Consider Young's Modulus in the case of a sphere, \(E_s\)
\(F_L=E_s\cfrac{A_o}{L_o}\Delta L\)
when \(L_o=1\) and \(A_o=4\pi(1)^2\)
\(F_L=4\pi E_s\Delta L=F_T=\sqrt{3}\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}\)
for the case of such a sphere under high temperature. If we have,
\(E_s=\sqrt{3}E\)
where \(E\) is the linear Young's Modulus and \(\sqrt{3}\) the 3D constant, then
\(4\pi E\Delta L=\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}\)
We have,
\(D.\cfrac{\partial (d_s)}{\partial T}=4\pi E\cfrac{\alpha}{\rho}\Delta L\) --- (*)
where \(\Delta L\) is the change in radius of the sphere, \(\rho\) is the density of the element, \(\alpha\)
is the linear thermal coefficient, and \(E\) the Young Modulus.
If both \(D\) and \(\cfrac{\partial (d_s)}{\partial T}\) are instead independent of the specific element, then the last formula provides a simple way of finding,
\(D.\cfrac{\partial (d_s)}{\partial T}\) in general.
\(\cfrac{\partial (d_s)}{\partial T}\) is a function in \(T\) and \(\Delta L\) changes with \(T\). In which case \(D\) can be found by plotting \(f(T)\) vs \(\Delta L\). Where \(f(T)\) is a function in \(T\) and \(f(T)\) vs \(\Delta L\) is a line through the origin with gradient,
\(gradient=4\pi E\cfrac{\alpha}{\rho D}\)
It should be noted that expression (*) is within the frame work of a solid lattice, an array of atoms in 3D. The limiting case of an infinitely sparse lattice will then approximate the situation of free space.
Have a nice day.
