Question

A beam is embedded at its left end and simply supported at its Right end. Find the deflection y(x)when the load is as given inProblem 77.

Short answer

The deflection of the beam is

$\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\left(\frac{9}{128}\frac{w_0}{EI}{L}^2\right)x^2+\frac{1}{3!}\left(-\frac{57}{128}\frac{w_0}{EI}L\right)x^3$

Step-by-step solution

Define Laplace transform.

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. The Laplace transform $\mathcal{L}$of a function f(t) is defined as

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

In words, we can describe this expression as the Laplace transform of F(t) equals the function F of s .

$\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$

Find the piecewise function

The piecewise function is defined 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 functioncan 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.

$$\begin{gathered} EI\mathcal{L}\left\{\frac{{\text{d}}^{4}y}{\text{d}{x}^4}\right\}=w_0\mathcal{L}\left\{1\right\}-w_0\mathcal{L}\left\{U\left(x-\frac{L}{2}\right)\right\} \\ =w_0\frac{1}{s}-w_0\frac{e^{-\frac{L}{2}s}}{s} \\ \left(\mathcal{L}\left\{1\right\}=\frac{1}{s},\mathcal{L}\left\{U\right(t-a\left)\right\}=\frac{e^{-as}}{s}\right) \end{gathered}$$

Hence the piecewise function after Laplace transform is$w_0\frac{1}{s}-w_0\frac{e^{-\frac{L}{2}s}}{s}$

Find inverse Laplace transform of both side

We know that,

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

So,

$y\left(t\right)=\frac{w_0}{4!EI}{x}^4-\frac{w_0}{4!EI}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3$

The inverse Laplace transform is$y\left(t\right)=\frac{w_0}{4!EI}{x}^4-\frac{w_0}{4!EI}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3$

Find the constants.

Find the constants using the initial conditions. $y\text{'}\text{'}\left(L\right)=0\text{,}y\left(L\right)=0$

The first derivative is

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

The second derivative is

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

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

$$\begin{gathered} y\text{'}\text{'}\left(L\right)=\frac{w_0}{EI2!}{L}^2-\frac{w_0}{EI2!}{\left(L-\frac{L}{2}\right)}^{2}U\left(L-\frac{L}{2}\right)+c_1+c_{2}L \\ 0=\frac{w_0}{EI2!}{L}^2-\frac{w_0}{EI2!}{\left(\frac{L}{2}\right)}^2+c_1+c_{2}L \\ 0=\frac{w_0}{EI2!}\frac{3}{4}+c_1+c_{2}L \\ {c}_1=-\left(\frac{3}{4}\frac{w_0}{EI2!}{L}^2+c_{2}L\right) \end{gathered}$$

Applying the condition $y\left(L\right)=0$we get,

role="math" localid="1663934286254" $$\begin{gathered} y\left(L\right)=\frac{w_0}{EI4!}{L}^4-\frac{w_0}{EI4!}{\left(L-\frac{L}{2}\right)}^{4}U\left(L-\frac{L}{2}\right)+\frac{c_1}{2!}{L}^2+\frac{c_2}{3!}{L}^3 \\ 0=\frac{w_0}{EI4!}{L}^4-\frac{w_0}{EI4!}{\left(\frac{L}{2}\right)}^4+\frac{c_1}{2!}{L}^2+\frac{c_2}{3!}{L}^3 \\ 0=\frac{5}{128}\frac{w_0}{EI}{L}^4-\frac{1}{2}{L}^2\left(\frac{3}{4}\frac{w_0}{EI2!}{L}^2+c_{2}L\right)+\frac{c_2}{3!}{L}^3 \\ 0=\frac{5}{128}\frac{w_0}{EI}{L}^4-\frac{3}{8}\frac{w_0}{EI2!}{L}^4-\frac{1}{2}{L}^3{c}_2+\frac{c_2}{6}{L}^3 \\ 0=\frac{5}{128}\frac{w_0}{EI}{L}^4-\frac{3}{16}\frac{w_0}{EI}{L}^4-\frac{1}{2}{L}^3{c}_2+\frac{1}{6}{L}^3{c}_2 \\ \frac{1}{3}{c}_2{L}^3=-\frac{19}{128}\frac{w_0}{EI}{L}^4 \\ {c}_2=-\frac{57}{128}\frac{w_0}{EI}L \end{gathered}$$

Hence the constants$c_1=-\left(\frac{3}{4}\frac{w_0}{EI2!}{L}^2+c_{2}L\right)$and$c_2=-\frac{57}{128}\frac{w_0}{EI}L$

Substituting the constant to find final equation.

Substitute the value of $c_{2}in{c}_1=-\left(\frac{3}{4}\frac{w_0}{EI2!}{L}^2+c_{2}L\right)$

$$\begin{gathered} c_1=-\left(\frac{3}{4}\frac{w_0}{EI2!}{L}^2+c_{2}L\right) \\ {c}_1=-\frac{3}{8}\frac{w_0}{EI}{L}^2-\left(-\frac{57}{128}\frac{w_0}{EI}L\right)L \\ {c}_1=\frac{9}{128}\frac{w_0}{EI}{L}^2 \end{gathered}$$

Substitute the value of the constants in

$$\begin{gathered} y\left(t\right)=\frac{w_0}{4!EI}{x}^4-\frac{w_0}{4!EI}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3 \\ y\left(x\right)=\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\left(\frac{9}{128}\frac{w_0}{EI}{L}^2\right)x^2+\frac{1}{3!}\left(-\frac{57}{128}\frac{w_0}{EI}L\right)x^3 \end{gathered}$$

$$\begin{gathered} y\left(t\right)=\frac{w_0}{4!EI}{x}^4-\frac{w_0}{4!EI}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3 \\ y\left(x\right)=\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\left(\frac{9}{128}\frac{w_0}{EI}{L}^2\right)x^2+\frac{1}{3!}\left(-\frac{57}{128}\frac{w_0}{EI}L\right)x^3 \end{gathered}$$

Hence, the value of deflection of the beam is

$y\left(x\right)=\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\left(\frac{9}{128}\frac{w_0}{EI}{L}^2\right)x^2+\frac{1}{3!}\left(-\frac{57}{128}\frac{w_0}{EI}L\right)x^3$