How Is Young's Modulus Calculated for a Cantilever in SHM Experiments?

[joelio36]

Youngs modulus of a cantilever, help please!

[1] BACKGROUND:
I am doing an experiment determining the young's modulus of a cantilever through principles of SHM (i.e. hypothesis: higher youngs modulus = higher frequency of oscillation.

See the attached Pictures for the diagram (strobe is used to find frequency of cantilever with use of beat frequencies).

We were given the attached formula to manipulate to find young's modulus.

T=Period of oscillation
M=Load at end of cantilever (Forget about mass of cantilever, negligible)
L=Length between cantilever's fixed pivot and the load's centre of mass
E=Youngs Modulus
I=Moment of Inertia

[2] Problem:
I have done everything a thousand times over, and i consistently get a value of 15Gpa for youngs modulus! isn't that far too high for a classroom wooden ruler?

I have used SI units and everything has been done right, the only thing i can think of is the formula is wrong is the formula we were given is flawed.

I personally think the problem lies in the moment of inertia sub-formula, I was given bd³, where b= width of beam d=depth of beam.

I don't know ANYHTING about moments of inertia, but surely that is wrong?? That would mean if the beam was 10cm or 10000km long, the moment of inertia would be equal.
I know its asking a lot but please help, this is my final report and I'm pretty screwed.

 

[tiny-tim]

I personally think the problem lies in the moment of inertia sub-formula, I was given bd³, where b= width of beam d=depth of beam.

I don't know ANYHTING about moments of inertia, but surely that is wrong??

Hi joelio36!

(there were no pictures)

Yes, you're right … moment of inertia has dimensions of distance squared times mass (or distance to the fifth times density).

So it can't be bd³.

 

[JohnSimpson]

I can't see any picture, but I assume you are dealing with the standard beam equation solved for some boundary values, in which case the "I" in the equation is actually "I_zz", the second moment of cross sectional area, and not the moment of inertia.

http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia

Assuming I'm correct, the correct dimensions are in fact [Length]^4

 

[tiny-tim]

second moment of area

I can't see any picture, but I assume you are dealing with the standard beam equation solved for some boundary values, in which case the "I" in the equation is actually "I_zz", the second moment of cross sectional area, and not the moment of inertia.

http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia

Assuming I'm correct, the correct dimensions are in fact [Length]^4

Thanks John!

Well, I never heard of that before!

hmm … from http://en.wikipedia.org/wiki/Second_moment_of_area" :

Many engineers refer to the second moment of area as the moment of inertia and use the same symbol I for both, which may be confusing.

The second moment of area depends only on shape, not on mass or even density.

So "moment of inertia" or "second moment of inertia" are really stupid names for this , because "inertia" means "mass", and mass isn't involved in the second moment of area.

The simplest-case operative formula seems to be a vector cross-product:

normal stress = σ_z =(M_x/I_x, M_y/I_y) x (C_x,C_y) = (bending moment/second moment of inertia) x (offset of centroid from axis). ​

I personally think the problem lies in the moment of inertia sub-formula, I was given bd³, where b= width of beam d=depth of beam.

The formula for a complete rectangle about its centroid seems to be bd³/12, not bd³.

But perhaps your beam is a T-section or similar, and the axis is off-centroid?

The picture you promised would really help!

 

[tiny-tim]

Hi joelio36!

Is this the equation: T = 2π√(4ML³/Ebd³)?

And are you working from this link … http://www.practicalphysics.org/go/Experiment_430.html ?