Matlab 二维杆单元有限元模型编程求解代码

Matlab 二维杆单元有限元模型编程求解代码

信息：The truss example to be solved using the completed code. The vertical and horizontal segments of the crane are made of aluminum (Young’s modulus E=70 GPa, and have a cross-section of 2 cm2. The diagonal truss elements are made of steel (Young’s modulus E=210 GPa, and have a cross-section of 3 cm2. The structure is subjected to a load P=6000 N applied as illustrated in the figure. The two support nodes are assumed fixed (i.e., x- and y-displacements are 0).

代码：
input代码

%读取input文件按顺序读取节点坐标，单元信息，受力节点信息，固定节点
fid = fopen('input.txt','rt'); %打开文件
NodeCoordinate = fscanf(fid,'%g',[3,25]); %读取节点坐标
NodeCoordinate = NodeCoordinate';
ElementMsg = fscanf(fid,'%g',[5,47]);%读取单元信息
ElementMsg = ElementMsg';
ForceNodeMsg = fscanf(fid,'%g',[3,1]);%读取受力节点信息（节点，方向，载荷）
FixNode = fscanf(fid,'%g',[2,1]);%读取固定节点信息
fclose(fid); 

总体刚度矩阵的计算函数

function K = TotalStiffness_2DTRUSS(ElementMsg,NodeCoordinate)
%输入单元信息，节点坐标，输出总体刚度矩阵
Num_Node = size(NodeCoordinate,1);%计算节点数
Num_Element = size(ElementMsg,1);%计算单元数
K = zeros(2*Num_Node,2*Num_Node);%预设总体刚度矩阵
for k = 1:Num_ElementN1 = ElementMsg(k,2);%提取节点编号N2 = ElementMsg(k,3);x1 = NodeCoordinate(N1,2);%提取节点坐标x2 = NodeCoordinate(N2,2);y1 = NodeCoordinate(N1,3);y2 = NodeCoordinate(N2,3);L = sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));%计算单元长度C = (x2-x1)./L;%计算单元转角cos值S = (y2-y1)./L;%计算单元转角sin值A = ElementMsg(k,4);%提取单元截面积E = ElementMsg(k,5);%提取单元弹性模量Ke = E*A/L*[C*C C*S -C*C -C*S; %计算单元刚度矩阵C*S S*S -C*S -S*S;-C*C -C*S C*C C*S;-C*S -S*S C*S S*S];Locate = [2*N1-1,2*N1,2*N2-1,2*N2];%定位单元刚度矩阵坐标%组装总体刚度矩阵Kfor i = 1:4for j = 1:4K(Locate(i),Locate(j)) = K(Locate(i),Locate(j)) + Ke(i,j);endend
end
end

求解函数

Num_Node = size(NodeCoordinate,1);%计算节点数
Num_Element = size(ElementMsg,1);%计算单元数
K = TotalStiffness_2DTRUSS(ElementMsg,NodeCoordinate);%计算K
F = zeros(2*Num_Node,1);%生成力初始边界条件
F(ForceNodeMsg(1)*2-2+ForceNodeMsg(2),1) = ForceNodeMsg(3);%生成力边界条件
%用置0置1处理边界条件
Num_FixNode=size(FixNode,1);
K0 = K;
for i = 1:Num_FixNode%非主对角线元素全为0K(FixNode(i)*2-1,:)