KaBoom

If from the post "Oops! Smooth Operator",

$d_s(x)-d_n=B.h(ln(T))$

when    $T=T_o$,    $d_s=d_s(L)$,  this is to say at a distance    $x=L$   from the maximum,  $T_{max}$,    at the furthest extend of the body,    $T=T_o$.  In the case of a shell, the opposite wall of the body,  in the case of a rod, the length of the rod.

So we consider,

$B=\cfrac{d_s(L)-d_n}{h(ln(T_o))}$

and also,

$B=\cfrac{d_s(0)-d_n}{h(ln(T_{max}))}$

this implies,

$\cfrac{d_s(L)-d_n}{d_s(0)-d_n}\cfrac{h(ln(T_{max}))}{h(ln(T_o))}=1$

$\cfrac{h(ln(T_{o}))}{h(ln(T_{max}))}=\cfrac{d_s(L)-d_n}{d_s(0)-d_n}$

and also,

$\cfrac{\partial (d_s)}{\partial T}=\cfrac{d_s(L)-d_n}{ln(T_o)}.h^{'}(ln(T))\cfrac{1}{T}$

From the post "Young And On Heat"

$F_T=\sqrt{3}\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}$

For the case of a sphere of radius    $L$,

$F_T=\sqrt{3}\rho.\cfrac{4}{3}\pi.L^3 D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{L\alpha}$

$F_T=\cfrac{4\sqrt{3}}{3}\pi\rho L^2 D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}$

$F_T=\cfrac{4\sqrt{3}}{3}\pi\rho L^2 D.\cfrac{d_s(L)-d_n}{ln(T_o)}.h^{'}(ln(T)).\cfrac{1}{T}.\cfrac{1}{\alpha}$

With the help of superposition,

at    $x=L_i$,    $T=T_{max}$,

$F_{T_{max}}=\cfrac{4\sqrt{3}}{3}\pi\rho L^2_i D.\cfrac{d_s(L)-d_n}{ln(T_o)}.h^{'}(ln(T_{max})).\cfrac{1}{T_{max}}.\cfrac{1}{\alpha}$

At    $x=L_e$,    $T=T_o$,

$F_{T_{o}}=\cfrac{4\sqrt{3}}{3}\pi\rho L^2_e D.\cfrac{d_s(L)-d_n}{ln(T_o)}.h^{'}(ln(T_{o})).\cfrac{1}{T_{o}}.\cfrac{1}{\alpha}$

$\cfrac{F_{T_{max}}}{F_{T_{o}}}=\cfrac{L^2_i}{L^2_e}\cfrac{h^{'}(ln(T_{max}))}{h^{'}(ln(T_o))}\cfrac{T_{o}}{T_{max}}$

From which we formulate,

$F_{T_{max}}-F_{T_{o}}={F_{T_{max}}}\left\{1-\cfrac{L^2_e}{L^2_i}\cfrac{h^{'}(ln(T_o))}{h^{'}(ln(T_{max}))}\cfrac{T_{max}}{T_{o}}\right\}$

Since,    $h(ln(T))$    is monotonously decreasing and gradient at     $T_{max}$    is greater than at   $T_{o}$,  because  when    $T\rightarrow0$,    $g_{T}=0=D.\cfrac{\partial d_s}{\partial T}.\cfrac{dT}{dx}$

$0<\cfrac{h^{'}(ln(T_o))}{h^{'}(ln(T_{max}))}\cfrac{T_{max}}{T_{o}}<1$

But    ${L^2_i}<{L^2_e}$,    if

$\cfrac{L^2_e}{L^2_i}\cfrac{h^{'}(ln(T_o))}{h^{'}(ln(T_{max}))}\cfrac{T_{max}}{T_{o}}<1$

then

$F_{T_{max}}>F_{T_{o}}$

and

$0<F_{T_{max}}-F_{T_{o}}={F_{T_{max}}}\left\{1-\cfrac{L^2_e}{L^2_i}\cfrac{h^{'}(ln(T_o))}{h^{'}(ln(T_{max}))}\cfrac{T_{max}}{T_{o}}\right\}<F_{T_{max}}$

This is the stress experienced by the shell of thickness    $L_e-L_i$.    To achieve containment, the shell must have a yield point higher than     $F_{T_{max}}$.    The shell should disintegrate  when    $T_{max}$  is at the melting point of the shell.   At that moment, the shell melts instantaneously and all energy is released in a spherical wave front.   If the shell were to yield, since  the stress inside the shell is greater, the interior surface will begin to crack first.

If however,

$\cfrac{L^2_e}{L^2_i}\cfrac{h^{'}(ln(T_o))}{h^{'}(ln(T_{max}))}\cfrac{T_{max}}{T_{o}}>1$

$F_{T_{max}}<F_{T_{o}}$

In this case if the shell were to yield it will crack on the exterior surface first.    To achieve containment, the shell must have a yield point higher than     $F_{T_{o}}$.  So it seems that the containment shell cannot be too thick.

When

$\cfrac{L^2_e}{L^2_i}\cfrac{h^{'}(ln(T_o))}{h^{'}(ln(T_{max}))}\cfrac{T_{max}}{T_{o}}=1$

then

$F_{T_{max}}=F_{T_{o}}$

Stress on the internal and external surface of the shell are then equal.  It is likely that the shell will deform   ($L_i$   and   $L_e$   changes) until   $F_{T_{max}}=F_{T_{o}}$  given an initial geometry subjected to   $T_{max}$ inside, unless the yield point of the material is reached first, and the containment shell breaks.

Note that the expansion along the thickness of the shell is not uniform as temperature decreases from    $T_{max}$    to    $T_{o}$.

Together with the post "Have A Heart, While It's Hot", have a big kaBoom.

Note:  It is possible to make this discussion without the assumption of   $h(ln(T))$  by simply using the term  $\cfrac{\partial (d_s)}{\partial T}|_{T}$  to denote the value of  $\cfrac{\partial (d_s)}{\partial T}$  at the temperature  $T$.  And we would have,

$F_{T_{max}}-F_{T_{o}}={F_{T_{max}}}\left\{1-\cfrac{L^2_e}{L^2_i}.{\cfrac{\partial (d_s)}{\partial T}|_{T_o}\over\cfrac{\partial (d_s)}{\partial T}|_{T_{max}}}\right\}$.