A problem on thermal stress Here's the question:

There are 2 rods placed between rigid supports. Where $Y_{i}, \alpha_{i}$ and $A_{i}$ are the Young's Modulus, Coefficient of linear expansion and Cross sectional area of the rods. When the system is heated to temperature $\theta_{2} $ from $\theta_1$, find the relation between $A_1$ and $A_2$ such that there lengths remain constant.

Now I know that if the length does not change I could consider that a thermal stress of $Y\alpha\Delta\theta $ is developed in the rods. The problem is the area that I have to consider to equate the force. My teacher tells me that I have to consider this equation:
$$A_{1}Y_{1}\alpha_{1}\Delta\theta=A_{2}Y_{2}\alpha_{2}\Delta\theta$$But I don't understand this equation. According to me, it must be $A_2$ on both the sides instead of $A_1$ and $A_2$. As the stress developed in the rods could only be transferred through the common area (I think...). Could someone explain me what's wrong?  
 A: What is going on there is that there is an equal and opposite force acting on the interface 

This force results in stress of $\sigma_1 = \frac{F}{A_1}$ on one side and $\sigma_2 = \frac{F}{A_2}$ on the other side far away from the interface.
The engineering assumption being that the contact pressure $P=\frac{F}{A_2}$ spreads to the full cross-section $A_1$ over a short distance compared to the overall length of the part. 
The two parts also have the strain due to this force equal the strain due to thermal expansion.
$$ \epsilon_1 = \frac{\sigma_1}{Y_1} = \alpha_1 \Delta \theta $$
$$ \epsilon_2 = \frac{\sigma_2}{Y_2} = \alpha_2 \Delta \theta $$
Combine the stress from above into the strain and solve for the common force $F$
$$ \frac{F}{A_1 Y_1} = \alpha_1 \Delta \theta $$
$$ \frac{F}{A_2 Y_2} = \alpha_2 \Delta \theta $$
$$ \boxed{ F = A_1 Y_1 \alpha_1 \Delta \theta = A_2 Y_2 \alpha_2 \Delta \theta } $$

All this is good in theory, but in real-life it is a bit more complex as you suspect since you are raising valid doubts about the use of the larger cross-section.
Doing an FEA example you see below that the thin section has uniform stress (1). But the larger section only has uniform stress (2) away from the interface. Just next to the interface (3) the stress is $F/(A_2 Y_1)$ and away from the contact (4) it is much lower than predicted.

A: The confusion is in part the difference between making a theoretical force balance and determining in practice force from stress. A theoretical free-body force balance is made at a point. Stress $\sigma$ is converted in practice to force $F$ using area $F = \sigma A$. To convert the latter (area) reference frame to the former (force balance reference frame), you must assume that the full force generated over the area of the (larger diameter) left rod and the full force generated over the area of the (smaller diameter) right rod both collapse to a common point. In practice, this is equivalent to stating that the face of the larger diameter rod is perfectly rigid.
An alternative view is to consider the equivalent of a hand pushing against a wall. The entire hand pushes with a force that is countered by the entire wall. Otherwise, the hand punches a hole through the wall.
A: The reason you take the complete area on $A_{1}$ is because its an approximation: no matter how locally the stress on its surface is being applied, the whole object feels the stress as if it were applied evenly across its cross-section. Otherwise there would be local deformation.
This isn't true in reality though.
