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0488 1.710 X121100 1.000 71 0.358 72 =0.156s 要求:用底部剪力法计算结构在多遇地震作用下各层的层间地震剪力 。提示 7=0.1s~Tg时 T=7~57时,a=am 4 71>1478且7<0.35s时 6n=0.087+0.07 0.35~0.55s时 n=0.087+0.01 0.25 解:7<7=0.35857,a1= 0.16=0.116 0.358 Ga=085×(60+50)×98=916.3kN Fs=a1Ga=0.116×9163=10629kN 71>147=14×025=0358,7=025<0.355 6n=0.087+0.07=008×0.358+0.07=0.10 斤=yGB1(- 60×98×4 0×9.8×4+50×98×8 ×10629×(1-0.10)=3587kN F F(1-n) 60×9.8×4+50×98× ×10629×(1-0.10)=5979KN      =       1.000 0.488 12 11 X X      − =       1.000 1.710 22 21 X X T1 = 0.358s T2 = 0.156s 要求:用底部剪力法计算结构在多遇地震作用下各层的层间地震剪力 Vi 。提示: T1 = 0.1s~Tg 时,  =  max ; T1 = Tg ~5 Tg 时, max 0.9 1           = T Tg ; T 4Tg 1. 1  且 0.35 T g  s时,  n = 0.08T1 + 0.07 ; 0.35 ~ 0.55 T g = s时, 1 0.08 0.01  n = + T 解: g 1 g T T T  = 0.358s<5 , 0.9 0.9 1 max 1 0.25 0.16=0.116 0.358 T g T       = =          eq G =  +  0.85 (60 50) 9.8=916.3kN Ek 1 eq F G = =  =  0.116 916.3 106.29kN 1 1.4 1.4 0.25 0.35s T T  =  = g , 0.25s 0.35s T g =  1 0.08 0.07 0.08 0.358 0.07 0.10  n = + =  + = T 1 1 1 Ek n i i 60 9.8 4 (1 ) 106.29 (1 0.10) 35.87kN 60 9.8 4 50 9.8 8 G H F F G H    = − =   − =    +   2 2 2 Ek n i i 50 9.8 8 (1 ) 106.29 (1 0.10) 59.79kN 60 9.8 4 50 9.8 8 G H F F G H    = − =   − =    +   k2 k1 m2 m1 4m 4m
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