k1=2k,k2=k,横梁为刚性梁。 解:k1X1+k12X2=m1O2X1 K2x1+k2x2=m,0'X k1=3k,k2=k,k12=k21=-k; 1a k22-m2a 2n2-5mm2+2k2=0 a2=0.k/m;01=0707√k/m 02=2k/m;a2=1.414k/m 1=1/2;=1=-1; X 八、试求图14示体系的自振频率和振型。已知mm2=m,各杆E 常数。 解:S1m12X1+612m2O2X2 621m1o2X1+2m2a2X2=X2 设 (1-4)X1+2x2=0 1 δ, 2X1+(1-4)X2=0 61=82=2P/E/; 612=S21=P3/6EI; 1/4 l/41-=04=5/42=3/4 6 El 5m/1=1095E/m1;,o2=2-;O2=144√Em/k1=2k，k2=k，横梁为刚性梁。 解： 1 2 k11X1 + k12X2 = m1 X 2 2 k21X1 + k22X2 = m2 X 3 , , ; 11 22 12 21 k = k k = k k = k = −k 0 2 21 22 2 12 2 11 1 = − −   k k m k m k 2 5 2 0 2 4 2 2 m  − mk + k = 0.5k / m; 1 0.707 k / m 2 1 =  = 2k / m; 2 1.414 k / m 2 2 =  = ; X X / ; X X 1 2 1 22 12 21 11 = = − 八、试求图 14 示体系的自振频率和振型。已知 m1=m2= m，各杆 EI= 常数。 解： 2 1 2 1 12 2 2  11m1 X + m  X = X 2 2 2 1 22 2 2  21m1 X + m  X = X 设 2 11 1 1    m = 1 0 1 0 1 2 11 21 2 11 12 1 + − = − + = X ( )X ( )X X       l / EI; l / EI; 6 2 3 12 21 3 11 22 = = = =     0 1 4 1 1 1 4 = − −   / / 1 = 5/ 4;2 = 3/ 4 3 3 2 2 2 3 3 1 2 1 ; 1.095 / ; 2 ; 1.414 / 5 6 EI ml ml EI EI ml ml EI  =  =  =  =