Chapter 9.5, Problem 112P

(a)

To determine

Find the slope $\left({\theta }_A\right)$ at end A.

Answer to Problem 112P

The slope $\left({\theta }_A\right)$ at end A is $\underset{\_}{-\frac{w{a}^2}{2EI}\left(\frac{3L-2a}{12}\right)}$.

Explanation of Solution

Calculation:

Use moment area method:

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

Determine the reaction of the support by taking the vertical equilibrium condition:

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ R_A=R_E=wa\end{array}$

Determine the moment over the span AB by taking the moment:

$M=R_{A}x-\left(wa\times \frac{a}{2}\right)$

Substitute $wa$ for $R_A$.

$M=wax-\left(wa\times \frac{a}{2}\right)$

Determine the moment over the span BD by taking the moment:

$\begin{array}{c}M=\left(wa\times \frac{a}{2}\right)\\ =\frac{w{a}^2}{2}\end{array}$

Show the moment $\left(\frac{M}{EI}\right)$ diagram for the loading beam as in Figure 2.

Due to symmetric beam with symmetric loading, consider the left side of the beam.

Calculate the moment $\left(M_1\right)$ due to reaction of the support using the formula:

$M_1=R_A\times a$

Substitute $wa$ for $R_A$.

$\begin{array}{c}{M}_1=wa\times a\\ =w{a}^2\end{array}$

Calculate the ratio of $\frac{M_1}{EI}$:

$\text{Ratio}=\frac{M_1}{EI}$

Substitute $w{a}^2$ for $M_1$.

$\frac{M_1}{EI}=\frac{w{a}^2}{EI}$

Calculate the area $\left(A_1\right)$ due to the moment $\left(M_1\right)$ using the formula:

$A_1=\frac{1}{2}{b}_1{h}_1$

Here, $b_1$ is the width of the triangle in area $\left(A_1\right)$ and $h_1$ is the height of the triangle in area $\left(A_1\right)$.

Substitute a for $b_1$ and $\frac{w{a}^2}{EI}$ for $h_1$.

$\begin{array}{c}{A}_1=\frac{1}{2}\times a\times \frac{w{a}^2}{EI}\\ =\frac{w{a}^3}{2EI}\end{array}$

Calculate the moment $\left(M_2\right)$ due to loading of the beam using the formula:

$M_2=-\frac{w{a}^2}{2}$

Calculate the ratio of $\frac{M_2}{EI}$:

$\text{Ratio}=\frac{M_2}{EI}$

Substitute $-\frac{w{a}^2}{2}$ for $M_2$.

$\frac{M_2}{EI}=-\frac{w{a}^2}{2EI}$

Calculate the area $\left(A_2\right)$ due to the moment $\left(M_2\right)$ using the formula:

$A_2=\frac{1}{3}{b}_2{h}_2$

Here, $b_2$ is the width of the triangle in area $\left(A_2\right)$ and $h_2$ is the height of the triangle in area $\left(A_2\right)$.

Substitute a for $b_1$ and $-\frac{w{a}^2}{2EI}$ for $h_1$.

$\begin{array}{c}{A}_2=\frac{1}{3}\times a\times -\frac{w{a}^2}{2EI}\\ =-\frac{w{a}^3}{6EI}\end{array}$

Calculate the moment $\left(M_3\right)$ due to loading of the beam using the formula:

$M_3=\frac{w{a}^2}{2}$

Calculate the ratio of $\frac{M_3}{EI}$:

$\text{Ratio}=\frac{M_3}{EI}$

Substitute $\frac{w{a}^2}{2}$ for $M_3$.

$\frac{M_3}{EI}=\frac{w{a}^2}{2EI}$

Calculate the area $\left(A_3\right)$ due to the moment $\left(M_3\right)$ using the formula:

$A_3=b_3{h}_3$

Here, $b_3$ is the width of the rectangle in area $\left(A_3\right)$ and $h_3$ is the height of the rectangle in area $\left(A_3\right)$.

Substitute $\left(\frac{L}{2}-a\right)$ for $b_3$ and $\frac{w{a}^2}{2EI}$ for $h_3$.

$\begin{array}{c}{A}_3=\left(\frac{L}{2}-a\right)\times \frac{w{a}^2}{2EI}\\ =\left(\frac{L-2a}{2}\right)\frac{w{a}^2}{2EI}\\ =\frac{w{a}^2}{4EI}\left(L-2a\right)\end{array}$

Show the tangent slope and deflection related to reference tangent as in Figure 3.

Place the reference tangent at C then the slope $\left({\theta }_C\right)$ and deflection $\left(y_C\right)$ is zero respectively.

Calculate the slope at the end A related to the point C $\left({\theta }_{C/A}\right)$ using the formula:

${\theta }_{C/A}=A_1+A_2+A_3$

Substitute $\frac{w{a}^3}{2EI}$ for $A_1$, $-\frac{w{a}^3}{6EI}$ for $A_2$ and $\frac{w{a}^2}{4EI}\left(L-2a\right)$ for $A_3$.

$\begin{array}{c}{\theta }_{C/A}=\frac{w{a}^3}{2EI}+\left(-\frac{w{a}^3}{6EI}\right)+\frac{w{a}^2}{4EI}\left(L-2a\right)\\ =\frac{w{a}^2}{2EI}\left(a-\frac{a}{3}+\frac{L-2a}{4}\right)\\ =\frac{w{a}^2}{2EI}\left(\frac{12a-4a+3\left(L-2a\right)}{12}\right)\end{array}$

$\begin{array}{c}=\frac{w{a}^2}{2EI}\left(\frac{12a-4a+3L-6a}{12}\right)\\ =\frac{w{a}^2}{2EI}\left(\frac{3L-2a}{12}\right)\end{array}$

Calculate the slope at the point A $\left({\theta }_A\right)$ using the formula:

${\theta }_A={\theta }_C-{\theta }_{C/A}$

Substitute 0 for ${\theta }_C$ and $\frac{w{a}^2}{2EI}\left(\frac{3L-2a}{12}\right)$ for ${\theta }_{C/A}$.

$\begin{array}{c}{\theta }_A=0-\left(\frac{w{a}^2}{2EI}\left(\frac{3L-2a}{12}\right)\right)\\ =-\frac{w{a}^2}{2EI}\left(\frac{3L-2a}{12}\right)\end{array}$

Thus, the slope $\left({\theta }_A\right)$ at point A is $\underset{\_}{-\frac{w{a}^2}{2EI}\left(\frac{3L-2a}{12}\right)}$.

(b)

To determine

Find the deflection $\left(y_C\right)$ at point C.

Answer to Problem 112P

The deflection $\left(y_C\right)$ at point C is $\underset{\_}{\frac{w{a}^2}{48EI}\left(3L^2-2a^2\right)(\downarrow )}$.

Explanation of Solution

Calculation:

Calculate the deflection at end C related to the left end A $\left(t_{C/A}\right)$ using the formula:

$t_{C/A}=\left(A_1\times \frac{2}{3}a\right)+\left(A_2\times \frac{3}{4}a\right)+\left[\left(A_3\right)\left(a+\frac{1}{2}\left(\frac{L}{2}-a\right)\right)\right]$

Substitute $\frac{w{a}^3}{2EI}$ for $A_1$, $-\frac{w{a}^3}{6EI}$ for $A_2$ and $\frac{w{a}^2}{4EI}\left(L-2a\right)$ for $A_3$.

$\begin{array}{c}{t}_{C/A}=\left(\frac{w{a}^3}{2EI}\times \frac{2}{3}a\right)+\left(-\frac{w{a}^3}{6EI}\times \frac{3}{4}a\right)+\left[\frac{w{a}^2}{4EI}\left(L-2a\right)\left(a+\frac{1}{2}\left(\frac{L}{2}-a\right)\right)\right]\\ =\frac{w{a}^3}{3EI}-\frac{w{a}^3}{8EI}+\frac{w{a}^2}{4EI}\left(L-2a\right)\left(a+\frac{1}{4}\left(L-2a\right)\right)\\ =\frac{w{a}^3}{3EI}-\frac{w{a}^3}{8EI}+\frac{w{a}^2}{4EI}\left(L-2a\right)\left(\frac{4a+L-2a}{4}\right)\end{array}$

$\begin{array}{c}=\frac{w{a}^3}{3EI}-\frac{w{a}^3}{8EI}+\frac{w{a}^2}{4EI}\left(L-2a\right)\left(\frac{L+2a}{4}\right)\\ =\frac{w{a}^3}{3EI}-\frac{w{a}^3}{8EI}+\frac{w{a}^2}{16EI}\left(L-2a\right)\left(L+2a\right)\\ =\frac{w{a}^3}{3EI}-\frac{w{a}^3}{8EI}+\frac{w{a}^2}{16EI}\left(L^2-4a^2\right)\end{array}$

$\begin{array}{c}=\frac{w{a}^2}{EI}\left(\frac{L^2}{16}-\frac{a^2}{24}\right)\\ =\frac{w{a}^2}{EI}\left(\frac{24L^2-16a^2}{384}\right)\\ =\frac{8w{a}^2}{384EI}\left(3L^2-2a^2\right)\\ =\frac{w{a}^2}{48EI}\left(3L^2-2a^2\right)\end{array}$

Calculate the deflection at the point C $\left(y_C\right)$ using the formula:

$y_C=-t_{A/C}$

Substitute $\frac{w{a}^2}{48EI}\left(3L^2-2a^2\right)$ for $t_{A/C}$.

$\begin{array}{c}{y}_C=-\frac{w{a}^2}{48EI}\left(3L^2-2a^2\right)\\ =\frac{w{a}^2}{48EI}\left(3L^2-2a^2\right)(\downarrow )\end{array}$

Thus, the deflection $\left(y_C\right)$ at point C is $\underset{\_}{\frac{w{a}^2}{48EI}\left(3L^2-2a^2\right)(\downarrow )}$.