Force Applied onto a Mounting Hole

AI Thread Summary
To determine the force applied to the four mounting holes of a 0.1" thick aluminum plate under a torque of 35 lb.in, the force can be calculated using the formula Force = Torque / (r*4 holes) for equal distribution, though a more accurate approach assumes only two screws react to the torque, leading to Force = Torque / (r*2 holes). After calculating the force, it is essential to evaluate the plate's thickness and hole placement by computing the bearing stress and shear tear-out stress. The bearing stress is calculated as force divided by the product of screw diameter and plate thickness, while shear tear-out stress is determined using the force divided by twice the plate thickness and the distance from the hole edge to the plate edge. It is recommended to limit these stresses to 40% of the yield strength of the aluminum plate to ensure safety and integrity. Proper calculations will confirm if the plate is adequately designed for the application.
numenor260
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Hi everyone:

I have an 0.1" thick aluminum plate. I am mounting a component to the plate using four 8-32 screws.

During use, this component will see a torque of about 35 Ib.in applied onto a shaft that is connected to it.

How do I determine the force that is applied onto the four mounting holes?

I really want to make sure the plate thickness is sufficient.

Thank you for any guidance.
 
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Assuming the screw holes are symmetrically arranged around the shaft's center-line at a radius of 'r', then the load on each would be Force = Torque / (r*4 holes). That assumes the forces on the screws will be equally divided, which is not likely. A more reasonable assumption would be to assume that only 2 screws react the torque, then the Force = Torque / (r*2 holes).
 
Once you have found this force, you can check if the plate is thick enough (and the holes are far enough away from the edge) by computing the "bearing stress" and the "shear tear out" stress. Let me know if you need any help with these.
 
edgepflow said:
Once you have found this force, you can check if the plate is thick enough (and the holes are far enough away from the edge) by computing the "bearing stress" and the "shear tear out" stress. Let me know if you need any help with these.

Thanks very much for your response. Yes, I would like help with this.
 
DickL said:
Assuming the screw holes are symmetrically arranged around the shaft's center-line at a radius of 'r', then the load on each would be Force = Torque / (r*4 holes). That assumes the forces on the screws will be equally divided, which is not likely. A more reasonable assumption would be to assume that only 2 screws react the torque, then the Force = Torque / (r*2 holes).

Thanks for your response. The hole are only symmetric about the horizontal axis.
 
numenor260 said:
Thanks very much for your response. Yes, I would like help with this.

Compute the bearing stress as follows:

bearing stress = force / ( screw diameter X plate thickness)

Compute the shear tear out stress as follows:

shear tear out stress = force / (2 X plate thickness X distance from hole edge to plate edge)

Then limit these calculated stresses to 40% of the yield stength of the plate (typical for shear).
 
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