Does the normal reaction on pull up bar change during the pull ups? Intuitively, I know the answer but I can't think of the right math.
I found this question but none of the answers were satisfying enough for me. Human body is not a rigid body so do how do we even apply $\Sigma F=ma_{net}$ over it?
 A: 
Does normal reaction on pull up bar changes during the pull ups?

The normal reaction of the bar changes while the body moves upwards (or downwards) because body does not move at a constant acceleration. You are right to say that human body is a complex system which cannot be modelled as a simple particle, but Newton's laws of motion still apply. For the body to accelerate there must be a net force which will provide the acceleration. In your example this comes from the bar
$$F_\text{bar} = m(a + g)$$
where $m$ is total mass of the body, and $a$ is its vertical acceleration.
Another interesting example similar to this would be doing squats on a scale. As body accelerates downwards the scale shows lower weight, and as the body slows down to rest the scale shows larger weight. Once the body is at rest, the scale shows the normal weight.

Although human body is a complex system that has many particles, it can be considered as a particle with all the mass concentrated at the center of mass.

When a collection of particles is acted on by external forces, the center of mass moves as though all the mass were concentrated at that point and acted on by the net external force.

Center of mass of a collection of particles can be calculated as
$$\vec{r}_\text{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2 + \dots}{m_1 + m_2 + \dots}$$
The sum in denominator is total mass of the object $M$, and the above equation becomes
$$M \vec{r}_\text{cm} = m_1 \vec{r}_1 + m_2 \vec{r}_2 + \dots$$
The second time-derivative of the above equation gives
$$M \vec{a}_\text{cm} = m_1 \vec{a}_1 + m_2 \vec{a}_2 + \dots$$
The forces acting on a complex object can be divided to (i) internal forces between the particles and (ii) external forces
$$\sum \vec{F}_\text{ext} + \sum \vec{F}_\text{int} = M \vec{a}_\text{cm}$$
By Newton's third law of motion, internal forces between the particles cancel (you cannot lift yourself by pulling your own belt) and we are left with
$$\boxed{\sum \vec{F}_\text{ext} = M \vec{a}_\text{cm}}$$
In your example external forces would be (i) gravitational force, and (ii) normal force by the bar.
A: 
Human body is not a rigid body so do how do we even apply $\Sigma
> F=ma_{net}$ over it?

It doesn't matter that the human body is not a rigid body. You only you need to apply Newton's second law to the acceleration of the center of mass (COM) of the human body. To simplify the problem, consider only vertical acceleration of the COM of body.
See the figures below showing free body diagrams of the body and the pull up bar, assuming symmetrical loading, while hanging and starting to accelerate upwards during the pull up.
While hanging, the total reaction force of the bar is simply the weight of the man, $Mg$ per Newton's 3rd law.
In order for the COM of the body to accelerate upward, there needs to be an upward force on the COM greater than $Mg$. To do this the man pulls down on the bar with a force $F>Mg$. Per Newton's third law the bar pulls up on the man with an equal and opposite reaction force of $F>Mg$ so that there is a net upward force on the COM of $F-Mg$ and an upward acceleration of $a=(F-Mg)/M$ on the COM.
Bottom line: The reaction force on the bar increases during the pull up.
Hope this helps.

