Short Answer:

When a thin-walled cylinder is under internal pressure, two principal membrane stresses develop in the wall: the hoop (circumferential) stress and the longitudinal (axial) stress. The hoop stress acts around the circumference and tends to split the cylinder along its length, while the longitudinal stress acts along the axis and tends to pull the end caps off.

For a thin cylinder (wall thickness  much smaller than radius , ), the standard formulas are:

where  is internal pressure,  internal radius and  internal diameter. Hoop stress is twice the longitudinal stress.

Detailed Explanation :

Hoop and Longitudinal Stresses in Thin Cylinder

Assumptions for thin-walled cylinder theory
Before deriving stresses we state common simplifying assumptions used for thin cylinders:

1. Wall thickness  is small compared with radius  ().
2. Stress through the wall thickness is constant (plane stress in the wall).
3. Radial stress is negligible compared with membrane stresses (hoop and longitudinal).
4. Material is linear elastic, homogeneous and isotropic; deformations are small.

These assumptions allow treating the cylinder wall as a membrane carrying in-plane forces only.

Derivation of hoop (circumferential) stress
Consider a thin cylindrical shell of internal radius , wall thickness  and length , subjected to internal pressure . To obtain hoop stress, make a longitudinal cut and look at the equilibrium of half the cylinder (cut by a plane parallel to the axis). The internal pressure acts on the projected area of the cut half.

1. Pressure force acting to separate the two halves: the projected area of the internal pressure on the cut half equals  (width = diameter , length = ), so the total outward force due to pressure is

1. This pressure force is resisted by the tensile forces in the two longitudinal edges (the shell walls on either side of the cut). The resisting tensile force on each longitudinal edge is , so both edges give

1. Equilibrium (force balance) in the radial direction gives

Cancel  and solve for :

If diameter  is used, this becomes .

Derivation of longitudinal (axial) stress
To derive axial (longitudinal) stress, consider the equilibrium of the cylinder cut by a plane perpendicular to the axis (i.e., consider the free body of the cylindrical shell and its end cap). The internal pressure acts on the circular cross-section and tends to blow off the end; this pressure is resisted by the axial tensile force in the cylindrical wall.

1. Pressure force on the circular end is the internal pressure times the area:

1. The resisting force is provided by the axial membrane stress acting over the hoop length around the circumference. The total axial resisting area of the wall is the circumferential length times thickness: . So resisting force is

1. Equilibrium (axial force balance) gives

Cancel  and solve:

Using diameter : .

Relationship between the two stresses
From the formulas it follows directly:

so

Thus the hoop stress is twice the longitudinal stress for a thin, closed cylinder under internal pressure. This is why failure (e.g., longitudinal split) often occurs along the length of a pressurized cylinder.

Sign convention and nature of stresses
Both stresses are tensile for internal pressure. If external pressure (vacuum inside, higher pressure outside) is applied, these stresses reverse sign (compressive) and buckling may govern the design instead of yielding.

When formulas are valid and limitations
The membrane formulas above neglect radial stress and assume uniform distribution through thickness. They are accurate when . For thicker walls or very high pressure where stress varies significantly across the thickness, use Lame’s thick-wall solutions. Also, the derivation assumes closed ends; for very long open cylinders or special end conditions the axial stress distribution and boundary effects may modify the longitudinal stress locally.

Design implications
Because hoop stress is larger, design against burst usually checks the hoop stress against allowable material strength first. Common safety checks compute required thickness:

or include a factor of safety. Welding, corrosion, and stress concentrations must also be considered.

Simple numeric example
A cylinder with internal pressure , internal diameter , thickness :

 

Hoop stress is 50 MPa and longitudinal 25 MPa.

Conclusion

For a thin-walled cylinder under internal pressure, simple force equilibrium gives the well-used membrane results: hoop stress  and longitudinal stress . These follow from balancing pressure forces with in-plane membrane forces and are valid under thin-wall assumptions. Since hoop stress is twice the axial stress, it usually controls strength checks for thin pressure vessels and pipes.