Calculating the Diameter of a Breakaway Bolt for 40 ft-lbs Torque Requirement

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Discussion Overview

The discussion revolves around calculating the diameter of a breakaway bolt designed to break at a specified torque of 40 ft-lbs. Participants explore theoretical and practical aspects of shear bolts, including material properties and failure theories.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks an equation to determine the torque required to break the head of a breakaway bolt, specifically at 40 ft-lbs.
  • Another participant suggests that the problem involves a shear bolt and provides a relationship between tensile and shear moduli, referencing Mohr's Circle and various failure theories like Von Mises and Maximum Shear theory.
  • A different participant expresses that the provided information does not directly help in calculating the required torque to break the bolt.
  • One participant proposes a formula for calculating the diameter of a solid circular breakaway shaft based on installation torque, tensile ultimate strength, and a dimensionless factor, suggesting that testing different bolt alloys is necessary to determine the factor accurately.
  • The same participant offers an example calculation using specific values for torque, tensile strength, and the dimensionless factor to illustrate the application of the proposed formula.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to calculate the required torque or the diameter of the breakaway bolt. Multiple viewpoints and methods are presented, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants note the importance of testing different bolt alloys to determine the dimensionless factor used in calculations, highlighting that assumptions about material properties may vary.

rsk2mc
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I am trying to design a breakaway Bolt, is there an equation to calculate how much torque is required to break the head of the bolt. Typically this bolt has two heads, one that breaks off and the permanent one. These are connected with a small neck that breaks at a given torque. That's what I need to calculate, the diameter needed to break at 40 ft-lbs.
 
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Sounds like a shear bolt.
So this is what i found

The relationship between the tensile and shear moduli is

E = 2*G*(1+v)

E = Elastic modulus
G = Shear modulus
v = Poisson's ratio

so for common steels assuming E = 30,000,000 psi and v = .33 Then G = 11,278,196 psi

You should also use Mohr's Circle.

There are several failure theories that can be used such as the Von Mises and the Maximum Shear theory.

Remember in shear bolt the failure (breakage) occurs at the ultimate strength not at the yield strength.
 
I saw the same post, however this does not help me figure out the required torque to break the bolt.
 
rsk2mc: In your case, the bolt breakaway head will not be preloaded, which is different from a typical bolt head. Therefore, in your particular case, if your breakaway shaft is a solid circular cross section, you could use the following.

d = 2.04[T/(Rsu*Stu)]^0.3333,​

where d = solid circular breakaway shaft diameter, T = breakaway bolt installation torque, Stu = bolt material tensile ultimate strength, and Rsu = dimensionless factor, such as 1.00 or 1.40.

You would need to test each different bolt alloy, to determine the value of Rsu for that particular alloy. Fortunately, there is probably no test easier than this test. Once you determine Rsu for a particular alloy, then you can plug it into the above formula, for that particular alloy.

For an initial guess at Rsu, I might arbitrarily guess 1.30 or 1.35 (?). Regardless of your initial guess, after you test it, you can then solve for Rsu, using the above formula, to obtain the correct value of Rsu, for that particular bolt alloy.

If you do determine your above Rsu value, then if you want to post your Rsu value, and bolt alloy, here, that would be great.

Here is an example. T = 54 240 N*mm, Stu = 515 MPa, Rsu = 1.45. Therefore, d = 2.04[(54 240 N*mm)/(1.45*515 MPa)]^0.3333 = 8.51 mm.
 
Last edited:

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