Chapter 9, Problem 5P

To determine

Find the reactions and sketch the shear and moment curves.

Locate the point of maximum deflection of the beam and repeat the computation considering uniform I over the length of the beam.

Answer to Problem 5P

Non-Uniform I:

The vertical reaction at C is $\underset{\_}{6.71\text{kips}(\downarrow )}$.

The moment and vertical reaction at A are $\underset{\_}{40.65\text{kip}\cdot \text{ft}\left(\text{Anticlockwise}\right)}$ and $\underset{\_}{6.71\text{kips}(\uparrow )}$

Uniform I:

The vertical reaction at C is $\underset{\_}{6\text{kips}(\downarrow )}$.

The moment and vertical reaction at A are $\underset{\_}{30\text{kip}\cdot \text{ft}\left(\text{Anticlockwise}\right)}$ and $\underset{\_}{6\text{kips}(\uparrow )}$.

Explanation of Solution

Case 1: Non uniform I.

The moment M is 60 kip ft.

Show the free body diagram of the beam as shown in Figure 1.

Refer Figure 1.

Consider the vertical reaction at C as the redundant.

Remove the redundant at C to get the released structure.

Sketch the released structure with applied loading as shown in Figure 2.

Refer Figure 2.

Find the value of $\frac{M}{EI}$ at A, B, C as follows:

$\begin{array}{c}{\left(\frac{M}{EI}\right)}_{AB}=\frac{60}{E\left(2I\right)}\\ =\frac{30}{EI}\end{array}$

${\left(\frac{M}{EI}\right)}_{BC}=\frac{60}{EI}$

Sketch the $\left(\frac{M}{EI}\right)$ for the beam as shown in Figure 3.

Refer Figure 3.

Find the deflection ${\Delta }_{CO}$ as follows:

$\begin{array}{c}{\Delta }_{CO}=t_{C/A}\\ =\frac{30}{EI}\times 6\left(12\right)+\frac{60}{EI}\times 9\left(4.5\right)\\ =\frac{4590}{EI}\end{array}$

Sketch the released structure with unit loading as shown in Figure 4.

Refer Figure 4.

Find the value of $\frac{M}{EI}$ at A, B, C as follows:

$\begin{array}{c}{\left(\frac{M}{EI}\right)}_A=\frac{1\times 15}{E\left(2I\right)}\\ =\frac{7.5}{EI}\end{array}$

$\begin{array}{c}{\left(\frac{M}{EI}\right)}_{B,left}=\frac{1\times 9}{E\left(2I\right)}\\ =\frac{4.5}{EI}\end{array}$

$\begin{array}{c}{\left(\frac{M}{EI}\right)}_{B,right}=\frac{1\times 9}{E\left(I\right)}\\ =\frac{9}{EI}\end{array}$

Sketch the $\left(\frac{M}{EI}\right)$ for the beam as shown in Figure 5.

Refer Figure 5.

Find the deflection ${\delta }_{CC}$ as follows:

$\begin{array}{c}{\delta }_{CC}=\left[\begin{array}{l}\left(\frac{1}{2}\times 9\times \frac{9}{EI}\right)\times 6+\\ \left(6\times \frac{4.5}{EI}\right)\times \left(9+3\right)+\\ \left(\frac{1}{2}\times 6\times \frac{3}{EI}\right)\times \left(9+4\right)\end{array}\right]\\ =\frac{684}{EI}\end{array}$

Show the compatibility Equation as follows:

$\begin{array}{c}{\Delta }_C={\Delta }_{CC}+{\delta }_{CC}{R}_C\\ 0=\frac{4590}{EI}+\frac{684}{EI}{R}_C\\ R_C=-\frac{4590}{EI}\times \frac{EI}{684}\\ =-6.71\text{kips}\\ =6.71\text{kips}(\downarrow )\end{array}$

Thus, the vertical reaction at C is $\underset{\_}{6.71\text{kips}(\downarrow )}$.

Find the moment at A as follows:

$\begin{array}{c}{M}_A=60-\left(6.71\times 15\right)\\ =40.65\text{kip}\cdot \text{ft}\left(\text{Anticlockwise}\right)\end{array}$

Find the vertical reaction at A as follows:

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_A+R_C=0\\ R_A-6.71=0\\ R_A=6.71\text{kips}(\uparrow )\end{array}$

Thus, the moment and vertical reaction at A are $\underset{\_}{40.65\text{kip}\cdot \text{ft}\left(\text{Anticlockwise}\right)}$ and $\underset{\_}{6.71\text{kips}(\uparrow )}$.

Find the shear force at A and C.

$V_A=6.71\text{kips}$

$V_C=-6.71\text{kips}$

Find the bending moment at A and C.

$\begin{array}{c}{M}_A=-6.71\times \text{15+60}\\ =\text{40}\text{.65}\text{kip}\cdot \text{ft}\left(\text{Clockwise}\right)\end{array}$

$M_C=\text{60}\text{kip}\cdot \text{ft}\left(\text{Anticlockwise}\right)$

Sketch the shear force and bending moment diagram as shown in Figure 6.

Refer Figure 6.

Find the distance $x_1$ from left support.

Find the distance $x_1$ as follows:

$\begin{array}{l}\frac{x_1}{40.65}=\frac{15-x_1}{60}\\ 60x_1=609.75-40.65x_1\\ 100.65x_1=609.75\\ x_1=6.058\text{ft}\end{array}$

Consider the point of maximum deflection occurs at a distance x from left support.

Find the deflection of AC as follows:

$\begin{array}{c}{\Delta }_{AC}=\left\{\begin{array}{l}\left(-\frac{1}{2}\times \frac{40.65}{2EI}\times 6.056\right)\left(\frac{1}{3}\times 6.056\right)+\\ \left(\frac{1}{2}\times 8.944\times \frac{60}{EI}\right)\left(6.056+\frac{2}{3}\times \left(8.944\right)\right)\end{array}\right\}\\ =\frac{-62.1}{EI}+\frac{3224.85}{EI}\\ =\frac{3162.75}{EI}\end{array}$

Find the slope at C as follows:

$\begin{array}{c}{\theta }_C=\frac{{\Delta }_{AC}}{L}\\ =\frac{3162.75}{15EI}\\ =\frac{210.85}{EI}\end{array}$        (1)

Calculate the area of the shaded region of M/EI diagram as follows:

$Area=\left(\frac{60+\frac{60}{8.944}}{2}\right)x_m$        (2)

Equate Equation (1) and (2).

$\begin{array}{l}{\theta }_C=Area\\ \frac{210.85}{EI}=\left(\frac{60+\frac{60}{8.944}{x}_m}{2EI}\right)x_m\\ 421.7=\left(60+6.7084x_m\right)x_m\\ 6.7084x_m^2+60x_m-421.7=0\end{array}$

Solve the above Equation.

$x_m=4.63\text{ft}$

Find the location of maximum deflection from the right support as follows:

$\begin{array}{c}\text{Distance}=8.944-x_m\\ =8.944-4.63\text{ft}\\ \text{=4}\text{.31}\text{ft}\end{array}$

Thus, the point of maximum deflection occurs at a distance $4.31\text{ft}$ from right support.

Case 2: Uniform I.

The moment M is 60 kip ft.

Show the free body diagram of the beam as shown in Figure 7.

Refer Figure 7.

Consider the vertical reaction at C as the redundant.

Remove the redundant at C to get the released structure.

Sketch the released structure with applied loading as shown in Figure 8.

Refer Figure 8.

Show the deflection ${\Delta }_{CO}$ at the free end of the beam as follows:

$\begin{array}{c}{\Delta }_{CO}=\frac{M{L}^2}{2EI}\\ =\frac{60\times {15}^2}{2EI}\\ =\frac{6750}{EI}\end{array}$

Sketch the released structure with unit loading as shown in Figure 9.

Refer Figure 9.

Show the deflection ${\delta }_{CC}$ due to unit load at free end as follows:

$\begin{array}{c}{\delta }_{CC}=\frac{P{L}^3}{3EI}\\ =\frac{1\times {15}^3}{3EI}\\ =\frac{1125}{EI}\end{array}$

Show the compatibility Equation as follows:

$\begin{array}{c}{\Delta }_C={\Delta }_{CC}+{\delta }_{CC}{R}_C\\ 0=\frac{6750}{EI}+\frac{1125}{EI}{R}_C\\ R_C=-\frac{6750}{EI}\times \frac{EI}{1125}\\ =-6\text{kips}\\ =6\text{kips}(\downarrow )\end{array}$

Thus, the vertical reaction at C is $\underset{\_}{6\text{kips}(\downarrow )}$.

Find the moment at A as follows:

$\begin{array}{c}{M}_A=60-\left(6\times 15\right)\\ =30\text{kip}\cdot \text{ft}\end{array}$

Find the vertical reaction at A as follows:

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_A+R_C=0\\ R_A-6=0\\ R_A=6\text{kips}\end{array}$

Thus, the moment and vertical reaction at A are $\underset{\_}{30\text{kip}\cdot \text{ft}\left(\text{Anticlockwise}\right)}$ and $\underset{\_}{6\text{kips}(\uparrow )}$.

Find the shear force at A and C.

$V_A=6\text{kips}$

$V_C=-6\text{kips}$

Find the bending moment at A and C.

$\begin{array}{c}{M}_A=-6\times \text{15+60}\\ =30\text{kip}\cdot \text{ft}\left(\text{Clockwise}\right)\end{array}$

$M_C=\text{60}\text{kip}\cdot \text{ft}\left(\text{Anticlockwise}\right)$

Sketch the shear force and bending moment diagram as shown in Figure 10.

Refer Figure 10.

Find the distance x as follows:

$\begin{array}{l}\frac{x_1}{30}=\frac{15-x_1}{60}\\ 60x_1=450-30x_1\\ 90x_1=450\\ x_1=5\text{ft}\end{array}$

Consider the point of maximum deflection occurs at a distance x from left support.

Find the deflection of AC as follows:

$\begin{array}{c}{\Delta }_{AC}=\left\{\begin{array}{l}\left(-\frac{1}{2}\times \frac{30}{EI}\times 5\right)\left(\frac{1}{3}\times 5\right)+\\ \left(\frac{1}{2}\times 10\times \frac{60}{EI}\right)\left(5+\frac{2}{3}\times 10\right)\end{array}\right\}\\ =\frac{-125}{EI}+\frac{3500}{EI}\\ =\frac{3375}{EI}\end{array}$

Find the slope at C as follows:

$\begin{array}{c}{\theta }_C=\frac{{\Delta }_{AC}}{L}\\ =\frac{3375}{15EI}\\ =\frac{225}{EI}\end{array}$        (1)

Calculate the area of the shaded region of M/EI diagram as follows:

$\text{Area}=\left(\frac{60+\frac{60}{10}}{2}\right)x_m$        (2)

Equate Equation (1) and (2).

$\begin{array}{l}{\theta }_C=\text{Area}\\ \frac{225}{EI}=\left(\frac{60+\frac{60}{10}{x}_m}{2EI}\right)x_m\\ 450=\left(60+6x_m\right)x_m\\ 6x_m^2+60x_m-450=0\end{array}$

Solve the above Equation.

$x_m=5\text{ft}$

Find the location of maximum deflection from the right support as follows:

$\begin{array}{c}\text{Distance}=10-x_m\\ =10-5\\ =\text{5}\text{ft}\end{array}$

Thus, the point of maximum deflection occurs at a distance $5\text{ft}$ from right support.