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 <u>equals or exceeds</u> 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
