Chapter 10, Problem 18P

To determine

Find the force in each member of the truss using structural symmetry.

Answer to Problem 18P

The force in the member AC and BE is $\underset{\_}{62.5\text{kN}\left(\text{C}\right)}$.

The force in the member CD is $\underset{\_}{32.5\text{kN}\left(\text{C}\right)}$.

The force in the member AD is $\underset{\_}{0}$.

The force in the member DE is $\underset{\_}{17.5\text{kN}\left(\text{T}\right)}$.

The force in the member BD is $\underset{\_}{130\text{kN}\left(\text{C}\right)}$.

Explanation of Solution

Given information:

The structure is given in the Figure.

The young’s modulus E and area A is constant.

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:

Refer the given structure.

The structure is symmetric with respect to the s axis.

Divide the magnitudes of forces and moments of the given loading by 2 to obtain the half loading.

Sketch the half loading for the given structure as shown in Figure 1.

Draw the reflection of half loading about the specified axis s.

Sketch the reflection of half loading as shown in Figure 2.

Add the half loading and reflection of half loading to find the symmetric component.

Sketch the symmetric loading component as shown in Figure 3.

Subtract the symmetric loading component from the given loading to obtain the antisymmetric component.

Sketch the antisymmetric loading component as shown in Figure 4.

Find the member forces due to symmetric loading component:

Sketch the substructure with symmetric boundary conditions as shown in Figure 5.

Find the reactions and member end forces of substructure using equilibrium equations and to the left of s axis.

The member end forces to the right of s axis are obtained by the reflection.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ -60-60+A_y=0\\ A_y=120\text{kN}\uparrow \end{array}$

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -25\times 12+60\times 3.5-60\times 5-D_x\times 12=0\\ -12D_x=390\\ D_x=32.5\text{kN}\leftarrow \end{array}$

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 25+A_x-D_x=0\\ 25+A_x-32.5=0\\ A_x=7.5\text{kN}\to \end{array}$

Find the angle $\theta$ made by the member AC with respect to the horizontal axis using the given Figure.

$\begin{array}{l}\tan \theta =\frac{12}{3.5}\\ \theta =73.74°\end{array}$

Consider joint C, find the force in the member AC and CD:

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ -60-F_{AC}\sin \left(73.74°\right)=0\\ F_{AC}=-62.5\text{kN}\\ F_{AC}=62.5\text{kN}\left(\text{C}\right)\end{array}$

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 25+F_{CD}+F_{AC}\cos \left(73.74°\right)=0\\ 25+F_{CD}-62.5\cos \left(73.74°\right)=0\\ F_{CD}=-7.5\text{kN}\end{array}$

$F_{CD}=7.5\text{kN}\left(\text{C}\right)$

Find the angle $\theta$ made by the member AD with respect to the horizontal axis using the given Figure.

$\begin{array}{l}\tan \theta =\frac{12}{5}\\ \theta =67.38°\end{array}$

Consider joint A, find the force in the member AD:

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ 120+F_{AD}\sin \left(67.38\right)+F_{AC}\sin \left(73.74°\right)=0\\ 120+F_{AD}\sin \left(67.38\right)-62.5\sin \left(73.74°\right)=0\\ F_{AD}=-64.9\text{kN}\end{array}$

$F_{AD}\approx 65\text{kN}\left(\text{C}\right)$

Sketch the substructure with antisymmetric boundary conditions as shown in Figure 6.

Find the reactions and member end forces of substructure using equilibrium equations and to the left of s axis.

The member end forces to the right of s axis are obtained by reflecting the negatives of computed forces and moments about the axis of symmetry.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 25-A_x=0\\ A_x=25\text{kN}\leftarrow \end{array}$

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -25\times 12+D_y\times 5=0\\ 5D_y=300\\ D_y=60\text{kN}\uparrow \end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+D_y=0\\ -A_y+60=0\\ A_y=60\text{kN}\downarrow \end{array}$

Consider joint C, find the force in the member AC and CD:

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_{AC}=0\end{array}$

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ 25+F_{CD}=0\\ F_{CD}=-25\text{kN}\\ F_{CD}=25\text{kN}\left(\text{C}\right)\end{array}$

Consider joint A, find the force in the member AD:

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ -60+F_{AD}\sin \left(67.38\right)=0\\ F_{AD}=65\text{kN}\left(\text{T}\right)\end{array}$

The total member end forces are obtained by superposing the member forces due to symmetric and antisymmetric components of loading.

Sketch the member end forces due to total loading as shown in Figure 7.

Therefore,

The force in the member AC and BE is $\underset{\_}{62.5\text{kN}\left(\text{C}\right)}$.

The force in the member CD is $\underset{\_}{32.5\text{kN}\left(\text{C}\right)}$.

The force in the member AD is $\underset{\_}{0}$.

The force in the member DE is $\underset{\_}{17.5\text{kN}\left(\text{T}\right)}$.

The force in the member BD is $\underset{\_}{130\text{kN}\left(\text{C}\right)}$.