Chapter 7.2, Problem 7.51P

To determine

Draw the shear force and bending moment diagrams.

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

Answer to Problem 7.51P

The maximum absolute shear force is $\underset{\_}{\left|V_{\max }\right|=90\text{lb}}$.

The maximum absolute bending moment is $\underset{\_}{\left|M_{\max }\right|=1,400\text{lb-in}\text{.}}$

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 value of angle $\theta$.

$\begin{array}{c}\tan \theta =\frac{\text{opposite}\text{side}}{\text{adjacent}\text{side}}\\ =\frac{14}{20}\\ \theta =34.99°\end{array}$

Find the force in the cable CD by taking moment about point H.

$\begin{array}{l}{\displaystyle \sum M_H=0}\\ 50\left(26\right)+F_{CD}\cos 34.99°\left(4\right)-F_{CD}\sin 34.99°\left(20\right)+100\left(10\right)-50\left(6\right)=0\\ 1,300+3.277F_{CD}-11.469F_{CD}+1,000-300=0\\ 2,000-8.192F_{CD}=0\end{array}$

$F_{CD}=244.141\text{lb}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -50-100+F_{CD}\sin 34.99°+H_y-50=0\\ -200+244.141\sin 34.99°+H_y=0\\ H_y=60\text{lb}(\uparrow )\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ H_x-F_{CD}\cos 34.99°=0\\ H_x-244.141\cos 34.99°=0\\ H_x=200\text{lb}(\leftarrow )\end{array}$

Show the calculated reaction values of the beam as in Figure 2.

Consider a section x from left end of the beam.

Section AE $\left(0\le x<6\text{in}\text{.}\right)$:

Show 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.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -50-V=0\\ V=-50\text{lb}\end{array}$ (1)

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ M+50\left(x\right)=0\\ M=-50x\end{array}$ (2)

When $x=0$;

Refer to Equation (1);

$V=-50\text{lb}$

Substitute 0 for x in Equation (2).

$\begin{array}{c}M=-50\left(0\right)\\ =0\end{array}$

When $x=6\text{in}\text{.}$;

Refer to Equation (1);

$V=-50\text{lb}$

Substitute 6 in. for x in Equation (2).

$\begin{array}{c}M=-50\left(6\right)\\ =-300\text{lb-in}\text{.}\end{array}$

Section EF $\left(6\text{in}\text{.}<x<16\text{in}\text{.}\right)$:

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.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -50+140-V=0\\ V=90\text{lb}\end{array}$ (3)

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ M+50\left(x\right)-140\left(x-6\right)-800=0\\ M=140\left(x-6\right)-50x+800\end{array}$ (4)

When $x=6\text{in}\text{.}$;

Refer to Equation (3);

$V=90\text{lb}$

Substitute 6 in. for x in Equation (4).

$\begin{array}{c}M=140\left(6-6\right)-50\left(6\right)+800\\ =0-300+800\\ =500\text{lb}\cdot \text{in}.\end{array}$

When $x=16\text{in}\text{.}$;

Refer to Equation (3);

$V=90\text{lb}$

Substitute 16 in. for x in Equation (4).

$\begin{array}{c}M=140\left(16-6\right)-50\left(16\right)+800\\ =1,400-800+800\\ =1,400\text{lb-in}\text{.}\end{array}$

Section FG $\left(16\text{in}\text{.}<x<26\text{in}\text{.}\right)$:

Show the free body diagram of the section as in Figure 5.

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -50+140-100-V=0\\ V=-10\text{lb}\end{array}$ (5)

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ 50\left(x\right)-140\left(x-6\right)-800+100\left(x-16\right)+M=0\\ M=140\left(x-6\right)+800-50x-100\left(x-16\right)\end{array}$ (6)

When $x=16\text{in}\text{.}$;

Refer to Equation (5).

$V=-10\text{lb}$

Substitute 16 in. for x in Equation (6).

$\begin{array}{c}M=140\left(16-6\right)+800-50\left(16\right)-100\left(16-16\right)\\ =1,400+800-800-0\\ =1,400\text{lb-in}\text{.}\end{array}$

When $x=26\text{in}\text{.}$;

Refer to Equation (5).

$V=-10\text{lb}$

Substitute 26 in. for x in Equation (6).

$\begin{array}{c}M=140\left(26-6\right)+800-50\left(26\right)-100\left(26-16\right)\\ =2,800+800-1,300-1,000\\ =1,300\text{lb-in}\text{.}\end{array}$

Section GB $\left(26\text{in}\text{.}<x\le 32\text{in}\text{.}\right)$:

Show the free body diagram of the section as in Figure 6.

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ V-50=0\\ V=50\text{lb}\end{array}$ (7)

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -M-50\left(32-x\right)=0\\ M=-50\left(32-x\right)\end{array}$ (8)

When $x=26\text{in}\text{.}$;

Refer to Equation (7).

$V=50\text{lb}$

Substitute 26 in. for x in Equation (8).

$\begin{array}{c}M=-50\left(32-26\right)\\ =-300\text{lb-in}\text{.}\end{array}$

When $x=32\text{in}\text{.}$;

Refer to Equation (7).

$V=50\text{lb}$

Substitute 32 in. for x in Equation (8).

$\begin{array}{c}M=-50\left(32-32\right)\\ =0\end{array}$

Tabulate the calculated shear force values as in Table 1.

Distance, x (in.)	Shear force, V (lb)
0	–50
6 (Just left)	–50
6(Just right)	90
16 (Just left)	90
16 (Just right)	–10
26 (Just left)	–10
26 (Just right)	50
32	50

Plot the shear force diagram as in Figure 7.

Tabulate the calculated bending moment values as in Table 2.

Distance, x (in.)	Bending moment, M (lb-in.)
0	0
6 (Just left)	–300
6 (Just right)	500
16	1,400
26 (just left)	1,300
26 (Just right)	–300
32	0

Plot the bending moment diagram as in Figure 8.

Refer to the Figure (7);

The maximum absolute shear force is $\underset{\_}{\left|V_{\max }\right|=90\text{lb}}$.

Refer to the Figure (8);

The maximum absolute bending moment is $\underset{\_}{\left|M_{\max }\right|=1,400\text{lb-in}\text{.}}$