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Posted: 1 decade ago Feb 15, 2013, 11:42 a.m. EST

Hi Ivan, Your analysis is correct. The factor (1+ alpha_solid(Tref)(Tref-293)) is however very small under all normal circumstances, taking into account possible actual values of temperatures and coefficients of thermal expansion. So this correction is currently ignored. Even with an extremely high reference temperature, it will hardly exceed 1.01 and for more common cases it would be much closer to unity. Ignoring this factor essentially means that the initial "reference" length of the object is taken to be the same at the reference temperature as at 293 K. Still, since the expression is already complicated, there is no need to ignore the correction term, and we will include it in the next version of COMSOL. For those interested in how these expressions are derived: Let [math]T_m[/math] be the reference temperature for the measured data (here 293 K). By definition [math]L(T)=(1+\alpha_m(T) (T-T_m)) L(T_m) \quad(1)[/math] When we have a strain free temperature [math]T_{ref}[/math] ('reference temperature') , which may not be the same temperature as the one on which the material data is based, then the same type of relation is expected to hold [math]L(T)=(1+\alpha_r(T) (T-T_{ref}) )L(T_{ref})[/math] Now [math]\alpha_r[/math] is the 'strain free based' coefficient of thermal expansion. [math]\alpha_r(T) = \frac{L(T)-L(T_{ref})}{(T-T_{ref}) L(T_{ref})} \quad(2) [/math] Using (1) with [math]T = T_{ref}[/math]: [math]L(T_{ref})=(1+\alpha_m(T_{ref}) (T_{ref}-T_m)) L(T_m) \quad(3)[/math] Inserting (1) and (3) into (2) we will arrive at the expression you presented: [math]\alpha_r(T) =\Big( \alpha_m(T)+ \frac{(T_{ref}-T_m)(\alpha_m(T)-\alpha_m(T_{ref}))}{T-T_{ref}} \Big)\Big(1+\alpha_m(T_{ref})(T_{ref}-T_m)\Big)^{-1} [/math] Regards, Henrik