Cantilever Beam Stiffness Calculation

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SUMMARY

The discussion focuses on the calculation of cantilever beam stiffness, specifically in the context of an underdamped 1 DOF mass/spring/damper system. Key formulas include the damping ratio defined as c/2sqrt(m/k) and the natural frequency wnatural = sqrt(k/m). The equivalent stiffness is calculated using kequivalent = kbeam + kshaker, with kshaker given as 45 N/m. The second moment of area for a rectangular beam is determined using I = bh^3/12, leading to a value of 15.83 x 10^-12 m.

PREREQUISITES
  • Understanding of cantilever beam mechanics
  • Familiarity with mass/spring/damper systems
  • Knowledge of damping ratios and natural frequencies
  • Ability to calculate the second moment of area for rectangular sections
NEXT STEPS
  • Learn about the derivation of the characteristic equation for damped systems
  • Study the implications of damping ratios on system behavior
  • Explore the calculation of deflection in beams using deflection x = Fl^3/3EI
  • Investigate LaTeX markup for better presentation of mathematical equations
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Mechanical engineers, structural analysts, and students studying dynamics and vibration analysis will benefit from this discussion on cantilever beam stiffness calculations.

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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