Chapter 1, Problem 1.4.13P

An L-shaped reinforced concrete slab 12 Ft X 12 ft, with a 6 Ft X 6 ft cut-out and thickness t = 9.0 in, is lifted by three cables attached at O, B, and D, as shown in the figure. The cables are are combined at point Q, which is 7.0 Ft above the top of the slab and directly above the center of mass at C. Each cable has an effective cross-sectional area of A_e= 0.12 in².

(a) Find the tensile force T_r(i = 1, 2, 3) in each cable due to the weight W of the concrete slab

(ignore weight of cables).

(b) Find the average stress ov in each cable. (See Table I-1 in Appendix I for the weight density of reinforced concrete.)

(c) Add cable AQ so that OQA is one continuous cable, with each segment having Force T, which is connected to cables BQ and DQ at point Q. Repeat parts (a) and (b). Hini: There are now three Forced equilibrium equations and one constrain equation, T₁= T₄.

  

To determine

Tensile force T in each cable due to weight.

Answer to Problem 1.4.13P

Tensile force T is:

  $T=\left(\begin{array}{l}5877\\ 4679\\ 7159\end{array}\right)\text{ lb}\text{.}$

Explanation of Solution

Given Information:

You have the following figure with all relevant information:

  

Thickness (t) is 9 in.

Calculation:

Consider the free body diagram as:

  

To calculate external reaction force F, take equilibrium of forces as:

  $\begin{array}{l}{\displaystyle \sum F_z}=0\\ \Rightarrow F=W=150\times t\times A\\ \Rightarrow F=150\times t\times A\\ \Rightarrow F=150\times \frac{9}{12}\times [12\times 6+6\times 6]=12150\text{ lb}\text{.}\end{array}$

The following force acts on point Q:

  $\begin{array}{l}\overrightarrow{F}=(0,0,12150)\text{ lb}\\ {\overrightarrow{T}}_1=\frac{T_1}{\sqrt{{(-5)}^2+{(-5)}^2+{(-7)}^2}}(-5,-5,-7)\text{ lb}\\ {\overrightarrow{T}}_2=\frac{T_2}{\sqrt{{(7)}^2+{(-5)}^2+{(-7)}^2}}(7,-5,-7)\text{ lb}\\ \overrightarrow{T_3}=\frac{T_3}{\sqrt{{(0)}^2+{(7)}^2+{(-7)}^2}}(0,7,-7)\text{ lb}\end{array}$

Take equilibrium of forces at Q as:

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}}=0\\ \Rightarrow \overrightarrow{F}+\overrightarrow{T_1}+\overrightarrow{T_2}+\overrightarrow{T_3}=0\\ \Rightarrow (0,0,12150)\text{ +}\frac{T_1}{\sqrt{{(-5)}^2+{(-5)}^2+{(-7)}^2}}(-5,-5,-7)\\ +\frac{T_2}{\sqrt{{(7)}^2+{(-5)}^2+{(-7)}^2}}(7,-5,-7)\\ +\frac{T_3}{\sqrt{{(0)}^2+{(7)}^2+{(-7)}^2}}(0,7,-7)=0\end{array}$

Solve the above equation to get:

  $T=\left(\begin{array}{l}{T}_1\\ T_2\\ T_3\end{array}\right)=\left(\begin{array}{l}5877\\ 4679\\ 7159\end{array}\right)\text{ lb}\text{.}$

Conclusion:

Therefore, the correct answers are $T=\left(\begin{array}{l}{T}_1\\ T_2\\ T_3\end{array}\right)=\left(\begin{array}{l}5877\\ 4679\\ 7159\end{array}\right)\text{ lb}\text{.}$

To determine

Average stress in each cable.

Answer to Problem 1.4.13P

Stress in each cable is $\sigma =\left(\begin{array}{l}48,975\\ 38,992\\ 59,658\end{array}\right)\text{ psi}\text{.}$

Explanation of Solution

Given Information:

You have following figure with all relevant information:

  

Thickness (t) is 9 in. and cross-sectional area of each cable is $0.12\text{ in}{\text{.}}^{\text{2}}$

Calculation:

Consider the free body diagram as,

  

To calculate external reaction force F, take equilibrium of forces as,

  $\begin{array}{l}{\displaystyle \sum F_z}=0\\ \Rightarrow F=W=150\times t\times A\\ \Rightarrow F=150\times t\times A\\ \Rightarrow F=150\times \frac{9}{12}\times [12\times 6+6\times 6]=12150\text{ lb}\text{.}\end{array}$

The following forces acts on point Q:

  $\begin{array}{l}\overrightarrow{F}=(0,0,12150)\text{ lb}\\ {\overrightarrow{T}}_1=\frac{T_1}{\sqrt{{(-5)}^2+{(-5)}^2+{(-7)}^2}}(-5,-5,-7)\text{ lb}\\ {\overrightarrow{T}}_2=\frac{T_2}{\sqrt{{(7)}^2+{(-5)}^2+{(-7)}^2}}(7,-5,-7)\text{ lb}\\ \overrightarrow{T_3}=\frac{T_3}{\sqrt{{(0)}^2+{(7)}^2+{(-7)}^2}}(0,7,-7)\text{ lb}\end{array}$

Take equilibrium of forces at Q as:

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}}=0\\ \Rightarrow \overrightarrow{F}+\overrightarrow{T_1}+\overrightarrow{T_2}+\overrightarrow{T_3}=0\\ \Rightarrow (0,0,12150)\text{ +}\frac{T_1}{\sqrt{{(-5)}^2+{(-5)}^2+{(-7)}^2}}(-5,-5,-7)\\ +\frac{T_2}{\sqrt{{(7)}^2+{(-5)}^2+{(-7)}^2}}(7,-5,-7)\\ +\frac{T_3}{\sqrt{{(0)}^2+{(7)}^2+{(-7)}^2}}(0,7,-7)=0\end{array}$

Solve the above equation to get,

  $T=\left(\begin{array}{l}{T}_1\\ T_2\\ T_3\end{array}\right)=\left(\begin{array}{l}5877\\ 4679\\ 7159\end{array}\right)\text{ lb}\text{.}$

Calculate stress as,

  $\sigma =\frac{T}{A}=\left(\begin{array}{l}5877\\ 4679\\ 7159\end{array}\right)\text{ lb}\times \frac{1}{0.12\text{ in}{\text{.}}^2}=\left(\begin{array}{l}48,975\\ 38,992\\ 59,658\end{array}\right)\text{ psi}\text{.}$

Conclusion:

Therefore, the correct answer is:

  $\sigma =\left(\begin{array}{l}48,975\\ 38,992\\ 59,658\end{array}\right)\text{ psi}\text{.}$

To determine

Tensile force and average stress in each cable.

Answer to Problem 1.4.13P

Stress in each cable is:

  $T=\left(\begin{array}{l}4278\\ 6461\\ 3341\\ 4278\end{array}\right)\text{ lb, }\sigma =\left(\begin{array}{l}35,650\\ 53,842\\ 27,842\\ 35,650\end{array}\right)\text{ psi}\text{.}$

Explanation of Solution

Given Information:

You have following figure with all relevant information:

  

Thickness (t) is 9 in. and cross-sectional area of each cable is $0.12\text{ in}{\text{.}}^{\text{2}}$

Calculation:

Consider the free body diagram as,

  

To calculate external reaction force F, take equilibrium of forces as:

  $\begin{array}{l}{\displaystyle \sum F_z}=0\\ \Rightarrow F=W=150\times t\times A\\ \Rightarrow F=150\times t\times A\\ \Rightarrow F=150\times \frac{9}{12}\times [12\times 6+6\times 6]=12150\text{ lb}\text{.}\end{array}$

The following force acts on point Q:

  $\begin{array}{l}\overrightarrow{F}=(0,0,12150)\text{ lb}\\ {\overrightarrow{T}}_1=\frac{T_1}{\sqrt{{(-5)}^2+{(-5)}^2+{(-7)}^2}}(-5,-5,-7)\text{ lb}\\ {\overrightarrow{T}}_2=\frac{T_2}{\sqrt{{(7)}^2+{(-5)}^2+{(-7)}^2}}(7,-5,-7)\text{ lb}\\ \overrightarrow{T_3}=\frac{T_3}{\sqrt{{(0)}^2+{(7)}^2+{(-7)}^2}}(0,7,-7)\text{ lb}\\ \overrightarrow{T_4}=\frac{T_4}{\sqrt{{(-5)}^2+{(7)}^2+{(-7)}^2}}(-5,7,-7)\text{ lb}\end{array}$

Take equilibrium of force at Q as:

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}}=0\\ \Rightarrow \overrightarrow{F}+\overrightarrow{T_1}+\overrightarrow{T_2}+\overrightarrow{T_3}+\overrightarrow{T_4}=0\\ \Rightarrow (0,0,12150)\text{ +}\frac{T_1}{\sqrt{{(-5)}^2+{(-5)}^2+{(-7)}^2}}(-5,-5,-7)\\ +\frac{T_2}{\sqrt{{(7)}^2+{(-5)}^2+{(-7)}^2}}(7,-5,-7)\\ +\frac{T_3}{\sqrt{{(0)}^2+{(7)}^2+{(-7)}^2}}(0,7,-7)\\ +\frac{T_4}{\sqrt{{(-5)}^2+{(7)}^2+{(-7)}^2}}(-5,7,-7)=0\end{array}$

Solve the above equation along with $T_1=T_4$to get,

  $T=\left(\begin{array}{l}{T}_1\\ T_2\\ T_3\\ T_4\end{array}\right)=\left(\begin{array}{l}4278\\ 6461\\ 3341\\ 4278\end{array}\right)\text{ lb}\text{.}$

Calculate stress as:

  $\sigma =\frac{T}{A}=\left(\begin{array}{l}4278\\ 6461\\ 3341\\ 4278\end{array}\right)\text{ lb}\times \frac{1}{0.12\text{ in}{\text{.}}^2}=\left(\begin{array}{l}35,650\\ 53,842\\ 27,842\\ 35,650\end{array}\right)\text{ psi}\text{.}$

Conclusion:

Therefore, the correct answers are:

  $T=\left(\begin{array}{l}4278\\ 6461\\ 3341\\ 4278\end{array}\right)\text{ lb, }\sigma =\left(\begin{array}{l}35,650\\ 53,842\\ 27,842\\ 35,650\end{array}\right)\text{ psi}\text{.}$