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