THE AXIAL COMPRESSION OF TALL SOLID CYLINDERS
WITH SMALL INTERFACIAL FRICTION
Gennady Mishuris1, Wiktoria Miszuris1, Sergei Alexandrov2,
1
Department of Mathematics
Rzeszow University of Technology, Poland
2
Institute for Problems in Mechanics
Russian Academy of Sciences
101-1 Prospect Vernadskogo, 119526 Moscow, Russia.
ABSTRACT: A regular perturbation technique is applied to determine an effect of
small friction on deformation of tall solid cylinders in the axial compression. The
profile of the free surface is illustrated for the rigid perfectly plastic constitutive law
and two conventional frictional laws after different amounts of deformation.
1. INTRODUCTION
Axial and plane-strain compression tests are widely used for the determination of
stress-strain data, ductile fracture conditions and friction laws, for example (1–3). In
the case of the determination of stress-strain curves, one of the principal limitations of
the test is caused by friction (3). Also, one of the ideal basic tests for determining
workability diagram assumes the axial compression with no friction (4). To approach
the ideal test conditions, recessed specimens with the groove filled with lubricant (5)
or special solid lubricant (6) are used. In either case, the thickness of the lubricant film
is sufficient to prevent or minimize breakdown of the film throughout the process. In
such conditions, the friction stress is very small. For example, assuming Coulomb’s
frictional law the coefficient of friction has been evaluated to less than 0.01 (6) or
even 0.001 (7). Nevertheless, barreling develops even under such frictional conditions
and has an effect on the interpretation of experimental data. For instance, the
triaxiality ratio should be corrected to accurately predict the corresponding point of
the workability diagram (5). Even though experimental techniques are available for
determining the shape of the free surface with a high accuracy (6,8), theoretical
methods are needed to find other quantities, for example the stress distribution. Of
such methods, the finite element method has been used in most cases, for example
(9,10). However, in the case of a very low level of friction stress, it is natural to adopt
a perturbation technique for solving the problem. In the theory of rigid/plastic solids,
perturbation techniques have been mainly used in combination with the method of
characteristics for finding instantaneous stress and velocity distributions (11-13). This
technique is inapplicable in the case under consideration because the equations are not
hyperbolic and continued deformation should be considered.
In the present paper, a first order solution is found for axisymmetric compression of a
solid cylinder. Even though the zero order solution (upsetting with no friction) is
trivial and can be obtained for quite a general constitutive law with no difficulty, the
first order solution is restricted to perfectly plastic material. This assumption is
reasonable for some practical applications (6) and significantly simplifies the solution.
The simplest possible constitutive behaviour, the Newtonian fluid, is assumed for the
lubricant film. Such a constitutive law is used for a class of lubricants (14). The
overall objective of the solution based on the two simplest models of the workpiece
2 - 65
material and lubricant is to understand major features of the solution behaviour that
can be useful for studying the problem with the use of more complicated constitutive
equations.
2. STATEMENT OF THE PROBLEM
Consider a solid cylinder of initial radius R0 and the height 2H0 subject to
compression between two parallel, well-lubricated plates, and introduce a cylindrical
coordinate system rz (Fig.1a). The side surface of the cylinder is stress free. The
radial and axial components of the velocity will be denoted by ur and uz, respectively,
and the non-zero components of the stress tensor in the cylindrical coordinates by rr,
, zz, and rz. The plates move with a velocity of the magnitude u0 which may
depend on the current half-height of the cylinder, H. Because of symmetry, it is
sufficient to find the solution at z  0 .
Obviously,
dH
 u 0
dt
(1)
uz  u0
(2)
where t is the time, and
at z  H . At the axis of symmetry, r = 0,
ur  0
and
 rz  0
(3)
and
 rz  0
(4)
Also, at the plane of symmetry, z = 0,
uz  0
Figure 1: Schematic shapes of cylinders compressed
without (a) and with (b) friction
The boundary conditions Eqns. (2) through (4) are valid independently of barreling of
the free surface. Barreling develops because of friction. It is supposed in the present
paper that the friction stress,  f , is small and is given by
 rz   f   (r , H ) ,
2 - 66
  1
(5)
at z = H. In general, the function  depends on physical quantities such as the tangent
velocity at the friction surface or the normal stress acting on this surface. However, it
will be seen later that for the case under consideration it can be represented in the
form of (5). If   0 barreling develops (Fig.1b) and the conditions at the free
boundary are
n  0
and
n  0
(6)
where  n and  n are the components of the traction vector. In the case of deformation
with no barreling, Eqn. (6) reduces to
 rr  0
and
 rz  0
(7)
at r = R, where R  R(H ) is the current radius of the cylinder. The shape of the free
surface can be described as
r  R    z, H 
(8)
A direct problem consists of prescribing the function  involved in Eqn. (5). Then,
the function  involved in (8) should be found from the solution. A great difficulty
here is that the function  is in fact unknown since the direct measurement of the
friction stress is very difficult (if even possible) and many methods of experimental
determination of the friction stress are based of measurements of geometric
parameters (for example, (1)). On the other hand, the current shape of the free surface
can be found experimentally with a high accuracy (for example (6,8)). In this case,
however, typical mathematical difficulties inherent to inverse problems arise.
Moreover, in the case of rate formulation of plasticity, it is not sufficient to have the
shape but is also necessary to have the rate of shape change. The latter is a difficult
experimental problem and, probably, the accuracy of such data is much low compared
to the data for the shape itself. In the present paper, the direct problem is solved.
Quite a general rigid/plastic constitutive law has the following form
ij   sij ,
(9)
2
sij sij   Y2
3
(10)
where  ij are the components of the strain rate tensor, sij are the components of the
stress deviator tensor,  is non-negative multiplier, and  Y is the tensile yield stress.
 Y may depend on the equivalent strain rate  eq , the equivalent strain  eq defined
respectively by
eq 
d eq
2
ijij ,
3
dt
 eq ,
(11)
and other internal variables. Equation (9) includes the incompressibility equation.
Equations (9) and (10) should be complemented with the equilibrium equations. In the
cylindrical coordinates, the non-trivial equations have the following form
 rr  zr 1

  rr      0 ,
r
z
r
2 - 67
 rz  zz 1

  rz  0 .
r
z
r
(12)
3. ZERO ORDER SOLUTION.
The zero order solution,   0 , is trivial for the constitutive equations in their general
form. The stress distribution is given by
 rz   rr     0
 zz   Y
and
(13)
This representation satisfies the stress boundary conditions of Eqns. (3) and (4), the
condition (5) at   0 , and the boundary conditions (7). Equation (10) is also
satisfied. The distribution (13) is compatible with the equilibrium equations (12) if  Y
is independent of z. Assume the velocity field in the form
u z  u 0
z
H
and
ur 
ru0
2H
(14)
Then, the components of the strain rate tensor are
 rr   
u0
,
2H
 zz  
u0
,
H
rz  0 .
(15)
It follows from Eqn. (15) the incompressibility equation is satisfied and from Eqns.
(15) and (11) that  eq and  eq are independent of the space coordinates. The latter
means that  Y is independent of the space coordinates and thus the equilibrium
equations are satisfied. It is possible to verify by substitution of Eqns. (13) and (15)
into (9) that the latter is also satisfied. Using (10) and the solution (15) the multiplier
in the flow rule (9) is determined in the form:

3u0
.
2 H Y ( eq ,  eq )
(16)
4. FIRST ORDER APPROXIMATION
The first order approximation is restricted to the perfectly plastic material:
 Y (eq ,  eq )   0  const .
(17)
All values of the order O( ) will be noticed by upper tilde. Thus, the flow rule and
incompressibility equation take now the form:
~
~
~
~ ~
~ ~ ~
 rz   ~srz ,  r     (~
s z  2~
sr ) ,  r   z    0 .
(18)
sr  ~
s ) ,  z  2 r   ( ~
The yield criterion (10) in the first approximation gives:
~
s ~
s  2~
s  0.
r

z
(19)
As a result, one can obtain the following linear relationships between the components
of the stress deviator and strain rate tensor:
~ ~
~
(20)
2 ~
sr  2 ~
s   r   ,
sz  0 .
Substituting the aforementioned formulae in the equilibrium conditions one can
obtain, after some algebra, the equation in terms of strain rate components:
2 - 68


 1 2
1   ~ ~  2
2
  1  ~

  r     2  2     zr  0 .

r  r  
r
 2 rz r z 
 z
(21)
At the top of the specimen, the boundary conditions follow from (2), (14) and (5):
~
srz (r, z) z H   (r, H ) ,
u~z (r, z ) z H  0 ,
(22)
At the symmetry plane, z  0 , and the symmetry axis, r  0 , the respective boundary
conditions (3) and (4) transform to:
~
(23)
u~ (r, z )  0 ,
s ( r, z )  0 .
z
z 0
rz
z 0
and:
u~r (r, z) r 0  0 ,
~
srz (r, z) r 0  0 .
Finally, the free boundary conditions (6) can be written in the main terms as:
~  ~
s
 0,
r r R
~
srz
r R
  z z ( z, H ) ,
(24)
(25)
(26)
where ~ is the first order approximation for hydrostatic pressure. Note that the
boundary condition (25) has to be considered as the initial condition in order to solve
Cauchy problem for the following equilibrium equation:
 ~  ~ 1   1  ~ 1   u~z u~r 
(27)
    s z    u r 


,
r
r
r  r r 
2 z  r
z 
when the first term approximation for the velocities, strain rate tensor and the stress
deviator is found, so that the right-hand side of (27) is known. Note also that the
function  ( z , H ) satisfies the following additional condition due to material
incompressibility:
H
 ( z, H )dz  0 .
(28)
0
To solve the boundary value problem (21) – (24), Eqn. (21) should be transformed,
taking into account incompressibility condition:

  1 ~
  u r  u~z  0 ,
z
 r r 
(29)
4 ~
2 ~
u

Mu r  M 2 u~r  0 ,
4 r
2
z
z
(30)
to the form:
Here we have introduced a new notation for the differential operator M by the
formula:
M
2 1 
1

 2 .
2
r r r
r
2 - 69
(31)
Incorporating a well known property of the Bessel functions (e.g. (15)),
MJ 1 (mr)  m 2 J 1 (mr) , it is natural to seek for the solution to the problem by Fourier
decomposition method in form:
(32)
u~ (r, z)  J (mr)U ( z) .
r
1
r
For the function U r (z ) one immediately obtains the following ordinary differential
equation:
2
4
2 
Ur  m
U r  m 4U r  0 ,
4
2
z
z
(33)
which has four linearly independent solutions of the form U r ( z )  e az . Here
parameters a j ( m)  m j  im j , j  1,...,4 can be easily calculated:
 3 1 
 3 1 

 3 1 
3 1 
a1  m
 i  , a 2   m
 i  , a 3  m 
 i  , a 4  m
 i  .
2
2
2
2
2
2 
2
2







As a result, the solution to equation (24) has the form:
mz 3
 mz 3 
mz
mz   2 
mz
mz 
u~r ( r, z )  J 1 (mr) e 2  C1 cos
 C2 sin

e
 C4 sin

 C3 cos
 .
2
2 
2
2 



Equation (29) together with boundary conditions (22)1 and (23)1 allows us to find the
velocity component:
m z 3
 mk z 3 ( k )
 k
mk z
m z
~
2
u z ( r, z )  Fk ( r ) e
D1 sin
 e 2 D2( k ) sin k  ,
2
2 

(34)
where
mk 
2k
, k  1,2,....
H
(35)
Incorporating all other homogeneous boundary conditions (22) - (24) one can finally
obtain the solution to the problem in terms of velocities:

 m z 3   mk z   1 
 sin 
u~z (r , z )   Dk cosh  k
   J 1 (mk r ),

k 1
 2   2  r r 
(36)


 m z 3   mk z 
 mk z 3  mk z 
m

 cos


u~r (r , z )   k Dk  cosh  k

3
sinh

  2 
 2  sin 2  J 1 (mk r ).

2
k 1 2






where unknown up to now constants D k have to be found from the given friction law
(22)2 using the following representation for the shear stress:

 m z 3   mk z 
m z 3
m k2 3
m z
 sin 
 cos k  J 1 (m k r )
Dk  3 cosh  k
  sinh  k

 2   2 
 2 

2
2 
k 1





or in term of the given function  :
1
~
s rz 
2

2 - 70
 (r , H ) 
m H 3
3 
 J 1 (mk r ) .
(1) k mk2 Dk sinh  k



4 ( H ) k 1
2


(37)
When the constants D k are calculated, the deviation of the spaceman shape can be
found from (26) with the use of (28):
 ( z, H )  
H
u0
1
2u0

 (1)
3
k 1
k
m H 3

Dk J 1 (mk R ( H )) sinh  k
2 

 m z 3   mk z 
 sin 
mk Dk sinh  k
 J 1 (mk R ( H ))


3 k 1
 2   2 

(38)
Note that one can also control in experiments the compression force deviation
~
P  P  P from that found in the frictionless experiment ( P  R 2 0 ):
R
~
P  2R  z ( H , H )  2  r~z ( r, H )dr
(39)
0
5. NUMERICAL EXAMPLES AND DISCUSSION
Full film lubrication regime. In this case there is no “metal – metal” contact and the
simplest conventional law is:
srz (r, H )  1ur (r, H ) .
(40)
where 1 is an arbitrary constant and should be given. However, due to our initial
assumptions:
(41)
srz (r, z)  ~
srz (r, z) and ur (r, H )  ur (r, H )  u~r (r, z) ,
or, comparing (40) and (41), we immediately conclude that the obtained solution
makes sense if and only if 1 ~  . In other words, for the friction law chosen, our
solution represents a perturbation with respect to the small friction parameter, which
results in a small deviation of the free surface from cylindrical shape (Fig.1b). Later
we will assume:
  1
(42)
Taking this fact into account condition (40) can be rewritten in the form:
~
s (r, H )  u (r, H ) .
rz
r
(43)
Then equation (22)2 determining unknown constants Dk can be rewritten in the
following manner:

mk2 3
1
Dk

2 ( H ) k 1 2






ru
 2k 
3 cosh k 3 sin k   sinh k 3 cos(k ) J 1 
r   0
2H
 H 
From the theory for the series of the Bessel function (see for example (15)), one can
show that:

4
 k ( 1) J kx   x
k
k 1
1
for all
2 - 71
x  (0,  ) .
(44)
As a result, the following representation of the shear stress can be found for the
friction law under consideration:






1u0
 3 cosh  mk z 3  sin  mk z   sinh  mk z 3  cos mk z   J 1 mk r 
 2   2 
 2   2 

k 1 k sinh k 3 






srz (r , z )  


while the shape deviation from the perfect cylinder is given by:
 ( z, H )  
2 1 ( H )  (1) k
J 1 (mk R ( H )) 

2
3
k 1 kmk
4  1 ( H ) H
3
 m z 3   mk z 
1
 sin 
sinh  k

  2  J 1 (mk R ( H )).
2
k 1 kmk sinh k 3




(45)

Tresca and Coulomb friction laws. Other specific friction laws can be combined
with the theory developed. For the Tresca law:
srz ( r, H )  2 0 , or ~
(46)
srz ( r, H )   0 .
The respective shear stress distribution is given by the following representation:
s rz 
(47)

2  2 0

 k sinh  (2k  1) 3  

k 1
 m z 3   m 2 k 1 z 
 m 2 k 1 z 3 
m z
 sin 

 cos 2 k 1  J 1 ( m 2 k 1 r ),
3 cosh  2 k 1

sinh

  2 


2
2
2 




while the shape deviation is:
 ( z, H ) 
8  2 0  H
u0 3
4  2 0 
u0 3


k 1
 (2k  1)m
k 1
1
 (2k  1)m
2
2 k 1
J 1 ( m2 k 1 R ( H )) 
(48)
 m z 3   m2 k 1 z 
1
 sin 
sinh  2 k 1
  2  J 1 ( m2 k 1 R ( H )).
2
sinh

(
2
k

1
)
3
2 k 1




In case of the Coulomb friction law one can write:
srz (r, H )  3  z (r, H ) , or
~
srz (r , H )   0 .
(49)
In main terms (zero and first order approximations), the respective solution takes the
same form (47), (48) as for the Tresca law and the only change is that the friction
parameter  2 should be replaced with  3 .
A crucial point for the analysis is that it makes sense only under the following strong
restriction:
2R ( H )

H
or
R (H ) 1
 .
H
2
(50)
This means that the solution obtained is applicable only for tall specimens.
As a numerical example, we consider a tall specimen with the following initial
dimensionless parameters: H 0  2 , R0  0.5 ( R0 / H 0  0.25 ). Several stages defined
by H  1.9; 1.7; 1.5; 1.3; 3 2 are considered. The last one corresponds to the limit of
2 - 72
applicability of the solution obtained. In Figure 2 the normalized deviations from the
initially perfect cylindrical shape are depicted for several stages of deformation and
two frictional laws. All shapes shown have a point of maximum in a vicinity of the
friction surface. In general, such behavior is consistent with experimental results for a
class of specimens (9). However, these experimental results have been obtained for
hollow cylinders with high friction. It seems that the presence of a hole makes such
deformation possible. On the other hand, the profile of the free surface of cylindrical
specimens found experimentally in (8) was a circular or parabolic arc with a high
accuracy. However, the specimens tested were short. In particular, the condition (50)
is not satisfied for these specimens. The exact reason for the appearance of the
maximum (Fig.2) in the compression of solid cylinders is unclear and will be a subject
of subsequent investigation. A possible reason (or one of reasons) is that there can be
necessary to use a singular perturbation technique since, in general, a singular point
can appear at the intersection of the free boundary and the friction surface where the
state of stress may depend on the path of approaching to the point. The singularity has
been excluded in the present analysis.
Figure 2: Shape of the free surface of a specimen at several consecutive stages of
plastic deformation in the case of the full film lubrication regime (a) and Tresca
frictional law (b).
CONCLUSIONS
A perturbation technique has been applied to account for small friction in the axial
compression of tall solid cylinders. The complete (up to the first order) theory has
been developed for rigid/perfectly plastic materials but the basis for considering more
realistic constitutive laws, such as rigid viscoplastic and rigid plastic hardening laws,
has been formulated. Numerical examples have been obtained for two different
friction laws and showed an unexpected maximum on the perturbed free surface.
ACKNOWLEDGMENTS
G.M. is very thankful to the MMT-2004 Conference Organizers for a support. S.A.
acknowledges support from the Center for Iron and Steel Research (South Korea), the
Russian Foundation for Basic Research (grant 02-01-00419) and the program State
Support of Leading Scientific Schools, Russia (grant NSH-1849.2003.1).
2 - 73
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