Question

When a uniform beam is supported by an elastic foundation, the differential equation for its deflection y(x) is

$\mathbf{E}\mathbf{I}\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{4}}}\mathbf{+}\mathbf{k}\mathbf{y}\mathbf{=}\mathbf{w}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}$

where k is the modulus of the foundation and -ky is the restoring force of the foundation that acts in the direction opposite to that of the load w(x). See Figure 7.R.11. For algebraic convenience suppose that the differential equation is written as

$\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{4}}}\mathbf{+}\mathbf{4}{\mathit{a}}^{\mathbf{4}}\mathit{y}\mathbf{=}\frac{\mathbf{w}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{E}\mathbf{I}}\mathbf{.}$

where$\mathit{a}\mathbf{=}\mathbf{(}\mathbf{k}\mathbf{/}\mathbf{4}\mathbf{E}\mathbf{I}{\mathbf{)}}^{\mathbf{1}\mathbf{/}\mathbf{4}}\mathbf{.}$ Assume$\mathit{L}\mathbf{=}\mathit{\pi }$ and a=1 Find the deflection y(x) of a beam that is supported on an elastic foundation when

(a) the beam is simply supported at both ends and a constant load W₀is uniformly distributed along its length,

(b) the beam is embedded at both ends and w(x). is a concentrated load W₀applied at $\mathit{x}\mathbf{=}\mathit{\pi }\mathbf{/}\mathbf{2}\mathbf{.}$[Hint: In both parts of this problem use the table of Laplace transforms in Appendix C and the fact that ${\mathit{s}}^{\mathbf{4}}\mathbf{+}\mathbf{4}\mathbf{=}\left(s^2-2s+2\right)\left(s^2+2s+2\right)$.

Short answer

A brief answer to the question. (Max 1-2 lines)

(a)

$$\begin{gathered} y\left(x\right)=\frac{w_0}{4EI}(1-\cos x\cosh x)+\frac{1}{2}\frac{w_0}{4EI}\frac{\cosh \pi }{\sinh \pi }+\frac{w_0}{4EI\sinh \pi }(\cos x\sinh x+\sin x\cosh x) \\ +\frac{1}{4}-\frac{w_0}{2EI}\frac{\cosh \pi }{\sinh \pi }-\frac{w_0}{2EI\sinh \pi }(\sin x\cosh x-\cos x\sinh x) \end{gathered}$$

(b)

$$\begin{gathered} y\left(x\right)=\frac{w_0}{2EI}\frac{\sinh \frac{\pi }{2}}{\sinh \pi }\left(\sin x\sinh x\right)-\frac{w_0}{4EI}\frac{\cosh \frac{\pi }{2}}{\sinh \pi }(\sin x\cosh x-\cos x\sinh x) \\ +\frac{w_0}{4EI}\sin x-\frac{\pi }{2}\cosh x-\frac{\pi }{2}-\cos x-\frac{\pi }{2}\sinh x-\frac{\pi }{2}Ux-\frac{\pi }{2} \end{gathered}$$

Step-by-step solution

Step 1:

(a) Take Laplace transform of both side of the given equation.

$\star \mathcal{L}\left\{\frac{{\text{d}}^{4}y}{\text{d}{x}^4}\right\}+4\mathcal{L}\left\{y\right\}=\frac{w_0}{EI}\mathcal{L}\left\{1\right\}$

$=\frac{w_0}{EI}\frac{1}{s}\left(\mathcal{L}\left\{1\right\}=\frac{1}{s}\right)$

We know that

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

So,

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

$s^{4}Y\left(s\right)-s{y}^{\text{'}\text{'}}\left(0\right)-y^{\text{'}\text{'}\text{'}}\left(0\right)+4Y\left(s\right)=w_0\frac{1}{EIs}\left(y\left(0\right)=0,y^{\text{'}\text{'}}\left(0\right)=0\right)$

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

$s^{4}Y\left(s\right)-s^2{c}_1-c_2+4Y\left(s\right)=w_0\frac{1}{EIs}$

$\left(s^4+4\right)Y\left(s\right)=w_0\frac{1}{EIs}+s^2{c}_1+c_2$

$Y\left(s\right)=w_0\frac{1}{EIs\left(s^4+4\right)}+\frac{s^2{c}_1}{\left(s^4+4\right)}+\frac{c_2}{\left(s^4+4\right)}----\left(2\right)$

Decompose the fraction$\frac{1}{s\left(s^2+2s+2\right)\left(s^2-2s+2\right)}$using the partial fraction technique.

$\star \frac{1}{s\left(s^2+2s+2\right)\left(s^2-2s+2\right)}=\frac{A}{s}+\frac{Bs+C}{s^2+2s+2}++\frac{Ds+E}{s^2-2s+2}\text{- - - - (1)}$

$=\frac{A{s}^2+2s+2s^2-2s+2+(Bs+C)s{s}^2-2s+2+(Ds+E)s{s}^2+2s+2}{s\left(s^2+2s+2\right)\left(s^2-2s+2\right)}$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$1=A{s}^2+2s+2s^2-2s+2+(Bs+C)s{s}^2-2s+2+(Ds+E)s{s}^2+2s+2$

$1=A{s}^4+4A+B{s}^4+C{s}^3-2B{s}^3-2C{s}^2+2B{s}^2+2Cs+D{s}^4+E{s}^3+2D{s}^3+2E{s}^2+2D{s}^2+2Es$

Comparing coefficient on both sides, we get

$(*)\left\{\begin{array}{l}\mathrm{cosktsinhkt}+\mathrm{sinktcoshkt}={\mathcal{L}}^{-1}\left\{\frac{2{\mathrm{ks}}^2}{{\mathrm{s}}^4+4{\mathrm{k}}^4}\right\}\begin{array}{l}4\mathrm{A}=1\\ 2\mathrm{C}+2\mathrm{E}=0\\ -\mathrm{C}+\mathrm{B}+\mathrm{E}+\mathrm{D}=0\\ \mathrm{C}-2\mathrm{B}+\mathrm{E}+2\mathrm{D}=0\\ \mathrm{A}+\mathrm{B}+\mathrm{D}=0\end{array}\\ \mathrm{sinktcoshkt}-\mathrm{cosktsinhkt}={\mathcal{L}}^{-1}\left\{\frac{4{\mathrm{k}}^3}{{\mathrm{s}}^4+4{\mathrm{k}}^4}\right\}\\ {\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}}\right\}=1\\ {\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}-\mathrm{a}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{k}}^2}\right\}={\mathrm{e}}^{\mathrm{at}}\mathrm{coskt}\end{array}\right.$

Based on the previous system we get$\mathrm{A}=\frac{1}{4},\mathrm{B}=-\frac{1}{8},\mathrm{C}=-\frac{1}{8},\mathrm{D}=-\frac{1}{8},\mathrm{E}=\frac{1}{8}$

Substituting A, B, C, D and E in the equation (1), we get.

$\frac{1}{s\left(s^2+2s+2\right)\left(s^2-2s+2\right)}=\frac{1}{8}\frac{2}{s}-\frac{1}{8}\frac{s+1}{s^2+2s+2}-\frac{1}{8}\frac{s-1}{s^2-2s+2}$

$=\frac{1}{8}\frac{2}{s}-\frac{1}{8}\frac{s+1}{s^2+2s+1+1}-\frac{1}{8}\frac{s-1}{s^2-2s+1+1}$

$=\frac{1}{8}\frac{2}{s}-\frac{1}{8}\frac{s+1}{{(s+1)}^2+1}-\frac{1}{8}\frac{s-1}{{(s-1)}^2+1}$

Take inverse Laplace transform of both side

The equation (2) becomes.

$Y\left(s\right)=w_0\frac{1}{EI8}\left(\frac{2}{s}-\frac{s+1}{{(s+1)}^2+1}-\frac{s-1}{{(s-1)}^2+1}\right)+\frac{c_1}{2}\frac{2s^2}{\left(s^4+4\right)}+\frac{c_2}{4}\frac{4}{\left(s^4+4\right)}$

$\star y\left(x\right)=w_0\frac{1}{EI8}\left[2{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{s+1}{{(s+1)}^2+1}\right\}-{\mathcal{L}}^{-1}\left\{\frac{s-1}{{(s-1)}^2+1}\right\}\right]+\frac{c_1}{2}{\mathcal{L}}^{-1}\left\{\frac{2s^2}{s^4+4}\right\}+\frac{c_2}{4}{\mathcal{L}}^{-1}\left\{\frac{4}{s^4+4}\right\}$

$\stackrel{(*)}{=}\frac{w_0}{EI}\frac{1}{8}\left(2-e^{-x}\cos x-e^x\cos x\right)+\frac{c_1}{2}(\cos x\sinh x+\sin x\cosh x)+\frac{c_2}{4}(\sin x\cosh x-\cos x\sinh x)$

$=\frac{w_0}{8EI}(2-2\cosh x\cos x)+\frac{c_1}{2}(\cos x\sinh x+\sin x\cosh x)+\frac{c_2}{4}(\sin x\cosh x-\cos x\sinh x)$

$=\frac{2w_0}{8EI}(1-\cosh x\cos x)+\frac{c_1}{2}(\cos x\sinh x+\sin x\cosh x)+\frac{c_2}{4}(\sin x\cosh x-\cos x\sinh x)$
So the solution is

$\begin{array}{rcl}y\left(x\right)& =& \frac{w_0}{4EI}(1-\cos x\cosh x)+\frac{c_1}{2}(\cos x\sinh x+\sin x\cosh x)\\ & & +\frac{c_2}{4}(\sin x\cosh x-\cos x\sinh x)----\left(3\right)\end{array}$

Find the derivative

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

$$\begin{gathered} y\left(x\right)=\frac{w_0}{2EI}\frac{\sinh \left(\frac{\pi }{2}\right)}{\sinh \pi }\left(\sin x\sinh x\right)-\frac{w_0}{4EI}\frac{\cosh \left(\frac{\pi }{2}\right)}{\sinh \pi }(\sin x\cosh x-\cos x\sinh x) \\ +\frac{w_0}{4EI}\left[\sin \left(x-\frac{\pi }{2}\right)\cosh \left(x-\frac{\pi }{2}\right)-\cos \left(x-\frac{\pi }{2}\right)\sinh \left(x-\frac{\pi }{2}\right)\right]U\left(x-\frac{\pi }{2}\right) \end{gathered}$$

The first derivative is

$$\begin{gathered} y^{\text{'}}\left(x\right)=\frac{w_0}{4EI}(\sin x\cosh x-\cos x\sinh x) \\ +\frac{c_1}{2}(-\sin x\sinh x+\cos x\cosh x+\cos x\cosh x+\sin x\sinh x) \\ +\frac{c_2}{4}(\sin x\sinh x+\cos x\cosh x-\cos x\cosh x+\sin x\sinh x) \end{gathered}$$

$=\frac{w_0}{4EI}(\sin x\cosh x-\cos x\sinh x)+\frac{c_1}{2}\left(2\cos x\cosh x\right)+\frac{c_2}{4}\left(2\sin x\sinh x\right)$

Simplifying, the first derivative is

$y^{\text{'}}\left(x\right)=\frac{w_0}{4EI}(\sin x\cosh x-\cos x\sinh x)+c_1\left(\cos x\cosh x\right)+\frac{c_2}{2}\left(\sin x\sinh x\right)$

The second derivative is

$$\begin{gathered} y^{\text{'}\text{'}}\left(x\right)=\frac{w_0}{4EI}(\cos x\cosh x+\sin x\sinh x+\sin x\sinh x-\cos x\cosh x) \\ +c_1(-\sin x\cosh x+\cos x\sinh x)+\frac{c_2}{2}(\cos x\sinh x+\sin x\cosh x) \end{gathered}$$

Simplifying, the second derivative is

$y^{\text{'}\text{'}}\left(x\right)=\frac{w_0}{2EI}\left(\sin x\sinh x\right)+c_1(\cos x\sinh x-\sin x\cosh x)+\frac{c_2}{2}(\cos x\sinh x+\sin x\cosh x)$

Applying the condition

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

$$\begin{gathered} y^{\text{'}\text{'}}\left(\pi \right)=\frac{w_0}{2EI}\left(\sin \pi \sinh \pi \right)+c_1(\cos \pi \sinh \pi -\sin \pi \cosh \pi ) \\ +\frac{c_2}{2}(\cos \pi \sinh \pi +\sin \pi \cosh \pi ) \end{gathered}$$

$0=\frac{w_0}{2EI}\left(0\right)+c_1(-\sinh \pi -0)+\frac{c_2}{2}(-\sinh \pi +0)$

$0=-c_1\sinh \pi -\frac{c_2}{2}\sinh \pi$

$c_1\sinh \pi =-\frac{c_2}{2}\sinh \pi$

$c_1=-\frac{c_2}{2}$

Applying the condition of $y\left(\pi \right)=0,$we have

$$\begin{gathered} y\left(\pi \right)=\frac{w_0}{4EI}(1-\cos \pi \cosh \pi )+\frac{c_1}{2}(\cos \pi \sinh \pi +\sin \pi \cosh \pi ) \\ +\frac{c_2}{4}(\sin \pi \cosh \pi -\cos \pi \sinh \pi ) \end{gathered}$$

$0=\frac{w_0}{4EI}(1+\cosh \pi )+\frac{c_1}{2}(-\sinh \pi +0)+\frac{c_2}{4}(0+\sinh \pi )$

$0=\frac{w_0}{4EI}(1+\cosh \pi )+\frac{1}{2}\left(-\frac{c_2}{2}\right)(-\sinh \pi )+\frac{c_2}{4}\left(\sinh \pi \right)$

$0=\frac{w_0}{4EI}+\frac{w_0}{4EI}\cosh \pi +\frac{c_2}{4}\left(\sinh \pi \right)+\frac{c_2}{4}\left(\sinh \pi \right)$

$0=\frac{w_0}{4EI}+\frac{w_0}{4EI}\cosh \pi +\frac{c_2}{2}\left(\sinh \pi \right)$

$\frac{c_2}{2}\left(\sinh \pi \right)=-\frac{w_0}{4EI}\cosh \pi -\frac{w_0}{4EI}$

$\frac{c_2}{2}=-\frac{w_0}{4EI}\frac{\cosh \pi }{\sinh \pi }-\frac{w_0}{4EI\sinh \pi }$

$c_2=-\frac{w_0}{2EI}\frac{\cosh \pi }{\sinh \pi }-\frac{w_0}{2EI\sinh \pi }$

So C₁ is

$c_1=-\frac{c_2}{2}$

$=-\frac{1}{2}\left(-\frac{w_0}{2EI}\frac{\cosh \pi }{\sinh \pi }-\frac{w_0}{2EI\sinh \pi }\right)$

$=\frac{w_0}{4EI}\frac{\cosh \pi }{\sinh \pi }+\frac{w_0}{4EI\sinh \pi }$

$$\begin{gathered} y\left(x\right)=\frac{w_0}{4EI}(1-\cos x\cosh x)+\frac{1}{2}\left(\frac{w_0}{4EI}\frac{\cosh \pi }{\sinh \pi }+\frac{w_0}{4EI\sinh \pi }\right)(\cos x\sinh x+\sin x\cosh x) \\ +\frac{1}{4}\left(-\frac{w_0}{2EI}\frac{\cosh \pi }{\sinh \pi }-\frac{w_0}{2EI\sinh \pi }\right)(\sin x\cosh x-\cos x\sinh x) \end{gathered}$$

Take Laplace transform

$\frac{{\text{d}}^{4}y}{\text{d}{x}^4}+4y=\frac{w_0}{EI}\delta \left(x-\frac{\pi }{2}\right)$

Take Laplace transform of both side of the above equation

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

$=\frac{w_0}{EI}{e}^{-s\frac{\pi }{2}}\left(\mathcal{L}\left\{\delta \left(t-t_0\right)\right\}=e^{-s{t}_0}\right)$

We know that

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

So,

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

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

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

$s^{4}Y\left(s\right)-s{c}_1-c_2+4Y\left(s\right)=w_0\frac{1}{EI}{e}^{-s\frac{\pi }{2}}$

$\left(s^4+4\right)Y\left(s\right)=w_0\frac{1}{EI}{e}^{-s\frac{\pi }{2}}+c_1+c_2$

$Y\left(s\right)=w_0\frac{4}{4EI\left(s^4+4\right)}{e}^{-s\frac{\pi }{2}}+\frac{s{c}_1}{\left(s^4+4\right)}+\frac{c_2}{\left(s^4+4\right)}$

Take Inverse Laplace transform

Take inverse Laplace transform of both side

$\star y\left(x\right)=w_0\frac{1}{EI4}\left[{\mathcal{L}}^{-1}\left\{\frac{4}{s^4+4}\right\}\right]+\frac{c_1}{2}{\mathcal{L}}^{-1}\left\{\frac{2s}{s^4+4}\right\}+\frac{c_2}{4}{\mathcal{L}}^{-1}\left\{\frac{4}{s^4+4}\right\}$

$$\begin{gathered} \stackrel{(**)}{=}\frac{w_0}{4EI}\left[\sin \left(x-\frac{\pi }{2}\right)\cosh \left(x-\frac{\pi }{2}\right)-\cos \left(x-\frac{\pi }{2}\right)\sinh \left(x-\frac{\pi }{2}\right)\right]U\left(x-\frac{\pi }{2}\right) \\ +\frac{c_1}{2}\left(\sin x\sinh x\right)+\frac{c_2}{4}(\sin x\cosh x-\cos x\sinh x)----\left(4\right) \end{gathered}$$

$(**)\left\{\begin{array}{l}\mathrm{sinktsinhkt}={\mathcal{L}}^{-1}\left\{\frac{{\displaystyle 2{\mathrm{k}}^2\mathrm{s}}}{{\displaystyle {\mathrm{s}}^4+4{\mathrm{k}}^4}}\right\}\\ \mathrm{sinktcoshkt}-\mathrm{cosktsinhkt}={\mathcal{L}}^{-1}\left\{\frac{{\displaystyle 4{\mathrm{k}}^3}}{{\displaystyle {\mathrm{s}}^4+4{\mathrm{k}}^4}}\right\}\\ \text{If}{\mathcal{L}}^{-1}\left\{\mathrm{F}\right(\mathrm{s}\left)\right\}=\mathrm{f}\left(\mathrm{t}\right)\text{and}\mathrm{a}>0,\text{then}\mathrm{f}(\mathrm{t}-\mathrm{a})\mathrm{U}(\mathrm{t}-\mathrm{a})={\mathcal{L}}^{-1}\left\{{\mathrm{e}}^{-\mathrm{as}}\mathrm{F}\left(\mathrm{s}\right)\right\}\end{array}\right.$

Find the derivative

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

The first derivative is

$$\begin{gathered} y^{\text{'}}\left(x\right)=\frac{w_0}{4EI}\left(2\sin \left(x-\frac{\pi }{2}\right)\sinh \left(x-\frac{\pi }{2}\right)\right)U\left(x-\frac{\pi }{2}\right) \\ +\frac{c_1}{2}(\cos x\sinh x+\sin x\cosh x)+\frac{c_2}{4}\left(2\sin x\sinh x\right) \end{gathered}$$

Simplifying, the first derivative is

$y^{\text{'}}\left(x\right)=\frac{w_0}{2EI}\left(\sin \left(x-\frac{\pi }{2}\right)\sinh \left(x-\frac{\pi }{2}\right)\right)U\left(x-\frac{\pi }{2}\right)+\frac{c_1}{2}(\cos x\sinh x+\sin x\cosh x)+\frac{c_2}{2}\left(\sin x\sinh x\right)$

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

$$\begin{gathered} y^{\text{'}}\left(\pi \right)=\frac{w_0}{2EI}\left(\sin \left(\pi -\frac{\pi }{2}\right)\sinh \left(\pi -\frac{\pi }{2}\right)\right)U\left(\pi -\frac{\pi }{2}\right)+\frac{c_1}{2}(\cos \pi \sinh \pi +\sin \pi \cosh \pi ) \\ +\frac{c_2}{2}\left(\sin \pi \sinh \pi \right) \end{gathered}$$

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

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

$c_1=\frac{w_0}{EI}\frac{\sinh \left(\frac{\pi }{2}\right)}{\sinh \pi }$

Applying the condition ofwe have

$$\begin{gathered} y\left(\pi \right)=\frac{w_0}{4EI}\left[\sin \left(\pi -\frac{\pi }{2}\right)\cosh \left(\pi -\frac{\pi }{2}\right)-\cos \left(\pi -\frac{\pi }{2}\right)\sinh \left(\pi -\frac{\pi }{2}\right)\right]U\left(\pi -\frac{\pi }{2}\right) \\ +\frac{c_1}{2}\left(\sin \pi \sinh \pi \right)+\frac{c_2}{4}(\sin \pi \cosh \pi -\cos \pi \sinh \pi ) \end{gathered}$$

$0=\frac{w_0}{4EI}\cosh \left(\frac{\pi }{2}\right)+\frac{c_2}{4}\left(\sinh \pi \right)$

$c_2=-\frac{w_0}{EI}\frac{\cosh \left(\frac{\pi }{2}\right)}{\sinh \pi }$

Substituting in the constants into equation (4), we have

$$\begin{gathered} y\left(x\right)=\frac{w_0}{2EI}\frac{\sinh \left(\frac{\pi }{2}\right)}{\sinh \pi }\left(\sin x\sinh x\right)-\frac{w_0}{4EI}\frac{\cosh \left(\frac{\pi }{2}\right)}{\sinh \pi }(\sin x\cosh x-\cos x\sinh x) \\ +\frac{w_0}{4EI}\left[\sin \left(x-\frac{\pi }{2}\right)\cosh \left(x-\frac{\pi }{2}\right)-\cos \left(x-\frac{\pi }{2}\right)\sinh \left(x-\frac{\pi }{2}\right)\right]U\left(x-\frac{\pi }{2}\right) \end{gathered}$$