习题课结构的极限载荷

1、习题课 12结构的极限载荷一已知截面极限弯矩为Mu，求极限荷载并画出破坏机构。FPuC(1)AMFqBuPuBAMuClq2l/2l/2l/2l/2F× l q = 2M q= 4MuFPuu2Pul4M u lMuFMuPuMuAqCBMuMulq2M 图（满足屈服条件）l/2l/2(2)quqBAlqqMuMuABlq 2lql/2l/2q l × lq= M q + M quuu2Mu12q l q2M q=2MuuuM 图（满足屈服条件）4M= u l 2qu(3)FPuFPABCD1.5qMqDMu2aa2a2aqu3aq2aa2a1.25 MuF× 2

2、aq = M q +1.5M qPuuuaMuMu2.5M= u 2aFPuMu / 2M 图（满足屈服条件）= 1.25MuaACB在题(3)中，若使A、B、D同时出现塑性铰，则铰C位置如何确定？(7)FPuMuMuABCMu2a1.5a1.5a如上图示，铰C平分DB 。= FPu= 6FPu aM= M´ 2a ´ 3a - M2MDuuu5a5= 5 Mu= 1.667 MuFPu3aaD(4)qquMuABCq1.5qMu3a2a3aq3a2aq × 1 × 5a × 3aq = M× (q +1.5q )MMuuuu21Mu

3、7.5q a q = 2.5M q224uuM 图（满足屈服条件）= Muqu3a2若使梁的正负弯矩均达最大值，则铰C位置如何确定？qu(8)M= M= 1 q × 25a2MuA- MMuBuuu中83.125q a2 = 2Mx=0.732aMuuu= 0.64 Mu2.5a2.5aqua2quå M= 0 CMuC1.6 Mu x - M- 0.5 × (0.64 Mux) × x2= 0ua2aM1.6u a- 5ax + 3.125a2 = 0x = 0.732ax2假设： B、C右塑性铰可能位置：B、C左、C右(5)M0uM0CMuqMAAB

4、uB2ql/3l/32l/33q2l/3´ 2q = Mu´ 3q + MuqM 0uMuMu= 2MMA0uuCB= 3MuM0u l/3FMyAlu2l/3M 图（满足屈服条件）1l3M=+ Mu ) =FyA(M 0u u l(6)M0CAABBCl/32l/3l/32l/3解：1) 虚3dq2ldqABM0uuu´ 3dq2l´q = 2M q + M= 2MMM0uuu0uu略去qqMCM2) 静力法M0uCMBMuMMMuuuABFyB3M= F3Mu u 2l=yA2l3Mu3Mu2l/3l/32l2l= 2MuM 0u- 3Mu´

5、; l3= 1 M2M= M(下拉)Buu2lMuM 图（满足屈服条件）CABMu0.5Mu2l/3l/3三已知链杆的极限轴力为FNu，求梁和链杆同时破坏时梁的极限弯矩Mu。CC0.6l0.6lFPFPuDMuMuqABABqD0.4l0.4l2q0.4lFNu0.4l解： 虚FPu× 0.4lq = 3MuqMDu= 7.5MuABFMuPul考虑DB平衡0.6FNuMuMu = 0.6FN u × 0.4l = 0.24FN ul= 4.167u DB0.4lMFNu l= 7.5 0.24FN ul = 1.8FFPuN ul四若使刚架中B、C、D、E各截面同时出现塑

6、性铰而破坏，求极限载荷和FP与q 的关系。qqFPFPMuMuCBDBDCMullAE EMuA0.5l0.5l0.5l0.5lq静力法1) 由横梁弯矩图可得：FPMuMuCBD1 ql 2q = 16Mu- M= MMuuul 282 M2) DE杆E端剪力l2MuulMuMAB杆A端剪力为 -3) 整体水平平衡ullAElMu0.5l0.5l+ Mu= 2MuFPllMu = FPlq = 16Mu= 16FPl = 16FP4)l 2l 2l另一种弯矩图qFPMuMuCBl 0.5l0.5lBC段M大于Mu不满足屈服条件DEMuAMu五图示刚架可能的破坏机构有哪几种？求极限载荷FPu 并

7、画出M 图。q = 2FP/ lCB2mFP2mA4m解：(1)q = 2F + / l = F + / 2P1P1F +CBq + (× 2 ×q ) × 2+F× 2 P1P12Mu2q2m= 3M qF +2qP1u4F + = 3M2mqP1uF += 0.75MMAP1uu2m2m破坏机构 10.375Mu作结构M图。考虑BC杆平衡，求最大弯矩的截面位B0.5CMu置。Mu2m0.75Mu0.5Mu2mF= 1 ( 1 M+ 0.375M´ 4 ´ 2)0.75Mu4mM 图0.375MuyCuu42MuA= 0.875M=

8、 0uFQx x = 0.875M= 2.333m/ 0.375MuuBCFyCx0.5Mu4m0.375MuCMmax2.333FyC=0.875Mum= 0.875Mu ´ 2.333 - 0.5´ 0.375Mu= 2.0417Mu -1.0208Mu = 1.021Mu´ 2.3332M maxF += 0.75M/1.021 = 0.735MP1uuF += (0.75M+ 0.735M) / 2 = 0.743MP1uuu(精确解为0.74025Mu )= F + = 0.743MFPuP1u(2)q = F + /2 0.75MP 2uMuCDBCB

9、MM2qu1.25Mu2mu2mqMu+FP 21.5MuMu2m2mAA4Mu4m2m2m破坏机构 2M 图（不满足屈服条件）= 1.5MuF +´ 2 ´q ) = 3Muq+2(P 2 2FP 2考虑DC平衡：0.75MuDCMuF= 1 (MF+ 0.75M´ 2 ´1)2myCyCuu2= 1.25Mu0.75Mu考虑BC平衡：BCMB4mFyC=1.25MuMB = (-1.25Mu ´4 + 0.75Mu ´4´2)= Mu于是求得AB杆A端弯矩为：M A= 4Mu ( 左拉 )(3).q = F +0.5Mu/2 P3BBCMCuMuq2m1.5625M2mu1.25MuF +MuP32qMuq2m2m2.5mMuAMuA4m4mM 图（不满足屈服条件）破坏机构3F +´ 2q = 2M q