上海交通大学：《结构概念设计 Conceptual design》教学资源_Chapter 8 Horizontal Linear Components

平 娱凝赴核筒 30m

Answer: Vertical total load:W=20x30x30x10=180000kN Stress of vertical load=W/A= 180000/0.3x4x4.7=31.9MPa Wind induced Mmax=1x30x60x30=54000kNm Earthquake induced Mmax= 5%Wx60x2/3=360000kNm>54000kNm Maximal stress of bending moment= (3/4)x360000/4.7x4.7x0.3)=+-40.7MPa Maximal stress in pressure is 31.9+40.7=72.6Mpa>40Mpa. The thickness of tube is not enough. 40.7-31.9=8.8 tensile stress,it's not a balance design

Answer: Vertical total load: W=20x30x30x10=180000kN Stress of vertical load=W/A= 180000/(0.3x4x4.7)= 31.9MPa Wind induced Mmax=1x30x60x30=54000kNm Earthquake induced Mmax= 5%Wx60x2/3=360000kNm>54000kNm Maximal stress of bending moment= (3/4)x360000/(4.7x4.7x0.3)=+-40.7MPa Maximal stress in pressure is 31.9+40.7=72.6Mpa>40Mpa. The thickness of tube is not enough. 40.7-31.9=8.8 tensile stress, it’s not a balance design

A M 40m 40m

Answer: (Assuming each column has same sectional area) Total vertical load W=50x3x40x40=240000kN, Each column take W'=0.5W/32=3750kN Mmax=1x50x3x40x75=450000kNm Average column load by Mmax=(3/4450000/(40x8)=1054.7kN According“Shear lag'”, the load in corner column =1.5x1054.7+3750=5332.05kN, Ac=5332050/315=16927mm2 the load in middle column=0.5x1054.7+3750=4277.35kN Am=4277350/315=13579mm2

Answer: (Assuming each column has same sectional area) Total vertical load W=50x3x40x40=240000kN, Each column take W’=0.5W/32=3750kN Mmax=1x50x3x40x75=450000kNm Average column load by Mmax=(3/4)450000/(40x8)=1054.7kN According “Shear lag”, the load in corner column =1.5x1054.7+3750=5332.05kN, Ac=5332050/315=16927mm² the load in middle column=0.5x1054.7+3750=4277.35kN Am=4277350/315=13579mm²

A two-story parking structure is shown in lecture book Fig.4-13,Fig.4-14 and Fig.4-15.All the calculation conditions are same as lecture book example 4-2.Please check the shear wall A and C against seismic overturning moment and propose what to do for them to resist seismic overturning moment(using metric units)

A two-story parking structure is shown in lecture book Fig.4-13, Fig.4-14 and Fig.4-15. All the calculation conditions are same as lecture book example 4-2. Please check the shear wall A and C against seismic overturning moment and propose what to do for them to resist seismic overturning moment(using metric units)

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=144.5Mpa200Mpa 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=144.5Mpa200Mpa 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