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On the application of complex variables in plane plastic deformations
ON TEE APPLICATION OF COMPLEX VARIABLES IN PLANE PLASTIC DEFORMATIONS (0 PBIMENENII KOMPLEKSNYKH PEREMENNYKH PLASTICBESKOI DEFORMATSII) PMW
Vo 1. 24, No. 5,
v.
L.
1960,
pp.
K PLOSKOI
955-958
DOBROSOL' SKI1 (Moscow)
The conditions of the stress function being real, applicable to the soluA few such solutions are given as tion of inverse problems, are derived. illustrative examples. The conditions of the stress function being biharmonic in the plastic range are also determined. 1. Consider a homogeneous isotropic material in the plastic stress components satisfy the equilibrium conditions
and the Huber-Von Mises (or St.
Venant-Tresca)
Plasticity
state.
The
conditions
(1.2) where k is the yield tion F: ar=
limit
in shear Ill.
k a2F 2ay2’
k Gy= y-
We introduce a2F ax2 9
the stress
k a2F z ry = - -2 axay
func-
(1.3)
Equation (1.1) is satisfied identically, and (1.2) is left for the determination of function F. This equation can be represented in the complex variable I = x + iy as a2F pr=i
From physical
considerations
asp
F has clearly F (1, i) = F (2, i)
Now (1.4)
can be written
(1.4) to be real,
i.e. (1.5)
as SF T
= exp [ if31
1447
(1.6)
V. L.
1448
where @ = @(r,
t)
is a real
function.
*s
F (z, Z) == Theorem
F(z, t)
1.
defined
Integration
of
(1.6)
results
in
-dr) L exp ti@ (b ;)I dE + ZTJ(z) + x (4 s i La no
The necessary by (1.7)
Applying
‘skii
rl
and sufficient
to be real -
Proof.
Dobrosol
condition
for
(1.7) the function
is
azei@ a22
aze-iQ _ -
(1.8)
a22
the operator
a4 az2ai2
(1.9)
to the functions F(z, T) and F(EI- I) and taking into account (1.5) we obtain the proof of the necessity of the condition (1.8). To prove that this condition is also sufficient we write (1.8) as
aw -=a22aiz
a43 a22a22
The operator (1.9) is biharmonic, and consequently F by a biharmonic additive term which can be selected the theorem is proved. The condition (1.8). for practical presented in the Cartesian coordinates
(
cos@ g-g or in the polar
-zEE ax ay,1
(
__2
$sin@
coordinates
(r,
purposes,
can differ - from so that F = F. Thus F
may be conveniently
+$-(+$T+($+)
(1.10)
$1
+sinCO+2~)(%~+~i~)z-_~-(~)2)=0 Introducing
the notation
re-
exp [ i@ 1 = u + iv.
(1.8)
(1.11)
can be represented
as (1.12)
A particular solution @ = @(z, z) of (1.8), which does not contain arbitrary parameters. determines the stress components within a constant hydrostatic pressure. Theorem
2.
Application
Proof.
0 = 8(x, term pzi
of
complex
variables
in
plane
plastic
1449
deforaations
Theorem 1 and Expression (1.7) show that the solution y) determines a real stress function F(z, I) within an additive (p = const) which represents constant hydrostatic pressure.
Special interest attaches to stress functions harmonic equation in the plastic range.
which satisfy
the bi-
Theorea 3. The necessary and sufficient condition for the stress tion F to be a biharmonic function is that the function @ = @(z, I) satisfies the following system of equations:
f320 -_a2(3_2ao 8x2 By* Proof.
conjunction
ax
(ig-(g,"-2$&0
-!z!%=(),
ay
If F, represented by (1.7). with (1.6) the following SF
az2&2
=
0,
is a biharmonic holds: or
(1.13) function,
then in
a2ei@
--0 azz
(1.14)
Representation of this relationship in Cartesian coordinates (1.13). and thus to the proof of the necessity condition. Inverse sufficiency
arguments lead from (1.13) condition.
It is obvious or (1.10).
that the solutions
The class of solutions of (1.13) not empty. As an example we propose 1. @ = const. 2.
0 = -2
corresponding
tan-‘(y/r)
+ const.
to (1.14)
of
(1.13)
func-
leads
and to the proof
are also
is apparently the following
to a uniform stress
solutions
very small, solutions:
to
of the
of
but it
(1.8)
is
field;
This case was utilized
by Galin [ 2 I .
Note 2. Equations (1.10) and (1.12) possess the following features. If 8 = 0(x, y) is a solution of (1. lo), then v, = - @( - x, y) and Yz = - 0(x, -y), and consequently ys = 0(-x, -y) are also solutions. If w = u + iv is a solution of (1.12), then w1 = u1 + iv1 is also a solution of the same equation. Note 2. If some solution of (1.8) is known. then it is not necessary to find the stress function from (1.7) and to calculate the stresses from this function. It is easier to utilize the equilibrium equation (1.1) and the fact that @ determines uniquely the difference between the normal and shear stresses.
Z. We shall now determine a few particular solutions of the equilibrium equations in the plastic range by solving (1.10) and (1.11).
1450
V.L.
1. We seek
a solution
Dobrosol
in the
‘skii
form @ = a(y).
Equation
(1.10)
can thus
be written d’Q
Solving a case
this
of 5,
where the tion
=
2.
r1.c
-
stresses
2 1/i
are
dy
% = sin-‘(Ay
by two rough (Z4y + B)“,
-
represented
coordinate
+ B).
plates
Qv = p -
relative
to
form 8 = 8(a x + p y ) reduces
Seeking
(1.12)
we find
compressed
p -
in the
ing the
equation
a strip
dQ 2
( j-0
tall8
-_ dy”
to
This
corresponds
to
[ 1,3 1 AZ,
ZXU= ny+
the magnitude
the
previous
B k.
(2.1) A solu-
one by rotat-
axes.
a solution
in the
form 0 = - 24 + R(r)
reduces
Equation
to
Solving
this
equation
we find
R=arcsin(a-$),
and the
corresponding
6,
or
stress
components
a2 In [l/l
=-2a(p+d1-
(2.2)
@=-2rp+sin-1(~-~,)
-
are
a2 1/(1 -
-a2) r4 + 2abr2-
b2 $.
% + (1 -
F +s jf( 1 -
a2) r2 + ab] + a sin-’
T Pm =P--_r2
The displacement theory
[l
1
components
and independent C
a?=79 Here b f
0.
If
b =
of
The solution of
among all the
represented entering
solutions.
(2.3)
from a simplified
angle
Hencky-Von
Mises
4 are az) r* + 2abr” -
(2.4)
b”
0 then
% =dv$& by (2.3)
known closed-form
parameters
particular
b2 + p
b
U = dr - & 1/(1 v
C Ur = 7,
tion
found the
~2) r4 + 2abr2 -
into
(2.5)
and (2.4) is the most general soluBy specializing the values
solutions. this
solution,
we obtain
a series
of
Application
a) For
of
a = 0,
complex
variables
b) For a = 1, b = 0,
stress
%
for
1=-
a plastic
2arp + VI -
d) For a = 0, b = metric loading [ 1 1 :
E there
with the corresponding
(2.7)
z,, = 1
a2 (In r2 T 1) + p, results
loaded cases
faces.
results
in
TrQ = a
a most general
a new solution
case of axisym-
of a problem pre-
displacements
I$ =
Cl F
r
y$,==-dr
-
Along the contours
r = r0 the stresses
3=Pr2*Tm
z rrp ==i--$
ur = Sz t r0
(2.11)
~1/2Cr2J
are
and displacements (PI, p3 = const)
UT = dr, - ;;
0
~_ T/2CQ -
C2
d = 0 and r -+ by we obtain
=r %
[ 1] .
of the two previous
o = 1. b = c we obtain in [ 3 ] :
Letting
(2.s)
wedge with uniformly
e) Letting
sented
field
P*
c) For b = 0 the superposition c 11) 6,
1451
we obtain
-2n,+
(cf.
deforaations
T9J =O
fp,
an axisymmetrical
which is a solution
plastic
we obtain
b = 0,
Inr2T1 which represents
in plane
1
=--2g,+n+p,
‘d,, =I,
up = 0,
up =
--Cl
J
‘5
-T--
_2
c
(2.12)
1452
V. L.
Dobrosol
‘skii
BIBLIOGRAPHY 1.
Sokolovskii, V.V., Teoriia plastichnosti Izd-vo Akad. Nauk SSSR, 1940.
2.
uprugo-plasticheskaia Galin, L. A. , Ploskaia plastic problem). PMM Vol. 10, No. 3,
3.
(Theory
of
zadacha
PZosticity).
(Plane
elasto-
1946.
deformatsii ideal’no plastiDobrovolskii. V. L., Zadacha o ploskoi cheskogo tela v kompleksnykh peremennykh (On a problem of plane deformations of an ideally plastic body in complex variables). PMM Vol. 24, No. 2, 1960.
Translated
by
R.M. E.-I.
Vo 1. 24, No. 5,
v.
L.
1960,
pp.
K PLOSKOI
955-958
DOBROSOL' SKI1 (Moscow)
The conditions of the stress function being real, applicable to the soluA few such solutions are given as tion of inverse problems, are derived. illustrative examples. The conditions of the stress function being biharmonic in the plastic range are also determined. 1. Consider a homogeneous isotropic material in the plastic stress components satisfy the equilibrium conditions
and the Huber-Von Mises (or St.
Venant-Tresca)
Plasticity
state.
The
conditions
(1.2) where k is the yield tion F: ar=
limit
in shear Ill.
k a2F 2ay2’
k Gy= y-
We introduce a2F ax2 9
the stress
k a2F z ry = - -2 axay
func-
(1.3)
Equation (1.1) is satisfied identically, and (1.2) is left for the determination of function F. This equation can be represented in the complex variable I = x + iy as a2F pr=i
From physical
considerations
asp
F has clearly F (1, i) = F (2, i)
Now (1.4)
can be written
(1.4) to be real,
i.e. (1.5)
as SF T
= exp [ if31
1447
(1.6)
V. L.
1448
where @ = @(r,
t)
is a real
function.
*s
F (z, Z) == Theorem
F(z, t)
1.
defined
Integration
of
(1.6)
results
in
-dr) L exp ti@ (b ;)I dE + ZTJ(z) + x (4 s i La no
The necessary by (1.7)
Applying
‘skii
rl
and sufficient
to be real -
Proof.
Dobrosol
condition
for
(1.7) the function
is
azei@ a22
aze-iQ _ -
(1.8)
a22
the operator
a4 az2ai2
(1.9)
to the functions F(z, T) and F(EI- I) and taking into account (1.5) we obtain the proof of the necessity of the condition (1.8). To prove that this condition is also sufficient we write (1.8) as
aw -=a22aiz
a43 a22a22
The operator (1.9) is biharmonic, and consequently F by a biharmonic additive term which can be selected the theorem is proved. The condition (1.8). for practical presented in the Cartesian coordinates
(
cos@ g-g or in the polar
-zEE ax ay,1
(
__2
$sin@
coordinates
(r,
purposes,
can differ - from so that F = F. Thus F
may be conveniently
+$-(+$T+($+)
(1.10)
$1
+sinCO+2~)(%~+~i~)z-_~-(~)2)=0 Introducing
the notation
re-
exp [ i@ 1 = u + iv.
(1.8)
(1.11)
can be represented
as (1.12)
A particular solution @ = @(z, z) of (1.8), which does not contain arbitrary parameters. determines the stress components within a constant hydrostatic pressure. Theorem
2.
Application
Proof.
0 = 8(x, term pzi
of
complex
variables
in
plane
plastic
1449
deforaations
Theorem 1 and Expression (1.7) show that the solution y) determines a real stress function F(z, I) within an additive (p = const) which represents constant hydrostatic pressure.
Special interest attaches to stress functions harmonic equation in the plastic range.
which satisfy
the bi-
Theorea 3. The necessary and sufficient condition for the stress tion F to be a biharmonic function is that the function @ = @(z, I) satisfies the following system of equations:
f320 -_a2(3_2ao 8x2 By* Proof.
conjunction
ax
(ig-(g,"-2$&0
-!z!%=(),
ay
If F, represented by (1.7). with (1.6) the following SF
az2&2
=
0,
is a biharmonic holds: or
(1.13) function,
then in
a2ei@
--0 azz
(1.14)
Representation of this relationship in Cartesian coordinates (1.13). and thus to the proof of the necessity condition. Inverse sufficiency
arguments lead from (1.13) condition.
It is obvious or (1.10).
that the solutions
The class of solutions of (1.13) not empty. As an example we propose 1. @ = const. 2.
0 = -2
corresponding
tan-‘(y/r)
+ const.
to (1.14)
of
(1.13)
func-
leads
and to the proof
are also
is apparently the following
to a uniform stress
solutions
very small, solutions:
to
of the
of
but it
(1.8)
is
field;
This case was utilized
by Galin [ 2 I .
Note 2. Equations (1.10) and (1.12) possess the following features. If 8 = 0(x, y) is a solution of (1. lo), then v, = - @( - x, y) and Yz = - 0(x, -y), and consequently ys = 0(-x, -y) are also solutions. If w = u + iv is a solution of (1.12), then w1 = u1 + iv1 is also a solution of the same equation. Note 2. If some solution of (1.8) is known. then it is not necessary to find the stress function from (1.7) and to calculate the stresses from this function. It is easier to utilize the equilibrium equation (1.1) and the fact that @ determines uniquely the difference between the normal and shear stresses.
Z. We shall now determine a few particular solutions of the equilibrium equations in the plastic range by solving (1.10) and (1.11).
1450
V.L.
1. We seek
a solution
Dobrosol
in the
‘skii
form @ = a(y).
Equation
(1.10)
can thus
be written d’Q
Solving a case
this
of 5,
where the tion
=
2.
r1.c
-
stresses
2 1/i
are
dy
% = sin-‘(Ay
by two rough (Z4y + B)“,
-
represented
coordinate
+ B).
plates
Qv = p -
relative
to
form 8 = 8(a x + p y ) reduces
Seeking
(1.12)
we find
compressed
p -
in the
ing the
equation
a strip
dQ 2
( j-0
tall8
-_ dy”
to
This
corresponds
to
[ 1,3 1 AZ,
ZXU= ny+
the magnitude
the
previous
B k.
(2.1) A solu-
one by rotat-
axes.
a solution
in the
form 0 = - 24 + R(r)
reduces
Equation
to
Solving
this
equation
we find
R=arcsin(a-$),
and the
corresponding
6,
or
stress
components
a2 In [l/l
=-2a(p+d1-
(2.2)
@=-2rp+sin-1(~-~,)
-
are
a2 1/(1 -
-a2) r4 + 2abr2-
b2 $.
% + (1 -
F +s jf( 1 -
a2) r2 + ab] + a sin-’
T Pm =P--_r2
The displacement theory
[l
1
components
and independent C
a?=79 Here b f
0.
If
b =
of
The solution of
among all the
represented entering
solutions.
(2.3)
from a simplified
angle
Hencky-Von
Mises
4 are az) r* + 2abr” -
(2.4)
b”
0 then
% =dv$& by (2.3)
known closed-form
parameters
particular
b2 + p
b
U = dr - & 1/(1 v
C Ur = 7,
tion
found the
~2) r4 + 2abr2 -
into
(2.5)
and (2.4) is the most general soluBy specializing the values
solutions. this
solution,
we obtain
a series
of
Application
a) For
of
a = 0,
complex
variables
b) For a = 1, b = 0,
stress
%
for
1=-
a plastic
2arp + VI -
d) For a = 0, b = metric loading [ 1 1 :
E there
with the corresponding
(2.7)
z,, = 1
a2 (In r2 T 1) + p, results
loaded cases
faces.
results
in
TrQ = a
a most general
a new solution
case of axisym-
of a problem pre-
displacements
I$ =
Cl F
r
y$,==-dr
-
Along the contours
r = r0 the stresses
3=Pr2*Tm
z rrp ==i--$
ur = Sz t r0
(2.11)
~1/2Cr2J
are
and displacements (PI, p3 = const)
UT = dr, - ;;
0
~_ T/2CQ -
C2
d = 0 and r -+ by we obtain
=r %
[ 1] .
of the two previous
o = 1. b = c we obtain in [ 3 ] :
Letting
(2.s)
wedge with uniformly
e) Letting
sented
field
P*
c) For b = 0 the superposition c 11) 6,
1451
we obtain
-2n,+
(cf.
deforaations
T9J =O
fp,
an axisymmetrical
which is a solution
plastic
we obtain
b = 0,
Inr2T1 which represents
in plane
1
=--2g,+n+p,
‘d,, =I,
up = 0,
up =
--Cl
J
‘5
-T--
_2
c
(2.12)
1452
V. L.
Dobrosol
‘skii
BIBLIOGRAPHY 1.
Sokolovskii, V.V., Teoriia plastichnosti Izd-vo Akad. Nauk SSSR, 1940.
2.
uprugo-plasticheskaia Galin, L. A. , Ploskaia plastic problem). PMM Vol. 10, No. 3,
3.
(Theory
of
zadacha
PZosticity).
(Plane
elasto-
1946.
deformatsii ideal’no plastiDobrovolskii. V. L., Zadacha o ploskoi cheskogo tela v kompleksnykh peremennykh (On a problem of plane deformations of an ideally plastic body in complex variables). PMM Vol. 24, No. 2, 1960.
Translated
by
R.M. E.-I.