MecMovies 4.0 : M5.6: Coaxial Tube and Core

scene 11 of 13

Having determined the internal forces, the normal stresses in aluminum tube $(1)$ and brass core $(2)$ can now be computed:

$σ_{1} = \frac{F_{1}}{A_{1}}$

$= \frac{(- 10.93 kN) (1000 N / kN)}{326.726 {mm}^{2}} = - 33.5 MPa$

$σ_{2} = \frac{F_{2}}{A_{2}}$

$= \frac{(- 19.07 kN) (1000 N / kN)}{380.133 {mm}^{2}} = - 50.2 MPa$