Pressure Vessels

*Example: A spherical pressure vessel has an I.D. of 220 mm, a wall thickness of

5 mm, and an internal pressure of 4.0 MPa. Find the maximum in-plane shearing

stress and the maximum absolute shearing stress in the pressure vessel.

Strategy: Use the expression σ₁ = σ₂ = pr/2t to determine the normal components of stress

on an element of the sphere as shown below. Then use Mohr’s circle to find the components

of stress at any point on the sphere.

σ₁ = σ₂ = pr/2t = (4.0)(110)/(2)(5) = 44.0 MPa

Mohr’s Circle for this element is shown below.

For “in-plane” stress Mohr’s Circle is just the common point with concident normal stresses

σ₁ and σ₂ . So the “in-plane” shearing stress is (σ₁ – σ₂ )/ 2 = 0 MPa.

For an element with internal pressure acting on a face of the element, the Mohr’s Circle has a

diameter of σ₁ – ( – p) = σ₁ + p . So Mohr’s Circle for “out-of-plane” surfaces is a circle

with radius r = (σ₁ + p )/ 2 . Thus the maximum absolute shearing stress

τ_max = ( 44 + 4 ) / 2 = 24 MPa. (result)

Note: If the “in-plane” surface is the xy-plane, then the “out-of-plane” surfaces are the

planes in xz and yz surfaces.

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