Chapter 6, Problem 1FP

Determine the force in each member of the truss. State if the members are in tension or compression.

Prob. F6-1

To determine

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

Answer to Problem 1FP

The magnitudes and the state of members are as follows:

• The magnitude of force in the member AD is $\underset{\_}{318.20\text{lb}\left(\text{C}\right)}$.
• The magnitude of force in the member AB is $\underset{\_}{225\text{lb}\left(\text{T}\right)}$.
• The magnitude of force in the member BD is $\underset{\_}{\text{zero}}$.
• The magnitude of force in the member BC is $\underset{\_}{225\text{lb}\left(\text{T}\right)}$.
• The magnitude of force in the member CD is $\underset{\_}{318.20\text{lb}\left(\text{T}\right)}$.

Explanation of Solution

Method adopted:Method of Joints

Assumptions:

• Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member.
• Consider the force indicating right side as positive and left side as negative in horizontal components of forces.
• Consider the force indicating upside is taken as positive and downside as negative in vertical components of forces.
• Consider clockwise moment as positive and anti-clock wise moment as negative wherever applicable.

Find the reactions.

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

Using Figure (1),

Take moment at joint A and equate to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ 450\left(4\right)-C_y\left(8\right)=0\\ C_y=225\text{lb}\end{array}$

Along the vertical direction:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y-C_y=0\end{array}$

Substitute 225 lb for $C_y$.

$\begin{array}{l}{A}_y-225=0\\ A_y=225\text{lb}\end{array}$

Along the horizontal direction:

Determine the horizontal reaction $C_x$ at the joint C by resolving the horizontal component.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ C_x-450=0\\ C_x=450\text{lb}\end{array}$

Joint A:

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

Using Figure (2),

Determine the value of $\tan \theta$.

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

Substitute 4 ft for opposite side and 4 ft for adjacent side in Equation (I).

$\begin{array}{l}\tan {\theta }_1=\frac{4}{4}\\ {\theta }_1=45°\end{array}$

Along the vertical direction:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y-F_{AD}\sin {\theta }_1=0\end{array}$ (II)

Along the horizontal direction:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{AB}-F_{AD}\cos {\theta }_1=0\end{array}$ (III)

Show the calculation of force in the members as follows:

Conclusion:

Substitute 225 lb for $A_y$ and $45°$ for ${\theta }_1$ in Equation (II).

$\begin{array}{l}225-F_{AD}\sin 45°=0\\ F_{AD}=318.20\text{lb}\left(\text{C}\right)\end{array}$

Thus, the magnitude of force in the member AD is $\underset{\_}{318.20\text{lb}\left(\text{C}\right)}$.

Substitute 318.20 lb for $F_{AD}$ and $45°$ for ${\theta }_1$ in Equation (III).

$\begin{array}{l}{F}_{AB}-318.20\cos 45°=0\\ F_{AB}=225\text{lb}\left(\text{T}\right)\end{array}$

Thus, the magnitude of force in the member AB is $\underset{\_}{225\text{lb}\left(\text{T}\right)}$.

Joint B:

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

Using Figure (3),

Along the vertical direction:

Determine the force in the member BD by resolving the vertical component of force.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{BD}=0\end{array}$ (IV)

Along the horizontal direction:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{BC}-F_{AB}=0\end{array}$ (V)

Show the calculation of force in the members as follows:

Conclusion:

Refer to Equation (IV).

There is no vertical force acting in the member BD. Therefore the force in the member $F_{BD}$ is zero.

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

Substitute 225 lb for $F_{AB}$ in Equation (V).

$\begin{array}{l}{F}_{BC}-225=0\\ F_{BC}=225\text{lb}\left(\text{T}\right)\end{array}$

Thus, the magnitude of force in the member BC is $\underset{\_}{225\text{lb}\left(\text{T}\right)}$.

Joint C:

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

Using Figure (4),

Substitute 4 ft for opposite side and 4 ft for adjacent side in Equation (I).

$\begin{array}{l}\tan {\theta }_2=\frac{4}{4}\\ {\theta }_2=45°\end{array}$

Along the vertical direction:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{CD}\sin {\theta }_2-C_y=0\end{array}$ (VI)

Show the calculation of forces in the members as follows:

Conclusion:

Substitute 225 lb for $C_y$ and $45°$ for ${\theta }_2$ in Equation (VI).

$\begin{array}{l}{F}_{CD}\sin 45°-225=0\\ F_{CD}=318.20\text{lb}\left(\text{T}\right)\end{array}$

Thus, the magnitude of force in the member CD is $\underset{\_}{318.20\text{lb}\left(\text{T}\right)}$.