Longitudinal Stress Calculator

How it is calculated

For a thin-walled cylinder closed at both ends, the internal pressure pushes on the end caps and stretches the wall along its length. The tool uses σ_L = P × d ÷ (4 × t) for the longitudinal (axial) stress and σ_hoop = P × d ÷ (2 × t) for the hoop stress around the circumference. P is the internal pressure in MPa (1 MPa = 1 N/mm²), d the inner diameter in mm and t the wall thickness in mm, so the stresses come out in MPa. The hoop value is always exactly twice the longitudinal one. Diameter and thickness must be above zero and the pressure cannot be negative (0 gives 0 MPa). A thickness of half the diameter or more would be a solid bar, not a cylinder, so it gives no result. Below that limit the tool does not check how thin the wall really is (a common rule of thumb is a diameter of about 20 times the thickness or more), and it ignores stress concentrations and external loads.

Common mistakes

• Entering the pressure in bar or kPa. The field expects MPa, so 20 bar must be entered as 2.
• Using the outer diameter or the mean diameter. The tool expects the inner diameter, and a different diameter shifts the stresses.
• Comparing the longitudinal figure alone with the allowable stress. The hoop stress is twice as large, so it is the one that normally governs the wall thickness.
• Applying the formula to a thick-walled vessel. With a thickness of 100 mm on a 500 mm bore the tool still prints 2.50 MPa for the same 2 MPa, although the stress is then not uniform through the wall and the thin-wall result is unreliable.
• Using it for a cylinder with open ends, such as a pipe with no end load. The axial formula assumes the pressure acts on closed end caps.