Internal stresses of accelerating body 
Refer to pics above.
I have an accelerating body that has a fin extruded from the side.
I want to analyze the stresses at the base of the fin that result from high accelerations.
So, the body is accelerating at $a$. I begin by taking the sum of the moments at the cut (base of fin) and get the following:
$$-M-F_{air}\frac{w}{2}=m_{fin}a\frac{w}{2}$$
where $M$ is the moment, $F_{air}$ is the total air drag force distributed over the entire face.
I want to solve for the moment, $M$ as a function of d so that I can find the bending stresses and determine the thickness $d$ that prevent material failure.
However, when I solve this equation and analyze the bending stress with:
$$\sigma = \frac{Md}{2I_{zz}}$$
I get strange answers.. I must be doing something wrong. 
It would be helpful if someone can walk me through this problem or suggest another way to analyze stresses as a result of shock / high accelerations
 A: What do you mean by "strange answers"? Please be more specific.
If there is constant acceleration in empty space, and the force is applied to the main body, then there is a constant inertial force on the fin, causing a torque.
If the object is moving in air, drag force is proportional to $v^2$. If the body is accelerating then $v$ will be constantly increasing, so the drag force will also be constantly increasing. 
I think you need to be looking at the constant velocity case when there is  a constant drag force. This could be when the body reaches terminal velocity - ie the propulsive force equals the drag force.
A: The first thing to realize is that not all fin parts accelerate with $a$. Only the base and the remaining shape are subject to the system's dynamics.
Look at a small section of the fin with mass ${\rm d}m = \tfrac{m}{\ell} {\rm d}x$. It's a kind of bad drawing, but I have the important variables defined below.

The shape of the fin at each time and position is defined by $y(x,t)$ as a displacement from some datum at $t=0$, and the base of the fin seems constant acceleration of magnitude $a$.
This means that at $x=0$ at all times you have the boundary condition $$ \ddot{y}(0,t) = a \tag{1}$$
If you want to consider the air resistance as uniform along the fin then
$${\rm d}F_{\rm air} = F_{\rm air} \tfrac{1}{\ell} {\rm d}x \tag{2}$$
Now the balance of forces on this small segment are as follows
$$ \begin{gathered}M(x,t)=EI\frac{\partial^{2}}{\partial x^{2}}y(x,t)\\
\left(\frac{\partial^{2}}{\partial x^{2}}M(x,t)+\frac{m}{\ell}\frac{\partial^{2}}{\partial t^{2}}y(x,t)-\frac{1}{\ell}F_{{\rm air}}\right){\rm d}x=0
\end{gathered}
 \tag{3}$$
subject to the boundary condition in (1).
We can form an inhomogeneous 2nd-order differential equation if we combine the air resistance and acceleration to as total weight $W=m\,a+F_{{\rm air}}$ and divide up for each slice for the new force balance
$$\left(EI\frac{\partial^{4}}{\partial x^{4}}y(x,t)+\frac{m}{\ell}\frac{\partial^{2}}{\partial t^{2}}y(x,t)-\frac{1}{\ell}W\right){\rm d}x=0 \tag{4}$$
but with the better boundary condition(s) $y(0,t) = 0$ and $ \frac{\partial}{\partial x} y(0,t) = 0$
My best approximation to the solution of the above inhomogeneous DE is
$$ \small y=\frac{W\ell^{3}}{3\,EI}\left(\frac{x^{4}}{8\ell^{4}}-\frac{x^{3}}{2\ell^{3}}+\frac{3x^{2}}{4\ell^{2}}-\frac{1}{\Phi^{2}}\sin\left(\omega\,t\right)\left(\left(\cos\left(\Phi\tfrac{x}{\ell}\right)-\cosh\left(\Phi\tfrac{x}{\ell}\right)\right)-\lambda\left(\sin\left(\Phi\tfrac{x}{\ell}\right)-\sinh\left(\Phi\tfrac{x}{\ell}\right)\right)\right)\right) \tag{5}$$
where for the first harmonic we have values of $$\begin{aligned}
\omega&=\sqrt{\frac{EI\,\Phi^{4}}{m\ell^{3}}}\\\lambda&=\frac{\sinh\Phi-\sin\Phi}{\cosh\Phi+\cos\Phi}\\\Phi&=1.875104068711961
\end{aligned}$$
For a beam in the definition of internal moment is $M = EI \frac{\partial^2}{\partial x^2}y(x)$ and thus the support moment at $x=0$ is
$$ \boxed{ M_0=\left.EI\frac{\partial^{2}}{\partial x^{2}}y\right|_{x=0}=\frac{7}{6}W\ell } \tag{7} $$
The above value is higher than the static case of $M_0 = \frac{W \ell}{2}$ by a bit more than 2×. This is the dynamic factor as the fin swings due to the acceleration.


I think air resistance should be a function of the velocity at each point.  If you consider only small Reynolds numbers then something like the following is valid
$$ {\rm d}F_{\rm air} = \beta\, \dot{y}(x,t) \tfrac{1}{\ell} {\rm d}x  \tag{2b}$$

A: Dear Sir:  I am not a physicist so my comment may be misguided however the problem as I see it is that your problem does not account for the mass of the fin compared to the mass of the block.
Let's say the acceleration on the block is a constant $g$ or $10\ \mathrm{m/s^2}$.  The block is being pushed to the back of the block is going to want to crunch forward into the front of the block.  If the block were a hollow cube made of the plastic of a soda bottle for instance filled with air the constant movement forward would compress the air and the block from front to back would be shorter.
The fin however is attached to the front of the block.  It is not being pressed forward at all since the acceleration or engine if you will is at the back of the block.  So as the front of the block goes forward the fin is pulled forward via its rigid attachment to the front.  Thus there will be a stress on the line where the fin attaches to the block as the fin tends to try to keep up (not a good metaphor).  So what is the mass of the fin.  This is important because Force equals mass time acceleration $F=ma$.  In our example we set acceleration to $g$ or approximately $10\ \mathrm{m/s^2}$ so if we know the mass we have the force.  Then you can determine the length of the fin attached to the block as well as the thickness.  You then have to calculate the stress over the length since a larger fin up and down with the same mass would spread that force over a larger area.  
To me this is what I think you are trying to solve.  The physicists would have to develop the formula.  Just my thoughts.
