CHAPTER 1Introduction Concrete Concrete is a mixture of water, cement, sand, gravel crushed rock, or other aggregates. The aggregates (sand, gravel, crushed rock) are held together in a rocklike mass with a paste of cement and water. REINFORCED CONCRETE As with most rocklike mass, concrete has very high compressive strength but have a very low tensile strength. As a structural member, concrete can be made to carry tensile stresses (as in beam in flexure). In this regard, it is necessary to provide steel bars to provide the tensile strength lacking in concrete. The composite member is called reinforced concrete. AGGREGATES Aggregates used in concrete may be fine aggregates (usually sand) and coarse aggregates (usually gravel or crushed stone). Fine aggregates are those that passes through a No. 4 sieve (about 6 mm in size). Materials retained are coarse aggregates. The nominal maximum sizes of coarse aggregate are specified in Section 5.3.3 of NSCP. These are follows: 1/5 the narrowest dimension between sides of forms, 1/3 the depth of slabs, or 3/4 the minimum clear spacing between individual reinforcing bars or wires, bundles of bars, or prestressing tendons or ducts. These limitations may not be applied if, in the judgment the Engineer, workability and methods of consolidation are such that concrete can be placed without honeycomb or voids. WATER According to Section 5.3.4, water used in mixing concrete shall be clean and free from injurious of oils, acids, alkalis, salts organic materials or other substances that may be deleterious to concrete or reinforcement. Mixing water for prestressed concrete or for concrete that will contain aluminum embedment’s, including that portion of mixing water contributed in the form of free moisture on aggregates, shall not be used in concrete unless the following are satisfied: (a) Selection of concrete proportions shall be based on concrete mixes using water from the same source and (b) mortar test cubes made with non-portable mixing water shall have 7-days and 28 day strengths equal to at least 90 MODULUS OF ELASTICITY Unlike steel and other materials, concrete has no definite modulus of elasticity. Its value is dependent on the characteristics of cement and aggregates used, age of concrete and strengths. According to NSCP (Section 5.8.5), modulus of elasticity E c for concrete for values of wc, between 1500 and 2500 kg/ m3 may be taken as. Eq. 1-1 EC =W c1.50 0.043 √ f ' c (¿ MPa) f 'c Wc Where is the day 28-day compressive strength of concrete in MPa is the unit weight on concrete in kg /m3 . For normal weight concrete, E c=4700 √ f ' c . Modulus of elasticity E for nonprestressed reinforced may be s taken as 200,000 MPa. DETAILS OF REINFORCEMENT STANDARD HOOKS Standard hooks refer to one of the following: 4 db 1. 180-degree bend plus extension but not less than 60 mm at free end of bar. 12 d b 2. 90-degree bed plus extension at free end of bar. 3. For stirrups and tie hooks: 6 db a) 61 mm diameter bar and smaller, 90-degree bend plus extension at free end bar, or 12 d b b) 20 and 25 mm diameter bar, 90-degree bend, plus extension at free end of bar, or 6 db c) 25mm diameter bar and smaller, 135-degree bend d plus extension at free end of bar. MINIMUM BEND DIAMETERS (SECTION 407.3) Diameter of bend measured on the inside of the bar, other than for stirrups and ties in sizes 10mm through 15 mm, shall not be less than the values in Table 1.1. 4 db Inside diameter of bend for stirrups and ties shall not be less than 16 mm bar and smaller. For bars larger than 16 mm, diameter of bend shall be in accordance with Table 1.1 Inside diameter of bend in welded wire fabric /9plain or deformed) for stirrups and ties 4 db 2 db shall not be less than for deformed wire larger than D56 and for all other wires. 4 db Bends with inside diameter of less than 8db shall not be less than from nearest welded intersection. Table 1.1- Minimum Diameters of Bend Bar Size Minimum Diameter 6 db 10 mm to25 mm 8 db 28 mm, 32 mm, and 36 mm PLAIN REINFORCEMENT (407.6) Reinforcement, prestressing tendons, and ducts shall not be accurately placed and adequately before concrete is placed, and shall be secured against displacement within tolerance permitted. Unless otherwise specified by the Engineer, reinforcement prestressing tendons, and prestressing ducts shall be placed within the following tolerances: Tolerance for depth d, and minimum concrete over a flexural members walls and compression members shall be as follows: Effective depth, d Tolerance on d Tolerance on minimum concrete cover d ≤200 mm ±10 mm -10 mm d ¿ 200 mm ±12 mm -12 mm Except that tolerance for the clear distance to formed soffits shall be minus 6 mm and tolerance for cover shall not exceed minus 1/3 the minimum concrete cover required in the design drawings or specifications. Tolerance for longitudinal location of bends and ends of reinforcement shall be ± 50 mm except at discontinuous ends of members where tolerance shall be ±12 mm. SPACING LIMITS FOR REINFORCEMENT According for Section 5.7.6 of NSCP, the minimum clear spacing between parallel bars in a layer should be db but not less than 25 mm. Where parallel reinforcement is placed in two or more layers, bars in the upper layers should be placed directly above bars in the bottom layer with clear distance between layers not less than 25mm. In spirally reinforced or tied reinforced compression members, clear distance between longitudinal bars shall be not less than 1.5 db nor 40mm. In walls and slabs other than concrete joist construction, primary flexural reinforced shall be spaced not for farther apart than three times the wall or slab thickness, nor 450 mm. BUNDLED BARS Groups of parallel reinforcing bars bundled in contact to act as unit shall be limited to four in any one bundle. Bundled bars shall be enclosed within stirrups or ties and bars Figure 1. Since spacing limitations and minimum concrete cover of most members are based on a single diameter db. D π π 4 (25)2 x 3 4 D2 . = 3-25mm Equivalent diameter.1 Bundled-bar arrangement Diameter of single bar equivalent to bundled bars according to NSCP to be used for spacing limitation and concrete cover.larger than 32 mm shall not be bundle in beams. bundled bars shall be treated as a single bar of a diameter derived from the equivalent total area. The individual bars within the span of flexural members should terminate at different points with at least 40 d b stagger. CONCRETE PROTECTION FOR REINFORCEMENT (SECTION 407. joists: 32 mm bar and smaller 40 Beams. The following minimum concrete cover shall be provided for reinforcement: Minimum cover. walls.1) Cast-in –place Concrete (nonprestressed). and smaller Precast concrete (Manufactured Under Plant Conditions). stirrups.The Following minimum concrete shall be provided for reinforcement Minimum cover. mm (a) Concrete cast against permanently exposed to earth 75 (b) Concrete exposed to earth or weather: 50 20 mm through 36 mm bars 40 16 mm bar. W31 or D31 wire. folded place members: 20 mm bar and larger 15 16 mm. mm (a) Concrete exposed to earth or weather: Wall panels: 20 32 mm bar and smaller Other members: 40 . columns Primary reinforcement. spirals 20 Shells. Wr1 or D31 wire. ties. and smaller (C Concrete not exposed to weather or in contact with ) ground: 20 slabs.8. Minimum cover. walls. and smaller (b) Concrete not exposed to weather or in contact with ground: 15 slabs. & 32 mm bar and smaller need not exceed 40 Beams. . Wr1 or D31 wire. folded plate members: 20 mm bar and larger 16 mm. stirrups. joists: Beams. joists: db but not less 15. mm (a) Concrete cast against permanently exposed to earth 75 (b) Concrete exposed to earth or weather: Wall panels. 20 mm through 32 mm bars 30 16 mm bar. columns: 40 Primary reinforcement. W31 wire. slabs joists 25 other members 40 (C) Concrete not exposed to weather or in contact with ground: 20 slabs. ducts and end fittings. walls. spirals 10 Shells. columns 10 Primary reinforcement 15 Ties. and smaller Prestressed Concrete The following minimum concrete cover shall be provided for prestressed and nonprestressed reinforcement. 3. Reinforcement for shrinkage and temperature stresses normal to flexural reinforcement shall be provided in structural slabs where the flexural reinforcement extends in one direction only. and smaller db but not less than Other Reinforcement 20 Bundled Bars For bundled bars. SHRINKAGE AND TEMPERATURE REINFORCEMENT (2010 NSCP) Shrinkage and temperature reinforcement is required at right angles to the principles reinforcement to minimize cracking and to tie the structure together to ensure its acting as assumed in the design. the minimum concrete cover shall be equal to the equivalent diameter of the bundle.4 and 408. The provisions of this section are intended for structural slabs only. but need to be greater than 50 mm. folded plate members: 10 16 mm. except for concrete cast against and permanently exposed to earth. spirals Shells. 25 Ties.3 shall be considered.6. Wr1 or D31 wire. stirrups. Shrinkage and temperature reinforcement shall be provided in accordance with either of the following: a) Where shrinkage and temperature movements are significantly restrained. b) Deformed reinforcement conforming to 43. they are not intended for soil-supported slabs on grade. the requirements of 408. the minimum cover shall be 75 mm.3 used for shrinkage and temperature reinforcement shall be provided in accordance with the following: .3. 0018x415/ Shrinkage and temperature reinforcement shall be spaced not farther apart than five times the slab thickness.. rain.0018 c) Slabs where reinforcement with stress exceeding 420 MPa measured at a yield strain of 0. It is permitted by the code to assume the following arrangement of live loads: (a) Factored dead load on all spans with full factored live load on two adjacent spans. and temperature changes.0.0. earthquakes. and the worst possible combination of these loads that might occur simultaneously.014: a) Slabs where Grade 280 or 350 deformed bars are used…….…. DEAD LOAD Dead loads are loads of constant magnitude that remain in one position.. Loads on structure may be classified as dead loads or live loads.0.35 percent is fy used……………………………………………………. nor farther apart than 450 mm. .. Live loads that move under their own power called moving loads.. Other Live loads are those caused by wind. soils. and (b) Factored dead load on all spans with full factored live load on alternative spans. Wind and earthquake loads are called lateral loads..Areas of shrinkage and temperature reinforced shall be provided at least the following rations of reinforcement area to gross concrete area. but no less than 0.0020 b) Slabs where Grade 420 deformed bars or welded wire reinforcement are used………………………………………………………………. LOADS The most important and most critical task of an engineer is the determination of the loads that can be applied to a structure during its life. LIVE LOAD Live loads are loads that may change in magnitude and position. ARRENGMENTS OF LIVE LOAD Live loads may be applied only to the floor or roof under consideration... and the far ends of columns built integrally with the structure may be considered fixed. This consists mainly of the weight of the structure and other permanent attachments to the frame . 1D + 1.3W But for any combination of D. required strength U shall not be less than Eq.9D + 1. 1-5 U=1.9D + 1. 1-2 U=1.4D + 1. 1-3 U=0. the following combination of D.4D + 1.3L + 1. and Eq. 1-6 U=0.7L If resistances to structural effects of a specified wind load W are included in design.7W) Where load combinations shall be include both full value and zero value of L to determine the more severe condition. required strength U shall not be less than Eq. L and E shall be investigated to determine the greatest required strength U: Eq. required strength U shall be at least equal to: . and Eq. and E. 1- 2 If resistance to earth pressure H is included in design. L.REQUIRED STRENGHT (FACTIRED LOAD).1E Where load combinations shall included both full value and zero value of L to determine the more severe condition. 1-4 U=0.1- 2 If resistance to specified earthquake loads of forces E is included in design.7L + 1.1E But for any combination of D. the following combinations of D. and W shall be investigated to determine the greatest required strength U: Eq. L. L. and W. U Required strength U to resist dead load (D) and live load (L) shall be at least equal to: Eq.75(1. and axial tension with flexure…………… 0. Where structural effects T of differential settlement.4T + 1.4(D + T) Estimations of differential settlement. required strength U shall be equal to Eq. shear.4 and to be added to all loading combinations that include live load. in terms of flexure.7L + 1.7 H Except where D or L reduces the effect of H. If resistance to impact effects is taken into account in design. L and H. and shrinkage expansion of shrinkage-compensating concrete or temperature change may be significant in design. such effects shall be included with live load L. 1-7 U=1. axial load. its connections to other members. STRENGTH REDUCTIONS FACTORS. and its cross sections. required strength U shall not be less than. 1-8 U=1.75(1. creep. (a) Flexure without axial load………………………………… 0. and shrinkage expansion of shrinkage compensating concrete or temperature change shall be based on a realistic assessment of such effects occurring in service.9D shall be substituted for 1.90 (c)Axial tension and axial tension with flexure: . For any combination of D. 0. and torsion shall be taken as the nominal strength multiplied by a strength reduction factor φ having following values. If resistance to loadings due to weight and pressure of fluids with well defined densities and controllable maximum heights F is included in design. creep.4D and zero value of L shall be used to determine the greatest required strength U. such loading shall have a factor of 1. φ (PHI) The design strength provided by a concrete member.7L) But required strength U shall not be less than Eq.4D +1.4D + 1.Eq. 1-9 U=1.90 (b) Axial tension. 0.70 (f) Post-tensioned anchorage zones…………………………… 0. mm D = dead loads. For a hollow section. is the area of the concrete only and does not include the area of the void(s) A v =¿ 2 area of shear reinforcement spacing. 0. Spiral reinforcement…………………………………….85 (e) Bearing on concrete…………………………………………. or diameter of circular section. or related internal moments and forces..75 (d) Shear and torsion…………………………………………….. h = overall thickness or height of member. mm b w =¿ web width.85 ACI-318-05 (NSCP C101-10-210) Notations A g=¿ 2 Ag gross of concrete sections . or related internal moments and forces d = distance from extreme compression fiber to centroid of longitudinal tension reinforcement. The reinforcement & other reinforced members……. 0. 1. or related internal moments and forces f yt =¿ fy specified yield strength of transverse reinforcement.... mm . 0. MPa F = loads due to weight and pressures of fluids with well-defined densities and controllable maximum heights. mm E = load effects of earthquake.75 2. mm . and shrinkage-compensating concrete. creep. to be taken as positive for compression and negative for tension. creep. Lr=¿ roof live loads or related internal moments and forces. excluding strains due to effective prestress. N W = wind load. differential settlement. or related internal moments and forces. M u=¿ factored moment at section. or related internal moments and forces. L = live loads or related internal moments and forces. N R = rain load. N-mm N u=¿ Vu factored axial force normal to cross section occurring simultaneously with Tu or . and temperature . Vs = nominal shear strength provided by shear reinforcement N Vu = factored shear force at section. N Vn = nominal shear strength. or other materials. Vc = nominal shear strength provided by concrete.H = loads due to weight and pressure of soil water in soil. related internal moments and forces εt = net tensile strain in extreme layer of longitudinal tension steel at nominal strength. U = required strength to resist factored loads or related internal moments and forces. T = cumulative effect of temperature. shrinkage. shrinkage . 3 Design of structures and structural members using the load factor combinations and strength reduction factors of Appendix C shall be permitted.2(D+F+T) + 1. φ = strength reduction factor ρw As bw d = ratio of to CHAPTER 9 – STRENGTH AND SERVVICEABILITY REQUIREMENTS 9.0L or 0.2 Required strength 9.1.6(L+H) + 0.1.2D + 1. 9.1.4 (D+F) (9-1) lR U = 1. 9. 9.5( or R) (9-2) Lr U = 1.GENERAL 9.1 Structures and structural members shall be designed to have design strengths at all sections at least equal to the required strengths calculated for the factored loads and forces in such combinations as are stipulated in this code.1. U = 1.8W) (9-3) .6( or R) + (1. Use of load factor combinations from this chapter in conjunction with strength reduction factors of appendix C shall be permitted.1 Required strength U shall be at least to the effects of factored loads in Eq.2. The effect of one or more loads not acting simultaneously shall be investigated.2 Members also shall meet all other requirements of this code to ensure adequate performance at service load levels. (9-1) through (9-7). 8N/ m . Where lateral earth pressure provides resistance to structural actions from other forces.6H (9-8) Except as follows: a) The load factor on the live load L in Eq. such effects shall be included with L. a load factor of 1.9D + 1. loads due to weight and pressure of soil. d) The load factor on H.4 If structure is in a flood zone. creep.0L (9-5) U = 0. 9.3 W in Eq. it shall be not be included in H but shall be included in the design resistance. 9.9D + 1. (9-5) and (9-7). or is subjected to forces from atmospheric ice loads.2.2.2D + 1. (9-3) to (9-5) shall be permitted to be reduced to 0. is based on service-level seismic forces. or temperature change shall be based on a realistic assessment of such effects occurring in service. 2 and all where L is greater than 4. 9. it shall be permitted to use 1. 9.6H (9-6) U = 0.6W+ 1. Lr U = 1.2D + 1.0E+ 1.6W + 1. shall be set equal to zero in Eq. expansion of shrinkage-compensating concrete.3 Design strength .2 If resistance to impact effects is taken into account id design.2 shall be applied to the maximum prestressing steel jacking force. 1. c) Where E. the flood or ice loads and the appropriate load combinations of SEI/ASCE7 shall be used.0L + 0.4E shall be used in place of 1.3 Estimations of differential settlement. areas occupied as places of public assembly. (9-4) and (9-6). shrinkage. 9. the load effects of earthquake.5 except for garages.0E+ 1.2. b) Where wind load W has not been reduced by a directionality factor. water in soil or other materials.0E Eq.2. (9-6) and (9-7) if the structural action due to H counteracts that due to W or E.5( or R) (9-4) U = 1.5 For post-tensioned anchorage zone design. 2.2.9.1 Tension-controlled sections as defined in 10.2. with symmetric reinforcement. and its cross sections. multiplied by the strength reduction factors φ in 9.10 (E=200.90 (See also 9.3.7) 9.2.2 Compression-controlled sections.3: a) Members with spiral reinforcement conforming to 10. 9. to zero.3. 9.. shall be taken as the nominal strength calculated in accordance with requirements and assumptions of this code. Transition and with (d-d’)/h not less than 0. φ shall be permitted to be increased from 0.3. φ shall Tension controlled be permitted to be increased linearly tocontrolled φ Pn f ' c Ag 0.3.3. 000 MPa) f ' c Ag φ Pb or .0. as defined 10.2.0.3….90 Spiral ( ) 0.0.70 ( ) 0. 9.3.4. axial load..3.1 through 9. φ shall be permitted to be linearly increase from that for compression-limit to 0.2 Strength reduction factor φ shall be as given in 9.75 . and 9. whichever is smaller.9.65 For sections in which the net tensile strain in the extreme tension steel at nominal εt strength is between the limits for compression-controlled and tension-controlled sections.3.3 –Shear and torsion…………………………………………0. for members in which does not exceed Compression 415 MPa.3.2 Strength reduction factor zero. its connections to other members.2.1 Design strength provided by a member.3.3. For other reinforced members.4………….3.65Other fy Alternatively.5.7: 9.3.2.70. when Appendix B is used. ( ) ( ) 0.90 as decreases from 0. in terms of flexure.70 b) Other reinforced members……………………………………. shear and torsion.10 to Figure 1.005. mm Ec =modulus of elasticity of concrete. MPa Es = modulus of elasticity of steel 200.4 – Bearing on concrete (except for post-tensioned and anchorage zones and struct-and-tie models)……………………. 9. MPa fy =specified yield strength of steel.……………0. mm d' =distance from extreme compression fiber to centroid of compression reinforcement. mm A s = area of tension reinforcement. mm d c = thickness of concrete cover measured from extreme tension fiber to center of bar or wire. MPa fs =calculated stress in reinforcement at service loads.2. mm d = distance from extreme compression fiber to centroid of tension reinforcement.3. mm Ig =moment of inertia of gross concrete section about centroidal axis.65 CHAPTER 1 Analysis and Design of Beam NOTAIONS AND SYMBOLS USED ∂ = depth of equivalent stress block. MPa h =overall thickness of member.000 MPa f 'c =specified compressive stress of concrete. mm 2/ m b = width of compression face of member. neglecting reinforcement . mm 2 A sk = area of skin reinforcement per unit height in one side face. mm c = distance from extreme compression fiber to neutral axis. Ec 2. 5. shall be assumed equal to 0. N-mm M u =factored moment at section. Maximum usable strain at extreme concrete compression fiber. Strain in reinforcement and concrete shall be based assumed directly proportional to the distance from the neutral axis.7). Tensile strength of concrete shall be neglected in axial and flexural calculations. h/L> 2/5 for continuous spans and h/L >4/5 for simple spans.10. = fy . 4.10. Expect for deep flexural members with overall depth to clear span to ratio.4 in Page 16 εc =strain in concrete (maximum = 0. . a nonlinear distribution of strain shall be considered (See Sec.2) 1.003 fs fy fs εs εs εs εs fs 3. be taken as x for > . I se =moment of inertia of reinforcement about centroidal axis of member cross- section M n =nominal moment. For below shall . N-mm β1 =factor defined in Section 410.003) εs f y /E s =strain in steel below yield point = εy =strain in steel at yield point ρ A s /bd =ration of tension reinforcement ρb =balance steel ratio Ø =strength reduction factor ASSUMPTION IN STRENGTH DESIGN IN FLEXURE (CODE SECTION 5. β1 f 'c =0. For rectangular distribution of stress: f 'c a) Concrete stress of 0.008( -30) but not shall be less than 0. but β1 s hall not be taken less than 0. parabolic. For > 30 MPa.65. or any other from that result in prediction of strength in substantial agreement with results of comprehensive tests.5.85 for ≤ 30 MPa and shall be reduced continuously at rate of 0.003 c a c d d-a/2 NA Mn As . 6.85 shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross-section and a straight line located parallel to the maximum compressive strain. For ≤ 30 MPa. Relationships between compressive stress distribution and concrete strain may be assumed rectangular.e f 'c β1 i.A. β1 f 'c β1 c) Factor shall be taken as 0.85-0.85 f 'c ii.85 0. trapezoidal. = 0. i.008 for each 1 MPa of strength in excess of 30 MPa.65 RECTANGULAR BEAM REINFORCED FOR TENSION ONLY (SINGLY REINFORCED) f 'c b 0. b) Distance c from fiber of maximum strain to the neutral axis hall is measured in the direction perpendicular to N. 2-1 f ' c ≤ 30 MPa .85 f ' c b Multiplying Eq.85 f ' c b d As f y d a= bd 0.05 f ' c ≤ 30 MPa . β 1=0. 2-2 a= 0.85 As f y Eq.85− f ' c −30 ¿ For 7 ( but shall not be less than 0. T= As f y y /¿ Es f¿ Stress Diagram Strain Diagram Figure 2. β 1=0.85 For 0.1: Stress and strain diagram for singly reinforced and rectangular beam a=β1 c Eq.85 f ' c .65 [ ∑ Fh =0 ] C=T f ' c ab= A s f y 0. 2-2 by d/d: As f y d a= x 0. 2-5 0.85 f ' c ρf y Let ω= f 'c ωd a= Eq.85 f ' c ab (d−a/2) ' ωd 1 ωd M n=0.85 Nominal Moment Capacity: From the stress diagram in Figure 2.85 f c b( d− ) 0. 2-3 ρ= bd and pfyd Eq.85 2 0.85 . As The term ρ .1: M n=Cx( d−a /2) M n=0. bd is called the ratio of steel reinforcement and is denoted as As Eq. 2-4 a= 0. 59 ω) Coefficient of Resistance Eq.003) and steel and (with strain of E s ) are reached simultaneously once he ultimate load is reached.2-10 ρ= fy [ √ 0.59 ω) Eq.85 f ' c 1− 1− 2 Rn 0. Eq.2-8 Rn=f ' c ω (1−0. UNDERREINFORCED DESIGN .59 ω) Ultimate Moment Capacity (Design Strength): M u=φ M u (where φ=0. yields the following formula the steel ratio ρ : Eq.2-6 M n=f ' c ωb d 2 (1−0.85 f ' c ] BALANCE DESIGN Balance design refers to a design so proportioned that the maximum stresses in concrete fy (with strain of 0. 2-8 and replacing it with.2-9 M u=φ Rn b d 2 ρf y Solving for an ω in Eq.2-7 M u=φ f ' c ωb d2 (1−0.90 for flexure) Eq. causing them to fail simultaneously. f ' c . b 0. and failure occurs suddenly without warning to the user of the structure.Underreinforced design is a design in which the steel reinforced is lesser than what is required for balance condition .000 MPa. the strain fy Es in concrete reached is maximum usable value of and the strain in steel is E s where =200. If the beam is overreinforced. that is why the Code limits the tensile steel percentage (P max=0. deflections are not noticeable although the compression concrete is highly stressed. Failure under this condition is ductile and will give warning to the user of the structure to decrease the load. the steel will begin to yield although the compression concrete is still understressed.00 3 c = c d d NA 0.00 3 Strain Diagram Figure 2. the concrete and steel yield simultaneously. In this condition. As the load is increased. the steel will continue to elongate. OVERREINFORCED DESIGN Overreinforced design is a design in which the steel reinforcement is more than what is required for balanced condition. the steel will not before failure. resulting in appreciable deflections and large visible crack in the tensile concrete. Overreinforced as well as balanced design should be avoided in concrete because of its brittle property. If the load is further increased.75pb) to ensure underreinforced beam with ductile type of failure to give occupants warning before occurs. If the ultimate load is approached. BALANCED STEEL RATIO ρb : In balanced condition.2-Balanced condition . 2-12 is for singly reinforced rectangular sections only.2: c 0.003+ Note: Es 0. Eq.85 f ' c β 1 p f yd 600 c=c = d 0.85 f ' c β 1 600 Eq.00 600 d c b= Eq.003+ 200.85 f ' c β1 c= β1 p f yd c= 0.00 0.003 = d fy Es =200.2-11 600+ f y β1 c But a = ρf y d a c= 0.003 c= d fy 0. .85 f ' c β 1 600+f y 0. 2-12 ρb= f y (600+ f y ) Note: Eq.By ratio and proportion in the triangle shown in Figure2. 2-11 is applicable to nay shape. 4 bw d Eq. the ratio of reinforcement that would produce balance strain condition for the section under flexure without axial.MAXIMUM STEEL REINFORCEMENT Section 410. MINIMUM REINFORCEMENT OF FLEXURAL MEMBERS 410.75 ρb Eq. ρmax =0. 2-15 A smin = 4 f y b w d 1. the portion of equalized by compression reinforcement need not be reduced by the0.61 At very section of flexural members where tensile reinforcement is required by As analysis. For members with compression ρb reinforcement.75 factor.4.75 ρb Eq.3: For flexural and for subject to combined flexure and compressive axial φ Pn 0. 2-13 and A smax =0. load. 2-14 This limitation is to ensure that the steel reinforcement will yield first to ensure ductile failure.10 f ' c A g load when the design axial load strength is less than the smaller of φ Pn ρb or . the area provided shall not be less than that given by: √ f 'c Eq.2-16 and not less than fy . Maximum spacing of this reinforcement shall not exceed three times the thickness and 450 mm. Approximate the weight of beam (DL) between 20% to 25% of (DL+LL). 410. 410.7L Mn IV. A smin 410. III. With a very small amount of tensile reinforcement. The provision for minimum amount of reinforcement applies to beams. 2-15 with set equal to the width of the flange.and VIII are the author’s recommendation based on his experience. V. STEPS IN DESIGNING A SINGLY REINFORCED RECTANGULAR BEAM FOR FLEXURE: Note: The assumptions made in steps II. which for architectural and other reasons are much larger in cross-section than required by strength consideration. I.62 For statically determinate T-section with flange in tension. 2-17 2fy w bw or Eq. Compute the factored moment to be resisted by the beam. the area shall be equal to or greater than the smaller value given either by: A smi n= √ f 'c b d Eq.1 and 410. the computed moment strength as a reinforced concrete section computed from its modulus of rapture. Identify the values of the dead load and live load to be carried by the beam.6.6.4 For structural slabs and footings of uniform thickness.3 The requirements of Sections 410.This weight is added to the de load. Compute the factored load and factored moment: Ex: factored Load =1. (DL & LL) II.13 (Shrinking and Temperature Reinforcement).6. .2 need to be applied if at every section the area of the tensile reinforcement is at least one-third greater than that required by analysis.4 DL+1. the minimum area of tensile reinforcement in the direction of span shall be the same as that required by Section 407. Failure in such a case can be quite sudden.6. 5 ρb but must not be less than ρmin . Solve for the required steel area and number of bars. V.85 for f ' c ≤30 MPa β 1=0.4 ρmin = fy ρf y VI.85 f ' c β 1 600 ρb= f y (600+ f y ) β 1=0. Solve for bd : ' 2 M u=φ f c bd ( 1−0. and solve for b. This value ρ will provided enough alloance in the actual value of ρ due to rounding-off of the number of bars to be used. After solving for d.008 ( f ' c −30 ) for f 'c >30 MPa 0.05ρb. for it not to exceed the maximum ρ of 0. ω= f 'c 2 VII.59 ω ) ¿ ¿ ¿ ¿ ¿ ¿ ¿ bd 2=¿ ¿ VIII. substitute its value to Step VII. and solve for d.1 in page 36. (round-off this value to reasonable dimension).85−00−. Check also the minimum thickness of beam required by the Code a given in Table 2. . Try a value of steel ratio ρ from 0. Try ratio b/ d ❑ ( from d=15b to d=2b). IX. Compute the weight of the beam and compare it to the assumption made in Step II. Compute the value of ω . 0. Solve for ρ : M u=φ Ru bd 2 ¿ ¿ ¿ ¿ Ru =¿ ¿ .85 f ' c β 1(600) Pmax =0.59 ω ¿ M u=M u max if design as singly reinforced (Step II) M u=M u max if design as doubly reinforced (Step III) II. Solve for Pmax =0.75 pb 0. A s =pbd Number of bars(diameter = D) π 2 D x number of bars = A s 4 AS STEPS IN COMPUTING THE REQUIRED TENSION STEEL AREA OF A BEM MU WITH KNOWN MOMENT NT AND OTHER PROPERTIES: Pmax ∧M u max I.75 f y (600+ f y ) ¿ ¿ ¿ ¿ ρf y ω= =¿ ¿ fc M u max=φ f c ω b d 2 (1-0. Solve for ρ : ρ= bd II. Check if steel yields by computing ρb ' 0. ` ρ ≤ ρb . Compression reinforcement is necessary.85 f c β1 (600) ρb= f y (600+ f y ) III. (See Chapter 3) MU STEPS IN COPUTING OF A BEAM WITH KNOWN TENSION STEEL AREA AS AND OTHER BEAM PROPERTIES: As I.85 f ' c ¿ ¿ ¿ 1−√¿=¿ ¿ 0. 2 Ru 1− 0.85 f ' c ρ= ¿ fy ¿ ¿ ¿ ¿ A s =ρbd=¿ ¿ III. 000MPa] .the given is not adequate for the beam dimension. ρ ≤ ρ min As Note: if .003 c=0.0 00 T= fs Es Solve for from the strain diagram: [Note: =200. IV.steel yields.85 ab 0. proceed to step IV. ω=ρ f y /f c φM u=φ f ' c ωb d 2(1−0.8 5 a c d d-a/2 d-c =200. proceed to III if ρ ≤ ρb . ρ> ρb b 0.steel does not yield.59 ω) if ρ ≤ ρb . 1 shall apply for one-way construction not supporting are attached to portions or other construction likely to be damaged by large deflections. fs Es 0. unless computation of deflection indicates a lesser thickness can be used without adverse effects. a=β 1 c c M u=Φ T d− ( a2 )=Φ A f ( d− a2 ) s s or M u=Φ C d− ( a2 )=Φ 0. .85 f ' ab(d − a2 ) c MINIMUM THICKNESS OF FLEXURAL MEMBERS According to Section 5. minimum thickness stipulated in Table 2.9.5 of NACP.85 β1 f c b c 2 fs Solve c by quadratic formula and solve for and a: d−c f s=600 . 2-18 c Σ F H =0 ¿ [ T=C A s f s=0.003 = d−c c d−c f s=600 Eq.85 f 'c (β 1 c)b c 600 A s ( d−c ) =0.85 f c ab β1c but a= d−c A s 600 =0. 2 Where deflections are to be computed. (b) For other than 415 MPa. For other conditions.005 ) but not less than 1.5 L/21 L/8 ribbed one-way slabs Span length L is in millimeters Values given shall be used directly for members with normal density concrete ( 3 ω=2300 kg /m ) and grade 415 reinforcement. considering effects of cracking and reinforcement on member stiffness.4 + BEAM DEFLECTION (SECTION 5. the values shall be multiplied by (0.2.65-0. . is the unit mass in fy f y /700 ¿ .9. 5. deflections that occur immediately on application of load shall be computed by usual methods or formulas for elastic deflections.Table 2.1 MINIMUM THICKNESS OF NON-PRESTRESSED BEAMS OR ONE-WAY SLABS UNLESS DEFLECTIONS ARE COMPUTED * Minimum thickness. the values shall be modified as follows: (a) For structural lightweight concrete having weights in the range 1500-2000 kg /m3 ωc values shall be multiplied by (1. h Simply One end Both ends Cantilever supported continuous continuous Members not supporting or attached to partitions or other construction Member likely to be damaged by large deflections Solid one-way L/20 L/24 L/28 L/10 slabs Beams or L/16 L/18.9.09.5 Sect.5. where ωc kg /m3 . 9. one of the following modifications shall apply: . Eq. Ec immediate deflection shall be computed with the modulus of elasticity for concrete Ig and with the effective moment of inertia as follows. When Lightweight aggregate is used.2-19 1− [ ]M cr Ma 3 ] I cr [ ] M I c = cr I g +¿ Ma Where f r Ig M cr = Yt Fr = modulus of rapture of concrete.7 √ f ' c M a = maximum moment in member at stage deflections is computed. MPa. but not greater than . for normal weight Concrete f r=0. I g = moment of inertia of gross concrete section about centroidal axis. I cr = moment of inertia of cracked section transformed to concrete Y 1 = distance from centroidal axis of gross section. neglecting reinforcement. to extreme fiber in tension. 5.3 Unless stiffness values are obtained by a more comprehensive analysis.2.5. neglecting reinforcement.Sect. 5.4: For continuous members. Sect. It is permitted to assume the time-dependent factor ε for sustained loads to be equal to: 5 years or more……………………2. 5. shall not be multiplied by 0.1. 2-10 1+50 ρ ' Where ρ' shall be taken the value of reinforcement ratio for non-prestressed compression reinforcement at midspan for simple and continuous spans. effective moment of inertia may be taken as the average of values obtained from Eq.2.85 for “ sand-lightweight” concrete.9.2. Linear interpolation is permitted if partial sand replacement is used. fr f ct √ f 'c 5. f ct fr (b) When is not specified. Sect. and at the support cantilevers.4 6 months…………………………….2 3 months………………………………1.5.75 for “all lightweight” concrete.. 2-19 for the critical positive and negative moment sections.5. ε λ= Eq.0 .8 f ct √ f 'c shall not exceed .5: Unless values are obtained by a more comprehensive analysis. 2-19 at midspan for simple and continuous spans.5. and 0. f ct (a) When is specified and concrete is proportioned in accordance with Sec.2.0 12 months…………………………. For prismatic members.9..8 for but the value of 1. effective moment of inertia may be taken as the value obtained from Eq. shall be modified by substituting 1.1. by the factor..a nd at support for cantilevers. additional long-term deflection resulting from creep and shrinkage of flexural members (normal weight or lightweight concrete) shall be determined by multiplying the immediate caused by the sustained load considered. 5.2 through Sec.5 shall not exceed limits stipulated in Table 2. including added deflections due to ponded water and considering long-term effects of all sustained loads.2.5. .5. This amount shall be determined on basis of accepted engineering. or attached to immediate deflection due to nonstructural elements not likely any additional live load)**** to be damaged by large deflections Limit not intended to safeguard against ponding.2: Maximum Permissible Computed Deflections Type of member Deflection to be considered Deflection limitation Flat roofs not supporting or Immediate deflection due to L/180* attached to nonstructural live load LL elements likely to be damage by large deflections Floors not supporting or Immediate deflection due to L/360* attached to nonstructural live load LL elements likely to be damaged by large deflections Roof or floor construction That part of the total L/480** supporting.5.2. data relating to time-deflection characteristics of members similar to those being considered. construction tolerances. camber. 5.Deflection computed in accordance with Sec.5 or Sec. or attached to deflection occurring after nonstructural elements not likely attachment of non structural to be damaged by large elements (sum of the long- deflections time deflection due to all Roof or floor construction sustained loads and the L/20**** supporting. Table 2.5.9. Limit may be exceeded if adequate measures are taken to prevent damage to supported or attached elements.2. Long=time deflections shall be determined in accordance with Sec.2. Attachment of nonstructural elements.9. and reliability of provisions for damage.9. Ponding should be cheated by suitable calculations of deflections. …… 2 Negative moment at face of all supports for: . Limit may be exceeded if camber is provided so that deflection minus camber does not exceeded limit. 2 ωu Ln /10 More than two spans……………………………………………….. and e) Members are prismatic. the following approximate moment and shears are permitted for design of continuous beams and one-way slabs (slabs reinforced to resist flexural stresses in only one direction)...3. c) Loads are uniformly distributed.. provided: a) There are two or more spans. with the larger of two adjacent spans not greater than uniformly than the shorter by more than 20 percent. But not greater than tolerance provided for nonstructural elements. Positive moment End spans ωu Ln /11 Discontinuous end unrestrained…………………… 2 ωu Ln /14 Discontinuous end integral with support………….8... 2 ωu Ln /11 Negative moment at other faces of interior supports………….3 of NSCP states that in lieu of frame analysis. 2 ωu Ln /16 Interior spans………………………………… 2 Negative moment at exterior face of first interior support ωu Ln /9 Two spans …………………………………………………………. b) Spans are approximately equal.... NSCP COEFFICICIENTS FOR CONTINUOUS BEAMS AND SLASBS Section 5. d) Unit live does not exceeded three times unit dead load. ... 2 Ln When =clear span positive moment or shear and average of adjacent clear spans for negative moment............ Column Column Column 1 ..... 2 Shear in end members at face of 1..... 2 ωu Ln /2 Shear at face of all other supports………………………………………..... and beams Where ratio of sum of column stiffness to beams ωu Ln /12 Stiffness exceeds eight at each end of the span…………………… 2 Negative moment at interior face of exterior Support members built integrally with ωu Ln / 24 Where support is a spandrel beam…………………………… 2 ωu Ln /16 When support is a column……………………………...…….5 ωu Ln /2 First interior support……………………………..15w 5 Shear w w w Moment - -w w -w ....... 1 w 1.….Slabs with spans not exceeding 3 m.... Figure 2.3: Shear and moment for continuous beam or slab with spans and discontinuous end integral with support Column Column Column Spandrel w Beam 1 w . 1.15 1 w w 5 w Shear w w Moment - - - - w - - w w w w w Column Column Column 1 w . w 1.15 1 w w 5 w Shear w w Moment - w - - - w w w Figure 2.5 Shear and moment for continuous beam or slab with more than two spans and discontinuous end unrestrained ACI-318-05 (NSCP C101-10-2010) 10.2 Design assumptions (410.3) 10.2.1 Strength design of members for flexure and axial loads shall be based on assumptions given in 10.2.2 through 10.2.7, and on satisfaction of applicable conditions of equilibrium and compatibility of strains. 10.2.2 Strain in reinforcement and concrete shall be assumed directly proportional to the distance from the neutral axis, except that, for deep beams as defined in 10.7.1, an analysis that considers a nonlinear distribution of strain shall be used alternatively, it shall be permitted to use a struct-and tie model. See 10.7,118, and Appendix A. 10.2.3 Maximum usable strain at extreme concrete compression fiber shall be assumed equal to 0.003. fy Es 10.2.4 Stress in reinforcement below shall be taken as times steel fy strain. For strains greater than that corresponding to , stress in reinforcement shall fy be considered independent of strain and equal to . 10.2.5 Tensile strength of concrete shall be neglected in axial and flexural calculations of reinforced concrete, except when meeting requirements of 18.4. 10.2.6 The relationship between concrete compressive stress distribution and concrete strain shall be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction of strength in substantial agreement with results of comprehensive tests. 10.2.7 Requirements of 10.2.6 are satisfied by an equivalent rectangular concrete stress distribution defined by the following: f 'c 10.2.7.1 Concrete stress of 0.85 shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross section β and a straight line located parallel to the neutral axis at distance a= 1 c form the fiber of maximum compressive strain. 10.2.7.2 Distance from the fiber of maximum strain to the neutral axis, c , shall be measured in direction perpendicular to the neutral axis. f 'c β1 10.2.7.3 For between 17 and 18 MPa, shall be taken as 0.85. f 'c β1 For above 28 MPa, shall not be taken less than 0.65 10.3 General principles and requirements (410.4) 10.3.1 Design of cross sections subject to flexure or axial loads, or to combined flexure and axial loads, shall be based on stress and strain compatibility using assumptions in10.2. 10.3.2 Balanced strain conditions exist at a cross section when tension fy reinforcement reaches the strain corresponding to just as concrete in compression reaches its assumed ultimate strain of 0.003. 10.3.3 Sections are compression-controlled if the next tensile strain in the extreme tension steel, εt , is equal to or less than the compression-controlled strain limit when the concrete in reaches its assumed strain limit of 0.003. The compression- controlled strain limit is the net tensile strain in the reinforcement at balanced strain conditions. For Grade 415 reinforcement, and for all prestressed reinforcement, it shall be permitted to set the compression-controlled strain limit equal to 0.002. 10.3.4 Sections are tension-controlled if the net tensile strain in the extreme tension steel εt is equal to greater than 0.005 when the concrete in compression reaches its assumed strain limit of 0.003. Sections with εt between the compression- controlled strain limit and 0.005 constitute a transition region between compression- controlled and tension-controlled sections. Derivation: for E = 200 GPa The beam is tension-controlled 0.003 0.008 fs When ε = 0.005 (or =1000MPa) c = d 0.005 c d = 0.003 0.008 3 c= d Eq. 2-21 8 3 a=β1 c=β 1 d 8 For rectangular beam: a ( ) φ M tn =φCc d− where φ=0.90 2 a φ M tn =0.90 x 0.85 f 'c ab( d− ) 2 3 β1 d ' 8 φ M tn =0.90 x 0.85 f c x β 1 3 /8 dxb(d − ) 2 459 3 φ M tn = β1 f 'c b d 2 (1− β 1) Eq. 2-22 1600 16 10.3.5 For nonprestressed flexural members and nonprestressed members with f c Ag factored axial compressive load less than 0.10 steel strain εt at nominal strength shall not be less than 0.004. 10.3.5.1 Use of compression reinforcement shall be permitted in conjunction with additional tension reinforcement to increase the strength of flexural members. Derivation: for E =200 GPa M n max Maximum steel area and when beam is singly reinforced: εt=0.004∨f s=εtx E=800 MPa 0.003 0.007 c = d 0.004 Strain diagram for minimum steel strain 2-23 7 For rectangular section: Cc A s max f y =0.007 7 3 Cmax = d Eq.85 f ' c ab where a=β 1 c T= 3 β1 x d 7 )b ρmax bd f y =0.85 f ' c ab ( ) 2 β 3 d 3 1 M n max=0. c= d 0.85 f c x β 1 dxb (d− 7 ) 7 2 .003 0.85 f c ¿ 3 0.85 f ' c β 1 Eq. c d 3 = . 2-14 ρmax 7 fy M n max=c c ( −a2 ) −a M n max=0. The factored axial force at given Mu eccentricity shall not exceed that given in 10.5: φ Pn max =0.3 For prestressed members.85 φ[0.4 or composite members conforming to 10. (10-1) or (10-2). 10.85 f 'c ( A g− A st )+ f y A st ] (10-1) 10.80 (for members with tie φ P0 reinforcement) of the design axial strength at zero capacity . 2-25 51 ) β c f 'c b d 2 ¿ 140 800−f y Eq.10.25 1000−f y φ Pn 10.85 (for members with spiral reinforcement) or 0.3. 2-26 φ=0.10 .6.6.85 φ[0. design axial shall not be taken greater than 0.6.3.6 Design axial strength of compression members shall not be taken greater φ Pn max than computed by Eq.7 Members subject to compressive axial load shall be designed for the maximum Pu moment that can accompany the axial load.3.1 For nonprestressed members with spiral reinforcement conforming to 7.10.2 For non nonprestressed members with spiral reinforcement conforming to 7.3. 10.3.85 f 'c ( A g− A st )+ f y A st ] (10-2) φ Pn 10.16: φ Pn max =0.3.6. 3 1− β 14 1 M n max=¿ Eq. The maximum factored moment shall be magnified for slenderness effects in accordance with 10.65+0. I. the least width of compression flange or face. Approximate the weight of beam (DL) as follows: .4.3. shall not bw 2 bw be less than the value given by eq.1 Minimum reinforcement of flexural members 10. STEPS IN THE DESIGN OF SINGLY REINFORCED RECTANGULAR BEAM FOR FLEXURE Note: The assumption made in steps II.5. V.5.5.4.5.4 Distance between lateral supports of flexural members 10. and VIII are the authors recommendation based on his experience. (10-3). LL and other loads II. whichever is smaller.4. 10.2 –For statically determinate members with a flange in tension.10.1 At every section of flexural members where tensile reinforcement is required by analysis.1 Spacing of lateral supports for a beam shall not exceed 50 times b. 10.2. except as provided in 10. as provided shall not be less than that given by √f 'c b d∨ ρmin = √f 'c b d 4fy w 4fy w (10-3) and not less than 1.2 Effects of lateral eccentricity of load shall be taken into account in determining spacing of lateral supports. Determine the values of loads. Dl. except that is replaced by either or the width of the flange. and 10.4 1.4 A smin = b w d∨ρmin fy fy (10-3) A smin 10. 10.5.5. Compute the factored load on different load combinations Example: Factored Load =1. ρ from 0.36 MPa .05 ' β 1=0.85− 7 ( f c −28 ) for f ' c > 28 MPa 3 0.8 but must not be ρmin ρ less than .5kN/m Large-sixed beams: 7kN/m 2 or Weight of beam in kN/m=24kN/ m x beam area in m2 III.7 ρmax ρmax V.2 DL + 1. 0.othewise ρ min= 1.6 LL Mu IV. Try a value of steel ratio to 0.85 f ' c β 1 600 ρb= f y ( 600+ f y ) β 1=0. This value of will provided enough allowance in the actual value of ρ due to rounding-off the numbers bars to be used so that it will not exceed the maximum ρ .85 for f ' c ≤ 28 MPa 0.85 f ' c ρmax = 7 fy ρmax = √ f ' c if f ' > 31. Small beams: 2kN/m Medium-sized beams: 3. Compute the factored moment to be resisted by the beam.4 c 4fy fy . 65+0. substitute its value to Step VII. Solve for bd : M u=φ f ' c ω b d 2 ( 1−0. Compute the value of ω . After solving for d. d−c f s=600 c f s ≥ 1000 MPa . (round-off this value to reasonable dimension) Check also the minimum thickness of beam required by the code as given in Table 2.90 if f s−f y if f s <1000 MPa . .1 in Page 26.59 ω ) ¿ ¿ ¿ ¿ 2 b d =¿ ¿ IX. Solve for the reduction factor φ: Solve for c: Note: For singly reinforced rectangular beam. and solve for b. φ=0.transition . ρ is directly proportional to c: 3 Cmax where C max = d c=(assumed factor) x 7 The assumed factor may range from 0.8 as suggested in step V.7 to 0. φ=0. Try a ratio d/b (from d= 1. ρ fy VI.25 1000−f y 2 VIII.tension−controlled .5b to d=2b). and solve for d. ω= f 'c VII. X. 3 0. A s =pbd Number of bars (diameter=D) π 2 D x number of bars = A s 4 AS STEPS IN FINDING THE REQUIRED TENSION STEEL AREA OF A BEAM MU WITH KNOW REQUIRE MOMENT AND OTHER BEAM PROPERTIES f 'c f y Mu Given b. Solve for and . d.65+0.85 f ' c β1 ρmax = 7 fy 51 3 M n max= 140 ( β 1 f ' c b d 2 1− β 1 14 ) 800−f y φ=φ=0. Compute the weight of the beam and it to the assumption made in Step II. Solve for the required steel area and number of bars. and : ρmax φ M n max I.25 1000−f y ¿ ¿ ¿ ¿ ¿ φ M n max =¿ ¿ . 4 c 4fy fy ¿ ¿ ¿ A s =pbd =¿ ¿ .90 if .proceed to step III f s >φ M n . Determine if the section in tension-controlled or transition 459 3 φ M tn = β1 f 'c bd 2(1− β 1) From Eq.transition if region. otherwise ρmin = 1.85 f 'c ρ= ¿ fy ρmin = √ f 'c if φ f ' >31. M u=φ Rn b d 2 ¿ ¿ Rn =¿ ¿ 2 Rn 1− 0. 2-11: 1600 16 f s <φ M n .85 f ' c ¿ ¿ 1−√ ¿=¿ ¿ 0.36 MPa . M u ≤ φ M n max if design as Singly Reinforced (Step II) M u> φ M n max if design as Doubly Reinforced (Step V) II. proceed to step IV III. φ=0.tension−controlled . 85 β 1 c b 0.4 c 4fy fy ¿ ¿ ¿ A s =pbd =¿ ¿ As : Solve for c and M u=φC c (d−a /2) f s−f y d−c φ=0.65+0.85 f ' c ab=0.36 MPa .25 where f s=600 1000−f y c a=β1 c C c =0. ρmin = √ f 'c if φ f ' >31.25( [ 600 d−c c −f y ] c )(0.65+0.85 f ' c β 1 c b)(d −β1 ) 1000−f y 2 M u =¿ . otherwise ρmin = 1.IV. 36 MPa . d. . : As I. . Solve for ρ= bd ρb II. otherwise ρ min= 1.85 f ' c a b ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ A s =¿ ¿ ρmin = √ f 'c f ' c >31. ¿ ¿ ¿ a ¿ ¿ ¿ c ¿ β 1 c =¿ ¿ A s f y =0.4 4fy if fy V. Compression reinforcement is necessary. Check if steel yields by computing .(See chapter 2) φ Mn STEPS IN FINDING OF A BEAM WITH KNOWN TENSION STEEL AREA AS AND OTHER BEAM PROPERTIES: As f 'c fy Given: b. otherwise ρmin = 1.c= a/ β 1 =_________ d−c f s=600 c f s ≥ 1000 MPa . proceed to step III ρ> ρb . Compression-controlled φ=0.36 MPa . ' 0.85 f c β 1(600) ρb= f y ( 600+ f y ) ρ ≤ ρb.85 f ' c a b(d− ) 2 ρ> ρb IV.65+0.85 As . ρ ≤ ρ min .90 if f s−f y if f s <1000 MPa . Solve for φ : A s f y =0. if steel dos not yield.transition .85 f ' c ab ¿ ¿ a=¿ ¿ . φ=0. As Note: if the given is not adequate for the beam dimension. proceed to step IV. φ=0.tension−controlled . ρmin = √ f 'c if f ' >31. if steel yields.25 1000−f y a φ M n=φ 0.4 c 4fy fy ρ ≤ ρb III.65 f 'c b 0. 85 f ' c ab .85 f ' c ( β 1 c ) b c d−c f s=600 c=__________ .85 f ' c ab β1c As f s but a= T= d−c A s 600 =0.65 φ M n=φT d − ( a2 )=φ A f (d− a2 ) s s or φ M n=φC d − ( a2 )=φ f ' ab(d − a2 ) c . d−c f s=600 a c= c 0. d d-a/2 T=C A s f s=0. c =__________ ¿ ¿ a= ¿ β 1 c =¿ ¿ φ=0. 1 A reinforced concrete rectangular beam 300 mm wide has an effective depth of 460 f ' c =21 MPa f y =345 MPa mm and is reinforced for tension only.85 ( 27.85 ) (600) ρb= 345(600+345) ρb=0. Concrete compressive strength f ' c =27.ILLUSTRATIVE PROBLEMS DESIGN PROBLEMS PROBLEM 2.85(600) ρb= ρ b= f y ( 600+f y ) 276 (600+276) ρb=0.(c) 485 kN-m.0495 .mm.85 ( 21 ) ( 0. Assuming and .85 f ' c β 1 600 ρb= β 1=0.(b)140 kN-m. determine the balance steel area in sq.85 f ' c β 1 600 0. and (d)620 kN-m.02792 A sb= ρb bd PROBLEM 2.6 ) 0.6 MPa f y =276 MPa and steel yield strength .85 since f ' c <30 MPa f y ( 600+f y ) 0. SOLUTION ρmax ∧M u max : Solve for 0.2 A rectangular beam has b = 300 mm and d =490 mm. SOLUTION 0. Calculate the required tension Mu steel area if the factored moment is (a) 20 kN-m. 001 ( 300 ) ¿ 6 M n max=576.6 ( 0.371 ) [1−0.03711 ( 276 ) ω max= ' ω max= f c 27.59 ( 0.0371 ρmax f y 0.37 ) ] Rn max =8.371 Rn max =f ' c ω(1−0.59 ω) Rn max =27.90 x 576.6 ω max=0.75 ρb ρmax =0.65 kN −m a) M u=20 kN −m< M u max=(singly reinforced ) 0 2 49 ¿ Mu=φ Rn bd 20 x ¿ 6 10 =0.001 MPa 490 2 ¿ M n max=R n max b d ¿ M n max=8.ρmax =0.309 MPa .75(0.279 M u max=518.0495) ρmax =0.90 R n ( 300 ) ¿ Rn=0.279 kN −mm M u max=φ M n max M u max=0.279 x 10 N −mm M n max=576. 00113< ρmin ρmin = √ f 'c if f ' >31.85 f ' c ] .4 ρmin = =0.00572 ( 300 )(490) 2 A s =746 mm M u=140 kN −m< M u max b) (singly reinforced) 0 49 ¿ M u=φ Rn bd 2 140 x 1 ¿ 6 0 =0.005072 fy A s =ρbd A s =0.4 c 4fy fy 1.85 f ' c fy [ √ 1− 1− 2 Rn 0. otherwise ρmin = 1.ρ= 0.309 ¿ 2¿ 1−¿ 1−√ ¿ 0.16 MPa ρ= fy [ √ 0.85(27.90 R n ( 300 ) ¿ Rn=2.6) ρ= ¿ 276 ρ=0.36 MPa .85 f ' c 1− 1− 2 Rn 0.085 f ' c ] 0. 85 f ' c 1− √¿ 0.6) 276 [1− 1− √ 2 ( 7.6 ) ] . ρ= 0.6 ) ] A s =ρbd ρ=0.48 MPa 2 Rn 1− 0.85 ( 27.90 Rn ( 300 ) ¿ Rn=7.85 f ' c ρ= ¿ fy ρ= 0.00822 ( 300 )(490) A s =1.85 ( 27.16 ) 0.00822> ρmin A s =0.209 mm2 M u=485 kN−m< M umax c) (singly reinforced) 0 2 49 ¿ M u=φ Rn b d 485 x ¿ 2 10 =0.6) 276 [ √ 1− 1− 2 ( 2.48 ) 0.85(27.85(27. 3 Unit weight concrete is 23. b) If the design ultimate moment capacity of the beam is 280 kN-m. PROBLEM 2. determine the required number of 20 mm tension bars.Other than the weight of the beam . Concrete cover is 70 mm from the centroid of the steel area. SOLUTION f ' c =300 MPa Given: b=300m β 1=0. c) If the beam will carry a factored load of 240 kN at midsoan.4 f y =415 MPa ρmin = =0. a) Determine the maximum factored moment on the beam. See Chapter 3. Use the strength design method. the beam carries a superimposed dead of 18 kN/m and a live load of 14 kN/m.5kN/ m .00337 fy .03384> ρmin A s =ρ b d A s =0. f y =415 The beam is simply supported over span of 5 m.3 (CE MAY 2012) A reinforced concrete beam has a width of 300 mm and an overall depth of 480 mm.975 m m2 M u=600 kN −m> M umax d) The beam will be doubly reinforced.03384 ( 300 ) ( 490) A s =4.85 d=480-70=410 mm 1. determine the required number of 20 mm tension bars. Steel strength MPa and f ' c =28 MPa concrete . ρ=0. 93 kN −m M u=280 kN −m b) M umax Solve for to determine whether compression steel is needed 0. d b=20 m Bar diameter .3 x 0. Maximum factored moment: 5 ¿ W u L2 ¿ M u= ¿2 8 53.48 )=3.02881 ρmax =0. W u=1.85 ) (600) ρb= ρb= f y ( 600+f y ) 415 (600+415) ρb=0. Factored load.85 f ' c β 1 600 0. kN w b=γ c A b=23.7(14) W u=53.5 ( 0.75 ρb ρmax =0. m a) Maximum factored moment on the beam.021261 .384 Weight of beam.738 kN /m Factored load.738 ¿ M u =¿ M u=167.4 ( 3.85 ( 28 ) ( 0.384+18 ) +. 619) 0.85 f ' c fy 1− [2 Rn 0.9 say 7 bars .274 2 M u max=φ R nmax b d =330.01755 ( 300 ) (410) 2 A s =2159 mm 0 2¿ ¿ π π A s = d b 2159= ¿ 2 4 4 N=6.01755> ρmin] A s =ρ b d A s =0. ρmax f y ω max= ω max=0.85 f ' c ] ρ= 0.59 ωmax )=7.85(28) =0.90 Rn (300)¿ Rn=6.14 kN −m M u=280 kN −m M u max Required < (singly reinforced) 0 41 ¿ ¿ 2 6 M u=φ Rn b d 280 x 10 =.85(28) 415 1− 1− [ √ 2(6.03203 f 'c Rn max =f ' c ωmax ( 1−0.169 MPa ρ= 0. 95 say 8 bars .02031 ( 300 )(410) 2 A s =2498 mm 0 π 2¿ As= db N ¿ N 2498 = 2 4 π ¿ 4 N=7. Pu=240 kN at midspan 3. W d =3.384 kN /m (weight of beam) P u L (1.85 f ' c ] =002031> ρmin A s =ρ b d A s =0.936 MPa ρ= 0.4 W d ) L2 M u= + =314.805 x 1 06 R n= ¿ Rn=6.90 ( 300 ) ¿ 314.85(28) 415 [ √ 1− 1− 2 Rn 0.85 f ' c fy [ √ 1− 1− 2 Rn 0.85 f ' c ] ρ= 0.805 kN−m< M u max (singly ) 4 8 0 41 ¿ Mu ¿ R n= φb d 2 0. 59 ω) 0 60 ¿ ¿ [1-0.85 ) (600) ρ b= 414 (600+414 ) ρb=0.59(0.3209 ) ( 300 ) ¿ .031 ) ρmax =0.85 f ' c β 1 600 ρb= β =0.02323 ρf y 0.4 (CE MAY 1993) A reinforced concrete beam has a width of 300 mm and an effective depth to tension bars of 600 mm.85 ( 30 ) ( 0.02323 ( 414 ) ω= ' ω= f c 30 ω=0.85 since f ' c <10 MPa f y ( 600+f y ) 1 0. determine the tension steel area if the beam is to resist an ultimate moment of 650 kN-m. PROBLEM 2.90 ( 30 )( 0. compression reinforcement if needed will be placed at a depth of 60 f ' c =30 MPa f y =414 MPa mm below the top.3209 M u max=φ f ' c ω b d 2 (1−0.15 ρb ρmax =0.75 ( 0. If and .031 ρmax =0. SOLUTION ρmax M umax Solve for and : 0.309) M u max=0. 90 R n ( 300 ) ¿ Rn=6. Assume and . the beam may be designed as singly reinforced.00338 fy A s =ρbd A s =0.85 f ' c ) ρ= 0. 0 Rn=6.5 (CE November 2000) A rectangular concrete beam has a width of 300 mm and an effective depth of 550 mm.85(30) ] =0.85 f ' c fy ( √ 1− 1− 2 Rn 0.687 MPa 60 ¿ 650 x 1 ¿ 06 =0.687) 0. The beam is simply supported over a span 6 m and is used to carry a uniform dead f ' c =21 MPa load of 25 kN/m and a uniform live load of 40 kN/m.85(30) 414 [ √ 1− 1− 2(6.0191 ( 300 ) (600) A s =3442m m2 PROBLEM 2.0191> ρmin 1.1 kN −m> M u M u< M u max Since .687 MPa Solve for ρ : ρ= 0. M u max=758.4 ρmin = =0. 85 ) (600) ρb= 312(312+ 600) ρb=0.75(0.4 ( 25 ) +1.7 ( 40 ) W u=103 kN /m Required strength: . a) Determine 80 mm from the outermost compression concrete.85 f ' c β 1 600 ρb= β =0.0319 9 ρmax =0. c) Determine the required number of 25-mm tension bars.4 D+ 1.02399 A s max =ρmax bd A s max =0.75 ρb ρmax =0. f y =312 MPa .85 ( 21 ) ( 0.03199) ρmax =0. Compression reinforcement if necessary shall be placed at a depth 80 mm from the outermost compression concrete. b) Determine the required tension steel area.959 mm2 b) Required tension steel area: Factored load: W u=1.7 L W u=1.85 since f c is less than 30 MPa f y ( 600+f y ) 1 0.02399 ( 300 ) (550) A s max =3. SOLUTION a) Maximum steel area: 0. 5 kN −m> M u singly reinforced 0 55 ¿ ¿ 2 M u=φ Rn b d M u =0.356 M u max=φ f ' c ωb d 2 (1−0.39 R n ( 300 ) ¿ 550 ¿ ¿ 6 463.59 ω) 0 55 ¿ ¿ M u max=0. 6 ¿ ¿ ¿2 103 ¿ W L2 M u= u M u=¿ 8 Mu =463.90(30)(0.0299(312) ω= ω= f 'c 21 ω=0.67 MPa .5kN-m M u max Solve for ρmax f y 0.5 x 10 =0.356)(300)¿ M u max=536.9 Rn (300) ¿ Rn=5. a) Determine the balanced steel ratio in percent.c ) ρ= 0.67) 0.85(21) 312 [ √ 1− 1− 2( 5.002269 ( 300 )( 550 ) A s =3743 mm2 c) Number of 25 mm bars: As Number of 25-mm bars= A s 25 25 ¿ ¿ ¿2 Number of 25-mm bars= π ¿ 4 3. Concrete strength f ' c =28 MPa f y =248 MPa and steel yield strength .743 ¿ PROBLEM 2. The beam will be design to carry a factored moment of 540kN-m.6 (CE MAY 2009) A reinforced concrete beam has a width of 300 mm and total depth of 600 mm.85 f .85(21) ] ρ=0. Solve using the strength design method.85 f ' c 1− 1− 2 Rn 0. b) Determine the minimum effective depth of the beam using a steel ratio ρ . ρ= fy ( √ 0.02269 A s =ρbd A s =0. 77 ρ=0.5 of balanced steel ratio.5 ( 0.0577 )=0.85 ( 28 )( 0.0289 ρf y 0. SOLUTION Given: f ' c =28 MPa b=300 mm f y =248 MPa h=600 mm M u=540 kN −m β 1=0.0289( 248) ω= ω= =0.85 f c β 1 600 0. equal to 0.0577=5.2556 f 'c 28 .5 ρb b) Effective depth using ρ=0. c) Determine the minimum effective depth of the beam using the maximum allowable steel ratio.85 ) 600 ρb= ρb= f y ( 600+ f y ) 248 ( 600+248 ) ρb=0.85 a) Balanced steel ratio: ' 0. Determine the required spacing 12 mm main bar if the total factored moment acting on 1-m width of slab is 23 kN-m width of slab is 23 kN-m.90 ( 8. The slab will be reinforced f y =275 MPa f ' c =21 MPa with 12-mm-diameter bars with . Rn=f ' c ω (1−0. mm s s cover=20 mm b = 1000 mm .0776 MPa M u=φ M n=φ R n b d2 6 540 x 1 0 =0.2556 ) [1−0.59 ω) Rn=28 ( 0.59 ( 0.Concrete strength .2556 ) ] Rn=6. .7 A concrete one-way slab has a total thickness of 120 mm.307 ) ( 300 ) d 2 d=491 mm PROBLEM 2. 12mm bars d h=120 . SOLUTION Note: Slabs are practically singly reinforced because of its small depths. Clear concrete cover is 20 mm. 1154 ( 1000 )( 94) A s =1085 mm2 Spacing of bars (for walls and slabs using unit width): 1000 b s= s= As N Ab .982) 0.90 R n(1000)¿ Rn=2.0284 f y (600+ f y ) 1.Effective depth.00509 fy A s =ρbd A s =0. b = 1000 mm 94 ¿ M u=φ Rn b d 2 23 x ¿ 6 10 =0.85(21) 275 1− 1−( √ 2(2.4 ρmin = =0.75 x 0.85(21) ) 0.85 f ' c 1− 1− 2 Rn 0.892 ρ= fy ( √ 0.85 f ' c β 1 600 ρmax = =0.85 f ' c ) ρ= 0. d= 120 -20-1/2(12)=94 mm Width. 1000 A b s= Eq. Assume and .85 f ' c β 1 600 ρmax =0.8 A 2. The factored moment at f ' c =21 MPa f y =275 MPa critical section for moment is 640 kN-m. Clear concrete cover is 75 mm. Maximum and minimum requirements: 0.0284 f y (600+ f y ) . 2- As 17 12 ¿ ¿ ¿2 π 1000 x ¿ 4 1000 A b s= s=¿ As s=100 mm PROBLEM 2. Determine the required number of 20 mm tension bars. b =2800 mm M u=640 kN −m Design strength.8 m square column fooring has a total thickness of 47 mm. d=470-75-1/2(20)=385 mm Width.75 x =0. SOLUTION Effective depth. 4 b w d A s min = =5488 mm2 fy Singly reinforced: 385 2 ¿ M u=φ Rn b d ¿ 640 x 106 =0.85 f ' c ρ= ) 0. M u max=2528 kN −m (Procedure is not shown anymore see Problem 2.00656 A s =ρ b d A s =0.2) 1.85(21) 275 ¿ ρ=0.85 f ' c fy ( √ 1− 1− 2 Rn 0.85(21) ¿ ¿) 1−¿ 1−√ ¿ ρ= 0.90 R n (2800)¿ Rn=1.713 ¿ 2(¿ 0.713 MPa 1.00656 ( 2800 ) (385) A s =7074 mm2 > A s min Number of 20 mm bars: . 85 ( 20. 20 ¿ ¿ π ¿ 4 As 7074 N= N= ¿ Ab N=22.85 f ' c β 1 600 0.6 kN −m (Note: this already includes the weight of beam) 0.7 MPa f y =276 MPa .85 ) (600) ρb= ρb= f y ( 600+f y ) 276 (600+276) .m.4 ( 60 ) +1.7( 48) M u=165.7 ) ( 0.7 M L M u=1. Use f ' c =20.4 M b +1.5 say 23 bars PROBLEM 2. and SOLUTION Required strength: M u=1.9 Design a rectangular beam reinforced for tension only to carry a dead load moment of 60 kN-m (including its own weight) and a live load moment of 48 kN. 7 ω=0.6 x 106=0.2968 ) ] Rn=5.90 ( 5.2968 ) [ 1−0.0371 )=0.00507 fy ρ=60 ρb Try Note: this is the author’s suggestion ρ=0.6 ( 0. ρb=0.59 ω ) Rn=20.068 M u=φ Rn bd 2 165.4 ρmin = =0.068 ) bd 2 bd 2=36.296 x 106 mm3 Try d = 1.59 ( 0.0371 1.2968 Rn=f ' c ω ( 1−0.049 mm2 .02226 ( 230 ) 2(400) A s =2.75 b b=228 mm say 230 mm d=399 say 30 mm A s =ρbd A s =0.02226(276) ω= ω= f 'c 20.7 ( 0.02226 ρf y 0. 7(44) W u=121.85 ( 34 )( 0.448 ¿ M u =¿ M u=546.821 ) (600) ρb= ρb= f y ( 600+f y ) 345 (600+345) .821 7 ( 34−30 ) 0. W b =24 x (0. oncrete with m 3 .448 kN /m 6 ¿ W u L2 ¿ . M u= 8 ¿2 121. Use SOLUTION Weight of beam: (this is the author’s assumption) Assuming a 300 mm x 600 mm.516 kN −m 0.05 β 1=0.4 ( 29+4.85 f ' c β 1 600 0.Summary: b = 230 mm d = 400 mm A s =2.10 Design a singly reinforced rectangular beam for a 6-m simple span to support a superimposed dead load of 29 kN/m and a live load of 44 kN/m.6)=4. f ' c =34 MPa .049 mm2 PROBLEM 2.85− =0.32kN /m W b =1.3 0. Assume normal weigth 24 kN γ= ρmax .∧f y =345 MPa .7 W L W u=1.32 ) +1 .4 W L +1. 4 mm∧d=489mm Use b = 280 mm.36 MPa c 4fy ρf y 0.6.2.087 )( b ) ¿ b=279. ¿ M u=φ Rn b d 546.332 ) [1−0.75 b (this is the author’s assumption) 75 b 2 1.087 MPa Assume d = 1.03277> ρmin ρmin = √ f 'c =0. d = 490 mm Minimum beam the thickness (Section 409.56 ω) Rn=34 ( 0.516 x ¿ 6 10 =0.1) . ρb=0.00423 since f ' >31.332 ) ] Rn=9.03277(345) ω= ω= f 'c 34 ω=0.332 Rn=f ' c ω (1−0.04369 ρ= ρmax=0.59 ( 0.90 ( 9.04369 ) ρ=0.75 ( 0. 4+ y h min= 700 6000 16 ( 0. . h6 - #10 d b∨25 mm ≥¿ .03277 ( 280 ) (490) A s =4496 mm 2 Using 32 mm bars (#100): 32 ¿ ¿ π ¿ 4 As 4496 N= N= ¿ Ab N=5.4 + 345 700 ) hmin =335 mm OK A s =ρ b d A s =0.6 say 6 bars 2 80 mm . f hmin = L 16( ) 0. 73 kN/m < 4.6 SOLUTION Given: f ' c =24 MPa f y =415 MPa .6 is made of reinforced concrete having a width of 290 mm overall depth of 490 mm. Concrete cover is 60 mm from the centroid of the bars.5mm> hmin Beam weight = 24 (0. f y =415 MPa . h=490+ ( 25 ) +32+20 h=554.5545) Beam weight = 3. and a uniform live load of 55 kN/m. Assume EI=constant. Given f ' c =24 MPa .11 A propped cantilever beam shown in Figure 2.28)(0. The beam is loaded with uniform dead load of 35 kN/m (including its own weight).32(OK) PROBLEM 2. 290mm 490 mm A 6m B 2m C Figure 2. Determine the required tension steel area for maximum positive moment. 7 W L W u=1. f yh=275 MPa b=290 mm D H=490 mm d ' =60 mm O Lo A L1=6 m B 2m C W D=35 kN /m x W L=55 kN /m MD R d=490−60=430 mm W u=1.5 kN /m MA Moment Diagram M8 Solve for moment reactions using the three-moment equation: M B=−142.4 W D +1.5 ( 2 ) (1 ) =−285 kN−m 6 A 0 á0 6 A1 b́0 Mo Lo + 2 M A ( Lo + L1 ) + M B L1 + L0 + L1 =0 .4 ( 35 ) +1.7(55) W u=142. 828 kN −m φ M n max Solve for : .5 ¿ 0+2 M A ( 0+6 ) + (−285 ) ( 6 ) +0+¿ M A=−498.75+¿ ¿ ¿ ¿2 ¿2 ¿ ¿ M D =R X −W u ¿ M D =676.75 kN−m A M A=M ¿ ¿ -489.676.75 )−142.5 ( 8 )−676.875 ( 2.875 = 0 x = 2.5(8)(4) R=676.5 ¿ M D =253.142.75 m 2 2 x +¿ 2. 6 ¿ ¿ ¿3 142.75 = R(6).125 kN Maximum positive moment: V D =0 W u ( 2+ x )−R=0 142.5(2 + x) .875 R A =463.875 kN R A =W u L−R R A =142. 0247 ρmax =0.75 ρb ρmax =0.90(334.316) φM n max=300.316 kN−m φ M n max=0.828 kN −m< φ M n max (Singly reinforced) .59 ω ) Rn max =415 ( 0.85 ) 600 ρb= ρb= f y ( 600+f y ) 415(600+ 415) ρb=0.59 ( 0.0247) ρmax =0.3203 ) ] 430 2 ¿ M n max=R n b d ¿ M n max=6.01852(415) ω max= ω max= f 'c 24 ω max=0.235 ( 290 ) ¿ M n max=334.3203 ) [ 1−0.85 ( 24 )( 0.75(0.884 kN −m At a point of maximum positive moment: M u=253.3203 Rn max =f ' c ω ( 1−0.85 f ' c β 1 600 0. 0.01852 ρmax f y 0. 430 2 ¿ M u=φ Rn b d 253.85(24 ) 415 [ √ 1− 1− 2( 5.26 MPa ρ= fy [ √ 0.01495 ( 290 ) (430) A s =1.01495 A s =ρ b d A s =0.828 x ¿ 106=.85( 24) ] ρ=0.85 f ' c 1− 1− Rn 0.90 Rn (290) ¿ Rn=5.864 mm2 .85 f ' c ] ρ= 0.26) 0. ANALYSIS OF RECTANGULAR BEAMS WHERE STEEL YIELDS ( f =f ¿ S Y PROBLEM 2.01023(400) ω= ω= =0.85 )(600) ρb= ρb = f y (600+ f y ) 400(600+ 400) ρb=0. b) The ultimate moment capacity of the beam. SOLUTION 0.12(CE MAY 1999) A reinforced concrete rectangular beam with b = 400 mm and d= 720 mm is reinforced ' for tension only with 6-25 mm diameter bars.85 ( 21 )( 0. determine the following : Rn a) The coefficient of resistance of the beam. If f c =21 MPa and f y =400 MPa .85 f ' c β 1 0.01023< ρb ( steel yields ) bd 400(720) ρf y 0.195 f 'c 21 .02276 25 ¿ ¿ π A s =6 x ¿ 4 As 2945 ρ= ρ= =0. 67 kN −m Answer PROBLEM 2.85 ( 27 ) ( 0. beam in kN-m.56 ω) Rn=21 ( 0.59 ( 0.13 A rectangular beam reinforced for tension only has b= 300 m.500 sq. Assume .85 f ' c β 1 600 0. mm. Rn=f ' c ω(1−0.62 )( 400 ) ¿ ( M u=675.02276 As 4.500 ρ= ρ= bd 300(490) .85 ) (600) ρb= ρb= f y ( 600+f y ) 275 (600+275) ρb=0. The tension steel area provided is 4.195 ) [1−0.90 3. y SOLUTION 0. Determine the ultimate moment capcity of the f ' c =27 MPa f =275 MPa .62 MPa Answer 720 2 ¿ M u=φ Rn b d ¿ M u=0. d = 490 mm.195 ) ] Rn=3. f y =414 MPa ¿ .2−30 )=0.85− ( 34.3 kN −m PROBLEM 2. ρf y 0.2 ) ( 0.3118 ) [1−0. grade 60 reinforcement ( Calculate the design moment SOLUTION 0.05 β 1=0.59 ω) Rn=27 ( 0.03407 .59 ( 0.85 f ' c β 1 600 0.90 ( 6.82 7 0.87 MPa 490 2 ¿ M u=φ Rn bd ¿ M u=0. f ' c =34.82 ) (600) ρb= ρb= f y ( 600+f y ) 414(600+ 414) ρb=0.2 MPa . A rectangular beam has b = 300 mm.87 ) ( 300 ) ¿ M u=445.0361(275) ω= ω= f 'c 27 ω=0. d = 500 mm. M u.3118 ) ] Rn=6.85 ( 34.3118 Rn=f ' c ω(1−0.14 A s =3−25 mm . 779 )( 300 ) ¿ M u=255.00353 OK 4fy ρf y 0. 25 ¿ ¿ π As= ¿ 4 As 1473 ρ= ρ= bd 300(500) ρ=0.779 MPa 500 ¿ ¿ M u=φ Rn bd 2 M u=0.2 ( 0.59 ω ) Rn=34.15 A 130-mm-thick-one-way slab is reinforced with 12-mm-diameter tension bars spaced f ' c =21 at 110 on centers.1188 ) [ 1−0.90 ( 3. Concrete cover is 20 mm.00982< ρb Steel yields Check if the beam satisfies the minimum requirement: ρmin = √ f 'c =0.2 Rn=f ' c ω ( 1−0.00982(414) ω= ω= f 'c 34.11 kN −m PROBLEM 2.1188 ) ] Rn=3. concrete strength MPa and .59 ( 0. b = 1000 mm db Effective depth: d = h – cover.1/2 d = 130-20-1/2(12)=104 mm 0.00989 Check if the beam satisfies the minimum steel requirement on flexures: . a) What is the ultimate moment capacity of the slab? b) If the slab is simply supported over a span of 4 m. f y =275 MPa 3 steel yield strength .85 ) (600) ρ b= ρb = f y (600+f y ) 275( 600+275) ρb=0.85 f ' c β 1 600 0.5 kN/ m . Unit weight of concrete is 23. what safe uniform live load pressure can the slab carry? SOLUTION a) Consider 1 m width of slab.85 ( 21 ) ( 0.0378 b A s =A b x N A s =A b x s 12 ¿ ¿ π As= ¿ 4 2 A s =1028 mm As 1028 ρ= ρ= bd 1000( 104) ρ=0. 1.90 ( 2.129 Rn=f ' c ω ( 1−0.443 kN −m 4 ¿ ¿ ¿2 b) W u¿ W u L2 M u= 24.055 kPa Dead load pressure.129 ) ( 1−0.511 ) ( 1000 ) ¿ M u=24.129 ) ] Rn=2.59 ω ) Rn =21 ( 0.00989( 275) ω= ω= f 'c 21 ω=0.222 kN /m ρ D=γ c Dead load pressure. .443=¿ 8 W u=12.4 ρmin = =0. ρ D=23.59 ( 0.13=3.5 x 0.00509 OK fy ρf y 0. x thickness of concrete.511 MPa 104 2 ¿ M u=φ Rn bd ¿ M u=0. 7 ) ( 0.5) ρb=0.7 W L W u=1. The beam is simply supported over a span of 6 m and carries a uniform dead load of 680 N/m including its own weight.01281< ρb ¿ ) . W u=1. Assume and .16 A rectangular beam with b = 250 mm and d = 460 m is reinforced for tension only with 3-25 mm bars.7 ( ρ L b ) 12.673 kPa PROBLEM 2.85 ) (600) ρb= ρb= f y ( 600+f y ) 276. SOLUTION 25 ¿ ¿ π A s =3 x ¿ 4 0.4 ( 3.473 ρ= ρ= bd 250 (460) steel yields ρ=0.4 W L +1. Calculate the uniform live load that the f y =276.5 MPa f ' c =20.03703 As 1.055 x 1 ) +1.7( ρL x 1) ρ L=4.222=1.7 MPa beam can carry.85 ( 20.5 (600+276.85 f ' c β 1 600 0.4 ( ρD b ) +1. 68=1.171 Rn=f ' c ω ( 1−0.01281(276.4 ρmin = =0.56 kN −m 6 ¿ ¿ ¿2 Wu ¿ W L2 M umax= u 151.5) ω= ω= f 'c 20.68 kN /m W u=1.7 ω=0.7 W ¿ .183 MPa 460 ¿ ¿ 2 M u=φ Rn bd M u=0.59 ( 0.183 )( 250 ) ¿ M u=151.4 W DL +1.4 ( 0.Check if the beam satisfies the minimum steel requirement on flexure: 1.171 ) ] Rn=3.90 ( 3.00506 OK fy ρf y 0.171 ) [1−0.56=¿ 8 W u=33.7 W ¿ 33.68 ) +1.59 ω ) Rn =27 ( 0. 028816 )( 300 )( 5) A s max =3.242 mm2 c) Nominal moment capacity Using 6-25 mm bars: .85 f ' c β 1 600 0.028816 ρb=2.88 b) Maximum steel area A s max =ρmax bd A s max=( 0.25 kN −m PROBLEM 2.75 x 0. a) What is the balanced steel ratio? b) What is the maximum steel area for singly reinforced? c) What is the nominal moment capacity of the beam? SOLUTION a) Balanced steel ratio: 0.85 ) 600 ρb= ρb= f y ( 600+f y ) 415(600+ 415) β 1=0. Steel yield fy f 'c is 415 MPa and concrete strength is 28 MPa. W ¿ =19.85 ( 28 ) ( 0.85 ρb=0.17 (CE JANUARY 2008) A reinforced concrete rectangular beam has a width of 300 mm and an effective depth of 55 mm. The beam is reinforced with six 25-mm-diameter tension bars.75 ρb ) bd A s max =( 0. 7494 MPa 500 ¿ ¿ 2 M n=Rn bd M n=6. 25 ¿ ¿ π As= ¿ 4 As 2.01963< ρb (tensio steel yields) ρf y 0.291 ) (1−0.7494 ( 300 ) ¿ M n=506.291 f 'c 28 ' Rn=f c ω ( 1+0.01963(415) ω= ω= =0.59 x 0.59 ω ) R n=28 ( 0.945 ρ= ρ= bd 300 (500) ρ=0.2 kN−m .291) Rn=6. a uniform live load of 3 kN/m. Calculate the following: a) The ultimate moment capacity of the section in kN-m.5 ) ( 0. and b) The maximum value of P in kN.5 kN/m (including its own weight). 2P P 2m 2m 2m Figure 2.5 MPa load of P and 2P as shown in Figure 2.7 SOLUTION 0.85− ( 34.818 ) (600) ρ b= ρb = f y (600+f y ) 414( 600+414) . PROBLEM 2.05 β 1=0.85 f ' c β 1 600 0. and concentrated dead f y =414 MPa . The beam has an effective depth of 446 mm.818 7 0.5−30 ) =0. The beam carries a uniform dead load of 4.7. f ' c =34.18 A 350 mm x 500 mm rectangular is reinforced for tension only with 5-28 mm bars. Assume .85 ( 34. 025 ) ( 300 ) ¿ .59 ( 0.00355 4fy ρ¿ ρf y 0.5 ( 0.01972< ρb Steel yields Check if the beam satisfies the minimum requirement: m∈¿= √ f 'c =0.01972(414) ω= ω= f 'c 34. ρb=0.2367 Rn=f ' c ω ( 1−0.2367 ) ] Rn=7.025 MPa 446 2 ¿ M u=φ Rn bd ¿ M u=0.2367 ) [1−0.5 ω=0.03428 28 ¿ ¿ π As= ¿ 4 As 3079 ρ= ρ= bd 300 (446) ρ=0.90 ( 7.59 ω ) Rn =34. the maximum moment can occur at B or C: M c =1.4P W u=1.7 ( 3 ) =11.4P(2) + 11.4(2)(1) P=149 kN .18 = 1.4(2P) 1.8 – Beam with factored loads For the given loads.4 ( 4.4 kN / m A Ra B C D 2m 2m 2m Figure 2. M u=440.5 ) +1.18 kN −m 1.4 P ( 2 )+11.4 (2)(1) At point C: M c =M u Set 440. SOLUTION 0.7 P ) −11.7 P ) (2 )−11.85 ( 27.7 P M B¿ M B = ( 17.1+0.1+ 0.27 kN Mu Thus the maximum value of P such that will not exceed 440.19 A s =6−32mm .18 kN-m is 149 kN. ANALYSIS OF RECTANGULAR BEAMS WHERE STEEL DOES NOT YIELDS ( f ≠ f ) S Y PROBLEM 2.4 ( 2 ) (1) Set P=306.4 (6)(1) R A =17.6 MPa .4 P ( 2 )=2.1+0. A rectangular beam has b = 300 mm.85 ) (600) ρb= ρb= f y ( 600+f y ) 414 (600+414 ) . f ' c =27.4 ( 2 ) ( 1) ∑¿ M B=M u 440.8 P ( 2 ) +11. grade 60 reinforcement ( Calculate the ultimate moment capacity of the beam. f y =414 MPa ¿ .85 f ' c β 1 600 0.6 ) ( 0.18=( 17. d = 500 mm. RA ¿ At point B: (First solve for ∑ M c=0 4 R A +1. 0285 32 ¿ ¿ π As= ¿ 4 As 4825 ρ= ρ= bd 300 (500) ρ=0. 2-18 d−c 500−c f s=600 f s =600 T= c c ∑ F H =0 .03217> ρb Steel does not yield b=.8 5 a d=500 500-a/2 =4825 From Eq. ρb=0.300 0.85 ab c=0. 85 f ' c a b .2 mm d−c 500−306 f s=600 f s =600 c 306 f s=379.85 c 500−c ( 4825 ) 600 =0.90 ( 4825 )( 379.2) a=260. a=β 1 c=0.8 kN −m .964=0 c=306.3 φM n=0.65 ) (500− ) 2 φ M n=609.3 mm ( a2 ) φM n=φ A s f s d− 260.85 (306.85 ( 27.6 ) ( 0.85 c ) (300) c c 2=484 c−241. T =C A s f s=0.65 MPa a=β1 c=0. Calculate the ultimate moment capacity of the beam.85 ( 21 ) ( 0.02337> ρb Steel does not yield b=300 0.8 5 a d=490 490-a/2 =3436 T= . The f y =415 MPa .85 f ' c β 1 600 0.02161 25 ¿ ¿ π As= ¿ 4 As 3436 ρ= ρ= bd 300 (490) ρ=0. SOLUTION 0. tension steel area provided is 7-25 mm diameter bars with f ' c =21 MPa .85 ab c=0. d = 490 mm.20 A rectangular beam reinforced for tension only has b=300 mm.PROBLEM 2.85 ) (600) ρ b= ρb = f y (600+f y ) 415( 600+415) ρb=0. 85 f ' c a b .24 f s=600 f s =600 c 296.43 MPa< f y a=β1 c=0. a=β 1 c=0.81mm .43) a=251.From Eq.85 c 490−c ( 3436 ) 600 =0.85 ( 221 ) ( 0.2-18: d−c 490−c f s=600 f =600 c s c ∑ F H =0 T =C A s f s=0.24 mm d−c 490−296.24 f s=392.85 c )(300) c c=296.85 (392. 9.81 φ M n=0. 125125 125 125 700mm 4-32mm 75 375mm Figure 2.9 .43 ) (490− ) 2 φM n=441.86 kN−m ANALYSIS & DESIGN OF SINGLY REINFORCED NON-RECTANGULAR BEAMS PROBLEM 2. ( a2 ) φ M n=φT d − ( a2 ) φ M n=φ A s f s d− 251.21 Compute the ultimate moment capacity of the beam shown in Figure 2.90 ( 3436 ) ( 392. Assume f y =345 MPa f ' c =21 MPa and . The moment can be computed using the assumptions in the Code and the conditions of equilibrium. Some formulas derived above (such as ρ . 2-11 125 c b= 600 d C= 600(625) a 625mm 600+ f y b 600+345 4-32mm 375mm . Cb =: 125125125 From Eq. 32 ¿ ¿ π As= ¿ 4 2 A s =3217 mm As Solve for the balanced to determine whether the given steel yield or not. n ) may not be applicable. ρb R .SOLUTION Note: This is not a rectangular beam. C=T 0.863) A sb=5.5 mm 125 125 125 b=375 a 1 2 5 C a 625mm 4.825 mm a=β1 c a=0.863 mm2 T =C A sb f y =0.8 f ' c Ac A sb ( 345 )=0.85 ( 21 ) (110.736 mm2 A s provided A sb Since < . d-a/2 32 N.3 ( 375 )−125 ( 125 ) =110. tension steel yields.A m m 375 mm .3 mm A c =337.85(396. Cb =396.217(345) a=207.825) a=337.85 f 'c ( ab−1252 )= A s f y 0.85 ( 21 ) ( a x 375−125 2) =3. 90( 567.03 kN −m φM n=0.5 M n=0. II I M n=M n 1−M n 2 M n=C 1 d−( a2 )−C (d− 1252 ) 2 207.22 Compute the ultimate moment capacity of the beam shown in Figure 2. Assume f y =345 MPa f ' c =21 MPa and .85 ( 21 ) (207.10 75 T 375mm .03) φM n=510. a 450mm c 375 450mm 450mm d-375 x d-(2/3)a 3-22mm 75 375mm 3-22mm Figure 2.10.33 kN −m PROBLEM 2.5 )( 375 ) (625− ) 2 M n=567. 140 mm As : Solve for 600 d Cb = 600+f y 600(375) Cb = 600+ 345 Cb =238 mm ab =β 1 C b ab =0.SOLUTION 22 ¿ ¿ π As= ¿ 4 2 A s =1.85(238) ab =202.7 mm 1 5 5 A c =1/2( x )(a) A c = x a x a= a2 2 6 a .4 mm x 375 5 = x= a a 450 6 x=168. 92 )=213.85 d−c 5 d −c f s=600 0. tension steel does not yield ( solve for c: CC =T f ' c Ac = A s f s 0.437 ¿ c=250. ¿ ¿ 7.85 f ' c Ac d− a 3 3 ( ) .85 ( 21 ) a2=1140 x 600 c 12 c 85 c a=β1 c 0.066) A sb=883 mm2 < A s A s provided > A sb f s <f y ¿ Since .85 ( 250.85 ( 21 ) (17.066 mm T =CC A sb f y =0.85 f ' c Ac A sb ( 345 )=0. 2 A c =17.92 mm a=β1 c a=0.3 mm 2 2 ( ) M n=C c x d− a M n=0. ¿ ¿ 5 M n=0.23 f ' c =28 MPa f y =345 MPa A hallow beam is shown in Figure 2.89 kN −m PROBLEM 2. b) What is the balanced moment capacity of the beam? c) What is the maximum steel area under singly reinforced condition? d) What is the maximum design moment strength under singly reinforced condition? M u=1200 kN −m e) Calculate the required tension steel area when . 500 mm 125 250 125 150 800 mm 500 150 75 mm Figure 2.77=70. 29 213.85 ( 21 ) ¿ 12 M n=78.90 x 78.11. Hallow beam SOLUTION . Assume and .11. M u=800 kN −m a) Calculate the required tension steel area when .77 kN −m φ M n=0. 85 ( 28 ) a ( 500 ) (725−0.90 x 0. let us solve the design moment when a=150 mm. Since the required Assuming tension steel yields: a M u=φM n M u =φ C c ( d− ) 2 a M u=φ 0.3 mm c β1 .90 x 0.85 ( 28 )( 150 ) 725− 150 2 ) φM n=1044. To guide us whether “a: will exceed 150 mm or not.85 f ' c a b (d− ) 2 800 x 106 =0. d = 800 – 75 = 725 mm ( a2 ) φ M n=φ C C d− ( φM n=0.5 a) a=111.225 kN −m M u=800 kN −m a) M u=800 kN −m<1044.25 kN−m.6 mm<150 mm Check is steel yields: d−c a f s=600 wherec = =131. a<150 mm . 85 ( 28 )( 111.3 mm z=a−150=241.6 ) (500) A s =3.27 ) =30.85 ( 460. 725−131.32 ) =391.712 MPa>f y steel yields 131.32mm 600+f y 600+345 a=β1 C b a=0.2 f s=600 =2.000 mm y 1=725− =650 mm 2 ( 150 ) 1 A 1=125 ( 241.85 f ' c ab A s ( 345 )=0.37 2 ( 241.3 T =C A s f y =0.12) 600 d 600(725) Cb = Cb = =460.27 ) M bn=C1 y 1+ 2C 2 y 2 .27 mm 2 1 A 1=500 ( 150 )=75.159 mm2 y 2=725−150− =454.850 mm2 b) Balanced condition (See Figure 2. Refer to Figure 2. A 1 y 1 +2 A 2 y 2 M bn=0.12: .12 T A s max c) Maximum steel area.000 x 650+ 2 x 30.159) A sb=9.37 ] M bn=1812.001 mm2 M u max : d) Maximum moment .52=1631.335 mm2 A s max=0.3 kN−m 5 1 02 1 05 2 2 1 150 5 m0 5 2 m 2 za 7 2 5 Figure 2.159 x 454.75 A sb A s max =0.00+ 2 x 30.90 x 1812.85 ( 28 ) [ 75. ' T =C1 +C 2 A sb f y =0.85 ( 28 ) (75.335 ) A s max =7.85 f c ¿ ) ' M bn=0.52 kN−m φ M n=0.85 f c ( A 1+2 A2 ) A sb ( 345 )=0.75 ( 9. 000+2 A 2 ]=7.000 mm y 1=650 mm A 2=125 z y2 =575−0.85 ( 28 ) [ 75.03 ] M n max=1489.5 z M u=φM n .244 mm2 A 2=125 z 13.95 mm 1501 y 2=725− =522.12 2 A 1=75.244 x 522.34=1340.95 ) M n max=C 1 + y 1+ 2C 2 y 2 M n max=0.4 kN −m M u=1200 kN −m< φ M n max ( Singly reinforced ) e) Refer to Figure 2.03 mm 2 ( 105. C1 +C 2=T 0.85 ( 28 ) [ 75.001(245) A 2=13.00 x 650+2 x 13.244=125 z z=105.90 x 1189.34 kN−m φM n max=0.85 f ' c (A 1 y 1 +2 A 2 y 2) M n max=0. 6 – 25 mm Ø d-c .24 A reinforced concrete beam is 350 mm wide and 600 mm deep. concrete and a) Calculate the maximum instantaneous deflection due to service loads.089 mm2 BEAM DEFLECTION PROBLEM PROBLEM 2.85 ( 28 ) (88.8 MPa . The beam is simply supported over a span of 8 m and carries a uniform dead load of 11 kN/m including its own weight and a uniform live load of 15 kN/m.2 mm2 T =C A s f y =0. Modulus of elasticity of Ec =21.90 x 0.7 MPa . The beam is reinforced tension bars of f ' c =20.90 x 0.2) A s =6.04 ) A c =88.5 z ) ] z=53.832 MPA .000+2 x 125 ( 53. 6 ' 1200 x 10 =0.A.85 ( 28 ) [ 75.259.04 mm A c = A 1+ A 2 A c =75.000 ( 650 ) +2 ( 125 z ) ( 575−0.650 MPa Es =200 GPa .259.85 f ' c A c A s ( 345 )=0. f r=2. 530 mm. f y =344. SOLUTION b = 350 mm b = 350 mm d = 530 mm h = 600 mm c N.85 f c ( A1 y 1+2 A2 y 2) 1200 x 106=0. b) Calculate the deflection for the same loads after five years assuming that 40% of the live load is sustained. Eq.Figure 2.13 Ie: Effective moment of inertia. 2-19 M cr 3 [ ( )] 3 M cr I e= ( ) Ma I g + 1− Ma I cr ≤ I g I g=moment of inertiaof gross section 600 ¿ ¿ ¿3 350 ¿ bh3 I g= I =¿ 12 g f r Ig M cr = where y t =1/2(600)=300 mm yt 10 600 x ¿ ¿ ¿6 2.832 ¿ M cr =¿ . 208(350-c) c = 219. n= =9.A.238 Ec 25 ¿ ¿ π n A s =9.A. 350 x c x c/2 = 27. = Moment of area below N.7 mm .13: Es Modular ratio.328 x 6 x ¿ 4 Solve for c: Moment of area above N. M a=Maximum monet ∈beam wL 2 M a= w=w D + w L=11 +15=26 kN / m 8 8 ¿ ¿ ¿2 26 ¿ M a=¿ I cr=¿ Moment of inertia of cracked section with steel transformed to concrete From Figure 2. 857 x 106 mm3 M cr 3 [ ( )] 3 M cr I e= Ma( ) I g + 1− Ma I cr 59.472 I e= ( 208 ) x 600 x 106 + 1− 208 6 I e=3.36 mm .7 ¿ ¿ ¿3 350 ¿ I cr=¿ I cr=3.650 ) (3.914 x 106 mm4 <I g (OK ) a) Instantaneous Deflection: 5 wL 4 2 ( 26 ) (8000)4 δ= δ= 384 Ec I e 384 ( 21.472 3 [ ( ) ] x 3.914 x 10 6) δ=16.857 x 10 3 59. d−c ¿ ¿ bc3 I cr=I NA = +n A s ¿ 3 219. 7 m δ + δ1 Long-term deflection = ξ λ= 1+50 ρ ' ξ=2 for 5 years∨more ' ρ =0 since there is no compression reinforcement 2 λ= =2 1+50(0) .7 mm Note: Since deflections are directly proportional to the load.914 x 10 6) δ =10.650 ) (3.4(15) = 17 kN/m 5 wL 4 Instantaneous deflection δ= 384 Ec I e 5 ( 17 ) (8)4 (1000)4 δ= 384 ( 21. δ 1 16. the instantaneous deflection due to sustained load can be found by ratio and proportion using the result in Part”a”.36 = 17 26 δ 1=10. b) Long-term Deflection Since only 40% of the live load was sustained: w = 11 + 0. resulting in the bending moment diagram shown. 7. Determine the following: a) The effective moment of inertia at the supports (maximum negative moment). Twenty percent of the live load will be sustained in nature. under the sustained loading if the instantaneous deflection due to the combined service dead and live load is 5 mm.14 is subjected to a uniform service dead load of 16 k/m and a service live load of 32 kN/m.2 MPa .Long-term deflection = 16.25 (CE NOVEMBER 2002) The continuous reinforced concrete beam shown in Figure 2. The concrete strength f c =17. while 80% will be applied only intermittently.76 mm PROBLEM 2. The modulus of elasticity of concrete is given by the expression Ec =4700 √(f ' c ) and the modulus of rapture is given by the expression f r=0. c) The additional deflection (in addition to the initial deflection) after 5 years. b) The effective moment of inertia for the continuous member.6 m 5-32 mmø 3-32 mmø 5-32 mmø 145 kN-m 202 kN-m 202 kN-m .7 √( f ' c ) .7) Long-term deflection = 37.36 + 2(10. y 560 mm y Gross Section Cracked Section I=0.14 SOLUTION Ec =4700 √ f ' c =4700 √17. I=0.0715 I=0.2=19.00573 AT SUPPORTS y=310 mm y=159 mm 1900 mm y 560 mm 620 mm y n As Gros Crack s ed AT MIDSPAN Sect Secti ion on I=0.492 MPA f r=0.903 MPa a) Effective moment of inertia at the supports .7 √ 17.70 √ f ' c =0.2=2.0 013 0573 8 y=10 y=1 7 mm 94 mm Figure 2. M u=202 kN −m Maximum moment.959 kN−m M cr 3 [ ( )] 3 M cr I e= Ma( ) I g + 1− Ma I cr [ ( ) ] x 0.00715+ 1− 66.0057817 m4 b) Effective moment of inertia for the continuous member Ie ¿ ¿ I ¿ I e=¿ .00573 3 3 I e= ( 66.00573 m4 f r Ig 2. I g=0.959 202 ) x 0.00715 m Moment of inertia of cracked section.00715 x 10004 ) M cr = M cr = yt 10 M cr =66. Y t =310 mm Distance from NA of gross section to extreme tension fiber. 4 Moment of inertia of gross section.903 ( 0. I g=0.959 202 I e=0. 00513 m f r Ig 2.At maximum negative moment (at support) I e=0.00715 x 10004 ) M cr = M cr = Yt 310 M cr =66.007932 4 I e= =0.0057817 m4 Ie Solving for at maximum positive moment (at midspan) 4 I g=0.903(0.0138 m Y t =620−194=246 mm ( bottom fibers∈tension ) 4 I cr=0.0057817+ 0.959 kN−m M cr 3 [ ( )] 3 M cr I e= Ma( ) I g + 1− Ma I cr Ie ¿ ¿ I ¿ I e=¿ 0.006857 m 2 . c) Additional long term deflection= long term deflection x λ ξ λ= 1+50 ρ ' ρ' =0 ( since there is no compression reinforcement at midspan ) ξ=2( after 5 years) 2 λ= =2 1+0 Solving for the instantaneous deflection under sustained loading: Instantaneous deflection = 5mm (given) Instantaneous loading = 16 kN/m + 32 kN/m Instantaneous loading = 48 kN/m Sustained loading = 16 + 20%(32) Sustained loading = 22.33 m .4 48 δ 1=2.4 kN/m Sine deflection is directly proportional to the load: δ1 5 = 22. 333 x 2 Additional long term deflection = 4. steel beams. they are referred to a one-way slabs since the bending occurs in one direction only.67 mm ONE-WAY SLAB Reinforced concrete design slabs are large flat plates that are supported at its sides by reinforced concrete beams. it is called two-way slab since the bending occurs in both direction. with bending occurring in the short direction. If a slab is supported on two opposite sides only.15: One-way slab on simple support . or by the ground. walls. If a rectangular slab is supported in all four sides but the long is two or more times the short side. Additional long term deflection = 2.333 x λ =2. for all practical purposes. the slab will. b = 1m h Figure 2. If the slab is supported on all four sides. columns. act as one way slab. 0.A one-way slab is considered as a wide.0018 x 415 fy Shrinkage and temperature reinforcement may not be spaced not farther apart than 5 times the slab thickness.7.13. This load may consist of: . Concrete shrinks as it hardens. rectangular beam. which is assumed independent of the adjacent strips. Identify the uniform floor pressure (Pa) to be carried by the slab.2. (where h is the slab thickness) but not less than 0.35% are used…………………………………………. the flexural reinforcement shall not be spaced farther apart than 3 times the slab thickness. the area of shrinkage reinforcement shall provide at least the following ratios of gross concrete area bh.. In this effect. swallow. One way-way slabs are analyzed by considering one-meter strip. ρT SHRINKAGE AND TEMPERATURE REINFORCEMENT. STEPS IN THE DESIGN OF ONE-WAY SLABS (FLEXURE) I. where flexural reinforcement extends in one direction only.0.0018 f y > 415 MPa c) Where reinforcement with measured at yield strain of 0. According to Section 407. nor 450 mm.5. temperature changes occur that causes expansion and construction of concrete.2). This method of analysis is somewhat conservative because we neglect the lateral restraint provided by the adjacent strips. MAXIMUM SPACING OF REINFORCEMENT According to Section 407.2.1. a) Where Grades 230 & 275 deformed bars are used………………. nor 450 mm (Section 407.0. In addition.0020 b) Where Grade 415 deformed bars or welded wire fabric (plain or deformed ) are used………………………………….132.. The reinforcing steel is usually spaced uniformly over its width.0014. the code (407.13) requires that one-way slab. should be reinforced for shrinkage and temperature stresses perpendicular to flexural reinforcement. If necessary adjust this value depending on your judgment. use ρ=ρmin If ρ is less than VII. A s =ρ b d=ρ ( 1000 ) d ≥ ρt b h A ¯¿ x 1000 As Spacing.85 f ' c 1− 1− 2 Rn 0. increase the depth of slab to ensure ductile failure ρmin . W u=Factored pressure x 1m Uniform load. V. Determine the minimum slab thickness “h” from Table 2. Compute the weight of slab (Pa) γ xh Weight = conc Mu¿ IV. Compute the required steel ratio ρ : 2 Rn M u=φ Rn b d where b=1000 mm Solve for from ρ= fy [ √ 0. d: d=h-covering (usually 20 mm)-1/2 (main bar diameter) VI. III. 1) Live load pressure 2) Dead load pressure 3) Ceiling load and other attachments below the slab II. use ρ ρmax If is greater than . . Calculate the factored moment ( to be carried by the slab. S 1=¿ . Compute the required main bar spacing. Compute the effective depth.85 f ' c ] ρmin Solve for ρ ρmax ρmin ρ If is less than and greater than .1. 36 . A st =ρt b h A ¯¿ x 1000 As S 2=¿ Use the smallest of the following for temperature bar spacing: S2 a) b) 5 xh c) 450 mm ILLUSTRATIVE PROBLEMS Problem 2. Temperature bars: See Page 81 for the required steel ratio. Use the smallest of the following for the main bar spacing: S1 a) b) 3 xh c) 450 mm ρt VIII. The slab is not exposed to earth or weather.5 KPa x 1 m=7. Bbars = 1000 mm d 120 mm h= 102mm main bars Cover + /2 .6 MPa f y =276 MPa load of 7. b= 1000 m w L =7.Design a one-way slab having a simple span 3 m.1: f hmin = L 20( ) 0.4+ 276 700 ) hmin =119 mm (use 120 mm) Effective depth: 10 mm temp.5 kN /m Uniform live load. The slab is to carry a uniform live f ' c =27. Minimum slab thickness from Table 2.500 Pa.5 kM /m .4+ y hmin = 700 3000 20 (0. Use unit weight of 3 concrete γ c =23. Assume and for main and temperature bars. SOLUTION Consider 1 m strip of slab. 85 ( 27.82 )+1.12) W s 2.5(1)(0.6 ) ] ρ=0.85 ( 27.785 x 106=0.362 MPa ρ= fy [ √ 0.90 R n ( 1000 ) (94)2 Rn=2.698(3)2 M u= M u= 8 2 M u=18.85 f 'c 1 − 1− Ru ' 0.362 ) 0.698 kN /m W u L2 16.6 ) 276 [ √ 1− 1− 2 ( 2.5) W u=16.009039 .4 ( 2.85 f c ] ρ= 0.785 kN −m M u=φ Rn b d 2 18.7 w L W u=1.82 kN /m Factored floor pressure load: W u=1.7 (7.4 w s+ 1.d = 120-20 mm (covering)-1/2 bar diameter (12mm) d=94 mm Weight of slab: W s =γ conc x b x h W s =23. 009039 ( 1000 ) (94) A s =850 mm2 per meter width of slab Using 12-mm main bars: 12 ¿ ¿ ¿2 π Spacing s = ¿ 4 A ¯¿ x 1000 s=¿ As ¿ s=138 mm say 135 mm Maximum spacing required by the Code: a) 3 ( h )=3 ( 120 ) =360 mmOK b) 450 mm .037> 0.009309(OK ) A s =ρbd A s=0.4 ρmin = =0. ρmin ρmax Check for and : 1.6 )( 0.75 0.85 ( 27.85 f ' c β 1 600 0.00507 OK fy 0.85 ) 600 ρmax = ρmax = f y (600+ f y ) 276 (600+276) ρmax =0.75 0. 002 ( 1000 ) ( 120 ) 2 A t =240 mm 10 ¿ ¿ ¿2 π Spacing = ¿ 4 A ¯¿ x 1000 s=¿ As ¿ s=327 mm say 325 mm Maximum spacing required by the Code: a) 5 h=5 ( 120 ) =600 mm b) 450 mm OK Thus. use 10 mm temperature bars at 325 mm o. Temperature bars: (Grade 275) A t =0. 20 mm L= 3m . use 12 mm main bars at 135 mm o.c. Thus.002 bh A t=0.c. mm o.c 325 .c. 12 mm main bars 10 mm temperature 120 mm bars @ 325 mm@ o. 1: L 4000 hmin = h min= 28 28 hmin =143 mm(use 150 mm) Weight of beam (DL): . Minimum slab thickness from Table 2. SOLUTION Consider 1 m strip.27 Design a one-way slab to carry a service live load of 4000 Pa. b = 1000 mm w L =4 kpa x 1 m=4 kN /m Uniform live load. The slab has a length of f ' c =21 MPa f y =415 MPa 4m with both ends continuous. Unit weight of 3 concrete is 23. Steel cover is 20 mm.PROBLEM 2. Assume and for main f y =276 MPa bars and for temperature bars.5 kN/ m . 735 kN /m Maximum factored moment.7( 4) w u=11. d = 1.525 kPa w u=1.525 ) +1.4 ( 3.4 (See Page 29) LL < 3 DL Column Column Column Spandrel Beam Shear Moment Effective depth.735(4)2 M u= M u= 16 16 .50 – 20 – 1/2 (12) Effective depth.5 (1 ) (0. Section 408. w D =γ conc x b x h w D =23. d = 124 mm At midspan: w u Ln 2 11.4 w D + 1.15) w D =3.4 w L w u=1. 4 ρmin = =0.85(21) ] ρ=0.00337 ( 1000 ) (124) 2 A s =418 mm π (12)2 A sb 4 Spacing.85 f ' c 1− 1− 2 Rn 0.0021 1.90 R n(1000)¿ Rn=0.00337> 0.735 kN −m 124 ¿ ¿ M u=φ Rn bd 2 11. M u=11.848 MPa ρ= fy [ √ 0.735 x 106=0. s = x 1000 s= x 1000 As 418 .85 f ' c]ρ= 0.00337 Use A s =ρbd A s=0.848) 0.0021 fy ρ= ρmin =0.85(21) 415 [ √ 1− 1− 2( 0. s=271 say 270 mm Maximum spacing required by the Code: a) 3 h=3 ( 150 ) =450 mm b) 450 mm Thus, use 12 mm bottom bars at 270 mm o.c. at midspan At support: w u L n2 11.735(4)2 M u= M u= 10 10 M u=18.776 kN −m M u=φ Rn bd 2 18.776 x 10 6=0.90 Rn ( 1000 ) (124)2 Rn=1.357 MPa ρ= fy [ √ 0.85 f ' c 1− 1−2 Rn 0.85 f ' c ] ρ= 0.85(21) 415 1− 1− [ √ 2(1.357) 0.85( 21) ] ρmax =0.0034> ρmin 0.85 f ' c β 1 600 0.85 ( 21 )( 0.85 ) 600 ρ max =0.75 ρmax =0.75 f y (600+ f y ) 415(600+415) ρmax =0.0162>0.0034 Use ρ=0.034 A s =ρbd A s =0.0034 ( 1000 ) (124) 2 A s =422 mm π 2 (12) A sb 4 Spacing, s= x 1000 s= x 1000 As 422 Spacing=268 say 265 mm Thus, use 12 mm top bars @ 265 mm o.c. at support ρt =0.002 ¿ Temperature bars (10 mm): ( A t =0.002 bh A t=0.002 ( 1000 ) (150) 2 A t =300 mm 10 ¿ ¿ ¿2 Spacing, s = π ¿ 4 A sb x 1000 s=¿ As s=261 say 260 mm Maximum spacing required by the Code: a) 5 h=5 ( 150 ) =750 mm b) 450 mm Thus, use 10 mm temperature bars @ 260 mm o.c. 10 mm temperature 150 mm bars @ 260 mm o.c. 12 mm main bars L/4 @ 265L/2 mm o.c. L/4 PROBLEM 2.28 A one-way slab having a simple span of 3 m is 160 mm thick. The slab is reinforced with 12 mm tension bars (f y =275 MPa) spaced at 140 mm o.c. Steel covering is 20 f ' c =20.7 MPa mm. Calculate the uniform live load pressure that a slab can carry. Use 3 . Unit weight of concrete is 23.5 kN/ m . SOLUTION Consider 1 m strip of slab, b = 1000 m w d=γ c b h Dead load: w d=23.5 ( 1 ) (0.16) w d=3.76 kN−m Effective depth: d = 160 – 20 – 1/2(12) d = 134 mm 12 ¿ ¿ Steel area, 1000 1000 π As= x A s A s= x ¿ s 140 24 2 A s =807.8 mm As 807.8 ρ= ρ= bd 1000( 134) ρ=0.006028 0.85 f ' c β 1 600 0.85 ( 20.7 ) ( 0.85 ) (600) ρb= ρb= f y ( 600+f y ) 275 (600+275) ρb=0.037> ρ( steel yields ) ρf y 0.006028( 275) ω= ω= f 'c 20.7 Rn=f ' c ω ( 1−0.59 ω ) Rn =20.7 ( 0.0801 ) [1−0.59 ( 0.0801 ) ] Rn=1.58 MPa 2 M u=φ Rn b d M u =20.7 ( 0.0801 ) [1−0.59 ( 0.0801 ) ] M u=25.5334 kN −m w u L2 wu (3)2 M u= 25.5334= 8 8 wu = 22.696 kN/m w u=1.4 w DL + 1.7 w¿ 22.696=1.4 ( 3.76 ) +1.7 wL w L =10.25 kN /m 05 ' 0. .836 a) Balanced steel area: ' 0. SOLUTION f ' c >28 MPa . f ' c =30 MPa f y =415 MPa .85 f c β 1 600 0.25 kPa Solved Problems Using 2010 NSCP PROBLEM 2.05 β 1=0. Since 0.836 )( 600 ) ρb= ρb= f y ( 600+ f y ) 415 ( 600+415 ) ρb=0.03036 . Determine the following: a) The balanced steel area b) The maximum steel area for singly reinforced condition c) The maximum design strength if the beam is singly reinforced d) The required steel area if the beam is subjected to dead load moment of 120 kN- m and live load moment of 170 kN-m.85− 7 ( f c −28 ) β 1=0. w ¿=Uniform pressure x b 10.85 ( 30 )( 0.29 A reinforced concrete beam has width of 310 mm and an effective depth of 490 mm.85− 7 (30−28) β 1=0.25 = Uniform pressure x 1 Uniform live load pressure = 10. 85 (30 )( 0.0221 A s max =ρmax b d A s max =0. 51 3 M n max= β 1 f ' c bd 2 (1− β 1 ) From Eq. 2-24: ρmax = 7 fy 3 0.836) 14 51 M n max= ( 0.343 mm2 φ M n max :ε =0. f s =800 MPa c) Maximum design strength.03036 ( 310 ) ( 490) 2 A sb =4. A sb = ρb b d A sb =0.611 mm b) Maximum steel area when beam is singly reinforced: ' 3 0.004.0221 ( 310 )( 490) A s max =3. 2-25 : 140 14 3 490 ¿2 (1− x 0.836 ) ρmax = 7 415 ( 600+415 ) ρmax =0.85 f c β 1 From Eq.836 )( 30 )( 310 ) ¿ 140 . 05 kN −m 800−f y From Eq.45 kN −m Thus.05) φ M n max=454.6 M L M u=1. Since the required the section is tension controlled.65+0.25 1000−f y 800−415 φ=0.6(170) d) M u=451. Determine if the beam is tension-controlled: 459 3 φ M tn = β1 f ' c bd 2 (1− β 1 ) From Eq. the beam is singly reinforced. 2-26: φ=0.90 .8145 φM n max=0.45 kN −m M u is less than M tn .2 ( 120 ) +1. M n max=558.65+0.8145(558.55 kN −m M u=1. 2-22: 1600 16 φM tn =451.2 M D +1. φ=0.25 1000−415 φ=0. 06 mm Check if it is really tension-controlled: a 139. Dead load moment M D =180 kN −m. a M u=φM n M n=φ x 0.85 f ' c a b(d − ) 2 a 416 x 106=0.06 c= = =166.2 ( 180 ) +1. Determine the required tension steel area .30 Given the following data for a rectangular beam: width b=320 mm .2 M D +1.167 MPa>1.4 mm β 1 0.6(167) .90 x 0. f ' c =27 MPa . effective depth d=520 mm . M L =167 kN −m.85(30)(a)(310)(490− ) 2 a=139.6 M L M u=1. f y =345 MPa .85 M U =1. Live load moment SOLUTION β 1=0.4 PROBLEM 2.000 MPa(OK ) c 166.4 f s=600 =600 =1.836 d−c 490−166. e 0. M u=483. the section is within “transition region’. 459 3 φM n= 1600 (16 ) β 1 f ' c bd2 1− β 1 =478.65+0.65 < φ<0.2 kN −m φM n max Solve for to determine if compression steel area is required.31 kN−m> M u ( singly reinforced ) φM tn Solve for to determine if the section is tension-controlled. 51 3 M n max= 140 ( β 1 f ' c bd 2 1− β 1 14 ) 520 ¿ ¿ 51 M n max= ( 0.85 )( 27 )( 320 ) ¿ 140 M n max=591.85 f ' c ab( d−a /2) .25 1000−f y =0.8237 φM n max=487.64 kN−m 800−f y φ=0. i.90 M u=φM n=φ x 0.9 kN −m M u> φM tn Since . 084 φ= +0.45 mm T =C A s f y =0.25 1000−f y 1000−345 119.2893) x 0.5 )=0.45 ) ( 320 ) 2 A s =3.2893 c φ=0.084 c + 0.85 c) c=208.2 x 106 = ( 119.65+0.777 mm . 520−c 600 −345 f −f c φ=0.850 ( 27 )( 177.85 c 483.8 mm a=β1 c=177.25 s y =0.85 f ' c ab A s ( 34.85 c)(320)(520−1/ 2 x 0.65+0.85( 27)(0. 85 )( 600 ) ρb= ρb= f y ( 600+ f y ) 415 ( 600+415 ) ρb=0.31 Given the following properties of a rectangular concrete beam: b = 280 mm. . SOLUTION β 1=0. d = 480 f ' c =21 MPa f y =415 MPa mm. c) When the beam is reinforced with seven 25 mm diameter bars. .85 ( 21 )( 0. a) When the beam is reinforced with three 25 mm diameter bars. Determine the design strength under the following conditions.01096< ρb ( steel yields) .85 f ' c β 1 600 0. b) When the beam is reinforced with four 25 mm diameter bars.85 since f ' c is less than28 MPa 0. The beam is reinforced for tension only.87 mm 2 4 a) A s =3 x Ab =1473 mm2 As 1473 ρ= ρ= bd 280 (480) ρ=0.0216 π A b = (25)2=490.PROBLEM 2. 87 kN −m φM n=0.86 mm β1 d−c 480−143.28 )( 280 ) (480−122.85 ( 21 )( a )( 280 )=1473( 415) a=122. φ=0. ' C=T 0.90( 255.000 MPa c 143.87) φM n=230.86 f s=600 =600 =1.85 f c a b=A s f y 0.28 mm a c= =143.28 /2) M n=255.85 f ' c a b(d−a /2) M n=0.402 MPa>1.86 The section is tension-controlled.85 ( 21 ) (122.90 M n=C c ( d−a/2 ) M n =0.28 kN −m b) A s =4 x A b=1963 mm 2 . 81 The section within” transition region”.65+0.85 f ' c a b= A s f y 0.25 φ=0. e 0.5−415 φ=0. i.65 < φ<0.90 f s−f y 901.65+0.04 mm a c= =191.81 f s=600 =600 =901.5 MPa< 1.85 ( 21 ) (163.81 mm β1 d−c 480−191.014961< ρb ¿ ) C=T 0.25 1000−f y 1000−415 φ=0. As 1963 ρ= ρ= bd 280 (480) steel yields ρ=0.04 ) (280 ) (480−163.85 ( 21 )( a )( 280 )=1963(415) a=163.04/2) .858 M n=C c ( d−a/2 ) M n =0.000 MPa c 191.85 f ' c a b(d−a /2) M n=0. 85 (21 )( 0.504 kN −m φM n=0.56 mm a=β1 c=252.91) .02557> ρb ( stel does not yield ) The section is compression-controlled.65( 446.92 ) ( 280 ) ( 480−252.858( 324. φ=0.85 c ) (280) c c=297. M n=324.92/ 2) φM n=0.85 f ' c a b 480−c 3436 x 600 =0.504) φM n=278.92 mm M n=C c ¿ d−a/2 ¿ M n =0.65 T =C A s f s=0.396 kN −m c) A s =7 x Ab =3436 mm2 As 3436 ρ= ρ= bd 280 (480) ρ=0.85 f ' c a b(d−a/2) M n=0.85 ( 21 ) (252. Assume and .16 . φM n=290.16.32 f ' c =28 MPa f y =345 MPa A hallow beam is shown in Figure 2.Hallow beam . 500 mm 125250125 500 150 800 mm 150 75 mm Figure 2. M u=800 kn−m a) Calculate the required tension steel area when b) What is the balanced moment capacity of the beam? c) What is the maximum steel area under singly reinforced condition? d) What is the maximum design moment strength under singly reinforced condition? M u=1200 kN −m e) Calculate the required tension steel area when .49 kN −m PROBLEM 2. let us solve the design moment when a =150 mm.865 MPa>1000 MPa Tension controls .225 kN−m. Since the required M u=φ M n M u =φ C c ( d−a/2) M u=φ 0.SOLUTION This problem is the same as Problem 2.47 mm β1 d−c f s=600 =1.85 ( 28 )( 150 ) ( 500 ¿(725−150/2) φM n=1044. a<150 mm .85 f ' c a b (d−a/2) .90 c φM n=φ C c ( d −a/2 ) φ M n=0.90 x 0. d=800−75=725 mm To guide us whether “a” will exceed 150 mm or not.φ=0. a c= =176.225 kN −m M u=800 kN −m a) M u=800 kN −m<1044.23. 3 T =C c A s f y =0.85 ( 28 ) a(500)(725−0.90 x 0. 800 x 106 =0.65 600 d 600(725) Cb = Cb = =460.6 ) (500) 2 A s =3.85 ( 460.6 mm<150 mm Stress in steel d−c a f s=600 wherec = =131.2 f s=600 =2.3 mm 500 mm 125 250 125 1 150 2 2 a N 725 T Figure 2.850 mm b) Balanced condition: φ=0.3 mm c β1 725−131.712 MPa>f y steel yields 131.5 a) a=111.32mm 600+f y 600+345 a=β1 c b a=0.32 )=391.17 .85 ( 28 )( 111.85 f ' c a b A s ( 345 )=0. 159 mm2 y 2=725−150−1/ 2 ¿ 241.71 mm a=β 1 c max=264.27 ) =30.52 kN−m φM bn=0.000 x 650+2 x 30.85 ( 28 ) [75.159 x 454.14 kN−m A s max Refer ¿ Figure 2.85 f ' c ( A1 y 1+ 2 A2 y 2) M bn=0.27) = 454.52 φ M bn=1178.11 mm 7 .65 x 1812.z=a−150=241. 3 Cmax = d=310.37 ] M bn=1812.37 M bn=C1 y 1+ 2C 2 y 2 M bn=0.000 mm y 1=725−1/2(150)=650 mm ¿ A 1=125 ( 241.17 c) Maximum steel area.27 mm 2 A 1=500 ( 150 )=75. 85 f ' c ( A1 +2 A 2) A s max ( 345 )=0.85 ( 28 ) [75.11 mm A 1=500 ( 150 )=75.11)=517.142 mm M n max d) Maximum moment.95 T =C1 +C 2 A s max f y =0.11 )=14.263 mm y 2=725−150−1/2 (114.85 f 'c +( A 1 y 1 +2 A 2 y 2 ) M n max=0. z=a−150=114.95] M n max=1511.000 x 650+ 2 x 14.9 kN−m .85 ( 28 ) [75.263 x 517.000 x 650+2 x 14.263] 2 A s max =7.000 mm 2 y 1=725−1/2(150)=650 mm 2 A 2=125 ( 114. : M n max=C 1 y1 +2 C2 y 2 M n max=0. 824 1000−f y φ M n max=0.85c-150) y 2=650−0.750 y 2=725−150−1/2 z =575-1/2(0.425 c .85 c−150 A 2=125 z=106.65+0.85 f 'c ( A 1 y 1 +2 A 2 y 2 ) f s+ f y d−c φ=0. 800−f y φ=0.25 =0.25 c−18.28 93 1000−345 c z=a−150=0.25 = +0.25 f s=600 1000−f y c 725−c 600 −345 c 166.65+0.824 x 1511.03 φ=0.65+0.3 kN−m e) M u=1200 kN −m< φ M n max (singly reinforced ) Refer to Figure 2.17 M u=φ 0.9 φM n max=1245. 7 mm 166. and take d=1.615) 2 A s =8. M u=φ 0.03 φ= +0.85 f 'c ( A 1 y 1 +2 A 2 y 2 ) 1200 x 106= ( 166.65 ρ b (including self weight) and live load moment of180 kN-m.750=23.25 ( 398.25 c−18.33 Design a singly reinforced rectangular beam to carry dead load moment of 110 kN-m ρ=0.7 ) −18.85 f ' c ( A 1+2 A2 ) A s ( 345 )=0.2893=0.03 c +0.9 b .706 398.85 ( 28 ) [ 75.425 c)¿ c=398. Use steel ratio f y =276 MPa f ' c =21 MPa .7 2 A 2=106.2893 ) 0.750)(650−0.432 mm PROBLEM 2.000 ] (650) +2(106.85 ( 28 ) (75.000+2 x 23. Assume and SOLUTION .615 mm T =C A s f y =0. 685 d 600+f y b Note: For singly reinforced rectangular beam.65 ρ b=0.65 c b Thus.2 M D +1. c=0.2 ( 110 )+ 1. ρ is directly proportional to c.03765 Note : β1 =0.473 MPa 600 d Cb = C =0.445 d d−c d −0.59 ω)¿=5. c=0.6(180) M u=420 kN−m 0.6 M L M u=1.445 d f s=747.85 f ' c β 1 600 ρb= =0.445 d f s=600 f s =600 c 0. M u=1.7 MPa<1000 MPa transition .02447 ρf y ω= =0.85 since f ' c < 28 MPa f y ( 600+f y ) ρ=0.322 f 'c ' Rn=f c ω (1−0. 65+0.25 φ=0. f s−f y 747.59 ω ) =4.03765 ρ=0.33 using a steel ratio SOLUTION M u=420 kN−m ρb=0.34 ρ=0.7−276 φ=0.473 )( b ) (1.01883 ρf y ω= (1−0.1000 mm PROBLEM 2.65+0.25 1000−f y 1000−276 φ=0.5 ρb=0.9 b=564 mm A s =ρbd A s =0.438 MPa f 'c .9 b)2 b=297 mm d=1.5 ρb Repeat Problem 2.813 M u=φ Rn b d 2 420 x 106 =0.813 (5.02447 ( 297 ) ( 564 ) 2 A s =4. Determine (a) the maximum design moment if the beam is singly reinforced and (b) the .01883 ( 308 ) (585) A s =3. 600 d Cb = C =0.324247 d f s=1152 MPa>1000 MPa .9 b=585 mm A s =ρbd A s =0.473 ) ( b ) (1.90 ( 5.34247 d f s=600 f s =600 d− c 0.685 d 600+f y b c=0.9 b)2 b=308 mm d=1. f ' c =20.35 b=250 mm .5 c b =0. φ=0.7 MPa A rectangular beam has . d=350 mm .390 mm2 SUPPLEMENTARY PROBLEMS PROBLEM 2.90 M u=φ Rn b d 2 420 x 10 6=0.34247 d d−c d−0. f y =414 MPa . Answer :a ¿ φM n max=148.36 Repeat Problem 2.38 Repeat Problem 2.37 Design a rectangular beam reinforced for tension only carry dead load moment of 85 ρ=0.3 kNm 2 b ¿ A s =1075 mm PROBLEM 2.8 kN−m b ¿ A s =1056 mm2 PROBLEM 2.273 mm2 PROBLEM 2. Answer :a ¿ φM n max=130. Use the 2001 NSCP Answer :b=250 mm . Use and f y =276 MPa f ' c =28 MPa use d= 1.required steel area if the beam is required to carry a dead load moment of 50 kN-m and a live load moment of 30 kN-m.6 ρb kN-m (including its estimated weight) and a live load of 102 kN-m. Assume and . d=436 mm .35 using the 2010 NSCP. Use the 2001 NSCP.75b. A s=3.37 using the 2010 NSCP. . and the corresponding service live load moment.39 A reinforced concrete beam has the following properties: Use 2001 NSCP) beam with.000 MPa reinforcing steel modulus.44.40 Repeat Problem 2.φ M n=687.549mm 2 . service dead load moment 350=kN−m a) If the beam is to be designed for a balanced condition. Es =200. design balanced moment. Answer :b=246 mm . b=320 mm effective depth.162mm . find the required area of steel area reinforcement. d=640 mm f ' c =25 MPa concrete strength.963 mm . d=430 mm . and the corresponding service live load moment if the beam is to be designed as singly reinforced. M L =167.549 mm2 . φ M n max=677.46. b) Find the maximum steel area. A s=3182 mm2 PROBLEM 2. Answer :a ¿ A sb=5. M L =168 kN −m PROBLEM 2.7 M L=161 kN −m . the maximum design moment.42 kN−m 2 b ¿ A s max =3. φ M n max =775. M L =272 kN −m 2 b ¿ A s max =4.87 kN−m. Answer :a ¿ 5.39 using the 2010 NSCP. f y =400 MPa reinforcing steel. φM n=952. 44 Repeat Problem 2. d=450 mm .2 kN −m PROBLEM 2. A s =5−25 mm . Use 2010 NSCP. Answer :φ M n=582. Answer :φ M n=366. Answer :φM n=514.2 kN −m PROBLEM 2. d=540 mm .43 using the 2010 NSCP.42 Repeat Problem 2. f y =345 MPa .43 Calculate the ultimate moment capacity of a rectangular beam with b=350 mm . Assume f ' c =24 MPa f =345 MPa . Assume f ' c =24 MPa .41 using the 2010 NSCP. y .41 Calculate the ultimate moment capacity of a rectangular beam with b=350 mm .45 .3 kN −m PROBLEM 2.9 kN −m PROBLEM 2. Use 2001 NSCP Answer :φM n=366. A s =7−28 mm .PROBLEM 2. 6 kN −m PROBLEM 2.5 kN −m CHAPTER 3 Analysis and Design of T-Beams and Doubly Reinforced Beams T-Beams . Use 2010 NSCP Answer :φ M n=729. d=500 mm . Asnwer :φ M n=522. Assume f ' c =34 MPa .45 using the 2010 NSCP.Calculate the ultimate moment capacity of a rectangular beam with b=300 mm .46 Repeat Problem 2. A s =9−28 mm2 . f y =414 MPa . as discussed in Chapter 2 is given by: Figure 3. If however it cover part of the web as shown in Figure 3. and the resulting section is called a T- beam. the beam will have extra width on top (which is usually under compression) called flangers. In this effect. If it falls within the flange as shown in Figure 3. the section is usually tension-controlled (extreme tension yields).1 (b).1 (a). (a) STEEL AREA BALANCED AND MAXIMUM (b) AND MOMENT The balanced value of “c” for any beam shape. the compression concrete no longer consist of a single rectangle and thus the rectangular formulas do not apply. which are placed or poured monolithically. The beam may also be L-shaped if it is located at the end of slab. the rectangular beam formulas in Chapter 2 applies since the concrete below neutral axis is assumed to be cracked and its shape has no effect on the flexure calculations. ANALYSIS AND DESIGN OF T-BEAMS WITH FLANGE IN COMPRESSION Because of the huge amount of compression concrete when the flange of a T-beams is compression.1: Location of neutral axis 600 d Cb = 600+f y . The compression block of T-beam may fall within the flange only or partly in the web.Reinforced concrete floors usually consist of slab and beams. 85 f ' c a max b(d− ) 2 However.85 f ' c β 1 600 ρb= f y ( 600+f y ) A sb= ρb br d ab M bn=0. or 0. ab =β 1 C b and If ”a” is less than the slab thickness. if “a” is greater than the slab thickness.75 A sb amax =0.85 f ' c ab (d − ) 2 A s max =0. the formulas for rectangular beam may be used.75 a b amax M n max=0. t a Z d T T =C1 +C 2 A sb f y =C 1+ C2 . the following formula will be used. ' A sb f y =0. The intention of this section is to minimize the possibilities of flexural cracks that will occur at the top face of the flange due to negative moments.85 f c (b f t+b w z ) 0. . some longitudinal reinforcement shall be provided in the outer portions of the flange.85 f ' c [f ' c t + ( a−t ) b w ] Eq. whichever is smaller.11. the formulas for rectangular beams can be applied. 3-1 A sb= fy A s max =0.6: Where flangers of T-beam construction are in tension. 408.A When T-beams are resisting negative moments so that far their flangers are in tension and the bottom of their stems in compression.75 A sb Eq. part of the flexural tension reinforcement shall be distributed over an effective flange width as defined in Sec.7. 3-2 DESIGN OF T-BEAMS WITH NEGATIVE MOMENTS N. The following code requirements shall be applied for this case: 410. If the effective flange width exceeds 1/10 the span. or width equal to 1/10 the span. and c) 1/2 the clear distance to the next web. The width of slab effective as T-beam shall not exceed 1/4 of the span of the beam. b) 6 times the slab thickness. and b) 1/2 the clear distance to the next web. For beams with slab on one side only. 3. and the effective overhanging flange on each side of th web shall not exceed: a) 8 times the slab thickness.MINIMUM STEEL RATIO For statically determinate T-section with flange in tension. 3-4 4fy f CODE REQUIREMENTS ON T-BEAMS (SECTION 408. In T-beam construction. the minimum steel area is equal to or greater than the smaller value of Eq. 2. 3-4: A s min = √ f 'c b d Eq. 3-3 2fy w A s min = √ f 'c b d Eq.11) NOTE: THESE REQUIREMENTS ARE THE SAME WITH 2010 NSCP 1. t Interior Beam end Beam Figure 3. the effective overhanging flange shall not exceed: a) 1/12 the span length of the beam. the flange and web shall be built integrally or otherwise effectively bonded together.2: Effective flange width . 3-3 and Eq. + +b w 2 2 For End Beam b'f is the smallest of : b'w 1. S 1=S2=S ¿ For symmetrical interior beam ( bf is the smallest of: . L/12 + 6 t+ b ' w 2. S1 S2 3. S 3 /2+b ' w 3.For Interior Beam bf is the smallest of: 1. L/4 16 t +bw 2. nor 450 mm. only the effective overhanging slab needs to be considered. t ≥ bw /2 t b f ≤ 4 bw 5. b) Transverse reinforcement shall be spaced not further apart than five times the slab thickness. reinforcement perpendicular to the beam shall be provided in the top of the slab in accordance with the following: a) Transverse reinforcement shall be designed to carry the factored load on the overhanging slab with assumed to act as a cantilever. Isolated beams in which T-shape are used to provide a flange for additional compression area shall have a flange thickness not less than 1/2 the width of the web and an effective flange width not more than four times the width of the web. Transverse reinforcement To be provided Primary slab reinforcement . Where primary flexural reinforcement in a slab that is considered as a T-beam flange is parallel to the beam.1. For isolated beam. center-to-center spacing of beams 4. L/4 16 t +bw 2. the full width of the overhanging flange shall be considered. 3. For other T-beams. thena< t . thena> t . M n 1=φ C ( d−t/2 ) ¿ ¿ ' ( 2t )= ¿ φ M n 1=φ 0. proceed ¿ Step III if φM n 1> M u . Follow the procedure in Page 105. the beam is singly reinforced. a<t . AS STEPS IN FINDING THE TENSION STEEL AREA OF SINGLY MU REINFORCED T-BEAMS WITH GIVEN AND OTHER BEAM PROPERTIES: φ Mn I. the beam is doubly reinforced Note : SKIP stepif ∈ your judgment M u is small∧¿ compression steel is not needed . proceed ¿ Step IV if III. φM n 1 II. M u ≤ φ M n max If .85 f ' c b f t Compressive force in concrete. M u> φ M n max If .85 f c bf d− ¿ φM n 1> M u . Solve for max to determine of compression steel is necessary. proceed to Step II. Solve for when a = t C=0. 85 f ' c ab ¿ ¿ A s =¿ A s min A s min is the smaller value of: . t a d d -a/2 Solve for a: a M u=φ M n=φ∁ (d− ) 2 M u=φ 0.85 f ' c ab (d−a/2) ¿ ¿ ¿ ¿ a=¿ ¿ T =C A s f y =0. t a Z d T M u=φ M n M u =φ M n 1+ M n 2 Note :φ M n is∈Step ¿ ¿ ¿ ¿ M n 2=¿ ¿ M n 2=C2 y 2=0.85 f ' c b w z y 2 ¿ ¿ ¿ ¿ ¿ z=¿ ¿ T =C1 +C 2 A s f y =C 1 +C2 .A s min = √ f 'c b d A s min = √ f 'c b d w f 2fy 4fy IV. tension steel yields. C=T 0. Compute the area of compression concrete. tension does not yield Note : This step may skipped if ∈ your judgement A s is small∧¿ instead assume that f s =f y . Solve for balanced steel area to determine if tension steel yiel. Follow the procedure in Page 105. Proceed to step II As> Asb If . .85 f 'c (b f t+ bw z) ¿ ¿ ¿ A s =¿ ¿ A s min is the smallest value of: A s min = √ f 'c b d A s min = √ f 'c b d w f 2fy 4fy φM n As STEPS IN FINDING OF SINGLY REINFORCED T-BEAMS WITH GIVEN AND OTHER BEAM PROPERTIES: A sb I. A s f y =0. A s ≤ A sb If . f s=f y Ac II.85 f ' c Ac = A s f y ¿ ¿ A c =¿ ¿ . Tension steel yields. A s < A f . a<t Solve for a: t a A c =b f x a ¿ d d -a/2 a=¿ ¿ φM n=φT (d−a/2) a φM n=φ A s f y (d − ) 2 IV. a<t . Ac A f =b f t Compare with the area of compression flange. If proceed to Step III A c > Af . proceed ¿ Step IV If III. a>t . a>t : t a Z d T . Solve for z : A c = A 1+b w z ¿ ¿ ¿ z=¿ ¿ φM n=φM n 1 +φM n 2 φM n=φ (C1 y 1+ C2 y 2) φM n=φ 0. The beam width of web is 250 mm. clear spacing of beams + b f =1500 mm Therefore PROBLEM 3. b w =3000+250=3250 mm 3. 1/4 span = 6000/4 = 1500 mm 16 t+ bw =16 ( 120 ) +250=2170 mm 2.85 f ' c [ A 1 y 1 + A2 y 2 ] ILLUSTRATIVE PROBLEMS SOLVED PROBLEMS IN T-BEAMS USING 2001 NSCP PROBLEM 3. SOLUTION For symmetrical T-beam. the slab thickness is 120 mm. the effective flange width is the smallest of: 1.2 .1 Determine the effective flange with for symmetrical T-beam with a span of 6 m. and the clear distance to adjacent beams is 3m. 85 ( 30 ) (120 )( 130 ) ( 470− ) 2 M fn =1611 kN −m φ M n=0. SOLUTION a M n=0. determine the ultimate moment capacity when the depth of compression concrete flange equals the flange thickness or a=t . d=470 mm If the beam is reinforced for tension only.Given the following elements of a T-beam: b f =1200 mm Concrete strength f ' c =30 MPa Flange width.8 f ' c b f a( d− ) 2 When a=t Eq. t=130 mm Steel strength . Effective depth.3 Given the following elements of a T-beam: .85 f 'c b f t (d −t /2) 180 M fn =0.90 x 1611=1450 kn−m PROBLEM 3. 3-5 M n=0. f y =345 MPa Flange thickness. b w =290 mm Width of web. t=110 m Steel strength. b w =310 mm Width of web.4 mm>t = 900mm t=100 d = 460 mm C a z T =250 mm .85(272. f y =414 MPa Flange thickness. d=460 mm If the beam is reinforced for tension only. Effective depth.2 mm a=β1 c a=0. determine the following: a) The balanced steel area b) The nominal and ultimate balanced moment capacity c) The maximum steel area d) The nominal and ultimate maximum moment capacity SOLUTION β 1=0.2) a=231.7 MPa Flange width.85 since f ' c is less than30 MPa a) Balanced condition 600 d 600 ( 460 ) Cb = Cb = 600+f y 600+ 414 Cb =272. b f =900mm Concrete strength f ' c =20. 622mm2 2 A cb =A 1 + A2=136.4 )=37.85 f ' c ( A1 + A2 ) A sb ( 414 )=0.4 mm 2 A 1=b f x t =900 ( 110 )=99.622 ( 289.000 ( 405 ) +37.000 mm A 2=b w x z=310 ( 121.90(897) M bn=807.3 mm ' M bn=C2 y 1+ C2 y 2 M bn=0.7 ) [99. Refer to Figure 3.85 f c ( A 1 y 1 + A 2 y 2) M bn=0.622 mm T =C1 +C 2 A sb f y =0.85 ( 20.3 z=a−t=121.3 kN −m→ ultimate balanced moment b) Maximum steel area and moment.85 ( 20.806 mm → balanced steel area y 1=d−t/2=405 mm y 2=d −t−z /2=289.3. Figure 3.000+37.622 ¿ 2 A sb=5.7 ) 99. .3 ) ] M bn=597 kN −m→ nominal balanced moment M bn=0. 8 kN−m→ultimate maximum moment PROBLEM 3.7 MPa Concrete compressive strength and steel area for the following load conditions: M D =150 kn−m.75(136. M L =120 kN −m a) .466=99.4 b f =820 mm .41 mm 2 M n mnx=C 1 y1 +C 2 y 2 M n max=0.466 mm > A 1 . b w =250 mm .5 kN −m →nominal max moment M n max=0.90 ( 726.466 ( 289.622) 2 A c max =102.85 ( 20.t =100 mm . A T-beam has the following properties: d=470 mm .355 mm → maximum steel area A c max =0.75 A cb A c max =0. f ' c =20.85 f ' c ( A 1 y 1 + A2 y 2) M n max=0.000 ( 405 ) +3.2 mm A 2=102. A s max =0.7 ) [99.thus a>t A c max =A 1 + A 2 102.3 ) ] M n max=726.466 mm2 z y 2=d −t− =344.466−99.000 + 310(z) z=11.75(5806) 2 A s max =4.5 ) M n max=653.75 A sb A s max=0.000=3. M D =175 kN −m.39 mm>t = 820mm t=100 d = 470 mm C a z T =250 mm FIGURE 3.85 Solve for ↑ φ M n when a=t ( 2t )=545.85 f ' c b f t d− Solve for φM n max : Balanced condition: 600 d c b= =278.39 mm A 1=b f t=82.11 mm 600+ f y a< β 1 c b=236.4 z=a−t=136.000 mm2 . M L =190 kN −m b) SOLUTION β 1=0.375 kN −mm φ M fn =0. singly reinforced M u is less than M fn .28 kN−m M D =150 kN −m.073 mm2 > A1 A 2= A cmax −82.29 mm bw z y 2=d −t− =359.85 mm 2 φ M n max= M n + M n 2=M fn +0.85 f ' c A2 y 2 φM n max=574.000=5.098 mm2 Maximum condition: A c max =0.098 mm A cb =A 1 + A2=116. AM L=120 kN−m a) M u=1. a is less than t.75 A c b=87.4 M D +1. 2 A 2=b w z =34. Since t =100 =820 mm C d = 470 mm d -a/2 T .7 M L =414 kN−m< φM n max .073 mm2 As z= =20. 85 f ' c a b f A s =2.85 f ' c a b f (d−a /2) 414 x 106=0. M u=0.565 mm2 As Minimum is the smaller of: √f 'c b d=646 mm2 √f 'c b d=1059 mm2 w f 2f y 4fy 2 Thus. = 820mm t=100 d = 470 mm C a z T =250 mm .7 ) a ( 820 ) ( 470−a/2) a=73.6 mm T =C A s f y =0.90 ( 0.4 M D +1.7 M L =568 kN−m< φM n max .565 mm M D =175 kN −m. M L =190 kN −m b) M u=1. A s =2.85 )( 20. singly reinforced Mu M fn . a Since is more than is more than t. f y =415 MPa .5 m long and spaced at 3 mo.7 )( 82.946 mm2 T =C1 +C 2 A s f y =0. f ' c =27 MPa .85 ( 20.7 )( 250 ) z (470− ) 2 z=15.90 ( 0. M D =450 kN −m ( including its own weight ) .85 f ' c ( A 1+ A 2) A s ( 414 )=0. The slab thickness is 100 mm. M L =350 kN −m. SOLUTION β 1=0. The Design a T-beam for a floor system for which and beams are 4.000+3946) A s =3.c.5 b w =300 mm d=550 mm . M u=φM fn +φM n 2 100 z 568 x 106=545.85 .78 mm A 2=b w z =3.85 )( 20.653 mm2 PROBLEM 3.375+ 0. 4 ( 450 ) +1.7 M L M u=1. M u=1.90 Solve for t φ M fn =φ 0.125 mm Thus.355 mm> t = 1125mm t=100 d =550mm ac y C fn +φ M u=φM 2 2 z 450 T =300 mm .85 f ' c t bf (d− ) 2 φM n=1161. 3. center-to center spacing of beams = 3 m b f =1.900 mm 2.7(350) M u=1225 kN −m Solve for bf: b f is the smallest of : 1. 600 d c b= =325.125 m 16t + bw =16 ( 100 ) +300=1.844 kN −m φM n max Solve for to determine if compression steel is needed. L/4 = 1.4 M D +1.123 mm 600+ f y a=β1 c b=276. φ M n whena=t=100 mm . φ=0. 844 x 106 + 0. The beam is reinforced with six 28 mm bars. z 1225 x 106=1161.7 MPa depth d = 600 mm. web width .85 ( 27 ) (112.975.02) A s =6. SOLUTION = 1500mm As : t=100 Solve for balanced d =600mm C a z =250 mm . effective f ' c =20.85 f c ( A1 + A 2) A s ( 415 )=0. A s =6.25 mm A 2=b w z =6975.02mm 2 ' T =c 1+ c 2 A s f y =0.607 mm PROBLEM 3.6 Determine the ultimate moment capacity of reinforced concrete T-beam with the b w =250 mm following properties: Flange width b = 1500 mm.500+ 6.90 x 0. slab thickness t = 100 mm.85(27)(300 z)( 450− ) 2 z=23. Assume and f y =345 MPa .607 mm2 As Minimum is the smaller value of: √f 'c b d=1033 mm 2 √f 'c b d=1937 mm2 w f 2f y 4fy 2 Thus. 85 0.441< A1 therefore a is less than t t =100 =1500 mm C d = 600 mm d -a/2 T .503 π A s =6 x ( 28 )2=3. 4 f s=f y Therefore.000 A 2=b w z =55. C=T f ' c Ac = A s f y 0.81>t z=a−t=22381 mm A 1=b f t=150.85 ( 20.952mm2 T =C A s f y =0.95 mm 600+ f y a=β1 c b=323.695 mm2 > A sb steel yields Steel area provided. 600 d c b= =380.85 f 'c ( A1 + A 2) A sb ( 345 )=0.00+ 55.7) A c =3.952) A sb=10.695(345) A c =72.85(20.7 )(150. . Determine the safe service live load if the beam is reinforced for tension only with twelve (12) 28-mm-diameter bars.85 ( 20.7 Given the following properties of T-beam: b f =900mm f ' c =21 MPa Flange width.29 mm2 a M n=0.6 kN −m PROBLEM 3.7 ) ( 48.85 f ' c a b f (d− ) 2 48. t=1200 b w =400 mm Width of web.29 ) ( 1500 ) (600− ) 2 M n=733.99 kN −m φM n=0.441 = a (1500) a=48.90( 733.29 M n=0.99) φM n=660. f y =345 MPa Flange thickness. A c =a bf 72. Effective depth. d = 580 mm M D =410 kn−m Service deal load. 813 mm2 > A1 a >t = 900mm t=120 d =580mm C a z 460 T =400 mm .206 mm2 T =C A sb f y =0.85 ( 21 ) A c =7.000 mm2 As Solve for balance : =400 mm 600 d c b= =368.02 mm>t z=a=t=193.90 t=120 d =580mm 28 a C ¿ ¿ z π A s =12 x ¿ 4 A 1=b f t=108. SOLUTION = 900mm β 1=0.85 f ' c A c = A s f y 0. φ=.000+77.85 ( 21 ) (108.85 .582m m Steel area provided is less than the balanced steel area. C=T 0.389(345) A c =142.206) 2 A sb=9.85 f ' c ( A1 + A2 ) A sb ( 345 )=0.02 mm A 2=b w z =77. Steel yields.25 mm 600+ f y c=β 1 c b=313. 000 ( 520 ) +34.813 = 400z z=87.138=1.48 ) ] M n=1.3 kN −m φM n=0. The beam is f y =415 MPa reinforced with 10 32-mm-diameter tension bars with .5.813=108.813 mm2 A 2=b w z 34.85 ( 21 ) [108.261.000+ A 2=34.90( 1.138 kN −m φM n=M u M u=1.48 mm 2 M n=C 1 y 1 +C2 y 2 ' M n=0.813 ( 416.8 The section of a reinforced concrete T-beam is shown in Figure 3.4 ( 410 ) +1.03 mm t y 1=d− =520 mm 2 z y 2=d −t− =416.135. Concrete strength .85 f c ( A1 y 1+ A 2 y 2 ) M n=0. A c = A 1+ A 2 A2 142.0 kN −m PROBLEM 3.7 M L 1.7 M L M L =330.261.3) φM n=1135.4 M D +1. =500mm d=530mm t = 120mm 10-32 mm SOLUTION 32 ¿ t = 120mm =500mm ¿ d=530mm =320mm π a A s =10 x ¿ 4 z A s =8. determine the safe service live load moment.85− (32−30) 7 β 1=0. If the total service dead load moment on the beam is 330 kN-m. f ' c =32 MPa .836 As : Solve for balance 600 d Cb = 600+f y Cb =313.05 β 1=0.000 mm =320mm 0.042 mm2 2 Figure 3.83 mm>t .3 mm a=β1 C b=261.5 A 1=b f t=60. 5 mm2 T =C A sb f y =0.00+ 45.85 ( 21 ) (60. z=a−t=141.385.385.85 f ' c ( A 1+ A 2) .907 mm2 A s > A sb Since .83 mm A 2=b w z =45.85 f ' c ( A1 + A2 ) A sb ( 345 )=0.000 mm 2 =320mm A 2=b w z =bw ( a−t) A 2=b w ( β1 c−t ) d−c f s=600 c T =C1 +C 2 A s f s=0.5) A sb=6. tension steel does not yield t = 120mm =500mm d=530mm a z T A 1=60. 4 ( 330 ) +1.4 M D +1.83 mm y 2=d −t− =332.56 kN −m φ M n=0.97 2 mm y 1=d−t/ 2=470 M n=C 1 y 1 +C2 y 2 M n=0.092=1.092.213.213.56) φM n=1.000+ 320 ( 0.7 kN −m .85 ( 32 ) [60.303 mm2 z z=a−t=141.85 f 'c ( A1 y 1+ A 1 y 2 ) M n=0.042 x 600 =0.85 ( 32 ) [60. 530−c 8.303 ( 332.83 mm A 2=b w z =49.2 kN −m M u=φ M n M u=1.7 M L 1.836 c−120 ) ] c c=327.95 mm a=β1 c=261.97 ) ] M n=1.7 M L M L =370.90(1.000 ( 470 ) +49. 4 mm>t = 900mm t=110 d = 460 mm C a z T =310 mm .85 since f ' c is lesst h an 28 MPa a) Balanced condition.7 MPa Given: t=110 mm f y =414 MPa b w =3210 mm d=460 mm β 1=0. SOLUTION b f =900mm f ' c =20. φ=0.2) a=231.9 Repeat Problem 3.2 mm a=β1 c a=0.3 using the 2010 NSCP. SOLVED PROBLEMS IN T-BEAMS USING 2010 NSCP PROBLEM 3.85(272.65 600 d 600(460) c b= c b= 600+ f y 600+414 c b=272. 806 mm2 → balanced steel area y 1=d−t/2=405 mm y 2=d −t−z /2=289.4 mm A 1=b f x t =900 ( 110 )=99.000+37.65( 897) φM bn=583 kN −m→ ultimatebalanced moment b) Maximum steel area and moment.3 ) ] M bn=897 kN −m→ nominal balanced moment φM bn=0. Refer to Figure 3.622mm 2 A cb =A 1 + A2=136. .7 ) (99.622) A sb=5. Figure 3.7 ) [99.6.85 f ' c ( A1 y 1+ A 2 y 2 ) M bn=0.622 mm T =c 1+ c 2 A sb f y =0.85 ( 20.000 ( 405 ) 37.622 (289.85 f ' c ( A1 + A2 ) A sb ( 414 )=0.6 z=a−t=121.3 mm M bn=c 1 y 1 +c 2 y 2 M bn=0.000 mm 2 2 A 2=b w x z=310 ( 121.4 )=37.85 ( 20. 25 =0.10 Repeat Problem 3.815 7 1000−f y a=β1 c a=0.6 mm z=a−t=57.7 ) [99.21mm T =c 1+ c 2 A s max f y =0.9 kN−m→ultimate maximum moment PROBLEM 3.34) φM n max=656.65+0.000 ( 415 ) +17. 3 800−f y c= d=197.14) a=167.14 mm .34 kN −m→ nominal max moment φM n max=0.6 )=17. SOLUTION .85 ( 20.571 mm A 2=b w z =310 ( 57.2 using the 2010 NSCP.85 f ' c (99.85(197.000+17.85 f c (A 1 y 1 + A2 y 2) M n max=0.815(806.85 f ' c (A 1+ A 2) A s max ( 414 ) =0.2 ) ] M n max=806.847 mm2 y 2=d −t−z /2=321.847) A s max =4966 mm2 → maximum steel area M n max=c1 y 1+ c 2 y 2 ' M n max=0.847 ( 321. φ=0. 05 ' β 1=0. φ=0.85 ( 30 ) (1200 )( 130 ) ( 470− ) 2 M fn =1611 kN −m Solving for φ : a=130 mm 0. d) Find the maximum design moment so that section is tension-controlled if it is reinforced for tension only. b f =1200 mm b w =290 mm Given: t=130 mm f ' c =30 MPa d=470 mm f y =345 MPa t M fn =0.56 mm β1 d−c f s=600 =1213 MPa>1000 MP tension-controls.90 c φM fn =090(1611) φM fn =1450 kN −m PROBLEM 3.4 using the 2010 NSCP.836 a c= =155. .85 f ' c t bf (d− ) 2 130 M fn =0.11 Repeat Problem 3. Additional questions: M D =195 kN −m M L =210 kN −m c) Find the required steel area if and .85− 7 ( f c −28 )=0. 25 =0.97 kN−mm a c= =117.90 c φ M fn =545.85 since f ' c <28< MPa φM n a=t : Solve for when M fn =0.SOLUTION b f =820 mm f ' c =20.7 MPa Given: b w =250 mm f y =414 MPa d=470 mm t=100 mm β 1=0. φ=0.815 1000−f y .65+0.85 f ' c b f t ( d−t /2 )=605.43 mm 7 800−f y φ=0.65 mm β1 d−c f s=600 =1797 MPa>1000 MPa .375 kN −m φ M n max : Solve for 3 Cmax = d=201. ”a” is less than t. M L =120 kN −m a) M u=1.6 mm =250 mm y 2=d −t−z /2=334.39 mm M n max=M fn +0.72 kN −m φM n max=579 kN−m M D =150 kN −m.21mm> t = 820mm t=100 d = 470 mm C a z z=a−t=71.85 f ' c A2 y 2 M n max=710. a=β1 c max =171.21 mm T 2 A 2=b w z =17.803. singly reinforced Mu φM fn Since is less than . t =100 =820 mm C d = 470 mm d -a/2 T .2 M D +1.6 M L =372 kN −m< φ M n max . 7 ) a ( 820 ) ( 470− ) 2 a=65.85 ( 20.283 mm As Minimum is the smaller value of: f 'c b w d=646 mm2 √ f ' c b d =1059mm 2 2fy 4f y f .85 )( 20.85 f ' c a b f ( d− ) 2 6 a 372 x 10 =0.08 mm β1 d−c f s=600 =3.85 f ' c a b f A s ( 345 )=0.φ=0.52mm a c= =77. tension controls .90 ( 0.90 c T =C A s f y =0.52 ) (820) 2 A s =2.90 a M u=φ 0.058 MPa>1000 MPa .7 )( 65.Assume φ=0. 85 f ' c a b f ( d− ) 2 6 514 x 10 =0.963 MPa>1000 MPa .6 M L =514 kN−m<φ M n max .90 c .2 M D +1.tension controls.03 mm β1 t =100 =820 mm C d = 470 mm d -a/2 T d−c f s=600 =1.85 ) ( 20.90 ( 0. 2 Thus.283 mm M D =175 kN −m.53 mm a c= =110.a Since is less than t. singly reinforced M u is less than φM fn . M L =190 kN −m b) M u=1. Assume φ=0.7 ) a ( 820 ) (470−a /2) a=93. φ=0. A s =2.90 a M u=φ 0. 85 )( 20.53 ) (820) A s =3.6 M L =570 kn−m<φ M n max .85 f ' c a b f A s ( 345 )=0.375+ 0. T =C A s f y =0. M L =210 kN −m c) M u=1.7 )( 93.7 )( 250 ) z (470−100− ) 2 z=17. = 820mm t=100 d =470mm C a z T =250 mm Assume φ=0.259 mm2 M D =195 kN −m. singly reinforced M u is more than φM u .2 M D +1.05 mm .90 ( 0. a Since is more than t.85 ( 20.90 M u=φM fn +φM n 2 z 570 x 106=545. φ=0.7 mm d−c f s=600 =1448 MPa>1000 MPa . φ=0. a=t+ z=117.453mm 2 z y 2=d −t− =3450.85 f ' c A 2 y 2 . c=a /β 1=137.81 mm A 2=b w z =12.9 mm 2 M tn =M fn +0.85 f c ( A1 + A 2) A s =3.90 c A 2=b w z =3.908 mm2 ' T =C1 +C 2 A s f y =0.81 mm>t = 820mm t=100 d = 470 mm C a z T =250 mm z=a−t=49.666 mm2 3 c= d=176.tension controls .05 mm .90 d) b a=β1 c=149.25 mm . 4 kN −m Mu φM n Note: If is less than or equal to .95 mm z a=β1 c b=323.694 mm As Solve for balanced : t=100 = 1500mm d =600mm 600 d C a c b= 600+ f y =380.7 MPa Given: b w =250 mm f y =345 MPa d=600 mm β 1=0.952mm =400 mm T =C A sb f y =0. M tn =681.6 using the 2010 NSCP.000 2 A 2=b w z =55. the beam is tension-controlled. SOLUTION b f =1500 mm f ' c =20.59 kN−m φM tn =613.81>t A 1=b f t=150.85 2 A s =6−28 mm=3.12 Repeat Problem 3. PROBLEM 3.85 f ' c ( A1 + A2 ) . A sb ( 345 )=0.441< A1 therefore a is less than t t =100 =1500 mm C d = 600 mm d -a/2 T A c =a bf 72.441=a (1500) .000+ 55.503 28 ¿ ¿ Steel area provided. C=T 0. π A s =6 x ¿ 4 f s=f y therefore .7 ) A c =3.7 )( 150.85 f ' c A c = A s f y 0.952) A sb=10.85 ( 20.695(345) A c =72.85 ( 20. 85 f ' c a b f ( d−a /2) 48.7 ) ( 48.7 using 2010 NSCP.29 M n=0.29 ) ( 1500 ) (600− ) 2 M n=733.82 mm β1 d−c f s=600 =5.90( 733. .13 Repeat Problem 3.736 MPa>1000 MPa tension controls c therefore φ=0.85 ( 20.29 mm Solve for φ : a c= =56.90 M n=0.99 kN −m φM n=0.99) φ M n=660.6 kN −m PROBLEM 3. 2 a=48. SOLUTION Given the following properties of a T-beam: b f =900mm f ' c =21 MPa Flange width. t=120 mm f y =345 MPa Flange thickness. φ=0. d=580 mm M D =410 kN −m Service deal load.000 mm 2 z As Solve for balance : 600 d Cb = =368.206) 2 A sb=9.000+77.582mm Steel area provided is less than the balanced steel area.85 .85 ( 21 ) (108. Steel yields.90 2 2 28 ¿ =7.85 f ' c ( A1 + A2 ) A sb ( 345 )=0.02 mm>t z=a−t=193.389 mm t=120 = 900mm π A s=12 x ¿ 4 d =580mm C a A 1=b f t=108. β 1=0. Effective depth. .25 mm =400 mm 600+f y a=β1 c b=313.206 mm2 T =C A sb f y =0.389 mm Width of web. 2 b w =400 mm A s =7.02 mm A 2=b w z =77. C=T f ' c Ac = A s f y 0.85 0.85 ( 21 ) A c =7,389(345) A c =142,813 mm2 > A1 “a” >t = 900mm t=120 d = 580 mm C a z 460 T =400 mm A c = A 1+ A 2 142,813=108,000+ A 2 2 A 2=34,813 mm A 2=b w z 34,813=400 z z=87.03 mm t y 1=d− =520 mm 2 z y 2=d −t− =416.48 mm 2 M n=C 1 y 1 +C2 y 2 M n=0.85 f 'c ( A1 y 1+ A 2 y 2 ) M n=0.85 ( 21 ) [108,000 ( 520 ) +34,813 ( 416.48 ) ] M n=1,261.6 kn−m Solve for φ : a a=t+ z=203.03mm c= =243.57 mm β1 d−c f s=600 =828.76 MPa<1000 MPa c f y < f s<1000 MPa , Since Transition region f s−f y φ=0.65+0.25 =0.8346 1000−f y φM n=0.8346 (1,261.3) φM n=1,052.703 kN −m φM n=M u M u=1.2 M D +.6 M L 1,052.703=1.2 ( 410 ) +1.7 M L M L =350.44 kN −m PROBLEM 3.14 Repeat Problem 3.8 using 2010 NSCP. =500mm d=530mm t = 120mm 10-32 mm =320mm Figure 3.7 SOLUTION 32 ¿2 t =120 mm =500mm π A s =10 x ¿ 4 d=530mm a A s =8,042 mm2 Z A 1=b f t=60,000 mm2 0.05 β 1=0.85− (32−28) 7 β 1=0.821 =320mm As Solve for balanced : 600 d c b= 600+ f y a=313.3 mm a=β1 c b=257.35 mm> t z=a−t=137.35 mm A 2=b w z =43,953 mm2 T =C A sb f y =0.85 f ' c ( A1 + A2 ) A sb ( 345 )=0.85 ( 21 ) (60,000+ 43,953) A sb=6,813 mm2 A s > A sb ,tension steel does not yield Since t =120 mm =500mm d=530mm a Z T =320mm φ=0.65 cpmression controls A 1=60,000 mm2 A 2=b w z =bw ( a−t ) =bw ( β 1 c−t) d−c f s=600 c ' T =C1 +C 2 A s f s =0.85 f c ( A1 + A2 ) 530−c 8,042 x 600 =0.85 ( 32 ) [60,000+ 320(0.821 c−1200)] c c=329.27 mm a=β1 c=270.47 mm A 2=b w z =48,151 mm2 z z=a−t=150.47 mm y 2=d −t− =334.76 mm 2 y 1=d−t/2=470 mm M n=C 1 y 1 +c 2 y 2 M n=0.85 f 'c ( A1 y 1+ A 2 y 2 ) M n=0.85 ( 32 ) [60,000 ( 470 ) +48,151 ( 334.76 ) ] M n=1,205.48 kN −m φM n=0.65( 1,205.48) φ M n=783.56 kn−m M u=φM n M u=1.2 M D +1.6 M L According to Section 407.2 ( 330 ) +1. 78. Continuous compression bars are also helpful for positioning stirrups and keeping them in place during concrete placement and vibration. Deformed wire or welded wire fabric of equivalent area is allowed. compression steel in beams must be enclosed by lateral ties. Compression reinforcement is needed to increase the moment capacity of a beam beyond that of a tensilely reinforced makes beams tough and ductile and reduces long-time deflection of beams.8. beams are restricted in small sizes by space or aesthetic requirements to such extent that the compression concrete should be reinforced with steel to carry compression.56=1. at least 10 mm in size for longitudinal bars 32 mm or smaller.003 d’ a c = + .23 kN −m DOUBLY REINFORCED BEAM Occasionally. and at least 12 mm in size for 36 mm and bundled bars.12 of NSCP. M u2 the part of the tension steel is the couple due to the compression steel A 's As2 . 48 tie bar or wire diameters. is the couple due to compression concrete and As1 . ANALYSIS OF DOUBLY REINFORCED BEAM M n1 Doubly reinforced beam is analyzed by dividing the beam into two couples and Mn2 M n1 as shown in Figure 3. Various tests show that compression reinforcement also prevents the beam to collapse even if the compression concrete crushes especially if it is enclosed by stirrups. or least dimension of the compression member. Compression steel also helps the beam withstand stress reversals that might occur during earthquakes. and the other part of the tension steel area b 0. The spacing of these ties shall not exceed 16 longitudinal bar diameters.6 M L M L =242. then . Thus. otherwise . 3-7 A s max =0. Figure 3. A 's ¿ by On the other hand.75 ρb bd + A ' s fy . the maximum permissible is: f 's Eq.e. where f 's is the stress of compression steel is given and is given by the following equation. the stress in tension ( is always equal for . This stress must always be checked. (see derivation in page 137) c−d ' f ' s=600 Eq.3 of NSCP. ρb the portion of equalized by compression reinforcement need not be reduced by the As 0. For the reason. tension steel must As ¿ fy yield).8 Compression reinforcement is provided to ensure ductile failure (i. for members with compression reinforcement. therefore. A ' s =A s 2 A ' s =A s 2 f y /f ' s If the compression steel yields. 3-6 c According to Section 410.4.75 factor. stress of compression steel ( may either be or below fy . 59 ω) M u ≤ φ M n max If design Singly Reinforced (See Chapter 2) M u>φ M n max If design as Doubly Reinforced(proceed to step II) M u> φ M n max II. Solve for and ρmax =0.d’ d -a/2 = + .85 f ' c β 1 (600) ρmax =0.75 . b d’ a d d . ρb bd= A s 1 The expression 0. As A 's STEPS TO DETERMINE AND OF ADOUBLY REINFORCED MU RECTANGULAR BEAM. GIVEN AND OTHER BEAM PROPERTIES ρmax M u max I.75 ρb 0.75 =ρ f y (600+f y ) ¿ ¿ ¿ ρf y ω= ' =¿ ¿ fc M n max=φ f ' c ω b d 2 (1−0. Figure 3.003 0.85 ¿ d’ a=¿ ¿ a=β1 c c ¿ c =¿ ¿ c – d’ . and M n 1=M n max φM n 2=M u−φ M n max ¿ φ T 2 ( d−d ' ) ' φ M n 2=φ A s 2 f y (d−d ) ¿ ¿ A s 2 =¿ ¿ III.9 As1 Solve for A s 1= ρmax b d M n1 M n2 As2 Solve for . Solve for the stress of compression steel Solve for a and c: C1 =T 1 f ' c a b= A s1 f y 0. 3-8 c f 's≥ f y If proceed to IV f ' s <f y If proceed to V f 's≥ f y f ' s=f y IV. AND OTHER BEAM PROPERTIES There are three possible cases in doubly reinforced beams. then use (compression steel will not yield) fy A ' s =A s 2 f 's Mn STEPS IN FINDING OF A DOUBLY REINFORCED RECTANGULAR BEAM AS A 'S . f ' s /E s 0. WITH GIVEN .003 = c−d ' c c−d ' f ' s=600 Eq. f 's V. then use (compression steel yields) A ' s =A s 2 f ' s <f y . f s=f ' s =f y ¿ Case 1: Both tension and compression yields ( . f ' s < f y ¿ Case 2: Tension steel yields and compression steel does not ( Case 3: Tension steel does not yield compression steel yields. Assume compression steel yields ( ) ¿ ¿ A s 2= A ' s =¿ ¿ ¿ A s 1= A s− A s 2=¿ ¿ II.d’ d -a/2 = + f ' s=f y I. f s=f y . Solve for a and c (assuming tension steel yields): . s y Note: For doubly reinforced beams with effective depth d=250 mm or more. f =f f ' < f ¿ ( s y. it is not possible for both steels not to yield. b d’ a d d . yields. proceed to step IV III. Solve for the stress in compression steel c−d ' f ' s=600 c f 's≥ f y . Check: c If tension steel. Since M n=M n 1+ M n 2 ¿ T 1 d− ( a2 )+T (d−d ) 2 ' M n= A s1 f y d− ( a2 )+ A s2 f y (d−d ' ) f ' s <f y . C1 =T 1 f ' c ab= A s1 f y 0. If compression steel does not yield.compression steel yields IV. proceed to step III f s <f y . . If proceed to step IV f ' s <f y . If tension steel does not yield. If proceed to step V f ' s ≥ f y .85 ¿ a=¿ ¿ a=β1 c ¿ c =¿ ¿ d−c f s=600 f s≥ f y . V. f ' s =600 =¿ ¿ c ¿ Solve for a.85 f ' c β 1 c b+ A ' s 600 = As f y c Solve for c by equation formula. c−d ' f ' s=600 d’ c a c d From the stress diagram: d – d’ d -a/2 C1 +C 2=T 0. ¿ c−d ' Solve for f ' s .85 f ' c ab+ A ' s f ' s= A s f y c−d ' 0.a= β 1 c =¿ ¿ M n: Solve for M n=M n 1+ M n 2 ( a2 )+C (d−d ) ¿ C1 d− 2 ' M n=0.85 f ' c ab d− ( a2 )+ A f (d −d ) ' s ' s ' . The area of skin reinforcement per meter of height on each side face shall be Eq.0(d−750) .7 of NSCP.85 f ' c a b d− ( a2 )+ A f ( d−d ) ' s y ' DEEP BEAMS According to Section 410. T =C c +C 2 A s f s=0. 3-9 A sk ≥1.85 f ' c a b+ A ' s f y d−c A s x 600 =0.8 of the Code. if the depth of web exceeds 900 mm. beams with overall depth to clear span ratios greater than 2/3 for continuous spans. shall be designed as deep flexural members taking into account nonlinear distribution of strain and lateral buckling.85 f ' c ( β 1 c ) b+ A ' s f y c ¿ ¿ c =¿ . a=β 1=¿ ______ M n=M n 1+ M c2 ( a2 )+C (d−d ) ¿ C1 d− 2 ' M n=0. longitudinal skin reinforcement shall be uniformly distributed along both side faces of the member for distance d/2 nearest the flexural A sk tension. Beams with web depth that exceed 900 mm have a tendency to develop excessive wide cracks in the upper parts of their tension zones.. f s <f y but f ' s =f y VI. According to Section 410. or 4/5 for simple spans. 305-mm wide rectangular beam has an overall depth of 560 mm.66 mm .The maximum spacing of the skin reinforcement shall not exceed the lesser of d/b and 300 mm. ILLUSATIVE PROBLEMS DESIGN PROBLEMS PROBLEM 3. The total area of longitudinal skin reinforcement in both faces need not exceed one-half of the required flexural tensile reinforcement.15 a . b) The maximum tension steel area and the nominal and ultimate maximum moment. SOLUTION 70 mm 305 mm β 1=0. The centroid fiber.85 d=490 mm a 4-25 mm d-d’420 a) Balanced condition = + 600 d 600(490) c b= c b= 600+ f y 600+ 415 70 mm c b=289. Determine the following: a) The balanced tension steel area and the nominal and ultimate balanced moment. Assume f y =415 MPa f ' c =29 MPa and . The beam is reinforced with four 25-mm-diameter compression bars. Such reinforcement may be included in strength computations if a strain compatibility analysis is made to determine stresses in the individual bars or wires. 21 M nb=0.021.85 ( 29 )( 246.4 kN −m φM nb=0.21 ) (305) 2 A s 1=4.66 f sc =455 MPa> f y yield f sc =f y T 1 =Cc A s 1 f y =0.90 ( 1. A sb= A s 1 + A s 2=6.460 mm T 2 =C ' s A s 2 f y =A 's f y 2 A s 2=1.85 f ' c a b A s 1 ( 415 ) =0.21 )( 305 ) (490− ) 2 M nb=1.85 ( 29 ) ( 246.85 Tf ' c a b d− ' s y ' 246.66) a=246.24 kN −m .66−70 f sc =600 f sc =600 c 289.a=β1 c b a=0.21mm d−c 289.964 mm Balanced steel area.4 ) =919.85(289.021.242mm 2 M nb=C c ( d−a/2 ) +C' s ( d−d ' ) ( a2 )+ A f (d−d ) M nb=0. b) Maximum tension steel area: According to Section 410.85 ( 29 ) ( a ) (305 )=3. A s 1=0.2 mm c−d ' 217.75( 4. the ρb portion of equalized by compression reinforcement need not be reduced by the 0.309 mm2 C c =T 1 0.7 /0.75 factor.2 f ' s=406.345+1.964 mm2 A s max = A s 1max + A s 2 A s max =3.345( 415) a=184.85 c=217.460) A s 1 max =3.3. for members with compression reinforcement.85 f ' c a b=A s 1 max f y 0.7 mm c=a/ β1 c=184.345 mm2 A s 2=1.75 A s 1 A s 1 max =0.2−70 f ' s=600 f ' s=600 c 217.964 A s max =5.4.7 MPa<f y ( will not yield ) . a) What is the balanced steel area considering the contribution of the compression steel? b) What is the maximum tension steel area allowed by the code? SOLUTION .45) φM n max=798.85 f 'c a b d− ' s ' s (d−d ' ) 184.7 ) ( 305 ) (490− ) 2 +1964 ( 406.7 M n max=0.7 ) (490−70) M n max=887.7 kN−m PROBLEM 3. Concrete strength and steel strength . The beam is reinforced with 2-28 mm compression bars placed 70 mm from extreme f ' c =35 MPa f y =345 MPa concrete.85 ( 29 )( 184. ( a2 )+C ( d−d ) M n max=C c d − ' s ' ( a2 )+ A f M n max=0.90(887.16 (CE NOVEMBER 2009) A reinforced concrete beam has width of 300 mm and effective depth of 460 mm.45 kN−m φM n max=0. 75 c b =0.85− ( 35−30 ) =0.814 x 292 ) ( 300 )+ 1232 ( 345 ) =A s (345) A s =7.05 mm a=β1 cmax=178.05 .814 7 π A ' s = (28)2 x 2=1. 0.85 ( 35 ) ( 0.384 mm2 b) Maximum steel area: For rectangular beams: Cmax =0.37 mm c−d ' 219.232 mm2 4 a) Balanced condition considering compression steel: 600 d 600(460) c b= c b= 600+ f y 600+345 c−d ' 292−70 f ' s=600 f ' s=600 c 292 f ' s=456 MPa>f y .05−70 f ' s=600 f ' s=600 c 219.85 f ' c a b+ A ' s f ' s= A s f y 0.05 β 1=0. thus f ' s=f y =345 MPa C c +C s=T 0.75(292) Cmax =219. 37 ) (300) +1232 ( 345 )=A s (345) 3 A s =5.85 f 'c a b(d − ) 2 M n max=0.96 /2) M n max=576.85 ( 27.76 kN−m φM n max=0. f ' s=408 MPa> f y .76 )=519 kN −m .846 mm PROBLEM 3.85 ( 35 ) (178.6 MPa f y =276 MPa and steel yield strength . SOLUTION This is the same problem in Chapter 2.96 )( 300 ) (490−213.90 (576.85 f ' c a b+ A ' s f ' s= A s f y 0.17 A rectangular beam has b=300 mm and d= 490 mm. Compressive steel if required shall have its centroid 60 mm from extreme concrete fiber. φM n max Solve for : 600 d c b= =335. Concrete compressive strength f ' c =27.6 )( 213. thus f ' s =f y =345 MPa C c +C s=T 0. Calculate the required Mu tension steel area if the factored moment is 620 kN-m.616 mm 600+ f y ab =β 1 c b=285.27 mm a M n max=0. b d’ ca M n 1=M n max=576. M u=620 kn−m> φM n max .90−576.85 ( 27.13 kN −m a=213.6 )( 213. Since the beam must be doubly reinforced.96 mm A s 1 f y =0.96 ) (300) A s 1=5456 mm2 Note: A s 1= A s max f 's: Solve for a c= =251.76 kN−m = d – a/2+ d – d’ Mu 620 M n 2= −M n 1 M n 2= φ 0.76 M n 2=112.71 f ' s=457 MPa>f y Compression steel yields Use f ' s=f y .71−60 f ' s=600 f ' s=600 c 251.71 mm β1 c−d ' 251.85 f ' c a b A s 1 ( 276 )=0. 13 x 106= A s 2 ( 276 )( 490−60) A s 2=945 mm2 Tension steel area. A s =A s 1 + A s 2=6401 mm2 Compression steel: c ' s=T 2 A ' s f y =A s 2 f y A ' s =A s 2 A ' s =945 mm2 PROBLEM 3.7 M L M u=1.85 M u=1. Use and SOLUTION β 1=0.18 A rectangular beam has b=310 mm and d=460 mm. M n 2=T 2 ( d−d ' ) 112.7 (190) M u=645 kN −m φ M n max Solve for : . Determine the required steel area.4 M D +1.4 ( 230 )+1. The beam will be designed to carry a service dead load of 230 kN-m and service live load of190 kn-m. Compression reinforcement if necessary will have its centtoid 70 mm from extreme concrete fiber. f ' c =30 MPa f y =415 MPa . 58) φM n m ax =460.90(511.35 ) ( 310 ) (460−173.85 f ' c a b(d −a/2) M n max=0.75 c b Note : For rectangular beams. compression steel is necessary Since b d’ ca = d – a/2+ d – d’ M n 1=M n max=511.58 kN−m Mu 645 M n 2= −M n 1 M n 2= −511.42 kN −m M u=645 kN −m>φ M n max .90 . 600 d c max =0.85 ( 30 )( 173.58 φ 0.35/2) M n max=511.35 mm M n max=0.75 =203.58 kN−m φM n max=0.94 mm 600+ f y a=β1 c max =173. c max =0. 85 ( 30 )( 173.06 Pa compression steel does nt yield.302 mm2 M n 2=T 2 ( d−d ' ) 205.06 )=1.267 mm2 A s =A s 1 + A s 2 A s =3.35 ) (310) A s 1=3.569 mm Compression steel: c−d ' 203.088 x 106= A s 2 ( 415 ) (460−70) A s 2=1.267( 415) A ' s =1334 mm2 .94 f ' s=394.94 mm a=173.302+ 1.94−70 f ' s=600 f ' s=600 c 203. C ' s=T 2 A ' s f ' s= A s 2 f y A ' s ( 394.35 mm Tension Steel: T 1 =Cc A s 1 f y =0.85 f ' c a b A s 1 ( 415 ) =0.088 kN −m c=cmax =203. M n 2=205.267 2 A s =4. b) Using the tributary area method.10.PROBLEM 3. a) Calculate the required tension steel area at the point of maximum positive moment.85 d ' =70 mm . Web dimensions.5 kN/ m . f y =415 MPa maximum positive factored moment of 1080 kN-m. 1.19 A floor system consists of a 100-mm concrete slab supported by continuous T beam with 9 m span. as b w =280 mm . Use . 3 Unit weight of concrete is 23. The beam is subjected to a f ' c =21 MPa . what is the uniform service dead load acting on the beam? c) Calculate the uniform service live load acting on the beam. Concrete cover is 70 mm from the centroid of the bars. are and d=500 mm . determined by negative-moment requirements. A L=9m B L=9m C SOLUTION f ' c =21 MPa b w =280 mm f y =414 MPa d=500 mm β 1=0.2 m on centers as shown in Figure 3. 16t+ b w =16 ( 100 )+ 280=1.57) a=251.2 m 3. M u max=1080 kN −m Maximum factored moment.25 m 2.85(295. bf : Effective flange width.23100 . Balanced condition: 600 d 600(500) c= c= 600+ f y 600+415 a=β1 c a=0.88 m S oc =1. b f =1. 1. L/4=9/4=2.23 mm t=100 = 1200mm d =500mm C a z =280 mm z=a−t z=251.2 m Use φM n max Solve for to determine if compression steel is required. 28 /2 .345 mm2 2 A cb =A 1 + A2=162.759 mm2 T A 2=b w z 1. z=151.759−120.000 mm A 2=280 ( 151.23 mm 2 A 1=1200 x 100=120.23 )=42.759=280 z =280 mm z=6.75(162.75 A cb A c max =0.000 A 2=¿ 1.345 mm Maximum condition: A c max =0.28 mm y 1=d−t/ 2 y 1=500−100/2 y 1=450mm y 2=d −t−z /2 y 2=500−100−6.345) A c max =121.759 mm2> A 1 = 1200mm t=100 d =500mm C a z A 2= A cmax − A1 A 2=121. 85 ( 21 ) (120.86 mm M n max=C 1 y1 + c2 y 2 M n max=0.28 a=106. =1200mm d’=70 a d=500mm Z 500 mm d-d’ 430 d’=70 =820mm a=t+ z a=100+6.36) φM n max=878.90(976. y 2=396.36 kN−m φM n max=0.85 f ' c ( A 1 y 1 + A2 y 2) M n max=0.080 kN −m> φM n max .72 kN−m M u=1.000 x 450) +1. Since the compression reinforcement must be provided.759 x 396.86 M n max=976.28 mm . 1 MPa< f y M n 1=M n max=976.85 f ' c ( A1 + A2 ) A s 1 ( 415 ) =0.04 f ' s=264.04 c−d ' 125.36 0.36 kN−m A s 1= A s max T 1 =C1 +C 2 A s 1 f y =0.64 kN −m M n 2=T 2 ( d−d ' ) M n 2=A s 2 f y ( d−d ' ) 6 223.85 ( 21 ) (120.253 mm 2 .000+ 1.c=a/ β1 c=106.90 M n 2=223.28 /0.64 x 10 =A s 2 ( 415 ) (500−70) A s 2=1.237 mm2 Mu M n 2=M n−M n 1 M n 2= −M n 1 φ 1080 M n 2= −976.85 c=125.759) A s 1=5.04−70 f ' s=600 f ' s=600 c 125. A s =A s 1 + A s 2=6.490 mm2 Compression steel area: C ' s=T 2 A ' s f ' s= A s 2 f y A ' s ( 264.6 m 5-32 mmø 3-32 mmø 5-32 5-32mmø mmø 3-32 mmø 5-32 mmø A L=9m B L=9m C 145 kN-m 145 kN-m 202 kN-m .2516) w c =5.47)=0.5(0.Tension steel area.969 mm2 b) Dead load=weight of concrete: Area=1.6 m 7.9126 kN / m →dead load c) Uniform live load 7.28(0.2516 m3 w c =γ c x Area w c =23.1)+0.253(415) A ' s =1.2(0.1 )=1. Maximum positive moment (at midspan) 9 ¿2 ¿ wu ¿ w L2 w u= u 1,080=¿ 24 w u=320 kN /m w u=1.4 w D + 1.7 w L 320=1.4 (5.9126 )+ 1.7 w L w L =183.37 kN /m →live load INVESTIGATION (ANALYSIS) PROBLEMS PROBLEM 3.20 The beam shown in Figure 3.11 is subjected to a maximum service dead load moment of 230 kN-m. Determine the service live load that the beam can carry. Use f ' c =20.7 MPa∧f y =345 MPa . 350 mm 60 mm 2-28 mm 540 mm 600 mm 4-36 mm Figure 3.11 SOLUTION 36 ¿ ¿ π As= ¿ 4 28 ¿2 x 2=1,232 mm2 π A ' s= ¿ 4 b d’ ca Assume all steel yield: = d – a/2+ d – d’ f s=f ' s =f y 2 A s 2= A ' s=1,232 mm A s 1= A s− A s 2=2,840 mm2 C c =T 1 0.85 f ' c a b=A s 1 f y 0.85 ( 20.7 ) a ( 350 )=2,840(345) a=159.1mm a c= =187.18 mm β1 d−c 600−187.18 f s=600 f =600 c s 187.18 f s=1.323> f y tension steel yields c−d ' 187.18−60 f s=600 f s=600 c 187.18 f s=407.7> f y compression steel yields Assumption is correct, all steel yield. M n=M n 1+ M n 2 M n=T 1 d−( a2 )+T (d−d ) 2 ' ( a2 )+ A M n= A s1 f y d− s2 f y (d−d ' ) 159.1 ( M n=2,840 ( 345 ) 600− 2 ) +1,232 ( 345 ) (600−60) M n=739.4 kn−m φM n=0.90 ( 739.4 )=665.43 kN −m φM n=M u=1.4 M D +1.7 M L 665.43=1.4 (230 )+ 1.7 M L M L =202.02 kN−m PROBLEM 3.21 A rectangular beam has the following properties: f y =415 MPa Width, b=400 mm f ' c =22 MPa Effective depth, d=620 mm Tension bars, 3 pcs 25-mm-diameter d’=70 mm Determine the design strength of the beam and the safe service live load if the service dead load is 320 kN-m. SOLUTION 28 ¿2=6,158 mm2 π A s=10 x ¿ 4 2 2 25 ¿ =1,473 mm π A ' s=3 x ¿ 4 Assume all steel yields: A s 2= A ' s=1,473 mm 2 2 A s 1= A s− A s 2=4,685 mm b d’ ca = d – a/2+ d – d’ ' 0.85 f c a b= A s 1 f y 0.85 ( 22 ) a ( 400 )=4,685 ( 415 ) a=260 mm a c= =305.8 mm β1 d−c f s=600 =616.5 MPa> f y ( yield ) c c−d ' f ' s=600 =463 MPa> f y ( yield) c All steel yields. Assumption is correct M n=M n 1+ M n 2 M n=T 1 d− ( a2 )+T (d−d ) 2 ' ( a2 )+ A M n= A s1 f y d− s2 (d−d ' ) M n=4,685 ( 415 )( 620−260/ 2 ) +1,473 ( 415 ) ( 620−70) M n=1288.9 kN −m φM n=0.90 ( 1288.9 )=1,160 kN −m φM n=M u=1.4 M D +1.7 M L 1160=1.4 ( 320 )+1.7 M L M L =419 kN −m PROBLEM 3.22 A 12-m long rectangular reinforced concrete beam is simply supported at its ends. The including its own weight. SOLUTION β 1=0.75 a) Determine the depth of the compression block. four bars at the tension side and 2 bars at the compression side .Concrete protective f ' c =30 MPa coverings is 70 mm form the centroid of the bars. c) Determine the factored uniform load. Use 0.023.beam is provided with an addition support at the mid span. Width of beam is 300 mm and the overall depth is 450 mm. Concrete strength and f y =415 MPa ρb=0. steel yield . The beam is reinforced with 25-mm-diameter bars. b) Determine the nominal moment capacity of the beam.85 300 mm f ' c =30 MPa 70 mm f y =415 MPa 2-25 mm 310 mm 2 25 ¿ 380 mm 450 mm π A s =4 x ¿ 4 A s =1963 mm2 2 25 ¿ 4-25 mm π A ' s =2 x ¿ 4 70 mm 2 A ' s =982 mm Assuming all steel yields: A s 2= A ' s=982mm 2 A s 1= A s− A s 2=982 mm2 . the beam can carry. 85 f ' c a b=A s 1 f y 0.26 mm a c= =62.85 c)(300) 1963 ( 415 )=0.87 mm 98.66 mm<70 mm β1 compression steel does not yield Assuming tension steel yields and compression steel does not.85¿ c−70 +982 x 600 c c=98.03 mm → answer ∈Part a M n=C c d−( a2 )+C ( d−d ) s ' M n=0.17 MPa <f y 98.86−70 f ' s=600 =175.85 ( 30 ) a ( 30 )=982( 415) a=53.58 kN −m→ asnwer ∈part b .86 d−c f s=600 =1.706> f y ( yield ) c a=β1 c=84. T =C c +C ' s A s f y =0. C c =T 1 0.85 f ' c a b+ A ' s f ' s 30 (0.85 f ' c a b+ A ' s f ' s (d −d ' ) M n=270. By there-moment equation: A B 6 A á 1 1 6 A 2 á 2 C M A L1+2 M B ( L1 + L2 )+ M c L2+ + =0 L1 L2 M A=M C =0 6 A 1 á 1 w u L13 = L1 4 3 6 A 2 á 2 w u L1 = L2 4 + 6+¿ ¿ 6 ¿3 ¿ 6 ¿3 ¿ wu ¿ wu ¿ 0+2 M B ¿ w u=54.12 kN /m PROBLEM 3. φM n=0.90 M n φ M n=243.53 kN −m c) Maximum factored uniform load: Factored load.23 (CE NOVEMBER 2010) . 7 MPa f y =415 MPa . each located 70 mm from the extreme concrete f ' c =20. SOLUTION L=6 m f ' c =20. and steel yield strength Determine the following: a) Depth of compression blocks assuming both tension and compression steel yields. b) What is the ultimate moment capacity of the beam in kN-m? c) Determine the additional concentrated live load that can be applied at midspan if the dead load including the weight of the beam is 20 kN/m.85 φ=0. Concrete strength . fiber.90 Assuming tension & compression steel yields: A s 2= A ' s=1232 mm2 A s 1= A s− A s 2=1232 mm2 . π A ' s= ¿ 4 β 1=0.7 MPa Given : b=350 mm f y =415 MPa d=400 mm d b=28 mm d ' =70 mm 28 ¿2 x 4=2463 mm2 Tension steel area π A s= ¿ 4 2 2 28 ¿ x 2=132 mm Compression steel area. The beam is reinforced with 2-28 mm compression bars on top and 4-28 tension bars at the bottom.A 6 meter long simply supported reinforced concrete beam has a width of 350mm and an overall depth of 470 mm. c c =T 1 0. c Assuming tension steel yields: C c +C s=T s 0. c−d ' f ' s=600 Since compression steel does not yield.85 f ' c a b=A s 1 f y 0.64−70 f ' s=600 f ' s=600 c 97.85 ( 20.64 f ' s=170 MPa< f y Thus.08 .08 f s=600 f s =600 c 130.85 c )( 350 ) +1232 x 600 =2463( 415) c c=130.85 ( 20.7 ) ( 0.85 f ' c a b+ A ' s f ' s= A s f y c−70 0. compression steel does not yield.6 mm d−c 400−130.08 mm a=β1 c=110.7 ) a ( 350 )=1232( 415) a=83 mm →answer ∈Part a c=a/ β1 c=83 /0.85 c=97.64 mm c−d ' 97. 33 kN −m Ultimate moment capacity= φ M n=0.4 M D +1.11 ) ( 400−70 ) M n=347.7 M L M u=1.4 +1.90( 347.08−70 f ' s=600 f ' s=600 c 130.85 ( 20.6 /2 ) +1232 ( 277.33) φM n=312.85 f ' c a b d− ' s ' s ' M n=0. f s=1245> f y ( yield ) c−d ' 130.6 ) ( 350 ) ( 400−110.6 kN −m L=6m M D L2 P L M u=1.08 f ' s=277.6 kN −m →answer ∈ Par t b Ultimate moment capacity= 3m 3m M u=φ M n=312.7 ) ( 110.11 MPa< f y ( a2 )+C ( d−d ) M n=c c d − ' s ' ( a2 )+ A f (d −d ) M n=0.7 L 8 4 . 25mm 30 mm SOLUTION β 1=0.24 A beam section is shown in Figure 3.175 kN PROBLEM 3. What is the safe service live load moment for this f ' c =21 MPa∧f y =415 MPa .12.4 ¿ PL =73.85 .25mm 25 mm Figure 3. The beam will be subjected to a maximum service dead load of 215 kN-m.12 8 .6=1. 2 6¿ ¿ 6 ¿2 ¿ PL¿ 20 ¿ 312. beam? Use 360 mm 30 mm 650 mm 5. 454 mm2 360 mm 30 mm d’ 650 mm 5. π A ' s =5 x ¿ 4 A ' s =2. π A s =8 x ¿ 4 A s =3.473 mm2 .5 mm 2 ( 25 ) Effective depth (to centroid of tension bar) d=650−30−25−1/2(25) d=582.5 mm M D −215 kN −m f ' c =21 MPa f y =415 MPa Assume all steel yields: 2 A s 2= A ' s=2. 2 25 ¿ Tension steel.927 mm2 2 25 ¿ Compression steel.25mm Effective depth to extreme tension bar: 30 mm 1 d t =650−30− =607.454 mm A s 1= A s− A s 2=1.25mm 25 mm ' d =30+ 1 =42.5 mm d 2 (25 ) 8 . 473( 415) a=95.9 f ' s=372 MPa<f y Compression steel does not yield.85 f ' c a b=A s 1 f y 0.5 f ' s=600 f ' s=600 c 111. 360 mm d’ 5. thier stresses are bothset equal ¿ f y ∧thier cg is located at thier geometric centroid . Thus .85 f c ( β 1 c ) b+ A ' s x 600 c 3.1 mm c=a/ β1 c=95.927 ( 415 ) =0.1 /0.85 ( 21 ) a ( 360 )=1.9−42.85 f ' c a b+ A ' s f ' s ' c−d ' A s f y =0.85 c=111. C c =T 1 0.-25mm 25mm T Note : There are two lawyersof tension bars which obviously yiel .9 mm c−d ' 111. T =C c +C ' s A s f y =0.25mm c a 25 mm d d-a/2d-d’ 88.85 c ) (360) .85 ( 21 )( 0. 31=1.38−42.7 M L 786.38 mm c−d ' 122.68 kN −m φM n=0.03 )( 360 ) (582.5 f ' s=600 f ' s=600 c 122.4 M D +1.25 .03 M n=0.85 f ' c a b d−( a2 )+ A ' f ( d−d ) s ' s ' 104.5) M n=873.7 (M L ) M L =285.68) φM n=786.64 MPa< f y a=β1 c a=0.454 ( 391.5 kN −m PROBLEM 3.38 f s=391. c−42.85 ( 21 ) (104.31 kn−m M u=φ M n M u=1.03 mm ( a2 )+C (d −d ) M n=C c d− ' s ' M n=0.454 x 600 c c=122.85(122.4 ( 215 ) +1.5−42.90( 873.64 )(582.5 +2.5− ) 2 +2.38) a=104. 85 d Figure 28 ¿ 3.28mm 10 .158 mm2 30 mm 2 25 ¿ Compression steel.25mm 2 .A beam section is shown in Figure 3. What is the safe service live load moment for this beam? Use f ' c =21 MPa∧f y =415 MPa .5 mm Effective depth (to centroid of tension bars) d=650−30−28−1/2(28) d=578 mm M D =360 kN −m f ' c =21 MPa f y =415 MPa Assume all steel yields: .28mm 4 30 mm A s =6. 320 30 mmmm 30 mmd’ 2.13 2 Tension steel. The beam will be subjected to a maximum service dead load of 360 kN-m. A s =10 x π ¿ 10 .13.25mm 650 mm 650 mm 28 mm 28 mm β 1=0. π A ' s =2 x ¿ 4 A ' s =982 mm2 ' d =30+1/2 ( 25 )=42. 5 f ' s=600 f ' s=600 c 442.25mm ca 28 mm d-d’ d-a/2 Thus . d 10 8 -.85 c=442.176( 415) a=376.25mm 28mm d 1=650−30−14=606 mm d 2=650−30−28−28−14=550 mm 2 2 28 ¿ =3.4−42.85 ( 21 ) a ( 320 )=5.04 mm c=a/ β1 c=376.176 mm2 C c =T 1 0.4 f s=222 MPa< f y Tension steel does not yield. thier cg isnot located at thier geometric centroid .4 mm c−d ' 442.079 mm π A st 1 =A st 2=5 x ¿ 4 . 2 A s 2= A ' s=982mm A s 1= A s− A s 2=5.04 /0.85 f ' c a b=A s 1 f y 0. 320 mm d’ are two layers of tension bars which have different stresses less thanf y . Note : There 5. 9 f s 1=600 f s 1=600 c 363.079 x 600 c c ¿ 0.5 f ' s=600 f ' s=600 c 363.9 MPa<f y c−d ' 363. T 1 +T 2=C c +C ' s A st 1 f s 1+ A st 2 f s 2=0.85(345.9 f s 2=306.85 c ) ( 320 ) +982 ( 415 ) c=363.4) a=301.9 f ' s=530 MPa> f y a=β1 c a=0.079 x 600 +3.85 f ' c a b+ A ' s f y d1 −c d −c A st 1 600 + A st 2 600 2 =0.9 mm a=β1 c=309.9 f s 2=600 f s 2=600 c 363.9 f s 1=399.9−42.25 MPa<f y d 2−c 550−363.2mm Solve for d: .85 ( 21 )( 0.29 mm d1 −c 606−363.85 f ' c a b+ A ' s f y c c 606−c 550−c 3. 7 M L 876.2 ( 606 )+ 944.07) φM n=876.079(399.9 kN T x d=T 1 x d1 +T 2 x d 2 2.1 d=1.65 kN −m M u=φ M n M u =1.29/2) +982 ( 415 ) (581.07 kn−m φM n=0.29 )( 320 ) (578−309.85 f ' c a b d− s y ' M n=0. T 1 =A st 1 f s 1 T 1=3.5) M n=974.25) T 1 =1.65=1.21 kN−m PROBLEM 3.85 ( 21 ) (309.229.26 .90( 947.9(550) d=581.4 M D +1.174.66−42.7( M L ) M L =219.079 (306.229.9 ) T 2 =944.66 m ( a2 )+C ( d−d ) M n=c c d − ' s ' ( a2 )+ A ' f ( d−d ) M n=0.4 ( 360 )+ 1.2kN T 2 =A s 2 f s 2 T 2=3. =600mm 25mm t=100mm 3-23mm Figure 3. A f =600 ( 110 ) =66.140 mm 2 2 A s 1= A s− A s 2=3. Use f ' c =27 MPa∧f y =350 MPa .000 mm Assume all steel yields: d=110 +390−20−10−25−1/2 ( 25 )=432.14. SOLUTION β 1=0.14 10mm =390mm stirrup 10-25mm 25mm 20mm =300mm .5 mm d ' =25+10+1 /2 ( 22 )=46 mm A s 2= A ' s=1.140 mm2 π A ' s=3 x ¿ 4 2 Flange area.85 2 2 25 ¿ =4.768 mm .909 mm π A s =10 x ¿ 4 22 ¿2=1.Calculate the design flexural strength of the T-beam shown in Figure 3. 85 c=112.7 f s=600 f =600 c s 112.7 f ' s=355 MPa> f y ( yield ) =600mm t=110mm 25mm =390mm d-d’ d-a/2 3-22mm 3-23mm d 10mm 10-25mm stirrup 25mm 20mm =300mm d−c 432.8 mm<t c=a/ β1 c=95.768(350) A c =57.7 f s=1.468 mm2 < A f therefore a< t A c =a bf 57.8/0.7−46 f ' s=600 f ' s=600 c 112.85 ( 27 ) A c =3.85 f ' c A c = A s 1 f y 0.469=a(600) a=95.7 mm c−d ' 112.Area of compression concrete: C c =T 1 0.5−112.703> f y ( yield ) Verify if the upper layer of tension steel yields . 15.15 10mm stirrup =390mm 10-28mm 25mm 20mm =315mm . =600mm 25mm t=100mm 2-22mm Figure 3. assumption is correct: a ( ) M n=C c d− +C ' s (d −d ' ) 2 ( a2 )+ A f (d−d ) M n=0.140(350)(432.5 mm d 2−c f s 2=600 =1.90( 661. d 2=d −1/2 ( 25 ) −1/29 ( 25 )=407. Use f ' c =25 MPa∧f y =345 MPa .4 kN−m PROBLEM 3.5−46 ¿ +1.567 MPa>f y ( yield) c All steel yields.5−46) M n=661.5) φM n=595.8 ) ( 600 ) 432.5 kN −m φM n=0.85 ( 27 ) ( 95.85 f ' c a b f d− ' s y ' M n=0.27 Calculate the design flexural strength of the T-beam shown in Figure 3. A f =600 ( 100 )=60.000 mm 2 Assume all steel yields: d=100+390−−20−10−28−1 /2 ( 25 ) =419.626 mm > A f therefore a>t A c = A f + A w 87.626=60.5 mm d ' =25+10+1 /2 ( 22 )=46 mm A s 2= A ' s=760 mm 2 A s 1= A s− A s 2=5.SOLUTION β 1=0.397(345) 2 A c =87.85 ( 25 ) A c =5.397 mm2 Area of compression concrete: C c =T 1 0.626 mm .158 mm π A s=10 x ¿ 4 22 ¿2=760 mm2 π A ' s=2 x ¿ 4 Flange area.85 2 2 28 ¿ =6.00+ Aw 2 A w =27.85 f ' c A c = A s 1 f y 0. 83 mm c−d ' 220.626=315 z z=87.7 mm a=100+ z=187.83−46 f ' s=600 f ' s=600 c 220.A w =bw z 27.83 f ' s=475 MPa> f y ( yield) .7 mm c=a/ β1 =220.