Chapter 7, Problem 7.161RP

(a)

To determine

Plot the shear force and bending moment diagram for the beam.

Find the magnitude and location of the maximum absolute value of the bending moment.

Answer to Problem 7.161RP

The location and magnitude of the maximum absolute bending moment is $\underset{\_}{18\text{kips-ft, }\text{3}\text{ft}\text{from}A}$.

Explanation of Solution

Given information:

The moment applied at A is $M=0$.

Calculation:

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

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

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ B_y\left(8\right)-\left(4\times 4\right)\times \frac{4}{2}=0\\ 8B_y-32=0\\ B_y=4\text{kips}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-\left(4\times 4\right)+B_y=0\\ A_y-16+4=0\\ A_y=12\text{kips}\end{array}$

Resolve the horizontal component of forces.

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

Consider the section AC: $\left(0\le x<4\text{ft}\right)$.

Consider a section at a distance x from left end A.

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

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ 12-\left(4\times x\right)-V=0\\ V=12-4x\end{array}$ (1)

Take moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -12\left(x\right)+\left(4\times x\right)\times \frac{x}{2}+M=0\\ M=12x-2x^2\end{array}$ (2)

At $x=0$:

Substitute 0 for x in Equation (1).

$\begin{array}{c}V=12-4\left(0\right)\\ =12\text{kips}\end{array}$

Substitute 0 for x in Equation (2).

$\begin{array}{c}M=12\left(0\right)-2{\left(0\right)}^2\\ =0\end{array}$

At $x=4\text{ft}$:

Substitute 4 ft for x in Equation (1).

$\begin{array}{c}V=12-4\left(4\right)\\ =-4\text{kips}\end{array}$

Substitute 4 ft for x in Equation (2).

$\begin{array}{c}M=12\left(4\right)-2{\left(4\right)}^2\\ =16\text{kips-ft}\end{array}$

Consider the section CB: $\left(4\text{ft}<x\le 8\text{ft}\right)$.

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

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ V+4=0\\ V=-4\text{kips}\end{array}$

Take moment about the section.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -M+4\left(8-x\right)=0\\ -M+32-4x=0\\ M=32-4x\end{array}$ (3)

At $x=4\text{ft}$:

Substitute 4 ft for x in Equation (3).

$\begin{array}{c}M=32-4\left(4\right)\\ =16\text{kips-ft}\end{array}$

At $x=8\text{ft}$:

Substitute 8 ft for x in Equation (3).

$\begin{array}{c}M=32-4\left(8\right)\\ =0\end{array}$

Tabulate the shear force values as in Table 1.

Location, x ft	Shear force, kips
0	12
4	–4
8	–4

Plot the shear force diagram as in Figure 4.

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

$V=0$

Refer to the Figure 4, the shear force changes in the section AC.

Substitute 0 for V in Equation (1).

$\begin{array}{l}0=12-4x\\ 4x=12\\ x=3\text{ft}\end{array}$

Substitute 3 ft for x in Equaiton (2).

$\begin{array}{c}{M}_{\max }=12\left(3\right)-2{\left(3\right)}^2\\ =36-18\\ =18\text{kips-ft}\end{array}$

Tabulate the bending moment values as in Table 2.

Location, x ft	Bending moment, kips-ft
0	0
3	18
4	16
8	0

Plot the bending moment values as in Figure 5.

Therefore, the location and magnitude of the maximum absolute bending moment is $\underset{\_}{18\text{kips-ft, }\text{3}\text{ft}\text{from}A}$.

(b)

To determine

Plot the shear force and bending moment diagram for the beam.

Find the magnitude and location of the maximum absolute value of the bending moment.

Answer to Problem 7.161RP

The location and magnitude of the maximum absolute bending moment is $\underset{\_}{34.125\text{kips-ft, }2.25\text{ft}\text{from}A}$.

Explanation of Solution

Given information:

The moment applied at A is $M=24\text{kip}\cdot \text{ft}$.

Calculation:

Show the free-body diagram of the entire beam as in Figure 6.

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

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ B_y\left(8\right)-\left(4\times 4\right)\times \frac{4}{2}-24=0\\ 8B_y-32-24=0\\ B_y=7\text{kips}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-\left(4\times 4\right)+B_y=0\\ A_y-16+7=0\\ A_y=9\text{kips}\end{array}$

Resolve the horizontal component of forces.

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

Consider the section AC: $\left(0\le x<4\text{ft}\right)$.

Consider a section at a distance x from left end A.

Show the free-body diagram of the section as in Figure 7.

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ 9-\left(4\times x\right)-V=0\\ V=9-4x\end{array}$ (4)

Take moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -24-9\left(x\right)+\left(4\times x\right)\times \frac{x}{2}+M=0\\ -24-9x+2x^2+M=0\\ M=9x-2x^2+24\end{array}$ (5)

At $x=0$:

Substitute 0 for x in Equation (4).

$\begin{array}{c}V=9-4\left(0\right)\\ =9\text{kips}\end{array}$

Substitute 0 for x in Equation (5).

$\begin{array}{c}M=9\left(0\right)-2{\left(0\right)}^2+24\\ =24\text{kips}\cdot \text{ft}\end{array}$

At $x=4\text{ft}$:

Substitute 4 ft for x in Equation (4).

$\begin{array}{c}V=9-4\left(4\right)\\ =-7\text{kips}\end{array}$

Substitute 4 ft for x in Equation (5).

$\begin{array}{c}M=9\left(4\right)-2{\left(4\right)}^2+24\\ =36-32+24\\ =28\text{kips-ft}\end{array}$

Consider the section CB: $\left(4\text{ft}<x\le 8\text{ft}\right)$.

Show the free-body diagram of the section as in Figure 8.

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ V+7=0\\ V=-7\text{kips}\end{array}$

Take moment about the section.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -M+7\left(8-x\right)=0\\ -M+56-7x=0\\ M=56-7x\end{array}$ (6)

At $x=4\text{ft}$:

Substitute 4 ft for x in Equation (6).

$\begin{array}{c}M=56-7\left(4\right)\\ =56-28\\ =28\text{kips-ft}\end{array}$

At $x=8\text{ft}$:

Substitute 8 ft for x in Equation (6).

$\begin{array}{c}M=56-7\left(8\right)\\ =0\end{array}$

Tabulate the shear force values as in Table 3.

Location, x ft	Shear force, kips
0	9
4	–7
8	–7

Plot the shear force diagram as in Figure 9.

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

$V=0$

Refer to the Figure 4, the shear force changes in the section AC.

Substitute 0 for V in Equation (4).

$\begin{array}{l}0=9-4x\\ 4x=9\\ x=2.25\text{ft}\end{array}$

Substitute 2.25 ft for x in Equaiton (5).

$\begin{array}{c}{M}_{\max }=9\left(2.25\right)-2{\left(2.25\right)}^2+24\\ =20.25-10.125+24\\ =34.125\text{kips-ft}\end{array}$

Tabulate the bending moment values as in Table 4.

Location, x ft	Bending moment, kips-ft
0	0
2.25	34.125
4	28
8	0

Plot the bending moment values as in Figure 10.

Therefore, the location and magnitude of the maximum absolute bending moment is $\underset{\_}{34.125\text{kips-ft, }2.25\text{ft}\text{from}A}$.