Matlab program
% hibbeler_6_11.m
% Ricardo E. Avila Montoya, Feb. 18, 2014
clc
clear
disp(' ')
disp('Solution of Prob. 6-11, Hibbeler,')
disp('"Mechanics of Materials", 8th. ed')
disp('_________________________________')
% Solution of equilibrium equations.
% The unknown forces of the system are Cx, Cy, Dx, Dy
% Matrix of coefficients: distance is measured in feet
A = [-1 0 1 0; % Equation: -Cx + Dx = 0;
0 -1 0 1; % Equation: -Cy + Dy = 800;
0 0 -2 -4; % Equation: -2Dx -4Dy = -8000;
0 0 3 -4]; % Equation: 4Dx -3Dy = 0;
% Right Hand Side of equilibrium equations:
RHS = [0; 800; -8000; 0];
% Solve the linear system to find the unknown forces:
Force_vector = A \ RHS; % A Matlab algorithm is used here.
disp(' ')
disp('Forces at points C and D, Lb: ')
disp('_________________________________')
format compact
% Assign values of the equilibrium solution to the forces:
Cx = Force_vector(1)
Cy = Force_vector(2)
Dx = Force_vector(3)
Dy = Force_vector(4)
% From here on, distance measurement unit is inches
L = 120; % length of A-C main beam, inch
dx = .25; % discrete division of x-axis, along beam A-C
% Generate discrete values for the x-axis of beam A-C
x = [0 : dx: L]';
N = size(x, 1); % number of rows in column of x-axis values
% Initialize values of variables to be used in beam problem:
V = zeros(N, 1); % Shear force
M = V; % Bending moment
y = V; % elastic curve, y(x) displacement
% Solution of linear system to calculate integration constants:
A = [72 1; 120 1]; % From boundary conditions of beam
RHS = [49766400; 164044800]; % Right-hand side of equations
C = A \ RHS; % Solution of the linear system of equations
disp(' ') % display empty line
disp('First constant of integration: ')
disp(C(1))
disp('Second constant of integration: ')
disp(C(2))
% y(x) is obtained by the method of double integration
for j = 1 : N
% First section of beam, from A to B
V(j) = -800;
M(j) = -800 * x(j);
y(j) = -133.333 * x(j)^3 + C(1) * x(j) + C(2);
if x(j) >= 72; % 2nd. section of beam, from B to C
V(j) = V(j) + Dy;
M(j) = M(j) + Dy * (x(j) - 72) + 24 * Dx;
y(j) = y(j) + (Dy/6) * (x(j) - 72)^3 + 12 * Dx * (x(j) - 72)^2;
end
end
% Material properties of square hollow section 4" x 4" x 1/4" wall
E = 29E6; % Lb/in2, ASTM A-36 steel
I = 7.8; % in^4, from tables of structural materials
% Divide by the product E*I, to obtain elastic curve
y = y/(E * I);
% Post-processing section
disp('Maximum absolute value of bending moment (Lb-in): ')
disp(max(abs(M)))
figure(1)
plot(x, V, 'r', 'linewidth', 2)
axis([0 120 -850 450])
grid
xlabel('x, in')
ylabel('V(x), in')
title('Shear force diagram')
figure(2)
plot(x, M, 'b', 'linewidth', 2)
grid
xlabel('x, in')
ylabel('M(x), Lb-in')
title('Bending moment diagram')
% The beam is considered as simply soported at points B (x = 72)
% and C (x = 120), for the calculation of the elastic curve.
% The effect of the motion of point D under load, due to bending
% of the projected arm BD, can be added later, and that effect
% superimposed to the elastic curve as it has been calculated here.
figure(3)
plot(x, y, 'k', 'linewidth', 2)
grid
xlabel('x, in')
ylabel('y, in')
title('Elastic curve of beam')
disp('_________________________________')
disp('Successful run of Matlab program. ')
disp(' ')
% end of Matlab program
Figures
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