上海交通大学：《结构概念设计 Conceptual design》教学资源（课程讲义）水平与竖向构件

Homework 5 Check if the thickness of concrete square core tube with 300mm is enough for wind, earthquake and dead loads.Use "Footprint Method”,Tube Action”and assume the layout diagram as attached (The building is totally supported by the core tube).The height of building is 60m with 20 stories,the concrete strength is 40N/mm2,the wind load is 1kN/m 2, the floor DL is 10kN/m 2,earthquake load is 5%DL

Homework 5 Check if the thickness of concrete square core tube with 300mm is enough for wind, earthquake and dead loads. Use “Footprint Method”, “Tube Action” and assume the layout diagram as attached (The building is totally supported by the core tube). The height of building is 60m with 20 stories, the concrete strength is 40N/mm², the wind load is 1kN/m ² , the floor DL is 10kN/m ²,earthquake load is 5% DL

娱凝±核筒 卡5m头 ↓ 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

Homework 6 A 50-story (every story's height is 3 m) steel square framed tube building utilizes its exterior column to carry 1/2 of the floor loads (3kN/m 2)and resist wind forces(1kN/m 2).Attached is the column layout.Using "footprint concept""Tube action”ands“Shear lag action'”, determine the section areas of middle and corner steel columns.The strength of steel is 315N/mm2

Homework 6 A 50-story (every story's height is 3 m) steel square framed tube building utilizes its exterior column to carry 1/2 of the floor loads (3kN/m ²) and resist wind forces(1kN/m ²). Attached is the column layout. Using “footprint concept” “Tube action” and “Shear lag action”, determine the section areas of middle and corner steel columns. The strength of steel is 315N/mm²

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 N=(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

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 N=(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

Homework 7: 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). Homework 7:

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

Homework 9 Estimate the most economic cost of the one-way truss on truss subsystem (Pl.see the attached figure. Ignoring bucking problem in truss).Total vertical design load on the truss subsystem is 10kN/m2.You can only choice two types for steel tubes with section area of 50cm2 and 25cm2.The steel tubes in main span truss have the same section.The tubes in secondary span truss have the same section.The design strength of steel tube is 200N/mm2.The unit price of steel tube in truss is 10000RMB/T

Homework 9 Estimate the most economic cost of the one-way truss on truss subsystem (Pl. see the attached figure. Ignoring bucking problem in truss). Total vertical design load on the truss subsystem is 10kN/m². You can only choice two types for steel tubes with section area of 50cm²and 25cm². The steel tubes in main span truss have the same section. The tubes in secondary span truss have the same section. The design strength of steel tube is 200N/mm². The unit price of steel tube in truss is 10000RMB/T

5"//s/sm//5/5/ 日=30°~60° 10m 人日 日=30°≈60°

5m 30。~ 60。 。60 。~ 30 10m 5m 5m 5m 5m 5m 5m