Let us consider a ceramic material, $\mathrm{Mg_2SiO_4}$ - Forsterite, with a sphere shape.

At room temperature, the ceramic sphere is "trapped" in a matrix which does not expand (null thermal expansion coefficient).

If I raise the temperature of the system (matrix + ceramic), the sphere will not be able to expand and internal stress will grow due to the resulting compression.

How would you estimate the maximum temperature, as well as the maximum internal stress associated before the ceramic sphere fails ?

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    $\begingroup$ Let us see some working so far... $\endgroup$
    – Solar Mike
    Commented Jul 26, 2018 at 4:59

1 Answer 1


I assume here, the linear coefficient of thermal expansion of the matrix is lower than that of the ceramic so we land in compression, otherwise the sphere tries to expand in all directions.

The fracture criterium of isotrope materials and anisotrope materials are not the same but in general, fractures occurs:

if the greatest principal stress is equal to the fracture stress as measured in axial tensile test. This is true for isotrope materials.

So what can we do, theoretically, first we should construct the stress tensor and then find the eigen values of this matrix, the greatest eigen values indicates the greatest principal stress. Again all this is true if the material is isotope. For anisotrope materials you need a comprehensive knowledge of fracture mechanics, and it ain't as easy as decibed above.


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