What isdynamicloading in structures

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.

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

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.

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.

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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?

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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.

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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.

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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.

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).

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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.

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.