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Dynamic contact problem of steady vibrations of an elastic half-space
DYNAMIC CONTACT PROBLEM OF STEADY VIBRATIONS OF AN ELASTIC HALF-SPACE (K DINAMICHESKOI
KONTAKTNOI
KOLEBANH
UPRUQOGO
ZADACHE
STATSIONARNYKH
POLUPROSTRANSTVA)
29, No. 3. 1965, pp. 545-552
PMM Vol.
V.N. ZAKORKO and (Komsomol’sk-on-Amur, (Received
N.A.
ROSTOVTSEV Novosibirsk)
Februav
8,
1965)
the effect of a sinusoidal load (with Two problems are treated with a common formulation: a time factor of the form exp 4 WC ) transmitted through a weightless (and inertialess) rigid punch of circular planform lying on the homogeneous isotropic half-space (t > 0). The vibrations are steady-state, and the effect of the static load is not considered, it being assumed that its magnitude is sufficient to maintain contact in the case where the punch is not joined to the half-space. For brevity, henceforth when we speak of vectors (or their The actual values are obtained by multiprojections), we shall mean their amplitudes. plying by exp (i W: ). In the fundamental mixed problem it is required to find, for a known distribution of the diaplacemen vector w (w, ucr, u,. ) in the region of contact 0, the disthe tangential tribution of the force vector p t p, t x, t } b rs the pressure, tx, c In the other problem forces) in this region. In particular, if t h ere is bonding, ug = IA = J* rX to find only (a punch without friction and bonding) we have I, = t = 0, an rt is required Y p=p(x,y)foraknownw=w(x,y)inthe regionn. A unified treatment of these problems is given developing an approximate method for their solution problem). 1. The
proposed
fundamental
mixed
method problem
consists
of the
(in vector
notation,
w (r) =
cs K (r -
in deriving the integral equations (which is applied to the second
following.
The
of course)
r’) p (r’) dS,
system
takes
of equations
the
rE
and
of the
form
Q
(1.1)
‘n
Here K
applied that
it is possible problem. system
is the difference
the displacement
at the origin
the kernel
sequently
r’)
K (r -
(r) represent
and directed
K (r) is singular to reduce
be shown Therefam of equations
(1.1)
(with
second
the axes
P weak
to s system
.If the solution
kernel.
in K
z, x, y,
singularity).
columns
can
644
respectively.
the second singularities
problem
in fact
(K”, K r, K’)
of concentrated
is known,
be carried
out.
of the matrix
unit
forces
It is therefore
By separating
of equationsof
(r) containing
of the static kind
The
due to the effects
along
that the terms of the
matrix
vectors
kind.
It will
correspond then
clear
out the singularity sub-
to the static
the transition
In order
to find K
to a
(r) we
Elastic
solve
the following (h +
system
p) v div w +
of Lamd
PAW +
half-space
equations
@W
=
vibration
645
(in rectangular
coordinates)
(p is the density elastic medium)
0
of the (1.2)
for the conditions
The
calculations
lead
w=KP,
to the
P=
PT [ T;
following
1
-
iae
(yy,
-
$0
(r2),
The
The
i (ax +
By) X
iat3 h2h
w
W)
‘l2H (‘r) + V2 (a2 - fJ2)Z W),
aSz
(r*)
E , 0, H, Z of the the
(~2 -
are wave
=
argument
(r2),
li2H
w h ere R is the
(a*--@)
here
(1.4)
~(72)
represent
expressions
Rayleigh
4~2 (~2 -
kz2) -
(2ya -
(r2)-‘1,
appearing
y’
E, , @I,, H,, Z,, whose
functions
k,2)‘/zI? (~9,
R (r2) (k, and k,
da dfi exp
ss
--co --a3
1 (r= l/c@+ IPI
by dividing
by the product
+a2 +ca
K = 4-P
aSz
functions
obtained
1
’
E (r2), x
result:
the quotients
are
given
below,
function
Ic,?)~/z (~8 -
k&“’
(1.5)
numbers). for E+,@+, H,,
expressions
Z,
are (1.6)
E, (~2) = -
kc (r2 -
8,
(~2) = (2~2 -
722, W) = The radicals Riemann The a set [l]
are
of cuts
angular
joined
which
(p. 945).
transforming
The
with
the vector
tions
of them 2~‘~
Then
formulasare
a cut which
is drawn
by A. Sommerfeld
double
Fourier
integral
displacements
u +
iv,
in (1.4)
and
and
to a problem may
in the sequel
forces,
w2 = u -
iv,
ha)
kla)“’
functions
to the
four
in an appropriate
in the CL, p plane
calculation
kze) (ya -
single-valued
corresponding
applied
coordinates
2 (r2 -
4* (?’ -
(k, = o -f/p / (11+ 2~), kz = 0 ‘t/P /P)
k22)
of four sheets,
along
In this
=
k22)“’ -
was
to polar
variable.
kz2 (r2 in these
consisting
ka2)“‘, H, (y2) = R (-r*) -
kz2) (ya -
R (P) appearing
surface
sheets
klz)“’ (r2 -
but rather
combinations.
manner.
We will
of radio-wave
be reduced
and
the following
t, = t, + it,,
integral
the integration
it is convenient
to deal complex
t, = t, -
use
propagation
to a single
performing
with
of y on the
sign
it,
by over
the
not directly combina-
(1.7)
and N. A. Rostovtsev
L’. N. Zakorko
646
Expressing
the integrals
in polar
we obtain
coordinates,
m
2&1W
A (xv Y) =
0 where 2 =
There appears [II,
is no danger
in the
transform
relation
function
H(l)
(yr)
not write
out this
the asymptote.
with
the
We note
As a consequence
associated
with
culating each
the root
the integrals of these
residue
to this
integral
equation in (1.9).
is equal
corresponding
the
to the
pole.
00, -/- co)
the
that
Hence
the vector
of the pole In cal-
from above;
the product (1.1)
the
of in
reduces
and
these
shaped
two-dimensional
region,
however.
regions
equations.
With this
n,
there
is no effective
A simplification
aim we expand
the
analytical
is possible
force
vector
for the
thus the
to
(E, ‘1) dE dq 2. For arbitrarily
of
in the upper
quadrant.
is bypassed
equation
We will
in calculating
contribution
this
minus
by
treatment
remaining
h owever,
value
from (1.9)
P = (P, 0, 0 1 the asymp-
0 in the right-hand pole
in
the
the Hankel
detailed
R (~2) =
principal
differs
is deformed,
includes
using
in the integrands.
note
never
as shown
(yr),
whereas
force
We also
[Zl.
(-
of this
of the
appearing
integrals
by Lamb
H(r)
a complete
normal
the latter
one may,
which
appear
aim to present
for a concentrated
of integration
half-plane.
obtained
y2
z , since
functions
of 1/2np,
will
+
kernel,
instead
indices
we do not
Jfx2
coordinate of this
y) is then
1/4nu
investigated path
12 1 =
oo) in the Hankel
corresponding
that
r =
the spatial
for i\ (a,
since only
was
the original
00, +
.4 formula
iy,
the asymptote
of the integral:
formula,
displacement
(-
x -
with
To calculate
to the interval
in front
Z =
of confusion
of the circuits.
integrals
iy,
+
formulas.
the multiplier
totic
2
(1.10)
method case
of solving
of a circular
(p, tr, te) in a Fourier
series
in
eid @, $., 63) =
For
each
of the components
t 1m = We note
[g, further
w1 =
etc.
(r) .+ ih,
u +
iv =
Introducing
these
the integration
We give
t,,
=
(r),
gm (r), h,
(2.1)
gm (r) eime,
(r)] oifm+lJe,
(r)) eime
t?,
L?l =
[g,
=
(r) e into we have
h,
(r) -
ihm (r)l
f.?i(m-l)Q
(2.2)
that
we perform
[31 (p. 392).
5 (I,, --CC
values only
(ur +
iu,)
in (l.lO), over
the result
eie, where
the angle
70, = .z should $J with
of the calculation
u -
iu =
be replaced the help
(ur -
iu,)
by z -
of Graf’s
8’ i =
addition
reio theorem
pei*
Elastic
Rere matrix
the index
with
m denotes
the following 03
the
half-space
Fourier
647
vibrotion
coefficient
(of
eime).
The
Afm) is *
kernel
elements: co
s
s
AZ,(~) =
4,tjm)= 2 E W) Jr,, (~4 J,(YP) T d-r,
0 W) Jm_l hr) J, (w) T’ti
0
0
Azlcm) = -
2 h2) Jm+l
s
(TP) Jm+I
(Tr) ‘fa dr
0
(2.4) The remark integrals.
made
It would
Cf.gs h lrn, plicated.
but
The
at the
have
end
been
possible
the formulas notation
would
here
be even
series
to the real
(the
m = 0 is omitted K,,
section
form.
here). co
= -
holds
to express
for the elements
of the Fourier
subscript
of the first
for the calculation
(tu, ur, r&
of the kernel
would
directly then
more
complicatedif
we went
Only
in the case
m = 0 is there
In these
k12)“’
R (P)
0
in terms
have
been
of
com-
from the complex
form
a simplification
equations
kz2 (r2 -
5
of these
JO (v)
JO (YP) TM
cc
K,,
c
k2 W - b?)
= -
b
R (r2)
J, (‘r4 J, (TP) ‘r dy,
KOI =
s T2e
=
W)
Equations
(2.5)
the first torsion
equations
into
Lam;
equations
tions
(2.5).
consequence
may
be considered
of equations
describes
(the
31 -
which
Reissner-Sagoci
dr
systems
from the
(1.6)
physical
The
(rS -
of the punch,
to the
vanishing
point
of view,
@
(r2) =
2 (y2 -
decomposition
depends ks2)“’
problem, while
of the system
of R (ya)
we have H (r2) -
(YP) 7 dr k,a)‘j*
of the axisymmetric
The resolution
corresponds
case.
(2.6)
so
the indentation
problem).
axisymmetric
is clear
of formula
as the equations
d escribes
(2.5)
two independent in the
Jo h4 J1 (‘rp)
K _ O”JI (~4 JI
JO(w) JI 04 &,
0
where
(P)
0
03 Kt,
JPe
the
of the system
in the second on the
second
of integral
fact
of equathat
as a
of
and N. A. Rostovtseo
V. N. Zokorko
648
The
solution
of the system
on the absenceof quadraturea
will
will
the description
restrict
without
exact
unavoidably
friction
3.
(2.3)
any simple
and
even
in the simplest
solution
reduce
the accuracy
of any
of such
au algorithm
to the problem
g,,, = h,
{ 1,
(P) PdP TE
0
is clear
This help
that
Fourier equation
in this
reduce
solved
i.e.
may
also
function.
method
equation functions
Rorodachev
is used,
In order
(Yr) J,
algorithm.
Therefore
we
of a punch
(YP) YdY
(3.1)
which
contains
(3.1)
from
required
Rorodachev’s
directly
friction
Hankel
transforms
pair
of integral
of N.N.
directly
coefficients
[2] with
was
were
used
to
were
then
equations
for a certain
numerical in that
pressure
to an equation
the
and bonding
Lebedev)
additional method
the sought-for
is transformed
out the singularities
The
of Lamb
without
(by the method
of the pressure differs
paper
to he the
sin r&l.
from the results problem
equations. kind
below
to simplify
directly
[41. In that
of integral
Calculation
fm (r ). Equation
by separating
turns
approximate
of indentation
of cos m f? and
axisymmetric
of the second
developed
(3.1)
The
to a pair
to an equation
auxiliary The
numerical
wm (r ), fm (r ) may be considered
be obtained
by N.M.
(Y) J,,,
as the coefficients
theorem.
the problem
reduced
(2.5)
0
equation
series,
of the addition
effectively
case
Cumulative
= 0 we have
wm (r) = It
problem.
bonding.
In the case
of a real
axisymmetric
of the static
only
quadratures. the single
through
the unknown
of the second
kind
of the kernel.
the subsequent
calculations
we change
to dimensionless
quantities
by setting
r
=a2, Making
p =
a&, ka = a / a, h = k, / k,
the further
y = kp
substitution
9.m (Z) =
\ IIm (E) io!Er
f,, (r) = pII,,,
(r)
CT),
I(!,,,
(3.1).
we obtain
J,"
(ll:S) Cf.7
(3.2) inside
the integral
*%j+
SJ,
(USS)
(3.3)
0
j? (a) = (29 In the neighbourhood
- as
fz@)7n (x)
-
1) -
49
(r? _
1)“’ (.$? _
jiZ)‘/Z
of s = J) we have
v/s% - ha =
F (4 Here C is a regular
function. 1
2 (i -
2 (1 ” Equation
! ha) \ . &I
h*)
-
(3.3) cc
takes
(E) SdE \ J,,W)
0
(G (32) -= I,)
G (s)
(3.4)
the form
J,(EQ
dh =
0
(3.5) = The kernel
%I
(z) -e { Ku 0
on the left-hand
side
(5) go% TJ_
(czars) J,
(ags)
G (s) ds
0 is a special
case
of the Weber-Shafkhaitlin
integral
Elastic
For
the
half-space
649
vibration
0
equation 1
s
63 w,,(;) (x9 El dE = f (4
‘p
0
the inversion
formula
151 is known -
cp(r)
=
r (i/u (1 +
1
21-r a?Q
p +
4) r (l/e (1 +
p-P-9&
s (t2 _
d
z2)1/~(1-r+o-p)
dt
this
formula
into
we obtain
(3.5),
u’+~ f (u) du so (ta _
x
Substituting
q - p)) ’
r 3_
t
d
’ dr
r +
ua)‘/z(l-r+p-91
an equation
of the
second
kind
1
(3.6)
in which 4 (1 -
N) xm-’
-
d
‘m (2) = - r (m + V2) r (l/J dx
d”
2 (1 -
Km (x7Y)=
G (s) J,
r trn+ l/2) r (l/2)
(axs) J,,,+
(ast)(as)
(3.7)
‘A ds } dt
(3.8) 4.
In the particular
case
of a plane
pr cos 8 =
w z wg + In this
inclined
punch
we have
a (W, -!-
px cos cl)
case @ =
cPl =
W@,
(0
@u
is the angle
of inclination
of the punch) Here
it is convenient
to transform
x+ t
H, (x) = II, (x) (1 Noting
that
1 -ha
= ?/?(1 -
to the new
v),
unknown
functions
H1 (I) = II, (x) (1 -
we obtain
z$‘*
the equations 1
2 Ho (x) =
2wo 3T (1 -
Y) -
JL (1 -
v) s K,
(x, y) Hc, (y) dy
0
HI (4 = with
kernels
equal,
x (r!!
respectively,
,,) 4x -n
to
(1”
y) i KI (x, Y) HI (Y) dy
(4.1)
V.N. Zokorko
650
Ke
(I,
Y) =
+
(ej)‘”
2
i
and N.A.
Rostovtsev
&f&
i
G (a) J,
(aya)
cos (ato)
(4.3)
da
0
‘Ce first solution
consider
of this
equation
equation
(4.1)
with
the kernel
H,(O) =
This
is related
to the solution p,(“’
to find
the
first
1
s
of the corresponding
zeroth
approximation
to the
approximation
Y) a l/a2
-
we calculate
z
the order
problem
2tL~O
(r) =
0
Changing
static
ra
the integral
(ejh{-&i$&--i
K, (2, y) dy = j,
0
The
2we 7I (1 -v)
n (1 In order
(4.3).
is
G (0) Jo (ay) cos (ata) du}dy
0
we obtain
of integration,
1 v’-&t&~C(u)cos(a+u)~du}
\&,(qy)dy=
(4.5)
lz
0
b
the integral
To calculate
00 1 z we use
the theory 1 =
of residues.
We have
C
cos auo) co9 am, +
[nc sinau,
Now the function of t , whose respect
I (a, t) =
to this
i WI=0
IP,
co9 am, +
1 -- (WI res G (a)
c=
a0 is the root of the Rayleigh
powers
s
V (a) (1 -
cos au) cos (atu) da
0
( where
da
G (a) cos (am) y
1
nc (1 -
-I- i
with
c b
V (a) sin au cos (atu) do] Im G (a)
) v (0) = ~ au a=0.
1
F (01 = 0.
equation
coefficients,
+
1
I = 1 (a, t ) may be expanded (complex)
I
which
in a rapidly depend
converging
on U, may
series
be expanded
in even in series
parameter (a) +
i
Q,,, (a)1tarn9
P,n
(a) =
(-(~~~
i
k=l
akfm)aab
(4 .6)
651
Elastic half-space vibration
Qm ($ =
(-,;Z p”” 2
bk(rWa2k+l
)
k=l
Substituting
uk-l [nC(J62m+2k-1 + q*+2k_11’
(-
b,(m) =
(2k -
q,,-
I)!
for I (CL, t
(4.7)
1 @+?'du 0
) in (4.5) its expansion
(4.6)
and integrating,
we have
1
. K, (r, Y) dy =
i
c
t2mtl
4-d' M,
(4
[P,
=
1:
ZG s Jfp-
za
z
(VI The polynomials polynomials
in z.
i Qm (a)1 M,
d dt = yg
even powers
Ml (2) =
1,
in u and are consequently
2ua -
1,
(4.8)
M, (I)
8~’ + 6~2 -
1M,,,(I)I< const
side of equation
R'egive below the resultsof
even-power
calculating
PI (a)= -$
+
ha =
the formulas
Qn for h2 = 1/3, taken from reference [4], we find PO (a) = 1.0068~~ - 0.07913a4 + 0.002665ae 0.3186as
P,
the coefficients
3tc =
-
0.01622aS
-
-
0.0004206ar
+
+
0.0004761a’
-
ak Qz (a)= q-[1.9466a - 0.3533as+ 0.01996a6-
.
[0.9583a2- 0.09361a4
P, (a)= - $
[1.1232a2
Q3 (a)= - $
12.1199a
-
0.1068a4
to the solution
(1”:q {1 -
+
0.3858aS
a8 ' P4 (a)= -g- [1.2816a2- . . .], The first approximation
+
n (1 “_
+
.
+
. . .]
...
. .]
. . .] . . .I
of equation
(4.1)
y)
(a) +
i
..,
. .]
as Q4 (a)= m (2.3946a-
-0
of
. ,.
[1.9113a
-fi- 0.003560a6
= ~
I/,). values
-
0.3244a3
Qrnca) for
and a tableof
0.00005433$
-
P, (a)= $
te) and
[0.9495a2- O.O7986a*f 0.000312a6 -
on the
1/s (A = ~1, v =
(4.7)
0.01766as
Q1 (a)= - T
the series
series
(ao)2"
In this case
2.3243a
i
x2)
and consequently
up to terms in U lo for the case 1.0248.Using
4u= -
1 -
by the convergent
m = 0, I, 2, 3, 4, accurate 1.0877,
V,u4 f
(US =
v/m,
is majorized
=
1
; If; m=O (Pm)!
Ho(‘)(4
us sin% T)~ dq
, VI-P=7)
-$=u
Mm (x) areof
It may be shown that
Q. (a) =
(4.6)
'lc * u sin cp (1 s 0
M, (x) = la/GE -
o. =
(I)
We give the first few polynomials
M, (5) =
right-hand
(a) f
m=o
[p,
,
consequently i Q,,,
. .] takes
the form
(41 Mm (4) f4ag)
652
C’.N. Zakorko
The
second
the solution
term
of the
We consider approximation.
in the curly
and changing
brackets
corresponding
now
static
equation
Calculating
the order
and N.A.
(4.2) the
in equation
with
the kernel
of integration,
1/l -
%\
l/t2 -
x3.
x
the theory
correction
to
(4.4)
and
seek
the solution
in the
first
m
x2"\ dx.
0
Applying
is the dynamic
we obtain
1
(x, y) ydy =
(4.9)
problem.
integral
1
K,
Rostovtsev
G (0) sin (ato)(F
-
cy)
da (4.10)
0
of residues
to the calculation
of the inner
integral
M 11 0% r) =
cos ao a.
--‘;,$r (’
G (0)
s
sin (ata) ds
0
we find I, (a, t) = [rtc (sin
i-
au0 -
au -
i V (a) (sin
I -
z:
’ -,“,“”
““)
“)
sin (au&
sin (at@ da]
+
-/-
0
+
ijnc (00s
aoo-s~)sin
(ao,$) +
iv
(0) (~0, t.J_
e),i,
@a)
da]
0
Expanding of the resultant
now
the function
series
I, (at)
in powers
in powers
of t and the a-dependent
coefficients
of CL, we obtain
1, (a, r) =
5
[Pa,
(a) +
iQk
(a)]
t2m-1
(4.11)
m=1
,,w =
((2/( (-
dkcrn) =
(2k Substituting
I)“-’ 2)
2mt2k-2
+
! 2k (no%
iy-*
3)
! (2k -
1) Wr”
for I, (CL, t ) in (4.10)
its
2m+2k-3 +
expansion
Y
2m+2k-2)
Y2n1+2h._J,
(4.11)
C&l)
&cm)
z
and integrating,
0
we obtain
1 1 KI (I, y) ydy
0 We quote h2 =
‘13 (Y =
the expansionsof l/r)
=
fPf,(a) ;f ia",, (a)lM,_,
2
(x)
rn=l the first
few coefficients
PC
(a)
and
Q;(a)
for
Elastic
P”1 (a) = G
(1.0068a
-
Q"1(a)= $
[--0.6371az
half-space
0.2374as +
+
0.01331~~
0.06489a4
-
iO.94950 - 0.2398as + 0.01560a6
Q; (a)= $
[-
&
(a) = $-
Qi (a) =
Pi (a) The
[0.9583a
gr
= T-
first
-
+
0.2808~~
[ -0.7066aa
+
a7 [l.l232u
0.07066~~
-
+
-
-
0.00005279ae
0.000445Oa’ +
. . .]
-
...]
+
. . .]
. . .]
.. .], Qi (a)= -4
to the solution
...]
+
. . .]
+
0.00039OOa’
0.002851a6
0.01780~~
0.07982a’
0.3204as
-
approximation
+
-
0.002524~~~
Pi (a)= $
0.6489a2
653
vibration
of equation
(4.2)
[-0.7982aa
consequently
. . .]
+
takes
the
form
f&(l)(4 =
W a
(I
-
v)
t
1-
It
2_
(I
v)
[Pk (a) + iQ& WI Mm_,
i
CT)} z
mz_1
Here,
as in (4.9),
corresponding the same
static
integrals
the second problem.
will
amount
in principle
of calculation
is the dynamic that
be encountered,
I (u, t ) and f 1 (u, t ), there Thus
term
We note
the
will
increases
except
be .f,,,+,,,
subsequent
correction
in obtaining
to the solution
the subsequent
in place
of the
approximations,
of sin CLCT/QU in the integrals
(uu) / (r~a)“+‘ls.
approximations
are
foundin
a similar
manner,
but the
considerably.
BIBLIOGRAPHY R., Die Differential-
und lntegralgleichungen
der Uechanik
the surfaceof
solid.
1.
Frank, P., and von vises, und Physik, 1930.
2.
Lamb, Trans.
3.
Watson, G.N., A Treatise Cambridge, 1944.
4.
kontaktnaia zadacha dlia shtampa s ploski m Borodachev, N.M., Dinamicheskaia krugovym osnovaniem, lezhashchego na uprugom poluprostranstve (The dynamical tact problem for a punch with a plane circular base lying on an elastic half-space). lru. AN SSSR, OTN, ?vlekhanika i mashinostroenie, NO. 2, 1964.
5.
H., On the propagations of tremors over Roy. Sot. (Series A), Vol. 203, 1904. on the Theory
of Bessel
Funcrions.
an elastic
Second
Phil.
ed.,
con-
-nogo uravneniia teorii lineinogo Rostovtsev, N.A., 0 nekotorykh resheniiakh integral’ deformiruemogo osnovaniia (Some solutions of the integral equations of the theory of a linear deformable foundation). PUU, Vol. 28, NO. 1, 1964.
Translated
by F.A.L.
KONTAKTNOI
KOLEBANH
UPRUQOGO
ZADACHE
STATSIONARNYKH
POLUPROSTRANSTVA)
29, No. 3. 1965, pp. 545-552
PMM Vol.
V.N. ZAKORKO and (Komsomol’sk-on-Amur, (Received
N.A.
ROSTOVTSEV Novosibirsk)
Februav
8,
1965)
the effect of a sinusoidal load (with Two problems are treated with a common formulation: a time factor of the form exp 4 WC ) transmitted through a weightless (and inertialess) rigid punch of circular planform lying on the homogeneous isotropic half-space (t > 0). The vibrations are steady-state, and the effect of the static load is not considered, it being assumed that its magnitude is sufficient to maintain contact in the case where the punch is not joined to the half-space. For brevity, henceforth when we speak of vectors (or their The actual values are obtained by multiprojections), we shall mean their amplitudes. plying by exp (i W: ). In the fundamental mixed problem it is required to find, for a known distribution of the diaplacemen vector w (w, ucr, u,. ) in the region of contact 0, the disthe tangential tribution of the force vector p t p, t x, t } b rs the pressure, tx, c In the other problem forces) in this region. In particular, if t h ere is bonding, ug = IA = J* rX to find only (a punch without friction and bonding) we have I, = t = 0, an rt is required Y p=p(x,y)foraknownw=w(x,y)inthe regionn. A unified treatment of these problems is given developing an approximate method for their solution problem). 1. The
proposed
fundamental
mixed
method problem
consists
of the
(in vector
notation,
w (r) =
cs K (r -
in deriving the integral equations (which is applied to the second
following.
The
of course)
r’) p (r’) dS,
system
takes
of equations
the
rE
and
of the
form
Q
(1.1)
‘n
Here K
applied that
it is possible problem. system
is the difference
the displacement
at the origin
the kernel
sequently
r’)
K (r -
(r) represent
and directed
K (r) is singular to reduce
be shown Therefam of equations
(1.1)
(with
second
the axes
P weak
to s system
.If the solution
kernel.
in K
z, x, y,
singularity).
columns
can
644
respectively.
the second singularities
problem
in fact
(K”, K r, K’)
of concentrated
is known,
be carried
out.
of the matrix
unit
forces
It is therefore
By separating
of equationsof
(r) containing
of the static kind
The
due to the effects
along
that the terms of the
matrix
vectors
kind.
It will
correspond then
clear
out the singularity sub-
to the static
the transition
In order
to find K
to a
(r) we
Elastic
solve
the following (h +
system
p) v div w +
of Lamd
PAW +
half-space
equations
@W
=
vibration
645
(in rectangular
coordinates)
(p is the density elastic medium)
0
of the (1.2)
for the conditions
The
calculations
lead
w=KP,
to the
P=
PT [ T;
following
1
-
iae
(yy,
-
$0
(r2),
The
The
i (ax +
By) X
iat3 h2h
w
W)
‘l2H (‘r) + V2 (a2 - fJ2)Z W),
aSz
(r*)
E , 0, H, Z of the the
(~2 -
are wave
=
argument
(r2),
li2H
w h ere R is the
(a*--@)
here
(1.4)
~(72)
represent
expressions
Rayleigh
4~2 (~2 -
kz2) -
(2ya -
(r2)-‘1,
appearing
y’
E, , @I,, H,, Z,, whose
functions
k,2)‘/zI? (~9,
R (r2) (k, and k,
da dfi exp
ss
--co --a3
1 (r= l/c@+ IPI
by dividing
by the product
+a2 +ca
K = 4-P
aSz
functions
obtained
1
’
E (r2), x
result:
the quotients
are
given
below,
function
Ic,?)~/z (~8 -
k&“’
(1.5)
numbers). for E+,@+, H,,
expressions
Z,
are (1.6)
E, (~2) = -
kc (r2 -
8,
(~2) = (2~2 -
722, W) = The radicals Riemann The a set [l]
are
of cuts
angular
joined
which
(p. 945).
transforming
The
with
the vector
tions
of them 2~‘~
Then
formulasare
a cut which
is drawn
by A. Sommerfeld
double
Fourier
integral
displacements
u +
iv,
in (1.4)
and
and
to a problem may
in the sequel
forces,
w2 = u -
iv,
ha)
kla)“’
functions
to the
four
in an appropriate
in the CL, p plane
calculation
kze) (ya -
single-valued
corresponding
applied
coordinates
2 (r2 -
4* (?’ -
(k, = o -f/p / (11+ 2~), kz = 0 ‘t/P /P)
k22)
of four sheets,
along
In this
=
k22)“’ -
was
to polar
variable.
kz2 (r2 in these
consisting
ka2)“‘, H, (y2) = R (-r*) -
kz2) (ya -
R (P) appearing
surface
sheets
klz)“’ (r2 -
but rather
combinations.
manner.
We will
of radio-wave
be reduced
and
the following
t, = t, + it,,
integral
the integration
it is convenient
to deal complex
t, = t, -
use
propagation
to a single
performing
with
of y on the
sign
it,
by over
the
not directly combina-
(1.7)
and N. A. Rostovtsev
L’. N. Zakorko
646
Expressing
the integrals
in polar
we obtain
coordinates,
m
2&1W
A (xv Y) =
0 where 2 =
There appears [II,
is no danger
in the
transform
relation
function
H(l)
(yr)
not write
out this
the asymptote.
with
the
We note
As a consequence
associated
with
culating each
the root
the integrals of these
residue
to this
integral
equation in (1.9).
is equal
corresponding
the
to the
pole.
00, -/- co)
the
that
Hence
the vector
of the pole In cal-
from above;
the product (1.1)
the
of in
reduces
and
these
shaped
two-dimensional
region,
however.
regions
equations.
With this
n,
there
is no effective
A simplification
aim we expand
the
analytical
is possible
force
vector
for the
thus the
to
(E, ‘1) dE dq 2. For arbitrarily
of
in the upper
quadrant.
is bypassed
equation
We will
in calculating
contribution
this
minus
by
treatment
remaining
h owever,
value
from (1.9)
P = (P, 0, 0 1 the asymp-
0 in the right-hand pole
in
the
the Hankel
detailed
R (~2) =
principal
differs
is deformed,
includes
using
in the integrands.
note
never
as shown
(yr),
whereas
force
We also
[Zl.
(-
of this
of the
appearing
integrals
by Lamb
H(r)
a complete
normal
the latter
one may,
which
appear
aim to present
for a concentrated
of integration
half-plane.
obtained
y2
z , since
functions
of 1/2np,
will
+
kernel,
instead
indices
we do not
Jfx2
coordinate of this
y) is then
1/4nu
investigated path
12 1 =
oo) in the Hankel
corresponding
that
r =
the spatial
for i\ (a,
since only
was
the original
00, +
.4 formula
iy,
the asymptote
of the integral:
formula,
displacement
(-
x -
with
To calculate
to the interval
in front
Z =
of confusion
of the circuits.
integrals
iy,
+
formulas.
the multiplier
totic
2
(1.10)
method case
of solving
of a circular
(p, tr, te) in a Fourier
series
in
eid @, $., 63) =
For
each
of the components
t 1m = We note
[g, further
w1 =
etc.
(r) .+ ih,
u +
iv =
Introducing
these
the integration
We give
t,,
=
(r),
gm (r), h,
(2.1)
gm (r) eime,
(r)] oifm+lJe,
(r)) eime
t?,
L?l =
[g,
=
(r) e into we have
h,
(r) -
ihm (r)l
f.?i(m-l)Q
(2.2)
that
we perform
[31 (p. 392).
5 (I,, --CC
values only
(ur +
iu,)
in (l.lO), over
the result
eie, where
the angle
70, = .z should $J with
of the calculation
u -
iu =
be replaced the help
(ur -
iu,)
by z -
of Graf’s
8’ i =
addition
reio theorem
pei*
Elastic
Rere matrix
the index
with
m denotes
the following 03
the
half-space
Fourier
647
vibrotion
coefficient
(of
eime).
The
Afm) is *
kernel
elements: co
s
s
AZ,(~) =
4,tjm)= 2 E W) Jr,, (~4 J,(YP) T d-r,
0 W) Jm_l hr) J, (w) T’ti
0
0
Azlcm) = -
2 h2) Jm+l
s
(TP) Jm+I
(Tr) ‘fa dr
0
(2.4) The remark integrals.
made
It would
Cf.gs h lrn, plicated.
but
The
at the
have
end
been
possible
the formulas notation
would
here
be even
series
to the real
(the
m = 0 is omitted K,,
section
form.
here). co
= -
holds
to express
for the elements
of the Fourier
subscript
of the first
for the calculation
(tu, ur, r&
of the kernel
would
directly then
more
complicatedif
we went
Only
in the case
m = 0 is there
In these
k12)“’
R (P)
0
in terms
have
been
of
com-
from the complex
form
a simplification
equations
kz2 (r2 -
5
of these
JO (v)
JO (YP) TM
cc
K,,
c
k2 W - b?)
= -
b
R (r2)
J, (‘r4 J, (TP) ‘r dy,
KOI =
s T2e
=
W)
Equations
(2.5)
the first torsion
equations
into
Lam;
equations
tions
(2.5).
consequence
may
be considered
of equations
describes
(the
31 -
which
Reissner-Sagoci
dr
systems
from the
(1.6)
physical
The
(rS -
of the punch,
to the
vanishing
point
of view,
@
(r2) =
2 (y2 -
decomposition
depends ks2)“’
problem, while
of the system
of R (ya)
we have H (r2) -
(YP) 7 dr k,a)‘j*
of the axisymmetric
The resolution
corresponds
case.
(2.6)
so
the indentation
problem).
axisymmetric
is clear
of formula
as the equations
d escribes
(2.5)
two independent in the
Jo h4 J1 (‘rp)
K _ O”JI (~4 JI
JO(w) JI 04 &,
0
where
(P)
0
03 Kt,
JPe
the
of the system
in the second on the
second
of integral
fact
of equathat
as a
of
and N. A. Rostovtseo
V. N. Zokorko
648
The
solution
of the system
on the absenceof quadraturea
will
will
the description
restrict
without
exact
unavoidably
friction
3.
(2.3)
any simple
and
even
in the simplest
solution
reduce
the accuracy
of any
of such
au algorithm
to the problem
g,,, = h,
{ 1,
(P) PdP TE
0
is clear
This help
that
Fourier equation
in this
reduce
solved
i.e.
may
also
function.
method
equation functions
Rorodachev
is used,
In order
(Yr) J,
algorithm.
Therefore
we
of a punch
(YP) YdY
(3.1)
which
contains
(3.1)
from
required
Rorodachev’s
directly
friction
Hankel
transforms
pair
of integral
of N.N.
directly
coefficients
[2] with
was
were
used
to
were
then
equations
for a certain
numerical in that
pressure
to an equation
the
and bonding
Lebedev)
additional method
the sought-for
is transformed
out the singularities
The
of Lamb
without
(by the method
of the pressure differs
paper
to he the
sin r&l.
from the results problem
equations. kind
below
to simplify
directly
[41. In that
of integral
Calculation
fm (r ). Equation
by separating
turns
approximate
of indentation
of cos m f? and
axisymmetric
of the second
developed
(3.1)
The
to a pair
to an equation
auxiliary The
numerical
wm (r ), fm (r ) may be considered
be obtained
by N.M.
(Y) J,,,
as the coefficients
theorem.
the problem
reduced
(2.5)
0
equation
series,
of the addition
effectively
case
Cumulative
= 0 we have
wm (r) = It
problem.
bonding.
In the case
of a real
axisymmetric
of the static
only
quadratures. the single
through
the unknown
of the second
kind
of the kernel.
the subsequent
calculations
we change
to dimensionless
quantities
by setting
r
=a2, Making
p =
a&, ka = a / a, h = k, / k,
the further
y = kp
substitution
9.m (Z) =
\ IIm (E) io!Er
f,, (r) = pII,,,
(r)
CT),
I(!,,,
(3.1).
we obtain
J,"
(ll:S) Cf.7
(3.2) inside
the integral
*%j+
SJ,
(USS)
(3.3)
0
j? (a) = (29 In the neighbourhood
- as
fz@)7n (x)
-
1) -
49
(r? _
1)“’ (.$? _
jiZ)‘/Z
of s = J) we have
v/s% - ha =
F (4 Here C is a regular
function. 1
2 (i -
2 (1 ” Equation
! ha) \ . &I
h*)
-
(3.3) cc
takes
(E) SdE \ J,,W)
0
(G (32) -= I,)
G (s)
(3.4)
the form
J,(EQ
dh =
0
(3.5) = The kernel
%I
(z) -e { Ku 0
on the left-hand
side
(5) go% TJ_
(czars) J,
(ags)
G (s) ds
0 is a special
case
of the Weber-Shafkhaitlin
integral
Elastic
For
the
half-space
649
vibration
0
equation 1
s
63 w,,(;) (x9 El dE = f (4
‘p
0
the inversion
formula
151 is known -
cp(r)
=
r (i/u (1 +
1
21-r a?Q
p +
4) r (l/e (1 +
p-P-9&
s (t2 _
d
z2)1/~(1-r+o-p)
dt
this
formula
into
we obtain
(3.5),
u’+~ f (u) du so (ta _
x
Substituting
q - p)) ’
r 3_
t
d
’ dr
r +
ua)‘/z(l-r+p-91
an equation
of the
second
kind
1
(3.6)
in which 4 (1 -
N) xm-’
-
d
‘m (2) = - r (m + V2) r (l/J dx
d”
2 (1 -
Km (x7Y)=
G (s) J,
r trn+ l/2) r (l/2)
(axs) J,,,+
(ast)(as)
(3.7)
‘A ds } dt
(3.8) 4.
In the particular
case
of a plane
pr cos 8 =
w z wg + In this
inclined
punch
we have
a (W, -!-
px cos cl)
case @ =
cPl =
W@,
(0
@u
is the angle
of inclination
of the punch) Here
it is convenient
to transform
x+ t
H, (x) = II, (x) (1 Noting
that
1 -ha
= ?/?(1 -
to the new
v),
unknown
functions
H1 (I) = II, (x) (1 -
we obtain
z$‘*
the equations 1
2 Ho (x) =
2wo 3T (1 -
Y) -
JL (1 -
v) s K,
(x, y) Hc, (y) dy
0
HI (4 = with
kernels
equal,
x (r!!
respectively,
,,) 4x -n
to
(1”
y) i KI (x, Y) HI (Y) dy
(4.1)
V.N. Zokorko
650
Ke
(I,
Y) =
+
(ej)‘”
2
i
and N.A.
Rostovtsev
&f&
i
G (a) J,
(aya)
cos (ato)
(4.3)
da
0
‘Ce first solution
consider
of this
equation
equation
(4.1)
with
the kernel
H,(O) =
This
is related
to the solution p,(“’
to find
the
first
1
s
of the corresponding
zeroth
approximation
to the
approximation
Y) a l/a2
-
we calculate
z
the order
problem
2tL~O
(r) =
0
Changing
static
ra
the integral
(ejh{-&i$&--i
K, (2, y) dy = j,
0
The
2we 7I (1 -v)
n (1 In order
(4.3).
is
G (0) Jo (ay) cos (ata) du}dy
0
we obtain
of integration,
1 v’-&t&~C(u)cos(a+u)~du}
\&,(qy)dy=
(4.5)
lz
0
b
the integral
To calculate
00 1 z we use
the theory 1 =
of residues.
We have
C
cos auo) co9 am, +
[nc sinau,
Now the function of t , whose respect
I (a, t) =
to this
i WI=0
IP,
co9 am, +
1 -- (WI res G (a)
c=
a0 is the root of the Rayleigh
powers
s
V (a) (1 -
cos au) cos (atu) da
0
( where
da
G (a) cos (am) y
1
nc (1 -
-I- i
with
c b
V (a) sin au cos (atu) do] Im G (a)
) v (0) = ~ au a=0.
1
F (01 = 0.
equation
coefficients,
+
1
I = 1 (a, t ) may be expanded (complex)
I
which
in a rapidly depend
converging
on U, may
series
be expanded
in even in series
parameter (a) +
i
Q,,, (a)1tarn9
P,n
(a) =
(-(~~~
i
k=l
akfm)aab
(4 .6)
651
Elastic half-space vibration
Qm ($ =
(-,;Z p”” 2
bk(rWa2k+l
)
k=l
Substituting
uk-l [nC(J62m+2k-1 + q*+2k_11’
(-
b,(m) =
(2k -
q,,-
I)!
for I (CL, t
(4.7)
1 @+?'du 0
) in (4.5) its expansion
(4.6)
and integrating,
we have
1
. K, (r, Y) dy =
i
c
t2mtl
4-d' M,
(4
[P,
=
1:
ZG s Jfp-
za
z
(VI The polynomials polynomials
in z.
i Qm (a)1 M,
d dt = yg
even powers
Ml (2) =
1,
in u and are consequently
2ua -
1,
(4.8)
M, (I)
8~’ + 6~2 -
1M,,,(I)I< const
side of equation
R'egive below the resultsof
even-power
calculating
PI (a)= -$
+
ha =
the formulas
Qn for h2 = 1/3, taken from reference [4], we find PO (a) = 1.0068~~ - 0.07913a4 + 0.002665ae 0.3186as
P,
the coefficients
3tc =
-
0.01622aS
-
-
0.0004206ar
+
+
0.0004761a’
-
ak Qz (a)= q-[1.9466a - 0.3533as+ 0.01996a6-
.
[0.9583a2- 0.09361a4
P, (a)= - $
[1.1232a2
Q3 (a)= - $
12.1199a
-
0.1068a4
to the solution
(1”:q {1 -
+
0.3858aS
a8 ' P4 (a)= -g- [1.2816a2- . . .], The first approximation
+
n (1 “_
+
.
+
. . .]
...
. .]
. . .] . . .I
of equation
(4.1)
y)
(a) +
i
..,
. .]
as Q4 (a)= m (2.3946a-
-0
of
. ,.
[1.9113a
-fi- 0.003560a6
= ~
I/,). values
-
0.3244a3
Qrnca) for
and a tableof
0.00005433$
-
P, (a)= $
te) and
[0.9495a2- O.O7986a*f 0.000312a6 -
on the
1/s (A = ~1, v =
(4.7)
0.01766as
Q1 (a)= - T
the series
series
(ao)2"
In this case
2.3243a
i
x2)
and consequently
up to terms in U lo for the case 1.0248.Using
4u= -
1 -
by the convergent
m = 0, I, 2, 3, 4, accurate 1.0877,
V,u4 f
(US =
v/m,
is majorized
=
1
; If; m=O (Pm)!
Ho(‘)(4
us sin% T)~ dq
, VI-P=7)
-$=u
Mm (x) areof
It may be shown that
Q. (a) =
(4.6)
'lc * u sin cp (1 s 0
M, (x) = la/GE -
o. =
(I)
We give the first few polynomials
M, (5) =
right-hand
(a) f
m=o
[p,
,
consequently i Q,,,
. .] takes
the form
(41 Mm (4) f4ag)
652
C’.N. Zakorko
The
second
the solution
term
of the
We consider approximation.
in the curly
and changing
brackets
corresponding
now
static
equation
Calculating
the order
and N.A.
(4.2) the
in equation
with
the kernel
of integration,
1/l -
%\
l/t2 -
x3.
x
the theory
correction
to
(4.4)
and
seek
the solution
in the
first
m
x2"\ dx.
0
Applying
is the dynamic
we obtain
1
(x, y) ydy =
(4.9)
problem.
integral
1
K,
Rostovtsev
G (0) sin (ato)(F
-
cy)
da (4.10)
0
of residues
to the calculation
of the inner
integral
M 11 0% r) =
cos ao a.
--‘;,$r (’
G (0)
s
sin (ata) ds
0
we find I, (a, t) = [rtc (sin
i-
au0 -
au -
i V (a) (sin
I -
z:
’ -,“,“”
““)
“)
sin (au&
sin (at@ da]
+
-/-
0
+
ijnc (00s
aoo-s~)sin
(ao,$) +
iv
(0) (~0, t.J_
e),i,
@a)
da]
0
Expanding of the resultant
now
the function
series
I, (at)
in powers
in powers
of t and the a-dependent
coefficients
of CL, we obtain
1, (a, r) =
5
[Pa,
(a) +
iQk
(a)]
t2m-1
(4.11)
m=1
,,w =
((2/( (-
dkcrn) =
(2k Substituting
I)“-’ 2)
2mt2k-2
+
! 2k (no%
iy-*
3)
! (2k -
1) Wr”
for I, (CL, t ) in (4.10)
its
2m+2k-3 +
expansion
Y
2m+2k-2)
Y2n1+2h._J,
(4.11)
C&l)
&cm)
z
and integrating,
0
we obtain
1 1 KI (I, y) ydy
0 We quote h2 =
‘13 (Y =
the expansionsof l/r)
=
fPf,(a) ;f ia",, (a)lM,_,
2
(x)
rn=l the first
few coefficients
PC
(a)
and
Q;(a)
for
Elastic
P”1 (a) = G
(1.0068a
-
Q"1(a)= $
[--0.6371az
half-space
0.2374as +
+
0.01331~~
0.06489a4
-
iO.94950 - 0.2398as + 0.01560a6
Q; (a)= $
[-
&
(a) = $-
Qi (a) =
Pi (a) The
[0.9583a
gr
= T-
first
-
+
0.2808~~
[ -0.7066aa
+
a7 [l.l232u
0.07066~~
-
+
-
-
0.00005279ae
0.000445Oa’ +
. . .]
-
...]
+
. . .]
. . .]
.. .], Qi (a)= -4
to the solution
...]
+
. . .]
+
0.00039OOa’
0.002851a6
0.01780~~
0.07982a’
0.3204as
-
approximation
+
-
0.002524~~~
Pi (a)= $
0.6489a2
653
vibration
of equation
(4.2)
[-0.7982aa
consequently
. . .]
+
takes
the
form
f&(l)(4 =
W a
(I
-
v)
t
1-
It
2_
(I
v)
[Pk (a) + iQ& WI Mm_,
i
CT)} z
mz_1
Here,
as in (4.9),
corresponding the same
static
integrals
the second problem.
will
amount
in principle
of calculation
is the dynamic that
be encountered,
I (u, t ) and f 1 (u, t ), there Thus
term
We note
the
will
increases
except
be .f,,,+,,,
subsequent
correction
in obtaining
to the solution
the subsequent
in place
of the
approximations,
of sin CLCT/QU in the integrals
(uu) / (r~a)“+‘ls.
approximations
are
foundin
a similar
manner,
but the
considerably.
BIBLIOGRAPHY R., Die Differential-
und lntegralgleichungen
der Uechanik
the surfaceof
solid.
1.
Frank, P., and von vises, und Physik, 1930.
2.
Lamb, Trans.
3.
Watson, G.N., A Treatise Cambridge, 1944.
4.
kontaktnaia zadacha dlia shtampa s ploski m Borodachev, N.M., Dinamicheskaia krugovym osnovaniem, lezhashchego na uprugom poluprostranstve (The dynamical tact problem for a punch with a plane circular base lying on an elastic half-space). lru. AN SSSR, OTN, ?vlekhanika i mashinostroenie, NO. 2, 1964.
5.
H., On the propagations of tremors over Roy. Sot. (Series A), Vol. 203, 1904. on the Theory
of Bessel
Funcrions.
an elastic
Second
Phil.
ed.,
con-
-nogo uravneniia teorii lineinogo Rostovtsev, N.A., 0 nekotorykh resheniiakh integral’ deformiruemogo osnovaniia (Some solutions of the integral equations of the theory of a linear deformable foundation). PUU, Vol. 28, NO. 1, 1964.
Translated
by F.A.L.