Chapter 4, Problem 46P

To determine

Find the forces in the members of the truss.

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 and section.

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 E is $E_y$.

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x+10+10=0\\ A_x=-20\text{k}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+E_y=40+30\\ A_y+E_y=70\text{k}\end{array}$        (1)

Take the sum of the moment about point A as zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ \left[\begin{array}{l}-\left(10\times 20\right)-\left(10\times 40\right)-\left(40\times 30\right)\\ -\left(30\times 45\right)+\left(E_y\times 60\right)\end{array}\right]=0\\ E_y=\frac{3,150}{60}\\ E_y=52.5\text{k}\end{array}$

Substitute $52.5\text{k}$ for $E_y$ in Equation (1).

$\begin{array}{l}{A}_y+52.5=70\\ A_y=70-52.5\\ A_y=17.5\text{k}\end{array}$

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

$\begin{array}{l}\tan \theta =\left(\frac{40}{15}\right)\\ \theta ={\tan }^{-1}\left(\frac{40}{15}\right)\\ \theta =69.44°\end{array}$

Consider a section a-a passing through the members BC, CG, and JK.

Show the portion of the truss to the left of the section a-a as shown in Figure 2.

Refer Figure 2.

Take the sum of the moment about C as zero.

$\begin{array}{l}{\displaystyle \sum M_F}=0\\ -\left(10\times 20\right)-\left(10\times 40\right)-\left(17.5\times 30\right)-F_{JK}\times 40=0\\ F_{JK}=\frac{1125}{40}\\ F_{JK}=28.13\text{k}\left(\text{C}\right)\end{array}$

JOINT A.

Show the joint as shown in Figure 3.

Refer Figure 3.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{AF}\sin \left(69.44°\right)+17.5=0\\ F_{AF}=-\frac{17.5}{\sin \left(69.44°\right)}\\ F_{AF}=18.69\text{k}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -20+F_{AB}+F_{AF}\cos \left(69.44°\right)=0\\ -20+F_{AB}+\left(-18.69\right)\cos \left(69.44°\right)=0\\ F_{AB}=18.69\cos \left(69.44°\right)+20\\ F_{AB}=26.56\text{k}\left(\text{T}\right)\end{array}$

JOINT J.

Show the joint as shown in Figure 4.

Refer Figure 4.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{JF}\sin \left(69.44°\right)-F_{JG}\sin \left(69.44°\right)=0\\ F_{JF}=-F_{JG}\end{array}$        (2)

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ 10-F_{JF}\cos \left(69.44°\right)+F_{JK}+F_{JG}\cos \left(69.44°\right)=0\\ 10-F_{JF}\cos \left(69.44°\right)+\left(-28.13\right)+F_{JG}\cos \left(69.44°\right)=0\\ 10-\left(-F_{JG}\right)\cos \left(69.44°\right)+\left(-28.13\right)+F_{JG}\cos \left(69.44°\right)=0\end{array}$

$\begin{array}{l}10+0.3511F_{JG}-28.13+0.3511F_{JG}=0\\ F_{JG}=\frac{18.13}{0.7022}\\ F_{JG}=25.81\text{k}\left(\text{T}\right)\end{array}$

Substitute $25.81\text{k}$ for $F_{JG}$ in Equation (2).

$\begin{array}{l}{F}_{JF}=-25.81\text{k}\\ F_{JF}=25.81\text{k}\left(\text{C}\right)\end{array}$

JOINT F.

Show the joint as shown in Figure 5.

Refer Figure 5.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{FA}\sin \left(69.44°\right)-F_{FB}\sin \left(69.44°\right)+F_{FJ}\sin \left(69.44°\right)=0\\ -F_{FA}-F_{FB}+F_{FJ}=0\\ -\left(-18.69\right)-F_{FB}+\left(-25.81\right)=0\\ F_{FB}=7.12\text{k}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ 10+F_{FJ}\cos \left(69.44°\right)+F_{FB}\cos \left(69.44°\right)+F_{FG}-F_{FA}\cos \left(69.44°\right)=0\\ 10+\left(-25.81\right)\cos \left(69.44°\right)+\left(-7.12\right)\cos \left(69.44°\right)+F_{FG}-F_{FA}\cos \left(69.44°\right)=0\\ \left[\begin{array}{l}10+\left(-25.81\right)\cos \left(69.44°\right)+\\ \left(-7.12\right)\cos \left(69.44°\right)+F_{FG}-\left(-18.69\right)\cos \left(69.44°\right)\end{array}\right]=0\\ F_{FG}+5=0\\ F_{FG}=5\text{k}\left(\text{C}\right)\end{array}$

JOINT G.

Show the joint as shown in Figure 6.

Refer Figure 6.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{GF}-F_{GJ}\cos \left(69.44°\right)-F_{GB}\cos \left(69.44°\right)+F_{GC}\cos \left(69.44°\right)=0\\ -\left(-5\right)-25.81\cos \left(69.44°\right)-7.12\cos \left(69.44°\right)+F_{GC}\cos \left(69.44°\right)=0\\ -25.81-7.12+F_{GC}=-\frac{5}{\cos \left(69.44°\right)}\\ -25.81-7.12+F_{GC}=-14.237\\ F_{GC}=18.69\text{k}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{GB}\sin \left(69.44°\right)-F_{GC}\sin \left(69.44°\right)+F_{GJ}\sin \left(69.44°\right)=0\\ -F_{GB}-F_{GC}+F_{GJ}=0\\ -F_{GB}-18.69+25.81=0\\ F_{GB}=7.12\text{k}\left(\text{T}\right)\end{array}$

JOINT E.

Show the joint as shown in Figure 7.

Refer Figure 7.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ 52.5+F_{EI}\sin \left(69.44°\right)=0\\ F_{EI}=-\frac{52.5}{\sin \left(69.44°\right)}\\ F_{EI}=56.07\text{k}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{ED}=F_{EI}\cos \left(69.44°\right)\\ -F_{ED}=\left(-56.07\right)\cos \left(69.44°\right)\\ F_{ED}=19.69\text{k}\left(\text{T}\right)\end{array}$

JOINT K.

Show the joint as shown in Figure 8.

Refer Figure 8.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{KJ}-F_{KH}\cos \left(69.44°\right)+F_{KI}\cos \left(69.44°\right)=0\\ -\left(-28.13\right)-F_{KH}\cos \left(69.44°\right)+F_{KI}\cos \left(69.44°\right)=0\\ -F_{KH}+F_{KI}=-\frac{28.13}{\cos \left(69.44°\right)}\\ -F_{KH}+F_{KI}=-80.09\end{array}$        (3)

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{KH}\sin \left(69.44°\right)=F_{KI}\sin \left(69.44°\right)\\ -F_{KH}=F_{KI}\\ F_{KI}+F_{KH}=0\end{array}$        (4)

Add Equation (3) and (4).

$\begin{array}{l}2F_{KI}=-80.09\\ F_{KI}=40.045\text{k}\left(\text{C}\right)\end{array}$

Subtract Equation (4) and (3).

$\begin{array}{l}-2F_{KH}=-80.09\\ F_{KH}=40.045\text{k}\left(\text{T}\right)\end{array}$

JOINT I.

Show the joint as shown in Figure 9.

Refer Figure 9.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{ID}\sin \left(69.44°\right)-F_{IE}\sin \left(69.44°\right)+F_{IK}\sin \left(69.44°\right)=0\\ -F_{ID}-F_{IE}+F_{IK}=0\\ -F_{ID}-\left(-56.07\right)+\left(-40.05\right)=0\\ F_{ID}=16.02\text{k}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{IH}-F_{ID}\cos \left(69.44°\right)-F_{IK}\cos \left(69.44°\right)+F_{IE}\cos \left(69.44°\right)=0\\ -F_{IH}-16.02\cos \left(69.44°\right)-\left(-40.05\right)\cos \left(69.44°\right)+\left(-56.07\right)\cos \left(69.44°\right)=0\\ F_{IH}=\left[-16.02-\left(-40.05\right)+\left(-56.07\right)\right]\cos \left(69.44°\right)\\ F_{IH}=11.25\text{k}\left(\text{C}\right)\end{array}$

JOINT H.

Show the joint as shown in Figure 10.

Refer Figure 10.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{HD}\cos 69.44°+F_{HK}\cos 69.44°-F_{HC}\cos 69.44°+F_{HI}=0\\ F_{HD}\cos 69.44°+40.05\cos 69.44°-F_{HC}\cos 69.44°-11.25=0\\ F_{HD}+40.05-F_{HC}=\frac{11.25}{\cos 69.44°}\\ F_{HD}-F_{HC}=-8.015\end{array}$        (5)

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{HD}\cos 69.44°+F_{HK}\cos 69.44°-F_{HC}\cos 69.44°=0\\ -F_{HD}\cos 69.44°+40.05\cos 69.44°-F_{HC}\cos 69.44°=0\\ -F_{HD}+40.05-F_{HC}=0\\ -F_{HD}-F_{HC}=-40.05\end{array}$        (6)

Add Equation (5) and (6).

$\begin{array}{l}-2F_{HC}=-48.065\\ F_{HC}=24.03\text{k}\left(\text{T}\right)\end{array}$

Subtract Equation (5) from Equation (6).

$\begin{array}{l}-2F_{HD}=-32.035\\ F_{HD}=16.02\text{k}\left(\text{T}\right)\end{array}$

JOINT B.

Show the joint as shown in Figure 11.

Refer Figure 11.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{BF}\cos \left(69.44°\right)-F_{BA}=F_{BC}+F_{BG}\cos \left(69.44°\right)\\ -\left(-7.12\right)\cos \left(69.44°\right)-26.56=F_{BC}+\left(7.12\right)\cos \left(69.44°\right)\\ -26.56=F_{BC}\\ F_{BC}=26.56\text{k}\left(\text{C}\right)\end{array}$

JOINT D.

Show the joint as shown in Figure 12.

Refer Figure 12.

For Equilibrium of forces,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{DH}\cos \left(69.44°\right)+F_{DC}=F_{DE}+F_{DI}\cos \left(69.44°\right)\\ -16.02\cos \left(69.44°\right)+F_{DC}=19.69+16.02\cos \left(69.44°\right)\\ F_{DC}=19.69\text{k}\left(\text{T}\right)\end{array}$

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

Member	Force (k)
DC	$19.69\text{k}\left(\text{T}\right)$
BC	$26.56\text{k}\left(\text{C}\right)$
HD	$16.02\text{k}\left(\text{T}\right)$
HC	$24.03\text{k}\left(\text{T}\right)$
IH	$11.25\text{k}\left(\text{C}\right)$
ID	$16.02\text{k}\left(\text{T}\right)$
KH	$40.045\text{k}\left(\text{T}\right)$
KI	$40.045\text{k}\left(\text{C}\right)$
ED	$19.69\text{k}\left(\text{T}\right)$
EI	$56.07\text{k}\left(\text{C}\right)$
GB	$7.12\text{k}\left(\text{T}\right)$
GC	$18.69\text{k}\left(\text{T}\right)$
FG	$5\text{k}\left(\text{C}\right)$
FB	$7.12\text{k}\left(\text{C}\right)$
JF	$25.81\text{k}\left(\text{C}\right)$
JG	$25.81\text{k}\left(\text{T}\right)$
AB	$26.56\text{k}\left(\text{T}\right)$
AF	$18.69\text{k}\left(\text{C}\right)$
JK	$28.13\text{k}\left(\text{C}\right)$