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8 – Stress on Thin-Walled Pressure Vessels and Pipes, Slides of Fluid Mechanics

8 – Stress on Thin-Walled Pressure Vessels and Pipes Description: Covers hoop and longitudinal stress analysis Applies to tanks, pipes, and cylindrical vessels Includes pressure formulas and stress safety checks

Typology: Slides

2021/2022

Available from 06/06/2025

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ENGR. BON RYAN ANIBAN
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ENGR. BON RYAN ANIBAN

Cylindrical and spherical vessels are commonly used in industry to serve as boilers

or tanks. When under pressure, the material of which they are made is subjected to a

loading from all directions. Although this is the case, the vessels can be analyzed in a

simple manner provided it has a thin wall.

In general, “thin wall” refers to a vessel having an inner-

radius-to-wall-thickness ratio of 10 or more ( r / t ≥ 10 ).

Provided the vessel wall is “thin,” the stress distribution

throughout its thickness will not vary significantly, and so

assumed that it is uniform or constant.

Using this assumption, we will analyze the state of stress in

thin-walled cylindrical and spherical pressure vessels.

t

σt

σL

D

a

a

L

The tangential stress ( σt ) can be determined

by considering the vessel to be cut by plane a-a ,

other common names are circumferential stress, hoop stress, and girth stress.

The tangential stress ( σt ) can be determined

by considering the vessel to be cut by plane a-a ,

other common names are circumferential stress, hoop stress, and girth stress.

σt

σt

t

t

L

D

p

z

y

x

t

σt

σL

D

L

The longitudinal stress ( σL ) can be determined by

considering the vessel to be cut by plane b-b , and isolate

the left portion of plane b-b. As shown in the figure, σL acts

uniformly throughout the wall, and p acts on the section

of the contained fluid.

t

σt

σL

D

L

b

The longitudinal stress ( σL ) can be determined by

considering the vessel to be cut by plane b-b , and isolate

the left portion of plane b-b. As shown in the figure, σL acts

uniformly throughout the wall, and p acts on the section

of the contained fluid.

Consider the vessel to have a wall thickness t , inner diameter

D , and subjected to an internal pressure p. A small element of the vessel that is sufficiently removed and oriented as shown, is subjected to normal stress σS. If the spherical vessel is cut in

half by plane c-c , the resulting free-body diagram is shown.

σs

σs

D

t

σs

σs

D

t

c

c

Consider the vessel to have a wall thickness t , inner diameter

D , and subjected to an internal pressure p. A small element of the vessel that is sufficiently removed and oriented as shown, is subjected to normal stress σS. If the spherical vessel is cut in

half by plane c-c , the resulting free-body diagram is shown.

Consider the wooden tank having an inner diameter D , and subjected to an internal pressure p that developed within the vessel by a contained fluid. The staves of wooden tank is bound together by steel hoops having a cross-sectional area AH and allowable stress σH. The spacing s can be determined by considering the vessel to be sectioned by planes a-a , b-b , and c-c.

c

c

a

a

b

b

s s

D

Consider the wooden tank having an inner diameter D , and subjected to an internal pressure p that developed within the vessel by a contained fluid. The staves of wooden tank is bound together by steel hoops having a cross-sectional area AH and allowable stress σH. The spacing s can be determined by considering the vessel to be sectioned by planes a-a , b-b , and c-c.

σH

p D

σH

AH

AH

s /2 s /

𝐻

𝐻

Where: s = spacing of hoops, mm σH = allowable hoop stress, MPa AH = cross-sectional area of the hoop, mm^2 p = the internal pressure developed by the contained gas or fluid, MPa D = the inner diameter of the cylindrical vessel, mm

z

y

x

F 1

F 2

= σH ( AH )

= p^ ( Ds )

F 1 = σ

H ( AH )

Determine the required thickness of the 450 mm diameter steel pipe to carry a maximum pressure of 5500 kPa if the allowable stress of steel is 124 MPa. Solution 𝜎𝑡 = 𝑝𝐷 2𝑡 124 𝑀𝑃𝑎 = ( 5. 5 MPa)( 450 mm ) 2 (𝑡) 𝑡 = 9. 980 mm 𝜎𝐿 = 𝑝𝐷 4𝑡 124 𝑀𝑃𝑎 = ( 5. 5 MPa)( 450 mm ) 4 (𝑡) 𝑡 = 4. 990 mm

A thin walled hallow sphere 3.5 m in diameter holds helium gas at 1700 kPa. Determine the minimum wall thickness of the sphere if the allowable stress is 60 MPa. Solution 𝜎𝑆 = 𝑝𝐷 4𝑡 60 𝑀𝑃𝑎 = ( 1. 7 MPa)( 3500 mm ) 4 (𝑡) 𝑡 = 24. 792 mm