Chapter 11, Problem 31P

To determine

Find the deflection at mid span of the truss.

Sketch the deflected shape of the truss.

Sketch the shear and moment diagrams for the members AB and AF.

Answer to Problem 31P

The deflection at mid span is $\underset{\_}{3.56\text{mm}}$.

The vertical reaction at D and F are $\underset{\_}{50\text{kN}\uparrow }$ and $\underset{\_}{50\text{kN}\uparrow }$.

The moment at A is $\underset{\_}{-44.44\text{kN}\cdot \text{m}}$.

The moment at B is $55.56\text{kN}\cdot \text{m}$.

Explanation of Solution

Given information:

The value of E is $200\text{GPa}$ and I is $250\times {10}^6{\text{mm}}^4$.

Calculation:

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.

Let $F_y$ and $D_y$ be the vertical reactions at the supports F and D.

Summation of moments about F is equal to 0.

$\begin{array}{l}\sum M_F=0\\ D_y\left(8\right)-100\left(4\right)=0\\ D_y=50\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_y+D_y-100=0\\ F_y+50-100=0\\ F_y=50\text{kN}\downarrow \end{array}$

Sketch the deflected shape of the truss as shown in Figure 1.

Refer to Figure 1.

The structure is symmetric about vertical axis at mid span.

Also the structure is anti symmetric about A horizontal axis at the mid depth.

Analyze the half of the structure with the following conditions exist.

The member EB is straight. Therefore the slope at B and E is zero ${\theta }_B={\theta }_E=0$.

The joints B and E are treated as fixed supports. Therefore, the deflection

${\Delta }_B={\Delta }_E=\Delta$

The deflected shape of members AB, CB, FE, and DE are identical. Therefore,

${\theta }_A={\theta }_F=\theta$

${\theta }_C={\theta }_D=-\theta$

The shear and moment values are identical for members AB, FE, DE, and CB.

Axial force exists in both horizontal and vertical members.

Find the fixed end moments of the truss:

Fixed end moments of member BA is equal to fixed end moments of member EF, AB, and FE.

$\begin{array}{c}{\text{FEM}}_{BA}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_A-3\Psi \right)\\ =\frac{2EI}{4}\left(0+0-3\Psi \right)\\ =\frac{3EI}{2}\Psi \\ ={\text{FEM}}_{EF}={\text{FEM}}_{AB}={\text{FEM}}_{FE}\end{array}$

Let consider arbitrarily $\frac{3EI}{2}\Psi$ as $90\text{kN}\cdot \text{m}$.

Find the distribution factor as follows:

$\begin{array}{c}{K}_{AB}=\frac{I}{L}\\ =\frac{I}{4}\\ =K_{FE}\end{array}$

$\begin{array}{c}{K}_{AF}=\frac{3}{2}\frac{I}{L}\\ =\frac{3}{2}\frac{I}{3}\\ =\frac{I}{2}\end{array}$

The summation is,

$\begin{array}{c}\Sigma K=\frac{I}{4}+\frac{I}{2}\\ =\frac{3I}{4}\end{array}$

$\begin{array}{c}D{F}_{AB}=D{F}_{FE}=\frac{I/4}{3I/4}\\ =\frac{I}{4}\times \frac{4}{3I}\\ =\frac{1}{3}\end{array}$

$\begin{array}{c}D{F}_{AF}=\frac{I/2}{3I/4}\\ =\frac{I}{2}\times \frac{4}{3I}\\ =\frac{2}{3}\end{array}$

Sketch the moment distribution of the truss as shown in Figure 2.

Find the vertical force V in the member AB and FE as follows:

Sketch the free body diagram of member AB as shown in Figure 3.

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ V\left(4\right)-60-75=0\\ V=33.75\text{kN}\end{array}$

The value of V is considered initially as 25 kN.

Therefore, reduce all the forces by the ratio of,

$\frac{25}{33.75}=0.7407$

The moment $60\text{kN}\cdot \text{m}$ becomes

$60\text{kN}\cdot \text{m}\left(0.7407\right)=44.44\text{kN}\cdot \text{m}$

The moment $75\text{kN}\cdot \text{m}$ becomes

$75\text{kN}\cdot \text{m}\left(0.7407\right)=55.56\text{kN}\cdot \text{m}$

Find the force in the member AF as follows:

Sketch the free body diagram of member AF as shown in Figure 4.

Summation of moments about F is equal to 0.

$\begin{array}{l}\sum M_F=0\\ -V\left(3\right)+44.44+44.44=0\\ V=29.63\text{kN}\end{array}$

Sketch the moment in each member of the truss as shown in Figure 5.

Sketch the axial force in each member of the truss as shown in Figure 6.

Therefore,

The vertical reaction at D and F are $\underset{\_}{50\text{kN}\uparrow }$ and $\underset{\_}{50\text{kN}\uparrow }$.

The moment at A is $\underset{\_}{-44.44\text{kN}\cdot \text{m}}$.

The moment at B is $55.56\text{kN}\cdot \text{m}$.

Find the value of deflection at the mid span of the truss as follows:

Use slope deflection equations:

For member FE,

$\begin{array}{l}{M}_{FE}=\frac{2EI}{L}\left(2{\theta }_F+{\theta }_E-3\Psi \right)\\ -44.44=\frac{2EI}{L}\left(2\theta +0-3\Psi \right)\\ -44.44=\frac{2EI}{L}\left(2\theta -3\Psi \right)\end{array}$        (1)

For member EF,

$\begin{array}{l}{M}_{EF}=\frac{2EI}{L}\left(2{\theta }_E+{\theta }_F-3\Psi \right)\\ -55.55=\frac{2EI}{L}\left(\theta +0-3\Psi \right)\\ -55.55=\frac{2EI}{L}\left(\theta -3\Psi \right)\end{array}$        (2)

Solve Equation (1) and Equation (2).

$\begin{array}{l}\theta =\frac{11.11L}{2EI}\\ \Psi =\frac{-66.66L}{3\times 2EI}\\ \frac{\Delta }{L}=\frac{-66.66L}{3\times 2EI}\\ \Delta =\frac{-66.66L^2}{6EI}\end{array}$

$\begin{array}{c}\Delta =\frac{-66.66{\left(4\right)}^2}{6\left(200\text{GPa}\right)\left(250\times {10}^6{\text{mm}}^4\right)}\\ =\frac{-66.66{\left(4\right)}^2}{6\left(200\text{GPa}\times \frac{{10}^6{\text{kN/m}}^2}{1\text{GPa}}\right)\left(250\times {10}^6{\text{mm}}^4\times {\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}^4\right)}\\ =3.56\times {10}^{-3}\text{m}\\ =3.56\text{mm}\end{array}$

Therefore, the deflection at mid span is $\underset{\_}{3.56\text{mm}}$.