Chapter 7.2, Problem 7.29P

(a)

To determine

Draw the shear force and bending moment diagrams.

Explanation of Solution

Assumptions:

Apply the sign convention for calculating the equations of equilibrium as below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Apply the following sign convention for calculating the bending moment at any section x-x while approaching from the left hand side.

• Take clockwise moment as positive and anticlockwise moment as negative

Apply the following sign convention for calculating the shear force at any section x-x while approaching from the left hand side.

• Take downward force as negative and upward force as positive.

Calculation:

Draw the free-body diagram of the beam as in Figure (1).

Find the vertical reaction at point D by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ D_y\left(L\right)-w\times \frac{L}{2}\times \left(\frac{L}{4}+\frac{1}{2}\times \frac{L}{2}\right)=0\\ D_y=\frac{wL}{4}(\uparrow )\end{array}$

Find the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-\frac{wL}{2}+D_y=0\\ A_y-\frac{wL}{2}+\frac{wL}{4}=0\\ A_y=\frac{wL}{4}(\uparrow )\end{array}$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ A_x=0\end{array}$

Consider a section x from left end of the beam.

Section AB $\left(0\le x<\frac{L}{4}\right)$:

Draw the free body diagram of the section AB as in Figure (2).

Find the shear force at the section by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \frac{wL}{4}-V=0\\ V=\frac{wL}{4}\end{array}$ (1)

Find the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -\frac{wL}{4}\left(x\right)+M=0\\ M=\frac{wLx}{4}\end{array}$ (2)

When $x=0$:

Refer to Equation (1),

$V=\frac{wL}{4}$

Substitute 0 for x in Equation (2).

$\begin{array}{c}M=\frac{wL\left(0\right)}{4}\\ =0\end{array}$

When $x=\frac{L}{4}$:

Refer to Equation (1),

$V=\frac{wL}{4}$

Substitute $\frac{L}{4}$ for x in Equation (2).

$\begin{array}{c}M=\frac{wL\left(\frac{L}{4}\right)}{4}\\ =\frac{w{L}^2}{16}\end{array}$

Section BC $\left(\frac{L}{4}<x<\frac{3L}{4}\right)$:

Draw the free body diagram of the section as in Figure (3).

Find the shear force at the section by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \frac{wL}{4}-V-w\left(x-\frac{L}{4}\right)=0\\ \frac{wL}{4}-V-wx+\frac{wL}{4}=0\\ V=\frac{wL}{2}-wx\end{array}$ (3)

Find the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -\frac{wL}{4}\left(x\right)+M+w\left(x-\frac{L}{4}\right)\times \frac{1}{2}\left(x-\frac{L}{4}\right)=0\\ M=\frac{wLx}{4}-\frac{w}{2}{\left(x-\frac{L}{4}\right)}^2\end{array}$ (4)

When $x=\frac{L}{4}$:

Substitute $\frac{L}{4}$ for x in Equation (3).

$\begin{array}{c}V=\frac{wL}{2}-w\left(\frac{L}{4}\right)\\ =\frac{wL}{4}\end{array}$

Substitute $\frac{L}{4}$ for x in Equation (4).

$\begin{array}{c}M=\frac{wL\left(\frac{L}{4}\right)}{4}-\frac{w}{2}{\left(\frac{L}{4}-\frac{L}{4}\right)}^2\\ =\frac{w{L}^2}{16}\end{array}$

When $x=\frac{3L}{4}$:

Substitute $\frac{3L}{4}$ for x in Equation (3).

$\begin{array}{c}V=\frac{wL}{2}-w\left(\frac{3L}{4}\right)\\ =-\frac{wL}{4}\end{array}$

Substitute $\frac{3L}{4}$ for x in Equation (4).

$\begin{array}{c}M=\frac{wL\left(\frac{3L}{4}\right)}{4}-\frac{w}{2}{\left(\frac{3L}{4}-\frac{L}{4}\right)}^2\\ =\frac{3w{L}^2}{16}-\frac{w{L}^2}{8}\\ =\frac{w{L}^2}{16}\end{array}$

Section CD $\left(\frac{3L}{4}<x\le L\right)$:

Show the free body diagram of the section as in Figure (4).

Find the shear force at the section by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ V+\frac{wL}{4}=0\\ V=-\frac{wL}{4}\end{array}$ (5)

Find the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -M+\frac{wL}{4}\left(L-x\right)=0\\ M=\frac{wL}{4}\left(L-x\right)\end{array}$ (6)

When $x=\frac{3L}{4}$:

Refer to Equation (5).

$V=-\frac{wL}{4}$

Substitute $\frac{3L}{4}$ for x in Equation (6).

$\begin{array}{c}M=\frac{wL}{4}\left(L-\frac{3L}{4}\right)\\ =\frac{w{L}^2}{16}\end{array}$

When $x=L$:

Refer to Equation (5).

$V=-\frac{wL}{4}$

Substitute L for x in Equation (6).

$\begin{array}{c}M=\frac{wL}{4}\left(L-L\right)\\ =0\end{array}$

Summarize the calculated shear force values as in Table (1).

Distance, x	Shear force, V
0	$\frac{wL}{4}$
$\frac{L}{4}$	$\frac{wL}{4}$
$\frac{3L}{4}$	$-\frac{wL}{4}$
L	$-\frac{wL}{4}$

Draw the shear force diagram as in Figure (5).

Refer to Figure (5):

Find location of maximum bending moment;

The bending moment is maximum where the shear force changes the sign.

$V=0$

The shear force changes its sign in the BC region.

Substitute 0 for V in Equation (3).

$\begin{array}{l}0=\frac{wL}{2}-wx\\ x=\frac{L}{2}\end{array}$

Substitute $\frac{L}{2}$ for x in Equation (4).

$\begin{array}{c}{M}_{\max }=\frac{wL\left(\frac{L}{2}\right)}{4}-\frac{w}{2}{\left(\frac{L}{2}-\frac{L}{4}\right)}^2\\ =\frac{w{L}^2}{8}-\frac{w{L}^2}{32}\\ =\frac{3w{L}^2}{32}\end{array}$

Summarize the calculated bending moment values as in Table (2).

Distance, x	Bending moment, M
0	0
$\frac{L}{4}$	$\frac{w{L}^2}{16}$
$\frac{L}{2}$	$\frac{3w{L}^2}{32}$
$\frac{3L}{4}$	$\frac{w{L}^2}{16}$
L	0

Draw the bending moment diagram as in Figure (6).

(b)

To determine

The maximum absolute values of the shear force and bending moment.

Answer to Problem 7.29P

The maximum absolute shear force is $\underset{\_}{\left|V_{\max }\right|=\frac{wL}{4}}$.

The maximum absolute bending moment is $\underset{\_}{\left|M_{\max }\right|=\frac{3w{L}^2}{32}}$

Explanation of Solution

Refer to the Figure (5),

The maximum absolute shear force is $\underset{\_}{\left|V_{\max }\right|=\frac{wL}{4}}$.

Refer to the Figure (6),

The maximum absolute bending moment is $\underset{\_}{\left|M_{\max }\right|=\frac{3w{L}^2}{32}}$