What is the Shear Stress in Bolts for a Given Torque and Separation Distance?

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Discussion Overview

The discussion revolves around calculating shear stress in bolts given a specific torque and separation distance. Participants explore the relationships between torque, force, and shear stress, while addressing a homework problem related to bolts in a mechanical context.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents an initial attempt at solving the problem, calculating the separation of bolts based on the circumference of a shaft and expressing uncertainty about the next steps.
  • Another participant questions the relevance of the circumference calculation, suggesting that the torque must be resisted by the bolts without considering friction, prompting a reevaluation of the force on the bolts.
  • A participant calculates the force on each bolt using the torque and separation distance, arriving at a figure of 10 kN per bolt, which is later challenged by others.
  • Subsequent posts explore the possibility of the force being 20 kN per bolt, with participants confirming this figure.
  • Finally, a participant applies the force to calculate the area and subsequently the diameter of the bolts, arriving at a value of 11.3 mm, expressing satisfaction with the result.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correct force on each bolt, with some proposing 10 kN and others suggesting 20 kN. The discussion reflects multiple competing views on the calculations and assumptions involved, and no consensus is reached on the final values.

Contextual Notes

Participants rely on specific assumptions about the problem setup, including the neglect of friction and the interpretation of torque distribution among the bolts. There are unresolved steps in the calculations, particularly regarding the application of shear stress formulas and the determination of bolt dimensions.

Matthew Heywood
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Homework Statement


jan 2012 q3.jpg
I've solved everything up to part d). I generally struggle with questions on bolts and I'm really not sure how to tackle this problem.

Homework Equations


τ = Tr/J, τ = 200MPa, T=1kNm

The Attempt at a Solution


I worked out the separation of the bolts (I think).
Circumference around the inside: 2π(25mm) = 0.157m ∴ Separation: 0.157/2 = 0.0785m

Not sure where to go next, if I'm honest. I tried this:

T = 1kNm ⇒ F = 1kNm/0.0785m = 12.7kN
so σ = F/2A as double shear so A = F/2σ = 12.7kN/2(200MPa) = 3.19x10-5m2

so πr2 = 3.19-5 ⇒ r = 3.18mm ⇒ d = 6.37mm

so I went wrong somewhere as the answer is 11.28mm

any help appreciated. Thanks
 

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Matthew Heywood said:

Homework Statement


jan 2012 q3.jpg
I've solved everything up to part d). I generally struggle with questions on bolts and I'm really not sure how to tackle this problem.

Homework Equations


τ = Tr/J, τ = 200MPa, T=1kNm

The Attempt at a Solution


I worked out the separation of the bolts (I think).
Circumference around the inside: 2π(25mm) = 0.157m ∴ Separation: 0.157/2 = 0.0785m

It's not clear how the circumference of the solid shaft is useful here. You were instructed by the problem statement that friction between the shaft surface and the coupling was to be neglected. Therefore, all of the torque transmitted to the solid shaft must be resisted by the two bolts. What force on the bolts would produce an equal torque?

Not sure where to go next, if I'm honest. I tried this:

T = 1kNm ⇒ F = 1kNm/0.0785m = 12.7kN
so σ = F/2A as double shear so A = F/2σ = 12.7kN/2(200MPa) = 3.19x10-5m2

so πr2 = 3.19-5 ⇒ r = 3.18mm ⇒ d = 6.37mm

so I went wrong somewhere as the answer is 11.28mm

any help appreciated. Thanks
This approach is not correct for the reason given above.
 
Ah okay. So the for the force on each bolt:
T = Fm ⇒ F = 1kNm/50mm = 20kN
So 10kN on each bolt?
 
Matthew Heywood said:
Ah okay. So the for the force on each bolt:
T = Fm ⇒ F = 1kNm/50mm = 20kN
So 10kN on each bolt?
You might want to check that figure of 10 kN/bolt.
 
Oh okay. So would it be 20kN?
 
Matthew Heywood said:
Oh okay. So would it be 20kN?
Yes.
 
And then use σ = F/A?
 
so A = 20kN/200Mpa = 1x10-4
then r = √(1x10-4/π) = 5.64mm
so d = 11.3mm
Thanks :D
 

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