Since ${\mathrm{L}}_1={\mathrm{L}}_2$ for the coaxial bars, this equation simplifies to:

${\mathrm{F}}_1={\mathrm{F}}_2\left(\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}\right)\left(\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}\right)$

Recall that aluminum tube $\left(1\right)$ has an outer diameter of $\mathrm{D}=30 \mathrm{mm}$ and an inner diameter of $\mathrm{d}=22 \mathrm{mm}$. The elastic modulus of the aluminum is $70 \mathrm{GPa}$. The brass core $\left(2\right)$ has a diameter of $\mathrm{D}=22 \mathrm{mm}$ and an elastic modulus of $105 \mathrm{GPa}$.

Compute the areas ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$.

$$\begin{gathered} {\mathrm{A}}_1=\frac{\mathrm{\pi }}{4} \left[{\left(30 \mathrm{mm}\right)}^2-{(22 \mathrm{mm})}^2\right] \\ =326.726 {\mathrm{mm}}^2 \\ {\mathrm{A}}_2=\frac{\mathrm{\pi }}{4}{(22 \mathrm{mm})}^2 \\ =380.133 {\mathrm{mm}}^2 \end{gathered}$$

$$\begin{gathered} {\mathrm{F}}_1={\mathrm{F}}_2 \left(\frac{326.726 {\mathrm{mm}}^2}{380.133 {\mathrm{mm}}^2}\right)\left(\frac{70 \mathrm{GPa}}{105 \mathrm{GPa}}\right) \\ {\mathbf{F}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{573}\mathbf{ }{\mathbf{F}}_{\mathbf{2}} \end{gathered}$$