Chapter 7.1, Problem 7.16P

Fig. P7.15 and P7.16

7.16 Knowing that the radius of each pulley is 100 mm and neglecting friction, determine the internal forces at (a) point C, (b) point J that is 100 mm to the left of C.

To determine

The internal forces exerted at the point $C$ in a frame.

Answer to Problem 7.16P

The internal forces of shearing force is $V=540\text{N}$ acting in the upward direction, axial force $F=520\text{N}$ acting in the leftside of the frame, and the bending moment acting at that point is $M=0$.

Explanation of Solution

Sketch the free body diagram for the internal forces acting on the frame and pulley system as shown in the Figure 1.

Write the equation of the axial force exerted at the axial point $A$ of the frame from x direction.

$\begin{array}{c}{\displaystyle \sum F_x}=0\\ E_x=0\end{array}$ (I)

Here, the force exerted on the frame at the point $E$ from x direction in equilibrium condition is $F_x$.

Write the equation of the moment of couple formed in the bending moment of the frame and pulley system supported at the point $E$ rotated in counterclockwise moment (Refer fig 1).

$\begin{array}{c}{\displaystyle \sum M_E}=0\\ F_B\left(l\right)-F_B\left(l_1+r\right)+Ad=0\end{array}$ (II)

Here, the axial force exerted on the pulley at point $B$ is $F_B$, the radius of the each pulley is $r$, distance between the pulley $B$ on the frame to the point $C$ is $l$, distance between the point $A$ on the frame to the point $D$ is $l_1$, and total distance of the frame is $d$.

Write the equation of the axial force exerted at the axial point of the frame from y direction (Refer fig 1).

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ E_y+A=0\end{array}$ (III)

Here, the axial force exerted on the pulley at point $E$ in y direction is $E_y$.

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

The slope of the cable (Refer fig 2):

$\begin{array}{c}BC=CD=200\text{mm}\\ BG=DH=120\text{mm}\end{array}$

The angle formed in the slope of the cable:

$\begin{array}{c}\sin \alpha =\frac{100\text{mm}}{200\text{mm}}\\ =\frac{1}{5}\\ \sin \alpha =0.2\\ \alpha =30°\end{array}$

Write the equation of the axial force exerted at the axial point $AC$ of the given free body diagram from $x$ direction.

$\begin{array}{c}{\displaystyle \sum F_x}=0\\ F_B\cos \alpha -F=0\end{array}$ (IV)

Here, the angle between the pulley $B$ to the frame is $\alpha$ and the axial force exerted on the frame and pulley system is $F$.

Sketch the free body diagram for the cable for the point $AC$ as shown in the Figure 3.

Write the equation of the axial force exerted at the point $A$ of the frame from $y$ direction.

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ V+A-B_x\sin \alpha -F_B=0\end{array}$ (V)

Here, shearing force acting on the semicircular rod is $V$

At the pulley $B$ it is resolved into two components $B_x$ and $B_y$.

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

$\begin{array}{c}{\displaystyle \sum M_C}=0\\ M-x_{1}A+\left(B_x\sin \alpha +F_B\right)x_2=0\end{array}$ (VI)

Here, the moment of couple exerted at the point $C$ in the frame is represented as $M_C$ and moment of couple acting in the bending beam is $M$.

Conclusion:

Substitute $600\text{N}$ for $F_B$, $100\text{mm}$ for $r$, $200\text{mm}$ for $l$, $1000\text{mm}$ for $d$, and $700\text{mm}$ for $l_1$ in equation (II) to solve for $A$.

$\begin{array}{c}\left(600\text{N}\right)\left(200\text{mm}\right)-\left(600\text{N}\right)\left(700\text{mm}+100\text{mm}\right)+A\left(1000\text{mm}\right)=0\\ \left(600\text{N}\right)\left(200\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)-\left(600\text{N}\right)\left(800\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)+A\left(1000\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)=0\\ \left(600\text{N}\right)\left(0.2\text{m}\right)-\left(600\text{N}\right)\left(0.8\text{m}\right)+A\left(1\text{m}\right)=0\end{array}$

Solve the above equation for $A$.

$\begin{array}{c}\left(120\text{N}\cdot \text{m}\right)-\left(480\text{N}\cdot \text{m}\right)+A\left(1\text{m}\right)=0\\ \left(-360\text{N}\cdot \text{m}\right)=-A\left(1\text{m}\right)\\ A=360\text{N}\end{array}$

Substitute $360\text{N}$ for $A$ in equation (III) to solve for $E_y$.

$\begin{array}{c}{E}_y+360\text{N}=0\\ E_y=-360\text{N}\end{array}$

Substitute $600\text{N}$ for $F_B$ and $30°$ for $\alpha$ in equation (IV) to solve for $F$.

$\begin{array}{c}{F}_B\cos \alpha -F=0\\ \left(600\text{N}\right)\cos 30°-F=0\\ 520\text{N}-F=0\\ F=520\text{N}\end{array}$

Substitute $360\text{N}$ for $A$, $600\text{N}$ for $B_x$, $600\text{N}$ for $F_B$, and $30°$ for $\alpha$ in equation (V) to solve for $V$

$\begin{array}{c}V+A-B_x\cos \alpha -F=0\\ V+360\text{N}-\left(600\text{N}\right)\sin 30°-600\text{N}=0\\ V+360\text{N}-300\text{N}-600\text{N}=0\\ V-540\text{N}=0\\ V=540\text{N}\end{array}$

Substitute $360\text{N}$ for $A$, $600\text{N}$ for $B_x$, $600\text{N}$ for $F_B$, $500\text{mm}$ for $x_1$, $200\text{mm}$ for $x_2$,$30°$ for $\alpha$ in equation (VI) to solve for $M$.

$\begin{array}{c}M-x_{1}A+\left(B_x\mathrm{co}\alpha +F_B\right)x_2=0\\ M-\left(500\text{mm}\right)\left(360\text{N}\right)+\left(600\text{Nsin}30°+600\text{N}\right)\left(200\text{mm}\right)=0\\ M-\left(500\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)\left(360\text{N}\right)+\left(300\text{N}+600\text{N}\right)\left(200\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)=0\\ M-\left(0.500\text{m}\right)\left(360\text{N}\right)+\left(900\text{N}\right)\left(0.200\text{m}\right)=0\end{array}$

The above equation can be written as,

$\begin{array}{c}M-180\text{N}\cdot \text{m}+180\text{N}\cdot \text{m}=0\\ M=0\end{array}$

Therefore, The internal forces of shearing force is $V=540\text{N}$ acting in the upward direction, axial force $F=520\text{N}$ acting in the leftside of the frame, and the bending moment acting at that point is $M=0$.

To determine

The internal forces exerted at the point $J$ to the left point $C$ in a frame of $100\text{mm}$.

Answer to Problem 7.16P

The internal forces of shearing force is $V=540\text{N}$ acting in the upward direction, axial force $F=520\text{N}$ acting in the leftside of the frame, and the bending moment acting at that point is $M=54.0\text{N}\cdot \text{m}$ rotates in the counterclockwise direction.

Explanation of Solution

Sketch the free body diagram for the cable for the point $AJ$ as shown in the Figure 3.

Write the equation of the axial force exerted at the axial point $AJ$ of the frame from x direction.

$\begin{array}{c}{\displaystyle \sum F_x}=0\\ -F+F_B=0\end{array}$ (VII)

Here, the force exerted on the frame at the point $E$ from x direction in equilibrium condition is $F_x$.

Write the equation of the axial force exerted at the axial point of the frame from y direction (Refer fig 4).

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ V+A-B_x\sin \alpha -F_B=0\end{array}$ (VIII)

Here, the axial force exerted on the pulley at point $D$ is $F_D$.

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

$\begin{array}{c}{\displaystyle \sum M_J}=0\\ M-x_{3}A+\left(B_x\sin \alpha +F_B\right)x_4=0\end{array}$ (IX)

Here, the moment of couple exerted at the point $J$ in the frame is represented as $M_J$ and moment of couple acting in the bending beam is $M$.

Conclusion:

Substitute $600\text{N}$ for $F_B$ and $30°$ for $\alpha$ in equation (VII) to solve for $F$.

$\begin{array}{c}-F+\left(600\text{N}\right)\cos 30°=0\\ -F+520\text{N}=0\\ F=520\text{N}\end{array}$

Substitute $360\text{N}$ for $A$, $600\text{N}$ for $B_x$, $600\text{N}$ for $F_B$, and $30°$ for $\alpha$ in equation (VIII) to solve for $V$.

$\begin{array}{c}V+A-B_x\sin \alpha -F_B=0\\ V+360\text{N}-\left(600\text{N}\right)\sin 30°-600\text{N}=0\\ V+360\text{N}-300\text{N}-600\text{N}=0\\ V-540\text{N}=0\\ V=540\text{N}\end{array}$

Substitute $360\text{N}$ for $A$, $600\text{N}$ for $B_x$, $600\text{N}$ for $F_B$, $500\text{mm}$ for $x_1$, $200\text{mm}$ for $x_2$,$30°$ for $\alpha$ in equation (IX) to solve for $M$.

$\begin{array}{c}M-x_{3}A+\left(B_x\sin \alpha +F_B\right)x_4=0\\ M-\left(400\text{mm}\right)\left(360\text{N}\right)+\left(600\text{Nsin}30°+600\text{N}\right)\left(100\text{mm}\right)=0\\ M-\left(400\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)\left(360\text{N}\right)+\left(300\text{N}+600\text{N}\right)\left(100\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)=0\\ M-\left(0.400\text{m}\right)\left(360\text{N}\right)+\left(900\text{N}\right)\left(0.100\text{m}\right)=0\end{array}$

The above equation can be written as,

$\begin{array}{c}M-144\text{N}\cdot \text{m}+90\text{N}\cdot \text{m}=0\\ M=54\text{N}\cdot \text{m}\end{array}$

Therefore, the internal forces of shearing force is $V=540\text{N}$ acting in the upward direction, axial force $F=520\text{N}$ acting in the leftside of the frame, and the bending moment acting at that point is $M=54.0\text{N}\cdot \text{m}$ rotates in the counterclockwise direction.