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Shear stress in bolts help

  • #1

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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Answers and Replies

  • #2
SteamKing
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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
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.
 
  • #3
Ah okay. So the for the force on each bolt:
T = Fm ⇒ F = 1kNm/50mm = 20kN
So 10kN on each bolt?
 
  • #4
SteamKing
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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.
 
  • #5
Oh okay. So would it be 20kN?
 
  • #6
SteamKing
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  • #7
And then use σ = F/A?
 
  • #8
so A = 20kN/200Mpa = 1x10-4
then r = √(1x10-4/π) = 5.64mm
so d = 11.3mm
Thanks :D
 

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