Chapter 6, Problem 6.165RP

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

To determine

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

Answer to Problem 6.165RP

The force in member $AB$ is $7.83\text{ kN}$ and the member $AB$ is in compression, the force in member $AC$ is $7.00\text{ kN}$ and the member $AC$ is in tension, the force in member $BD$ is $6.34\text{ kN}$ and the member $BD$ is in compression, the force in member $BC$ is $1.886\text{ kN}$ and the member $BC$ is in compression, the force in member $CD$ is $1.491\text{ kN}$ and the member $CD$ is in tension, the force in member $CE$ is $5.00\text{ kN}$ and the member $CE$ is in tension, the force in member $DF$ is $3.35\text{ kN}$ and the member $DF$ is in compression, the force in member $DE$ is $2.83\text{ kN}$ and the member $BC$ is in compression, the force in member $FG$ is $4.24\text{ kN}$ and the member $FG$ is in compression, the force in member $EF$ is $2.75\text{ kN}$ and the member $EF$ is in tension, the force in member $GH$ is $5.30\text{ kN}$ and the member $GH$ is in compression, the force in member $EG$ is $1.061\text{ kN}$ and the member $EG$ is in compression, the force in member $EH$ is $3.75\text{ kN}$ and the member $EH$ is in tension.

Explanation of Solution

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

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

$\Sigma M_A=0$ (I)

Here, $\Sigma M_A$ is the sum of the moments about the point $A$ .

Write the equation for $\Sigma M_A$ .

$\Sigma M_A=H\left(18\text{ m}\right)-\left(2\text{ kN}\right)\left(4\text{ m}\right)-\left(2\text{ kN}\right)\left(8\text{ m}\right)-\left(1.75\text{ kN}\right)\left(12\text{ m}\right)-\left(1.5\text{ kN}\right)\left(15\text{ m}\right)-\left(0.75\text{ kN}\right)\left(18\text{ m}\right)$

Here, $H$ is the magnitude of the reaction at the point $H$.

Put the above equation in equation (I).

$\begin{array}{c}H\left(18\text{ m}\right)-\left(2\text{ kN}\right)\left(4\text{ m}\right)-\left(2\text{ kN}\right)\left(8\text{ m}\right)-\left(1.75\text{ kN}\right)\left(12\text{ m}\right)-\left(1.5\text{ kN}\right)\left(15\text{ m}\right)-\left(0.75\text{ kN}\right)\left(18\text{ m}\right)=0\\ H=4.50\text{ kN}\\ \overrightarrow{H}=4.50\text{ kN}\uparrow \end{array}$

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

$\Sigma F_x=0$ (II)

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

Write the expression for $\Sigma {\overrightarrow{F}}_x$ .

$\Sigma F_x=A_x$

Here, $A_x$ is the $x$ component of the reaction at the point $A$ .

Put the above equation in equation (II).

$A_x=0$

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

$\Sigma F_y=0$ (III)

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

Write the expression for $\Sigma F_y$ .

$\Sigma F_y=A_y+H-9$

Put the above equation in equation (III).

$\begin{array}{c}{A}_y+H-9=0\\ A_y=-H+9\end{array}$

Here, $A_y$ is the $y$ component of the reaction at the point $A$ .

Substitute $4.50\text{ kN}$ for $H$ in the above equation to find $A_y$ .

$\begin{array}{c}{A}_y=-4.50\text{ kN}+9\\ =4.50\text{ kN}\uparrow \end{array}$

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

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

$AB$ pushes on the joint so $AB$ is in compression and member $AC$ pulls on the joint so $AC$ is in tension.

Obtain the magnitudes of the two forces from proportion.

$\begin{array}{c}\frac{F_{AB}}{\sqrt{5}}=\frac{3.50\text{ kN}}{1}\\ F_{AB}=\frac{\sqrt{5}\left(3.50\text{ kN}\right)}{1}\\ =7.8262\text{ kN}\\ =7.83\text{ kN}\\ \frac{F_{AC}}{2}=\frac{3.50\text{ kN}}{1}\\ F_{AC}=\frac{2\left(3.50\text{ kN}\right)}{1}\\ =7.00\text{ kN}\end{array}$

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

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

Write the equilibrium equations.

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

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

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

$\to \Sigma F_x=\frac{2}{\sqrt{5}}{F}_{BD}+\frac{2}{\sqrt{5}}\left(7.8262\text{ kN}\right)+\frac{1}{\sqrt{2}}{F}_{BC}$

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

Put the above equation in equation (IV).

$\begin{array}{c}\frac{2}{\sqrt{5}}{F}_{BD}+\frac{2}{\sqrt{5}}\left(7.8262\text{ kN}\right)+\frac{1}{\sqrt{2}}{F}_{BC}=0\\ F_{BD}+0.79057F_{BC}=-7.8262\text{ kN}\end{array}$ (V)

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

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

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

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

$\uparrow \Sigma F_y=\frac{1}{\sqrt{5}}{F}_{BD}+\frac{1}{\sqrt{5}}\left(7.8262\text{ kN}\right)-\frac{1}{\sqrt{2}}{F}_{BC}-2\text{ kN}$

Put the above equation in equation (VI).

$\begin{array}{c}\frac{1}{\sqrt{5}}{F}_{BD}+\frac{1}{\sqrt{5}}\left(7.8262\text{ kN}\right)-\frac{1}{\sqrt{2}}{F}_{BC}-2\text{ kN=0}\\ F_{BD}-1.58114F_{BC}=-3.3541\end{array}$ (VII)

Multiply equation (V) by $2$ and add equation (VII).

$\begin{array}{c}3F_{BD}=-19.0065\\ F_{BD}=\frac{-19.0065}{3}\\ =-6.3355\text{ kN}\\ =6.34\text{ kN, compression}\end{array}$

Subtract equation (VII) from (V).

$\begin{array}{c}2.37111F_{BC}=-4.4721\\ F_{BC}=\frac{-4.4721}{2.37111}\\ =-1.8861\text{ kN}\\ =1.8861\text{ kN},\text{ compression}\end{array}$

The free-body diagram of the joint C is shown in figure 5.

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

$\uparrow \Sigma F_y=\frac{2}{\sqrt{5}}{F}_{CD}+\frac{1}{\sqrt{2}}\left(1.8861\text{ kN}\right)$

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

Put the above equation in equation (VI).

$\begin{array}{c}\frac{2}{\sqrt{5}}{F}_{CD}+\frac{1}{\sqrt{2}}\left(1.8861\text{ kN}\right)=0\\ F_{CD}=1.4911\text{ kN}\\ =1.491\text{ kN, tension}\end{array}$

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

$\to \Sigma F_x=F_{CE}-7.00\text{ kN}+\frac{1}{\sqrt{2}}\left(1.8861\text{ kN}\right)+\frac{1}{\sqrt{5}}\left(1.4911\text{ kN}\right)$

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

Put the above equation in equation (IV).

$\begin{array}{c}{F}_{CE}-7.00\text{ kN}+\frac{1}{\sqrt{2}}\left(1.8861\text{ kN}\right)+\frac{1}{\sqrt{5}}\left(1.4911\text{ kN}\right)=0\\ F_{CE}=5.00\text{ kN, tension}\end{array}$

The free-body diagram of the joint D is shown in figure 6.

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

$\to \Sigma F_x=\frac{2}{\sqrt{5}}{F}_{DF}+\frac{1}{\sqrt{2}}{F}_{DE}+\frac{2}{\sqrt{5}}\left(6.3355\text{ kN}\right)-\frac{1}{\sqrt{5}}\left(1.4911\text{ kN}\right)$

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

Put the above equation in equation (IV).

$\begin{array}{c}\frac{2}{\sqrt{5}}{F}_{DF}+\frac{1}{\sqrt{2}}{F}_{DE}+\frac{2}{\sqrt{5}}\left(6.3355\text{ kN}\right)-\frac{1}{\sqrt{5}}\left(1.4911\text{ kN}\right)=0\\ F_{DF}+0.79057F_{DE}=-5.5900\text{ kN}\end{array}$ (VIII)

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

$\uparrow \Sigma F_y=\frac{1}{\sqrt{5}}{F}_{DF}-\frac{1}{\sqrt{2}}{F}_{DE}+\frac{1}{\sqrt{5}}\left(6.3355\text{ kN}\right)-\frac{2}{\sqrt{5}}\left(1.4911\text{ kN}\right)-2\text{ kN}$

Put the above equation in equation (VI).

$\begin{array}{c}\frac{1}{\sqrt{5}}{F}_{DF}-\frac{1}{\sqrt{2}}{F}_{DE}+\frac{1}{\sqrt{5}}\left(6.3355\text{ kN}\right)-\frac{2}{\sqrt{5}}\left(1.4911\text{ kN}\right)-2\text{ kN=0}\\ F_{DF}-0.79057F_{DE}=-1.1188\text{ kN}\end{array}$ (IX)

Add equations (VIII) and (IX).

$\begin{array}{c}2F_{DF}=-6.7088\text{ kN}\\ F_{DF}=-3.3544\text{ kN}\\ F_{DF}=3.3544\text{ kN, compression}\end{array}$

Subtract equation (IX) from equation (VIII).

$\begin{array}{c}1.58114F_{DE}=-4.4712\text{ kN}\\ F_{DE}=-2.8278\text{ kN}\\ F_{DE}=2.8278\text{ kN, compression}\end{array}$

Consider the joint F. The free-body diagram of the joint F is shown in figure 7.

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

$\to \Sigma F_x=\frac{1}{\sqrt{2}}{F}_{FG}+\frac{2}{\sqrt{5}}\left(3.3544\text{ kN}\right)$

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

Put the above equation in equation (IV).

$\begin{array}{c}\frac{1}{\sqrt{2}}{F}_{FG}+\frac{2}{\sqrt{5}}\left(3.3544\text{ kN}\right)=0\\ F_{FG}=-4.243\text{ kN}\\ F_{FG}=4.24\text{ kN, compression}\end{array}$

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

$\uparrow \Sigma F_y=-F_{EF}-1.75\text{ kN}+\frac{1}{\sqrt{5}}\left(3.3544\text{ kN}\right)-\frac{1}{\sqrt{2}}\left(-4.243\text{ kN}\right)$

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

Put the above equation in equation (VI).

$\begin{array}{c}-F_{EF}-1.75\text{ kN}+\frac{1}{\sqrt{5}}\left(3.3544\text{ kN}\right)-\frac{1}{\sqrt{2}}\left(-4.243\text{ kN}\right)=0\\ F_{EF}=2.75\text{ kN, tension}\end{array}$

Consider the joint G. The free-body diagram of the joint G is shown in figure 8.

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

$\to \Sigma F_x=\frac{1}{\sqrt{2}}{F}_{GH}-\frac{1}{\sqrt{2}}{F}_{EG}+\frac{1}{\sqrt{2}}\left(4.243\text{ kN}\right)$

Here, $F_{GH}$ is the force exerted by $GH$ and $F_{EG}$ is the force exerted by $EG$.

Put the above equation in equation (IV).

$\begin{array}{c}\frac{1}{\sqrt{2}}{F}_{GH}-\frac{1}{\sqrt{2}}{F}_{EG}+\frac{1}{\sqrt{2}}\left(4.243\text{ kN}\right)=0\\ F_{GH}-F_{EG}=-4.243\text{ kN}\end{array}$ (X)

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

$\uparrow \Sigma F_y=-\frac{1}{\sqrt{2}}{F}_{GH}-\frac{1}{\sqrt{2}}{F}_{EG}+\frac{1}{\sqrt{2}}\left(4.243\text{ kN}\right)-1.5\text{ kN}$

Put the above equation in equation (VI).

$\begin{array}{c}-\frac{1}{\sqrt{2}}{F}_{GH}-\frac{1}{\sqrt{2}}{F}_{EG}+\frac{1}{\sqrt{2}}\left(4.243\text{ kN}\right)-1.5\text{ kN=0}\\ F_{GH}+F_{EG}=-6.364\text{ kN}\end{array}$ (XI)

Add equations (X) and (XI).

$\begin{array}{c}2F_{GH}=-10.607\text{ kN}\\ F_{GH}=-5.303\text{ kN}\\ F_{GH}=5.30\text{ kN, compression}\end{array}$

Subtract equation (X) from equation (XI).

$\begin{array}{c}2F_{EG}=-2.121\text{ kN}\\ F_{EG}=-1.0605\text{ kN}\\ F_{DE}=1.061\text{ kN, compression}\end{array}$

Consider the joint H. The free-body diagram of the joint H is shown in figure 9.

The joint H is subject to the forces exerted by $EH$ and $GH$ . Use a force triangle to determine $F_{EH}$ . The force triangle is shown in figure 10.

$EH$ pulls on the joint so $EH$ is in tension.

Obtain the magnitudes of the force from proportion.

$\begin{array}{c}\frac{F_{EH}}{1}=\frac{3.75\text{ kN}}{1}\\ F_{EH}=\frac{\left(3.75\text{ kN}\right)}{1}\\ =3.75\text{ kN}\end{array}$

Conclusion:

Thus, the force in member $AB$ is $7.83\text{ kN}$ and the member $AB$ is in compression, the force in member $AC$ is $7.00\text{ kN}$ and the member $AC$ is in tension, the force in member $BD$ is $6.34\text{ kN}$ and the member $BD$ is in compression, the force in member $BC$ is $1.886\text{ kN}$ and the member $BC$ is in compression, the force in member $CD$ is $1.491\text{ kN}$ and the member $CD$ is in tension, the force in member $CE$ is $5.00\text{ kN}$ and the member $CE$ is in tension, the force in member $DF$ is $3.35\text{ kN}$ and the member $DF$ is in compression, the force in member $DE$ is $2.83\text{ kN}$ and the member $BC$ is in compression, the force in member $FG$ is $4.24\text{ kN}$ and the member $FG$ is in compression, the force in member $EF$ is $2.75\text{ kN}$ and the member $EF$ is in tension, the force in member $GH$ is $5.30\text{ kN}$ and the member $GH$ is in compression, the force in member $EG$ is $1.061\text{ kN}$ and the member $EG$ is in compression, the force in member $EH$ is $3.75\text{ kN}$ and the member $EH$ is in tension.