Thin-Walled Pressure Vessels: Stress Analysis and Applications - Prof. Kweche, Schemes and Mind Maps of Economic Growth and Globalization

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THIN-WALLED

PRESSURE VESSELS

THIN-WALLED PRESSURE VESSELS

Assumptions:

1. Inner-radius-to-wall-thickness ratio ≥ 10
2. Stress distribution in thin wall is uniform or constant

• Cylindrical vessels:
• The cylindrical vessel in Fig. a has a wall thickness t , inner radius r , and

is subjected to an internal gas pressure p.

• Two types of stresses: circumferential or hoop stress , & longitudinal

stress

THIN-WALLED PRESSURE VESSELS

Hoop direction: 𝜎 1 =

• To find the circumferential or hoop stress , we can section the vessel by

planes a , b , and c (in the previous slide).

• Considering only loadings in the x- direction: shown in fig (b) below

2𝑡𝑑𝑦𝜎 1 = 2𝑝𝑟𝑑𝑦

෍ 𝐹𝑥 = 0 ; 2 [𝜎 1 𝑡 𝑑𝑦 ] − 𝑝 2𝑟 𝑑𝑦 = 0

Hoop direction:

THIN-WALLED PRESSURE VESSELS (cont)

LONGITUDINAL STRESS

• Comparing the two stresses, it can be seen that the hoop or

circumferential stress is twice as large as the longitudinal or

axial stress.

• This implies that:
  • when fabricating cylindrical pressure vessels from rolled-

formed plates, it is important that the longitudinal joints

be designed to carry twice as much stress as the

circumferential joints.

THIN-WALLED PRESSURE VESSELS (cont)

• Spherical vessels:

2

2 t

pr  =

෍ 𝐹𝑦 = 0 ; 𝜎 2 2 𝜋𝑟𝑡 − 𝑝 𝜋𝑟

2 = 0

This is the same result as that obtained for the longitudinal

stress in the cylindrical pressure vessel, although this stress

will be the same regardless of the orientation of the

hemispheric free-body diagram

EXAMPLE 1

A cylindrical pressure vessel has an inner diameter of 1.2 m

and a thickness of 12 mm.

• Determine the maximum internal pressure it can sustain

so that neither its circumferential nor its longitudinal stress

component exceeds 140 MPa.

• Under the same conditions, what is the maximum internal

pressure that a similar-size spherical vessel can sustain?

EXAMPLE 1 (cont)

• The stress in the longitudinal direction will be
  • The maximum stress in the radial direction occurs on the material at

the inner wall of the vessel and is

• The maximum stress occurs in the circumferential direction.

Solutions

𝜎 1 =

𝑝𝑟

𝑡

140 =

𝑝 600

12

𝑝 = 2. 8 MPa

( 140 ) 70 MPa

2

1  2 = =

𝜎 3 (max) = 𝑝 = 2. 8 MPa

EXAMPLE 2

A tall open-topped standpipe below has an inside diameter of

2,750 mm and a wall thickness of 6 mm. The standpipe

contains water, which has a mass density of 1,000 kg/m

3 .

(a) What height h of water will produce a circumferential

stress of 16 MPa in the wall of the standpipe?

(b) What is the axial stress in the wall of the standpipe due to

the water pressure?

(a) Height h of water

EXAMPLE 2 (cont)

• Circumferential or hoop stress:

Solutions

hoop

𝑝( 1375 mm)

( 6 mm)

= 16 Mpa

∴ 𝑝 = 0. 0698 MPa

3

3 2

3 2

69.818 10 MPa

69.818 10 N/m

7.122684 m

(1,000 kg/m )(9.81 m

7.12 m

/s )

p gh

h

−

STRESS CAUSED BY

COMBINED

LOADINGS

STRESS CAUSED BY

COMBINED LOADINGS

• So far, we’ve determined the stress in a member

subjected to either an internal axial force, a shear

force, a bending moment, or a torsional moment.

• Most often, however, the cross-section of a member

will be subjected to several of these loadings

simultaneously, and when this occurs, then the

method of superposition should be used to

determine the resultant stress.

• The following procedure for analysis provides a

method for doing this

REVIEW OF STRESS ANALYSES (cont)

• Torsional moment T leads to:
• Stresses in pressure thin-walled vessels

𝑠ℎ𝑒𝑎𝑟 − 𝑠𝑡𝑟𝑒𝑠𝑠 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛, 𝝉 =

𝑻𝝆

𝑱

(for circular shaft)

𝝉 =

𝑻

𝟐𝑨𝒎𝒕

(for closed thin−walled tube)

Circumferential or hoop stress, 𝝈𝟏 =

𝒑𝒓

𝒕

Longitudinal or axial stress, 𝝈𝟐=

𝒑𝒓

𝟐𝒕

RESULTANT STRESSES BY SUPERPOSITION

Once the normal and shear stress components for each

loading have been calculated, use the principal of

superposition to determine the resultant normal and shear

stress components.

Represent the results on an element of material located at a

point, or show the results as a distribution of stress acting over

the member’s cross-sectional area.