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

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SUMMARY

The discussion focuses on calculating shear stress in bolts given a torque of 1 kNm and a separation distance of 50 mm. The correct approach involves using the equation T = Fm to determine the force on each bolt, resulting in a force of 20 kN per bolt. The area is then calculated using σ = F/A, leading to a diameter of approximately 11.3 mm for each bolt. The initial miscalculations were clarified through peer feedback, emphasizing the importance of torque distribution among bolts.

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  • Understanding of shear stress and torque equations (τ = Tr/J)
  • Familiarity with basic mechanics of materials concepts
  • Knowledge of double shear calculations
  • Ability to perform unit conversions and area calculations
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  • Learn about double shear scenarios and their implications in bolt design
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Mechanical engineers, students studying mechanics of materials, and professionals involved in bolted joint design will benefit from this discussion.

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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