有限元钢架结构分析~手算+matlab+ansys模拟

由于单元①局部坐标系与整体坐标系的夹角为：??90，则单元①的局部坐标变换矩阵为：

?T1e?cos?sin?0000????sin?cos?0000???001000?????00cos?sin?0??0?000?sin?cos?0?????000001???12EI?l3??0???6EI?l2'e?K1?T1??12EI??3?l?0??6EI??2?l6EIl204EIl6EIl203EIl?0???1?0??0?0???010000100000000??000?000??

010??100??001??06EI?l2??0??3EI?l? 6EI??l2?0??4EI??l?可以得到在总体坐标系下的单元①的刚度矩阵：

0EAl00?EAl0??12EIl30?0EAl00EAl0?K1?TeT16EIl212EIl306EIl2对于单元②，求局部坐标系的单元刚度矩阵：

?EA0?2l?3EI?02l3?3EI?0?'2l2K2??EA0???2l3EI?0?3?2l?3EI0?2l2?03EI2l22EIl0?3EI2l2EIl?EA2l0003EI2l33EI?22l?03EI2l33EI?22l'EA2l00??0?3EI??22l?EI?l?

?0??3EI?2?2l?2EI??l?由于单元②局部坐标系与整体坐标系的夹角为??0,则K2?K2。 2、求单元节点等效载荷向量

将P等效在单元①两侧节点1,2上：

Fx11?Fx12?0

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Fb2?3a?b?44pF???? 3125L1y1Fa2?a?3b?81pF????

125L31y2Fab212plM??2??

125L11Fa2b18plM?2?

125L12将均布载荷等效在单元②两侧的节点2，3上：

Fx22?Fx23?0

Fy22?Fy23??1112pl?P 2Pl2PlM??M????

123与作用在节点上的力叠加为整体坐标系下的节点载荷：

Fx'1?44p 125Fy'1?0

M1'??12pl 12581pFx'2?

125Fy'2??P

67Pl?1811?' M2?????Pl??125103750??Fx'3?0

Fy'1??P

M3'?

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pl 3三、单元组装

将两个整体坐标系下的单元刚度矩阵组装为整体刚度矩阵：

?12EI?l3??0???6EI?l2?12EI??3?lK??0??6EI??2?l?0???0??0?0EAl00?EAl0000?6EIl20?12EIl30?0EAl00EA3EI?3l2l3EI2l20-3EI2l33EI2l2?6EIl20000-EA2l00EA2l0000003EI2l33EI-22l-03EI2l33EI-22l4EIl6EIl202EIl0006EIl212EIEA?l32l06EIl2EA-2l003EIl6EIl23EI2l26EIl0-3EI2l2EIl?0??0??0???0??3EI? 2?2lEI??l?0??3EI?-2?2l2EI??l?四、边界条件引入及组装总体方程

由于节点1、3为固定约束，所以节点1和3的x、y方向的位移以及转角均为0，节点2无位移约束，不存在支反力，所以力约束即为外力约束。

Qx1???U1?0???Qy1?V?0????1???M1??1?0??????Q?F'?81p?x2?U2??x2125?'K??V2???Qy2?Fy2??P?

???67Pl?'??2??M2?M2????U?0??750?U3?0?3????V3?0???V3?0???0????3????0?3??

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五、求解整体刚度方程，计算节点2的位移和转角

提取节点2位移的相关要素：

?12EIEA?l3?2l??0??6EI?l2?求得：

0EA3EI?3l2l3EI2l26EI?81p??'2?Q?F?x2x2l??U2??125?3EI??????V2???Qy2?Fy'2??P?2? 2l?67Pl?'6EI????2???M2?M2????750???l?

??14l3P（?316Al2?273I）??375E（8A2l4?105AIl2?72I2）??U2???7l3P（419Al2?4080I）?V?????

22422???375E（8Al?105AIl?72I）?????2???l2P（134A2l4?12831AIl2?31680I2）???24221125EI（8Al?105AIl?72I）??六、求节点1、3支撑反力

根据总体方程，提取求解节点1支撑反力所需方程：

?12EI??3l?Qx1??????Qy1???0?M1??6EI???2?l6EI?0?2?l?EA?0??l3EI?0?l??2P（394A2l4?20643?AIl2?13500I2）???2422'375（8Al?105AIl?72I）?U2??Fx1???227AlP（419Al?4080I）???????V2???Fy'1???2422??375（8Al?105AIl?72I）'??2??M1???2422????2Pl（298Al?32655AIl?1170I）??2422??1125（8Al?105AIl?72I）??

根据总体方程，提取求解节点2支撑反力所需方程：

????Qx3??????Qy3????M3??????EA2l000?3EI2l33EI2l2?0??3EI??22l?EI??l???7Al2P（?316Al2?273I）??2422'375（8Al?105AIl?72I）???U2?Fx3??2422??P（3067Al?50190AIl?54000I）?????V2???Fy'3???2422??375（8Al?105AIl?72I）'??2??M3??2422?????Pl（6268Al?130809AIl?247680I）???2422??2250（8Al?105AIl?72I）??

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