Temperature of Material Under Compression

[jp4294]

Homework Statement

This is one of the questions in my materials engineering assignment.
1. Thermal shock is the focus of this problem. Note that the resistance to thermal shock of a material, “R”, is defined here as the temperature change required causing fracture: (σc*λ*(1-v))/(α*E)
Where:
σc is the compressive strength
v is Poison’s ratio
α is the thermal expansion coefficient
E is the Young’s modulus
λ is the thermal conductivity
The Poison’s ratio term is necessary because thermal shock often results in a biaxial stress.
a) If the surface of a material is under compression, is it hot or cold relative to the body of the same piece of material? Explain why?

Homework Equations

The Attempt at a Solution

I have a few ways of thinking of this problem.
A) If a material's surface is to be under compression, it must be colder on the inside and as the warmer inside tries to expand, the outside will go under compression in an attempt to hold itself together.
B) The outside will be hotter than the inside and as the outside begins to cool due to the colder inside and the diffusion of heat, it will begin to contract and therefore come under compression.

I'm not sure which one is right if any at all and the question seems a bit vague to me...?

Any help would be great anyway.
Cheers,
JP

 

[LawrenceC]

Think about an I beam heated by a torch on on one flat. It will expand causing the beam to curve. So if you forcibly straighten it, the side that grew must go under compression for it to be straight once again.

Another way of thinking about it is if a blow torch is placed on a thick piece of material, the outer fibers get hot immediately but the material that is deeper in has not sensed the temperature change yet. The surface material wants to grow but the majority of it has yet to experience the higher temperature so it wants to remain in its original size and resists. This puts the material that has become hotter in compression because it has not been permitted to expand.