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

[Matthew Heywood]

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

 

[SteamKing]

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.

 

[Matthew Heywood]

Ah okay. So the for the force on each bolt:
T = Fm ⇒ F = 1kNm/50mm = 20kN
So 10kN on each bolt?

 

[SteamKing]

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.

 

[Matthew Heywood]

Oh okay. So would it be 20kN?

 

[SteamKing]

Oh okay. So would it be 20kN?

Yes.

 

[Matthew Heywood]

And then use σ = F/A?

 

[Matthew Heywood]

so A = 20kN/200Mpa = 1x10^-4
then r = √(1x10^-4/π) = 5.64mm
so d = 11.3mm
Thanks :D