Chapter 4, Problem 22P

To determine

Find the forces in the members of the truss by the method of joints.

Answer to Problem 22P

The forces in the members of the truss are

$\underset{\_}{F_{AE}=81.87\text{k}\left(\text{C}\right)}$, $\underset{\_}{F_{AC}=102.18\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{EG}=89.49\text{k}\left(\text{C}\right)}$, $\underset{\_}{F_{EC}=56.57\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{CG}=73.89\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{CD}=60\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{BF}=70.47\text{k}\left(\text{C}\right)}$, $\underset{\_}{F_{BD}=39.27\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{FG}=62.83\text{k}\left(\text{C}\right)}$, $\underset{\_}{F_{FD}=56.57\text{k}\left(\text{T}\right)}$, and $\underset{\_}{F_{DG}=10.94\text{k}\left(\text{T}\right)}$.

Explanation of Solution

Given information:

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as negative and the counter clockwise moment as positive.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates Tension (T).

Calculation:

Consider the forces in the member AE, EG, AC, CG, BF, FG, BD, DG, CD, CE, and DF are $F_{AE},F_{EG},F_{AC},F_{CG},F_{BF},F_{FG},F_{BD},F_{DG},F_{CD},F_{CE},F_{DF}$.

Show the free body diagram of the truss as shown in Figure 1.

Refer Figure 1.

Consider the horizontal and vertical reactions at A are $A_x$ and $A_y$.

Consider the vertical reaction at B is $B_y$.

Calculate the value of the angle $\theta$ as follows:

$\begin{array}{l}\tan \theta =\frac{25}{25}\\ \theta ={\tan }^{-1}\left(1\right)\\ \theta =45°\end{array}$

Calculate the value of the angle $\alpha$ as follows:

$\begin{array}{l}\tan \beta =\frac{10}{10}\\ \beta ={\tan }^{-1}\left(1\right)\\ \beta =45°\end{array}$

Calculate the value of the angle $\alpha$ as follows:

$\begin{array}{l}\tan \alpha =\frac{35}{15}\\ \alpha ={\tan }^{-1}\left(\frac{35}{15}\right)\\ \alpha =66.80°\end{array}$

Calculate the value of the angle $\gamma$ as follows:

$\begin{array}{l}\tan \gamma =\frac{15}{35}\\ \gamma ={\tan }^{-1}\left(\frac{15}{35}\right)\\ \gamma =23.198°\end{array}$

Take the sum of the forces in the vertical direction as zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+B_y=40\end{array}$        (1)

Take the sum of the forces in the horizontal direction as zero.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x+10+20+10=0\\ A_x=-40\text{k}\end{array}$

Take the sum of the moments at A is zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ \left(B_y\times 100\right)-\left(10\times 35\right)-\left(20\times 25\right)-\left(20\times 50\right)-\left(20\times 75\right)-\left(10\times 35\right)=0\\ B_y\times 100=3700\\ B_y=37\text{k}\end{array}$

Substitute $37\text{k}$ for $B_y$ in Equation (1).

$\begin{array}{l}{A}_y+37=40\\ A_y=3\text{k}\end{array}$

Show the joint A as shown in Figure 2.

Refer Figure 2.

Find the forces in the member AC and AE as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y+F_{AE}\sin \alpha +F_{AC}\sin \theta =0\\ 3+F_{AE}\sin \left(66.80°\right)+F_{AC}\sin \left(45°\right)=0\\ 3+0.9191F_{AE}+0.7071F_{AC}=0\\ 0.9191F_{AE}+0.7071F_{AC}=-3\end{array}$        (2)

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ A_x+F_{AE}\cos \alpha +F_{AC}\cos \theta =0\\ -40+F_{AE}\cos \left(66.80°\right)+F_{AC}\cos \left(45°\right)=0\\ -40+0.3939F_{AE}+0.7071F_{AC}=0\\ 0.3939F_{AE}+0.7071F_{AC}=40\end{array}$        (3)

Subtract Equation (2) from Equation (3).

$\begin{array}{l}-0.5252F_{AE}=43\\ F_{AE}=-\frac{43}{0.5252}\\ F_{AE}=81.87\text{k}\left(\text{C}\right)\end{array}$

Substitute $-81.87\text{k}$ for $F_{AE}$ in Equation (3).

$\begin{array}{l}0.3939\times \left(-81.87\right)+0.7071F_{AC}=40\\ 0.7071F_{AC}=40+0.3939\times \left(81.87\right)\\ F_{AC}=\frac{40+0.3939\times \left(81.87\right)}{0.7071}\\ F_{AC}=102.18\text{k}\left(\text{T}\right)\end{array}$

Show the joint E as shown in Figure 3.

Refer Figure 3.

Find the forces in the member EC and EG as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{EA}\sin \alpha -F_{EC}\sin \beta +F_{EG}\sin \gamma =0\\ -\left(-81.87\right)\sin 66.80°-F_{EC}\sin 45°+F_{EG}\sin 23.198°=0\\ 75.249-0.7071F_{EC}+0.3939F_{EG}=0\\ 0.3939F_{EG}-0.7071F_{EC}=-75.249\end{array}$        (4)

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ 10-F_{AE}\cos \alpha +F_{EC}\cos \beta +F_{EG}\cos \gamma =0\\ 10-\left(-81.87\right)\cos 66.80°+F_{EC}\cos 45°+F_{EG}\cos 23.198°=0\\ 10+32.252+0.7071F_{EC}+0.9191F_{EG}=0\\ 0.9191F_{EG}+0.7071F_{EC}=-42.252\end{array}$        (5)

Solve Equation (4) and (5).

$\begin{array}{l}{F}_{EG}=-89.49\text{k}\\ F_{EG}=89.49\text{k}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{F}_{EC}=56.57\text{k}\\ F_{EC}=56.57\text{k}\left(\text{T}\right)\end{array}$

Show the joint C as shown in Figure 4.

Refer Figure 4.

Find the forces in the member CG and CD as follows:

For Equilibrium of forces,

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{EC}\sin \beta +F_{CG}\sin \theta -F_{CA}\sin \theta -20=0\\ F_{EC}\sin 45°+F_{CG}\sin 45°-F_{CA}\sin 45°-20=0\\ 56.57\times 0.7071+0.7071F_{CG}-0.7071F_{CA}-20=0\end{array}$

$\begin{array}{l}20+0.7071F_{CG}-0.7071F_{CA}=0\\ F_{CG}=\frac{-20+0.7071F_{CA}}{0.7071}\\ F_{CG}=\frac{-20+0.7071\times \left(102.18\right)}{0.7071}\\ F_{CG}=73.89\text{k}\left(\text{T}\right)\end{array}$        (6)

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{EC}\cos \beta +F_{CG}\cos \theta -F_{CA}\cos \theta =0\\ -F_{EC}\cos 45°+F_{CG}\cos 45°-F_{CA}\cos 45°+F_{CD}=0\\ -0.7071F_{EC}+0.7071F_{CG}-0.7071F_{CA}+F_{CD}=0\end{array}$

$\begin{array}{l}-0.7071\times \left(56.57\right)+0.7071\times \left(73.89\right)-102.18+F_{CD}=0\\ -60+F_{CD}=0\\ F_{CD}=60\text{k}\left(\text{T}\right)\end{array}$        (7)

Show the joint B as shown in Figure 5.

Refer Figure 5.

Find the forces in the member BF and BD as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{BF}\cos \alpha -F_{BD}\cos \theta =0\\ -F_{BF}\cos 66.80°-F_{BD}\cos 45°=0\\ 0.394F_{BF}+0.707F_{BD}=0\end{array}$        (8)

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{BF}\sin \alpha +F_{BD}\sin \theta +B_y=0\\ F_{BF}\sin 66.80°+F_{BD}\sin 45°+37=0\\ 0.919F_{BF}+0.707F_{BD}=-37\end{array}$        (9)

Solve Equation (8) and (9).

$\begin{array}{l}{F}_{BF}=70.47\text{k}\left(\text{C}\right)\\ F_{BD}=39.27\text{k}\left(\text{T}\right)\end{array}$

Joint F

Show the joint F as shown in Figure 6.

Refer Figure 6.

Find the forces in the member EC and EG as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{FG}\cos \gamma -F_{FD}\cos \beta +10+F_{FB}\cos \alpha =0\\ -F_{FG}\cos 23.198°-F_{FD}\cos 45°+10+F_{FB}\cos 66.80°=0\\ -0.919F_{FG}-0.707F_{FD}+10+0.3939\left(-70.47\right)=0\\ -0.919F_{FG}-0.707F_{FD}=17.758\end{array}$        (10)

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{FG}\sin \gamma -F_{FD}\sin \beta -F_{FB}\sin \alpha =0\\ F_{FG}\sin 23.198°-F_{FD}\sin 45°-F_{FB}\sin 66.80°=0\\ 0.3939F_{FG}-0.707F_{FD}-\left(-70.47\right)\times 0.919=0\\ 0.3939F_{FG}-0.707F_{FD}=-64.761\end{array}$        (11)

Solve Equation (10) and (11).

$\begin{array}{l}{F}_{FG}=62.83\text{k}\left(\text{C}\right)\\ F_{FD}=56.57\text{k}\left(\text{T}\right)\end{array}$

Joint D:

Show the joint D as shown in Figure 7.

Refer Figure 7.

Find the forces in the member DG as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{DC}-F_{DG}\cos \theta +F_{DF}\cos \beta +F_{DB}\cos \theta =0\\ -F_{DC}-F_{DG}\cos 45°+F_{DF}\cos 45°+F_{DB}\cos 45°=0\\ -60-F_{DG}\cos 45°+56.57\cos 45°+39.24\cos 45°=0\\ F_{DG}=\frac{7.737}{\cos 45°}\\ F_{DG}=10.94\text{k}\left(\text{T}\right)\end{array}$

Thus, the forces in the members of the truss are

$\underset{\_}{F_{AE}=81.87\text{k}\left(\text{C}\right)}$, $\underset{\_}{F_{AC}=102.18\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{EG}=89.49\text{k}\left(\text{C}\right)}$, $\underset{\_}{F_{EC}=56.57\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{CG}=73.89\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{CD}=60\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{BF}=70.47\text{k}\left(\text{C}\right)}$, $\underset{\_}{F_{BD}=39.27\text{k}\left(\text{T}\right)}$, $\underset{\_}{F_{FG}=62.83\text{k}\left(\text{C}\right)}$, $\underset{\_}{F_{FD}=56.57\text{k}\left(\text{T}\right)}$, and $\underset{\_}{F_{DG}=10.94\text{k}\left(\text{T}\right)}$.