Chapter 4, Problem 12P

To determine

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

Answer to Problem 12P

The forces in the member are $\underset{\_}{F_{IJ}=F_{JH}=0\text{kN}}$, $\underset{\_}{F_{HI}=50\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{GI}=30\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{GH}=40\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{FH}=30\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{FG}=150\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{EG}=120\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{EF}=120\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{DF}=120\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{DE}=250\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{CE}=270\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{CD}=200\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{BD}=270\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{BC}=350\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{AC}=480\text{kN}\left(\text{T}\right)}$, and $\underset{\_}{F_{AB}=280\text{kN}\left(\text{T}\right)}$.

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 AC, CE, EG, GI, IJ, BD, DF, FH, HJ, AB, BC, CD, DE, EF, FG, GH, and HI, are $F_{AC},F_{CE},F_{EG},F_{GI},F_{IJ},F_{BD},F_{DF},F_{FH},F_{HJ},F_{AB},F_{BC},F_{CD},F_{DE},F_{EF},F_{FG},F_{GH},F_{HI}$.

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

Refer Figure 1.

Consider the member BC, DE, FG, and HI with the horizontal as $\theta$.

$\begin{array}{l}\tan \theta =\frac{6}{8}\\ \theta ={\tan }^{-1}\left(\frac{6}{8}\right)\\ \theta =36.86°\end{array}$

Consider the horizontal and vertical reactions at A are $A_x$ and $A_y$.

Consider the vertical reaction at B is $B_y$.

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+B_y=0\\ A_y=-B_y\end{array}$        (1)

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x+80+80+80+40=0\\ A_x=-280\text{kN}\end{array}$

Take the sum of the moments at A is zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -\left(80\times 6\right)-\left(80\times 12\right)-\left(80\times 18\right)-\left(40\times 24\right)+8B_y=0\\ -3840+8B_y=0\\ B_y=\frac{3840}{8}\\ B_y=480\text{kN}\end{array}$

Substitute $480\text{kN}$ for $B_y$ in Equation (1).

$A_y=-480\text{kN}$

Show the joint A as shown in Figure 2.

Refer Figure 2.

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

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x=-F_{AB}\\ F_{AB}=-\left(-280\text{kN}\right)\\ F_{AB}=280\text{kN}\left(\text{T}\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y=-F_{AC}\\ F_{AC}=-\left(-480\text{kN}\right)\\ F_{AC}=480\text{kN}\left(\text{T}\right)\end{array}$

Show the joint B as shown in Figure 3.

Refer Figure 3.

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

 $\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{BC}\cos \theta =F_{BA}\\ F_{BC}=-\frac{F_{BA}}{\cos \left(36.86°\right)}\\ F_{BC}=-\frac{280}{\cos \left(36.86°\right)}\\ F_{BC}=350\text{kN}\left(\text{C}\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ B_y+F_{BC}\sin \theta =-F_{BD}\\ B_y+\left(-350\right)\sin \left(36.86°\right)=-F_{BD}\end{array}$

Substitute $480\text{kN}$ for $B_y$

$\begin{array}{l}480+\left(-350\right)\sin \left(36.86°\right)=-F_{BD}\\ 480+\left(-350\right)\sin \left(36.86°\right)=-F_{BD}\\ F_{BD}=-270\text{kN}\\ F_{BD}=270\text{kN}\left(\text{C}\right)\end{array}$

Show the joint C as shown in Figure 4.

Refer Figure 4.

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

 $\begin{array}{l}{\displaystyle \sum F_x}=0\\ 80+F_{CB}\cos 36.86°+F_{CD}=0\\ 80+\left(-350\right)\cos 36.86°+F_{CD}=0\\ F_{CD}=350\cos 36.86°-80\\ F_{CD}=200\text{kN}\left(\text{T}\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{CE}-F_{CA}-F_{CB}\sin 36.86°=0\\ F_{CE}-\left(\text{480}\right)-\left(-350\right)\sin 36.86°=0\\ F_{CE}=270\text{kN}\left(\text{T}\right)\end{array}$

Show the joint D as shown in Figure 5.

Refer Figure 5.

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

 $\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{DE}\cos \theta =F_{DC}\\ F_{DE}=-\frac{200}{\cos \left(36.86°\right)}\\ F_{DE}=250\text{kN}\left(\text{C}\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{DB}+F_{DE}\sin \theta =-F_{DF}\\ -\left(-270\right)+\left(-250\right)\sin \left(36.86°\right)=-F_{DF}\\ 120\text{kN}=-F_{DF}\\ F_{DF}=120\text{kN}\left(\text{C}\right)\end{array}$

Show the joint E as shown in Figure 6.

Refer Figure 6.

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

 $\begin{array}{l}{\displaystyle \sum F_x}=0\\ 80+F_{ED}\cos 36.86°+F_{EF}=0\\ 80+\left(-250\right)\cos 36.86°+F_{EF}=0\\ F_{EF}=250\cos 36.86°-80\\ F_{EF}=120\text{kN}\left(\text{T}\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{EG}-F_{EC}-F_{ED}\sin 36.86°=0\\ F_{EG}-\left(270\right)-\left(-250\right)\sin 36.86°=0\\ F_{EG}=120\text{kN}\\ F_{EG}=120\text{kN}\left(\text{T}\right)\end{array}$

Show the joint F as shown in Figure 7.

Refer Figure 7.

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

 $\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{FG}\cos \theta =F_{EF}\\ F_{FG}=-\frac{120}{\cos \left(36.86°\right)}\\ F_{FG}=150\text{kN}\left(\text{C}\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{FD}+F_{FG}\sin \theta =-F_{FH}\\ -\left(-120\right)+\left(-150\right)\sin \left(36.86°\right)=-F_{FH}\\ 30\text{kN}=-F_{FH}\\ F_{FH}=30\text{kN}\left(\text{C}\right)\end{array}$

Show the joint G as shown in Figure 8.

Refer Figure 8.

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

 $\begin{array}{l}{\displaystyle \sum F_x}=0\\ 80+F_{GF}\cos 36.86°+F_{GH}=0\\ 80+\left(-150\right)\cos 36.86°+F_{GH}=0\\ F_{GH}=150\cos 36.86°-80\\ F_{GH}=40\text{kN}\left(\text{T}\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{GI}-F_{GE}-F_{GF}\sin 36.86°=0\\ F_{GI}-\left(120\right)-\left(-150\right)\sin 36.86°=0\\ F_{GI}=30\text{kN}\\ F_{GI}=30\text{kN}\left(\text{T}\right)\end{array}$

Show the joint H as shown in Figure 9.

Refer Figure 9.

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

 $\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{HI}\cos \theta =F_{HG}\\ F_{HI}=-\frac{40}{\cos \left(36.86°\right)}\\ F_{HI}=50\text{kN}\left(\text{C}\right)\end{array}$

The forces in the member IJ and JH is zero as no force is acts at the joint J.

Thus, the forces in the member are $\underset{\_}{F_{IJ}=F_{JH}=0\text{kN}}$, $\underset{\_}{F_{HI}=50\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{GI}=30\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{GH}=40\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{FH}=30\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{FG}=150\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{EG}=120\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{EF}=120\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{DF}=120\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{DE}=250\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{CE}=270\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{CD}=200\text{kN}\left(\text{T}\right)}$, $\underset{\_}{F_{BD}=270\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{BC}=350\text{kN}\left(\text{C}\right)}$, $\underset{\_}{F_{AC}=480\text{kN}\left(\text{T}\right)}$, and $\underset{\_}{F_{AB}=280\text{kN}\left(\text{T}\right)}$.