User:Eml4507.s13.team3.chen/Lecture 1

R1.4
On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Given
Consider a plane truss consisting of tree bars with the properties E = 200 GPa, A1 = 6.0 · 10-4 m2, A2 = 3.0 · 10-4 m2 and A3 = 10.0 · 10-4 m2, and loaded by a single force P = 80 kN. The corresponding finite element model consists of three elements and eight degrees of freedom.

Solution
The free body diagram is defined by the matrix Edof=[1 1 2 5 6; 2 5 6 7 8;         3 3 4 5 6]; Where the first number in each row is the element, and the four other numbers are the degrees of freedom at each node of the element.

Next, create the n x n stiffness matrix K, and the n x 1 load vector f >> K=zeros(8); f=zeros(8,1); f(6)=-80e3;

The axial rigidity of the elements are defined by the element property vectors >> E=2.0e11; >> A1=6.0e-4;   A2=3.0e-4;   A3=10.0e-4; >> ep1=[E A1];  ep2=[E A2];   ep3=[E A3];

and the element coordinate vectors ex1, ex2, ex3, ey1, ey2, ey3 >> ex1=[0 1.6];  ex2=[1.6 1.6];   ex3=[0 1.6]; >> ey1=[0 0];  ey2[0 1.2];   ey3=[1.2 0];

Using bar2e, compute Ke1, Ke2, and Ke3 >> Ke1=bar2e(ex1,ey1,ep1) Ke1 = 1.0e+007 * 7.5000   0   -7.5000    0        0    0         0    0  -7.5000    0    7.5000    0        0    0         0    0   >> Ke2=bar2e(ex2,ey2,ep2) Ke2 = 1.0e+007 * 0        0    0         0   0    5.0000    0   -5.0000   0   -5.0000    0    5.0000   >> Ke3=bar2e(ex3,ex3,ep3) Ke3 = 1.0e+007 * 6.4000  -4.8000   -6.4000   4.8000  -4.8000    3.6000    4.8000  -3.6000  -6.4000    4.8000    6.4000  -4.8000   4.8000   -3.6000   -4.8000   3.6000

Assemble the element stiffness matrices to create the global stiffness matrix

>> K=assem(Edof(1,:),K,Ke1); >> K=assem(Edof(2,:),K,Ke2); >> K=assem(Edof(3,:),K,Ke3) K = 1.0e+008 * 0.7500   0        0        0   -0.7500         0    0         0        0    0        0        0         0         0    0         0        0    0   0.6400   -4.800   -0.6400    0.4800    0         0        0    0        0        0         0         0    0         0        0    0  -0.6400    4.800    1.3900   -0.4800    0         0        0    0   0.4800   -3.600   -0.4800    0.8600    0   -0.5000        0    0        0        0         0         0    0         0        0    0        0        0         0   -0.5000    0    0.5000

Create a prescribed displacement matrix bc >> bc=[1 0;2 0;3 0;4 0;7 0;8 0]; Then solve the system of equations using solveq, which results in displacements a and support forces r >> [a,r]=solveq(K,f,bc) a = 1.0e-002 * 0        0         0         0   -0.0398   -0.1152         0         0   r = 1.0e+004 * 2.9845        0   -2.9845    2.2383    0.0000    0.0000         0    5.7617

From the displacement matrix, it can be seen that the vertical displacement at the point of loading is 1.15 mm. Obtain the element displacements ed1, ed2, and ed3 using extract. >> ed1=extract(Edof(1,:),a); >> ed2=extract(Edof(2,:),a); >> ed3=extract(Edof(3,:),a); Then calculate the normal forces N1, N2, and N3 using bar2s. >> N1=bar2s(ex1,ey1,ep1,ed1) N1 = -2.9845e+004 >> N2=bar2s(ex2,ey2,ep2,ed2) N2 = 5.7617e+004 >> N3=bar2s(ex3,ey3,ep3,ed3) N3 = 3.7306e+004 The normal forces are N1 = −29.84 kN, N2 = 57.62 kN and N3 = 37.31 kN.