4 Thick Cylinders

 

TOPIC 4 THICK CYLINDERS References: 1.

Boresi AP Boresi AP,, Schm Schmidt idt RJ RJ.. Adva Advance nced d Mech Mechani anics cs of Mat Materi erials als , 6th edition, John Wiley & Sons, USA, 2003.

2.

Hea earn rn,, E. J., Me Mech chan aniics of ma matter eria ialls 1, 3rd  edition, ButterworthHeinemann, UK, 1997.

 

INTRODUCTION

• Examples of the use of thick-wall cylinders in industry: Pressure vessels

http://www.envirosep.com/ http://www .envirosep.com/APVCS/asme-pressur APVCS/asme-pressure-vessels.html e-vessels.html

Pipes

http://www.chillzoneindia.com/Piping.html 2

 

STRESSES IN CYLINDERS

• For a cylinder subjected to external loadings, at any point on the cylinder wall, three types of stresses are induced on three mutually perpendicular planes: Circumferential erential Stress (σ H) –directed 1. Hoop or Circumf along the tangent to the circumference and tensile in nature. Tends to increase the diameter. 3

 

STRESSES IN CYLINDERS

2. Longitudinal Stress (σL ) – directed along the length of the cylinder. cylinder. This is also tensile in nature and tends to increase the length. 3. Radial Stress (σr ) – It is compressive in nature. Its magnitude is equal to the pressure on the inside wall and zero on the outer wall if it is open to atmosphere. 4

 

STRESSES IN CYLINDERS σL p

σL

p

σH

σH

1. Hoop Hoop Stress Stress (sH)

σr

p

σL

σL

2. Longitudinal Stress (sL)

3. Radial Radial Stress Stress (sr) σr

•Element on the cylinder wall subjected to these three stresses

σH

•No change in shape (no shear) σL

σL

r   z  θ 

σH

σr

5

 

THIN VS. THICK CYLINDERS • Thin cylinders

• Thick cylinders

 –  –  –

Ri/t > 10 Radial stress = 0 No pressure gradient across

 – Ri/t    10

 –

the wall Hoop stress is constant across the thickness of the cylinder wall State of stress is membrane/ biaxial/plane stress

across the wall  – Hoop stress varies across the cylinder wall

 –

 – Radial stress is considered  – pressure distribution varies

 – State of stress is triaxial

6

 

OPEN VS. CLOSED CYLINDERS • Open cylinders (no end caps) and unconstrained: axisymmetric deformations which ar are e independent of z.

• Closed cylinders (constrained by end caps): deformation and stress will depend on the axial deformation coordinate z in the vicinity of the supports or junction between the cylinder and end caps.

7

 

LIMITATIONS OF THE ANALYSIS

• Only at locations far from the end caps where the effects of the constraints constraints imposed by the end caps are negligible. • In other words, the analysis is also applicable to open cylinders at any location.

• A function of radial coordinate r , since only axially symmetrical loads and constraints are admitted (con (const stant ant in θ and L direc directions) tions).. 8

 

LAMÉ’s THEORY (1833)

VALID FOR BOTH OPEN AND CLOSED CYLINDERS

• Assumptions: 1. Plane Plane sect section ionss of the cylind cylinder er normal normal to its its axis axis remain plane and normal even under pressure. 2. Hence, σL and εL remain constant throughout the thickness of the wall. 3. Since σL and εL are constant, it follows that the σH+σr at any point on the cylinder wall is a constant.. constant 4. The mat materi erial is homogeneous, homogeneous, isotropic and obeys H Hooke’ ooke’ss law. law. 9

 

LAMÉ’s THEORY (1833)

dθ/2

dθ/2

10

 

LAMÉ’s THEORY (1833)

• Radial equilibrium • For small angles, • Neglecting second-order small quantities

11

 

LAMÉ’s THEORY (1833)

• Constant εL (plane sections remain plane)

(away y from the cap ends) • Constant σL (awa

• Substituting (10.2) in (10.1) for σH

12

 

LAMÉ’s THEORY (1833)

• Multiplying through by r and rearranging

Note: σr is a function of r.

Integrating ating yields • Integr

13

 

LAMÉ’s THEORY (1833)

• Radial stress

 s H   2 A s r  s H    A  s avg 

sr

2 Since s r and s H  are principal stresses,

• Hoop stress

 B 2

r 

is thus the radius of the Mohr's circle

14

 

GENERAL CASE: BOTH INTERNAL AND EXTERNAL PRESSURES t

• Boundary conditions: 



r = Ri, σr = -Pi r = Ro, σr = -Po

Ri P

Pi

o

R o

σ r 

σ  H 

• Note: pressure causes negative radial stress since it produces radial compression (thinning) of the cylinder walls. 15

 

Since  A   A 

s r  

 B 2 i

 R

 B 2 o

 R

 B

  A 

Substituting B into (1)

2

r 

  P i

[You may also choose (2)]

(1)

 A 

  P o   ( 2)



 B 2



B

2 i

  P i



( Pi  Po ) Ri2 Ro2  Pi ( Ro2  Ri2 ) Ri2



 Pi Ro2  Po Ro2  Pi Ro2  Pi Ri2



 Pi Ri2  Po Ro2

2 o

 R R  B 

2 i

( Ro2  Ri2 ) Ri2

2

 Ri 2 Ro 2  B ( Ro  Ri ) 2 i

( R  R ) R 2 o

Po ) Ri2 Ro2  P   A  ( Pi  i ( Ro2  Ri2 ) Ri2

( 2)  (1)  Pi  P o 

( Pi  Po ) Ri2 Ro2

( Pi  Po ) R R 2 i

( Ro2  Ri2 )

2 o

( Ro2  Ri2 ) 2

 Ro



2

 Ri

16

 

 s r    A 

 B r 2

 Pi Ri2  Po Ro2 ( Pi  Po ) Ri2 Ro2   2 2 ( Ro2  Ri2 ) r 2  Ro  Ri s r 



 Pi Ri2  Po Ro2  ( Po  Pi ) Ri2 Ro2 / r 2  R 2  R 2 o

i

 Note:  No te: If you use us e (1)-(2) (1 )-(2),, you will wi ll ge gett th thee fo follow llowing ing equ equati ation on::

 Pi Ri2  Po Ro2  ( Po  Pi ) Ro2 Ri2 / r 2 2 i

2 o

 R  R However, the above equation is less preferable since  Ro > Ri . Just multiply (-) to the denominator and nominator then s r 

as shown in the previous form is obtained. 17

 

Hoop stress s  H 

  A   B2 r 

 P R 2  P R 2 i

i

o

  R  R

s  H 



2 o

2 i

o

(P  P )R2 R2 i

o

i

o 2

 ( R  R ) r  2 o

2 i

 Pi R  Po R  ( Po  Pi ) R R / r  2 i

2 o

2 i

2 o

2

 Ro2  Ri2 18

 

LONGITUDINAL STRESS

• Exists in closed-end cylinders. • For open-end cylinders, σ  L=0. • Say, consider both internal pressure Pi and external pressure Po Po Ro Pi

i

19

 

LONGITUDINAL STRESS

• Horizontal Horizontal equilibrium equi librium  Pi   Ri2  Po   Ro2  s L   ( R o2  Ri2 ) • Longitudinal stress σL s  L

  s H

  A 

s  L



 Pi Ri2  Po Ro2  Ro2  Ri2

 s r 

2 • Note: END CYLINDERS.is ONLY VALID FOR CLOSED-

20

 

LONGITUDINAL STRESS

• For open-end cylinders, σ  L=0. But

  s H

 A 

 s r  2

is not equal to 0. In I n other words, for for open-end cylinders, s      A  L

• Nevertheless, it should be noted that based on Lamé’s theory theor y,  A  2 is always true, which applies to both open and closed-end cylinders.   s H

s r 

21

 

CLOSED-END CYLINDER

Case 1: For Po > Pi

0

σH

-50

    )    a    P -100    M     (    s    s    e -150    r    t    S

σL

σr  σ

σr

τ

σL σH

Note: Open cylinder

-200

-250 0

0.2

0.4

0.6

0.8

1

σH

A=σavg σr  σL σ

normalised radius

Inner radius

Outer radius

NOTE:

1.

Variations of σH and σr are parabolic across the cylinder wall.

2.

σH and σr are principal stresses, and σL is σavg.

3.

radius. Th The e llar arge gest st diff differ eren ence ce be betw twee een n σH and σr is at the inner radius.

τ

22

 

CLOSED-END CYLINDER Case 2: For Pi > Po 150

0

100

-5

    ) 50    a    P    M     ( 0    s    s    e    r    t    S -50

-10

σH σL σr

    )    a    P    M -15     (    s    s -20    e    r    t    S

σH σL σr

-25

-100

-30

-150

-35

0

0.2

Inner radius

0.4

0.6

normalised radius

0.8

1

0

Outer radius

Inner radius

0.2

0.4

0.6

normalised radius

0.8

1

Outer radius

NOTE:

1. 2.

Variations of σH and σr are parabolic across the cylinder wall. σH and σr are principal stresses, and σL is σavg.

3.

radius. Th The e llar arge gest st diff differ eren ence ce be betw twee een n σH and σr is at the inner radius.

23

 

SPECIAL CASE: INTERNAL PRESSURE ONLY

• Boundary conditions:

 A 





r = Ri, σr = -Pi



r = Ro, σr = 0

 Pi Ri2  Po Ro2  R  R 2 o

2 i

 Pi Ri2  R  R 2 o

2 i

 B 



s r 



 Pi Ri2 (1  Ro2 / r 2 )  Ro2  Ri2

 ( Ro / r ) 2  1    P   i  k 2  1  ( Pi  Po ) Ri2 Ro2 ( R  R ) 2 o

2 i

 Pi Ri2 Ro2 ( Ro2  Ri2 )

 s  L  (closed) For open-end cylinders, σ  L=0

s  H 



 Pi Ri2 (1  Ro2 / r 2 )  Ro2  Ri2

 ( Ro / r ) 2  1    P   i   2  k  1   Where k is the diameter ratio =

 D2 /D1=R2 /R1 24

 

SPECIAL CASE: INTERNAL PRESSURE ONLY

σ σr 

CLOSED-END CYLINDER

40 30

L

σH σ

τ

20

    )    a    P    M10     (    s 0    s    e    r    t    S

σH

Note:

σL σr

Open cylinder

-10 -20

σr  σL A

-30 0 Inner radius

0.2

0.4

0.6

normalised radius

0.8

σH

σ

1 Outer radius

τ

NOTE:

1.

Variations of σH and σr are parabolic across the cylinder wall.

2.

σH and σr are principal stresses, and σL is σavg.

3. 4.

At the inner edge (r minimum), minimum), the stresses (magnitude) are maximum maximum.. At the outer edge (r maximum), maximum), the stresses (magnitude) are minimum minimum..

25

 

SPECIAL CASE: EXTERNAL PRESSURE ONLY

• Boundary conditions:

 A 





r = Ri, σr = 0



r = Ro, σr = -Po

 Pi Ri2  Po Ro2 2 o

2 i

 R  R  Po Ro2  R  R 2 o

2 i

 B 



s r 

 

 Po Ro2 ( Ri2 / r 2  1)  Ro2  Ri2 2

 ( Ri / r ) 2  1 

 Po k  

( Pi  Po ) Ri2 Ro2 2 o

2 i

( R  R )  Po Ri2 Ro2 ( Ro2  Ri2 )

k 2  1



 Po Ro2 (1  Ri2 / r 2 ) s  H    Ro2  Ri2  ( R / r )  1    Po  k   i 2   k  1   2

2

Where k is the diameter ratio =

 D2 /D1=R2 /R1 26

 

SPECIAL CASE: EXTERNAL PRESSURE ONLY

At Ri

CLOSED-END CYLINDER

0

σH

σL

σr  σ

-10     ) -20    a    P    M     (    s    s -30    e    r    t    S-40

τ

σr σL σH

Note: Open cylinder

-50

σH

-60 0 Inner radius

0.2

0.4

0.6

normalised radius

0.8

σL σr  σ

A

1 Outer radius

τ

NOTE:

1.

Variations of σH and σr are parabolic across the cylinder wall.

2.

σH and σr are principal stresses, and σL is σavg.

3.

Th The e llar arge gest st diff differ eren ence ce be betw twee een n σH and σr is still at the inner radius radius..

27

 

EXAMPLE 1 A thick cylinder of 100 mm internal radius and 150 mm external radius is subjected to an internal pressure of 60 MN/m2 and an external pressure of 30 MN/m2. Determine the hoop and radial stres stresses ses at the inside and outside of the cylinder together with the longitudinal stresss if tthe stres he cylinder is assumed to have closed ends.

28

 

EXAMPLE 1

?

29

 

MAXIMUM SHEAR STRESS

• Take note that σH and σr are always the stresses, and σL and σavg. principal stresses, • Hence,

= radius of the Mohr’s circle Largest τmax at r = Ri (inner radius)

30

 

AILURE URE THEORIES – DUCTIL DUCTILE E MATERIA MATERIALS LS FAIL Note that Note that:: 1. The largest Mohr ’s circle always occur at Ri. Hence, to eval ev alua uate te the the cyli cylind nder er,, the the prin princi cipa pall str tres esse sess at Ri ar are e used used.. 2. The principal stresses must be σH,Ri and σr,Ri. 3.

σH,Ri could be +ve/-ve, but σr,Ri is alwa alway ys –ve. –ve.

1. Maximum-shear-stress theory (or Tresca criterion) 

s  H , Ri

r , Ri



s r ,R

i



s

s H ,R

i

s H , R

i

H , Ri



s r , R

s    s r , R    Y        i



s 

i

s Y    



s Y   

    

s H , R , s r , R i

s H ,R ,s r ,R h   ave i

i

i

have same signs

opposite signs

Internal pressure only case

31

 

AILURE URE THEORIES – DUCTIL DUCTILE E MATERIA MATERIALS LS FAIL 2. Maximum-distortion-energy theory (or Von Mises criterion) 2

s H , R

i

 s r2, R  s L2  s H , R s r , R i

i

i

 s    s H , R s L  s r , R s L    Y       i

2

i

3. Total-strain-energy theory (or Haigh criterion) 2 s H , R i



2 s r ,R i



2 sL

 2 (s H , R s r , R i

i

s   s H , R s L  s r , R s L )    Y       i

2

i

32

 

AILURE URE THEORIES – BRITTL BRITTLE E MATERIA MATERIALS LS FAIL 4. Maximum-normal-stress theory (or Rankine criterion) s H , R

i



s r ,R



i

s H , R

i



s ult   

s r ,R



s H ,R

i

i



s r , R

i



s ult   

5. Mohr’s failure criterion Stress Region

> 0, ssH,Ri  < 0 r,Ri 

sH,Ri  , s r,Ri 

≤0

Mohr ’s Circle

Failure

s H , R



i

sUT

i



i

sUC

s H , R s r , Ri

s r , R

 

1  

 s UC  /     UC   s   /   

33

 

AILURE URE THEORIES – BRITTL BRITTLE E MATERIA MATERIALS LS FAIL 6. Maximum-principal-strain criterion (or St. Venant criterion) s H , R

 (s r , R  s L )   

s ult 

s r ,R

 (s H , R  s L )   

s ult 

  (s H , R  s r , R  )  

s ult 

i

i

s L

i

i

i

i

 

 

 

Note: for open-end cylinders, σL = 0. Hence, failure failur e criteria 2, 3 and 6 could be further simplified to be plane-stress case. 34

 

CHANGE OF CYLINDER DIMENSIONS

(a) Change Change of diamet diameter er

35

 

CHANGE OF CYLINDER DIMENSIONS

(a) Change Change of diamet diameter er

36

 

CHANGE OF CYLINDER DIMENSIONS

(a) Change Change of diamet diameter er = diametr diametral al strain strain x original original diameter diameter = circumferential circumferential strain x original diameter  D   H  2r 

2r 

s H  v s r  s L     E 

(b) Change of length = longitudinal strain x original length  L   L  L 

 L

s L  v s r  s H        E  37

 

EXAMPLE 2 An external pressure pressure of 10 MN/m2 is applied to a thick cylinder of internal diameter 160 mm and external diameter of 320 mm. If the maximum hoop stress permitted2 on the inside wall of the cylinder is limited to 30 MN/m , what maximum internal pressure can be applied assuming the cylinder has closed ends? What will be the change in outside diameter when this pressure is applied? Given E = 207 MN/m2, ν = 0.29.

38

 

EXAMPLE 2

39

 

EXAMPLE 2

40

 

EXAMPLE 2

41

 

COMPARISON WITH THIN CYLINDER THEORY • Compare only σH since the σHmax is normally the limiting factor factor to determine the limit of D/t ratio within which is it safe to use the thin cylinder theory or not for a given pressure. THIN CYLINDER

THICK CYLINDER

s  H 



 PRi t



PDi   PK  2t 



at r  Ri , s H max

2

, where K  Di / t 

 2 2 i Ro2  2  P     R  Ro  Ri 

Substituting Substitutin g Ro=Ri+t and Di=2Ri s  H max

  K 2   P  2( K  1)  1

42

 

COMPARISON WITH THIN CYLINDER THEORY • Thin cylinder condition limit: K = Di/t = 20 THICK CYLINDER s  H max

2 2      K   P  2( K   1 =P  2(220  1 =10.52P    1 ) 0 1 )    

THIN CYLINDER  

 PRi

s  H 

t



PDi 2t 

 10 P 

%error

%error 

s H max

 s H 

s  H 

  10.52 P  10 P  100  100  5.2% 10 P 

43

 

COMPARISON WITH THIN CYLINDER THEORY

Error will be held within 5% if D/t  20 Thick cylinderwhen theoryD/t approaches thin cylinder theory becomes larger.

44

 

METHODS OF STRENGTHENING THICK CYLINDERS

 

COMPOUND CYLINDERS

• Large hoop stress variation across the cylinder wall when subjected to internal pressure. the refore ore • The material is theref not used to its best advantage. 46

 

COMPOUND UND CYLINDERS CYLINDERS – SAME MA MATERIAL TERIALS S COMPO • To obtain a more uniform hoop stress distribution, cylinders are often built up by shrinking one tube on to the outside of another. • When the outer tube contracts contracts on cooling, the inner tube is brought into a state of compression. The outer tube will conversely be brought into a state of tension. • A smaller total fluctuation on the resultant hoop stress is obtained.

47

 

COMPO COMPOUND UND CYLINDERS CYLINDERS – SAME MA MATERIAL TERIALS S

3 separate effects: i.

Shri Sh rink nkag age e pre press ssur ure e onl only y on on the the in insi side de cy cyli lind nder  er 

At r  R1,

s r 

0

At r  Rc ,

s r 

p

48

 

COMPO COMPOUND UND CYLINDERS CYLINDERS – SAME MA MATERIAL TERIALS S

3 separate effects: ii. Shrink Shrinkage age pressure only on the outside cylinder 

At r  R2 ,

s r 

0

At r  Rc ,

s r 

p

49

 

COMPO COMPOUND UND CYLINDERS CYLINDERS – SAME MA MATERIAL TERIALS S

3 separate effects: iii. Internal pressure only on the complet complete e cylinder 

At r  R2 ,

s r 

At r  R1 ,

s r 

0 1 P  

50

 

COMPO COMPOUND UND CYLINDERS CYLINDERS – SAME MA MATERIAL TERIALS S •

At any radius, apply the principle of superposition to obtain the corresponding hoop and radial stresses (total = internal pressure + shrinkage)

51

 

COMPO COMPOUND UND CYLINDERS CYLINDERS – SAME MA MATERIAL TERIALS S

• In practice this means that the compound cylinder is able to withstand greater internal internal pressure before failure occurs. Alternatively, a thinner compound cylinder • Alternatively, (with the associated reduction in material cost) may be used to withstand the same sa me internal pressure as the single thick cylinder it replaces.

52

 

COMPO COMPOUND UND CYLINDERS CYLINDERS – DIFFE DIFFERENT RENT MATERIA MATERIALS LS

• Diametra Diametrall strains strains must must be equal a att the common surface. • Ignoring longitudinal strain and stress,

53

 

HEATING OF COMPOUND CYLINDERS UNIFORM HEATING OF DIFFERENT MATERIALS • Two different materials upon uniform heating will expand at different rates and differential thermal stresses will be set up.

Both tubes being effectively effectively compr compressed essed radially (on their thickness) and an effective increase pt  in “shrinkage” pressure is thus introduced.

54

 

SHRINKAGE AND INTERFERENCE ALLOWANCE

• In the design of compound cylinders, it is important to relate the difference in diameter of the mating cylinders to the stresses it will produce.

• This difference in diameter at the “common” surface is normally termed the shrinkage or interference interference allowance.

55

 

SHRINKAGE AND INTERFERENCE ALLOWANCE Pressure at the junction of two cylinders owing to shrink fit

The material of the two cylinders not necessarily being the same

56

 

HUB AND SOLID SHAFT

• La Lamé mé equa equati tion onss

• In reality, reality, at r=0 (centre of the solid shaft), hoop and radial stresses cannot be infinite, infinite, the only solution to yield finite values for the stresses is B must be zero.

57

 

HUB AND SOLID SHAFT

• At the outer surface of the shaft

Therefore the σ H and σ r throughout a solid shaft are everywhere equal to the shrinkage or interference  pressure and both are compressive.

The τ max  = ½(σ1-σ2) is thus zero throughout the shaft.

58

 

FORCE FITS

• Another technique to form compound cylinders. Force-fitt technique: the interference allowance is • Force-fi small enough to allow the outer cylinder to be pressed over the inner cylinder with a large axial force the interference pressure set upofatthe thecontact • For common surface p and axial length surfaces L, the normal for force ce between the mating cylinders N = p(2πrL)

59

 

FORCE FITS

• Friction between the cylinders to be overcome by the applied for force ce cylinders F =μN  (where μ =coefficient of friction)  F

=2μπprL p

p Fapp

Fapp

60

 

AUTO-FRETT O-FRETTAGE” AGE” PLASTIC YIELDING – “AUT

• Inner radius is most highly stressed, thus yielding will start at this position. • For partial yielded inner wall, when the internal pressure is released, the elastic e lastic region tends to return to its original dimensions but is prevented by the permanent deformation of the yielded zone. • Hence, the elastic region is in residual tension whilst the plastic region is in residual compressio compr ession n – auto-fre auto-frett ttage. age. • Same effect as shrinking.

61

 

WIRE-WOUND THICK CYLINDERS

• Wound the cylinder with wire under tension • The resulting hoop and radial stresses developed in the cylinder will depend upon the way in which the tension in the wire varies.

62

 

CONCEPT TEST 1.

Which of the follo following wing Mohr’s circle represent the stress states at the inner radius of the thick cylin cylinder der which is subjected to internal pressure only?

σr

σL

σL σH

σr 

σH σ τ (a)

σr 

σH σ

σr 

σH σ

(b)

σr

σH

σr σ

τ

τ

σH

τ (c)

(d)

63

 

CONCEPT TEST 2.

Whi Whic ch o off th the e ffol ollo lowi wing ng Mo Mohr hr’’s cir circ cle represent the stress states at the inner radius of the thick cylin cylinder der which is subjected to external pressure only?

σr

σL

σL σH

σr 

σH σ τ (a)

σr 

σH σ

σr 

σH σ

(b)

σr

σH

σr σ τ

τ

τ

σH

(c)

(d)

64

 

CONCEPT TEST 3.

σr

Whi Whic ch o off th the e ffol ollo lowi wing ng Mo Mohr hr’’s cir circ cle represent the stress states at the inner radius of the thick cylin cylinder der which is subjected to both internal and external pressure, with P > P ? 2

σH σL

σL σH

σr

1

σ

σ σr 

σH τ (a)

σ

σr 

σH

σ

σH

r 

τ

σ

σH

r 

σ

τ

τ (b)

(c)

(d)

65

 

CONCEPT TEST 4.

σr

Whi Whic ch o off th the e ffol ollo lowi wing ng Mo Mohr hr’’s cir circ cle represent the stress states at the inner radius of the thick cylin cylinder der which is subjected to both internal and external pressure, with P > P ? 1

σH σL

σL σH

σr

2

σ σr 

σH τ (a)

σ

σr 

σH

σ

σr 

σH

σ

σH

r 

σ

τ

τ (b)

τ (c)

(d)