When we do pull-ups, does the bar takes more weight than when we hang down on the bar? When I do pull-ups, I feel I push down to the bar. But does the bar really take more weight than just hang down?
For people who don't know pull-ups and hang down, here is an illustration.
Left: Hang Down-----------------------Right: Pull ups

So, does in right picture the bar take more weight than the left one?
 A: Let's suppose you are pulling yourself up and down in approximately simple harmonic motion so your height above the ground will be give by:
$$ h = h_o + h' sin(\omega t) $$

Your acceleration is just $d^2h/dt^2$, and the force is just your mass times the acceleration, so the force due to your motion will be:
$$ F = - m h' \omega^2 sin(\omega t) $$
and the total force on the bar is:
$$ F = - m \left( g + h' \omega^2 sin(\omega t)\right) $$
So in this model the force is greatest at the bottom of your cycle as you are slowing your decent and accelerating yourself back up. It is lowest at the top where your ascent is slowing and you're allowing gravity to pull you back down.
A: Yes, you do put more weight on the bar.
Your mass here is $m$.
To hang, you simply put force $F=mg$ on the bar (and equivalently, the bar puts that same force on you, so the forces cancel and you don't move anywhere).
For you to move upwards at some acceleration $a$, now you need the net force on you to equal $ma$: $\sum F = ma = F_{bar}-mg$. So, $F_{bar} = m(g+a)$.
This is the force the bar must put on you to accelerate up, so it's the force you put on the bar.
Note that in the case of hanging ($a=0$), it reduces to the first case, $F_{bar} = mg$.
Edit: I'll add a tiny caveat because it can possibly be confusing. You'll notice that in this scenario, there's only more force on the bar if you're accelerating up. If you manage to go up at a totally constant speed, the force should be $mg$ still. I guess the reason this doesn't happen is because it's nearly impossible for a human to do.
A: If you are accelerating upward then the force on the bar will be greater.  F=ma.
A: The act of pulling oneself up (or holding oneself up) will require a change in force applied to your arms, or at least a redistribution of it when holding oneself up. This will change the force applied to a portion of the bar- specifically, where the hands are, as they will almost assuredly grip the bar harder than when hanging limply. The overall weight applied to the bar, as a whole, won't have changed, but by gripping tighter, additional force has been applied.
This, of course, in addition to the increase in force required to actively accelerate upwards.
