Chapter 9.2, Problem 29P

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

Fig. P9.29

To determine

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

Answer to Problem 29P

The reaction at the roller support A is $\underset{\_}{A_y=\frac{41wL}{128}(\uparrow )}$.

The deflection at the point C is $\underset{\_}{y_C=\frac{19w{L}^4}{6,144EI}(\downarrow )}$.

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.

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

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

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

Integrate the equation with respect to x;

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

Integrate the Equation (2) with respect to x.

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

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

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

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

Integrate the equation with respect to x;

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

Integrate the Equation (4) with respect to x.

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

Boundary condition 1:

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

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

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

Boundary condition 2:

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

Substitute L for x and 0 for $\frac{dy}{dx}$ in Equation (4).

$\begin{array}{l}EI\left(0\right)=\frac{A_y{\left(L\right)}^2}{2}-\frac{wL}{4}{\left(L-\frac{L}{4}\right)}^2+C_3\\ 0=\frac{A_y{L}^2}{2}-\frac{9w{L}^3}{64}+C_3\\ C_3=-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\end{array}$

Boundary condition 3:

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

Equate Equation (2) and (4).

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

Substitute $\frac{L}{2}$  and x and $\left(-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\right)$ for $C_3$.

$\begin{array}{c}-\frac{w{\left(\frac{L}{2}\right)}^3}{6}+C_1=-\frac{wL}{4}{\left(\frac{L}{2}-\frac{L}{4}\right)}^2-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\\ -\frac{w{L}^3}{48}+C_1=-\frac{w{L}^3}{64}-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\\ C_1=-\frac{3w{L}^3}{192}-\frac{A_y{L}^2}{2}+\frac{27w{L}^3}{192}+\frac{4w{L}^3}{192}\\ =-\frac{A_y{L}^2}{2}+\frac{28w{L}^3}{192}\end{array}$

Boundary condition 4:

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

Equate Equation (3) and (5).

$\begin{array}{l}\frac{A_y{x}^3}{6}-\frac{w{x}^4}{24}+C_{1}x+C_2=\frac{A_y{x}^3}{6}-\frac{wL{x}^3}{12}+\frac{w{L}^2{x}^2}{16}+C_{3}x+C_4\\ -\frac{w{x}^4}{24}+C_{1}x+C_2=-\frac{wL{x}^3}{12}+\frac{w{L}^2{x}^2}{16}+C_{3}x+C_4\end{array}$

Substitute $\frac{L}{2}$  for x, 0 for $C_2$, $\left(-\frac{A_y{L}^2}{2}+\frac{28w{L}^3}{192}\right)$ for $C_1$, and $\left(-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\right)$ for $C_3$.

$\begin{array}{c}-\frac{w{\left(\frac{L}{2}\right)}^4}{24}+\left(-\frac{A_y{L}^2}{2}+\frac{28w{L}^3}{192}\right)\frac{L}{2}+0=-\frac{wL}{12}{\left(\frac{L}{2}-\frac{L}{4}\right)}^3+\left(-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\right)\left(\frac{L}{2}\right)+C_4\\ -\frac{w{L}^4}{384}-\frac{A_y{L}^3}{4}+\frac{28w{L}^4}{384}=-\frac{w{L}^4}{768}-\frac{A_y{L}^3}{4}+\frac{9w{L}^4}{128}+C_4\\ -\frac{2w{L}^4}{768}+\frac{56w{L}^4}{384}=-\frac{w{L}^4}{768}+\frac{54w{L}^4}{768}+C_4\\ C_4=\frac{w{L}^4}{768}\end{array}$

Boundary condition 5:

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

Substitute L for x, 0 for y, $\left(-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\right)$ for $C_3$, and $\frac{w{L}^4}{768}$ for $C_4$ in Equation (5).

$\begin{array}{l}EI\left(0\right)=\frac{A_y{L}^3}{6}-\frac{wL}{12}{\left(L-\frac{L}{4}\right)}^3+\left(-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\right)\left(L\right)+\frac{w{L}^4}{768}\\ 0=\frac{A_y{L}^3}{6}-\frac{27w{L}^4}{768}-\frac{A_y{L}^3}{2}+\frac{9w{L}^4}{64}+\frac{w{L}^4}{768}\\ 0=\frac{A_y{L}^3}{6}-\frac{27w{L}^4}{768}-\frac{3A_y{L}^3}{6}+\frac{108w{L}^4}{768}+\frac{w{L}^4}{768}\\ \frac{2A_y{L}^3}{6}=\frac{82w{L}^4}{768}\end{array}$

$A_y=\frac{41wL}{128}(\uparrow )$

Therefore, the reaction at the roller support A is $\underset{\_}{A_y=\frac{41wL}{128}(\uparrow )}$.

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

Substitute $\frac{L}{2}$ for x, $y_C$ for y, $\frac{41wL}{128}$ for $A_y$, $\left(-\frac{A_y{L}^2}{2}+\frac{9w{L}^3}{64}\right)$ for $C_3$, and $\frac{w{L}^4}{768}$ for $C_4$ in Equation (5).

$\begin{array}{c}EI{y}_C=\frac{\frac{41wL}{128}{\left(\frac{L}{2}\right)}^3}{6}-\frac{wL}{12}{\left(\frac{L}{2}-\frac{L}{4}\right)}^3+\left(-\frac{\left(\frac{41wL}{128}\right)L^2}{2}+\frac{9w{L}^3}{64}\right)\left(\frac{L}{2}\right)+\frac{w{L}^4}{768}\\ =\frac{41w{L}^4}{6,144}-\frac{w{L}^4}{768}-\frac{41w{L}^4}{512}+\frac{9w{L}^4}{128}+\frac{w{L}^4}{768}\\ =\frac{41w{L}^4}{6,144}-\frac{8w{L}^4}{6,144}-\frac{492w{L}^4}{6,144}+\frac{432w{L}^4}{6,144}+\frac{8w{L}^4}{6,144}\\ y_C=-\frac{19w{L}^4}{6,144EI}(\downarrow )\end{array}$

Therefore, the deflection at the point C is $\underset{\_}{y_C=\frac{19w{L}^4}{6,144EI}(\downarrow )}$.