Question

A uniform cantilever beam of length L is embedded at its left end (x=0) and free at its right end. Find the deflection y(x) if the load per unit length is given by

$\mathbf{w}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}{\mathbf{w}}_{\mathbf{0}}}{\mathbf{L}}\left[\frac{L}{2}-x+\left(x-\frac{L}{2}\right)u\left(x-\frac{L}{2}\right)\right]$.

Short answer

$y\left(x\right)=\frac{{\omega }_0}{EI}\frac{1}{4!}{x}^4-\frac{2{\omega }_0}{LEI}\frac{1}{5!}{x}^5+\frac{2{\omega }_0}{LEI}\frac{1}{5!}{\left(x-\frac{L}{2}\right)}^{5}u\left(x-\frac{L}{2}\right)+\frac{1}{2}\frac{{\omega }_0}{EI}\frac{L^2}{24}{x}^2+\frac{1}{3!}\left(-\frac{{\omega }_0}{EI}\left(\frac{L}{4}\right)\right)x^3$

Step-by-step solution

given information

$w\left(x\right)=\frac{2w_0}{L}\left[\frac{L}{2}-x+\left(x-\frac{L}{2}\right)u\left(x-\frac{L}{2}\right)\right]$

 use laplace transform

$EI\frac{d^{4}y}{d{x}^4}=\frac{2{\omega }_0}{L}\left[\frac{L}{s}-x+\left(x-\frac{L}{2}\right)U\left(x-\frac{L}{2}\right)\right]\left(EI\frac{d^{4}y}{d{x}^4}=\omega \left(x\right)\right)$

Take laplace transform of both side

$\begin{array}{rcl}EIL\left\{\frac{d^{4}y}{d{x}^4}\right\}& =& \frac{2{\omega }_0}{L}\left[\frac{L}{s}L\left\{1\right\}-L\left\{x\right\}+L\left\{\left(x-\frac{L}{2}\right)U\left(x-\frac{L}{2}\right)\right\}\right]\\ & =& \frac{2{\omega }_0}{L}\left[\frac{L}{2}\frac{1}{s}-\frac{1}{s^2}+e^{-s\frac{L}{2}}\frac{1}{s^2}\right]\\ & =& {\omega }_0\frac{1}{s}-\frac{2{\omega }_0}{L}\frac{1}{s^2}+\frac{2{\omega }_0}{L}{e}^{-s\frac{L}{2}}\frac{1}{s^2}\end{array}$

$$\begin{gathered} \left(**\right)\left\{F\left(s\right)=L\left\{f\left(t\right)\right\}anda>0thenL\left\{f\left(t-a\right)U\left(t-a\right)\right\}{e}^{-as}F\left(s\right) \\ L\left\{t^n\right\}=\frac{n!}{s^{n+1}} \\ L\left\{1\right\}=\frac{1}{s}\right. \end{gathered}$$

inverse laplace transform

$L\left\{\frac{d^{4}y}{d{x}^4}\right\}=s^{4}Y-s^{3}y\left(0\right)-s^{2}y\text{'}\left(0\right)-sy\text{'}\text{'}\left(0\right)-y\text{'}\text{'}\text{'}\left(0\right)$

$$\begin{gathered} s^{4}Y\left(s\right)-s^{3}y\left(0\right)-s^{2}y\text{'}\left(0\right)-sy\text{'}\text{'}\left(0\right)-y\text{'}\text{'}\text{'}\left(0\right)={\omega }_0\frac{1}{EI}-\frac{2{\omega }_0}{L}\frac{1}{EI{s}^2}+\frac{2{\omega }_0}{L}{e}^{-s\frac{L}{2}}\frac{1}{EI{s}^2} \\ {s}^{4}Y\left(s\right)-sy\text{'}\text{'}\left(0\right)-y\text{'}\text{'}\text{'}\left(0\right)=\frac{{\omega }_0}{EI}-\frac{2{\omega }_0}{L}\frac{1}{EI{s}^2}+\frac{2{\omega }_0}{L}\frac{e^{-s\frac{L}{2}}}{EI{s}^2} \end{gathered}$$

Let$y\text{'}\text{'}\left(0\right)=c_1$and$y\text{'}\text{'}\text{'}\left(0\right)=c_2$then

$\begin{array}{rcl}{s}^{4}Y\left(s\right)-s{c}_1-c_2& =& \frac{{\omega }_0}{EIs}-\frac{2{\omega }_0}{L}\frac{1}{EI{s}^2}+\frac{2{\omega }_0}{L}\frac{e^{-s\frac{L}{2}}}{EI{s}^2}\\ Y\left(s\right)& =& \frac{{\omega }_0}{EI{s}^5}-\frac{2{\omega }_0}{L}\frac{1}{EI{s}^6}+\frac{2{\omega }_0}{L}\frac{e^{-s\frac{L}{2}}}{EI{s}^6}+\frac{c_1}{s^3}+\frac{c_2}{s^4}\end{array}$

Take inverse laplace transform of both side.

$\begin{array}{rcl}{L}^{-1}\left\{Y\left(s\right)\right\}& =& \frac{{\omega }_0}{4!EI{s}^5}{L}^{-1}\left(\frac{4!}{s^{4+1}}\right)-\frac{2{\omega }_0}{5!LEI}{L}^{-1}\left\{\frac{5!}{s^{5+1}}\right\}+\frac{2{\omega }_0}{5!LEI}{L}^{-1}\left\{\frac{5!e^{-\frac{L}{2}s}}{s^{5+1}}\right\}+\frac{c_1}{2!}{L}^{-1}\left\{\frac{2!}{s^{2+1}}\right\}+\frac{c_2}{3!}{L}^{-1}\left\{\frac{3!}{s^{3+1}}\right\}\\ & =& \frac{{\omega }_0}{EI}\frac{1}{4!}{x}^4-\frac{2{\omega }_0}{LEI}\frac{1}{5!}{x}^5+\frac{2{\omega }_0}{LEI}\frac{1}{5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3\end{array}$

So,

$y\left(t\right)=\frac{{\omega }_0}{EI}\frac{1}{4!}{x}^4-\frac{2{\omega }_0}{LEI}\frac{1}{5!}{x}^5+\frac{2{\omega }_0}{LEI}\frac{1}{5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2$(1)

find constants C1 and C2

Initial conditions$y\text{'}\text{'}\left(L\right)=0$and$y\text{'}\text{'}\text{'}\left(L\right)=0$

The first derivative

$\begin{array}{rcl}y\text{'}\left(t\right)& =& \frac{{\omega }_0}{EI}\frac{4}{4!}{x}^3-\frac{2{\omega }_0}{LEI}\frac{5}{5!}{x}^4+\frac{2{\omega }_0}{LEI}\frac{5}{5!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{2c_1}{2!}x+\frac{3c_2}{3!}{x}^2\\ & =& \frac{{\omega }_0}{EI}\frac{1}{3!}{x}^3-\frac{2{\omega }_0}{LEI}\frac{1}{4!}{x}^4+\frac{2{\omega }_0}{LEI}\frac{1}{4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+c_{1}x+\frac{c_2}{2!}{x}^2\\ & & \end{array}$

The second derivative

$\begin{array}{rcl}y\text{'}\text{'}\left(x\right)& =& \frac{{\omega }_0}{EI}\frac{3}{3!}{x}^2-\frac{2{\omega }_0}{LEI}\frac{4}{4!}{x}^3+\frac{2{\omega }_0}{LEI}\frac{4}{4!}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+c_1+\frac{2c_2}{2!}x\\ & =& \frac{{\omega }_0}{EI}\frac{1}{2!}{x}^2-\frac{2{\omega }_0}{LEI}\frac{1}{3!}{x}^3+\frac{2{\omega }_0}{LEI}\frac{1}{3!}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+c_1+c_2{x}^2\end{array}$

The third derivative

$$\begin{gathered} y\text{'}\text{'}\text{'}\left(x\right)=\frac{{\omega }_0}{EI}\frac{2}{2!}x-\frac{2{\omega }_0}{LEI}\frac{3}{3!}{x}^2+\frac{2{\omega }_0}{LEI}\frac{3}{3!}{\left(x-\frac{L}{2}\right)}^{2}U\left(x-\frac{L}{2}\right)+c_2 \\ =\frac{{\omega }_0}{EI}x-\frac{2{\omega }_0}{LEI}\frac{1}{2!}{x}^2+\frac{2{\omega }_0}{LEI}\frac{1}{2!}{\left(x-\frac{L}{2}\right)}^{2}U\left(x-\frac{L}{2}\right)+c_2 \end{gathered}$$

Apply the condition of $y\text{'}\text{'}\text{'}\left(L\right)=0$

$$\begin{gathered} y\text{'}\text{'}\text{'}\left(L\right)=\frac{{\omega }_0}{EI}L-\frac{2{\omega }_0}{LEI}\frac{1}{2!}{L}^2+\frac{2{\omega }_0}{LEI}\frac{1}{2!}{\left(L-\frac{L}{2}\right)}^{2}U\left(L-\frac{L}{2}\right)+c_2 \\ 0=\frac{{\omega }_0}{EI}L-\frac{{\omega }_0}{LEI}{L}^2+\frac{{\omega }_0}{LEI}{\left(\frac{L}{2}\right)}^2+c_2 \end{gathered}$$

$$\begin{gathered} 0=\frac{{\omega }_0}{EI}L-\frac{{\omega }_0}{EI}L+\frac{{\omega }_0}{LEI}\frac{L}{4}+c_2 \\ {c}_2=-\frac{{\omega }_0}{LEI}\left(\frac{L}{4}\right) \end{gathered}$$

Apply the condition of $y\text{'}\text{'}\left(L\right)=0$

$$\begin{gathered} y\text{'}\text{'}\left(L\right)=\frac{{\omega }_0}{EI}\frac{1}{2!}{L}^2-\frac{2{\omega }_0}{LEI}\frac{1}{3!}{L}^2+\frac{2{\omega }_0}{LEI}\frac{1}{3!}{\left(L-\frac{L}{2}\right)}^{3}U\left(L-\frac{L}{2}\right)+c_1+c_{2}L \\ 0=\frac{{\omega }_0}{EI}\frac{1}{2!}{L}^2-\frac{2{\omega }_0}{LEI}\frac{1}{3!}{L}^2+\frac{2{\omega }_0}{LEI}\frac{1}{3!}{\left(L-\frac{L}{2}\right)}^{3}U\left(L-\frac{L}{2}\right)+c_1+\left(-\frac{{\omega }_0}{EI}\left(\frac{L}{4}\right)\right) \end{gathered}$$

$$\begin{gathered} 0=-\frac{1}{24}\frac{{\omega }_0}{EI}{L}^2+c_1 \\ {c}_1=\frac{{\omega }_0}{EI}\frac{L^2}{24} \end{gathered}$$

Substituting in the constants into equation (1)

$y\text{'}\text{'}\left(x\right)=\frac{{\omega }_0}{EI}\frac{1}{4!}{x}^4-\frac{2{\omega }_0}{LEI}\frac{1}{5!}{x}^5+\frac{2{\omega }_0}{LEI}\frac{1}{5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\frac{{\omega }_0}{EI}\frac{L^2}{24}{x}^2$

conclusion

The final solution is

$y\left(x\right)=\frac{{\omega }_0}{EI}\frac{1}{4!}{x}^4-\frac{2{\omega }_0}{LEI}\frac{1}{5!}{x}^5+\frac{2{\omega }_0}{LEI}\frac{1}{5!}{\left(x-\frac{L}{2}\right)}^{5}u\left(x-\frac{L}{2}\right)+\frac{1}{2}\frac{{\omega }_0}{EI}\frac{L^2}{24}{x}^2+\frac{1}{3!}\left(-\frac{{\omega }_0}{EI}\left(\frac{L}{4}\right)\right)x^3$