Answer: The vertical load in wall A Wa=80x10x80x2+80000=208000Ib=925.6kN The vertical load in wall C Wc=80x60x60x2+80000=656000Ib=2919.2kN The horizontal load of each story =0.1x80x120x160=155kip=689.8kN Me=689.8x20x0.305+689.8x10x0.305=6311.7kNm Check wall C:Me/Wc=6311.7/2919.2=2.16m>(1/6x40x0.305=2.03m. Balance design is not satisfied. The width of wall C should be increased. Check wall A: Me/2Wa=6311.7/(2x925.6)=3.4m>(1/6)40x0.305=2.03m. Balance design is not satisfied. The width of wall A and B should be increased
Answer: The vertical load in wall A Wa=80x10x80x2+80000=208000Ib=925.6kN The vertical load in wall C Wc=80x60x60x2+80000=656000Ib=2919.2kN The horizontal load of each story =0.1x80x120x160=155kip=689.8kN Me=689.8x20x0.305+689.8x10x0.305=6311.7kNm Check wall C:Me/Wc=6311.7/2919.2=2.16m>(1/6)x40x0.305=2.03m. Balance design is not satisfied. The width of wall C should be increased. Check wall A: Me/2Wa=6311.7/(2x925.6)=3.4m>(1/6)40x0.305=2.03m. Balance design is not satisfied. The width of wall A and B should be increased
/"y5m/5/-5Y5y5mY5/ 日=30~60 10m 人Θ 日=30~60°
5m 30。~ 60。 。 60 。~ 30 10m 5m 5m 5m 5m 5m 5m
Answer: Load in sub-truss: w'=5x10=50kN/m,M'max=(1/8)wL2=(1/8)x50x10x10=625kNm If 0=60,h=1.73m,it is the first choice for sub-truss structure. N=625/1.73=361kN,stress=N/A=361000/2500=144.5Mpa<200Mpa,OKl Check shear,Qmax=250kN,Ns=250/0.866=289kN,OK! For most economical purpose,0 can be changed as 51,h=1.25m,so that the maximal stress in sub-truss can reach 200Mpa,h=1.25m,N=500kN Load in main truss:P=w'x5=250kN, Mmax=7.5x0.5P+2.5xP=6.25P=6.25x250=1562.5kNm If 0=60.h=4.325m,it is the first choice for main truss structure. N=1562.5/4.325=361kN,Stress=-N/A=361000/2500=144Mpa,OK! Check shear,Qmax=500kN,Ns=500/0.866=577kN,Stress=N/A=231>200Mpa Fail! The tube with 50cm2 can be used in main truss with 0=32,h=1.563m N=1562.5/1.563=1000kN,Stress=-1000/5000=200Mpa,OK Qmax=500kN,Ns=500x1.89=945kN,Stress=N/A=189<200Mpa OK Weight of sub-truss=-8(10+8+10x1.6)x0.0025x7850=5.338T Weight of main truss==2(35+30+14x2.95)0.005x7850=8.344T Cost=(5.338+8.344)10000=136820RMB
Answer: Load in sub-truss: w’=5x10=50kN/m, M’max=(1/8)w’L²=(1/8)x50x10x10=625kNm If θ=60, h=1.73m, it is the first choice for sub-truss structure. N=625/1.73=361kN, stress=N/A=361000/2500=144.5Mpa<200Mpa, OK! Check shear, Qmax=250kN,Ns=250/0.866=289kN, OK! For most economical purpose, θ can be changed as 51, h=1.25m, so that the maximal stress in sub-truss can reach 200Mpa, h=1.25m, N=500kN Load in main truss: P=w’x5=250kN, Mmax=7.5x0.5P+2.5xP=6.25P=6.25x250=1562.5kNm If θ=60, h=4.325m, it is the first choice for main truss structure. N=1562.5/4.325=361kN, Stress=N/A=361000/2500=144Mpa, OK! Check shear, Qmax=500kN, Ns=500/0.866=577kN, Stress=N/A=231>200Mpa Fail! The tube with 50cm² can be used in main truss with θ=32, h=1.563m N=1562.5/1.563=1000kN, Stress=1000/5000=200Mpa, OK Qmax=500kN, Ns=500x1.89=945kN, Stress=N/A=189<200Mpa OK Weight of sub-truss=8(10+8+10x1.6)x0.0025x7850=5.338T Weight of main truss=2(35+30+14x2.95) 0.005x7850=8.344T Cost=( 5.338+8.344)10000=13 6820RMB
Chapter 8 Horizontal Linear Components 8.1.Sectional Shapes and Proportions 8.2.Moment Diagrams 8.3.Internal Resisting Couple 8.4.Prestressing Design 8.5.Connections
Chapter 8 Horizontal Linear Components 8.1. Sectional Shapes and Proportions 8.2. Moment Diagrams 8.3. Internal Resisting Couple 8.4. Prestressing Design 8.5. Connections
8.1.Sectional Shapes and Proportions Horizontal linear components are variously named as slabs,beams,joist or girders,with shear,bending,torsion and axial force in these components
8.1. Sectional Shapes and Proportions Horizontal linear components are variously named as slabs, beams, joist or girders, with shear, bending, torsion and axial force in these components