Chapter 9.2, Problem 30P

9.29 and 9.30 Determine the reaction at the roller support and the deflection at point C.

Fig. P9.30

To determine

Find the reaction at the roller support and the deflection at point C of the beam.

Answer to Problem 30P

The reaction at the roller support B is $\underset{\_}{B_y=\frac{17wL}{64}(\uparrow )}$.

The deflection at point C of the beam is $\underset{\_}{y_C=\frac{w{L}^4}{1,024EI}(\downarrow )}$.

Explanation of Solution

Consider a section at a distance x from left end A of the section AC.

Show the free-body diagram of the section AC as in Figure 1.

Determine the moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M=0}\\ M+M_A-A_y\left(x\right)-wx\times \frac{x}{2}=0\\ M=A_{y}x+\frac{w{x}^2}{2}-M_A\end{array}$

Write the second order differential equation as follows;

$\frac{d^{2}y}{d{x}^2}=\frac{M\left(x\right)}{EI}$ (1)

Here, the moment at the corresponding section is $M\left(x\right)$, the modulus of elasticity of the material is E, and the moment of inertia of the section is I.

Substitute $\left(A_{y}x+\frac{w{x}^2}{2}-M_A\right)$ for $M\left(x\right)$ in Equation (1).

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{A_{y}x+\frac{w{x}^2}{2}-M_A}{EI}\\ EI\frac{d^{2}y}{d{x}^2}=A_{y}x+\frac{w{x}^2}{2}-M_A\end{array}$

Integrate the equation with respect to x;

$EI\frac{dy}{dx}=\frac{A_y{x}^2}{2}+\frac{w{x}^3}{6}-M_{A}x+C_1$ (2)

Integrate the Equation (2) with respect to x.

$EIy=\frac{A_y{x}^3}{6}+\frac{w{x}^4}{24}-\frac{M_A{x}^2}{2}+C_{1}x+C_2$ (3)

Show the free-body diagram of the section BC as in Figure 2.

Determine the moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M=0}\\ M+M_A-A_y\left(x\right)-\frac{wL}{2}\left(x-\frac{L}{4}\right)+\frac{w}{2}{\left(x-\frac{L}{2}\right)}^2=0\\ M=-M_A+A_{y}x+\frac{wL}{2}\left(x-\frac{L}{4}\right)-\frac{w}{2}{\left(x-\frac{L}{2}\right)}^2\end{array}$

Substitute $\left(-M_A+A_{y}x+\frac{wL}{2}\left(x-\frac{L}{4}\right)-\frac{w}{2}{\left(x-\frac{L}{2}\right)}^2\right)$ for $M\left(x\right)$ in Equation (1).

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{-M_A+A_y\left(x\right)+\frac{wL}{2}\left(x-\frac{L}{4}\right)-\frac{w}{2}{\left(x-\frac{L}{2}\right)}^2}{EI}\\ EI\frac{d^{2}y}{d{x}^2}=-M_A+A_y\left(x\right)+\frac{wL}{2}\left(x-\frac{L}{4}\right)-\frac{w}{2}{\left(x-\frac{L}{2}\right)}^2\end{array}$

Integrate the equation with respect to x;

$EI\frac{dy}{dx}=-M_{A}x+\frac{A_y{x}^2}{2}+\frac{wL}{4}{\left(x-\frac{L}{4}\right)}^2-\frac{w}{6}{\left(x-\frac{L}{2}\right)}^3+C_3$ (4)

Integrate the Equation (4) with respect to x.

$EIy=-\frac{M_A{x}^2}{2}+\frac{A_y{x}^3}{6}+\frac{wL}{12}{\left(x-\frac{L}{4}\right)}^3-\frac{w}{24}{\left(x-\frac{L}{2}\right)}^4+C_{3}x+C_4$ (5)

Boundary condition 1:

At the point A; $x=0;y=0$.

Substitute 0 for x and 0 for y in Equation (3).

$\begin{array}{l}EI\left(0\right)=\frac{A_y{\left(0\right)}^3}{6}+\frac{w{\left(0\right)}^4}{24}-\frac{M_A{\left(0\right)}^2}{2}+C_1\left(0\right)+C_2\\ 0=0+0-0+0+C_2\\ C_2=0\end{array}$

Boundary condition 2:

At the point A; $x=0;\frac{dy}{dx}=0$.

Substitute 0 for x and 0 for $\frac{dy}{dx}$ in Equation (2).

$\begin{array}{l}EI\left(0\right)=\frac{A_y{\left(0\right)}^2}{2}+\frac{w{\left(0\right)}^3}{6}-M_A\left(0\right)+C_1\\ 0=0+0-0+C_1\\ C_1=0\end{array}$

Boundary condition 3:

At the point C; $x=\frac{L}{2};\frac{dy}{dx}=\frac{dy}{dx}$.

Equate Equation (2) and (4).

$\begin{array}{l}\frac{A_y{x}^2}{2}+\frac{w{x}^3}{6}-M_{A}x+C_1=-M_{A}x+\frac{A_y{x}^2}{2}+\frac{wL}{4}{\left(x-\frac{L}{4}\right)}^2-\frac{w}{6}{\left(x-\frac{L}{2}\right)}^3+C_3\\ \frac{w{x}^3}{6}+C_1=\frac{wL}{4}{\left(x-\frac{L}{4}\right)}^2-\frac{w}{6}{\left(x-\frac{L}{2}\right)}^3+C_3\end{array}$

Substitute $\frac{L}{2}$  and x and 0 for $C_1$.

$\begin{array}{c}\frac{w{\left(\frac{L}{2}\right)}^3}{6}+0=\frac{wL}{4}{\left(\frac{L}{2}-\frac{L}{4}\right)}^2-\frac{w}{6}{\left(\frac{L}{2}-\frac{L}{2}\right)}^3+C_3\\ \frac{w{L}^3}{48}=\frac{w{L}^3}{64}-0+C_3\\ C_3=\frac{4w{L}^3}{192}-\frac{3w{L}^3}{192}\\ =\frac{w{L}^3}{192}\end{array}$

Boundary condition 4:

At the point C; $x=\frac{L}{2};y=y$.

Equate Equation (3) and (5).

$\begin{array}{l}\frac{A_y{x}^3}{6}+\frac{w{x}^4}{24}-\frac{M_A{x}^2}{2}+C_{1}x+C_2=-\frac{M_A{x}^2}{2}+\frac{A_y{x}^3}{6}+\frac{wL}{12}{\left(x-\frac{L}{4}\right)}^3-\frac{w}{24}{\left(x-\frac{L}{2}\right)}^4+C_{3}x+C_4\\ \frac{w{x}^4}{24}+C_{1}x+C_2=\frac{wL}{12}{\left(x-\frac{L}{4}\right)}^3-\frac{w}{24}{\left(x-\frac{L}{2}\right)}^4+C_{3}x+C_4\end{array}$

Substitute $\frac{L}{2}$  for x, 0 for $C_1$, 0 for $C_2$, and $\frac{w{L}^3}{192}$ for $C_3$.

$\begin{array}{l}\frac{w{\left(\frac{L}{2}\right)}^4}{24}+\left(0\right)\left(\frac{L}{2}\right)+0=\frac{wL}{12}{\left(\frac{L}{2}-\frac{L}{4}\right)}^3-\frac{w}{24}{\left(\frac{L}{2}-\frac{L}{2}\right)}^4+\frac{w{L}^3}{192}\left(\frac{L}{2}\right)+C_4\\ \frac{w{L}^4}{384}=\frac{w{L}^4}{768}-0+\frac{w{L}^4}{384}+C_4\\ C_4=-\frac{w{L}^4}{768}\end{array}$

Boundary condition 5:

At the point B; $x=L;y=0$.

Substitute L for x, 0 for y, $\frac{w{L}^3}{192}$ for $C_3$, and $-\frac{w{L}^4}{768}$ for $C_4$ in Equation (5).

$\begin{array}{l}EI\left(0\right)=-\frac{M_A{L}^2}{2}+\frac{A_y{L}^3}{6}+\frac{wL}{12}{\left(L-\frac{L}{4}\right)}^3-\frac{w}{24}{\left(L-\frac{L}{2}\right)}^4+\frac{w{L}^3}{192}\left(L\right)-\frac{w{L}^4}{768}\\ 0=-\frac{M_A{L}^2}{2}+\frac{A_y{L}^3}{6}+\frac{27w{L}^4}{768}-\frac{w{L}^4}{384}+\frac{w{L}^4}{192}-\frac{w{L}^4}{768}\\ 0=-\frac{M_A{L}^2}{2}+\frac{A_y{L}^3}{6}+\frac{27w{L}^4}{768}-\frac{2w{L}^4}{768}+\frac{4w{L}^4}{768}-\frac{w{L}^4}{768}\\ 0=-\frac{M_A{L}^2}{2}+\frac{A_y{L}^3}{6}+\frac{28w{L}^4}{768}\end{array}$ (6)

Show the free-body diagram of the beam AB as in Figure 3.

Resolve the vertical component of forces as follows;

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y+B_y+\frac{wL}{2}-\frac{wL}{2}=0\\ A_y=-B_y\end{array}$ (7)

Take moment about the point A as follows;

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ M_A+\frac{wL}{2}\left(\frac{L}{4}\right)-\frac{wL}{2}\left(\frac{3L}{4}\right)+B_y\left(L\right)=0\\ M_A+\frac{w{L}^2}{8}-\frac{3w{L}^2}{8}+B_{y}L=0\\ M_A=\frac{w{L}^2}{4}-B_{y}L\end{array}$ (8)

Substitute $\left(\frac{w{L}^2}{4}-B_{y}L\right)$ for $M_A$ and $\left(-B_y\right)$ for $A_y$ in Equation (6).

$\begin{array}{l}-\frac{\left(\frac{w{L}^2}{4}-B_{y}L\right)L^2}{2}+\frac{-B_y{L}^3}{6}+\frac{28w{L}^4}{768}=0\\ -\frac{w{L}^4}{8}+\frac{B_y{L}^3}{2}-\frac{B_y{L}^3}{6}+\frac{7w{L}^4}{192}=0\\ -\frac{24w{L}^4}{192}+\frac{3B_y{L}^3}{6}-\frac{B_y{L}^3}{6}+\frac{7w{L}^4}{192}=0\end{array}$

$\begin{array}{l}\frac{2B_y{L}^3}{6}=\frac{17w{L}^4}{192}\\ B_y=\frac{17wL}{64}(\uparrow )\end{array}$

Therefore, the reaction at the roller support B is $\underset{\_}{B_y=\frac{17wL}{64}(\uparrow )}$.

Substitute $\frac{17wL}{64}$ for $B_y$ in Equation (7).

$\begin{array}{c}{A}_y=-\frac{17wL}{64}\\ =\frac{17wL}{64}(\downarrow )\end{array}$

Substitute $\frac{17wL}{64}$ for $B_y$ in Equation (8).

$\begin{array}{c}{M}_A=\frac{w{L}^2}{4}-\frac{17wL}{64}\left(L\right)\\ =\frac{16w{L}^2}{64}-\frac{17w{L}^2}{64}\\ =-\frac{w{L}^2}{64}\left(\text{Clockwise}\right)\end{array}$

At point C; $x=\frac{L}{2};y=y_C$.

Substitute $\frac{L}{2}$ for x, $y_C$ for y, $\frac{w{L}^3}{192}$ for $C_3$, $-\frac{17wL}{64}$ for $A_y$, $-\frac{w{L}^2}{64}$ for $M_A$, and $-\frac{w{L}^4}{768}$ for $C_4$ in Equation (5).

$\begin{array}{c}EI{y}_C=\left[\begin{array}{l}-\frac{\left(-\frac{w{L}^2}{64}\right){\left(\frac{L}{2}\right)}^2}{2}+\frac{\left(-\frac{17wL}{64}\right){\left(\frac{L}{2}\right)}^3}{6}+\frac{wL}{12}{\left(\frac{L}{2}-\frac{L}{4}\right)}^3\\ -\frac{w}{24}{\left(\frac{L}{2}-\frac{L}{2}\right)}^4+\frac{w{L}^3}{192}\left(\frac{L}{2}\right)-\frac{w{L}^4}{768}\end{array}\right]\\ =\frac{w{L}^4}{512}-\frac{17w{L}^4}{3,072}+\frac{w{L}^4}{768}-0+\frac{w{L}^4}{384}-\frac{w{L}^4}{768}\\ =\frac{6w{L}^4}{3,072}-\frac{17w{L}^4}{3,072}+\frac{8w{L}^4}{3,072}\\ =-\frac{3w{L}^4}{3,072}\end{array}$

$y_C=\frac{w{L}^4}{1,024EI}(\downarrow )$

Therefore, the deflection at point C of the beam is $\underset{\_}{y_C=\frac{w{L}^4}{1,024EI}(\downarrow )}$.