有限元大作业

有限元大作业姓名：学院：机械工程学院

题目要求：简支梁解:一、将简支梁分成5个单元e1-e5，相对应产生6个节点n1-n6，每个节点有2个自由度，如下列图所示：求形函数对单元e1分析令〔1〕选择局部坐标，其中是单元长度。用矩阵的形式表示位移为〔2〕〔3〕得到形函数〔4〕矩阵形式：〔5〕三、求单元的质量矩阵和刚度矩阵动能表达式：〔6〕将〔5〕式带人〔6〕式得：〔7〕所以：〔8〕由matlab编程得：Me=[0.0032091,0.000036206,0.0011109,-0.000021394][0.000036206,5.2663e-7,0.000021394,-3.9497e-7][0.0011109,0.000021394,0.0032091,-0.000036206][-0.000021394,-3.9497e-7,-0.000036206,5.2663e-7]势能表达式：〔9〕将〔5〕式带入得：〔10〕所以：〔11〕由matlab编程得：Ke=[21875.0,875.0,-21875.0,875.0][875.0,46.667,-875.0,23.333][-21875.0,-875.0,21875.0,-875.0][875.0,23.333,-875.0,46.667]三、求总体质量矩阵和刚度矩阵〔12〕我们可以得到：〔13〕由公式：〔14〕

得:(15)由matlab编程得总质量阵:M=[0.00321,3.62e-5,0.00111,-2.14e-5,0,0,0,0,0,0,0,0][3.62e-5,5.27e-7,2.14e-5,-3.95e-7,0,0,0,0,0,0,0,0][0.00111,2.14e-5,0.00642,0,0.00111,-2.14e-5,0,0,0,0,0,0][-2.14e-5,-3.95e-7,0,1.05e-6,2.14e-5,-3.95e-7,0,0,0,0,0,0][0,0,0.00111,2.14e-5,0.00642,0,0.00111,-2.14e-5,0,0,0,0][0,0,-2.14e-5,-3.95e-7,0,1.05e-6,2.14e-5,-3.95e-7,0,0,0,0][0,0,0,0,0.00111,2.14e-5,0.00642,0,0.00111,-2.14e-5,0,0][0,0,0,0,-2.14e-5,-3.95e-7,0,1.05e-6,2.14e-5,-3.95e-7,0,0][0,0,0,0,0,0,0.00111,2.14e-5,0.00642,0,0.00111,-2.14e-5][0,0,0,0,0,0,-2.14e-5,-3.95e-7,0,1.05e-6,2.14e-5,-3.95e-7][0,0,0,0,0,0,0,0,0.00111,2.14e-5,0.00321,-3.62e-5][0,0,0,0,0,0,0,0,-2.14e-5,-3.95e-7,-3.62e-5,5.27e-7]总刚度阵：K=[2.19e4,875.0,-2.19e4,875.0,0,0,0,0,0,0,0,0][875.0,46.7,-875.0,23.3,0,0,0,0,0,0,0,0][-2.19e4,-875.0,4.38e4,0,-2.19e4,875.0,0,0,0,0,0,0][875.0,23.3,0,93.3,-875.0,23.3,0,0,0,0,0,0][0,0,-2.19e4,-875.0,4.38e4,0,-2.19e4,875.0,0,0,0,0][0,0,875.0,23.3,0,93.3,-875.0,23.3,0,0,0,0][0,0,0,0,-2.19e4,-875.0,4.38e4,0,-2.19e4,875.0,0,0][0,0,0,0,875.0,23.3,0,93.3,-875.0,23.3,0,0][0,0,0,0,0,0,-2.19e4,-875.0,4.38e4,0,-2.19e4,875.0][0,0,0,0,0,0,875.0,23.3,0,93.3,-875.0,23.3][0,0,0,0,0,0,0,0,-2.19e4,-875.0,2.19e4,-875.0][0,0,0,0,0,0,0,0,875.0,23.3,-875.0,46.7]施加约束条件并计算各阶固有频率由于梁为简支梁，所以节点1和节点6的挠度v=0。划去总质量阵和总刚度阵的第一行第一列和第11行第11列，可由matlab编程得到。由公式〔16〕可知：由matlab编程得各阶固有频率：表1各阶固有频率阶数固有频率12345678910各阶主振型为矩阵[L]的各列向量：表2各阶主振型阶数12345678910主振型00000000Matlab程序：clearclcsymsxldenbhEnLfL=0.4;n=5;l=L/n;b=0.02;h=0.002;den=2700;A=b*h;E=70*10^9;I=(b*h^3)/12;%参数定义N1=((x-l)^2)*(l+2*x)/l^3;N2=(x*(x-l)^2)/l^2;N3=(x^2*(3*l-2*x))/l^3;N4=(x^2*(x-l))/l^2;N=[N1,N2,N3,N4];%构造形函数Me0=int(N'*N,x,0,l);Me=den*A*Me0;Me=vpa(Me,5);%求单元质量阵Ke0=int(diff(N.',x,2)*diff(N,x,2),x,0,l);Ke=E*I*Ke0;Ke=vpa(Ke,5);%求单元刚度阵M=zeros(12,12);K=zeros(12,12);fori=1:1:5a=zeros(4,12);forj=1:1:4a(j,2*i+j-2)=1;endM=M+a.'*Me*a;K=K+a.'*Ke*a;endM=vpa(M,3);K=vpa(K,3);%求总体质量阵和刚度阵for