Question

A uniform beam of length carries a concentrated load ${\mathit{w}}_{\mathbf{0}}$at $\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{L}$. See Figure 7.5.5 (Problem 15) .Use the Laplace transform to solve the differential equation$\mathbf{El}\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{4}}}\mathbf{=}{\mathbf{w}}_{\mathbf{0}}\mathbf{\delta }\left(\mathbf{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{L}\right),\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{L}$subject to the given boundary conditions.

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

Beam embedded at its left end and free at its right end.

Short answer

The solution for the differential equation $El\frac{d^{4}y}{d{x}^4}=w_0\delta \left(x-\frac{1}{2}L\right),0<x<L$using the Laplace transform isrole="math" localid="1663908917907" $*y\left(x\right)=\left\{\begin{array}{l}\frac{w_{0}L}{EI}\left[\frac{1}{4}{x}^2-\frac{1}{6}{x}^3\right],0⩽x<\frac{L}{2}\\ \frac{w_0{L}^2}{4EI}\left(\frac{1}{2}x-\frac{L}{12}\right),\frac{L}{2}⩽x⩽L\end{array}\right.$

Step-by-step solution

Define the Dirac Delta Function theorem:

For ${\mathbf{t}}_{\mathbf{0}}\mathbf{>}\mathbf{0}$, $\mathbf{L}\left\{\mathbf{\delta }\left(\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{0}}\right)\right\}\mathbf{=}{\mathbf{e}}^{\mathbf{-}{\mathbf{xt}}_{\mathbf{0}}}$

Derive the differential equation by applying the Laplace transform:

Consider the equation,

$El\frac{d^{4}y}{d{x}^4}=w_0\delta \left(x-\frac{1}{2}L\right),0<x<L$

Apply Laplace transform on both sides,

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

Use Dirac Delta Function theorem,

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

Hence, we know

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

Thus,

$*s^{4}Y\left(s\right)-s^{3}y\left(0\right)-s^{2}y\text{'}\left(0\right)-sy\text{'}\text{'}\left(0\right)-y\text{'}\text{'}\text{'}\left(0\right)=\frac{1}{EI}{w}_0{e}^{-\frac{L}{2}s}$

$s^{4}Y\left(s\right)-sy\text{'}\text{'}\left(0\right)-y\text{'}\text{'}\text{'}\left(0\right)=\frac{1}{EI}{w}_0{e}^{-\frac{L}{2}s}\left(y\left(0\right)=0,y\text{'}\left(0\right)=0\right)$

Assume$y\text{'}\text{'}\left(0\right)=c_{1}andy\text{'}\text{'}\text{'}\left(0\right)=c_2$, then

$s^{4}Y\left(s\right)-s{c}_1-c_2=\frac{1}{EI}{w}_0{e}^{-\frac{L}{2}s}$

$Y\left(s\right)=\frac{1}{EI{s}^4}{w}_0{e}^{-\frac{L}{2}s}+\frac{c_1}{s^3}+\frac{c_2}{s^4}$

Use Inverse Laplace transform,

${\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}=\frac{w_0}{3!EI}{\mathcal{L}}^{-1}\left\{\frac{3!e^{-\frac{L}{2}s}}{s^{3+1}}\right\}+\frac{c_1}{2!}{\mathcal{L}}^{-1}\left\{\frac{2!}{s^{2+1}}\right\}+\frac{c_2}{3!}{\mathcal{L}}^{-1}\left\{\frac{3!}{s^{3+1}}\right\}$

$\stackrel{\left(**\right)}{=}\frac{w_0}{3!EI}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3$

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

Thus,

$y\left(t\right)=\frac{w_0}{3!EI}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3$

Determine the derivatives and apply the condition in the equation:

Hence, use the initial conditions$y\text{'}\text{'}\left(L\right)=0andy\text{'}\text{'}\text{'}\left(L\right)=0$, to find the constants$c_{1}and{c}_2$ .

Therefore, the first derivative is,

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

$=\frac{w_0}{EI2!}{\left(x-\frac{L}{2}\right)}^{2}U\left(x-\frac{L}{2}\right)+c_{1}x+\frac{c_2}{2!}{x}^2$

The second derivative,

$y\text{'}\text{'}\left(x\right)=\frac{2w_0}{EI2!}\left(x-\frac{L}{2}\right)U\left(x-\frac{L}{2}\right)+c_1+\frac{2c_2}{2!}x$

$=\frac{w_0}{EI}\left(x-\frac{L}{2}\right)U\left(x-\frac{L}{2}\right)+c_1+c_{2}x$

The third derivative is,

$y\text{'}\text{'}\text{'}\left(x\right)=\frac{w_0}{EI}U\left(x-\frac{L}{2}\right)+c_2$

Thus, apply the condition $y\text{'}\text{'}\text{'}\left(L\right)=0$,

$y\text{'}\text{'}\text{'}\left(L\right)=\frac{w_0}{EI}U\left(L-\frac{L}{2}\right)+c_2$

$0=\frac{w_0}{EI}+c_2\left(U\left(L-\frac{L}{2}\right)=1\right)$

$c_2=-\frac{w_0}{EI}$

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

$y\text{'}\text{'}\left(L\right)=\frac{w_0}{EI}\left(L-\frac{L}{2}\right)U\left(L-\frac{L}{2}\right)+c_1+c_{2}L$

$0=\frac{w_0}{EI}\left(\frac{L}{2}\right)+c_1+\left(-\frac{w_0}{EI}\right)L$

$0=-\frac{w_{0}L}{EI2}+c_1$

$c_1=\frac{w_{0}L}{EI2}$

Substitute the values of $c_{1}and{c}_2$in the equation (1),

$y\left(x\right)=\frac{w_0}{3!EI}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\frac{w_{0}L}{EI2}{x}^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\right)x^3$

Thus,

$*y\left(x\right)=\left\{\begin{array}{l}\frac{w_{0}L}{EI}\left[\frac{1}{4}{x}^2-\frac{1}{6}{x}^3\right],0⩽x<\frac{L}{2}\\ \frac{w_0{L}^2}{4EI}\left(\frac{1}{2}x-\frac{L}{12}\right),\frac{L}{2}⩽x⩽L\end{array}\right.$

Therefore, the solution for the differential equation by using Laplace transform is $*y\left(x\right)=\left\{\begin{array}{l}\frac{w_{0}L}{EI}\left[\frac{1}{4}{x}^2-\frac{1}{6}{x}^3\right],0⩽x<\frac{L}{2}\\ \frac{w_0{L}^2}{4EI}\left(\frac{1}{2}x-\frac{L}{12}\right),\frac{L}{2}⩽x⩽L\end{array}\right.$,