Question

A cantilever beam is embedded at its left end and free at its right end. Use the Laplace transform to find the deflection $y\left(x\right)$whenthe load is given by

$w\left(x\right)=\left\{\begin{array}{ll}{w}_0,& 0<x<L/2\\ 0& L/2⩽x<L\end{array}\right.$

Short answer

The deflection of$y\left(x\right)$ when the load is given by,

$y\left(x\right)=\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\frac{w_0}{EI}\frac{L^2}{8}{x}^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{2}\right)\right)x^3$

Step-by-step solution

Definition of Laplace transform.

• The integral transform of a given derivative function with real variable t into a complex function with variable s is known as the Laplace transform.
• Let f (t) be supplied for t (0), and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f (t), which is indicated by Lf (t) or F (s)

$F\left(s\right)={\int }_0^{+\infty }f\left(t\right)e^{-s·t}·dt$

Find the value

A piecewise function defined as:

$f\left(t\right)=\left\{\begin{array}{ll}g\left(t\right),& 0⩽t<a\\ h\left(t\right)& t⩾a\end{array}\right.$

The equation can be expressed as:

$f\left(t\right)=g\left(t\right)-g\left(t\right)U(t-a)+h\left(t\right)U(t-a)$

So, the given function$w\left(x\right)$can be expressed as:

$$\begin{gathered} w\left(x\right)=w_0-w_{0}U\left(x-\frac{L}{2}\right)\left(EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w\left(x\right)\right) \\ EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w_0-w_{0}U\left(x-\frac{L}{2}\right) \end{gathered}$$

Take Laplace transform of both side of the above equation.

$EIL\left\{\frac{{\text{d}}^{4}y}{\text{d}{x}^4}\right\}=w_{0}L\left\{1\right\}+w_{0}L\left\{U\left(x-\frac{L}{2}\right)\right\}$

$=w_0\frac{1}{s}+w_0\frac{e^{-\frac{L}{2}s}}{s}$

We know that

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

So,

$$\begin{gathered} s^{4}Y\left(s\right)-s^{3}y\left(0\right)-s^2{y}^{\text{'}}\left(0\right)-s{y}^{\text{'}\text{'}}\left(0\right)-y^{\text{'}\text{'}\text{'}}\left(0\right)=w_0\frac{1}{EIs}+w_0\frac{e^{-\frac{L}{2}s}}{EIs} \\ {s}^{4}Y\left(s\right)-s{y}^{\text{'}\text{'}}\left(0\right)-y^{\text{'}\text{'}\text{'}}\left(0\right)=w_0\frac{1}{EIs}+w_0\frac{e^{\frac{L}{2}s}}{EIs}\left(y\left(0\right)=0,y^{\text{'}}\left(0\right)=0\right) \end{gathered}$$

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

$$\begin{gathered} s^{4}Y\left(s\right)-s{c}_1-c_2=w_0\frac{1}{EIs}+w_0\frac{e^{\frac{L}{2}s}}{EIs} \\ Y\left(s\right)=w_0\frac{1}{EI{s}^5}+w_0\frac{e^{\frac{L}{2}s}}{EI{s}^5}+\frac{c_1}{s^3}+\frac{c_2}{s^4} \end{gathered}$$

Find the derivatives by initial conditions.

Next, we find the constantsandby using the initial conditions$y^{\text{'}\text{'}}\left(L\right)=0$and$y^{\text{'}\text{'}\text{'}}\left(L\right)=0$

The first derivative is

$$\begin{gathered} y^{\text{'}}\left(x\right)=\frac{4w_0}{EI4!}{x}^3-\frac{4w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+\frac{2c_1}{2!}x+\frac{3c_2}{3!}{x}^2 \\ {y}^{\text{'}}\left(x\right)=\frac{w_0}{EI3!}{x}^3-\frac{w_0}{EI3!}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+c_{1}x+\frac{c_2}{2!}{x}^2 \end{gathered}$$

The second derivative is

$$\begin{gathered} y^{\text{'}\text{'}}\left(x\right)=\frac{3w_0}{EI3!}{x}^2-\frac{3w_0}{EI3!}{\left(x-\frac{L}{2}\right)}^{2}U\left(x-\frac{L}{2}\right)+c_1+\frac{2c_2}{2!}x \\ {y}^{\text{'}\text{'}}\left(x\right)=\frac{w_0}{EI2!}{x}^2-\frac{w_0}{EI2!}{\left(x-\frac{L}{2}\right)}^{2}U\left(x-\frac{L}{2}\right)+c_1+c_{2}x \end{gathered}$$

The third derivative is

$$\begin{gathered} y^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{2w_0}{EI2!}x-\frac{2w_0}{EI2!}\left(x-\frac{L}{2}\right)U\left(x-\frac{L}{2}\right)+c_2 \\ {y}^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{w_0}{EI}x-\frac{w_0}{EI}\left(x-\frac{L}{2}\right)U\left(x-\frac{L}{2}\right)+c_2 \end{gathered}$$

Apply the conditions in the given equation.

Applying the condition of $y^{\text{'}\text{'}\text{'}}\left(L\right)=0$, we have

$$\begin{gathered} y^{\text{'}\text{'}\text{'}}\left(L\right)=\frac{w_0}{EI}L-\frac{w_0}{EI}\left(L-\frac{L}{2}\right)U\left(L-\frac{L}{2}\right)+c_2 \\ 0=\frac{w_0}{EI}L-\frac{w_0}{EI}\left(\frac{L}{2}\right)+c_2 \\ 0=\frac{w_0}{EI}\left(\frac{L}{2}\right)+c_2 \\ {c}_2=-\frac{w_0}{EI}\left(\frac{L}{2}\right) \end{gathered}$$

Applying the condition of $y^{\text{'}\text{'}}\left(L\right)=0$, we have

$$\begin{gathered} y^{\text{'}\text{'}}\left(L\right)=\frac{w_0}{EI2!}{L}^2-\frac{w_0}{EI2!}{\left(L-\frac{L}{2}\right)}^{2}U\left(L-\frac{L}{2}\right)+c_1+c_{2}L \\ 0=\frac{w_0}{EI2!}{L}^2-\frac{w_0}{EI2!}{\left(\frac{L}{2}\right)}^2+c_1+\left(-\frac{w_0}{EI}\left(\frac{L}{2}\right)\right)L \\ 0=\frac{w_0}{EI}\left[\frac{L^2}{2}-\frac{L^2}{8}-\frac{L^2}{2}\right]+c_1 \\ {c}_1=\frac{w_0}{EI}\frac{L^2}{8} \end{gathered}$$

Substituting in the constants into $y\left(t\right)$,we have

$y\left(x\right)=\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\frac{w_0}{EI}\frac{L^2}{8}{x}^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{2}\right)\right)x^3$

Hence, the deflection of $y\left(x\right)$when the load is given by,

$y\left(x\right)=\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\frac{w_0}{EI}\frac{L^2}{8}{x}^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{2}\right)\right)x^3$