Chapter 6.1, Problem 6.38P

To determine

The force in each of the members of the truss for the given loading.

Answer to Problem 6.38P

The force in member $AD$  is $2500\text{ lb}$ and the member $AD$ is in tension, the force in members $AC\text{ and }AB$ is $1061\text{ lb}$ and these members are in compression, the force in members $BE,CD\text{ and }BD$ is $1250\text{ lb}$ and these members are in compression, the force in member $BC$ is $2100\text{ lb}$ and it is in tension, the force in member $DE$ is $1500\text{ lb}$ and it is in tension.

Explanation of Solution

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

Refer to figure 1 and use symmetry.

$A_z=B_z=0$

Here, $A_z$ is the $z$ component of the reaction at the point A, $B_z$ is the $z$ component of the reaction at the point B.

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

$\begin{array}{c}\Sigma {\overrightarrow{F}}_x=0\\ \Rightarrow A_x=0\end{array}$

Here, $\Sigma {\overrightarrow{F}}_x$ is the $x$ component of the net force and $A_x$ is the $x$ component of the reaction at the point A.

Sum of the moments must be equal to zero.

$\Sigma M_{BC}=0$ (I)

Here, $\Sigma M_{BC}$ is the sum of the moments of force.

Write the equation for $\Sigma M_{BC}$ .

$\Sigma M_{BC}=A_y\left(6\text{ ft}\right)+\left(1600\text{ lb}\right)\left(7.5\text{ ft}\right)$

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

Put the above equation in equation (I).

$\begin{array}{c}{A}_y\left(6\text{ ft}\right)+\left(1600\text{ lb}\right)\left(7.5\text{ ft}\right)=0\\ A_y=-2000\text{ lb}\end{array}$

Write the expression for the reaction at the point A.

$\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}+A_z\widehat{k}$

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

Substitute $0$ for $A_x,A_y$ and $-2000\text{ lb}$ for $A_y$ in the above equation to find $\overrightarrow{A}$ .

$\begin{array}{c}\overrightarrow{A}=0+\left(-2000\text{ lb}\right)\widehat{j}+0\\ =-\left(2000\text{ lb}\right)\widehat{j}\end{array}$

Use symmetry.

$B_y=C$

Here, $B_y$ is the $y$ component of the reaction at the point B and $C$ is the reaction at the point C.

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

$\Sigma {\overrightarrow{F}}_y=0$ (II)

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

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

$\Sigma {\overrightarrow{F}}_y=2B_y-2000\text{ lb}-1600\text{ lb}$

Put the above equation in equation (II).

$\begin{array}{c}2B_y-2000\text{ lb}-1600\text{ lb}=0\\ B_y=1800\text{ lb}\end{array}$

Write the expression for the reaction at the point A.

$\overrightarrow{B}=B_y\widehat{j}+B_z\widehat{k}$

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

Substitute $0$ for $B_z$ and $1800\text{ lb}$ for $B_y$ in the above equation to find $\overrightarrow{B}$ .

$\begin{array}{c}\overrightarrow{B}=\left(1800\text{ lb}\right)\widehat{j}+0\widehat{k}\\ =\left(1800\text{ lb}\right)\widehat{j}\end{array}$

Consider the free body $A$ . The diagram is shown in figure 2.

The net force must be equal to zero.

$\Sigma \overrightarrow{F}=0$ (III)

Here, $\Sigma \overrightarrow{F}$ is the net force.

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

$\Sigma \overrightarrow{F}={\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{AC}+{\overrightarrow{F}}_{AD}-\left(2000\text{ lb}\right)\widehat{j}$

Put the above equation in equation (III).

${\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{AC}+{\overrightarrow{F}}_{AD}-\left(2000\text{ lb}\right)\widehat{j}=0$ (IV)

Here, ${\overrightarrow{F}}_{AB}$ is the force exerted by the member $AB$ , ${\overrightarrow{F}}_{AC}$ is the force exerted by member $AC$ and ${\overrightarrow{F}}_{AD}$ is the force exerted by $AD$ .

Write the expression for ${\overrightarrow{F}}_{AB}$ .

${\overrightarrow{F}}_{AB}=F_{AB}\frac{\widehat{i}+\widehat{k}}{\sqrt{2}}$ (V)

Here, $F_{AB}$ is the magnitude of ${\overrightarrow{F}}_{AB}$ .

Write the expression for ${\overrightarrow{F}}_{AC}$ .

${\overrightarrow{F}}_{AC}=F_{AC}\frac{\widehat{i}-\widehat{k}}{\sqrt{2}}$ (VI)

Here, $F_{AC}$ is the magnitude of ${\overrightarrow{F}}_{AC}$ .

Write the expression for ${\overrightarrow{F}}_{AD}$ .

${\overrightarrow{F}}_{AD}=F_{AD}\left(0.6\widehat{i}+0.8\widehat{j}\right)$ (VII)

Here, $F_{AD}$ is the magnitude of ${\overrightarrow{F}}_{AD}$ .

Put equations (V), (VI) and (VII) in equation (IV).

$F_{AB}\frac{\widehat{i}+\widehat{k}}{\sqrt{2}}+F_{AC}\frac{\widehat{i}-\widehat{k}}{\sqrt{2}}+F_{AD}\left(0.6\widehat{i}+0.8\widehat{j}\right)-\left(2000\text{ lb}\right)\widehat{j}=0$ (VIII)

Factorize $\widehat{i},\widehat{j},\widehat{k}$ and equate their coefficient to zero.

Equate the coefficient of $\widehat{i}$ to zero.

$\frac{1}{\sqrt{2}}{F}_{AB}+\frac{1}{\sqrt{2}}{F}_{AC}+0.6F_{AD}=0$ (IX)

Equate the coefficient of $\widehat{j}$ to zero.

$\begin{array}{c}0.8F_{AD}-2000\text{ lb}=0\\ F_{AD}=2500\text{ lb, tension}\end{array}$

Equate the coefficient of $\widehat{k}$ to zero.

$\begin{array}{c}\frac{1}{\sqrt{2}}{F}_{AB}-\frac{1}{\sqrt{2}}{F}_{AC}=0\\ F_{AC}=F_{AB}\end{array}$ (X)

Put equation (X) in equation (IX).

$\begin{array}{c}\frac{1}{\sqrt{2}}{F}_{AB}+\frac{1}{\sqrt{2}}{F}_{AB}+0.6F_{AD}=0\\ \frac{2}{\sqrt{2}}{F}_{AB}+0.6F_{AD}=0\end{array}$

Substitute $2500\text{ lb}$ for $F_{AD}$ in the above equation.

$\begin{array}{c}\frac{2}{\sqrt{2}}{F}_{AB}+0.6\left(2500\text{ lb}\right)=0\\ F_{AB}=-1060.7\text{ lb}\\ F_{AB}=1061\text{ lb, compression}\end{array}$

Put the above equation in equation (X).

$F_{AC}=1061\text{ lb, compression}$

Consider the free-body joint B. The free-body diagram of joint B is shown in figure 3.

Refer to figure (3) and write the expression for the forces.

${\overrightarrow{F}}_{BA}=F_{AB}\frac{\overrightarrow{BA}}{BA}$

Here, ${\overrightarrow{F}}_{BA}$ is the force exerted by member BA, $BA$ is the magnitude of the vector $\overrightarrow{BA}$ .

Substitute $1060.7\text{ lb}$ for $F_{AB}$ , $\widehat{i}+\widehat{k}$ for $\overrightarrow{BA}$ and $\sqrt{2}$ for $BA$ in the above equation to find the expression for ${\overrightarrow{F}}_{BA}$ .

$\begin{array}{c}{\overrightarrow{F}}_{BA}=\left(1060.7\text{ lb}\right)\frac{\widehat{i}+\widehat{k}}{\sqrt{2}}\\ =\left(750\text{ lb}\right)\left(\widehat{i}+\widehat{k}\right)\end{array}$ (XI)

Write the expression for ${\overrightarrow{F}}_{BC}$ .

${\overrightarrow{F}}_{BC}=-F_{BC}\widehat{k}$ (XII)

Here, ${\overrightarrow{F}}_{BC}$ is the force exerted by member BC.

Write the expression for ${\overrightarrow{F}}_{BD}$ .

${\overrightarrow{F}}_{BD}=F_{BD}\left(0.8\widehat{j}-0.6\widehat{k}\right)$ (XIII)

Here, ${\overrightarrow{F}}_{BD}$ is the force exerted by member BD.

Write the expression for ${\overrightarrow{F}}_{BE}$ .

${\overrightarrow{F}}_{BE}=F_{BE}\frac{\overrightarrow{BE}}{BE}$

Here, ${\overrightarrow{F}}_{BE}$ is the force exerted by member BE, $BE$ is the magnitude of the vector $\overrightarrow{BE}$ .

Substitute $7.5\widehat{i}+8\widehat{j}+6\widehat{k}$ for $\overrightarrow{BE}$ and $12.5$ for $BE$ in the above equation to find the expression for ${\overrightarrow{F}}_{BE}$ .

${\overrightarrow{F}}_{BE}=F_{BE}\frac{\left(7.5\widehat{i}+8\widehat{j}+6\widehat{k}\right)}{12.5}$ (XIV)

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

$\Sigma \overrightarrow{F}={\overrightarrow{F}}_{BA}+{\overrightarrow{F}}_{BC}+{\overrightarrow{F}}_{BD}+{\overrightarrow{F}}_{BE}+\overrightarrow{B}$

Put the above equation in equation (III).

${\overrightarrow{F}}_{BA}+{\overrightarrow{F}}_{BC}+{\overrightarrow{F}}_{BD}+{\overrightarrow{F}}_{BE}+\overrightarrow{B}=0$

Put equations (XI), (XII), (XIII) , (XIV) and substitute $\left(1800\text{ lb}\right)\widehat{j}$ for $\overrightarrow{B}$ in the above equation.

$\left(750\text{ lb}\right)\left(\widehat{i}+\widehat{k}\right)-F_{BC}\widehat{k}+F_{BD}\left(0.8\widehat{j}-0.6\widehat{k}\right)+F_{BE}\frac{\left(7.5\widehat{i}+8\widehat{j}+6\widehat{k}\right)}{12.5}+\left(1800\text{ lb}\right)\widehat{j}=0$ (XV)

Equate the coefficient of $\widehat{i}$ in equation (XV) to zero.

$\begin{array}{c}750\text{ lb}+\left(\frac{7.5}{12.5}\right)F_{BE}=0\\ F_{BE}=-1250\text{ lb}\\ =1250\text{ lb},\text{ compression}\end{array}$

Equate the coefficient of $\widehat{j}$ in equation (XV) to zero.

$0.8F_{BD}+\left(\frac{8}{12.5}\right)F_{BE}+\left(1800\text{ lb}\right)=0$

Substitute $-1250\text{ lb}$ for $F_{BE}$ in the above equation.

$\begin{array}{c}0.8F_{BD}+\left(\frac{8}{12.5}\right)\left(-1250\text{ lb}\right)+\left(1800\text{ lb}\right)=0\\ F_{BD}=-1250\text{ lb}\\ =1250\text{ lb},\text{ compression}\end{array}$

Equate the coefficient of $\widehat{k}$ in equation (XV) to zero.

$750\text{ lb}-F_{BC}-0.6\left(F_{BD}\right)-\frac{6}{12.5}\left(F_{BE}\right)=0$

Substitute $-1250\text{ lb}$ for $F_{BD}$ and $F_{BE}$ in the above equation.

$\begin{array}{c}750\text{ lb}-F_{BC}-0.6\left(-1250\text{ lb}\right)-\frac{6}{12.5}\left(-1250\text{ lb}\right)=0\\ F_{BC}=2100\text{ lb, tension}\end{array}$

Use symmetry.

$F_{BD}=F_{CD}$

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

Substitute $1250\text{ lb, compression}$ for $F_{BD}$ in the above equation.

$F_{CD}=1250\text{ lb, compression}$

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

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

$\Sigma \overrightarrow{F}={\overrightarrow{F}}_{DA}+{\overrightarrow{F}}_{DB}+{\overrightarrow{F}}_{DC}+{\overrightarrow{F}}_{DE}\widehat{i}$

Put the above equation in equation (III).

${\overrightarrow{F}}_{DA}+{\overrightarrow{F}}_{DB}+{\overrightarrow{F}}_{DC}+{\overrightarrow{F}}_{DE}\widehat{i}=0$

Only ${\overrightarrow{F}}_{DA}$ and ${\overrightarrow{F}}_{DE}$ in the above equation contains $\widehat{i}$ .

Equate the coefficient of $\widehat{i}$ in the above equation to zero.

$\begin{array}{c}{F}_{DA}+F_{DE}=0\\ -0.6F_{AD}+F_{DE}=0\end{array}$

Substitute $2500\text{ lb}$ for $F_{AD}$ in the above equation.

$\begin{array}{c}-0.6\left(2500\text{ lb}\right)+F_{DE}=0\\ F_{DE}=1500\text{ lb, tension}\end{array}$

Conclusion:

Thus, the force in member $AD$  is $2500\text{ lb}$ and the member $AD$ is in tension, the force in members $AC\text{ and }AB$ is $1061\text{ lb}$ and these members are in compression, the force in members $BE,CD\text{ and }BD$ is $1250\text{ lb}$ and these members are in compression, the force in member $BC$ is $2100\text{ lb}$ and it is in tension, the force in member $DE$ is $1500\text{ lb}$ and it is in tension.