2
$\begingroup$

I want to build myself a pull-up bar. I am confident enough that I can guesstimate how much material I need for it to be strong enough, but this time I would like to challenge myself to properly calculate all the strengths needed.

However I am stuck at the first point -- how do I properly estimate the loading on such a thing? By this I mean only the maximal instant force that my body would apply to the bar, safety factors will be added later.

Is there some standard that tells you what dynamic forces to expect from a human body?

My best guess so far goes something like this: I can hang on single hand, but not with another person pulling me down, so the max total force that I can apply to the bar will be somewhere between 2x and 4x my body weight, anything more and my hands will certainly slip. Therefore I design for 4x my body weight and add safety factors on top of that. Does that make sense?

$\endgroup$
7
  • $\begingroup$ Yes, mass times acceleration. So allow your body mass to accelerate for the length of your arms and then a sudden stop... $\endgroup$
    – Solar Mike
    Feb 10 at 21:16
  • $\begingroup$ Sudden stop is a bit of a problem -- how sudden? And I won't be able to hold if it is too sudden, limiting the force. $\endgroup$
    – cube
    Feb 10 at 22:07
  • $\begingroup$ re: sudden stop - yes I think (de)acceleration at the bottom of the motion would be most intense. Also your hands might be stronger than you think. If you can do 6-8 pull ups, you can probably do one or two holding the bar with one hand and holding the same wrist with your other hand, and doing so won't fully challenge your hand grip if you have been fairly active. Catching yourself from a drop would use the muscles and tendons etc in the body and inside the shoulders (with arms extended, weak point will feel like the elbows probably). all of those typically stronger than pull up muscles... $\endgroup$
    – Pete W
    Feb 10 at 23:11
  • 1
    $\begingroup$ 20 years or so ago when I was taking a suspended platform set up course, we were told that the life line that would support a worker if they fell was to be designed to carry a minimum of 10X their body weight. So the rope, buckles, connectors and attachment point each had to be designed to carry 2500 lbs (assumed weight of worker with attached gear: 200-250 lbs). Meanwhile, the platform support lines were designed only to carry 4X the supported load. $\endgroup$
    – Forward Ed
    Feb 14 at 7:45
  • 1
    $\begingroup$ From what I recall. If you have a motorized winch to lift the platform, the winch was capable of exerting 1000 lbs on the steel line. so you had to make sure the the line could handle 4000 lbs, The outriggers up top had to have enough counter weight to resist 4X the over turning. I have never had to set one up and I have not had to inspect a set up in the past 18 years so I may have a few details mixed up. $\endgroup$
    – Forward Ed
    Feb 14 at 18:25
2
$\begingroup$

Like alephzero already mentioned the pull up part is usually the less stresses part. The problem is with the deceleration forces when you are at the top of the pull up bar and suddenly you let your self down. Although that seldom happen (i.e. there is a constant force for deceleration), it is best to assume the worst case scenario.

I'll make the following assumptions:

  • the fully extended arm is h = 1[m]
  • the weight is 100 [kg] ~ 1000N

Assuming you are at the top and you let go, the velocity can be calculated by the following formula:

$$v= \sqrt{2 g h}\approx 4.47[m/s]$$

That velocity is the free fall velocity of your body (arms should be excluded because they will not be travelling with the same speed as the torso but that's a detail, so I'll just keep it simple).

Now assuming that the duration of deceleration is 0.1[s], from the relationship betwen impulse and momentum you obtain:

$$ m\cdot v + \int F\cdot dt = 0 \rightarrow$$

$$ \int F\cdot dt = -m\cdot v = 447 [Ns]$$

From that point it depends what sort of approximation you make for F.

enter image description here

If you assume that F is: -constant then you can estimate that:

$$F_c= 4472 [N]$$

  • if you assume it has a triangular profile, then

$$F_{tri}= 8944 [N]$$

There are other approximations (sinusoidal), but for the purposes of this don't make much sense. I would go for the triangular.

It is noteworthy that the deceleration force, is added to the force of weight, but you can see that is almost another order of magnitude (just like alephzero proposed).

I would take it a step further and suggest that you should add something extra (possibly double it although that is starting to be an overkill). However, its something you don't want it to bend nor break while you are at it.

things to note.

This is very dependent on the duration of the deceleration. So doubling $dt$ results in smaller forces.

However, given that the duration of free fall is about 0.45[s], I don't think you can go much higher that 0.1[s].

comparison with pull up weight

As you can see, the comparing the deceleration force, compared to the pull up force (which is quasi static), makes it easy to understand that the deceleration scenario is much more punishing to the beam.

$\endgroup$
2
  • $\begingroup$ What I don't like about this answer is the 0.1s, but in comparison to the total free fall time it really does look reasonable. I still think that the grip strength will be a limiting factor here, but this helps me get another POV on the problem $\endgroup$
    – cube
    Feb 11 at 8:54
  • $\begingroup$ even if you doubled $dt$, you would still have approx 5kN for the triangular load. If you added a decent safety factor for a person carrying structure, you would be well over 10x the body weight as alephzero proposed. Also keep in mind, that if you compare that time with the time it takes to do a pull up you'll see that the deceleration will be more punishing. $\endgroup$
    – NMech
    Feb 11 at 8:58
2
$\begingroup$

Yes, That is a good estimate.

Another reasonable way is to measure how fast you can pull yourself up. Say if you can time your lift acceleration and it is $12m.s^2$ you are applying, $mg+1.2mg=2.2mg \ , $ 2.2 times your weight.

Of course, If you have access to one of those benches with lift up weights, the ones that have rubber bars locking you in, you can directly measure your maximum lift.

$\endgroup$
3
  • 1
    $\begingroup$ The pull up calculation is the easy part. The hard part is estimating the "shock load" when you release the tension in your arms but don't let go of the bar so you are still hanging from it. A reasonable safety factor might be 10x your body weight (but that is NOT professional design advice!). The maximum load for that scenario will depend on the flexibility of the bar, not just on your weight. $\endgroup$
    – alephzero
    Feb 10 at 23:01
  • $\begingroup$ @alephzero, when I was a kid, 18, i got into a head-on accident riding my Husquarna dirt bike with a Hillman assembled in Iran. My foot in a cotton shoe totally ripped the side of the car thru door panels and the column. the shoe was torn to shreds, but no bone breakage, no immediate severe injury but pain. later it started to turn blue and black, my skin. Flesh striking correctly can tear the steel. $\endgroup$
    – kamran
    Feb 11 at 2:46
  • $\begingroup$ @alephzero: When using 4x body weight as peak force, and 1.5x safety factor, it comes out to 12x body weight of "average" load capacity for the whole bar (the max force could be all applied to only one of the bar mounts, so another 2x in the calculation) -- that's pretty close to your 10x :) $\endgroup$
    – cube
    Feb 11 at 8:38
0
$\begingroup$

Since you want to challenge yourself with the design process, it would be good to use the force information not just for strength of the bar, but also the deflection. For example when you design a bow for archery you want deflection, and when you design a bridge, you don't. Olympic weightlifting bars are somewhere in between So it is worth considering how flexibility will influence your use of the bar.

$\endgroup$

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.