Repairs... My garage door has counterweights, and the weight of the door is connected to the counterweights with a steel cable which runs over two pulleys. The original (cast iron) pulleys had needle roller bearings, but lacked any lubrication, and probably incorrect termination/hardening of the outer ring caused excessive wear, and I had to replace them.

I found size-compatible replacements, but they have 'normal' ball bearings, which should be able to bear the load. I've been trying to find a way to tell if the axis will stand hold the narrower load though. The simplified configuration is now:

Pulley mounting

White is the axle, blue is the fixed hardware, and green is the bearing of the new pulley. All measures are in mm. The extra space is because the original roller bearing was wider.

This is the path the cable follows:

Path of the cable

I've looked over many pages, and I'm not sure how to tackle this. The problem is not completely shearing stress. I've found precious little info on 'bending' limits of round bars (axle). The weight of the door is about 60 kg (same as weights, of course) and the load is distributed at each side of the door.

Intuitively I suspect this will hold, but then my mechanical intuition isn't as developed (I'm an electronics engineer). In short, will the (mild steel) 10 mm axis bear the weight? Or do I need some more hardened material?

  • $\begingroup$ Does this pulley hold the whole 60kg weight, or are there two pulleys so it only sees about 30 kg of tension? Also, does the cable leave the pulley 180 degrees from where it came in or 90 degrees? That impacts the resultant force on the axle/pulley assembly. FWIW, in general bending will not govern for such a short span and shear is the only likely failure mode. If the motor starts and stops suddenly, it would be wise to add a multiplier to the static load to represent that. doubling the load is a reasonably conservative rule of thumb. $\endgroup$
    – Ethan48
    Commented Mar 4, 2016 at 18:13
  • 1
    $\begingroup$ Yes, it holds the complete weight (well, only halt the weight, as I commented the load is distributed over two counterweights at each side of the door). I'll add a drawing to the question, but the pulleys do not reduce the load. The cable follows the pulley for 90°. $\endgroup$
    – jcoppens
    Commented Mar 4, 2016 at 18:35
  • $\begingroup$ OK, so with 30 kg worth of tension in the cable, you'll have ~1.4* 30kg of resultant load on each axle. $\endgroup$
    – Ethan48
    Commented Mar 4, 2016 at 19:06
  • $\begingroup$ Yes. With the safety factor you proposed, and taking into account friction of the counterweights in their enclosure, peaking at maybe a little more. $\endgroup$
    – jcoppens
    Commented Mar 4, 2016 at 21:54

1 Answer 1


A good start would be to disregard the fact that the load is distributed over a small part of the shaft and look at the problem as a basic "Three Point Bending" of a simply supported beam with a center load.

For that load case, the Max Stress ($\sigma_{max}$) is given by:

$$\sigma_{max} = \frac{P*L} {4Z} $$

Where $P$ is the load at the center of the shaft, $L$ is the distance between the supports and $Z$ is the section modulus. For a circular beam/shaft:

$$Z = 0.78r^3$$

For mild steel, the yield strength is probably about 240MPa

If this analysis gives an acceptable result, you should be OK, since distributing the load over the bearing areas will result in lower maximum stresses.

  • $\begingroup$ This is only the bending check right? $\endgroup$
    – Ethan48
    Commented Mar 4, 2016 at 19:07
  • 1
    $\begingroup$ @Ethan48 Yes. The OP indicated that analysis of bending in a round shaft was their main problem. $\endgroup$
    – DLS3141
    Commented Mar 4, 2016 at 19:16
  • $\begingroup$ Ok... With approximate values, this would give me 600*0.026/(4*0.78*0.005^3)=40MPa, correct? (26mm beam length, 60 kg load, 10mm diameter shaft) So I'd be more than OK... $\endgroup$
    – jcoppens
    Commented Mar 4, 2016 at 19:36
  • $\begingroup$ I see practical applications like this every day and that arrangement looks absolutely fine to me. Your main concern is fitting the bearings which will be more or less easy depending on the condition and tolerances of the shaft an pulleys. $\endgroup$ Commented Mar 4, 2016 at 21:05
  • $\begingroup$ Thanks Chris. I consider myself a fairly practical guy, and though most of the time at the computer, I've done plenty of mechanical/construction work at home. Once in a while though, a scenario appears where intuition fails... $\endgroup$
    – jcoppens
    Commented Mar 4, 2016 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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