What Is the Correct Bolt Diameter for a Rigid Flange Coupling?

[ilahac]

A rigid flange coupling, secured by 3 bolts on a PCD of 200mm, is subjected to a torque of 10kN/m. Determine the bolt diameter (to the nearest mm)

Bolt material shear stress = 225N/mm^2
Factor of safety = 2

Hi, I'm chris, I'm an aeronautical engineering student and I'm a little puzzled by a step in the above problem. I don't understand why I have to figure out the bolt area then reverse the step in the following calculation. I'm really just looking for some clarity, y'know, someone to say "yeah, that's correct". Any help appreciated, my work is below.

Working stress = 225N/mm^2 ÷ 2 = 112.5N/mm^2

Torque = Force*Radius => Force = Torque÷Radius
= 10*10^3÷0.1
= 10*10^4 kN

Stress=Force÷Area => Area = Force÷Stress
= 10*10^4÷112.5
= 888.8mm^2

Bolt area = 888.8÷3
= 296.2mm^2

Diameter = √(3*296.2 ÷ ᴨ)
= 16.81mm
= 17mm

 

[mgb_phys]

Looks correct except for the '3' in the bolt diamter - you have already takne into account that there are 3bolts, by dividing 888.8mm^2 by 3.
So I get a bolt radius of 9.7, so a diameter of 20mm

 

[piercas]

(rounded to nearest mm)

Hi Chris,

Thank you for sharing your work and questions. It seems like you have approached the problem correctly, but there is a slight misunderstanding in the calculation of bolt area. The reason for dividing by 3 is to account for the three bolts, but the resulting area should be multiplied by 3 instead of divided by 3. This is because each bolt will have the same area and will contribute to the total area of the coupling.

So, the correct calculation for bolt area would be:

Bolt area = 888.8 * 3 = 2666.4 mm^2

Then, using the same formula for diameter:

Diameter = √(4 * 2666.4 ÷ ᴨ) = 23.2 mm = 23 mm (rounded to nearest mm)

I hope this helps clarify the calculation and provides the confirmation you were looking for. Keep up the good work!