Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 147P from Chapter 4 from Hibbeler's Engineering Mechanics.
Problem 147P
Step-by-Step Solution
We are given the equivalent resultant force is $F = 0$.
We are given the resultant couple moment is $M = 8\;{\rm{kN}} \cdot {\rm{m}}$.
We are asked the length b of the triangular load and its position a on the beam.
Step 2
The following is the diagram of the beam:
We have the length of the beam is $l = 9\;{\rm{m}}$.
We have the magnitude of force per unit length acting at upper part of the beam is ${F_1} = 4\;{\rm{kN/m}}$.
We have the magnitude of force per unit length acting at lower part of the beam is ${F_2} = 2.5\;{\rm{kN/m}}$.
Step 3
The equivalent resultant force acting over the beam is,
\[F = \frac{1}{2}{F_2}l - \frac{1}{2}{F_1}b\]Step 4
Substitute the values in the above expression.
\begin{array}{c} 0 = \left[ {\frac{1}{2}\left( {2.5\;{\rm{kN/m}}} \right)\left( {9\;{\rm{m}}} \right)} \right] - \left[ {\frac{1}{2}\left( {4\;{\rm{kN/m}}} \right)b} \right]\\ b = 5.625\;{\rm{m}} \end{array}Step 5
The couple moment acting at point A is,
\[M = \frac{1}{2}{F_2}l\left( {\frac{2}{3} \times l} \right) - \frac{1}{2}{F_1}b\left[ {a + \left( {\frac{2}{3} \times b} \right)} \right]\]Step 6
Substitute the values in the above expression.
\begin{array}{c} 8\;{\rm{kN}} \cdot {\rm{m}} = \left[ {\frac{1}{2}\left( {2.5\;{\rm{kN/m}}} \right)\left( {9\;{\rm{m}}} \right) \times \left( {\frac{2}{3} \times 9\;{\rm{m}}} \right)} \right] - \left[ {\frac{1}{2}\left( {4\;{\rm{kN/m}}} \right)\left( {5.625\;{\rm{m}}} \right)\left( {a + \frac{2}{3} \times 5.625\;{\rm{m}}} \right)} \right]\\ 8\;{\rm{kN}} \cdot {\rm{m}} = 67.5\;{\rm{kN}} \cdot {\rm{m}} - 11.25\;{\rm{kN}}\left( {a + 3.75\;{\rm{m}}} \right)\\ a = 1.54\;{\rm{m}} \end{array}