Chapter 4, Problem 14P

To determine

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

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 AB, BC, CD, DE, EF, FG, HI, IJ, JK, KL, AH, HB, HC, IC, JC, JD, JE, KE, LE, LF, LG, are $F_{AB},F_{BC},F_{CD},F_{DE},F_{EF},F_{FG},F_{HI},F_{IJ},F_{JK},$$F_{KL},F_{AH},F_{HB},F_{HC},F_{IC},F_{JC},F_{JD},F_{JE},F_{KE},F_{LE},F_{LF},F_{LG}$.

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 horizontal reaction at G is $G_y$.

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

$\begin{array}{l}\tan \theta =\frac{6}{8}\\ \theta ={\tan }^{-1}\left(\frac{6}{8}\right)\\ \theta =36.869°\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+G_y=30+30+60+60+60\\ A_y+G_y=240\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x+40=0\\ A_x=-40\text{kN}\end{array}$        (1)

Take the sum of the moment about A as zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ \left[\begin{array}{l}-\left(30\times 8\right)-\left(30\times 16\right)-\left(60\times 24\right)\\ -\left(60\times 32\right)-\left(60\times 40\right)+\left(40\times 6\right)+G_y\times 48\end{array}\right]=0\\ -6,240+G_y\times 48=0\\ G_y=\frac{6,240}{48}\\ G_y=130\text{kN}\end{array}$

Substitute $130\text{kN}$ for $G_y$ in Equation (1).

$\begin{array}{l}{A}_y+130=240\\ A_y=110\text{kN}\end{array}$

Show the joint A as shown in Figure 2.

Refer Figure 2.

Find the force in the member AH as follows:

For the Equilibrium of forces,

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ 110-F_{AH}\sin \left(36.869°\right)=0\\ F_{AH}=\frac{110}{\sin \left(36.869°\right)}\\ F_{AH}=183.3\text{kN}\left(\text{T}\right)\end{array}$

Find the force in the member AB as follows:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{AB}+F_{AH}\cos 36.869°-40=0\\ F_{AB}=-183.3\cos 36.869°+40\\ F_{AB}=106.6\text{kN}\left(\text{C}\right)\end{array}$

Show the joint B as shown in Figure 3.

Refer Figure 3.

Find the force in the member BH as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -30-F_{BH}=0\\ F_{BH}=-30\text{kN}\\ F_{BH}=30\text{kN}\left(\text{C}\right)\end{array}$

Find the force in the member BC as follows:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{BC}=F_{BA}\\ F_{BC}=-106.6\\ F_{BC}=106.6\text{kN}\left(\text{C}\right)\end{array}$

Show the joint H as shown in Figure 4.

Refer Figure 4.

Find the force in the member HC as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{HA}\sin \left(36.869°\right)+F_{HC}\sin \left(36.869°\right)+F_{HB}=0\\ 183.3\sin \left(36.869°\right)+F_{HC}\sin \left(36.869°\right)-30=0\\ F_{HC}=-\frac{80}{\sin \left(36.869°\right)}\\ F_{HC}=133.3\text{kN}\left(\text{C}\right)\end{array}$

Find the force in the member HI as follows:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{HA}\cos \left(36.869°\right)+F_{HC}\cos \left(36.869°\right)+F_{HI}=0\\ -183.3\cos \left(36.869°\right)+\left(-133.3\right)\cos \left(36.869°\right)+F_{HI}=0\\ F_{HI}=253.3\text{kN}\left(\text{T}\right)\end{array}$

Show the joint I as shown in Figure 5.

Refer Figure 5.

Find the force in the member IC as follows:

Take the sum of the forces in the vertical direction as follows:

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{IC}=0\text{kN}\end{array}$

Find the force in the member IJ as follows:

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

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{IJ}=F_{IH}\\ F_{IJ}=253.3\text{kN}\left(\text{T}\right)\end{array}$

Show the joint C as shown in Figure 6.

Refer Figure 6.

Find the force in the member CJ as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{CH}\sin \left(36.869°\right)-F_{CJ}\sin \left(36.869°\right)-F_{CI}-30=0\\ -\left(-133.3\right)\sin \left(36.869°\right)-F_{CJ}\sin \left(36.869°\right)-0-30=0\\ F_{CJ}=\frac{50}{\sin \left(36.869°\right)}\\ F_{CJ}=83.3\text{kN}\left(\text{T}\right)\end{array}$

Find the force in the member CD as follows:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{CB}+F_{CD}-F_{CH}\cos \left(36.869°\right)+F_{CJ}\cos \left(36.869°\right)=0\\ -\left(-106.6\right)+F_{CD}-\left(-133.3\right)\cos \left(36.869°\right)+83.3\cos \left(36.869°\right)=0\\ 280+F_{CD}=0\\ F_{CD}=280\text{k}\left(\text{C}\right)\end{array}$

Show the joint D as shown in Figure 7.

Refer Figure 7.

Find the force in the member DJ as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -60-F_{DJ}=0\\ F_{DJ}=60\text{kN}\left(\text{C}\right)\end{array}$

Find the force in the member DE as follows:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{DE}=F_{CD}\\ F_{DE}=-280\\ F_{DE}=280\text{kN}\left(\text{C}\right)\end{array}$

Show the joint J as shown in Figure 8.

Refer Figure 8.

Find the force in the member JE as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{JC}\sin \left(36.869°\right)+F_{JE}\sin \left(36.869°\right)+F_{JD}=0\\ 83.3\sin \left(36.869°\right)+F_{JE}\sin \left(36.869°\right)+\left(-60\right)=0\\ F_{JE}=\frac{10}{\sin \left(36.869°\right)}\\ F_{JE}=16.67\text{kN}\left(\text{T}\right)\end{array}$

Find the force in the member JK as follows:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{JC}\cos \left(36.869°\right)+F_{JE}\cos \left(36.869°\right)-F_{JI}+F_{JK}=0\\ -83.3\cos \left(36.869°\right)+16.7\cos \left(36.869°\right)-253.3+F_{JK}=0\\ -306.6+F_{JK}=0\\ F_{JK}=306.6\text{kN}\left(\text{T}\right)\end{array}$

Show the joint K as shown in Figure 9.

Refer Figure 9.

Find the force in the member KE as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{KE}=0\text{kN}\end{array}$

Find the force in the member KL as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{KL}=F_{KJ}\\ F_{KL}=306.6\text{kN}\left(\text{T}\right)\end{array}$

Show the joint E as shown in Figure 10.

Refer Figure 10.

Find the force in the member EL as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{EJ}\sin \left(36.869°\right)-F_{EL}\sin \left(36.869°\right)-F_{EK}-60=0\\ -16.7\sin \left(36.869°\right)-F_{EL}\sin \left(36.869°\right)-0-60=0\\ F_{EL}=-\frac{70}{\sin \left(36.869°\right)}\\ F_{EL}=116.7\text{kN}\left(\text{C}\right)\end{array}$

Find the force in the member EF as follows:

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{ED}+F_{EF}-F_{EJ}\cos \left(36.869°\right)+F_{EL}\cos \left(36.869°\right)=0\\ -\left(-280\right)+F_{EF}-16.7\cos \left(36.869°\right)+\left(-116.7\right)\cos \left(36.869°\right)=0\\ -\left(-280\right)+F_{EF}-16.7\cos \left(36.869°\right)+\left(-116.7\right)\cos \left(36.869°\right)=0\\ F_{EF}=173.3\text{kN}\left(\text{C}\right)\end{array}$

Show the joint F as shown in Figure 11.

Refer Figure 11.

Find the force in the member FL as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{FL}-60=0\\ F_{FL}=60\text{kN}\left(\text{C}\right)\end{array}$

Find the force in the member FC as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{FC}=F_{FE}\\ F_{FC}=173.3\text{kN}\left(\text{C}\right)\end{array}$

Show the joint G as shown in Figure 12.

Refer Figure 12.

Find the force in the member LG as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ 130-F_{GL}\sin \left(36.869°\right)=0\\ F_{GL}=\frac{130}{\sin \left(36.869°\right)}\\ F_{GL}=216.7\text{kN}\left(\text{T}\right)\end{array}$

Show the forces in the members of the truss as follows:

Member	Force
AB	$106.6\text{kN}\left(\text{C}\right)$
BC	$106.6\text{kN}\left(\text{C}\right)$
CD	$280\text{k}\left(\text{C}\right)$
DE	$280\text{kN}\left(\text{C}\right)$
EF	$173.3\text{kN}\left(\text{C}\right)$
FG	$173.3\text{kN}\left(\text{C}\right)$
HI	$253.3\text{kN}\left(\text{T}\right)$
IJ	$253.3\text{kN}\left(\text{T}\right)$
JK	$306.6\text{kN}\left(\text{T}\right)$
KL	$306.6\text{kN}\left(\text{T}\right)$
AH	$183.3\text{kN}\left(\text{T}\right)$
HB	$30\text{kN}\left(\text{C}\right)$
CH	$133.3\text{kN}\left(\text{C}\right)$
CI	$0\text{kN}$
CJ	$83.3\text{kN}\left(\text{T}\right)$
DJ	$60\text{kN}\left(\text{C}\right)$
EJ	$16.7\text{kN}\left(\text{T}\right)$
EK	$0\text{kN}$
EL	$116.7\text{kN}\left(\text{C}\right)$
LF	$60\text{kN}\left(\text{C}\right)$
LG	$216.7\text{kN}\left(\text{T}\right)$