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 members AB, BC, CD, DE, EF, FG, AH, HI, IJ, JK, KL, LG, BH, CI, DJ, EK, FL, CH, DI, DK, and EL are $F_{AB},F_{BC},F_{CD},F_{DE},F_{EF},F_{FG},F_{AH},F_{HI},F_{IJ},F_{JK},F_{KL},F_{LG},F_{BH},F_{CI},F_{DJ},F_{EK},F_{FL},F_{CH},F_{DI},F_{DK},F_{EL}$

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 G is $G_y$.

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

$\begin{array}{l}\tan \theta =\frac{3}{4}\\ \theta ={\tan }^{-1}\left(\frac{3}{4}\right)\\ \theta =36.86°\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+G_y=60+60+60+30+30\\ A_y+G_y=240\end{array}$        (1)

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

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

Take the sum of the moments at A is zero. Then,

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -\left(60\times 4\right)-\left(60\times 8\right)-\left(60\times 12\right)-\left(30\times 16\right)-\left(30\times 20\right)+G_y\times 24=0\\ -240-480-720-480-600+24G_y=0\\ G_y=\frac{2520}{24}\\ G_y=105\text{kN}\end{array}$

Substitute $105\text{kN}$ for $G_y$ in Equation (1).

$\begin{array}{l}{A}_y+105=240\\ A_y=135\text{kN}\end{array}$

Show the joint A as shown in Figure 2.

 

Refer Figure 2.

Find the forces in the member AH and AB as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-F_{AH}\sin \left(36.86°\right)=0\\ 135-0.6F_{AH}=0\\ F_{AH}=\frac{135}{0.6}\\ F_{AH}=225\text{kN}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{AH}\cos \left(36.86°\right)+F_{AB}=0\\ F_{AB}=-F_{AH}\cos \left(36.86°\right)\\ F_{AB}=-225\cos \left(36.86°\right)\\ F_{AB}=180\text{kN}\left(\text{C}\right)\end{array}$

Show the joint B as shown in Figure 3.

Refer Figure 3.

Find the forces in the member BH and BC as follows:

For Equilibrium of forces,

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

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{BC}-F_{AB}=0\\ F_{BC}=F_{AB}\\ F_{BC}=-180\text{kN}\\ F_{BC}=180\text{kN}\left(\text{C}\right)\end{array}$

Show the joint H as shown in Figure 4.

 

Refer Figure 4.

Find the forces in the member HI and HC as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{HC}\sin \left(36.86°\right)+F_{HA}\sin \left(36.86°\right)+F_{HB}=0\\ F_{HC}\sin \left(36.86°\right)+225\sin \left(36.86°\right)-60=0\\ 0.6F_{HC}=-75\\ F_{HC}=125\text{kN}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{HC}\cos \left(36.86°\right)-F_{HA}\cos \left(36.86°\right)+F_{HI}=0\\ -125\cos \left(36.86°\right)-225\cos \left(36.86°\right)+F_{HI}=0\\ -280+F_{HI}=0\\ F_{HI}=280\text{kN}\left(\text{T}\right)\end{array}$

Show the joint C as shown in Figure 5.

 

Refer Figure 5.

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

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{CH}\sin \left(36.86°\right)-F_{CI}-60=0\\ 125\sin \left(36.86°\right)-F_{CI}-60=0\\ -F_{CI}+15=0\\ F_{CI}=15\text{kN}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{CB}+F_{CD}-F_{CH}\cos \left(36.86°\right)=0\\ -\left(-180\right)+F_{CD}-\left(-125\right)\cos \left(36.86°\right)=0\\ 280+F_{CD}=0\\ F_{CD}=-280\text{kN}\\ F_{CD}=280\text{kN}\left(\text{C}\right)\end{array}$

Show the joint I as shown in Figure 6.

Refer Figure 6.

Find the forces in the member ID and IJ as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{ID}\sin \left(36.86°\right)+F_{CI}=0\\ F_{ID}\sin \left(36.86°\right)+15=0\\ F_{ID}=-\frac{15}{\sin \left(36.86°\right)}\\ F_{ID}=25\text{kN}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{HI}+F_{ID}\cos \left(36.86°\right)+F_{IJ}=0\\ -280+\left(-25\right)\cos \left(36.86°\right)+F_{IJ}=0\\ -300+F_{IJ}=0\\ F_{IJ}=300\text{kN}\left(\text{T}\right)\end{array}$

Show the joint J as shown in Figure 7.

Refer Figure 7.

Find the forces in the member JD and JK as follows:

For Equilibrium of forces,

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

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{JK}=F_{JI}\\ F_{JK}=300\text{kN}\left(\text{T}\right)\end{array}$

Show the joint G as shown in Figure 8.

 

Refer Figure 8.

Find the forces in the member GL and GF as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ G_y-F_{GL}\sin \left(36.86°\right)=0\\ F_{GL}=\frac{G_y}{\sin \left(36.86°\right)}\\ F_{GL}=\frac{105}{\sin \left(36.86°\right)}\\ F_{GL}=175\text{kN}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{GL}\cos \left(36.86°\right)-F_{FG}=0\\ F_{FG}=-F_{GL}\cos \left(36.86°\right)\\ F_{FG}=-175\cos \left(36.86°\right)\\ F_{FG}=140\text{kN}\left(\text{C}\right)\end{array}$

Show the joint F as shown in Figure 9.

Refer Figure 9.

Find the forces in the member FE and FL as follows:

For Equilibrium of forces,

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

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{FE}+F_{FG}=0\\ F_{FE}=-140\\ F_{FE}=140\text{kN}\left(\text{C}\right)\end{array}$

Show the joint L as shown in Figure 10.

Refer Figure 10.

Find the forces in the member LE and LK as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{LF}+F_{LG}\sin \left(36.86°\right)+F_{LE}\sin \left(36.86°\right)=0\\ -30+175\sin \left(36.86°\right)+F_{LE}\sin \left(36.86°\right)=0\\ 75+F_{LE}\sin \left(36.86°\right)=0\\ F_{LE}=-\frac{75}{\sin \left(36.86°\right)}\\ F_{LE}=125\text{kN}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{LK}-F_{LE}\cos \left(36.86°\right)+F_{LG}\cos \left(36.86°\right)=0\\ -F_{LK}-\left(-125\right)\cos \left(36.86°\right)+175\cos \left(36.86°\right)=0\\ -F_{LK}+240=0\\ F_{LK}=240\text{kN}\left(\text{T}\right)\end{array}$

Show the joint E as shown in Figure 11.

 

Refer Figure 11.

Find the forces in the member EK and ED as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{EK}-30-F_{EL}\sin \left(36.86°\right)=0\\ -F_{EK}-30-\left(-125\right)\sin \left(36.86°\right)=0\\ -F_{EK}+45=0\\ F_{EK}=45\text{kN}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{ED}+F_{EF}+F_{EL}\cos \left(36.86°\right)=0\\ -F_{ED}+\left(-140\right)-125\cos \left(36.86°\right)=0\\ -F_{ED}-140-100=0\\ F_{ED}=240\text{kN}\left(\text{C}\right)\end{array}$

Show the joint K as shown in Figure 12.

Refer Figure 12.

Find the forces in the member KD and KL as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{KE}+F_{KD}\sin \left(36.86°\right)=0\\ F_{KD}\sin \left(36.86°\right)=-F_{KE}\\ F_{KD}=-\frac{45}{\sin \left(36.86°\right)}\\ F_{KD}=75\text{kN}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{KJ}-F_{KD}\cos \left(36.86°\right)+F_{KL}=0\\ -F_{KJ}-\left(-75\right)\cos \left(36.86°\right)+240=0\\ -F_{KJ}+300=0\\ F_{KJ}=300\text{kN}\left(\text{T}\right)\end{array}$

Show the forces in the members of truss as shown in Table 1.

Member	Force (kN)
AB	180 kN (C)
BC	180 kN (C)
CD	280 kN (C)
DE	240 kN (C)
EF	140 kN (C)
FG	140 kN (C)
AH	225 kN (T)
HI	280 kN (T)
IJ	300 kN (T)
JK	300 kN (T)
KL	240 kN (T)
LG	175 kN (T)
BH	60 kN (C)
HC	125 kN (C)
CI	15 kN (T)
ID	25 kN (C)
DJ	0 kN
DK	75 kN (C)
KE	45 kN (T)
EL	125 kN (C)
LF	30 kN (C)