Thermal Strain in Two Plates Held by Bolt

In the 13th edition of the MERM companion problem books, question number 6 in chapter 51 (page 51-2) asks for the tensile load on a bolt that is holding a steel and aluminum plate when the temperature is raised 250 deg F. In the solution, one of the lines reads "The unrealized elongation of the bolt is:"

I don't understand this: how can the bolt have an "unrealized elongation" when its elongation was less than the total of the elongation of the aluminum and steel plates? It seems to me that instead it is the steel and aluminum plates that have "unrealized elongation".

With that being said, the balance of the solution does not make sense to me. I mean, I get what they're doing but I don't understand how the strain in the bolt can be calculated using this "unrealized elongation".

The closest I can get is that the thermal strain in the bolt must be equal to the sum of the thermal strain in the aluminum and steel plates. That being said, if the aluminum and steel plates have a total "unrealized elongation" of X, then the bolt must have an "equivalent unrealized elongation" of the same amount? i.e. the thermal strain Mohr's law should hold for the bolt even though it was allowed to fully elongate?

I think you might be getting a little hung up on the wording in the explanation. I will try to give a different perspective.

The unconstrained thermal expansion of the members is greater than the unconstrained thermal expansion of the bolt because one of the members is aluminum and has a higher coefficient of thermal expansion than the steel bolt. This causes a force on the bolt since the members get bigger than the bolt and they are clamped together by the bolt. Since the members are clamped the bolt has to stretch more than it's own thermal expansion. I drew out a simple diagram for you of the unconstrained thermal expansion of the bolt and each member (see attached).

The difference between the thermal expansion of the bolt and the thermal expansion of the members is the change in length of the bolt that causes a thermal strain. That thermal strain can be used to calculate the force in the bolt.

If both members had been steel, all the members and the bolt would have the same coefficient of thermal expansion and there would be no thermal strain. If the bolt had been aluminum instead of steel, it would have expanded more than the members and there would also be no thermal strain.

Hopefully I am making sense, but I don't think there is anything wrong numerically. How they got there and explanation may be a little confusing, but in my opinion the end result is correct.

View attachment 9852

J_MEC said:

Since the members are clamped the bolt has to stretch more than it's own thermal expansion.

Click to expand...

This is the part that I am struggling with. I understand that the members have expanded more than the bolt due to thermal expansion. When working this problem, I considered the bolt head and nut to act as 'walls' that were constraining the plates. Since the bolt expands due to thermal strain, it makes sense that the 'gap' between these two 'walls' increases. This allows for the plates to expand, again due to thermal expansion.

Once the plates expand sufficiently that they reach the bounds of the 'walls', though, I did NOT assume that the additional thermal strain from the plates would cause additional, non thermal strain in the bolt. What's more, had I made this assumption, I still would not have assumed that the additional mechanical strain in the bolt would be equal to the non-realized thermal strain in the plates.

How can these assumption be valid? It just seems like such a long jump: at best, I wanted to take the residual thermal strain in the plates and equate that to a force being applied to the bolt head and nut face. This approach would have required assuming some sort of ratio between the forces due to each of the plates, as well as finding the surface area of the bolt head and nut face.