EN0175 11 / 14 / 06

Linear elasticity solution in polar coordinates Typical problems: Stress around a circular hole in an elastic solid.

0σ

0σa

Boundary conditions:

Traction free @ ar = : 0=+= θθσσσ eee rrrrrvvv

, i.e. 0=rrσ , 0=θσ r

pa

b

Boundary conditions:

prr −=σ , 0=θσ r @ ar =

0=rrσ , 0=θσ r @ br =

Governing equation: 022 =∇∇ φ

In Cartesian coordinates: 02

2

2

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂ φ

yx

rθ

revθev

xev

yev

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EN0175 11 / 14 / 06

Proposition: use ( )θ,r instead of , ( )yx, ( )θφφ ,r=

θθ ∂∂

+∂∂

=∂∂

+∂∂

=∇r

er

ey

ex

e ryxvvvv

⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

⋅⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

=∇⋅∇=∇θθ θθ r

er

er

er

e rr112 vvvv

yxr eee vvv θθ sincos +=

yx eee vvv θθθ cossin +−=

θθθθ

eeeeyx

r vvvv

=+−==∂∂ cossin

ryx eeee vvvv

−=−−==∂∂ θθθθ sincos

It follows from above that 2

2

22

22 11

θ∂∂

+∂∂

+∂∂

=∇rrrr

Governing equation in polar coordinates: ( )θφφ ,r=

0112

2

2

22

222 =⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∇∇ φθ

φrrrr

Stress components in Cartesian coordinates:

2

2

yxx ∂∂

=φσ ,

yxxy ∂∂∂

−=φσ

2

, 2

2

xyy ∂∂

=φσ

φσσ 2∇=+ yyxx

Stress components in polar coordinates:

2

2

r∂∂

=φσθθ , ⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂

−=θφσ θ rrr

1, 2

2

2

11θφφσ

∂∂

+∂∂

=rrrrr

Equilibrium equations in polar coordinates:

01=+

−+

∂∂

+∂∂

rrrrrr f

rrrθθθ σσ

θσσ

021=++

∂∂

−∂∂

θθθθθθ σ

θσσ f

rrrr

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EN0175 11 / 14 / 06

Hooke’s law in polar coordinates:

( )θθνσσε −= rrrr E1

( )rrEνσσε θθθθ −=

1

θθ σνε rr E+

=1

Strain-displacement relations in polar coordinates:

rur

rr ∂∂

=ε , ruu

rr+

∂∂

=θ

ε θθθ

1,

ru

ruu

rr

rθθ

θ θε −

∂∂

+∂∂

=12

Example 1: Thick-walled pressure vessel

ab ip

0p

Since the problem is axisymmetric,

( )rφφ =

⎟⎠⎞

⎜⎝⎛=

∂∂

+∂∂

=∇r

rrrrrr d

ddd11

2

22

Boundary conditions:

irr prr

−=∂∂

=φσ 1

, 0=θσ r @ ar =

01 p

rrrr −=∂∂

=φσ , 0=θσ r @ br =

In mathematical description, the problem becomes an ordinary differential equation

0dd

dd1

dd

dd1

=φr

rrrr

rrr

with boundary conditions

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EN0175 11 / 14 / 06

( ) iapa −='φ

( ) 0' bpb −=φ

The above differential equation can be directly integrated and has the solution

DCrrBrrA +++= 22 lnlnφ

The constant term is nothing but a rigid body motion and can be neglected in stress analysis, i.e. .

D0=D

The tangental displacement associated with the term rBr ln2 comes out to be θθ rEBu 4

=

plus a rigid body motion, which is not a single-valued function. Actually, the term rBr ln2 represents a so-called disclination (think of gluing a cut-opened ring back into a circle). For the present problem, take . Therefore, the solution to thickwalled cylinder is 0=B

2ln CrrA +=φ

CrA

rrrr 212 +=

∂∂

=φσ

CrA

r222

2

+−=∂∂

=φσθθ

The constants are determined from the boundary conditions: CA,

iarrr pCaA

−=+==

22σ

02 2 pCbA

brrr −=+==

σ

The results are

( )22

022

abppbaA i

−−

=

22

20

2

2ab

bpapC i

−−

=

In the special case of : 00 =p

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−= 2

2

22

2

1rb

abpa i

rrσ ( 0< , compressive)

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EN0175 11 / 14 / 06

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−= 2

2

22

2

1rb

abpa i

θθσ ( 0> , tensile)

θθσ

rrσ

Consider the tensile hoop stress,

@ , br = 22

22abpa i

−=θθσ

@ ar = , ii pSCFpabba

⋅=−+

= 22

22

θθσ

Maximum stress occurs @ ar = .

22

22

abbaSCF

−+

= is called stress concentration factor.

In the case of a pressurized circular hole in an infinite medium, i.e. 00 =p and : ∞=b

2

2

rapirr −=σ , 2

2

rapi=θθσ

Example 2: Pressurized underground tunnel

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EN0175 11 / 14 / 06

ipa

P

Qd

The solution is discussed in Timoshenko’s book (Timoshenko and Goodier, 1987). The interesting features are that the maximum stress occurs at two potential sites

@ point P : ixx pad

a22

24−

=σ

@ point : Q ipadad

22

22

−+

=θθσ

For ad 3= , QPxx θθσσ =

If ad 3< , maximum stress occurs at ground point . P

If ad 3> , maximum stress occurs at the hole boundary point . Q

Example 3:

a∞σ

This is a special case of example 1. Take 0=ip , ∞−= σ0p , ∞→b . We find

∞−= σ2aA

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EN0175 11 / 14 / 06

∞=σC2

The stress fields are:

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∞ 2

2

1ra

rr σσ

⎟⎟⎠

⎞⎜⎜⎝

⎛+= ∞ 2

2

1raσσθθ

The maximum stress occurs at ar = , ∞== σσθθ 2

ar with a stress concentration factor of 2.

The general solution of in polar coordinates (i.e. for any 2D elasticity problem) can

be expressed as:

022 =∇∇ φ

( ) ( )( ) ( )

( ) (∑∑∞

=

+−−+∞

=

+−−+

−−

++++++++

++++++

⎟⎠⎞

⎜⎝⎛ −+++++=

2

2''2

2

2''2

'1

1'1

31

'1

1'1

31

11'0

20

20

200

sincos

sinlncosln

cos2

sin2

lnln

n

nn

nn

nn

nn

n

nn

nn

nn

nn nrdrcrdrcnrbrarbra

rrdrcrdrrbrarb

rcraardrrcrbra

θθ

θθ

θθθθθθφ

) The general solution can be conveniently used to solve boundary value problems. Example 4: Circular hole under uniaxial tension (remote)

a∞σ

Governing equation: 022 =∇∇ φ

Boundary conditions: irr prr

−=∂∂

=φσ 1

, 0=θσ r @ ar =

∞=σσ xx , 0=yyσ , 0=xyσ @ ∞=r

First, let us transform the remote stresses into polar coordinates

xx ee vv ⊗= ∞σσ

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EN0175 11 / 14 / 06

( )θσθσσσ 2cos12

cos2 +==⋅= ∞∞rrrr ee vv

( )θσθσσσ θθθθ 2cos12

sin 2 −==⋅= ∞∞ee vv

θσσσ θθ 2sin2∞−=⋅= eerr

vv

The above expressions suggest that the Airy stress function should have the form of

. Pick the corresponding expression in the general solution

associated with

( ) θφ 2cosln 221 rfrCrC ++=

θ2cos , we see ( ) '2

2'2

22 brararf ++= − (the 4r term is discarded since it

generates infinite stress at large r ). Applying the boundary conditions allow all the parameters to be determined. The solution is

θσσσφ 2cos2444

ln2

2

2

422

2

∞∞∞

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−++−=

ar

arrra

The associated stress fields are

θσσσ 2cos3112

12 2

2

2

2

2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−= ∞∞

ra

ra

ra

rr

θσσσθθ 2cos312

12 4

4

2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛+= ∞∞

ra

ra

θσσ θ 2sin3112 2

2

2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−−= ∞

ra

ra

r

∞σ

∞σ3

The maximum tensile stress occurs at ar = , 2πθ =

∞= σσθθ 3max

Therefore, the stress concentration factor is 3. The above problem can also be directly treated without knowing the general solution (next lecture).

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EN0175 11 / 14 / 06

For an elliptic hole under uniaxial tension (remote),

∞σ

a2

b2

⎟⎠⎞

⎜⎝⎛ += ∞ a

b1max σσθθ

The stress concentration factor is ⎟⎠⎞

⎜⎝⎛ +

ab21 , which depends on the aspect ratio of the elliptic

hole.

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