Thin-walled pressure vessel, hoop stress Question

[kidsasd987]

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfmaccording to the text above, we assume that force acting on the thin wall is p*2pi*r^2
whcih is the internal area of vertical slice.

But I am not sure why the area has to be 2*pi*r^2. I mean, why pressure on the crossectional area all applied to the thin wall?

 

[benny_91]

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfmaccording to the text above, we assume that force acting on the thin wall is p*2pi*r^2
whcih is the internal area of vertical slice.

But I am not sure why the area has to be 2*pi*r^2. I mean, why pressure on the crossectional area all applied to the thin wall?

I did not see a p*2pi*r^2 anywhere. Is this a case of a cylinder or a sphere?

 

[Chestermiller]

They split the cylinder in half (including the air inside), and determine the forces on half the cylinder, either axially or in the hoop direction. Axially, the forces on the half-cylinder are $p\pi R^2$ and $2\pi R\sigma_{axial} t$, where $\sigma_{axial}$ is the axial stress in the shell. So, for equilibrium, $$2\pi R t\sigma_{axial}=p\pi R^2$$
Similarly, for the hoop direction, the forces on half the half-cylinder are $p(2R)L$ and $2Lt\sigma_{hoop}$. So, for equilibrium,
$$2Lt\sigma_{hoop}=p(2R)L$$