大工20秋高层建筑结构大作业题目及要求10VIP免费

远远程与^号远程与继续教育学院'?高层建筑结构?大作业题 目：题目二、题目四内容仅供参考题目二：底部剪力法.钢筋混凝土 8 层框架结构各层高均为 3m,经质量集中后,各楼层的重力荷 载代 表值分 别为：G1 = 1000kN,G2 = G3 = G4 = G5 = G6 = G7 = 800kN , -=700kN.结构阻尼比 0.05,自振周期为* = 0.80s ,11 类场地类别,设计 地震分组为第二组,抗震设防烈度为 7 度.按底部剪力法计算结构在多遇地震时 的水平地震作用及地震剪力.解答:设防烈度相对应的多遇地震和罕遇地震的匕映值地震设防烈度6789多遇地震0.040.08 (0.12)0.16 (0.24)0.32罕遇地震0.50 (0.72)0.90 (1.20)1.4场地特征周期 Tg (S)设计地震 分组场地类别IIIIIIW第一组0.250.350.450.65第二组0.300.400.550.75第三组0.350.450.650.90(1)抗震设防烈度为 7 度,在多遇地震下,查表可得 a 谈「0.12,在 H 类场地类别,设计地震分组为第二组,查表可得 T =0.40S,因此,T =0.40<,=0.08<5 T =2.00.⑵ 计算地震影响系数 ai计算结构等效总重力为 GeqG =0.85 G =0.85 X( 1000+800X6+700) =5525KNeq(3)计算结构底部剪力 FEKFEK = a. G =0.064 X5525=353.6KN(4)计算各质点的水平地震作用 FiT =0.04s, T =0.80s>1.4T =0.56s.该结构为钢筋混凝土房屋结构,需要考g1虑结构顶部附加集中作用.8 =0.08 T +0.01=0.08 X 0.80+0.01=0.074那么 4 F = 8 FEK =0.074 X353.6=26.166KN(T )-gTYn a2 max0.9X 1X0.12 7 0.064F = G1H1(1-8 ) F =1000X3 X327.4336=11.611KN1 ?n EK 84600GH iii =1⑸ 计算各楼层地震剪力V= F+△ F =65.022+26.166=91.188KNV7 = F + V=65.022+91.188=156.21KNV6 = F + V7 =55.733+156.21=211.943KN(1-8 ) FK =327.4336KNX GH =1000X3+800X(6+9+12+15+18+21) +700X24=84600KN*mi i i =1口 GHF =—8~8— (1- 8 ) F8 EGHE EKi i i =1700 x 24X327.4336=65.022KN84600F = G7H 7(1-8 ) F91Vn EK乙 GHii800 X 21 X327.4336=65.022KN84600F = G 6H 6(1-8 ) F6n EK乙 GHii800 X18 X327.4336=55.733KN84600F = G5H 5(1-8 ) F5 1Vn EK乙 GHii800 X15 X327.4336=46.444KN84600F = G4H 4(1-8 ) F4n EK乙 GHii800 X12 X327.4336=37.156KN84600F = G 3H 3(1-8 ) F3 1Vn EK乙 GHii800 x 9—X327.4336=27.867KN 84600...