Cantilever Beam Stiffness Calculation

AI Thread Summary
A cantilever beam at low frequency operates as an underdamped one-degree-of-freedom mass/spring/damper system. The discussion focuses on calculating the roots of the characteristic equation and determining the damping ratio, natural frequency, and equivalent stiffness of the beam and shaker system. Key formulas include the second moment of area for a rectangular beam and the relationship between force, deflection, and stiffness. Participants emphasize the importance of using proper notation and formulas, particularly for the second moment of area and equivalent stiffness. The conversation highlights the need for clarity in calculations and reasoning to solve the stiffness calculation problem effectively.
kev.thomson96
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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
 
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kev.thomson96 said:
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)
##\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}|##

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
... I don't follow.
Am I reading that right: ##I=bh^3/12## ?? Where are these numbers coming from?
Please explain your reasoning?
 
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Simon Bridge said:
Or, you can learn LaTeX markup - which is what the rest of us do:
Apologies for not using that.
Simon Bridge said:
Am I reading that right: I=bh3/12I=bh3/12I=bh^3/12 ?? Where are these numbers coming from?
The second moment of area of the beam, which is a rectangular prism.
 
OK - so how are you thinking of the problem? Please show your best attempt with reasoning.
 
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.
 
kev.thomson96 said:
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.
... 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 :) )
 
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