Unit III – Thick Cylinders€™s theory • Assumptions: • The material is homogeneous and isotropic. • Plane sections of the cylinder perpendicular to the longitudinal axis

Unit III – Thick Cylinders

Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE

2

• References:

Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi 2004.

Rattan.S.S., "Strength of Materials", Tata McGraw Hill Education Pvt. Ltd., New Delhi, 2011.

Rajput R.K., "Strength of Materials (Mechanics of Solids)", S.Chand & company Ltd., New Delhi, 2010.

Ramamrutham S., “Theory of structures” Dhanpat Rai & Sons, New Delhi 1990.

Contents

• Thick cylinders

• Compound cylinders


3

Difference between Thick cylinder and Thin cylinder

Thick Cylinder Thin Cylinder

1.Circumferential stress varies

along the thickness of the shell.

2. Radial stress is no longer

negligible since a thick cylinder is

required to have a heavy

internal pressure.

1. Circumferential stress ‘f’ is

constant throughout the

thickness of the shell.

2.Radial stress ‘p’ is negligible in

comparison of ‘f’ and ‘f0’.


4

Lame’s theory

• Assumptions:

• The material is homogeneous and isotropic.

• Plane sections of the cylinder perpendicular to the longitudinal axis remain plane under pressure.

That is longitudinal strain is the same at all points in the cylinder

wall. (i.e., it is independent of the radius)

Hence to satisfy the requirements of uniform longitudinal strain, we

have e0 = 1

𝐸𝑓0 −

𝑓𝑥

𝑚+

𝑝𝑥

𝑚= Constant

• 𝑓𝑥 − 𝑝𝑥 = Constant = 2A (say)

𝑓𝑥 = Circumferential stress (tensile)

𝑝𝑥 = radial pressure.

Dr.P.Venkateswara Rao, Associate Professor,

Dept. of Civil Engg., SVCE 5

• Figure 1a shows a thick cylinder subjected to internal and

external radial pressure.

Lame’s theory


6

𝑓𝑥 𝑓𝑥

𝑝𝑟 𝑟

𝑅

𝑝𝑥

𝑝𝑥+𝛿𝑝𝑥

Figure 1 (a)

𝛿𝑥

𝑥

𝑝𝑥


𝑓𝑥𝑙𝛿𝑥 𝑓𝑥𝑙𝛿𝑥

(b)

• Consider an angular ring of the cylinder, of internal radius x and thickness δ𝑥.

• Let the internal radial pressure on this ring be 𝑝𝑥 and external pressure 𝑝 + 𝛿𝑝𝑥 .

• On any small element of this ring, 𝑓𝑥 is circumferential stress.

Lame’s theory


7

𝛿𝑥

𝑥

𝑝𝑥


𝑓𝑥𝑙𝛿𝑥 𝑓𝑥𝑙𝛿𝑥

(b)

• The condition for equilibrium of one-half of thin ring are similar to those in the case of thin cylinder. Thus we have,

• the bursting force = (𝑝𝑥 2𝑥𝑙) − (𝑝𝑥 + 𝛿𝑝𝑥 ) 2 (x+ δ𝑥)𝑙

= 2𝑙 – 𝑝𝑥𝛿𝑥 − 𝑥𝛿𝑝𝑥 − 𝛿𝑥. 𝛿𝑝𝑥

= −2𝑙 𝑝𝑥𝛿𝑥 + 𝑥𝛿𝑝𝑥 ( Neglecting the products of small quantities)

• The resisting force = 2𝑓𝑥 𝑙 𝛿𝑥.

• For equilibrium we have , 2𝑓𝑥 𝑙 𝛿𝑥 = −2𝑙 𝑝𝑥𝛿𝑥 + 𝑥𝛿𝑝𝑥

• Or, 𝑓𝑥 =−[𝑝𝑥 + 𝑥𝛿𝑝𝑥

𝛿𝑥 ]

Lame’s theory


8

• In the limit when thickness of the element is reduced indefinitely,

𝑓𝑥 + 𝑝𝑥+𝑥𝑑𝑝𝑥

𝑑𝑥= 0 ------------(i)

• Another relation is obtained from the assumption that the longitudinal strain is independent of 𝑥. Thus from the equation 𝑓𝑥 − 𝑝𝑥 = 2𝐴 ---------(ii)

• And hence, 𝑓𝑥 = 𝑝𝑥 + 2𝐴 and by substituting it in equation (i), one can obtain the following relation;

(𝑝𝑥 + 2𝐴) + 𝑝𝑥+𝑥𝑑𝑝𝑥

𝑑𝑥= 0


9

Lame’s theory

• (𝑝𝑥 + 2𝐴) + 𝑝𝑥+𝑥𝑑𝑝𝑥

𝑑𝑥= 0

• or 𝑑𝑝𝑥

𝑑𝑥 =−

2 𝑝𝑥+𝐴

𝑥

•𝑑𝑝𝑥

𝑝𝑥+𝐴= −2

𝑑𝑥

𝑥


10

Lame’s theory

Integrating, log𝑒 𝑝𝑥 + 𝐴 =− log𝑒 𝑥2 + log𝑒 𝐵

Where log𝑒 𝐵 is a constant of integration.

Therefore, log𝑒 𝑝𝑥 + 𝐴 = log𝑒𝐵

𝑥2

𝑝𝑥 = 𝐵

𝑥2 – 𝐴

𝑓𝑥= 𝐵

𝑥2 + 𝐴 we know, 𝑓𝑥 − 𝑝𝑥 = 2𝐴 -----(ii) From (ii),

Problems

Problem 1.

The internal and external diameter of a thick hollow cylinder are 80 mm and 120 mm respectively. It is subjected to an external pressure of 40 N/mm2 and an internal pressure of 120 N/mm2. Calculate the circumferential and radial stresses at the mean radius.


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• Solution:

At 𝑥 = r, 𝑝𝑟 = 120 N/mm2

i.e., at r = 40mm, 𝑝𝑟 = 120 𝑁/𝑚𝑚2

At 𝑥 = 𝑅, 𝑝𝑅 = 40 N/mm2

i.e., at R= 60 mm, 𝑝𝑅 = 40 𝑁/𝑚𝑚2


12

Problems

• Lame’s equations are:

• 𝑝𝑥 = 𝐵

𝑥2 – 𝐴 −− − (1)

• 𝑓

𝑥 =

𝐵

𝑥2 + 𝐴 −− −(2)

Where 𝑝𝑥 = radial stress at a radius x from the centre of the cylinder.

𝑓𝑥 = circumferential stress at a radius ‘x’ from the centre of the cylinder


13

Problems

From equation (1),

• At 𝑥 = 𝑟, 𝑝𝑟 = 𝐵

𝑟2 – 𝐴 -----------(1a)

• At 𝑥 = 𝑅, 𝑝𝑅= 𝐵

𝑅2 – 𝐴 ----------------(1b)

• From (1a), 120 = 𝐵

402 − 𝐴

• From (1b), 40 = 𝐵

602 − 𝐴

• ------------------------------

• (1a) – (1b), 80 = B 1

402 −1

602


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Problems

• B= 80∗ 402∗602

602−402 = 230400

• From (1a), 120 = 230400

402 − 𝐴

• 120 = 144 − 𝐴

• A= 24

• Radial stress at the mean radius ‘50 mm’ is:

• 𝑝50 =𝐵

502 – 𝐴

= 230400

502 − 24 = 68.16 N/mm2


15

Problems

• Circumferential stress at mean radius ’50 mm’ is,

• 𝑓50 = 𝐵

502 +𝐴

= 230400

502 + 24 = 116.16 N/mm2


16

Problems

∴ 𝑝50 = 68.16 N/mm2 and 𝑓50=116.16 N/mm2

120 N/mm2

40 N/mm2

88 N/mm2

168 N/mm2

116.16 N/mm2

68.16 N/mm2

𝑝𝑥

𝑓𝑥 𝑝𝑟 = 120 𝑁/𝑚𝑚2 𝑝𝑅 = 40 𝑁/𝑚𝑚2

𝑓𝑟 = 168 𝑁/𝑚𝑚2 𝑓𝑅 = 88 𝑁/𝑚𝑚2

Problem 2

A cylinder has an internal radius of 200 mm and external radius of 300 mm. Permissible stress for the material is 15.5 N/mm2. If the cylinder is subjected to an external pressure of 4 N/mm2, find the internal pressure that can be applied.


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Problems

• Solution: r =20 mm; R= 300mm,

• 𝑓𝑟 = 15.5 N/mm2 ; 𝑝𝑅 = 4𝑁/𝑚𝑚2

• Lame’s equations are:

𝑝𝑥 = 𝐵

𝑥2 – 𝐴 −− − (1)

𝑓

𝑥 =

𝐵

𝑥2 + 𝐴 −− −(2)

From (1), 𝑝𝑅 = 𝐵

𝑅2 – 𝐴

4 = 𝐵

3002 − 𝐴 ---------------(1a)



Problems

• From (2), 𝑓

𝑟 =

𝐵

𝑟2 + 𝐴

15.5 = 𝐵

2002 + 𝐴 ----------------(2a)

4 = 𝐵

3002 − 𝐴 ---------------(1a)

________________________________

(1a) + (2a), 19.5 = B 1

3002 +1

2002

• B= 19.5∗ 3002∗2002

2002+3002 = 540000



Problems

• From (1a), 4= 540000

3002 − 𝐴

• A = 6-4 = 2

• Internal pressure, 𝑝𝑟 = 𝐵

𝑟2 – 𝐴

= 540000

2002 − 2 = 11.5 𝑁/𝑚𝑚2


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Problems

Problem 3

A pipe with internal diameter 400 mm is to carry a fluid pressure of 12 MPa. If the maximum stress in the material of the pipe is restricted to 110 MPa, calculate the minimum thickness of the pipe required.


21

Problems

• Solution:

• d= 400 mm, r = 200 mm

• 𝑝𝑟 = 12 𝑀𝑃𝑎 = 12 𝑁/𝑚𝑚2

• 𝑓𝑟 = 110 𝑀𝑃𝑎 = 140 𝑁/𝑚𝑚2

• Lame’s equations:

• 𝑝𝑥 = 𝐵

𝑥2 – 𝐴 −− − (1)

• 𝑓

𝑥 =

𝐵

𝑥2 + 𝐴 −− −(2)

• From (1), 𝑝𝑟 = 𝐵

𝑟2 – 𝐴



Problems

• 12 = 𝐵

2002 − 𝐴 ---------------(1a)

• From (2), 𝑓

𝑟 =

𝐵

𝑟2 + 𝐴

110 = 𝐵

2002 + 𝐴 ----------------(2a)

12 = 𝐵

2002 − 𝐴 ---------------(1a)

___________________________________

(2a) +(1a) is, 122= 2𝐵

2002


23

Problems

B = 122×2002

2 = 2440000

From (1a), 12 = 2440000

2002 − 𝐴

A = 49.

For minimum thickness, 𝑝𝑅 = 𝐵

𝑅2 – 𝐴 =0

2440000

𝑅2 − 49 = 0

𝑅2 = 2440000

49 = 49795.9

R= 223.15 mm.

• Thickness of the pipe is R-r = 223.15-200 = 23.15 mm.


24

Problems

• Problem 4

A pipe with internal diameter 400 mm is to carry a fluid at a pressure of 10 MPa. If the maximum stress in the material of the pipe is restricted to 150 MPa, calculate the minimum thickness of the pipe required.


25

Problems

• Solution:

d= 400 mm

r = 200 mm

𝑝𝑟 = 10 𝑀𝑃𝑎 = 10 𝑁/𝑚𝑚2

𝑓𝑟 = 150 𝑀𝑃𝑎 = 150 𝑁/𝑚𝑚2

• Lame’s equations:

𝑝𝑟 = 𝐵

𝑟2 – 𝐴 −− − (1)

𝑓

𝑟 =

𝐵

𝑟2 + 𝐴 −− −(2)


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Problems

400 mm

Pr=10 MPa

fr = 150 MPa

From (1), 10 = 𝐵

2002 − 𝐴 ---------------(1a)

From (2), 150 = 𝐵

2002 + 𝐴 ----------------(2a)

_____________________________________

(1a) + (2a), 160 = 2𝐵

2002

and hence, B = 160×2002

2 = 320× 104

From (1a), 10 = 320 ×104

2002 − 𝐴

A = 70.

Since PR = 0,



Problems

• Since PR = 0,

𝐵

𝑅2 – 𝐴 =0

•320 ×104

𝑅2 − 70 = 0

• R2 = 320 ×104

𝑅2 = 45714.2

• R=213.8

• Thickness of pipe = R-r = 213.8-200 = 13.8 mm.


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Problems

• In the thick cylinders when the cylindrical shells subjected to internal pressure, the circumferential stress (hoop stress) is maximum at inner circumference and it is decreases towards the outer circumference.

• Hence the maximum pressure inside the shell is limited corresponding to the condition that the hoop stress at the inner circumference reaches the permissible value.



Compound Cylinders

𝑝𝑥

𝑓𝑥

• But suppose the shell is made of shrinking one tube over the other. This will initially introduce hoop compressive stresses in the inner tube and hoop tensile stresses in the outer tube.

• If now the compound tube is subjected to internal pressure, both the inner and outer tubes will be subjected to hoop tensile stress, due to internal pressure alone.


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Compound Cylinders

𝑟1

𝑟2 𝑟3

• Adding the internal stresses caused while shrinking and the stresses due to internal pressure alone, the final hoop stresses in both the tubes can be determined .

• By this arrangement the hoop stresses throughout the metal will be more or less uniform.


31

Compound Cylinders

𝑟1

𝑟2 𝑟3

• Let r1 and r2 be the outer and inner radii of the compound tube. Let the radius at the junction of the two tubes be r3 .

• Let pj be the radial pressure intensity at the junction of the two tubes after shrinking the outer tube over the inner tube.


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Compound Cylinders

𝑟1

𝑟2 𝑟3

• Let Lamme’s relation for the outer tube be given by,

• 𝑝𝑥 =𝑏1

𝑥2 − 𝑎1

• and 𝑓𝑥 =𝑏1

𝑥2 + 𝑎1

• At 𝑥 = 𝑟1, 𝑝𝑥 = 0

• ∴ 0 =𝑏1

𝑟12 − 𝑎1 -----(1)

and at 𝑥 = 𝑟3,

𝑝𝑗 =𝑏1

𝑟32 − 𝑎1--------(2)

The constants 𝑎1 and 𝑏1 can be

determined from equations (1) and (2).


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Compound Cylinders

𝑟1

𝑟2 𝑟3

• Let Lamme’s relation for the inner tube be given by,

• 𝑝𝑥 =𝑏2

𝑥2 − 𝑎2


𝑥2 + 𝑎2

• At 𝑥 = 𝑟2, 𝑝𝑥 = 0

• ∴ 0 =𝑏2

𝑟12 − 𝑎2 -----(3)


𝑝𝑗 =𝑏2

𝑟32 − 𝑎2--------(4)

The constants 𝑎2 and 𝑏2 can be

Determined from equations (3) and (4).

Now the hoop stresses for the outer and inner tube can be easily determined.


34

Compound Cylinders

𝑟1

𝑟2 𝑟3

• Suppose the compound tube is subjected to an internal fluid pressure 𝑝0. For this analysis , the inner and the outer tubes will together be considered as one thick shell. The stresses due to internal fluid pressure alone can now be determined. For this condition let Lame’s relations be,

• 𝑝𝑥 =𝐵

𝑥2 − 𝐴

• and 𝑓𝑥 =𝐵

𝑥2 + 𝐴

• At 𝑥 = 𝑟1, 𝑝𝑥 = 0

• ∴ 0 =𝐵

𝑟12 − 𝐴 -----(5)


𝑝0 =𝐵

𝑟22 − 𝐴--------(6)

The constants 𝐴 and 𝐵 can now be evaluated. Dr.P.Venkateswara Rao, Associate Professor,


Compound Cylinders

𝑟1

𝑟2 𝑟3

𝑝0

The hoop stresses across the section can now be easily determined.

By algebraically adding, the hoop stresses caused due to shrinking to the hoop stresses caused by internal fluid pressure, the final hoop stresses may be determined.


36

Compound Cylinders

𝑟1

𝑟2 𝑟3

Problem:

A compound tube is composed of a tube 25 cm internal diameter and 2.5 cm thick shrunk on a tube of 25 cm external diameter and 2.5 cm thick. The radial pressure at the junction is 80 kg/cm2 . The compound tube is subjected to an internal fluid pressure of 845 kg/cm2. Find the variation of the hoop stress over the wall of the compound tube.


37

Compound Cylinders - Problems

Solution:

Stresses due to shrinking the outer tube to the inner tube:

Outer tube: Let Lamme’s relations for the outer tube be given by,

• 𝑝𝑥 =𝑏1

𝑥2 − 𝑎1


𝑥2 + 𝑎1

• At 𝑥 = 15 𝑐𝑚, 𝑝𝑥 = 0

• ∴ 0 =𝑏1

152 − 𝑎1 -----(1)

and at 𝑥 = 12.5 𝑐𝑚, 𝑝𝑗 = 80 kg/cm2

80 =𝑏1

12.52 − 𝑎1--------(2)

Solving equations (1) and (2), we get 𝑎1=181.8 and 𝑏1=40910




12.5 𝑐𝑚

Hoop stresses for the outer tube are given by,

• 𝑓12.5 =40910

12.52 + 181.8 = 443.6 𝑘𝑔/𝑐𝑚2(tensile)

• 𝑓15 =40910

152 + 181.8 = 363.6 𝑘𝑔/𝑐𝑚2(tensile)

Inner Tube:

Let Lamme’s relations for the inner tube

be given by, 𝑝𝑥 =𝑏2

𝑥2 − 𝑎2

and 𝑓𝑥 =𝑏2

𝑥2 + 𝑎2

• At 𝑥 = 12.5 𝑐𝑚, 𝑝𝑥 = 80 kg/cm2

• ∴ 80 =𝑏2

12.52 − 𝑎2 -----(3)

and at 𝑥 = 10 𝑐𝑚,

0 =𝑏2

102 − 𝑎2--------(4)


39


12.5 𝑐𝑚

Solving equations (3) and (4) we get 𝑎2 = −222 and 𝑏2 = −22220

Hence the hoop stresses for the inner tube are given by,

• 𝑓12.5 = −22220

12.52 − 222 = −364.2 𝑘𝑔/𝑐𝑚2(Compressive)

• 𝑓10 = −22220

102 − 222 = −444.2 𝑘𝑔/𝑐𝑚2(Compressive)

Stresses due to internal fluid pressure alone:

For this condition both the tubes together will

Be considered as acting as one cylinder.

Let Lamme’s relations for this condition be

• 𝑝𝑥 =𝐵

𝑥2 − 𝐴


𝑥2 + 𝐴




12.5 𝑐𝑚

• 𝑝𝑥 =𝐵

𝑥2 − 𝐴


𝑥2 + 𝐴

• At 𝑥 = 15𝑐𝑚, 𝑝𝑥 = 0

• ∴ 0 =𝐵

152 − 𝐴 -----(5)

and at 𝑥 = 10 𝑐𝑚, 𝑝𝑥 =845 kg/cm2

845 =𝐵

102 − 𝐴--------(6)

Solving equations (5) and (6), we get

A=676.1 and B= 152200


41


12.5 𝑐𝑚

Hence the hoop stresses due to internal

fluid pressure alone given by

• 𝑓10 =152200

102 + 676.1 = 2198.1𝑘𝑔/𝑐𝑚2(Tensile)

• 𝑓12.5 =152200

12.52 + 676.1 = 1650.1 𝑘𝑔/𝑐𝑚2(Tensile)

• 𝑓15 =152200

152 + 676.1 = 1352.2 𝑘𝑔/𝑐𝑚2(Tensile)

Hence due to the combined effect of shrinking the outer tube on the innner tube and internal fluid pressure the final hoop stresses will be as follows:

Outer tube: 𝐹15 = 363.6 + 1352.2 = 1715.8 kg/cm2 (tensile)

𝐹12.5 = 443.6 + 1650.1 = 2093.7 kg/cm2 (tensile)

Inner tube: 𝐹12.5 = −364.2 + 1650.1 = 1285.9 kg/cm2 (tensile)

𝐹10 = −444.2 + 2198.1 = 1753.9 kg/cm2 (tensile)


42


Initial difference in radii at junction is 𝜹𝒓′: 𝛿𝑟′

𝑟′=

1

𝐸

𝐵′

𝑟32

+ 𝐴′ −𝐵

𝑟32

+ 𝐴

Let the Lame’s equations for inner tube be

𝑝𝑥 =𝐵

𝑥2− A; 𝑓𝑥 =

𝐵

𝑥2+ A

and for the outer tube be

𝑝𝑥 =𝐵′

𝑥2 − 𝐴′; 𝑓𝑥 =𝐵′

𝑥2 + 𝐴′


43

Compound Cylinders

𝑟1

𝑟2 𝑟3

A thick cylinder of external diameter 40 cm and internal diameter 30 cm is shrunk on to another cylinder of external diameter 30 cm and 5 cm thick. If the radial pressure at the junction to shrink fit is 15 Mpa, calculate the initial difference in radii at the junction.


44


Unit III – Thick Cylinders€™s theory • Assumptions: • The material is homogeneous and isotropic. • Plane sections of the cylinder perpendicular to the longitudinal axis

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