Chapter 4, Problem 26P

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 reactions at A are $A_x$ and $A_y$.

Consider the vertical reaction at F is $F_y$.

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+F_y=30+30+30+30\\ A_y+F_y=120\end{array}$        (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.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -\left(30\times 16\right)-\left(30\times 48\right)-\left(30\times 64\right)-\left(30\times 96\right)+F_y\times 80=0\\ F_y\times 80=6720\\ F_y=84\text{k}\end{array}$

Substitute $84\text{k}$ for $F_y$ in Equation (1).

$\begin{array}{l}{A}_y+84=120\\ A_y=36\text{k}\end{array}$

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

$\begin{array}{l}\tan \alpha =\left(\frac{12}{16}\right)\\ \alpha ={\tan }^{-1}\left(\frac{12}{16}\right)\\ \alpha =36.869°\end{array}$

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

$\begin{array}{l}\tan \beta =\left(\frac{8}{16}\right)\\ \beta ={\tan }^{-1}\left(\frac{8}{16}\right)\\ \beta =26.565°\end{array}$

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

$\begin{array}{l}\tan \gamma =\left(\frac{4}{16}\right)\\ \gamma ={\tan }^{-1}\left(\frac{4}{16}\right)\\ \gamma =14.036°\end{array}$

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

$\begin{array}{l}\tan \theta =\left(\frac{20}{16}\right)\\ \theta ={\tan }^{-1}\left(\frac{20}{16}\right)\\ \theta =51.340°\end{array}$

Show the joint A as shown in Figure 2.

Refer Figure 2.

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

For the Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{AH}\sin 36.869°=-36\\ F_{AH}=-\frac{36}{\sin 36.869°}\\ F_{AH}=60\text{k}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{AH}\cos 36.869°=-F_{AB}\\ -60\cos 36.869°=-F_{AB}\\ F_{AB}=48\text{k}\left(\text{T}\right)\end{array}$

Show the joint B as shown in Figure 3.

Refer Figure 3.

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

For the Equilibrium of forces,

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{BC}=F_{BA}\\ F_{BC}=48\text{k}\left(\text{T}\right)\end{array}$

Show the joint H as shown in Figure 4.

Refer Figure 4.

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

For the Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{HI}\cos 26.565°+F_{HC}\cos 36.869°-F_{HA}\cos 36.869°=0\\ F_{HI}\cos 26.565°+F_{HC}\cos 36.869°-\left(-60\right)\cos 36.869°=0\\ 0.894F_{HI}+0.8F_{HC}=-48\end{array}$        (2)

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{HI}\sin 26.565°-F_{HC}\sin 36.869°-F_{HB}-F_{HA}\sin 36.869°=0\\ F_{HI}\sin 26.565°-F_{HC}\sin 36.869°-30-\left(-60\right)\sin 36.869°=0\\ 0.447F_{HI}-0.6F_{HC}=-6\end{array}$        (3)

Solve Equation (2) and (3).

$\begin{array}{l}{F}_{HI}=37.57\text{k}\left(\text{C}\right)\\ F_{HC}=18\text{k}\left(\text{C}\right)\end{array}$

Show the joint C as shown in Figure 5.

Refer Figure 5.

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

For the Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{CI}=-F_{CH}\sin 36.869°\\ F_{CI}=-\left(-18\right)\sin 36.869°\\ F_{CI}=10.8\text{k}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{CB}+F_{CD}-F_{CH}\cos 36.869°=0\\ -48+F_{CD}-\left(-18\right)\cos 36.869°=0\\ F_{CD}=33.6\text{k}\left(\text{T}\right)\end{array}$

Show the joint I as shown in Figure 6.

Refer Figure 6.

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

For the Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{IH}\sin 26.565°-F_{IC}-F_{ID}\sin 51.340°+F_{IJ}\sin 14.036°=0\\ -\left(-37.57\right)\sin 26.565°-10.8-F_{ID}\sin 51.340°+F_{IJ}\sin 14.036°=0\\ -0.780F_{ID}+0.242F_{IJ}=-6\end{array}$        (4)

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{IH}\cos 26.565°+F_{ID}\cos 51.340°+F_{IJ}\cos 14.036°=0\\ -\left(-37.57\right)\cos 26.565°+F_{ID}\cos 51.340°+F_{IJ}\cos 14.036°=0\\ 0.624F_{ID}+0.97F_{IJ}=-33.60\end{array}$        (5)

Solve Equation (4) and (5).

$\begin{array}{l}{F}_{ID}=2.56\text{k}\left(\text{C}\right)\\ F_{IJ}=32.98\text{k}\left(\text{C}\right)\end{array}$

Show the joint G as shown in Figure 7.

Refer Figure 7.

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

For the Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{GL}\sin 36.869°=30\\ F_{GL}=\frac{30}{\sin 36.869°}\\ F_{GL}=50\text{k}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{GL}\cos 36.869°=F_{GF}\\ -\left(50\right)\cos 36.869°=F_{GF}\\ F_{GF}=40\text{k}\left(\text{C}\right)\end{array}$

Show the joint F as shown in Figure 8.

Refer Figure 8.

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

For the Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{FL}=84\text{k}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{EF}=F_{FG}\\ F_{EF}=40\text{k}\left(\text{C}\right)\end{array}$

Show the joint L as shown in Figure 9.

Refer Figure 9.

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

For the Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{LK}\cos 26.565°-F_{LE}\cos 36.869°+F_{LG}\cos 36.869°=0\\ -F_{LK}\cos 26.565°-F_{LE}\cos 36.869°+50\cos 36.869°=0\\ -0.894F_{LK}-0.8F_{LE}=-40\end{array}$        (6)

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{LK}\sin 26.565°-F_{LE}\sin 36.869°-F_{LF}-F_{LG}\sin 36.869°=0\\ F_{LK}\sin 26.565°-F_{LE}\sin 36.869°-\left(-84\right)-50\sin 36.869°=0\\ 0.447F_{LK}-0.6F_{LE}=-54\end{array}$        (7)

Solve Equation (2) and (3).

$\begin{array}{l}{F}_{LE}=74\text{k}\left(\text{T}\right)\\ F_{LK}=21.47\text{k}\left(\text{T}\right)\end{array}$

Show the joint E as shown in Figure 10.

Refer Figure 10.

Find the force in the members EK and ED as follows:

For Equilibrium of the forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{ED}+F_{EF}+F_{EL}\cos 36.869°=0\\ -F_{ED}+\left(-40\right)+74\cos 36.869°=0\\ F_{ED}=19.2\text{k}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{EK}-30+F_{EL}\sin 36.869°=0\\ F_{EK}-30+74\sin 36.869°=0\\ F_{EK}=14.4\text{k}\left(\text{C}\right)\end{array}$

Show the joint K as shown in Figure 11.

Refer Figure 11.

Find the force in the members KJ and KD as follows:

For Equilibrium of the forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{KJ}\cos 14.036°-F_{KD}\cos 51.340°+F_{KL}\cos 26.565°=0\\ -F_{KJ}\cos 14.036°-F_{KD}\cos 51.340°+\left(-21.47\right)\cos 26.565°=0\\ -0.970F_{KJ}-0.624F_{KD}=19.203\end{array}$        (8)

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{KJ}\sin \left(14.036°\right)-F_{KD}\sin \left(51.340°\right)-F_{KL}\sin \left(26.565°\right)-F_{KE}=0\\ F_{KJ}\sin \left(14.036°\right)-F_{KD}\sin \left(51.340°\right)-\left(-21.47\right)\sin \left(26.565°\right)-\left(-14.4\right)=0\\ 0.242F_{KJ}-0.780F_{KD}=-24\end{array}$        (9)

Solve Equation (8) and (9).

$\begin{array}{l}{F}_{KJ}=32.98\text{k}\left(\text{C}\right)\\ F_{KD}=20.49\text{k}\left(\text{T}\right)\end{array}$

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

Member	Force
AB	48 k (T)
BC	48 k (T)
CD	33.6 k (T)
DE	19.2 k (T)
EF	40 k (T)
FG	40 k (T)
AH	60 k (T)
HI	37.57 k (T)
IJ	32.98 k (T)
JK	32.98 k (T)
KL	21.47 k (T)
LG	50 k (T)
BH	30 k (T)
HC	18 k (C)
CI	10.8 k (T)
ID	2.56 k (C)
DJ	16 k (T)
DK	20.49 k (T)
KE	14.4 k (C)
EL	74 k (T)
LF	84 k (C)