This document discusses stresses and strains in thin-walled pressure vessels such as cylinders and spheres. It describes how internal pressure leads to hoop and axial stresses, and how these stresses can be calculated for thin cylindrical and spherical shells. It also discusses techniques for reinforcing thin shells, such as adding wire winding or using hemispherical ends to reduce hoop stresses. Thick cylinders are also briefly mentioned as being used for higher internal pressures compared to thin-walled vessels.

1. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 1
UNIT-IV
Thin Cylinders & Spheres: Introduction
Difference between Thin Walled and Thick Walled Pressure Vessels
Thin walled spheres and cylinders
Hoop and Axial Stresses and Strain
Volumetric Strain.
Thick Cylinders:
Radial, axial and circumferential stresses in thick cylinders subjected to internal or external
pressures,
Compound Cylinders
Stresses in Rotating Shaft and Cylinders,
Stresses due to Interference Fits

2. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 2
Thin Cylindrical and Spherical Shells:
Thin cylindrical and spherical shells are used mainly for storage of gas, petrol, liquid, chemicals, grains and so on.
Some are subjected to internal/external pressures and the order of pressure is low (10–30 atmospheres).
Their D/t ratio, that is, the ratio of diameter to wall thickness, is large, that is, D/t is greater than 20. Because in
comparison to diameter, thickness is very small, so the variation of stresses along the thickness is taken
negligible.
Thin pressure vessels are subjected to internal uniform pressure or hydrostatic pressure of low magnitude.
Development of axial and hoop stresses in the wall thickness of the pressure vessel provides useful data for the
designer. Stresses in thin shells are dependent on the D/t ratio, that is, the ratio of shell diameter to wall
thickness.
Thin Cylinder Subjected to Internal Pressure:
Consider a thin cylindrical shell of inner diameter D, length L, with wall thickness equal to t. The shell is full of
liquid. The volume of the liquid inside the shell is [π/4 (D2L)] and the pressure of the liquid is atmospheric. Now,
the additional volume of the liquid is pumped under pressure inside the shell; as a result, there is overall
expansion in the shell, that is, expansion in both diameter and length. However, the expansion of liquid is partly
prevented by the wall of the metallic cylinder, which offers equal and opposite reaction and compresses the
liquid inside the cylinder (Fig 5.1).
Figure. 5.1 Small element of cylinder subjected to internal pressure
Mathematically
δV = additional volume of the liquid pumped inside
= δv1 + δv2 = expansion in the volume of cylinder + contraction in the volume of the liquid.
Since the liquid exerts pressure on cylinder wall, the length and diameter of the cylinder get increased
introducing axial and circumferential stresses, in the cylinder. Figure 5.2 shows a small element of the cylinder
subjected to internal pressure p; σc is the circumferential a hoop stress developed (along circumference)
and σa is the axial stress developed (along the axes); at outer surface pa is atmospheric pressure, such
that p >>pa.
Axial Stress:
Diameter of cylinder = D
Wall thickness = t
Axial stress = σa
Axial bursting force on cylinder in axial direction, [Pa = π/4(D2p)] acting on the end plates as shown in Fig. 5.3
Resisting area = πDt
Figure. 5.2 Small element under internal pressure
Figure. 5.3 Axial bursting force

3. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 3
Circumferential stress (σc):
Consider a small portion along length δL and inner radius D/2, subjected to internal pressure p, Fig 5.4 Take a
small element at an angle θ from x-axis, subtending an angle dθ at inner surface.
The horizontal component of the force is cancelled
out when the force is integrated over the semicircular
portion of the cylinder.
Therefore, total diametral bursting force,
Area of cross-section resisting the diametral bursting force = 2t dl
Strains:
Consider Fig. 5.2, σc, σa and p stresses are perpendicular to each other. Note that p acts as a compressive radial
stress on inner surface of cylinder. Stresses σc and σa are much larger than p; Therefore in calculation of strains,
the effect of internal pressure p is neglected. Say E is Young’s modulus and v is Poisson’s ratio of the material.
Note that εc = circumferential strain ; εa = axial strain
Change in the diameter of cylinder, δD = εcD
Change in the length of cylinder, δL = εaL
Volumetric Strain:
Volume of cylinder [v = π/4 (D2L)], taking partial derivative
Figure. 5.4 Diametral bursting force

4. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 4
Change in Volume:
Change in volume of cylinder, δV1 = ev V
where K is bulk modulus of liquid.
Additional volume of liquid pumped inside the cylinder, δv = δv1 + δv2.
Note that if, in any problem, K for the liquid is not given that δv2, that is, reduction in volume of liquid can be
taken as negligible.
Thin Spherical Shell:
Consider a thin spherical shell of diameter D and wall thickness t subjected to internal fluid pressure p as shown
in Fig. 5.5(a). The internal pressure p acts throughout the inner curved surface and the diametral bursting force
tends to break the thin shell in two halves, that is, two hemispherical halves, as shown in Fig. 5.5(b).
Figure. 5.5 Thin spherical shell
Diametral bursting force, PD = p × projected area of curved surface
Say σc = hoop stress developed in the wall of the shell.
Area resisting the diametral bursting force = π Dt
So
Strains:
Consider a small element of thin spherical shell subjected to internal radial stress p, due to which circumferential
stress σc is developed in shell, as shown in Fig. 5.6.
pa = atmosphere pressure on outer surface.
If E and v are Young’s modulus and Poisson’s ratio of the material, respectively, then
Internal pressure p << σc ; therefore, the effect of p is not considered in determining the strain εc.
Volume of the shell

5. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 5
Figure. 5.6 Small element of thin spherical shell
Note that circumferential stress in thin spherical shell is only PD/4t; but in thin cylindrical shell, circumferential
stress is PD/2t, that is, two times, the stress which is developed in a spherical shell, so it is advantageous to use a
thin spherical shell, but economically it is costlier to make a thin spherical shell than a thin cylindrical shell. To
make a thin spherical shell, two hemispherical portions are made and then joined. To make a hemispherical
portion, several segments are made and joined to develop the hemispherical part.
Cylindrical Shell with Hemispherical Ends:
For thin cylindrical shell subjected to internal pressure, we have derived expressions for axial strain, εa =
[(PD/4tE)(1 − 2v)] and we have assumed that along the length of the shell, axial strain remains constant,
meaning that there is no distortion of end plates. But in actual practice there is distortion of end plates.
Moreover, we have learnt that hoop stress developed in spherical shell is only half the hoop stress developed in
cylindrical shell for same values of p, D and t. Therefore, it is advantageous to use hemispherical ends for thin
cylindrical shell as shown in Fig. 5.7. Figure shows a cylindrical shell of diameter D, length Land wall thickness t,
with hemispherical ends of diameter D and wall thickness t2. Obviously, the thickness of hemispherical portion
will be less than the wall thickness of cylindrical part.
Figure. 5.7
Say, σc′ = hoop stress in cylindrical part; σc″ = hoop stress in hemispherical part

6. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 6
Similarly hoop strain
εc′ = hoop strain in cylindrical part
εc″ = hoop strain hemispherical part
There can be two criteria for design in this case:
1.Hoop stress developed in both portions is the same.
2. Hoop strain in both the portions is the same, so that there is no distortion at the junction of two portions.
Wire Winding of Thin Cylindrical Shells:
We have learnt that due to internal pressure, hoop stress developed in the cylindrical shell is double the axial
stress. Under pressure, there are chances of development of longitudinal crack in the cylinder, if circumferential
stress exceeds the safe limit. Moreover, due to high hoop stress, the pressure-bearing capacity of cylindrical shell
is reduced. Therefore, to reduce the intensity of hoop stress and to strengthen the cylinder in longitudinal
direction, a wire under tension is wound around the circumference of the cylinder, as shown in Fig. 5.8 tensile
stress in winding wire puts pressure on cylinder from outside and a negative hoop stress is developed in the
cylinder wall.
Figure. 5.8
Consider a thin cylinder of diameter D, length L, and wall thickness t. It is wound with a single layer of wire of
diameter dw under an initial tension of σw.
(Note that there are two sections)
Compressive force developed in cylinder = σc 2Lt
For equilibrium

7. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 7
This is the initial compressive hoop stress developed in cylinder wall due to wire winding. Figure 5.8 shows, the
development of σc, in cylinder
Internal pressure:
Now the wire-wound cylinder is subjected to internal pressure p. Due to this say the stresses developed in
cylinder are σa′, σc′ and σw′.
Axial stress developed,
Dimensional bursting force, PD = pDL
Say, the stresses due to internal pressure developed in cylinder and wire are, respectively, σc′, σw′. Then,
putting the value of n,
There are two unknowns σc′ and σw′ but only one equation, so let us consider compatibility condition, taking
strain into account.
Circumferential strain in wire = circumferential strain in cylinder
Let us take Ec and vc as Young’s modulus and Poisson’s ratio for cylinder, respectively, and Ew and vw as Young’s
modulus and Poisson’s ratio of wire, respectively.
From Eqs (5.9) and Eqs (5.10), σc′ in cylinder and σw′ in wire can be determined.
Resultant Stresses:
In cylinder, σcr = σc + σc′ (tensile)
In wire, σwr = σw + σw′ (tensile)
Thick Cylinders
For low pressures, thin pressure vessels are used where D/t ratio is greater than 20. For high internal pressure
as in the case of tubes used for high pressure gauge and bulk compression gauge, tubes with considerable
thickness are used where D/t << 20.
In a thin pressure vessel, the hoop and axial stresses are much greater than the intensity of fluid pressure and in
many applications such as tube (for high-pressure gauge) it is subjected to high internal pressure. The tube
section can be designed so that the ratio of hoop stress/internal pressure is not large.
Lame’s Equations:
Stresses in a thick cylinder are determined on the assumption that plane sections (perpendicular to the axes) of
the cylinder remain plane after the application of internal pressure. In other words, axial strain in cylinder
remains constant along its lengths and there is no distortion of end plates.

8. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 8
Figure 6.1 shows a section of a thick cylinder of which thickness t = R2 − R1 is considerable. It is subjected to
internal pressure p such that axial and circumstantial stresses develop in the cylinder. Radial stress σr varies
from p (compressive) at radius R1 to zero stress at radius R2, a nonlinear variation of σr. Consider a thin circular
cylinder of inner radius r and outer radius r +dr, that is, radial thickness dr subjected to radial stress σr at inner
radius r and σr + dσr at outer radius r + dr. Consider the equilibrium of half of the cylinder, that is, XAX as shown
in Fig. 6.1, σc is the circumstantial stress (taking force equal to stress × projected area with cylinder length equal
to unity).
2(σr + dσr) (r + dr) + 2σc dr = 2rσr
2σcdr + 2rdσr + 2σrdr + 2dσrdr = 0
But, 2dσrdr is a negligible quantity,
So, σcdr + σrdr + rdσr = 0
or,
Say E and v are elastic constants
Figure 6.1 Radial and hoop stresses on thin cylinder Figure 6.2 Stresses on a small element
Stresses on a small element of thick cylinder are shown in Fig. 6.2, that is, σr, σc and σa in radial, circumstantial
and axial directions. Note that σr is compressive stress, while σa and σc are tensile stresses.
But stress σa is constant axial stress and E and v are elastic constants; therefore, σc − σr = constant = 2A
From Eqs (6.3) and (6.4)
σc − σr = 2A (6.4)
From Eqs (6.1)
Subtracting Eq. (6.4) from Eq. (6.5)
or ln (A + σr) = −2 ln r + ln B, where ln B is another constant

9. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 9
From Eq. (6.4)
Boundary Conditions:
Radial stress at
or
putting the value of A in Eq. (6.9)
putting the values of A and B in equation for σc we get,
Thick Cylindrical Shell Subjected to External Pressure:
In the previous chapter, we have studied about the thick cylinder subjected to an internal pressure which causes
tensile hoop and axial stresses in the cylinder. Yet, in many applications, a cylinder may be subjected to an
external pressure as in the case of the submarine tank subjected to external pressure and the inner cylinder of a
compounded cylinder subjected to external pressure which causes compressive hoop and axial stresses in the
cylinder. Figure 6.5 shows a thick cylinder of inner radius R1, outer radius R2 and length L subjected to an
external pressure p.
σr = p at r = R2
σr = 0 = at r R1
Using Lame’s equations,
Figure 6.3 Variation of radial and
circumferential stresses

10. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 10
or, A = B/R1
2, putting the value in Eq.(6.12)
at r = R1, σr = 0
at r = R2 ,
Hoop Stress:
Note that 2R2
2 > (R2
2 + R1
2)
Therefore, the maximum hoop stress occurs at inner radius.
as the pressure acts on the outer surface.
Stresses on an element of a cylinder subjected to pressure p are shown in Fig. 6.6. Note that all the
stresses σc, σr and σa are compressive. The maximum hoop stress (though compressive in this case) occurs at
inner radius.
Figure 6.6
Compound Cylinder:
To increase the pressure-bearing capacity of a cylinder and to decrease the variation in hoop stress developed in
cylinder, another cylinder is force fitted or shrink fitted over the cylinder. This process is called compounding of
cylinders. Therefore, for compounding, the inner diameter of the outer cylinder is slightly less than the outer
diameter of the inner cylinder so as to provide the force fit allowance or shrinkage allowance.
In shrinkage fitting, the outer cylinder is heated such that its inner radius R3″ expands to R3′ and then the outer
cylinder is slipped over the inner cylinder. When the outer cylinder cools, it exerts a radial pressure p′ on the
outer surface of the inner cylinder. In reaction, the inner cylinder exerts a pressure p′ on the inner surface of the
outer cylinder as shown in Fig. 6.8(c). The outer pressure p′ on the inner cylinder develops compressive hoop
stresses in the inner cylinder. Similarly the inner pressure p′ on the outer cylinder develops tensile hoop stresses
Figure 6.5 Thick cylinder subjected to external
pressure

11. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 11
in the outer cylinder. The final junction radius between two cylinders is R3 such that R3 < R3′ but R3 > R3″ The
difference R3″ ‒ R3′ is termed as shrinkage allowance.
Stresses Due to Junction Pressure:
Inner Cylinder:
Outer Cylinder:
Figure 6.8
Compound Cylinder:
Say the compound cylinder is now subjected to an internal pressure p. Say A and B are Lame’s constants, thus
or
Stresses at R1, R2 and R3 due to internal pressure
Figure 6.9
Figure 6.10 Resultant hoop stress in the
inner and outer cylinders of a compound
cylinder

12. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 12
Resultant Stresses:
Inner Cylinder:
Outer Cylinder:
Figure 6.10 shows the variation of resultant hoop stresses developed due to junction pressure p′ and internal
pressure p on the inner and outer cylinders.
Shrinkage Allowance:
In a compound cylinder at the junction radius due to junction pressure, tensile hoop stress in the outer cylinder
and compressive hoop stress in the inner cylinder are developed. At common radius R3
Inner Cylinder:
Junction pressure, p′ (compressive)
When vi and Ei are the Poisson’s ratio and the elastic modulus of the inner cylinder, respectively
Outer Cylinder:
Junctions pressure, p′ (compressive)
where v0 and E0 are the Poisson’s ratio and the elastic modulus of the outer cylinder.
Shrinkage allowance (R′3 − R′′3) from Eqs (16) and (18)
If both the cylinders are of different materials, then the shrinkage allowance is calculated from Eq. (6.19).
If Ei = E0 = E,vi = v0 = v, that is, both the cylinders are of same material. The shrinkage allowance on common
radius R3 is

13. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 13
where R1, R3 and R2 are inner radius, junction radius and outer radius of compound cylinder. Remember that
(numerical sum of hoop stresses developed at the common radius of two cylinders).
Thick Spherical Shell:
To determine radial and hoop stresses in a thick spherical shell subjected to an internal pressure, let us first
learn about the radial and circumferential strains in such axi-symmetric cases. Consider a thin disc of inner
radius R1 and outer radius R2 subjected to an internal pressure p. A small element abcd is subtending an
angle δθ at the centre is considered as shown in Fig. 6.12.
Radius oa = r
Radius od = r + δr before the application of internal pressure.
After the application of internal pressure, say
r → changes to r + u
δr → changes to δr + δu
These strains are tensile if positive.
On the element of a spherical shell, say stresses are σr + δσr at radius r + δr
Circumferential stress on element = σc
Bursting force P on the elementary shell (taking the projected area of hemispherical portion)
Figure 6.12
πr2σr − π (r + δr)2 (σr + δσr)
Resisting force = σc × 2 πrδr , where δr is the radial thickness.
For equilibrium
πr2σr − π (r + δr)2 (σr + δσr) = σc × 2πrδr
2σrδr − rδσr = 2σc δr (neglecting the small quantities of higher order terms)
Three principal stresses at any point in the elementary shell are:
1.radial pressure, σr (compressive),
2.hoop stress, σc (tensile), and

14. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 14
3.hoop stress, σc on another plane (at right angle to the first plane).
From initial Eqs. (22) and (23)
Substituting for εr and εc from Eqs. (25) and (26)
After simplification we get
From Eq. (6.24)
Differentiating Eq. (6.29) we get
Substituting for σc and dσc/dr in Eq. (6.24) and after multiplication we get
Let us put dσr/dr = k, then we have
So ln k = −4 ln r + ln C1
where C1 is a constant of integration
or
Integrating Eq. (6.30),
or

15. Mechanics of Solids (NME-302) Unit-V
Unit-V Page 15
Let us put constants C1 = 6B and C2 = ‒A
Then,
For a thick spherical shell subjected to an internal pressure p, the boundary conditions are