Involute Spline Design Generator - Inventor Calculations
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I was generating a spline connection and wanted to review the calculations so that I would better understand how the formulas that Inventor uses were derived. This way I can understand better how Inventor is checking if the spline will work or not. There were a couple of things that I found that I would like clarification on. I am using a ANSI B92.1 - 30 deg, Fillet root, Side fit 10/20 - Class 5 spline in english units. The size of the spline is 3.300 in - 10ul/in x 32 ul that is 4.0 in long.
=> What reference did Inventor use to get the Lmin equation? I can find the stress equations in my machinery's handbook but I don't see any minimum length equations. When trying to derive this equation I started with pressure is equal to force times a safety factor divided by area.
Pmin=F*SF/A
The force is the tangential force created by the applied torque.
F=2*T/ds
The area is half the groove height times the length times the number of teeth.
A=(hst/2)*Lmin*N
Substituting the force and area back into the pressure equation I get.
Pmin=[2*T*SF]/[ds*(hst/2)*Lmin*N]
Rearranging this equation to solve for Lmin I get.
Lmin=[2*T*SF]/[ds*(hst/2)*Pmin*N]
This equation is very similar to one that Inventor is using except they have a factor of 12 on the torque were I have a factor of 2 on the torque. If the torque is in lb-ft then you would need to multiple by 12 to get lb-in so that Lmin comes out in inches. So I believe if Inventor is using the torque in lb-ft then the equation should have a factor of 24 on the torque.
Where does Inventor get the allowable pressure for the materials? If I want to create a new material I can only enter the allowable shear stress, the allowable compressive stress and the allowable tensile stress.
Please correct me if I have missed something, I just want to understand how Inventor is working so that I can trust the numbers it gives me. If my derivation is correct then the strength check that Inventor uses will also need to be corrected, as it uses the same equation.
=> When calculating the radial tensile stress I have found an problem with the way Inventor does this calculation. In the equations for these calculations Inventor states the the wall thick is the outside diameter of the hub minus the major diameter of the internal spline. The wall thickness should be this but then divided by two. For my application I have a hub diameter of 120mm or 4.724 inches, and a torque of 11750 lb-ft or 141000 lb-in. Inventor calculates (Sr) to be 2493.117 psi. Using the Inventor wall thickness I get 1506.258 psi. Using the wall thickness that has been divided by two I get 3012.516 psi, which is still not the value the Inventor gives. When I use the MINOR diameter of the internal spline for the wall thickness and divided by two then I get 2493.117 psi. So I believe that Inventor is calculating this value incorrectly, please correct me if I have missed something.
=> When calculating the beam tensile stress I found that Inventor is using the same equation as what is in the machinery's handbook but with a (Ks) factor in the denominator. The equation from the machinery's handbook states the the factor of 4 is from the assumption that only half the teeth are engaged. If I use Ks = 0.5 which is what Inventor says for half the teeth engaged then I would have 8 times the torque instead of 4 times the torque. I believe the equation that Inventor should be using is:
SB=[2*T]/[D^2*Lf*Y*Ks]
Please correct me if I have missed something.
Thank you in advance for all your help.
