Considering that the half sphere must be in equilibrium, the net force in the z direction can only be balanced by the normal stress in the wall s_s.

 

S F_z = 0 :  s_s 2 p r_avg t = p (r_i)² p_i

 

Because the wall is thin, we can make an approximation and replace r_avg by r_i. Then

 

s_s = (p_i r_i)/(2 t).

 

Note that we can cut the sphere in half with a vertical plane and we have the exact same situation that we have just considered except rotated 90^o.  Therefore, there must be a normal stress in  the horizontal direction equal to s_s as well.  In fact, we recognize that we can cut the sphere in half in any orientation and the normal stress is the same.  The spherical pressure vessel then has a state of stress in the wall which can be represented as

 

 

And the element could be oriented in any way on the surface of the sphere.

 

 

 

 

 

Now, lets consider the thin wall cylindrical pressure vessel.

 

Again, because of the internal pressure, there will be normal stresses in the wall of the cylinder.  To determine the stresses, lets first pass a cutting plane through the cylinder perpendicular to its length.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Again, there are forces created by the internal pressure on the inside wall.  Considering an element dA on the side of the cylinder, we see that the force only has components in the x and y directions. 

Lecture #21, page 2                 Back                        Lecture #21, page 3