西北工业大学(王生楠)结构力学课后题答案第五章位移法

第五章位移法5-5-5-5-1111求装配图如图所示桁架结构刚度矩阵K。各杆截面刚度均为EA。45°12345678aaa(a)(a)(a)(a)(a)解：以铰点5为原心，建立如图所示的总体坐标系。建立元素刚度矩阵。杆元1-2，�0=θ，1cos==θλ，.,0sinal===θµ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−000001010000010121aEAK同理：877665433221−−−−−−=====KKKKKK杆元2-6，.,1sin,0cos,270al=−=====θµθλθ�⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−101000001010000062aEAK同理：847362−−−==KKK杆元1-6，.2,22sin,22cos,315al=−=====θµθλθ�⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−−−−=−11111111111111112261aEAK同理：837261−−−==KKK建立总体刚度矩阵，并进行删行删列：⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡+−−+−−−−−+−−+−−−−+−−+−−−−+−−−+−−−−+=24040000220002420000402242280400002200024200004000422800000000000400000000000404002400000242002000004228040224000002420220000042284aEAK60°1246357aaa(b)解：以节点1为原点，建立如图所示的总体坐标系。建立元素刚度矩阵。杆元1-3，.0sin,1cos,0,======θµθλθ�al⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−000001010000010131aEAK同理：6442755331−−−−−====KKKKK杆元1-2，.23sin,21cos,60,======θµθλθ�al⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−−=−3333313133333131421aEAK同理：654321−−−==KKK杆元2-3，.23sin,21cos,300,−======θµθλθ�al⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−−=−3333313133333131432aEAK同理：765432−−−==KKK建立总体刚度矩阵K，并进行删行删列：⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−=2304343000000023434101000043432304343000043410254341010000434323043430001434102543410100004343230434300014341025434100000043432300000014341023aEAK5-25-25-25-2平面桁架的形状及尺寸如图所示，各杆的截面刚度EA均相同，,45,30==βα�用位移法求结点位移及各杆内力。Pl123654(a)αααα(a)(a)(a)(a)解：以1为原点，建立如图所示的总体坐标系。结点力和结点位移列阵。[].,,,,,,,,,,,665544332211TyxyxyxyxyxyxPPPPPPPPPPPPP=[]Tvuvuvuvuvuvu66554,4332211,,,,,,,,,,=∆建立元素刚度矩阵。杆元1-2，长度为.21sin,23cos,150,2==−===θµθλθ�l⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−−=−1313333313133333821lEAK杆元1-3，长度为,23sin,21cos,120,32==−===θµθλθ�l⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−−=−33333131333331318331lEAK杆元1-4，长度为.1sin,0cos,90,=====θµθλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−101000001010000041lEAK杆元1-5，长度为,23sin,21cos,60,32=====θµθλθ�l⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−−=−33333131333331318351lEAK杆元1-6，长度为.21sin,23cos,30,2=====θµθλθ�l⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−=−1313333313133333861lEAK建立总体刚度矩阵，并删行删列后。⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡++=⎥⎦⎤⎢⎣⎡−1136100032680vulEAPEAPlvu3923.0,011−==PPPPPPPPvlEAPPvlEAPxyxx14711.036103,39230.03610814711.03610338,08494.083541312=+==+=−=+−=×=−==对于结点2：300N1-22P2yP2xPPNPNxx15470.132030cos221221−=−==+×−−�对于结点3：N3-13P3yP3x600PPNNPx29423.036106,02113133−=+−==+−−对于结点4：P4yN4-14PPNy39230.0414==−对于结点5：5P5yP5x600PPNx29423.02515==−利用对称性得：PNN09808.01216==−−⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡−−−−−09808.029423.039230.029423.009808.06151413121PNNNNNPl12543(b)βαα0.5P(b)(b)(b)(b)解：以结点1为原点，建立如图所示的总体坐标系。结点力与结点位移列阵。[]TyxyxyxyxyxPPPPPPPPPPP5544332211,,,,,,,,,=[]Tvuvuvuvuvu554,4332211,,,,,,,,=∆建立元素刚度矩阵，杆元1-2，长度为.22sin,22cos,135,2==−===θµθλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−−−−=−11111111111111112221lEAK杆元1-3，长度为.1sin,0cos,90,=====θµθλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−101000001010000031lEAK杆元1-4，长度为.23sin,21cos,60,332=====θµθλθ�l⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−−=−33333131333331318341lEAK杆元1-5，长度为.21sin,23cos,30,2=====θµθλθ�l⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−−=−1313333313133333851lEAK建立总体刚度方程，并删行删列之后，⎥⎦⎤⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++−+++−++=⎥⎦⎤⎢⎣⎡11118339221221833833221221833vulEAPPyxEAPlvEAPlu544.0,665.011−==()()()PPPPPPPPPPPPPPxxyx13160.0544,083665.08306002.0544.083665.083544.042745.022544.022665.05432−=−×−×−==−×−×−==−=−−=对于结点2：2N2-1P2yP2x450PPNx60450.02221=−=−对于结点3：P3yN1-33PPNy544.0331==−对于结点4：4N1-4P4yP4x600PPNx12005.02441==−对于结点5：5N1-5P5yP5x300PPNx15195.032551==−⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−15195.012005.0544.060450.051413121PNNNNα14232PPl(c)(c)(c)(c)(c)解：以结点4为原心，建立如图所示的总体坐标系。结点力和结点位移列阵。[]TyxyxyxyxPPPPPPPPP44332211,,,,,,,=[]Tvuvuvuvu4,4332211,,,,,,=∆建立元素刚度矩阵。杆元4-1，长度为,1sin,0cos,90,=====θµθλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−101000001010000014lEAK同理：2314−−=KK杆元1-2，长度为.0sin,1cos,0,30=====θµθλθl⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−0000010100000101321lEAK同理：3421−−=KK杆元4-2，长度为.21sin,23cos,30,2=====θµθλθ�l⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−−−−−−=−1313333313133333824lEAK建立总体刚度矩阵，并删行删列后：⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡−−−−+=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡4228981838181838383318302vvulEAPP⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡329.0671.1745.0422EAPlvvuPPPPPPPyxyx671.1,0,329.0,43013.03311−===−=对于结点1：P1yN1-41N1-2P1xPPNPPNyx329.0,43013.0141121−===−=−−对于结点3：P3yN2-33N4-2P3xPPNPNyx671.1,0332334=−===−−对于结点4：N2-44N4-3P4x300N1-4PNN658.024142=−=−−⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡−=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡−−−−−658.0671.1329.00430.02432413421PNNNNN2314Plββ(d)(d)(d)(d)(d)解：利用对称性化简，其结构为：xy231P/2选取如图所示的总坐标系。结点力与结点位移列阵。[][]TTxyxxvuvuvuPPPPPP332211,322,1,,,,,0,,,2=∆−=建立元素刚度矩阵。杆元2-1，长度为.22sin,22cos,315,−=====θµθλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−−−−=−1111111111111111212lEAK杆元2-3，长度为.22sin,22cos,45,=====θµθλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−−−−=−1111111111111111232lEAK杆元1-3，长度为.1sin,0cos,90,2=====θµθλθ�l13000001012200000101EAKl−⎡⎤⎢⎥−⎢⎥=⎢⎥⎢⎥−⎣⎦建立总体刚度矩阵，并删行删列之后：131112222111022PvEAlv⎡⎤+−⎡⎤−⎡⎤⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥−+⎣⎦⎣⎦⎢⎥⎣⎦1322212vPlEAv⎡⎤−⎢⎥⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦−⎢⎥⎣⎦以下计算可能有错PPPxx1423,142413+−=−−=对于结点3：3450P3xN2-3N1-3PNNNPPNPNxx724,0271222,022313231332332−−==+−−===−×−−−−−对于结点1：1P1xN1-3N1-2P/2PNPNx5.0,02221121==+×−−利用对称性知，41214332,−−−−==NNNN.⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡−−=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡−−−−−5.0207.0293.0207.05.04143313221PNNNNN5-35-35-35-3如图所示刚架结构，在点5作用有刚架平面内的集中力矩M，用位移法求结点5处的转角5θ.M23145解：建立如图所示的总体坐标系。x14325M结点力与结点位移列阵。[][]TTvuvuvuvuvuMQNMQNMQNMQNMQNP555444333222111555444333222111,,,,,,,,,,,,,,,,,,,,,,,,,,,,θθθθθ=∆=建立总刚度矩阵，55444333222111,,,,,,,,,,,,,vuvuvuvuvuθθθθ均为0.⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡++MvulEJlEAlEJlEAlEJ001600022400022455533θ故删行删列之后，EJMlMlEJlEJi161645540=∴=×=×∑=θθ5-45-45-45-4求如图所示桁架结构的位移，已知杆2-3及杆1-3的长度为l,各杆的截面刚度为EA，斜支座和水平倾角为�45，左右对称。P1234解：以2为原点，建立如图所示的总体坐标系。Pxy1234结点力和结点位移列阵。[][]TTyxyxyxyxvuvuvuvuPPPPPPPPP4433221144332211,,,,,,,,,,,,,,=∆=建立元素刚度矩阵。杆元2-1.长度为.22,22,45,2===µλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−−−−=−11111111111111112212lEAK杆元2-3，长度为.0,1,0,===µλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−000001010000010132lEAK同理：4332−−=KK杆元3-1，长度为.1,0,90,===µλθ�l⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=−101000001010000013lEAK杆元4-1，长度为.22,22,135,2=−==µλθ�⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−−−−=−11111111111111112214lEAK建立总体刚度矩阵，并删行删列之后：⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡−−+=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−33111010020010211000021000vuvulEAP44223311,12020vuvuEAPlvuvu===⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡