Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 51P from Chapter 7 from Hibbeler's Engineering Mechanics.
Problem 51P
Step-by-Step Solution
We are given uniform distributed load on beam is $w$.
We are asked the shear and moment diagrams for the beam.
Step 2
The free body diagram of the system is:
We have the distance between point A and point B is $AB = a$.
We have the distance between point B and point C is $BC = a$.
We have the distance between point A and point C is $AC = 2a$.
Step 3
The moment of forces at point A is,
\[\begin{array}{c} \Sigma {M_A} = 0\\ w\left( {AC} \right)\left( {AB} \right) - {B_y}\left( {AB} \right) = 0 \end{array}\]Step 4
Substitute the values in the above expression.
\[\begin{array}{c} w\left( {2a} \right)\left( a \right) - {B_y}\left( a \right) = 0\\ {B_y} = 2wa \end{array}\]Step 5
The forces acting in vertical direction is,
\[w\left( {AC} \right) = {A_y} + {B_y}\]Step 6
Substitute the values in the above expression.
\[\begin{array}{c} w\left( {2a} \right) = {A_y} + 2wa\\ {A_y} = 0 \end{array}\]Step 7
Consider a section at a distance x from point A. For $0 \le x < a$.
Step 8
The forces acting over the component in vertical direction is,
\[\begin{array}{c} - V - wx = 0\\ V = - wx \end{array}\]Step 9
The moment about the section is,
\[\begin{array}{c} \Sigma M = 0\\ M + wx\left( {\frac{x}{2}} \right) = 0\\ M = - \frac{{w{x^2}}}{2} \end{array}\]Step 10
Consider a section at a distance x from point A. For $a < x \le 2a$.
Step 11
The forces acting over the component in vertical direction is,
\[\begin{array}{l} - V + 2wa - wx = 0\\ V = w\left( {2a - x} \right) \end{array}\]Step 12
The moment about the section is,
\[\begin{array}{l} \Sigma M = 0\\ M + wx\left( {\frac{x}{2}} \right) - 2wa\left( {x - a} \right) = 0\\ M = 2wax - 2w{a^2} - \frac{w}{2}{x^2} \end{array}\]Step 13
The shear force and bending moment diagrams of the system are shown as:
