bending moment of inertia for multiple step shaft

[Cheetos]

I created a shaft with 3 steps and using CAD program I was able to find out the Ix, Iy, Iz on the shaft.
L1/D1 = 2"/5"
L2/D2 = 4"/7"
L3/D3 = 3"/6"
density (rho) = 0.283 lb/in3

From CAD, I got Ix = 65.74 Lb-in2 and Iy = Iz = 785.7 lb-in2

How do I get Iy and Iz analytically? I was able to get Ix by summing up rho*L*pi*D^4/32 for all sections, but this doesn't work for Iy and Iz.

 

[rb1957]

do you want mass moments of inertia or area moments of inertia ?

i think Iy (and Iz) are based on the plan view of the shaft (like areas of D*L) about the center of mass of the shaft ?

Quando Omni Flunkus Moritati

 

[Cheetos]

yes, you're right. The center line through the shaft is the x-axis, so I'm looking for the bending mass moment of inertia

 

[rb1957]

so Iy is the mass moment of three cyclinders stacked on top of one another about the CG of the stack

Quando Omni Flunkus Moritati

 

[Cheetos]

Yes, the cylinders are right next to each other. If the all the diameters were the same, I can use standard formula to calculate Ix, Iy and Iz.

Ix = m*D^2/8
Iy = Iz = m/48*(3*D^2+4*L^2)

For Ix, I can calculate each section separately and sum all the sections to obtain the Ix for the entire shaft. I can't do that for Iy and Iz.

 

[rb1957]

no, when i mentioned the CG of the assembly you have to take into account the stacking of the cylinder. the cylinders have an Iy about their base, when the axis don't align (note how they align for Ix) you need to use the parallel axis theroem.

Quando Omni Flunkus Moritati

 

[electricpete]

Note you don't need a single value of Iy and Iz to do bending calculations using Euler Bernoulli beam theory. For example if you know the forces w(x):
V(x) = Int {w(x)}dx
M(x) = Int {V(x)}dx
Slope(x) = Int {M(x)}dx / [E*I(x)]
displacement y(x) = Int{Slope(x)}dx
Note that I(x) denotes I varying... different in the three regions.

Sorry if I missed your point and am rambling about irrelevant and obvious things.



=====================================
(2B)+(2B)' ?

 

[GregLocock]

Many of the above answers are corect, i fear none address the exact issue directly.

Your software has calculated the rotational moment of inertia, as if it were a 3 dimensional flywheel. That's fine and dandy, if you are designing 3 dimensional flywheels.

I suspect from the thread title you are interested in the bending behaviour of your shaft considered as a beam, in which case the software has led you up the proverbial garden path, taken you into the wood shed, and is about to have its wicked way with you.

So go back to elastic beam theory. If you describe the problem you need solving, I expect somebody here can help.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[rb1957]

guys, the OP has repeatedly posted using mass moments of inertia ... possibly he knows what he wants ?

you've posted an expression of mass moment of a cylinder about the y- (and z-) axis through the centroid of the cylinder (i think so) or through the base (i could look up mass moments of inertia but i don't feel inclined to) ? just summing the x-axis mass moments for the individual cylinders works because they share the same x-axis and it's through the CG of the assembly, but doesn't work for y- and z- 'cause these axes are not colinear (they're parallel). Research parallel axes theorem.

alternatively, since he can't do (or recognise) the simple parallel axis maybe he doesn't ??

Quando Omni Flunkus Moritati

 

[GregLocock]

so why did he give the thread the title he did?

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[Cheetos]

thanks for the reminder, rb1957. i totally forgot about the parallel axis theorem since i haven't used it in a long time. thanks for the help.