Question

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est

Answer 1

Question:

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.

Answer:

See explanation below

Explanation:

Given:

d = 2m = 2*10³ = 2000

thickness, t = 10 mm

Length of strain guage = 20 mm

i) Let's calculate d/t

$\frac{d}{t} = \frac{2000}{10} = 200$

Since $\frac{d}{t}$ is greater than length of strain guage, the pressure vessel is thin.

For the minimum normal stress, we have:

$\sigma max= \frac{pd}{4t}$

$\sigma max= \frac{2000p}{4 * 20}$

= 50p

For the minimum normal strain due to pressure, we have:

$E_max= \frac{change in L}{L_g}$

$= \frac{0.012}{20} = 0.60*10^-^3$

The minimum normal stress for a thin pressure vessel is 0.

$\sigma _min = 0$

i) Let's use Hookes law to calculate the pressure causing this deformation.

$E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)]$

Substituting figures, we have:

$0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)]$

$120 * 10^6 = 35p$

$p = \frac{120*10^6}{35}$

$p = 3.429 * 10^6$

p = 3.4 MPa

ii) Calculating the maximum in-plane shear stress, we have:

$\frac{\sigma _max - \sigma _int}{2}$

$= \frac{50p - 50p}{2} = 0$

Max in plane shear stress = 0

iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:

$\frac{\sigma _max - \sigma _min}{2}$

$= \frac{50p - 0}{2} = 25p$

since p = 3.429 MPa

25p = 25 * 3.4 MPa

= 85.71 ≈ 85.7 MPa

The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa