Chapter 9, Problem 159RP

For the beam and loading shown, determine (a) the equation of the elastic carve, (b) the slope at end A, (c) the deflection at the midpoint of the span.

Fig. P9.159

To determine

The equation of the elastic curve (y) of the beam.

Answer to Problem 159RP

The equation of the elastic curve (y) of the beam is $\underset{\_}{y=\frac{w_0}{EI{L}^2}\left(\frac{1}{90}{x}^6-\frac{1}{90}L{x}^5+\frac{1}{18}{L}^3{x}^3-\frac{1}{30}{L}^{5}x\right)}$.

Explanation of Solution

Calculation:

Write the load equation (w).

$\begin{array}{c}w=4w_0\left[\frac{x}{L}-\frac{x^2}{L^2}\right]\\ =\frac{4w_{0}x}{L}-\frac{4w_0{x}^2}{L^2}\\ =\frac{w_0}{L^2}\left(4Lx-4x^2\right)\end{array}$ (1)

Integrate the Equation (1).

$\begin{array}{l}\frac{dV}{dx}=-w=\frac{w_0}{L^2}\left(4x^2-4Lx\right)\\ \frac{dM}{dx}=V=\frac{w_0}{L^2}\left(\frac{4}{3}{x}^3-2L{x}^2\right)+C_1\\ M=\frac{w_0}{L^2}\left(\frac{1}{3}{x}^4-\frac{2}{3}L{x}^3\right)+C_{1}x+C_2\end{array}$ (2)

Substitute the boundary condition 0 for x and 0 for M in Equation (2).

$\begin{array}{c}\left(0\right)=\frac{w_0}{L^2}\left(\frac{1}{3}{\left(0\right)}^4-\frac{2}{3}L{\left(0\right)}^3\right)+C_1\left(0\right)+C_2\\ C_2=0\end{array}$

Substitute the boundary condition L for x, 0 for $C_2$ and 0 for M in Equation (2).

$\begin{array}{l}\left(0\right)=\frac{w_0}{L^2}\left(\frac{1}{3}{\left(L\right)}^4-\frac{2}{3}L{\left(L\right)}^3\right)+C_1\left(L\right)+C_2\\ 0=\frac{w_0{L}^4}{3L^2}+C_1\left(L\right)\end{array}$

$\begin{array}{l}{C}_1\left(L\right)=\frac{w_0{L}^4}{3L^2}\\ C_1=\frac{\left(\frac{w_0{L}^4}{3L^2}\right)}{L}\\ C_1=\frac{w_{0}L}{3}\end{array}$

Substitute $\frac{w_{0}L}{3}$ for $C_1$ and 0 for $C_2$ in Equation (2).

$\begin{array}{c}M=\frac{w_0}{L^2}\left(\frac{1}{3}{x}^4-\frac{2}{3}L{x}^3\right)+\frac{w_{0}L}{3}x+0\\ =\frac{w_0}{L^2}\left(\frac{1}{3}{x}^4-\frac{2}{3}L{x}^3+\frac{1}{3}{L}^{3}x\right)\end{array}$ (3)

Calculate the equation of the elastic curve (y) by integrating the Equation (3).

$EI\frac{d^{2}y}{d{x}^2}=M=\frac{w_0}{L^2}\left(\frac{1}{3}{x}^4-\frac{2}{3}L{x}^3+\frac{1}{3}{L}^{3}x\right)$

$EI\frac{dy}{dx}=\frac{w_0}{L^2}\left(\frac{1}{15}{x}^5-\frac{1}{6}L{x}^4+\frac{1}{6}{L}^3{x}^2\right)+C_3$ (4).

$EIy=\frac{w_0}{L^2}\left(\frac{1}{90}{x}^6-\frac{1}{30}L{x}^5+\frac{1}{18}{L}^3{x}^3\right)+C_{3}x+C_4$ (5).

Substitute 0 for x and 0 for $y$ in Equation (4).

$\begin{array}{l}EI\left(0\right)=\frac{w_0}{L^2}\left(\frac{1}{90}{\left(0\right)}^6-\frac{1}{30}L{\left(0\right)}^5+\frac{1}{18}{L}^3{\left(0\right)}^3\right)+C_3\left(0\right)+C_4\\ C_4=0\end{array}$

Substitute the boundary condition L for x, 0 for $C_4$ and 0 for y in the Equation (5).

$\begin{array}{l}EI\left(0\right)=\frac{w_0}{L^2}\left(\frac{1}{90}{\left(L\right)}^6-\frac{1}{30}L{\left(L\right)}^5+\frac{1}{18}{L}^3{\left(L\right)}^3\right)+C_3\left(L\right)+0\\ 0=\frac{w_0}{L^2}\left(\frac{1}{90}{\left(L\right)}^6-\frac{1}{30}{L}^6+\frac{1}{18}{L}^6\right)+C_{3}L\end{array}$

$\begin{array}{l}{C}_{3}L=-\frac{w_0{L}^4}{30}\\ C_3=\frac{-\frac{w_0{L}^4}{30}}{L}\\ C_3=-\frac{w_0{L}^3}{30}\end{array}$

Substitute the boundary condition 0 for $C_4$ and $-\frac{w_0{L}^3}{30}$ for $C_3$ in the Equation (5).

$\begin{array}{l}EIy=\frac{w_0}{L^2}\left(\frac{1}{90}{x}^6-\frac{1}{30}L{x}^5+\frac{1}{18}{L}^3{x}^3\right)+\left(-\frac{w_0{L}^3}{30}\right)x+\left(0\right)\\ EIy=\frac{w_0}{L^2}\left(\frac{1}{90}{x}^6-\frac{1}{30}L{x}^5+\frac{1}{18}{L}^3{x}^3-\frac{1}{30}{L}^{5}x\right)\\ y=\frac{w_0}{EI{L}^2}\left(\frac{1}{90}{x}^6-\frac{1}{30}L{x}^5+\frac{1}{18}{L}^3{x}^3-\frac{1}{30}{L}^{5}x\right)\end{array}$

Thus, the equation of the elastic curve (y) of the beam is $\underset{\_}{y=\frac{w_0}{EI{L}^2}\left(\frac{1}{90}{x}^6-\frac{1}{90}L{x}^5+\frac{1}{18}{L}^3{x}^3-\frac{1}{30}{L}^{5}x\right)}$.

To determine

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

Answer to Problem 159RP

The slope $\left({\theta }_A\right)$ the free end at point A is $\underset{\_}{\frac{1}{30}\frac{w_0{L}^3}{EI}}$.

Explanation of Solution

Calculation:

Calculate the slope $\left({\theta }_A\right)$ the free end at point A in Equation (4).

Substitute $-\frac{w_0{L}^3}{30}$ for $C_3$ and L for x.

$\begin{array}{c}EI\frac{dy}{dx}=\frac{w_0}{L^2}\left(\frac{1}{15}{\left(L\right)}^5-\frac{1}{6}L{\left(L\right)}^4+\frac{1}{6}{L}^3{\left(L\right)}^2\right)-\frac{w_0{L}^3}{30}\\ =\frac{w_0}{EI{L}^2}\left(\frac{1}{15}{\left(L\right)}^5-\frac{1}{6}{L}^5+\frac{1}{6}{L}^5-\frac{1}{30}{L}^5\right)\\ =\frac{w_0}{EI{L}^2}\left(\frac{1}{30}{L}^5\right)\\ =\frac{1}{30}\frac{w_0{L}^3}{EI}\end{array}$

Thus, the slope $\left({\theta }_A\right)$ the free end at point A is $\underset{\_}{\frac{1}{30}\frac{w_0{L}^3}{EI}}$.

To determine

The deflection $\left(y_C\right)$ at the midpoint of the span.

Answer to Problem 159RP

The deflection $\left(y_C\right)$ at the midpoint of the span is $\underset{\_}{\frac{61}{5760}\frac{w_0{L}^4}{EI}\left(\text{downward}\right)}$.

Explanation of Solution

Calculation:

Calculate the deflection $\left(y_C\right)$ at the midpoint of the span using the equation (5).

Substitute  $\frac{L}{2}$ for x, 0 for $C_4$ and $-\frac{w_0{L}^3}{30}$ for $C_3$ in Equation (5).

$\begin{array}{c}EIy=\frac{w_0}{L^2}\left(\frac{1}{90}{\left(\frac{L}{2}\right)}^6-\frac{1}{30}L{\left(\frac{L}{2}\right)}^5+\frac{1}{18}{L}^3{\left(\frac{L}{2}\right)}^3-\frac{1}{30}{L}^5\left(\frac{L}{2}\right)\right)\\ y=\frac{w_0}{EI{L}^2}\left(\frac{L^6}{5760}-\frac{L^6}{960}+\frac{L^6}{144}-\frac{L^6}{60}\right)\\ =\frac{w_0{L}^6}{EI{L}^2}\left(-\frac{61}{5760}\right)\\ =-\frac{61}{5760}\frac{w_0{L}^6}{EI{L}^2}\end{array}$

Thus, the deflection $\left(y_C\right)$ at the midpoint of the span is $\underset{\_}{\frac{61}{5760}\frac{w_0{L}^4}{EI}\left(\text{downward}\right)}$.