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1、20000 +10000 +1001000020000 10000、2 +50 J10000( 2 x 20000 +10000100102x10000)50建筑结构抗震（清华大学出版社）计算题及例题解答1.某两层房屋计算简图如图1所示。已知楼层集中质量为m1 = 100t，m2 =50t，每层层高均为h，楼板平面内刚度无限大，沿某抗震主轴方向的层间剪切刚度为k = 20000kN.m，1k2 = 10000kN m。求该结构体系在该抗震主轴方向的自振周期、振型和振型参与系数。叫f&h图1动力模型计算简图【解】m】=1001，m2 = 50f，匕=20000kN/m，k2 = 10000kN/
2、m（1）自振圆频率I21,2% + k2 *m1k 匕-kT+ 匕(2k1 + k2 + 色) m m mJ m m m=-(300 + 200 300)2J100400=10rad/s， aw = 20rad/s（2）自振周期T =四=M14 = 0.6281 也 10(3)振型第一主振型:Xu _ % + k? - m】气2 _ 20000 +10000 -100 x 102X2 _2k2 1 110000第二主振型:X k + k - m w 220000 +10000 -100 x 20222 = 1 2 1=X21k210000(4)振型参与系数EmXY =叫1+巴上1 mX2 m
3、X2 + m X2i 1ji=1100x1 + 50x2 _2100 x 12 + 50 x 22 = 3EmXi 2iY = -4=12 EmX2i 2ii=1mX* + mX” _ 100x 1 + 50x(-1)_ 1 m X2 + m X2 100x 12 + 50x(-1)2 32.某三层钢筋混凝土框架，如图2和图3所示。已知:设防烈度为8度，设计基本地震加速度为0.20g, 设计地震分组为第一组，建造于I类场地，结构阻尼比&= 0.05。各楼层层高均为4m,各层重力荷载代表值分别为 G广G = 800kN ,电=520kN。各层抗侧刚度分别 匕=21400kN/m，k = 1920
4、0 kNm，3 = 17600 kN m，结构沿某抗震主轴方向的自振圆频率分别为w = 8.92rad/s、w = 24.74 rad，s、w=38.71rad s，相应的振型分别为0.445、X = 0.92011.000)-0.654、X = -0.58921.000 )1.654、X = -1.28931.000 )。试用振型分解反应谱法确定多遇地震作用下在该抗震主轴方向的框架层间地震剪力和层间位移角，并绘出层间地震剪力分布图。&-LI 4r由：GiG J 、血:7T7777Z 1rZ /图2框架图图3动力模型计算简图【解】G = G = 800kN , G = 520kN ,k = 1
5、7600 kN mk = 21400 kNm , k = 19200 kNm ,w = 8.92rad/s,X =1x 11 (0.4450.920=24.74 rad s,x 2=38.71rads ,r x 21、-0.654、X22=-0.589k X /23k 1.000 /r x 31， 1.654、X32=-1.289X33k 1.000 /k 1.000)X12k X13X =32 兀2 x 3.14=0.704s 8.92=0.254s24.7422兀_ 2 x 3.142 x 3.1438.713=0.162s设防烈度8度，0.20g,设计地震分组第一组，I类场地，查表知，a
6、 =0.16 ,Tg = 0.25(s)当阻尼比& = 0.05时，得门=1.0 , y = 0.92由于T = 0.65s T 5T = 3.25s ,故 a =TT9n a= L0.21 T9 x1.0x0.16 = 0.0631 T J *2 max 0.704J又由于T T = 0.65s，且T T = 0.65s ,故 a2=a3 = n2amax = L X 0.16=0.16=叩+ G?X2 + G X = 800 x 0.445 + 800 x 0.920 + 520 x 1 = 口於71一 G；X： + G：X12 + G；X13 800 x 0.4452 + 800 x 0
7、.9202 + 520 x 12 .GX + G X + GX800x(-0.654) + 800x(-0.589) + 520x 1八八/7 = 1_212_223_23 = = -0.4162 G X ； + G2 X 22 + G3 X 23800 x (-0.654)2 + 800 x (-0.589)2 + 520 x 12GX + G X + GX 800x 1.654 + 800x(-1.289) + 520x 1 n7 3 = G X31 + G X32 + G X33 800 x 1.6542 + 800 x (-1.289)2 + 520 x 12 =131232333(1
8、) 求各振型下的地震作用:F =a 7 X G = 0.063 x 1.1 89 x 0.445 x 800 = 26.67kN F1 =ay 1 X11 = 0.063x 1.189x0.920x800 = 55.13kNF = a171X12G2 = 0.063 x 1.189 x 1 x 520 = 38.95kN131 113 3F =a 7 X G = 0. 1 6x(-0.4 1 6) x(-0.654) x800 = 34.82kN F =a7 X G = 0.16x(-0.416) x(-0.589) x800 = 3 1.36kNF =a7 X G = 0.16x(-0.41
9、6)x 1 x520 = -34.61kN231 123 3F =a 7 X G = 0. 1 6x0.20 1 x 1 .654x800 = 42.55kN F =a7 X G = 0.16x0.201 x(-1.289) x800 = -33.1 6kNF =以7 X G = 0.16x0.201 x 1 x520 = 16.72kNv 331 1 33 3(2) 求各振型下的地震作用效应：32VIVIVI厂 =F = 38.95kN=F + F = 38.95 + 55.13 = 94.08kN=F + F + F = 38.95 + 55.13 + 26.67 = 120.75kN13
10、1211V = F =-34.61kNV = F + F =-34.61 + 31.36 = -3.25kNV = F + F + F = -34.61 + 31.36 + 34.82 = 31.57kN 21232221V = F = 16.72kNV = F + F = 16.72 - 33.16 = -16.44kNV = F + F + F = 16.72 - 33.16 + 42.55 = 26.11kN 31333231（3）求体系总的地震作用效应:层间剪力：（见图4）54.72V 气V 2 + V 2 + V 2 =*120.752 + 31.572 + 26.112 127.5
11、1kN1 112131.94.082 + (-3.25)2 + (-16.44)2 = 95.56kNV3 = J咋 + 气 + 匕；=J38.952+ (-34.61)2 +16.722 = 54.72kN127.11层间位移:图4层间剪力匕分布图（kN）=6.0mm=3.1mmA u1A u2层间位=5.0mmAu3移转角:Au1h16 = 1.5 x 10-34 x 103AuI hi )AuI hi )Aw5.0 =1.3 x 10-34 x 103性2I h1 )、2Au -T23I hi )Au3 h33.1 =0.8 x 10-34 x 1033. 某三层钢筋混凝土框架，已知条件
12、和计算要求同第2题，试用底部剪力法计算框架层间地震剪力和结构顶层总位移，并绘出层间地震剪力分布图。【解】（1）求地震影响系数由表知，在多遇地震下，设防烈度为8度，设计基本地震加速度为0.20g时，ama= 0.16由表知，当设计地震分组为第一组，建造于11类场地时，V0.252 丸 2 x 3.14由w = 8.92rad，s得，T = gg? = 0.704s1当阻尼比& = 0.05时，得门2 1.0，Y = 0.9由于T = 0.65s T 1.4T = 0.35s，所以由表得，5 = 0.08“+ 0.07 = 0.126顶点附加地震作用：AF = 0.126 x 113.53 = 1
13、4.305kN H G =HG +H G +H G = 4 x 800 + (4 + 4) x 800 + (4 + 4 + 4) x 520 = 15840kN m k k 1 12 23 3k =1F =3G3 F (1 -5 ) + AF = 12 x520 x 113.53 x (1 -0.126) +14.305 = 53.39kN3Ek/ n 15840H Gk kk =1F = H2G2 F (1 -5 ) = 8 x800 x 113.53 x (1 -0.126) = 40.09kN2 Ek n 15840HGkkk=1F =H1G1 F (1 -5 ) = 4 x 800 x 113.53 x (1 - 0.126) = 20.05kN 1H G Ek/15840kkk=1(4) 求
