Chapter 3, Problem 36P

To determine

Calculate the reactions at support A, B, C, and D and the force in the brace CH.

Answer to Problem 36P

The horizontal reaction at A is $\underset{\_}{A_x=0}$.

The vertical reaction at A is $\underset{\_}{A_y=60\text{kips}(\uparrow )}$.

The horizontal reaction at B is $\underset{\_}{B_x=0}$.

The vertical reaction at B is $\underset{\_}{B_y=120\text{kips}(\uparrow )}$.

The horizontal reaction at C is $\underset{\_}{C_x=6\text{kips}(\leftarrow )}$.

The vertical reaction at C is $\underset{\_}{C_y=116.8\text{kips}(\uparrow )}$.

The horizontal reaction at D is $\underset{\_}{D_x=0}$.

The vertical reaction at D is $\underset{\_}{D_y=63.2\text{kips}(\uparrow )}$.

The force acting on the brace CH is $\underset{\_}{F_{CH}=6.8\text{kips}}$.

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as positive and the counter clockwise moment as negative.

Calculation:

Sketch the free body diagram of the member EF as shown in Figure 1.

Refer to Figure 1.

Summation of moments about E is equal to 0.

$\begin{array}{l}\sum M_E=0\\ F_y\times 30-4\times 30\times \frac{30}{2}=0\\ F_y=60\text{kips}\end{array}$

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 6-F_x=0\\ F_x=6\text{kips}\end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ E_y-4\times 30+60=0\\ E_y=60\text{kips}\end{array}$

Sketch the free body diagram of the member AE as shown in Figure 2.

Refer to Figure 2.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 6-6-A_x=0\\ A_x=0\end{array}$

Hence, the horizontal reaction at A is $\underset{\_}{A_x=0}$.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y-60=0\\ A_y=60\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at A is $\underset{\_}{A_y=60\text{kips}(\uparrow )}$.

Sketch the free body diagram of the member FG as shown in Figure 3.

Refer to Figure 3.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 6-G_x=0\\ G_x=0\end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ G_y-4\times 30+60=0\\ G_y=60\text{kips}\end{array}$

Sketch the free body diagram of the member BF as shown in Figure 4.

Refer to Figure 4.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 6-6-B_x=0\\ B_x=0\end{array}$

Hence, the horizontal reaction at B is $\underset{\_}{B_x=0}$.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ B_y-60-60=0\\ B_y=120\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at B is $\underset{\_}{B_y=120\text{kips}(\uparrow )}$.

Sketch the free body diagram of the frame as shown in Figure 5.

Refer to Figure 5.

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ 60\times 90+6\times 16-4\times 90\times \frac{90}{2}+120\times 60+C_y\times 30=0\\ C_y=116.8\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at C is $\underset{\_}{C_y=116.8\text{kips}(\uparrow )}$.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ D_y+60+120+116.8-4\times 90=0\\ D_y=63.2\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at D is $\underset{\_}{D_y=63.2\text{kips}(\uparrow )}$.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 6-D_x-C_x=0\\ D_x+C_x=6\end{array}$        (1)

Sketch the free body diagram of the member DH without considering the brace as shown in Figure 6.

Refer to Figure 6.

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ 6\times 16-H_x\times 16=0\\ H_x=6\text{kips}\end{array}$

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 6-6-D_x=0\\ D_x=0\end{array}$

Hence, the horizontal reaction at D is $\underset{\_}{D_x=0}$.

From Equation (1).

$\begin{array}{l}0+C_x=6\\ C_x=6\text{kips}(\leftarrow )\end{array}$

Hence, the horizontal reaction at C is $\underset{\_}{C_x=6\text{kips}(\leftarrow )}$.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ -H_y+63.2=0\\ H_y=63.2\text{kips}\end{array}$

Sketch the free body diagram of the member GH with the brace as shown in Figure 7.

Refer to Figure 7.

Calculate the angle $\left(\theta \right)$ as shown below.

$\begin{array}{l}\tan \theta =\frac{16}{30}\\ \theta =28.07°\end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{CH}\sin 28.07°+60+63.2-4\times 30=0\\ F_{CH}=6.8\text{kips}\end{array}$

Therefore, the force acting on the brace CH is $\underset{\_}{F_{CH}=6.8\text{kips}}$.