Transcript Slide 1
Introduction to
Contact Mechanics
18.07.2015
Contact mechanics
Contact mechanics is the study of the deformation of solids that touch each other at one
or more points. The physical and mathematical formulation of the subject is built upon the
mechanics of materials and continuum mechanics. The original work in this field dates
back to the publication of the paper “On the contact of elastic solids” (“Ueber die
Berührung fester elastischer Körper”) by Heinrich Hertz in 1882.
Hertz was attempting to understand how the optical properties of multiple, stacked lenses
might change with the force holding them together. Results in this field have since been
extended to all branches of engineering, but are most essential in the study of tribology and
indentation hardness.
2
Heinrich Rudolf Hertz
Born
February 22, 1857
Hamburg
Died
January 1, 1894 (aged 36)
Bonn, Germany
Heinrich Hertz was buried in Ohlsdorf,
Hamburg at the Jewish cemetery.
His nephew Gustav Ludwig Hertz
was a Nobel Prize winner
(inelastic electron collisions in
gases), and Gustav's son Carl
Hellmuth Hertz invented medical
ultrasonography.
3
Memorial of Heinrich Hertz on the campus of
the University of Karlsruhe
4
Motivation of Hertz’s research
Hertz's combination of theoretical and experimental work was so remarkable that during
the next spring he was allowed to present his research as his doctoral dissertation, "On
Induction in Rotating Spheres," and received the degree magna cum laude ("with great
honor" (direct translation: "with great praise") ). He became Helmholtz's assistant at the
Physikalisches Institut at the University of Berlin, and during his 3 years there Hertz's
most remarkable achievement was his work on the pressure arising between two plates
in contact. The influence of his conclusions on the construction of precision instruments
was so great that his paper "On the Contact of Elastic Solids" was simultaneously
published in a scientific and a technical magazine. But the promise of the future lay with
his work on electric and cathode-tube discharge, published in 1883 in two papers.
5
Worm drive
Worm and worm gear
(worm wheel).
A worm drive
controlling a gate.
6
A double bass
features worm
gears as tuning
mechanisms.
Gearing
Bevel gear
used to lift
floodgate by
means of
central screw.
Helical gears from a Meccano construction set.
Top: parallel configuration.
Bottom: crossed configuration .
7
Contact of rail tracks and rail wheel
Two wheelsets in a North American (Bettendorf-style)
freight bogie displayed at the Illinois Railway Museum.
8
Driving wheel of steam locomotive
Tyre of the Wheel in Contact With
the Road
9
Tyre of the Wheel in Contact With
the Road
10
The area of contact mechanics
•
The area of contact mechanics focuses on computations involving elastic,
viscoelastic, and plastic bodies in static or dynamic contact. Contact mechanics is
foundational to the field of mechanical engineering; it provides necessary information
for the safe and energy efficient design of technical systems.
•
Principles of contacts mechanics can be applied in areas such as locomotive wheel-
rail contact, coupling devices, braking systems, tires, bushings, ball bearings,
combustion engines, mechanical linkages, gasket seals, metalworking, metal forming,
ultrasonic welding, electrical contacts, and many others. Current challenges faced in
the field may include stress analysis of contact and coupling members and the
influence of lubrication and material design on friction and wear. Applications of
contact mechanics further extend into the micro- and nanotechnoligical realm.
11
Conforming and non-conforming
problems
A contact is said to be conforming if the surfaces of the two bodies
‘fit’ exactly or even closely together without deformation. Flat slider
bearings and journal bearings are examples of conforming contact.
12
Conforming contact
13
Conforming contact
14
Non-conforming contact
Self-aligning ball bearing
15
Cylindrical Roller Bearing
Bodies which have dissimilar profiles are said to be non-conforming.
When brought into contact without deformation they will touch first at
a point – ‘point contact’ – or along a line – ‘line contact’.
Conforming or non-conforming
contact?
Deep Groove Ball Bearings
16
Thrust Ball Bearing
Non-conforming contact
Line contact arises when the profiles of the bodies are conforming in
one direction and non-conforming in the perpendicular direction. The
contact area between non-conforming bodies is generally small
compared with the dimensions of the bodies themselves; the stresses
are highly concentrated in the region close to the contact zone and are
not greatly influenced by the shape of the bodies at a distance from the
contact area. These are the circumstances with which Contact
Mechanics is mainly concerned.
17
Nonlinearity of non-conforming
contact
P
M
q
The Pierre Pflimlin bridge being
constructed over the river Rhine
between Germany and France.
18
This concrete balanced cantilever
bridge under construction employs
movable cantilevers to support
formwork.
u; v;
P;M; q
Nonlinearity of non-conforming
contact
P
P
u; v;
u; v;
P
19
u; v; ; ; P
P
ij
1
(ui, j u j,i )
2
εσ
Schematic of non-conforming
contact
Non-conforming surfaces in contact at O
20
Forces and moments
acting on contact area S
Equilibrium equations in Contact
Mechanics
21
Axisymmetric problem of the
Theory of Elasticity
(two aspects: geometry of elastic medium; loading)
u u r e r u e θ u 3e 3
u 0
r
0
r 3 r 3 0
ur
rr
;
r
u3
33
;
x3
22
x3
ur
;
r
1 u r u3
r 3
2 x3 r
Axisymmetric problem of the
Theory of Elasticity
rr 1 r r 3 rr
0
r
r
x
r
3
r 1 3 2 r
0
r
r
x
r
3
r 3 1 3 33 r 3
0
r
r
x
r
3
r
Body forces are absent.
In thecase
0
r 3 r 3 0 2nd equation is satisfiedidentically
rr r 3 rr
0
r x
r
3
r 3 33 r 3 0
r
x3
r
23
x3
Love’s stress function
It may be verified by substitution that Equilibrium equations are satisfied if we take
2
2
2
rr
x3
r
2
1
x3
r r
2
2
2 2
33
x3
x3
2
2
r 3 1 2
r
x3
Body forces are absent.
Provided that the stress function (r, x3) satisfies biharmonic equation:
24
2 1 2 2 1 2
2
2 2
2 2 2 0
r r x3 r
r r x3
r
Displacements corresponding to
Love’s stress function
1 2
ur
2G r x3
2
1
2
21 2
u3
2G
x3
A particular problem is solved if we can find such a biharmonic function (r, x3) that
also satisfies the boundary conditions.
25
Polynomial solutions of biharmonic
equation
0 A0
1 A1 x3
1
2 A2 x32 r 2 x32
3
3
3 A3 x33 x3 r 2 x32
2 B2 r 2 x32
5
3 2
4 6 2 2
2
2 2
3 B3 x3 r 2 x32
r x3
4 A4 x3 x3 r x3
7
35
4 B4 2 x32 r 2 r 2 x32
5
5 10 3 2
2
2
2 2
3
2
2
2
A
x
x
r
x
x
r
x
5 3
3
3
3
3
5
B
2
x
3
r
x
r
x
5
5
3
3
3
9
21
.......................................................................
....................................................................
26
Kelvin problem (1848)
Concentrated force acting at a point in an
infinite solid
P
r
O
27
x3
Kelvin problem solution
O P
r
1
2
2 2
1 A1 r x3
3
2
2 2
2 A2 x3 r x3
5
3
1
2
2
2
2
2
2
3 A3 x3 r x3 r x3 2
3
.......................................................................
x3
B r2 x
28
1
2 2
3
1
2
2 2
1 B1 r x3
1
2
2 2
2 B2 x3 r x3
.......................................................................
Kelvin problem solution
Small spherical cavity around the singular point O.
O P
r
R
O
x3
n
Br x
2
1
2 2
3
dS
f3
r
rr
fr
r3
fdS P
33
S
x3
29
n cos e3 sin er
Stress Field in Kelvin problem
2
2
2 5 2
2
2
2 3 2
2
rr x r 2 B 1 2 x3 r x3 3r x3 r x3
3
1
2
2 3 2
2
x
r
x
2
1
B
3
3
x
r r
3
2
2 5 2
2
3
2 3 2
2
2
2 2 B 1 2 x3 r x3 3 x3 r x3
33
x3
x3
2
2 5 2
2
2
2 3 2
2
2
r 3 r 1 x 2 B 1 2 r r x3 3rx3 r x3
3
30
O
r
dS
n
f3
rr
fr
r3
x3
31
33
dS 2r Rd r 2 x32 2rd
From the condition of equilibrium of a ring-shaped element,
adjacent to the cavity, the component of surface forces in the x3
direction is
Fi 3 0
i
f 3 dS 33dS cos r 3 dS sin 0
f 3 33 cos r 3 sin
sin
x3
x3
r
r
;
cos
r 2 x32 R
r 2 x32 R
3 x32
1
f 3 B 1 2 2
2
2
2 2
r
x
r
x
3
3
fdS P P f 3 dS
S
S
2
P 2 f 3 r 2 x32 2rd 8 B(1 )
0
B
32
P
81
33 0; r 3
B1 2
P1 2
@ x3 0
2
2
r
81 r
Center of compression
O P
P
O
r
x3
P
r
x3
RR
33
O
P
41 2 A
3
R
P
r
P
x3
Force on boundary of a semi-infinite
body (Boussinesq problem, 1885)
P
r
O
B.C. 33 r 3 0 @ x3 0
r0
?
rr 0
@ x3 0
R
O P
r
x3
34
Kelvin problem:
33 0; r 3
x3
B1 2
P1 2
@ x3 0
2
2
r
81 r
Boussinesq problem (Cont’d.)
For bottom semi-infinite space
O P
r
x3
Kelvin problem:
33 0; r 3
B1 2
P1 2
@ x3 0
r2
81 r 2
r 3
B1 2
P/2
r2
O
r
R
35
x3
Boussinesq problem (Cont’d.)
P
O
r
=
R
x3
O
+
r
O
r
36
x3
x3
Boussinesq problem (Cont’d.)
For bottom semi-infinite space
r
O
O
r
x3
x3
37
Boussinesq problem (Cont’d.)
P
O
r
=
R
x3
O
r
38
x3
+
O
r
x3
Boussinesq problem solution
P
1 x3 2
2 1 2
2
2
2 5 2
1
2
r
x
3
r
x
r
x
3
3
3
2
2
rr 2
r
r
3P 3 2
2 5 2
x
r
x
3
3
33
2
P 1 2 1 x3 r 2 x 2 1 2 x r 2 x 2 3 2
3
3
3
r 2 r 2
2
r 3 3P rx32 r 2 x32 5 2
2
1 2 1 P 2 2 1 2
1
2
2
2 3 2
u
x
r
x
1
r
x
r
x
3
3
3
r
3
2Er
1
2
u P 1 x 2 r 2 x 2 3 2 21 2 r 2 x 2 1 2
3
3
3
3 2E
1 2 1 P
u
r
2Er
@
x3 0
2
P
1
u
3
Er
39
Boussinesq problem solution
(semi-non-dimensional form)
2
1 x3 2
2 1 2
2
2
2 5 2
1
2
r
x
3
r
x
r
x
rr
3
3
3
rr P
r 2 r 2
5 2
33 3 x33 r 2 x32
1 2 1 x3 r 2 x 2 1 2 x r 2 x 2 3 2
3
3
3
r 2 r 2
5 2
r 3 3rx32 r 2 x32
1 2 1 2 2 1 2
1
2
2
2 3 2
u
x
r
x
1
r
x
r
x
3
3
3
r
3
Er
1
2
u 1 1 x 2 r 2 x 2 3 2 21 2 r 2 x 2 1 2
3
3
3
3 E
1 2 1
u
r
Er
@
x3 0
2
u 21
3
Er
40
Boussinesq problem solution
(semi-non-dimensional form; x3=0, r > 0)
1 2
2
rr P rr r 2
33 0
1 2
r2
0
r3
41
1 2 1
ur
Er
2
2
1
u
3
Er
Boussinesq problem solution
(semi-non-dimensional form; r = 0, x3> 0)
2
1 x3 2
2 1 2
1
2
r
x
rr
3
rr P
r 2 r 2
3
2
33
x
3
1 x
1
1 2
1 2 2 23 r 2 x32 2
x3
r r
0
r3
ur 0
1
1
1
2 1
2
u
1
2
1
3
2
3 E
x3
x3 Ex3
42
Boussinesq problem solution
(semi-non-dimensional form; r = 0, x3> 0)
(Cont’d.)
1 x3 2
2 1 2
r
x
3
2
2
r r
r2
1 1
2
2
2
x3
r x3 x3
x3
2 2
r2
r r x32
r 2 x32
1 r2 1 r4
1 2 4 ... 1
2
2
x
8
x
x3
1
1
r
1
3
3
2
2
3 ...
r 0
2
2
2
2
r
2
x
8
x
2
x
r x3
r x3 3
3
3
43
Boussinesq problem solution
(semi-non-dimensional form (Cont’d.)
x3 0;
44
r 0
1 2
2
rr P rr r 2
33 0
1 2
r2
0
r3
1 2 1
ur
Er
2
2
1
u
3
Er
r 0;
x3 0
1
2
rr P rr 1 2 2 x 2
3
3
33 2
x3
r 3 0
ur 0
1
2
u
3
2
3 Ex
3
Load distributed over a part of the
boundary of a semi-infinite solid
q
r
O
x3
a
r
45
Solution for displacements
@ x3 0;
ra
2
a 2 2
41 2 qr 2
d
a
2
u3
1 sin d 1
2
0
E
r
r 0
a
2
1
sin
r
@ x3 0;
ra
41 2 qa 2
r
2
u3
1 sin d
0
E
a
2
@ x3 0;
ra
@ x3 0;
r 0
41 2 qa
1 2 qa
u3
1.27
E
E
46
1 qa u
2
2
u3
E
|
3 max
Solution for stresses
@ r 0;
x3 0
x33
33 q 1 2
2 32
a
x
3
3
21 x3
x3
q
rr 1 2
2
2
2
2
2
a x3 a x3
@ r 0;
x3 0
If 0.3
q1 2
2
rr 0.8q
max |r 0; x 0 0.1q
@ r 0;
x3 0
33 q;
rr
3
3
x3
x3
1
q 1 2
3
max |r 0; x 0 rr 33
1 2
2
2
2
2
2 2
a x3 2 a x3
q 1 2 2
21
max |r 0; x 0|
1 21 @ x3 a
2 2
9
7 2
If 0.3
max |r 0; x 0| 0.33q @ x3 0.638a
3
3
max
3
or
47
max
1
max |r 0; x 0| q
3
3
max
2
@ x3 a
3
@ r 0;
x3
33 rr max 0
x
r
r ; x3 3 ;
a
a
q
@ r 0;
x3 0
x33
1
33 1
1
32
2 32
1 x3
1
2 1
x3
21 x3
1
rr 1 2
2
2
1
x
3
@ r 0;
x3 0
33 1;
rr
If 0.3
1 2
2
rr 0.8
@ r 0;
x3 0
3
3
x3 1
21 1
1 2
1
1
1 x32 2
1
1
2
x32
x3
max |r 0; x 0 0.1
3
3
x3
1
1 1 2
3 x3
max |r 0; x 0 rr 33
1
2
2
2
2 2
2
1
x
1
x
3
3
3
1 1 2
1
3 1
1
2 2
2 1
1
1
2
2
x
x
3
3
1 1 2 2
21
max |r 0; x 0|
1 21 @ x3
2 2
9
7 2
If 0.3
max |r 0; x 0| 0.33 @ x3 0.638
48
1
2
or
max |r 0; x 0|
@ x3
3
3
3
3
max
3
max
3
max
@ r 0;
x3
33 rr max 0
Load distributed over a part of the
boundary of a semi-infinite solid
Radial (Circumferential), Axial and Maximum Shear
Stresses
Dimensionless stress
Dimensionless coordinate
-1.2
0.0
49
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.5
1.0
1.5
2.0
2.5
3.0
Axial Stress
Radial Stress
Maximum Tangential Stress
0.4
Yield Criteria
Stress deviator tensor
The stress tensor ij can be expressed as the sum of two other stress tensors:
1. A mean hydrostatic stress tensor or volumetric stress tensor or mean
normal stress tensor, pdij, which tends to change the volume of the stressed
body; and
2. A deviatoric component called the stress deviator tensor, sij ,which tends to
distort it .
ij = pdij+ sij
where p is the mean stress given by p=kk/3=(11+22+33)/3=I1/3
The deviatoric stress tensor can be obtained by subtracting the mean hydrostatic
stress tensor from the stress tensor:
sij ij
s11
s
21
s31
50
s12
s22
s32
kk
dij
3
s13 11 12
s23 21 22
s33 31 32
13 p
23 0
33 0
0
p
0
0 11 p
12
13
0 21
22 p
23
p 31
32
33
p
Invariants of the stress deviator
tensor sij
As it is a second order tensor, the stress deviator tensor also has a set of
invariants, which can be obtained using the same procedure used to calculate
the invariants of the stress tensor. It can be shown that the principal directions
of the stress deviator tensor sij are the same as the principal directions of the
stress tensor ij. Thus, the characteristic equation is
where J1, J2 and J3 are the first, second, and third deviatoric stress
invariants, respectively. Their values are the same (invariant) regardless of the
orientation of the coordinate system chosen. These deviatoric stress invariants
can be expressed as a function of the components of sij or its principal values
s1, s2, s3 or alternatively, as a function of ij or its principal values 1, 2, 3
51
Invariants of the stress deviator
tensor sij
Because ,
the stress deviator tensor is in a state of pure shear.
A quantity called the equivalent stress or von Mises stress is commonly used in solid
mechanics. The equivalent stress is defined as
52
Isotropic yield criteria
Maximum Shear Stress Theory
Distortion Energy Theory
53
Pressure between two spherical bodies
in contact (Hertz’s problem, 1881)
Geometry of non-conforming contacting surfaces
z2
R2
r
N
z2
O
z1
M
Assumptions:
R1
54
z1
r R1 ;
r R2
z1 R1 ;
z 2 R2
Hertz’s problem (Cont’d.)
z2
r
r 2 2 R2 z 2 z 22
r 2 R2 z 2
z 2 R2 z 22 2 R2 z 2 r 2 2 R2 z 2
O2
r2
z2
2 R2
r
z2
O
55
N
By analogy
r2
z1
2 R1
Hertz’s problem (Cont’d.)
r2
z1
;
2 R1
z2
r2
z2
2 R2
1
1 r 2 R1 R2
MN z1 z 2 r
2 R1 R2
2 R1 2 R2
2
R2
r
N
z2
O
z1
M
R1
Particular cases:
(a) Contact between a sphere and a plane
56
1
r2
0 MN
R1
2 R2
r 2 R1 R2
(b) Contact between a ball and a spherical seat R1 0 MN
2 R1 R2
z1
Hertz’s problem (Cont’d.)
z2
P
R2
r
N
z2
O
z1
M
R1
P
57
z1
If the bodies are pressed together along the
normal at O by a force P, there will be a local
deformation near the point of contact producing
contact over a small surface with a circular
boundary, called the surface of contact.
Assuming that the radii of curvature R1 and R2
are very large in comparison with the radius of
the boundary of the surface of contact, we can
apply, in discussing local deformation, the
results obtained before for semi-infinite bodies
(Boussinesq problem). In other words, the
contact area lies approximately in the
separating plane and slight warping is
neglected.
Hertz’s problem (Cont’d.)
z2 O
2
w2
The tangent plane at O is held immovable
during local compression.
P O2΄
N
O
N΄
M΄
M
Contact interference
O1O2 O1O2 w w1 w2
w1
58
P O1΄
z1 O1
Hertz’s problem (Cont’d.)
M΄
M
M
z1
z1
z1
Final state: MM ' w1 uz
Rigid body motion: w1
Origin state
1
The distance between points M
and N diminishes by
z2
N
O
59
M
z1
N΄
M΄
M
MN M ' N ' MM ' NN '
w1 u z w2 u z
1
w uz uz
1
2
2
Hertz’s problem (Cont’d.)
If, due to local compression, the points M and N come inside the surface of contact
uz
w1
M
M
z1
Origin state
Rigid body motion: w1
N
O
60
z1
M΄
z1
M
z1
z1
z2
1
M
N΄
M΄
Final state:
MM ' z1 w1 uz
1
w1 z1 u z
1
w2 z 2 u z
2
w w1 w2 z1 z 2 u z u z
1
2
Hertz’s problem (Cont’d.)
w w1 w2 z1 z 2 u z u z
z1 z 2 w u z u z r 2
1
1
where
2
R1 R2
2 R1 R2
For any point of the contact area:
uz uz w r
1
61
2
2
2
Hertz’s problem (Cont’d.)
The intensity q of pressure between the contacting bodies and the corresponding
deformation are symmetrical with respect to the center O of the contact area
P 1 2
u3
- Boussinesq problem result at x3 0
Er
1 12 q s ds d
duz1
E1
s
uz1
1
q ds d
u z2
1
q ds d
2
1
E1
2
2
E 2
uz1 uz2 k1 k2 q ds d
where
1 12
k1
;
E1
k1 62
k2 q ds d w r 2
1 22
k2
E 2
Integral equation for q. Contact area
is unknown.
Although the integration is extended
over the entire contact area, integral
varies with the current point
coordinates.
Hertz’s problem (Cont’d.)
k1 k2 q ds d w r
2
Thus we must find an expression for q to satisfy the integral equation at each
point of the contact area. It may be shown that this requirement is satisfied by
assuming the semi-ellipsoidal distribution of contact pressure :
q q0
r2
1 2
a
or
q q0
a
2
r2 a
where q0 is the pressure at the center O
of the contact area.
Along a chord mn the pressure q varies by the dotted
semi-ellipse with the semi-principal axes:
a
2
r 2 sin 2 and q0
a
2
r 2 sin 2 a
The area enclosed by an ellipse is ab
1
2
2
2
63q ds
π
q
a
r
sin
a
0
The area of semi-ellipse indicated by the dotted line 2
m
n
Hertz’s problem (Cont’d.)
q ds
q0
A
a
1
A a 2 r 2 sin 2
2
k1 k2 q ds d w r 2
k1 k 2 q0 2 2 2 2
2
a r sin d w r
0
a
q0 2
k1 k2
2a 2 r 2 w r 2
4a
2 q0
2 a
w k1 k 2 q0
;
a k1 k 2
2
4
64
Hertz’s problem (Cont’d.)
The volume of an ellipsoid is given by V=(4/3)abc,
where a; b; c are the semi-principal axes of an ellipsoid.
The value of the maximum pressure q0 is obtained by equating the sum of
the pressures over the contact area to the compressive force P. For the
semi-ellipsoidal distribution of contact pressure this gives:
2
π q0a 2 P
3
q0
3P
2a 2
It means, the maximum pressure is 1.5 times the
average pressure on the contact area.
R1 R2
2 R1 R2
3 Pk1 k 2 R1 R2
a3
a 3 P ; q0 3 P
R1 R2
4
9 2 P 2 k1 k 2 R1 R2
w3
w 3 P2
65 16
R1 R2
2
Hertz’s problem (Cont’d.)
3 Pk1 k 2 R1 R2
a3
R1 R2
4
z2 O
2
w2
P O2΄
9 2 P 2 k1 k 2 R1 R2
w3
16
R1 R2
2
NN
N΄
M΄
O
M
Assuming that both spheres have
the same elastic properties and
taking =0.3, these
a 1.109
a
w 1.23
w1
66
P O1΄
z1 O1
3
3
P R1 R2
E R1 R2
P 2 R1 R2
E 2 R1 R2
3 P
q0
0.388
2
2 a
3
PE
2
R1 R2 2
R12 R22
Hertz’s problem (Cont’d.)
3 Pk1 k 2 R1 R2
a3
R1 R2
4
z2 O
2
w2
P O2΄
9 2 P 2 k1 k 2 R1 R2
w3
16
R1 R2
2
N
N΄
For a ball pressed into a plane
surface (1/R1=0), and the same
elastic constants for both bodies:
O
a 1.109
a
w 1.23
w1
67
P O1΄
z1 O1
PR2
E
3
3
q0 0.388
P2
E 2 R2
3
PE 2
R22
Hertz’s problem (Cont’d.)
z2
P
w2
O
a
w1
P
z1
3 Pk1 k2 R1 R2
a3
;
R1 R2
4
92 P 2 k1 k2 R1 R2
w3
16
R1 R2
2
For a ball pressed into a plane surface (1/R1=0), and the different
elastic properties of the contacting bodies:
3 Pk1 k 2 3 3
a
Pk1 k 2 R2
4
1
3 4 1
R1 R2
2
2
3
1
w 3 Pk1 k 2
3
R2
4
R2
68
6
1
a
3
a 3 R2 a w 3 3 a wR2
R2
R2
Hertz’s problem (Cont’d.)
z2
P
w2
O
w1
3 Pk1 k2 R1 R2
a3
;
R1 R2
4
2
a
P
z1
If a plane surface (1/R1=0) is rigid (E1 → ∞; k1→0):
3 Pk1 k 2 3 3
a
Pk2 R2 ;
4
1
3 4 1
R1 R2
3 P1 22 R2
a3
;
4
E2
69
92 P 2 k1 k2 R1 R2
w3
16
R1 R2
2
3
1
w 3 Pk2
4
R2
3 P 1 22 1 a 2 3 P1 22
w 3
4
E
R
R
4 E2 a
2
2
2
R2 q0
q0 a
3P
2
2
Substitution of q0
leads
to
a
(
1
);
w
(
1
2
2)
2
2a
2 E2
2 E2
2
Hertz’s problem (Cont’d.)
The contact pressures by the Hertz theory over the frictionless contact area
between two elastic solids of revolution are represented by the semi-ellipsoidal
distribution :
q r q 0
a
2
r2 a
At the external surface of the contacting spheres the displacements are given by:
1 2 q0
2
2
u
r
2
a
r
z
E 4a
2
1
2
1
a
2 3/ 2
u r
q
1
1
(
r
/
a
)
0
r
3E
r
1 2 q0
2
2
1
2
2 1/ 2
u
r
2
a
r
sin
(
a
/
r
)
r
(
a
/
r
)
1
(
a
/
r
)
z
E 2a
2
1
2
1
a
u r
q0
r
3E
r
r a; z 0
r a; z 0
70 over bar is used to denote values of the variable at the surface z = 0
The
Stress analysis in Hertz’s problem
Inside the loaded circle:
rr r 1 2 a 2
2 3/ 2
2 1/ 2
1
1
(
r
/
a
)
1
(
r
/
a
)
3 r
q0
2
r
1
2
a
2 3/ 2
2 1/ 2
1
1
(
r
/
a
)
2
1
(
r
/
a
)
3 r
q0
zz r 1 (r / a ) 2 1 / 2
q0
zr r z 0
r a; z 0
Outside the loaded circle:
71
2
rr r
r 1 2 a
q
q
3
r
0
0
0
zz
zr
r
z
r a; z 0
Stress analysis in Hertz’s problem
(Cont’d.)
The stresses along the z-axis:
1
rr r r
1
1
1
(
z
/
a
)
tan
(
a
/
z
)
2
q
q
2
1
(
z
/
a
)
0
0
1
zz r
2
q
1
(
z
/
a
)
0
zr r z 0
72
r 0; z 0
Comparison of the normalized stresses in
Hertz problem with the stresses produced
by a uniform pressure acting on a circular
area. The maximum shear stresses
1=0.5|zz- | are also plotted. = 0.3.
For the uniform pressure distribution:
1max 0.33 p @ z 0.64a
For the Hertz pressure distribution:
73
1max 0.31q0 0.47
P
0.47 pm @ z 0.57a
2
a
Stress analysis in Hertz’s problem
(Cont’d.)
In the case of brittle materials failure is produced by maximum tensile stress. This
stress occurs at the circular boundary of the surface of contact. It acts in a radial
direction and has the magnitude:
rr r a; z 0 1 2
q0
3
The other principal stress, acting in the circumferential direction, , is numerically
equal to the above radial stress but of opposite sign. Hence, along the boundary of the
contact area, where normal pressure (zz) on the free surface becomes equal to zero,
we have pure shear of the ammount q0(1-2)/3. Taking =0.3, this shear becomes
equal to 0.133q0. This stress is much smaller than the maximum shearing stress
calculated along the symmetry axis, but it is larger than the shearing stress at the
center of the contact area, where the normal pressure is the largest.
74