Mos unit iv

Mechanics of Solids (NME-302) Unit-V

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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


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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


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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


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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

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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


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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

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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.


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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


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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


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or, A = B/R12, putting the value in Eq.(6.12)

at r = R1, σr = 0

at r = R2 ,

Hoop Stress:

Note that 2R22 > (R2

2 + R12)

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


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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


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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


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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


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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

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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

Mos unit iv

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