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1、2012年有限元作业对于三角形单元有：1、形函数其中：相乘后得：也可写成：简写为：其中2、应变有弹性力学知道：，可得3、应力4、刚度矩阵题目：如图所示是一平面梁。载荷沿梁的上边均匀分布，其单位长度上的分布载荷P=100N/cm。假定μ=0，墙梁的厚度t=0.1cm。在不计自重情况下，试求其位移和应力。MATLAB程序：%FiniteElementAnalysisE=1;u=0;%对单元1进行分析xy1=[0,0;6,6;0,6];A1=0.5*det([1,xy1(1,1),xy1(1,2);1,xy1(2,1),xy1(2,2);1,
2、xy1(3,1),xy1(3,2)]);B1=zeros(3,6);B1(1,1)=xy1(2,2)-xy1(3,2);B1(1,3)=xy1(3,2)-xy1(1,2);B1(1,5)=xy1(1,2)-xy1(2,2);B1(2,2)=xy1(3,1)-xy1(2,1);B1(2,4)=xy1(1,1)-xy1(3,1);B1(2,6)=xy1(2,1)-xy1(1,1);B1(3,1)=xy1(2,2)-xy1(3,2);B1(3,2)=xy1(3,1)-xy1(2,1);B1(3,3)=xy1(3,2)-xy1(1,2);B1(
3、3,4)=xy1(1,1)-xy1(3,1);B1(3,5)=xy1(1,2)-xy1(2,2);B1(3,6)=xy1(2,1)-xy1(1,1);%对单元2进行分析xy2=[0,0;6,0;6,6];A2=0.5*det([1,xy2(1,1),xy2(1,2);1,xy2(2,1),xy2(2,2);1,xy2(3,1),xy2(3,2)]);B2=zeros(3,6);B2(1,1)=xy2(2,2)-xy2(3,2);B2(1,3)=xy2(3,2)-xy2(1,2);B2(1,5)=xy2(1,2)-xy2(2,2);B2(
4、2,2)=xy2(3,1)-xy2(2,1);B2(2,4)=xy2(1,1)-xy2(3,1);B2(2,6)=xy2(2,1)-xy2(1,1);B2(3,1)=xy2(2,2)-xy2(3,2);B2(3,2)=xy2(3,1)-xy2(2,1);B2(3,3)=xy2(3,2)-xy2(1,2);B2(3,4)=xy2(1,1)-xy2(3,1);B2(3,5)=xy2(1,2)-xy2(2,2);B2(3,6)=xy2(2,1)-xy2(1,1);%单元1的刚度矩阵K1=zeros(6,6);fori=1:3forj=1:3K
5、1(2*i-1,2*j-1)=((0.1*E)/(4*(1-u^2)*A1))*(B1(1,2*i-1)*B1(1,2*j-1)+(1-u)/2*B1(2,2*i)*B1(2,2*j));K1(2*i-1,2*j)=((0.1*E)/(4*(1-u^2)*A1))*(u*B1(1,2*i-1)*B1(1,2*j-1)+(1-u)/2*B1(2,2*i)*B1(1,2*j-1));K1(2*i,2*j-1)=((0.1*E)/(4*(1-u^2)*A1))*(u*B1(2,2*i)*B1(1,2*j-1)+(1-u)/2*B1(1,2*i
6、-1)*B1(2,2*j));K1(2*i,2*j)=((0.1*E)/(4*(1-u^2)*A1))*(B1(2,2*i)*B1(2,2*j)+(1-u)/2*B1(1,2*i-1)*B1(1,2*j-1));endendK1Ka=zeros(8,8);Ka(1:6,1:6)=K1;temp=Ka(1:1,:);Ka(1:1,:)=Ka(5:5,:);Ka(5:5,:)=temp;temp=Ka(2:2,:);Ka(2:2,:)=Ka(6:6,:);Ka(6:6,:)=temp;temp=Ka(3:3,:);Ka(3:3,:)=Ka(
7、7:7,:);Ka(7:7,:)=temp;temp=Ka(4:4,:);Ka(4:4,:)=Ka(8:8,:);Ka(8:8,:)=temp;temp=Ka(3:3,:);Ka(3:3,:)=Ka(5:5,:);Ka(5:5,:)=temp;temp=Ka(4:4,:);Ka(4:4,:)=Ka(6:6,:);Ka(6:6,:)=temp;temp=Ka(:,1:1);Ka(:,1:1)=Ka(:,5:5);Ka(:,5:5)=temp;temp=Ka(:,2:2);Ka(:,2:2)=Ka(:,6:6);Ka(:,6:6)=temp
8、;temp=Ka(:,3:3);Ka(:,3:3)=Ka(:,7:7);Ka(:,7:7)=temp;temp=Ka(:,4:4);Ka(:,4:4)=Ka(:,8:8);Ka(:,8:8)=temp;te
