Cantilever Investigation

  • Thread starter ash1098
  • Start date
  • #1
5
0
[SOLVED] Cantilever Investigation

1.Hi,

I am writing a report on cantilever oscillations, my experiment involves fixing different cantilvers e.g. a ruler to the end of a table then measuring the period and height of oscillations while varing the mass attached to the cantilever, and other varients e.g. lenght of cantilever.


2. I have found these two formulae: (Shown much more clearly in attachments)

T= 2(pi)*[(4ML^3)/(bd^3E)]^1/2

and:

h= 4MgL^3/Ebd^3

where:
b= width of cantilever
d= thickness of cantilever
E= Youngs Modulus
M=Mass
L=Lenght of cantilever
T=period of oscillations
h=height of oscillation


3. I have looked at eqn's involving Hooke's and simple harmonic motion but cannot work out how these formulae have been derived.

Does anyone know how these formulae where derived, or where I can find information on this in general?

Thanks.

ash.

p.s. I have written out the formulae using math open office and attached them in pdf if it helps make them easier to read.
 

Attachments

  • h.pdf
    40.1 KB · Views: 302
  • T.pdf
    41.1 KB · Views: 275

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Well I have done a lab to find the Young's Modulus of a loaded cantilever and this is theory which is written down on the paper:

The depression,s,due to a load W(=Mg) at the end of a cantilever of length,l, is

[tex]s=\frac{Wl^3}{3IE}[/tex]

This strain brings into play internal stresses which produce a restoring force equal to W, i.e. equal to [itex]\frac{3IEs}{l^3}[/itex].

If the acceleration of the load [itex]\frac{d^2s}{dt^2}[/itex] when the cantilever is displaced to produce vertical oscillations,then

[tex]M\frac{d^2s}{dt^2}=\frac{-3IE}{l^3}s[/tex]

OR
[tex]\frac{d^2s}{dt^2}+\frac{3IE}{Ml^3}s=0[/tex]

Hence the motion is Simple harmonic and the periodic time,T, is

[tex]T=2\pi \sqrt{\frac{Ml^3}{3IE}}[/tex]

from which

[tex] E=\frac{4\pi^2Ml^3}{3IT^2}[/tex]

For a beam of rectangular section:
[tex]I=\frac{bd^3}{12}[/tex]



I hope that helps in some way
 
Last edited:
  • #3
5
0
Thanks for your reply rock.freak it's really really useful, just one question:

why does [tex]T=2\pi \sqrt{\frac{Ml^3}{3IE}}[/tex]

Thanks, ash.
 
  • #4
rock.freak667
Homework Helper
6,230
31
Well from
[tex]M\frac{d^2s}{dt^2}=\frac{-3IE}{l^3}s[/tex]

[tex]\frac{d^2s}{dt^2}=\frac{-3IE}{Ml^3}s[/tex]

and that is of the form [itex]a=-\omega^2 s[/itex] where [itex]a=\frac{d^2s}{dt^2}[/itex]

so from that

[tex]\omega=\sqrt{\frac{3IE}{Ml^3}}[/tex]

and since it moves with SHM, the period,T, is given by

[tex]T=\frac{2\pi}{\omega}[/tex]
 
  • #5
5
0
Oh yeah I see it now, Thanks alot.

Ash.
 

Related Threads on Cantilever Investigation

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
11
Views
874
Top