实用标准文案
2011级南京理工大学研究生有限元理论作业
1．已知平面应力问题下三节点三角形单元的节点坐标i?6,0?、j?8,5?和m?3,4?；单元的节点位移分量ui?uj?1、vi?vj?um?vm?0；材料弹性模量E，泊松比?。试求：（1）单元的形函数Ni，Nj和Nm；（2）单元内应变和应力分量？
1xi1解：(1)A?1xj21xmai?xjxm1yi160123yj?185?22ym134yj?xjym?xmyj?8?4?3?5?17;aj??24;am?3ymyj?yj?ym?5?4?1;bj?4;bm??5bi??ci??Ni?1ym1xj?xm?xj?3?8??5;cj?3;cm?21xm11(ai?bix?ciy)?(17?x?5y)2A2311Nj?(aj?bjx?cjy)?(?24?4x?3y)2A2311Nm?(am?bmx?cmy)?(30?5x?2y)2A23?bi0bj0bm0?040?5?11??1?(2)应变矩阵?B????0ci0cj0cm??0?50302A23??cibicjbjcmbm????51342???1??0??5?040?50?????1?23??11????????B????e??0?50302????0???23?0???2????51342?5???0??23???????0???1??0???0??1040?50????1??1?E?Ee????????D??B??????100?50302?????01??2?1??2?1????51342?5????00????0?2?????0???0?2???5???5??5????????1??
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实用标准文案
2．图示为一个三个节点的杆单元，O为坐标原点，其位移模式取为
U?C1?C2?x?C3?x2。设其E?A为常数，试求其单元的刚度矩阵?K?。
1 2 O L/2 L/2 3 x 解： 节点坐标X1??l/2,X2?0和X3?l/2,则代入位移模式U?C1?C2?x?C3?x2有?U1?C1?C2l/2?C3l2/4U3?U12(U1?U3?2U2)?U?C?C?U,C?,C?,则?2112232ll?2?U3?C1?C2l/2?C3l/4x2x24x2x2x2位移模式可写成U?(??2)U1?(1?2)U2?(?2)U3,取位移模式lllllx2x24x2U(x)?N1(x)U1?N2(x)U2?N3(x)U3,其中N1(x)=??2，N2(x)=1?2lllx2x28x14x??14xN3(x)=?2，由此可的单元的应变矩阵?B?????2?2?2?lllll??ll杆单元的?D??E,则单元的刚度矩阵?K??????B??D??B?dv?EA??B??D??B?dxveTT代入?B?有：?14x???l?l2???l/28xe?14x?2????2?K?=EA??l/2??l??ll??14x???2??ll???7?81?EA????816?8?3l???1?87???8xl214x??dxll2??3．已知图示正方形薄板的边长为a，厚度为t，弹性模量为E，泊松比?。现将其分成两个三角形单元，设节点2、3和4为不动点，在节点1处受到向上的集中载荷P。试求节点位移，支座反力以及单元A和单元B内的应力？
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y P 实用标准文案 a 1 i m B A j 2 3 j m i 4 x 解：单元的节点编号如下： 单元节点号 A B i 1 4 a j 3 2 m 4 1
单元A的节点坐标为：i(0，a) ; j?0,0? ;m?a,0?
ai?0,aj?a2,am?0;bi?0,bj??a,bm?a;ci?a,cj??a,cm?0由此可计算出单元A的刚度矩阵?K??1???2??0?1????Et?2?2(1??2)?1?????2?0?1????2A?K?A1??0?21??3????21???12???10?1??21???2?11??23??2???1??21??02?01???1?21?????2100????????????1????2?单元B的节点坐标为：i(0，a) ; j?0,0? ;m?a,0?
ai?a2,aj??a2,am?a2;bi?0,bj?a,bm??a;ci??a,cj?a,cm?0
由此可计算出单元A的刚度矩阵?K?B1??1??1????1??0??0?2222???01???1?0???1??3??1??1??????1????EtA2222???K??21??3??1???2(1??)?1????1?????2222?????1??10??0?1??1??1??1?????0??0?2222?由A、B单元的刚度矩阵可写出整体刚度矩阵?K?精彩文档
实用标准文案
?3-?0-1?2?3-??-1?0?22??-13-??-122???-1?+1?-?Et22??K??2(1??2)??-1-10?2???-1-10?2??+1?-1?0?22??+1?0-??2采用主对角线元素为\的方法?3???2??0??0Et?2(1??2)?0??0?0??0?0?
-??-12-?00?-12-1000?+1??-12?+123-?200?+12?-12?-12-?-?3-?20?+13-?22?+13-?22-1-??-12-1?-12?-122??0???-????-1???-1??2??-1?2??0??3-???2?03??200000003-?2?-12?-12-1-1?000000??u1??0??????vP000000??1?????0??0?????100000?4(1??2)P??0??0?010000??0?=?0??u1=0,v1?(3??)Et?001000??0??0??????000100?0??0??????000010??0??0??????000001?-1-??3-??2??0???-1???-?Et?22(1??)??-1??2??-1?2??0???+1??2?-12-?00?-12-1000?+1??-123-?2?+1200?-12?+123-?200?+12?-12?-12-?-?3-?20?+13-?22?+13-?22-1-??+120?-12-??-12-1?-12?-122??0????u1??0?-???v1??P????????0??R2x?-1????R???0???2y??-1??0??R3x??????2??0??R2y??-1??0??R4x????2?????0????R2y??0??3-???2?精彩文档
实用标准文案
将u1，v1带入上式可求的各支座的约束反力R2x?P(??1)P(??1)2P2PP(??1),R2y?;R3x?,R3x?;R4x?,R4y?03??3????3??33????cm??cm?1???bm?2?单元内应力{?}=?S?{?}e??bi?cibj?cj?E?bici?bjcj?S??2(1??2)A??1??1??1??1??cibicjbj??2222代A、B单元的相关参数计算其单元内应力?4?P??(3??)ta???x??????4P?单元A{?}=??y????(3??)ta?????xy????0????bm?bm1??2cm?????x??0??????单元B{?}=??y???0?????2P(??1)??xy?????(3??)ta??
4．已知集中载荷P，试求图示六节点三角形单元的等效节点载荷列阵。
i(2,2) y 2 1 30o m(5,6)
P (5,4) 3 j(6,3) x 精彩文档
2011级南京理工大学有限元理论作业
