# Cantilever Beam problem

1. Oct 20, 2016

### kev.thomson96

A cantilever beam at low frequency behaves like an underdamped 1 DOF mass/spring/damper system.

We are trying to find the roots of the Characteristic equation which are lambda1,2 = -dampingRatio x wnatural /sqrt(1-dampingRatio)

Relevant formulas and given values:

damping ratio = c/2sqrt(m/k)

wnatural = sqrt(k/m)

wdamped = wnatural x sqrt(1-dampingRatio^2)

log decrement = (1/n)ln(x1/xn+1) = 2pi(damping ratio)/sqrt(1-dampingRatio)

beam width = 50 mm

beam depth = 3.8 mm

Modulus of elasticity = 200 GPa

density = m/v = 7800 kg/m^3

deflection x = Fl^3/3EI, l is length and I is second moment of area

kequivalent = kbeam + kshaker, where kshaker = 45 N/m.

I've only found I = bh^3/12 = 15.83 x 10^-12 m and from then on I'm stuck

2. Oct 20, 2016

### Simon Bridge

$\lambda_{1,2} = -\zeta \omega_0/\sqrt{1-\zeta}$ ... you mean like that?

It's a heroic effort but still hard to read. You could just use roman characters ... L=-zw0/√(1-z) because you can define any variable names you like.
Or, you can learn LaTeX markup - which is what the rest of us do:
https://www.physicsforums.com/help/latexhelp/

So:
damping ratio = c/2sqrt(m/k) $\zeta = c/2\sqrt{km}$ https://en.wikipedia.org/wiki/Damping_ratio#Definition
wnatural = sqrt(k/m) $\omega_0 = \sqrt{k/m}$
wdamped = wnatural x sqrt(1-dampingRatio^2) $\omega = \omega_0\sqrt{1-\zeta^2}$
log decrement = (1/n)ln(x1/xn+1) = 2pi(damping ratio)/sqrt(1-dampingRatio) $\delta = \frac{1}{n}\ln|x_1/x_{n+1}|$

... I don't follow.
Am I reading that right: $I=bh^3/12$ ?? Where are these numbers coming from?

3. Oct 20, 2016

### kev.thomson96

Apologies for not using that.
The second moment of area of the beam, which is a rectangular prism.

4. Oct 20, 2016

### Simon Bridge

OK - so how are you thinking of the problem? Please show your best attempt with reasoning.

5. Oct 20, 2016

### kev.thomson96

We can find the damping ratio from the log decrement.

Shaker and beam have the same ground, therefore they are in parallel in terms of stiffness, so the equivalent stiffness formula is $k_equivalent = k_beam + k_shaker$

To find k_beam, we need to equate it to $\frac F x$, and equate $\frac F x$ to $\frac 3EI l^3$(just rearranging the deflection formula).

To plug I into the deflection formula, we find it by using the appropriate formula for a rectangular prism.

And then I'm stuck.

6. Oct 23, 2016

### Simon Bridge

... did you figure it out?
(BTW: a_bcd gets you $a_bcd$ while a_{bcd} gets $a_{bcd}$ same with \frac ab cd vs \frac{ab}{cd} ... but better this time :) )