Chapter 7.3, Problem 7.92P

(a)

To determine

Find the value of P and Q.

Answer to Problem 7.92P

The value of the unknown forces is $\underset{\_}{P=0.540\text{lb}(\downarrow )\text{;}Q=1.860\text{lb}(\downarrow )}$.

Explanation of Solution

Given information:

The bending moment at point D is $M_D=+5.96\text{kip-ft}$.

The bending moment at point E is $M_E=+6.84\text{kip-ft}$.

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 section DE as in Figure 1.

Find the shear force at point E by taking moment about point D.

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ -5.96\text{kip-ft}\times \frac{\text{1,000}\text{lb}}{\text{1}\text{kip}}-250\times 4\times \frac{4}{2}+V_E\left(4\right)+6.84\text{kip-ft}\times \frac{\text{1,000}\text{lb}}{\text{1}\text{kip}}=0\\ -5,960-2,000+4V_E+6,840=0\\ V_E=280\text{lb}(\uparrow )\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ V_D-250\times 4+V_E=0\\ V_D-1,000+280=0\\ V_D=720\text{lb}(\uparrow )\end{array}$

Show the free-body diagram of the section AD as in Figure 2.

Find the value of force P by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ 5.96\text{kips-ft}\times \frac{\text{1,000}\text{lb}}{\text{1}\text{kip}}-720\left(4\right)-P\left(2\right)-250\times 4\times \frac{4}{2}=0\\ 5,960-2,880-2P-2,000=0\\ P=540\text{lb}\times \frac{\text{1}\text{kip}}{\text{1,000}\text{lb}}\end{array}$

$P=0.540\text{lb}(\downarrow )$

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

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

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-250\times 4-P-720=0\\ A_y-1,000-540-720=0\\ A_y=2,260\text{lb}(\uparrow )\end{array}$

Show the free-body diagram of the section EB as in Figure 3.

Find the value of force Q by taking moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -6.84\text{kips-ft}\times \frac{\text{1,000}\text{lb}}{\text{1}\text{kip}}+280\left(4\right)+Q\left(2\right)+250\times 4\times \frac{4}{2}=0\\ -6,840+1,120+2Q+2,000=0\\ Q=1,860\text{lb}\times \frac{\text{1}\text{kip}}{\text{1,000}\text{lb}}\end{array}$

$Q=1.860\text{lb}(\downarrow )$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ B_y-250\times 4-Q-280=0\\ B_y-1,000-1,860-280=0\\ B_y=3,140\text{lb}(\uparrow )\end{array}$

Show the free-body diagram of the beam as in Figure 4.

Therefore, the value of the unknown forces is $\underset{\_}{P=0.540\text{lb}(\downarrow )\text{;}Q=1.860\text{lb}(\downarrow )}$.

(b)

To determine

Draw the shear force and bending moment diagrams.

Explanation of Solution

Show the calculation of shear force as follows;

Shear force at $x=0$;

$\begin{array}{l}{V|}_{x=0\left(\text{Just}\text{left}\right)}=0\\ {V|}_{x=0\left(\text{Just}\text{right}\right)}=2,260\text{lb}\end{array}$

Shear force at $x=2\text{ft}$;

$\begin{array}{c}{V|}_{x=2\text{ft}\left(\text{Just}\text{left}\right)}=2,260-250\times 2\\ =1,760\text{lb}\\ {V|}_{x=2\text{ft}\left(\text{Just}\text{right}\right)}=1,760-540\\ =1,220\text{lb}\end{array}$

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

$\begin{array}{c}{V|}_{x=10\text{ft}\left(\text{Just}\text{left}\right)}=2,260-250\times 10-540\\ =-780\text{lb}\\ {V|}_{x=10\text{ft}\left(\text{Just}\text{right}\right)}=-780-1,860\\ =-2,640\text{lb}\end{array}$

Shear force at $x=12\text{ft}$;

$\begin{array}{c}{V|}_{x=12\text{ft}\left(\text{Just}\text{left}\right)}=2,260-250\times 12-540-1,860\\ =-3,140\text{lb}\\ {V|}_{x=12\text{ft}\left(\text{Just}\text{right}\right)}=-3,140+3,140\\ =0\end{array}$

Tabulate the calculated shear force values as in Table 1.

Distance, x (ft)	Shear force, V (lb)
0 (Just left)	0
0 (Just right)	2,260
2 (Just left)	1,760
2 (Just right)	1,220
10 (Just left)	–780
10 (Just right)	–2,640
12 (Just left)	–3,140
12 (Just right)	0

Plot the shear force diagram as in Figure 5.

Location of maximum bending moment;

The maximum bending moment occurs where the shear force changes it’s sign.

$V=0$

Refer to the shear force diagram the shear force changes in the region CF.

Show the similar triangle of the shear force diagram for the region CF as in Figure 6.

Use the similar triangle method.

$\begin{array}{l}\frac{x}{8-x}=\frac{1,220}{780}\\ 780x=9,760-1,220x\\ 2,000x=9,760\\ x=4.88\text{ft}\end{array}$

Therefore, the maximum bending moment occurs at a distance 6.88 ft from the left end A.

Show the calculation of bending moment as follows;

Bending moment at $x=0$;

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

Bending moment at $x=2\text{ft}$;

$\begin{array}{c}{M|}_{x=2\text{ft}}=2,260\times 2-250\times 2\times \frac{2}{2}\\ =4,520-500\\ =4,020\text{lb-ft}\end{array}$

Bending moment at $x=6.88\text{ft}$;

$\begin{array}{c}{M|}_{x=6.88\text{ft}}=2,260\times 6.88-250\times 6.88\times \frac{6.88}{2}-540\times 4.88\\ =15,548.8-5,916.8-2,635.5\\ =6,996.5\text{lb-ft}\end{array}$

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

$\begin{array}{c}{M|}_{x=10\text{ft}}=2,260\times 10-250\times 10\times \frac{10}{2}-540\times 8\\ =22,600-12,500-4,320\\ =5,780\text{lb-ft}\end{array}$

Bending moment at $x=12\text{ft}$;

$\begin{array}{c}{M|}_{x=12\text{ft}}=2,260\times 12-250\times 12\times \frac{12}{2}-540\times 10-1,860\times 2\\ =27,120-18,000-5,400-3,720\\ =0\end{array}$

Tabulate the calculated bending moment values as in Table 2.

Distance, x (ft)	Bending moment, M (lb-ft)
0	0
2	4,020
6.88	6,996.5
10	5,780
12	0

Plot the bending moment diagram as in Figure 7.