Chapter 7, Problem 1RP

The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E. Neglect the thickness of both the beam and column in the calculation.

To determine

Find the resultant internal loadings acting on cross sections located at D and E.

Answer to Problem 1RP

The resultant internal loadings at cross section at D are $\underset{\_}{N_D=-2.16\text{kip},V_D=0}$, and $\underset{\_}{M_D=2.16\text{kip}\cdot \text{ft}}$.

The resultant internal loadings at cross section at E are $\underset{\_}{N_E=4.32\text{kip},V_E=0.54\text{kip}}$, and $\underset{\_}{M_E=2.16\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is $120\text{lb}/\text{ft}$.

The weight of the column FC is $180\text{lb}/\text{ft}$.

Calculation:

Find the loading at the center of the beam AB $\left(P_{AB}\right)$:

$P_{AB}=\text{Weight of beam}AB\times \text{Length}\text{of}\text{beam}AB$

Substitute $120\text{lb}/\text{ft}$ for the weight of beam AB and 12 ft for the length of beam AB.

$\begin{array}{c}{P}_{AB}=120\times 12\\ =1,440\text{lb}\end{array}$

Convert the unit from lb to kip.

$\begin{array}{c}{P}_{AB}=1,440\text{lb}\times \frac{1\text{kip}}{1,000\text{lb}}\\ =1.44\text{kip}\end{array}$

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

Refer to Figure 1.

Find the angle of cable BC to the horizontal $\left(\theta \right)$:

$\begin{array}{l}\sin \theta =\frac{1}{\sqrt{1^2+3^2}}\\ \sin \theta =\frac{1}{3.1623}\\ \theta ={\sin }^{-1}\left(0.3162\right)\\ \theta =18.43°\end{array}$

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ \left(F_{BC}\sin 18.43°\times 12\right)+1.44\times 6=0\\ 3.794F_{BC}+8.64=0\\ 3.794F_{BC}=-8.64\end{array}$

$F_{BC}=-2.277\text{kip}$

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x-2.277\cos 18.43°=0\\ A_x-2.16=0\\ A_x=2.16\text{kip}\end{array}$

Summation of forces along vertical direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y-1.44+2.277\sin 18.43°=0\\ A_y-0.72=0\\ A_y=0.72\text{kip}\end{array}$

Find the loading at the center of the beam AD $\left(P_{AD}\right)$:

$P_{AD}=\text{Weight of beam}AD\times \text{Length}\text{of}\text{beam}AD$

Substitute $120\text{lb}/\text{ft}$ for the weight of beam AD and 6 ft for the length of beam AD.

$\begin{array}{c}{P}_{AD}=120\times 6\\ =720\text{lb}\end{array}$

Convert the unit from lb to kip.

$\begin{array}{c}{P}_{AD}=720\text{lb}\times \frac{1\text{kip}}{1,000\text{lb}}\\ =0.72\text{kip}\end{array}$

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ N_D+2.16=0\\ N_D=-2.16\text{kip}\end{array}$

Summation of forces along vertical direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ V_D+0.72-0.72-=0\\ V_D=0\end{array}$

Take moment about D is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_D}=0\\ M_D-0.72\times 3=0\\ M_D-2.16=0\\ M_D=2.16\text{kip}\cdot \text{ft}\end{array}$

Hence, the resultant internal loadings at cross section at D are $\underset{\_}{N_D=-2.16\text{kip},V_D=0}$, and $\underset{\_}{M_D=2.16\text{kip}\cdot \text{ft}}$.

Find the loading at the center of the column FC $\left(P_{FC}\right)$:

$P_{FC}=\text{Weight of column}FC\times \text{Length}\text{of}\text{column}FC$

Substitute $180\text{lb}/\text{ft}$ for the weight of column FC and 16 ft for the length of column FC.

$\begin{array}{c}{P}_{FC}=180\times 16\\ =2,880\text{lb}\end{array}$

Convert the unit from lb to kip.

$\begin{array}{c}{P}_{FC}=2,880\text{lb}\times \frac{1\text{kip}}{1,000\text{lb}}\\ =2.88\text{kip}\end{array}$

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

$\begin{array}{l}\sin \theta =\frac{3}{\sqrt{4^2+3^2}}\\ \sin \theta =\frac{3}{5}\\ \theta ={\sin }^{-1}\left(0.6\right)\\ \theta =36.87°\end{array}$

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{CG}\cos 36.87°-2.277\sin 18.43°=0\\ 0.8F_{CG}-0.72=0\\ 0.8F_{CG}=0.72\end{array}$

$F_{CG}=0.9\text{kip}$

Find the loading at the center of the column FE $\left(P_{FE}\right)$:

$P_{FE}=\text{Weight of column}FE\times \text{Length}\text{of}\text{column}FE$

Substitute $180\text{lb}/\text{ft}$ for the weight of column FE and 4 ft for the length of column FC.

$\begin{array}{c}{P}_{FE}=180\times 4\\ =720\text{lb}\end{array}$

Convert the unit from lb to kip.

$\begin{array}{c}{P}_{FE}=720\text{lb}\times \frac{1\text{kip}}{1,000\text{lb}}\\ =0.72\text{kip}\end{array}$

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ V_E-0.54=0\\ V_E=0.54\text{kip}\end{array}$

Summation of forces along vertical direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ N_E+0.72-5.04=0\\ N_E-4.32=0\\ N_E=4.32\text{kip}\end{array}$

Take moment about E is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_E}=0\\ -M_E+0.54\times 4=0\\ -M_E+2.16=0\\ M_E=2.16\text{kip}\cdot \text{ft}\end{array}$

Therefore, the resultant internal loadings at cross section at E are $\underset{\_}{N_E=4.32\text{kip},V_E=0.54\text{kip}}$, and $\underset{\_}{M_E=2.16\text{kip}\cdot \text{ft}}$.