Chapter 11: Thick-wall cylinders

• End caps or hemispherical ends. Solution far from end caps.

Closed cylinder with internal pressure, external pressure, and axial loads.

Governing equations

• Equations of equilibrium

Stresses in thick-wall cylinder. Thin annulus of thickness dz.

( ) θθθθ σσσσσ=−= rrrr

rr rdrd

drdr or

Governing equations - continued

• Strain displacement

• Compatibilityor

• Hooke’s law

ru

rr ∂∂

=∈ ru

=∈θθ zw

zz ∂∂

=∈

rrdr

drθθ

θθ∈

=∈ −∈ ( ) rrrdrd

=∈∈θθ

( )[ ]

( )[ ]

( )[ ] constantTvE

TvE

TvE

rrzzzz

zzrr

zzrrrr

=∆++−=∈

∆++−=∈

∆++−=∈

ασσσ

ασσσ

ασσσ

θθ

θθθθ

θθ

1

1

1

Cylinder with close ends• Combining equilbrium, compatibility and Hooke’s law

• With equilbrium

• Integration,

• Where

10 21 1rr rr

d E T E T Cdr v vθθ θθ

α ασ σ σ σ∆ ∆⎛ ⎞+ + = + + =⎜ ⎟− −⎝ ⎠

( ) rCvTEr

drd

rr 12 2

1+

−∆

−=σσ

( ) 22

12

2

2 11 r

CCraTrdr

vrE r

arr +⎟⎟

⎠

⎞⎜⎜⎝

⎛−+∆

−−= ∫

ασ

( ) 22

12

2

2 111 r

CCra

vTETrdr

vrE r

a

−⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−∆

−∆−

= ∫αασθθ

⎟⎟⎞

⎜⎜⎛

∆+−= ∫b

TrdrEbpapC 1 22

21221

α2apC −=

⎠⎝ −− avab 112

Z-stress and strain

• With some algebra

( ) ( ) ( )( ) ∫ ∆−−−

+−∆

−−

+−−

=b

aclosedendzz Trdr

abvaE

vTE

abP

abbpap

22

2

2222

22

21

12

1αα

πσ

( ) ( )( ) ( ) ∫ ∆−+

−+−

−−

=∈b

aclosedendzz Trdr

abEabPbpap

abEv

22222

22

122

221 απ

Axial equilibrium of closed-end cylinder

( ) ( )2 21 22

b

zza

r dr P p a p bσ π π= + −∫

Constant temperature

• Equations simplify to

( )

( )constant

abbpap

constantab

Pab

bpap

ppabr

baab

bpap

ppabr

baab

bpap

rr

zz

rr

=−−

=+

=−

+−−

=

−−

+−−

=

−−

−−−

=

22

22

21

2222

22

21

21222

22

22

22

21

21222

22

22

22

21

)(2

)()(

)(

θθ

θθ

σσ

πσ

σ

σ

Does it reduce to thin cylinder equations? How fast?

• Internal pressure with average radius R and thickness t

• Check the other two!

( )

( ) ( ) ( ) ( )( ) ( )

( )

( ) ( )

2 2 21 1

2 2 2 2 2

2 2 2 2 2

22 22 2 2 2 4 2 2

3 2 211

2

2 21

1 1max min

( )( )( ) 2 ( 0.5 ) 1 /

0.5 0.5 0.25 1 0.5 /

1 0.5 /1 /2 2

1 / /2

/ 0.2 1.02 0.81

p a p a bb a r b a

b a b a b a Rt a R t R t R

a b R t R t R t R t R

p R t Rp R t Rt r t

p R t R R rt

p R p Rt Rt t

θθ

θθ

θθ θθ

σ

σ

σ σ

= +− −

− = + − = = − ≈ −

= − + = − ≈ −

−−≈ +

≈ − +

= = =

Example 11.1• A thick-wall cylinder is made of steel (E = 200 GPa and v = 0.29), has an

inside diameter of 20mm, and has an outside diameter of 100mm. The cylinder is subjected to an internal pressure of 300 MPa. Determine the stress components and at r = a = 10mm, r = 25mm, and r = b = 50mm.

• The external pressure = 0. Equations 11.20 and 11.21 simplify to

,

Substitution of values for r equal to 10mm, 25mm, and 50mm, respectively, into these equations yields the following results:

Stress r = 10 mm r = 25 mm r = 50 mm

-300.0 MPa -37.5 MPa 0.0

325.0 MPa 62.5 MPa 25.0 MPa

rrσ

rrσ

θθσ

θθσ

2p

)((

222

)222

1 abrbraprr −

−=σ

)()(

222

222

1 abrbrap

−+

=θθσ

Compare to thin cylinder approximation

• Use average radius of 60mm and thickness of 80mm. Then

1 300 60 22580

p r MPatθθσ ×

= = =

Problem 11.4• A long closed cylinder has an internal radius a = 100mm

and an external radius b = 250mm. It is subjected to an internal pressure = 80.0 MPa ( = 0). Determine the maximum radial, circumferential and axial stresses in the cylinder.

• By Eqs. (11.20) – (11.22), with a = 100mm, b = 250 mm, = 80.0 MPa, and = 0, we have at r = a = 100mm,

Mpaabbap

Mpapabbaprr

5.11010025025010080

80

22

22

22

22

1

122

22

1

=−+

=−+

=

−=−=−−

=

θθσ

σ

Mpaab

apzz 2.15100250

10080 22

2

22

2

1 =−

=−

=σ

1p 2p

1p 2p

Reading assignmentSections 11.4-11.5: Question: What is “ideal” about ideal

residual stress distributions?

Source: www.library.veryhelpful.co.uk/ Page11.htm