材料力学：第七章弯曲变形2

1、F1F2F1F2wC1F1F2CwC2 EIlMwEIlMAA22ee EIFlwEIFlAA3232 EIqlwEIqlAA8643 AlFAlMeAlq EIFlwEIFlCA481632 EIqlwEIqlCA38452443 EIlMEIlMBA36ee BA2l2lCqBA2l2lCMeBA2l2lCF EIlMwC162e例题例题1已知：已知：q , a , EI = 常数常数求：求： A, wC解：解： 1. 分解分解qaBACaaq EIaqawCF4823 EIaqwCq384254qaBACaaqwCqaBACaawCFBACaaqwCqqaBACaaqwCqaBACaaw

2、CFBACaaqwCqAq EIaqa4823= EIaq384254 EIqa834AqAFA EIaqEIaqa24216232 EIqa1273AAFCFlABF2l2lCABF2l2lwBFB2lBCFlAB2l2lwCM EIFlwlwwCMBBFC481923e EIFlEIFwlBF243)(332 EIFlB2)(22 EIlFlwCM22CFlABF2l2lCABF2l2lwBFB2lB CFlAB2l2lwCMBA2aqa分析：分析：AB段段B截面转角引起截面转角引起 B a （） EIqa324EI=常数，常数，求求wC 注意：引起注意：引起 B的有两项的有两项: q 和

3、和qa2，他们的转向不同，他们的转向不同，叠加时注意正负号。叠加时注意正负号。qaBACa2aqqa2qaBCaFBvCF结果结果 wC= wCF - B a BC段弹性弯曲引起段弹性弯曲引起 wCF()CFB悬臂梁悬臂梁AC 如图，如图，EI=常数常数wAPBC段段B截面向下位移截面向下位移wB ,转角引起转角引起B a求求: A , wA AB段弹性弯曲引起段弹性弯曲引起 wAF分析：分析： 分为分为悬臂梁悬臂梁AB和和BCBAaFCEI2EIaABFEIBFaMB FBAaC2EIawBwBB aCFBwAP求求: A , wA BAaFCEI2EIaABFEIBFaMB FBAaC2E

4、IavBwBB aEIFaEIaFaEIFaB432)2( 222 ( ) EIFaEIaFaEIFawB125)2(2)2(3223aEIFaEIFaEIFaawwwBBAFA431253233 EIFa233EIFaEIFaBAFA43222EIFa452( )CAB2l2l CwCqCAB2l2l CwCqAB2l2lqAB2l2lqAB2l2lqCAB2l2lwCqbdF=qdxdxx)43(48)d()43(48dd2222xlEIxxqxlEIFxwC xxxlEIqwwbCCd)43(48d220 )23(48222blEIqb lwlwmax max lw 1000015000

5、1 75014001 100012501 BACAFBAFBFBFAYFAXMAAFBAFBAFBlABlABlFFB静定基静定基解除所有外力和多余约束解除所有外力和多余约束相当系统相当系统静定基加全部载荷和多余静定基加全部载荷和多余未知力未知力M 82ql12892ql EIqlqwB84 EIlFFwBBB33 03834 EIlFEIqlB qlFB83AlqBAlqBxyFB超静超静EIqlwBq84 EIlFwBF33N llABCqEAlFlN BClqABEIEAFNFNM 1652ql51292qlEIqlwBq84 EIlFwBF33N EAlFEIlFEIqlN3N438 qlF163N llABCqEAlFlN 32AlI BClqABEIEAFNFN 1.应用应用可以推知某些未知量：可以推知某些未知量：FXAFXB= 0AlqBFYAFXAMAFYBFXB对称性对称性AlqBFYAFXAMAFYBFXBMBAlqBAlqB 0 BByBBBBMFq FyBFxBMB 0BByBBBBMwFwqww三次静不定三次静不定0 BA 或 0 BAAAAAMMq AlqBBACqlMAMBN AqC2lFSMCCB2lqFSMCAlqB CCC