Chapter 10, Problem 20P

To determine

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

Answer to Problem 20P

The force in the member AC and CG is $\underset{\_}{10.34\text{k}\left(\text{T}\right)}$.

The force in the member AE is $\underset{\_}{16.78\text{k}\left(\text{C}\right)}$.

The force in the member CE is $\underset{\_}{7.07\text{k}\left(\text{C}\right)}$.

The force in the member EG is $\underset{\_}{23.86\text{k}\left(\text{C}\right)}$.

The force in the member BD and DG is $\underset{\_}{20.68\text{k}\left(\text{T}\right)}$.

The force in the member DF is $\underset{\_}{14.14\text{k}\left(\text{C}\right)}$.

The force in the member DG is $\underset{\_}{20.68\text{k}\left(\text{T}\right)}$.

The force in the member FG is $\underset{\_}{44.18\text{k}\left(\text{C}\right)}$.

The force in the member FB is $\underset{\_}{58.32\text{k}\left(\text{C}\right)}$

Explanation of Solution

Given information:

The structure is given in the Figure.

The young’s modulus E and area A are 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 loading 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.

$\begin{array}{l}+\uparrow \sum F_y=0\\ -10-10+A_y=0\\ A_y=20\text{k}\uparrow \end{array}$

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ G_x\times 24-10\times 24-10\times 12-5\left(12\right)=0\\ G_x=17.5\text{k}\leftarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ 5+A_x-G_x=0\\ 5+A_x-17.5=0\\ A_x=12.5\text{k}\to \end{array}$

Consider joint A, find the force in the member AC and AE:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 20+F_{AE}\sin \left(45°\right)+F_{AC}\sin \left(25°\right)=0\\ F_{AE}\sin \left(45°\right)+F_{AC}\sin \left(25°\right)=-20\end{array}$        (1)

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

$\begin{array}{l}+\to \sum F_x=0\\ 12.5+F_{AE}\cos \left(45°\right)+F_{AC}\cos \left(25°\right)=0\\ F_{AE}\cos \left(45°\right)+F_{AC}\cos \left(25°\right)=-12.5\end{array}$        (2)

Solve Equation (1) and Equation (2).

$\begin{array}{l}{F}_{AE}=-37.55\text{k}\\ F_{AC}=15.51\text{k}\end{array}$

Force in the member $CG$ is equal to $F_{AC}$ with the value of $15.51\text{k}$.

Consider joint E, find the force in the member CE and EG:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -10+F_{EG}\sin \left(45°\right)-F_{EC}\sin \left(45°\right)-F_{AE}\sin 45°=0\\ -10+F_{EG}\sin \left(45°\right)-F_{EC}\sin \left(45°\right)-\left(-37.55\right)\sin 45°=0\\ F_{EG}\sin \left(45°\right)-F_{EC}\sin \left(45°\right)=-16.55\end{array}$        (3)

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

$\begin{array}{l}+\to \sum F_x=0\\ 5+F_{EG}\cos \left(45°\right)+F_{EC}\cos \left(45°\right)-F_{AE}\cos 45°=0\\ 5+F_{EG}\cos \left(45°\right)+F_{EC}\cos \left(45°\right)-\left(-37.55\right)\cos 45°=0\\ F_{EG}\cos \left(45°\right)+F_{EC}\cos \left(45°\right)=-31.55\end{array}$        (4)

Solve Equation (3) and Equation (4).

$\begin{array}{l}{F}_{EG}=-34.02\text{k}\\ F_{EC}=-10.61\text{k}\end{array}$

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 moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ G_y\times 24-5\times 24-5\times 12+10\times 12=0\\ G_y=2.5\text{k}\uparrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -A_x+5+5=0\\ A_x=10\text{k}\leftarrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+10+G_y=0\\ -A_y+10+2.5=0\\ A_y=12.5\text{k}\downarrow \end{array}$

Consider joint A, find the force in the member AC and AE:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -12.5+F_{AE}\sin \left(45°\right)+F_{AC}\sin \left(25°\right)=0\\ F_{AE}\sin \left(45°\right)+F_{AC}\sin \left(25°\right)=12.5\end{array}$        (5)

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

$\begin{array}{l}+\to \sum F_x=0\\ -10+F_{AE}\cos \left(45°\right)+F_{AC}\cos \left(25°\right)=0\\ F_{AE}\cos \left(45°\right)+F_{AC}\cos \left(25°\right)=10\end{array}$        (6)

Solve Equation (5) and Equation (6).

$\begin{array}{l}{F}_{AE}=20.77\text{k}\\ F_{AC}=-5.17\text{k}\end{array}$

Force in the member $CG$ is equal to $F_{AC}$ with the value of $-5.17\text{k}$.

Consider joint E, find the force in the member CE and EG:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 10+F_{EG}\sin \left(45°\right)-F_{EC}\sin \left(45°\right)-F_{AE}\sin 45°=0\\ 10+F_{EG}\sin \left(45°\right)-F_{EC}\sin \left(45°\right)-\left(20.77\right)\sin 45°=0\\ F_{EG}\sin \left(45°\right)-F_{EC}\sin \left(45°\right)=4.69\end{array}$        (7)

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

$\begin{array}{l}+\to \sum F_x=0\\ 5+F_{EG}\cos \left(45°\right)+F_{EC}\cos \left(45°\right)-F_{AE}\cos 45°=0\\ 5+F_{EG}\cos \left(45°\right)+F_{EC}\cos \left(45°\right)-\left(20.77\right)\cos 45°=0\\ F_{EG}\cos \left(45°\right)+F_{EC}\cos \left(45°\right)=9.69\end{array}$        (8)

Solve Equation (7) and Equation (8).

$\begin{array}{l}{F}_{EG}=10.16\text{k}\\ F_{EC}=3.54\text{k}\end{array}$

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 CG is $\underset{\_}{10.34\text{k}\left(\text{T}\right)}$.

The force in the member AE is $\underset{\_}{16.78\text{k}\left(\text{C}\right)}$.

The force in the member CE is $\underset{\_}{7.07\text{k}\left(\text{C}\right)}$.

The force in the member EG is $\underset{\_}{23.86\text{k}\left(\text{C}\right)}$.

The force in the member BD and DG is $\underset{\_}{20.68\text{k}\left(\text{T}\right)}$.

The force in the member DF is $\underset{\_}{14.14\text{k}\left(\text{C}\right)}$.

The force in the member DG is $\underset{\_}{20.68\text{k}\left(\text{T}\right)}$.

The force in the member FG is $\underset{\_}{44.18\text{k}\left(\text{C}\right)}$.