Chapter 4, Problem 16P

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 horizontal reaction at point B by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ B_x\left(6\right)-60\left(9\right)+60\left(6\right)=0\\ 6B_x-540+360=0\\ B_x=30\text{kN}(\to )\end{array}$

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

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ A_y-60=0\\ A_y=60\text{kN}(\uparrow )\end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -A_x+B_x+60=0\\ -A_x+30+60=0\\ A_x=90\text{kN}(\leftarrow )\end{array}$

Find the zero force bars in the truss.

By observation, the zero force bars are HG, GF, IF, KD, and LB.

Pass a section 1-1 as in Figure 1.

Consider the lower portion of the section 1-1.

Show the free-body diagram of the lower portion 1-1 as in Figure 2.

Find the force in the member EF by taking moment about the joint L.

$\begin{array}{l}{\displaystyle \sum M_L=0}\\ 30\left(3\right)+60\left(3\right)-60\left(9\right)+F_{EF}\left(9\right)=0\\ 90+180-540+9F_{EF}=0\\ F_{EF}=30\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint E;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ Y_{KE}+30-60=0\\ Y_{KE}=30\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{KE}}{1}=\frac{Y_{KE}}{1}=\frac{F_{KE}}{\sqrt{2}}\\ \frac{X_{KE}}{1}=\frac{30}{1}=\frac{F_{KE}}{\sqrt{2}}\\ X_{KE}=30\text{kN}\\ F_{KE}=42.43\text{kN}\left(\text{T}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -F_{DE}-X_{KE}+60=0\\ -F_{DE}-30+60=0\\ F_{DE}=30\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint D;

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

Resolve the vertical component of forces.

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

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -F_{CD}+30=0\\ F_{CD}=30\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint F;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ Y_{HF}-30=0\\ Y_{HF}=30\text{kN}\end{array}$

Use the proportion:

$\begin{array}{l}\frac{X_{HF}}{1}=\frac{Y_{HF}}{1}=\frac{F_{HF}}{\sqrt{2}}\\ \frac{X_{HF}}{1}=\frac{30}{1}=\frac{F_{HF}}{\sqrt{2}}\\ X_{HF}=30\text{kN}\\ F_{HF}=42.43\text{kN}\left(\text{T}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -X_{HF}+F_{KF}=0\\ -30+F_{KF}=0\\ F_{KF}=30\text{kN}\left(\text{C}\right)\end{array}$

Consider the joint H;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ Y_{HJ}-30=0\\ Y_{HJ}=30\text{kN}\end{array}$

Use the proportion;

$\begin{array}{l}\frac{X_{HJ}}{1}=\frac{Y_{HJ}}{1}=\frac{F_{HJ}}{\sqrt{2}}\\ \frac{X_{HJ}}{1}=\frac{30}{1}=\frac{F_{HJ}}{\sqrt{2}}\\ X_{HJ}=30\text{kN}\\ F_{HJ}=42.43\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ X_{HJ}+30-F_{IH}=0\\ 30+30-F_{IH}=0\\ F_{IH}=60\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint I;

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

Resolve the vertical component of forces.

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

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -F_{AI}+60=0\\ F_{AI}=60\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint K;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ Y_{CK}-30=0\\ Y_{CK}=30\text{kN}\end{array}$

Use the proportion;

$\begin{array}{l}\frac{X_{CK}}{1}=\frac{Y_{CK}}{1}=\frac{F_{CK}}{\sqrt{2}}\\ \frac{X_{CK}}{1}=\frac{30}{1}=\frac{F_{CK}}{\sqrt{2}}\\ X_{CK}=30\text{kN}\\ F_{CK}=42.43\text{kN}\left(\text{C}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ X_{CK}+30-30-F_{KJ}=0\\ 30-F_{KJ}=0\\ F_{KJ}=30\text{kN}\left(\text{T}\right)\end{array}$

Consider the joint C;

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

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ Y_{CL}-30=0\\ Y_{CL}=30\text{kN}\end{array}$

Use the proportion;

$\begin{array}{l}\frac{X_{CL}}{1}=\frac{Y_{CL}}{1}=\frac{F_{CL}}{\sqrt{2}}\\ \frac{X_{CL}}{1}=\frac{30}{1}=\frac{F_{CL}}{\sqrt{2}}\\ X_{CL}=30\text{kN}\\ F_{CL}=42.43\text{kN}\left(\text{T}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -X_{CL}+F_{BC}-30+30=0\\ -30+F_{BC}=0\\ F_{BC}=30\text{kN}\left(\text{C}\right)\end{array}$

Consider the joint J;

Show the forces acting at the joint J as in Figure 10.

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ Y_{AJ}-30=0\\ Y_{AJ}=30\text{kN}\end{array}$

Use the proportion;

$\begin{array}{l}\frac{X_{AJ}}{1}=\frac{Y_{AJ}}{1}=\frac{F_{AJ}}{\sqrt{2}}\\ \frac{X_{AJ}}{1}=\frac{30}{1}=\frac{F_{AJ}}{\sqrt{2}}\\ X_{AJ}=30\text{kN}\\ F_{AJ}=42.43\text{kN}\left(\text{T}\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -X_{AJ}+F_{LJ}-30+30=0\\ -30+F_{LJ}=0\\ F_{LJ}=30\text{kN}\left(\text{C}\right)\end{array}$

Consider the joint L;

Show the forces acting at the joint L as in Figure 11.

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ F_{AL}-30=0\\ F_{AL}=30\text{kN}\left(\text{T}\right)\end{array}$

Show the forces in the bars of the truss as in Figure 12.