Chapter 4, Problem 19P

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:

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

Refer Figure 1.

Consider the horizontal and vertical reaction at A are denoted by $A_x$ and $A_y$.

Consider the vertical reaction at G is denoted by $G_y$.

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

$\begin{array}{l}\tan \theta =\frac{24}{48}\\ \theta ={\tan }^{-1}\left(\frac{24}{48}\right)\\ \theta =26.565°\end{array}$

Consider the triangles ABH, ACI, ADJ.

$\begin{array}{l}\frac{BH}{AB}=\frac{CI}{AC}=\frac{DJ}{AD}\\ \frac{BH}{16}=\frac{CI}{32}=\frac{24}{48}\end{array}$

Solve the above relation.

$\begin{array}{l}BH=\frac{24}{48}\times 16\\ BH=8\text{ft}\end{array}$

$\begin{array}{l}CI=\frac{24}{48}\times 32\\ CI=16\text{ft}\end{array}$

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

$\begin{array}{l}\tan \alpha =\frac{CI}{CD}\\ \tan \alpha =\frac{16}{16}\\ \alpha ={\tan }^{-1}\left(1\right)\\ \alpha =45°\end{array}$

Consider the triangle GDJ.

$\begin{array}{c}GJ=\sqrt{G{D}^2+D{J}^2}\\ =\sqrt{{48}^2+{24}^2}\\ =53.66\text{ft}\end{array}$

Consider the triangles GFL, GEK, and GDJ.

$\begin{array}{l}\frac{GL}{GF}=\frac{GK}{GE}=\frac{GJ}{GD}\\ \frac{GL}{16}=\frac{GK}{32}=\frac{53.66}{48}\end{array}$

Solve the above relation.

$\begin{array}{l}GL=\frac{53.66}{48}\times 16\\ GL=17.88\text{ft}\end{array}$

$\begin{array}{l}GK=\frac{53.66}{48}\times 32\\ GK=35.77\text{ft}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ A_x-\left(10+20+20\right)\sin 26.565°=0\\ A_x=22.36\text{k}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y+G_y-10-\left(10+20+20\right)\cos 26.565°=0\\ A_y+G_y=54.721\end{array}$        (1)

Take the sum of the moment about G as zero.

$\begin{array}{l}{\displaystyle \sum M_G=0}\\ -10\times 80-10\times GJ-20\times GK-20\times GL+A_y\times 96=0\\ -10\times 80-10\times 53.66-20\times 35.77-20\times 17.88+A_y\times 96=0\\ A_y=\frac{2409.6}{96}\\ A_y=25.1\text{k}\end{array}$

Substitute $25.1\text{k}$ for $A_y$ in Equation (1).

$\begin{array}{l}25.1+G_y=54.721\\ G_y=29.621\text{k}\end{array}$

Show the joint A as shown in Figure 2.

Figure 2

Find the forces in the members AH as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y+F_{AH}\sin 26.565°=0\\ 25.10+F_{AH}\sin 26.565°=0\\ F_{AH}=-\frac{25.10}{\sin 26.565°}\\ F_{AH}=-56.12\text{k}\\ F_{AH}=56.12\text{k}\left(\text{C}\right)\end{array}$

Find the forces in the members AB as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ A_x+F_{AB}+F_{AH}\cos 26.565°=0\\ 22.36+F_{AB}-56.12\cos 26.565°=0\\ F_{AB}=-22.36+56.12\cos 26.565°\\ F_{AB}=27.83\text{k}\left(\text{T}\right)\end{array}$

Show the joint B as shown in Figure 3.

Refer Figure 3.

Find the forces in the members BC as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{AB}+F_{BC}=0\\ F_{BC}=F_{AB}\\ F_{BC}=27.83\text{k}\left(\text{T}\right)\end{array}$

Find the forces in the members BH as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{BH}=0\end{array}$

Show the joint H as shown in Figure 4.

Refer Figure 4.

Find the forces in the members CH and IH as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -10+F_{IH}\sin 26.565°-F_{AH}\sin 26.565°-F_{CH}\sin 26.565°=0\\ F_{IH}\sin 26.565°-F_{CH}\sin 26.565°=10-F_{AH}\sin 26.565°\\ F_{IH}-F_{CH}=\frac{10-56.12\sin 26.565°}{\sin 26.565°}\\ F_{IH}-F_{CH}=\frac{10-F_{AH}\sin 26.565°}{\sin 26.565°}\\ F_{IH}-F_{CH}=-33.759\end{array}$        (2)

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{AH}\cos 26.565°+F_{CH}\cos 26.565°+F_{IH}\cos 26.565°=0\\ -F_{AH}+F_{CH}+F_{IH}=0\\ F_{CH}+F_{IH}=F_{AH}\\ F_{CH}+F_{IH}=-56.12\end{array}$        (3)

Solve Equation (2) and (3).

$\begin{array}{l}{F}_{IH}=44.94\text{k}\left(\text{C}\right)\\ F_{CH}=11.18\text{k}\left(\text{C}\right)\end{array}$

Show the joint C in Figure 5.

Refer Figure 5.

Find the forces in the members IC as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{IC}+F_{CH}\sin 26.565°=0\\ F_{IC}=-\left(-11.18\right)\sin 26.565°\\ F_{IC}=5\text{k}\left(\text{T}\right)\end{array}$

Find the forces in the members CD as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{CD}-F_{BC}-F_{CH}\cos 26.565°=0\\ F_{CD}=F_{BC}+F_{CH}\cos 26.565°\\ F_{CD}=27.85+\left(-11.18\right)\cos 26.565°\\ F_{CD}=17.85\text{k}\left(\text{T}\right)\end{array}$

Show the joint I as shown in Figure 6.

Refer Figure 6.

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

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{IC}-F_{ID}\sin 45°+F_{IJ}\sin 26.565°-F_{IH}26.565°=0\\ -5-0.707F_{ID}+0.447F_{IJ}+44.94\sin 26.565°=0\\ -5-0.707F_{ID}+0.447F_{IJ}+44.94\sin 26.565°=0\\ -0.707F_{ID}+0.447F_{IJ}=15.10\end{array}$        (4)

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{IH}\cos 26.565°+F_{IJ}\cos 26.565°+F_{ID}\cos 45°=0\\ 44.94\cos 26.565°+0.894F_{IJ}+0.707F_{ID}=0\\ 0.894F_{IJ}+0.707F_{ID}=40.20\end{array}$        (5)

Solve Equation (4) and (5).

$\begin{array}{l}{F}_{IJ}=41.24\text{k}\left(\text{C}\right)\\ F_{ID}=4.712\text{k}\left(\text{C}\right)\end{array}$

Show the joint J as shown in Figure 7.

Refer Figure 7.

Find the forces in the members JK as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{IJ}\cos 26.565°-10\sin 26.565°+F_{JK}\cos 26.565°=0\\ 41.24\cos 26.565°-10\sin 26.565°+F_{JK}\cos 26.565°=0\\ F_{JK}=-\frac{32.414}{\cos 26.565°}\\ F_{JK}=36.24\text{k}\left(\text{C}\right)\end{array}$

Find the forces in the members DJ as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -10\cos 26.565°-F_{IJ}\sin 26.565°-F_{JK}\sin 26.565°-F_{DJ}=0\\ -10\cos 26.565°+41.24\sin 26.565°+36.24\sin 26.565°-F_{DJ}=0\\ F_{DJ}=25.70\text{k}\left(\text{T}\right)\end{array}$

Show the joint D as shown in Figure 8.

Refer Figure 8.

Find the forces in the members KD as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{ID}\sin 45°+F_{DJ}+F_{KD}\sin 45°=0\\ -F_{KD}=\frac{F_{ID}\sin 45°+F_{DJ}}{\sin 45°}\\ -F_{KD}=\frac{-4.712\sin 45°+25.70}{\sin 45°}\\ F_{KD}=31.60\text{k}\left(\text{C}\right)\end{array}$

Find the forces in the members DE as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{CD}+F_{DE}-F_{ID}\cos 45°+F_{KD}\cos 45°=0\\ -17.85+F_{DE}+4.712\cos 45°-31.60\cos 45°=0\\ -17.85+F_{DE}+4.712\cos 45°-31.60\cos 45°=0\\ F_{DE}=36.86\text{k}\left(\text{T}\right)\end{array}$

Show the joint K as shown in Figure 9.

Refer Figure 9.

Find the forces in the members KL as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{JK}\cos 26.565°-20\times \sin 26.565°-F_{KD}\cos 45°+F_{KL}\cos 26.565°=0\\ 36.24\cos 26.565°-20\times \sin 26.565°+31.60\cos 45°+F_{KL}\cos 26.565°=0\\ F_{KL}=51.22\text{k}\left(\text{C}\right)\end{array}$

Find the forces in the members KE as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{KE}-20\cos 26.565°+F_{JK}\sin 26.565°-F_{KL}\sin 26.565°-F_{KD}\sin 45°=0\\ -F_{KE}-20\cos 26.565°-36.24\sin 26.565°-\left(-51.22\right)\sin 26.565°-\left(-31.60\right)\sin 45°=0\\ F_{KE}=11.15\text{k}\left(\text{T}\right)\end{array}$

Show the joint E as shown in Figure 10.

Refer Figure 10.

Find the forces in the members EL as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{EK}+F_{EL}\sin 26.565°=0\\ F_{EL}=-\frac{11.15}{\sin 26.565°}\\ F_{EL}=-\frac{11.15}{\sin 26.565°}\\ F_{EL}=25\text{k}\left(\text{C}\right)\end{array}$

Find the forces in the members EF as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -F_{DE}+F_{EF}+F_{EL}\cos 26.565°=0\\ F_{EF}=F_{DE}-F_{EL}\cos 26.565°\\ F_{EF}=36.86-\left(-25\right)\cos 26.565°\\ F_{EF}=36.86-\left(-25\right)\cos 26.565°\\ F_{EF}=59.22\text{k}\left(\text{T}\right)\end{array}$

Show the joint F as shown in 11.

Refer Figure 11.

Find the forces in the members FL as follows:

For Equilibrium of forces,

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

Find the forces in the members FG as follows:

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{EF}+F_{FG}=0\\ F_{FG}=F_{EF}\\ F_{FG}=59.22\text{k}\left(\text{T}\right)\end{array}$

Show the joint G as shown in Figure 12.

Refer Figure 12.

Find the force in the member GL as follows:

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ G_y+F_{GL}\sin 26.565°=0\\ F_{GL}=-\frac{G_y}{\sin 26.565°}\\ F_{GL}=-\frac{29.621}{\sin 26.565°}\\ F_{GL}=66.23\text{k}\left(\text{C}\right)\end{array}$

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

Members	Force
AH	56.12 k (C)
AB	27.83 k (T)
BC	27.83 k (T)
BH	0 k
IH	44.94 k(C)
CH	11.18 k (C)
IC	5 k (T)
CD	17.85 k (T)
IJ	41.24 k (C)
ID	4.712 k (C)
JK	36.24 k (C)
DJ	25.70 k (T)
KD	31.60 k (C)
DE	36.86 k (T)
KL	51.22 k (C)
KE	11.15 k (T)
EL	25 k (C)
EF	59.22 k (T)
FL	0 k
FG	59.22 k (T)
GL	66.23 k (C)