Question

Find the deflection y(x) of a cantilever beam embedded at its left end and free at its right end when the load is as given in Example 10.

Short answer

The deflection of a of a cantilever beam embedded at its left end and free at its right end is

$$\begin{gathered} y\left(x\right)=\frac{w_0}{EI5!}{x}^5-\frac{2}{L}\frac{w_0}{EI6!}{x}^6-\frac{w_0}{EI5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right) \\ +\frac{w_0}{EI6!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{6}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\left(\frac{w_0}{EI}\frac{L^3}{64}\right)x^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{12}\right)\right)x^3 \end{gathered}$$

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. Laplace transform $\mathcal{L}$of a function $f\left(t\right)$is defined as

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

In words we can describe this expression as the Laplace transform of equals function F of S .

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

Find the piecewise function.

A piecewise function is defined as

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

It can be expressed as

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

The function $w\left(x\right)$is expressed as

$$\begin{gathered} w\left(x\right)=w_0\left(1-\frac{2}{L}x\right)-w_0\left(1-\frac{2}{L}x\right)U\left(x-\frac{L}{2}\right)\left(EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w\left(x\right)\right) \\ EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w_0\left(1-\frac{2}{L}x\right)-w_0\left(1-\frac{2}{L}x\right)U\left(x-\frac{L}{2}\right) \end{gathered}$$

Hence the piecewise function of $w\left(x\right)$ is$EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w_0\left(1-\frac{2}{L}x\right)-w_0\left(1-\frac{2}{L}x\right)U\left(x-\frac{L}{2}\right)$

Find the Laplace transform for the given equation.

Take the Laplace transform on both sides of $EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w_0\left(1-\frac{2}{L}x\right)-w_0\left(1-\frac{2}{L}x\right)U\left(x-\frac{L}{2}\right)$

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

Hence, the Laplace transform of the equation

$$\begin{gathered} EI\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=w_0\left(1-\frac{2}{L}x\right)-w_0\left(1-\frac{2}{L}x\right)U\left(x-\frac{L}{2}\right)is \\ {w}_0\left(\frac{1}{s}-\frac{2}{L}\frac{1}{s^2}\right)-w_0{e}^{-\frac{L}{2}s}\left(\frac{1}{s}-\frac{2}{L}\frac{1}{s^2}\right) \end{gathered}$$

Find the value

We know that,

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

So,

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

Substitute $y\text{'}\text{'}\left(0\right)=c_1\text{,}y\text{'}\text{'}\text{'}\left(0\right)=c_2$in the above equation.

$$\begin{gathered} s^{4}Y\left(s\right)-s{c}_1-c_2=\frac{w_0}{EI}\left(\frac{1}{s}-\frac{2}{L}\frac{1}{s^2}\right)-\frac{w_0}{EI}{e}^{-\frac{L}{2}s}\left(\frac{1}{s}-\frac{2}{L}\frac{1}{s^2}\right) \\ Y\left(s\right)=\frac{w_0}{EI}\frac{1}{s^6}-\frac{2}{L}\frac{w_0}{EI}\frac{1}{s^7}-\frac{w_0}{EI}{e}^{-\frac{L}{2}s}\frac{1}{s^6}+\frac{w_0}{EI}\frac{2}{L}{e}^{-\frac{L}{2}s}\frac{1}{s^7}+\frac{c_1}{s^3}+\frac{c_2}{s^4} \end{gathered}$$

Hence the value of $Y\left(s\right)=\frac{w_0}{EI}\frac{1}{s^6}-\frac{2}{L}\frac{w_0}{EI}\frac{1}{s^7}-\frac{w_0}{EI}{e}^{-\frac{L}{2}s}\frac{1}{s^6}+\frac{w_0}{EI}\frac{2}{L}{e}^{-\frac{L}{2}s}\frac{1}{s^7}+\frac{c_1}{s^3}+\frac{c_2}{s^4}$

Find the inverse of the Laplace value.

Take inverse The Laplace transform on both sides of

$$\begin{gathered} Y\left(s\right)=\frac{w_0}{EI}\frac{1}{s^6}-\frac{2}{L}\frac{w_0}{EI}\frac{1}{s^7}-\frac{w_0}{EI}{e}^{-\frac{L}{2}s}\frac{1}{s^6}+\frac{w_0}{EI}\frac{2}{L}{e}^{-\frac{L}{2}s}\frac{1}{s^7}+\frac{c_1}{s^3}+\frac{c_2}{s^4} \\ {\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}=\frac{w_0}{EI5!}{\mathcal{L}}^{-1}\left\{\frac{5!}{s^6}\right\}-\frac{2}{L}\frac{w_0}{EI6!}{\mathcal{L}}^{-1}\left\{\frac{6!}{s^7}\right\}-\frac{w_0}{EI5!}{\mathcal{L}}^{-1}\left\{e^{-\frac{L}{2}s}\frac{5!}{s^6}\right\} \\ +\frac{w_0}{EI6!}\frac{2}{L}{\mathcal{L}}^{-1}\left\{e^{-\frac{L}{2}s}\frac{6!}{s^7}\right\}+\frac{c_1}{2!}{\mathcal{L}}^{-1}\left\{\frac{2!}{s^{2+1}}\right\}+\frac{c_2}{3!}{\mathcal{L}}^{-1}\left\{\frac{3!}{s^{3+1}}\right\} \end{gathered}$$

$$\begin{gathered} =\frac{w_0}{EI5!}{x}^5-\frac{2}{L}\frac{w_0}{EI6!}{x}^6-\frac{w_0}{EI5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right) \\ +\frac{w_0}{EI6!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{6}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3 \end{gathered}$$

$$\begin{gathered} t^n={\mathcal{L}}^{-1}\left\{\frac{n!}{s^{n+1}}\right\} \\ y\left(t\right)=\frac{w_0}{EI5!}{x}^5-\frac{2}{L}\frac{w_0}{EI6!}{x}^6-\frac{w_0}{EI5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right) \\ +\frac{w_0}{EI6!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{6}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3 \end{gathered}$$

The inverse Laplace transform is $$\begin{gathered} y\left(t\right)=\frac{w_0}{EI5!}{x}^5-\frac{2}{L}\frac{w_0}{EI6!}{x}^6-\frac{w_0}{EI5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right) \\ +\frac{w_0}{EI6!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{6}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3 \end{gathered}$$

Find the first derivative

Next we find the constants. The first derivative is

$$\begin{gathered} y\text{'}\left(x\right)=\frac{5w_0}{EI5!}{x}^4-\frac{2}{L}\frac{6w_0}{EI6!}{x}^5-\frac{5w_0}{EI5!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right) \\ +\frac{6w_0}{EI6!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right)+\frac{2c_1}{2!}x+\frac{3c_2}{3!}{x}^2 \\ =\frac{w_0}{EI4!}{x}^4-\frac{2}{L}\frac{w_0}{EI5!}{x}^5-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right) \\ +\frac{w_0}{EI5!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right)+c_{1}x+\frac{c_2}{2!}{x}^2 \end{gathered}$$

Hence the first derivative is$$\begin{gathered} \frac{w_0}{EI4!}{x}^4-\frac{2}{L}\frac{w_0}{EI5!}{x}^5-\frac{w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right) \\ +\frac{w_0}{EI5!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right)+c_{1}x+\frac{c_2}{2!}{x}^2 \end{gathered}$$

Find the second and third derivative.

The second derivative is

$$\begin{gathered} y\text{'}\text{'}\left(x\right)=\frac{4w_0}{EI4!}{x}^3-\frac{2}{L}\frac{5w_0}{EI5!}{x}^4-\frac{4w_0}{EI4!}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+\frac{5w_0}{EI5!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+c_1+\frac{2c_2}{2!}x \\ =\frac{w_0}{EI3!}{x}^3-\frac{2}{L}\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI3!}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+\frac{w_0}{EI4!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+c_1+c_{2}x \end{gathered}$$

The third derivative is

$$\begin{gathered} y\text{'}\text{'}\text{'}\left(x\right)=\frac{3w_0}{EI3!}{x}^2-\frac{2}{L}\frac{4w_0}{EI4!}{x}^3-\frac{3w_0}{EI3!}{\left(x-\frac{L}{2}\right)}^{2}U\left(x-\frac{L}{2}\right)+\frac{4w_0}{EI4!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+c_2 \\ =\frac{w_0}{EI2!}{x}^2-\frac{2}{L}\frac{w_0}{EI3!}{x}^3-\frac{w_0}{EI2!}{\left(x-\frac{L}{2}\right)}^{2}U\left(x-\frac{L}{2}\right)+\frac{w_0}{EI3!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+c_2 \end{gathered}$$

Hence the second and third derivative is

$$\begin{gathered} \frac{w_0}{EI3!}{x}^3-\frac{2}{L}\frac{w_0}{EI4!}{x}^4-\frac{w_0}{EI3!}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+\frac{w_0}{EI4!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{4}U\left(x-\frac{L}{2}\right)+c_1+c_{2}x \\ \frac{w_0}{EI2!}{x}^2-\frac{2}{L}\frac{w_0}{EI3!}{x}^3-\frac{w_0}{EI2!}{\left(x-\frac{L}{2}\right)}^{2}U\left(x-\frac{L}{2}\right)+\frac{w_0}{EI3!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{3}U\left(x-\frac{L}{2}\right)+c_{2}respectively \end{gathered}$$

Finding the value of deflection.

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

$$\begin{gathered} y\text{'}\text{'}\text{'}\left(L\right)=\frac{w_0}{EI2!}{L}^2-\frac{2}{L}\frac{w_0}{EI3!}{L}^3-\frac{w_0}{EI2!}{\left(L-\frac{L}{2}\right)}^{2}U\left(L-\frac{L}{2}\right) \\ +\frac{w_0}{EI3!}\frac{2}{L}{\left(L-\frac{L}{2}\right)}^{3}U\left(L-\frac{L}{2}\right)+c_2 \\ 0=\frac{w_0}{EI2!}{L}^2-\frac{2}{L}\frac{w_0}{EI3!}{L}^3-\frac{w_0}{EI2!}{\left(\frac{L}{2}\right)}^2+\frac{w_0}{EI3!}\frac{2}{L}{\left(\frac{L}{2}\right)}^3+c_2\left(U\left(L-\frac{L}{2}\right)=1\right) \\ 0=\frac{1}{2}\frac{w_0}{EI}{L}^2-\frac{2}{6}\frac{w_0}{EI}{L}^2-\frac{1}{8}\frac{w_0}{EI}{L}^2+\frac{1}{24}\frac{w_0}{EI}{L}^2+c_2 \\ {c}_2=-\frac{w_0}{EI}\left(\frac{L}{12}\right) \end{gathered}$$

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

$$\begin{gathered} y\text{'}\text{'}\left(L\right)=\frac{w_0}{EI3!}{L}^3-\frac{2}{L}\frac{w_0}{EI4!}{L}^4-\frac{w_0}{EI3!}{\left(L-\frac{L}{2}\right)}^{3}U\left(L-\frac{L}{2}\right) \\ +\frac{w_0}{EI4!}\frac{2}{L}{\left(L-\frac{L}{2}\right)}^{4}U\left(L-\frac{L}{2}\right)+c_1+c_{2}L \\ 0=\frac{1}{6}\frac{w_0}{EI}{L}^3-\frac{2}{24}\frac{w_0}{EI}{L}^3-\frac{1}{48}\frac{w_0}{EI}{L}^3+\frac{1}{24·8}\frac{w_0}{EI}{L}^3+c_1+\left(-\frac{1}{12}\frac{w_0}{EI}{L}^2\right)L \\ 0=\frac{1}{6}\frac{w_0}{EI}{L}^3-\frac{2}{24}\frac{w_0}{EI}{L}^3-\frac{1}{48}\frac{w_0}{EI}{L}^3+\frac{1}{24·8}\frac{w_0}{EI}{L}^3+c_1-\frac{1}{12}\frac{w_0}{EI}{L}^3 \\ {c}_1=\frac{w_0}{EI}\frac{L^3}{64} \end{gathered}$$

Substitute the value of the constants in

$$\begin{gathered} y\left(t\right)=\frac{w_0}{EI5!}{x}^5-\frac{2}{L}\frac{w_0}{EI6!}{x}^6-\frac{w_0}{EI5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right) \\ +\frac{w_0}{EI6!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{6}U\left(x-\frac{L}{2}\right)+\frac{c_1}{2!}{x}^2+\frac{c_2}{3!}{x}^3 \\ y\left(x\right)=\frac{w_0}{EI5!}{x}^5-\frac{2}{L}\frac{w_0}{EI6!}{x}^6-\frac{w_0}{EI5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right) \end{gathered}$$

$+\frac{w_0}{EI6!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{6}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\left(\frac{w_0}{EI}\frac{L^3}{64}\right)x^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{12}\right)\right)x^3$

Hence the value of deflection is

$$\begin{gathered} y\left(x\right)=\frac{w_0}{EI5!}{x}^5-\frac{2}{L}\frac{w_0}{EI6!}{x}^6-\frac{w_0}{EI5!}{\left(x-\frac{L}{2}\right)}^{5}U\left(x-\frac{L}{2}\right) \\ +\frac{w_0}{EI6!}\frac{2}{L}{\left(x-\frac{L}{2}\right)}^{6}U\left(x-\frac{L}{2}\right)+\frac{1}{2!}\left(\frac{w_0}{EI}\frac{L^3}{64}\right)x^2+\frac{1}{3!}\left(-\frac{w_0}{EI}\left(\frac{L}{12}\right)\right)x^3 \end{gathered}$$