Force Applied onto a Mounting Hole

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SUMMARY

The discussion focuses on calculating the force applied to four mounting holes in a 0.1" thick aluminum plate when a torque of 35 Ib.in is applied to a component attached via 8-32 screws. The load on each screw is determined using the formula Force = Torque / (r*2 holes), assuming only two screws react to the torque. To ensure the plate's adequacy, participants recommend calculating bearing stress and shear tear out stress, limiting these values to 40% of the yield strength of the aluminum plate.

PREREQUISITES
  • Understanding of torque and its effects on mechanical components
  • Familiarity with bearing stress and shear tear out stress calculations
  • Knowledge of screw specifications, specifically 8-32 screws
  • Basic principles of material strength, particularly yield strength of aluminum
NEXT STEPS
  • Research how to calculate bearing stress in mechanical applications
  • Study shear tear out stress and its implications in mounting scenarios
  • Learn about the yield strength of various aluminum alloys
  • Explore torque distribution among multiple fasteners in mechanical assemblies
USEFUL FOR

Mechanical engineers, product designers, and anyone involved in the mounting and securing of components to aluminum plates will benefit from this discussion.

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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