Chapter 7, Problem 40P

To determine

Sketch the slope and deflection diagrams of the beam.

Explanation of Solution

Given information:

The flexural rigidity of segment ABCD is 2I and DEF is I.

Calculation:

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.

Use conjugate beam method.

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.

$\begin{array}{l}\sum M_D^{DEF}=0\\ -R_F\times 16-10\times 8=0\\ R_F=5\text{ kips}(\downarrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_{DR}+10-5=0\\ R_{DR}=5\text{ kips}(\downarrow )\end{array}$

The vertical reaction at the left of the hinge at D is $R_{DL}=5\text{ kips}(\uparrow )$.

Consider span ABCD.

Use equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A^{ABCD}=0\\ -5\times 20-R_C\times 8+25\times 4=0\\ R_C=0\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_A+5-25=0\\ R_A=20\text{kips}(\uparrow )\end{array}$

Calculate the moment at each point as shown below.

$\begin{array}{c}{M}_A=0\\ M_B=20\times 4\\ =80\text{ kip}\cdot \text{ft}\\ M_C=20\times 8-25\times 4\\ =60\text{ kip}\cdot \text{ft}\end{array}$

$\begin{array}{c}{M}_D=20\times 20-25\times 16\\ =0\\ M_E=20\times 28-25\times 24\\ =-40\text{ kip}\cdot \text{ft}\\ M_F=20\times 36-25\times 32+10\times 8\\ =0\end{array}$

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

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

Change the hinged support at A as hinged support, the roller at F as roller support, the roller at C as internal hinge, and the internal hinge at 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 A is equal to 0.

$\begin{array}{l}\sum M_C=0\\ R_A\times 8+\frac{1}{2}\times 4\times \frac{40}{EI}\times \left(\frac{1}{3}\times 4+4\right)+\frac{1}{2}\times 4\times \frac{10}{EI}\times \left(\frac{2}{3}\times 4\right)+4\times \frac{30}{EI}\times \frac{4}{2}=0\\ R_A=\frac{90}{EI}(\downarrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_{CL}+\frac{1}{2}\times 4\times \frac{40}{EI}+\frac{1}{2}\times 4\times \frac{10}{EI}+4\times \frac{30}{EI}-\frac{90}{EI}=0\\ R_{CL}=\frac{130}{EI}(\downarrow )\end{array}$

Consider span CDEF.

Use equilibrium equations:

Summation of moments about F is equal to 0.

$\begin{array}{l}\sum M_F=0\\ \left(\begin{array}{l}\frac{130}{EI}\times 28+\frac{1}{2}\times 12\times \frac{30}{EI}\times \left(\frac{2}{3}\times 12+16\right)+R_D\times 16\\ -\frac{1}{2}\times 8\times \frac{40}{EI}\times \left(\frac{1}{3}\times 8+8\right)-\frac{1}{2}\times 8\times \frac{40}{EI}\times \left(\frac{2}{3}\times 8\right)\end{array}\right)=0\\ R_D=\frac{337.5}{EI}(\downarrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_F-\frac{337.5}{EI}+\frac{130}{EI}+\frac{1}{2}\times 12\times \frac{30}{EI}-\frac{1}{2}\times 16\times \frac{40}{EI}=0\\ R_F=\frac{347.5}{EI}\end{array}$

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

$\begin{array}{c}{V}_A=-\frac{90}{EI}\\ V_B=-\frac{90}{EI}+\frac{1}{2}\times 4\times \frac{40}{EI}\\ =-\frac{10}{EI}\\ V_C=-\frac{90}{EI}+\frac{1}{2}\times 4\times \frac{40}{EI}+\frac{1}{2}\times 4\times \frac{10}{EI}+4\times \frac{30}{EI}\\ =\frac{130}{EI}\end{array}$

$\begin{array}{l}{V}_{@\text{ left of }D}=-\frac{90}{EI}+\frac{1}{2}\times 4\times \frac{40}{EI}+\frac{1}{2}\times 16\times \frac{40}{EI}\\ =\frac{310}{EI}\\ V_D=-\frac{90}{EI}+\frac{1}{2}\times 4\times \frac{40}{EI}+\frac{1}{2}\times 16\times \frac{40}{EI}-\frac{337.5}{EI}\\ =-\frac{27.5}{EI}\end{array}$

$\begin{array}{l}{V}_E=-\frac{90}{EI}+\frac{1}{2}\times 4\times \frac{40}{EI}+\frac{1}{2}\times 16\times \frac{40}{EI}-\frac{337.5}{EI}-\frac{1}{2}\times 8\times \frac{40}{EI}\\ =-\frac{187.5}{EI}\\ V_{@\text{ left of }F}=-\frac{90}{EI}+\frac{1}{2}\times 4\times \frac{40}{EI}+\frac{1}{2}\times 16\times \frac{40}{EI}-\frac{337.5}{EI}-\frac{1}{2}\times 16\times \frac{40}{EI}\\ =-\frac{347.5}{EI}\\ V_E=-\frac{90}{EI}+\frac{1}{2}\times 4\times \frac{40}{EI}+\frac{1}{2}\times 16\times \frac{40}{EI}-\frac{337.5}{EI}-\frac{1}{2}\times 16\times \frac{40}{EI}+\frac{347.5}{EI}\\ =0\end{array}$

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

$\begin{array}{c}{M}_A=0\\ M_B=-\frac{90}{EI}\times 4+\frac{1}{2}\times 4\times \frac{40}{EI}\times \left(\frac{1}{3}\times 4\right)\\ =-\frac{253.3}{EI}\\ M_C=-\frac{90}{EI}\times 8+\frac{1}{2}\times 4\times \frac{40}{EI}\times \left(\frac{1}{3}\times 4+4\right)+\frac{1}{2}\times 4\times \frac{10}{EI}\times \left(\frac{2}{3}\times 4\right)+4\times \frac{30}{EI}\times \frac{4}{2}\\ =0\end{array}$

$\begin{array}{c}{M}_D=-\frac{90}{EI}\times 20+\frac{1}{2}\times 4\times \frac{40}{EI}\times \left(\frac{1}{3}\times 4+16\right)+\frac{1}{2}\times 16\times \frac{40}{EI}\times \left(\frac{2}{3}\times 16\right)\\ =\frac{3000}{EI}\\ M_E=\left(\begin{array}{l}-\frac{90}{EI}\times 28+\frac{1}{2}\times 4\times \frac{40}{EI}\times \left(\frac{1}{3}\times 4+24\right)+\frac{1}{2}\times 16\times \frac{40}{EI}\times \left(\frac{2}{3}\times 16+8\right)-\frac{337.5}{EI}\times 8\\ -\frac{1}{2}\times 8\times \frac{40}{EI}\times \left(\frac{1}{3}\times 8\right)\end{array}\right)\\ =\frac{2353.3}{EI}\\ M_F=\left(\begin{array}{l}-\frac{90}{EI}\times 36+\frac{1}{2}\times 4\times \frac{40}{EI}\times \left(\frac{1}{3}\times 4+32\right)+\frac{1}{2}\times 16\times \frac{40}{EI}\times \left(\frac{2}{3}\times 16+16\right)-\frac{337.5}{EI}\times 16\\ -\frac{1}{2}\times 8\times \frac{40}{EI}\times \left(\frac{1}{3}\times 8+8\right)-\frac{1}{2}\times 8\times \frac{40}{EI}\times \left(\frac{2}{3}\times 8\right)\end{array}\right)\\ =0\end{array}$

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