Chapter 6, Problem 6.164RP

Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

To determine

The force in each of the member of the truss using method of joints and whether each member is in tension or compression.

Answer to Problem 6.164RP

The force in member $BE$ is $5.00\text{ kN}$ and the member $BE$ is in tension, the force in member $DE$ is $4.00\text{ kN}$ and the member $DE$ is in compression, the force in member $AB$ is $4.00\text{ kN}$ and the member $AB$ is in tension, the force in member $BD$ is $9.00\text{ kN}$ and the member $BD$ is in compression, the force in member $AD$ is $15.00\text{ kN}$ and the member $AD$ is in tension, the force in member $CD$ is $16.00\text{ kN}$ and the member $CD$ is in compression.

Explanation of Solution

The free-body diagram of the truss is shown in figure 1.

The sum of the moments about the point $C$ must be equal to zero.

$\Sigma M_C=0$

Here, $\Sigma M_C$ is the sum of the moments about the point $C$ .

The above equation gives,

${\overrightarrow{A}}_x=16\text{ kN}\leftarrow$

Here, ${\overrightarrow{A}}_x$ is the $x$ component of the reaction at the point $A$ .

The $y$ component of the net force must be equal to zero.

$\Sigma F_y=0$

Here, $\Sigma F_y$ is the $y$ component of the net force.

The above equation gives,

${\overrightarrow{A}}_y=9\text{ kN}\uparrow$

Here, ${\overrightarrow{A}}_y$ is the $y$ component of the reaction at the point $A$ .

The $x$ component of the net force must be equal to zero.

$\Sigma F_x=0$

Here, $\Sigma {\overrightarrow{F}}_x$ is the $x$ component of the net force.

The above equation gives,

$\overrightarrow{C}=16\text{ kN}\to$

Here, $\overrightarrow{C}$ is the reaction at the point C.

The free-body diagram of joint E is shown in figure 2.

The joint E is subject to the forces exerted by $BE$ and $DE$ . Use a force triangle to determine $F_{BE}$ and $F_{DE}$ . The force triangle is shown in figure 3.

$BE$ pulls on the joint so $BE$ is in tension and member $DE$ pushes on the joint so $DE$ is in compression.

Obtain the magnitudes of the two forces from proportion.

$\begin{array}{c}\frac{F_{BE}}{5}=\frac{3\text{ kN}}{3}\\ F_{BE}=\frac{5\left(3\text{ kN}\right)}{3}\\ =5.00\text{ kN}\\ \frac{F_{DE}}{4}=\frac{3\text{ kN}}{3}\\ F_{DE}=\frac{4\left(3\text{ kN}\right)}{3}\\ =4.00\text{ kN}\end{array}$

Here, $F_{BE}$ is the force exerted by $BE$ and $F_{DE}$ is the force exerted by $DE$ .

The free-body diagram of the joint B is shown in figure 3. Write the equilibrium equations.

$\to \Sigma F_x=0$ (I)

Here, $\to \Sigma F_x$ is the $x$ component of the net force.

Write the expression for $\to \Sigma F_x$ .

$\to \Sigma F_x=\frac{4}{5}\left(5\text{ kN}\right)-F_{AB}$

Here, $F_{AB}$ is the force exerted by $AB$.

Put the above equation in equation (I).

$\begin{array}{c}\frac{4}{5}\left(5\text{ kN}\right)-F_{AB}=0\\ F_{AB}=4.00\text{ kN, tension}\end{array}$

The $y$ component of the net force must be equal to zero.

$\uparrow \Sigma F_y=0$ (II)

Here, $\uparrow \Sigma F_y$ is the $y$ component of the net force.

Write the expression for $\uparrow \Sigma F_y$ .

$\uparrow \Sigma F_y=-6\text{ kN}-\frac{3}{5}\left(5\text{ kN}\right)-F_{BD}$

Here, $F_{BD}$ is the force exerted by $BD$.

Put the above equation in equation (II).

$\begin{array}{c}-6\text{ kN}-\frac{3}{5}\left(5\text{ kN}\right)-F_{BD}=0\\ F_{BD}=-9.00\text{ kN}\\ =9.00\text{ kN, compression}\end{array}$

The free-body diagram of the joint D is shown in figure 4.

Write the expression for $\uparrow \Sigma F_y$ .

$\uparrow \Sigma F_y=-9\text{ kN+}\frac{3}{5}{F}_{AD}$

Here, $F_{AD}$ is the force exerted by $AD$.

Put the above equation in equation (II).

$\begin{array}{c}-9\text{ kN+}\frac{3}{5}{F}_{AD}=0\\ F_{AD}=15.00\text{ kN, tension}\end{array}$

Write the expression for $\to \Sigma F_x$ .

$\to \Sigma F_x=-4\text{ kN}-\frac{4}{5}\left(15\text{ kN}\right)-F_{CD}$

Here, $F_{CD}$ is the force exerted by $CD$.

Put the above equation in equation (I).

$\begin{array}{c}-4\text{ kN}-\frac{4}{5}\left(15\text{ kN}\right)-F_{CD}=0\\ F_{CD}=-16\text{ kN}\\ =16\text{ kN},\text{ compression}\end{array}$

Conclusion:

Thus, the force in member $BE$ is $5.00\text{ kN}$ and the member $BE$ is in tension, the force in member $DE$ is $4.00\text{ kN}$ and the member $DE$ is in compression, the force in member $AB$ is $4.00\text{ kN}$ and the member $AB$ is in tension, the force in member $BD$ is $9.00\text{ kN}$ and the member $BD$ is in compression, the force in member $AD$ is $15.00\text{ kN}$ and the member $AD$ is in tension, the force in member $CD$ is $16.00\text{ kN}$ and the member $CD$ is in compression.