Disclaimer: amateur mechanic, not an engineer of any kind.

I just bought a spring compressor for my car. It's the factory recommended type so I'm assuming it's strong enough, but I'm wondering exactly how strong - since it looks (to my untrained eye) like there's a clear weakest link. The spring gets sandwiched between cast iron plates. The bottom plate rests on a threaded handle that looks really thick and sturdy. The top plate however is held in place by a hemispherical piece ('Upper Ball' in Figure 1), which is in turn held onto the main shaft by a small dowel pin ('Pin' in Figure 1). The upper ball fills a tapered hole in the upper plate, which presses down onto the coil. Everything seems like a well-machined fit - the pin slides in easily with no play.

Here's an image of the actual pieces:

The pin is 7.75mm in diameter and 56.0mm in length. The upper ball has an ID of 18.6mm and an OD of 44.3mm, so during use the pin sticks out both sides as per the image. I can't tell what it's made of, but I'm assuming either grade 5 or grade 8 steel or the equivalent metric grade.

Two questions:

(a) am I correct in assuming that this is the weakest part of the assembly?

(b) is it possible to calculate an approximate yielding/breaking strength of this part of the assembly, assuming either grade 5 of grade 8 steel?

For reference, the total force I would routinely put on this compressor is 1000 lbs (~4500 N).


I would have thought the nut and screw screw thread was more likely to be the failure point, but you are looking at the part and I'm only looking at a sketch of it.

The yield strength of grade 5 steel is around 600 MPa. The area of the pin is about 47 mm^2. So to permanently bend the pin by shearing it would take about 28000 N or about 6300 lbf. (Or arguably twice that, since the load goes to both sides of the pin).

Nothing much to worry IMO.

  • $\begingroup$ Woah that's nuts. My intuition about how strong that little steel nub is was totally off. Plus if it actually is grade 8 like most automotive suspension parts, and assuming an extra factor of 2, it would take over 18,000 lbf to shear the pin. $\endgroup$ – user28400 Feb 11 at 6:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.