Chapter 7, Problem 36P

To determine

Calculate the deflection and the slopes on both sides of the hinge at B using conjugate beam method.

Answer to Problem 36P

The slope on the right of hinge at B is $\underset{\_}{{\theta }_{BR}=\frac{83.3}{EI}}$.

The slope on the left of hinge at B is $\underset{\_}{{\theta }_{BL}=0}$.

The deflection on the both side of the hinge at B is $\underset{\_}{{\Delta }_B=0}$.

Explanation of Solution

Given information:

Apply the sign conventions for calculating reactions 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 positive and the counter clockwise moment as negative.

Sketch the Free Body Diagram of the beam as shown in Figure 1.

Refer to Figure 1.

Consider span DEF.

Use equilibrium equations:

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ -R_F\times 10+40\times 5=0\\ R_F=20\text{ kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_{DR}-40+20=0\\ R_{DR}=20\text{ kN}\end{array}$

The vertical reaction at the left of the hinge at D is $R_{DL}=20\text{ kN}(\downarrow )$.

Consider span BCD.

Use equilibrium equations:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ 20\times 10-R_C\times 5=0\\ R_C=40\text{ kN}\end{array}$

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

$\begin{array}{c}+\uparrow \sum F_y=0\\ R_{BR}+40-20=0\\ R_{BR}=20\text{kips}\\ R_{BL}=20-20\\ =0\end{array}$

The vertical reaction at the left of the hinge at B is $R_{BL}=0$.

Consider span AB.

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

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

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ M_A=0\end{array}$

Calculate the moment at each point as shown below.

$\begin{array}{c}{M}_A=0\\ M_B=0\\ M_C=-20\times 5\\ =-100\text{ kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{M}_D=-20\times 10+40\times 5\\ =0\\ M_E=-20\times 15+40\times 10\\ =100\text{ kN}\cdot \text{m}\\ M_C=-20\times 20+40\times 15-40\times 5\\ =0\end{array}$

Sketch the $\frac{M}{EI}$ diagram as shown in Figure 2.

In the real beam A is the fixed end, F and C are the roller support, and B and D are the internal hinge but in conjugate-beam method the end conditions has to be change.

Change fixed end at A as free end, the roller at F as roller support, the roller at B as internal hinge, and the internal hinges at B and D as roller support.

Sketch the conjugate beam of the loading from M/EI diagram as shown in Figure 3.

Refer to Figure 3.

Consider span ABC.

Use equilibrium equations:

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ R_B\times 5-\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{1}{3}\times 5\right)=0\\ R_B=\frac{83.3}{EI}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_{CL}+\frac{83.3}{EI}-\frac{1}{2}\times 5\times \frac{100}{EI}=0\\ R_{CL}=\frac{166.7}{EI}\end{array}$

The vertical reaction at the right of the hinge at C is $R_{CR}=\frac{166.7}{EI}(\downarrow )$.

Consider span CDEF.

Use equilibrium equations:

Summation of moments about F is equal to 0.

$\begin{array}{l}\sum M_F=0\\ \left(\begin{array}{l}-\frac{166.7}{EI}\times 15+R_D\times 10-\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{2}{3}\times 5+10\right)\\ +\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{1}{3}\times 5+5\right)+\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{2}{3}\times 5\right)\end{array}\right)=0\\ R_D=\frac{333.3}{EI}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_F-166.7+\frac{333.3}{EI}-\frac{1}{2}\times 5\times \frac{100}{EI}+\frac{1}{2}\times 10\times \frac{100}{EI}=0\\ R_F=\frac{416.7}{EI}(\downarrow )\end{array}$

Calculate the shear (slope) of conjugate beam at each point as shown below.

$\begin{array}{c}{V}_A=0\\ V_B=\frac{83.3}{EI}\\ V_C=\frac{83.3}{EI}-\frac{1}{2}\times 5\times \frac{100}{EI}\\ =-\frac{166.7}{EI}\end{array}$

$\begin{array}{l}{V}_{@\text{ left of }D}=\frac{83.3}{EI}-\frac{1}{2}\times 10\times \frac{100}{EI}\\ =-\frac{416.7}{EI}\\ V_D=\frac{83.3}{EI}-\frac{1}{2}\times 10\times \frac{100}{EI}+\frac{333.3}{EI}\\ =-\frac{83.3}{EI}\end{array}$

$\begin{array}{l}{V}_E=\frac{83.3}{EI}-\frac{1}{2}\times 10\times \frac{100}{EI}+\frac{333.3}{EI}+\frac{1}{2}\times 5\times \frac{100}{EI}\\ =\frac{166.7}{EI}\\ V_{@\text{ left of }E}=\frac{83.3}{EI}-\frac{1}{2}\times 10\times \frac{100}{EI}+\frac{333.3}{EI}+\frac{1}{2}\times 10\times \frac{100}{EI}\\ =\frac{416.7}{EI}\\ V_E=\frac{83.3}{EI}-\frac{1}{2}\times 10\times \frac{100}{EI}+\frac{333.3}{EI}+\frac{1}{2}\times 10\times \frac{100}{EI}-\frac{416.7}{EI}\\ =0\end{array}$

Calculate the moment (deflection) of conjugate beam at each point as shown below.

$\begin{array}{l}{M}_A=0\\ M_B=0\\ M_{\text{@ left of}C}=\frac{83.3}{EI}\times 5\\ =\frac{416.7}{EI}\\ M_C=\frac{83.3}{EI}\times 5-\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{1}{3}\times 5\right)\\ =0\end{array}$

$\begin{array}{c}{M}_D=\frac{83.3}{EI}\times 10-\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{1}{3}\times 5+5\right)-\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{2}{3}\times 5\right)\\ =-\frac{1666.7}{EI}\\ M_E=\left(\begin{array}{l}\frac{83.3}{EI}\times 20-\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{1}{3}\times 5+15\right)-\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{2}{3}\times 5+10\right)\\ +\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{1}{3}\times 5+5\right)+\frac{1}{2}\times 5\times \frac{100}{EI}\times \left(\frac{2}{3}\times 5\right)+\frac{333.3}{EI}\times 10\end{array}\right)\\ =0\end{array}$

Sketch the slope (shear) and deflection (moment) of the beam as shown in Figure 4.

Refer to Figure 4.

The slope on the right of hinge at B is $\underset{\_}{{\theta }_{BR}=\frac{83.3}{EI}}$.

The slope on the left of hinge at B is $\underset{\_}{{\theta }_{BL}=0}$.

The deflection on the both side of the hinge at B is $\underset{\_}{{\Delta }_B=0}$.