Chapter 8, Problem 8.130RP

To determine

Tension on each rope supporting steel plate

Answer to Problem 8.130RP

The tension force $T_A$, $T_C$ at point A and C is $16.267lb$.

The tension force $T_B$ at point B is $13.13lb$.

Explanation of Solution

Given information:

The specific weight of plate is $0.284lb/i{n}^3$.

The weight of plate is defined as:

  $W=\gamma \times V$

  $V$ - Volume

Steps to follow in the equilibrium analysis of a body are:

1. Draw the free body diagram.

2. Write the equilibrium equations.

3. Solve the equations for the unknowns.

Calculation:

Assume $t,b$ as the thickness and side length of the plate.

The weight $W_1$ of bottom part 1

  $\begin{array}{l}{W}_1=\gamma t{b}^2\\ =\left(0.284lb/i{n}^3\right)\left(0.5in\right){\left(24in\right)}^2\\ =81.792lb\end{array}$

The point of action $\overline{y_1}$ of center of gravity in y direction

  $\overline{y_1}=12in$

The weight $W_2$ of arc part 2

  $\begin{array}{l}{W}_2=-\gamma t{\left(\frac{\pi r}{4}\right)}^2\\ =-\left(0.284lb/i{n}^3\right)\left(0.5in\right){\left(\frac{\pi \left(18in\right)}{4}\right)}^2\\ =-36.13lb\end{array}$

The point of action $\overline{y_2}$ of center of gravity in y direction

  $\begin{array}{l}\overline{y_2}=b-\frac{4r}{3\pi }\\ =24in-\frac{4\left(18in\right)}{3\pi }\\ =16.361in\end{array}$

The weight $W$ of whole plate

  $\begin{array}{l}W={\displaystyle \sum W_i}\\ =W_1-W_2\\ =81.792lb-36.13lb\\ =45.66lb\end{array}$

The sum of $W_i\overline{y_i}$

  $\begin{array}{l}{\displaystyle \sum W_i\overline{y_i}}=W_1\overline{y_1}+W_2\overline{y_2}\\ =\left(81.792lb\right)\left(12in\right)+\left(-36.13lb\right)\left(16.361in\right)\\ =390.4lbin\end{array}$

FBD of plate

Assume $T_A,T_B,T_C$ as the tension in ropes at point A,B and C.

For the equilibrium of plate, the moment about x axis is equal to zero.

  ${\displaystyle \sum M_x=0}$

  $T_{A}b-W\overline{y}=0$

Substitute and solve

  $\begin{array}{l}{T}_A\left(24in\right)-390.4lbin=0\\ T_A=16.267lb\end{array}$

The plate is symmetrical.

Therefore

  $T_A=T_C=16.267lb$

Write equilibrium equation in z direction.

  $\uparrow {\displaystyle \sum F_z=0}$

  $T_A+T_B+T_C-W=0$

Substitute and solve

  $\begin{array}{l}16.267lb+T_B+16.267lb-45.66lb=0\\ T_B=13.13lb\end{array}$

Conclusion:

The tension force $T_A$, $T_C$ at point A and C is $16.267lb$.

The tension force $T_B$ at point B is $13.13lb$.