Chapter 10, Problem 19P

To determine

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

Answer to Problem 19P

The force in the member AB and BC is $\underset{\_}{180\text{kN}\left(\text{C}\right)}$.

The force in the member AH is $\underset{\_}{225\text{kN}\left(\text{T}\right)}$.

The force in the member HI is $\underset{\_}{280\text{kN}\left(\text{T}\right)}$.

The force in the member BH is $\underset{\_}{60\text{kN}\left(\text{C}\right)}$.

The force in the member CH is $\underset{\_}{125\text{kN}\left(\text{C}\right)}$.

The force in the member CI is $\underset{\_}{15\text{kN}\left(\text{T}\right)}$.

The force in the member CD is $\underset{\_}{280\text{kN}\left(\text{C}\right)}$.

The force in the member DI is $\underset{\_}{25\text{kN}\left(\text{C}\right)}$.

The force in the member IJ and JK is $\underset{\_}{300\text{kN}\left(\text{T}\right)}$.

The force in the member DK is $\underset{\_}{75\text{kN}\left(\text{C}\right)}$.

The force in the member EK is $\underset{\_}{45\text{kN}\left(\text{T}\right)}$.

The force in the member EF and FG is $\underset{\_}{140\text{kN}\left(\text{C}\right)}$.

The force in the member GL is $\underset{\_}{175\text{kN}\left(\text{T}\right)}$.

The force in the member FL is $\underset{\_}{30\text{kN}\left(\text{C}\right)}$.

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

The force in the member EL is $\underset{\_}{125\text{kN}\left(\text{C}\right)}$.

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 right of s axis.

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

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -30-45-45+G_y=0\\ G_y=120\text{kN}\uparrow \end{array}$

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ 120\times 12-45\times 8-45\times 4-J_x\times 3=0\\ -3J_x=-900\\ J_x=300\text{kN}\leftarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ D_x-J_x=0\\ D_x-300=0\\ D_x=300\text{kN}\to \end{array}$

Consider joint J, find the force in the member JK and DJ:

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

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

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

$\begin{array}{l}+\to \sum F_x=0\\ -300+F_{JK}=0\\ F_{JK}=300\text{kN}\left(\text{T}\right)\end{array}$

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

$\begin{array}{l}\tan \theta =\frac{3}{4}\\ \theta =36.87°\end{array}$

Consider joint D, find the force in the member DE and DK:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -30-F_{DK}\sin \left(36.87°\right)=0\\ F_{DK}=-50\text{kN}\\ F_{DK}=50\text{kN}\left(\text{C}\right)\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ 300+F_{DE}+F_{DK}\cos \left(36.87°\right)=0\\ 300+F_{DE}-50\cos \left(36.87°\right)=0\\ F_{DE}=-260\text{kN}\end{array}$

$F_{DE}=260\text{kN}\left(\text{C}\right)$

Consider joint K, find the force in the member KL and EK:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_{EK}+F_{DK}\sin \left(36.87°\right)=0\\ F_{EK}-50\sin \left(36.87°\right)=0\\ F_{EK}=30\text{kN}\end{array}$

$F_{EK}=30\text{kN}\left(\text{T}\right)$

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{JK}+F_{KL}-F_{DK}\cos \left(36.87°\right)=0\\ -300+F_{KL}+50\cos \left(36.87°\right)=0\\ F_{KL}=260\text{kN}\end{array}$

$F_{KL}=260\text{kN}\left(\text{T}\right)$

Consider joint E, find the force in the member EF and EL:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -45-F_{EK}-F_{EL}\sin 36.87°=0\\ -45-30-F_{EL}\sin 36.87°=0\\ F_{EL}=-1\text{25}\text{kN}\end{array}$

$F_{EL}=1\text{25}\text{kN}\left(\text{C}\right)$

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{DE}+F_{EF}+F_{EL}\cos \left(36.87°\right)=0\\ 260+F_{EF}-125\cos \left(36.87°\right)=0\\ F_{EF}=-160\text{kN}\end{array}$

$F_{EF}=160\text{kN}\left(\text{C}\right)$

Consider joint F, find the force in the member FL and FG:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -45-F_{FL}=0\\ F_{FL}=-45\text{kN}\\ F_{FL}=4\text{5}\text{kN}\left(\text{C}\right)\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{EF}+F_{FG}=0\\ 160+F_{FG}=0\\ F_{FG}=-160\text{kN}\end{array}$

$F_{FG}=160\text{kN}\left(\text{C}\right)$

Consider joint G, find the force in the member GL:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 120-F_{GL}\sin 36.87°=0\\ F_{GL}=200\text{kN}\\ F_{GL}=200\text{kN}\left(\text{T}\right)\end{array}$

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 right of s axis.

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

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ -G_y\times 12+15\times 8+15\times 4=0\\ -12G_y=-180\\ G_y=15\text{kN}\downarrow \end{array}$

The vertical reaction at joint J is 0.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ J_y-D_y+15+15-G_y=0\\ 0-D_y+15+15-15=0\\ D_y=15\text{kN}\downarrow \end{array}$

Consider joint J, find the force in the member JK:

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

$\begin{array}{l}+\to \sum F_x=0\\ F_{JK}=0\end{array}$

Consider joint D, find the force in the member DE and DK:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -15-F_{DK}\sin \left(36.87°\right)=0\\ F_{DK}=-25\text{kN}\\ F_{DK}=25\text{kN}\left(\text{C}\right)\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ F_{DE}+F_{DK}\cos \left(36.87°\right)=0\\ F_{DE}-25\cos \left(36.87°\right)=0\\ F_{DE}=20\text{kN}\end{array}$

$F_{DE}=20\text{kN}\left(\text{T}\right)$

Consider joint K, find the force in the member KL and EK:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_{EK}+F_{DK}\sin \left(36.87°\right)=0\\ F_{EK}-25\sin \left(36.87°\right)=0\\ F_{EK}=15\text{kN}\end{array}$

$F_{EK}=15\text{kN}\left(\text{T}\right)$

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{JK}+F_{KL}-F_{DK}\cos \left(36.87°\right)=0\\ 0+F_{KL}+25\cos \left(36.87°\right)=0\\ F_{KL}=-20\text{kN}\end{array}$

$F_{KL}=-20\text{kN}\left(\text{C}\right)$

Consider joint E, find the force in the member EF and EL:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 15-F_{EK}-F_{EL}\sin 36.87°=0\\ 15-15-F_{EL}\sin 36.87°=0\\ F_{EL}=0\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{DE}+F_{EF}+F_{EL}\cos \left(36.87°\right)=0\\ -20+F_{EF}-0=0\\ F_{EF}=20\text{kN}\left(\text{T}\right)\end{array}$

Consider joint F, find the force in the member FL and FG:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 15-F_{FL}=0\\ F_{FL}=15\text{kN}\\ F_{FL}=1\text{5}\text{kN}\left(\text{T}\right)\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{EF}+F_{FG}=0\\ -20+F_{FG}=0\\ F_{FG}=20\text{kN}\left(\text{T}\right)\end{array}$

Consider joint G, find the force in the member GL:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -15-F_{GL}\sin 36.87°=0\\ F_{GL}=-25\text{kN}\\ F_{GL}=25\text{kN}\left(\text{C}\right)\end{array}$

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 AB and BC is $\underset{\_}{180\text{kN}\left(\text{C}\right)}$.

The force in the member AH is $\underset{\_}{225\text{kN}\left(\text{T}\right)}$.

The force in the member HI is $\underset{\_}{280\text{kN}\left(\text{T}\right)}$.

The force in the member BH is $\underset{\_}{60\text{kN}\left(\text{C}\right)}$.

The force in the member CH is $\underset{\_}{125\text{kN}\left(\text{C}\right)}$.

The force in the member CI is $\underset{\_}{15\text{kN}\left(\text{T}\right)}$.

The force in the member CD is $\underset{\_}{280\text{kN}\left(\text{C}\right)}$.

The force in the member DI is $\underset{\_}{25\text{kN}\left(\text{C}\right)}$.

The force in the member IJ and JK is $\underset{\_}{300\text{kN}\left(\text{T}\right)}$.

The force in the member DK is $\underset{\_}{75\text{kN}\left(\text{C}\right)}$.

The force in the member EK is $\underset{\_}{45\text{kN}\left(\text{T}\right)}$.

The force in the member EF and FG is $\underset{\_}{140\text{kN}\left(\text{C}\right)}$.

The force in the member GL is $\underset{\_}{175\text{kN}\left(\text{T}\right)}$.

The force in the member FL is $\underset{\_}{30\text{kN}\left(\text{C}\right)}$.

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

The force in the member EL is $\underset{\_}{125\text{kN}\left(\text{C}\right)}$.