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Torque Capacity of Straight Splines 4

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PhilBW

Mechanical
Feb 17, 2004
112
I am trying to find out the maximum torque that can be transmitted by a shaft on one of our existing machines. The shaft has 10 straight sided splines. The major diameter is 5", minor diameter 4.3".

In a Machine Design textbook (Faires)it gives a formula: T=s*A*rm where s is side pressure, A is area of splines and rm is mean radius. It also states that for splines with axial sliding under load, s should be 1000 psi, but if there is no axial sliding under load, s will be "much greater". He doesn't say how much greater. Since there will be no axial sliding under load, I need a realistic value for s.

The torque values for straight sided splines in Machinery's Handbook are approximately the same as what I calculated using 1000 psi side load, but the torque we currently have on this shaft is much higher than the calculation or what Machinery's Handbook shows.

Can anyone help me out? Thanks.
 
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What is the material and core hardness
of the material or physical
strength of the material?
The formula that I have is similar
but must be multiplied by N or the
Number of splines.

T = (1000) (hL) (rm) (N) inchpounds
Design of Machine Elements (1935)
by Virgil Morino Faires, M.E., M.S.

Does this give you better results?
(hL) is equal to the area in your
equation. h is the height of the
spline and L is the Length.


 
The one that I came up with (from the Norton book) is:

T = 1/16 * Ashear * (.67Sut) * FOS

Ashear = pi* dia(root)* Lenght of engagement

FOS - factor of safety

Sut - Ultimate Tensil Strength.



Wes C.
 
diamondjim - Thanks for your suggestion. I had already included number of splines in the area calculation


wes616 - Thanks to you also. I'll give this a try

Phil
 
Phil;

might I suggest getting the rod tested? Don't know if you have that possibility, but it could reduce the headaches. I know there are lots of test frames out there in the world that can do torsion testing under either constant axial stress or constant axial strain (or zero axial stress or strain). Having the test data could help, I would think!

-Ron
 
Ron:

Thanks for the suggestion. It would be great to get the torsion testing done and see how the results compare to the calculated values.

Phil
 
Lets assume the Safety factor was 20
for this part.

20,000psi would be a typ. max shear stress
for soft steels and result in an allowable
shear stress of 1000psi. 20,000/20 = 1000.

A one time fracture test should give you
a near 20,000psi value. For the endurance
limit it would still be 1000psi if the Safety
factor is 20.

You would still have to estimate how the Safety
factor decreases if no sliding takes place.
So you are back to square one after the testing.
 
israelkk, that is an interesting paper and a great web site. Thanks.
 
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