At temperature T_o, two metal strips of length ${\mathrm{l}}_{\mathrm{o}}$ and thickness d is bolted so that their ends coincide. The upper strip is made up of metal A and have coefficient of expansion ${\mathrm{\alpha }}_{\mathrm{A}}$and lower strip is made up of metal B with coefficient of expansion ${\mathrm{\alpha }}_{\mathrm{B}}\left({\mathrm{\alpha }}_{\mathrm{A}}>{\mathrm{\alpha }}_{\mathrm{B}}\right)$. When temperature of their blastic strip is increased from T_o to $\left({\mathrm{T}}_0+\mathrm{\Delta T}\right)$, one stop becomes longer than the other and the blastic strip is bend in the form of a circle as shown in .
 

The radius of curvature R of the strip is

At temperature $\left({\mathrm{T}}_0+\mathrm{\Delta T}\right)$, the increase in length
                          ${\mathrm{L}}_{\mathrm{A}}={\mathrm{l}}_0\left(1+{\mathrm{\alpha }}_{\mathrm{A}}\mathrm{\Delta T}\right) ....\left(1\right)$
and                    ${\mathrm{L}}_{\mathrm{B}}={\mathrm{l}}_0\left(1+{\mathrm{\alpha }}_{\mathrm{B}}\mathrm{\Delta T}\right) ....\left(2\right)$
Let R_A and R_B the radius of the strips, then
                         ${\mathrm{L}}_{\mathrm{A}}={\mathrm{\theta R}}_{\mathrm{A}} ....\left(3\right)$
  and                        ${\mathrm{L}}_{\mathrm{B}}={\mathrm{\theta R}}_{\mathrm{B}} .....\left(4\right)$
where $\mathrm{\theta }$ = common angle as shown in the fig. (2).
From eqs. (1) and (2), we get
      ${\mathrm{L}}_{\mathrm{A}}-{\mathrm{L}}_{\mathrm{B}}={\mathrm{l}}_0\mathrm{\Delta T}\left({\mathrm{\alpha }}_{\mathrm{A}}-{\mathrm{\alpha }}_{\mathrm{B}}\right) ....\left(5\right)$
From eqs. (3) and (4), we get
   ${\mathrm{L}}_{\mathrm{A}}-{\mathrm{L}}_{\mathrm{B}}=\mathrm{\theta }\left({\mathrm{R}}_{\mathrm{A}}-{\mathrm{R}}_{\mathrm{B}}\right)=\mathrm{\theta d} ....\left(6\right)$
From eqs. (5) and (6), we get   
                 $\mathrm{\theta }=\frac{{\mathrm{l}}_0\mathrm{\Delta T}\left({\mathrm{\alpha }}_{\mathrm{A}}-{\mathrm{\alpha }}_{\mathrm{B}}\right)}{\mathrm{d}} .....\left(7\right)$
From eqs. (1), (2), (3) and (4), we get
 ${\mathrm{R}}_{\mathrm{A}}+{\mathrm{R}}_{\mathrm{B}}=\frac{2{\mathrm{l}}_0+{\mathrm{l}}_0\left({\mathrm{\alpha }}_{\mathrm{A}}+{\mathrm{\alpha }}_{\mathrm{B}}\right)\mathrm{\Delta T}}{\mathrm{\theta }} ....\left(8\right)$
$\therefore$Mean radius R is given by 
$\mathrm{R}=\frac{{\mathrm{R}}_{\mathrm{A}}+{\mathrm{R}}_{\mathrm{B}}}{2}=\frac{2{\mathrm{l}}_0+{\mathrm{l}}_0\left({\mathrm{\alpha }}_{\mathrm{A}}+{\mathrm{\alpha }}_{\mathrm{B}}\right)\mathrm{\Delta T}}{2\mathrm{\theta }} ....\left(9\right)$
Substituting the value of $\mathrm{\theta }$ from eq. (7) in eq. (9), we get
             $\mathrm{R}=\frac{2{\mathrm{l}}_0+{\mathrm{l}}_0\left({\mathrm{\alpha }}_{\mathrm{A}}+{\mathrm{\alpha }}_{\mathrm{B}}\right)\mathrm{\Delta T}}{2\left[{\mathrm{l}}_0\mathrm{\Delta T}\left({\mathrm{\alpha }}_{\mathrm{A}}-{\mathrm{\alpha }}_{\mathrm{B}}\right)/\mathrm{d}\right]}$
or,        $\mathrm{R}=\frac{\left[2+\left({\mathrm{\alpha }}_{\mathrm{A}}+{\mathrm{\alpha }}_{\mathrm{B}}\right)\mathrm{\Delta T}\right]\mathrm{d}}{2\left({\mathrm{\alpha }}_{\mathrm{A}}-{\mathrm{\alpha }}_{\mathrm{B}}\right)\mathrm{\Delta T}}$