Chapter 6, Problem 1P

To determine

Calculate the reaction forces at the supports.

Calculate the magnitude of cable sag at joints B and E.

Calculate the tension force in every segment of cable and total length of the cable.

Answer to Problem 1P

The vertical reaction at A and F is $\underset{\_}{A_y=\text{60}\text{k}}$ and $\underset{\_}{F_y=\text{60}\text{k}}$ respectively and the horizontal reaction is $\underset{\_}{H=\text{75}\text{k}}$.

The magnitude of cable sag at joint B and E is $\underset{\_}{h_B=16\text{ft}}$ and $\underset{\_}{h_E=16\text{ft}}$ respectively.

The tension force in the cable segment AB and EF is $\underset{\_}{96\text{k}}$.

The tension force in the cable segment BC and DE is $\underset{\_}{80.7\text{k}}$.

The tension force in the cable segment CD is $\underset{\_}{75\text{k}}$.

The total length of the cable is $\underset{\_}{114.3\text{ft}}$.

Explanation of Solution

Given information:

The cable supporting concentrated load as shown in given Figure P6.1.

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 positive and counter clockwise moment as negative.

Calculation:

Draw the free body diagram as shown in Figure 1.

Calculate the support reactions.

Take moment about F.

$\begin{array}{l}{\displaystyle \sum M}=0\\ \left(A_y\times 100\right)-\left(30\times 80\right)-\left(30\times 60\right)-\left(30\times 40\right)-\left(30\times 20\right)=0\\ 100A_y=2400+1800+1200+600\\ A_y=60\text{k}\end{array}$

Calculate the vertical reaction at F.

$\begin{array}{l}{A}_y+F_y=\left(4\times 30\text{k}\right)\\ 60+F_y=120\text{k}\\ F_y=60\text{k}\end{array}$

Therefore, the vertical support reaction at A and F is $\underset{\_}{A_y=\text{60}\text{k}}$ and $\underset{\_}{F_y=\text{60}\text{k}}$ respectively.

Find the horizontal reaction using general cable theorem.

Take moment about C from support A.

$\begin{array}{l}\left(A_y\times 40\right)-\left(H\times 24\right)-\left(30\times 20\right)=0\\ \left(60\times 40\right)-\left(H\times 24\right)-\left(30\times 20\right)=0\\ 24H=1,800\text{k}\\ H=\text{75}\text{k}\end{array}$

Therefore, the horizontal support reaction is $\underset{\_}{H=\text{75}\text{k}}$.

Calculate the magnitude of cable sag at B.

Consider the vertical distance at B is $h_B$.

Take moment at B from support A.

$\begin{array}{l}\left(A_y\times 20\right)-\left(H\times h_B\right)=0\\ \left(60\times 20\right)-\left(75\times h_B\right)=0\\ h_B=16\text{ft}\end{array}$

Calculate the magnitude of cable sag at E.

Consider the vertical distance at E is $h_E$.

Take moment at E from support F.

$\begin{array}{l}-\left(F_y\times 20\right)+\left(H\times h_E\right)=0\\ -\left(60\times 20\right)+\left(75\times h_E\right)=0\\ h_E=16\text{ft}\end{array}$

Therefore, the magnitude of cable sag at joint B and E is $\underset{\_}{h_B=16\text{ft}}$ and $\underset{\_}{h_E=16\text{ft}}$ respectively.

Calculate the tension force in the segment of cable AB $\left(T_{AB}\right)$ and EF $\left(T_{EF}\right)$.

Take joint A as shown in Figure 2.

Find the horizontal angle for $T_{AB}$ as follows:

$\begin{array}{l}\tan \theta =\frac{16\text{ft}}{20\text{ft}}\\ \theta =38.65°\end{array}$

Apply horizontal equilibrium Equation.

$\begin{array}{l}{\displaystyle \sum H}=0\\ T_{AB}\cos 38.65°-75\text{k}=\text{0}\\ T_{AB}=96\text{k}\end{array}$

The cable with supports are symmetrical about vertical axis. Hence, the tension force in the segment AB is equal to EF. So, $T_{EF}=96\text{k}$

Therefore, the tension force in the cable segment AB and EF is $\underset{\_}{96\text{k}}$.

Calculate the tension force in the segment of cable BC $\left(T_{BC}\right)$ and DE $\left(T_{DE}\right)$

Take joint B as shown in Figure 3.

Find the horizontal angle for $T_{BC}$ as follows:

$\begin{array}{l}\tan \theta =\frac{8\text{ft}}{20\text{ft}}\\ \theta =21.80°\end{array}$

Apply horizontal equilibrium Equation.

$\begin{array}{l}{\displaystyle \sum H}=0\\ T_{BC}\cos 21.80°-T_{AB}\cos 38.65=\text{0}\\ T_{BC}\cos 21.80°=96\cos 38.65\\ T_{BC}=80.7\text{k}\end{array}$

The cable with supports are symmetrical about vertical axis. Hence, the tension force in the segment BC is equal to DE. So, $T_{DE}=80.7\text{k}$.

The tension force in the cable segment BC and DE is $\underset{\_}{80.7\text{k}}$.

Calculate the tension force in the segment of cable CD $\left(T_{CD}\right)$.

Take the joint C as shown in Figure 4.

Apply horizontal equilibrium Equation.

$\begin{array}{l}{\displaystyle \sum H}=0\\ T_{CD}-T_{BC}\cos 21.80°=0\\ T_{CD}-80.7\cos 21.80°=0\\ T_{CD}=75\text{k}\end{array}$

Therefore, the tension force in the cable segments CD is $\underset{\_}{75\text{k}}$.

Show the vertical and horizontal length of each segment of cable as shown in Figure 5.

Calculate the total length of the cable (L) using Figure 4.

$\begin{array}{c}L=L_{AB}+L_{BC}+L_{CD}+L_{DE}+L_{EF}\\ =\sqrt{{20}^2+{16}^2}+\sqrt{{20}^2+8^2}+20+\sqrt{{20}^2+8^2}+\sqrt{{20}^2+{16}^2}\\ =114.3\text{ft}\end{array}$

Therefore, the total length of the cable is $\underset{\_}{114.3\text{ft}}$.