《结构力学》习题集下册) 23、T=1656√(W/EAg) 24、c=√1/m6=√EA/10.5m O=(E/8m/2)=242551,y=1(1-02/o2)=3127, Momax =AMst, YMax=Stp=1. 3029cm 26、O=√1/m=√l/m(4/3E+1/4k)=3416s-1u=1/(1-/a2)=1.522 YDmax=Astp=0.006m,, MDmx=HMsp=7.6IkN m Ys=P/mo, uo=1.04067, Y Asin( ar ) Bcos(ot)+ m2人Dcos(bh) A=D-,B=//1000 Y=000lcos(o)-0.20833xsm(on)+1.04167sn(n) =√27EAm) 29、O=3892s1,0=157s,4=1.19,y=m=209/10m 30、O=52.36s1,B=1378,,yt=1.9610mA=y=0.27mm D= BFM=2.756M 31、O=71.50s,0=62.83s5,:B=4.389:A=BFδ=3.37mm ym=(+)6=528m 32、b=57596,B=1496,M=PFM=748M, Mm=Ms+Mp= 15.48 0.52 M 3EI k Wml 运动方程:m+ky=k△1p,y+O2y 16m 特征解y sin At=0.0595-sin 6 t 16mo2,O2《结构力学》习题集 （下册） —— 47 —— 23、T =16.56 (W / EAg) 24、 = 1/ m = EA/10.5m 25、 M Y Ystp cm EI ml , 1.3029 ( / 8 / 2) 24.25s , 1/(1 / ) 3.127, Dmax Mstp Max -1 2 2 = = = = = = − =       26、 =  = + = − 1/ m 1/ m(4 / 3EI 1/ 4k) 34.16s 1  = 1 1−  = 1 522 2 2 /( / ) . 0.006m, YD max = ystp = ， 7.61kN m, M Dmax = M stp = 27、 0.001 cos( ) 0.20833 sin( ) 1.04167 sin( ), , /1000, / , 1.04067, sin( ) cos( ) cos( ), s t s t s t 2 2 s t Y l t Y t Y t A Y B l t m P Y P m Y A t B t D D D              = − + = = = = = + + 28、 27 /( ) 3  = EI ml 29、 -1  = 38.92s ， -1  = 15.71s ，  = 1.19 ， 2.09 /10 m 3 ymax = 30、 52.36s , 1.378, -1  =  = ， 1.9610 m, st 0.27mm 4 st = = = − y A y M D = FM = 2.756M 31、 71.50s , 62.83s , -1 -1  =  = ； = 4.389 ； A = F = 3.37mm ； ymax = (w+ F) = 5.28mm 32、 = 57.596s ,  =1.496 -1 ， MD = F M = 7.48M ， M M M   M T max = st + D = 15.48 0.52 33、 3 3 3 , 3 l EI k ml EI  = = ， 运动方程： m P my ky k y y P 16 5 , 2  + = 1  + = 特征解 y * ： y P m t P m t * = sin . sin − = 5 16 0 0595 0 2 2 2 0      1 1