Hibbeler Problem 6-35. Shear force, bending moment and elastic curve diagrams.

Problem 6-35, 8th. edition of Hibbeler's "Mechanics of Materials".

A Matlab program to calculate and plot shear force, bending moment and elastic curve
of the loaded beam.

Length of beam is 6 m. Establish the x axis with its origin (x = 0) at the left end of the beam.

The beam is simply supported at its ends (x = 0 at left end, x = 6 at right end).

The load on the beam can be divided to form two loads, and thus simplify the problem:

1. A distributed uniform load of 200 N/m is located on the right half, 3 to 6 m of the beam.

2. A distributed uniformly increasing load starts from 0 N/m at x = 3 m, and it increases
to 200 N/m at the right end of the beam, x = 6 m.

Calculation of reaction forces at the supports: lump the loads, then there is a concentrated
load of 600 N located at x = 4.5, and another concentrated force of 300 N at x = 5.

The reactions are found to be RA = 200N, and RB = 700 N.

The figures below show the calculations to find bending moment M(x), shear force V(x) and
the elastic curve y(x).

Click on the figures to make them larger.

 

MATLAB PROGRAM:

% hibbeler_6_35.m
% Irma Iveth Martinez Huerta / Ricardo E. Avila
% mechanicsofmaterials.blogspot.com / ricardo_avila@hotmail.com

clc

clear

disp(' ')

disp('Problem 6-35, Hibbeler''s ')
disp('"Mechanics of Materials", 8th. ed.')
disp(' ')

% Pre-processing section

L = 6;              % Total length of beam
dx = L/600;         % x-axis division is 1 centimeter
x = [0 : dx : L]';  % x-axis is defined
n = size(x, 1);     % Size of the x-axis column of data

% Initialize data structures

V = zeros(n, 1);   
M = V;             
Y = V;

% Processing section of Matlab program

% Since no values of E (elastic modulus) and I (moment of inertia)
% are given, we use E = 1 and I = 1 to plot the elastic curve.

% First constant of integration, calculated from boundary condition

% x = 0 @ y = 6:
C1 = -1065;

for index = 1 : n

     % First portion of the beam, 0 < x < 3

     V(index) = 200;
     M(index) = 200 * x(index);
     Y(index) = 33.3333 * x(index)^3 + C1 * x(index);

     % Second portion of the beam, 3 < x < 6

      if  x(index) >= 3
          V(index) = V(index) - 200 * (x(index)-3) ...
                              - 33.3333 *((x(index)-3)^2);

          M(index) = M(index) - 100 *((x(index)-3)^2) ...

                              - 11.1111 * ((x(index)-3)^3);

          Y(index) = Y(index) -  8.3333 * ((x(index)-3)^4) ...

                              -  0.5556 * ((x(index)-3)^5);

      end

end

% Material and geometric properties are given unit value:

E = 1;      % Elastic constant is unit value (not calculated)
I = 1;      % Moment of Inertia is unit value (not calculated)

% Post-processing

figure(1)
plot(x, V, 'r', 'linewidth', 3)
xlabel('x, m')
ylabel('V(x), N')
grid
title('Shear Force Diagram')

figure(2)

plot(x, M, 'b', 'linewidth', 3)
grid
xlabel('x, m')
ylabel('M(x), N-m')
title('Bending Moment Diagram')

figure(3)

plot(x, Y, 'k', 'linewidth', 3)
grid
xlabel('x, m')
ylabel('y(x), m')
title('Elastic Curve Diagram')

disp(' ')

disp('_______________________________________')
disp('Successful execution of Matlab program.')

% end of Matlab program

Figure 1.

 

Figure 2.

 

Figure 3.