Question

Solve Problem 77 when the load is given by

$w\left(x\right)=\left\{\begin{array}{ll}0,& 0<x<L/3\\ W_{0,}& L/3<x<2L/3\\ 0,& 2L/3<x<L\end{array}\right.$

Short answer

The deflection of when the load is given by,

$y\left(x\right)=\frac{w_0}{4!EI}{\left(x-\frac{L}{3}\right)}^{4}U\left(x-\frac{L}{3}\right)-\frac{w_0}{4!EI}{\left(x-\frac{2L}{3}\right)}^{4}U\left(x-\frac{2L}{3}\right)+\frac{1}{2!}\left(\frac{w_0}{EI}\frac{L^2}{6}\right)x^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{3}\right)\right)x^3$

Step-by-step solution

 Step 1: 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

The function of the type

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

The equation can be written as

$f\left(t\right)=g\left(t\right)\left[U\right(t-a)-U(t-b\left)\right]$

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

$$\begin{gathered} w\left(x\right)=w_{0}U\left(x-\frac{L}{3}\right)-w_{0}U\left(x-\frac{2L}{3}\right)\left(EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w\left(x\right)\right) \\ EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w_{0}U\left(x-\frac{L}{3}\right)-w_{0}U\left(x-\frac{2L}{3}\right) \end{gathered}$$

Take the Laplace transform of both sides of the above equation.

$$\begin{gathered} EIL\left\{\frac{{\text{d}}^{4}y}{\text{d}{x}^4}\right\}=w_{0}L\left\{U\left(x-\frac{L}{3}\right)\right\}-w_{0}L\left\{U\left(x-\frac{2L}{3}\right)\right\} \\ =w_0\frac{e^{\frac{L}{3}s}}{s}-w_0\frac{e^{\frac{2L}{3}s}}{s}\left(L\left\{U\right(t-a\left)\right\}=\frac{e^{-as}}{s}\right) \end{gathered}$$

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,

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

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

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

Take inverse Laplace transform of both side.

$$\begin{gathered} L^{-1}\left\{Y\right(s\left)\right\}=\frac{w_0}{4!EI}{L}^{-1}\left\{\frac{4!e^{-\frac{L}{3}s}}{s^{4}1}\right\}-\frac{w_0}{4!EI}{L}^1\left\{\frac{4!e^{\frac{2L}{3}s}}{s^{4+1}}\right\}+\frac{c_1}{2!}{L}^{-1}\left\{\frac{2!}{s^2-1}\right\}+\frac{c_2}{3!}{L}^{-1}\left\{\frac{3!}{s^{3+1}}\right\} \\ y\left(t\right)=\frac{w_0}{4!EI}{\left(x-\frac{L}{3}\right)}^{4}U\left(x-\frac{L}{3}\right)-\frac{w_0}{4!EI}{\left(x-\frac{2L}{3}\right)}^{4}U\left(x-\frac{2L}{3}\right)+\frac{1}{2!}\left(\frac{w_0}{EI}\frac{L^2}{6}\right)x^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{3}\right)\right)x^3 \end{gathered}$$

Find the derivatives by initial conditions

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

The first derivative is

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

The second derivative is

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

The third derivative is

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

Apply the conditions in the given equation.

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

$$\begin{gathered} y^{\text{'}\text{'}\text{'}}\left(L\right)=\frac{w_0}{EI}\left(L-\frac{L}{3}\right)U\left(L-\frac{L}{3}\right)-\frac{w_0}{EI}\left(L-\frac{2L}{3}\right)U\left(L-\frac{2L}{3}\right)+c_2 \\ 0=\frac{w_0}{EI}\left(\frac{2L}{3}\right)-\frac{w_0}{EI}\left(\frac{L}{3}\right)+c_2 \\ 0=\frac{w_0}{EI}\left(\frac{L}{3}\right)+c_2 \\ {c}_2=-\frac{w_0}{EI}\left(\frac{L}{3}\right) \end{gathered}$$

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

$$\begin{gathered} y\left(x\right)=\frac{w_0}{4!EI}{\left(x-\frac{L}{3}\right)}^{4}U\left(x-\frac{L}{3}\right)-\frac{w_0}{4!EI}{\left(x-\frac{2L}{3}\right)}^{4}U\left(x-\frac{2L}{3}\right)+\frac{1}{2!}\left(\frac{w_0}{EI}\frac{L^2}{6}\right)x^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{3}\right)\right)x^3 \\ Hence,thedeflectionofy\left(x\right)whentheloadisgivenby, \\ y\left(x\right)=\frac{w_0}{4!EI}{\left(x-\frac{L}{3}\right)}^{4}U\left(x-\frac{L}{3}\right)-\frac{w_0}{4!EI}{\left(x-\frac{2L}{3}\right)}^{4}U\left(x-\frac{2L}{3}\right)+\frac{1}{2!}\left(\frac{w_0}{EI}\frac{L^2}{6}\right)x^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{3}\right)\right)x^3 \end{gathered}$$