The finite element method linear static and dynamic finite element analysis
The global stiffness matrix Assignment
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Question 1
- Part a: The global stiffness matrix
- Part b: Displacements of Nodes 2 and 3
- Part c: The reaction force at nodes 1 and 4
- Part d: The force in the spring 2
K_2=200;
K_3=100;
Number of nodes is 4 and number of elements is 3 Global stifffness matrix, [K] is given by: [K]=[K_1]+[K_2]+[K_3] where [K_1] the local stiffness matrix of element 1 [K_2] the local stiffness matrix of element 2 and [K_3] the local stiffness matrix of element 3
K_1m=[100 -100; -100 100]; %local stiffness matrix of element 1
%updating the elements of the global stiffness matrix using the local
%stiffness matrices
K =
100.0000e+000 -100.0000e+000 0.0000e+000 0.0000e+000
From equilibrium equation [K][d]=[F] where [d] is the displacement vector [F] the force vector
%We know that:
eqn1=K(2,2)*u_2+K(2,3)*u_3==500;
eqn2=K(3,2)*u_2+K(3,3)*u_3==0;
3
d3 =
u_1=0; u_2=2; u_3=3;u_4=0;
d=[u_1; u_2; u_3; u_4]; %displacement vector
R_1 =
-200.0000e+000
F_2=K_2*(d3-d2)
F_2 =
Part a
E=30*10^6; %psi
% we know that at x=0, u(x)=0, which gives
% a_0=0;
% and u(x)=-6*a_2*x+a_2*x^2;
% u(x)=a_2*(x^2-6x)
T_2(x)=300*x/x; %T(x)=300 lb/in
P_e=0.5*(int(((E*A)*(diff(u,x))^2),x,0,60)-int((T_1(x)*u*A), x, 0,30)-int((T_2(x)*u*4*A), x, 30,60));
a_2=solve(diff(P_e,a_2)==0,a_2) %solving for a_2
%using a_2 in u(x) expression
u_2 =
(499*x^2)/109760000 - (1497*x)/54880000
%plot of u versus x
figure (1)
title('plot of u versus x')
%plot of sigma_xx versus x
ylabel('\sigma')
title('plot of \sigma versus x')
- Using force displacement method
- Nodal elongations
- Element stress calculations
- Calculating the support reactions
%Elongation is given
% Elong=PL/AE
L3=0.2;A3=400*10^-6; P3=600*10^3;
L4=0.2;A4=400*10^-6; P4=0;
% Hence there is a compression force at ends less than the deformation of
% the bars
%Reaction at unfixed end P5
P5=vpa(soln, 5)
u1=vpa(((P1-P5)*L1)/(A1*E),3)%node 1
u2=vpa(((P2-P5)*L2)/(A2*E),3) %node 2
0.00202
u2 =
-5.68e-4
u5 =
S3=vpa((P3-P5)/A3, 5) %Element 3
S4=vpa((P4-P5)/A4, 5) %Element 4
S3 =
9.3182e+8
P_start=vpa((600+300)*10^3-P5, 5)
P_end =
- Obtaining the element stiffness matrices
- Obtaining the global stiffness matrix
- Obtaining the nodal displacements and Support reactions from the Force displacement equation
- Element stresses
clc;
clear;
%same material, the stiffness matrices are equal:
E=200*10^9; %N/m^2
K1=((A1*E)/L1).*M %element 1 stiffness matrix
K2=((A2*E)/L2).*M %element 2 stiffness matrix
-333.3333e+009 333.3333e+009
K2 =
-400.0000e+009 400.0000e+009
K4 =
%updating the elements of the global stiffness matrix using the local
%stiffness matrices
disp("The Global stiffness matrix is")
disp(K)
0.0000e+000 -333.3333e+009 733.3333e+009 -400.0000e+009
0.0000e+000 0.0000e+000 -400.0000e+009 800.0000e+009
0.0000e+000
-400.0000e+009
u=[0 u2 u3 u4 3.5*10^-3]';
F=[F1 900*10^3 600*10^3 600*10^3 F5]';
u3=vpa(soln.u3, 3) %m
u4=vpa(soln.u4, 3) %m
9.58e-4
u3 =
-3.1925e+8
F5 =
FS=[F1 F2 F3 F4]';
Stress=[s1 s2 s3 s4]';
s3=vpa(sol.s3, 5) %Pa
s4=vpa(sol.s4, 5) %Pa
s3 =
1500.0
Felton, Lewis P, and Richard B Nelson. Matrix Structural Analysis. New York, J. Wiley, 2018.
Hughes, Thomas J R. The Finite Element Method : Linear Static and Dynamic Finite Element Analysis. New York, Dover Publication, Inc, 2017.
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