Chapter 6.1, Problem 6.36P

To determine

The force in each of the members of the truss for $\overrightarrow{P}=\left(-2184\text{ N}\right)\widehat{j}$ and $\overrightarrow{Q}=0$ .

Answer to Problem 6.36P

The force in member AC is $676\text{ N}$ and the member AC is in compression, the force in members $AB\text{ and }AD$ is $861\text{ N}$ and these members are in compression, the force in members $BC\text{ and }CD$ is $162.5\text{ N}$ and these members are in tension, the force in member $BD$ is $244\text{ N}$ 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.

$\begin{array}{c}{D}_x=B_x\\ D_y=B_y\end{array}$ (I)

Here, $D_x$ is the $x$ component of the reaction at the point D, $D_y$ is the $y$ component of the reaction at the point D, $B_x$ is the $x$ component of the reaction at the point B and $B_y$ is the $y$ component of the reaction at the point B.

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

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

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

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

$\Sigma {\overrightarrow{F}}_x=2B_x\widehat{i}$

Put the above equation in equation (II).

$\begin{array}{c}2B_x\widehat{i}=0\\ B_x=0\end{array}$

Put equation (I) in the above equation.

$D_x=0$

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

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

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

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

$\Sigma F_z=B_z$

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

Put the above equation in equation (III).

$B_z=0$

Write the equilibrium equations taking the moments about the point C in the $z$ direction.

$\Sigma M_{cz}=0$ (IV)

Here, $\Sigma M_{cz}$ is the sum of the moments about the point C in the $z$ direction.

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

$\Sigma M_{cz}=-2B_y\left(2.8\text{ m}\right)+\left(2184\text{ N}\right)\left(2\text{ m}\right)$

Put the above equation in equation (IV).

$\begin{array}{c}-2B_y\left(2.8\text{ m}\right)+\left(2184\text{ N}\right)\left(2\text{ m}\right)=0\\ B_y=780\text{ N}\end{array}$

Write the expression for the reaction at the point B.

$\overrightarrow{B}=B_y\widehat{j}$

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

Substitute $780\text{ N}$ for $B_y$ in the above equation to find $\overrightarrow{B}$ .

$\overrightarrow{B}=\left(780\text{ N}\right)\widehat{j}$

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

Refer to figure (2) and write the expression for the forces.

${\overrightarrow{F}}_{AB}=F_{AB}\frac{\overrightarrow{AB}}{AB}$ (V)

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

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

$\overrightarrow{AB}=-0.8\widehat{i}-4.8\widehat{j}+2.1\widehat{k}$

Find the magnitude of $\overrightarrow{AB}$ .

$\begin{array}{c}AB=\sqrt{{\left(-0.8\right)}^2+{\left(-4.8\right)}^2+{\left(2.1\right)}^2}\\ =5.30\end{array}$

Substitute $-0.8\widehat{i}-4.8\widehat{j}+2.1\widehat{k}$  for $\overrightarrow{AB}$ and $5.30$ for $AB$ in equation (V) to find the expression for ${\overrightarrow{F}}_{CA}$ .

${\overrightarrow{F}}_{AB}=F_{AB}\frac{\left(-0.8\widehat{i}-4.8\widehat{j}+2.1\widehat{k}\right)}{5.30}$ (VI)

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

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

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

Substitute $2\widehat{i}-4.8\widehat{j}$ for $\overrightarrow{AC}$ and $5.20$ for $AC$ in the above equation to find the expression for ${\overrightarrow{F}}_{AC}$ .

${\overrightarrow{F}}_{AC}=F_{AC}\frac{\left(2\widehat{i}-4.8\widehat{j}\right)}{5.20}$ (VII)

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

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

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

Substitute $0.8\widehat{i}-4.8\widehat{j}-2.1\widehat{k}$ for $\overrightarrow{AD}$ and $5.30$ for $AD$ in the above equation to find the expression for ${\overrightarrow{F}}_{AD}$ .

${\overrightarrow{F}}_{AD}=F_{AD}\frac{\left(0.8\widehat{i}-4.8\widehat{j}-2.1\widehat{k}\right)}{5.30}$ (VIII)

The net force must be equal to zero.

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

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

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

$\Sigma \overrightarrow{F}={\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{AC}+{\overrightarrow{F}}_{AD}-\left(2184\text{ N}\right)\widehat{j}$

Put the above equation in equation (IX).

${\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{AC}+{\overrightarrow{F}}_{AD}-\left(2184\text{ N}\right)\widehat{j}=0$

Put equations (VI), (VII) and (VIII) in the above equation.

$F_{AB}\frac{\left(-0.8\widehat{i}-4.8\widehat{j}+2.1\widehat{k}\right)}{5.30}+F_{AC}\frac{\left(2\widehat{i}-4.8\widehat{j}\right)}{5.20}+F_{AD}\frac{\left(0.8\widehat{i}-4.8\widehat{j}-2.1\widehat{k}\right)}{5.30}-\left(2184\text{ N}\right)\widehat{j}=0$ (X)

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

$-\frac{0.8}{5.30}\left(F_{AB}+F_{AD}\right)+\frac{2}{5.20}{F}_{AC}=0$ (XI)

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

$-\frac{4.8}{5.30}\left(F_{AB}+F_{AD}\right)-\frac{4.8}{5.20}{F}_{AC}-2184\text{ N=0}$ (XII)

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

$\begin{array}{c}\frac{2.1}{5.30}\left(F_{AB}-F_{AD}\right)=0\\ F_{AD}=F_{AB}\end{array}$ (XIII)

Multiply equation (XI) by $-6$ and add equation (XII).

$\begin{array}{c}-\left(\frac{16.8}{5.20}\right)F_{AC}-2184\text{ N}=0\\ F_{AC}=-676\text{ N}\\ F_{AC}=676\text{ N, compression}\end{array}$

Put equation (XIII) in equation (XI).

$\begin{array}{c}-\frac{0.8}{5.30}\left(F_{AB}+F_{AB}\right)+\frac{2}{5.20}{F}_{AC}=0\\ -\left(\frac{0.8}{5.30}\right)2F_{AB}+\frac{2}{5.20}{F}_{AC}=0\end{array}$

Substitute $-676\text{ N}$ for $F_{AC}$ in the above equation.

$\begin{array}{c}-\left(\frac{0.8}{5.30}\right)2F_{AB}+\frac{2}{5.20}\left(-676\text{ N}\right)=0\\ F_{AB}=-861.25\text{ N}\\ F_{AB}=861\text{ N, compression}\end{array}$

Put the above equation in equation (XIII).

$F_{AD}=861\text{ N, compression}$

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.

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

Substitute $861.25\text{ N}$ for $F_{AB}$ , $-0.8\widehat{i}-4.8\widehat{j}+2.1\widehat{k}$ for $\overrightarrow{AB}$ and $5.30$ for $AB$ in the above equation to find the expression for ${\overrightarrow{F}}_{AB}$ .

$\begin{array}{c}{\overrightarrow{F}}_{AB}=\left(861.25\text{ N}\right)\frac{\left(-0.8\widehat{i}-4.8\widehat{j}+2.1\widehat{k}\right)}{5.30}\\ =-\left(130\text{ N}\right)\widehat{i}-\left(780\text{ N}\right)\widehat{j}+\left(341.25\text{ N}\right)\widehat{k}\end{array}$ (XIV)

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

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

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

Substitute $2.8\widehat{i}-2.1\widehat{k}$ for $\overrightarrow{BC}$ and $3.5$ for $BC$ in the above equation to find the expression for ${\overrightarrow{F}}_{BC}$ .

$\begin{array}{c}{\overrightarrow{F}}_{BC}=F_{BC}\frac{\left(2.8\widehat{i}-2.1\widehat{k}\right)}{3.5}\\ =F_{BC}\left(0.8\widehat{i}-0.6\widehat{k}\right)\end{array}$ (XV)

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

${\overrightarrow{F}}_{BD}=-F_{BD}\widehat{k}$ (XVI)

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

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

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

Put the above equation in equation (IX).

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

Put equations (XIV), (XV) and (XVI) in the above equation.

$-\left(130\text{ N}\right)\widehat{i}-\left(780\text{ N}\right)\widehat{j}+\left(341.25\text{ N}\right)\widehat{k}+F_{BC}\left(0.8\widehat{i}-0.6\widehat{k}\right)-F_{BD}\widehat{k}+\overrightarrow{B}=0$

Substitute $\left(780\text{ N}\right)\widehat{j}$ for $\overrightarrow{B}$ in the above equation.

$-\left(130\text{ N}\right)\widehat{i}-\left(780\text{ N}\right)\widehat{j}+\left(341.25\text{ N}\right)\widehat{k}+F_{BC}\left(0.8\widehat{i}-0.6\widehat{k}\right)-F_{BD}\widehat{k}+\left(780\text{ N}\right)\widehat{j}=0$ (XVII)

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

$\begin{array}{c}-130\text{ N}+0.8F_{BC}=0\\ F_{BC}=162.5\text{ N, tension}\end{array}$

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

$\begin{array}{c}341.25\text{ N}-0.6F_{BC}-F_{BD}=0\\ F_{BD}=341.25\text{ N}-0.6F_{BC}\end{array}$

Substitute $162.5\text{ N}$ for $F_{BC}$ in the above equation.

$\begin{array}{c}{F}_{BD}=341.25\text{ N}-0.6\left(162.5\text{ N}\right)\\ =244\text{ N}\end{array}$

From symmetry,

$F_{CD}=F_{BC}$

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

Substitute $162.5\text{ N, tension}$ for $F_{BC}$ in the above equation to find $F_{CD}$ .

$F_{CD}=162.5\text{ N, tension}$

Conclusion:

Thus, the force in member AC is $676\text{ N}$ and the member AC is in compression, the force in members $AB\text{ and }AD$ is $861\text{ N}$ and these members are in compression, the force in members $BC\text{ and }CD$ is $162.5\text{ N}$ and these members are in tension, the force in member $BD$ is $244\text{ N}$ and it is in tension.