1一、平面刚架力学模型(1)(2)(3)(4)(5)(6)123456xyo6xP6yP6Mq▲杆件处于同一平面内,结点刚性连接。▲单元的线位移和转角在结点处都连续。▲每个杆件可看作梁单元。▲基本未知量为:结构的结点位移:T12T{}{{}{}......{}}{}{}nssssddddduv()T{}{}eiiijjjFXYmXYmxyiX,jjXu,iiYv,jjYv,iim,jjmij()e,iu平面刚架结构示意图梁单元示意图整体结构有限个梁单元平面刚架有限元分析单元的内力:2二、局部坐标系中的单元分析平面刚架有限元分析()T{}{}eiiijjjFXYmXYm梁单元示意图T{}{}iiijjjduvuv()()(){}[]{}eeeFkd结点载荷向量:结点位移向量:刚度方程:基本参数:,,,,lAIExyiX,jjXu,iiYv,jjYv,iim,jjmij()e,ius323223()323222000012612600646200[]000012612600626400eEAEAllEIEIEIEIllllEIEIEIEIllllkEAEAllEIEIEIEIllllEIEIEIEIllll局部坐标系中的单元刚度矩阵3梁单元示意图xyiX,jjXu,iiYv,jjYv,iim,jjmij()e,iursk刚度矩阵的系数的物理意义:第s个位移对第r个力的贡献。或1sdrF当第s个位移分量等于1,即,其余分量为零时需要施加的力()323223323222000012612600646200000012612600626400eiiiiiijjjjjjEAEAllEIEIEIEIuXllllvYEIEIEIEImlllluXEAEAllvYEIEIEIEImllllEIEIEIEIllll()e(r,s=1,2,..,6)平面刚架有限元分析4二、整体坐标系中的单元刚度矩阵平面刚架有限元分析1.矢量的坐标转换xxyyoij()eiXiYiXiYimimcossinsincosiiiiiiiiXXYYXYmmcossin0sinsin0001iiiiiiXXYYmm{}[t]{}iiFF同理:{}[t]{}jjFF()()(){}[T]{}eeeFF杆端力的坐标转换:()[][0][T][0][]ett坐标转换矩阵★只取决于()e5二、整体坐标系中的单元刚度矩阵平面刚架有限元分析1.矢量的坐标转换1T[T][T]()()(){}[T]{}eeeFF杆端力的坐标转换:()()(){}[T]{}eeedd杆端位移坐标转换:反变换:整体坐标局部坐标整体坐标局部坐标()T(){}[T]{}eeFF杆端力的坐标转换:()T(){}[T]{}eedd杆端位移坐标转换:可证:2.整体坐标系中的刚度方程()T(2)()()(){}[T]{}[T][]{}[T][][T]{}[]{}eTTeeeFFkdkdkd其中:为整体坐标系中的单元刚度矩阵,()()[][T][][T]eTekkxyiX,jjXu,iiYv,jjYv,iim,jjmij()e,iu整体坐标系中的单元的力和位移6二、整体坐标系中的单元刚度矩阵平面刚架有限元分析单元刚度矩阵的性质:(1)对称性(2)奇异性(3)分块性质T[][]kkrssrkk或T[][T][][T]kk位移互等定理由的对称性可知也对称。[]k给定一组平衡的杆端力,不等位移确定杆端位移。(){}eF(){}ed方程的解不唯一()()(){}[]{}eeeFkd()det[]0ek物理意义?7平面刚架有限元分析()(){}{},{}eiejFFF杆端力:()(){}{},{}eiejFFF()(){}eieiiiXFYm()(){},eieiiiXFYm()(){}{},{}eiejddd杆端位移:()(){}{},{}eiejddd()(){}eieiiiudv()(){},eieiiiudv()()()()()()[][]{}{},{}{}[][]eeeeiiijiieejjjijjkkFdFdkk刚度方程:()()()()()()[][]{}{}{}{}[][]eeeeiiijiieejjjijjkkFdFdkk(3)分块性质()()()()()()()()()(){}[]{}[]{}{}[]{}[]{}eeeeeiiiiijjeeeeejjiijjjFkdkdFkdkd或:8平面刚架有限元分析三、整体刚度矩阵的形成整体结点载荷矢量:T1234{}{{}{}{}{}}FFFFFT{}{}ssxsysFFFM其中:(1,2,3,4)sT1234{}{{}{}{}{}}ddddd整体结点位移矢量:其中:T{}{}ssssduv(1,2,3,4)s结构的刚度方程1211212121{}[]{}FKd1112131411212223242233313233344441324344[][][][]{}{}[][][][]{}{}{}{}[][][][]{}{}[][][][]KKKKFdKKKKFdFdKKKKFdKKKK分块形式:问题:[]?rsKxyo1234(1)(2)(3)44,xFu44,yFv44,m9平面刚架有限元分析xyo1234(1)(2)(3)44,xFu44,yFv44,m刚度集成法:考虑单元刚度对结构刚度的贡献。将单元刚度矩阵扩充为于整体刚度矩阵同阶的矩阵。三个杆件组成的平面刚架表1单元的局部码和整体码的对应关系两种编码的关系(1)(2)(3)i132j344()()[][]eekK()()1[][]eeKK(1)(1)(1)(1)(1)[][0][][0][0][0][0][0][][][0][][0][0][0][0][0]iiijjijjkkKkk(2)(2)(2)(2)(2)[0][0][0][0][0][0][0][0][][0][0][][][0][0][][]iiijjijjKkkkk10平面刚架有限元分析xyo1234(1)(2)(3)44,xFu44,yFv44,m三个杆件组成的平面刚架(1)(1)(1)(1)(1)[][0][][0][0][0][0][0][][][0][][0][0][0][0][0]iiijjijjkkKkk(2)(2)(2)(2)(2)[0][0][0][0][0][0][0][0][][0][0][][][0][0][][]iiijjijjKkkkk(3)(3)(3)(3)(3)[0][0][0][0][0][][0][][][0][0][0][0][0][][0][]iiijjijjkkKkk(1)(1)(3)(3)(1)(1)(2)(2)(3)(2)(2)(3)[][0][][0][0][][0][][][][0][][][][0][][][][]iiijiiijjijjiiijjijijjijkkkkKkkkkkkkk11平面刚架有限元分析思考题:1、如何推导?(虚位移原理,转角位移法)2、如何推导?()[]ek[]K(结点平衡法,能量法,虚位移原理)12平面刚架有限元分析▲用插值函数推导局部坐标系中的单元刚度矩阵:T{}{}iijjdvv231234vxxx0d,dixvxddjxlvx(0),()ijvvvvl(0)(1)(0)(1)[]{}iiiijjjjvHvHHvHHd(0)23(1)23(0)23(1)32132(2)32()iijjHHlHHlHermite插值多项式/xl(01)ijjiiujuivjvxyxyiXjXiYjYimjmij()e(1)由上面方程求得系数代入(1)后可得i13平面刚架有限元分析▲用插值函数推导局部坐标系中的单元刚度矩阵:*T*T*T0T22*T2220T221*T3220{}{}{}{}dd{}{}dddd{}{}dddddd{}d{}ddlAVlAdFxAxAHHdEydxAxxEIHHddl2222dd,ddvHyyxx22ddHEyEx虚位移原理:令T2213220dd[]dddEIHHkl由虚位移的任意性,可得*{}d{}[]{}Fkd14平面刚架有限元分析223221261266462[]1261266264llllllEIklllllll于是得到弯曲刚度矩阵在叠加拉伸刚度矩阵11[k]11EAl最终得到拉弯变形梁的刚度矩阵:323223323222000012612600646200[]000012612600626400EAEAllEIEIEIEIllllEIEIEIEIllllkEAEAllEIEIEIEIllllEIEIEIEIllll刚度系数:130dddddsrrssNEINkl(0)(1)(0)(1)1234,,,iijjNHNHNHNH15四、非结点载荷等效虚功相等条件进行等效。iqjabxy0iX0iY0im0jX0jY0jm*T*T00*TT0{}{}{}{}d{}[]{}dlldFvqxdHqx*T00T0{}{}{}d[]dllFvqxHqx000,0,ijXX230023231,222ijaaqaaYqaYllll323002683,431212ijqaaaqaammllll单元的等效结点载荷为:()T00{}[T]{}epF(1)均布载荷:平面刚架有限元分析-非结点载荷等效固端力:(a)均布载荷T0{}{,,,,,}iiijjjFXYmXYm虚功方程:16*T*T00*TT0{}{}{()}d{}[(/)]dlldFvaPxdHalPxijabxy0iX0iY0im0jX0jY0jmPT00{}[(/)]dlFHalPx(2)横向集中载荷:000,0,ijXX2200221,122ijbaaaYPYPllll220022,ijababmPmPll平面刚架有限元分析-非结点载荷等效(b)横向集中载荷虚功方程:固端力:17*T*T00T*T0{}{}{()}dd{}ddllxadFaMxHdMxxijabxy0iX0iY0im0jX0jY0jmMT00d{}ddlxaHFM