Chapter 5, Problem 1RP

All the problems solutions must include FBD’s.

R5-1. Determine the force in each member of the truss and state if the members are in tension or compression.

Prob. R5-1

To determine

Find the force in each member of truss and the state of members are in tension or compression.

Answer to Problem 1RP

The magnitudes and the state of members are as follows:

• The magnitude of force in the member EF is $\underset{\_}{\text{zero}}$.
• The magnitude of force in the member ED is $\underset{\_}{13.125\text{kN}\left(\text{C}\right)}$.
• The magnitude of force in the member DF is $\underset{\_}{5.21\text{kN}\left(\text{T}\right)}$.
• The magnitude of force in the member CD is $\underset{\_}{4.17\text{kN}\left(\text{C}\right)}$.
• The magnitude of force in the member CF is $\underset{\_}{3.13\text{kN}\left(\text{C}\right)}$.
• The magnitude of force in the member AF is $\underset{\_}{4.17\text{kN}\left(\text{T}\right)}$.
• The magnitude of force in the member AC is $\underset{\_}{1.45\text{kN}\left(\text{C}\right)}$.
• The magnitude of force in the member BC is $\underset{\_}{3\text{kN}\left(\text{C}\right)}$.
• The magnitude of force in the member AB is $\underset{\_}{8\text{kN}\left(\text{C}\right)}$.

Explanation of Solution

Calculation:

Find the reactions.

Show the free body diagram of the truss as in Figure (1).

Using Figure (1):

Take moment at joint A and equate to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ E_y\left(4\right)-10\left(4\right)-4\left(2\right)-3\left(1.5\right)=0\\ E_y=13.125\text{kN}\end{array}$

Along the vertical direction:

Determine the vertical reaction at the joint A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+E_y-8-4-10=0\\ A_y+E_y=22\end{array}$ (1)

Substitute 13.125 kN for $E_y$ in Equation (1).

$\begin{array}{l}{A}_y+13.125=22\\ A_y=8.875\text{kN}\end{array}$

Along the horizontal direction:

Determine the horizontal reaction at the joint A by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x=0\end{array}$

Joint E:

Show the free body diagram of the Joint E in Figure (2).

Along the vertical direction:

Determine the force in the member ED by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ E_y-F_{ED}=0\end{array}$ (2)

Along the horizontal direction:

Determine the force in the member EF by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{EF}=0\end{array}$

Thus, the magnitude of force in the member EF is $\underset{\_}{\text{zero}}$.

Conclusion:

Substitute 13.125 kN for $E_y$ in Equation (2).

$\begin{array}{l}13.125-F_{ED}=0\\ F_{ED}=13.125\text{kN}\left(\text{C}\right)\end{array}$

Thus, the magnitude of force in the member ED is $\underset{\_}{13.125\text{kN}\left(\text{C}\right)}$.

Joint D:

Show the free body diagram of the Joint D in Figure (3).

Using Figure (3),

Determine the value of $\tan \theta$.

$\tan \theta =\frac{\text{Opposite}\text{side}}{\text{Adjacent}\text{side}}$ (3)

Substitute 1.5 m for opposite side and 2 m for adjacent side in Equation (3).

$\begin{array}{l}\tan {\theta }_1=\frac{1.5}{2}\\ {\theta }_1=36.87°\end{array}$

Along the vertical direction:

Determine the force in the member DF by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{ED}-10-F_{DF}\sin {\theta }_1=0\end{array}$ (4)

Along the horizontal direction:

Determine the force in the member CD by resolving the horizontal component of force.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{CD}-F_{DF}\cos {\theta }_1=0\end{array}$ (5)

Show the calculation of force in the members as follows.

Conclusion:

Substitute 13.125 kN for $F_{ED}$ and $36.87°$ for ${\theta }_1$ in Equation (4).

$\begin{array}{l}13.125-10-F_{DF}\sin 36.87°=0\\ F_{DF}=5.21\text{kN}\left(\text{T}\right)\end{array}$

Thus, the magnitude of force in the member DF is $\underset{\_}{5.21\text{kN}\left(\text{T}\right)}$.

Substitute 5.21 kN for $F_{DF}$ and $36.87°$ for ${\theta }_1$ in Equation (5).

$\begin{array}{l}{F}_{CD}-5.21\cos 36.87°=0\\ F_{CD}=4.17\text{kN}\left(\text{C}\right)\end{array}$

Thus, the magnitude of force in the member CD is $\underset{\_}{4.17\text{kN}\left(\text{C}\right)}$.

Joint F:

Show the free body diagram of the Joint F in Figure (4).

Using Figure (4),

Substitute 1.5 m for opposite side and 2 m for adjacent side in Equation (3).

$\begin{array}{l}\tan {\theta }_2=\frac{1.5}{2}\\ {\theta }_2=36.87°\end{array}$

Along the vertical direction:

Determine the force in the member CF by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{DF}\sin {\theta }_2-F_{CF}=0\end{array}$ (6)

Along the horizontal direction:

Determine the force in the member AF by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{EF}+F_{DF}\cos {\theta }_2-F_{AF}=0\end{array}$ (7)

Show the calculation of force in the members as follows:

Conclusion:

Substitute 5.21 kN for $F_{DF}$ and $36.87°$ for ${\theta }_2$ in Equation (6).

$\begin{array}{l}5.21\sin 36.87°-F_{CF}=0\\ F_{CF}=3.13\text{kN}\left(\text{C}\right)\end{array}$

Thus, the magnitude of force in the member CF is $\underset{\_}{3.13\text{kN}\left(\text{C}\right)}$.

Substitute 0 for $F_{EF}$, 5.21 kN for $F_{DF}$, and $36.87°$ for ${\theta }_2$ in Equation (7).

$\begin{array}{l}0+5.21\cos 36.87°-F_{AF}=0\\ F_{AF}=4.17\text{kN}\left(\text{T}\right)\end{array}$

Thus, the magnitude of force in the member AF is $\underset{\_}{4.17\text{kN}\left(\text{T}\right)}$.

Joint C:

Show the free body diagram of the Joint C in Figure (5).

Using Figure (5),

Substitute 1.5 m for opposite side and 2 m for adjacent side in Equation (3).

$\begin{array}{l}\tan {\theta }_2=\frac{1.5}{2}\\ {\theta }_2=36.87°\end{array}$

Along the vertical direction:

Determine the force in the member AC by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{AC}\sin {\theta }_3+F_{CF}-4=0\end{array}$ (8)

Along the horizontal direction:

Determine the force in the member BC by resolving the horizontal component of force.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{CD}-F_{BC}-F_{AC}\cos {\theta }_3=0\end{array}$ (9)

Conclusion:

Substitute $36.87°$ for ${\theta }_2$ and 3.13 kN for $F_{CF}$ in Equation (8).

$\begin{array}{l}{F}_{AC}\sin 36.87°+3.13-4=0\\ F_{AC}=1.45\text{kN}\left(\text{C}\right)\end{array}$

Thus, the magnitude of force in the member AC is $\underset{\_}{1.45\text{kN}\left(\text{C}\right)}$.

Substitute 4.17 kN for $F_{CD}$, 1.45 kN for $F_{AC}$, and $36.87°$ for ${\theta }_2$ in Equation (9).

$\begin{array}{l}4.17-F_{BC}-1.45\cos 36.87°=0\\ F_{BC}=3\text{kN}\left(\text{C}\right)\end{array}$

Thus, the magnitude of force in the member BC is $\underset{\_}{3\text{kN}\left(\text{C}\right)}$.

Joint B:

Show the free body diagram of the Joint B in Figure (6).

Using Figure (6):

Along the vertical direction:

Determine the force in the member AB by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{AB}-8=0\\ F_{AB}=8\text{kN}\left(\text{C}\right)\end{array}$

Thus, the magnitude of force in the member AB is $\underset{\_}{8\text{kN}\left(\text{C}\right)}$.