Chapter 6.1, Problem 6.38P

The truss shown consists of nine members and is supported by a ball and socket at A, two short links at B, and a short link at C. Determine the force in each of the members for the given loading.

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.