Pressure Vessels and Failure Theories

Thin-walled pressure vessel stresses

A thin-walled pressure vessel is a shell that holds a fluid at internal gauge pressure p. The shell is called thin when its radius r (or mean diameter d) is large compared with the wall thickness t, with the common working limit t \le d/20. Under this approximation the stresses are taken as uniform through the wall and the radial stress is neglected, so the wall is treated as a plane-stress membrane.\n\nFor a closed cylinder there are two membrane stresses. The hoop (circumferential) stress acts perpendicular to the axis and equals \sigma_h = \frac{pd}{2t} = \frac{pr}{t}. The longitudinal (axial) stress acts parallel to the axis and equals \sigma_l = \frac{pd}{4t} = \frac{pr}{2t}, which is half the hoop value. For example, with p = 1 MPa, d = 1 m and t = 10 mm, the hoop stress is 50 MPa and the longitudinal stress is 25 MPa. On a surface element whose sides lie along and across the axis, the normal stress across the axis is the hoop stress and the normal stress along the axis is the longitudinal stress.\n\nThe radial stress varies from -p at the inner surface to zero at the outer surface. Because r \gg t, it is far smaller than the membrane stresses and is dropped, so the principal stresses are \sigma_1 = \sigma_h, \sigma_2 = \sigma_l and \sigma_3 \approx 0. The shear stress in the wall must be read from this principal set. The in-plane shear (\sigma_h - \sigma_l)/2 = \frac{pd}{8t} is not the largest shear once the zero radial stress is included. The absolute maximum shear is \tau_{max} = \sigma_1/2 = \frac{pd}{4t}, and this is the value to use in a shear-based check.\n\nPressure also pushes the two halves of a split cylinder apart. Over a longitudinal plane the separating force is the pressure times the projected area, F = p\,D\,L. If the end caps carry no radial load, the whole separating force is taken by the flange bolts, so the load per bolt is F divided by the number of bolts that share it.\n\nWhen a distortion-energy check is required, the equivalent stress for the biaxial membrane state is \sigma_e = \sqrt{\sigma_h^2 + \sigma_l^2 - \sigma_h\sigma_l}. For a closed cylinder this reduces to \sigma_e = \frac{\sqrt{3}\,pd}{4t}, and yielding occurs when \sigma_e = \sigma_y. The allowable pressure is therefore p = \frac{4t\sigma_y}{\sqrt{3}\,d}.\n\nProblems in this set often combine vessel stresses with a quick unit or RMS conversion, so keep that bookkeeping in mind. For a sinusoidal quantity the RMS value is the peak divided by \sqrt{2}, so a_{max} = \sqrt{2}\,a_{rms}. With \omega = 2\pi f and a_{max} = \omega^2 X for harmonic motion, the amplitude is X = a_{max}/\omega^2. Convert all lengths to metres before substituting.