Determining the neutral axis of an I-shaped cross section with dissimilar materials Determine the neutral axis of the  I- shaped cross section with dissimilar materials.
The top rectangle of the cross section is made out of aluminum and the second and third part are titanium ( these are respectively the rectangle that connects the upper and lower part and the lower rectangle itself ).
I do not think the dimensions are needed since I only want to find out in what way this has to be solved.
How do I regard the difference in materials in this case? I have never done this before and could not find anything useful on the site nor the internet.
Kenny
 A: The shape has to share strain across the cross section. If the centroid is at $y=0$ then assume linear strain in the section of the form $\epsilon(y) = C_0 + C_1 y$ which would yield a neutral axis at $y_0 = -\frac{C_0}{C_1}$. Make material [1] the Ti and material [2] the Al with top and bottom bar width $w$, overall height $h$ and thickness $t$.
The stresses on the two materials are $\sigma_1(y) = E_1 \epsilon(y)$ and $\sigma_2(y) = E_2 \epsilon(y)$. Now we integrate the stress over the area to get internal force and moment. 
$$ F = \int_{-\frac{h}{2}}^{-\frac{h}{2}+t} w \sigma_1 \,{\rm d}y + \int_{-\frac{h}{2}+t}^{\frac{h}{2}-t} t \sigma_1 \,{\rm d}y + \int_{\frac{h}{2}-t}^{\frac{h}{2}} w \sigma_2 \,{\rm d}y $$
 $$ M = \int_{-\frac{h}{2}}^{-\frac{h}{2}+t} w\,y \sigma_1 \,{\rm d}y + \int_{-\frac{h}{2}+t}^{\frac{h}{2}-t} t\,y \sigma_1 \,{\rm d}y + \int_{\frac{h}{2}-t}^{\frac{h}{2}} w\,y \sigma_2 \,{\rm d}y $$
For pure bending we have $F=0$ and $M\ne 0$ yielding some rather lengthy expressions for $C_0$ and $C_1$. But we only need the ratio
$$ y_0 = -\frac{C_0}{C_1} = \frac{ \frac{h-t}{2} ( E_1 - E_2 ) }{ E_1 \left( \frac{h}{w} \left( \frac{2 t}{h}-1 \right)-1\right)-E_2} $$
I would like to confirm this value with an FEA before using it anywhere.
NOTE:I used CAS software to do the integrals and simplification.
ADDENDUM: You can set the applied moment, to be an eccentric axial load with $M=z\,F$ and then find for what values of $z$ the neutral axis goes to infinity (pure axial elongation). It turns out that $z=y_0$, confirming that a load through the neutral axis yields pure axial deformation (similar to a load through the a center of mass yields a pure translation).
A: For a section with different materials, you need to transform the section into a section of one material. The way to do it is to first get the Young's modulus $E$ of the different materials. For your problem, let's say that the modulus of aluminum is $E_a$, and the modulus of titanium is $E_t$. You may then proceed in either of the following ways. If the section is transformed into aluminum, then the widths of the second and third rectangles should be multiplied by a factor of $E_t/E_a$. If, otherwise, the section is transformed into titanium, then the width of the first rectangle should be multiplied by a factor of $E_a/E_t$. The neutral axis is at the geometric centroid of the transformed section.
