Contents

Aaron Klapheck

% Home Work Ch 3: Problem #23 22-Sep-08
clear, clc, home
fprintf('The date and time: %s \n', datestr(now))

The date and time: 29-Sep-2008 10:01:19 

Givens and Setup

% See 3-23 in the book for the diagram on page 131.
% Given: l, a, b, w, and F in units of m and N.
% Find: Reaction forces, maximum shear, and maximum moment.

% Given
l = 1; a = .4; b = .6; w = 200; F = 500;

% Solution

% State reactions
By = (w*a.^2./2 + F*b)/l;
Ay = F + w*a - By;


% State range
x = [0:0.001*l:l];

Shear

% State V:
% (x>=a) is the same as heaviside(x-a) except that it is defined for x=a.
V = Ay*(x>=0).*(x-0).^0 - w*(x>=0).*(x-0).^1 + w*(x>=a).*(x-a).^1 - ...
    F*(x>=b).*(x-b).^0 + By*(x>=l).*(x-l).^0;

plot(x,V), grid

maxV = max(abs(V));

fprintf('\nThe Maximum Shear is %g N \n', maxV)

The Maximum Shear is 316 N 

Moment

M = Ay*(x>=0).*(x-0).^1 - w/2*(x>=0).*(x-0).^2 + w/2*(x>=a).*(x-a).^2 - ...
    F*(x>=b).*(x-b).^1 + By*(x>=l).*(x-l).^1;

plot(x,M), grid

maxM = max(abs(M));

fprintf('\nThe Maximum Moment is %g N \n', maxM)

The Maximum Moment is 126.4 N 

Published with MATLAB® 7.5