h' M y = 250 + 0.446 × × 150 = 356 kNm 340 b' 356 × 10 6 M = = 2. Assume a 4-bar column.5 5 5.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 8 2 8. N 4000000 = = 0.5 3 3. M x = 250 kNm. the solution process will be very tedious and not possible for irregular section. by balancing axial load and the 2 bending moments.T32 Extra T20 Figure 5. However. neutral axis depth and the required reinforcement. 3 equations can be obtained for solution of the neutral axis orientation. Theoretically. Such approach can be applied to section of irregular shape. 64 5. Nevertheless. various combinations of the neutral axis depth and orientation will be presumed for calculation of axial load and moment carrying capacities of the section.5 . Practical approach by computer method may be based on trial and error method.3. If the reinforcements are pre-determined. a designer can actually solve the biaxial bending problem by locating the orientation of the extent of the neutral axis (which generally does not align with the resultant moment except for circular section) by balancing axial load and the bending in two directions.3. Though the Code has provision for converting the biaxial bending problem into a uniaxial bending problem by (i) searching for the controlling bending axis and (ii) aggravate the moments about the controlling bending axis as appropriate to account for effects of bending in the non-controlling axis. Direct sectional analysis to Biaxial Bending without the necessity of converting the biaxial bending problem into a uniaxial bending problem. spread sheets have been prepared and 2 samples are enclosed at the end of Appendix F. the design of reinforcements can be based on formulae derived in Appendix F. However.5 – Design Chart and Worked Re-bar details for Example 5.4 5. The section will be considered adequate if a combination of neutral axis depth and orientation produce axial load and moments of resistance in excess of the applied loads and moments. as the algebraic manipulations are complicated (may involve solution of 4th polynomial equations) and cases are many.4 Alternatively. the approach may only be practical by computer methods. the smallest bar diameter used shall not be less than 2/3 of the largest bar diameter used (D). 9.Figure 5.2% (D). 9.8 f cu as per (v) (vi) (vii) fy (Ceqn 9. For grade 35 concrete and based on high yield bar.5 and Cl.2. the limiting bar diameter is simply 65 .6 illustrates the method of solution. The total sectional resistance of the section under the stress strain profile resists the applied axial loads and moments Neutral axis Strain profile on section strain profile across section concrete stress profile Figure 5. In any row of bars.2h 0. At laps.6 – General Biaxial Bending on irregular section 5. the bar diameter should be limited to φ ≤ 1. the ductility requirements are marked with “D”) (i) (ii) (iii) (iv) Minimum longitudinal steel based on gross area of a column is 0.25 × 3. In columns having a polygonal cross-section. Longitudinal bars should have a diameter > 12 mm.8%. at least one bar be placed at each corner.7) of the Code where h is the beam depth. Where the columns bars pass through the beams at column beam joints. the sum of reinforcement sizes in a particular layer should not exceed 40% of the breadth at that section.9.1(a) of the Code. Maximum longitudinal steel based on gross area of a column is 4% except at lap which can be increased to 5. The minimum number of longitudinal bars should be four in rectangular columns and six in circular columns.4 Detailing Requirements for longitudinal bars in columns (generally by Cl. (ix) No column bar should be terminated in a joint area without a horizontal 90o standard hook or an equivalent device as near as practically possible to the far side of the beam and not closer than 3/4 of the depth of the beam to the face of entry (D).2. 9. the anchorage of the column bars into the joint region should commence at 1/2 of the depth of the beam/foundation member or 8 times the bar diameter from the face at which the bars enter the beam or the foundation member (D) and the bend should be “inwards”.5D or 8ø D Beam or foundation element Figure 5.7 – Column Bar anchorage requirements as per 5. 20 mm and aggregate size + 5 mm.25 can be removed if the column is not intended to form a plastic hinge (D).φ ≤ 0. i.4 (viii) and (ix) (x) Minimum clear spacing of bars should be the minimum of bar diameter.1(a) of the Code.6 implying maximum bar size is 25 mm.9. 5. Items (i) to (iv) below are requirements for columns not within “critical regions” as defined in item (v) : (i) (ii) (iii) Diameter of transverse reinforcements > the greater of 6 mm and 1/4 of longitudinal bar diameter. (viii) Where column bars terminate in a beam-column joints or joints between columns and foundation members where a plastic hinge in the column could be formed.5. The spacing of transverse reinforcement shall not exceed 12 times the diameter of the smallest longitudinal bar.e. the ductility requirements are marked with “D”. every corner bar and each alternate bar (or bundle) shall be laterally supported by a link passing 66 .2 and Cl.75D >0. if beam depth is 600 mm. For rectangular or polygonal columns. Anchorage should commence at this point >0.5 Detailing Requirements for transverse reinforcements in columns (generally by Cl.046h . 9. φ ≤ 27. The factor 1. (b) 6 times the diameter of the longitudinal bar to be restrained (D). For circular columns. (c) 200 mm.(iv) (v) around the bar and having an included angle < 135o. Links be adequately anchored by hooks through angles < 135o. 9.8 (Re Cl.2 of the Code) shall have additional requirements as for the maximum spacing of transverse reinforcements as : (a) 1/4 of the least lateral column dimension (D) in case of rectangular or polygonal column and 1/4 of the diameter in case of a circular column.2. Loops (circular links) should be anchored with a mechanical connection or a welded lap by terminating each end with a 135o hook bent around a longitudinal bar after overlapping the other end of the loop by a minimum length. Spiral should be anchored either by welding to the previous turn or by terminating each end with a 135o hook bent around a longitudinal bar and at not more than 25 mm from the previous turn. which causes spalling of the concrete cover. loops or spiral reinforcement satisfying (i) to (iii) should be provided. Transverse reinforcements in “Critical regions” within columns as defined in Figure 5. Loops and spirals should not be anchored by straight lapping. 67 .9. No bar within a compression zone be further than 150 mm from a restrained bar. 5h or 1.65 H > hm H > 2h or 2D hm M max h b D Figure 5.85 H > hm H > h or D (b) 0. x = 0. x = 0.8 – “Critical Regions (Potential Plastic Hinge Regions)” in Columns 68 .Height of “critical region” . Critical regions with enhanced transverse reinforcements Normal transverse reinforceme xM max H.3.5D (c) 0.1. depends on N/Agfcu ratio : H.1 <N/Agfcu ≤ 0.75 H > hm H > 1.6. x = 0.3 <N/Agfcu ≤ 0. Critical regions with enhanced transverse reinforcements (a) 0<N/Agfcu ≤ 0. h. 1(b) indicating typical loading arrangement acting on the joint. the aim of design comprise minimization of (a) the risk of concrete cracking and (b) spalling near the beam-column interface. reversing the hogging moment on the beam on the left side of the joint. To be specific.2 The phenomenon of “diagonal splitting” of joint Diagonal crushing or splitting of column-beam joints is resulted from “shears” and unbalancing moment acting on the joints as illustrated in Figure 6. providing vertical and horizontal shear links within the joint and confinement to the longitudinal reinforcements of the columns adjacent to the joint.1(a) shows a large horizontal shear from the right. and (c) checking provisions against diagonal crushing or splitting of the joint and where necessary.1(a) and 6. Vc1 Potential failure surface (tension) CBL TBR sagging moment in beam TBL Vb2 Vb1 CBR hogging moment in beam Vc2 Figure 6.1(a) – Phenomenon of Diagonal Joint Splitting by moments of opposite signs on both sides of joint 69 . 6.1 General The design criteria of a column-beam joint comprise (i) performance not inferior than the adjoining members at serviceability limit state.0 Column-Beam Joints 6.6. Figure 6. and (ii) sufficient strength to resist the worst load combination at ultimate limit state. Generally the load due to unbalanced forces by the beams on both sides of the joints need be considered.1(b) – Phenomenon of Diagonal Joint Splitting by moments of same sign on both sides of joint The unbalanced forces due to unbalanced flexural stresses by the adjoining beams on both sides of the joint tend “tear” the joint off with a potential tension failure surface. By the load capacity concept. 6.2(a) and 6.2(b) indicates how to determine V jh by this criterion. V jh should be worked out by considering equilibrium of the joint.3 Design procedures : (i) Work out the total nominal horizontal shear force across the joint V jh in X and Y directions generally. Though generally there are shears in the column usually tend to act oppositely. The horizontal shears by the columns can generally be ignored as they are comparatively small and counteract the effects by bending. Reinforcements in form of links may therefore be necessary if the concrete alone is considered inadequate to resist the diagonal splitting. producing “diagonal splitting”.Potential failure surface (tension) Vc1 T BR TBR > TBL TBL hogging moment in beam Vb2 CBL Vb1 CBR hogging moment in beam Vc2 Figure 6. the reinforcing bars in the beam will be assumed to have steel stress equal to 125% yield strength of steel if such assumption will lead to the most adverse conditions. Figure 6. 70 . the effects are usually less dominant than that of moment and therefore can be ignored. 71) in the Code.25 f y AsR T BL ML = 0. The vector sum of v jh for both the X and Y directions should be worked out and check that it does not exceed 0. v jh = V jh b j hc where hc is the overall depth of the column in the direction of shear b j = bc or b j = bw + 0.25 f y AsR + ML 0.5hc whichever is the smaller when bc ≥ bw b j = bw or b j = bc + 0. same sign beam moments on both sides (ii) With the V jh determined.2(b) – Calculation of V jh .25 f cu . opposite sign beam moments on both sides Potential failure surface (tension) TBR > TBL T BL T BR = 1.87 f y z L zL CBR hogging moment in beam ML CBL hogging moment in beam V jh = T BL + T BR = 1.25 f y AsR sagging moment in beam V jh = TBL + TBR = 1.25 f cu 71 2 2 .5hc whichever is the smaller when bc ≥ bw where bc is the width of column and bw is the width of the beam.2(a) – Calculation of V jh .25 f y ( AsL + AsR ) hogging moment in beam Figure 6.25 f y AsL TBL CBR TBR T BR = 1.87 f y z Figure 6. the nominal shear stress is determined by (Ceqn 6.Potential failure surface (tension) C BL = T BL TBL = 1. v jhx + v jhy ≤ 0. no horizontal shear reinforcements will be required.(iii) Horizontal A jh = reinforcements based on (Ceqn 6. i. 6.87 f yv should be worked out in both directions and be provided in the joint as horizontal links. (e) Reduced by half to that provisions required in (a) if the joint is connected to beams in all its 4 faces.4(hb / hc )V jh * −C j N * 0. Concrete grade is 40.5. (v) Notwithstanding the provisions arrived at in (iii) for the horizontal reinforcements. (iv) Similarly vertical reinforcements based on (Ceqn 6.2 of the Code. If the numerical values arrived by (Ceqn 6.72) reading V jh *  C N *  0.87 f yh  Ag f cu    be provided in the joint as horizontal links. 72 .73) reading A jh = 0. It may be more convenient to use close links which can serve as horizontal reinforcements in both directions and confinements as required by (v).5 − j  should be worked out in both directions and 0.3 in the X-direction and Y-directions. no horizontal shear reinforcements will be required. (f) Link spacing ≤ 10φ (diameter of smallest column bar) and 200 mm.1: Consider the column beam joints with columns and beams adjoining as indicated in Figure 6.5 (i) to (iv) of this Manual if the joint has a free face in one of its four faces. confinements in form of closed links within the joint should be provided as : (d) Not less than that in the column shaft as required by Cl. Section 5.4 Worked Example 6. If the numerical values arrived by (Ceqn 6.73) becomes negative. 9.72) becomes negative.e. It may be more convenient to use close links which can serve as horizontal reinforcements in both directions. The top steel provided on the left beam is 3T32.2(a) is applicable. 73 .3 – Design Example for Column Beam Joint (i) Check nominal shear stress : X-direction The moments on the left and right beams are of opposite sign. So Figure 6. with Ast1 = 2413 mm2 whilst the bottom steel provided on the right beam is 4T20 with Ast1 = 1257 mm2.X-direction N * = 6000 kN Beam moment 550 kNm (hogging) Beam size 700 × 500 (effective depth 630) Beam moment 300 kNm (hogging) Beam size 700 × 500 (effective depth 630) Column size 900 × 800 900 800 Column size 900 × 800 Y-direction N * = 6000 kN Beam moment 550 kNm (hogging) Beam size 700 × 500 (effective depth 630) Beam moment 300 kNm (hogging) Beam size 700 × 500 (effective depth 630) Column size 900 × 800 900 800 Figure 6. = 0.2(b) is applicable. 2 bd d TBL = 0. M z = 1.49 mm2. As the moment on the right beam is higher. b j = 900 .25 × 460 ×1257 = 723 kN. C BR = TBR TBL is to be determined by conventional beam design method. TBL = 1. the potential plastic hinge will be formed on the right beam.5hc = 900 + 0.25 × 460 × 2413 = 1445 kN.CBL TBR TBL CBR TBR = 1. similarly.25 × 460 × 2413 = 1445 kN.948 .87 f y AsL = 502.512 MPa.5 × 800 = 1300 .01 MPa 900 × 800 Y-direction The moments on the left and right beams are of equal sign. TBR > TBL T BL zL CBR CBL TBR = 1.05 kN So V jy = 1445 − 502 = 943 kN In the Y-direction hc = 900 . 74 . C BR = TBR C BL = TBL So the total shear is V jx = TBL + C BR = TBR + C BL = 2168 kN In the X-direction hc = 800 As bc = 900 > bw = 500 . AsL = 1254.the effective joint width is the smaller of bc = 900 and bw + 0. so b j = 900 So v jx = V jx b j hc = 2168000 = 3. So Figure 6. 87 × 460    = 1922 mm2 A jhx = Y-direction V jy 943 = = 0.87 × 460 = −8. A jvx = = Y-direction 75 0.So v jy = V jy b j hc = 943000 = 1.303 C jx y = V jx + V jy 2168 + 943 V jhx *  C N *  0.87 f yh 0.697 × 5000000   0.4(700 / 900)2168 − 0.87 × 460  0. reading A jh = 0.303 × 5000000     0.5 − 0.5 − jx =  0.4(hb / hc )V jh * −C j N * . 900 × 800 Combining X and Y direction.87 f yh  900 × 800 × 40  Ag f cu  0.87 f yh  900 × 800 × 40  Ag f cu   = 1029 mm2 A jhx = Use 5T16 close stirrups (Area provided = 2011 mm2) which can adequately cover shear reinforcements in both directions (iii) To calculate the vertical joint reinforcement by (Ceqn 6.31 MPa. 2 2 (ii) To calculate the horizontal joint reinforcement by (Ceqn 6.012 + 1.87 f yh X-direction 0.697 × 6000 0.697 C jx = V jx + V jy 2168 + 943 V jhx *  C N *  2168000  0. So no vertical shear reinforcement is required.28 < 0.8 .5 −  0.5 − j  Code).73 of the Code). 0.25 f cu = 10 MPa.87 f yh  Ag f cu    V jh where C j = V jx + V jy X-direction V jx 2168 = = 0. the total shear stress is v jh = v jx + v jy = 3.5 − jx  = 943000  0.4(hb / hc )V jh * −C j N * reading A jv = 0.312 = 3.72 of the V jh *  C N *  0. 303 × 6000 0. A jvy = = (iv) The provision of outermost closed stirrups in the column shaft is T 10 at 200 which is in excess of the confinement requirements.4(hb / hc )V jh * −C j N * Closed links 5T16 Figure 6.4(700 / 800)943 − 0.8 . 0.87 f yh 0.4 – Details of Column Beam Joint Detail 76 .0.87 × 460 = −3. So no additional confinement requirement. So no vertical shear reinforcement is required. 7.4.3 Other than 7.2. 7.2 Limits of slender ratio (Re Table 6.2.3. 6. (Re Cl. 6.1.2 Categorization of Walls Walls can be categorized into (i) slender walls.1 Similar to column by design to resist axial loads and biaxial moments.2(a) and Cl.1 As similar to column. (ii) stocky walls.3(a) of the Code).1 and 7.4 Stocky Wall 7. (iii) 30 for unbraced wall.1 Determination of effective height l e (of minor axis generally which controls) – (i) in case of monolithic construction. same as that for plain wall.2. 7. and (ii) in case of simply supported construction. 7. 7. (iii) reinforced concrete walls.3 Minimum eccentricity for transverse moment design is the lesser of 20 mm or h / 20 . as similar to columns.15 of the Code) – (i) 40 for braced wall with reinforcements < 1%. and (iv) plain walls.1.3. 7.1. (ii) 45 for braced wall with reinforcements ≥ 1%. reinforced concrete design is similar to that of columns.1 Design Generally 7.2. 7.3. same as that for column. 7.2 The design ultimate axial force may be calculated on the assumption that the beams and slabs transmitting force to it are simply supported.2.3 Slender Wall Section Design 7. stocky walls are walls with slenderness ratio < 15 for braced walls and slenderness ratio < 10 for unbraced walls. 77 .3.3.0 Walls 7. For detailed design criteria including check for concentrated load. 7. load carrying capacities etc. refer to Cl. 7. (b) l e = 2.8 and 3.5l 0 when one end support restraint movements only and the other is free.4.1 Effective height of unbraced plain wall.35 f cu Ac + 0.Plain wall are walls the design of which is without consideration of the presence of the reinforcements.7 Sectional Design The sectional design of wall section is similar to that of column by utilizing stress strain relationship of concrete and steel as indicated in Figure 3. (Alternatively.67 f y Asc 7.2 The load carrying capacity of plain wall will depend on concrete strength alone.2.75l 0 when the two end supports restraint movements and rotations.2 and the design for deflection induced moment M add . Effective height ratio for braced plain wall is determined by (a) l e = 0.2. n w ≤ 0. design of stocky wall is similar to slender walls.59) of the Code provided that the walls support approximately symmetrical arrangement of slabs with uniformly distributed loads and the spans on either side do not differ by more than 15%. (b) l e = 2. (b) l e = 2.3 Other than 7.6. is determined by : (a) l e = 1.4.4. where l 0 is the clear height of the wall between support.5 Reinforced Concrete Walls design is similar to that of columns with categorization into slender walls and stocky walls.9 of the Code.6. 7.0l 0 for other cases. 7.6 7. Plain Wall .2 Stocky reinforced wall may be designed for axial load n w only by (Ceqn 6. 6.7.5l 0 when it is supporting a floor slab spanning at right angles to it. (c) l e = l 0 when the two end supports restraint movements only. shear.3 of the Code. the simplified stress block of concrete as 78 .0l 0 when one end support restraint movements and rotations and the other is free. d = 0.1) which is considered as a more realistic idealization. Conventionally.3.75h Shear Wall Section Current idealization based on 4-bar column design chart Proposed idealization with reinforcing bars as continuum with areas equal to the bars Figure 7-1 – Idealization of Reinforcing bars in shear wall 7.1 can also be used.3. Nevertheless. walls with uniformly distributed reinforcements along its length can be treated as if the steel bars on each side of the centroidal axis are lumped into two bars each carrying half of the steel areas as shown in Figure 7. Concrete grade : 35 Connection conditions at both ends of the wall : connected monolithically with floor structures shallower than the wall thickness. Derivation of the formulae for the design with reinforcements idealized as continuum is contained in Appendix G. together with design charts also enclosed in the Appendix. it is suggested in the Manual that the reinforcements can be idealized as a continuum (also as shown in Figure 7. Wall Height : 3.6 m.8 Worked Example 7.indicated in Figure 6. Check for slenderness 79 .) ε 0 should be shifted as discussed in Section 5.1 and design is carried out as if it is a 4 bar column. plan length : 2000 mm.1 Wall Section : thickness : 200 mm. 48 and (4) N × emin = 7200 × 0. le / h = 3060 / 2000 = 1.2 = 0.48 = 192. M x will be the greatest of (1) M 2 = 110 .2 = 0. (2) M i + M add where M i = 0. For bending about the minor axis.08 kNm for design.6M 2 ≥ 0.117 × 1 × 0.6 × 25 + 168.) M 1 + M add / 2 in which M 1 is the smaller initial end moment due to design ultimate load.08 .4 × 100 + 0.2) For bending about the major axis. the greater initial end moment due to design ultimate load. So M x = 144 kNm for design.48 = 193.02 = 144 .49 of the Code) in the initial trial.117 × 1 × 0. 1  le  1  3060  βa =   =   = 0. So M y = 193. (3) M 1 + M add = 24 + 168. M x = 110 kNm at top and 110 kNm at bottom .4 × 24 + 0.0234 M add = Nau = 7200 × 0.6 = 3.02 = 144 . so M add = 0 .117 2000  b  2000  200  2 2 au = β a Kh = 0. (2) M i + M add = 0.0234 = 168.4M 2 (with M 2 as (3) (4) positive and M 1 negative. (3) M 1 + M add = 100 + 0 = 100 and (4) N × emin = 7200 × 0. l e = 0. 1  le  1  3060  βa =   =   = 0. N × emin (discussed in 5. End conditions are 2 for both ends. le / b = 3060 / 200 = 15.2.48 kNm.6 × 110 + 0 = 106 .53 < 15 . So the factored axial load and moments for design are 80 .117 2000  b  2000  200  2 2 au = β a Kh = 0.4M 1 + 0. (2) M i + M add = 0.3 > 15 .0234 (i) Axial Load : N = 7200 kN. β = 0.85 (by Table 6. M y = 25 kNm at top and 24 kNm Final design moment M t about the major and minor axes will be the greatest of : (1) M 2 .11 of the Code). M y will be the greatest of (1) M 2 = 25 .85 × 3.06 m Taken K = 1 (Ceqn 6.Only necessary about the minor axis. 096 < = 1. M x = 144 kNm.5 3 3.025 × 200 × 2000 = 10000 mm or 5000 mm /m Use T20 – 125 (B.8 50 45 40 35 30 25 20 15 10 5 0 0 0.46 kNm h' 1500 M 199. = 2.825 = 18 .4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 6.4933 . = 2 2 hb b 200 bh 2000 × 200 Choosing the most appropriate chart (Chart F-7) in Appendix F which is Grade 35 with d/h = 0.N = 7200 kN.8. and plot N M = 18 and = 2.5 7 7.46 ×10 6 d 165 N = = 0. bottom .5% 2 2 0. β = 0.514 .403 from Table 6. the greater initial end moment due to design ultimate load.5 6 2 0. f cu bh 35 × 200 × 2000 My Mx = 0.75 = 1500 N 7200000 = = 0.5 8 8.5 4 4.F. 81 .5 2 2.14.08 kNm b' = 200 − 25 − 10 = 165 .5 1 1.403 × ×144 = 199. the bh hb 2 which is estimated steel percentage is 2.5 12 M/bh N/mm 2 (ii) Axial Load : N = 7200 kN. 4-bar column.5 10 10.5 11 11. h' b' ∴M y '= M y + β b' 165 M x = 193. d/h = 0.08 + 0. M x = 1800 kNm at top and 1200 kNm at M y = 25 kNm at top and 24 kNm Final design moment M t about the major and minor axes will be the greatest of : (1) M 2 .5 9 9.5 5 5.494 on it.17 . h' = 2000 × 0.) Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35. M y = 193. 6 kNm b' 165 N M 2415. h' = 2000 × 0.75 = 1500 N 7200000 = = 0.4 × 24 + 0.08 kNm for design. 1  le  1  3060  βa =   =   = 0.6 × 1800 + 0 = 1560 .48 kNm.403 × ×139. M y will be the greatest of (1) M 2 = 25 .6 × 25 + 168. M x = 1800 kNm.02 = 144 . le / h = 3060 / 2000 = 1.02 = 144 . β = 0.6 × 10 6 = 18 . (2) M i + M add = 0.14.4 × 1200 + 0. For bending about the minor axis.48 = 193. So M x = 1800 kNm for design.8.2 < = 1.08 kNm b' = 200 − 25 − 10 = 165 .08 .2) For bending about the major axis. h' b' ∴M x '= M x + β h' 1500 M y = 1800 + 0.117 2000  b  2000  200  2 2 au = β a Kh = 0.53 < 15 .48 and (4) N × emin = 7200 × 0.2 = 0. = = 3.514 . M y = 193.3 > 15 . M x will be the greatest of (1) M 2 = 1800 .) M 1 + M add / 2 in which M 1 is the smaller initial end moment due to design ultimate load. and plot N = 18 and bh 82 . N × emin (discussed in 5. (3) M 1 + M add = 24 + 168.2.08 = 2415. So M y = 193.117 × 1 × 0. (2) M i + M add = 0.17 . (3) M 1 + M add = 1200 + 0 = 1200 and (4) N × emin = 7200 × 0.(2) (3) (4) M i + M add where M i = 0. f cu bh 35 × 200 × 2000 My Mx = 1.4M 1 + 0. So the factored axial load and moments for design are N = 7200 kN. le / b = 3060 / 200 = 15.6M 2 ≥ 0. bh hb 2 200 × 2000 2 Choosing the wall design chart for grade 35 for wall (Chart G-2) in Appendix G which is Grade 35 with d/h = 0.48 = 192. so M add = 0 .0234 = 168.4M 2 (with M 2 as positive and M 1 negative.0195 .403 from Table 6.0234 M add = Nau = 7200 × 0. 5 2 6 6.5 9 9.1% amounting to 0. Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35.031× 200 × 2000 = 12400 mm2 or 6200 mm2/m which is slightly higher. the estimated steel percentage is 3% which is hb 2 0.494 on it. Concrete Grade 35 50 45 40 35 30 25 20 15 10 5 0 0 0. However.5 M/bh N/mm Alternatively.F.5 2 2.75 50 45 40 35 30 25 20 15 10 5 0 0 0.5 4 4.5 5 2 0.5 4 4.03 × 200 × 2000 = 12000 mm2 or 6000 mm2/m Use T20 – 100 (B.5 1 1.5 2 2.5 8 8.5 3 3. use Column Design Chart F-6 in Appendix F by treating it as column of d/h = 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 5.5 2 6 6.5 7 7. the calculations are too tedious and cases to try 83 .5 7 7. d/h = 0. the estimated steel percentage is 3.75.) Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004.5 9 9.M = 2.5 3 3.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 5.5 1 1.5 8 8. 4-bar column.5 M/bh N/mm Design can be performed by calculations with the formulae derived in Appendices F and G.5 5 2 0. links be provided as : (g) to anchor every vertical compression longitudinal bar. 84 . When this reinforcement controls the design. ≤ 200 ≤ 200 (ii) bars not required to take compression bars required to take compression Figure 7. (ii) Maximum steel percentage : 4%.9 Detailing Requirements There are no ductility requirements in Cl.9 of the Code for walls. Spread sheets have been devised to tackle the problem and a sample is enclosed in Appendix G. maximum spacing not to exceed 16 times the vertical bar diameter in the vertical direction. 9. 7.3% for f y = 250 MPa if the required vertical reinforcement does not exceed 2%.4%. In addition. (iii) Maximum distance between bars : the lesser of 3 times the wall thickness and 400 mm.2 – Anchorage by links on vertical reinforcements (i) (j) minimum diameter : the greater of 6 mm and 1/4 of the largest compression bar. maximum spacing : twice the wall thickness in both the horizontal and vertical directions. (h) no bar be at a distance further than 200 mm from a restrained bar at which a link passes round at included angle ≤ 90o. Horizontal and transverse reinforcements for reinforced concrete walls (i) Minimum horizontal reinforcements : 0. 9. If the required vertical reinforcement > 2%.6 of the Code : Vertical reinforcements for reinforced concrete walls (i) Minimum steel percentage : 0. half of the steel area be on each side. The detailing requirements are summarized from Cl.25% for f y = 460 MPa and 0.are too many without the use of computer methods. 25% for f y = 460 MPa and 0.Plain walls If provided. 85 . minimum reinforcements : 0.3% for f y = 250 MPa in both directions generally. 9 x v Concrete ultimate strain ε ult = 0.2.0 Corbels 8. 6. The strut action (compressive) is carried out by concrete and the tensile force at top is carried by the top steel.1 General – A corbel is a short cantilever projection supporting a load-bearing member with dimensions as shown : av < d Applied Load ≥ 0 .8.1 According to Cl.5 h h d Top steel bar Figure 8. the basis of design method of a corbel is that it behaves as a “Tie-and-strut” model as illustrated in Figure 8.0035 xv β Concrete stress block at corbel support Figure 8.2 – Strut-and-Tie Action of a Corbel 86 .5.2 of the Code.1 – Dimension requirement for a Corbel 8. Steel strain to be determined by linear extrapolation av Tie action by reinforcing bar Applied Load Vu Ft d neutral axis Strut action by concrete Vu Fc Balancing force polygon 0.2.2 Basis of Design 8. 2. av cos β = av a v + (d − 0.405 f cu bx cos β By the force polygon.9 x = 0. Fc sin β = Vu ⇒ 0.2 In addition to the strut-and tie model for the determination of the top steel bars. The net bearing width is obtained by ultimate load effective bearing length × ultimate bearing stress The Precast Concrete Code has specified that the effective bearing length be the least of (i) physical bearing length.8.45 f cu × b × 0. 8.8 f cu for contact face of a steel bearing plate cast on the corbel with each of the bearing width and length not exceeding 40% of the width and length of the corbel.4 f cu for dry bearing.9. shear reinforcements should be provided in form of horizontal links in the upper two thirds of the effective depth of the corbel.405 f cu bx a v (d − 0. In short. 2.45 x ) a v + (d − 0. (ii) one half of the physical bearing length plus 100 mm. (iii) 600 mm. (iii) 0.7. the design ultimate bearing pressure to ultimate loads should not exceed (i) 0.2.405 f cu bx where b is the length of the corbel The force in the compressive strut is therefore Fc = 0.405 f cu bx sin β cos β = Vu As tan β = d − 0.3 Bearing pressure from the bearing pad on the corbel should be checked and properly designed in accordance with “Code of Practice for Precast Concrete Construction 2003” Cl.45 x ) 2 So 0.405 f cu bxa v (d − 0.45 x .45 x ) 2 2 a v + (d − 0. (ii) 0.45 x ) 2 2 . The horizontal links should not be less than one half of the steel area of the top steel.45 x ) a v + (d − 0. 8.6 f cu for bedded bearing on concrete.3 Design Formulae for the upper steel tie The capacity of concrete in providing lateral force as per Figure 8.45 x ) 2 = Vu ⇒ Vu = 87 0.45 x ) 2 2 sin β = (d − 0.2 is 0. the top steel force is T = Vu cot β = (Eqn 8-2) The strain at the steel level is.18225 f cu bav . However. C = Vu av + d 2 x= − B − B 2 − 4 AC 2A B = −0.45 f cu bav ) ( 2 ) (Eqn 8-1) Vu av d − 0.45 x By the equilibrium of force. (v) Solve the neutral axis depth x by the equation (Eqn 3-1).45 f cu bav )x + Vu (av 2 + d 2 ) = 0 Putting A = 0.0035 x x (Eqn 8-3) 8. (iv) Check bearing pressures.2025Vu + 0. estimate the size of the corbel and check that the estimated dimensions comply with Figure 8. work out the strain at the top steel level as ε s .Expanding and grouping (0.2 is ε s = d−x d−x ε ult = × 0. (viii) Obtain the force in the top steel bar T by (Eqn 8-2) (ix) Obtain the required steel area of the top steel bars Ast by Ast = (x) Check the shear stress by v = T σs Vu .0035 . However.18225 f cu bav )x 2 − 0. the total shear area provided which is 88 .9d (Vu + 0.2025Vu + 0. If v > v c (after enhancement as (2 / 3)bd A b(v − v c ) over the applicable). (vii) Obtain the steel stress as σ s = E s ε s where E s = 205× 10 6 kPa.87 f y upper 2 d where Asv is the cross sectional area of each link and s v 3 is the link spacing. by extrapolation of the strain diagram in Figure 8. (vi) By the assumption plane remains plane and that the linear strain at the base of the corbel is the ultimate strain of concrete ε ult = 0.002 . the stress should be limited to 0.4 Design Procedure : (iii) Based on the design ultimate load and av . provide shear reinforcements by sv = sv 0.87 f y at which ε s = 0.9d (Vu + 0.1. 3(b) Vu Top main bar av >c c.e. (ii) Typical detailings are shown in Figure 8. the bearing area of the load should stop short of the transverse bar by a distance equal to the cover of the tie reinforcement. sv 3 2 8. In the latter case.Asv 2 × d should not be less than half of the top steel area.5 Detailing Requirements (i) Anchorage of the top reinforcing bar should (a) either be welded to a bar of equivalent strength or diameter. sv 3 Asv 2 1 × d ≥ Asl . In the former case. or (b) bent back to form a closed loop. the bearing area of the load should not project beyond the straight portion of the bars forming the tension reinforcements. Shear reinforcements be provided in the upper two thirds of the effective depth and total area not less than half of the top bars.3(a) and 8. cover to transverse bar 2 d 3 d transverse bar welded to the main tension bar of equal diameter or strength Additional bar for shear link anchorage Shear reinforcements Figure 8.3(a) – Typical Detailing of a Corbel 89 . i. av = 200 < d = 450 Vu = 600 250 ≥ 0. 6. i.4 – Worked Example. 8.5.Vu Top main bar av >0 2 d 3 d Shear reinforcements Figure 8. The load is transmitted from a bearing pad of length 300 mm.3(b) – Typical Detailing of a Corbel 8.e. av = 200 mm. Concrete grade is 40.1 Design a corbel to support an ultimate load of 600 kN at a distance 200 mm from a wall support.1 (a) The dimensions of the corbel are detailed as shown which comply with the requirement of Cl.1 of the Code with length of the corbel 90 .6 Worked Example 8. Vu = 600 kN.5h h = 500 d = 450 Net bearing width Figure 8. b = 300 mm.87 × 460 sv So use T12 closed links – 100 ( Asv provided is 2.33 MPa > vc = 0.002 276. Net bearing width is (c) With the following parameters.87 × 460 (g) Steel area required is (h) Check shear stress over the upper two thirds of d . Total steel area is 113 × 2 × 3 = 679 mm2 > 0. Vu = 600 kN.45 x 440 − 0.32 mm2.0035 = 0.5) = = 2. 91 .5 × 921 = 461 mm2. (b) Check bearing stress : Design ultimate bearing stress is 0.45 f cu bav )x + Vu (av 2 + d 2 ) = 0 Solving x = 276. d − 0.9d (Vu + 0.77 d−x ε ult = × 0.8 × 40 = 32 MPa 600 ×10 3 = 62.33 − 0. 600000 = 3.87 f y as ε s > 0.77 368710 = 921. 300 × 32 So use net bearing width of bearing pad 70 mm.26) over the top 300 sv mm.8 f cu = 0.e. The force in the top steel is T= Vu av 600 × 200 = = 368. 0.12 mm So shear reinforcement sv = 0.002 .77 mm. av = 200 mm d = 450 mm substituted into (Eqn 8-1) b = 300 mm.71 kN. (d) The strain at steel level εs = (e) (f) 450 − 276. f cu = 40 MPa.5 mm.87 f y 0.2025Vu + 0. provide 3T20.77 x The stress in the top steel is 0. i. (2 / 3)450 × 600 A b(v − vc ) 300(3.5 MPa.45 × 276.18225 f cu bav )x 2 − 0.00219 > 0. 300 mm. (0. 1 92 .Vu 3T20 av 300 450 T20 anchor bar T12 closed links Figure 8.5 – Detailing of Worked Example 8. it will nevertheless create out-of-plane loads in the transfer structures which need to be taken into account. transfer structures are horizontal elements which redistribute vertical loads where there is a discontinuity between the vertical structural elements above and below. Sidesway of the transfer structures under lateral loads and unbalanced gravity loads should also be taken into account. Local effects of shear walls on transfer structures – shear walls will stiffen up transfer structures considerably and the effects should be taken into account.5 of the Code.1 According to Cl.0 Transfer Structures 9. behaving somewhat between (a) flexible floor structures on hard columns. Care should be taken if the structural model above the transfer structure is analyzed separately with the assumption that the supports offered by the transfer structures are rigid. consideration should be given to the followings : (i) Construction and pouring sequence – the effects of construction sequence can be important in transfer structures due to the comparative large stiffness of the transfer structure and sequential built up of stiffness of structures above the transfer structure as illustrated in Figure 9.2 In the analysis of transfer structures.9. 93 (ii) (iii) (iv) (v) (vi) (vii) . Temporary and permanent loading conditions – especially important when it is planned to cast the transfer structures in two shifts and use the lower to support the upper shift as temporary conditions. and (b) rigid structures (like rigid cap) on flexible columns. Deflection of the transfer structures – will lead to redistribution of loads of the superstructure. The phenomenon is more serious as the transfer structure usually possesses large flexural stiffness in comparison with the supporting structural members. thus creating locked-in stresses. 5.1. Re-examination of the load redistribution should be carried out if the deflections of the transfer structures are found to be significant. Varying axial shortening of elements supporting the transfer structures – which leads to redistribution of loads. Lateral shear forces on the transfer structures – though the shear is lateral. The effects should be considered if the transfer structure is analyzed as a 2-D model. 9. .S. + G/F + 1/F Stress/force in T.....S.S. due to weight of 2/F structure and stiffness of the T.. only G/F Stress/force in T.S. Final force induced on T.S. + G/F Stage (4) : 1/F hardened and 2/F wet concrete just poured Stage (5) : 2/F hardened and 3/F wet concrete just poured 3/F 2/F 1/F G/F 2/F 1/F G/F Stage (6) and onwards Structure above transfer structure continues to be built.Stage (1) : Transfer Structure (T. {F2} being force induced in transfer structure due to weight of G/F structure and stiffness of the T.. and stiffness of the T.) just hardened Stage (2) : Wet concrete of G/F just poured Stage (3) : G/F hardened and 1/F wet concrete just poured 1/F G/F G/F Stress/force in T. being {F1} + {F2} + {F3} + {F4} + {F5}.. becomes {Fn} + {Fn-1} + {Fn-2} + . + G/F + 1/F + 2/F Figure 9. being {F1} + {F2}.S.S.1 – Diagrammatic illustration of the Effects of Construction Sequence of loads induced on transfer structure 94 . due to weight of 1/F structure and and stiffness of the T.S.S. {F5} being force induced in T.S.S.. {F3} being force induced in transfer structure due to weight of 1/F structure and stiffness of the T.S. {F4} being force induced in T. being {F1} due to own weight of T. being {F1} + {F2} + {F3}. + {F2 } + {F1} Stress/force in T.S.S. being {F1} + {F2} + {F3} + {F4}..S. Stress/force in T. Lateral loads have to be analyzed separately.9. The vertical settlement support stiffness should take the length of the column/wall support down to level of adequate restraint against further settlement such as pile cap level. However.e. Reference can be made to Appendix H discussing the method of “Compounding” of vertical stiffness and the underlying assumption. Correction to allow for sidesway effects is necessary. It is therefore necessary to check such settlements. Walls which are constructed monolithically with the supporting transfer structures may help to stiffen up the transfer structures considerably. Four or five storeys of walls may be used for multi-storey buildings. 95 (ii) (iii) (iv) (v) . It should be noted that the conventional use of either 4 EI / L or 3EI / L have taken the basic assumption of no lateral movements at the transfer structure level. especially under unbalanced applied moments such as wind moment. i. Fuller discussion and means to assess such effects are incorporated in Appendix H. vertical loads and moments.3 Mathematical modeling of transfer structures as 2-D model (by SAFE) : The general comments in mathematical modeling of transfer structures as 2-D model to be analyzed by computer methods are listed : (i) The 2-D model can only examine effects due to out-of-plane loads. care should be taken to incorporate such stiffening effect in the mathematical modeling of the transfer structures which is usually done by adding a stiff beam in the mathematical model. It is a basic requirement that the transfer structure must be adequately stiff so that detrimental effects due to settlements of the columns and walls being supported on the transfer structure are tolerable. Furthermore. Effects of construction sequence may be taken into account in checking. It is not advisable to take the full height of the wall in the estimation of the stiffening effect if it is of many storeys as the stiffness can only be gradually built up in the storey by storey construction so that the full stiffness can only be effected in supporting the upper floors. loads induced in these stiffening structures (the stiff beams) have to be properly catered for which should be resisted by the wall forming the stiff beams. Care should be taken in assigning support stiffness to the transfer structures. 4 Modeling of the transfer structure as a 3-dimensional mathematical model can eliminate most of the shortcomings of 2-dimensional analysis discussed in section 9. If large differences occur. as most of these softwares may not have the sub-routines for detailed design. the designer should review his approach. as indicated by the differences in recovery of nodal forces in the 2-D model.9. The designer should check consistencies in reactions acting on the 2 models. Only the displacements of the nodes with external loads (applied loads and reactions) should be transported. However.2 – 3-D model to 2-D with transportation of nodal displacements 96 . as the finite element meshing of the 2-D model is usually finer than that of the 3-D one. However. including the effects of construction sequence if the software has provisions for such effects. A 2-D structure will be re-formulated in the 2-D software for re-analysis by which the structure is re-analyzed by forced displacements (the transported displacements) with recovery of the external loads (out-of-plane components only) and subsequently recovery of the internal forces in the structure. the designer may need to “transport” the 3-D model into the 2-D model for detailed design.3. there are differences incurred between the 2 models. External nodal force is {F3D} External nodal force is {F2D} ≠ {F3D} after re-analysis 3-D model (usually coarser meshing) with displacements at nodes with external loads marked with 2-D model (usually finer meshing) with nodal forces recovered by forced displacement analysis at nodes marked with Figure 9. Theoretically the results of the two models should be identical if the finite element meshing and the shape functions adopted in the 2 models are identical. For such “transportation”. revealing lesser loads on the structure. two approaches can be adopted : (i) Transport the structure with the calculated displacements by the 3-D software (after omission of the in-plane displacements) into the 2-D software for re-analysis and design. However.c. no effects due to this support will be incurred by this support because the support reactions should be zero as the transported loads from the 3-D model are in equilibrium. the sectional design and r. detailing The structural sectional design and r.5 Structural Sectional Design and r.c. Though not so common in Hong Kong. in the re-analysis of the 2-D structure. i. if simulation as strut-and-tie model is employed. Reference to Appendix D can be 97 . detailing of a transfer structure member should be in accordance with the structural element it simulates.(ii) Transport the out-of-plane components of the external loads (applied loads and reactions) acting on the 3-D model to the 2-D model for further analysis. Nevertheless. This type of transportation is simpler and more reliable as full recovery of loads acting on the structure is ensured. twisting moments and out-of-plane shears should be designed for.c. it should be designed and detailed as a beam if simulated as a beam and be designed and detailed as a plate structure if simulated as a plate structure. The out-of-plane components of all loads acting on the structure including reactions be transported 3-D model with external loads obtained by analysis 2-D model with out-of-plane components of external forces transported from 3-D model and re-analyzed with a fixed support at any point Figure 9. a fixed support has to be added on any point of the structure for analysis as without which the structure will be unstable. detailing should be accordingly be based on tie and strut forces so analyzed. As such.e. The commonest structural simulation of a transfer plate structure is as an assembly of plate bending elements analyzed by the finite element method.3 – 3-D model to 2-D with transportation of nodal displacements 9. the analytical results comprising bending. made for the principles and design approach of the plate bending elements. 98 . 7.5d lc2 2lc1 Plan Figure 10.2 – Distribution of reinforcing bars when lc > (3c/4 + 9d/4) 99 .1 of the Code allows a footing be analyzed as a “rigid footing” provided it is of sufficient rigidity with uniform or linearly varying pressures beneath. critical sections for design Footing under pure axial load creating uniform pressure beneath Footing under eccentric load creating linearly varying pressure beneath Figure 10.5d c 1. As suggested by the Code.1.5d 1. the Code sets the some rules in requiring concentration of reinforcing bars in areas with high concentration of stresses. the critical section for design is at column or wall face as marked in Figure 10. Re Figure 10. 6.1 Analysis and Design of Footing based on the assumption of rigid footing Cl.1 – Assumed Reaction Pressure on Rigid Footing As it is a usual practice of treating the rigid footing as a beam in the analysis of its internal forces.5d c 1.0 Footings 10.10.2 : area with 2/3 of the required reinforcements area with 2/3 of the required reinforcements lc is the greater of lc1 and lc2 d is the effective depth 1. structural depth 500 mm Concrete grade of footing : grade 35 400 D. Column: O.4 × 600 + 1.L.3) 2 × 80 = 160 kNm Factored load : Axial load 1.L. 300kNm L. Design data are as follows : Column Loads (for each): Axial Load: D.6 × 1200 = 9169.5 × 20 = 4800 kN. (Re Example 4.4 of the Code requires checking of shear be based on (i) section through the whole width of the footing (as a beam). 600 kN Moment D.2 Worked Example 10. 300 kNm L.1 (i) Loading Summary : D.1 Consider a raft footing under two column loads as shown in Figure 10. and (ii) local punching shear check as if it is a flat slab.5 in Section 4).L.5 × 2 ×1.5 × 24 = 108 kN L. 2400 kN L.7. Total 5178 kN Moment (bending upwards as shown in Figure 10.Cl.2.L.4 × 5178 + 1. Overburden Soil 2 × 2400 = 4.5 × 2. 80kNm for each column 400 1000 400 1000 1000 2500 1000 Plan Figure 10.6 ×160 = 1096 kNm (ii) The pressure beneath the footing is first worked out as : 100 . 270 kN 4.3 – Footing layout for Worked Example 10.L.L.L.3) 2 × 300 = 600 kNm Column 2 × 600 = 1200 kN. 10. 6.3. Moment (bending upwards as shown in Figure 10.0 × 0.5 m deep Footing dimensions : plan dimensions as shown.W.L. 80 kNm Overburden soil : 1.2 kN Moment 1. 13 kN/m2 400 Section for critical design 400 1091.5 2 By the formulae in Section 3 for Rigorous Stress Approach.87 kN/m2 3 4.5 +   × 0 . bd 2 1000 × 412.125 .5 × 2 9169.5 × 2 / 12 The pressures are indicated in Figure 10.At the upper end : At the lower end : At the critical section 9169.5 = 412. p 0 = 0.3(a) 1384.5 × 2 4 .5 2 2 3   2 (iv) Design for bending : Moment per m width is : 1809.5 × 0.66 = 402.1 (iii) At the critical section for design as marked in Figure 10.13 − 1091.8 = 1809. the total shear is due to the upward ground pressure minus the weight of the overburden soil which is  1384.5 d = 500 − 50 − 12.5 − 20 × 1.2 1096 × 0.47 kN/m2 Figure 10.8 × 4.8 × 4.87    × 0.655 %.66 kNm 2  1384.5 = 4348.2 6 × 1096 − = 653.47 kN/m2 2 4 .13 + 1091.15 ×10 6 = = 2.87 kN/m2 400 1000 2000 1000 653.5 × 2 9169.5d 101 .8 × × 4 .87  2 × 4 .5 × 2 4.13 kN/m2 2 4 .5 mm M 402. 4.8 kN 2   The total bending moment is (1091.15 kNm/m. two K= thirds of the reinforcements have to be distributed within a zone of 1.3(a).5)× 0.2 6 × 1096 + = 1384.3(a) – Bearing Pressure for Worked Example 10.2 + = 1091.5 × 2 4 .5 / 4 = 1228.87 − 20 ×1.363 . Ast = 2702 mm2/m As l c = 1250 > 3c / 4 + 9d / 4 = 3 × 400 / 4 + 9 × 412. Asv provided is 5.4 2 × 20 × 1.475 = 3275 mm2/m within the critical zone. So provide T25 – 150.475) = 2001 mm2/m.e. Total flexural reinforcements over the entire width is 2702 × 4.5 the Code. i. reinforcements per metre width is 12159 / 3 / (4. 2/3 of which in 1.1 Factored load by the column is 1.343 − 0.4 × 2400 + 1.475 m.343 N/mm2 > v c = 0. 1. Other than the critical zone.2375 . The first critical perimeter is at 1.3(b) – checking punching shear for Worked Example 10.6 × 600 = 4320 kN Weight of overburden soil is 1.Ws).5d×2=1637.5d from the column face. So shear reinforcement required is Asv b(v − v c ) 1000(2.6375 2 − 0.32 mm/m width 0.3 of 1000 × 412. Provide T25 – 225 (v) Design for Strip Shear : Shear per m width at the critical section is 4348.64 kN ( ) 102 .5 = 75.5 400+1.5 × 2 = 1.614) = = = 4. Critical perimeter for punching shear checking 400 400 1000 400 1000 400+1.5 Shear stress at the critical section is v= 966400 = 2. So 12159 × 2 / 3 / 2.5 − 2.87 f yv 0.5 = 12159 mm2.5 1000 Plan Figure 10.5 × 412.87 × 460 sv Use T12 – 150 (B.614 N/mm2 as per Table 6.8 = 966.5d×2=1.637.from the centre of column.4 kN 4.2375 × 2 = 2.02 sv (vi) Check punching shear by first locating the first critical perimeter for punching shear checking. one can raise a comment that the design has to some extent be duplicated as checking of bending has been carried out in the perpendicular direction.81 − 0.64 + 30.5 = = 1323 mm2 Asv ≥ 0.6375 2 = 2731.6375   stress is 2196.2 × 1. As unlike shear where enhancement can be adopted with “shear span” less than 2d or 1.Weight of concrete is 1. So shear reinforcements in form of links required.4 2 × 24 × 0.614)× 4 ×1637.614 N/mm2. Furthermore.4 2 = 12670 mm2. the “beam” should have a free length of beam stirrup width + depth to develop the torsion (as illustrated in Figure 3.44) of the Code.5 = 30.6375 2 − 0.5 × 4 × 412. However.81 N/mm2 1637.2 So shear reinforcements in (v) controls (vii) Checking of bending and shear in the direction parallel to the line joining the columns can be carried out similarly.2 × 0.5 × 2 ( ) Net load along the critical perimeter is 4320 + 75.06 kN   1694. 0.5M t Veff = Vt 1 +  Vt x sp  Punching shear  1. by (Ceqn 6.5d .87 f yv 0. To be on the conservative side.08 kN Effective shear force for punching shear stress checking is. (v − v c )ud (0. by (Ceqn 6.81 < 1.12 in Section 3) which is generally not possible for footing of considerable width.26 kN Upthrust by ground pressure is 9169.5 × 412. for full torsion to be developed for design in accordance with (Ceqn 6.40) of the Code :  1.5 × 548   = 1694. As 0.06 ×10 3 = 0.5 > v c = 0.6375 2 − 0. Total shear reinforcement is. shear arising due to this torsion should be checked and designed accordingly as a beam as necessary. it should be noted that there is a net “torsion acting on any section perpendicular to the line joining the two columns due to linearly varying ground pressure.68) of the Code. Nevertheless.6v c = 0. full reduction by v c can be carried out.98 N/mm2.87 × 460 Comparing with the shear reinforcements obtained in (v) by which the total reinforcements within the perimeter is 201 × 1.08 ×1.82 kN 4. 103 ( ) .26 − 2731.081 +   = 2196.82 = 1694.65) to (Ceqn 6. though by the same phenomenon there should be some shear strength enhancement. flexible footings are often modeled as plate bending 104 .5 400 400 T25 – 225(B1) 400 T25 – 150(B1) Plan Shear links T12 – 150 BWs in the whole footing Figure 10.no similar strength enhancement is allowed in Code. Though it is comparatively easy to model the cap structure. it is difficult to model the surface supports provided by the ground because : (i) the stiffness of the ground with respect to the hardness of the subgrade and geometry of the footing are difficult to assess. we are currently lacking computer softwares to solve the problem. (viii) The flexural and shear reinforcements provisions for the direction perpendicular to the line joining the columns is 1237. However.1 (in the direction perpendicular to the line joining the two columns only) 10. the analysis of footing under the assumption of its being a flexible structure will also take the stiffness of the structure into account by which the deformations of the structure itself will be analyzed.5 1237. The deformations will affect the distribution of the internal forces of the structure and the reactions which are generally significantly different from that by rigid footing analysis. As the out-of-plane deformations and forces are most important in footing analysis and design. So full design for bending in both ways together with torsion will likely result in over-design.3(c) – Reinforcement Details for Worked Example 10. (ii) the supports are interacting with one another instead of being independent “Winkler springs” supports.3 Flexible Footing Analysis and Design As contrast to the footing analyzed under the rigid footing assumption. it would be reasonable to “spread” the stresses over certain width for design. care must be taken of not taking widths too wide for “spreading” as local effects may not be well captured. The design at any location in the footing can be based the calculated stresses directly. twisting moments and out-of-plane shears for consideration in design. Nevertheless.4 – Spreading of peak stress over certain width for design (iii) The design against flexure should be done by the “Wood Armer 105 .elements analyzed by the finite element method discussed in 10.4 Analysis and Design by Computer Method The followings are highlighted for design of footing modeled as 2-D model (idealized as assembly of plate bending elements) on continuous surface supports: The analytical results comprise bending. However.4 in more details. (i) peak stress width over which the peak stress is designed for Figure 10. if “peak stresses” (high stresses dropping off rapidly within short distance) occur at certain locations which are often results of finite element analysis at points with heavy loads or point supports. the rules governing distribution of reinforcements in footing analyzed as a beam need not be applied. 10. (ii) As local “stresses” within the footing are revealed in details. The checking of shear and design of shear reinforcements can be based directly on the shear stresses revealed by the finite element analysis. it is a practice of “lumping” the design reinforcements of a number of nodes over certain widths and evenly distributing the total reinforcements over the widths. It should be noted that as the finite element analysis give detailed distribution of shear stresses on the structure. as illustrated in Figure D5a to D5c. 106 . together with discussion of its underlying principle. as is done by the popular software “SAFE”. The principle together with a worked example for design against shear is included in Appendix D. care must be taken of not taking widths too wide for “lumping” as local effects may not be well captured. The design of reinforcements by SAFE is illustrated on the right portion of Figure 10. Again.3. As the finite element mesh of the mathematical model is often very fine. it is not necessary to calculate shear stress as done for flat slab under empirical analysis in accordance with the Code.(iv) Equations” listed in Appendix D. 1 – Pile load profile under rigid cap assumption Upon solution of the pile loads. the reactions of the piles follow a linearly varying pattern. Appendix I contains derivation of formulae for solution of pile loads under rigid cap assumption. The conventional assumption is to consider the cap as a beam structure spanning in two directions and perform analysis and design separately. 107 . 6. 6.7.7. pile loads.1 Rigid Cap analysis Cl. As the designer can only obtain a total moment and shear force in any section of full cap width. If it is assumed that the piles are identical.11. magnitude follows linear profile under assumption of equal pile stiffness Figure 11. Cl.7. As the deformations of the piles are governed.5 of the Code requires checking of torsion based on rigid body theory.3. The Code (Cl.3.3) therefore imposes a condition that shear enhancement due to closeness of applied load and support cannot be applied.3 of the Code allows a pile cap be analyzed and designed as a “rigid cap” by which the cap is considered as a perfectly rigid structure so that the supporting piles deform in a planar structure.0 Pile Caps 11. the internal forces of the pile cap structure can be obtained with the applied loads and reactions acting on it. 6. the reactions by the piles can be found. there may be under-design against heavy local effects in areas having heavy point loads or pile reactions. It is also a requirement under certain circumstances that some net torsions acting on the cap structure (being idealized as a beam) need be checked. 108 .11. of Cap Weight of overburden soil 11 × 9 × 2 × 24 = 4752 kN 11× 9 × 1.W. of Cap and soil is 1.1 (Rigid Cap Design) The small cap as shown in Figure 11.2 is analyzed by the rigid cap assumption and will then undergo conventional designed as beams spanning in two directions.2 – Pile cap layout of Worked Example 11.4(4752 + 2970 ) = 10811 kN So total axial load is 50000 + 10811 = 60811 kN.5 × 20 = 2970 kN Factored load due to O.2 Worked Example 11.W. Design data : Pile cap plan dimensions : as shown Pile cap structural depth : 2 m Pile diameter : 2 m Column dimension : 2 m square Factored Load from the central column : P = 50000 kN M x = 2000 kNm (along X-axis) M y = 1000 kNm (along Y-axis) 1500 3000 P1 P2 P3 Y 3000 P4 P5 P6 X 400 critical sections for shear checking 1500 1500 1500 4000 4000 Figure 11.1 (i) Loads : from Columns : P = 50000 kN M x = 2000 kNm (along X-axis) M y = 1000 kNm (along Y-axis) O. and soil is 109 .33 kN Downward shear by cap’s O.65 kN 6 32 54 60811 2000 × 0 1000 × 3 − − = 10116.69 kN 6 32 54 60811 2000 × 4 1000 × 3 − − = 9866. 6. b = 9000 M 76423.65)× 4 = 83081.69 kN 6 32 54 60811 2000 × 0 1000 × 3 − + = 10153.2 Total shear at the critical section is : Upward shear by P3 and P6 is 10403. B1 and B3) (iv) Design for shear in the X-direction By Cl.69 + 10366. requiring T40 – 200 (2 layers.58 kNm d = 2000 − 75 − 60 = 1865 .65 = 20770. the critical section for shear checking is at 20% of the diameter of the pile inside the face of the pile as shown in Figure 11.65 kN 6 32 54 (iii) Design for bending in the X-direction The most critical section is at the centre line of the cap Moment created by Piles P3 and P6 is (10403.69 + 10366.2 of the Code.33 − 10617.75 = 10617.3.69 kN 6 32 54 60811 2000 × 4 1000 × 3 + + = 10403.65 kN 6 32 54 60811 2000 × 4 1000 × 3 + − = 10366.33 kNm Counter moment by O. of cap and soil is (4752 + 2970) ÷ 2 × 2.75 = 76423.75 kNm The net moment acting on the section is 83081.441 .W.668% bd 2 9000 × 1865 2 Ast = 112070 mm2.W.7.(ii) Analysis of pile loads – assume all piles are identical (Reference to Appendix I for analysis formulae) I x of pile group = 6 × 3 2 = 54 I y of pile group = 2 × 4 2 + 2 × 0 = 32 Pile Loads on P1 : P2: P3: P4: P5: P6: 60811 2000 × 4 1000 × 3 − + = 9903. p = 0.58 × 10 6 K= = = 2. 1 = 1801.8 kN 9 110 .17 N/mm2 > v c = 0.87 × 460 sv Use T12 links – 200 in X-direction and 400 in Y-direction by which Asv provided is 1.25 = 8687.13 × 103 v= = 1. and soil is (4752 + 2970)× 2.(4752 + 2970)× 2. B2 and B4) (vi) Checking for shear in the Y-direction By Cl.06 kN Downward shear by cap’s O.92 kNm d = 2000 − 75 − 60 − 40 = 1825 . No shear enhancement can be effected.257 .W.13 kN 19299.3.69 + 10403.2 = 19299.69)× 3 = 30461. P2 and P3 is 30461.2 of the Code.17 kNm Counter moment by O. P2 and P3 (9903.613% bd 2 9000 × 1825 2 Ast = 122422 mm2.06 × 3 = 91383.7. 6. b = 11000 K= M 82695.17 − 0.41.618) = = = 1.87 f yv 0. sv (v) Design for bending in the Y-direction The most critical section is at the centre line of the cap Moment created by Piles P1.33 − 1474.69 + 10153.379 0.92 × 10 6 = = 2. of cap and soil is (4752 + 2970) ÷ 2 × 2. requiring T40 – 200 (2 layers.1 = 1474.25 = 82695.W. the critical section for shear checking is at 20% of the diameter of the pile inside the face of the pile as shown in Figure 11.618 N/mm2 by Table 6.2 kN 11 Net shear on the critical section is 20770.3 of 9000 × 1865 the Code. p = 0. Shear reinforcements in form of links per metre width is Asv b(v − v c ) 1000(1.25 kNm The net moment acting on the section is 91383.17 − 8687.2 Total shear at the critical section is : Upward shear by P1. it may not be necessary to design the torsion as for that for beams.3 11000 ×1825 of the Code.601 N/mm2 by Table 6.87 f yv 0.097 N/mm2.12 kNm (537.04 is the difference in pile loads between P3 and P4). the net torsion is this example is small.82. 2T (ix) Reinforcement details are as shown in Figure 11. sv for shear reinforcement design Use T12 links – 200 BWs by which (vii) Punching shear : As the critical perimeter for punching shear checking of the piles and column overlap with each other. so adopt this sv Asv provided is 2. creating torsional shear stress in the order of 2 × 1611.12 × 10 6 vt = = = 0. punching shear cannot be effected. (viii) Checking for torsion : There are unbalanced torsions in any full width sections at X-Y directions due to differences in the pile reactions. However. as discussed in sub-section 10.26 × 10 3 v= = 1.Net shear on the critical section is 30461.2(vii) of this Manual for footing. h  2000   2 hmin  hmax − min  2000 2  9000 −  3  3    the torsional shear effects should be negligible.87 × 460 sv As Asv in Y-direction is greater than that in X-direction. No shear enhancement can be effected.04 × 3 = 1611.26 kN 28259. being 537. So shear checking as a beam in (iv) and (vi) is adequate.417 − 0.417 N/mm2 < v c = 0.06 − 1801.601) = = = 2.8 = 28259. Anyhow.038 0. Shear reinforcements in form of links per metre width is Asv b(v − v c ) 1000(1.3 So 111 . a strut is a compression member whose strength is provided by concrete compression and a tie is a tension member whose strength is provided by added reinforcements.B1. The Code has specified the following requirements : (i) Truss be of triangular shape.3. (ii) Nodes be at centre of loads and reinforcements. In the analysis of a S&T model.7. or more commonly known as “Strut-and-Tie Model” (S&T Model) in which a concrete structure is divided into a series of struts and ties which are beam-like members along which the stress are anticipated to follow. (iii) For widely spaced piles (pile spacing exceeding 3 times the pile diameter).1500 3000 P1 P2 T40-200 (T1. (iii) Strut members cannot cross each other while a tie member can cross another tie member. In a S&T model.. 6.3 Strut-and-Tie Model Cl. only the reinforcements within 1. the following basic requirements must be met (Re ACI Code 2002): (i) Equilibrium must be achieved.B4) P4 P5 X P6 Shear links T12 – 200 BWs on the whole cap 1500 4000 4000 1500 Figure 11.1 of the Code allows pile cap be designed by the truss analogy. (iv) The smallest angle between a tie and a strut joined at a node should not exceed 25o.3 – Reinforcement Design of Worked Example 11. B2.1 11. (ii) The strength of a strut or a tie member must exceed the stress induced on it.B3) P3 Y 3000 1500 T40-200 (T2.5 times the pile diameter 112 . 2 (i) Determine the dimension of the strut-and-tie model Assume two layers of steel at the bottom of the cap. (ii) A simple force polygon is drawn and the compression in the strut can be simply worked out as 2C sin 36.62 0 = 6000 ⇒ C = 5028.2 (Strut-and-Tie Model) Consider the pile cap supporting a column load of 6000kN supported by two piles with a column of size 1m by 1 m. 113 .4.5 m. 6000kN 2500 Elevation 3000 1000 dia. The dimension of the cap is as shown in Figure 11.75 kN. 3000 1000 dia. with the thickness of cap 1.5.4 – Pile Cap Layout of Worked Example 11. 1500 Plan Figure 11.4 Worked Example 11. the centroid of both layers is at 75 + 40 + 20 = 135 mm from the base of the cap. So the effective width of the tension tie is 135 × 2 = 270 mm. 11.from the centre of pile can be considered to constitute a tension member of the truss. The tie and strut dimension is drawn in Figure 11. the reinforcement steel required is (iv) 4035.62 0 = 4035.62o 36. the strut width at bottom is 1000 sin 36.62 0 = 813.26 mm As the bottom part is in tension. strength be provided by steel 3000 3000 Elevation 6000 kN 1000 dia. there is a reduction of compressive strength of concrete to 1.87 ×10 3 = = 10085 mm2.87 kN.2 (iii) To provide the bottom tension of 4035. 2230 36.87 kN.And the tension in the bottom tie is T = C cos 36.8 35 = 10. 0. 1000 top strut width = 596. Use 9-T40.57mm 6000kN concrete strut 2230 2500 270 Bottom strut width = 813.87 f y 0.08 MPa as suggested 114 . bottom tie.8 f cu = 1.87 ×10 3 4035.26mm 1000 dia.87 × 460 Check stresses in the struts : Bottom section of the strut.62o 4000 kN 3000 3000 4000 kN Figure 11.4 – Analysis of strut and tie forces in Worked Example 11.62 0 + 270 cos 36. 62 0 = 596. Figure 11. – 200 2500 5T40 B1 & 4T40 B2 3000 1000 dia.5 Flexible Cap Analysis A pile cap can be analyzed by treating it as a flexible structure..8 f cu as stated in BS8110.s. the flexible pile cap structure will deform and the 115 . As a conservative approach.45 f cu = 15. it is conservative to assume a sectional area of 1000mm × 596. 3000 Elevation 1000 dia.69 MPa < 10.2 11.57 mm.57 1000 5T32 T1 T16 s.75 MPa 1000 × 59. assuming a circular section at the base of the strut since the pile is circular.by OAP.5 – Reinforcement Details of Worked Example 11. which is an implied value of the ultimate concrete shear strength of 0.75 × 10 3 is = 9. the sectional width is 2 × 500 sin 36. The compressive stress of the strut at top section is 5028.439 MPa < 0. i.08MPa 813 2 π / 4 For the top section of the strut.e.75 × 10 3 = 8. as in contrast to the rigid cap assumption in which the cap is a perfectly rigid body undergoing rigid body movement only with no deformation upon the application of loads. the stress at the base of the strut 5028.57 mm As the sectional length of the column is 1 m. 6 – Effective width for shear enhancement in pile cap around a pile (ii) averaging of shear force shall not be based on a width > the effective depth on either side of the centre of a pile. 116 . Cl. Effects by soil restraints on the piles can be considered as less significant in end-bearing piles such large diameter bored piles.deformations will affect the distribution of internal forces of the structure and the reactions. as analysis by computer methods can often account for load distribution within the pile cap structure. most of the flexible cap structures are modeled as plate structures and analyzed by the finite element method. Analysis of the flexible cap structure will require input of the stiffness of the structure which is comparatively easy. 11. Similar to the flexible footing. or pile width plus 2 × least dimension of pile. or as limited by the actual dimension of the cap.7. as the out-of-plane loads and deformation are most important in pile cap structures. 6. especially for the friction pile which will interact significantly with each other through the embedding soil. as similar to that of footing.3. Area where shear enhancement may apply φ φ av av av φ B B B av Figure 11.6 Analysis and Design by Computer Method Analysis and design by computer method for pile cap are similar to Section 10. Nevertheless. the stiffness of the pile cap itself which is mainly offered by the supporting pile is often difficult.3 for footing.3 of the Code has specified the followings which are particularly applicable for pile cap in respect of design : (i) shear enhancement may be applied to a width of 3φ for circular pile. However. 4(ii) of this Manual.7 – Width in cap over which shear force at pile can be averaged for Design Figure 11.7 can be a guideline for “effective widths” adopted in averaging “peak stresses” as will often be encountered in finite element analysis for pile cap structure modeled as an assembly of plate bending elements under point loads and point supports.φ φ Width over which the shear force can be averaged in the cap for design ≤φ peak shear at pile centre Shear force diagram Figure 11. 117 . as in the same manner as that for footing as discussed in 10. φ is the bar diameter.89 40   1 + 2 ×  100   . If the above requirement is not fulfilled. (ii) avoid overstress by bearing on concrete in the bend.2 12. 12.8 > rφ 120 × 40  φ  1 + 2 a b  So the (Ceqn 8.12.1 In this section. (iii) 20 mm. Permissible bent radii of bars.1) is not fulfilled. Fbt = 0.3 Fbt 2 f cu ≤ rφ  φ 1 + 2  ab      where Fbt is the tensile force in the bar.87 × 460 × 1257 = 503051 N 2 f cu Fbt 503051 = = 104. Take an example of a layer of T40 bars of centre to centre separation of 100 mm and internal bend radii of 120mm in grade 35 concrete. The purpose of requiring minimum bend radii for bars are (i) avoid damage of bar.0 General Detailings 12. Requirements marked with (D) are ductility requirements. They are mainly taken from Section 8 of the Code. (ii) maximum aggregate size (hagg) + 5 mm. bearing pressure inside the bend should be checked so that (Ceqn 8. Minimum spacing of reinforcements – clear distance is the greatest of (i) maximum bar diameter. Generally.1) listed as below is fulfilled 12. the provisions of detailing requirements are general ones applicable to all types of structural members.4 Anchorage of longitudinal reinforcements 118     = 2 × 35 = 38. a b is centre to centre distance between bars perpendicular to the plane of the bend and in case the bars are adjacent to the face of the member. the minimum bend radii are 3φ for φ ≤ 20 mm and 4φ for φ > 20 mm. r the internal bend radius of the bar. a b = φ + cover. 87 f y   4    . columns and walls.3) of the Code. In addition to (i).8φ ≈ 34φ which agrees (ii) with Table 8.7 of the Code. resulting in increase of bond and anchorage lengths for beam should be increased by 1 / 0.1.2 of the Code displays bend of links of bend angles from 90o to 180o.5 of the Code.1) of the Code if anchorage of bar requires a length more than 4φ beyond the bend Code (Re Figure 8. However. 12.8 of the Code. For example.87 f y   4  =β   f cu πφL ⇒ L = 0.1 of the Code) (iv) With the minimum support width requirements as stated in Cl.4. f bu = β f cu where for high yield bars β = 0.5 Anchorage of links – Figure 8.” The steel bars should therefore be carrying stress of f y instead of 0. bends of bars can start beyond the centre line of supports offered by beams.96 MPa for grade 35.2. 9. (D) 119 . The required bond length L will then be related by    φ 2π  0.1. f bu = 0.87 f y  φ 2π is 0.(i) Anchorage is derived from ultimate anchorage bond stress with concrete assessed by the (Ceqn 8. (D) (iii) Bearing stress on concrete by bend must be checked by (Ceqn 8.5 35 = 2. the total force up to 0. it should be noted that the Code requires anchorage links in beams and columns to have bend angles not more than 135o as ductility requirements.9.87 − 1 = 15% . By the same clause the requirement can be considered as not confining to simply supported beam as stated in Cl.1(c) of the Code which contains ductility requirements for beams that “For the calculation of anchorage length the bars must be assumed to be fully stressed.5 for tension and β = 0.87 f y φ 4β f cu = 33. 8.87 f y in determination of anchorage (and bond) lengths. For a bar of diameter φ . it has been stated in 9.65 for compression. 12.4 of the Code reproduced in Figure 12.7 Lap Length – the followings should be noted : (i) Absolute minimum lap length is the greater of 15φ and 300 mm.6 Laps – the followings should be noted (i) (ii) If sum of reinforcement sizes in a particular layer exceeds 40% of the breadth of the section at that level.1 for ease of reference. Details of requirements in bar lapping are indicated in Figure 8.3l0 l0 ≤50mm ≤4ø ≥20mm >4ø Figure 12.5 of the Code. the laps must be staggered. Tension lap length should be at least equal to the design tension anchorage length and be based on the diameter of the smaller bar. (ii) 120 . ≥0.0 as indicated in Figure 8.4 or 2. 12.1 – Lapping arrangement for tension laps (iii) Compression and secondary reinforcements can be lapped in one section. (iii) Lap length be increased by a factor of 1. Appendix A Clause by Clause Comparison between “Code of Practice for Structural Use of Concrete 2004” and BS8110 . 2.0 for the beneficial case. The clause in CoPConcrete2004 affirms engineers to ignore 2. In addition Pt. 2 3. Clause No. 1. Pt.4.5 – Design working life – Nil No similar statement in BS8110.1 of BS8110 except that the partial factor for load due to earth and water is 1.2 stipulates a limit on inter-storey drift of Storey height/500 for avoidance of damage to non-structural elements. BS8110 does not apply to high strength concrete.2.3 & 2.2.4 – The clauses explicitly state that these effects need only be considered when they are significant for – No similar clauses in BS8110 A-1 .0 in CoPConc2004 may not be adequately conservative as there may even be over-estimation in the determination of the unfactored soil load.3 – Response to wind loads Reference to specialist literature is required. precast concrete (under the separate code for precast concrete). In addition. 2.2 for the usual limits of H/500 to lateral deflection at top of the building and accelerations of 1-in-10 year return period of 10 minutes duration of 0.2.1 is generally reproduced from Table 2.25m/sec2 for office. Where design working life differs from 50 years. 1 2.1. 1 2. the recommendations should be modified. shells. though the draft steel code has a requirement of storey height/400.2. lightweight aggregate concrete etc.3 – Response to wind loads Pt. CoPConc2004 is more specific.2. 1 1. Table 2. with the exclusion of (i) no fines.3. 2. However.3.3.1 – Scope HK CoP Structural Use of Concrete 2004 Contents The clause has explicitly stated that the Code applies only to normal weight concrete.1 – Scope BS8110:1997 Contents The clause only explicitly excludes bridge structures and structural concrete of high alumina cement. However.3. 2.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. The clause states that the Code assumes a design working life of 50 years.1. 1.3. ICU has raised this comment during the comment stage when the draft Code has exactly the content of BS8110. no distinction should be made between adverse and beneficial loads. (ii) bridge and associated structures.15m/sec2 for residential and 0.3.3. and (iii) particular aspects of special types of structures such as membranes.2 – Partial factors for earth pressures It is stated in the clause that when applying the load factor.1 – Loads for ultimate limit state Pt. aerated. Appendix A Remark The exclusion of CoPConc2004 should also be applied to BS8110.2. It is even a practice to set the load to zero in beneficial case. method for determination of the acceleration is not given in the Code and in the HKWC-2004. The clause refers to clause 7. there is no requirement on the inter-storey drift. 3.95fy.87fy even if BS8110 was used before the promulgation of the new concrete code The coverage of CoPConc2004 is wider.4.87fy. Table 2.1.1 BS 8110 has included temperature effects be a minor one that can be ignored in calculation.1 – Values of γm for ULS Table 2. Nevertheless. Pt. 1 Figure 2. Differential settlement of foundations.15. concrete grades covered by the design charts in Part 3 of the Code range from grade 25 to 50 whilst other provisions such as vc (Pt. Contents Appendix A Remark consideration of these effects in normal cases which are the usual practices.5 – Elastic deformation Table 3. shrinkage.1. The determination of short term static Young’s Modulus of concrete is given by the slope gradient in the figure which is 5. and no specific calculation are required.2 gives γm for ULS for concrete and re-bars. γm for re-bars is 1. 1 Table 3. Values in the CoPConc2004 should be used as it is based on local research and concrete E values are affected by constituents which are of local materials. Pt.1 states concrete strength grades from 20 MPa up to 100 MPa which is the range covered by the Code.4.27) are up to grade 40.2 gives γm for ULS for concrete and re-bars. The clause states that “For the ULS. 3.2 – Values of γm for ULS HK CoP Structural Use of Concrete 2004 Contents ULS.3.21 in MPa derived from local research.46√fcu+3.4. BS8110:1997 Clause No. 3. creep. γm for re-bars is 1. these effects will usually be minor and no specific calculations will be necessary.2 stipulates short term static Young’s Modulus of concrete of various grades based on the formula 3. However.3 – creep. shrinkage and temperature effects Pt. temperature effects 2.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. lap lengths (Pt. creep and shrinkage are minor. implying strength of re-bars for design remain as 0.05. 1 Table 3. BD has been insisting on the use of 0. 1 2. 3. – BS8110 has not explicitly stated the concrete grades covered by the BS.5√(fcu/γm). In most other cases they need not be considered provided ductility and rotational capacity of the structure sufficient.4.8). 1 2. it should be noted that E values in the new Code are slightly A-2 .3 – Strength grades Table 3.4 – Deformation of Concrete It is stated in the 1st paragraph of the clause that for ULS.1. implying strength of re-bars for design remain as 0. 5. The short term design stress-strain curve of concrete (Fig.4. 2. 1 Fig.1. 3. 2.3. 3.1. The approach is identical to the “Structural Design Manual” issued by Highways Department and the charts are extracted from BS5400:Pt 4:1990. though there are differences in the values of the Young’s Moduli. the ultimate strain is limited to below 0.3.1. 3. as stated in 2. 1 2.4. together with the decrease of ultimate strain for grade above C60.9 – Thermal Expansion – No account has been given for temperature.4. effects need only be considered if they are significant.4.7 & 3. A-3 .3. – No account has been given for creep and shrinkage. Furthermore.0035 for concrete grade exceeding C60.1). as stated in 2.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No.3. HK CoP Structural Use of Concrete 2004 Contents BS8110:1997 Clause No.3.3 – Analysis of section for ULS Pt. these clauses contain detailed formulae and charts for prediction of creep and shrinkage strain.8 – Creep and shrinkage Though it is stated in 3.4. (i) The Young’s Moduli of concrete stipulated in CoPConc2004 should be adopted as they are based on local data and cover up to grade 100 concrete. (ii) “Smooth” connection between the parabolic curve and the straight line portion of the stress-strain curve cannot be effected if ε0 = 2. The linear coefficient of thermal expansion given in the Code for normal weight concrete is 10×10-6/oC whilst that stated in the “Structural Design Manual” issued by Highways Department is 9×10-6/oC in Clause 2. Contents Appendix A Remark higher than the previous ones in “The Structural Use of Concrete – 1987” (Table 2.2.1 shows stress-strain relation of normal weight concrete. the “plastic strain”. The linear coefficient of thermal expansion given by both CoPConc2004 is slightly higher than SDM and both are independent of concrete grades. Nevertheless. As stated in the CoPConc2004 Cl.10 – Stress -strain relationship for design Pt.1.4×10-4√fcu/γm is kept 3.1. strain beyond which stress is constant remains identical as BS8110.8) follows closely the traditional one in BS8110. to account for the brittleness of high strength concrete.4 that for ULS creep and shrinkage are minor and require no specific calculations. Further provisions for welding are given in 10.3.3.2 – Load cases and Pt.3.4 of BS8110.2 – Design for Durability Pt. 1 3.4.1. Pt. Contents Appendix A Remark and the Young’s moduli in Table 3. detailed descriptions are different except the last one “abrasive”. 1 7. Pt. CoPConc2004 is more related to local practice. 1 2. 1 3. Pt.2. the user has to refer to the current fire code for additional requirements against fire resistance. though not truly adequate. Pt. 2 Section 4 The approaches of the two codes are quite different.2 states that when cement content > 400 kg/m3 and section thicker than 600 mm. The followings are highlighted :(i) In 4.5.2 2 simplified load cases are considered sufficient for design of spans and beams (with DL always present): (i) all spans loaded with LL. However. outlining methods of determining fire resistance of structural elements and with reference to BS476 Pt.3 – Concrete cover to rebars Pt. 1 3. Nevertheless.4 – Durability.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. the followings are highlighted : (i) 3. However.12. 5.5.1.2. it is stated that requirements are based on design working life of 50 years. Section 4 – Durability and fire resistance The requirements are general.16. (iii) In 4.2. Steel complying CS2 is likely weldable as CS2 does not differ significantly from BS4449. measures for temperature control should be implemented.16 7. exposure conditions 1 to 5 are classified with headings similar to that in Pt. 1 3. the current A-4 .2 for various concrete grades and exposure conditions. The requirements are general. and general welding requirements in Pt.7 – Weldability (of re-bars) The clause states that re-bars can be welded provided the types of steel have the required welding properties given in acceptable standards and under Approval and inspection by competent person. 3. (ii) A full description for fire resistance control is given in Pt.3. Nevertheless.6.67fcu/γm.2. For smooth connection.7. (ii) alternate spans CoPConc2004 more reasonable. control of AAR has been incorporated from the previous PNAP 180.6 – Fire resistance. However.1.1.3.6.1. 1 2. 1 3.2 of CoPConc2004 are used.2. (ii) alternate spans loaded with LL. HK CoP Structural Use of Concrete 2004 Contents BS8110:1997 Clause No. another BS7123 titled “Metal arc welding of steel for concrete reinforcement” requires the re-bars be in compliance with BS4449 or BS4482. 2 Section 4. ε0 should be revised 2σult/E where σult=0. 1 3. (iv) Concrete covers are given in Table 4.12.8.2. (v) By 4. 8.1.8.6 There are provisions for lapping re-bars by welding in Pt.2.1 under 4. 3 simplified load cases for Dead + Live loads are recommended (with DL always present): (i) all spans loaded with LL. (ii) In Table 4.2. shears. It is also stated that the effective flange can be adopted in structural analysis.and Clause No. one-way or two-ways). sagging moment and the 3rd case is likely to seek for max. (iii) Clearer definition of effective spans of beams is also included in 5.1. 5. A-5 . column in accordance with their geometries. slab (included ribbed. waffle.1. the effective span of a simply supported beam. But true max hogging moment should also include alternate spans from the support being loaded.2. (ii) By definition.2 – Analysis of Structures 3.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. (iii) Method of analysis in both codes are old-fashioned ones that can be performed by hand calculations. combinations for beams and slabs HK CoP Structural Use of Concrete 2004 Contents (iii) adjacent spans loaded with LL. The following differences with BS8110 are highlighted : (i) Conditions in relations to rib spacings and flange depths for analysis of a ribbed or waffle slab as an integral structural unit are given in 5.2(b). transfer structures are not given. (ii) Consideration for high-rise buildings (second order effects) and structures such as shear walls. (iii) The effective span of a simply supported beam is the clear span plus the lesser of half effective depth and half of support width whilst that of continuous beam and cantilever are span between supports (presumably mid-support width) except at end span in case of continuous beam and a cantilever forming the end of a continuous beam.2 (a) are slightly greater than that in BS8110 by 0. The clause contains definition of beam. Appendix A Remark softwares mostly can account for the load case to search for max.2 – Analysis of Structure 3. together with reduction of span moments due to support width in 5.4 – Beams The followings are highlighted : (i) Generally provisions are applicable to normal strength concrete.1.2. (v) The definition of effective flange for T.2(a) with illustration by diagrams.2.2. (ii) Extended use of effective flange in beam in structural analysis and moment reduced to support shear are explicitly stated in CoPConc2004. hogging moment. hogging moment at support. transfer structures. high rise buildings. The following remarks are made : (i) CoPConc2004 is more suitable for use in Hong Kong as it cover high strengths concrete.1(d). (iv) Furthermore reduction in support moments due to support width is allowed which is not mentioned in BS8110.1. the effective flange widths in T and L-beams included in 5. BS8110:1997 Contents loaded with LL. the effective span is the clear span plus mid-support width should be used. In principle. Use of computer methods (extensively adopted currently in Hong Kong) are not mentioned. continuous beam or cantilever is the clear span plus the minimum of half effective depth and half support width except that on bearing where the span should be up to the centre of the bearing. The 2nd case is to seek for max.1×clear rib spacing unless for narrow flange width controlled by rib spacing. More stringent limit on neutral axis is added for concrete grade higher than C40 as per 6. The followings are highlighted : (i) Provisions are made in CoPConc2004 for concrete grades higher than 40 including limitation of neutral axis depths etc.4.4 which if redistribution of moment exceed 10%. 1 3. 3.4 – Beams In the design aspects. (ii) Neutral axis depths limited to less than 0.1. HK CoP Structural Use of Concrete 2004 Contents L-beams are slightly different from BS8110 though the upper bound values are identical to that of BS8110. (vi) Moment redistribution is restricted to grade below C70 as per 5.2.0035 for concrete grade exceeding C60 as per Fig. as different from BS8110 which makes not differences among concrete grades in Cl 3. 6.2.9.5d for concrete grade higher than C40 as per 6.5 though the provisions are too general.1 relation to checking of neutral axis depths of beam in adopting moment re-distribution is different from Cl.4(b). (viii) Provision for second order effects with axial loads in 5.2.1.1.4(b). (iii) The limitations on neutral axis depth ratio under redistribution of moments are further restricted for grade > 40.2. 70<fcu≤100) concrete BS8110:1997 Clause No. the Code is limited to grade 40.1 – Members in Flexure Pt.2. it is also noted that the ultimate stress in Fig. A-6 .4.1. (ix) Provisions for shears walls and transfer structures are added in 5.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No.1 remains unchanged for high concrete grade though the ultimate strain is reduced. So more stringent. (ii) Ultimate concrete shear strengths increased in CoPConc2004. (vii) Condition 2 in 5. (iv) Different design formulae are used for the higher grades (40<fcu≤70. Contents Appendix A Remark 6. The provisions are quite general except the statement “second order effects can be ignored if they are less than 10%”.2.4 and 5.3 is added. 6.1 b) of BS8110. However.9. The followings differences with BS8110 are identified : (i) Ultimate strains are reduced below 0. 2. the required shear strength is increased by factor (fcu/40)2/3 as per Table 6. Contents Appendix A Remark (iii) Due to the more stringent limitation on neutral axis depth for high grade concrete in CoPConc2004. Design charts similar to that BS8110 based on stress-strain relationship in figure 3. (iv) 6.2.2.8√fcu is identical to BS8110.3. for both flexure and shear.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No.4 MPa is for concrete grade below C40 (requirement by BS8110). The minimum amount of shear reinforcements required is reduced to that can provide shear resistance of 0.87fy in accordance with BS8110:1985 instead of 0. BS8110:1997 Clause No. ribbed slabs (6. similar formulae are not given for flanged beams. (vi) Partial strength factor for steel remains as 0. 1 Table 3. the minimum shear reinforcement to cater for shear strength of 0. BS8110 does not explicitly require equal structural properties in mutually perpendicular directions in adopting design method as flat slab (BS8110.95fy as in accordance with BS8110:1997. (viii) By 6.1.1.6.1.8 are not available.1. The other limitation of 0.4(c). different formulae have been devised for x > 0. A-7 .5.4.1.5d for rectangular beams.4.4. (vii) Ultimate shear strength of concrete increased to 7. Above grade C40. 1 3. HK CoP Structural Use of Concrete 2004 Contents based on simplified stress blocks as per 6.8 of BS8110 except that applicability is extended to C80 whereas BS8110 limits to grade 40.4 MPa. However.2 in relation to vc of concrete as related to tensile reinforcements is identical to Pt. it is ICU’s comment to qualify that if ribbed slabs are to be designed as two-ways spanning as similar to a flat slab. it should be qualified that the slab has equal structural properties in two mutually perpendicular directions.0 MPa as compared 5. (ix) By 6. (v) Table 6. However.2.2 of CoPConc2004 is similar to BS8110 Pt.4(fcu/40)2/3 MPa for grade over 40 whilst that for concrete at and below grade 40 remains to that can provide 0.2) can be designed as two-ways spanning as similar to flat slab if they have equal structural properties in two mutually perpendicular directions.0 MPa in BS8110. 2 2. (ii) A more tedious approach for determination of effective height of column by consideration stiffness of the connecting beams and columns is outlined in Pt.4. Strictly speaking. the design charts in BS8110 are not applicable (a) the Young’s Moduli of concrete are different.4 – Design for robustness Pt 2 2. The followings are highlighted : Table 6. the Code provides values for vtmin and vt for different grades up to 80.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No.2 2. HK CoP Structural Use of Concrete 2004 Contents 3. Above grade 40. So the Code is more stringent as consideration to loss of elements is required for all buildings.2. 2 2.3 contains specific values for vtmin and vt up to grade 40. it should be noted that members with axial loads creating axial stress < 0.6.1fcu may be regarded as beam in design.17 in relation to limitation of torsional shear stresses contains specific values for grades 25 to 80.8 – Columns Pt 2 2.6. The followings are highlighted: (i) The CoP contains no design charts for column design with moments.4 The followings are highlighted (i) Table 2. 1 3.56) are lower than BS8110 (Eqn 28.2). CoPConcrete 2004 is more stringent.3 in relation to design of “bridging elements” applies to buildings of five or more storeys. However. Contents Appendix A Remark 6.1. the factors on steel in equations for strengths of column sections (6.3 – Torsion and Combined Effects Pt.15). (iii) The provisions for more accurate assessment of effective column height in accordance with BS8110 Pt. For values below grade 40. (ii) CoPConc2004 contains more specific values for ultimate concrete shear stress.2 in relation to design of “bridging elements” which is identical to BS8110 Pt. 2 2. they are identical to BS8110. 6. the provisions are limited to grade 40. (ii) Due to the use of lower partial strength factor of steel (γm = 1.3.4 of CoPConc2004. BS8110:1997 Clause No.55. the words “where required in buildings of five or more storeys” have been deleted.2 – Members axially loaded with or without flexure Pt. 2 2. A-8 . The values remain constant from grade 40 thereon. 6.5 – Effective column height The followings are highlighted : (i) In the design aspects.3 The followings are highlighted : (i) Pt. Re 6. 6.6. The followings are highlighted : (i) The limitation of neutral axis depth as for beam is clearly not applicable to columns. The followings are highlighted : (i) In 6. and (b) the ultimate strain for concrete grade > C60 are reduced. beyond which the values remain constant whilst BS8110 set the values to that of grade 40 for grades above 40.6.5.5 are not incorporated in the Code. 29). Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. if any.11. The followings are highlighted : (i) In Pt 1 3. it is explicitly stated that no shear reinforcement is required if v<vc. There is no stipulation that the staircase may include landing.3. as similar to direct stress. So a pre-requisite for the use of these assumptions is stated at the end of the clause which reads “if a base or pile cap is considered be of sufficient rigidity. the assumption of uniform or linearly varying assumption is without the pre-requisite that the footing or pile cap is sufficiently rigid. (iii) In 6. This is not too sound as the formulae for beam are under the assumption that torsional cracks can be fully developed for a beam length of b+d. 6.4. the assumptions of uniform reaction or linearly varying reaction of footing and pile cap are based on use of rigid footings or pile caps. Appendix A Remark It is more reasonable to assume staircase should include landing. there is no mention of pile cap rigidity which affects design.11 The followings are highlighted : (i) CoPConc2004 is generally more reasonable as it makes provision for the modern analysis by treating the pile cap as a flexible structure by computer methods. there is no mention that shear enhancement shall not be applied to where shear distribution across section has not been considered.7.2.1.11.3. determination of effective width of section for resisting shear is included.7 – Foundations The following differences with BS8110 are highlighted : (i) In 6. it is stated that “where the shear distribution across section has not been considered.1 Contents The followings are highlighted : (i) A note in 3. (v) In 6.7.4 b). shear enhancement in pile cap can only be applied where due consideration has been given to shear distribution across section.3.1 is in relation to design of staircase.10.3. No similar provision is found in BS8110. BS8110:1997 Clause No.4. (ii) Apparently CoPConc2004 forces the designer to check torsion as if the cap or footing is a beam under the rigid cap (footing).3 3rd dot. 2nd dot. This is not good enough as significant errors may arise if the footing or pile cap is flexible. No similar provision is found in BS8110. Pt 1 3. (iv) In 6.” No similar provision is found in BS8110. (iii) In Pt 1 3. No similar provision is found in BS8110. no A-9 . (vi) In 6. (ii) In Pt 1 3.3.3 5th dot.7.1 in relation to design of staircase has stipulated that a staircase also include a section of landing spanning in the same direction and continuous with flight.1. as in CoPConc2004. (ii) In 6. If such length is not available which is very common for cap or footing structures which are usually wide structures. shear enhancement shall not be applied.10. (vii) No explicit statement that torsion is required to be checked.”.7.1.11.11.4.7. there is no mention that shear enhancement in pile cap shall be applied to under the condition that shear distribution across section has not been duly considered. a statement has been inserted that a pile cap may be designed as rigid or flexible. (iv) In Pt 1 3. depending on the structural configuration. 6.6.1.3.7. There should be shear enhancement if the full length cannot mobilized. No Pt 1 3.3. (v) There is no explicit stipulation on the limit of effective width for checking shear.3 4th dot.1. (vi) No explicit statement that shear reinforcement is required if v<vc though it is a normal practice of not providing shear stirrups in pile caps.6 – Staircase HK CoP Structural Use of Concrete 2004 Contents The followings are highlighted : (i) 6. However. 2.7. (iii) 7.2 of BS8110 Pt. The followings are highlighted : (i) This section contains provisions in BS8110 Pt.2 where the difference in temperatures is divided into 2 parts and a factor of 0.7.3 in 7. The omitted portion is in relation to estimated limiting temperatures changes to avoid cracking. HK CoP Structural Use of Concrete 2004 Contents similar stipulation is found in BS8110.1 is identical to BS8110 Pt.3mm as compared with BS8110 except water retaining structures and pre-stressed concrete (0.2 is different from Equation 14 of BS8110 Pt. Contents Appendix A Remark study data is available for torsional shear enhancement.4. (viii) In 6. 2 3. it is stated that torsion for a rigid pile cap should be checked based on rigid body theory and where required.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. Pt. the ultimate shear stress is increased to 7 MPa.5.8.2.1 as a guide.8.4. A-10 .2 appears to be a refined version of Equation 14 of BS8110 Pt.4. cracks (including thermal cracking) are given.4.4. 2 3.8 is employed in the estimation of thermal strain.1 have been omitted.3. The factor 0. (ii) Equation 7. (ii) Some of the provisions may be applicable to UK only. 2 3.1 except that the last paragraph and Table 3. 2 Section 3 The followings are highlighted : (i) General provisions for determination of deflections. 2 Section 3 and the deem-to-satisfy requirements (for deflections) in BS8110 Pt.4.2 of BS8110 Pt.4.8 – Beam Column Joints Section 7 – Serviceability Limit States – No similar provision. (iii) Limit of calculated crack widths is given as 0.2 mm). 2 3. 6.4. as compared with BS8110.8 accounts for discount of strain due to the long term strain. 1 Section 3.2.3.2. Also it is allowed in the clause to take reinforcements into BS8110:1997 Clause No. torsional reinforcements be provided. The Code contains detailed provisions for design of beam column joints. No similar provision found in BS8110.3mm in 3.2. (ii) Limits of calculated crack widths are given in 7. (vii) In 6.4. (iv) Equation 7.8. The following remarks are made : (i) The omission of the last paragraph and Table 3.8.1 is likely because of the different climate and material properties in Hong Kong.1 for member types and exposure conditions. The limited values are mostly 0.8.4. 2 3. Design checking on beam-column shall be done if CoPConc2004 is used.2.3 in 7. (viii) Table 7.3.2.3. Also. HK CoP Structural Use of Concrete 2004 Contents consideration. (ix) Table 7.4.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. However.2 indicates limit of h/500 as inter-storey drift for avoidance of damage to non-structural members. Furthermore.2. 2 3.3.2. deflection limit after construction for avoidance of damage to structure is limited to span/500. (vii) In 7. Contents Appendix A Remark A-11 . pre-camber is limited to span/250 to compensate excessive deflection.1.2.9 of BS8110 Pt 1. duration) are given as avoidance of “excessive response” to wind loads whilst no numerical values are given in BS8110 Pt. 2 3.2. In addition.3. whilst BS8110 Pt. 2 3. deflection limit due to wind load is given as H/500 whilst BS8110 Pt. (v) In 7. there is extra requirement in 7.15 and 0. requirements for “two-way slabs” and “end spans” are included.25 m/s2) on accelerations (10 years return period on 10 min.1.2 in relation to deem-to-satisfy requirement for basic span/depth ratios of beam and slab contains. (vi) In 7.4 in relation to modification factor to effective span depth ratio by tensile BS8110:1997 Clause No. as in comparison with Table 3. 2 3.3. excessive vibration should be avoided as similar in BS8110 Pt.2.3.2 specifies span/500 or span/350 in accordance with brittleness of finishes for avoidance of damage to non-structural elements. The limit is not given in BS8110 Pt.3 that dynamic analysis be carried out in case structural frequency less than 6 Hz for structural floor and footbridge less than 5 Hz.2.1. 2 3. 20 mm as an absolute value is also imposed in BS8110.2.3 under 7. limits (0.3. 5 is identical to BS8110 Pt 2 Cl.3 etc have requirements for bend of bars. zinc coated bars etc. Contents Appendix A Remark Section 8 – Reinf’t requirements Pt.6 is not identical to equation 9 in BS8110 Pt.22. 3.7 in 7.2. 1 Section 3 Pt. (ii) Consideration for ductility is not adequate.1.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. HK CoP Structural Use of Concrete 2004 Contents reinforcement is identical to Table 3. machine vibration.2 requires standard be in accordance with BS4466. it is stated that the distance between horizontal layer of bars should be sufficient without quantification whilst BS8110 Pt.3.2. (xii) Equation 7. The followings are highlighted : (i) CoPConc2004 requires bar scheduling to acceptable standards. However. (iii) In 8. it is stated that bar scheduling should be in accordance with acceptable standards whilst BS8110 Pt.12. No similar exclusion is found in BS8110. except that a single Table 8. 1 3.8 of the Code seems to incorporate the comment. 1 3. the minimum spacing of bars should be the lesser of bar diameter.8. 8.12.4(a) should extend to all support conditions.7.7 except that the last sentence in BS8110 is deleted.6. 1 3. 1 3.12.4.2.10 in relation to anchorage of tendons in prestressed concrete The followings are highlighted : (i) The provisions are general.2. it is declared that the rules given in the Section for re-bars detailings do not apply to seismic. fatigue etc and epoxy.11. The followings are highlighted : (i) In 8.10 of BS8110 except that the row with service stress 307 (fy = 460) has replaced that of 333 (fy = 500). 1 4. (x) 7.3.3.3. (v) 8.11. hagg+5 mm and 20 mm.1. BS8110 Pt. (xi) The provision of deflection calculation in 7.1 requires minimum be 2hagg/3.4. (iv) In 8.1 does not include 20 mm and bar diameter is only required when bar size > hagg+5 mm.6 is identical to BS8110 Pt.3 is essentially identical to BS8110 Pt.4. (ii) In 8. 24. 1 3.4. 9.9.12. 25.2 (in relation to minimum bend radii of re-bars) BS8110:1997 Clause No. provisions in 8. 1 3. (ii) ICU has commented that requirement in BS8110 Pt.12.1. 2 in the derivation of shrinkage curvature. (iii) ICU has also suggested the use of torsional links similar to that in ACI code (135o hood) which is of shape other than shape code 77 of BS4466. A-12 . This effectively extend the requirement of BS8110 Pt.12.6 is added with Figure 8.12.6. Contents Appendix A Remark A-13 .1 that when mechanical device is used.4. (ix) Except for the requirement that the sum of re-bar sizes at lapping < 40% of the breadth of BS8110:1997 Clause No.8. (2) the greater of 10φ or 70 mm for 90o bend in the Code but only 8φ in BS8110. (c) 8. Provisions for 135o bend and welded bars are also added. 1 3.8 except (a) It is mentioned in 8. (d) 8. (b) Type 1 bars are not included in the Code.4. (vi) 8.6 contains requirements for anchorage by welded bars which is not found in BS8110. No similar provision is found in BS8110. (vii) 8.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. (viii) 8. – (1) the greater of 4φ or 50 mm for 180o bend in the Code but only 4φ in BS8110.5 in relation to anchorage of links and shear reinforcements contains more stringent requirement for the length of the link beyond bends of bars than BS8110 Pt. As such. no distinction is made between mild steel bars and HY bars – both adopting minimum radii of HY bars. 1 3.8 in relation to minimum support widths requires any bend inside support be beyond the centre line of the support.4 is essentially identical to BS8110 Pt.12.4(a) to support conditions other than simply support. their effectiveness has to be proven.4. HK CoP Structural Use of Concrete 2004 Contents to replace Table 3 of BS4466 to which BS8110 is making reference. Provisions to the newer BS – BS8666 where the minimum bend radii are generally smaller is not adopted.9.1 for illustration of bend anchorage. 1 3. 8 requiring 30% of the calculated BS8110:1997 Clause No.2. 1 Section 3 and 5.1. (xi) 8.2 tabulates minimum steel percentage equal to that of BS8110.8 and 8. 9. (ii) Though ductility is not a design aid explicitly stated in the BS.16 for butt joints of re-bars is not found in the Code.12. Furthermore.2.2. minimum longitudinal and transverse distances between adjacent lapping of bars which are not found in BS8110 in case that sum of re-bar sizes at lapping > 40% of the breadth of the section in not using staggered laps.4.11.7.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. the stipulation in BS8110 that demonstration of crack width < 0. 8.1 under Cl.2. 1 3.12.12. HK CoP Structural Use of Concrete 2004 Contents the section.1. (ii) Ductility is more emphasized in CoPConc2004 9.1.2.2). 1 3.12. a requirement of 15% span moment be used to design beam support even under simply supported assumption is not found in BS8110. 1 3. A-14 . (ii) In 9. (iv) 9.9 in relation to large diameter bars (>40φ) and bundle bars which are not found in BS8110. the BS does requires 135o bend of links in anchoring compression bars in columns and beams (Pt.7 contains more requirements in terms of “staggering laps”. The followings are highlighted : (i) The stress reduction in design of cantilevered projecting structures in PNAP173 is not incorporated is likely because the PNAP is based on working stress design method. it is also stated in the clause that total tension re-bars of a flanged beam over intermediate supports can be spread over the effective width of the flange provided that half of the steel within the web width.1 contains different and more detailed requirements for transverse reinforcements in lapped zone than BS8110 Pt.4. The followings are highlighted : (i) Table 9. There is also no such provision in BS8110.7 The followings are highlighted : (i) The analysis procedures are largely old-fashioned relying on old theories of Johansen.3 mm can be accepted is omitted.5.8. So there should be some other approaches and this is not mentioned in the CoPConc2004.9 which largely stem from seismic design. Detailings to cater for behaviours not well understood or quantified are thus provided.7. Contents Appendix A Remark Section 9 – Detailing of Members and particular rules Pt. (xii) The provision in BS8110 Pt.4 in relation to maximum distance of bars in tension as similar to BS8110 Pt. 2 3. (x) 8.8.12. though the determination of which are largely empirical or from past experiences. The requirements are also schematically indicated in Fig. (iii) In 9. 8. Hillerborg. Contents Appendix A Remark A-15 .4. (viii) In 9. HK CoP Structural Use of Concrete 2004 Contents mid-span re-bars be continuous over support appears to be adopted from Fig. (c) more stringent requirements are added for slabs with concentrated loads or areas of maximum moments whilst no similar requirements are found in BS8110.3 of the Code provides an alternative of using closed links of 135o bend. maximum spacing of bent-up bars is stipulated whilst no such requirement is found in BS8110. in 9. (b) for secondary re-bars 3. 1 3. 2 2.1. 3. The provisions are. The requirement is identical to BS8110 Pt.2.2 of the BS is not quoted.9 requires top steel of cantilever to extend beyond the point of contraflexure of the supporting span whilst Fig.5h ≤ 450 mm whilst no provision in BS8110.1 (b) in relation to maximum spacing of re-bars in slab is more detailed than BS8110 Pt.3 requires half of the area of the calculated span re-bars be provided at the simply supported slabs and end support of continuous slabs.12. the provision in BS8110 is under the condition BS8110:1997 Clause No. 3. However.2.3. 1 3.24 c) requires at least half of the top steel to extend beyond half span of the cantilever or 45φ.24 a) of BS8110. (vi) In 9. 3h ≤ 400 mm whilst BS8110 is 3h ≤ 750 mm.2.7 and appears more reasonable. 9.1. more stringent : (a) for principal re-bars.11.2. (ix) The first para.2.12.1.12. However.3.3.2. (vii) Torsional links has to be closed links (shape code 77 of BS4466) as required by BS8110 Pt. in fact.10. the circumstances by which the Figure is applicable as listed in 3.10. (v) 9.8.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. However. 3 as the termination of links should be at 135o hook.l.3. there is also a provision that if the ultimate shear stress < 0. anchorage length from PNAP173.10.5vc at support. (xii) 9.3.9.6 requiring closed loops of longitudinal rebars at free edge of slab is not found in BS8110.12.9.3.3.1.4 in relation to cantilevered projecting structures has incorporated requirements for minimum thickness.7. 3. However. (xiii) 9.4. (xv) There is an extra requirement in the maximum spacing of traverse reinforcement in wall in 9.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. and in case of continuous slabs.6.23 in relation to plain walls BS8110:1997 Clause No.2.5 in relation to reinforcement provisions to plain walls include BS8110 Pt.3. other requirements such as design with reduced stresses are not found. No similar provision is found in BS8110.4.3 which is 400 mm.12.1. 1 3. Contents Appendix A Remark A-16 . 1 3.12. (xi) 9. maximum bar spacing.3 contains more stringent requirements of links in circular columns than that in BS8110 Pt.1 that the slabs are designed predominantly to carry u. straight length of bar beyond effective anchorage for 1/3 of support width or 30 mm (whichever is the greater) is considered effective anchorage. 1 3. minimum steel areas. However.d. (xvi) 9.6. as in comparison with BS8110 Pt. (xiv) 9. These conditions are not mentioned in the Code.5.7. approximately equal span.2 is in relation to shear in slabs which should be identical to that for beams.19 to 22. However it is stated that shears should be avoided in slabs < 200 mm thick. HK CoP Structural Use of Concrete 2004 Contents listed in 3.4. (x) In 9. 1 is under the heading “Beam”.5.9.3 in relation to tie beams has included (xviii) a requirement that the tie beam should be designed for a load of 10 kN/m if the action of compaction machinery can cause effects to the tie beams. 1 5. The followings are highlighted : (a) 9. (xvii) In 9. this is not important as the wall should not be regarded as plain wall.9.1(c) requires anchorage of beam bar into exterior column to commence beyond centre line of column or 8φ instead of column face unless the moment plastic hinge can be formed at 500 mm or half beam depth from column face.2.4 and is identical to BS8110 Pt.1 and 9.8 in relation to design of corbels carries the same content as 6.7.1(d) imposes restriction in locations of laps and mechanical couplers – (i) not within column/beam joints.7.2 in relation to pile caps and footings. (b) 9.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. So the minimum steel percentage of 0.1 though Table 9.9 contains requirements more stringent than BS8110 in detailing with the aim to enhance ductility.3% and percentage of tensile steel < 2.13% for HY bars should be observed.2. Contents Appendix A Remark A-17 . (xx) 9.1.1(a) requires steel percentage in beam > 0.5. This is not found in BS8110. 9. There is no explicit provision in BS8110 for minimum steel in pile caps and footings. (ii) not within BS8110:1997 Clause No. (xix) 9.1. it is stipulated that the minimum steel percentage should refer to Table 9.2 to 6. (c) 9.7.2.7.5%.9. Anyhow. HK CoP Structural Use of Concrete 2004 Contents with more than 1/10 in tension to resist flexure is not included.1. 1(b) requires spacing of links to longitudinal bars not be spaced further than 1/4 of the adjacent column dimension or 200 mm.2. bars be terminated by a 90o bend or equivalent to the far face of column.2% at lap)in column.2. dia.2(b) states that links be adequately anchored by means of 135o or 180o hooks. (iii) reversing stresses exceeding 0. of column re-bar through beam by (eqn 9.1(a) states min.9.8%) and max.2(e) requires distribution and curtailment of flexural re-bars be attained in critical sections (potential plastic hinge regions).2.2.9. (e) 9.1. (i) 9.2(a) states the link spacing in beam < the lesser of 16φ and beam width or depth and corner and alternate compression re-bars be anchored by links.9. (h) 9.6fy unless with specified confinement by links.9. (g) 9.9. In addition.1(a) requires the smallest dia. Contents Appendix A Remark A-18 .1.Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. HK CoP Structural Use of Concrete 2004 Contents 1 m from potential plastic hinge locations. (0. of any bars in a row > 2/3 of the largest bar.1.9.9. (d) 9. steel% (4% with increase to 5.7) dependent on beam depth. (f) 9. (k) 9.1(c) requires anchorage of column bar into exterior beam or foundation to commence beyond centre line of beam or foundation or 8φ instead of interface unless BS8110:1997 Clause No. (j) 9. Anchorage by 90o hooks or welded cross bars not permitted.9.2.1(a) limits max. with increase by 25% if not forming plastic hinge. Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. HK CoP Structural Use of Concrete 2004 Contents the moment plastic hinge can be formed at 500 mm or half beam depth from column face; (l) 9.9.2.1(d) states restrictions in locations of laps (m) 9.9.2.2 describes the establishment of “critical regions” in columns where there are extra requirements on links – (i) link spacing in column < the lesser of 6φ and least 1/4 of column lateral dimension; (ii) each longitudinal bar be laterally supported by a link passing around the bar and having an included angle < 135o. (Regions other than “critical regions” fallow 9.5.2) The followings are highlighted : (i) 10.2 lists figures for construction tolerances whilst BS8110 refers most of the requirements to other BS; (ii) 10.3.4 in relation to sampling, testing and compliance criteria of concrete. They are extracted from HKB(C)R but with incorporation of 100 mm test cubes. Such provision is not found in BS8110; (iii) The sub-clause on “Concreting in cold weather” in BS8110 is not incorporated. 10.3.7 on “Concreting in hot weather” is modified from BS8110 Pt. 1 6.2.5 (reference to BS deleted); (iv) Table 10.4 is similar to BS8110 Pt. 1 Table 6.1. However the parameter t (temperature) is deleted and the categorization of cement is OPC and “others” instead of the few types of cement in BS8110; (v) 10.3.8.1 contains general requirements for BS8110:1997 Clause No. Contents Appendix A Remark Section 10 Pt. 1 2.3, Section 6, Section 7, Section 8 The followings are highlighted : (i) Pt. 1 2.3 lists general requirements for inspection of construction; (ii) References to other BS are often stated in Section 6 and 7; (iii) Provisions of works in extreme temperatures are given which are deleted in CoPConc2004. The followings are highlighted : (i) The first part of Section 10 of CoPConc2004 mainly stems from HKB(C)R, CS1, CS2 whilst the second part incorporates workmanship requirements listed in BS8110 Pt. 1 Section 6; (ii) (iii) A-19 Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. HK CoP Structural Use of Concrete 2004 Contents “Formwork and falsework” similar (but not identical) to BS8110 Pt. 1 6.2.6.1; (vi) 10.3.8.2 lists criteria for striking of formwork identical to that in BS8110 Pt. 1 6.2.6.3.1. In addition, provisions for using longer or shorter striking time for PFA concrete and climbing formwork are included; (vii) Minimum striking time in 10.3.8.2 are in accordance with HKB(C)R Table 10 (with the addition of props to cantilever requiring 10 days striking time) instead of BS8110 Pt. 1 Table 6.2. Furthermore, BS8110 Pt. 1 Table 6.2 gives temperature dependent striking time whilst the striking times in CoPConc2004 are not temperature dependent; (viii) The contents of 10.3.9 in relation to surface finish are extracted from BS8110 Pt. 1 6.2.7. However, the general requirements are differently written and the “classes of finish” have been deleted; (ix) 10.3.10 and 10.3.11 in relation to joints are identical to BS8110 Pt. 1 6.2.9 and 6.2.10 though the wordings are different, except the last sentence of 6.2.9 last para. in relation to treating vertical joint as movement joint; (x) 10.4.1 contains general requirements on re-bars to standards CS2 and other acceptable standards whilst BS8110 Pt. 1 7.1 requires conformance to other BS; (xi) 10.4.2 in relation to cutting and bending of re-bars is identical to BS8110 Pt. 1 7.2 except (a) conformance is not restricted to BS but to acceptable standards; and (b) the requirement of pre-heating re-bars at temperatures below 5oC is deleted; BS8110:1997 Clause No. Contents Appendix A Remark A-20 Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. HK CoP Structural Use of Concrete 2004 Contents (xii) 10.4.3 is effectively identical to BS8110 Pt. 1 7.3 except that the requirement for spacer blocks be of concrete of small aggregates of equal strength to the parental concrete is replaced by spacer blocks to acceptable standards; (xiii) 10.4.6 is effectively identical to BS8110 Pt. 1 7.6 except (a) conformance to BS changed to acceptable standards; (b) detailed descriptions of the types of welding omitted; and (c) requirement to avoid welding in re-bar bends omitted; (xiv) 10.5.1 is identical to BS8110 Pt. 1 8.1 except conformance to BS is changed to acceptable standards; (xv) 10.5.5.3 in relation to tensioning apparatus of prestressing tendons is effectively identical to BS8110 Pt. 1 8.7.3 except that CoPConc2004 has an additional requirements that apparatus be calibrated within 6 months; (xvi) 10.5.5.4 in relation to pre-tensioning of deflected tendons, compressive and tensile stresses should be ensured not to exceed permissible limits during transfer of prestressing force to the concrete with the release of holding-up and down forces. BS8110 Pt. 1 8.7.4.3 has omitted the compressive forces; (xvii) 10.5.5.5(b) requires anchorage of post tensioning wires to conform to acceptable standards whilst BS8110 Pt. 1 8.7.5.2 requires compliance with BS4447 (xviii) 10.5.5.5(d) in relation to tensioning procedures which is identical to BS8110 Pt. 1 8.7.5.4, the requirement of not carrying out BS8110:1997 Clause No. Contents Appendix A Remark A-21 Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. HK CoP Structural Use of Concrete 2004 Contents tensioning below 0oC is omitted. Further, the paragraph in BS8110 stipulating that when full force cannot be developed in an element due to breakage, slip during the case when a large no. of tendons is being stressed is omitted in CoPConc2004; (xix) 10.5.7 contains detailed provisions for grouting of prestressed tendons whilst BS8110 Pt. 1 8.9 requires compliance to BS EN 445, 446. This section outlines measures and procedures in general quality assurance and control, with reference to local practice. The followings are highlighted : (i) Control are on design, construction and completed products; (ii) Control can be by independent organization; (iii) Concrete must be from supplier certified under the Quality Scheme for the Production and Supply of Concrete (QSPSC); (iv) Control on construction includes surveillance measures. This section is basically identical to Section 4 of BS8110 Pt. 1. The followings are highlighted : (i) 12.1.5 in relation to durability and fire resistance makes reference to previous recommendations in Sections 4 and 10 whilst BS8110 makes reference also to Part 2 of BS8110; (ii) 12.2.3.1 in relation to redistribution of moments is restricted to concrete grade C70, as in consistency with reinforced concrete. BS8110 Pt. 1 4.2.3.1 does not have this limitation. But the BS covers grades up to 40; (iii) The first loading arrangement in 12.3.3 for BS8110:1997 Clause No. Contents Appendix A Remark Section 11 – Quality Assurance and Control – No similar provisions in BS8110. The control in CoPConc2004 are summaries of local good practice. Section 12 – Prestressed Concrete Pt. 1 Section 4 The provisions are general. CoPConc2004 follows quite closely the provisions in BS8110 except for minor changes. A-22 Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (or 1985) Clause No. HK CoP Structural Use of Concrete 2004 Contents continuous beam is not found BS8110 Pt. 1 4.3.3. The loading arrangement is in consistency with 5.1.3.2 for reinforced concrete beams. Though not truly adequate (per similar argument as above), CoPConc2004 is more conclusive than BS8110; (iv) 12.3.8.2 gives ultimate concrete stress 7.0 MPa, as similar to r.c. works; (v) 12.8.2.2 in relation to 1000 h relaxation value which is identical to BS8110 Pt. 1 4.8.2.2, “UK” has been deleted in description of manufacturer’s appropriate certificate; (vi) 12.8.4 and 12.8.5 in relation to shrinkage and creep of concrete make reference to 3.1.8 and 3.1.7 whilst BS8110 Pt 1. 4.8.4 and 4.8.5 list UK data; (vii) 12.10 makes reference to 8.10.2.2 for transmission lengths in pre-stressed members which is titled “transfer of prestress” which is identical to BS8110 Pt. 1 4.10.1 except that the 2nd paragraph of the BS in relation to the difficulty of determination of transmission length has been deleted; (viii) 12.12.3.1(a) is identical to BS8110 Pt. 1 4.12.3.1.1 except that not only protection against corrosion is added. In 12.12.3.1(c), reference for protection against fire is not identical to BS8110; This section contains testing of structures during construction stage under circumstances such as sub-standard works are suspected and visible defects are identified. BS8110:1997 Clause No. Contents Appendix A Remark Section 13 – Load Tests of Structures or parts of structures – No similar provisions in BS8110. A-23 Appendix B Assessment of Building Accelerations . ω is the circular frequency of the building where ω = 2πf ( f is the natural frequency of the building) can be used. However. The formula G2 shows clearly it utilizes the dynamic resonant component Furthermore. Thus the equation of && = −ω 2 x where && x x is the acceleration and x is the displacement of the motion. ω of the building can either be obtained by detail dynamic analysis or by some empirical formula such as 460/h.Appendix B Assessment of acceleration of Buildings Underlying principles : Two Approaches are outlined in this Appendix : (i) The first one is based on the assumption that the building will undergo simple harmonic motion under wind loads. it should also be noted that g R SE ζ for assessment of acceleration. Nevertheless. the acceleration that will be causing discomfort to the occupants is the “dynamic resonant component” of the motion. The approach is also based on the same principle as (i) though the formula G2 listed in the Australian Code has incorporated empirical parameters m 0 and h for assessment of the dynamic properties of the building. (b) the dynamic background component which is 2 I h g v B . So it is only necessary to calculate the dynamic resonant component. there is a denominator of 1 + 2 g v I h in the formula in the B-1 . and (c) the dynamic resonant g f SE 2 2 component which is 2 I h ζ . It is the last displacement that should be multiplied to ω 2 to obtain the acceleration by which the building is responding to wind loads.2:2002 Appendix G2 as extracted at the end of this Appendix. If the G factor which is equal to G = 1 + 2 I h g v B + 2 g f SE 2 ζ in Appendix F of the Wind Code 2004 is used to arrive at a total displacement. it can be considered that the total displacement is made of up of three components : (a) the static part which is 1 in the equation. (ii) The second approach is that listed in Australian Wind Code AS/NSZ 1170. 00 h Lh = 1000   10  0.11  h I h = 0. In addition.384 Hz Total displacement at the uppermost floor by the approach listed in Appendix G of the Hong Kong Wind Code is 74 mm Total dead load is 251555 kN and total live load is 48171 kN (A) To compute the dynamic resonant component of the total force Computation of the G factor.3 m Building natural frequency n a = 0. the reason being that the Australian Code is based on V des .11 n a = 0.θ which is 3 second gust whilst Hong Kong Code is based on hourly mean wind speed.1047 g f = 2 ln(3600n a ) = 4.4  = 0.4 m b = 37. For conversion.1055   90  = 0. Building width b = 37.) Worked Example Data : Building height h = 96. the dynamic background and dynamic resonant components in accordance with HKWC 2004: h = 96.4 m. the formula listed in Appendix B can be used (as confirmed by some experts that the formula can be used for downward conversion from 1-in-50 year to 1-in-10 year return periods). The 10 minutes mean speed can also be taken as identical to that of hourly mean speed (also confirmed by some experts. it should be noted that the Concrete Code requires the wind load for assessment of acceleration to be 1-in-10 year return period of 10 minutes duration whilst the wind load arrived for structural design in the Hong Kong Wind Code is based on 1-in-50 year return period of hourly duration.7  96.Appendix B Australian Code as different from Hong Kong Wind Code.25 = 1762.05 B-2 . So this denominator can be omitted.3 m −0.384 Hz −0.1055   90  g v = 3. 607 (B) To scale the wind loads from 50 years return period 1 hour duration to 10 years return period 10 minutes duration The formula in Appendix B of the Wind Code can be used for scaling to wind loads of shorter return period.898 The dynamic background component is G = 2 I h g v B = 0.5n a h   4n a b  1 +  1 +  Vh   Vh   0.73 V h = 49.1285  3. B-3 .Appendix B B= 1 36h 2 + 64b 2 1+ Lh = 0.662 2 The dynamic resonant component is G dynresonant = 2 I h g f SE 2 ζ = 0. Thus no adjustment needs be carried out.6 m/sec n L N = a h = 13. Thus the factor in scaling down the wind load from 50 years to 10 years is  5 + ln 10    = 0.02 So G = 1 + 2 I h g v B + 2 g f SE 2 ζ = 1.47 N E= = 0. as confirmed by some experts that the formula could be used for period shorter than 50 years.64 Vh 1 S= = 0.6714  5 + ln 50  2 The wind pressures of 10 minutes duration and 1 hour duration can be taken as identical.0816 (2 + N 2 )5 / 6 ζ = 0. By a = 3 ˆ M b in accordance with the Australian Code m0 h 2 (II) As total dead load is 251555 kN and total live load is 48171 kN The weight used is full dead load + 0.138 m/sec2 which is for 50 year return period and one hour mean wind speed. based on assumption of simple harmonic motion.093 m/sec2 < 0.4 2734000 = 279022 kg/m 9. The dynamic resonant component is x = 0.6714 = 120708 kNm B-4 . For the dynamic portion 562163 × 0. the acceleration for 10 years return period and 10 minute duration as required by Clause 7.2 of the Concrete Code 2004 is 0.607 = 0. x Fundamental natural frequency = 0.898 Scaled to 10 minutes duration and 10 years return period ˆ M b = 179786 × 0.25 live load = 263598 kN h = 96.074 × 0.607 = 179786 kNm 1.6714 = 0.898 Total Displacement = 0.024 m 1.074 m.898 The acceleration is ω 2 x = 0.412 /sec Dynamic Magnification factor = 1.138 × 0. Scaling down to 10 year return period.15 m/sec2.8 ∴m0 = 279022 kg/m The total bending moment about X axis is 562163 kNm for full static + dynamic load.4 m The weight per m height is which is equal to 263598 = 2734 kN/m 96.3.Appendix B Assessment of accelerations By the two approaches (I) By && = −ω 2 x .384 Hz Circular frequency = ω = 2πf = 2. 14 m/sec2 < 0.θ used in the wind pressure calculation in the same formula is based on 3 second gust.15 m/sec2.Appendix B So the acceleration is 3 × 120708 × 10 3 = 0. downscaling by the factor is not necessary. 2 279022 × 96. As our code is based on hourly mean speed. B-5 .4 Explanatory Note : The existence of the denominator 1 + 2 g v I h in the formula G2 listed in the Australian Code is because Vdes . . Beam sections against Flexure .C.Appendix C Derivation of Basic Design Formulae of R. cover to compressive reinforcements are x. ε ult = 0.Appendix C Derivation of Basic Design Formulae of R. Beam sections against Flexure The stress strain relationship of a R. as used in BS8110 and the Code. To derive the contribution of force and moment by the concrete stress block.34 f cu ⇒ = ⇒ ε0 = γ mε 0 Ec γ m 2ε 0 2 γ mε 0 2 ε0 (Eqn (C-4) (accords with 3. assume the parabolic portion of the concrete stress block be represented by the equation σ = Aε 2 + Bε (where A and B are constants) (Eqn C-1) dσ So = 2 Aε + B (Eqn C-2) dε dσ = Ec ⇒ B = Ec where Ec is the tangential Young’s Modulus of As dε ε =0 concrete listed in Table 3. and d’.14 of the Concrete Code Handbook) C-1 .67 ∴ 0. the symbols for the neutral axis depth. effective depth. d.2 of the Code. Also dσ dε = 0 ⇒ 2 Aε 0 + B = 0 ⇒ A = − ε =ε 0 E B =− c 2ε 0 2ε 0 (Eqn C-3) As σ = 0.67 f cu γm − f cu Ec when ε = ε 0 =− Ec 0.0035 d’ x d neutral axis Stress Diagram Strain Diagram Figure C-1 – Stress Strain diagram for Beam In Figure C-1 above.C.67 f cu Ec 1.C. beam section is illustrated in Figure C-1. A plot for grade 35 is included for illustration : Stress Strain Profile for Grade 35 18 16 14 Stress (MPa) 12 10 8 6 4 2 0 0 0.8 1 Distance Ratio from Neutral axis 0.34 f cu Ec γ m So the equation of the parabola is σ =− Ec 2 ε + Ecε for ε ≤ ε 0 2ε 0 Consider the linear strain distribution xε 0 / ε ult ε = ε0 ε ult = 0. the stress strain profiles can be determined. ε = ε ult So stress at u from the neutral axis up to x 2 u x ε0 is ε ult 2 E E  Eε Eε u u  σ = − c ε 2 + Ecε = − c  ε ult  + Ec  ε ult  = − c ult2 u 2 + c ult u (Eqn C-5) x x x 2ε 0 2ε 0  2ε 0 x  Based on (Eqn C-5).001319 Figure C-3 – Stress strain profile of grades 35 C-2 .4 0.2 0.0035 x u h Figure C-2 – Strain diagram across concrete section At distance u from the neutral axis.Appendix C A=− Ec 2ε 0 where ε 0 = 1.3769 where ε0 = 0.6 0. 34ε 0 f cu 2 ε0 x0.67 f cu  ε x− x 0 γm  ε ult   0.67 f cu bx  ε b = 1 − 0   ε γm ult       (Eqn C-8) The moment offered by the constant part about the centre line of the whole section is h  ε M c 2 = Fc 2  − 1 − 0  ε ult 2   x   2   (Eqn C-9) The compressive force by concrete as stipulated in (Eqn C-6) and (Eqn C-8) is Fc = Fc1 + Fc 2 = ε 1.Appendix C Sectional Design of rectangular Section to rigorous stress strain profile Making use of the properties of parabola in Figure C-4 offered by the parabolic section as Fc1 given by centre of mass b Area = 2 ab 3 a 3 a 8 Figure C-4 – Geometrical Properties of Parabola Fc1 = b f 1.67 f cu bx  1 − 0 bx +  ε γm 3γ m ε ult ult   0.34ε 0 f cu 0.67 f cu bx  ε 3 − 0 =  ε  3γ m ult       For singly reinforcing sections.67 cu = bx 3 ε ult γm 3γ m ε ult (Eqn C-6) and the moment exerted by Fc1 about centre line of the whole section h  ε M c1 = Fc1  − x1 − 0  ε ult 2  h  5 ε0  3 ε0  − x  = Fc1  − x1 −  8 ε  ult    2  8 ε ult     (Eqn C-7) The force by the straight portion is Fc 2 = 0. by (Eqn 2-7) and (2-9) C-3 . moment by concrete about the level of the tensile steel is. Consider grade 100 where C-4 .1.152 for grade 35 K ' = 0.113.4 for singly d reinforcing sections under moment distribution not greater than 10% (Clause 6.1199 for grade 70 again less than more significantly less than 0.5 for singly reinforcing sections for grades up to 40 under As d moment distribution not greater than 10% (Clause 6.151 for grade 40 which are all smaller than 0.67 f cu  1 1 ε 0 0.1.34ε 0 f cu x  x  5 ε 0  0.67 f cu x  1 ε 0   1 1 ε 0 1  ε   x M   1 −  + − + ⇒ = −  0     2  ε  d γ m d  3 ε ult   2 3 ε ult 12  ult    bd     2 2 0.34ε 0 f cu    + 1 − 0   d − 1 − 0   bx d − x1 − =   ε   ε 2 γm 3γ mε ult  ult   ult    8 ε ult        1. again by (Eqn 3-1) bd 2 f cu However. x is limited to 0. K ' = 0. M will be limited to K values as in bd 2 f cu K ' = 0.2.2.126 for grade 45 K ' = 0.125 for grade 50 K ' = 0. by (Eqn C-9). The implication is that design based on the rigorous stress block approach requires compression steel at smaller moments than the simplified stress approach.4 M will be limited to of the Code).67 f cu x  ε  1  ε  x  M 1 − 0  1 − 1 − 0   ⇒ 2 = 1 − 1 −     8 ε  + γ  ε  3γ m ε ult d  d  d bd ult   m ult    2  ε ult  d     2 0.154 for grade 30 K ' = 0.156 under the simplified stress block. again more d significantly less than that arrived by the simplified stressed block which is 0.132 under the simplified stress block.Appendix C     5 ε 0  ε0  x M = M c1 + M c 2 = Fc1 d − x1 −  8 ε  + Fc 2 d − 1 − ε  2        ult  ult         5 ε 0  0.0958 .33. x is limited to 0. for grades 45 and 50 and 70 where K ' = 0.67 f cu  1 ε 0  x M 1  ε   x   1 −  − − + ⇒ −  0    + =0 ε γ m  2 3 ε ult 12  ult   d  γ m  3 ε ult  d bd 2     (Eqn C-9) x which is a quadratic equation in d x is limited to 0.67 f cubx  ε   ε  x  1.4 of the Code). the forces in concrete d 1. Furthermore.87 f y 1 −   d 0.67 f cu  1 ε 0  x 1 −   γ n  3 ε ult  d   1 ε0 1 −  3ε ult  x  d  ≤ 0.67 f cu  1 ε 0  x A 0. thus A 0.Appendix C x analyzed by (Eqn C-9). there is a limitation of lever arm ratio not to exceed 0.87 f y γ m  3 ε ult  d γ m  3 ε ult  d     (Eqn C-10) With the When M exceeds the limited value for single reinforcement. Compression bd 2 f cu reinforcements at d ' from the surface of the compression side should be added.67 f cu bx  0.34ε 0 f cu 0.   M   2  bd f − K '  f cu  1 ε0  cu    1 − 3 ε η +   d' ult   0. The compression reinforcements will take up the difference between the applied moment and K 'bd 2  M   2   bd f − K '  f cu  Asc  cu  = − K ' ⇒  bd  d'  0.87 f y Asc  d '   M 1 −  =   bd  d   bd 2 f cu (Eqn C-11) And the same amount of steel will be added to the tensile steel.87 f y st = bd bd 0.87 f y γ m (Eqn C-12) where η is the limit of x ratio which is 0.67 f cu  1 1 ε 0 0.67 f cu  1 ε 0  x 1 1 −  ⇒ st = 1 −  0.67 f cu  1 ε 0  x ε  F 1 − 0  ⇒ c = 1 −  Fc = Fc1 + Fc 2 = bx +  ε  bd 3γ m ε ult γm γ m  3 ε ult  d ult     can be calculated which will be equal to the required force to be provided by steel.5 for grade 40 and below and 0.95 C-5 .4 for d grades up to and including 70.87 f y 1 −   d Ast 1 0.67 f cu 1  ε 0   x     + − + −  γ m  2 3 ε ult 12  ε ult   d  γm     0.67 f cu = bd 0.95 which requires 2 2 0. 051 −  3ε ult      ⇒ x ≥ d 1 1 ε 1 ε 0  − +  0   2 3 ε ult 12  ε ult      2     (Eqn C-13) Thus the lower limits for the neutral axis depth ratio are 0.112.13% in accordance with Table 9. As illustration for comparison between the rigorous and simplified stress block approaches. 40 and 45 respectively which are in excess of the minimum of 0. With such lower limits of neutral axis depth ratio. plots of M against steel percentages is plotted as bd 2 Comparison of Reinforcement Ratios for Grade 35 according to the Rigorous and Simplified Stress Block Ast/bd . 45 respectively.1 of the Code.Simplified Approach 14 12 10 M/bd2 8 6 4 2 0 0 1 2 3 4 5 Steel Percentage (%) It can be seen that the differences are very small. 0. 0.114 and 0.Appendix C  1 ε0 0. 0. C-6 . 40.44% and 0.Rigorous Approach Ast/bd .Rigorous Stress Approach Asc/bd . 35.33%. 0. 35. maximum error is 1%.113.Simplified Approach Asc/bd .115 for grades 30.39%. the tensile steel required will be 0.5% for grades 30. 9 ≤ . taking moment about the level of the reinforcing steel.45  2  b − 1 d  1 − 2 d  + γ γm  w  d  d bw d  m   C-7 . If the 90% of neutral axis depth is within the x df depth of the flange.67 f cu   (b f − bw )d f  d − f  + 0.1 of the Code is adopted. The exercise is first carried out by treating the width of the beam as b f and analyze the beam as if it is a rectangular section. however.9 x )(d − 0.67 f cu bw (0.9  x K  df ≤ = 2 0. 0.9 C-5. the reinforcement so arrived is adequate for the d d x df is section.45 x ) (Eqn C-15) M =   2  γm γm   d f  1 d f  0.9 ≤ d d 0.or L-Sections For simplicity.9 x x bw Figure C-5 – Analysis of a T or L beam section For singly reinforced sections.9 1 − 0.67 f cu γm df d 0. 0.67 f cu  0.e. x df > . i. d 0. The requirement for 0. the section has to be reconsidered with reference to Figure d d bf 0.67 f cu  b f M x  x     ⇒ =  0.Appendix C Determination of reinforcements for Flanged Beam Section – T.5 − 0.9  d   (Eqn C-14) If.25 −  d 0. only the simplified stress block in accordance with Figure 6. 67 f cu d f γm d 0.67 f cu = bw d 0.9 d 1 − 0.87 f y st = γm γm bw d ⇒ Ast 1 0.67 d f  b f    =  b − 11 − 2 d  + 0. the maximum moment of resistance is.45 × 0.4 depends on concrete grade in accordance with (Eqn 6. The compression reinforcements will take up the difference between the applied moment and Kbd 2 Asc bd d'  M  1 −  =  2 d   bd f cu    M   − K f '  f cu b d2 f   A w cu  − K  ⇒ sc =   d' bd   0.1) to (Eqn 6.Appendix C Putting ⇒ Mf bw d 2 = 0. For grades 40 and below :  1 d f  0.67 d f  b f  − 11 −  + 0.67 f cu (b f − bw )d f + 0.132 K f '= γ m d  bw  2 d     For tensile steel ratio : A 0.87 f y 1 −  d  C-8 0.87 f y γ m  b f df x + 0.67 f cu bw (0.9 x ) 0. the equation becomes  w   2  x  0.9 f cu x M − M f + =0   − d γm bw d 2 d  2 x M −M f x ⇒ 0.67 × 0.45 d  γ m f cu d  w  f cu bw d   m cu    1 d f  0.67 f cu d f  b f M x  x  − 11 −  = 2 b  2 d  + γ f  0.87 f y (Eqn C-17) .3) of the Code.402 f cu + =0 d bw d 2 d  (Eqn C-16) As the neutral axis depth ratio is limited to 0.1809 f cu   − 0.5 or 0.156 γ m d  w   Mf plus K ' of the rectangular section which is simply f cu bw d 2 K f '= Similarly for grade above 40 and up to 70  1 d f  0.67 × 0.67 f cu  0.9 f cu γm  bf  1 d f      b − 11 − 2 d  . by (Eqn C-15).9   − 1  d  d  bw    (Eqn C-16) When M exceeds the limited value for single reinforcement. Compression bw d 2 f cu reinforcements at d ' from the surface of the compression side should be added. 87 f y 1 −   d 0.45 +  d'  w     0.67 f cu Ast 1 = bd 0.Appendix C And the same amount of steel will be added to the tensile steel.  M     b d 2 f − K f '  f cu  b f   w df cu     b − 1 d + 0.87 f y γ m (Eqn C-18) C-9 . 05 Grade 30 Ast/bd Grade 40 Asc/bd 14 13 12 11 10 9 M/bd 2 Grade 30 Asc/bd Grade 45 Ast/bd Grade 35 Ast/bd Grade 45 Asc/bd Grade 35 Asc/bd Grade 40 Ast/bd 8 7 6 5 4 3 2 1 0 0 0.6 1.4 3.4 0.8 3 3.6 3.8 2 2.2 3.4 1.2 0.6 2.4 2.2 1.6 0.Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.8 4 Reinforcement ratios (%) Chart-C1 .2 2.8 1 1. Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.1 Grade 30 Ast/bd Grade 40 Asc/bd 14 13 12 11 10 9 M/bd 2 Grade 30 Asc/bd Grade 45 Ast/bd Grade 35 Ast/bd Grade 45 Asc/bd Grade 35 Asc/bd Grade 40 Ast/bd 8 7 6 5 4 3 2 1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) Chart-C2 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.15 Grade 30 Ast/bd Grade 40 Asc/bd 14 13 12 11 10 9 M/bd 2 Grade 30 Asc/bd Grade 45 Ast/bd Grade 35 Ast/bd Grade 45 Asc/bd Grade 35 Asc/bd Grade 40 Ast/bd 8 7 6 5 4 3 2 1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) Chart-C3 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.2 Grade 30 Ast/bd Grade 40 Asc/bd 14 13 12 11 10 9 M/bd 2 Grade 30 Asc/bd Grade 45 Ast/bd Grade 35 Ast/bd Grade 45 Asc/bd Grade 35 Asc/bd Grade 40 Ast/bd 8 7 6 5 4 3 2 1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) Chart-C4 Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.25 Grade 30 Ast/bd Grade 40 Asc/bd 14 13 12 11 10 9 M/bd 2 Grade 30 Asc/bd Grade 45 Ast/bd Grade 35 Ast/bd Grade 45 Asc/bd Grade 35 Asc/bd Grade 40 Ast/bd 8 7 6 5 4 3 2 1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Reinforcement ratios (%) Chart-C5 Appendix D Underlying Theory and Design Principles for Plate Bending Element each of which comprises two bending moments and a shear force can be obtained. D-1 . Balancing of moments is similar. F3 and F4 which are node vertical shear of 4 elements at the common node will sum up to balance the externally applied force FA such that F1+ F2+ F3 + F4 = FA. FA.Appendix D Underlying Theory and Design Principles for Plate Bending Element By the finite element method. F2 Figure D-1b – Diagrammatic illustration of balancing of Node shear forces at a common node to 2 or more adjoining elements. F2. the summation of which will balance the applied load at the node. external load applied at the common node F4 F1 F3 For equilibrium. F1. “node forces” at each node of an element. Y axes and the force in the Z axis at the nodes of the plate bending element Figure D-1a – Diagrammatic illustration of the Node Forces at the four Nodes of a Plate Bending Element 1234. The four elements joined at the common node are displaced diagrammatically for clarity. a plate bending structure is idealized as an assembly of discrete elements joined at nodes. Figures D-1a and D-1b illustrates the phenomena. Through the analysis. FZ at Node 1 FZ at Node 4 FZ at Node 3 MX at Node 2 Note : MY at Node 3 Z Y 3 FZ at Node 2 2 MY at Node 2 X 1 MX at Node 1 MX at Node 2 MX. MY. FZ represents respectively the bending moments about X. They represent the actual internal forces within the plate structure in accordance with the plate bending theory. M T is therefore termed the “twisting moment”. As the hypotenuse can be in any directions of the plate structure. The existence of the twisting moment and its nature are discussed in the followings. it follows that at any point in the plate bending structure. M Y and M B which produces flexural stresses. The equations relating M B . twisting moments and shear forces per unit width in plate bending element. a twisting moment can be proved to be in existence by the plate bending theory. in order to achieve rotational equilibrium about an axis out of plane. as the resultant of M X and M Y does not necessarily align with M B . Woods (1968) has developed the famous Wood-Armer Equations to convert the bending moments and twisting moments (both are moments per unit width) at any point to “design moments” in two directions for structural design purpose. it “twists” the element and produce shear stress in the in-plane direction. M T with M X . Outline of the plate bending theory Apart from bending moment in two mutually perpendicular directions as well known by engineers. the shear stress will have to be “complementary”.Appendix D The finite element method goes further to analyze the “stresses” within the discrete elements. The vector direction of M T is normal to the face of the hypotenuse. So instead of “bending” the element like M X . M Y . R. M XY and θ derived from equilibrium conditions are stated as follows: D-2 .H. say M X and M Y in two mutually perpendicular directions coupled with a complementary twisting moment M XY as indicated in Figure 11a. Furthermore. A moment M B will generally be acting on the hypotenuse making an angle of θ with the X-axis as shown to achieve equilibrium. there will generally be two bending moments. It should be noted that “stresss” is a terminology of the finite element method which refer to bending moments. Consider a triangular element in a plate bending structure with two of its sides aligning with the global X and Y directions as shown in Figure D-2 where moments M X and M Y (both in kNm per m width) are acting respectively about X and Y. However. The bending and twisting moments constitutes a “moment field” which represents the actual structural behaviour of a plate bending structure. The phenomenon is in exact analogy to the in-plane stress problem where generally two direct stresses coupled with a shear stress exist and these components vary with directions. The shear stress will follow a triangular pattern as shown in Figure D-2 for stress-strain compatibility. so there will generally be a moment acting in the perpendicular direction of M B to achieve equilibrium which is denoted as M T . if θ is so varied that M T vanishes when θ = φ . again in exact analogy with the in-plane stress problem having principal stresses at orientations where shear stresses are zero. M 2 . The 2 M XY 1 angle φ can be worked out by φ = tan −1 (Eqn D-2) 2 (M X − M Y ) Plate Structure MT MY Y MX X θ MB Complementary shear stress pattern Figure D-2 – Derivation and nature of the “Twisting Moment” D-3 . then the element will be having pure bending in the direction.Appendix D MB = 1 (M X + M Y ) + 1 (M X − M Y ) cos 2θ + M XY sin 2θ 2 2 1 M T = (M X − M Y ) sin 2θ − M XY sin 2θ 2 (Eqn D-1) In addition. The moments will be termed the “principal moments” and denoted as M 1 . The effects of the twisting moments have been taken into account in the formulae. as similar to the in-plane stress problem. For bottom steel reinforcement provisions: ∗ M X = M X + M XY if ∗ MX ≥ 0. the derivation of which is based on the “normal yield criterion” which requires the provided reinforcing steels at any point to be adequate to resist the normal moment which is the bending moment M B in any directions as calculated from (Eqn D-1). ∗ M Y = M Y + M XY if 2 ∗ MY ≥ 0 If ∗ M X < 0 . However. it will be adequate if the designer designs for these principal moments in the principal directions which generally vary from point to point.Appendix D MXY Y X MX MY M1 M2 MXY MY MX MXY M2 φ M1 MXY Figure D-3a – General co-existence of bending moments and twisting moment in a plate bending structure Figure D-3b – Principal moment in a plate bending structure Again. The “stress” approach for design against flexure would therefore involve formulae for providing reinforcing steels in two directions (mostly in orthogonal directions) adequate to resist the “moment field” comprising the bending moments and twisting moments. Theoretically. The Wood Armer Equations are listed as follows. one may view that the plate bending structure is actually having principal moments “bending” in the principal directions which are free of “twisting”. then ∗ MX =0 ∗ and M Y = M Y + M XY MX D-4 . practically this is not achievable for reinforced concrete structures as we cannot vary the directions of the reinforcing steels from point to point and from load case to load case. The most popular one is the “Wood Armer” Equations by Woods (1968). Q max . more accurate design or checking can be carried out which is based on finite element analysis by which an accurate shear stress distribution in the slab structure can be obtained.Appendix D ∗ M Y < 0 . then M Y = 0 ∗ and M X = M X − M XY MY (Eqn D-3) The equations have been incorporated in the New Zealand Standard NZS 3101:Part 2:1995 as solution approach for a general moment field. The “stress” approach is therefore based on the actual structural behaviour of the plate bending structure which is considered as a direct and realistic approach.7 of the Code by which the punching shear load created by column (or pile in pile cap) is effectively averaged over a perimeter.1.5. It can be easily shown that the maximum shear after “compounding” these two components will occur in a plane at an orientation θ = tan −1    Q XZ  QYZ     on plan and the value of the maximum shear is D-5 . The finite element analysis outputs the “shear stresses” (shear force per unit width) in accordance with the general plate bending theory at the “X-face” and “Y-face” of an element which are ∂M Y ∂M XY ∂M X ∂M XY respectively Q XZ = − + QYZ = − and . then ∗ MX =0 M XY MX 2 ∗ ∗ M Y < 0 . Design has to cater for all these components to ensure structural adequacy. Design against shear As an alternative to checking or designing punching shear in slab in accordance with 6. The approach is particularly suitable for structures analyzed by the finite element method which produces a complete set of components of the internal forces of the plate bending structures including twisting moments. then ∗ MY = 0 ∗ and M X = M X + If M XY MY 2 For top steel reinforcement provisions: ∗ M X = M X − M XY if ∗ MX ≤ 0’ ∗ M Y = M Y − M XY ∗ and M Y = M Y − if 2 ∗ MY ≤ 0 If If ∗ M X > 0 . as ∂x ∂y ∂y ∂y diagrammatically illustrated in Figure D-4. So the designer needs to check or design for Qmax at the spot. Derivation A rectangular element extracted from a plate structure showing shear per unit width in the X and Y directions For vertical equilibrium. The actual shear stress is Qmax . potential shear failure and Derivation of the magnitude and orientation of the design shear stress Following the usual practice of designing against shear in accordance with the Code. Plate structure Out-of plane shear QXZ.Appendix D Qmax = Q XZ + QYZ 2 2 as per the illustration in the same Figure. Thus one can view that both Q XZ and QYZ are components of the actual shears in a pre-set global axis system. There is no necessity to design for Q XZ and QYZ separately. QYZ and Qθ Z Y X Q XZ QYZ Qθ θ Plan θ Q XZ QYZ The triangular element formed from the rectangular element by cutting into half. the maximum shear on the hypotenuse is obtained. By varying the angle θ. if the Qmax does not exceed allowable shear strength of concrete based on vc (the D-6 . the action of which tends to produce shear failure at the angle θ on plan as shown in Figure D-3. Qθ can be expressed as Qθ × 1 = Q XZ sin θ + QYZ cos θ For Qθ to be maximum set dQθ Q = 0 ⇒ Q XZ cos θ − QYZ sin θ = 0 ⇒ tan θ = XZ dθ QYZ Q max θ ⇒ Qθ = Qmax = Q XZ 2 Q XZ 2 + QYZ 2 + QYZ 2 Q XZ 2 + QYZ 2 X = Q XZ 2 + QYZ 2 Q max Potential shear failure at the orientation θ = tan −1 (QXZ / QYZ ) Figure D-4 – Diagrammatic illustration of shear “stresses” in the X and Y faces of an element in Plate bending structure. The “stress” approach for shear design based on Qmax can best be carried out by graphical method. (iii) locations where the stresses exceed the enhanced strengths be reinforced by shear links as appropriate in accordance with established code requirements as shown in Figure D-5(c). reinforcements will be required to cater for the difference.Appendix D design concrete shear stress with enhancement as appropriate) no shear reinforcements will be required. An illustration of the method for a raft footing is indicated in Figure D-5 as : (i) an enveloped shear stress (shear force per unit width) contour map of a structure due to applied loads is first plotted as shown in Figure D-5(a). so as to avoid handling the large quantity of data obtainable in the finite element analysis. Otherwise. (ii) the concrete shear strength contour of the structure which is a contour map indicating the shear strength of the concrete structure after enhancement of the design concrete shear stress (vc) to closeness of supports in accordance with established code requirements (say BS8110) is plotted as shown in Figure D-5(b). Figure D-5a – Stress contour of enveloped shear “stresses” of a raft footing due to applied load D-7 . Appendix D Figure D-5b – Strength contour of the raft footing with enhancement Figure D-5c – Arrangement of shear reinforcements D-8 . Appendix E Extracts of Design Charts for Slabs . . . Appendix F Derivation of Design Formulae for Rectangular Columns to Rigorous Stress Strain Curve of Concrete . 3.8 of HKCoP2004 be represented by the equation σ = Aε 2 + Bε (where A and B are constants) (Eqn F-1) dσ So = 2 Aε + B (Eqn F-2) dε dσ = Ec ⇒ B = Ec where Ec is the tangential Young’s Modulus of As dε ε =0 concrete listed in Table 3.34 f cu ⇒ = ⇒ ε0 = γ mε 0 Ec γ m 2ε 0 2 γ mε 0 2 ε0 (Eqn (F-4) (accords with 3.0035 x u h Figure F-1 – Strain diagram across concrete section F-1 .14 of the Concrete Code Handbook) E 1.34 f cu A=− c where ε 0 = 2ε 0 Ec γ m So the equation of the parabola is σ =− Ec 2 ε + Ecε for ε ≤ ε 0 2ε 0 Consider the linear strain distribution xε 0 / ε ult ε = ε0 ε ult = 0.67 ∴ 0.Appendix F Derivation of Design Formulae for Rectangular Columns to Rigorous Stress Strain Curve of Concrete (I) Computing stress / force contribution of concrete stress block Assuming the parabolic portion of the concrete stress block as indicated in Fig. Also dσ dε = 0 ⇒ 2 Aε 0 + B = 0 ⇒ A = − ε =ε 0 E B =− c 2ε 0 2ε 0 (Eqn F-3) As σ = 0.67 f cu Ec 1.67 f cu γm − f cu Ec when ε = ε 0 =− Ec 0.2 of the Code. we can formulate total force offered by the parabolic section as Fc1 given by centre of mass b Area = 2 ab 3 a 3 a 8 Figure F-3 – Geometrical Properties of Parabola F-2 .6 0.8 1 Distance Ratio from Neutral axis 0. Stress Strain Profile for Grade 35 18 16 14 Stress (MPa) 12 10 8 6 4 2 0 0 0.3769 where ε0 = 0.Appendix F At distance u from the neutral axis.2 0. the stress strain profiles for grade 35 within the concrete compression section are plotted in Figure F2 for illustration.4 0. ε = ε ult So stress at u from the neutral axis up to x 2 u x ε0 is ε ult 2 E E  Eε Eε u u  σ = − c ε 2 + Ecε = − c  ε ult  + Ec  ε ult  = − c ult2 u 2 + c ult u (Eqn F-5) x x x 2ε 0 2ε 0  2ε 0 x  Based on (Eqn F-5).001319 Figure F-2 – Stress strain profile of grades 35 By the properties of parabola as shown in Figure F-3. 67 cu = bx 3 ε ult γm 3γ mε ult h  5 ε0  3 ε0  − x  = Fc1  − x1 −  8 ε  ult    2  8 ε ult     (Eqn F-6) and the moment exerted by Fc1 about centre line of the whole section h  ε M c1 = Fc1  − x1 − 0  ε ult 2  (Eqn F-7) The force by the straight portion is Fc 2 = 0.34ε 0 f cu  x   1  x  5 ε 0  0.67 f cu = + = bh bh bh γm 0. the F-3 . column sections Cases 1 to 7 with different stress / strain profile of concrete and steel across the x .67 f cu  x  ε  h  ε  x 1 ε 0   ε0  x M c 2 0. Pursuant to the derivation of the stress strain relationship of concrete and steel.34ε 0 f cu  h  5 ε 0  1 1.Appendix F Fc1 = b 1.34ε 0 f cu f 2 ε0 x0.34ε 0 f cu  x   1  x  5 ε 0   bx  − x1 − =    −  1 − 2   8 ε  bh 2 = 3γ ε 3γ m ε ult bh  h   2  h  8 ε ult  2  ult   m ult      0.C. d' . The section is reinforced by continuous reinforcements Ash along its length h idealized as continuum and reinforcements at its end faces Asb with cover column section due to the differences in the neutral axis depth ratios.67 f cu  ε x− x 0  γm  ε ult  0.67 f cu  x  = =  1 − 0  1 − 1 − 0      −  1 − 2 2  ε   ε h  8 ε  + 2γ  3γ mε ult  h   2  h  bh bh  h  ult   m ult   ult    Fc Fc1 Fc 2 0.67 f cu bx  ε b = 1 − 0   ε γm ult       (Eqn F-8) The moment offered by the constant part about the centre line of the whole section is h  ε M c 2 = Fc 2  − 1 − 0  ε ult 2   x   2    1 ε0 1 −  3ε ult  (Eqn F-9) Thus if full section of concrete in compression exists in the column section x  (Eqn F-10) h  M c1 1.67 f cu bx  1 − 0   − 1 − 0   2 = =  1 − 2     ε  1 − 1 − ε  h   ε  2  ε  2 bh 2γ m  h  γm bh ult   ult   ult   ult          ε   ε  x  M c M c1 + M c 2 1.67 f cu  x  1 1 ε 0  1 1 ε 0 1ε  = + − + −  0   − γ m  h  2 6 ε ult  2 3 ε ult 12  ε ult        2  x     h   (Eqn F-11) (II) Derivation of Basic Design Formulae of R. are h investigated. the stress strain diagram of concrete and steel for Case 1 is as indicated in Figure F-4.1 : Steel compressive force in the portion steel elastic zone by Asb is 7  d'  x − d' Fsc1 =   × 0.87 f y × sh   × = 0.87 f y × sh   = 0.87 f y × sh  h −  = 0.87 f y × Ash   h  7  7 h Steel compressive force in the portion steel elastic zone by Ash is A  4x  1 2 x Fsc 3 = 0. = + + − − bh bh bh bh bh bh bh bh By (Eqn 10) F-4 .5 Asb (Eqn F-12) (Eqn F-13) (Eqn F-14) (Eqn F-15) Steel tensile force in the portion steel plastic zone by Ash is A  11x   11 x  Fst 2 = 0.87 f y × 0.87 f y × 0.87 f y × 0.5 Asb 4 x  4x / 7  Steel compressive force in the portion steel plastic zone by Ash is A  3x  3 x Fsc 2 = 0.87 f y × Ash 1 −  h  7  7 h  Steel tensile force in the portion steel elastic zone by Ash is Fst 3 = 0.Appendix F stress strain diagram of concrete and steel for Cases 1 to 7 are as follows.87 f y × Ash   h  7  2 7 h (Eqn F-16) (Eqn F-17) To balance the external load N u N u Fc1 Fc 2 Fsc1 Fsc 2 Fsc 3 Fst1 Fst 2 Fst 3 = + + + + − − − bh bh bh bh bh bh bh bh bh N u Fc Fsc1 Fsc 2 Fst1 Fst 2 F Fsc 3 as and st 3 are equal and opposite.87 f y × Ash   h  7  2 7 h Steel tensile force in the portion steel plastic zone by Asb is Fst1 = 0.87 f y × Ash  4 x  1 2 x   × = 0. under the definition of symbols as : b: h: x: Asb : d' : Ash : width of the column length of the column neutral axis depth of the column total steel area at the end faces of the column concrete cover to the centre of the end face steel total steel area along the length of the column Case 1 – where x/h ≤ 7/3(d’/h) Pursuant to the derivation of the stress strain relationship of concrete and steel.5 Asb = 1 −  × 0. 002 ε = ε0 4x/7 4x/7 3x/7 ε ult = 0.67 f cu γm Concrete stress Block Fsc1 h–11x/7 4x/7 Fsc3 Fst2 Fst3 4x/7 Fsc2 0.0035 u h x Strain diagram across whole section Fc1 Fc2 0.87 f y sb h bh  h  bh  8 8 h x   (Eqn F-18) d' xε 0 / ε ult d' εs = 0.87fy 3x/7 Steel stress Block Fst1 Figure F-4.5 bh + 0.87 f y  2 − 1 sh +  −    0.87 f y × 0.Appendix F ⇒  x 7  d' Asb Ash  3 x    h + 4 1 − x  × 0.67 f cu = bh γm  1 ε0 1 −  3ε ult  x A A x 3 7 d' h   + 0.87 f y × bh  7 h       A A  11 x  − 0.87 f y × 0.87 f y × sh 1 −  bh bh  7 h N u 0.5 sb − 0.67 f cu bx  1 ε 0 1 − =  3ε bh γm ult  ⇒ N u 0.1 – Concrete and steel stress strain relation for Case 1 Re-arranging (Eqn F-18) F-5 . 87 f y   −  (Eqn F-23) 2 bh  7 h  2 21 h  bh  7 h  2 21  bh M st1 0.87 f y sh   −  u + 0.87 f y sb = 0 bh h 8 x (Eqn F-19) can be used for solve for h To balance the external load M u M u M c1 M c 2 M sc1 M sc 2 M sc 3 M st1 M st 2 M st 3 = + + + + + 2 + + (Eqn F-20) bh 2 bh 2 bh 2 bh 2 bh 2 bh 2 bh bh 2 bh 2 By (Eqn F-11) 2 M c M c1 M c 2 0.87 f y sh   −  + 0.87 f y × 1 −   2 bh  bh  7 h  2 14 h  7 h  14 h  bh 2 Ash  x  65  x   = 0.87 f y × 0.87 f y × 0.5  −  h x bh  2 h  bh  2 h  4 bh 2 A  1 d '  11 7 d ' h  = 0.5 Asb  h Asb  1 d '   = (Eqn F-24)  − d '  = 0.87 f y  sh −      3ε  γm  bh  h   bh  bh 8 bh  h ult   A d' 7 (Eqn F-19) − 0.87 f y × 0.67 f cu  x  1 1 ε 0  1 1 ε 0 1  ε0   x   = 2 + 2 = + − + −    − 2  ε   h  γ m  h  2 6 ε ult  2 3 ε ult 12  ult    bh bh bh     M sc1 7  d '  A h A  1 d'  7  d' h  = 1 −  × 0.87 f y × Ash    bh  7 h  21 h 2  bh 2  7 h  21 2  bh (Eqn F-26) M sc1 + M st1 7  d ' h  Asb  1 d '  Asb  1 d '  = 1 −  × 0.87 f y sb  −  − (Eqn F-27)  bh  2 h  8 8 h x  Ash  11 x  11 x  A  3 x  1 3 x  M sc 2 + M st 2 = 0.5 sb  − d '  = 1 −  × 0.87 f y × 1 −   2 2 bh  7 h  14  7 h  14 h  bh  bh A  2 x  29 x 1  M st 3  2 x  29 x h  1 −  −  2 = 0.Appendix F 2  0.87 f y × sh  = 0.87 f y × Ash   −  2 = 0.5  −  + 0.87 f y × Ash   −  2 = 0.87 f y ×   2 bh  7 h  21 h 2  bh  7 h  2 21 h  bh A 32  x  = 0.87 f y × sh 1 − (Eqn F-25)   = 0.87 f y × 0.5 sb  −  2 2 x h x bh  2 h  4 bh bh  2  4 (Eqn F-21) M sc 2 Ash  3 x  1 3 x   3 x  h 3x  1 = 0.87 f y   −  (Eqn F-22) 2 bh  7 h  2 14 h  bh  7 h  2 14  bh Ash  2 x  1 13 x  M sc 3  2 x  h 13x  1 = 0.87 f y × 0.87 f y sh   bh 147  h  2 (Eqn F-29) F-6 .87 f y × 0.87 f y sh   −  + 0.5  −  bh  2 h  bh 2 bh 2 2  Ash  11 x  11 x  A  11 x  11x  M st 2 = 0.67 f cu  1 ε 0  N A  x 3 Asb  x A 1 −  + 2 × 0.87 f y (Eqn F-28)   −    bh  h  49  h     Ash  2 x  29 x 1  A  2 x  1 13 x  M sc 3 + M st 3 −  = 0. i. Case 2(a) – d' 3 d' 3 ≥ and Case 2(b) – < h 14 h 14 For Case 2(a). where doesn’t exist.87 f y  sb  −  −  + sh   −     bh  2 h  8 8 h x  bh  h  147  h       (Eqn F-30) (Eqn F-31) Case 2 – where 7/3(d’/h) < x/h ≤ 7/11 (1 – d’/h) There are two sub-cases to be considered in Case 2.87 f y sh   −     bh 2 bh  2 h  8 8 h x  bh  h  147  h     Total ⇔  A  1 d '  11 7 d ' h  A  x  163  x  2   Ms   = 0.87 f y sh x bh bh ⇒ = h  0.5 Asb and M sc1 = 0. so cancel out.87 f y  2 − 1 sh    3ε h  h  bh ult   Nu A + 0.87 f y × 0. By formulation similar to the above. However.67 f cu  1 ε 0  A  1 −  + 2 × 0.67 f cu = bh γm F-7 (Eqn F-32) (Eqn F-33) .Appendix F 2 Ms A  1 d '  11 7 d ' h  A  x  163  x   = 0.e.87 f y sh     bh   γ m  3 ε ult  N u 0.5 Asb  −  2 h  It can be seen that Fst1 and Fsc1 are identical but opposite in direction.  1 ε0  x A x 1 −  + 0.87 f y  sb  −  −  + sh   −    2 bh  bh  2 h  8 8 h x  bh  h  147  h       Substituting into (Eqn F-20) 2 M u 0. For Case 2(b). ≥ and 1 −  < .67 f cu  x  1 1 ε 0  1 1 ε 0 1  ε0   x   = + − + −    − 2  ε   h  γ m  h  2 6 ε ult  2 3 ε ult 12  ult    bh      A  1 d '  11 7 d ' h  A  x  163  x  2     + 0. 7 d' 1 7  d' 1 d' 3 ≥ .87 f y × 0. where Figure 3. So this case h 14 h 2 11  3 h 2 d' 3 < both Asc1 and Ast1 are in the plastic zone as shown in h 14 The various components of stresses in concrete and steel are identical to that of Case 1 except that of Asc1 where  1 d' Fsc1 = 0.87 f y sb  −  − + 0.2. 0035 ε = ε0 4x/7 u h x Strain diagram across whole section Fc1 Fc2 0.67 f cu = γm bh 2     2  x     h   (Eqn F-34) d' xε 0 / ε ult d' εs = 0.2 – Concrete and steel stress strain relation for Case 2(b) Case 3 – where 7/11(d’/h) < x/h ≤ 7/11 for d’/h > 3/14 and 7/11(1 – d’/h) < x/h ≤ 7/11 for d’/h < 3/14 The concrete / steel stress / strain diagram is worked out as shown in Figure F-4.67 f cu γm Concrete stress Block h–11x/7 4x/7 Fsc3 Fst2 Fst3 4x/7 Fsc2 0.002 ε ult = 0.87fy 3x/7 Steel stress Block Figure F-4.Appendix F  1ε  x  1 1 ε 0  1 1 ε 0 + − + −  0   −  h  2 6 ε ult  2 3 ε ult 12  ε ult     A  1 d '  A  x  163  x  2     + 0.87 f y  sb  −  + sh   −     bh  2 h  bh  h  147  h       M u 0. It F-8 .3. 87 f y (h − x − d ') 4x / 7 × 0. N u 0.3 – Concrete and steel stress strain relation for Case 3 By derivation similar to the above in balancing axial load.Appendix F should be noted that the components of stresses are identical to that of Case 2 except that by Ast1 which is Fst1 = 0.87 f y  − − 1 Asb  x x 8 (Eqn F-35) d' xε 0 / ε ult d' εs = 0.87 f y (h − x − d ') 7 A x 8 sb  h d'  7 = 0.0035 ε = ε0 4x/7 u h x Strain diagram across concrete section Fc1 Fc2 0.67 f cu  1 ε 0  x Asb  11 7 h 7 d '   x  Ash 1 − =  3 ε  h + 0.5 Asb = 0.67 f cu γm Concrete stress Block h–11x/7 4x/7 Fsc3 Fst3 4x/7 Fsc2 0.87fy 3x/7 Fst2 Steel stress Block Figure F-4.002 ε ult = 0.87 f y  2 h − 1 bh  bh γm      ult  (Eqn F-36) F-9 .87 f y bh  8 − 8 x + 8 x  + 0. + 0.Appendix F Re-arranging (Eqn F-36) 2  0.87 f y bh  8 − 8 x + 8 x  + 0. N u 0.87 f y  −     −    −  +   −     (Eqn F-38)  2 h   8  x  8  h  x  8  bh  h  147  h   bh      Case 4 – where x < h < 11x/7.87 f y  bh 8 + bh 4  − bh  h       ult   γm     7  Asb  d '  Ash  =0  − 1 − 8  bh  h  bh    which can be solved as a quadratic equation. It should be noted that the components of stresses are identical to that of Case 3 except that by Ast 2 which vanishes.67 f cu  1 ε 0  x Asb  11 7 h 7 d '  Ash  9 x 7 h 7  1 − =  3 ε  h + 0.67 f cu  x  1  x  1 ε 0 1 ε 0  x  1  ε 0   x    = +    1 −  −  −  2  ε   h     γ m  h  2  h  6 ε ult 3 ε ult  h  12  ult  bh    2  1 d '   7  h  7  d '  h  3  A 7 h 9 x 9  x   Ash    sb + − + + 0.87 f y bh  h  + 0.87 f y bh  h  + 0.e.4.67 f cu  1 ε 0  9 Ash  x    Asb 11 Ash 7  N u  x  1 −   3 ε  − 56 × 0.87 f y (Eqn F-37)  − 1 = 0 8 bh  h  By balancing moments : 2 M u 0.67 f cu  x  1 1 ε 0  1 1 ε 0 1  ε0   x   = + − + −    − 2  ε   h  γ m  h  2 6 ε ult  2 3 ε ult 12  ult    bh      1 d '   7  h  7  d '  h  3  A  x  163  x  2  Ash    sb + 0.87 f y  bh 8 − bh  − bh  h       ult   γm     7 Asb  d '  + 0.67 f cu  1 ε 0  Ash  x    Asb 11 Ash  N u  x  1 −   3 ε  + 2 × 0.87 f y  −     −    −       2 h   8  x  8  h  x  8  bh  48 x 112 h 392  h   bh      (Eqn F-41) F-10 . i. 7/11 < x/h ≤ 1 The concrete / steel stress / strain diagram is worked out as shown in Figure F-4. Hence by derivation similar to the above by balancing axial load.87 f y (Eqn F-40) By balancing moment : 2 M u 0.87 f y bh  − 56 h − 8 x + 4   bh γm      ult  (Eqn F-39) Re-arranging 2  0. F-11 . Fst 3 vanish.4 – Concrete and steel stress strain relation for Case 4 Case 5 – where x>h>(1 – ε0/εult)x.67 f cu γm Concrete stress Block h–x Fsc3 Fst3 4x/7 Fsc2 0.0035 ε = ε0 ε1 = 0.0035(h − x ) x 4x/7 u h x Strain diagram across concrete section Fc1 Fc2 0.87fy 3x/7 Steel stress Block Figure F-4. i. The whole section is in compression so that the neutral axis depth ratio is greater than unity and hence becomes a hypothetical concept.Appendix F d' xε 0 / ε ult d' εs = 0.e 1 < x/h ≤ 1/(1 – ε0/εult) The concrete / steel stress / strain diagram is worked out as follows. The contribution by Fst1 become compression Fsc1' and Fst 2 .002 ε ult = 0. The contribution by Fc1 becomes partial and need be calculated by integration to partial limits. 5 – Concrete and steel stress strain relation for Case 5 Concrete compressive stresses and forces xε 0 / ε ult 2 Ec ε ult 2 Ec ε ult u + u Fc1 = ∫ σ bdu where σ = − 2ε 0 x 2 x x−h xε 0 / ε ult (Eqn F-42) xε / ε 2 xε / ε  Ec ε ult 2 2 Ec ε ult  Ec ε ult b 0 ult2 Ec ε ult b 0 ult Fc1 = ∫ − u + u bdu = − u du + ∫ u du x x 2ε 0 x 2 2ε 0 x 2 x∫h x −h  x−h −  2 2  E c ε ult 2  ε 0 3  x  Fc1  E c ε ult  ε 0 Eε   −   +  E c ε ult − c ult  ⇒ = − 1 − 1  6ε 0  ε ult 3 2ε 0  bh  2  ε ult 2     h     F-12 .87fy 4x/7 3x/7 Steel stress Block Figure F-4.0035(h − x ) x d' 4x/7 u x h Strain diagram across concrete section Fc1 Fc2 0.002 ε ult = 0.0035 ε = ε0 ε1 = 0.Appendix F d' xε 0 / ε ult εs = 0.67 f cu γm Concrete stress Block Fsc3 Fsc2 0. 87 f y 9 sh  x  56 bh  h   F-13 .87 f y 1 − × h − × 4x / 7 h  7  7  2  4x / 7  h  F  7 7 h 33 x  Ash ⇒ sc 3 = 0.5 sb bh bh 0.87 f y × (Eqn F-46) (Eqn F-47) (Eqn F-48) (Eqn F-49) 3x  3x  1 x − h Ash  x − h  Ash   ×  h −  + 0.87 f y  − − (Eqn F-50)  bh  4 8 x 56 h  bh Fsc 2 Fsc 3  3 x  Ash  7 7 h 33 x  Ash  7 7 h 9 x  Ash = 0.Appendix F  E ε 2 E ε  h E ε 2  h 2 +  c ult − c ult  − c ult    2 x 6ε 0  x    2ε 0 0.67 f cu  ε x 1 − 0  (same as the previous cases) Fc 2 =  ε h γm  ult  2  Ec ε ult 2  ε 0 3  Eε  ε Fc Fc1 Fc 2  Ecε ult  ε 0  2 − 1 −  3 − 1 + c 0  1 − 0 = + =  ε ε  ε  6ε 0  ult 2  bh bh bh  2  ult ult    Eε 2 Eε +  c ult − c ult  2  2ε 0  h Ecε ult 2  h  2  −   x 6ε 0  x   (Eqn F-43) (Eqn F-44) 2  x  Ec ε ult    +  Ec ε ult −  h  2ε 0      (Eqn F-45) Steel compressive force in the portion steel plastic zone by Asb is F A Fsc1 = 0.87 f y × Fsc 3 = 0.87 f y   h  7  bh  7 h  bh Steel compressive force in the portion steel elastic zone by Ash is Fsc 2 = 0.87 f y  + 0.87 f y × 0.87 f y  − 1 −   h  bh 2ε 0 2 x 6ε 0  x   8 8 x   7 7 h 9 x  Ash + 0.5 Asb ⇒ sc1' = 0.87 f y  − −   4 8 x 56 h  bh ⇔  1 ε 0 2 1 ε 0 1 1 ε ult Nu  =  Ec ε ult  −  6ε 2 +2ε −2+6 ε bh  0 ult ult    A   − 0.87 f y × 0.87 f y 1 − 1 −  Fsc1' = 4x / 7 bh h  8 bh  x Fsc1 Fsc1' 11 7 h  d '  Asb + = 0.5 Asb ⇔ sc1 = 0.87 f y  − − + = 0.87 f y ( x − h + d ') F  h  d '  7 Asb × 0.87 f y  − 1 −   bh bh h  bh  8 8 x Steel compressive force in the portion steel plastic zone by Ash is Ash  3 x  F  3 x  Ash   ⇔ sc 2 = 0.87 f y  − −    bh bh  7 h  bh  4 8 x 56 h  bh  4 8 x 56 h  bh (Eqn F-51) N u Fc Fsc1 Fsc1' Fsc 2 Fsc 3 As = + + + + bh bh bh bh bh bh 2 2 2  Eε   Eε ε 3  ε  N u  Ec ε ult  ε 0 Eε   2 − 1 − c ult  0 3 − 1 + c 0 1 − 0  x +  Ec ε ult − c ult  =   bh  2  ε ult 6ε 0  ε ult 2  ε ult  h  2ε 0            E ε 2 E ε  h E ε 2  h 2 11 7 h  d '  Asb   +  c ult − c ult  − c ult   + 0. Appendix F 2  Eε 11 Asb +  Ec ε ult − c ult + 0.87 f y 2 8 bh  2ε 0 + 0.87 f y − 2ε 0 8 bh 4 bh bh  h    2 2  E c ε ult E c ε ult Asb 7  d '  7 Ash  x E c ε ult + − + 0.87 f y 1 +  − 1  −  sb 8   h  x  2 h  bh xε 0 / ε ult F-14 .67 f cu  x  1 − 0   − 1 − 0   2 = =  1 − 0  1 − 1 − 0   2  ε  2  ε  2 bh  ε   ε h γm 2γ m  h  bh ult   ult   ult   ult      M sc1 0.67 f cu bx  ε  h  ε  x  1 ε   ε  x  0.87 f y × 0.87 f y =0  − 1 − 0.87 f y 9 sh  x     56 bh  h    2 2  Eε 11 Asb 7 Ash N u  x     Ec ε ult − c ult + 0.87 f y  2ε 0 8 bh  E ε 2 E ε 7 Asb +  c ult − c ult + 0.87 f y + + 0.87 f y 1 − + 2 8  x h x  2 h  bh bh h  x x  8 bh  2 7   d '  h  1 d '  A = −0. bh bh Summing the Moments as follows : Concrete compressive stresses and moments xε 0 / ε ult 2 Eε Eε h  M c1 = ∫ σ bdu  − x + u  where σ = − c ult2 u 2 + c ult u 2ε 0 x x 2  x−h  E c ε ult 2 2 E c ε ult   h  (Eqn F-54) ∫h − 2ε 0 x 2 u + x u b 2 − x + u du     x−   Integrating and expanding 3  1  1 x   ε 2 x  x  2 M c1 h  1  ε 0 x h  0 = E c ε ult   −   2 − 1 + 2 −  +  3 − 1  + 3 − 3 +    h  h  x  3  ε ult h x  bh 2  2  2 h   ε ult        M c1 = − 2 2 2 3 4 x  x  2 E c ε ult  1  1 x   ε 0 h  h   1  ε x h  h     −   3 − 1 + 3 − 3 +    +  0 4 − 1  + 4 − 6 + 4 −     h  h  2ε 0  3  2 h   ε ult x  x   4  ε ult h x  x          M c 2 0.87 f y −   8 bh  x 6ε 0 h  Re-arranging (Eqn F-52)   1 ε 0 2 1 ε 0 1 1 ε ult Ec ε ult  −   6ε 2 +2ε −2+6 ε 0 ult  ult   3  A   − 0.5 sb  −  2 2 bh  2 h  bh bh  2 M sc1' 7  h d ' h  1 d '  Asb 1  h d '  7 Asb  h = −0.87 f y 7 Ash   4 bh   h   (EqnF-52)  x 2 2 7 Ash  h Ec ε ult  d'  − 1 − 0. sb and under applied N u .5 Asb  h A  1 d'  =  − d '  = 0.87 f y 1 − +   −   − d '  = −0.87 f y × 0.87 f y  − bh 8  h  2 8 bh  h 6ε 0  2ε 0   (Eqn F-53) x that can be solved by formulae or by numerical which is a cubic equation in h A A method with pre-determined sh . e. i.87 f y  − +     48 x 112 h 392  h   bh   2 3 2  1 ε0 Mu 1 ε0 1 ε 0 1 1 ε ult  x    − ⇒ 2 = Ec ε ult  + − + − 3 6 ε ult 2 4 ε ult 6 24 ε 0  h  bh   24 ε ult 2  1 ε0  1 1 ε ult 1 ε 0 1 1 ε ult  x + E c ε ult − + − +  + E c ε ult  − 2  12 12 ε 4 ε ult 4 12 ε 0  h  12 ε ult 0    h 1 ε ult  h   + E c ε ult   x 24 ε 0  x   2 2 7 h 9 x Asb  1 d '   7  d '  h 3  9  x   Ash + 0.87 f y sb ×  −   1 −  −  2 bh  2 h   8  h  x 8 bh bh  3 x 9  x  2  Ash M sc 2  3 x  1 3 x  Ash = 0.87 f y  −    − = 0.5 sb  −  − 0.87 f y 1 +  − 1  −  sb 2 2 bh  2 h  8  h bh bh  x  2 h  bh M M A  1 d '  7  d '  h 3  (Eqn F-55) ⇒ sc + sc21' = 0.87 f y ×  −  1 −  −  + 0. 1/(1 – ε0/εult) < x/h ≤ 7/3 The concrete / steel stress / strain diagram is worked out as follows : F-15 .67 f cu 2γ m ε  x   1 − 0  ε  h  ult   ε  1 − 1 − 0   ε ult    x A  1 d '  7 d ' h 3   + 0.87 f y    bh 2  7 h  2 14 h  bh 14 h 98  h   bh   M sc 3 A 5 7 h 3 x 3 x A  7 h 3 9 x  2 x 1  −  − 0.87 f y  − +    bh  2 h   8  h  x 8   48 x 112 h 392  h   bh   (Eqn F-57) Case 6 – where (1 – ε0/εult)x>h>3x/7.87 f y × sh ×  − −   2 bh  2 4 x 4 h  14 h bh  8 x 4 56 h  7 h 6  bh 2 A  7 h 33 x 45  x   = 0.87 f y × 0.87 f y sh  − +    bh  48 x 112 h 392  h     2 M sc 2 M sc 3 Ash  7 h 9 x 9  x  ∴ 2 + (Eqn F-56) = 0.Appendix F A  1 d' M sc M sc1' 7   d '  h  1 d '  A + = 0.87 f y × sh ×  − + = −0.87 f y sb ×  −   1 −  −    h bh  2 h   8  h  x 8   2 7 h 9 x 9  x   Ash + 0.87 f y − +     bh bh 2 bh  48 x 112 h 392  h     Total Moment M M M M M M M = c21 + c22 + sc1 + sc21' + sc22 + sc 3 2 2 bh bh bh bh bh bh bh 2 2  1  1 x   ε 2   3    Mu   0 − 1 x + 2 − h  + 1   ε 0 − 1 x  + 3 x − 3 + h  = Ec ε ult   −   2    bh 2 x  x  3   ε ult 3  h  h  2  2 h   ε ult  h      2 2 2 3 4  x  2 x Ec ε ult  1  1 x   ε 0 h  h   1  ε x h  h     −   3 − 1 + 3 − 3 +    +   0 4 − 1  + 4 − 6 + 4 −     h  h 2ε 0  3  2 h   ε ult x  x   4   ε ult h x  x          − + 0. 0035 x h Strain diagram across concrete section Fc2 0. So formulation as similar to above : N u 0.87 f y     + 56 bh  h   bh γm  8 bh 4 bh  h  F-16 .Appendix F d' εs = 0.67 f cu 11 7 h  d '  Asb  7 7 h 9 x  Ash = + 0.87 f y  − − 1 −    bh h  bh γm  4 8 x 56 h  bh  8 8 x (Eqn F-58) Re-arranging (Eqn F-58) 2 9 Ash  x   N u 0.67 f cu  11 Asb 7 Ash  x  − − 0.002 d' 4x/7 ε ult = 0.67 f cu γm Concrete stress Block Fsc3 Fsc2 0.6 – Concrete and steel stress strain relation for Case 6 Comparing with Case 5.87 f y  + 0. contribution by Fsc1 vanishes.87 f y  − + 0.87fy 4x/7 3x/7 Steel stress Block Figure F-4. The process involves : bh A (i) For the 7 cases discussed in the foregoing.87 f y − +     bh  2 h   8  h  x 8 bh  48 x 112 h 392  h     (Eqn F-60) Case 7 – where x/h > 7/3 In this case. eliminate sb between equations bh F-17 . The axial load will be simply N u 0. whilst that by steel is identical to Case 5 bh 2 M M M M M Total Moment = sc1 + sc21' + sc22 + sc 3 2 2 bh bh bh bh bh 2 2 Asb  1 d '   7  d '  h 3  Ash  7 h 9 x 9  x  = 0.87 f y ×  −   1 −  −  + 0.87 f y  sb + sh  (Eqn F-61) bh γm  bd bd  and the moment is zero. sh are set to zero.67 f cu A  A = + 0.87 f y   − 1 sb − =0  8  h  bh 8 bh  x which can be solved which is a quadratic equation in h In moment balancing moment (Eqn F-59) Mc = 0 .Appendix F 7  d'  A 7 Ash  − 0. In bh A the following derivations. Mu =0 (Eqn F-62) bh 2 (III) Design formulae for 4-bar column sections for determination of reinforcement ratios b h It is the aim of the section of the Appendix to derive formulae for the determination of Asb against applied axial load and moment under a pre-determined sectional size. the concrete and steel in the entire column section are under ultimate stress. 87 f y bh h = 0   γm    ult  0.87 f y sb = 7 d' 3 x bh − 8 h 8h A Substituting into (Eqn F-34) again with sh = 0 bh M u 0. x into the equation obtained by Back substitute the accepted value of h Nu A balancing to solve for sb . bh bh obtained from balancing Case 1 – where x/h ≤ 7/3(d’/h) By (Eqn F-18) with Ash =0 bh 2 Asb d ' 0.67 f cu 7  d '   1 ε 0  M u  x 7 d '  M u N u  1 d '    + u  − −   1 −  3 ε  − bh 2  h + 8 h  bh 2 − bh  2 − h  = 0  γm 8 h    bh 8  2 h      ult     0.67 f cu 3  1 ε 0 1 1  ε 0   x      ⇒ − −  γ m 8  3 ε ult 2 12  ε ult   h      2 2 0.Appendix F (ii) Nu Mu A and by making sb subject of formulae 2 bh bh bh Nu substitute into the equation for balancing in the equation for balancing of bh Mu x which can be solved of .87 f y 8 bh  h − 8 0.67 f cu  1 ε 0  x   N u 3 Asb  x 7 1 −  3 ε  h  −  bh − 0.67 f cu  1 29 d ' 7 ε 0 3 ε 0 d ' 7  ε 0  d '  x     + + − +  − +   γ m  2 16 h 24 ε ult 4 ε ult h 96  ε ult  h  h      2  N 11  1 d '  0.67 f cu 3γ m 2  ε  x N x  3 − 0   − u   ε  h  bh h A ult   ⇒ 0.67 f cu  x  1 1 ε 0   1 ε 0 1 1  ε 0  + = − −    − 2  γ m  h  2 6 ε ult   3 ε ult 2 12  ε ult bh     (Eqn F-63)     2  x      h    2  1 ε 0  x  N u x  1 −   −    bh h  1 d '  11 7 d ' h   3 ε ult  h  − −  2 h  8 8 h x  7 d' 3 x    −  8 h 8h   2 3 0. x will be valid if the value arrived at agree with the Solution in h pre-determined range of the respective case. The equation obtained in a polynomial in 2 bh h by equations (if quadratic or cubic or even 4th power) or by numerical methods.67 f cu  γm +     F-18 . 87 f y   8 h 8 h 7/3(d’/h) < x/h ≤ 7/11 – 7/11(d’/h) (applicable to d’/h > 3/14 only) Ash = 0.87 f y sb  − +    3ε h bh  8 8 x 8 x  ult   N u 0.87 f y  − +   8 8 x 8 h x 0.67 f cu  x   1  x  1 ε 0 1 ε 0  x  1  ε 0  = 2 − +    1 −  −  −  bh  bh γ m  h   2  h  6 ε ult 3 ε ult  h  12  ε ult    By (Eqn F-33) with Case 3 – where 7/11(d’/h) < x/h ≤ 7/11 for d’/h > 3/14 and 7/11(1 – d’/h) < x/h ≤ 7/11 for d’/h < 3/14 and Case 4 – where 7/11 < x/h ≤ 1 By (Eqn F-36) and setting N u 0.Appendix F The equation can be solved for x and substituted into (Eqn F-18) h 2 0.67 f cu  1 ε 0  x 1 −  − γ m  3 ε ult  h Asb bh   = bh  11 7 h 7 d ' h  0.87 f y  − +   8 8 x 8 h x A Substituting into (Eqn F-38) and again setting sh = 0 bh 2 M u 0.87 f y  −     −    −  11 7 h 7 d ' h  2 h   8  x  8  h  x  8    0.67 f cu = bh γm Ash =0. bh x N u  0. bh 2   x    1 d'       ÷ 0.67 f cu = ÷ h bh  γ m Case 2 – where  1 ε 0  1 −   3 ε  ult    A x Substituting into (Eqn F-34) obtained and calculate sb as h bh Asb  M u 0.67 f cu  ε 0  x  N u x 3 −   − bh h 3γ m  ε ult  h  Asb   = bh  7 d' 3 x  − 0.67 f cu  x  1  x  1 ε 0 1 ε 0  x  1  ε 0   x    = +    1 −  −  −  2  ε   h     γ m  h  2  h  6 ε ult 3 ε ult  h  12  ult  bh    N u 0.67 f cu 11  1 ε 0 1 1  ε 0  ⇔ − −  γ m 8  3 ε ult 2 12  ε ult       2  x  3    h   F-19 .87 f y  −     h     2 h    (Eqn F-64)  1 ε0  x A 11 7 h 7 d '  1 −  + 0.67 f cu  1 ε 0  x 1 −  − γ m  3 ε ult  h  1 d '   7  h  7  d '  h  3  bh   + 0. 67 f cu 7  d '  1 ε 0 d ' 1 d ' ε 0  3  1 d '  N u 11 M u  x  −  −  + − + −  − 11 −   γ m 8  h  3 ε ult h 3 h ε ult  8  2 h  bh 8 bh 2  h     7  1 d '  d '  N 7 M u  d'  −  −  − 1 u − − 1 = 0 (Eqn F-65) 2  8  2 h  h  bh 8 bh  h   x which is a cubic equation in   h 7  d' A  x Upon solving   lying between 1 −  and 1.87 f y sb ×  −   1 −  −  bh  2 h   8  h  x 8 2γ m  h  ε ult    ε ult  h     x to solve for . sb can be obtained by 11  h bh h N u 0. h A Back-substituting into the previous equation to solve for sb by trial and error bd F-20 .67 f cu  1 ε 0  x 1 −  − γ m  3 ε ult  h Asb bh   = bh  11 7 h 7 d ' h  0.87 f y  − +   8 8 x 8 h x (Eqn F-66) Case 5 – where 1 < x/h ≤ 1/(1 – ε0/εult) Solving Asb by (Eqn F-52) bd 2 Asb  N u  1 1 ε 0 1 ε ult 1 ε 0 − − = − + + 0.87 f y 2 bh  bh  2 6 ε ult 6 ε 0 2 ε ult   x  11 7  d '  h    ÷  −  − 1  h    8 8  h  x  2 2 2 2   E ε  E ε E ε  h E ε  h   11 7  d '  h  + −  Ec ε ult − c ult  −  c ult − c ult  + c ult    ÷  −  − 1    2ε 0   2ε 0 2 x 6ε 0  x    8 8  h  x        Substituting into (Eqn F-57) 2  1  1 x   ε 2  ε 3     M   0 2 − 1 x + 2 − h  + 1  0 3 − 1 x  + 3 x − 3 + h   = Ec ε ult   −       h  h bh 2 x  3  ε ult h x   2  2 h   ε ult        2 2  2 2 3  ε 4   x  Ec ε ult  1  1 x   ε 0   + 3 − 3 h +  h   + 1   0 4 − 1 x  + 4 x − 6 + 4 h −  h   −   −   3 − 1          h  x  x   4   ε ult h x  x   2ε 0  3  2 h   ε ult h      ε   ε  x  A  1 d '  7  d '  h 3  0.67 f cu  21 7 ε 0 13 d ' 7  ε 0  5 ε 0 d ' 7  ε 0  d '  x   +   −   + − +  −  γ m 16 12 ε ult 16 h 96  ε ult  12 ε ult h 96  ε ult  h  h         0.67 f cu  x   +  1 − 0  1 − 1 − 0   + 0.Appendix F 2 2 2 0. 67 f cu  11 7 h  d '  ÷ ⇒ 0.67 f cu  1 d '  M   −  + − 1 −     h   bh γ m  2 h  bh 2  x 8  ⇒ = h 11 M  N u 0.Appendix F Case 6 – where (1 – ε0/εult)x > h > 3x/7 i. calculate sb by (Eqn F-58) With h bh N u 0. 1/(1 – ε0/εult) < x/h ≤ 7/3(1–d’/h) Referring to (Eqn F-58) and setting Ash =0 bd N u 0.87 f y  N 0.87 f y sb =  u − − 1 −   bh bh  γ m   8 8 x  h     Ash = 0 and re-arranging bd 7  d '   N u 0.e.67 f cu 11 7 h  d '  Asb = + 0.67 f cu − Asb bh γm ⇒ = (Eqn F-67) bh 11 7 h  d '  0.67 f cu N u 0.87 f y sb ⇒ sb =  u −  bh bh γm bh bh  γm Mu =0 bh 2 (Eqn F-68) F-21 .87 f y  − 1 −   bh γm h  bh 8 8 x N A 0.67 f cu  1 d '  3   −   + −  γ m  2 h  8  8 bh 2  bh    A x determined.87 f y  − 1 −   h  8 8 x Substituting into (Eqn F-60) and again setting Case 7 – where x/h > 7/3  1   0.67 f cu A A = + 0.67 f cu 11 7 h  d '  Asb = + 0.87 f y  − 1 −   bh γm h  bh 8 8 x N u 0. 5 1 1.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30.5 8 8.5 2 2.5 9 M/bh N/mm Chart F-1 . 4-bar column.5 7 7. d/h = 0.75 50 45 40 35 30 25 20 15 10 5 0 0 0.5 6 6.5 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 5 2 5.5 4 4.5 3 3. 5 8 8.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30. d/h = 0.5 1 1.8 50 45 40 35 30 25 20 15 10 5 0 0 0. 4-bar column.5 4 4.5 2 0.5 6 6.5 5 5.5 M/bh 2 N/mm Chart F-2 .5 2 2.5 9 9.5 10 10.5 3 3.5 11 11.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 7 7. 5 11 11.5 2 0.5 8 8.5 9 9.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 7 2 7.5 4 4.5 1 1. 4-bar column.5 10 10.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30.5 12 12.5 13 M/bh N/mm Chart F-3 . d/h = 0.5 6 6.5 5 5.5 3 3.85 50 45 40 35 30 25 20 15 10 5 0 0 0.5 2 2. 4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 7.5 9 9.5 M/bh N/mm Chart F-4 .5 11 11.5 12 12.5 8 2 8.5 4 4.5 1 1.5 14 14.5 10 10. 4-bar column.5 7 2 0.5 5 5. d/h = 0.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30.5 2 2.9 50 45 40 35 30 25 20 15 10 5 0 0 0.5 3 3.5 13 13.5 6 6. 5 16 16.5 15 15.5 2 2.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 30. 4-bar column.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 8.5 11 11.5 9 2 9.5 6 6.95 50 45 40 35 30 25 20 15 10 5 0 0 0. d/h = 0.5 4 4.5 M/bh N/mm Chart F-5 .5 10 10.5 8 2 0.5 3 3.5 14 14.5 1 1.5 12 12.5 7 7.5 5 5.5 13 13. 5 7 7.5 8 8.5 4 4.5 1 1. d/h = 0. 4-bar column.5 M/bh N/mm Chart F-6 .75 50 45 40 35 30 25 20 15 10 5 0 0 0.5 5 2 0.5 2 2.5 3 3.5 9 9.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 5.5 2 6 6. 5 3 3.8 50 45 40 35 30 25 20 15 10 5 0 0 0. 4-bar column.5 2 2.5 10 10.5 1 1.5 5 5.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 6.5 8 8.5 4 4.5 12 M/bh N/mm 2 Chart F-7 .5 9 9.5 6 2 0.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35.5 11 11. d/h = 0.5 7 7. 5 3 3.85 50 45 40 35 30 25 20 15 10 5 0 0 0.5 1 1. d/h = 0.5 2 8 8.5 11 11. 4-bar column.5 4 4.5 6 6.5 12 12.5 10 10.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 7 7.5 M/bh N/mm Chart F-8 .Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35.5 2 0.5 5 5.5 2 2.5 9 9.5 13 13. 5 2 2.5 7 7.5 11 11.5 13 13.5 5 5.9 50 45 40 35 30 25 20 15 10 5 0 0 0. d/h = 0.5 2 0.5 12 12.5 4 4.5 1 1.5 15 M/bh N/mm Chart F-9 .4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 8 2 8.5 3 3.5 14 14. 4-bar column.5 6 6.5 10 10.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35.5 9 9. 5 7 7.5 8 2 0.5 11 11.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 8. d/h = 0.5 9 2 9.5 1 1.5 4 4.5 10 10.95 50 45 40 35 30 25 20 15 10 5 0 0 0.5 6 6.5 12 12.5 14 14.5 16 16.5 2 2.5 M/bh N/mm Chart F-10 .5 15 15.5 3 3.5 13 13.5 5 5.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 35. 4-bar column. 5 8 8.75 50 45 40 35 30 25 20 15 10 5 0 0 0.5 3 3. d/h = 0. 4-bar column.5 2 6 6.5 7 7.5 4 4.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40.5 M/bh N/mm Chart F-11 .5 2 2.5 9 9.5 5 2 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 5.5 1 1. 5 9 9.5 8 8.5 2 2.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40.5 10 10. 4-bar column.5 6 2 0.8 50 45 40 35 30 25 20 15 10 5 0 0 0.5 4 4.5 3 3. d/h = 0.5 12 M/bh N/mm 2 Chart F-12 .5 11 11.5 1 1.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 6.5 5 5.5 7 7. d/h = 0.5 2 8 8.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 7 7. 4-bar column.5 11 11.5 13 13.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40.5 M/bh N/mm Chart F-13 .5 2 0.5 9 9.5 1 1.85 50 45 40 35 30 25 20 15 10 5 0 0 0.5 3 3.5 2 2.5 5 5.5 10 10.5 4 4.5 12 12.5 6 6. 4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 8 2 8.5 9 9.5 5 5.5 6 6.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40. d/h = 0.9 50 45 40 35 30 25 20 15 10 5 0 0 0.5 1 1.5 2 2. 4-bar column.5 15 M/bh N/mm Chart F-14 .5 13 13.5 12 12.5 3 3.5 11 11.5 14 14.5 7 7.5 4 4.5 2 0.5 10 10. 5 16 16.5 12 12.5 8 2 0.5 13 13.95 50 45 40 35 30 25 20 15 10 5 0 0 0.5 10 10.5 5 5.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 40.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 8. 4-bar column.5 1 1.5 6 6.5 11 11.5 4 4. d/h = 0.5 3 3.5 9 9.5 2 2.5 17 2 M/bh N/mm Chart F-15 .5 14 14.5 15 15.5 7 7. 4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 5 2 N/bh N/mm 2 5.5 2 2.75 55 50 45 40 35 0.5 6 6.5 8 8.5 4 4.5 10 M/bh N/mm 2 Chart F-16 .5 7 7.5 3 3.5 1 1. d/h = 0.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45. 4-bar column.5 9 9. 8 55 50 45 40 35 0.5 1 1.5 8 8.5 6 2 N/bh N/mm 2 6.5 11 11.5 M/bh N/mm Chart F-17 .5 9 9.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 12 12.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45.5 5 5.5 4 4.5 10 10. 4-bar column.5 7 2 7. d/h = 0.5 3 3.5 2 2. 5 10 10.5 4 4.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45. d/h = 0. 4-bar column.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 6 6.5 12 12.5 9 9.85 55 50 45 40 35 0.5 8 2 8.5 13 13.5 1 1.5 2 2.5 5 5.5 3 3.5 14 M/bh N/mm Chart F-18 .5 7 2 N/bh N/mm 2 7.5 11 11. 5 2 9 9. d/h = 0.5 7 7.5 11 11.5 5 5.5 13 13.5 10 10.5 M/bh N/mm Chart F-19 .5 2 2.5 3 3.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45.5 6 6.5 15 15.5 14 14.5 1 1.5 2 N/bh N/mm 2 8 8.5 12 12. 4-bar column.9 55 50 45 40 35 0.5 4 4.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0. 5 12 12.5 10 10.5 3 3.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 45.5 13 13.5 8 2 N/bh N/mm 2 8.5 15 15.5 6 6.5 5 5.5 17 2 M/bh N/mm Chart F-20 .5 4 4.5 2 2.5 14 14.5 9 9. 4-bar column.5 11 11.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 16 16. d/h = 0.5 7 7.5 1 1.95 55 50 45 40 35 0. 5 7 7.5 10 M/bh N/mm 2 Chart F-21 .75 55 50 45 40 35 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0. d/h = 0.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50.5 6 6.5 3 3.5 9 9.5 2 2.5 5 2 N/bh N/mm 2 5.5 8 8. 4-bar column.5 4 4.5 1 1. Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50.8 55 50 45 40 35 0.5 7 2 7.5 11 11.5 4 4.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 12 12.5 5 5.5 1 1.5 M/bh N/mm Chart F-22 .5 3 3.5 8 8. d/h = 0.5 10 10.5 9 9.5 2 2. 4-bar column.5 6 2 N/bh N/mm 2 6. 5 9 9.5 10 10.5 8 2 8.5 14 M/bh N/mm Chart F-23 .5 1 1.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 7 2 N/bh N/mm 2 7.5 13 13. 4-bar column.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50. d/h = 0.5 4 4.5 3 3.5 11 11.5 5 5.85 55 50 45 40 35 0.5 6 6.5 12 12.5 2 2. 5 8 2 N/bh N/mm 2 8. d/h = 0.5 14 14.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50.5 10 10.5 6 6.5 12 12.5 4 4.5 16 M/bh N/mm Chart F-24 .5 11 11. 4-bar column.5 13 13.5 1 1.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 7 7.9 55 50 45 40 35 0.5 3 3.5 9 2 9.5 5 5.5 2 2.5 15 15. 5 2 2.5 2 M/bh N/mm Chart F-25 .5 7 7.5 13 13.5 16 16.5 11 11.5 14 14.5 6 6.5 2 N/bh N/mm 2 9 9.5 5 5. 4-bar column.5 8 8.5 1 1.5 10 10.5 12 12.5 3 3.95 55 50 45 40 35 0.5 4 4.5 15 15.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 50.5 17 17. d/h = 0. 5 5 5.5 10 10. d/h = 0.5 M/bh N/mm Chart F-26 .5 8 8.75 60 55 50 45 40 2 0.5 4 4. 4-bar column.5 7 7.5 3 3.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55.5 2 6 2 6.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0.5 9 9.5 1 1.5 2 2. 5 1 1.5 4 4.5 11 11. 4-bar column.5 8 8.5 5 5.5 13 M/bh N/mm Chart F-27 .5 10 10.5 2 7 2 7.5 2 2.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0. d/h = 0.5 9 9.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55.5 12 12.5 6 6.5 3 3.8 60 55 50 45 40 2 0. 5 14 14.5 10 10. d/h = 0.5 12 12.5 4 4.5 11 11.5 6 6. 4-bar column.5 5 5.5 9 9.5 13 13.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0.5 8 2 8.5 M/bh N/mm Chart F-28 .Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55.5 3 3.5 7 2 7.85 60 55 50 45 40 2 0.5 1 1.5 2 2. 5 12 12. 4-bar column.5 7 7.5 6 6.5 14 14.5 4 4.5 8 2 8.5 9 2 9.5 11 11.5 10 10.5 1 1.9 60 55 50 45 40 2 0.5 3 3.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55.5 13 13.5 15 15. d/h = 0.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0.5 5 5.5 16 M/bh N/mm Chart F-29 .5 2 2. d/h = 0.5 11 11.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 55.5 10 10. 4-bar column.5 2 M/bh N/mm Chart F-30 .5 13 13.5 1 1.95 60 55 50 45 40 2 0.5 17 17.5 2 2.5 12 12.5 16 16.5 7 7.5 15 15.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0.5 14 14.5 5 5.5 6 6.5 3 3.5 4 4.5 2 9 9.5 8 8. 4-bar column.5 3 3.75 60 55 50 45 40 2 0.5 2 2.5 9 9.5 2 6 2 6.5 1 1.5 4 4. d/h = 0.5 M/bh N/mm Chart F-31 .5 5 5.5 10 10.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0.5 8 8.5 7 7.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60. 4-bar column.5 10 10.5 6 6.5 2 2.5 9 9.8 60 55 50 45 40 2 0.5 11 11.5 12 12.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0.5 2 7 2 7.5 3 3.5 5 5.5 4 4.5 13 M/bh N/mm Chart F-32 .5 1 1.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60.5 8 8. d/h = 0. Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60.5 3 3.5 11 11.5 1 1.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0.5 4 4.5 10 10.5 14 14.5 7 2 7. d/h = 0.5 M/bh N/mm Chart F-33 .5 12 12.5 13 13. 4-bar column.85 60 55 50 45 40 2 0.5 9 9.5 5 5.5 2 2.5 6 6.5 8 2 8. 5 12 12.5 6 6.5 M/bh N/mm Chart F-34 .Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60.5 7 7.5 8 2 8.5 16 16.5 13 13.5 1 1.5 11 11.5 5 5.5 15 15. 4-bar column.5 3 3.9 60 55 50 45 40 2 0. d/h = 0.5 10 10.5 9 2 9.5 14 14.5 2 2.5 4 4.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 35 30 25 20 15 10 5 0 0 0. 5 12 12.4% steel 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 35 30 25 20 15 10 5 0 0 0.5 2 9 9.5 14 14.5 5 5.5 18 2 M/bh N/mm Chart F-35 .5 15 15.5 7 7.Appendix F Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004 Concrete Grade 60.5 13 13.5 6 6.95 60 55 50 45 40 0. d/h = 0.5 16 16.5 3 3.5 11 11. 4-bar column.5 1 1.5 8 8.5 17 17.5 10 10.5 2 2.5 4 4. 4 -6265.4(D-Wy) 1.4D+1.89 976.4(D+W135) 1.2(D+L+Wy) 1.1 -2611.8 6446.0D+1.9 3825.3 5477.3 2042.4W135 Mx' Mx' Mx' Mx' My' My' My' My' Mx' Mx' Mx' Mx' My' My' My' My' Mx' Mx' Mx' Mx' My' My' My' Mx' Mx' = = = = = = = = = = = = = = = = = = = = = = = = = 476.1 4045.2 3865. Ultimate Load = 76069 mm2/mm kN Total Steel Area = 72885 mm2 Area of Steel per mm length for the long sides bars (including corner bars) = 21.46 1 D.3 8209.3 4892.9 5.0D-1.4Wx 1.4 -24.My Actual Loads Mx control Actual Loads My control P (kN) 50000 40000 30000 20000 10000 0 0 5000 10000 15000 20000 25000 30000 35000 M (kNm) .55 179.36 Area of Steel along long sides (excluding corner bars) = Area of Steel along short sides (excluding corner bars) = Basic Load Case Load Case No.31 cover= 50 b = 1500 Steel provided : h = 2000 15 12 4 Y Y Y 40 40 40 b' = 1285.4(D-Wx) 1.L.4W135 1.01 6999.4(D-W45) 1.88 -959.Rectangular Column R.8 10685 387.3 3298.2 11689 Mux = Mux = Mux = Mux = Muy = Muy = Muy = Muy = Mux = Mux = Mux = Mux = Muy = Muy = Muy = Muy = Mux = Mux = Mux = Mux = Muy = Muy = Muy = Mux = Mux = 16452 23474 18743 22861 14871 18945 13130 11590 26076 20575 14185 19771 11560 16951 9164 7132 23910 27896 23206 27278 18343 22460 16626 19771 29921 Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK Section OK P versus Mx and My of the Column Section P .4(D+Wx) 1.8 4772.7 -2741.8 2159.77 1090.2(D+L-W45) 1.09 P 54782 42413 51121 43634 49900 39936 53598 56618 36916 47941 58099 49365 56676 45051 60989 64513 41527 32792 42951 34216 41527 29902 45841 49364 26379 3 4 5 6 Wx Wy W45 W135 -3628.2 4884.5 251.1 3736.2(D+L+W45) 1.2(D+L-Wy) 1.2 -7257.3 -3900.3 4707.4 -4972. excluding corner bars) (Along each short sides b.2 6558.4 4911.9 -4959.4(D-W135) 1.17 Mx -467. 1101 -37.2(D+L-Wx) 1.8 7805.72 8512.33 2 L.0D-1.7 -3258.624 -51.3 -31.2(D+L-W135) 1.4W45 1.2 170.7 -3811.7 -3913.9 -2494.Mx 80000 70000 60000 P .2 Load Comb 1 Load Comb 2 Load Comb 3 Load Comb 4 Load Comb 5 Load Comb 6 Load Comb 7 Load Comb 8 Load Comb 9 Load Comb 10 Load Comb 11 Load Comb 12 Load Comb 13 Load Comb 14 Load Comb 15 Load Comb 16 Load Comb 17 Load Comb 18 Load Comb 19 Load Comb 20 Load Comb 21 Load Comb 22 Load Comb 23 Load Comb 24 Load Comb 25 1.L.2 973.0D+1.06 -4834.1 -4242.6 6442.5 -3335.2 5179.4(D+Wy) 1.7 -7141.6L 1.5 1706.3 273.2(D+L+W135) 1.1 -5692.3 3838.0D+1.2(D+L+Wx) 1.81 -3700 -1750.6 4639 4352. Load Case Axial Load P (kN) Moment Mx (kNm) Moment My (kNm) Area of Steel per mm length for the short sides bars (including corner bars) = 23.4W45 1.2 9507.2 5808.084 -24.5 5236.3 2700 My -18.9 4416.6 4206 -36.43 -5471.3 4888.492 3221.31 -1067 -5587.568 3748.4Wx 1.0D-1.9 367.17 (Along each long sides h.118 -12.4Wy 1.C.092 -38.0D-1. Design to Code of Practice for Structural Use of Concrete 2004 Project : Column fcu = 35 N/mm2 fy = Floor 460 N/mm2 Ec = 23700 N/mm2 h' = 1684.2 470.0D+1.6 4897 2764 -3520.4(D+W45) 1. excluding corner bars) (Corner bars) Steel Percentage = 2.11 16.43 % 37699 mm2 mm2/mm 30159 mm2 Max.83 -950.4Wy 1.1 -3823. 37872 -291.8 9502.6 5801.2 -2858. 7 2719.964 -80.624 5 W45 56.12 44.5 0.06 49.6151 1.85 0.852 -46.317 19.4(D-Wy) 1.12 -96.2 0. 2304.0D+1.07 -36.038 134.9637 0.4Wy 1.85 1.531 15.922 11.63 191.4Wx 1.0771 4154.93 35.7078 15.58 191.7 3306.1087 2 b = 400 Basic Load Case Load Case No.328 74.8125 0.33 2 L.2(D+L+Wy) 1.4W135 Mx' Mx' My' My' Mx' Mx' My' Mx' My' Mx' My' Mx' Mx' Mx' My' Mx' My' Mx' My' Mx' Mx' Mx' My' Mx' My' = = = = = = = = = = = = = = = = = = = = = = = = = 99.8125 1.3 166.05 20.81 -76.8543 1708.4015 0.8 1600 0.09 8.Rectangular Column R.158 -64.2 0.79 15.6 3067.85 0.7 0.47 -108.13 -31.0602 0.13 55. Load Case Axial Load P (kN) Moment Mx (kNm) Moment My (kNm) b' = 325.85 1.0902 1.3577 0.85 0.5679 0.0214 1.5644 1.118 -15.42 215.9764 0.8 1600 0.872 18.85 1.3 3146.8755 0.964 -60.85 0.0205 2.9538 3907.591 17.8 2810 4118.4W45 1.0304 2.949 16.494 14.098 10.85 0.3179 0.21 76.6 0.6L 1.2945 1.6773 0.0D-1.182 78.2(D+L+Wx) 1.12 17.3331 4666.8 2384.8125 1.31 1.8125 0.15 25.284 28.9181 1.4(D-Wx) 1.628 4 Wy -545.233 192.3 3532.4(D-W45) 1.9 3562.158 130.4265 2852.482 192.85 0.386 145.1 2.85 1.8 1600 0.8 1600 0.Mx 6000 5000 4000 P .0401 0.5185 1.662 16.4(D+Wx) 1.0D+1.2 0.4(D+Wy) 1.4 3989.00 cover= 55 bar size = Mx 1 D.37 125.7208 0.54 -40.232 84.9306 0.85 0.9885 0.L.5 Load Comb 1 Load Comb 2 Load Comb 3 Load Comb 4 Load Comb 5 Load Comb 6 Load Comb 7 Load Comb 8 Load Comb 9 Load Comb 10 Load Comb 11 Load Comb 12 Load Comb 13 Load Comb 14 Load Comb 15 Load Comb 16 Load Comb 17 Load Comb 18 Load Comb 19 Load Comb 20 Load Comb 21 Load Comb 22 Load Comb 23 Load Comb 24 Load Comb 25 1.2 My (kNm) -18.7 3365.3 0.68 23.C.8125 1.432 -93.7 1797.4Wy 1.8 1600 0.437 140.0D-1.029 81.3 0.77 54.56 121.01 16.714 135.15 Mx (kNm) 92.76 56.148 19.1 32.4Wx 1. Design to Code of Practice for Structural Use of Concrete 2004 .708 -20.0D-1.L.6 0.74 63.1 -110.8125 0.00 40 h My b d/h / d/b x/h / y/h Steel Steel area (%) (mm2) 0.72 -25.558 8.2(D+L-W45) 1.708 15.6862 3372.9883 14.339 11.85 0.09 N (kN) 4157.3137 1.614 88.4(D+W135) 1.1 -75.6 3898.6 2189.2341 2468.77 226.558 53.8 3 Wx -362.7798 0.0D+1.9 3341.3345 0.9675 1935 0.0D+1.1601 1.8 1600 0.828 13.92 98.4W135 1.5 3395.7268 3453.34 30.21 -44.156 -28.8 1600 0.95 178.8 1600 0.85 1.1552 2310.9377 2.5083 5016.17 47.8125 1.2111 2422.734 16.99 6 W135 82.4D+1.924 -151.456 -139.693 P versus Mx and My of the Column Section P .8 1600 Steel required = 2.9671 0.11 16.3891 0.7 1541.9 3029.2(D+L-Wy) 1.83 193.4(D+W45) 1.44 97.0583 1.5 3733.431 41.4 0.668 12.059 7.9292 2.21 41.1502 1.78 186.9381 2.9944 1.772 106.125 12.6 0.9242 2.1578 2.32 -34.5 3111.949 M/bh (N/mm2) 0.0D-1.1 76.2(D+L-W135) 1.204 62.868 34.0402 0.8 1600 0.8125 0.4(D-W135) 1.My Actual Loads Mx control Actual Loads My control P (kN) 3000 2000 1000 0 0 100 200 300 400 500 600 M (kNm) .8125 0.65 5.9703 1.4W45 1.85 0.2(D+L-Wx) 1.5083 5016.752 -71.813 16.9731 1.207 105.28 95.2(D+L+W135) 1. 582.55 -93.7229 0.92 231.632 16.6 2463.2106 0.598 18.2658 2.1402 0.8 1600 0.7 2811.979 17.85 0.1989 2397.968 -16.4 bar column Project : Column Mark fcu = 35 N/mm2 h = 500 fy = Floor 460 N/mm2 Ec = 23700 N/mm2 h' = 425.4 2225 2419.444 103.656 163.7 29.9268 1.8125 1.585 0.9 0.018 -105.5561 1.7823 1.8694 1.8 1600 0.2(D+L+W45) 1.21 N/bh (N/mm2) 20. Appendix G Derivation of Design Formulae for Walls to Rigorous Stress Strain Curve of Concrete . 87 f y  2 − 1  h  x  h  (Eqn G-1) Substituting into (Eqn F-31) or (Eqn F-34) or (Eqn F-38) and putting Asb =0 bh 2 2 M u 0. x/h ≤ 7/11 Cases 1 to 3 – where By (Eqn F-18) or (Eqn F-32) or (Eqn F-36) of Appendix F ε x N u 0.87 f y sh   − = +       1 −  −  − γ m  h  2  h  6 ε ult 3 ε ult  h  12  ε ult   h  bh  h  147  h   bh 2       2 2 3 2 0. back substituting into (Eqn G-1) to calculate sh h bh Case 4 – where 7/11 < x/h ≤ 1 G-1 .67 f cu  1 1 ε 0 1  ε 0   N u 163  x         +   −  − ⇒ + −  +     γ m 147 441 ε ult 6  ε ult   h   γ m  2 3 ε ult 12  ε ult   bh 147  h            N 2M u 0. using the bh equations summarized in Appendix F.67 f cu  16 131 ε 0 1  ε 0   x   0.67 f cu  x  1  x  1 ε 0 1 ε 0  x  1  ε 0   x  A  x  163  x         + 0.67 f cu − 3γ m bh  ε 3 − 0  ε ult   x  0.67 f cu   x A  3 − 0  + 0.67 f cu  1 1 ε 0  x M u  +  − + u − − =0 (Eqn G-2) bh bh 2 γ m  2 6 ε ult  h bh 2    A x Upon solving .87 f y  2 − 1 sh = h  ε ult  3γ m  bh  h  bh Ash = bh N u 0.Appendix G Derivation of Design Formulae for Shear Walls to Rigorous Stress Strain Curve of Concrete b h As similar to the exercise in Appendix F for columns. the exercise in this Appendix is A repeated for walls by which sb are set to zero in the various cases 1 to 7. Appendix G Asb =0 bh ε0  x Ash  9 x 7 h 7  N u 0.67 f cu 7  (Eqn G-4) + + − =0 3 −  + − 2 4 bh 3γ m 12  ε ult  h  8 bh 2 48 bh    A x Upon solving .67 f cu    3 − ε  h + 0.67 f cu  45 9 ε0 3  ε 0   x      − ⇒ + − γ m  784 196 ε ult 224  ε ult   h      2 3  0.5 + + − + −  2 γm 8  3 ε ult 12  ε ult   112 bh  h   56 bh       ε 0  x  7 M u 7 Nu  7 M u 0.67 f cu  3 − 0  − 3γ m  ε ult  h bh A   ⇒ sh = bh  9 x 7 h 7 0.67 f cu 7  2 ε0 1  ε 0   9 N u  x  u    + − 1.67 f  7 7 ε 7  ε 0   9 N u  x  cu 0     −  − + +   γ m  8 12 ε ult 48  ε ult   392 bh  h        2 2  9 M 0.67 f cu  x  1  x  1 ε 0 1 ε 0  x  1  ε 0   x    = +    1 −  −  −  2  ε   h     γ m  h  2  h  6 ε ult 3 ε ult  h  12  ult  bh    2 7 h 9 x 9  x   Ash + 0.67 f cu  x  1 1 ε 0   1 ε 0 1 1  ε 0   x  7 9 x 7 h    ⇒ 2 − − =   −        2 6 ε  +  3 ε − 2 − 12  ε   h  4 − 56 h − 8 x  γ m  h  bh  4 56 h 8 x    ult ult   ult        2 7 h ε  x 9 x 9  x    N u 0. back-substitute into (Eqn G-3) to solve for sh h bh Case 5 – where 1 < x/h ≤ 1/(1 – ε0/εult) A Referring to (Eqn F-52) and setting sb = 0 bd 2  1 1 ε0  1 ε ult Nu 1 ε ult 1 ε 0  x = Ec ε ult − − + +  + Ec ε ult 1 − 2  2ε 6 ε 0 2 ε ult  h bh 0   2 6 ε ult     G-2 .87 f y bh  − 56 h − 8 x + 4  = bh  3γ m    ult  ε x N u 0.87 f y  − +     48 x 112 h 392  h   bh   Referring to (Eqn F-39) of Appendix F and setting (Eqn G-3) 2 M u  7 9 x 7 h  0.87 f y  − − +   56 h 8 x 4  A Substituting into (Eqn F-41) with sb = 0 bh 2 M u 0.67 f cu  3 − 0   + − + −    3γ m  ε ult  h   48 x 112 h 392  h    bh     2 4 0. is to be solved by numerical h h + G-3 .Appendix G  1 ε ult 1  h E c ε ult  h   7 7 h 9 x  Ash + E c ε ult   2 ε − 2  x − 6ε  x  + 0. again setting sb = 0 bd bh 2 2 3  1 ε0 1 ε0 1 ε 0 1 1 ε ult  x  M  = Ec ε ult  − 2  24 ε 3 + 6 ε 2 − 4 ε + 6 − 24 ε  h  bh ult 0   ult ult   1 ε 0 2 1 ε 0 1 1 ε ult  x  1 1 ε ult  h 1 ε ult + E c ε ult − + − +  + E c ε ult  − 2  12 12 ε  x + E c ε ult 24 ε  4 ε ult 4 12 ε 0  h  12 ε ult 0  0    2 2 h    x 2 2 7 h 9 x 9  x   Ash + 0.87 f y  − +     48 x 112 h 392  h   bh    1 ε 0 3 1 ε 0 2 1 ε 0 1 1 ε ult  7 7 h 9 x  x  2 M 7 7 h 9 x  − ⇒ 2 − − = E c ε ult  − + − + − −     24 ε 3 6 ε 2 4 ε 6 24 ε 0  4 8 x 56 h  h  bh  4 8 x 56 h  ult ult ult    1 ε 0 2 1 ε 0 1 1 ε ult  7 7 h 9 x  x  1 1 ε ult  7 7 h 9 x  h + E c ε ult − + − + −  + E c ε ult  −  − 2  12 12 ε  4 − 8 x − 56 h  x  4 ε ult 4 12 ε 0  4 8 x 56 h  h   12 ε ult  0    2 2 Nu  7 h 9 x 1 ε ult  7 7 h 9 x  h  9  x  − + + E c ε ult −  −   +     24 ε 0  4 8 x 56 h  x  bh  48 x 112 h 392  h     2  1 1 ε 0 2 1 ε ult 1 ε 0  x  7 h 9 x 9  x  − E c ε ult − − + + − +      2 6 ε 0 2 ε ult  h  48 x 112 h 392  h     2 6 ε ult    2 2  1 ε ult 1  h  7 h 9 x  1 ε ult   7 h 9 x 9  x  9  x   −   − + + − E c ε ult 1 − −       − E c ε ult        2 ε 0 2  x  48 x 112 h 392  h    2 ε 0   48 x 112 h 392  h      2 2 2 E c ε ult  h   7 h 9 x 9  x  − +       6ε 0  x   48 x 112 h 392  h     3 3 2  3 ε0 9 ε0 45 ε 0 3 9 ε ult  x    ⇒ E c ε ult  − + − +  448 ε 3 392 ε 2 1568 ε 196 3136 ε 0  h  ult ult ult   2 3 2   −7 ε0 7 ε0 7 ε0 79 289 ε ult  9 N u  x   + +  E c ε ult + − + −    96 ε 3 24 ε 2 16 ε 294 4704 ε 0  392 bh  h   ult ult ult     3 2   7 ε0 7 ε0 21 ε 0 289 1229 ε ult  9 M 9 N u  x  + +  E c ε ult  − + − + −   3 2 2  193 ε 24 ε ult 32 ε ult 588 9408 ε 0  56 bh 112 bh  h   ult     2 2   7 ε0 7 ε0 5 281 ε ult  7 M   E c ε ult 125 7 M 7 Nu  h −  +  E c ε ult  − + − + +  + −  72 ε 2 24 ε ε 0 1344 8 bh 2 48 bh  x 21 7056 ε 0  4 bh 2    ult ult       2 3 2 2 Eε E c ε ult 7  h  7 h + c ult (Eqn G-6)   −   =0 ε 0 96  x  ε 0 576  x  x x which is in fact an equation of 6th power in .87 f y sh into (Eqn F-57).87 f y  4 − 8 x − 56 h  bh (Eqn G-5)      0 0   A A Substituting for 0. 87 f y  (Eqn G-9) G-4 .67 f cu Ash bh γm  7 7 h 9 x  Ash = + 0.67 f cu A A + 0.87 f y  − −   4 8 x 56 h  (Eqn G-7) Substituting into (Eqn F-60) 2 Mu Ash  7 h 9 x 9  x  = 0.67 f cu = bh γm Mu =0 bh 2 Asb =0 bh N 0.67 f cu − N u 0.87 f y  −  6ε 0  x   4 8 x 56 h  bh  2 ε0 2  x Ash = bh  1 1 ε 0 2 1 ε ult 1 ε 0  x  1 ε ult Nu − E c ε ult − − + +  − Ec ε ult 1 − 2  2 ε bh 6 ε 0 2 ε ult  h  2 6 ε ult 0    7 7 h 9 x − 0.67 f cu  x    +     − − 2  h   γ m  h     56 bh 112  bh 7 M u x  7  N u 0.87 f y − +     bh  48 x 112 h 392  h   bh 2   3 2  x   9 M u 9  N u 0.87 f y  − − ⇒ =  bh bh γm 7 7 h 9 x  4 8 x 56 h  bh 0.67 f cu  7 M u  + − +  − (Eqn G-8) =0 γ m  8 bh 2  4 bh 2 h  48  bh   A x and substituting into (Eqn G-7) to solve for sh Solving the cubic equation for h bh 9  N u 0.87 f y  −   4 8 x 56 h    1 ε ult 1  h Ec ε ult 2  h   − Ec ε ult    2 ε − 2  x + 6ε  x     0 0    2 Case 6 – where 1/(1 – ε0/εult) < x/h ≤ 7/3 Asb =0 bh Consider (Eqn F-58) and substituting N u 0. By back-substituting  1 1 ε 02 Nu = Ec ε ult − − 2 bh  2 6 ε ult x into (Eqn G-5) h  1 ε ult 1 ε ult 1 ε 0  x + +  + Ec ε ult 1 −  2ε 6 ε 0 2 ε ult  h 0  2      1 ε ult 1  h Ec ε ult 2  h   7 7 h 9 x  Ash   + Ec ε ult  −  − −   + 0.67 f cu  ⇒ −  γm 392  bh Case 7 – where x/h > 7/3 By (Eqn F-61) and setting N u 0.87 f y sh ⇒ sh =  u −  bh bh bh  γm  1   0.Appendix G method. 4% steel 45 40 35 30 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 25 20 15 10 5 0 0 0.5 2 2.5 7 7.5 8 8.5 9 M/bh N/mm Chart-G1 . Concrete Grade 30 50 0.5 4 4.5 2 5 2 5.5 1 1.5 6 6.Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004.5 3 3. 4% steel 45 40 35 30 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 25 20 15 10 5 0 0 0.5 1 1.5 3 3.5 4 4.5 7 7.5 2 2.5 9 9. Concrete Grade 35 50 0.Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004.5 M/bh N/mm Chart-G2 .5 8 8.5 2 6 6.5 5 2 5. 5 3 3.5 7 7.4% steel 45 40 35 30 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 25 20 15 10 5 0 0 0.5 2 6 6.5 9 9.5 1 1.5 M/bh N/mm Chart-G3 .5 4 4.5 8 8. Concrete Grade 40 50 0.5 5 2 5.Design Chart of Rectangular Shear Wall with Uniform Vertical reinforcements to Code of Practice for Structural Use of Concrete 2004.5 2 2. 5 7 7.5 9 9. Concrete Grade 45 55 0.5 5 2 2 N/bh N/mm 5.5 4 4.4% steel 50 45 40 35 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 8 8.5 3 3.5 2 2.5 10 M/bh N/mm 2 Chart-G4 .5 6 6.Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004.5 1 1. 5 8 8.5 4 4.5 2 2.5 5 2 2 N/bh N/mm 5.5 7 7.5 6 6.5 1 1.4% steel 50 45 40 35 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel 30 25 20 15 10 5 0 0 0.5 9 9.Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004.5 10 M/bh N/mm 2 Chart-G5 .5 3 3. Concrete Grade 50 55 0. 5 2 6 2 6.5 7 7. Concrete Grade 55 60 0.5 10 10.5 1 1.5 9 9.Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004.4% steel 55 50 45 40 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 35 30 25 20 15 10 5 0 0 0.5 8 8.5 4 4.5 3 3.5 M/bh N/mm Chart-G6 .5 5 5.5 2 2. Concrete Grade 60 60 0.5 8 8.4% steel 55 50 45 40 1% steel 2% steel 3% steel 4% steel 5% steel 6% steel 7% steel 8% steel N/bh N/mm 2 35 30 25 20 15 10 5 0 0 0.Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice for Structural Use of Concrete 2004.5 2 6 2 6.5 4 4.5 10 10.5 M/bh N/mm Chart-G7 .5 1 1.5 3 3.5 5 5.5 2 2.5 7 7.5 9 9. 8 0.2 -1744. 1582.396 14.424 11.74 245.6025 3.4 1826 1578.0D-1.7255 1.9842 0.4Wx 1.8 2473.2065 0.54 435.708 -20.4 44.7 1755.2(D+L-Wy) 1.0311 0.5972 0.6439 0.4(D+W135) 1.279 6.5 6 W135 82.2839 9135.5491 7.825 0.4Wy 1.964 -60.1 -2836.6 -1215.C.4(D+W45) 1.0079 12032 0.7 2387.7 5765.2 -993.1 1.9 1918.4Wx 1.2 1273.3 5932.4 3050.6 3189.7152 3.5706 Steel required = Steel Steel area (%) (mm2) 0.9046 3618.7 3811. 3304.11 16.8 5570 6158.4(D+Wy) 1.9942 9.13 -31.8 3150.6673 2.8 -1756 1814.574 15.2 2.7599 11040 3.2(D+L-W135) 1.8314 3.93 35.1 2.0D+1.34 30.7417 2966.17 -245.158 2530 -2383 -1456.4(D-Wx) 1.8 3 4 5 Wx Wy W45 -362. Design to Code of Practice for Structural Use of Concrete 2004 .00 cover= 25 My 1 D.7727 0.1 1731.92 2047.9 5962.2(D+L+W135) 1.8667 0.4305 13722 0.5 5133.624 206.7 2797.5051 3.7913 7165.834 10.3848 9539.747 13.L.8274 2.2 1624.3059 9223.8 3384.5941 4.7508 4.4 2.0D-1.8328 0.2 381.5637 10255 1.2 2.9 2.52 1249.835 0.2 1140.607 2428.964 -80.4W135 My' Mx' Mx' Mx' Mx' My' My' Mx' Mx' Mx' Mx' Mx' Mx' My' My' Mx' Mx' Mx' Mx' Mx' Mx' My' My' Mx' Mx' = = = = = = = = = = = = = = = = = = = = = = = = = 22.33 2 L.6 1285.4 4969.9149 11660 0.1 32.9 5429.825 0.2(D+L+Wx) 1.6995 2.414 10.6 1391.4 350.7384 0.4576 0.278 0.2(D+L-Wx) 1.7031 4.1 1098.8122 3248.6743 2697.2(D+L+W45) 1.Uniform Reinforcements Project : Wall Mark fcu = 35 N/mm2 fy = Floor 460 N/mm2 Ec = 23700 N/mm2 h' = 1500.0D-1.7772 2.8126 0.1196 8.709 12.96 5.7 -2825.My Actual Loads Mx control Actual Loads My control P (kN) 8000 6000 4000 2000 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 M (kNm) .7 4119.9744 M/bh2 d/h / d/b x/h / y/b (N/mm2) 0.5843 2.62 1.4 1600 3.0D+1.55 -93.752 -71.626 10504 3.3 Plot of P (kN) versus M (kNm) P .23 2906.5 5795.5 -1508.5266 0.24 -332.7861 7144.6 3093.9032 3.4(D-W135) 1.0928 4371.L.1902 8760.1 56.825 0.51 -266.925 15.4Wy 1.2 -1244.Mx 14000 12000 10000 P .825 0.2597 0.21 N/bh (N/mm2) 17.3 4546.4D+1.7 2961.4(D-Wy) 1.82 442.21 257.461 8.5 1277.1 2.00 Load Comb 1 Load Comb 2 Load Comb 3 Load Comb 4 Load Comb 5 Load Comb 6 Load Comb 7 Load Comb 8 Load Comb 9 Load Comb 10 Load Comb 11 Load Comb 12 Load Comb 13 Load Comb 14 Load Comb 15 Load Comb 16 Load Comb 17 Load Comb 18 Load Comb 19 Load Comb 20 Load Comb 21 Load Comb 22 Load Comb 23 Load Comb 24 Load Comb 25 1.1 1.6 3647.766 11.9384 0.28 3.2(D+L-W45) 1.1 -1496.868 34.4(D-W45) 1.3272 5308.432 -93.54 -40.118 -15.1941 0.5293 7.6 6298.2(D+L+Wy) 1.1645 0.825 0.8 2.831 14.2 My (kNm) -18.4 1600 0.4039 9.9 306.0D-1.7 2895.367 11.9 2.1 1966 2405.4 3225 3419.3986 0.299 12.8365 2.5456 1.1 -1275.0915 0.895 13.852 -46.7 1.4W45 1.0D+1.4W45 1.5307 3.09 888.4 1600 0.5618 0.09 23.3 -1203.5596 5.5 2.7373 3.328 229.09 N (kN) 7157.77 -320.9185 1.825 0.Shear Wall R.7541 3016.6 4283.0065 0.0625 8.7 4706.4384 0.825 1.13 1390.7 29.5 4511.7996 11199 0.07 1315.907 14.3 1566.4W135 1.49 14.3121 0.3818 0.15 Mx (kNm) 92.32 -34.4(D+Wx) 1.156 -28.6436 2.854 11.7 3068 1849.62 336. Load Case Axial Load P (kN) Moment Mx (kNm) Moment My (kNm) h = 2000 b' = 165.238 2607.43 17.6532 5.9 4741.4305 13722 bar size = 20 Mx b h b = 200 Basic Load Case Load Case No.6L 1.018 -105.0D+1.872 18.5143 10057 2.7 1603.1 2.4 0. Appendix H Extracts of HKIE paper for plate bending analysis and design . Cheng1 and C. By ignoring such transfer. K θX . Simulation of Support Stiffness in Plate Bending Structure For support stiffness. if the column / wall is more than one storey high.Appendix H Some Problems in Analysis and Design of Thin/Thick Plate (Extracts) Y. the settlement AE L simply measures the ‘elastic shortening’ of the H-1 . if the linking beams (usually floor beams) in the structural frame are ‘flexible’.W. and rotational stiffnesses about X and Y directions. the expression AE L is only correct if the column / wall is one storey high and restrained completely from settlement at the bottom. E is the Young’s Modulus of the construction material and L is the free length of the column / wall. Law2 Department of Civil and Structural Engineering. Hong Kong Polytechnic University. However. Nevertheless. the transfer of loads from one column / wall through the linking beams to the rest of the frame will generally be negligible. Hong Kong SAR Government2 III. the value is simply AE L where A is the cross sectional of the support which is either a column or a wall. ‘interacting’ with that of others through the structural frame linking them together by transferring the axial loads in the column / wall to others through shears in the linking beams. It will not even be a constant value. The column / wall. Hong Kong1 Housing Department. in fact. Strictly speaking. the settlement stiffness estimation can be very complicated. These stiffnesses are independent parameters which can interact only through the plate structure.M. we are referring to the force or moment required to produce unit vertical movement or unit rotation at the support which are denoted by K Z . Most softwares allow the user either to input numerical values or structural sizes and heights of the support (which are usually walls or columns) by which the softwares can calculate numerical values for the support stiffnesses as follows : (i) For the settlement stiffness K Z . The settlement of the support is. K θY for settlement stiffness along the Z direction. the errors can be minimized by finding the force that will be required to restrain the slab structure from sideswaying and applying a force of the same magnitude but of opposite direction to nullify this force. the formulae L L the column / wall are restrained from lateral movement (sidesway)..Appendix H stiffness of a column / wall can be obtained by ‘compounding’ the settlement stiffness of the individual settlement stiffness at each floor as 1 1 KZ = = L L Li L1 L + 2 + 3 + .. This approach is adopted by some local engineers and the procedure for analysis is illustrated in Figure 13.. most of the existing softwares calculate the numerical 4 EI 3EI values either by or . n ∑AE A1 E1 A2 E 2 A3 E3 An E n i i (ii) For the rotational stiffness. IV.. apart from the assumed fixity at the far 4 EI 3EI or are also based on the assumption that both ends of end. It is obvious that the assumption will not be valid if the out-of-plane load or the structural layout is unsymmetrical where the plate will have lateral movement. If there are moment connections between the plate and the supporting walls/columns/piles. However. such effects have to be assessed separately by plane frame or space frame mathematical H-2 . the analysis of finding the effects on the plate by the “nullifying force” has to be done on a plane frame or a space frame structure as the 2-D plate bending model cannot cater for lateral in-plane loads. Effects of Lateral Shears If the plate structure is also subjected to an in-plane shear. Nevertheless. However. The errors may be significant if the structure is to simulate a transfer plate under wind load which is in the form of an out-of-plane moment tending to overturn the structure. depending on whether the far end of the supporting L L column / wall is assumed fixed or pinned (where I is the second moment of area of the column / wall section).. it will sway under the shear force (typically wind load) unless it is effectively restrained from lateral movement. As the 2-D plate model cannot cater for the in-plane load. there will be frame actions creating extra loads on the plate when it sways. This magnitude of this restraining force or nullifying force is the sum of the total shears produced in the supporting walls / columns due to the moments induced on the walls / columns from the plate analysis. However. the reactions on the added support should also be zero. generally (i) the forces at the joining nodes of every element and. V. (ii) “stresses” within every element can be obtained. However. Design based on “stresses” is however not popular as many commercial programs generate the design forces from nodal forces. The 2-D model is then re-analyzed and effects can be conveniently superimposed onto the original one which has been used to solve for the out-of-plane loads. It is however well known that the average stresses from all the adjoining elements of a node are reasonably close to the exact values and can be based on for design. H-3 . Out-ofplane support reactions as derived from the plane frame or space frame models due to the lateral shear can be added onto the 2-D plate model with all support stiffnesses removed but adding a single support with restraints in settlement and rotations at any point of the plate structure.Appendix H models. It should be perfect if the designer relies on “stresses” for structural design. The effects due to lateral shears as obtained from such analysis should be superimposed onto that of the 2-D plate mathematical model for design. thus creating no effects on the 2-D model. General discussions on design based on “Nodal Forces” and “Stresses” As pursuant to completion of analysis by the finite element method based on discretized elements joined at nodes. as the out-of-plane support reactions (on the actual supports) derived from the plane frame or space frame models should have a net effect equal to zero. The differences can be accounted for as errors in the finite element analysis and it should diminish as the meshing of elements is getting fine. he will soon find that the “stresses” at any nodes joined by more than one element (which should have unique values except under point loads or supports) are generally different when the results are generated from different adjoining elements. The purpose of adding the single support is to prevent the stiffness matrix of the 2-D plate model from being singular. in a plate bending problem where there should be “stresses” in form of bending. Basically. VI. there exist bending moments in two mutually perpendicular directions and a twisting moment (all per unit widths) which are termed “stresses”. The problems that are mentioned are actually very common for recent design (though the error is not great). Actually. The problems encountered in design based on nodal forces will be discussed in fuller details in Section VII. Twisting moment does not appear as a component in the nodal force and the out-of-plane nodal shear cannot be resolved into its original components at the X and Y faces. Such design approach possesses the beauty of ensuring structural adequacy (but not necessarily an accurate distribution of forces in the structure) as perfect equilibrium among all elements is achieved.Appendix H A popular approach in design in Hong Kong is based on nodal forces of the elements. it is worthwhile to conduct a brief review in the structural behaviour of a plate bending structure. design based on “stresses” instead of nodal reactions is very popular in the past when finite element analysis is first introduced to Hong Kong for plate analysis. A perfect analogy can be drawn to an “in-plane” load problem where at any point there exist normal stresses in two mutually perpendicular H-4 . However. Some recent popular commercial programs adopt nodal reactions instead of “stresses” in the design and most of the local engineers just simply adopt this approach without greater thought. For example. one may view that the design forces so chosen are not representing the actual behaviour of the structure. the “nodal force” approach simply reduces the structure into an assembly of discretized elements joined at nodes with nodal out-of-plane shears and moments. Structural Design in Plate Bending Structure Before carrying out more detailed discussion in the structural design of a plate bending structure. one may view that at any point inside the plate bending structure. twisting moments and out-of-plane shears per unit width at every point in the structure. The problem relating to bending and twisting moment is first discussed. The nodal force at each node is to be resisted by the “tributary” structural width extending from the node to mid-way to the adjacent node of the same element. he will face the problem of how to design adequately for the net torsion on the section due to the generally unbalanced effects of the out-of-plane nodal forces and nodal moments bending in the plane of the section. On the contrary. design by nodal moments (current local practice) can never capture the effects of twisting correctly. this approach is impractical for the engineers to design as the reinforcements have to follow the principal directions which change from point to point and load case to load case and is therefore not adopted in practice. The “components” of the optimum moments in any orientation.Appendix H directions and a shear stress. likewise the “principal moments” can be derived by compounding the bending and twisting moment in certain directions (known as the principal directions) free of twisting. Wood-Armer equations are recently introduced to H-5 . Whilst the normal stresses and the shear stress at the point can be compounded to the “principal stresses” in the directions where the shear stress is zero. which are derived by Johansen’s Yield Criterion [10] can adequately cover the bending moments normal and perpendicular to the plane of the orientation. if the designer goes further to sum up the nodal forces across a number of adjacent nodes and design the forces against the section formed from the summed tributary length of the nodes (currently the most popular approach in Hong Kong as provided by some commercial softwares). One can then view that the plate is under “pure bending” in the principal directions without twisting. Theoretically this net torsion is not a real force in the structure. but preferably mutually perpendicular) in a “moment field”. During the process. However. Furthermore. The Wood-Armer Equations developed by Wood [19] serve to calculate the optimum design moments at any point in two directions (not necessarily. So it is neither correct to design the torsion for the section by treating the section as a beam section. Of course the principal directions do not necessarily align with the artificial X or Y directions pre-determined by the designer. as there is no twisting moment in the nodal forces. Theoretically the designer can perform his design in the principal directions as the twisting moment is not involved. the twisting moments have been catered for and the design is structurally adequate. the designer will easily find that the vertical shear force in a node cannot be split into H-6 . the orientation of the shear links arrangement on plan does not really matter so long as the same total cross sectional area is provided. 14(d). 14(b). as the required shear resistance depends only on the total cross sectional area of links per unit area. However. Otherwise. Turning back to the current practice in design of shear by nodal forces. Thus it will be at the designer’s discretion to arrange the shear links on plan in any convenient orientations.Appendix H some commercial programs but nearly all the local designs are still based on the nodal moments in the bending reinforcement design. 14(c) which is practically impossible to be achieved. if the Qmax does not exceed allowable shear stress of concrete vc. reinforcements will be required. Following the usual practice of designing against shear. So the designer needs to check or design for the maximum shear of magnitude Qmax at the point. it is required to calculate the design shear based on the components of “stresses” in relation to shear derived from structural analysis. By the general plate bending theory. 14(a). as it is the shear “stress” under which the plate will most likely to fail with failure surface at the orientation θ. the shear force per unit width at the “X-face” and “Y-face” of an element are respectively Q XZ = − ∂M Y ∂M XY ∂M X ∂M XY and QYZ = . The designer may initially wonder that he has to arrange the vertical shear links in the direction of the orientation θ as shown in Fig. To provide a sound design approach against out-of-plane shears. as similar to that in establishing the WoodArmer Equations. say in the global Xdirection as shown in Fig. It can be easily shown that the maximum shear will occur in a plane at an orientation θ Q on plan such that θ = tan −1  XZ Q  YZ 2 2   and the value of the   maximum shear is Qmax = Q XZ + QYZ as per the illustration in Fig. as diagrammatically illustrated + − ∂x ∂y ∂y ∂y in Fig. no shear reinforcements will be required. So. in the first place. in a slab simply supported on two opposite sides. the slab structure does not behave as discretized elements with nodal forces. the moment at the lower end will be M Li = 0. A practical way for design appears to be lumping a series of adjacent nodal shear forces along a cut-section and design the summed shear force against the section. M + M Li The shear on the wall / column will be S i = Ui where M Ui is hi obtained from plate bending analysis and the total restraining shear is S = ∑ Si Figure 13. the moment at it will be zero. And yet. Diagrammatic illustration of the restraining shear or nullifying shear H-7 . Lateral force. problems again arise as to the determination of the orientation of the cut-section on plan. The designer will find that.5M Ui (carry-over from the top). However. The summed shears along any sections in the direction of the span of the slab are small and that in the direction perpendicular to the span are close to the actual shear values in the slab structure. S . apparently the designer has to identify the main span direction for his determination of the orientation of the cut-sections. it is difficult to determine appropriate guidelines for adequate design as. If the wall / column is prismatic and the lower end is restrained from rotation.Appendix H components in the adjacent faces of the element. 2. to prevent sidesway S1 M U1 S2 MU2 h2 S3 MU3 h3 S3 h1 S1 S2 Note : 1. if the lower end is assumed pinned. Diagrammatic illustration of shear “stresses” in the X and Y faces of an element Figure 14(b).Appendix H Qθ θ QYZ QXZ Z QXZ Plan Y QYZ For vertical equilibrium. Diagrammatic illustration of convenient arrangement of shear links on Plan that will produce the identical shear capacity against Qmax H-8 . Derivation of the direction and magnitude of the maximum shear “stress” sv sv Asv total link area in one row θ Qmax QXZ Asv total link area in one row Qmax sv sv QXZ QYZ θ QYZ θ Plan Figure 14(c). Diagrammatic illustration of arrangement of theoretical shear links on Plan against Qmax Plan Figure 14(d). Qθ can be expressed as Qθ × 1 = Q XZ sin θ + QYZ cos θ For Qθ to be maximum set dQθ Q = 0 ⇒ Q XZ cos θ − QYZ sin θ = 0 ⇒ tan θ = XZ dθ QYZ X ⇒ Qθ = Qmax = Q XZ 2 Q XZ 2 + QYZ 2 + QYZ 2 Q XZ 2 + QYZ 2 = Q XZ 2 + QYZ 2 Figure 14(a). Appendix I Derivation of Formulae for Rigid Cap Analysis . taking into consideration of the connection conditions of the piles. The cap itself may settle. translate or rotate. the settlement of Pile i denoted by ∆ iZ can be defined by bO + b1 xi + b2 y i which is the equation for a plane in ‘co-ordinate geometry’ where bO . b1 and b2 are constants. i. but as a rigid body. The deflections of a connecting pile will therefore be well defined by the movement of the cap and the location of the pile beneath the cap.Derivation of Formulae for Rigid Cap Analysis Underlying Principles of the Rigid Cap Analysis The “Rigid Cap Analysis” method utilizes the assumption of “Rigid Cap” in the solution of pile loads beneath a single cap against out-of-plane loads. the cap is a perfectly rigid body which does not deform upon the application of loads. The upward reaction by Pile Summing all pile loads : Balancing the applied load Balancing the applied Moment i is K iZ (bO + b1 xi + b2 y i ) ⇒ P = bO ∑ K iZ + b1 ∑ K iZ xi + b2 ∑ K iZ y i P = ∑ K iZ (bO + b1 xi + b2 y i ) M X = −∑ K iZ (bO + b1 xi + b2 y i ) y i I-1 . Consider a Pile i situated from a point O on the pile cap as shown in Figure I-1 with settlement stiffness K iZ Y Pile i +ve M X and M Y yi O xi X Figure I-1 – Derivation of Pile Loads under Rigid Cap As the settlement of all piles beneath the Cap will lie in the same plane after the application of the out-of-plane load.e. So the three equations become P = bO ∑ K iZ ⇒ 2 bO = P ∑ K iZ M X = −b2 ∑ K iZ y i M Y = b1 ∑ K iZ xi 2 ⇒ ⇒ b2 = b1 = ∑K ∑K −MX iZ yi xi 2 MY 2 iZ The load on Pile i is then P = ∑ K iZ (bO + b1 xi + b2 y i )  P  MY MX = K iZ  + xi − yi  2  ∑ K iZ ∑ K x 2 ∑ K iZ yi  iZ i   PK iZ M Y K iZ M X K iZ x − = + yi 2 i ∑ K iZ ∑ K iZ xi ∑ K iZ yi 2 To effect ∑K iZ xi = ∑ K iZ y i = ∑ K iZ xi y i = 0 . and Y-Y must then be the “principal axes” of the Conventionally. then the pile load become PiZ = PK iZ M U K iZ M Y K iZ u − + vi 2 i ∑ K iZ ∑ K iZ ui ∑ K iZ vi 2 If all piles are identical. all K iZ are equal. then only ∑K iZ xi = ∑ K iZ y i = 0 and the moment balancing equations becomes I-2 . the location of O and the orientation of the axes X-X pile group.e.⇒ M X = −bO ∑ K iZ y i − b1 ∑ K iZ xi y i − b2 ∑ K iZ y i Balancing the applied Moment 2 M Y = ∑ K iZ (bO + b1 xi + b2 y i )xi 2 ⇒ M Y = bO ∑ K iZ xi + b1 ∑ K iZ xi + b2 ∑ K iZ xi y i It is possible to choose the centre O such that ∑ K iZ xi = ∑ K iZ yi = ∑ K iZ xi yi = 0 . 2 i 2 i N ∑ ui vi ∑ Or if we do not wish to rotate the axes to U and V . designers may like to use moments along defined axes instead of moments about defined axes. then the formula is reduced MU MV P PiZ = + u − v where N is the number of piles. If we rename the axes and U-U and V-V after translation and rotation of the axes X-X and Y-Y such that the condition ∑K iZ u i = ∑ K iZ vi = ∑ K iZ u i vi = 0 can be satisfied. i. then the formula is reduced − M Y ∑ xi y i − M X ∑ xi M X ∑ xi y i + M Y ∑ y i P PiZ = + xi + yi 2 2 2 N − (∑ xi y i )2 + ∑ xi 2 ∑ y i 2 − (∑ xi y i ) + ∑ xi ∑ y i 2 ) ( ) For a symmetrical layout where ∑x y i i = 0 .e.M X = −bO ∑ K iZ y i − b1 ∑ K iZ xi y i − b2 ∑ K iZ y i ⇒ M X = −b1 ∑ K iZ xi y i − b2 ∑ K iZ y i and 2 2 M Y = bO ∑ K iZ xi + b1 ∑ K iZ xi + b2 ∑ K iZ xi y i 2 2 ⇒ M Y = b1 ∑ K iZ xi + b2 ∑ K iZ xi y i Solving P = bO ∑ K iZ ⇒ bO = P ∑ K iZ 2 2 b1 = − (∑ K iZ xi y i ) + 2 M X ∑ K iZ xi y i + M Y ∑ K iZ y i (∑ K iZ xi 2 ∑K iZ yi 2 ) ) b2 = − (∑ K iZ xi y i ) + 2 − M Y ∑ K iZ xi y i − M X ∑ K iZ xi (∑ K iZ xi 2 ∑K iZ yi 2 So the pile load becomes M X ∑ K iZ xi y i + M Y ∑ K iZ y i PK iZ + K iZ xi PiZ = ∑ K iZ − (∑ K iZ xi yi )2 + ∑ K iZ xi 2 ∑ K iZ yi 2 2 ( ) + − (∑ K iZ xi y i ) + 2 − M Y ∑ K iZ xi y i − M X ∑ K iZ xi 2 2 (∑ K ( iZ xi 2 ∑K 2 iZ yi ) K iZ y i . the equation is further reduced to PiZ = MY −MX P x + yi + 2 i N ∑ xi ∑ yi 2 I-3 . If all piles are identical. all KiZ are equal. i.