Chapter 4, Problem 43P

To determine

Find the force in each bar and mention the force is tension or compression of the bars in the truss.

Explanation of Solution

Assumptions:

• Consider the state of bars as tension (T) where the force is pulling the bar and as compression (C) where the force is pushing the bar.
• The sign of the force in the bar is positive when the force is in tension and negative when the force is in compression.
• Consider the force indicating right side as positive and left side as negative in horizontal components of forces.
• Consider the force indicating upward is taken as positive and downward as negative in vertical components of forces.
• Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable.

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

Find the vertical reaction at point E by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ E_y\left(20\right)-8\left(15\right)-16\left(10\right)-16\left(5\right)=0\\ 20E_y-120-160-80=0\\ E_y=18\text{kN}(\uparrow )\end{array}$

Find the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ A_y-16-16-8+E_y=0\\ A_y-40+18=0\\ A_y=22\text{kN}(\uparrow )\end{array}$

Find the horizontal reaction at point E by resolving the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ E_x=0\end{array}$

Consider the joint E;

Show the forces acting at the joint E as in Figure 2.

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ 18-Y_{IE}=0\\ Y_{IE}=18\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{IE}}{5}=\frac{Y_{IE}}{2}=\frac{F_{IE}}{\sqrt{29}}\\ \frac{X_{IE}}{5}=\frac{18}{2}=\frac{F_{IE}}{\sqrt{29}}\\ X_{IE}=45\text{kN}\\ F_{IE}=48.47\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -F_{DE}+X_{IE}=0\\ -F_{DE}+45=0\\ F_{DE}=45\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint D;

Show the forces acting at the joint D as in Figure 3.

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ F_{ID}=0\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -F_{CD}+45=0\\ F_{CD}=45\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint A;

Show the forces acting at the joint A as in Figure 4.

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ 22-Y_{AF}=0\\ Y_{AF}=22\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{AF}}{5}=\frac{Y_{AF}}{2}=\frac{F_{AF}}{\sqrt{29}}\\ \frac{X_{AF}}{5}=\frac{22}{2}=\frac{F_{AF}}{\sqrt{29}}\\ X_{AF}=55\text{kN}\\ F_{AF}=59.24\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ F_{AB}-X_{AF}=0\\ F_{AB}-55=0\\ F_{AB}=55\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint B;

Show the forces acting at the joint B as in Figure 5.

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ F_{FB}=0\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -55+F_{BC}=0\\ F_{BC}=55\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint F;

Show the forces acting at the joint F as in Figure 6.

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ 22+Y_{FC}-16-Y_{FG}=0\\ Y_{FG}-Y_{FC}=6\end{array}$        (1)

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ 55-X_{FG}-X_{FC}=0\\ X_{FG}+X_{FC}=55\end{array}$        (2)

Use the proportion:

$\begin{array}{l}\frac{X_{FG}}{5}=\frac{Y_{FG}}{2}=\frac{F_{FG}}{\sqrt{29}}\\ X_{FG}=0.928F_{FG}\\ Y_{FG}=0.371F_{FG}\end{array}$

$\begin{array}{l}\frac{X_{FC}}{5}=\frac{Y_{FC}}{2}=\frac{F_{FC}}{\sqrt{29}}\\ X_{FC}=0.928F_{FC}\\ Y_{FC}=0.371F_{FC}\end{array}$

Substitute $0.371F_{FG}$ for $Y_{FG}$ and $0.371F_{FC}$ for $Y_{FC}$ in Equation (1).

$0.371F_{FG}-0.371F_{FC}=6$        (3)

Substitute $0.928F_{FG}$ for $X_{FG}$ and $0.928F_{FC}$ for $X_{FC}$ in Equation (2).

$0.928F_{FG}+0.928F_{FC}=55$        (4)

Solve equation (3) and (4).

$\begin{array}{l}{F}_{FG}=37.70\text{kN}\left(\text{C}\right)\\ F_{FC}=21.54\text{kN}\left(\text{C}\right)\end{array}$

Consider the joint C;

Show the forces acting at the joint C as in Figure 7.

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -55+45+20-X_{CI}=0\\ X_{CI}=10\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{CI}}{5}=\frac{Y_{CI}}{2}=\frac{F_{CI}}{\sqrt{29}}\\ \frac{10}{5}=\frac{Y_{CI}}{2}=\frac{F_{CI}}{\sqrt{29}}\\ Y_{CI}=4\text{kN}\\ F_{CI}=10.77\text{kN}\left(\text{C}\right)\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -8-Y_{CI}+F_{CG}=0\\ -8-4+F_{CG}=0\\ F_{CG}=12\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint H;

Show the forces acting at the joint H as in Figure 8.

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ X_{GH}=0\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{GH}}{5}=\frac{Y_{GH}}{2}=\frac{F_{GH}}{\sqrt{29}}\\ \frac{0}{5}=\frac{Y_{GH}}{2}=\frac{F_{GH}}{\sqrt{29}}\\ Y_{GH}=0\\ F_{GH}=0\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ Y_{GH}+F_{IH}-8=0\\ 0+F_{IH}-8=0\\ F_{IH}=8\text{kN}\left(\text{C}\right)\end{array}$

Consider the joint G;

Show the forces acting at the joint G as in Figure 9.

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ 35-X_{GI}=0\\ X_{GI}=35\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{GI}}{5}=\frac{Y_{GI}}{2}=\frac{F_{GI}}{\sqrt{29}}\\ \frac{35}{5}=\frac{Y_{GI}}{2}=\frac{F_{GI}}{\sqrt{29}}\\ F_{GI}=37.7\text{kN}\left(\text{C}\right)\end{array}$

Show the forces in the bars of the truss as in Figure 10.