Question

In Problems 15 and 16 a uniform beam of length L carries a concentrated load w₀ at$\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{L}$. See Figure 7.5.5 (Problem 15) andFigure 7.5.6 (Problem 16). Use the Laplace transform to solve thedifferential

$\mathit{E}\mathit{I}\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{4}}}\mathbf{=}{\mathit{w}}_{\mathbf{0}}\mathit{\delta }\mathbf{(}\mathit{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{L}\mathbf{)}\mathbf{,}\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathit{L}$,

equationsubject to the given boundary conditions.

$\mathbf{16}\mathbf{.}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathit{L}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathit{L}\mathbf{\right)}\mathbf{=}\mathbf{0}$

FIGURE 7.5.6 Beam embedded at both ends

Short answer

The solution of the given system of differential equations is,$y\left(t\right)=\frac{w_0}{6EI}{\left(t-\frac{L}{2}\right)}^3+\frac{w_{0}L}{16EI}{t}^2-\frac{w_0}{12EI}{t}^3$

Step-by-step solution

Step 1:Definition of Laplace Transform

Let $f$be a function defined for $t⩾0$. Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of $f$, provided that the integral converges.

Step 2:Use Laplace Transform

Deflection of$y\left(x\right)$from,

$EI\frac{d^{4}y}{d{x}^4}=w_0\delta (x-\frac{1}{2}L)$

Where$y\left(0\right)=0,y\text{'}\left(0\right)=0,y\left(L\right)=0,$,and$y\text{'}\left(L\right)=0$. Here, we apply Laplace transform operator both side of expression we get,

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

Since we know,

$L\left\{f^n\right(t\left)\right\}=s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-....s{f}^{n-2}\left(0\right)-f^{n-1}\left(0\right)$

And as we know from property of Dirac Delta function is,

$$\begin{gathered} \underset{0}{\overset{\infty }{\int }}f\left(t\right)\delta (t-t_0)=f(t_0\left) \\ EI\right[s^{4}Y\left(s\right)-s^{3}y\left(0\right)-s^{2}y\text{'}\left(0\right)-sy\text{'}\text{'}\left(0\right)-y\text{'}\text{'}]=w_0{e}^{-\frac{L}{2}s} \\ Y(s)=\frac{w_0}{EI}\frac{e^{-\frac{L}{2}s}}{s^4}+\frac{y\text{'}\text{'}\left(0\right)}{s^3}+L^{-1}\left\{\frac{y\text{'}\text{'}\text{'}\left(0\right)}{s^4}\right\} \end{gathered}$$

Using inverse Laplace transform we evaluate$y\left(t\right)$ ,

$L^{-1}\left\{Y\left(s\right)\right\}=\frac{w_0}{EI}{L}^{-1}\left\{\frac{e^{-\frac{L}{2}s}}{s^4}\right\}+\frac{1}{2}{L}^{-1}\left\{\frac{2!y\text{'}\text{'}\left(0\right)}{s^3}\right\}+\frac{1}{3!}{L}^{-1}\left\{\frac{3!y\text{'}\text{'}\text{'}\left(0\right)}{s^4}\right\}$

$y\left(t\right)=\frac{w_0}{6EI}{\left(t-\frac{L}{2}\right)}^{3}u\left(t-\frac{L}{2}\right)+\frac{y\text{'}\text{'}\left(0\right)}{2}{t}^2+\frac{1}{6}y\text{'}\text{'}\text{'}\left(0\right)t^3$

using boundary conditions.

Evaluate$y\text{'}\text{'}\left(0\right)$and $y\text{'}\text{'}\text{'}\left(0\right)$by using boundary condition$y\left(L\right)=0$

$y\left(t\right)=\frac{w_0}{6EI}{\left(t-\frac{L}{2}\right)}^{3}u\left(t-\frac{L}{2}\right)+\frac{y\text{'}\text{'}\left(0\right)}{2}{t}^2+\frac{1}{6}y\text{'}\text{'}\text{'}\left(0\right)t^3$

$0=\frac{w_0}{6EI}{\left(L-\frac{L}{2}\right)}^3+\frac{y\text{'}\text{'}\left(0\right)}{2}{L}^2+\frac{1}{6}y\text{'}\text{'}\text{'}\left(0\right)L^3$...........(1)

And from boundary condition

$y\text{'}\left(t\right)=\frac{w_0}{2EI}{\left(t-\frac{L}{2}\right)}^2+y\text{'}\text{'}\left(0\right)t+\frac{1}{2}y\text{'}\text{'}\text{'}\left(0\right)t^2$

$0=\frac{w_0}{2EI}{\left(L-\frac{L}{2}\right)}^2+y\text{'}\text{'}\left(0\right)L+\frac{1}{2}y\text{'}\text{'}\text{'}\left(0\right)L^2$.............(2)

From equation (1) and (2) we get,

$$\begin{gathered} y\text{'}\text{'}\left(0\right)=\frac{w_{0}L}{8EI} \\ y\text{'}\text{'}\text{'}\left(0\right)=-\frac{w_0}{2EI} \end{gathered}$$

Now the final solution become,

$y\left(t\right)=\frac{w_0}{6EI}{\left(t-\frac{L}{2}\right)}^3+\frac{w_{0}L}{16EI}{t}^2-\frac{w_0}{12EI}{t}^3$