So I'm given this question with the following diagram:

Two sections of a pipe are joined in a flanged joint as shown in the Figure below. The difference in pressure from interior to exterior is 2 MPa.

The joint is connected with 12 off bolts, equally pitched around a pitch circle diameter of 350 mm. If the maximum allowable tensile stress on the bolts is 300 MPa, what diameter do the bolts need to be?

enter image description here

My working out

enter image description here

First I find the area of the section that the bolt's diameter takes up:

$A = \pi (0.175+r)^2 - \pi(0.175-r)^2 $

$A = 0.7\pi r\ $

Then I find out the force acting on that area of section and divide by twelve to obtain the force for a section that has one bolt.

$F_{12} = (2\times 10^6)(0.7\pi r) = 14 \times 10^5 \pi r$

$F_1 = 366519.1429r$

I then sub this into the stress equations since I know the maximum tensile stress for 1 bolt:

$A = {F_1 \over \sigma_{max}}$

$\pi r^2 = {366519.1429r \over 300 \times 10^6}$

$Diameter = 0.039 m$

But the answer I'm given says it is supposed to be 5.9mm. Is there anything wrong in my thinking?


1 Answer 1


Your calculation for the maximum forces to the bolt appears flawed.

The forces resisted by the bolts in tension is equal to the total force of the area on either side of the tube.

You have the tube diameter and thickness, you can calculate the interior area.

Multiple the pressure by this area to find the total force trying to pull these two sides apart.

Divide that by 12 and you have your force in one bolt. Then divide by the allowable bolt stress to get the minimum required area and then diameter.

When I do this I get 5.8mm this is the root area (doesn't include the threads).

  • $\begingroup$ So far in my course we haven't touched on factor of safety so unfortunately I'm not familiar with the factor of safety. The maximum stress came from the question. If it helps, I was made to calculate the hoop and axial stress of the same setup in a previous question. Do I make use of that? $\endgroup$
    – Kyzen
    Apr 22, 2019 at 21:47
  • $\begingroup$ Okay... Can you demonstrate how you find the force that goes to the bolt? It seems to me that you have 2MPa over the area on each end that needs to be resisted by the 12 bolts. The total force across the flange is the the product of the interior area x the pressure. This is the force that should go over the bolts. I dont think this should be a function of the bolt diameter. The bolts are outside the pressurized area. $\endgroup$
    – ShadowMan
    Apr 22, 2019 at 21:52
  • $\begingroup$ So to correctly calculate the force acting on all 12 bolts, I calculate the interior area which would be found by ignoring the flange bit (so π x 0.125^2 x 2MPa)? I'm not too sure why we would need the tube diameter or thickness since the bolts aren't on the tube itself? I had assumed that I needed to make use of the pitch circle diameter somehow. Thank you for your time for helping me. $\endgroup$
    – Kyzen
    Apr 22, 2019 at 22:08
  • $\begingroup$ The pressure doesnt act over the area where the thickness lies. So the radius of the inside of the vessel is 125 - 1.5. As I said the calculation is for the minimum root area of the bolt. The threads are not included in the root area. $\endgroup$
    – ShadowMan
    Apr 22, 2019 at 22:13
  • $\begingroup$ Thanks for your help, I now understand how to get the answer. My confusion came from my assumption that as you move outwards from the tube (outside the 250mm where the pressure no longer acts) the force acting on the parts of the tube would decrease. Would you happen to know if this is correct in real life and that the questions is under ideal conditions? $\endgroup$
    – Kyzen
    Apr 22, 2019 at 22:21

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