Q:

Finding Neutral Axis of a simple shape undergoing axial load

I know that the neutral axis occurs where the sum of stresses (bending + axial) is equal to 0. I have found the stresses and drawn a diagram to represent forces. How do I determine the neutral axis?

A:

In order to calculate the neutral axis under this definition (see my comment under the OP), you need to figure out the stress distribution for each of the applied loads and then superpose them.
So, for this we need to calculate the properties:
$$\begin{align}
A &= 80\cdot80 - 60\cdot60 = 2800 \\
I &= \dfrac{80\cdot80^3 - 60\cdot60^3}{12} = 2333333
\end{align}$$
So, under the axial force, the stress distribution is uniform and equal to (positive for tension)
$$\sigma = \dfrac{N}{A} = \dfrac{72000}{2800} = +25.71$$
Now, under the moment, the stress distribution is linear, and the maximum stresses are equal to
$$\sigma = \dfrac{My}{I} = \dfrac{(72000\cdot(60+40))\cdot(\pm40)}{2333333} = \pm123.43$$
where the stresses on the left side are negative (compression) and on the right side are positive (tension). The stress profile due to bending can therefore be described by the equation
$$\sigma = -123.43 + \dfrac{2\cdot123.43}{80}y$$
where $y=0$ is on the left side.
Now, adding the effect of the axial load's stress, this equation becomes
$$\sigma = -123.43 +25.71 + \dfrac{2\cdot123.43}{80}y$$
So, to find the neutral axis, you just need to find the zero of this equation:
$$\begin{gather}
\sigma = 0 = -97.72 + \dfrac{2\cdot123.43}{80}y \\
y = \dfrac{97.72}{\left(\dfrac{2\cdot123.43}{80}\right)} = 31.7
\end{gather}$$
So there you have it, the neutral line is 31.7 mm from the left side.

