Chapter 4, Problem 20P

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 G by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ G_y\left(8\right)-100\left(4\right)=0\\ 8G_y-400=0\\ G_y=50\text{kN}(\uparrow )\end{array}$

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

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ A_y-100+G_y=0\\ A_y-100+50=0\\ A_y=50\text{kN}(\uparrow )\end{array}$

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

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

Consider the joint G;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -Y_{FG}+50=0\\ Y_{FG}=50\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{FG}}{1}=\frac{Y_{FG}}{3}=\frac{F_{FG}}{\sqrt{10}}\\ \frac{X_{FG}}{1}=\frac{50}{3}=\frac{F_{FG}}{\sqrt{10}}\\ X_{FG}=16.67\text{kN}\\ F_{FG}=52.70\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -F_{GH}+X_{FG}=0\\ -F_{GH}+16.67=0\\ F_{GH}=16.67\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint A;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ 50-Y_{AB}=0\\ Y_{AB}=50\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{AB}}{1}=\frac{Y_{AB}}{3}=\frac{F_{AB}}{\sqrt{10}}\\ \frac{X_{AB}}{1}=\frac{50}{3}=\frac{F_{AB}}{\sqrt{10}}\\ X_{AB}=16.67\text{kN}\\ F_{AB}=52.70\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -X_{AB}+F_{AH}=0\\ -16.67+F_{AH}=0\\ F_{AH}=16.67\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint H;

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

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ X_{HB}-X_{HF}+16-16=0\\ X_{HB}-X_{HF}=0\end{array}$        (1)

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -Y_{HB}-Y_{HF}=0\end{array}$        (2)

Use the proportion:

$\begin{array}{l}\frac{X_{HB}}{1}=\frac{Y_{HB}}{2}=\frac{F_{HB}}{\sqrt{5}}\\ X_{HB}=0.447F_{HB}\\ Y_{HB}=0.894F_{HB}\end{array}$

$\begin{array}{l}\frac{X_{HF}}{1}=\frac{Y_{HF}}{2}=\frac{F_{HF}}{\sqrt{5}}\\ X_{HF}=0.447F_{HF}\\ Y_{HF}=0.894F_{HF}\end{array}$

Substitute $0.447F_{HB}$ for $X_{HB}$ and $0.447F_{HF}$ for $X_{HF}$ in Equation (1).

$0.447F_{HB}-0.447F_{HF}=0$        (3)

Substitute $0.894F_{HB}$ for $Y_{HB}$ and $0.894F_{HF}$ for $Y_{HF}$ in Equation (2).

$-0.894F_{HB}-0.894F_{HF}=0$        (4)

Solve the Equation (3) and (4);

$\begin{array}{l}{F}_{HB}=0\\ F_{HF}=0\end{array}$

Consider the joint B;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -Y_{BC}+50=0\\ Y_{BC}=50\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{BC}}{1}=\frac{Y_{BC}}{3}=\frac{F_{BC}}{\sqrt{10}}\\ \frac{X_{BC}}{1}=\frac{50}{3}=\frac{F_{BC}}{\sqrt{10}}\\ X_{BC}=16.67\text{kN}\\ F_{BC}=52.70\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ 16.67-X_{BC}+F_{BI}=0\\ 16.67-16.67+F_{BI}=0\\ F_{BI}=0\end{array}$

Consider the joint F;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -Y_{EF}+50=0\\ Y_{EF}=50\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{EF}}{1}=\frac{Y_{EF}}{3}=\frac{F_{EF}}{\sqrt{10}}\\ \frac{X_{EF}}{1}=\frac{50}{3}=\frac{F_{EF}}{\sqrt{10}}\\ X_{EF}=16.67\text{kN}\\ F_{EF}=52.70\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -16.67+X_{EF}-F_{IF}=0\\ -16.67+16.67-F_{IF}=0\\ F_{IF}=0\end{array}$

Consider the joint I;

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

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -X_{CI}-X_{EI}=0\end{array}$        (5)

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ Y_{CI}-Y_{EI}=0\end{array}$        (6)

Use the proportion:

$\begin{array}{l}\frac{X_{CI}}{1}=\frac{Y_{CI}}{2}=\frac{F_{CI}}{\sqrt{5}}\\ X_{CI}=0.447F_{CI}\\ Y_{CI}=0.894F_{CI}\end{array}$

$\begin{array}{l}\frac{X_{EI}}{1}=\frac{Y_{EI}}{2}=\frac{F_{EI}}{\sqrt{5}}\\ X_{EI}=0.447F_{EI}\\ Y_{EI}=0.894F_{EI}\end{array}$

Substitute $0.447F_{CI}$ for $X_{CI}$ and $0.447F_{EI}$ for $X_{EI}$ in Equation (5).

$-0.447F_{CI}-0.447F_{EI}=0$        (7)

Substitute $0.894F_{CI}$ for $Y_{CI}$ and $0.894F_{EI}$ for $Y_{EI}$ in Equation (6).

$-0.894F_{CI}-0.894F_{EI}=0$        (8)

Solve the Equation (7) and (8);

$\begin{array}{l}{F}_{CI}=0\\ F_{EI}=0\end{array}$

Consider the joint C;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -Y_{CD}+50=0\\ Y_{CD}=50\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{CD}}{1}=\frac{Y_{CD}}{3}=\frac{F_{CD}}{\sqrt{10}}\\ \frac{X_{CD}}{1}=\frac{50}{3}=\frac{F_{CD}}{\sqrt{10}}\\ X_{CD}=16.67\text{kN}\\ F_{CD}=52.70\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ 16.67-X_{CD}+F_{CE}=0\\ 16.67-16.67+F_{CE}=0\\ F_{CE}=0\end{array}$

Consider the joint E;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -Y_{DE}+50=0\\ Y_{DE}=50\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{DE}}{1}=\frac{Y_{DE}}{3}=\frac{F_{DE}}{\sqrt{10}}\\ \frac{X_{DE}}{1}=\frac{50}{3}=\frac{F_{DE}}{\sqrt{10}}\\ F_{DE}=52.70\text{kN}\left(\text{C}\right)\end{array}$

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