Chapter 7.3, Problem 7.76P

(a)

To determine

Draw the shear force and bending moment diagrams.

Explanation of Solution

Assumption:

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:

Show the free-body diagram of the beam as in Figure 1.

Refer to Figure 1:

Find the vertical reaction at point G by taking moment about point C.

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ 16\times 10\times \frac{10}{2}-100\left(6\right)-150\left(12\right)-100\left(18\right)+G_y\left(24\right)-16\times 10\times \left(24+\frac{10}{2}\right)=0\\ 800-600-1,800-1,800+24G_y-4,640=0\\ G_y=335\text{lb}(\uparrow )\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -16\times 10+C_y-100-150-100+G_y-16\times 10=0\\ -160+C_y-350+335-160=0\\ C_y=335\text{lb}(\uparrow )\end{array}$

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

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

Show the calculation of shear force as follows;

Shear force at $x=0$;

${V|}_{x=0}=0$

Shear force at $x=10\text{in}\text{.}$;

$\begin{array}{c}{V|}_{x=10\text{in}\text{.}\left(\text{Just}\text{left}\right)}=-16\times 10\\ =-160\text{lb}\\ {V|}_{x=10\text{in}\text{.}\left(\text{Just}\text{right}\right)}=-160+335\\ =175\text{lb}\end{array}$

Shear force at $x=16\text{in}\text{.}$;

$\begin{array}{c}{V|}_{x=10\text{in}\text{.}\left(\text{Just}\text{left}\right)}=175\text{lb}\\ {V|}_{x=10\text{in}\text{.}\left(\text{Just}\text{right}\right)}=175-100\\ =75\text{lb}\end{array}$

Shear force at $x=22\text{in}\text{.}$;

$\begin{array}{c}{V|}_{x=22\text{in}\text{.}\left(\text{Just}\text{left}\right)}=75\text{lb}\\ {V|}_{x=22\text{in}\text{.}\left(\text{Just}\text{right}\right)}=75-150\\ =-75\text{lb}\end{array}$

Shear force at $x=28\text{in}\text{.}$;

$\begin{array}{c}{V|}_{x=28\text{in}\text{.}\left(\text{Just}\text{left}\right)}=-75\text{lb}\\ {V|}_{x=28\text{in}\text{.}\left(\text{Just}\text{right}\right)}=-75-100\\ =-175\text{lb}\end{array}$

Shear force at $x=34\text{in}\text{.}$;

$\begin{array}{c}{V|}_{x=34\text{in}\text{.}\left(\text{Just}\text{left}\right)}=-175\text{lb}\\ {V|}_{x=34\text{in}\text{.}\left(\text{Just}\text{right}\right)}=-175+G_y\\ =-175+335\\ =160\text{lb}\end{array}$

Shear force at $x=44\text{in}\text{.}$;

$\begin{array}{c}{V|}_{x=44\text{in}\text{.}}=160-16\times 10\\ =0\end{array}$

Tabulate the calculated shear force values as in Table 1.

Distance, x (in.)	Shear force, V (lb)
0	0
10 (Just left)	–160
10 (Just right)	175
16 (Just left)	175
16 (Just right)	75
22 (Just left)	75
22 (Just right)	–75
28 (Just left)	–75
28 (Just right)	–175
34 (Just left)	–175
34 (Just right)	160
44	0

Plot the shear force diagram as in Figure 2.

Show the calculation of bending moment as follows;

Bending moment at $x=0$;

${M|}_{x=0}=0$

Bending moment at $x=10\text{in}\text{.}$;

$\begin{array}{c}{M|}_{x=10\text{in}\text{.}}=-16\times 10\times \frac{10}{2}\\ =-800\text{lb-in}\text{.}\end{array}$

Bending moment at $x=16\text{in}\text{.}$

$\begin{array}{c}{M|}_{x=16\text{in}\text{.}}=-16\times 10\times \left(\frac{10}{2}+6\right)+A_y\left(6\right)\\ =-1,760+335\left(6\right)\\ =250\text{lb-in}\text{.}\end{array}$

Bending moment at $x=22\text{in}\text{.}$

$\begin{array}{c}{M|}_{x=22\text{in}\text{.}}=-16\times 10\times \left(\frac{10}{2}+12\right)+A_y\left(12\right)-100\left(6\right)\\ =-2,720+335\left(12\right)-600\\ =700\text{lb-in}\text{.}\end{array}$

Bending moment at $x=28\text{in}\text{.}$

$\begin{array}{c}{M|}_{x=28\text{in}\text{.}}=-16\times 10\times \left(\frac{10}{2}+18\right)+A_y\left(18\right)-100\left(12\right)-150\left(6\right)\\ =-3,680+335\left(18\right)-1,200-900\\ =250\text{lb-in}\text{.}\end{array}$

Bending moment at $x=34\text{in}\text{.}$

$\begin{array}{c}{M|}_{x=34\text{in}\text{.}}=-16\times 10\times \left(\frac{10}{2}+24\right)+A_y\left(24\right)-100\left(18\right)-150\left(12\right)-100\left(6\right)\\ =-4,640+335\left(24\right)-1,800-1,800-600\\ =-800\text{lb-in}\text{.}\end{array}$

Bending moment at $x=44\text{in}\text{.}$

$\begin{array}{c}{M|}_{x=44\text{in}\text{.}}=\left(\begin{array}{l}-16\times 10\times \left(\frac{10}{2}+34\right)+A_y\left(34\right)-100\left(28\right)-150\left(22\right)-100\left(16\right)+G_y\left(10\right)\\ -16\times 10\times \frac{10}{2}\end{array}\right)\\ =-6,240+335\left(34\right)-2,800-3,300-1,600+335\left(10\right)-800\\ =0\end{array}$

Tabulate the calculated bending moment values as in Table 2.

Distance, x (in.)	Bending moment, M (lb-in.)
0	0
10	–800
16	250
22	700
28	250
34	–800
44	0

Plot the bending moment diagram as in Figure 3.

(b)

To determine

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

Answer to Problem 7.76P

The maximum absolute shear force is $\underset{\_}{\left|V_{\max }\right|=175\text{lb}}$.

The maximum absolute bending moment is $\underset{\_}{\left|M_{\max }\right|=800\text{lb-in}\text{.}}$

Explanation of Solution

Refer to the Figure (2);

The maximum absolute shear force is $\underset{\_}{\left|V_{\max }\right|=175\text{lb}}$.

Refer to the Figure (3);

The maximum absolute bending moment is $\underset{\_}{\left|M_{\max }\right|=800\text{lb-in}\text{.}}$