The outer diameter of the hollow cylindrical roller is $2b = 4000 \mathrm{mm}$, and the diameter of the inner circular hole is $2a = 80 \mathrm{mm}$. The coefficient of linear expansion is $\alpha = 12 \times 10^{-6} / \mathbf{°C}$, the modulus of elasticity is $E = 2.16 \times 10^{5} \mathrm{N/mm}^{2}$, and Poisson's ratio is $\nu = 0.3$. The temperature on the outer surface of the roller is $T_{b}$, while the temperature on the surface of the inner hole is $T_{a}$. Try to determine the transient thermal stress of the hollow cylindrical roller is ().

Options:
A. $$
\begin{array} {l} {{{\sigma_{θ}=1. 2 5 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{2 2 0} {r}+0. 1 1 0 \left( \frac{2 2 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 2 5 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{2 2 0} {r}-0. 9 7 2 \right)}}} \\ {{{\sigma_{r}=1. 2 5 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{2 2 0} {r}-0. 1 1 0 \left( \frac{2 2 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
B. $$
\begin{array} {l} {{{\sigma_{θ}=1. 1 8 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{3 5 0} {r}+0. 0 9 5 \left( \frac{3 5 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 1 8 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{3 5 0} {r}-0. 8 8 0 \right)}}} \\ {{{\sigma_{r}=1. 1 8 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{3 5 0} {r}-0. 0 9 5 \left( \frac{3 5 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
C. $$
\begin{array} {l} {{{\sigma_{θ}=1. 1 0 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{2 5 0} {r}+0. 1 0 0 \left( \frac{2 5 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 1 0 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{2 5 0} {r}-0. 9 0 0 \right)}}} \\ {{{\sigma_{r}=1. 1 0 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{2 5 0} {r}-0. 1 0 0 \left( \frac{2 5 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
D. $$
\begin{array} {l} {{{\sigma_{θ}=1. 2 7 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{2 6 0} {r}+0. 1 2 5 \left( \frac{2 6 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 2 7 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{2 6 0} {r}-0. 7 0 5 \right)}}} \\ {{{\sigma_{r}=1. 2 7 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{2 6 0} {r}-0. 1 2 5 \left( \frac{2 6 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
E. $$
\begin{array} {l} {{{\sigma_{θ}=1. 0 5 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{3 0 0} {r}+0. 0 5 5 \left( \frac{3 0 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 0 5 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{3 0 0} {r}-0. 9 5 0 \right)}}} \\ {{{\sigma_{r}=1. 0 5 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{3 0 0} {r}-0. 0 5 5 \left( \frac{3 0 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
F. $$
\begin{array} {l} {{{\sigma_{θ}=1. 2 2 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{5 0 0} {r}+0. 0 7 0 \left( \frac{5 0 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 2 2 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{5 0 0} {r}-0. 8 0 0 \right)}}} \\ {{{\sigma_{r}=1. 2 2 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{5 0 0} {r}-0. 0 7 0 \left( \frac{5 0 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
G. $$
\begin{array} {l} {{{\sigma_{θ}=1. 1 5 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{2 0 0} {r}+0. 0 6 7 \left( \frac{2 0 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 1 5 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{2 0 0} {r}-0. 8 6 6 \right)}}} \\ {{{\sigma_{r}=1. 1 5 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{2 0 0} {r}-0. 0 6 7 \left( \frac{2 0 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
H. $$
\begin{array} {l} {{{\sigma_{θ}=1. 3 0 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{4 0 0} {r}+0. 0 9 0 \left( \frac{4 0 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 3 0 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{4 0 0} {r}-0. 6 5 0 \right)}}} \\ {{{\sigma_{r}=1. 3 0 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{4 0 0} {r}-0. 0 9 0 \left( \frac{4 0 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
I. $$
\begin{array} {l} {{{\sigma_{θ}=1. 2 0 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{1 0 0} {r}+0. 0 8 0 \left( \frac{1 0 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 2 0 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{1 0 0} {r}-0. 7 5 0 \right)}}} \\ {{{\sigma_{r}=1. 2 0 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{1 0 0} {r}-0. 0 8 0 \left( \frac{1 0 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
J. $$
\begin{array} {l} {{{\sigma_{θ}=1. 0 8 \, ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{1 5 0} {r}+0. 0 7 5 \left( \frac{1 5 0^{2}} {r^{2}}+1 \right)-1 \right]}}} \\ {{{\sigma_{s}=1. 0 8 \, ( T_{b}-T_{a} ) \left( 2 \operatorname{l n} \frac{1 5 0} {r}-0. 7 8 5 \right)}}} \\ {{{\sigma_{r}=1. 0 8 ( T_{b}-T_{a} ) \left[ \operatorname{l n} \frac{1 5 0} {r}-0. 0 7 5 \left( \frac{1 5 0^{2}} {r^{2}}-1 \right) \right]}}} \\ \end{array}
$$
Answer: G