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Bolt calculation 1

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TGK07

Student
Sep 28, 2021
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My colleague calculated a safety factor of 1.31. Is my calculation correct?
 
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You have a bending moment and axial compression on the bolt which is going to generate tensile stress on one side of the bolt (M/S - P/A) and compression stress on the other edge of the bolts (M/S + P/A) , but I don't see a section modulus calculated, so at a glance, I would say there's no way it's right.

Rod Smith, P.E., The artist formerly known as HotRod10
 
Wait, is this supposed to be supported at point "M"?

Rod Smith, P.E., The artist formerly known as HotRod10
 
The "M" is confusing. Not sure why you'd take a moment about there instead of along the vertical axis of the CG. Doesn't make physical sense to me.
 
Point M is the assumed pivot point under headwind when the fastener starts to loosen.
 
Assuming the area exposed to wind is correct, and it's only about the size of a small tissue box (0.0144 m^2 = 22.32 in^2), and your support assumptions are valid (which is what I would be scrutinizing closely), the calculations appear to be correct; strength of the bolt shouldn't be an issue. The fatigue on the supporting material at Point M (it will experience millions of stress cycles), is another matter entirely, but since you haven't given us any information about that, there's no way for us to assess it.

Rod Smith, P.E., The artist formerly known as HotRod10
 
TGK07 said:
Point M is the assumed pivot point under headwind when the fastener starts to loosen.

What is your basis for that assumption?

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You need to address boundary conditions and equilibrium before you can solve the problem.
1. F[sub]A[/sub] has no force to balance it, so does not satisfy equilibrium.
2. The second equation (for F[sub]A[/sub]) is incorrect as it neglects the contribution of F[sub]S[/sub].

To answer your initial question, your solution is not correct. Given the magnitude of the applied force, it seems unlikely that your colleague's solution is even in the ballpark.

BA
 
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