If you're using a 3D CAD pkg. you've got the information already. We're using SolidEdge...a great tool. It will explode drawings, allow you to change them, creat .jpg's, etc....great stuff.
B
To all....looks like, and I calculate (w/ everyone's help) we're looking at about 13k in-lbs. That's great, and will work well w/ our components! Many thanks to all.....you're all very smart.....& I'm VERY grateful!
Regards,
Bill
pj...you're right!
only thing is that 36 is the diameter, not the radius. So if you take 18 for "r"...I get 12577 in lbs.
Only one more ? though...how do you go from multiplying (sl-in) X (rad/s^2) to get (in-lbs) for the resulting units?
Bill
Also...in calculating this, I get 1.62million in lbs for "I", and consequently 305million in lbs for torque at an acceleration of 188.5 rad/s/s.
Geez this seems huge enough to not be real!
Bill
Prex...could you please expand on:
"the peak acceleration will be higher, especially if you don't want to brake the mass and can't accept an overspeed".
I don't need to brake it....and overspeed is not an issue.
Why is this important?
Thanks,
Bill
To all,
I have a 10,000 lb mass that is about 36 inches in diameter that is going to have to be accelerated from rest to about 300 rpms in roughly 10 seconds.
My question is...I guess I can figure it's "I" value (ie 1/2*m*r^2), but where does the acceleration of this body come into...
The calc should be easy enough now that I will have the correct data. Though, there are still issues such as the one CoryPad presented w/ the tightening procedure. Sometimes things like this are almost impossible to direct for us. Our customers are going to do whatever's easiest. Thus, we...
Metalguy,
Could you please explain how to relate the hardness that I would find, with the UTS...(that I would then take 60% of to get the shear strength).
Thanks
I think you're probably right. I doubt the rod is fully hardened through. We need the outside to be hardened b/c of its abusive environment.
Also, the rod is actually 2 1/2" OD (furthering the arguement that it's not going to be 45RC all the way thru), and the bolt I'm proposing to use in...
Metalguy,
Actually, I'm calc.ing that the 8 bolts will be weaker than the one big one. I'm using the equation in the other tread that you and I are chatting on "tensile capacity/min engagement", given by CoryPad. Maybe I'm missing something here, but I show the eight 1" bolts...
Thank you all for the help.
The 'cross talk' would be an issue here b/c there will be a fair amount of disassembly and reassembly with this devise...it's rather high maintenance. Thus, one of the things we always try to incorporate into our designs is simplicity.
Just for clarification...shear strength, T, is AKA Yield Strength???
I have data on UTS, Tensile Yield, Bulk Modulus, & Shear Modulus...I'm getting confused.
Which is the number I'm looking for?
