Question

Rotating of a Shaft

Suppose the x-axis on the interval$[0,L]$is the geometric center of a long straight shaft, such as the propeller shaft of a ship. See Figure 5.2.12. When the shaft is rotating at a constant angular speed about this axis the deflection$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}$of the shaft satisfies the differential equation

$\mathbf{EI}\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{4}}}\mathbf{-}\mathit{\rho }{\mathit{\omega }}^{\mathbf{2}}\mathit{y}\mathbf{=}\mathbf{0}$

Where is its density per unit length. If the shaft is simplify supported, or hinged , at both ends the boundary conditions are then,

$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{n}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathit{L}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{n}}\mathbf{\left(}\mathit{L}\mathbf{\right)}\mathbf{=}\mathbf{0}$

(a) If $\mathit{\lambda }\mathbf{=}{\mathit{\alpha }}^{\mathbf{4}}\mathbf{=}\frac{\mathbf{\rho }{\mathbf{\omega }}^{\mathbf{2}}}{\mathbf{EI}}$, then find the eigenvalues and eigenfunctions for this boundary – value problem.

(b) Use the eigenvalues ${\mathit{\lambda }}_{\mathbf{n}}$in part (a) to find corresponding angular speeds ${\mathit{\omega }}_{\mathbf{n}}$. The values ${\mathit{\omega }}_{\mathbf{n}}$are called critical speeds. The value is ${\mathit{\omega }}_{\mathbf{1}}$called the fundamental critical speed and analogous to example 4, at this speed the shaft changes shape from $\mathit{y}\mathbf{=}\mathbf{0}$to a deflection given by ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$.

Short answer

Finally we get

(a)

${\lambda }_n={\left(\frac{n\pi }{L}\right)}^4,n=1,2,3,...$

And

$y_n\left(x\right)=c_4\sin \frac{n\pi x}{L},n=1,2,3,...$

Where is arbitrary.

(b)

${\omega }_n=\sqrt{\frac{\mathrm{E}l}{\rho }}{\left(\frac{n\pi }{L}\right)}^2,n=1,2,3,...$

Step-by-step solution

Given Information

The given equation is:

$\mathrm{EI}\frac{d^{4}y}{d{x}^4}-\rho {\omega }^{2}y=0$

For this boundary – value problem, find the eigenvalues and eigenfunctions.

(a). The characteristic equation of the given differential equation is

$m^4-{\alpha }^4=0$

Where${\alpha }^4=\rho {\omega }^2/{\rm E}l$

Whose roots are $\pm \alpha$and

We get the general solution is

$y\left(x\right)=c_1\cosh \alpha x+c_2\sinh \alpha x+c_3\cos \alpha x+c_4\sin \alpha x$

The initial conditions

$y\left(0\right)=y^n\left(0\right)=0$

Which gives$c_1=c_3=0$

and so

$y\left(x\right)=c_2\sinh \alpha x+c_4\sin \alpha x$

The conditions

$y\left(L\right)=y^n\left(L\right)=0$ gives

$c_2\sinh \alpha L+c_4\sin \alpha L=0$

And

$c_2\sinh \alpha L-c_4\sin \alpha L=0$

So that $c_2=0$.

To get a nonzero solution, we chose$c_4=0$ and

$\sin \alpha L=0$

Hence , $\alpha L=n\pi$and so the values are

${\lambda }_n={\left(\frac{n\pi }{L}\right)}^4,n=1,2,3,...$

The Eigen functions are

$y_n\left(x\right)=c_4\sin \frac{n\pi x}{L},n=1,2,3,...$

Where$c_4$ is arbitrary.

Find corresponding angular speed

(b) Indeed

$\rho {{\omega }_n}^2/{\rm E}l={\left(\frac{n\mathrm{\pi }}{L}\right)}^4\mathbb{S}$

And so we get

${\omega }_n=\sqrt{\frac{\mathrm{E}l}{\rho }}{\left(\frac{n\pi }{L}\right)}^2,n=1,2,3,...$