Shearing force and bending moment diagram: aircraft wing problem.

Problem 6-38 of "Mechanics of Materials", by Hibbeler, 8th. edition,
in US system of customary units (Lb, in).

The problem is to draw the diagrams for shearing force and
bending moment for a beam, which is the wing of an aircraft,
while it is static on the ground. The forces applied on the structure
are a concentrated weight of 3000 Lb corresponding to an engine,
the weight carried by the landing gear is 15000 Lb, and the
weight of the structure itself (this may include the fuel inside
the wings), which is a distributed uniformly varying load.

The values of shear force and bending moment have to be
calculated at point A, which is the attachment point of the
wing to the fuselage of the aircraft.
________________________________________________________

% prob6_38.m

% Ricardo Ernesto Avila Montoya, Sep. 9, 2013.
% Universidad Autonoma de Ciudad Juarez, Mexico

clc

clear
disp(' ')
disp('Problem 6-38, Hibbeler, "Mechanics of Materials", 8th. ed.')
disp(' ')

% Data input section

% The units of the problem are US customary system (Lb, inch)

% The wing of an aircraft is loaded with a 3000 Lb engine,    
% the reaction from the ground through the landing gear,  
% and the distributed load of the structure itself.

delta_x = .5;           % Increment for discretization of x axis.

x = (0: delta_x: 156)'; % Generate column array for x-axis (156 inch). 

N = size(x, 1);         % Size of the x-axis discretization array

V = zeros(N, 1);        % Shear force function of x.

M = V;                  % Bending moment function of x.

% Data processing section

for index = 1: N;       % For every discrete point along x.

   % First portion of the beam, 0 < x < 96, inch
   V(index) = -125 * x(index)^2 / (8*12^2);
   M(index) = V(index) * x(index)/3;

   if x(index) > 96;    % Engine weight minus distributed load 

   V(index) = V(index)-3000 + 6.25* (x(index)-96)^2 / (10*12^2);
   M(index) = M(index)-3000*(x(index)-96)...
        +6.25* (x(index)-96)^3 / (30*12^2);
   end

   if x(index) > 120;   % Last portion of wing, next to fuselage

       V(index) = V(index) + 15000;
       M(index) = M(index) + 15000*(x(index)-120);
   end   

end

% Data post-processing section

% Shear force diagram

figure(1)

plot(x,V, 'b', 'linewidth', 2.5)
grid
xlabel('x, inch')
ylabel('Shear force, Lb')
title('Shear force diagram')
axis tight

% Bending moment diagram

figure(2)
plot(x,M, 'r', 'linewidth', 2.5)
grid
xlabel('x, inch')
ylabel('Bending moment, Lb-in')
title('Bending Moment diagram')
axis tight

format long e

disp('Shear force at point A: ')
disp(V(N))
disp(' ')

disp('Bending moment at point A: ')

disp(M(N))
disp('_______________________________________')
format short

disp('*** Successful execution of program ***')

% end of Matlab program

figure(1)

figure(2)

 

Beam with uniformly varying distributed load: Matlab

Matlab program solves problem 6-73 of the textbook
"Mechanics of Materials", 8th. edition, by Hibbeler.

Applying the symmetry of the problem, only half of the beam
is calculated.

% prob6_73.m

% Ricardo E. Avila, Sep. 6, 2013
% Universidad Autonoma de Ciudad Juarez, Mexico

clc

clear
disp(' ')
disp('Problem 6-73, Hibbeler, "Mechanics of Materials", 8th. ed.')

% Data input section

% The units of the problem are US customary, not SI (Lb, inch)

% A beam supports a uniformly varying load applied on 24 ft. span length.   
% At midpoint of beam load is 500 Lb/ft, going down to 0 at both ends.
% Only half the problem is solved (12 ft.), using symmetry of beam.

I = 152.34;             % Moment of inertia, in^4, steel beam section.

E = 29e6;               % Elastic modulus, Lb/in^2, ASTM A-36 steel
delta_x = .5;           % Increment for discretization of x axis.
x = (0: delta_x: 144)'; % Generate column array for x-axis. 

N = size(x, 1);         % Size of the x-axis discretization

V = zeros(N, 1);        % Shear force function of x.

M = V;                  % Bending moment function of x.
y = V;                  % Coordinates of elastic curve.
C1 = -1250 * 12^4;      % Integration constant. (dy/dx) = 0 @ x = 144 in.

 

% Data processing section

for index = 1: N;   % For every discrete point along x.

   % First half of the beam, 0 < x < 144 (inch)
   V(index) = 3000 - 1.446759259259259e-001 * x(index)^2;
   M(index) = 3000 * x(index) - 4.822530864197531e-002 * x(index)^3;
   y(index) = 500 * x(index)^3 ...
       - 2.411265432098765e-003 * x(index)^5 + C1 * x(index);
end

y = y / (E*I); % E and I are constants, through the beam length.

% Data post-processing section

% Shear force diagram

figure(1)
plot(x,V, 'b', 'linewidth', 2.5)
grid
xlabel('x, inch')
ylabel('Shear force, Lb')
title('Shear force diagram')
axis tight

% Bending moment diagram

figure(2)
plot(x,M, 'r', 'linewidth', 2.5)
grid
xlabel('x, inch')
ylabel('Bending moment, Lb-in')
title('Bending Moment diagram')
axis tight

% Beam deflection: elastic curve

figure(3)
plot(x,y, 'k', 'linewidth', 2.5)
grid
xlabel('x, inch')
ylabel('Elastic Curve, in')
title('Bending Deflection')
axis tight

format long e

disp('Maximum deflection of beam: ')
disp(min(y))
disp(' ')
format short

disp('*** Successful execution of program ***')

% end of Matlab program

_____________________________________________________________________

Problem 6-73, Hibbeler, "Mechanics of Materials", 8th. ed.
Maximum deflection of beam:
   -5.407113851502764e-001

*** Successful execution of program ***
_____________________________________________________________________

figure(1)

figure(2)

 

figure(3)