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.

 

Shear force and bending moment diagrams for hyperstatic beam problem

Matlab program for problem 12-121 of Hibbeler's
"Mechanics of Materials", 8th. edition
 

% hibbeler12_121.m

% Luis Adan Villa Chaparro / Ricardo E. Avila Montoya
% 3 Mar. 2014 / ricardo_avila@hotmail.com

clc

clear

disp(' ')

disp('Problem 12-121, Hibbeler''s ')
disp('"Mechanics of Materials", 8th. ed.')
disp(' ')

% The beam is hyperstatic, so the method of superposition

% has to be used in order to find the reactions at the
% supports of the shaft.
L = 4;              % Total length of beam 
dx = L/400;         % Division is 1 centimeter
x = [0 : dx : L]';  % x-axis is defined
n = size(x, 1);     % Size of the x-axis column of data
V = zeros(n, 1);    % Initialize structures to data
M = V;             
Y = V;

% Each half of the beam exerts a moment of 150 N-m

% on the other half, in order to achieve dy/dx = 0 @ x = 2,
% otherwise the shaft will break in two parts, if the
% derivative were not countinuous at the middle of the shaft.

% The necessary moment of 150 N-m is found by superposition,

% using the formulas at appendix C of the textbook.

% Support force on left end: 125 N

% Support force on middle of shaft: 550 N
% Support force on right end: 125 N

% Data processing

for index = 1 : n
  V(index) = 125 ;
  M(index) = 125*x(index);
  Y(index) = 20.83333*x(index)^3 - 50*x(index);

  if  x(index) >= 1

      V(index) = V(index) - 400;
      M(index) = M(index) - 400*(x(index)-1);
      Y(index)= Y(index)  - 66.66666*(x(index)-1)^3;

  if  x(index) >= 2

      V(index) = V(index) + 550;
      M(index) = M(index) + 550*(x(index)-2);
      Y(index) = Y(index) + 91.66666*(x(index)-2)^3;

  if  x(index) >= 3

      V(index) = V(index)-400;
      M(index) = M(index) - 400*(x(index)-3);
      Y(index) = Y(index) - 66.66666*(x(index)-3)^3;
  end
  end
  end
end

% Material and geometric properties are temporarily made 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), m')
title('Bending Moment Diagram')

figure(3)

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

disp(' ')

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

% end of Matlab program

figure(1)

figure(2)

 

figure(3)
The deflection of the beam is exaggerated because E=1, I=1. It is necessary to design
the beam, in order to have a diameter determined, and to select a proper material,
in order to determine proper values for E and I.

 

Shear force, bending moment and elastic curve: Matlab program for cantilever beam

Problem 6.31, Hibbeler's "Mechanics of Materials", 8th. edition.

Solution of a cantilever beam, supported on its left end.
___________________________________________________________________

% hibbeler_6_31.m

% Luis Adan Villa Chaparro / Ricardo E. Avila, 24 Feb. 2014
% 

% Problem 6.31 of Hibbeler's "Mechanics of Materials", 8th. edition.

% A cantilever beam of unit length, supported only at its left end,
% has a uniformly distributed unit load w0 on the left half
% of its length; then the load decreases uniformly from w0 to 
% become zero load on the right half of the beam.

clc

clear
disp('Problem 6.31, Hibbeler''s "Mechanics of Materials"')
disp(' ')

L = 1;      % The length of the beam is set to be unit length

w0 = 1;     % The unit load, which may be increased at will.
dx = L/100; % Division of the beam length into segments.
x = [0 : dx : L]';  % Generate values of the x-axis.

n = size(x, 1);     % Size of the x-axis column of data

V = zeros(n, 1);    % Initialize column of shear force data.
M = V;              % Initialize column of bending moment data.
Y = V;              % Initialize column of elastic curve data.

% Processing of the data: shear force, bending moment

% and elastic curve equation, obtained by double integration.
% The two constants of integration are found to be zero, for a
% cantilever beam supported on its left end, at x = 0.

for index = 1 : n

  % First half of the beam length
  V(index) = .75*w0*L - w0*x(index);
  M(index) = (-7*w0*L^2)/24 + .75*w0*L*x(index) - .5*w0*x(index)^2;
  Y(index) = (-7*w0*L^2*(x(index))^2)/48 + ...
             (3*w0*L*(x(index))^3)/24 - (w0*x(index)^4)/24;

  % Second half of the beam length:
  length_2 = x(index) - L/2;  % Define this variable to save time
  if x(index) >= L/2
      V(index) = V(index) + w0*length_2^2/L;
      M(index) = M(index) + w0*length_2^3/(3*L);
      Y(index) = Y(index) + w0*length_2^5/(60*L) ;
  end
end

E = 1; % Elastic constant is unit value

I = 1; % Moment of Inertia is unit value

figure(1)

plot(x, V, 'r', 'linewidth', 3)
grid
xlabel('x, length units')
ylabel('V(x), force units')
title('Shear Force Diagram')

figure(2)

plot(x, M, 'b', 'linewidth', 3)
grid
xlabel('x, length units')
ylabel('M(x), force times distance')
title('Bending Moment Diagram')

figure(3)

plot(x, Y, 'b', 'linewidth', 3)
grid
xlabel('x, length units')
ylabel('y(x), length units')
title('Elastic Curve Diagram')

disp('________________________________ ')

disp('Successful run of Matlab program ')
disp(' ')

% end of Matlab program

Figure(1)

Figure(2)

Figure(3)

Shear force and bending moment diagrams using Matlab, Hibbeler's Mechanics.

Problem 6-11 of "Mechanics of Materials", Hibbeler, 8th. edition is solved using a Matlab program.

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