Structural Design of G+5 Building (Final Year Project for BSC in Civil Engineering)

MEKELLE UNIVERSITYFACULTY OF SCIENCE AND TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING In Partial Fulfillment of B.Sc. Degree in Civil Engineering Structural Design of A G+4 Commercial Building with Solid & Pre-cast Slab and Cost Comparison Prepared by:- Keralem Adane Osman Giragn Samuel Tesfaye Tewodros Kassa Tigistu Fisseha Advisor:- Kibrealem Mebratu June 2008 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 1 Table of content Acknowledgment…………………………………………………………..3 Introduction…………………………………………………………….…..4 Specification………………………………………………………………..5 1. Roof design...............................................................................................6 1.1. Wind load analysis………………………………………………...6 1.2. Analysis of lattice purlin………………………………………… 14 1.3. Design of truss……………………………………………………15 1.4. Design of lattice purloin………………………………………… 21 1.5. Slab roof design………………………………………………… 23 1.6. Weld design……………………………………………………… 25 2. Design of slab…………………………………………………………...26 2.1 Solid slab design…………………………………………………. 26 2.1.1 Depth determination……………………………………… 26 2.1.2 Loading…………………………………………………… 28 2.1.3 Analysis…………………………………………………… 29 2.1.4 Reinforcement design……………………………………… 38 2.2 Pre-cast slab design……………………………………………….39 2.2.1 Loading…………………………………………………… 40 2.2.2 Analysis and design……………………………………… 41 3. Design of stair………………………………………………………… 48 4. Frame analysis………………………………………………………… 55 4.1 Vertical load analysis…………………………………………… 55. 4.1.1 Solid slab………………………………………………… 55 4.1.2 Pre-cast slab……………………………………………… 59 4.2 Lateral load analysis…………………………………………… 62. 4.2.1 Earth quake analysis……………………………………… 62 4.2.2 Wind load analysis……………………………………… 64 4.3 Distribution of storey shear…………………………………… 67 4.4 Load combination……………………………………………… 83 5.Design of beam and column…………………………………………… 85 5.1 Beam design……………………………………………………85 5.1.1 Solid slab beams…………………………………………86 5.1.2 Pre-cast beams………………………………………… 93 5.1.3 Design of beams for shear and torsion………………… 99 5.2 Column design…………………………………………………104 5.2.1 Design procedure………………………………………. 104 5.2.2 Design of isolated columns…………………………… 111 5.2.3 Reinforcement design………………………………… 120 5.2.3.1 Solid slab column……………………………… 120 5.2.3.2 Pre-cast slab column…………………………… 121 6. Foundation design…………………………………………………… 124 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 2 6.1 Footing design……………………………………………… 125 6.2 Mat foundation design…………………………………………128 7. Cost estimation…………………………………………………………135 Conclusion and recommendation…………………………………… 138 References…………………………………………………………… 139 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 3 Acknowledgment We would like to express our sincere gratitude to our advisor Ato Kibrealem Mebratu for giving us critical advices throughout the project work. We would also like to thank our friend Fasil Meles for his material support. At last but not the least, we would like to thank to our parents for giving their countless material & moral support. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 4 Introduction Now a day’s it is being practical to choose types of structural members for different criteria’s especially economy after assuring safety. So in this particular project we have determined the cost variation for solid and pre-cast floor system. To do this we have passed many steps. This paper is prepared in partial fulfillment for the B.Sc. degree in civil engineering. The project is a structural design of a G+4 commercial building with solid and pre-cast slab systems with cost estimation & comparison for each staff. The building is located in Mekelle city. The structural design consists of the design of roof truss, Slabs, Staircase, Beams, Columns and foundation. The cost estimation comprises of cost for slabs, beams, columns and footings with their respective formwork. As can be seen from the architectural drawing, the floor arrangement is typical for all floors except some modification on the cantilever parts. For the solid type we first determine depth for deflection and made analysis for dead and live load as EBCS recommends. During the design of beams, columns and footings we have grouped them in reference to the stress they are bearing i.e. each group is taking relatively similar stress. Limit state design method has been adopted for the whole components. Ethiopian building code of standard EBCS 1, EBCS 2, EBCS 7 and EBCS 8, are referred for the design of the building. The roof truss, ribs and frame are analyzed using SAP 2000 V 9 for different combination of loads. And a combination with a critical effect is taken for sizing members and determination of rebar. Working drawings for beams, columns, footings stair case and floor slabs are prepared. And finally Bill of Quantity for Concrete work, Rebar and Footing are prepared. Generally in this project we have shown the basic steps for analysis and design of frame structures and we believe we have done a fabulous work which is almost accurate and we deserve a big hug. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 5 Specification Purpose – commercial building Approach- Limit state design method Material – Concrete – 25, class – I works Steel S – 300 deformed bars RHS for roof truss and purlin EGA- 300 for roof cover is used. Partial safety factors – concrete γc=1.5 Steel γs=1.15 Unit weight of concrete γc=24KN/m 3 Supporting ground condition = sandy gravel with bearing capacity of 560KPa Design Data and Materials Concrete fck = 0.8*25MPa =20MPa fctk= 0.21*fck 2/3 =1.547MPa fcd= 0.85*fck =11.33MPa γc fctd= fctk=1.032MPa γc Steel fyk= 300MPa fcd= 260.87MPa Design loads Fd= γf*Fk Where Fk = characteristics loads γf = partial safety factor for loads = 1.3 for dead loads = 1.6 for live loads Seismic condition Mekelle – Zone 4 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 6 1. Roof Design Roof layout 1.1 Wind load Analysis The roof is categorized according to EBCS-1,1995 table 2.1.3 under category-H :- roof not accessible except normal maintenance, repair, painting and minor repairs. From table 2.1.4 the roof will be sloping roof with category-H q k =0.25KN/m 2 Q k =1.0KN Characteristics wind load Wind pressure (EBCS-1, 1995 Art. 3.5) I. External wind pressure Wind pressure acting on external surface of the structure will be obtained from W e =q ref .C e (z e ) (C pe ) Where q ref =reference wind pressure = ρ/2 V ref 2 …………… (Art. 3.7.1) ρ= air density (from site with altitude above sea level) > 2000m (Mekelle) ρ=0.94 Kg/m 3 V ref = reference wind velocity V ref = C dir . Ctem .Calt .V ref o = 1*1*1*22m/s q ref = ρ/2 V ref 2 =0.5* 0.94Kg/m 3 *22 2 =227.48N/m 2 Ce=pressure coefficient that accounts the effect of terrain roughness, topography and height above the ground on the mean wind speed is defined as Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 7 C e (z) = ( ¸ ( ¸ + ) ( * ) ( * 7 1 * ) ( * ) ( 2 Z C Z C K Z C Z C t r T t r Where K T = terrain factor - For urban area in which at least 15° of the surface is covered with buildings by their average height is 15m. K T = 0.24 Cr(ze)= roughness coefficient = K T. ln.(z/z 0 ) for Z min ≤ Z Cr(ze)=Cr(Z min ) for Z<Z min Z 0 is the roughness length ( from table 3.2 Art. 3.8.3) Z min is the minimum height Z min =16m , Z0 =1m ( from table 3.2 Art. 3.8.3) Z= the height of the building at roof level Z=16.91m C r (Z e )=K T *ln | | ¹ | \ | o e Z Z = 0.24*ln(16.91/1) C r (Z e = 0.679 Ct(ze)= is the topography coefficient (Art. 3.8.4) and is unity or 1 for not topography unaffected zone Ce(ze)= 0.679 2 *1 2 *[ 1+(7* 0.24/(.679 *1))]= 1.602 C pe is the external pressure coefficient derived from appendix .A of EBCS-1, 1995 (Art. A.2.5 for duo pitch roofs) - the roof should be divided in to zones as shown below Case -1 when wind direction 0 0 I J H F G F 6.8 0.8m 26m 6.5m 13m 6.5m 2.6m 0.8m 2.6m Wind θ=0 0 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 8 Reference height Ze=h=16.91m e=b=26m Area F=16.9m 2 H= 20.8m 2 J=67.6m 2 G= 33.8m 2 I=20.8m 2 For duo pitch roof the external pressure coefficient is given on table A.4 EBCS 1, 1995 Wind ward side ( upwind face ) Pitch angle =15 0 Lee ward side α= 15 0 Since for A ≥ 10m 2 Cpe= Cpe10 (EBCS 1, 1995 Art. A.2.1) Case-2 when wind direction θ = 90 0 e/4=1.7m 1.7m 1.7m Θ=90 0 e/4 =1.7m Pitch angle Zone for wind ward direction θ=0 0 F G H I J Cpe, 10 Cpe, 10 Cpe, 10 Cpe, 1 0 Cpe, 1 0 15 o -0.9 -0.8 -0.3 -0.4 -1.0 0.2 0.2 0.2 _ _ F H I G G H I F Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 9 e/10=0.68m 2.72m 3.4m b=6.8m e= min b=6.8m e=6.8m 2h=33.82m Area: F = 1.156 m 2 G = 1.156m 2 H = 9.248 m 2 I = 11.56 m 2 External pressure coefficient Pitch angle Zone for wind direction θ = 90 0 F G H I Cpe, 10 Cpe, 1 Cpe, 10 Cpe, 1 Cpe, 10 Cpe, 10 15 0 -1.3 -2.0 -1.3 -2.0 -0.6 -0.5 W e = q ref Ce(z) cpe = 227.48*1.602 Cpe = 0.3644Cpe KN/m 2 = 0.3644 cpe K/m 2 Cpe = Cpe 1 , a <1 m 2 Cpe = Cpe 1 + ( Cpe 10 – Cpe 1 ) log A 10 for 1m 2 < A <10m 2 Cpe = Cpe 10 ; a<10 m 2 The calculation for the external pressure for each zones are calculated below Case1 θ=0 0 Zone F = 0.3644*-0.9= - 0.328KN/m 2 = 0.3644*0.2 = 0.073 KN/m 2 Zone G = 0.3644*- 0.8 = -0.292 KN/m 2 =0.3644*0.2 = 0.073 KN/m 2 Zone H =0.3644*-0.3 = -0.109 KN/m 2 = 0.3644*0.2 = 0.073 KN/m 2 Zone J = 0.3644*-1.0 = -0.3644 KN/m 2 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 10 Zone I = 0.3644 *-o.4 = -0.146 KN/m 2 Case 2 θ=90 0 Cpe for F Cpe = Cpe 1 + (Cpe 10 – Cpe 1 ) log A 10 = -2+ (-1.3+2) log 1.156 = -1.96 For G = -1.96 For H=-0.62 Zone F = 0.3644*-1.96 = -0.714 KN/m 2 Zone G = 0.3644*-1.96 = -0.714KN/m 2 Zone H = 0.3644*-0.62 = -0.226 KN/m 2 Zone I = 0.3644 *-0.5 = -0.184 KN/m 2 II. Internal wind pressure W i = q ref Ce(z i ) Cpi Where Cpi is the internal pressure coefficient obtained from appendix of EBCS 1-1995 (Art 1.2.9) -For closed buildings with internal partitions and opening windows the extreme values Cpi = 0.8 or Cpi = -0.5 - therefore the critical wind load on the roof Wi =0.3644 *0.8 = 0.292 K/m 2 or Wi =0.3644 *-0.5 = -0.182 K/m 2 Critical wind load Critical external wind pressure occurs On zone For G of case-2 =-0.714K/m 2 (-ve pressure) On zone F,H or G of case-1=0.073K/m 2 (+ve pressure) Net -ve wind pressure = -0.714-0.292 =-1.006 K/m 2 Net +ve wind pressure = 0.073+0.182 = 0.255K/m 2 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 11 Cpe calculation for the flat roof The roof is divided as shown below e/4=0.675 m wind 1.35m e/4 =0.675m e/10=0.27m 1.08m 5.3m b=2.7m e=min{b=2.7, 2h=2(19)=38 } e=2.7 Areas F=.675*.27=.182m 2 G=1.35*.27=.364m 2 H=1.08*2.7=2.916 I=2.7*5.3=14.31m 2 External pressure coefficients F G H I Sharp eves Cpe 10 Cpe 1 Cpe 10 Cpe 1 Cpe 10 Cpe 1 Cpe 10 -1.8 -2.5 -1.2 -2 -0.7 -1.2 ±0.2 Cr(z) roughness coefficient Cr(z)=K t ln(z/z 0 ) for z min < Z Z o is the roughness length z 0 =1m (EBCS-1,1995 table 3.2 Art 3.8.3 ) Z min is the minimum height =16m Z= the height of the building at roof level F H I G F Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 12 Z=19m Cr(z)=K t ln(z/z 0 ) =0.24*ln(19/1)=0.707 C t (z) is topography coefficient (Art 3.8.4) C t (z)=1 for not topography affected zone C e (Z e )=0.707 2 *1 2 *(1+(7*0.24)/(0.707*1)) =1.688 W e = q ref Ce(z) cpe =227.48N/m 2 (1.688)Cpe =384Cpe N/m 2 The calculation for the external pressure for each zones are calculated below Zone F = Cpe1= -2.5 Zone G = Cpe1=-2 Zone H = Cpe = Cpe 1 + ( Cpe 10 – Cpe 1 ) log A 10 for 1m 2 < A <10m 2 = -1.432 Zone F = 0.384*-2.5 = -0.86 KN/m 2 Zone G = 0.384*-o.2 = -0.768 KN/m 2 Zone H = 0.384*-1.432 = -0.55 KN/m 2 Zone I = 0.384*-o.2 = -0.768 KN/m 2 Internal wind pressure W i = 0.384*0.8=0.3072 W i = 0.384*-0.5=0.192 Critical wind load Occurs on zone F=-0.96KN/m 2 Occurs on zone I=0.077 KN/m 2 Net -ve wind pressure =-0.96 – 0.3072=-1.267 K/m 2 Net +ve wind pressure =0.077 + 0.192= 0.269 K/m 2 Point load calculation and determination of purloin spacing By looking into the wind pressures and referring “MAUAL OF COLD FORMED WELDED STRUCTURAL AD FURITURE STEEL TUBIG “(from kaliti steel industry) we have selected EGA-300 with thickness of 0.4mm. The possible loads on purloin -wind load - Self weight of EGA-300 - distributed imposed load Using EGA-300 of thickness 0.4mm from the previous table weight =3.14Kg/m or 0.0314 KN/m To determine per area load the weight is divided by width of the sheet =(0.0314KN/m)/0.823 DL=0.0382KN/m 2 (dead load) Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 13 40mm ST-20 Ø8 deformed bars The LL=0.25KN/m 2 Case-1 wind load (+ve pressure) WL=0.255KN/m 2 * cos15=0.246 KN/m 2 Load combination P d= 1.3DL + 1.6LL + WL P d =1.3(0.0382) + 1.6(0.25) + 0.246 =0.696 KN/m 2 So from table taking this load and thickness of EGA the purloin spacing is found to be = 1.75 m Case-2 Wind load (-ve) (suction) P d =1.3(0.0382) + 1.6(0.25) – 1.006 cos 15 = -0.522 KN/m 2 From table for this loading purloin spacing will be 2m Use purloin spacing of 1.75m. Total design load on purloin (lattice girder purloin) To get the total design load three cases are considered and compiled. Case-1 Dead Load + Live Load (concentrated) 205mm Assuming a ST-30 from table which is having a weight =0.0203 P d = 0.038cos15 *1.25+ 1KNcos15 *1 = 0.045KN/m + 0.966KN We have only considered for design the vertical perpendicular load to the EGA sheet. We neglect the effect of the load which is parallel since its effect is counter balance by its weight and the wind pressure. Case -2 Dead Load + Live Load (distributed) P d =0.038cos15 + 0.25*1m*cos15 = 0.036KN/m + 0.24KN/m Case -3 Dead Load + Wind Load 11@410m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 14 P d (+ve) =0.036KN/m + 0.255KN/m 2 *1m =0.036KN/m + 0.255KN/m P d (-ve) = 0.036KN/m – 1.006KN/m 2 *1m = 0.036KN/m – 1.006KN/m Reaction determination For case -1 0.966KN 0.0375KN/m 1.75m R R R=0.0394KN and 1KN at center 2R=.0788KN For case- 2 R=0.249 2R=0.498 For case- 3 R (+ve) = 0.218 2R = 0.436 R (-ve) = -0.665 2R = -1.33 We have used factor of safety 1- for live load 0.8 – for wind load 1.25 – for dead load, according to EBCS 1, 2, 3, 1995 1.2 Analysis of lattice purloin The analysis is done for each case and we include the effect of self weight when we do SAP analysis by defining the sections. And also we apply the distributed load on the purloin which is calculated as reaction from the EGA sheet. To apply this to the nodes we concentrate the distributed load to the center and divide it to the no of nodes, but the concentrated 1KN load is applied at the middle of the purloin as the code requires. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 15 Load Combination where:- COMB 1 = 1(DL) + 1.3(DL p ) + LL c DL r = dead load of roof COMB 2 = 1(DL + LLD) + 1.3DL P DL P = dead load of purloin COMB 3 = 1(DL r + WL +) + 1.3DL P LL c = live load concentrated COMB 4 = 1(DL r + WL - ) + 1.3DL P WL + = wind load +ve WL - = wind load -ve SAP result(max result) COMB 1 Top = -7.67 Diag = -1.27 Bott = 8.16 COMB 2 Top =-8.14 Diag = -1.9 Bott = 8.25 COMB 3 Top = -7.31 Diag tension = 1.79 Diag compression = -1.7 Bott = 7.4 Critical loads Top = -8.14 Diag tesion = 1.99 Diag compress = -1.9 Bott = 8.25 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 16 1.3 Design of truss The load on truss comprises of reaction from the purloins and the dead load of the truss itself. Determination of reactions of purloin Case -1 wind load i. positive wind Reaction from purloin R= 0.89 KN R x = 0.23KN R y = 0.86KN 2R x = 0.46KN 2R y = 1.72KN ii. egative (suction) wind Reaction from purloin R= -3.51 KN R x = -0.91KN R y = -3.39 KN 2R x = -1.82KN 2R y = -6.78KN Case-2 Live load (distributed) R= 1.09 KN 2R= 2.18KN Case -3 Dead load *EGA sheet R= .167KN 2R= 0.334KN * Purloin R=0.15KN 2R= 0.3 KN Total dead load = 0.334 + 0.3 = 0.634KN Load combinations COMB -1, 1.25DL + 0.8 WL + COMB – 2, 1.25DL + LL COMB – 3, 1.25DL + WL - Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 17 Analysis result COMB -1 upper chord max. comp.=10.96 KN Max. tens. = 6.42 KN Diagonal max. comp.= 3.04 KN Max. tens. =12.46 KN Bottom chord max. comp = 10.42KN Max. tens. = 11.11KN COMB -2 upper chord max. comp. = 10.22 KN Max. tens. = 7.3 KN Diagonal max. comp. = 3.38 KN Max. tens. = 11.27 KN Bottom chord max. comp = 11.09KN Max. Tens. = 10.63KN COMB -3 upper chord max. comp. = 11.66 KN Max. tens. = 22.78KN Diagonal max. comp.= 25.43 KN Max. tens. = 5.19 KN Bottom chord max. comp = 22.63KN Maximum tension upper chord = 22.78 K Diagonal = 12.46K Bottom chord = 20.09K Post = 13.88K Maximum compression upper chord = 22.78 K Diagonal = 25.43K Bottom chord = 20.09K Post = 6.99K DESIG OF TRUSS MEMBERS Material: - Fe 430 F y = 275 MPA F u = 430 MPA Diagonal steel member design Design actions Nsd =12.46 KN Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 18 Ncd = 25.43 KN Design is made for the longer member of the diagonal member i.e 1.13 1.13 1.13 tan 15 0 = y/2.26m y= 0.61m tan α = 0.61/1.13 α = 28.36 0 α y l = √1.13 + .61 = 1.28 m design is done for compression action and checked for tension actions section selection N brd = χβ A f y γ M1 where β A =1 For class 1-3 χ is reduction factor for the relevant buckling mode Assuming the reduction factor to be χ = 0.4 25.43= 0.4*1*A*275*10 3 /1.1 A= 2.543cm 2 So lets use ST-30 with t = 3mm SECTIOAL PROPERTIES H= 30mm b= 30mm 30mm t = 3mm A= 3.01 cm 2 I x =I z =3.5 cm 4 Wplx = wplz=2.34cm 3 30mm r x = r z =1.08 cm check class of the x-section d/t ≤ 90 E 2 24/3 = 8 ≤ 90*0.92 2 = 76.17 So the section is at least class 3 ⇒ β A =1 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 19 Determination of λ λ 1 = 86.4 λ x = λ z = l/r =1.28/0.0108 = 118.52 λ - x= λ - z= (118.52/86.4) * 1 1/2 =1.372 using curve c, x=0.354 check for N b,Rd b,Rd = 26.64K ≥ sd = 25.43K ⇒ ⇒⇒ ⇒OK! Check for tension Our design tension force N sd =12.46KN since our connection is welding A eff =A gross Resistance capacity N pl,Rd =Af y / γ M0 =3.01*10 -4 m 2 *275*10 6 N/m 2 /1.1=75.25KN N u,Rd =0.9Ae ff fy/ γ M2 =0.9*3.01*10 -4 m 2 *275*10 6 N/m 2 /1.25=59.59KN sd =25.43K< u,Rd =59.59K ⇒ ⇒⇒ ⇒ OK! Therefore the section is capable to carry both tension and compression forces Post steel member design Design actions Nsd =13.88 KN Ncd = 6.99 KN Check for tension for rolled section first we calculate the gross area N u,Rd =0.9Ae ff fy/ γ M2 ⇒ Ae ff = N u,Rd γ M1 /0.9fu=13.88*10 3 *1.25/0.9*275 =70mm 2 since our connection is welding A eff =A gross Let’s assume a square section ST-25 with the following sectional properties A=0.93 cm 2 I x =0.88cm 4 25mm S=0.71cm 3 r=0.97cm w=0.97Kg 25mm t=1mm In the absence of any information check for slenderness ratio λ≤ 180 ⇒ l e /r ≤ 180 ⇒ r≥910mm/180 =5.056mm =.5056 cm ⇒ r min =0.5056 For r min ≥0.5056 take nominal size of 25 * 25 r y =0.97 A= 0.93 cm 2 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 20 ⇒ 0.93cm 2 > 0.7cm 2 Therefore no need to check for N pl,Rd and N u,Rd Check for compression N sd =6.99KN l= 0.91m taking the above trial section d/t≤90E 23/1≤90*0.92 ⇒ 23≤82.8 OK! Therefore the class is at least class-3 and has no problem of local buckling thus β A =1 Determine λ - λ 1 = 86.4 λ x = λ z = l/r =910/97 =9.381 λ - x= λ - z= (9.381/86.4) * 1 1/2 =0.109 for cold formed RHS we use curve C and the value of the reduction factor x=2.784 calculate the design buckling resistance N brd = χβ A f y = 2.784*1*93*275/1.1 = 64.73KN>>6.99KN γ M1 Compression design Member design for upper and lower chords N sd =22.78KN use F e 430 f y =275 Determination of buckling length L=1.17m Selection of trial section by assuming trial value of reduction factor x=0.8 N brd = χβ A f y 22.78=0.8*1*A*275/1.1 A=113.9mm 2 γ M1 Trial section use ST-30 with cross sectional properties t=2mm I=2.72cm 4 30mm r=1.13cm 30mm Class of x-section d/t ≤ 90 E 2 24.5/2 = 12.25 ≤ 90*0.92 2 = 76.17 Therefore the class is at least class-3 and has no problem of local buckling thus β A =1 Determine λ - λ 1 = 86.4 λ x = λ z = l/r =1.17*10 2 /1.13 =103.54 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 21 λ - x= λ - z= (103.54/86.4) * 1 1/2 =1.2 for cold formed RHS we use curve C and the value of the reduction factor x=0.4338 calculate the design buckling resistance N brd = χβ A f y = 0.4338*1*214*275/1.1 = 23.21KN >22.78 OK! γ M1 1.4 Design of members for the lattice purlion CHECK FOR TESIO N u,Rd =0.9Ae ff fy/ γ M2 =0.9*214*275*/1.25 =42.37K >22.78K OK ! Therefore use ST-30 with thickness 2mm for lower and upper chords Top members N sd =8.14KN use F e 430 f y =275 Determination of buckling length L=0.41m Selection of trial section by assuming trial value of reduction factor x=0.45 N brd = χβ A f y 8.14=0.45*1*A*275/1.1 A=72.36mm 2 γ M1 Trial section use ST-20 with thickness t=1mm A=0.73cm 2 I=0.73 cm 4 20mm r=0.77cm t=1.2mm 20mm Class of x-section d/t ≤ 90 E 2 17/1.2 = 14.17 ≤ 90*0.92 2 = 76.17 Therefore the class is at least class-3 and has no problem of local buckling thus β A =1 Determine λ - λ 1 = 86.4 λ x = λ z = l/r =0.41*10 2 /0.77 =53.25 λ - x= λ - z= (53.25/86.4) * 1 1/2 =0.616 for cold formed RHS we use curve C and the value of the reduction factor x=0.7757 calculate the design buckling resistance N brd = χβ A f y = 0.7757*1*73*275/1.1 = 14.16K >8.14K OK! γ M1 Bottom members N sd =8.25KN(tension) take asection of ST-20 with thickness t=1mm A=0.73 I=0.73 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 22 r=0.77 N u,Rd =0.9Ae ff fy/ γ M2 =0.9*73*275/1.25 =14.45KN>8.25KN OK! DESIG ACTIOS FOR THE DIAGOAL MEMBERS N sd =1.99KN N cd =1.9KN Material Ø 8 ,S-300 f y =260.8 Check is done for both compression and tension actions . Check for compression resistance Φ8 deformed bar buckling length for pin –pin support l= 289.91mm class determination :-the class of the section is taken as class -2⇒βa=1 determination of slenderness ratio (λ) λ 1 =93.9є=93.9*0.92=86.4 λ x = λ y =l/r,r=√(I/A), A=πD 2 /4=π*0.8 2 /4=0.503cm 2 I=πr 4 /4=0.02cm 4 r= √(0.02/0.503)=0.199≅0.2cm λ x = λ y =28.91cm/0.2cm=144.55 λ - =[144.55/86.4]* βa 1/2 =1.673 using curve C , χ =0.2667 determination of buckling resistance N brd = χβ A f y =0.2677*0.503*10 -4 m 2 *260.87*10 6 N/m 2 γ M1 1.1 N brd= 3.193KN > Ncd=1.9KN OK ! Check for tensile capcity ,Nsd=1.99KN A eff =A gross N pl,rd =Af y =0.503*10 -4 m 2 *260.87*10 6 N/m 2 γ M1 1.1 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 23 =11.93KN >1.99KN OK! Nu,rd= 0.9A eff* f y = 0.9*0.503*10 -4 m 2 *260.87*10 6 N/m 2 γ M2 1.25 =9.45 KN > 1.99 KN OK! So the section can carry both tensile and compressive action 1.5 Slab roof Design For the design purpose we take the max positive pressure. The suction pressure can be easily counter balanced by the weight of the slab. Net positive wind pressure = 0.269 KN Changing into equivalent rectangular section using Bare’s equation Equivalent rectangular section c/a= 3.9/5= 0.78 > 0.25 ar =2/3[(2c +a)*a/(a +c)] = 4.79 4.79 br = b-{a*(a – c)} 6*(a + c) br= 2.599≅2.6 2.6 Depth determination [From EBCS 2-1995 Art. 5.2.3] for ps d= [0.4 + 0.6*f yk ]le =0.85*2600/24= 108.32 mm 400 βa For pc d= 0.85*1650 =116.87 mm 12 Use for both slabs overall depth, D= d +15 +5= 116.87 + 15 +5=136.87 Use D= 150 mm Loading :- From water tank having a capacity of 5000 lit = 5m 3 ps p c Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 24 unit wt of water= 9.81KN/ m 3 P= 5*9.81= 49.05 KN Distributed load =49.05 KN =4.08 KN/ m 2 12.015 m 2 Self weight wt of slab = 0.15* 24KN/ m 3 =3.6 KN/m 2 Live load =0.5KN/ m 2 Wind load =0.269 KN/ m 2 P d =1.25DL+ LL + 0.8WL = 1.25*7.68 +0.5 + 0.8*.269 P d =10.31 KN/ m 2 Design of slab For p s ly/lx= 1.84 s.c =8 α xf =0.0922 α ys = 0.0584 α yf = 0.045 Mi = α i *Pd*lx 2 M ys =0.0584* 10.31*2.7 2 = 4.36 KN-m M yf =0.044*10.31*2.7 2 =3.31 KN-m M xf = 0.0922*10.31*2.7 2 = 6.85 KN-m Moment for cantilever slab ps ,pd=5.21KN/m M= 0.5*wl 2 = 5.21*1.65^2*0.5 =7.099 KN-m Moment adjustment Reinforcement calculation Moment k m k s As 7.091 20.35 3.96 216 A s min =ρmin*b*d= 0.002*1000*130=260mm 2 A s min<As ⇒ use all slab minimum reinforcement =260mm 2 S=1000*50.3 = 193.46mm 260 Use Ø8 c/c 190mm Reinforcement detail See the reinforcement detail in the auto cad file Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 25 Design of roof slab 2 is the same as that of the previous roof slab of spacing Ø8 c/c 190mm 1.3 1.65 3.9 1.6 Weld design For the upper chord Design actions Nsd=22.78KN Ncom=22.7KN Arrangement of the weld : section selection to be upper chord ST-30 with 2mm thickness Fe=430 s 10mm,Ls available =30mm so lets take S=10mm a=0.70750.707*10mm=7.07mm lets use fy=270Mpa fu=430Mpa Fw, Rd =fvw,da fvw,d=0.63fy/gmw 0.65fu/gmw where Fw,Rd=design strength of the fillet weld per unit 0f length fvw,d=design shear strength gmw=1.25 fvw,d =0.63*270Mpa/1.25 =136.08Mpa 0.65fu/gmw Fw,Rd= 136.08 N/mm2*7.07mm=962.085N/mm But Nsd/2Fw,Rd =22.78*10^3N/2*962.085N/mm =11.84mm So the available Ls=30mm >the needed Ls=11.84mm So use fillet weld with length Ls=30mm and also use connected plate with Fe-430 and thickness 2mm. and also use this conection for all other joints since their design action is less than this. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 26 2. DESIG OF SLAB 2.1-SOLID SLAB DESIG Sample floor layout 2.1.1 Depth determination The effective depth requirement for deflection can be calculated by using the following formula (EBCS- 2; 1995 Art 5.2.3) D ≥ (0.4+0.6 fyk) Le 400 βa Where fyk = is the characteristics strength of the reinforcement Le = is the effective span and for two span the shorter span Βa = is the appropriate constant from table 5.1 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 27 Fourth floor slab Panel 1 Ly/Lx = 1 Lx = 5000mm Ly = 5000 mm βa= 45 d ≥ (0.4+0.6*300) 5000 = 94.44mm 400 45 Panel 4 Lx = 5000mm Ly = 5000 mm Ly/Lx = 1 βa= 40 d ≥ (0.4+0.6*300) 5000 = 106.25mm Panel 8 400 40 Ly/Lx = 5/1.65 = 3.03 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 28 βa= 12 d ≥ (0.4+0.6*300) 1650 = 116.88mm 400 12 First and second floor A = ½ ЛD 2 = 0.5*Л*3.8 2 = 5.67m 2 4 4 βa= 12 Equivalent rectangle d ≥ (0.4+0.6*300) 1530 = 108.37mm 400 12 Comparing all the above critical panels d ≥ 116.88mm Over all depth D = 116.88 + 10 + 15 = 141.88mm Use D = 150 mm Actual depth d =150 – 10 – 15 = 125mm For first and second floor slabs 2.1.2 Loading Unit weight -pvc – 16 KN/m 3 -Cement – 23 KN/m 3 -Terrazzo - 23 KN/m 3 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 29 - Concrete - 24 KN/m 3 Dead load -pvc tile = 16 *0.008 = 0.128 KN/m 2 -Cement screed = 23 *0.03 = 0.69 KN/m 2 -terrazzo tile = 23 *0.02 = 0.46 KN/m 2 -slab concrete = 24*0.15 = 3.6 KN/m 2 -plaster = 23*0.03 = 0.69 KN/m 2 -partition load = 1,2 KN/m 2 DL 1 = (PVC floor finish) = (0.128 + 0.69 + 3.6 + 0.69 + 1.2) KN/m 2 = 6.31 KN/m 2 DL 2 = (Terrazzo tile floor finish = (0.46 + 0.69 + 3.6 + 0.69 + 1.2) KN/m 2 = 6.64 KN/m 2 Live load (from EBCS-1-1995 Table 2.9 and 2.10) Category D 1 (Areas in general retail shops) = 5 KN/m 2 For stairs = 3 KN/m 2 Balconies = 4 KN/m 2 2.1.3 Analysis Design moment calculation The support and span moment of simply supported (external edges) or fully fixed (contentious edges) are calculated as Mi = αi Pd L X 2 Where; Mi = the design moment per unit width at point of reference Pd = the uniformly distributed load αi = the coefficient given in table A-1 as function of aspect ratio (Ly/Lx) and Support condition Lx = shorter span of the panel Ly = longer span of the panel Fig. Where; s = support f = field (span) x = direction of shorter span y = direction of longer span Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 30 Panel 1 Ly/Lx =1 αxs = 0.039 αys = 0.039 αxs = 0.029 αyf = 0.029 DL 1 = 6.31 KN/m 2 LL = 5 KN/m 2 Pd = ((1.3* DL 1 ) + (1.6*LL)) ((1.3* 6.31) + (1.6*5)) KN/m 2 = 16.203 KN/m 2 Mi = αi Pd L X 2 Mys = Mxs = 0.039*16.203 KN/m 2 *(5m) 2 = 15.8 KN.m Myf = Mxf = 0.029*16.203 KN/m 2 *(5m) 2 = 11.75 KN.m Panel 2,3 and 4 Ly/Lx =1 αxs = 0.032 αys = 0.032 αxs = 0.024 αyf = 0.024 DL 1 = 6.31 KN/m 2 LL = 5 KN/m 2 Pd =16.203KN/m 2 Mys = Mxs = 0.032*16.203 KN/m 2 *(5m) 2 = 12.96 KN.m Myf = Mxf = 0.024*16.203 KN/m 2 *(5m) 2 = 9.72 KN.m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 31 Panel 5 Ly/Lx =1 αxs = αys = 0.039 αxs = αyf = 0.03 Pd =16.203KN/m 2 Mys = Mxs = 0.039*16.203 KN/m 2 *(5m) 2 = 15.8 KN.m Myf = Mxf = 0.03*16.203 KN/m 2 *(5m) 2 = 12.15 KN.m Panel 6 Equivalent rectangular section Let’s take equivalent rectangular area Fig Ly/Lx = 5/0.905 = 5.52m one way slab Pd =16.203KN/m 2 * 1m = 16.203KN/m Fig Mxs = Pd Lx 2 2 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 32 = 16.203KN/m * (0,905m) 2 = 6.64KN.m 2 Panel 7 Equivalent rectangular section A = ½ ЛD 2 = 0.5*Л*3.8 2 = 5.67m 2 4 4 Equivalent rectangular area Ly*Lx = 5.67m 2 Lx = 5.67m 2 = 1.49m 3.8m Ly/Lx = 3.8/1.49 = 2.55 one way slab Pd = 16 203 KN/m Fig Mxs = Pd Lx 2 = 16.203KN/m* (1.49m) 2 = 17.986KN.m 2 2 Panel 8 Equivalent rectangular area using Bales theorem a r =2*((2c + a) * a ) 3 (a + c) b r = b – a(a – c) b(a + c) so, a r =2*((2* 1.2m + 1.9m) * 1.9 ) = 1.756 m 3 (1.9m + 1.2m) b r = 5 m – 1.9m(1.9m – 1.2m) = 4.57 m 5(1.9m + 1.2 m) Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 33 Ly/Lx = 4.57/ 1.756 = 2.6 > 2 one way slab Pd = 16.203KN/m Mys = Pd Lx 2 = 16.203KN/m *(1.756m) 2 = 24.81KN.m 2 2 Panel 9 Equivalent rectangle a r =2*((2c + a) * a ) 3 (a + c) b r = b – a(a – c) b(a + c) so, a r =2*((2* 0.525m + 1.2m) * 1.2 ) = 1.043 m 3 (1.2m + 0.525m) b r = 3.1 m – 1.2m(1.2m – 0.525m) = 2.95 m 3.1(1.2m + 0.525 m) Ly/Lx = 2.95m/ 1.043m = 2.8 > 2 one way slab Pd = 16.203KN/m Mys = Pd Ly 2 = 16.203KN/m *(1.043m) 2 = 8.8132KN.m 2 Panel 10 Lx/Ly = 3.4m/ 1.35m = 2.52 > 2 one way slab Pd = 16.203KN/m Mys = Pd Ly 2 = 16.203KN/m *(1.35m) 2 = 14.76KN.m 2 2 Panel 11 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 34 Equvalent rectangular area Ly/Lx = 3.25m/ 1.4m = 2.62 > 2 one way slab Pd = 16.203KN/m Mys = Pd Ly 2 = 16.203KN/m *(1.4m) 2 = 15.88 KN.m 2 2 Panel 12’ 13,14 and 15 Ly/Lx = 5m/ 1.55m = 3.23 > 2 one way slab Pd 2 = 16.632 KN/m 2 Mys = Pd 2 Ly 2 = 16.632KN/m *(1.55m) 2 = 19.98 KN.m 2 2 Panel 17 Pd 2 = 16.632 KN/m 2 Mys = Pd 2 Ly 2 = 16.632KN/m *(1.55m) 2 = 19.98 KN.m 2 2 Panel 18 Ly/Lx = 4.65m/ 1.1m = 4.23 > 2 one way slab DL 2 = 6.31 KN/m 2 LL = 5 KN/m 2 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 35 Pd = ((1.3* DL 1 ) + (1.6*LL)) ((1.3* 6.31) + (1.6*5)) KN/m 2 = 16.203 KN/m 2 For one meter strip Pd =16.203KN/m Mys = Pd Lx 2 = 16.203KN/m *(1.1m) 2 = 2.45 KN.m 8 8 Panel 1 Is cantilever slab with length of 1.03m Pd 2 = 16.632 KN/m 2 Mys = Pd Ly 2 = 16.632KN/m *(1.03m) 2 = 8.82KN.m 2 2 Adjustment of support and span moment Support adjustment Between 1&4 and 4&5 M R = 15.8 KN.m M = 12.96 KN.m ΛM = 15.8 – 12.96 = 2.84 ΛM = 2.84 *100% = 17.97% < 20% ; Then M d =15.8 +12.86 = 14.38KN/m 2 Between 5&6 M R = 6.64 KN.m M L = 15.8 KN.m M d = 15.8 KN. m ; Since it is cantilever Span moment Panel 1&5 Mxfd = Mxf + Cx ΛM = 11.75 + 0.38*(2.84) =12.83 KN.m Panel 5 Mxfd = Mxf + Cx ΛM = 12.15 + 0.38*(2.84) = 13.58 KN.m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 36 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 37 After moment adjustment check depth for flexure, taking the maximum moment M max =24.81 KN-m From material used the design constants f cd =11.33Mpa c 1 =0.0858 f yd = 260.87Mpa c 2 =3074 m= 28.78 ρ b = 0.8E cu * f cd ( E cu + E yd ) f yd b= 1000mm, E cu = 0.0035, E yd = (260.87/(200*10 3 ) = 0.0013 ρ b =0.8*0.0035*11.33*0.75 = ( 0.0035 + 0.0013) *260.87 Check for effective depth d=79.89< 125 mm OK!! Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 38 Reinforcement design Reinforcement detail NB: The full reinforcement details are attached with AutoCAD files. Moment Km Ks As Asmin S calculated S provided 19.98 35.76 4.108 656.62 212.5 119.55 Φ10c/c110 12.83 28.66 4.03 413.64 212.5 189.78 Φ10c/c180 17.986 33.92 4.09 588.50 212.5 133.39 Φ10c/c130 24.81 39.85 4.16 825.68 212.5 95.07 Φ10c/c90 14.38 30.34 4.05 465.91 212.5 168.49 Φ10c/c160 9.72 24.94 4 311.04 212.5 252.38 Φ10c/c250 12.96 28.8 4.04 418.87 212.5 187.41 Φ10c/c180 14.76 30.73 4.06 479.40 212.5 163.74 Φ10c/c160 15.88 31.88 4.07 517.05 212.5 151.82 Φ10c/c150 13.58 29.48 4.05 439.99 212.5 178.41 Φ10c/c170 2.45 12.52 3.87 75.85 212.5 236.71 Φ8c/c220 8.813 23.75 3.982 280.75 212.5 179.16 Φ8c/c170 6.64 20.65 3.961 210.41 212.5 236.71 Φ8c/c220 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 39 2.2Pre-cast slab design General Precast beam slab system is a system of slab construction in which reinforced concrete precast beam elements, with their latticed reinforcement bars projected out, are used. During construction, these beam elements will be placed at certain intervals, to accommodate hollow concrete block. These blocks of specified dimensions are placed along these prefabricated beams and across the span of these elements in a similar fashion as in the case of ribbed slab construction. Concrete will be casted over the blocks and the beam elements. The projected reinforcement bars from the beam elements are used as an anchorage for the concrete, in addition to their main purpose, i.e. shear resistance. The beam elements, together with the blocks, act as formwork for the concrete casted. In addition the beam elements will acts as flexural members to carry the loads until the cast in-situ concrete attains its full strength The pre-cast beam span is 5m and it is only one type therefore from the GTZ technical manual- II the section is recommended as follows And also the cross-sectional dimensions of the HCB is given below 220mm 550mm Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 40 Design constants and assumptions Material properties C-25 fcd = 0.85 fck in compression fctd=fck/gc in tension Therefore fcd=11.33 Mpa fctd=1.0 Mpa steel S-300 fyd=fyk/gs Hollow concrete block (HCB) g=14KN/m 2 Design of pre-cast beam element For starting use Ø10 for top members and Ø8 for diagonal members 2.2.1Loading A) Initial condition Dead load – pre-cast beam and Hollow Block Pre-cast beam = 1.3*0.12*0.08*24=0.299KN/m Hollow block = 1.3*0.069*14=1.256 KN/m Live load = 1.6*2*0.625= 2KN/m q d = 0.299+1.256+2= 3.6K/m B)Final condition Dead load – hf 625mm 60 220mm 280 625mm Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 41 Pre-cast beam 1.3*.12*.08*24 = .299 KN/m Hollow Block 1.3*.069*14 = 1.256 KN/m Cast In situ Concrete 1.3(.625*.06+2(.14*0.06))*24= 1.69KN/m Concrete Floor finishes and partition wall Partition load=0.625*1.2*1.3=0.975 KN/m Plastering = 1.3*23*0.03*0.625=0.561KN/m PVC floor finish = 1.3*16*0.002*0.625=0.104KN/m Terrazzo floor finish=1.3*23*0.02*0.625=0.374KN/m Live load =5*0.625*1.6=5KN/m q d1 = 0.299+1.256+1.69+0.975+0.561+0.104+5=9.884K/m (for PVC floor finish) q d2 = 0.299+1.256+1.69+0.975+0.561+0.374+5=10.154K/m (for terrazzo floor finish) 2.2.2Analysis and design Final condition Mmax=wl 2 /8=10.154*5 2 /8=31.73KNm 10.154K/m The minimum depth required for deflection d≥ (.4+.6*300/400)5000/20=212.5 Actual d=D-Ø/2 –cover = 280-16/2-15=257>212.5 OK!!! be ≤ bw + le/5=120+5000/5=1120 actual=625 be = 625mm Determination of neutral axis ρ= ½ c 1 -√(c 1 2 -4M/(c 2 bd 2 )) ρ= 0.00305 m=fyd/fcd*0.8=260.87/0.8*11.33=28.78 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 42 x= ρmd=0.00305*28.78*257=22.56 y=0.8X=0.8*22.56=18.05<60 the section is rectangular with b=be=625 M=31.73 using design table Km=(√(m/b))/d = 27.72 Ks=4.024 As=KsM/d=4.024*31.73/0.257=496.81mm 2 Therefore use 2Ø20 Check for shear Acting shear Vsd=pdl/2=10.173*5/2=25.385KN The shear force V C carried by the concrete in members with out significant axial forces shall be taken as V C = 0.25f ctd K 1 K 2 b w d Where K 1 = 1.6-d =1.6-0.257=1.343 K 2 = (1+50ρ)=1.1` fctd= 1 Mpa Vc=0.25*1*1.343*1.1*120*257=11.41 KN For T-section Vc=1.1*11.41=12.55KN V RD =0.25 *fcd *b w *d=0.25*11.33*120*257=87.354KN Vs=Vsd-Vc=25.385-12.55=12.84KN S= Assume spacing =150mm S=2*50.3*260.87*(257-20)(cos72.6+sin72.6)/12.84 S=289.69mm 240mm Assume s=200mm 67.3 0 β= α=67.38 0 200mm S=2*50.3*260.87*(257-20)(cos67.38+sin67.38)/12.84 S=372.66mm Let’s use Ø8 c/c 200mm Check for initial condition 3-dimentional truss model analysis Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 43 q=3.6KN/m ∆l=200mm p=q*∆l/2 = 3.6KN/m*0.2m/2=0.36KN p/2 =0.18K Location of neutral axis Assume the neutral axis depth x to be above the concrete section (i.e x<180) Moment about N.A As1*x= As2*(235-X) = 628*(235-x) 78.5x = 147.58-628.x X=208.89mm So the concrete part that carry the compression load is (120*(208.89-200)=1066.8mm 2 Force resisted by concrete section is =11.33Mpa*1066.8mm 2 =12.086KN To get the compressive force acting on steel members we have modeled the pre-cast beam as 3-D truss whose members are pin connected and also we reduced the acting compressive force which is resisted by the concrete section. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 44 SAP result (1) Max. Tension =6.13 KN (2) Max. Compression = 11.91 KN (3) Max. compression =2.72KN Max. Tension = 1.99KN (4) Max. Tension 0.3KN Capacity check for members According to EBCS-3,1995 N t,rd ≤ A*f y / γ M1 γ M1 = 1.1 0.9 A eff f U / γ M2 γ M2 = 1.25 Since the Top reinforcement bar is ф10 using Fe430 N t,rd 78.5*260.87/1.1=18.62 KN 0.9*78.5*300/1.25=16.96 KN So take Ntrd = 16.96KN 2 1 3 4 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 45 And also the buckling resistance of axial loaded compression members is Nb,rd= N brd = χβ A f y / γ M1 Design axial load Nsd=11.91KN Buckling length l=0.2m Determination of λ λ 1 = 89.12 λ x = λ z = l/r =80 β A =1 λ - x= λ - z= (80/89.12) * 1 1/2 =0.898 using curve c, x=0.5998 check for N b,Rd N b,Rd = 11.16KN is alittle bit less than Nsd=11.91 Since the centroid of the section is below the concrete section the concrete section carrys alittle bit so the total Nsd doesn’t apply on the steel so it’s safe. But for the safety purpose we insert the formwork at the mid span. Case 1 when all the LL at the 1 st span and to get the maximum span moment. 2.5m 2.5m Case 2 when all live loads are distributed over the whole span For case 1 P 1 =1/2*(3.6*0.2)=0.36 KN P 1 ’=1/2(1.6*0.2)=0.16KN For case 2 P 2 =0.36KN Case 1 The maximum tension for(2)=1.94KN Compression for (2) =2.37KN Diagonal (3) comp.= 1.25KN Tension= 1.15KN Bott (1) comp.=1.5KN Tens. =0.57KN For case 2 For (2) max. tens. 2.57KN Max.comp. 2.02KN For bott (1) =1.2KN tension Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 46 For diagonal comp. (3) =1.02 KN Tens. =0.76KN For bottom horizontal comp.=0.1KN Critical actions from the two cases Tension = 2.57KN Compression = 2.37KN Ntrd = 0.9A eff *f yd / γ M2 = 0.9*78.5*300/1.25 = 16.96 Ntsd=2.57< Ntrd=16.96 KN For compression x=0.5998 Nbrd= χβ A f y / γ M1 =0.5998*78.5*260.87/1.1 Nbrd=11.17 KN Nsd=2.37<11.17KN OK! For diagonal members Maximum tension =1.15 Maximum compression=1.25 Ntrd = 0.9A eff *f yd / γ M2 =0.9*50.3*300/1.25=10.865KN Ntsd=1.15<Ntrd=10.865KN OK! For compression Buckling length =0.26 λ 1 = 89.92 λ x = λ z = l/r =130 β A =1 λ - x= λ - z= (130/89.92) * 1 1/2 =1.446 using curve c, x=0.332 Nbrd= χβ A f y / γ M1 =0.332*50.3*260.87/1.1 Nbrd=3.96 KN Nsd=1.25<3.96KN =Ntrd OK! For bottom longitudinal reinforcement the max. Compression and tension are in significant, thus no need of checking. Hence use ф10 fo the top members and ф8 for the diagonal members. 1Ф10 Ф8 2 Ф20 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 47 NB: The irregular solid slabs are already designed in the solid slab design part. Sample pre-cast slab layout and reinforcement detail B:- the remaining details are attached with AutoCAD files Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 48 3. Design of stair DESIG OF STAIR 1 No of risers in one flight =9 Risers height =0.17 Length of thread =0.3 tanø = 0.17 =29.5 0 0.3 1.65m 2.4m 2.15m Depth required for deflection From survisability limt state(EBCS-2 ,1995-Art 5-2-3) d = (0.4+0.6 *fyk)/400 (le/βa ) fyk=300mpa, βa =24 (end span) le=5400 d=0.85*4550/24=161.1 D =161.1 + 15 +8 =184.1 take D =190 for end slab Actual d=167mm Loading (taking 1m strip) Dead load on the stair RC slab (inclined) = 0.91*1*24 =5.24KN/m Cos 29.5 Due to steps per meter width = 0.3*0.17*1*24*1 = 2.04KN/m 2 0.3 Floor Finishing Terrazzo tile (2cm) = 0.02*23 = 0.46KN/m 2 Cement Screed (2 cm) = 0.03*23 = 0.69 KN/m 2 Plastering =23*0.03=0.69 KN/m 2 On the tread = (0.46+0.69)*1=1.15KN/m On the riser = 1.15*0.17*1*9/2.4 =0.73KN/m Dead load on stair = (5.24+2.04+1.15+0.73+0.69/cos29.5) = 9.95 KN/m Dead load on the landing portion Landing Slab = 0.19*1*24 = 4.56KN/m Plastering = 0.3*1*23 = 0.69KN/m Cement screed = 0.69*1 = 0.69KN/m Terrazzo tile =23*0.02*1 = 0.46KN/m 6.4KN/m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 49 Total load on the landing = 6.4KN/m Live Load On stair category A, qt = 3 KN/m 2 (EBCS-2,1995 table 2.10) Total live load on stair case = 3 KN/m 2 * 1m = 3 KN/m Design Load On stair, DL 1 = 1.3*9.95+1.6*3 = 17.735 KN/m On landing, DL 2 = 1.3*6.4+1.6*3 = 13.12 KN/m Shear Force Diagram Bending moment Diagram Reinforcement Design Reinforcements are calculated using design table At support Mdes =17.86KNm Km = √(M/b) = √(17.86/1) = 25.3 d 0.167 Ks =3.994 Hence As = Ks * M = 3.994 * 17.86 = 427.12 mm 2 d 0.167 Asmin=0.002*1000*167=334 mm 2 Spacing S = as * b = 113 * 1000 = 264.56mm As 427.12 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 50 Spacing provided Ф12 c/c 260 Transverse reinforcement As=0.2As=0.2*427.12=85.424 S = as * b = 50.3 * 1000 = 588.8mm As 85.424 Smax 2D =380 350 Use Ф8 c/c 350mm Span moment m=31.65KN.m Km= =33.68 Ks=4.087 As=Ks m/d =4.087*31.63/0.167 =774.05 mm 2 S= = =145.98 use ø12 c/c 140 Distribution reinforcement A=0.2As=154.81 S= = =324.91mm use ø8 c/c 200 Design of Stair 2 No of risers in one flight =14 Risers height =2.3/14=0.165 Length of thread =0.3 tanø = 0.165 =28.81 0 0.3 3.9m 2.15 Depth required for deflection d= (0.4+0.6fyk) le/ba fyk=300mpa, ba=24 400 le=6050 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 51 d=(0.4+0.6*300)*6050 =214.27mm 400 24 D=214.27+15+8 = 237.27mm Use D=240mm d=240-15-8=217mm Loading (taking 1m strip) Dead load on the stair RC slab (inclined) = 0.24*1*24 =6.574KN/m Cos 28.81 Due to steps per meter width = 0.3*0.165*1*24*1 = 1.98KN/m 2 0.3 Floor Finishing Terrazzo tile (2cm) = 0.02*23 = 0.46KN/m 2 Cement Screed (2 cm) = 0.03*23 = 0.69 KN/m 2 Plastering =23*0.03=0.69 KN/m 2 On the tread = (0.46+0.69)*1=1.15KN/m On the riser = 1.15*0.165*1*14/3.9 =0.681KN/m Dead load on stair = (6.574+1.98+0.681+1.15+0.69/cos28.81) = 11.17 KN/m Dead load on the landing portion Landing Slab = 0.24*1*24 = 5.76KN/m Plastering = 0.3*1*23 = 0.69KN/m Cement screed = 0.69*1 = 0.69KN/m Terrazzo tile =23*0.02*1 = 0.46KN/m 7.6KN/m Total load on the landing = 7.6KN/m Live Load On stair category A, qt = 3 KN/m 2 (EBCS-2,1995 table 2.10) Total live load on stair case = 3 KN/m 2 * 1m = 3 KN/m Design Load On stair, DL 1 = 1.3*11.17+1.6*3 = 19.321 KN/m On landing, DL 2 = 1.3*7.6+1.6*3 = 14.68 KN/m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 52 Shear Force Diagram Bending moment Diagram Reinforcement Design Reinforcements are calculated using design table At span Mdes =82.93KNm Km = √(M/b) = √(82.93/1) = 41.97 d 0.217 Ks =4.2094 Hence As = Ks * M = 4.2094 * 82.93 = 1608.69 mm 2 d 0.217 Spacing S = as * b = 201 * 1000 = 124.95mm As 1608.69 Spacing provided Ф16 c/c 120 Transverse reinforcement Asmin =0.002*1000*217= 434mm 2 Asmin =0.2As=0.2*1608.69=321.738mm 2 Asmin=434 mm 2 S = as * b = 50.3 * 1000 = 156.34mm As 434 Use Ф8 c/c 150mm Check for shear Vc = 0.25fctdK 1 K 2 bwd d=240.15-8=217mm Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 53 Where: K1 = (1+50ρ)=1.37 ≤ 2.0 K2=1.6-d=1.383 ≥ 1.0 (d in meter) Fctd=(0.1(25/1.25)^2/3)/1.5=1.032 Vc=0.25*1.032*1.383*1.37*1000*217 Vc=106.07 KN Vsd=56.67<106.07KN OK! Staircase for precast and solid slab which is continuous with the slab 2.4m 2.15m No of risers in one flight =9 Risers height =2.3/14=0.17 Length of thread =0.3 tanø = 0.165 =29.81 0 0.3 Depth required for deflection d= (0.4+0.6fyk) le/ba fyk=300mpa, ba=24 400 le=6050 d=(0.4+0.6*300)*4550 =161.1mm 400 24 D=161.1+15+8=184.1 take D=190mm Loading on one meter strip from the previous cases On stair =17.35 KN/m On landing = 13.12 KN/ m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 54 Shear force diagram Bending moment diagram Reinforcement design At the support m=39.63KN.m d=190-15-8=167mm Km= =37.69 Ks=4.15 As=Ks m/d =4.15*39.63/0.167 =984.82 mm 2 >Asmin=334mm 2 S= = =114.74 use ø12 c/c 110mm Since the moment on the span is minimum the design is governed by top reinforcement For span use ø12 c/c 110mm Transverse reinforcement Asmin Use ø8 c/c 350mm Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 55 4. FRAME AALYSIS 4.1Vertical load analysis The load on the beam includes self weight, load from the wall on it, transferred load from slab, Live load on the beam. According to EBCS-2, 1995(Art. 3.4) the load transferred to beams supporting a two way slab distribution on a supporting beam is shown in figure 4.5 of EBCS-2, 1995 4.1.1Solid slabs The design loads on beams supporting the solid slabs spanning in two directions right angles supporting uniformly distributed loads may be assessed by using EBCS-2 1995 Art A.3.8 The equn. Is given by the formula, V=β v P d L x β v is a constant that depends on the support condition of the panel and L x is the shortest span of the panel. The assumed distribution of the load on the supporting beam is shown. Which assures that the load is assumed to be transferred only to 75% of the beam length. To convert such a distribution to uniformly distributed load through out the span, we compared the fixed end moment equation value with the span moment equation value and we took the maximum. From fixed end moment Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 56 Fixed end moment: M F AB = -M F BA =W*S/24(3L 2 -S 2 ) Span moment: M S =0.75L/2*WL/2-0.076WL 2 + (0.75L/2) 2 *W/2. For the equivalent beam having UDL W , and length L W , Fixed end moment: M F AB =-M F BA = W , L 2 /12 Span moment: M S = W , L 2 /24 Equating the fixed end moment equation W*S/24(3L 2 -S 2 ) = W , L 2 /12 W , = 0.914W Equating the span moment equations: 0.75L/2*WL/2-0.076WL 2 + (0.75L/2) 2 *W/2 = W , L 2 /24 W , =0.984W Taking weighted moment distribution factor =0.914+0.984/2=0.949 W= 0.949Vs load transfer to beams 1st and 2nd βvx βvy panel type of panel Lx Ly Ly/LX βvxc βvxD βvyc βvyD Pd Lx vxc vxD vyc 1&5 3 5 5 1 0.36 0.24 0.36 16.2 5 27.68 18 27.68 2,3,4 1 5 5 1 0.33 0.24 0.33 16.2 5 25.37 25.37 p6 cantilever 16.2 0.9 14.66 p7 cantilever 16.2 1.5 24.14 p8 cantilever 16.2 1.8 28.45 p9 cantilever 16.2 1 16.9 p10 cantilever 16.2 1.4 21.87 p11 cantilever 16.2 1.4 22.68 p12 cantilever 16.63 1.6 25.78 p13,14,15 cantilever 16.63 1.6 25.78 p16 cantilever 13.12 1.7 20.25 p17 cantilever 16.63 1.6 25.78 p18 cantilever 16.2 1.1 17.82 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 57 Vertical load on the frames AXIS-3 AXIS-4 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 58 AXIS-A Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 59 4.1.2 Pre-cast slab The total loads on beams include slab load, wall loads and reactions from cantilevers. The load distribution from slab to beam is a serious of concentrated loads AXIS-A Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 60 AXIS-E Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 61 AXIS-3 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 62 4.2 Lateral Load Analysis The lateral loads to be considered are: a) earthquake analysis and b) wind load analysis The analysis for both will be done and the critical load condition will be considered for the design. 4.2.1 Earthquake Analysis Determination of base shear According to EBCS-8, 1995 static method of analysis is used The seismic base shear force F b for each main direction is determined by the equation. F b = S d (T) W T Where S d (T) = ordinate of the design spectrum at period T and is given by: = α β γ The parameter α is the ratio of the design bed rock acceleration to acceleration of gravity, g, and given by: α = α o I α 0 = the bed rock acceleration ratio for the site and depends on the seismic zone as given in table 1.1 of EBCS-8, 1995. For zone 4, building location in mekelle, αo=0.10 I = the importance factor of the building i.e. 1.4 for important buildings such as hospital and 1.2 for buildings whose seismic resistance is of importance in view of the consequences associated with a collapse e.g. schools, assembly halls, cultural institutions etc. (Table 2.4, EBCS-1995) use I = 1.2 The parameter β is the design response factor for the site and is given by: β = 1.2 S ≤ 2.5 T 1 2/3 S = site coefficient for soil characteristics, table 1.2 = 1.0 for subsoil class A.(includes rocks, stiff deposits of sand, gravel or over consolidated clay T 1 = the fundamental period of vibration of the structure (in sec) for translational motion in the direction of motion. For buildings with heights up to 80m, the value of T 1 may be approximated using: T 1 = C 1 H ¾ H=height of the building= 19.3m C 1 =0.075 for moment resisting concrete frames T 1 = 0.075* 19.3 ¾ =0.691 sec Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 63 β = 1.2 *1.2 =1.84 ≤ 2.5 ok! 0.691 2/3 The parameter, γ, is the behavior factor to account energy dissipation capacity given by: γ = γ o k D k R k W ≤ 0.7 γ o = basic type of the behavior factor, dependent on structural type (table 3.2) for frame system , γ o =0.20 k D = factor reflecting ductility class ,use k D =2.0 for DC “L” k R = factor reflecting the structural regularity in elevation use k R =1.0 k W = factor reflecting the prevailing failure mode in Structural system with walls. =1.0 for frame and frame equivalent dual systems. γ =0.2*2.0*1.0*1.0=0.4 ≤ 0.7 ok! S d (T) = α β γ = 0.12*1.84*0.4= 0.0883 W T = seismic dead load, obtained as the total permanent load =11360.26 The base shear will be F b =0.0883*11360.26 = 1003.11 KN Distribution of base shear over height of a building The base shear force shall be distributed over the height of a structure, concentrated at each floor level, ∑ = − = n j j j i i t b i h W h W F F F 0 ) ( Where n = number of stories F i = the concentrated lateral force acting at floor i F t = the concentrated extra force at the top of the building accounting whiplash effect for slender building, which is given by: F t =0.07*T 1 *F b (EBCS- 8,1995 Art. 2.3.3.2.3) =0.07*0.691*1002KN =48.47KN W i , W j =that portion of total weight W located at or assigned to level i or j, respectively. h i , h j = height above the base to level i or j, respectively. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 64 4.2.2 Wind Load Analysis According to EBCS, 1995, the wind pressure acting on external surface, W e & internal surface, W i are: W e =q ref C e (Z i )C pe W I= q ref C e (Z i )C pi Reference wind pressure, q ref q ref = 1/2*ρ*V ref 2 Where V ref =C DIR C TEMP C ALT V ref,O =1.0*1.0*1.0*22 m/s = 22 m/s ρ=air density = 0.94Kg/m 3 Hence, q ref =0.5*0.94*22 2 =227.48N/m 2 Exposure coefficient To get the critical wind load we calculate the wind acting on the longest side of the building. Long side elevation of the building level h i w i h i F i (K) Ground floor 1.50 1546.76 13.29 first floor 5.30 12276.74 105.45 second floor 8.35 19082.77 163.90 third floor 11.40 26566.53 228.18 fourth floor 14.45 35161.19 302.00 roof 17.50 12704.15 109.12 head room 20.55 3677.63 31.59 ∑ 111015.8 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 65 4.3m 16m 27.75m Ce(Z)=Cr(z) 2 Ct(z) 2 { 1+7Kt/ Cr(z)Ct(z)} Where Z is reference height write wind load analysis Reference height Cross wind width, b=28m h<b, hence the building is considered as one part. Terrain category: category 4, urban areas in which at least 15% of the surface is covered with buildings and their average height exceeds 15m. (table 3.1, EBCS-1995) K T =0.24 Z o =1m Z min =16m Roughness coefficient, Cr(z) K T ln(Z/ Z o ) ; Z ≥ Z min C r (z) = C r (Z min ) ; Z < Z min Case-1 Where Ze =16m= Z min =16m C r (16) = C r (16) = 0.24*ln16=0.67 Case-2 Where Ze=20.3m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 66 C r (20.3) = 0.24*ln20.3=0.723 Topography coefficient Ct = 1, assuming that it is topography unaffected. The exposure coefficient will be; Case-1 Ce(16)=Cr(16) 2 Ct(16) 2 { 1+7Kt/ Cr(16)Ct(16)} = 0.67 2 *1*{1+7*0.24/(0.67*1) } = 1.575 Case-2 Ce(20.3)=1.74 External pressure coefficient, C pe Wall zonation: e min= b= 27.75 e=27.75, d <e 2h=32 The zonation is according to EBCS-1 Determination of external pressures and internal pressure Face Cpe10 upper @ 20.3m lower @ 16m Net external internal upper lower D 0.8 0.32 0.29 0.52 E -0.3 -0.12 -0.11 -0.44 D -0.5 -0.2 -0.18 0.47 E 0.8 0.32 0.29 -0.4 We neglect the effect of wind on adjacent faces due to its possession of high moment of inertia. zone A B D E d/h=0.425 Cpe10 Cpe10 Cpe10 Cpe10 ≤1 -1 -0.8 0.8 -0.3 ≥4 -1 -0.8 0.6 -0.3 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 67 By looking in to the earth quake and wind pressures the earth quake pressure governs and design is done for the earth quake forces. 4.3 Distribution of Storey Shear The horizontal forces at each level, F i , determined in the above manner are distributed to lateral load resistive structural elements in proportion to their rigidities assuming rigid floor diaphragms. a) Determination of center of mass Center of mass is a point on a floor level where the whole mass and its inertial effects can be replaced using lumped equivalent mass. X m = ∑W i X i ; Ym = ∑W i Y i ∑ W i ∑ W i X m Y m = the coordinate of the point of application of Fi when the seismic action is parallel to the Y- directin and X – direction respectively. GROUND FLOOR BLDNG PART TOTAL WT TOTAL MOMENT WX WY wall 659.27 9487.16 2830.32 beam 209.09 2890.80 638.25 column 162.82 2362.87 761.16 TOTAL 1031.17 14740.83 4229.73 Xm,Ym 14.30 4.10 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 68 FIRST FLOOR BLDNG PART TOTAL WT TOTAL MOMENT WX WY wall 709.69 9876.74 2798.11 beam 294.42 3934.78 1001.36 column 218.40 3244.09 1045.84 slab 734.59 8986.83 2005.45 inclined slab 56.52 1198.14 415.39 landing 19.44 464.62 117.13 pvc finishing 33.19 407.79 170.40 screed finishing 250.13 3090.75 677.47 Total 2316.37 31203.74 8231.15 Xm,Ym 13.47 3.55 SECOND FLOOR BLDNG PART TOTAL WT TOTAL MOMENT WX WY wall 709.69 9876.74 2798.11 beam 294.42 3934.78 1001.36 column 187.39 2719.53 876.06 slab 734.59 8986.83 2005.45 inclined slab 56.52 1198.14 415.39 landing 19.44 464.62 117.13 pvc finishing 33.19 407.79 170.40 screed finishing 250.13 3090.75 677.47 Total 2285.36 30679.17 8061.37 Xm,Ym 13.42 3.53 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 69 THIRD FLOOR BLDNG PART TOTAL WT TOTAL MOMENT Wx Wy wall 732.78 9664.66 2992.85 beam 294.42 3934.78 1001.36 column 187.39 2719.53 876.06 slab 744.23 8981.63 2053.64 inclined slab 56.52 1198.14 415.39 landing 19.44 464.62 117.13 pvc finishing 40.31 422.13 157.06 screed finishing 255.30 3139.18 695.70 Total 2330.40 30524.65 8309.19 Xm,Ym 13.10 3.57 FOURTH FLOOR BLDNG PART TOTAL WT TOTAL MOMENT Wx Wy wall 845.74 10937.92 3060.36 beam 294.42 3934.78 1001.36 column 187.39 2719.53 876.06 slab 745.32 9053.85 2084.45 inclined slab 56.52 1198.14 415.39 landing 19.44 464.62 117.13 pvc finishing 40.04 414.53 156.94 screed finishing 244.43 2839.08 690.89 Total 2433.30 31562.45 8402.57 Xm,Ym 12.97 3.45 ROOF LEVEL BLDNG PART TOTAL WT TOTAL MOMENT Wx Wy wall 237.21 4578.80 1170.83 beam 212.25 3017.76 713.01 column 117.12 1904.78 520.01 slab 39.37 793.86 297.03 inclined slab 56.52 1198.14 415.39 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 70 landing 19.44 464.62 117.13 TRUSS+EGA+CHIPWOOD 44.05 577.53 149.76 Total 725.95 12535.49 3383.17 Xm,Ym 17.27 4.66 HEAD ROOM LEVEL BLDNG PART TOTAL WT TOTAL MOMENT Wx Wy beam 52.92 1138.04 330.90 column 23.42 518.84 148.74 slab 64.64 1379.37 407.03 inclined slab 28.26 599.07 207.70 landing 9.72 232.31 58.56 Total 178.96 3867.64 1152.92 Xm,Ym 21.61 6.44 Determination of center of stiffness Center of stiffness is a point where the stiffness or strength of the floor is concentrated. Xs = ∑ DiyXi ; Ys = ∑ DixYi ∑ Diy ∑ Dix Where Xi, Yi = coordinates of the shear center of the frame element Dix, Diy = lateral stiffness of a particular element along X and Y axes, respectively. The rigidity, D of a column has a relation with: - stiffness of the column itself - stiffness of upper and lower beams - heights of upper and lower columns - upper and lower shear forces - location of storey D = a Kc Kc = column stiffness a = factor depending on boundary conditions Computation of ‘a’ If K represents the total sum of stiffness ratios of beams above and below the column divided by 2Kc, the approximate formula to obtain ‘a’ for general cases is: a = K / (2+K) Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 71 k1 k2 k3 k4 Kc k =0.5*k1+k2+k3+k4 Kc k2 k4 Kc k =0.5*k2+k4 Kc k = I/ L The above expressions of ‘a’ for cases of fixed column base is a = (0.5+K) and K = ∑Ktop (2+k) Kc and for pin supported columns Calculation of shear center LEVEL Axis Dy Xi DyXi Axis Dx Yi DxYi GROUND A 0.000894 0 0 4 0.00299 0 0 B 0.000894 5 0.00447 3 0.003182 5 0.01590825 C 0.000894 10 0.00894 2 0.001377 7.7 0.01060269 D 0.000894 15 0.01341 1 0.001198 9 0.01078116 E 0.002393 20 0.04785 ∑ 0.008746 0.0372921 F 0.000576 23.54 0.01356 Ys 4.26372227 G 0.000103 24.02 0.00247 H 0.000913 25 0.02281 ∑ 0.00756 0.11352 xS 15.0157 FIRST Axis Dy Xi DyXi Axis Dx Yi DxYi A 0.00038 0 0 4 0.012172 0 0 B 0.00038 5 0.0019 3 0.012976 5 0.06488222 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 72 C 0.00038 10 0.0038 2 0.004799 7.7 0.03695594 D 0.00038 15 0.00571 1 0.004908 9 0.0441748 E 0.009963 20 0.19925 ∑ 0.034857 0.14601295 F 0.002886 23.54 0.06794 Ys 4.18896113 G 0.003373 24.02 0.08101 H 0.002142 25 0.05355 ∑ 0.019885 0.41317 Xs 20.7777 1 ST -4 th Axis Dy Xi DyXi Axis Dx Yi DxYi A 0.00038 0 0 4 0.001574 0 0 B 0.00038 5 0.0019 3 0.001678 5 0.00838865 C 0.00038 10 0.0038 2 0.00062 7.7 0.0047703 D 0.00038 15 0.00569 1 0.000634 9 0.00570805 E 0.001437 20 0.02875 ∑ 0.004505 0.018867 F 0.000369 23.54 0.00869 Ys 4.18777854 G 0.000429 24.02 0.0103 H 0.000442 25 0.01105 ∑ 0.004196 0.07018 Xs 16.7262 ROOF Axis Dy Xi DyXi Axis Dx Yi DxYi A 0.00028 0 0 4 0.001212 0 0 B 0.00028 5 0.0014 3 0.001293 5 0.00646251 C 0.00028 10 0.0028 2 0.000484 7.7 0.00372346 D 0.00028 15 0.0042 1 0.000491 9 0.00441705 E 0.001185 20 0.0237 ∑ 0.003479 0.01460303 F 0.000305 23.54 0.00718 Ys 4.19713009 G 0.000368 24.02 0.00885 H 0.000334 25 0.00836 ∑ 0.003312 0.05648 Xs 17.0533 H.ROOM Axis Dy Xi DyXi Axis Dx Yi DxYi E 0.000268 20 0.00535 3 0.000206 5 0.00103186 G 0.000116 24.02 0.0028 2 0.00024 7.7 0.00184569 H 0.000116 25 0.00291 0.000446 0.00287755 ∑ 0.0005 0.01106 ∑ Ys 6.45086338 Xs 22.0981 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 73 Direct shear force distribution For a given storey shear Q ix = D ix * Q, Q ix = D iy * Q ∑ D ix ∑ D iy Axes in x-dxn FLOOR GROUND FIRST SECOND THIRD FORCE 13.29 105.78 164.42 228.27 AXIS D- VALUE Qi D- VALUE Qi D- VALUE Qi D- VALUE Qi 4 0.003 4.543171 0.012 36.9382 0.002 57.44 0.001574 79.74 3 0.0032 4.83463 0.013 39.3783 0.002 61.23 0.001678 85.007 2 0.0014 2.092359 0.005 14.5645 6E-04 22.61 0.00062 31.39 1 0.0012 1.820261 0.005 14.8947 6E-04 23.15 0.000634 32.135 0.0087 0.035 0.005 0.004505 FOURTH ROOF HEADROOM FLOOR 302.12 109.16 31.60 FORCE D-VALUE Qi D- VALUE Qi D-VALUE Qi AXIS 0.00157377 105.5368 0.001212 38.03917 4 0.00167773 112.508 0.001293 40.55121 0.000206372 14.61945 3 0.00061952 41.54477 0.000484 15.17151 0.0002397 16.98038 2 0.00063423 42.53109 0.000491 15.39791 1 0.00450525 0.003479 0.000446072 Axes in y-dxn FLOOR GROUND FIRST SECOND THIRD FORCE 13.29042136 105.7757269 164.4221758 228.2715446 D- VALUE Qi D- VALUE Qi D- VALUE Qi D- VALUE Qi A 0.0009 1.57154 4E-04 2.02353 0.00038 14.87 0.00038 20.65 B 0.0009 1.57154 4E-04 2.02353 0.00038 14.87 0.00038 20.65 C 0.0009 1.57154 4E-04 2.02353 0.00038 14.87 0.00038 20.65 D 0.0009 1.57154 4E-04 2.02353 0.00038 14.87 0.00038 20.65 E 0.0024 4.206323 0.01 52.9951 0.00144 56.32 0.001437 78.197 F 0.0006 1.012809 0.003 15.3529 0.00037 14.47 0.000369 20.084 G 0.0001 0.180896 0.003 17.9401 0.00043 16.81 0.000429 23.338 H 0.0009 1.604234 0.002 11.3935 0.00044 17.33 0.000442 24.053 0.0076 0.02 0.0042 0.004196 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 74 FOURTH ROOF HEADROOM FLOOR 302.1207209 109.1597995 31.59983145 FORCE D-VALUE Qi D- VALUE Qi D-VALUE Qi AXIS 0.00037955 27.33061 0.00028 9.22275 4 0.00037955 27.33061 0.00028 9.22275 3 0.00037955 27.33061 0.00028 9.22275 2 0.00037955 27.33061 0.00028 9.22275 1 0.00143729 103.4949 0.001185 39.05695 0.00026757 16.89932 0 0.00036915 26.58108 0.000305 10.05581 0 0.00042895 30.88771 0.000368 12.13888 0.000116378 7.350258 0 0.0004421 31.8346 0.000334 11.01716 0.000116378 7.350258 0.00419571 0.003312 0.000500326 Calculation of eccentricities Eccentricity is the difference between the center of mass and center of stiffness of the floor. Actual eccentricities e x = Xm – Xs e y = Ys – Ym Accidental eccentricities For various sources of eccentricities in locating the masses and spatial variation of seismic motion, an additional accidental eccentricity, eli is considered in addition to the actual eccentricity. It is given by: e li = ± 0.05 li Where e li is the floor dimension perpendicular to the direction of seismic action. Design eccentricities e d,x = e x + e lx and e d,y = e y + e ly Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 75 eccentrities ground first second third fourth roof head room actual ex 0.72 7.22 3.21 3.63 3.76 -0.21 0.49 ey 0.16 0.62 0.64 0.62 0.73 -0.46 0.01 accidental ex ±1.25 ±1.25 ±1.25 ±1.25 ±1.25 ±1.25 ±0.25 ey ±0.45 ±0.45 ±0.45 ±0.45 ±0.45 ±0.45 ±0.135 DESIGN ECCENTRICITIES LEVEL ground first second third fourth roof head room edx1 1.97 8.47 4.46 4.88 5.01 1.04 1.74 edx2 -0.53 5.97 1.96 2.38 2.51 -1.46 -0.76 edy1 0.61 1.07 1.09 1.07 1.18 -0.01 0.46 edy2 -0.29 0.17 0.19 0.17 0.28 -0.91 -0.44 Calculation of shear correction factor When the shear center and mass center do not coincide, torsion will be developed due to the lateral forces. This is also somehow amplified by the inherent existence of accidental eccentricities. As a result the direct shear forces obtained above need to be corrected to take this effect. The shear correction factors are calculated using α ix = 1+(D ix )e dy *y i J r α iy = 1+(D iy )e dx *x i J r J r = Jx + J y where J x =∑ (D ix y i 2 ) J y =∑ (D iy x i 2 ) x i = X i – X s y i =Y s – Y i Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 76 Ground floor DI R AXI S Ys- Yi Xs- Xi DX DY DX(Y s-Yi) 2 DY(X i-Xs) 2 αx1 αx2 αy1 αy2 αx,m ax αy,m ax X 4 4.2 6 0.003 0.054 4 1.01 3 0.99 4 1.013 3 - 0.7 4 0.003 2 0.001 7 0.99 8 1.00 1 1.001 2 - 3.4 4 0.001 4 0.016 3 0.99 5 1.00 2 1.002 1 - 4.7 4 0.001 2 0.026 9 0.99 4 1.00 3 1.003 Y A 15.01 6 9E- 04 0.2 1.043 1 0.98 8 1.043 B 10.01 6 9E- 04 0.09 1.028 8 0.99 2 1.029 C 5.015 7 9E- 04 0.02 1.014 4 0.99 6 1.014 D 0.015 7 9E- 04 0 1 1 1 E - 4.984 0.00 2 0.06 0.961 7 1.01 1.01 F - 8.524 6E- 04 0.04 0.984 2 1.00 4 1.004 G - 9.004 1E- 04 0.01 0.997 1.00 1 1.001 H - 9.984 9E- 04 0.09 0.970 7 1.00 8 1.008 0.099 2 0.51 Jx Jy Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 77 First floor X 4 4.1 9 0.012 2 0.213 6 1.06 7 1.01 1 1.06 7 3 - 0.8 1 0.013 0.008 5 0.98 6 0.99 8 0.99 8 2 - 3.5 1 0.004 8 0.059 2 0.97 8 0.99 6 0.99 6 1 - 4.8 1 0.004 9 0.113 6 0.96 9 0.99 5 0.99 5 Y A 20.77 8 4E- 04 0.1 6 1.082 4 1.05 8 1.08 2 B 15.77 8 4E- 04 0.0 9 1.062 5 1.04 4 1.06 3 C 10.77 8 4E- 04 0.0 4 1.042 7 1.03 1.04 3 D 5.777 7 4E- 04 0.0 1 1.022 9 1.01 6 1.02 3 E 0.777 7 0.01 0.0 1 1.080 7 1.05 7 1.08 1 F -2.762 0.00 3 0.0 2 0.916 9 0.94 1 0.94 1 G -3.242 0.00 3 0.0 4 0.886 1 0.92 0.92 H -4.222 0.00 2 0.0 4 0.905 8 0.93 4 0.93 4 0.394 9 0.4 2 Jx Jy Second floor X 4 4.19 0.0016 0.0276 1.023 1.004 1.023 3 - 0.81 0.0017 0.0011 0.995 0.999 0.999 2 - 3.51 0.0006 0.0076 0.992 0.999 0.999 1 - 4.81 0.0006 0.0147 0.989 0.998 0.998 Y A 16.726 4E-04 0.11 1.0904 1.04 1.09 B 11.726 4E-04 0.05 1.0633 1.028 1.063 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 78 C 6.7262 4E-04 0.02 1.0363 1.016 1.036 D 1.7262 4E-04 0 1.0093 1.004 1.009 E -3.274 0.001 0.02 0.933 0.971 0.971 F -6.814 4E-04 0.02 0.9642 0.984 0.984 G -7.294 4E-04 0.02 0.9555 0.98 0.98 H -8.274 4E-04 0.03 0.9479 0.977 0.977 0.051 0.26 Jx Jy Third floor X 4.00 4.19 0.00 0.03 1.02 1.00 1.02 3.00 -0.81 0.00 0.00 1.00 1.00 1.00 2.00 -3.51 0.00 0.01 0.99 1.00 1.00 1.00 -4.81 0.00 0.01 0.99 1.00 1.00 Y A 16.73 0.00 0.11 1.10 1.05 1.10 B 11.73 0.00 0.05 1.07 1.03 1.07 C 6.73 0.00 0.02 1.04 1.02 1.04 D 1.73 0.00 0.00 1.01 1.00 1.01 E -3.27 0.00 0.02 0.93 0.96 0.96 F -6.81 0.00 0.02 0.96 0.98 0.98 G -7.29 0.00 0.02 0.95 0.98 0.98 H -8.27 0.00 0.03 0.94 0.97 0.97 0.05 0.26 Jx Jy Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 79 Fourth floor X 4.00 4.26 0.00 0.03 1.03 1.01 1.03 3.00 -0.81 0.00 0.00 0.99 1.00 1.00 2.00 -3.51 0.00 0.01 0.99 1.00 1.00 1.00 -4.81 0.00 0.01 0.99 1.00 1.00 Y A 16.73 0.00 0.11 1.10 1.05 1.10 B 11.73 0.00 0.05 1.07 1.04 1.07 C 6.73 0.00 0.02 1.04 1.02 1.04 D 1.73 0.00 0.00 1.01 1.01 1.01 E -3.27 0.00 0.02 0.93 0.96 0.96 F -6.81 0.00 0.02 0.96 0.98 0.98 G -7.29 0.00 0.02 0.95 0.98 0.98 H -8.27 0.00 0.03 0.94 0.97 0.97 0.05 0.26 Jx Jy Roof X 4 4.2 0.0012 0.0214 1 0.981 1 3 -0.8 0.0013 0.0008 1 1.004 1.004 2 -3.5 0.0005 0.0059 1 1.006 1.006 1 -4.8 0.0005 0.0113 1 1.009 1.009 Y A 17.053 3E-04 0.08 1.0207 0.971 1.021 B 12.053 3E-04 0.04 1.0146 0.979 1.015 C 7.0533 3E-04 0.01 1.0086 0.988 1.009 D 2.0533 3E-04 0 1.0025 0.996 1.002 E -2.947 0.001 0.01 0.9848 1.021 1.021 F -6.487 3E-04 0.01 0.9914 1.012 1.012 G -6.967 4E-04 0.02 0.9889 1.016 1.016 H -7.947 3E-04 0.02 0.9885 1.016 1.016 0.0394 0.2 Jx Jy Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 80 Head room X 3 1.4 5 0.000 2 0.0004 1.0 4 0.96 1 1.04 2 - 1.2 5 0.000 2 0.0004 0.9 6 1.03 9 1.03 9 Y E 2.098 1 3E- 04 0 1.287 0.87 4 1.28 7 G -1.922 1E- 04 0 0.885 6 1.05 1.05 H -2.902 1E- 04 0 0.827 3 1.07 6 1.07 6 0.0008 0 Jx Jy Corrected shear forces for torsion Corrected shear forces are given by = α max * Q LEVEL GROUND FIRST SECOND THIRD Axis αmax*Qi αmax αmax*Qi αmax αmax*Qi αmax αmax*Qi 4 4.60 1.07 39.42 1.02 58.76 1.02 81.54 3 4.84 1.00 39.29 1.00 61.18 1.00 84.94 2 2.10 1.00 14.51 1.00 22.58 1.00 31.35 1 1.83 1.00 14.82 1.00 23.10 1.00 32.08 A 1.64 1.08 2.19 1.09 16.22 1.10 22.69 B 1.62 1.06 2.15 1.06 15.82 1.07 22.08 C 1.59 1.04 2.11 1.04 15.41 1.04 21.47 D 1.57 1.02 2.07 1.01 15.01 1.01 20.86 E 4.25 1.08 57.27 0.97 54.67 0.96 75.40 F 1.02 0.94 14.45 0.98 14.24 0.98 19.70 G 0.18 0.92 16.50 0.98 16.48 0.98 22.78 H 1.62 0.93 10.64 0.98 16.93 0.97 23.39 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 81 FOURTH ROOF HEAD ROOM LEVEL αmax αmax*Qi Qi αmax αmax*Qi Qi αmax αmax*Qi Axis 1.03 108.20 1.00 38.03 4 1.00 112.37 1.00 40.71 1.04 15.21 3 1.00 41.46 1.01 15.27 1.04 17.64 2 1.00 42.41 1.01 15.54 1 1.10 30.09 1.02 9.41 A 1.07 29.27 1.01 9.36 B 1.04 28.44 1.01 9.30 C 1.01 27.62 1.00 9.25 D 0.96 99.61 1.02 39.89 1.29 21.75 E 0.98 26.05 1.01 10.18 F 0.98 30.12 1.02 12.33 1.05 7.72 G 0.97 30.91 1.02 11.20 1.08 7.91 H Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 82 Earth quake loading on frames AXIS-A Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 83 AXIS-E Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 84 AXIS-3 4.4 Load combination The final critical design bending moments, shear forces and axial forces will be obtained from whichever the following five combinations of loading that produces the critical effect. Case 1. Consider vertical load only Case 2. Consider 75% of vertical load with earthquake effect from positive X– direction Case 3. Consider 75% of vertical load with earthquake effect from negative X– direction Case 4. Consider 75% of vertical load with earthquake effect from positive Y– direction Case 5. Consider 75% of vertical load with earthquake effect from negative Y– direction The frame is analyzed using sap2000 version 9 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 85 NB:- The same procedure is performed for the pre-cast slab system. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 86 5. DESIG OF BEAMS AD COLUMS 5.1 Beam design Beams are flexural members which are used to transfer the loads from slab to columns. Basically beams should be designed for flexure (moment). Furthermore it is essential to check and design the beam sections for torsion and shear. Beams may be designed for flexural moment depending on the magnitude of the moment and the X- sectional dimensions. on the other hand the beam can be singly reinforced, doubly reinforced T or Г section. Style of beam reinforcement Singly reinforced cross section The moment capacity of a given singly reinforced beam is given by M=ρbd 2 fyd (1-0.4ρm) ρ is taken to be 0.75 ρ b to re assure ductility of the material. Afterward for the given sectional dimensions and material data, the total area of reinforcement required for the applied moment M is given by As=ρbd Where: ρ =1/2(c 1 + ) / 4 ( 2 2 2 1 bd C M C − M=moment m=f yd / (0.8f cd ) d=effective depth Doubly reinforced cross section Incase when the dimension of the section is limited, the concrete may be subjected to higher compression stress. Thus additional steel bars are placed in the compression zone of the section. Hence the design moment, M d is obtained by M d = M 1 +M 2 Where; M 1 =the moment resisted by concrete and partial steel A s M 2 = The moment resisted by steel in compression, A s ’, and the left over steel As 2 M 1 can be computed via in the manner of singly reinforced section M 1 =0.8ρbd 2 f cd *m(1-0.4ρm), ρ is stated above. As 1= ρbd but since M 2 =M-M 1 =As’fs’(d-d c’ )=As 2 fyd(d-d c’ ) At yielding the compression steel both area of steel becomes equal As’=As 2 = (M-M1)/ ( fyd(d-d c ’ ) = (M-M1)/( fs’(d-d c ’ )…….i.e. f yd = f s ’ f s ’=(x-d c ’)Es*Є c /x ,x= ρmb Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 87 Longitudinal reinforcement design In this specific project we have used design tables for the calculation of reinforcement area as per the provision of EBCS 2, 1995. 5.1.1 Solid slab beams AXIS-A moment LOCATION b d Km KS AS1 Ф(As1 ) 22.42 TTBeam 0.25 0.307 30.85 4.06 296.499 2Ф12 12.4 TTBeam 0.25 0.307 22.94 3.98 Asmin 2Ф12 44.97 TTBeam 0.25 0.307 43.69 4.252 622.8418 4Ф12+1Ф16 173.14 1st &2nd 0.25 0.446 59.01 2410.72 4Ф24+3Ф16 102.74 1st &2nd 0.25 0.446 45.45 4.296 989.6212 5Ф16 174.08 1st &2nd 0.25 0.455 58.00 2410.72 4Ф24+3Ф16 128.5 3rd&4th 0.25 0.446 50.83 4.425 1274.916 4Ф20 71.05 3rd&4th 0.25 0.455 37.05 4.12 643.3538 3Ф16+2Ф20 14.31 3rd&4th 0.25 0.455 16.63 3.963 124.6385 2Ф12 141.28 3rd&4th 0.25 0.455 52.25 4.47 1387.96 4Ф20+1Ф14 96.44 Gbeam 0.25 0.355 55.33 1285.81 4Ф20 102.44 Gbeam 0.25 0.355 57.02 1275.096 4Ф20 66.6 Gbeam 0.25 0.355 45.98 4.306 807.8299 4Ф16 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 88 AXIS-B moment LOCATION b d Km KS As1 Ф(As1 ) 17.17 TTBeam 0.25 0.307 26.99 4.015 224.5523 2Ф12 5.16 TTBeam 0.25 0.307 14.80 min 153.5 2Ф12 19.68 TTBeam 0.25 0.307 28.90 4.039 258.917 2Ф14 179.28 1st &2nd 0.25 0.445 60.18 4.27 1720.282 4Ф24 140.87 1st &2nd 0.25 0.445 53.34 4.3 1361.216 3Ф24 205.44 1st &2nd 0.25 0.445 64.42 2707.4 6Ф24 157.05 3rd&4th 0.25 0.445 56.32 4.52 1595.204 3Ф24&1Ф20 90.16 3rd&4th 0.25 0.445 42.68 4.22 855.0004 2Ф20 +2Ф12 158.5 3rd&4th 0.25 0.445 56.58 1595.204 3Ф24&1Ф20 100.71 Gbeam 0.25 0.357 56.22 4.52 1275.096 4Ф20 102.03 Gbeam 0.25 0.357 56.59 1275.096 4Ф20 69.23 Gbeam 0.25 0.357 46.61 4.32 837.7412 2Ф20 +2Ф12 AXIS -C moment LOCATION b d Km KS AS1 Ф(As1 ) 22.15 TTBeam 0.25 0.307 30.66 3.994 288.1664 3Ф12 9.89 TTBeam 0.25 0.307 20.49 3.97 127.8935 2Ф12 16.01 TTBeam 0.25 0.307 26.07 4 208.5993 2Ф12 177.98 1st&2nd 0.25 0.446 59.82 2416.4 5Ф24+1Ф16 114.32 1st&2nd 0.25 0.446 47.95 4.36 1117.568 4Ф20 35.44 1st&2nd 0.25 0.455 26.17 4.01 312.3393 2Ф16 182.2 1st&2nd 0.25 0.455 59.33 2241 4Ф24+3Ф16 166.48 3rd&4th 0.25 0.446 57.86 1941.5 3Ф24+3Ф16 95.32 3rd&4th 0.25 0.446 43.78 4.26 910.4556 2Ф20+2Ф14 150.1 3rd&4th 0.25 0.455 53.85 4.53 1494.402 2Ф24+3Ф16 115.1 Gbeam 0.25 0.355 60.44 1550 3Ф24+1Ф20 111.28 Gbeam 0.25 0.355 59.43 1550 3Ф24+1Ф20 61.8 Gbeam 0.25 0.355 44.29 4.19 729.4141 4Ф16 AXIS-D moment LOCATION b d Km KS AS1 Ф(As1 ) 24.86 TTBeam 0.25 0.307 32.48 4.07 329.5772 3Ф14 12.74 TTBeam 0.25 0.307 23.25 3.98 165.1635 2Ф12 14.44 TTBeam 0.25 0.307 24.76 3.988 187.5789 2Ф12 166.42 4th&3rd 0.25 0.446 57.85 1941.5 3Ф24+3Ф16 96.08 4th&3rd 0.25 0.446 43.96 4.26 917.7148 2Ф20+2Ф14 157.49 4th&3rd 0.25 0.455 55.16 1595.204 3Ф24&1Ф20 165.87 1st&2nd 0.25 0.446 57.75 1941.5 3Ф24+3Ф16 105.43 1st&2nd 0.25 0.446 46.04 4.31 1018.841 4Ф20 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 89 80.93 1st&2nd 0.25 0.455 39.54 4.16 739.9314 2Ф16+2Ф20 145.5 1st&2nd 0.25 0.455 53.02 4.51 1442.209 4Ф20+1Ф16 103.85 Gbeam 0.25 0.355 57.41 1398.54 3Ф24 99.45 Gbeam 0.25 0.355 56.18 1398.54 3Ф24 74.01 Gbeam 0.25 0.355 48.47 4.37 911.0527 2Ф14+2Ф20 AXIS-E moment LOCATION b d Km KS AS1 Ф(As1 ) 32.22 HEAD ROOM 0.25 0.307 36.98 4.12 432.3987 3Ф14 6.42 HEAD ROOM 0.25 0.307 16.51 Asmin 2Ф12 31.25 TTBeam 0.25 0.307 36.42 4.11 418.3632 4Ф12 21.98 TTBeam 0.25 0.307 30.54 4.06 290.6801 2Ф14 13.42 TTBeam 0.25 0.307 23.87 Asmin 2Ф12 10.41 TTBeam 0.25 0.307 21.02 Asmin 2Ф12 12.46 TTBeam 0.25 0.307 23.00 3.97 161.1277 2Ф12 3.9 TTBeam 0.25 0.307 12.87 Asmin 2Ф12 157.15 4th&3rd 0.25 0.446 56.22 1744.84 4Ф24 112.47 4th&3rd 0.25 0.446 47.56 4.349 1096.709 2Ф24+1Ф16 121.7 4th&3rd 0.25 0.446 49.47 4.396 1199.536 2Ф24+2Ф14 47.81 4th&3rd 0.25 0.455 30.39 4.04 424.5108 3Ф14 130.59 4th&3rd 0.25 0.455 50.23 4.415 1267.154 2Ф24+2Ф16 81.82 4th&3rd 0.25 0.455 39.76 4.166 749.1475 2Ф20+1Ф12 52.57 4th&3rd 0.25 0.455 31.87 4.054 468.3929 2Ф14+1Ф16 179.08 1st&2nd 0.25 0.446 60.01 1754.68 4Ф24 120.73 1st&2nd 0.25 0.446 49.27 4.392 1188.893 2Ф24+2Ф14 156.16 1st&2nd 0.25 0.446 56.04 1744.84 4Ф24 101.81 1st&2nd 0.25 0.446 45.25 4.29 979.2935 2Ф20+2Ф16 146.86 1st&2nd 0.25 0.446 54.34 1442.209 4Ф20+1Ф16 149.72 1st&2nd 0.25 0.446 54.87 1717.45 4Ф24 150.8 1st&2nd 0.25 0.446 55.07 1717.45 4Ф24 100.84 Gbeam 0.25 0.335 59.95 4.275 1286.839 3Ф20 59.44 Gbeam 0.25 0.335 46.03 4.09 725.7003 2Ф20 97.73 Gbeam 0.25 0.335 59.02 4.28 1248.61 3Ф20 85.69 Gbeam 0.25 0.335 55.27 4.2 1074.322 3Ф20 119.53 Gbeam 0.25 0.335 65.27 4.386 1564.951 2Ф24+2Ф14 106.97 Gbeam 0.25 0.355 58.27 4.32 1301.719 2Ф20+3Ф14 130.62 Gbeam 0.25 0.355 64.39 4.43 1629.99 3Ф20+1Ф24 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 90 AXIS-H moment LOCATION b d Km KS AS1 Ф(As1 ) 36.14 TTBeam 0.25 0.307 39.16 4.159 489.5969 3Ф16 12.3 TTBeam 0.25 0.307 22.85 Asmin 2Ф12 27.97 TTBeam 0.25 0.307 34.45 4.092 372.8119 2Ф16 151.31 4th&3rd 0.25 0.446 55.16 1717.45 4Ф24 117.03 4th&3rd 0.25 0.446 48.51 4.373 1147.471 3Ф20+2Ф12 83.215 4th&3rd 0.25 0.446 40.91 4.18 779.9074 2Ф20+1Ф14 173.14 1st&2nd 0.25 0.446 59.01 2412.33 4Ф24+2Ф20 154.14 1st&2nd 0.25 0.446 55.67 1744.84 4Ф24 96.66 1st&2nd 0.25 0.446 44.09 4.265 924.3383 3Ф20 100 Gbeam 0.25 0.355 56.34 1394.7 3Ф24 86.56 Gbeam 0.25 0.355 52.42 4.481 1092.607 2Ф24+1Ф16 83.43 Gbeam 0.25 0.355 51.46 4.435 1042.287 2Ф20+3Ф14 AXIS-4 moment LOCATION b d Km KS AS1 Ф(As1 ) 186.912 1st&2nd 0.25 0.446 61.31 4.51 2341 4Ф24+3Ф16 184.18 1st&2nd 0.25 0.446 60.86 2341 4Ф24+3Ф16 150.25 1st&2nd 0.25 0.455 53.88 4.53 1495.896 3Ф24+1Ф14 174 1st&2nd 0.25 0.446 59.15 4.51 1759.507 4Ф24 176.859 1st&2nd 0.25 0.455 58.46 4.51 1753.042 4Ф24 180.71 1st&2nd 0.25 0.455 59.09 1754.68 4Ф24 92.19 1st&2nd 0.25 0.455 42.20 4.25 861.1154 2Ф24 81.38 1st&2nd 0.25 0.455 39.65 4.2 751.2 4Ф16 71.86 1st&2nd 0.25 0.455 37.26 4.15 655.4264 3Ф14+1Ф16 85.69 1st&2nd 0.25 0.455 40.69 4.22 794.7512 4Ф16 104.94 1st&2nd 0.25 0.455 45.03 4.31 994.047 5Ф16 155.98 3rd&4th 0.25 0.446 56.01 4.52 1580.784 3Ф24+2Ф14 148.96 3rd&4th 0.25 0.446 54.73 1717.45 4Ф24 132.93 3rd&4th 0.25 0.455 50.68 4.427 1293.365 2Ф24+2Ф16 134.47 3rd&4th 0.25 0.455 50.97 4.44 1312.191 3Ф20+2Ф16 132.34 3rd&4th 0.25 0.455 50.57 4.42 1285.589 4Ф20 137.81 3rd&4th 0.25 0.455 51.60 4.45 1347.812 2Ф24+3Ф14 82.511 3rd&4th 0.25 0.455 39.93 4.21 763.4534 4Ф16 70.31 3rd&4th 0.25 0.455 36.86 4.14 639.7437 1Ф16+3Ф14 48.76 3rd&4th 0.25 0.455 30.69 4.01 429.731 3Ф14 67.357 3rd&4th 0.25 0.455 36.08 4.13 611.3943 2Ф16+2Ф12 73.61 3rd&4th 0.25 0.455 37.71 4.16 673.0057 2Ф12+3Ф14 42.44 TTBeam 0.25 0.307 42.44 4.25 587.5244 3Ф12+2Ф14 37.65 TTBeam 0.25 0.307 39.97 4.19 513.855 3Ф16 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 91 33.94 TTBeam 0.25 0.307 37.95 4.15 458.798 3Ф16 25.63 TTBeam 0.25 0.307 32.98 4.13 344.7945 2Ф16 17.05 TTBeam 0.25 0.307 26.90 4.014 222.9274 2Ф12 25.89 TTBeam 0.25 0.307 33.15 4.082 344.2442 2Ф12+1Ф14 17.73 TTBeam 0.25 0.307 27.43 min 2Ф12 13.94 TTBeam 0.25 0.307 24.32 min 2Ф12 9.27 TTBeam 0.25 0.307 19.83 min 2Ф12 8.61 TTBeam 0.25 0.307 19.12 min 2Ф12 5.69 TTBeam 0.25 0.307 15.54 min 2Ф12 97.3 Gbeam 0.25 0.357 55.26 4.598 1253.18 4Ф20 78.96 Gbeam 0.25 0.357 49.78 4.405 974.2824 2Ф20+2Ф16 81.53 Gbeam 0.25 0.357 50.58 4.425 1010.561 2Ф20+2Ф16 83.34 Gbeam 0.25 0.357 51.14 4.43 1034.163 2Ф20+2Ф16 81.92 Gbeam 0.25 0.357 50.71 4.425 1015.395 2Ф20+2Ф16 22.74 Gbeam 0.25 0.357 26.72 4 254.7899 2Ф14 75.44 Gbeam 0.25 0.357 48.66 4.37 923.4532 3Ф20 66.48 Gbeam 0.25 0.357 45.68 4.29 798.8773 4Ф16 68.82 Gbeam 0.25 0.357 46.47 4.33 834.7076 2Ф20+2Ф12 69.32 Gbeam 0.25 0.357 46.64 4.34 842.7137 2Ф20+2Ф12 73.37 Gbeam 0.25 0.357 47.99 4.34 891.949 2Ф24 AXIS-1 moment LOCATION b d Km KS As1 Ф(As1 ) 21.36 TTBeam 0.25 0.307 30.11 3.97 276.2189 2Ф14 11.92 TTBeam 0.25 0.307 22.49 3.96 153.7564 2Ф12 158.02 1st&2nd 0.25 0.455 55.26 4.52 1748 4Ф24 177.72 1st&2nd 0.25 0.455 58.60 1857 5Ф20+2Ф16 164.47 1st&2nd 0.25 0.455 56.37 4.52 1895 5Ф20+2Ф16 117.51 3rd&4th 0.25 0.455 47.65 4.39 1133.778 3Ф20+2Ф12 104.24 3rd&4th 0.25 0.455 44.88 4.282 981.0015 2Ф24+1Ф12 71.82 3rd&4th 0.25 0.455 37.25 4.14 653.4831 3Φ14+2Φ12 122.93 Gbeam 0.25 0.355 62.46 4.63 1817 4Ф24 122.06 Gbeam 0.25 0.355 62.24 1817 4Ф24 95.75 Gbeam 0.25 0.355 55.13 4.57 1232.613 4Ф20 AXIS-2 moment LOCATION b d Km KS As1 Ф(As1 ) 30.11 Head Room 0.25 0.307 35.75 4.11 403.1013 2Ф16 14.5 Head Room 0.25 0.307 24.81 3.97 187.5081 2Ф12 11.27 TTBeam 0.25 0.307 21.87 3.96 145.372 2Ф12 28.93 TTBeam 0.25 0.307 35.04 4.1 386.3616 2Ф16 32.82 TTBeam 0.25 0.307 37.32 4.15 443.658 4Ф12 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 92 177.03 1st&2nd 0.25 0.455 58.48 4.7 1857 5Ф20+2Ф16 171.09 1st&2nd 0.25 0.455 57.50 2412.33 4Ф24+2Ф20 121.87 1st&2nd 0.25 0.455 48.53 4.41 1181.202 2Ф24+2Ф14 115.06 3rd&4th 0.25 0.455 47.15 4.38 1107.611 2Ф24+2Ф12 103.82 3rd&4th 0.25 0.455 44.79 4.28 976.5925 2Ф20+2Ф16 57.36 3rd&4th 0.25 0.455 33.29 4.05 510.567 2Ф16+1Ф12 118.84 Gbeam 0.25 0.355 61.42 4.65 1712 3Ф24+2Ф16 116.89 Gbeam 0.25 0.355 60.91 1550 3Ф24+1Ф20 87.01 Gbeam 0.25 0.355 52.55 4.51 1105.395 2Ф24+2Ф12 AXIS-3 moment LOCATION b d Km KS As1 Ф(As1 ) 15.34 Head Room 0.25 0.307 25.52 3.99 199.37 2Ф12 42.74 Head Room 0.25 0.307 42.59 4.23 588.8932 2Ф20 14.28 TTBeam 0.25 0.307 24.62 4.09 190.245 2Ф12 22.64 TTBeam 0.25 0.307 31.00 4.434 326.9894 3Ф12 16.53 TTBeam 0.25 0.307 26.49 4.02 216.4515 2Ф12 17.64 TTBeam 0.25 0.307 27.36 4.03 231.5609 2Ф14 27.43 TTBeam 0.25 0.307 34.12 4.09 365.4355 2Ф16 38.55 TTBeam 0.25 0.307 40.45 4.19 526.1384 2Ф12 +2Ф14 199.64 1st&2nd 0.25 0.446 63.36 4.51 2569 5Ф24+2Ф16 178.245 1st&2nd 0.25 0.446 59.87 4.51 2170 4Ф24+2Ф16 182.77 1st&2nd 0.25 0.446 60.62 2241 4Ф24+3Ф16 182.33 1st&2nd 0.25 0.446 60.55 4.51 2241 4Ф24+3Ф16 148.37 1st&2nd 0.25 0.455 53.54 4.53 1477.178 3Ф24+1Ф14 193.54 1st&2nd 0.25 0.455 61.15 2415.64 4Ф24+2Ф20 117.14 1st&2nd 0.25 0.455 47.57 4.35 1119.91 3Ф20+2Ф12 87.28 1st&2nd 0.25 0.455 41.07 4.2 805.6615 2Ф20+2Ф12 75.44 1st&2nd 0.25 0.455 38.18 4.14 686.4211 3Ф14+2Ф12 94.43 1st&2nd 0.25 0.455 42.71 4.23 877.8877 4Ф16+1Ф12 72.55 1st&2nd 0.25 0.455 37.44 4.15 661.7198 3Ф14+2Ф12 144.86 3rd&4th 0.25 0.446 53.97 4.53 1471.336 2Ф24+2Ф16 154.125 3rd&4th 0.25 0.446 55.67 4.5 1555.073 3Ф24+2Ф12 154.971 3rd&4th 0.25 0.446 55.82 4.51 1567.083 3Ф24+2Ф12 142.89 3rd&4th 0.25 0.446 53.60 4.53 1451.327 4Ф20+1Ф16 145.66 3rd&4th 0.25 0.446 54.12 4.47 1459.866 3Ф24+1Ф12 107.87 3rd&4th 0.25 0.446 46.57 4.28 1035.165 2Ф20+2Ф16 76.46 3rd&4th 0.25 0.455 38.44 4.18 702.4237 2Ф16+2Ф14 77.375 3rd&4th 0.25 0.455 38.67 4.18 710.8297 2Ф16+2Ф14 77.025 3rd&4th 0.25 0.455 38.58 4.18 707.6143 2Ф16+2Ф14 75.474 3rd&4th 0.25 0.455 38.19 4.17 691.7068 2Ф16+2Ф14 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 93 48.39 3rd&4th 0.25 0.455 30.58 4.01 426.4701 3Ф14 95.54 Gbeam 0.25 0.355 55.07 4.52 1253 4Ф20 93.45 Gbeam 0.25 0.355 54.46 4.52 1239 4Ф20 92.83 Gbeam 0.25 0.355 54.28 4.52 1230 4Ф20 114.01 Gbeam 0.25 0.355 60.16 1550 3Ф24+1Ф20 94.49 Gbeam 0.25 0.355 54.76 4.52 1252 4Ф20 100.71 Gbeam 0.25 0.355 56.54 4.52 1416 3Ф24+1Ф12 68.65 Gbeam 0.25 0.355 46.68 4.37 845.0718 2Ф20+2Ф12 56.89 Gbeam 0.25 0.355 42.49 4.27 684.2825 3Ф14+2Ф12 57.4 Gbeam 0.25 0.355 42.68 4.27 690.4169 2Ф14+2Ф16 55.78 Gbeam 0.25 0.355 42.08 4.26 669.36 3Ф14+2Ф12 61.71 Gbeam 0.25 0.355 44.26 4.31 749.2115 3Ф16+1Ф14 Inclined beam LOCATION AS Ф(AS) landing beams Gbeam 1332.63 3Ф20+2Ф16 LOCATION AS Ф(AS) Gbeam 1171.31 3Ф20+2Ф12 G-floor 2179.44 4Ф24+2Ф16 Gbeam 964.61 2Ф20+2Ф16 G-floor 1898.36 3Ф24+3Ф16 Gbeam 1240.9 4Ф20 G-floor 2221.63 4Ф24+3Ф14 Gbeam 924.75 2Ф20+2Ф14 1st&2nd 1560.77 3Ф24+2Ф12 1st&2nd 1645.39 3Ф24+2Ф14 1st&2nd 1357.31 3Ф24 1st&2nd 1194.29 2Ф24+2Ф14 1st&2nd 1728.52 3Ф24+2Ф16 1st&2nd 1201.26 2Ф24+2Ф14 3rd&4th 790.18 4Ф16 1st&2nd 1409.64 4Ф20+1Ф14 3rd&4th 615.56 4Ф14 1st&2nd 972.08 2Ф24+1Ф14 3rd&4th 915.1 3Ф16+2Ф14 3rd&4th 796.81 3Ф14+3Ф12 roof 356.15 2Ф16 3rd&4th 668.66 3Ф14+2Ф12 roof 225 2Ф12 3rd&4th 639.49 3Ф14+2Ф12 roof 225 2Ф12 3rd&4th 696.39 2Ф14+2Ф16 3rd&4th 525.17 2Ф14+2Ф12 TTBeam 157.5 2Ф12 TTBeam 187.61 2Ф12 TTBeam 159.75 2Ф12 TTBeam 157.5 2Ф12 Head Room 157.5 2Ф12 Head Room 157.5 2Ф12 Head Room 157.5 2Ф12 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 94 5.1.2 Pre-cast slab beams AXIS-A moment LOCATION b d Km KS AS1 Ф(As1 ) 15.41 TTBeam 0.25 0.307 25.57 4.007 201.1331 2Ø12 49.82 TTBeam 0.25 0.307 45.98 4.31 699.4274 2Ø16+2Ø14 170.54 1st&2nd 0.3 0.455 52.40 1829.2 3Ø24+2Ø20 66.22 1st &2nd 0.3 0.455 32.65 4.0765 593.2875 3Ø16 156.53 1st &2nd 0.3 0.455 50.20 1599.4 5Ø20 197.12 1st &2nd 0.3 0.455 56.34 2396.13 4Ø24+2Ø20 129.72 3rd&4th 0.3 0.455 45.70 130.95 3rd&4th 0.3 0.455 45.92 4.308 1239.852 4Ø20 177.07 3rd&4th 0.3 0.455 53.39 2014.8 2Ø16+3Ø24+2Ø14 88.64 Gbeam 0.3 0.355 48.42 4.36 1088.649 3Ø14+2Ø20 93.91 Gbeam 0.3 0.355 49.84 1168.59 3Ø20+2Ø12 85.39 Gbeam 0.3 0.355 47.52 AXIS-B moment LOCATION b d Km KS AS1 Ф(As1 ) 10.83 TTBeam 0.25 0.307 21.44 3.966 139.9081 2Ø12 23.12 TTBeam 0.25 0.307 31.32 4.067 306.2835 2Ø14 194.49 1st&2nd 0.3 0.455 55.96 4Ø24+2Ø20 168.55 1st &2nd 0.3 0.455 52.09 1829.2 3Ø24+2Ø20 245.89 1st &2nd 0.3 0.455 62.92 3260.34 5Ø24+5Ø16 178.456 3rd &4th 0.3 0.455 53.60 2086.4 3Ø24+2Ø16+3Ø14 198.58 3rd&4th 0.3 0.455 56.55 2396.13 4Ø24+2Ø20 154.49 3rd&4th 0.3 0.455 49.87 1654.94 3Ø24+2Ø14 84.374 Gbeam 0.3 0.355 47.24 4.341 1031.74 3Ø14+2Ø20 92.296 Gbeam 0.3 0.355 49.41 1158.51 3Ø20+2Ø12 89.94 Gbeam 0.3 0.355 48.77 AXIS -C moment LOCATION b d Km KS AS1 Ф(As1 ) 6.52 TTBeam 0.25 0.307 16.63 min ####### 2Ø12 18.94 TTBeam 0.25 0.307 28.35 4.032 248.7494 1Ø14+1Ø12 196.44 1st&2nd 0.3 0.455 56.24 2371.5 4Ø24+2Ø20 174.54 1st&2nd 0.3 0.455 53.01 1958.3 3Ø24+2Ø20 228.84 1st&2nd 0.3 0.446 61.93 3060.63 4Ø24+4Ø20 161.03 3rd&4th 0.3 0.455 50.92 1744.8 3Ø24+3Ø14 146.9 3rd&4th 0.3 0.455 48.63 1427.68 4Ø20+2Ø12 189.22 3rd&4th 0.3 0.455 55.20 2222.2 4Ø24+3Ø14 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 95 83.01 Gbeam 0.3 0.355 46.86 4.324 1011.085 2Ø20+2Ø16 91.4 Gbeam 0.3 0.355 49.17 1180.58 3Ø16+2Ø20 92.33 Gbeam 0.3 0.355 49.42 1195.44 3Ø16+2Ø20 AXIS-D moment LOCATION b d Km KS AS1 Ф(As1 ) 4.67 TTBeam 0.25 0.307 14.08 min 153 2Φ12 17.43 TTBeam 0.25 0.307 27.20 4.018 228.1229 2Ø12 165.37 4th&3rd 0.3 0.455 51.60 1791.79 3Ø24+3Ø14 143.17 4th&3rd 0.3 0.455 48.01 4.36 1371.915 2Ø24+3Ø14 179.26 4th&3rd 0.3 0.455 53.72 2086.4 3Ø24+2Ø16+3Ø14 196.46 1st&2nd 0.3 0.455 56.24 2371.5 4Ø24+2Ø20 171.23 1st&2nd 0.3 0.455 52.51 1901.71 4Ø24+1Ø12 218.24 1st&2nd 0.3 0.455 59.28 2765.05 4Ø24+3Ø20 81.81 Gbeam 0.3 0.355 46.52 4.323 996.2384 2Ø20+3Ø14 91.12 Gbeam 0.3 0.355 49.09 1180.58 3Ø16+2Ø20 94.02 Gbeam 0.3 0.355 49.87 1205.57 3Ø16+2Ø20 AXIS-E moment LOCATION b d Km KS AS1 Ф(As1 ) 30.99 HEAD ROOM 0.25 0.307 36.27 4.113 415.1852 2Ø14+1Ø12 27.88 HEAD ROOM 0.25 0.307 34.40 4.094 371.7939 2Ø16 28.95 TTBeam 0.25 0.307 35.05 4.101 386.723 2Ø16 18.28 TTBeam 0.25 0.307 27.85 4.026 239.724 1Ø12+1Ø14 1.22 TTBeam 0.25 0.307 7.20 min 153.5 2Ø12 8.45 TTBeam 0.25 0.307 18.94 min 153.5 2Ø12 14.04 TTBeam 0.25 0.307 24.41 3.986 182.2913 2Ø12 2.71 TTBeam 0.25 0.307 10.72 min 153 2Ø12 10.07 TTBeam 0.25 0.307 20.67 3.961 129.926 2Ø12 158.53 4th&3rd 0.3 0.455 50.52 1654.94 3Ø24+2Ø14 123.65 4th&3rd 0.3 0.455 44.62 4.276 1162.038 3Ø20+2Ø12 96.97 4th&3rd 0.3 0.455 39.51 4.162 887.0091 3Ø16+2Ø14 83.29 4th&3rd 0.3 0.455 36.62 4.114 753.088 5Ø14 139.69 4th&3rd 0.3 0.455 47.43 4.346 1334.27 3Ø20+2Ø16 36.74 4th&3rd 0.3 0.455 24.32 3.986 321.8585 3Ø12 87.24 4th&3rd 0.3 0.455 37.48 4.141 793.9799 4Ø16 181.92 1st&2nd 0.3 0.455 54.12 2086.4 3Ø24+2Ø16+3Ø14 135.14 1st&2nd 0.3 0.455 46.65 4.326 1284.87 3Ø24 166.35 1st&2nd 0.3 0.455 51.75 4.463 1631.692 3Ø24+2Ø14 111.05 1st&2nd 0.3 0.455 42.29 4.217 1029.226 2Ø20+2Ø16 110.87 1st&2nd 0.3 0.455 42.25 4.216 1027.314 2Ø20+2Ø16 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 96 173.46 1st&2nd 0.3 0.455 52.85 4.498 1714.776 4Ø24 112.42 Gbeam 0.3 0.455 42.55 4.224 1043.653 2Ø20+2Ø16 47.2 Gbeam 0.3 0.355 35.33 4.103 545.5256 2Ø16+1Ø14 98.49 Gbeam 0.3 0.355 51.04 4.436 1230.709 4Ø20 107.95 Gbeam 0.3 0.355 53.43 4.5315 1377.959 4Ø20+1Ø14 108.6 Gbeam 0.3 0.355 53.60 4.54 1388.856 4Ø20+1Ø14 136.06 Gbeam 0.3 0.355 59.99 1926.9 4Ø24+1Ø12 AXIS-H moment LOCATION b d Km KS AS1 Ф(As1 ) 35.57 TTBeam 0.25 0.307 38.85 4.151 480.9481 2Ø16+1Ø12 26.93 TTBeam 0.25 0.307 33.81 4.088 358.5988 2Ø16 123.6 4th&3rd 0.3 0.455 44.61 4.275 1161.297 3Ø20+2Ø12 107.526 4th&3rd 0.3 0.455 41.61 4.202 993.0203 5Ø16 144.544 4th&3rd 0.3 0.455 48.24 4.36 1385.081 4Ø20+1Ø14 156.45 1st&2nd 0.3 0.455 50.19 1599.4 5Ø20 122.3 1st&2nd 0.3 0.455 44.38 1161.297 3Ø20+2Ø12 164.56 1st&2nd 0.3 0.455 51.47 1631.692 3Ø24+2Ø14 100.8 Gbeam 0.3 0.355 51.63 4.45 1263.549 2Ø24+2Ø16 83.79 Gbeam 0.3 0.355 47.08 4.334 1022.946 2Ø20+2Ø16 89.25 Gbeam 0.3 0.355 48.59 4.375 1099.912 2Ф20+3Ф14 AXIS-4 moment LOCATION b d Km KS AS1 Ф(As1 ) 144.44 1st&2nd 0.25 0.455 52.83 1604.53 4Ø20+2Ø16 104.14 1st&2nd 0.25 0.455 44.86 4.304 985.0957 5Ø16 60.13 1st&2nd 0.25 0.455 34.09 4.091 540.6414 2Ø16+1Ø14 118.76 1st&2nd 0.25 0.455 47.90 4.353 1136.181 3Ø20+2Ø12 100.73 1st&2nd 0.25 0.455 44.12 4.263 943.7626 3Ø20 120.93 1st&2nd 0.25 0.455 48.34 4.36 1158.802 3Ø20+2Ø12 97.79 1st&2nd 0.25 0.455 43.47 4.248 912.9932 3Ø20 124.1 1st&2nd 0.25 0.455 48.97 1206.11 4Ø20 93.55 1st&2nd 0.25 0.455 42.51 4.223 868.2674 3Ø16+2Ø14 125.01 1st&2nd 0.25 0.455 49.15 1214.93 4Ø20 131.28 1st&2nd 0.25 0.455 50.36 1340 3Ø24 140.096 1st&2nd 0.25 0.455 52.03 1516.28 3Ø24+2Ø12 116.75 3rd&4th 0.25 0.455 47.49 4.35 1116.181 2Ø24+2Ø12 81.84 3rd&4th 0.25 0.455 39.77 4.16 748.2514 2Ø20+1Ø14 91.44 3rd&4th 0.25 0.455 42.03 4.211 846.2722 2Ø20+2Ø12 62.82 3rd&4th 0.25 0.455 34.84 4.098 565.7942 3Ø16 30.14 3rd&4th 0.25 0.455 24.13 3.984 263.9072 2Ø14 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 97 83.27 3rd&4th 0.25 0.455 40.11 4.172 763.5218 2Ø20+1Ø14 59.01 3rd&4th 0.25 0.455 33.77 4.088 530.1822 2Ø16+1Ø14 93.14 3rd&4th 0.25 0.455 42.42 4.22 863.8479 2Ø20+2Ø12 33.75 3rd&4th 0.25 0.455 25.54 3.994 296.2582 2Ø14 64.06 3rd&4th 0.25 0.455 35.18 4.101 577.3847 3Ø16 66.05 3rd&4th 0.25 0.455 35.72 4.107 596.192 3Ø16 91.94 3rd&4th 0.25 0.455 42.15 4.214 851.5058 2Ø20+2Ø12 95.02 3rd&4th 0.25 0.455 42.85 4.231 883.5816 3Ø16+2Ø14 39.71 TTBeam 0.25 0.307 41.05 4.19 541.9704 2Ø16+1Ø14 17.62 TTBeam 0.25 0.307 27.35 4.019 230.667 2Ø12 39.53 TTBeam 0.25 0.307 40.96 4.189 539.3849 2Ø16+1Ø14 14.15 TTBeam 0.25 0.307 24.51 3.987 183.7656 2Ø12 35.45 TTBeam 0.25 0.307 38.79 4.15 479.2101 2Ø14+1Ø16 9.23 TTBeam 0.25 0.307 19.79 3.955 118.9077 2Ø12 24.97 TTBeam 0.25 0.307 32.55 4.078 331.6862 1Ø16+1Ø14 18.67 TTBeam 0.25 0.307 28.15 4.032 245.2034 2Ø14 8.32 TTBeam 0.25 0.307 18.79 min 153 2Ø12 22.34 TTBeam 0.25 0.307 30.79 4.058 295.2955 2Ø14 4.23 TTBeam 0.25 0.307 13.40 min 153 2Ø12 96.69 Gbeam 0.25 0.357 55.09 4.593 1243.97 4Ø20 75.31 Gbeam 0.25 0.357 48.62 4.379 923.7605 3Ø20 81.12 Gbeam 0.25 0.357 50.46 4.434 1007.524 2Ø20+2Ø16 68.79 Gbeam 0.25 0.357 46.46 4.32 832.4168 2Ø20+2Ø12 83.48 Gbeam 0.25 0.357 51.19 4.456 1041.98 2Ø20+3Ø14 70.06 Gbeam 0.25 0.357 46.89 4.332 850.1398 2Ø20+2Ø12 83.07 Gbeam 0.25 0.357 51.06 4.452 1035.932 2Ø20+2Ø16 76.76 Gbeam 0.25 0.357 49.08 4.392 944.3415 3Ø20 80.23 Gbeam 0.25 0.357 50.18 4.455 1001.189 2Ø20+2Ø16 99.24 Gbeam 0.25 0.357 55.81 4.614 1282.614 2Ø24+2Ø16 AXIS-1 moment LOCATION b d Km KS AS1 Ф(As1 ) 2.74 TTBeam 0.25 0.307 10.78 min 154 2Ø12 13.23 TTBeam 0.25 0.307 23.70 3.981 171.5591 2Ø12 22.64 TTBeam 0.25 0.307 31.00 4.06 299.4085 2Ø14 175.83 1st&2nd 0.25 0.455 58.29 4.481 1853 4Ф24 142.2 1st&2nd 0.25 0.455 52.42 4.483 1401.061 1Ø16+4Ø20 185.74 1st&2nd 0.25 0.446 61.12 2126.2 4Ø24+2Ø16 79.896 3rd&4th 0.25 0.455 39.29 4.158 730.1265 2Ø20+1Ø12 87.07 3rd&4th 0.25 0.455 41.02 4.19 801.8095 4Ø16 108.94 3rd&4th 0.25 0.455 45.88 4.307 1031.219 2Ø20+2Ø16 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 98 123.09 Gbeam 0.25 0.355 62.50 2003.4 4Ø24+2Ø12 97.46 Gbeam 0.25 0.355 55.62 4.609 1265.333 3Ø24 124.68 Gbeam 0.25 0.355 62.91 2003.4 4Ø24+2Ø12 AXIS-2 moment LOCATION b d Km KS AS1 Ф(As1 ) 5.87 Head Room 0.25 0.307 15.78 min 153 2Ф12 14.54 Head Room 0.25 0.307 24.84 3.989 188.9253 2Ф12 30.95 Head Room 0.25 0.307 36.24 4.112 414.5485 2Ø14+1Ø12 11.59 TTBeam 0.25 0.307 22.18 3.971 149.915 2Ф12 33.42 TTBeam 0.25 0.307 37.66 4.131 449.7004 3Ø14 133.38 1st&2nd 0.25 0.455 50.76 1340 3Ø24 117.64 1st&2nd 0.25 0.455 47.68 4.352 1125.207 2Ø24+2Ø12 176.79 1st&2nd 0.25 0.446 59.62 1853 4Ø24 77.46 3rd&4th 0.25 0.455 38.69 4.148 706.1628 2Ø16+2Ø14 81.29 3rd&4th 0.25 0.455 39.63 4.164 743.9375 2Ø20+1Ø12 91.79 3rd&4th 0.25 0.455 42.11 4.213 849.9149 2Ø20+2Ø12 100.9 Gbeam 0.25 0.355 56.59 1429.3 1Ø16+4Ø20 94.9 Gbeam 0.25 0.355 54.88 1220.01 4Ø20 108.26 Gbeam 0.25 0.355 58.62 1345.67 4Ø20+1Ø12 AXIS-3 moment LOCATION b d Km KS As1 Ф(As1 ) 16.27 Head Room 0.25 0.307 26.28 4.006 212.305 2Ø12 20.93 Head Room 0.25 0.307 29.80 4.048 275.976 2Ø14 41.28 Head Room 0.25 0.307 41.86 4.207 565.6839 3Ø16 18.53 TTBeam 0.25 0.307 28.04 4.028 243.1233 2Ø14 9.5 TTBeam 0.25 0.307 20.08 3.957 122.4479 2Ø12 11.38 TTBeam 0.25 0.307 21.98 3.97 147.1616 2Ø12 8.83 TTBeam 0.25 0.307 19.36 3.952 113.6683 2Ø12 16.86 TTBeam 0.25 0.307 26.75 4.012 220.3333 2Ø12 17.25 TTBeam 0.25 0.307 27.06 4.016 225.6547 2Ø12 9.25 TTBeam 0.25 0.307 19.81 3.955 119.1653 2Ø12 29.58 TTBeam 0.25 0.307 35.43 4.104 395.4278 2Ø16 35.82 TTBeam 0.25 0.307 38.99 4.153 484.5618 2Ø14+1Ø16 11.77 TTBeam 0.25 0.307 22.35 3.972 152.2816 2Ø12 126.82 1st&2nd 0.25 0.455 49.50 1214.93 4Ø20 110 1st&2nd 0.25 0.455 46.10 4.312 1042.462 2Ø24+1Ø14 110.44 1st&2nd 0.25 0.455 46.19 4.315 1047.36 2Ø24+1Ø14 107.01 1st&2nd 0.25 0.455 45.47 4.297 1010.598 2Ø24+1Ø12 98.47 1st&2nd 0.25 0.455 43.62 4.25 919.7747 2Ø14+2Ø20 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 99 106.94 1st&2nd 0.25 0.455 45.46 4.3 1010.642 2Ø24+1Ø12 101.23 1st&2nd 0.25 0.455 44.23 4.267 949.3372 3Ø20 109.9 1st&2nd 0.25 0.455 46.08 4.312 1214.45 4Ø20 103.44 1st&2nd 0.25 0.455 44.71 4.278 972.5633 3Ø12+2Ø20 119.22 1st&2nd 0.25 0.455 47.99 4.36 1142.416 3Ø20+2Ø12 82.61 3rd&4th 0.25 0.455 39.95 4.17 757.107 2Ø20+1Ø14 88.42 3rd&4th 0.25 0.455 41.33 4.198 815.796 2Ø20+1Ø16 58.23 3rd&4th 0.25 0.455 33.54 4.085 522.7902 2Ø16+1Ø14 33.26 3rd&4th 0.25 0.455 25.35 4.001 292.4687 2Ø14 76.67 3rd&4th 0.25 0.455 38.49 4.147 698.7923 2Ø16+2Ø14 56.91 3rd&4th 0.25 0.455 33.16 4.082 510.564 2Ø16+1Ø12 38.72 3rd&4th 0.25 0.455 27.35 4.019 342.0125 2Ø16 76.07 3rd&4th 0.25 0.455 38.34 4.142 692.4878 2Ø16+2Ø14 60.62 3rd&4th 0.25 0.455 34.22 4.092 545.1803 2Ø16+1Ø14 81.59 3rd&4th 0.25 0.455 39.70 4.165 746.8623 2Ø20+1Ø14 51.57 3rd&4th 0.25 0.455 31.57 4.066 460.8431 3Ø14 99.4 3rd&4th 0.25 0.455 43.82 4.256 929.7723 3Ø20 89.94 Gbeam 0.25 0.355 53.43 1317.8 3Ø20+2Ø16 87.35 Gbeam 0.25 0.355 52.65 1246.82 4Ø20 86.16 Gbeam 0.25 0.355 52.29 1246.82 4Ø20 83.54 Gbeam 0.25 0.355 51.49 1160.19 3Ø20+2Ø12 85.72 Gbeam 0.25 0.355 52.16 1246.82 4Ø20 89.12 Gbeam 0.25 0.355 53.19 1317.8 3Ø20+2Ø16 71.8 Gbeam 0.25 0.355 47.74 4.354 880.6118 3Ø16+2Ø14 70.2 Gbeam 0.25 0.355 47.20 4.34 858.2197 2Ø20+2Ø12 68.24 Gbeam 0.25 0.355 46.54 4.324 831.1824 2Ø16+2Ø14 73.15 Gbeam 0.25 0.355 48.18 4.36 898.4056 3Ø16+2Ø14 Inclined beam(SAP) landing beams LOCATION AS Ф(AS) LOCATION AS Ф(AS) Gbeam 1377.25 4Ф20+1Ø14 G-floor 2172.62 4Ф24+2Ф16 Gbeam 1235.31 4Ф20 G-floor 2002.28 4Ф24+2Ø12 Gbeam 1176.7 2Ø24+1Ф20 G-floor 2291.24 4Ø24+2Ø20 Gbeam 961.86 5Ø16 1st&2nd 1500.52 4Ø20+2Ø14 Gbeam 966.54 5Ø16 1st&2nd 1475.75 4Ø20+2Ø12 1st&2nd 1723.22 4Ø24 1st&2nd 1839 3Ø20+2Ø24 1st&2nd 1426.32 4Ø20+1Ø16 3rd&4th 729.35 4Ø16 1st&2nd 1183.39 2Ø24+1Ф20 3rd&4th 697.65 2Ø16+2Ø14 1st&2nd 991.29 5Ø16 3rd&4th 984.77 5Ø16 1st&2nd 1269.18 3Ø24 roof 343.9 2Ø16 3rd&4th 847.42 2Ø20+2Ф12 roof 225 2Ф12 3rd&4th 745 4Ø16 roof 225 2Ф12 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 100 3rd&4th 692.88 2Ø16+2Ø14 roof 225 2Ф12 3rd&4th 520.69 2Ø14+2Ø12 3rd&4th 678.21 2Ø16+2Ø14 TTBeam 159.65 2Ф12 TTBeam 162.19 2Ф12 TTBeam 204.96 2Ф12 TTBeam 157.5 2Ф12 Head Room 157.5 2Ф12 Head Room 157.5 2Ф12 Head Room 157.5 2Ф12 corridor beams added for prec 2.4 0.25 0.455 6.81 min 227 2Ø12 6.15 0.25 0.455 10.90 min 227 2Ø12 3.21 0.25 0.455 7.88 min 227 2Ø12 3.9 0.25 0.455 8.68 min 227 2Ø12 5.65 0.25 0.455 10.45 min 227 2Ø12 16.36 0.25 0.455 17.78 min 227 2Ø12 6.62 0.25 0.455 11.31 min 227 2Ø12 5.1.3 Design of beams for shear and torsion Shear reinforcement design Beam sections are subjected to shear forces in addition to flexural actions. Shear is resisted by the combined actions of the following Shear resistance of concrete in compression zone Shear reinforcements or stirrups Dowel action in tension bars across crack Aggregate interlocking across the inclined crack in tension zone ominal reinforcement The shear force V C carried by the concrete in members with out significant axial forces shall be taken as V C = 0.25f ctd K 1 K 2 b w d Where K 1 = (1+50ρ) < 2.0 K 2 = 1.6-d > 1.0 (d in meters). For members where more than 50%of the bottom reinforcement is curtailed, K 2 = 1. ρ max =0.04 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 101 A s is the area of tensile reinforcement The ultimate limit state in shear is characterized by either diagonal compression failure of concrete or failure of shear reinforcement due to diagonal tension. Diagonal compression failure of concrete To avoid compression failure of concrete the shear resistance of the section , V RD shall not be less than the applied shear force V d . Where: V RD =0.25 *fcd *b w *d If V RD < V d ………….increase the concrete section Diagonal Tension failure of web reinforcement If the applied shear force V d < V RD > the shear resistance of the section Vc Shear reinforcement need be provided. The spacing in this case is given by S=(d-d’)Asv*fyk/(Vd-Vc) ( section 4.5 EBCS-2,1995) The shear between the critical section which is at a distance d from the face of the support, and the point beyond which maximum spacing is used is reinforced by the difference of shear capacity of concrete. The region beyond this is reinforced with maximum spacing.( section 4.5- 6,EBCS-2,1995,page 43-46) The maximum spacing S max between stirrups, in the longitudinal direction should be = 800 = d Shear design sample Axis –D 1 st & 2nd floor level As, actual =1941.5 mm 2 , d=446mm ,b= 250 mm ρ=As/bd=0.0174 V C = 0.25f ctd K 1 K 2 b w d K1 =1+50ρ=1.87 k2=1.6-d=1.154 Then Vc=62.1KN V RD =0.25 *fcd *b w *d=0.25*11.33*250*446=311.1 KN Taking the design shear at d distance from the face of the column Vd=112.6KN S=(d-d’)Asv*fyk/(Vd-Vc) = 2*50.3*260.87*(446-43)/(112.6-62.1)=209.7 mm But Vd< 2/3Vrd as per EBCS 2-1995 Smax= 0.5d=220mm Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 102 Use Φ8 c/c 200mm.Since the minimum reinforcement &the design result are all most similar , so use Φ8 c/c 200mm through out the length The reimaing shear designs are displayed in the reinforcement detail Torsional reinforcement Torsion results from the monolithic character of construction; and any assymetric in the loading of the floor slab produces torsion to the supporting beams. Torsion shear stresses create diagonal tension resulting in diagonal crack. Thus, we need to provide both closed stirrups and longitudinal steel to avoid brittle fracture. Torsional resistance of concrete Torsional effect may be disregarded whenever the design torque T SD is less than T C given by ( section 4.6.4 EBCS-2, 1995,page 48) T C = 1.2f ctd A ef h ef however, minimum reinforcement may be provided in such away that ρ min = 0.4 /f yk , and the spacing of stirrups shall not exceed U ef /8 More ever, at least one longitudinal bar shall be placed at each corner of the closed stirrup with spacing not exceeding 350mm. Limiting value of ultimate shear In order to prevent diagonal compression failure in the concrete the torsional resistance, T RD , of a section shall not be less than the applied torque T Sd . Where T RD = 0.8f cd A ef h ef > T sd . Design for torsional reinforcement When T sd >T c ……..high torsional moment. Hence ,both longituidinal bars Asl & closed stirrups A str must be provided. Combined actions a) Torsion and bending or torsion and axial forces Simple super position by separately determining area of reinforcements may be applied. b) Torsion and shear limiting values for torsion and shear are T Rd, com = β t T Rd and V Rd,com = β v V Rd in which Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 103 Further the torsional and shear resistance of the concrete shall be T C, com = β tc T C and V C,com = β vc V C in which (section 4.6.6 EBCS-2,1995,page 49-50.) AXIS-4 (TORSIOAL DESIG) Applied shear actions , Tsd=6.01kn.m Vsd=32.19kn 350 Equivalent hallow section 250 A=250*350=87.5*10 3 mm 2 U=2(250+350)=1200mm Hef=def/5=(350-43)/5=61.4<A/U=72.92 250 Aef=[350-61.4]8[250-61.4]=54429.96mm 2 Uef={[350-61.4]+[250-61.4]}*2=954.4mm Check shear capacity V C = 0.25f ctd K 1 K 2 b w d ρ=As/bd=[3*201]/[250*307]=0.00689 K 1 =1+50ρ=1.3446 K 2 =1.6-d=1.291 Vc=0.25*1.032*1.291*1.3446*250*309=34.6kn V RD =0.25 *fcd *b w *d=218.81kn Torsion T C = 1.2f ctd A ef h ef =4.14kn.m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 104 T RD = 0.8f cd A ef h ef =30.29kn.m Using combined action Factor for combined action Using the above formula βt=0.8033 βtc=0.842 βv=0.5956 βvc=0.540 Combined section capacity Vc=βvc*Vc=0.540*34.6=18.63kn Vrd,com=Vrd*βv=0.5956*218.81=130.41kn Tc,com=Tc*βtc=0.842*4.41=3.48kn.m Trd,com=Trd*βt=0.8033*30.29=24.32kn.m Check Vsd with VC,com Vsd=32.19>Vc,com=18.63 It requires shear reinforcement S=(d-d’)Asv*fyk/(Vd-Vc) = [2*50.3*260.87*(309-41)]/[13.56*10 3 ] =518.67mm 2/3Vrd=86.94 Smax=0.5d=0.5*309=154.5 UseΦ8c/c 150mm Tc=3.486kn.m<Tsd=6.01kn.m needs torsional reinforcement Tef=Tsd-Tc=2.614kn.m Longitudinal reinforcement Al= =87.85mm 2 Al/2=43.93mm 2 Top reinforcement=602.1+43.93=646.03mm 2 So use 2Φ20 Bottom reinforcement=235.5+43.93=279.43 So use 2Φ14 Spacing S= =546.455mm Smax=0.5d hence use Φ8c/c 150mm The same procedure is applied for the reimaing torsional action. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 105 Sample reinforcement detail AXIS-4 5.2 Column design Columns are axially loaded vertical members, which carry their load primarily in compression. The majority of compression or tension members carry a portion of the load in bending which may arise due to the unbalanced moments in the members connected to their ends. The result of such bending moments in axially loaded members into reduces the range of axial force that the member can carry. For this reason, it is essential to note that the effect of bending in axially loaded members should be considered in designing these columns. 5.2.1 Design Procedure 1. To design a column in a particular frame first the frame is classified wheather it is sway or non sway. 2. To determine the nature of the frame we substitute the beams and columns by one substitute frame 3. The value of the axial force on each substitute frame column is obtained by adding the axial load each column for the story including self weight. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 106 4. the values of the stiffness coefficients of the substitute frame is given by For beams=2*∑Kbi For column =∑Kci Where: Kbi stiffness coefficient of beam Where: Kci stiffness coefficient of column 5. The effective length of the substitute frame is computed for each storey assuming as sway frame as shown below. The effective length buckling Le of a column in a given plane is obtained from the following approximate equation provided that certain restriction is complied with. a) Non sway mode Le/L= (αm+0.4)/ (αm+0.8) ≥ 0.8 b) Sway mode: Conservatively: Le/L=√ (1+0.8 αm) ≥ 1.15 Where: αm is a stiffness coefficient which will be discussed Using the following theoretical model. K 12 K c K 11 K 21 K 22 K c1 K c2 α1 =Kc1+Kc K11+K12 α2 =Kc2+Kc K21+K22 α1 = α1 + α2 2 • Kc1 and Kc2 are column stiffness coefficients (EI/l) • Kc is the stiffness coefficient of the column being designed α =1.0 if opposite end elastically or rigidly restrained α= 0.5 if opposite ends are free to rotate α= 0 for cantilever beam Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 107 The above approximate equation for effective length calculation is applicable for values of α1 and α 2 not exceeding 10. If a base is designed to resist the column moment may be taken as 1.0. 6. the dimension of the substitute column is computed to find the moment of inertia of the section (Ic) 7. The amount of reinforcement required by the substitute column is computed and the moment of inertia of the reinforcement with respect to the centroid of the concrete section is determined. In lieu of more accurate determination, the first order moment, Mdl, at critical section of the substitute may be determined using: Mdl = α 2 +3 HL α 1 + α 2 +6 Where: H= the total horizontal reaction at the bottom of the story. L= the story shear 8. The buckling load of a story may be assumed to be equal to that of the substitute Beam-column frame, and may be determined as Ncr= π 2 EIe Le 2 Where: EIe is the effective stiffness of the substitute column designed Le is the effective length. In lieu of more accurate determination, the effective stiffness of a column may be taken as: EIe= 0.2Ec Ic + Es Is Where Ec= 1100fcd Es is the modulus of elasticity of steel Ic, Is are the moment of inertia of the concrete and reinforcement sections, respectively of the substitute column, with respect to the centroid of the concrete section. Computation of moment of inertia of reinforced concrete section with respect to the centroid of the concrete. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 108 Is = (n (П * r 4 )/4 + П * r 2 *d 2 ) Where n = number of bars As/2 As/2 S S d Reinforcing the substitute column using the biaxial chart using the following formulas: ν = N sd * 10 3 A c *f cd µ = M dl * 10 6 A c * f cd *S w = A s , tot *f yd A c *f cd We have designed five columns for solid and six columns for the pre-cast slab systems. Substitute column Determination of substitute column for each axis. Sample substitute frame is shown below. AXIS -4 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 109 The result of the substitute frames are tabulated as follows AXIS- 4 LEVE L α1 α2 H L Le Nsd HL Md1 µ v w Found 8.92 1.00 301.09 1.50 3.34 1975.36 451.64 113.48 0.04 0.4 4 mi n Groun d 2.9 8.92 224.05 3.80 9.10 1903.78 851.39 569.50 0.20 0.4 3 0.2 1 3.22 2.90 158.88 3.05 5.66 1446.68 484.58 235.89 0.08 0.3 3 mi n 2 3.22 3.22 98.67 3.05 5.77 999.10 300.94 150.47 0.05 0.2 3 mi n 3 3.22 3.22 50.43 3.05 5.77 502.31 153.81 76.91 0.03 0.1 1 mi n 4 4.69 3.22 15.67 3.05 6.23 101.84 47.79 21.37 0.01 0.0 2 mi n As Ǿ Is Ic EIe Ncr Nsd/Nc r condition 3135.008 10Ǿ20 196957002 1.28E+1 0 7.129E+13 63007705.1 8 0.03 non-sway 3403.96 8Ǿ24 226854001 1.28E+1 0 7.727E+13 9210012.38 4 0.21 ''sway'' 3135.008 10Ǿ20 196957002 1.28E+1 0 7.129E+13 21917624.0 2 0.07 non-sway 3135.008 10Ǿ20 196957002 1.28E+1 0 7.129E+13 21126915.3 7 0.05 non-sway 3135.008 10Ǿ20 196957002 1.28E+1 0 7.129E+13 21126915.3 7 0.02 non-sway 3135.008 10Ǿ20 196957002 1.28E+1 0 7.129E+13 18127122.2 9 0.01 non-sway AXIS-2 LEVEL α1 α2 H L Le Nsd HL Md1 µ v w Found 11.9 1.00 253.85 1.50 3.73 833.87 380.78 80.42 0.07 0.32 min Ground 4.02 11.94 197.14 3.80 10.33 778.22 749.13 509.75 0.42 0.30 0.79 1 4.46 4.02 140.73 3.05 6.39 627.80 429.23 208.06 0.17 0.24 0.2 2 4.46 4.46 91.53 3.05 6.52 471.96 279.17 139.58 0.11 0.18 0.1 3 4.46 4.46 51.78 3.05 6.52 315.78 157.93 78.96 0.06 0.12 0.02 4 12.6 4.46 23.18 3.05 8.53 166.29 70.70 22.86 0.02 0.06 min Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 110 5 6.31 12.61 17.49 3.05 8.93 72.19 53.34 33.42 0.03 0.03 min As Ǿ Is Ic EIe Ncr Nsd/Ncr condition 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 17262278.72 0.05 non-sway 7774.04 10Ǿ32 291981824 4.278E+09 6.906E+13 6385878.969 0.12 ''sway'' 1968.111 4Ǿ26 77089269 4.278E+09 2.608E+13 6293839.196 0.10 non-sway 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 5647044.625 0.08 non-sway 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 5647044.625 0.06 non-sway 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 3294931.883 0.05 non-sway 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 3009827.243 0.02 non-sway AXIS-3 LEVEL α1 α2 H L Le Nsd HL Md1 µ v w Found 8.91 1 324.3 1.5 3.34 2061.48 486.41 122.29 0.04 0.46 min Ground 2.9 8.91 270.6 3.8 9.09 1915.2 1028.2 687.59 0.25 0.43 0.33 1 3.22 2.9 200.7 3.05 5.66 1452.15 612.14 297.99 0.11 0.33 min 2 3.22 3.22 123.1 3.05 5.77 1014.74 375.46 187.73 0.07 0.23 min 3 3.22 3.22 59.96 3.05 5.77 583.79 182.88 91.44 0.03 0.13 min 4 6.26 3.22 12.6 3.05 6.68 155.85 38.43 15.44 0.01 0.04 min 5 7.82 6.26 5.44 3.05 7.85 59.35 16.592 7.65 0.00 0.01 min As Ǿ Is Ic EIe Ncr Nsd/Ncr condition 3135.008 10Ǿ20 196957002 1.28E+10 7.129E+13 63007705.18 0.00 non-sway 5531.435 8Ǿ30 354950030 1.28E+10 1.029E+14 12277159.55 0.00 non-sway 3135.008 10Ǿ20 196957002 1.28E+10 7.129E+13 21940864.41 0.00 non-sway 3135.008 10Ǿ20 196957002 1.28E+10 7.129E+13 21112271.86 0.00 non-sway 3135.008 10Ǿ20 196957002 1.28E+10 7.129E+13 21112271.86 0.00 non-sway 3135.008 10Ǿ20 196957002 1.28E+10 7.129E+13 15751926.29 0.01 non-sway 3135.008 10Ǿ20 196957002 1.28E+10 7.129E+13 11406365.47 0.03 non-sway AXIS- B LEVEL α1 α2 H L Le Nsd HL Md1 µ v w Found 14.9 1 319.3 1.5 4.07 2252.2 478.92 87.63 0.07 0.88 0.08 Ground 4.84 14.86 269 3.8 11.32 2108.1 1022.2 710.34 0.26 0.47 0.35 1 5.37 4.84 209.9 3.05 6.88 1566.1 640.1 309.51 0.25 0.61 0.4 2 5.37 5.374 134.4 3.05 7.02 1032.92 409.8 204.90 0.17 0.40 0.13 3 5.37 5.374 72.69 3.05 7.02 540.28 221.7 110.85 0.09 0.21 0.02 4 7.82 5.374 18.75 3.05 7.64 44.99 57.188 24.95 0.02 0.02 0.03 As Ǿ Is Ic EIe Ncr Nsd/Ncr condition Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 111 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 14518736.79 0.00 non-sway 3444.195 8Ǿ24 226854001 1.28E+10 7.727E+13 5945286.218 0.00 non-sway 3936.223 8Ǿ26 154178539 4.278E+09 4.15E+13 8644145.467 0.00 non-sway 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 4868281.781 0.00 non-sway 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 4868281.781 0.00 non-sway 1812.608 6Ǿ20 68346169 4.278E+09 2.433E+13 4110202.46 0.02 non-sway AXIS-E LEVEL α1 α2 H L Le Nsd HL Md1 µ v w Found 4.44 1 473 1.5 2.67 2061.48 709.43 248.05 0.12 0.57 min Ground 1.45 4.44 368.2 3.8 6.96 1915.2 1399 875.39 0.31 0.43 0.47 1 1.61 1.45 290.6 3.05 4.55 1452.15 886.36 435.35 0.21 0.40 0.23 2 1.61 1.61 195.9 3.05 4.61 1014.74 597.62 298.81 0.15 0.28 0.13 3 1.61 1.61 113.7 3.05 4.61 583.79 346.82 173.41 0.08 0.16 0.04 4 1.33 1.61 44.02 3.05 4.5 155.88 134.26 69.23 0.03 0.04 0.03 5 4.23 1.33 14.49 3.05 5.48 59.35 44.195 16.55 0.01 0.02 min As Ǿ Is Ic EIe Ncr Nsd/Ncr condition 2562.848 6Ǿ24 139106542 8.552E+09 4.914E+13 67961309.81 0.00 non-sway 7999.306 10Ǿ32 504623744 1.28E+10 1.328E+14 27034263.05 0.00 non-sway 3200.122 6Ǿ26 163439197 8.552E+09 5.401E+13 25720151.2 0.00 non-sway 2562.848 6Ǿ24 139106542 8.552E+09 4.914E+13 22797247.4 0.00 non-sway 2562.848 6Ǿ24 139106542 8.552E+09 4.914E+13 22797247.4 0.00 non-sway 2562.848 6Ǿ24 139106542 8.552E+09 4.914E+13 23925401.56 0.01 non-sway 2562.848 6Ǿ24 139106542 8.552E+09 4.914E+13 16133297.64 0.05 non-sway Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 112 5.2.2Design of isolated columns For buildings, a design method may be used which assumes the compression members to be isolated and adopts a simplified shape for the deformed axis of the column. Total eccentricity The total eccentricity to be used for the design of columns of constant cross section at the critical section is given by: e tot = e e +e a +e 2 Where e e is equivalent constant first-order eccentricity of the design axial load - for first-order eccentricity e o is equal at both ends of a column e e = e o for first-order moment varying linearly along the length, the equivalent eccentricity is the higher of the following two values e e =0.6e o2 + 0.4 e o1 e e =0.4 e o2 e o1 and e o2 are first-order eccentricities at the ends e o2 being positive and greater in magnitude than e o1. e a is the additional eccentricity to account for geometric imperfection, introduced by increasing the eccentricity of the longitudinal force acting in the most unfavorable direction. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 113 e a = Le/ 300 ≥20mm Le is the effective length of isolated column. e 2 is the second order eccentricity According to EBCS-2, 1995 Art. 4.4.6(2) second order effects in compressive members need not be taken into account in the following cases a) for sway frames the greater of λ ≤ 25 ; which ever is maximum 15/√ﬠ d ﬠ d = N sd /A c f cd b) for non sway frames λ ≤ 50 – 25 (M 1 /M 2 ) Where M 1 and M 2 are the 1 st order (calculated moments at the ends, M 2 being always positive and greater than M 1 , and M 1 being positive if member is bent in single curvature and negative if bent in double curvature. For sway frames in the absence of rigorous method the amplified sway moments method can be employed to obtain the sway contribution by multiplying the 1 st order moment by magnification factor given by: σ s = 1/(1 – Nsd/Ncr) provided; Nsd/Ncr ≤ 0.25 Determination of first order moment col A-3 Mdy level M1 M2 P eo1 e02 0.6eo2+0.4eo1 0.4eo2 eo M found -10.62 11.17 1368.42 - 0.0077608 0.00816 0.0017933 0.003265 0.003265 4.468 ground - 115.16 118.57 1076.01 -0.107025 0.11019 0.0233065 0.044078 0.044078 47.428 first -99.46 111.08 806.45 - 0.1233306 0.13774 0.0333114 0.055096 0.055096 44.432 second -37.56 42.01 557.52 - 0.0673698 0.07535 0.018263 0.030141 0.030141 16.804 third -41.94 44.65 314.86 - 0.1332021 0.14181 0.0318046 0.056724 0.056724 17.86 fourth -16.04 43.3 113.28 -0.141596 0.38224 0.1727048 0.152895 0.172705 19.564 Mdx found -10.87 17.56 1368.42 - 0.0079435 0.01283 0.004522 0.005133 0.005133 7.024 ground -13.75 15.94 1076.01 - 0.0127787 0.01481 0.0037769 0.005926 0.005926 6.376 first -18.35 21.51 806.45 -0.022754 0.02667 0.0069019 0.010669 0.010669 8.604 second -70.31 91.01 557.52 - 0.16324 0.0474996 0.065296 0.065296 36.404 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 114 0.1261121 third -62.69 80.18 314.86 - 0.1991044 0.25465 0.07315 0.101861 0.101861 32.072 fourth -10.99 35.68 113.28 - 0.0970162 0.31497 0.1501766 0.125989 0.150177 17.012 column B-3 Mdy level M1 M2 P eo1 e02 0.6eo2+0.4eo1 0.4eo2 eo M found 0.42 0.89 2252.23 0.0001865 0.0004 0.0003117 0.000158 0.000312 0.702 ground - 120.88 131.3 1602.82 - 0.0754171 0.08192 0.018984 0.032767 0.032767 52.52 first -11.14 12.12 1278.28 - 0.0087148 0.00948 0.002203 0.003793 0.003793 4.848 second -2.3 5.93 1032.91 - 0.0022267 0.00574 0.0025539 0.002296 0.002554 2.638 third -1.48 2.82 429.55 - 0.0034455 0.00657 0.0025608 0.002626 0.002626 1.128 fourth -6.08 7.08 44.99 - 0.1351411 0.15737 0.0403645 0.062947 0.062947 2.832 Mdx found -5.96 9.68 2252.23 - 0.0026463 0.0043 0.0015203 0.001719 0.001719 3.872 ground -24.34 36.7 1602.82 - 0.0151857 0.0229 0.007664 0.009159 0.009159 14.68 first - 113.43 116.88 1278.28 - 0.0887364 0.09144 0.0193666 0.036574 0.036574 46.752 second -56.35 56.89 1032.91 - 0.0545546 0.05508 0.0112246 0.022031 0.022031 22.756 third -71.73 92.79 429.55 - 0.1669887 0.21602 0.0628146 0.086407 0.086407 37.116 fourth -6.39 44.02 44.99 - 0.1420316 0.97844 0.5302512 0.391376 0.530251 23.856 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 115 column E-3 Mdy level M1 M2 P eo1 e02 0.6eo2+0.4eo1 0.4eo2 eo M found -0.67 4.15 1790.15 - 0.0003743 0.00232 0.0012412 0.000927 0.001241 2.222 ground -0.75 5.18 1633.3 - 0.0004592 0.00317 0.0017192 0.001269 0.001719 2.808 first -36.87 37.55 949.05 - 0.0388494 0.03957 0.0081998 0.015826 0.015826 15.02 second -32.36 33.38 671.98 - 0.0481562 0.04967 0.010542 0.01987 0.01987 13.352 third -30.2 32.22 387.46 - 0.0779435 0.08316 0.0187168 0.033263 0.033263 12.888 fourth -16.38 23.42 90.38 - 0.1812348 0.25913 0.082983 0.103651 0.103651 9.368 H.room -13.77 25.07 28.97 - 0.4753193 0.86538 0.3290991 0.346151 0.346151 10.028 Mdx found -32.76 97.68 1790.15 - 0.0183001 0.05457 0.0254191 0.021826 0.025419 45.504 ground - 101.76 119.28 1633.3 - 0.0623033 0.07303 0.0188967 0.029212 0.029212 47.712 first -141.7 147.18 949.05 - 0.1493072 0.15508 0.033326 0.062033 0.062033 58.872 second - 120.83 129.96 671.98 - 0.1798119 0.1934 0.0441144 0.077359 0.077359 51.984 third - 100.46 112.08 387.46 - 0.2592784 0.28927 0.0698498 0.115707 0.115707 44.832 fourth -34.9 55.16 90.38 - 0.3861474 0.61031 0.2117283 0.244125 0.244125 22.064 H.room -14.95 29.25 28.97 - 0.5160511 1.00967 0.3993787 0.403866 0.403866 11.7 column E-2 Mdy level M1 M2 P eo1 e02 0.6eo2+0.4eo1 0.4eo2 eo M found 1.28 8.38 833.97 0.0015348 0.01005 0.0066429 0.004019 0.006643 5.54 ground 1.7 4.34 456.7 0.0037224 0.0095 0.0071907 0.003801 0.007191 3.284 first -1.24 1.47 461.38 - 0.0026876 0.00319 0.0008366 0.001274 0.001274 0.588 second -3.93 5.69 401.311 - 0.0097929 0.01418 0.00459 0.005671 0.005671 2.276 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 116 third -9.19 9.78 297.53 - 0.0308876 0.03287 0.0073673 0.013148 0.013148 3.912 fourth -5.13 9.34 157.38 - 0.0325963 0.05935 0.0225696 0.023739 0.023739 3.736 H.room -9.89 17.57 68.19 - 0.1450359 0.25766 0.0965831 0.103065 0.103065 7.028 Mdx found -62.05 105.98 833.97 - 0.0744032 0.12708 0.0464861 0.050832 0.050832 42.392 ground - 140.04 144.6 456.7 - 0.3066346 0.31662 0.0673177 0.126648 0.126648 57.84 first - 122.54 126.93 461.38 - 0.2655945 0.27511 0.0588279 0.110044 0.110044 50.772 second -97.43 104.25 401.311 - 0.2427793 0.25977 0.0587524 0.103909 0.103909 41.7 third -68.94 76.62 297.53 - 0.2317077 0.25752 0.0618291 0.103008 0.103008 30.648 fourth -33.02 33.98 157.38 - 0.2098106 0.21591 0.0456221 0.086364 0.086364 13.592 H.room -24.69 32.2 68.19 - 0.3620766 0.47221 0.1384954 0.188884 0.188884 12.88 column H-3 Mdy level M1 M2 P eo1 e02 0.6eo2+0.4eo1 0.4eo2 eo M found -4.61 19.84 1557.47 - 0.0029599 0.01274 0.0064592 0.005095 0.006459 10.06 ground -1.89 29.48 560.41 - 0.0033725 0.0526 0.0302136 0.021042 0.030214 16.932 first -16.62 22.94 514.99 - 0.0322725 0.04454 0.0138177 0.017818 0.017818 9.176 second 16.68 18.59 434.68 0.0383731 0.04277 0.0410095 0.017107 0.041009 third 9.17 23.29 321.27 0.028543 0.07249 0.0549133 0.028997 0.054913 17.642 fourth -2.9 23.43 240.88 - 0.0120392 0.09727 0.0535453 0.038907 0.053545 12.898 H.room -13.21 16.42 124.59 - 0.1060278 0.13179 0.0366643 0.052717 0.052717 6.568 Mdx found -15.38 92.96 1557.47 -0.009875 0.05969 0.0318619 0.023875 0.031862 49.624 ground - 116.33 165.83 560.41 - 0.2075802 0.29591 0.0945129 0.118363 0.118363 66.332 first -68.08 123.76 514.99 - 0.1321967 0.24032 0.0913105 0.096126 0.096126 49.504 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 117 Moment due to imperfection column E- 2 x-dxn level Le eacalc 20 ea Nsd Ma foundation 1.39253 0.004642 0.02 0.02 833.97 16.6794 ground 10.2917 0.034306 0.02 0.0343056 456.7 15.667383 first 2.80135 0.009338 0.02 0.02 461.38 9.2276 second 2.81176 0.009373 0.02 0.02 401.311 8.02622 third 2.81176 0.009373 0.02 0.02 297.53 5.9506 fourth 2.91835 0.009728 0.02 0.02 157.38 3.1476 H.room 2.93109 0.00977 0.02 0.02 68.19 1.3638 y-dxn level Le eacalc 20 ea Nsd Ma foundation 1.23737 0.004125 0.02 0.02 833.97 16.6794 ground 5.36735 0.017891 0.02 0.02 456.7 9.134 first 2.22377 0.007413 0.02 0.02 461.38 9.2276 second 2.24308 0.007477 0.02 0.02 401.311 8.02622 third 2.24308 0.007477 0.02 0.02 297.53 5.9506 fourth 2.49388 0.008313 0.02 0.02 157.38 3.1476 H.room 2.74134 0.009138 0.02 0.02 68.19 1.3638 column E- 3 x-dxn 0 level Le eacalc 20 ea Nsd Ma foundation 1.34109 0.00447 0.02 0.02 1790.15 35.803 ground 3.42767 0.011426 0.02 0.02 1633.3 32.666 first 2.56249 0.008542 0.02 0.02 949.05 18.981 second 2.57922 0.008597 0.02 0.02 671.98 13.4396 third 2.57922 0.008597 0.02 0.02 387.46 7.7492 fourth 2.76646 0.009222 0.02 0.02 90.38 1.8076 H.room 2.88328 0.009611 0.02 0.02 28.97 0.5794 y-dxn level Le eacalc 20 ea Nsd Ma second -54 97.89 434.68 - 0.1242293 0.2252 0.0854284 0.09008 0.09008 39.156 third -37.45 74.51 321.27 - 0.1165686 0.23192 0.0925265 0.092769 0.092769 29.804 fourth -12.36 25.33 240.88 - 0.0513119 0.10516 0.0425689 0.042062 0.042569 10.254 H.room -17.28 17.41 124.59 -0.138694 0.13974 0.028365 0.055895 0.055895 6.964 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 118 foundation 1.29167 0.004306 0.02 0.02 1790.15 35.803 ground 3.27488 0.010916 0.02 0.02 1633.3 32.666 first 2.40302 0.00801 0.02 0.02 949.05 18.981 second 2.42191 0.008073 0.02 0.02 671.98 13.4396 third 2.42191 0.008073 0.02 0.02 387.46 7.7492 fourth 2.64814 0.008827 0.02 0.02 90.38 1.8076 H.room 2.78359 0.009279 0.02 0.02 28.97 0.5794 column A- 3 x-dxn level Le eacalc 20 ea Nsd Ma foundation 1.38032 0.004601 0.02 0.02 1368.42 27.3684 ground 3.53444 0.011781 0.02 0.02 1076.01 21.5202 first 2.68623 0.008954 0.02 0.02 806.45 16.129 second 2.70014 0.009 0.02 0.02 557.52 11.1504 third 2.70014 0.009 0.02 0.02 314.86 6.2972 fourth 2.75236 0.009175 0.02 0.02 113.28 2.2656 H.room 0 0 0 0 y-dxn 0 level Le eacalc 20 ea Nsd Ma foundation 1.43125 0.004771 0.02 0.02 1368.42 27.3684 ground 3.65724 0.012191 0.02 0.02 1076.01 21.5202 first 2.84348 0.009478 0.02 0.02 806.45 16.129 second 2.85241 0.009508 0.02 0.02 557.52 11.1504 third 2.85241 0.009508 0.02 0.02 314.86 6.2972 fourth 2.88509 0.009617 0.02 0.02 113.28 2.2656 H.room 0 0 0.02 0.02 0 0 column B- 3 x-dxn 0 level Le eacalc 20 ea Nsd Ma foundation 1.34109 0.00447 0.02 0.02 2252.23 45.0446 ground 3.42767 0.011426 0.02 0.02 1602.82 32.0564 first 2.56249 0.008542 0.02 0.02 1278.28 25.5656 second 2.57922 0.008597 0.02 0.02 1032.91 20.6582 third 2.57922 0.008597 0.02 0.02 429.55 8.591 fourth 2.64323 0.008811 0.02 0.02 44.99 0.8998 H.room 0 0 0.02 0.02 0 0 y-dxn 0 level Le eacalc 20 ea Nsd Ma Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 119 foundation 1.43125 0.004771 0.02 0.02 2252.23 45.0446 ground 3.65724 0.012191 0.02 0.02 1602.82 32.0564 first 2.84348 0.009478 0.02 0.02 1278.28 25.5656 second 2.85241 0.009508 0.02 0.02 1032.91 20.6582 third 2.85241 0.009508 0.02 0.02 429.55 8.591 fourth 2.88509 0.009617 0.02 0.02 44.99 0.8998 column H- 3 level ea Nsd Ma foundation 0.02 1557.47 31.1494 ground 0.02 560.41 11.2082 first 0.02 514.99 10.2998 second 0.02 434.68 8.6936 third 0.02 321.27 6.4254 fourth 0.02 240.88 4.8176 H.room 0.02 124.59 2.4918 Design moment column E- 2 level Mdx Ma Mdxtot Mdy Ma Mdytot foundation 42.392 16.6794 59.07 5.54 16.6794 22.22 ground 62.2952 9.134 71.43 3.53695 15.667383 19.20 first 50.772 9.2276 60.00 0.588 9.2276 9.82 second 41.7 8.02622 49.73 2.276 8.02622 10.30 third 30.648 5.9506 36.60 3.912 5.9506 9.86 fourth 13.592 3.1476 16.74 3.736 3.1476 6.88 H.room 12.88 1.3638 14.24 7.028 1.3638 8.39 colmnE-3 level Mdx Ma Mdxtot 0 Mdy Ma Mdytot foundation 45.504 35.803 81.31 2.222 35.803 38.03 ground 47.712 32.666 80.38 2.808 32.666 35.47 first 58.872 18.981 77.85 15.02 18.981 34.00 second 51.984 13.4396 65.42 13.352 13.4396 26.79 third 44.832 7.7492 52.58 12.888 7.7492 20.64 fourth 22.064 1.8076 23.87 9.368 1.8076 11.18 H.room 11.7 0.5794 12.28 10.028 0.5794 10.61 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 120 Column A-3 level Mdx Ma Mdxtot 0 Mdy Ma Mdytot foundation 7.024 27.3684 34.39 4.468 27.3684 31.84 ground 6.376 21.5202 27.90 47.428 21.5202 68.95 first 8.604 16.129 24.73 44.432 16.129 60.56 second 36.404 11.1504 47.55 16.804 11.1504 27.95 third 32.072 6.2972 38.37 17.86 6.2972 24.16 fourth 17.012 2.2656 19.28 19.564 2.2656 21.83 colmn B-3 level Mdx Ma Mdxtot 0 Mdy Ma Mdytot foundation 3.872 45.0446 48.92 0.702 45.0446 45.75 ground 14.68 32.0564 46.74 52.52 32.0564 84.58 first 46.752 25.5656 72.32 4.848 25.5656 30.41 second 22.756 20.6582 43.41 2.638 20.6582 23.30 third 37.116 8.591 45.71 1.128 8.591 9.72 fourth 23.856 0.8998 24.76 2.832 0.8998 3.73 column H- 3 level Mdx Ma Mdxtot 0 Mdy Ma Mdytot foundation 49.624 31.1494 80.77 10.06 31.1494 41.21 ground 66.332 11.2082 77.54 16.932 11.2082 28.14 first 49.504 10.2998 59.80 9.176 10.2998 19.48 second 39.156 8.6936 47.85 17.826 8.6936 26.52 third 29.804 6.4254 36.23 17.642 6.4254 24.07 fourth 10.254 4.8176 15.07 12.898 4.8176 17.72 H.room 6.964 2.4918 9.46 6.568 2.4918 9.06 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 121 5.2.3 Reinforcement Design 5.2.3.1 solid slab columns Longitudinal Reinforcement As min = 0.008Ac As max = 0.08Ac Colmn A-3 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundn 1368.42 34.39 31.84 1.0 0.07 0.07 0.28 1489.704 980 1489.70 8Ф16 ground 1076.01 27.90 68.95 0.8 0.06 0.14 0.29 1542.907 980 1542.91 8Ф16 first 806.45 24.73 60.56 0.6 0.05 0.12 0.1 532.037 980 980 4Ф12&4Ф14 second 557.52 47.55 27.95 0.4 0.10 0.06 0.06 319.2222 980 980 4Ф12&4Ф14 third 314.86 38.37 24.16 0.2 0.08 0.05 0.045 239.4167 980 980 4Ф12&4Ф14 fourth 113.28 19.28 21.83 0.1 0.06 0.07 0.12 469.0612 720 720 4Ф16 colmn B-3 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundn 2252.23 48.92 45.75 1.2 0.07 0.06 0.46 3196.565 1280 3196.6 8Ф24 ground 1602.82 46.74 84.58 0.9 0.06 0.12 0.325 2258.443 1280 2258.4 4Ф24&4Ф14 first 1278.28 72.32 30.41 0.9 0.15 0.06 0.32 1702.518 980 1702.5 4Ф20& 4Ф12 second 1032.91 43.41 23.30 0.7 0.09 0.05 0.055 292.6204 980 980 4Ф14&4Ф12 third 429.55 45.71 9.72 0.4 0.15 0.02 0.12 469.0612 720 720 4Ф16 fourth 44.99 24.76 3.73 0.0 0.08 0.01 0.2 781.7687 720 781.77 4Ф16 colmn E-3 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundn 1790.15 81.31 38.03 1.0 0.11 0.05 0.36 2501.66 1280 2501.7 4Ф24&8Ф12 ground 1633.3 80.38 35.47 0.9 0.11 0.05 0.275 1910.99 1280 1911 4Ф20&4Ф16 first 949.05 77.85 34.00 0.7 0.16 0.07 0.35 1862.13 980 1862.1 4Ф20&4Ф14 second 671.98 65.42 26.79 0.5 0.13 0.06 0.115 611.8426 980 980 4Ф12&4Ф14 third 387.46 52.58 20.64 0.3 0.11 0.04 0.06 319.2222 980 980 4Ф12&4Ф14 fourth 90.38 23.87 11.18 0.1 0.08 0.04 0.115 449.517 720 720 4Ф16 H.room 28.97 12.28 10.61 0.0 0.04 0.03 0.1 390.8843 720 720 4Ф16 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 122 5.2.3.2 pre-cast slab columns Following the same procedure the final out put for the pre-cast slab columns is tabulated below colum A-3 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundation 1422.5 33.11 36.90 1.0 0.07 0.08 0.3 1596 980 1596.11 8Ф16 ground 1141.5 70.57 26.12 0.8 0.15 0.05 0.32 1703 980 1702.52 4Ф20+4Ф16 first 850.75 61.16 21.39 0.6 0.13 0.04 0.11 585.2 980 980.00 4Ф12+4Ф14 second 626.26 51.90 17.72 0.5 0.11 0.04 0.02 106.4 980 980.00 4Ф12+4Ф14 third 340.59 40.05 13.36 0.3 0.13 0.04 0.18 703.6 720 720.00 4Ф16 fourth 89.96 13.92 7.08 0.1 0.03 0.01 0.035 136.8 720 720.00 4Ф16 column B- 3 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundation 1751.2 83.70 38.10 1.0 0.12 0.05 0.37 2571 1280 2571.15 4Ф24+4Ф16 ground 1606.5 79.52 36.57 0.9 0.11 0.05 0.27 1876 1280 1876.24 4Ф20+4Ф14 first 1195.3 68.90 28.16 0.7 0.10 0.04 0.065 451.7 1280 1280.00 4Ф16+4Ф14 second 797.91 54.77 18.70 0.4 0.08 0.03 0.01 39.09 1280 1280.00 4Ф16+4Ф14 colmn E-2 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundn 833.97 59.07 22.22 0.6 0.12 0.05 0.09 478.8333 980 980 4Ф12&4Ф14 ground 456.7 71.43 19.20 0.3 0.15 0.04 0.16 851.2592 980 980 4Ф12&4Ф14 first 461.38 60.00 9.82 0.3 0.12 0.02 0.06 319.2222 980 980 4Ф12&4Ф14 second 401.311 49.73 10.30 0.3 0.10 0.02 0.03 159.6111 980 980 4Ф12&4Ф14 third 297.53 36.60 9.86 0.3 0.12 0.03 0.07 273.619 720 720 4Ф16 fourth 157.38 16.74 6.88 0.2 0.05 0.02 0 0 720 720 4Ф16 H.room 68.19 14.24 8.39 0.1 0.05 0.03 0.06 234.5306 720 720 4Ф16 colmn H-3 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundn 1557.47 80.77 41.21 0.9 0.11 0.06 0.28 1945.735 1280 1945.7 4Ф20&4Ф16 ground 560.41 77.54 28.14 0.4 0.16 0.06 0.23 1223.685 980 1223.7 4Ф20 first 514.99 59.80 19.48 0.4 0.12 0.04 0.05 266.0185 980 980 4Ф20 second 434.68 47.85 26.52 0.4 0.16 0.09 0.3 1172.653 720 1172.7 4Ф20 third 321.27 36.23 24.07 0.3 0.12 0.08 0.13 508.1497 720 720 4Ф16 fourth 240.88 15.07 17.72 0.2 0.05 0.06 0 0 720 720 4Ф16 H.room 124.59 9.46 9.06 0.1 0.03 0.03 0.04 156.3537 720 720 4Ф16 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 123 third 413 45.04 9.70 0.4 0.15 0.03 0.13 508.1 720 720.00 4Ф16 fourth 48.43 20.25 1.66 0.0 0.07 0.01 0.18 703.6 720 720.00 4Ф16 column E- 3 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundation 1526 75.72 33.08 0.8 0.10 0.05 0.16 1112 1280 1280.00 4Ф16+4Ф14 ground 968.71 83.51 25.31 0.5 0.12 0.03 0.045 175.9 1280 1280.00 4Ф16+4Ф14 first 751.83 81.73 23.87 0.4 0.11 0.03 0 0 1280 1280.00 4Ф16+4Ф14 second 545.71 69.51 19.15 0.3 0.10 0.03 0.035 136.8 1280 1280.00 4Ф16+4Ф14 third 331.26 59.24 13.59 0.3 0.19 0.04 0.315 1231 720 1231.29 4Ф16+4Ф12 fourth 96.61 28.24 7.54 0.1 0.09 0.02 0.14 547.2 720 720.00 4Ф16 H.room 30.85 11.95 11.54 0.0 0.04 0.04 0.11 430 720 720.00 4Ф16 column E- 2 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundation 334.26 50.00 12.67 0.2 0.10 0.03 0.08 425.6 980 980.00 4Ф14+4Ф12 ground 413.92 71.97 17.44 0.3 0.15 0.04 0.19 1011 980 1010.87 4Ф14+4Ф12 first 428.13 60.28 10.68 0.3 0.12 0.02 0.075 399 980 980.00 4Ф14+4Ф12 second 377.7 49.76 8.38 0.4 0.16 0.03 0.17 664.5 720 720.00 4Ф16 third 283.8 36.44 7.40 0.3 0.12 0.02 0.075 293.2 720 720.00 4Ф16 fourth 154.59 15.53 5.71 0.2 0.05 0.02 0 0 720 720.00 4Ф16 H.room 66.65 13.81 8.08 0.1 0.05 0.03 0.06 234.5 720 720.00 4Ф16 columnH-3 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundation 1521.23 80.19 40.40 0.8 0.11 0.06 0.2 1390 1280 1389.81 4Ф16+4Ф14 ground 1437.09 94.95 46.87 0.8 0.13 0.06 0.27 1876 1280 1876.24 4Ф20+4Ф14 first 471.89 64.01 11.33 0.5 0.21 0.04 0.375 1466 720 1465.82 4Ф16+4Ф14 second 448.71 61.02 17.51 0.4 0.20 0.06 0.36 1407 720 1407.18 4Ф16+4Ф14 third 321.21 53.96 22.38 0.3 0.18 0.07 0.345 1349 720 1348.55 4Ф16+4Ф14 fourth 146.81 15.39 10.26 0.1 0.05 0.03 0.06 234.5 720 720.00 4Ф16 H.room 63.26 14.99 25.21 0.1 0.05 0.08 0.13 508.1 720 720.00 4Ф16 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 124 column E- 1 level Nsd Mdx Mdy ν µh µb ω As Asmin Asprvd Ф foundation 237.74 51.4 6.267 0.2 0.17 0.02 0.26 1016 720 1016.30 4Ф14+4Ф12 ground 239.7 51.31 12.35 0.2 0.17 0.04 0.3 1173 720 1172.65 4Ф16+4Ф12 first 208.25 27.51 9.225 0.2 0.09 0.03 0.05 195.4 720 720.00 4Ф16 second 161.85 9.725 27.44 0.2 0.03 0.09 0.05 195.4 720 720.00 4Ф16 third 103.27 8.393 18.76 0.1 0.03 0.06 0.075 293.2 720 720.00 4Ф16 fourth 28.77 4.683 6.647 0.0 0.02 0.02 0.05 195.4 720 720.00 4Ф16 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 125 6. Foundation design Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 126 6.1 Footing Design A) Interior column footing design (column at (b3)) Y a b Using comb-1 My Given L Mx = 5.96 kN.m L’ Mx X My =-0.42 kN.m -Pd =2252.23 kN Assume L = B c d And an allowable bearing capacity for the soil as δall =560 KPa C-25 and S-300 Proportioning Using unfactored load; Pd = 73 . 1608 4 . 1 23 . 2252 = m P M e m P M e d x y d y x 00026 . 0 73 . 1608 96 . 5 0037 . 0 73 . 1608 42 . 0 = = = = − = = 0.89m 1.5m Since a surface footing, B = L 0.61m ) 0037 . 0 * 6 00026 . 0 * 6 1 ( * 73 . 1608 560 ) * 6 * 6 1 ( * 2 B B B B e B e A p y x d all ± ± = ± ± = σ Solving the equation by trial and error, we get B =1.7m We use B = L = 1.7m 47 . 562 ) 00094 . 0 013 . 0 1 ( * 7 . 1 73 . 1608 2 = + + = a ult σ kn/m 2 42 . 561 ) 00094 . 0 013 . 0 1 ( * 7 . 1 73 . 1608 2 = − + = b ult σ kn/m 2 ) 00094 . 0 013 . 0 1 ( * 7 . 1 73 . 1608 2 + − = c ult σ =548.00 kn/m 2 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 127 ) 00094 . 0 013 . 0 1 ( * 7 . 1 73 . 1608 2 − − = d ult σ =546.96 kn/m 2 562.47 561.43 1.7m 1.7m 548 546.96 ult δ avg = 554.7 Kpa I . Punching shear According to EBCS-2, 1995 article 4.7.6 the resistance of footing without punching shear reinforcement is give by: Vrd = 0.25 * fctd * k1 * k2 * u * d Where: K1 = 1+50ρ ≤ 2.0 K2 = 1.6 – d ≥ 1.0 (d in meter) Assume ρ = 0.002, k1 = 1.0 + 50 * 0.002 = 1.1 0.4+d L’ U = 4* ( 0.4 + d) = 1.6 + 4d Vrd = 0.25*1043*1.1*1.0*(1.6 + 4*d)*d B’ Vacating = (1.7*1.7 – (0.4 + d) 2 )*554.7 0 .4+d Then Vrd > Vacting (for safe condition) B’ Hence solving for d d ≥ 0.545 Therefore , D = 549 + 50 + 10 =605mm So use over all depth of 610mm II.wide beam shear Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 128 According to EBCS-2, 1995, article 4.5.3, the shear force Vc carried by the concrete is given by : Vc = 0.3*fctd*k1*k2*bw*d Using 50mm clear cover and φ 16 longitudinal reinforcement bar d= 610 – 50– 8 = 560 mm Vc = 0.3*1043*1.1*1*1.7*0.552 =327.6 kN Vacting = (B/2 - B’/2 - d)*qu*B = 0.09 *561.42*1.7=85.89kN Since Vc > Vacting ..... the section is safe !! So punching governs and the overall depth is found to be 610 mm III.Reinforcement M = 364.18*1*1.05*1.05/2 = 200.75 kN.m Using table No.1 For design d b M K m = = 19.46 Ks = 3.96 As = Ks * d M = 3.96* 2 22 . 2428 56 . 0 99 . 201 mm = Amin = ρmin * b*d = .002*1700*560 = 1904 2 mm s s A b a s * = taking φ 16 mm s 140 22 . 2428 1700 * 201 = = Take φ 16 c/c 140 mm (both direction) No of reinforcement = 12 140 ) 100 1700 ( 1 = − + Hence provide 12φ 16 c/c 140 mm Similarly the remaining footing and the precast slab footing is done as that of the above procedure. The result is found in the auto cad detail for both cases. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 129 6.2 Mat foundation design The purpose of designing this mat foundation is due to irregularity of our building and occurrence of small amount of tension at the edge columns. For a building of G+4 with bearing capacity of 560 it is not recommended to use mat foundation but we do it to be safer. And also it is recommended to use cantilever structures to balance the tension by discussing with the architect. since we don’t have this chance we opted to design mat foundation to the local areas in which tension was developed. The design procedure for the precast floor system is presented as follows and for the solid case the results are found in the reinforcement detail and are also tabulated in table. E G F G 0.5 1 1.3 2 2.7 0.5 m 3 1.5 0.5m 3.54m 0.48m 0.98m 0.53 Case 1 when E-1& F-1 are in tension Column P Factored p mx My E-1 -143.77 -102.7 -95.52 3.15 F-1 -279.59 -199.71 -93.38 -10.64 E-2 803.13 573.7 -108.53 5.47 G-2 599.8 428.43 -99.51 -12.49 E-3 1525.96 1089.97 -99.8 2.91 H-3 1521.23 1086.6 -94.01 -4.65 R=-102.7-199.71+573.7+428.43+1086.6+1089.97=2876.3 KN Taking moment about axis E - Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 130 e’x =3.15-10.64+5.47-12.49+2.91-4.65-199.71*3.54+428.43*4.02+1086.6*5 2876.3 =2.24 ex=0.5+2.24= 2.74m similarly taking moment about axis 3 ey=1.81m centroid of the area is found to be x= 2.797 m y=2.82m Eccentricities Ex= x-ex= 2.797-2.74 =0.057m Ey =2.82-1.81= 1.01m Then M x = R*ey= 2876.3*1.01=2905.1 KN-m My =R*ex= 2876.3*0.057= 164.01KN-m Second moment of inertia of the area Ix= 130.76 m 4 Iy=52.76 m 4 d= ± ± =105.6 ±3.142x ±22.22y B C A A A D da=105.6 +3.142*2.797+ 22.22*2.82 =177.05 KN/m 2 db= 105.6+3.142*2.797-22.22*3.18 =43.73 KN/ m 2 dc= 105.6-3.142*1.743-22.22*3.18 =29.5 KN/ m 2 dd= 105.6-3.142*3.733+22.22*2.82 =156.53 KN/ m 2 Case 2 When E-1&F-1 are in compression The same procedure is followed to calculate eccentricities & moment Column P Factored p mx My E-1 1064.74 760.53 93.85 -5.91 F-1 1087.33 776.7 91.5 3.55 E-2 803.13 573.7 -108.53 5.47 G-2 857.76 612.7 -5.87 -86.22 E-3 1525.96 1089.97 -99.8 2.91 H-3 1521.23 1086.6 -94.01 -4.65 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 131 d= ± ± co-ordinates B G C F (-2.797,2.03) ,J (-0.527,-2.83) F H E (-2.797,0.03) H (2.16,2.03) E I I (2.449,0.03) J G (-0.527,3.18) A D ey=0.56m Mx=Rey=2744.11KN.m ex=0.137 m My=Rex=671.33KN.m p/A=179.9KN/m 2 Mx/Ix=20.99KN/m 2 My/Ix=12.86KN/m 2 d A= 179.9+12.86*2.797-20.99*2.82=156 KN/m 2 d B= 179.9+12.86*2.797+20.99*3.18=282.62 KN/m 2 d C= 179.9-12.86*1.743+20.99*3.18=224.23 KN/m 2 d D= 179.9-12.86*3.73-20.99*2.82=72.7 KN/m 2 d F= 179.9+12.86*2.797+20.99*2.03=258.48 KN/m 2 d E= 179.9+12.86*2.797+20.99*0.03=216.5 KN/m 2 d H= 179.9-12.86*2.16+20.99*2.03=194.73 KN/m 2 d I= 179.9-12.86*2.449+20.99*0.33=194.04 KN/m 2 d G= 179.9+12.86*0.527+20.99*3.18=253.43 KN/m 2 d C= 179.9+12.86*0.527-20.99*2.82=127.5 KN/m 2 Since the stress in case 2 greater than that of case 1 so the mat is designed for the compression case X-direction Strip 1 ∑Pu=760.53+776.7=1537.23 KN A=5.46m 2 Q avg =240.015 KN/m 2 Q=240.015 KN/m 2 *5.46m 2 =1310.48 KN Average load =1310.48 KN+1537.23KN/2=1423.86 KN Modified soil rxn =Q*( ) STRIP 2 STRIP 3 STRIP 1 S T R I P 5 S T R I P 4 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 132 = 240.02KN/m 2 *[ ]= 260.79 KN/m 2 Load correction factor = =1423.86/1537.23= 0.93 Corrected load for E-1,p= 707.3 KN F-1,P=722.33KN The same procedure is followed for the next strips and the results are tabulated as follows. Strip ∑pu Average soil rxn Average load Modified soil rxn Load correction factor 2 1186.4 204.7 1637.41 160.5 1.38 3 2176.57 148.73 2336.11 139.22 1.07 4 2424.2 205.1 2608.85 191.54 1.08 5 2476 169.5 2897.9 147.96 1.17 .B. strip 4&5 are spanning in the y- direction Analysis of individual strips SAP result for the solid case is presented for sample STRIP 1 SFD BMD Strip 2 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 133 SFD STRIP 3 Depth determination (precast case) Case 1 when one of the edge has no resistance Punching check for the column E-3 which is located at the edge with maximum axial load Vac=p-d(0.4+d)(0.7+0.5d) 0.4+d =1166.3-139.22(0.4+d)(0.7+0.5d) =1127.32-125.3d-69.61d 2 0.4m Vres=0.5 f ctd (1+50ρ)*(1.8+d)d =1021.68d+1135.2 *d 2 0.5 0.2+0.5d equating v act ≤ v res 0.7+0.5d then d = 0.602m 0.7+d 0.5 0.2+d For strip 4 case d=191.54KN/m 2 Vact=1166.3-191.54(0.4+d)*(0.7+0.5d) Vact= 112.67-172.4d-95.77d 2 Vres ==1021.68d+1135.2 *d 2 Equating v act ≤ v res then d= 0.6m Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 134 Case when the edge has resistance 0.4+d 0.4+d Check for the wide beam shear Maximum wide beam shear =306 KN for strip -1 Vres = 0.3fctd (1+50ρ) bw *d =0.3*1.043*10^3*1.1*1.15*0.788=312 KN Vres < vact ok So use 0ver all depth of 850mm d=850-50-12=78 Reinforcement design moment b d km ks As spacing s.provided Strip-1 364.06 1.15 0.788 22.58 3.974 1835.70 196.71 Φ20c/c190 65.72 1.15 0.788 9.59 min 1812.4 199.24 Strip-2 255.71 2.00 0.788 14.35 min 3152 199.24 Ø20c/c190 113.22 2.00 0.788 9.55 min 3152 199.24 Ø20c/c190 Strip-3 Strip-3 417.34 2.85 0.788 15.36 min 4491.6 199.24 Ø20c/c190 20.31 2.85 0.788 3.39 min 4491.6 199.24 Ø20c/c190 Strip-4 215.48 2.27 0.788 12.36 min 3577.52 199.24 Ø20c/c190 51.95 2.27 0.788 9.585 min 3577.52 199.24 Ø20c/c190 129.49 2.27 0.788 9.585 min 3577.52 199.24 Ø20c/c190 75.58 2.27 0.788 7.323 min 3577.52 199.24 Ø20c/c190 Strip-5 166.46 3.265 0.788 9.06 min 5145.64 199.24 Ø20c/c190 71.22 3.265 0.788 5.93 min 5145.64 199.24 Ø20c/c190 89.81 3.265 0.788 6.66 min 5145.64 199.24 Ø20c/c190 70.72 3.265 0.788 4.91 min 5145.64 199.24 Ø20c/c190 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 135 Reinforcement design for solid case moment b d Km Ks As spacing s.provided Strip-1 436.78 1.15 0.788 24.73 3.988 2210.62 163 Ø20c/c160 79.06 1.15 0.788 10.52 Min 1812.4 199.24 Ø20c/c190 Strip-2 596.5 2.00 0.788 21.92 3.9695 3004.302 209 Ø20c/c190 142.75 2.00 .788 8.02 Min 3152 199.24 Ø20c/c190 Strip-3 1296.38 2.85 0.788 27.066 4.0158 6606.64 135.45 Ø20c/c130 113.62 2.85 0.788 8.01 Min 4491.6 199.24 Ø20c/c190 Strip-4 590.95 2.27 0.788 20.48 3.96 2969.65 240 Ø20c/c190 91.57 2.27 0.788 8.66 Min 3577.52 199.24 Ø20c/c190 96.62 2.27 0.788 8.28 Min 3577.52 199.24 Ø20c/c190 Strip-5 494.10 3.265 0.788 15.61 Min 5145.64 199.24 Ø20c/c190 169.26 3.265 0.788 9.17 Min 5145.64 199.24 Ø20c/c190 The reinforcement detail is attached with AutoCAD Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 136 7. Cost estimation Solid slab A-SUB STRUCTURE 1. COCRETE WORK 1.10 Reinforced concrete quality C-25, 360 kg of cement/m3, filled into formwork and vibrated around rod reinforcement (formwork and reinforcement measured separately) a) Footing , Grade beam &footing columen m 2 53.74 1300.00 69,856.80 1.20 Provide, cut and fix in position sawn zigba wood or steel formwork which ever appropriate. a) Footing , Grade beam &footing columen m 2 384.26 75.00 28,819.50 1.30 Mild steel reinforcement according to structural drawings. Price includes cutting, bending, placing in position and tying wire and required spacers. a) Φ8-Φ24 mm deformed bar kg 110.20 26.00 2,865.20 b) Φ6 mm plain bar kg 812.44 25.00 20,311.05 TOTAL CARRIED TO SUMMARY .................... 121,852.55 B-SUPER STRUCTURE 1. COCRETE WORK 1.10 Reinforced concrete quality C-25, 360 kg of cement/m3, filled into formwork and vibrated around rod reinforcement (formwork and reinforcement measured separately) a) floor slabs m 3 121.97 1250.00 152,462.50 b) beams . m 3 62.19 1250.00 77,742.75 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 137 c) columns . m 3 26.61 1250.00 33,262.50 1.20 Provide, cut and fix in position sawn zigba wood or steel formwork which ever appropriate. a) Elevation column, beams & floor slabs m 2 1250.40 70.00 87,528.00 1.30 Mild steel reinforcement according to structural drawings. Price includes cutting, bending, placing in position and tying wire and required spacers. a) Φ8-Φ24 mm deformed bar kg 28873.64 25.00 721,841.00 b) Φ6 mm plain bar kg 357.56 26.00 9,296.56 TOTAL CARRIED TO SUMMARY .................... 1,082,133.31 Grand total of solid 1,203,989.96 pre-cast ribbed slab A. super structure 1. COCRETE WORK 1.10 Reinforced concrete quality C-25, 360 kg of cement/m3, filled into formwork and vibrated around rod reinforcement (formwork and reinforcement measured separately) a) floor slabs a1)solid part m3 17.51 1250.00 21,891.25 a2)cast in situ m3 51.36 1250.00 64,193.75 a3)beam element m3 7.76 1250.00 9,700.75 b) beams . m 3 76.17 1250.00 95,212.50 c) columns . m 3 26.29 1250.00 32,862.50 1.20 Class C 200mm thick HCB wall which can satisfy the designed strength , bedded in cement mortar (1:3).Price shall include mortar bed. pcs 4992.00 9.60 47,923.20 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 138 1.30 Provide, cut and fix in position sawn zigba wood or steel formwork which ever appropriate. a) Elevation column, beams & floor slabs m 2 681.86 70.00 47,730.20 1.40 Mild steel reinforcement according to structural drawings. Price includes cutting, bending, placing in position and tying wire and required spacers. a) Φ8-Φ24 mm deformed bar kg 26465.95 25.00 661,648.75 b) Φ6 mm plain bar kg 1429.20 26.00 37,159.20 TOTAL CARRIED TO SUMMARY 1,018,322.10 B.sub structure 1. COCRETE WORK 1.10 Reinforced concrete quality C-25, 360 kg of cement/m3, filled into formwork and vibrated around rod reinforcement (formwork and reinforcement measured separately) a) Footing , Grade beam &footing columen m 2 52.78 1300.00 68,608.02 1.20 Provide, cut and fix in position sawn zigba wood or steel formwork which ever appropriate. a) Footing , Grade beam &footing columen m 2 338.65 75.00 25,398.75 1.30 Mild steel reinforcement according to structural drawings. Price includes cutting, bending, placing in position and tying wire and required spacers. a) Φ8-Φ24 mm deformed bar kg 800.69 25.00 20,017.25 b) Φ6 mm plain bar kg 110.70 26.00 2,878.20 TOTAL CARRIED TO SUMMARY 116,902.22 Grand total of pre-cast 1,135,224.32 Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 139 Conclusion and recommendation Conclusion The main aim of the project is to compare the total cost variation of pre-cast and solid slab systems. To attain this we have done a detail analysis and design for each case. And finally we have come out with the total cost of each system as shown below. Total cost(birr) pre-cast solid difference concrete work 340,391.97 333,333.95 -7,058.02 reinforcement 721,703.40 754,308.51 32,605.11 form-work 73128.95 116347.5 43,218.55 total 1,135,224.32 1,203,989.96 68,765.64 Total difference = (1,203,989.96-1,135,224.32)*100/(1,203,989.96)=5.7115% As the above data shows the pre-cast slab type is less-costier than that of solid floor type. And also it is clear that the major difference is caused by the cost of form-work and reinforcement instead of concrete cost. Generally using pre-cast slabs is advantageous by minimizing the construction time, work man- ship, construction equipment and attaining quality of materials. Recommendation The overall works of the building should be inspected and supervised throughout the entire construction time in order to achieve the design strength. Care should be taken when handling, casting and placing the precast beam element. Especially the support condition on construction time must be the same as that of previously determined arrangement on the design part. And also it is better to remove application of concentrated force for the precast floor system, which can cause falling out of the hallow block concrete. The quality of materials should fulfill the design strength. Senior project, structural design & cost comparison June 2008 Mekelle University, Department of Civil Engineering Page 140 References Ethiopian Building Code Standard 1, 2, 3, 7, & 8 – 1995. Gtz technical manual Kality steel industry manual of cold framed welded structural & furniture steel tubing