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• # Calculating torque on hardened pins

Discussion in 'Calculations' started by Vector Victor, Mar 29, 2012.

1. ### Vector VictorMember

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Hey!

This seems a simple enough task to determine, but I don't think I'm doing it correctly. I would like to determine how much torque a pair of pins can handle.
I am using the pins to transmit torque from one rotating mass to another. The pins are pressed into one mass and assume a close tolerance fit in the other and the two masses are flush against each other.
I'm given the double shear load rating of one pin at 40,000 pounds. The pins are each 3" from the center of rotation. Now torque being radius x force, would not one pin be rated at 3" x 45,000 pounds? But that seems too high a max torque, so I don't think I'm working that out correctly. How do I need to go about this calculation then?

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3. ### VirguleActive Member

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Could you provide a drawing?

This would help me understand.

4. ### Vector VictorMember

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Sorry, I don't have access to my drawings until next week. But if it's helpful, the pins are part of a coupling that join the two rotating masses. I just tried to keep it simple in description the first time around. One half of the coupling has the pins pressed in it and the other half of the coupling accepts the pins, then the two halves are securely held together by other means. So the pins allow one half to turn the other half by transferring the torque. I'm trying to calculate the max allowable torque on the pins. Does that help?

5. ### dons3dNew Member

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You're right, that's too much.

For a better understanding, i would just make it 2 pins in the same location (or close)--then you have your center for torque.

Now, the force the pins can withstand is still the same, whether it's direct or torque--it's just tricky with torque. You either need to multiply or divide your value depending on how far out you are from the center. You are 3" from the center, and torque (normally lbs.-ft.) is 12" out.

Looking at your application with that in mind, if your pins were 12" out, they could withstand 40,000 lbs.-ft. of torque. That is because the torque is the same as a direct force. Your pins are at 3" out from center. The mechanical advantage is 4X the amount. If you apply that same 40,000 lbs-ft, the force at that 3" is 160,000 lbs., so you have to divide by 4 to get your value. So 10,000 lbs.-ft would give 40,000 lbs. of force at 3". You have to pins, so they can withstand 20,000 lbs.-ft.

Hope this helps! (and hope it's right, anybody concur?)
Don

6. ### Dan PohlyMember

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If the rating you have for the pins is double shear, you will need to cut that value in half because the arrangement as you describe it will put each pin in single shear. The calculation in general would be shear strength X distance from the center of rotation = torque strength. So you have 3" X 20,000lb = 60,000 in-lb per pin. Divide this by 12 for the torque in ft-lb. However, I stress that this is just the general calculation method. Other factors may apply, such as how the "double shear load rating" was calculated, whether the arrangement truly puts the pins in shear or perhaps bending stresses become significant, whether the load will be shared equally, even maybe the speed of rotation might be something to consider. Also a safety factor needs to be applied to establish the torque load "rating". Lots of pins are designed based on shear strength and then fail due to bending stresses, so one has to be careful.

7. ### Vector VictorMember

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Thanks for the input. Actually, I'm dealing with in-lb anyway as the gearbox it is connected to is rated in in-lb. It seems as if I'm on the right track then with the simple torque = r x f, but I hadn't considered the single shear aspect. Maybe my understanding of double shear is not correct. Can you better define single versus double shear? I did get that double shear rating from the Machinery Handbook by the way. I understand where you're coming from regarding bending stresses and other factors, so for now the general calculation is the right place to start.

8. ### Dan PohlyMember

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Double shear is the typical case when a pin is used in a clevis, such as at the end of a typical hydraulic cylinder. There is one plate in the middle and two ears of a clevis around it and the pin passes through all three. So effectively there are three plates acting on the pin. One plate in the center applies the full applied force; the two ears of the clevis each apply half that force in reaction. In double shear, there are two shear planes in the pin, so it is at least twice as strong as single shear. Single shear occurs for example in a bolted connection where only two plates act on the pin. There is only one shear plane in the pin. Single shear is also more likely to induce bending in addition to shear so it can be much weaker.

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10. ### Dan PohlyMember

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One final thought: Machinery's Handbook had to make certain assumptions for the "double shear load rating" you found. That calculation has to involve the material (particular steel alloy) and its temper, as well as assumptions about the loading conditions - not just the diameter of the pin. If your material and conditions don't match then those ratings may not be applicable.

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