Chapter 7.1, Problem 7.17P

A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members as shown. Knowing that the combined weight of the pipe and its contents is 10 lb/ft and neglecting the effect of friction, determine the magnitude and location of the maximum bending moment in member AC.

Fig. P7.17

To determine

The bending moment of the couple exerted at the point $AC$ in a pipe and also its location.

Answer to Problem 7.17P

The magnitude of the bending moment of couple exerted at the point $AC$ in a pipe is $M_{\max }=150.0\text{lb}\cdot \text{in}$ and also its location is at $D$.

Explanation of Solution

Sketch the free body diagram for the $10\text{ft}$ section of the pipe as shown in the Figure 1.

Write the equation of the axial force exerted at the axial point of the pipe from x direction.

$\begin{array}{c}{\displaystyle \sum F_x}=0\\ D-\frac{4}{5}F=0\end{array}$ (I)

Here, the pipe is supported by a small frame on the member is $D$ and the force exerted on the frame.

Write the equation of the axial force exerted at the point on the pipe from y direction.

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ E-\frac{3}{5}F=0\end{array}$ (II)

Here, the force exerted on the pipe at y direction in equilibrium condition is $F_y$ and the pipe is supported by a small frame on the member is $E$.

Sketch the free body diagram for the frame as shown in the Figure 2.

Write the equation of the moment of couple formed in the bending moment of the frame at the point $B$ rotated in counterclockwise moment (Refer fig 2).

$\begin{array}{c}{\displaystyle \sum M_B}=0\\ -A_{y}d+Dr+El=0\end{array}$ (III)

Here, the force exerted on the member of the pipe at point $A$ is $A_y$, the distance between the two member of the frame is $d$, radius of the pipe, distance of the frame $BC$ and force exerted on the axial point is $E$.

Write the equation of the axial force exerted at the point on the pipe from y direction (Refer fig 2).

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ B_y+A_y-\frac{4}{5}D-\frac{3}{5}E=0\end{array}$ (IV)

Here, the force exerted on the member of the pipe on frame at the point $B$ is $B_y$,

Write the equation of the axial force exerted at the axial point of the pipe from x direction (Refer fig 2).

$\begin{array}{c}{\displaystyle \sum F_x}=0\\ A_x+B_x-\frac{3}{5}D+\frac{4}{5}E=0\end{array}$ (V)

Sketch the free body diagram for the member of the frame from the point $AC$ as shown in the Figure 3.

Write the equation of the moment of couple formed in the bending moment supported at the point $C$ rotated in counterclockwise moment.

$\begin{array}{c}{\displaystyle \sum M_C}=0\\ Dr-A_y{d}_{AC}-A_x{d}_{BC}=0\end{array}$ (VI)

Here, the distance of the frame $AC$ is $d_{AC}$ and the distance of the frame $BC$ is $d_{BC}$.

Sketch the free body diagram for the member of the frame from the portion $\left(AJ\le AD\right)$ as shown in the Figure 4.

Write the equation of the moment of couple formed in the bending moment supported at the point $J$ rotated in counterclockwise moment for $x\le 12.5\text{in}\left(AJ\le AD\right)$.

$\begin{array}{c}{\displaystyle \sum M_J}=0\\ A_x\frac{3}{5}x-A_y\frac{4}{5}x+M=0\end{array}$ (VII)

Sketch the free body diagram for the member of the frame from the portion $\left(AJ>AD\right)$ as shown in the Figure 5.

Write the equation of the moment of couple formed in the bending moment supported at the point $J$ rotated in counterclockwise moment for $x>12.5\text{in}\left(AJ>AD\right)$.

$\begin{array}{c}{\displaystyle \sum M_J}=0\\ A_x\frac{3}{5}x-A_y\frac{4}{5}x+D\left(x-12.5\right)+M=0\end{array}$ (VIII)

Conclusion:

Substitute $90\text{lb}$ for $F$ in equation (I) to solve for $D$.

$\begin{array}{c}D-\frac{4}{5}\left(90\text{lb}\right)=0\\ D-72\text{lb}=0\\ D=72\text{lb}\end{array}$

Substitute $90\text{lb}$ for $F$ in equation (II) to solve for $E$.

$\begin{array}{c}E-\frac{3}{5}\left(90\text{lb}\right)=0\\ E-54\text{lb}=0\\ E=54\text{lb}\end{array}$

Substitute $72\text{lb}$ for $D$, $54\text{lb}$ for $E$, $18.75\text{in}$ for $d$, $2.5\text{in}$ for $r$, and $8.75\text{in}$ for $l$ in equation (III) to solve for $A_y$.

$\begin{array}{c}-A_y\left(18.75\text{in}\right)+\left(72\text{lb}\right)\left(2.5\text{in}\right)+\left(54\text{lb}\right)\left(8.75\text{in}\right)=0\\ -A_y\left(18.75\text{in}\right)+180\text{lb}\cdot \text{in}+472.5\text{lb}\cdot \text{in}=0\\ -A_y\left(18.75\text{in}\right)=-\left(652.5\text{lb}\cdot \text{in}\right)\\ A_y=34.8\text{lb}\end{array}$

Substitute $34.8\text{lb}$ for $A_y$, $72\text{lb}$ for $D$, and $54\text{lb}$ for $E$ in equation (IV) to solve for $B_y$.

$\begin{array}{c}{B}_y+\left(34.8\text{lb}\right)-\frac{4}{5}\left(72\text{lb}\right)-\frac{3}{5}\left(54\text{lb}\right)=0\\ B_y+34.8\text{lb}-57.6\text{lb}-32.4\text{lb}=0\\ B_y-55.2\text{lb}=0\\ B_y=55.2\text{lb}\end{array}$

Substitute $72\text{lb}$ for $D$ and $54\text{lb}$ for $E$ in equation (V).

$\begin{array}{c}{A}_x+B_x-\frac{3}{5}\left(72\text{lb}\right)+\frac{4}{5}\left(54\text{lb}\right)=0\\ A_x+B_x-43.2\text{lb}+43.2\text{lb}=0\\ A_x+B_x=0\end{array}$ (IX)

Substitute $72\text{lb}$ for $D$, $34.8\text{lb}$ for $A_y$, $12\text{in}$ for $d_{AC}$, $2.5\text{in}$ for $r$, and $9\text{in}$ for $d_{BC}$ in equation (VI) to solve for $A_x$.

$\begin{array}{c}\left(72\text{lb}\right)\left(2.5\text{in}\right)-\left(34.8\text{lb}\right)\left(12\text{in}\right)-A_x\left(9\text{in}\right)=0\\ 180\text{lb}\cdot \text{in}-417.6\text{lb}\cdot \text{in}-A_x\left(9\text{in}\right)=0\\ -237.6\text{lb}\cdot \text{in}-A_x\left(9\text{in}\right)=0\\ A_x=-26.4\text{lb}\end{array}$

Substitute $-26.4\text{lb}$ for $A_x$ in equation (IX).

$\begin{array}{c}-26.4\text{lb}+B_x=0\\ -26.4\text{lb}=-B_x\\ B_x=26.4\text{lb}\end{array}$

Substitute $26.4\text{lb}$ for $A_x$, $12.5\text{in}$ for $x$, and $34.8\text{lb}$ for $A_y$ in equation (VII) for $x\le 12.5\text{in}\left(AJ\le AD\right)$.

$\begin{array}{c}\left(26.4\text{lb}\right)\frac{3}{5}x-\left(34.8\text{lb}\right)\frac{4}{5}x+M=0\\ \left(15.84\text{lb}\right)x-\left(27.84\text{lb}\right)x+M=0\\ -\left(12\text{lb}\right)\left(12.5\text{in}\right)=-M\\ M=150.0\text{lb}\cdot \text{in}\end{array}$

Substitute $26.4\text{lb}$ for $A_x$, $72\text{lb}$ for $D$, and $34.8\text{lb}$ for $A_y$ in equation (VIII) for $x>12.5\text{in}\left(AJ>AD\right)$.

$\begin{array}{c}{A}_x\frac{3}{5}x-A_y\frac{4}{5}x+D\left(x-12.5\right)+M=0\\ \left(26.4\text{lb}\right)\frac{3}{5}x-\left(34.8\text{lb}\right)\frac{4}{5}x+\left(72\text{lb}\right)\left(x-12.5\right)+M=0\\ \left(15.84\text{lb}\right)x-\left(27.84\text{lb}\right)x+\left(72\text{lb}\right)x-900\text{lb}\cdot \text{in}+M=0\\ 900\text{lb}\cdot \text{in}-\left(60\text{lb}\right)x=M\end{array}$

Substitute $12.5\text{in}$ for $x$ in the above equation can be written as,

$\begin{array}{c}900\text{lb}\cdot \text{in}-\left(60\text{lb}\right)\left(12.5\text{in}\right)=M\\ 150\text{lb}\cdot \text{in}=M\end{array}$

Therefore, the magnitude of the bending moment of couple exerted at the point $AC$ in a pipe is $M_{\max }=150.0\text{lb}\cdot \text{in}$ and also its location is at $D$.