Chapter 3, Problem 31P

(a)

To determine

Calculate the reactions of the supports.

Answer to Problem 31P

The horizontal force at AB is $\underset{\_}{F_{ABx}=5.6\text{kips}(\to )}$.

The vertical force at AB is $\underset{\_}{F_{ABy}=5.6\text{kips}(\uparrow )}$.

The horizontal reaction at C is $\underset{\_}{C_x=25.6\text{kips}(\leftarrow )}$.

The vertical reaction at C is $\underset{\_}{C_y=38.4\text{kips}(\uparrow )}$.

The horizontal reaction at E is $\underset{\_}{E_x=20\text{kips}(\to )}$.

The vertical reaction at E is $\underset{\_}{E_y=40\text{kips}(\uparrow )}$.

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 counter clockwise moment as positive and the clockwise moment as negative.

Calculation:

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

Refer to Figure 1.

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

Refer to Figure 2.

Calculate the angle of the member DE as shown below.

$\begin{array}{l}\tan \theta =\frac{2}{1}\\ \theta =63.43°\end{array}$

Use equilibrium equations.

Find the vertical reaction at E:

Summation of moment about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ -E_y\times 60+60\times 40=0\\ E_y=40\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at E is $\underset{\_}{E_y=40\text{kips}(\uparrow )}$.

Calculate the horizontal reaction $E_x$ as shown below.

$\begin{array}{l}\frac{E_x}{1}=\frac{E_y}{2}\\ E_x=20\text{kips}(\to )\end{array}$

Hence, the horizontal reaction at E is $\underset{\_}{E_x=20\text{kips}(\to )}$.

Find the horizontal force at B:

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

$\begin{array}{l}+\to \sum F_x=0\\ -B_x+20=0\\ B_x=20\text{kips}(\leftarrow )\end{array}$

Hence, the horizontal force at B is $\underset{\_}{B_x=20\text{kips}(\leftarrow )}$.

Find the vertical force at B:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ B_y-60+40=0\\ B_y=20\text{kips}(\uparrow )\end{array}$

Hence, the vertical force at B is $\underset{\_}{B_y=20\text{kips}(\uparrow )}$.

Sketch the free body diagram of the truss 1 as shown in Figure 2.

Refer to Figure 3.

Use equilibrium equations.

Find the horizontal force at AB:

Summation of moments about c is equal to 0.

$\begin{array}{l}\sum M_C=0\\ -F_{ABx}\times 50+20\times 20+24\times 20-20\times 30=0\\ F_{ABx}=5.6\text{kips}(\to )\end{array}$

Hence, the horizontal force at AB is $\underset{\_}{F_{ABx}=5.6\text{kips}(\to )}$.

Calculate the horizontal reaction $F_{ABx}$ as shown below.

$\begin{array}{l}\frac{F_{ABx}}{1}=\frac{F_{ABy}}{1}\\ F_{ABy}=5.6\text{kips}(\uparrow )\end{array}$

Hence, the vertical force at AB is $\underset{\_}{F_{ABy}=5.6\text{kips}(\uparrow )}$.

Refer to Figure 3.

Use equilibrium equations.

Find the horizontal reaction at C:

Summation of moment about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -C_x\times 50+60\times 60+24\times 20+20\times 60-40\times 100=0\\ C_x=25.6\text{kips}(\leftarrow )\end{array}$

Hence, the horizontal reaction at C is $\underset{\_}{C_x=25.6\text{kips}(\leftarrow )}$.

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

$\begin{array}{l}+\to \sum F_y=0\\ C_y+5.6-24-60+40=0\\ C_y=38.4\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at C is $\underset{\_}{C_y=38.4\text{kips}(\uparrow )}$.

(b)

To determine

Calculate the forces applied to the trusses at joints C, B, and D.

Answer to Problem 31P

The horizontal force at B is $\underset{\_}{B_x=20\text{kips}(\leftarrow )}$.

The vertical force at B is $\underset{\_}{B_y=20\text{kips}(\uparrow )}$.

The horizontal reaction at C is $\underset{\_}{C_x=25.6\text{kips}(\leftarrow )}$.

The vertical reaction at C is $\underset{\_}{C_y=38.4\text{kips}(\uparrow )}$.

The horizontal force at D is $\underset{\_}{E_x=20\text{kips}(\to )}$.

The vertical force at D is $\underset{\_}{E_y=40\text{kips}(\uparrow )}$.

Explanation of Solution

Given information:

The structure is given in the Figure.

Calculation:

Refer to part (a).

The horizontal force at B is $\underset{\_}{B_x=20\text{kips}(\leftarrow )}$.

The vertical force at B is $\underset{\_}{B_y=20\text{kips}(\uparrow )}$.

The horizontal reaction at C is $\underset{\_}{C_x=25.6\text{kips}(\leftarrow )}$.

The vertical reaction at C is $\underset{\_}{C_y=38.4\text{kips}(\uparrow )}$.

The horizontal force at D is $\underset{\_}{E_x=20\text{kips}(\to )}$.

The vertical force at D is $\underset{\_}{E_y=40\text{kips}(\uparrow )}$.