ES240
Solid Mechanics
Fall 2007
2.3.2 Solution of 2D problems in polar coordinates
1. Transformation of stress components due to change of coordinates. A material particle is
in a state of plane stress. If we represent the material particle by a square in the (x, y ) coordinate
system, the components of the stress state are " xx ," yy ,! xy . If we represent the same material
particle under the same state of stress by a square in the (r ,! ) coordinate system, the
components of the stress state are # rr ,# !! ," r! . From the transformation rules, we know that the
two sets of the stress components are related as
# xx + # yy # xx $ # yy
+
cos 2! + " xy sin 2!
2
2
# + # yy # xx $ # yy
# !! = xx
$
cos 2! $ " xy sin 2!
2
2
# $ # yy
" r! = $ xx
sin 2! + " xy cos 2!
2
# rr =
2. Equations in polar coordinates. The Airy stress function is a function of the polar
coordinates, " (r ,! ). The stresses are expressed in terms of the Airy stress function:
$ rr =
! 2"
1 !"
+
,
2
2
r !#
r !r
# $$ =
' & '* #
! 2"
, + r) = ( $
!
2
'r % r') "
!r
The biharmonic equation is
& '2
'
' 2 #& ' 2) ')
' 2) #
$$ 2 +
+ 2 2 !!$$ 2 +
+ 2 2 !! = 0 .
r'r r '( "% 'r
r'r r '( "
% 'r
The stress-strain relations in polar coordinates are similar to those in the rectangular coordinate
system:
$ rr =
"
!
" rr
!
2(1 + # )
" r!
% # !! , # $$ = $$ % " rr , $ r! =
E
E
E
E
E
The strain-displacement relations are
# rr =
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%ur
u
%u
%u
%u
u
, # !! = r + ! , " r! = r + ! $ ! .
%r
r r%!
r%! %r
r
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Fall 2007
3. A stress field symmetric about an axis. Let the Airy stress function be ! (r ). The biharmonic
equation becomes
& d 2 1 d #& d 2' 1 d' #
$$ 2 +
!$
!=0 .
+
r dr !"$% dr 2 r dr !"
% dr
Each term in this equation has the same dimension in the independent variable r. Such an ODE is
known as an equi-dimensional equation. A solution to an equi-dimensional equation is of the
form
! = rm .
Inserting into the biharmonic equation, we obtain that
m 2 (m ! 2 ) .
2
The fourth order algebraic equation has a double root of 0 and a double root of 2. Consequently,
the general solution to the ODE is
! (r ) = A log r + Br 2 log r + Cr 2 + D .
where A, B, C and D are constants of integration. The components of the stress field are
$ rr =
$ %%
! 2"
1 !" A
+
= + B(1 + 2 log r )+ 2C ,
2
2
r !#
r !r r 2
" 2#
A
= 2 = ! 2 + B(3 + 2 log r )+ 2C ,
"r
r
+ r) = (
' & '* #
$
!=0.
'r % r') "
The stress field is linear in A, B and C.
The contributions due to A and C are familiar: they are the same as the Lame problem. For
example, for a hole of radius a in an infinite sheet subject to a remote biaxial stress S, the stress
field in the sheet is
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& , a )2 #
& , a )2 #
/ rr = S $1 - * ' !, / .. = S $1 + * ' ! .
$% + r ( "!
%$ + r ( "!
The stress concentration factor of this hole is 2. We may compare this problem with that of a
spherical cavity in an infinite elastic solid under remote tension:
/ rr
& , a )3 #
& 1 , a )3 #
= S $1 - * ' !, / .. = S $1 + * ' ! .
$% + r ( !"
$% 2 + r ( !"
A cut-and-weld operation. How about the contributions due to B? Let us study the stress field
(Timoshenko and Goodier, pp. 77-79)
! rr = B(1 + 2 log r ),
" !! = B(3 + 2 log r ), " r! = 0 .
The strain field is
# rr =
1
(" rr % !" $$ ) = B [(1 % 3! )+ 2(1 % ! )log r ]
E
E
# $$ =
1
(" $$ % !" rr ) = B [(3 % ! )+ 2(1 % ! )log r ]
E
E
" r! = 0
To obtain the displacement field, recall the strain-displacement relations
# rr =
%ur
u
%u
%u
%u
u
, # !! = r + ! , " r! = r + ! $ ! .
%r
r r%!
r%! %r
r
Integrating ! rr , we obtain that
ur =
B
[2(1 # " )r log r # (1 + " )r ]+ f (! ),
E
where f (! ) is a function still undetermined. Integrating " !! , we obtain that
u# =
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4 Br#
! " f (# )d# + g (r ),
E
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where g (r ) is another function still undetermined. Inserting the two displacements into the
expression
$ r# =
"u r "u# u#
+
!
=0,
r"#
"r
r
and we obtain that
f ' (# )+ " f (# )d# = g (r )! rg ' (r ).
In the equation, the left side is a function of ! , and the right side is a function of r.
Consequently, the both sides must equal a constant independent of r and ! , namely,
f ' (# )+ " f (# )d# = G
g (r )! rg ' (r ) = G
Solving these equations, we obtain that
f (! ) = H sin ! + K cos !
g (r ) = Fr + G
Substituting back into the displacement field, we obtain that
B
[2(1 # " )r log r # (1 + " )r ]+ H sin ! + K cos !
E
.
4 Br!
u! =
+ Fr + H cos ! # K sin !
E
ur =
Consequently, F represents a rigid-body rotation, and H and K represent a rigid-body translation.
Now we can give an interpretation of B. Imagine a ring, with a wedge of angle ! cut off. The
ring with the missing wedge was then weld together. This operation requires that after a rotation
of a circle, the displacement is
v(2" )# v(0 ) = !r
This condition gives
B=
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"E
.
8!
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Fall 2007
This cut-and-weld operation clearly introduces a stress field in the ring. The stress field is
axisymmetric, as given above.
4. A circular hole in an infinite sheet under remote shear. Remote from the hole, the sheet is
in a state of pure shear:
" xy = S , ! xx = ! yy = 0 .
The remote stresses in the polar coordinates are
# rr = S sin 2! ,
# !! = $ S sin 2! , " r! = S cos 2! .
Recall that
$ rr =
! 2"
1 !"
+
,
2
2
r !#
r !r
# $$ =
' & '* #
! 2"
, + r) = ( $
!.
2
'r % r') "
!r
We guess that the stress function must be in the form
" (r ,! ) = f (r )sin 2! .
The biharmonic equation becomes
& d2
d
4 #& ( 2 f
(f
4f
$$ 2 +
' 2 !!$$ 2 +
' 2
rdr r "% (r
r(r r
% dr
#
!! = 0 .
"
A solution to this equi-dimensional ODE takes the form f (r ) = r m . Inserting this form into the
ODE, we obtain that
(( m ! 2 )
2
)(
)
! 4 m2 ! 4 = 0 .
The algebraic equation has four roots: 2, -2, 0, 4. Consequently, the stress function is
C
(
%
" (r ,! ) = & Ar 2 + Br 4 + 2 + D # sin 2! .
r
'
$
The stress components inside the sheet are
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+ 2"
1 +"
6C 4 D &
)
# rr = 2 2 +
= *' 2 A + 4 + 2 $ sin 2!
r +!
r +r
r
r %
(
# !! =
* 2" )
6C &
= ' 2 A + 12 Br 2 + 4 $ sin 2!
2
*r
r %
(
# r! = *
+ ) +" & )
6C 2 D &
2
'
$ = ' * 2 A * 6 Br + 4 + 2 $ cos 2! .
+r ( r+! % (
r
r %
To determine the constants A, B, C, D, we invoke the boundary conditions:
1. Remote from the hole, namely, r " ! , # rr = S sin 2! , " r! = S cos 2! , giving
A = ! S / 2, B = 0 .
2. On the surface of the hole, namely, r = a, # rr = 0, " r! = 0 , giving D = Sa 2 and
C = ! Sa 4 / 2 .
The stress field inside the sheet is
4
2
(
.a+
.a+ %
" rr = S &1 + 3, ) / 4, ) # sin 2!
-r*
- r * #$
&'
" !!
4
(
.a+ %
= / S &1 + 3, ) # sin 2!
- r * #$
&'
4
2
(
.a+
.a+ %
" r! = S &1 / 3, ) + 2, ) # cos 2!
-r*
- r * $#
'&
5. A hole in an infinite sheet subject to a remote uniaxial stress. Use this as an example to
illustrate linear superposition. A state of uniaxial stress is a linear superposition of a state of pure
shear and a state of biaxial tension. The latter is the Lame problem. When the sheet is subject to
remote tension of magnitude S, the stress field in the sheet is given by
/ rr
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& , a )2 #
& , a )2 #
= S $1 - * ' !, / .. = S $1 + * ' ! .
$% + r ( !"
$% + r ( !"
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Solid Mechanics
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Illustrate the superposition in figures. Show that under uniaxial tensile stress, the stress around
the hole has a concentration factor of 3. Under uniaxial compression, the material may split in
the loading direction.
6. A line force acting on the surface of a half space. A half space of an elastic material is
subject to a line force on its surface. Let P be the force per unit length. The half space lies in
x > 0 , and the force points in the direction of x. This problem has no length scale. Linearity and
dimensional considerations requires that the stress field take the form
" ij (r , ! ) =
P
g ij (! ),
r
where g ij (! ) are dimensionless functions of ! . We guess that the stress function takes the form
" (r ) = rPf (! ),
where f (! ) is a dimensionless function of ! . (A homework problem will show that this guess
is not completely correct, but it suffices for the present problem.)
Inserting this form into the biharmonic equation, we obtain an ODE for f (! ):
f +2
d2 f d4 f
+
= 0.
d! 2 d! 4
The general solution is
" (r , ! ) = rP(A sin ! + B cos ! + C! sin ! + D! cos ! ).
Observe that r sin ! = y and r cos ! = x do not contribute to any stress, so we drop these two
terms. By the symmetry of the problem, we look for stress field symmetric about ! = 0 , so that
we will drop the term ! cos ! . Consequently, the stress function takes the form
" (r , ! ) = rPC! sin ! .
We can calculate the components of the stress field:
# rr =
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2CP cos !
, # !! = " r! = 0 .
r
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Solid Mechanics
Fall 2007
This field satisfies the traction boundary conditions, # !! = " r! = 0 at ! = 0 and " = ! . To
determine C, we require that the resultant force acting on a cylindrical surface of radius r balance
the line force P. On each element rd! of the surface, the radial stress provides a vertical
component of force " rr cos !rd! . The force balance of the half cylinder requires that
P+
# /2
!%
rr
cos $rd$ = 0 .
"# / 2
Integrating, we obtain that C = "1 / ! .
The stress components in the x-y coordinates are
$ xx = %
2P
2P 2
2P
cos 4 ! , $ yy = %
sin ! cos 2 ! , # xy = %
sin ! cos 3 !
"x
"x
"x
The displacement field is
(1 $ # )P ! sin !
2P
cos ! log r $
"E
"E
(1 $ # )P ! cos ! $ (1 $ # )P sin !
2#P
2P
u! = $
sin ! +
sin ! log r $
"E
"E
"E
"E
ur = $
7. Separation of variable. One can obtain many solutions by using the procedure of separation
of variable, assuming that
" (r, ! ) = R(r )#(! ).
Formulas for stresses and displacements can be found on p. 205, Deformation of Elastic Solids,
by A.K. Mal and S.J. Singh.
A real-life example.
From: S. Ho, C. Hillman, F.F. Lange and Z. Suo, " Surface cracking in layers under biaxial,
residual compressive stress," J. Am. Ceram. Soc. 78, 2353-2359 (1995).
In previous treatment of laminates, we have ignored edge effect. However, we also know that
edges are often the site for failure to initiate. Here is a phenomenon discovered in the lab of Fred
Lange at UCSB. A thin layer of material 1 was sandwiched in two thick blocks of material 2.
Material 1 has a smaller coefficient of thermal expansion than material 2, so that, upon cooling,
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material 1 develops a biaxial compression in the plane of the laminate. The two blocks are nearly
stress free. Of course, these statements are only valid at a distance larger than the thickness of the
thin layer. It was observed in experiment that the thin layer cracked, as shown in Fig. 1.
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It is clear from Fig. 2 that a tensile stress ! yy can develop near the edge. We would like to know
its magnitude, and how fast it decays as we go into the layer.
We analyze this problem by a linier superposition shown in Fig. 3. Let ! M be the magnitude of
the biaxial stress in the thin layer far from the edge. In Problem A, we apply a compressive
traction of magnitude ! M on the edge of the thin layer, so that the stress field in thin layer in
Problem A is the uniform biaxial stress in the thin layer, with no other stress components. In
problem B, we remover thermal expansion misfit, but applied a tensile traction on the edge of the
thin layer. The original problem is the superposition of Problem A and Problem B. Thus, the
residual stress field ! yy in the original problem is the same as the stress ! yy in Problem B.
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With reference to Fig. 4, let us calculate the stress distribution ! yy (x,0 ). Recall that when a half
space is subject to a line force P, the stress is given by
# yy = $
2P 2
sin ! cos 2 ! .
"x
We now consider a line-force acting at y = ! . On an element of the edge, d! , the tensile
traction applied the line force P = #" M d! . Summing up over all elements, we obtain the stress
field in the layer:
& yy (x,0 ) =
2& M d%
sin 2 # cos 2 # .
$x
"t / 2
t/2
!
Note that " = x tan ! , and let tan ! = t / 2 x . Consequently,
d" =
x
d! ,
cos 2 !
and the integral becomes
& yy (x,0 ) =
2& M
%
#
2
! sin $d$ =
"#
2& M
%
#
1 " cos 2$
d$ .
2
"#
!
Integrating, we obtain that
* yy (x,0 ) =
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2* M &
1
#
$ ( ' sin 2 ( ! .
) %
2
"
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At the edge of the layer, x / t ! 0 and " = ! / 2 , so that ! yy (0,0 ) = ! M . Far from the edge,
t / x ! 0,
) yy (x,0) '
3
)M & t #
$ ! .
6( % x "
Thus, this stress decays as x !3 .
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