This document discusses stresses in thin cylindrical shells. It defines a thin cylindrical shell as having a thickness to inner radius ratio of not greater than 1/10. For thin shells, hoop stress and longitudinal stresses are constant through the thickness, and radial stresses are negligible. Equations are provided for calculating hoop stress (circumferential stress) and longitudinal stress in a thin cylindrical shell based on the internal pressure, internal diameter, thickness, and material properties. An example problem calculates the hoop stress, longitudinal stress, maximum shear stress, and changes in diameter and length for a given thin cylindrical shell subjected to internal pressure.

1. THIN CYLINDER ANALYSIS
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2. Thin Cylinder

3. • Introduction
• Cylindrical and spherical vessels are used in the engineering field to
store and transport the fluids.
• Such vessels are tanks, boilers, compressed air receivers, pipe line etc.
• These vessels, when empty, are subjected to atmospheric pressure
internally as well as externally and resultant pressure on the walls of the
shell is nil. The cylinders are subjected to fluid pressures.
• But, whenever a vessel is subjected to an internal pressure its walls are
subjected to tensile stresses.

4. Thin wall pressure vessels
• If the wall thickness of the cylinder is less than 1/20th of the internal
diameter (di), the variation of the tangential stresses through the
wall thickness is small and radial stresses may be neglected. The
solution can be then treated as statically determinate and the vessel
is said to be thin pressure vessel. Thus a pin pressure vessel is one
whose thickness to inner radius ratio is not greater than 1/10.

5. Thin Cylindrical shell
 When, t ≤ d/10 to d/15, It is called thin cylindrical shell.
t = thickness of the shell
d = internal diameter of shell

6. Stresses in thin cylindrical shell
In thin cylindrical shells hoop stress and longitudinal stresses are
constant over the thickness and radial stresses are negligible.
 This will essential focus attention on two stress components at any
point these stress components are:
1. Stress along the circumferential direction, called hoop or tangential
stress.
2. Longitudinal stress in the direction the axis of the cylinder. This stress
is perpendicular to the plane of the paper. So the longitudinal stress
will remain same for any section of the thick cylinder.

7. Longitudinal and hoop stresses

8. Hoop Stress
Consider a thin cylindrical shell
subjected to an internal pressure as
shown in figure.
σc = circumferential stress in shell
material
p = internal pressure
d = internal diameter of shell
t = thickness of the shell
Total pressure(Bursting force),
P = Pressure * Area = p.d.l.

9. .
Resisting Area = A= 2tl
Circumferential stress (σC),
σC = P/A
= p.d.l / 2tl
σC = p.d/2t

10. Longitudinal Stress
Total pressure along the length of
the shell,
P = p.π.d.d/4
Resisting Area,
A = π.d.t
Longitudinal stress,
σl = P/A
= ( p. π.d.d/4)/ π.d.t
σl = p.d/4t

11. Change in dimensions due to internal pressure
Consider a thin cylindrical shell subjected to internal pressure.
p = internal pressure
d = internal dia. Of shell
l = length of the shell
We know that ,
σc = p.d/2.t and σl = p.d/4.t
Let , δd = change in dia. Of shell
δl = change in length of shell
So, circumferential strain,
ε1 = δd/d = σc /E - σl /mE

12. .
ε1 = p.d/2tE – p.d/4tmE
ε1 = (1-1/2m)pd/2tE
Longitudinal Strain,
ε2 = δl/l = σl /E - σc /mE
= pd/4tE - pd/2t.mE
ε2 = (1/2-1/m)pd/2tE
Change in diameter = δd = ε1. d
Change in length = δl = ε2 .l

13. • Example: A cylindrical shell has 3.5 m length, 1.2 m diameter and 10 mm
thickness. The shell is subjected to internal pressure of 2 N/mm . Calculate the
maximum shear stress and change in dimension of the shell.
Diameter of the cylindrical shell (d) = 1.2 m = 1200 mm
Length (l) = 3.5 m = 3500 mm
Thickness (t) = 10 mm
Internal pressure (p) = 2 N/mm
Modulus of Elasticity = 2 * 10 N/ mm
μ = 0.3
Hoop stress
σc = p.d/2t = (2*1200)/(2*10) = 120 N/mm
2
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14. Longitudinal stress
σl = p.d/4t
= (2*1200)/(4*10)
= 60 N/ mm
ε1 = σc /E - σl /mE
= 120/( 2*10) - 60*0.3/(2*10)
= 5.1* 10
ε2 = σl /E - σc /mE
= 60/(2*10) – (120*0.3)/ (2*10)
= 1.2 * 10
2
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15. Maximum shear stress
ťmax = (σc - σl)/2
= (120-60)/2
= 30 N/ mm
ε1 = δd/d
δd = ε1. d
= 5.1*10 *1200 = 0.612 mm
ε2 = δl/l
δl = ε2 .l
= 5.1*10 *3500 = 0.42 mm
δV/V = (ε2 + 2 ε1 )
δV = (ε2 + 2 ε1 )*V
= (1.2*10 + 2*5.1 *10) *(3.14*1200*3500)
= 4512584 mm = 4. 51 liter
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16. Thank You