Textbook Problem

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4.33 through 4.45 Determine the forces in the members identified by “×” of the truss shown by the method of sections.

FIG. P4.42

To determine

Find the forces in the members DG, EG, and EH of the truss by the method of sections.

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 sections:

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 truss as shown in Figure 1.

Refer Figure 1.

Consider the section a-a passing through the members DG, DE, EF, and FH.

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

$\begin{array}{l}\tan \theta =\frac{27}{9}\\ \theta ={\tan }^{-1}\left(\frac{27}{9}\right)\\ \theta =71.565°\end{array}$

Consider the similar triangles IAC and IGH.

$\begin{array}{l}\frac{GH}{AC}=\frac{9}{27}\\ GH=\frac{9}{27}\times AC\\ GH=\frac{9}{27}\times 18\\ GH=6\text{ft}\end{array}$

Consider the similar triangles IDF and IGH.

$\begin{array}{l}\frac{DF}{AC}=\frac{18}{27}\\ GH=\frac{18}{27}\times AC\\ GH=\frac{18}{27}\times 18\\ GH=12\text{ft}\end{array}$

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

Refer Figure 2.

Consider the forces in the members DG, DE, EF and FH are denoted by $F_{DG}$, $F_{DE}$, $F_{EF}$, and $F_{FH}$.

Take the sum of the moments acting at F as zero.

$\begin{array}{l}{\displaystyle \sum M_F}=0\\ \left[\begin{array}{l}\left(50\times 6\right)-\left(30\times 18\right)-\left(30\times 9\right)\\ +F_{DG}\sin \left(71.565°\right)\times 9+F_{DG}\cos \left(71.565°\right)\times 9\end{array}\right]=0\\ -510+8.538F_{DG}+2.846F_{DG}=0\\ F_{DG}=\frac{510}{11.384}\\ F_{DG}=44.80\text{k}\left(\text{T}\right)\end{array}$

Take the sum of the vertical forces equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{DG}\sin (71\end{array}$