MECH 303 Chapter4
Chapter 4 Solution of Plane Problems in Polar Coordinates
4.1 Equilibrium equations in polar coordinates
O
x

r
d

P r
r
r
K
dr
A
Kr
 r 
B
y
 
 
d

C

 r  r d

 r
dr
r
r 
 r
dr
r
Equilibrium condition of the element in radial, tangential directions, and
the equilibrium of moment lead to
  r 1  r  r   
 Kr  0
 r  r  
r

 1     r  2 r  K  0

 r 
r
r
and  r   r , Kr and K are the radial and circumferential
components of body force.
* how to remember and to derive the above equilibrium equations
4 -1
MECH 303 Chapter4
4.2 Geometrical and Physical equations in polar coordinates
O
x

r P
d
O
dr
ur
x

r
d’
P
P’
B
u
A
B”
y
A
P”
A’
B’
dr
B
A”
y
(b)
(a)
Definition of strains: r ,  , r , displacements ur , u (r, ).

If only radial displacement takes place as shown in (a), then
r 
u r
r
,  
ur
r
,  r 
1 ur
r 
 if only the circumferential displacement takes place in (b), then we have
 r  0 ,  
1 u
r 
,  r 
u u

r
r
 In the general case , we obtain the geometric equations:
r 
u r
u r 1 u
1 ur u u



,  
,  r 
r 
r
r
r
r r 
 : Starting from that ur and u are continuous functions of r and , derive
the compatibility equations between r ,  , r .
4 -2
MECH 303 Chapter4
Physical equations: since polar coordinates r and  are orthogonal , just as
the rectangular coordinates x and y , the physical equations between stress
and strain must have the same form, i.e. ,
r 
1
 r    
E
 
1
    r 
E
 r 
1
21   
 r 
 r
G
E
For a plane strain problem, simply replace E by
E

,  by
2
1 
1 
4 -3
MECH 303 Chapter4
4.3 Stress function and compatibility equations in polar coordinates.
 x, y  in rectangular coordinates and  r, 
Stress function    (.)
in polar coordinates
 The relations between polar and rectangular coordinates:
r 2  x 2  y 2 ,   arctan
x  r cos
   r 
 x  r x  
   r 



 y r y 
, y  r sin 
y
x
( x , y)  (r ,  )
(r ,  )  ( x , y)

 sin 
 cos 

x
r
r

 cos 
 sin 

y
r
r




    x, y    r  x, y ,  x, y 
~
 2
 2
 2

......

.....
,
,
 ...... functions of r, 
y 2
x 2
xy
Derive:  x 
 2
 2
 2







f
r
,


f
r
,

,
,



 f 2 r , 
y
1
xy
y 2
x 2
xy
Prove that:

 2
1  1  2


0
,





 2
r
x  0

2  0
r

r

y
r  2

 2
 2



0
,







y  0
 0
x 2
r 2


  1  

  0, r   xy  0   
r  r  

: It is easy to verify that the above stress expressions satisfy the equations
of equilibrium when Kr = K = 0.
4 -4
MECH 303 Chapter4
: On the other hand, we can prove that
 2  2  2 1  1  2
,




x 2 y 2 r 2 r r r 2  2
So the compatibility equation in rectangular coordinates
 2
2
 2  2
y
 x
2

   0

becomes that ( in polar coordinates)
 2 1 
1 2
 2 
 2
r r r  2
 r
2

   0

: In summary : In solving a plane problem in polar coordinates, it is
necessary to solve only the above compatibility equation for stress
function , then obtain the stress components which satisfy the boundary
condition and condition of single-valued displacements. Most of case, the
inverse or semi-inverse methods are employed.
:
4 -5
MECH 303 Chapter4
4.4 Coordinate transformation of stress components
O


x
r
r
r
r
B
x
y
y
c
r
xy
a
x
y
(a)
A
r b

O

y
y
y
x
x
B
r

xy
yx
a
xy
x
c
(b)
A
x
r
r
b
y
By equilibrium analysis, we have following relations between stress
components in two-coordinates systems:
:  r 
 
 r
:
 x  y
2
 x  y
 x  y
2
 x  y
cos 2   xy sin 2
cos 2   xy sin 2
2
 x  y
  xy cos 2 
sin 2
2
x 
2
 r  
2
y 
 xy 
or



 r  
2
 r  
2
 r 
2

cos 2   r sin 2
 r  
2
cos 2   r sin 2
sin 2   r cos 2
 x   r cos 2     sin 2   2 r sin  cos
4 -6
MECH 303 Chapter4
 y   r sin 2     cos 2   2 r sin  cos
 xy   r    sin  cos   r cos 2   sin 2  
4 -7
MECH 303 Chapter4
4.5 Axisymmetrical stress and corresponding displacements
 Now we apply inverse method to assume that  =  (r), then we have:
1 d
r 
,
r dr
d 2
  2 ,
dr
 r   r  0
The compatibility Equation reduces to
2
 d2 1 d 
 2 
  0
r dr 
 dr
The general solution of this ordinary differential equation is
 = A ln r + B r2 ln r + C r2 + D,
where A, B,C, D are arbitrary constants, the stress then becomes
A
 B (1  2 ln r )  2C
r2
A
    2  B(3  2 ln r )  2C
r
 r   r  0
r 
It is seen that the stress distribution is symmetrical with respect to any
plane passing through the z axis, and this is said to be symmetrical about
the axis.
 By using physical equations, we obtain strain of axisymmetrical
problems:
u r
1
A

  r  1    2  1  3 B  21   B ln r  21   C 
r
E
r

u r 1 u
1
A


     1    2  3   B  21   B ln r  21   C 
r r 
E
r

1 u r u u

   r  0
   
4 -8
MECH 303 Chapter4
By integration, we obtain displacements: (using above 3 equations)
1
A

 1     21   Br ln r  1  1  3 Br  21   Cr   f  ,

E
r

4 Br 
u 
  f  d  f1 r 
E
ur 
 r  0  f1 r   r
df1 r  df  

  f  d
dr
d
where f() and f1(r) are respectively arbitrary functions of  and r
df1 r 

 f1 r   r dr  F

 df    f  d  F

 d
F is a constant
from which we obtain:
f1(r) = Hr+F, H is an arbitrary constant.
f() = I cos + K sin, I,K are arbitrary constants.
Finally, we have the displacement solutions for the axisymmetric
problems
ur 
1
A

 1     21   Br ln r  1  1  3 Br  21   Cr  I cos   K sin 

E
r

u 
4 Br 
 Hr  I sin   K sin 
E
The arbitrary constants A, B, C, H, I, K can be determined by known
(boundary) conditions. It is noted that the displacements are usually not
4 -9
MECH 303 Chapter4
symmetrical about the z axis.
For plane strain problem, the solution is obtained by simply replace
E

E
, 
2
1    .
1 


4.5 4.6 Hollow cylinder subjected to uniform pressures


qa
qa
r
r
qb
qb
a
b
(a)
(b)
(c)
 Evidently the stress distribution must be axisymmetrical, then the stress
components can be expressed by
A
 B1  2 ln r   2C ,
r2
A
    2  B3  2 ln r   2C ,
r
r 
 r   r  0.
The above arbitrary constants A, B, C are determined by
(1) boundary conditions:

 r r a  0;  r r b  0;


r
r a
 qa ,  r
r b
 qb
already satisfied
(2) single-valued displacement condition
4 -10
MECH 303 Chapter4
4 Br 
E

u 
 Hr  I sin   K sin 
,
 not sin gle valued , so we must take B  0

So finally, we have the stress components:
2
2
b
a
1  
  1
r
r q ,
 r   2
qa 
2 b
b
a
1  
  1
a
 
b
2
2
b
a
1  
  1
r
r q
   2
qa 
2 b
b
a
1  
  1
a
 
b
 r   r  0,  r ,  are principal stress.
If qb=0, only internal pressure qa acts,
~ give the stress expressions
~ give the stress expressions if b/a   (large body with a circular hole of
radius a)
If qa=0, only external pressure qb acts,
~ give the stress expressions
~ give the stress expressions when a/b  0.
4 -11