I am wondering whether a typical mechanic can break a bolt using a wrench.

For example, let's say the wrench is 15" long and the force is applied at a radius of 13". The mechanic applies a force of 40 pounds. Imagine the thread of the bolt is 3/8" in diameter and has 16 threads per inch. The bolt has an ultimate tensile strength of 150,000 PSI.

Will the bolt fail and break apart?

  • $\begingroup$ @AndyT I changed the title to make it more descriptive. $\endgroup$ – Wallace Park Nov 17 '16 at 12:35
  • $\begingroup$ You're underestimating the force applied by several factors. For back-of-napkin calculations, I'd swag that 250 lbs is a safer estimate. And realistically speaking, a mechanic would use an extension bar to increase the leverage applied. Personally, I've warped lower-quality lug nut wrenches while attempting to remove lug nuts. And I'm reasonably certain that I've sheared the head off of 3/8" bolts while tightening them down. $\endgroup$ – user16 Nov 17 '16 at 12:59
  • $\begingroup$ @GlenH7 The question is not so much is it possible. I know it is possible. I want to get a sense for how hard it would be, and whether it could happen accidentally with normal procedures and tools. My concern is somebody breaking a bolt while just trying to tighten it, not somebody deliberately trying to break it. $\endgroup$ – Wallace Park Nov 17 '16 at 14:36
  • $\begingroup$ @WallacePark The short answer would be yes. You did not limit the question by providing parameters one can use. Please share with us a) the bolt grade (if steel) or material (plastic (nylon, PVC etc.), brass, steel, etc.). b) are standard wrenches/spanners being considered only, or adjustable wrenches? The reason I ask is that it is quite easy to destroy a nylon bolt. Also, adjustable wrenches allow for a longer lever arm than standard wrenches. I believe standard wrench lengths were determined to limit the torque the average human can generate, thus protecting the bolt during tightening. $\endgroup$ – NamSandStorm Nov 19 '16 at 4:43

torsion T = 40 pounds * 13 inches = 520 pound-inches

shear stress = Tr/J

I would assume that threads would be ignored, and stress would only be calculated on the shank diameter, but seeing as you haven't provided that, I'll just use 3/8in as the available shaft to resist the torque.

Torsional constant J = pi * D4 / 32 = 3.1415 * (3/8)4 / 32 = 1.941E-03 in4

radius r = D/2 = 3/16in

Therefore stress = 520 * (3/16) / 1.941E-03 = 50,220psi

Shear strength is about 55% of tensile strength, giving you a limiting shear stress of 0.55 * 150,000 = 82500psi.

Based on your numbers, the bolt survives. But I reckon I could supply 100 pounds of force (and I'm not exactly a strong guy), which would be enough to shear the bolt.

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  • $\begingroup$ This does not take into account the wedging effect of the threads. Also, you seem to be confused about the relationship between tensile fracture and shear fracture. $\endgroup$ – Wallace Park Nov 17 '16 at 14:40
  • $\begingroup$ @WallacePark - Clearly you know more than me; do please feel free to share your knowledge by answering your own question. Also, if you can make your feedback more constructive it would be appreciated: what about the relationship between tensile and shear fracture have I misunderstood? (Although there is a hint of sarcasm in this comment, I do also mean it all genuinely.) $\endgroup$ – AndyT Nov 17 '16 at 15:51

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