Question

The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest millimeter using
(a) the maximum-shear-stress theory, and
b) maximumdistortion-energy theory. Apply a factor of safety of 1.5 against yielding.

Answer 1

Answer:

a. t = 23mm, b. t = 20mm

Explanation:

Obtain the value of yield strength in tension for A- 36 steel from table ‘Average Mechanical Properties of typical Engineering Materials’, which is σ(y) = 250 MPa.

a.

Assume that the thin wall analysis is valid, Calculate the hoop stress

σ(1) = pd/2t, where p is the pressure in the tank, d is the internal diameter of the tank and t is the thickness.

Substitute 5MPa for p and 1.5m for d

σ(1) = 5 x 10⁶x 1.5/2t

σ(1) = (3.75 x 10⁶)/t

Calculate the longitudinal stress

σ(2) = pd/4t

σ(2) = (5 x 10⁶ x 1.5)/4t

Apply Maximum Shear stress theory which states that failure occurs when the maximum shear stress from a combination of principal stresses equals or exceeds the value obtained for the shear stress at yielding in the uni-axial tensile test. Hence

τ(abs.max) ≤ τ (allowed)

τ (abs.max) ≤ σy /2FS, where FS is the factor of safety

Substitute σ(1)/2 for τ (abs.max) as both the principal stresses have same sign.

σ(1)/2 ≤ σy/2FS

3.75 x 10⁶/2t = 250 x 10⁶/2 x 1.5

T = 0.0225m = 22.5mm = 23mm to the nearest millimeter

Hence the required minimum thickness using the maximum shear stress theory is t = 23mm

b.

Apply maximum distortion energy theorem

σ²(allowed) =σ²(1) – σ(1) x σ(2) + σ²(2)

σ²(y)(allowed)/FS = (3.75 x 10⁶/t)² – (3.75 x 10⁶/t) x (1.875 x 10⁶/t) + (1.875 x 106/t)²

250 x 10⁶/1.5 = (3.2476 x 10⁶)/t

t = 3.2476/166.67

t = 0.0195 m = 19.50 mm = 20mm to the nearest millimeter

Hence, the required minimum thickness using the maximum distortion energy theory is t = 20 mm