STRESSES DUE TO THE PRESSURE OF ONE
ELASTIC SOLID UPON ANOTHER
I. INTRODUCTION
1. Practical and Historical Aspects.-The question as to what
stresses are produced within the area of contact when one elastic solid
is pressed against another has for a long time been recognized as one
of great importance. That a solution of the question has a definite
practical application may be realized when it is considered that the
calculation of the stresses encountered in ball and roller bearings,
bridge rollers, rolling lift bridges and wheels on rails is impossible by
the common formulas of mechanics. The design of such structural
parts is based largely on results of tests, or trial and error methods.
This problem was attacked mathematically by H. Hertz who in
1881 published a paper* "On the Contact of Elastic Solids," and in
the following year an extension of the same analysis under the title
"On the Contact of Rigid Elastic Solids and on Hardness." Previous
to Hertz, Winkler and Grashof had both attempted mathematical
solutions, but their results were either approximate or involved some
unknown empirical factor. Hertz's analysis has long been considered
the basic method of solution and probably will give accurate results,
provided the radius of curvature is large in comparison with the size
of the area of contact. However, Hertz's computations deal defi-
nitely only with surface stresses, and he fails to recognize the great
importance of components of stress at right angles to the direction
of pressure.
Even though by the use of a mathematical method of solution the
various stress components (principal stresses) at points within the
area of contact are calculated, still the interpretation of the signifi-
cance of the magnitude of the stress components requires further
explanation.
2. Theories of Failure.-When at a point in a body there exist
stress components in more than one direction, the conditions under
which failure will occur have been the subject of numerous tests and
speculations, and as a consequence there have been proposed four
principal theories of failure.t By failure, in this connection, is meant
*H. Hertz, Gesammelte Werke, Vol. 1 (Leipzig) 1895. English Translation in "Miscellaneous
Papers," H. Hertz, 1896.
tFor a more detailed, yet brief, discussion of the theories of strength see Timoshenko and Lessells,
"Applied Elasticity," first edition, 1925, Chapter XVII.
ILLINOIS ENGINEERING EXPERIMENT STATION
the occurrence of a plastic yielding or flow of the material in the case
of ductile materials or rupture in the case of brittle materials. The
four theories mentioned are:
Maximum Stress Theory
Under the maximum stress (Rankine) theory it is assumed that
failure will occur whenever the maximum stress component equals or
exceeds the yield point of the material for that kind of stress, without
regard to the effect of other stress components that may be present.
This is the common theory and is sufficiently accurate if the stress is
almost entirely in one direction.
The Maximum Strain Theory
By the maximum strain (Saint Venant) theory it is assumed that
the maximum strain (deformation) due to the stresses acting at a
point is the criterion of failure. This theory also has been found to
be limited in correctness.
The Maximum Shear Theory
By the maximum shear (Coulomb) theory it is assumed that
failure will occur when the shearing stress on some plane (usually
oblique) due to the stress components is equal to the shearing yield
point of the material. It is generally agreed that this theory is satis-
factory for ductile material having approximately equal yield points
in tension and compression.
Mohr Theory
The Mohr theory is an extension of the maximum shear theory
which makes it possible to deal with such materials as cast iron where
the limits in compression and tension are widely different. It be-
comes the equivalent of the maximum shear theory when the tensile
and compressive limits are equal.
It is not considered necessary to discuss Haigh's Strain-Energy
theory, which is more recent than those mentioned. In all of the
calculations and discussion which are to follow the maximum shear
theory will be taken as the criterion of strength.
In text books on mechanics of materials* will be found the proof
that at any point in a body subjected to stress components acting at
right angles to each other the maximum shearing stress is numerically
equal to one-half of the algebraic difference between the maximum
and minimum stress components, and that the plane of the maximum
shearing stress bisects the angle between the stress components.
*For example: Fred B. Seely, "Resistance of Materials," Wiley & Sons, 1925, page 53.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 9
3. Division of Work.-In the preparation of the material for this
bulletin, the work of the two authors has been in large measure dis-
tinctly independent, although each has critically read the other's
manuscript. To V. A. Hoersch belongs the credit for the whole of
Chapter II, "Mathematical Derivation," which is the mathematical
work upon which the bulletin is based.
The laboratory work involved in the verification of the mathemat-
ical analysis was under the direction of H. R. Thomas, who also has
written all except Chapter II. The stress-component factors for
Tables 1 and 2 were calculated by both authors.
4. Acknowledgment.-This study was made possible by funds pro-
vided by the UTILITIES RESEARCH COMMISSION, WM. L. ABBOTT,
Chairman, for an investigation of transverse fissures in rails now
under way in the Fatigue of Metals Laboratory at the University of
Illinois. The following members of the Commission were appointed
as an advisory committee for this investigation:
G. F. TEBBETTS, Engineer of Structures, Chicago North Shore
and Milwaukee R. R. Co., Chairman.
A. HERZ, Engineer of Tests, Public Service Company.
J. S. HYATT, Engineering Assistant to Vice President, Chicago
Rapid Transit Company.
A number of meetings have been held with the advisory com-
mittee during the course of this work and they have given much
helpful advice.
The investigation has been carried on as a part of the work of the
Engineering Experiment Station at the University of Illinois and has
been under the general administrative direction of DEAN M. S.
KETCHUM, director of the Engineering Experiment Station, and of
PROF. M. L. ENGER, head of the Department of Theoretical and
Applied Mechanics.
All work in the Fatigue of Metals Laboratory is under the direc-
tion and supervision of Prof. H. F. MOORE, and to him the authors
wish to acknowledge their indebtedness for many helpful suggestions
and for sympathetic encouragement during all stages of this study.
The careful and painstaking work of MR. N. H. Roy, Research
Graduate Assistant, was of great value in the laboratory work in-
volved in this problem.
The idea that the shearing stress in the interior of bodies in con-
tact might be a critical factor was suggested some years ago by MR.
T. M. JASPER, who was at that time Engineer of Tests in the Fatigue
of Metals Laboratory, University of Illinois.
ILLINOIS ENGINEERING EXPERIMENT STATION
II. MATHEMATICAL DERIVATION
5. Introduction to Theory.-Two bodies are considered to be orig-
inally in geometrical contact at a single point and then pressed to-
gether so that there is contact over a small area called the compressed
area, bounded by a curve called the curve of compression. The theory
of the stress set up in either body was first developed by H. Hertz.*
Hertz completely solved the problem but left his solution in terms of a
certain Newtonian potential function. The integral involved in this
function was not expressed in terms of standard elliptic integrals.
The innovation in the present workt consists in doing this for points
directly under the center of the compressed area. A rather condensed
account of Hertz's theory appears in Love, "Mathematical Theory of
Elasticity," 3rd edition, 1920, pp. 190-196. All references to Love
are to this book. The notation of Love will here be followed. The
author will present what he hopes is a simplified account of the Hertz
theory, followed by his own elaboration of it.
Consider the dimensions of the compressed area as small in com-
parison with the radii of curvature of bodies 1 and 2. Body 1 may
then be considered in the unstressed state as an infinite solid
occupying all of space on the positive side of a plane zl = 0. When
body 2 is pressed against body 1, the stress in body 1 may be regarded
as due to a pressure P' distributed over the compressed area regarded
as an area in the plane z, = 0. One system of coordinates xiyizl will
be used for body 1 and another system x2y2z2 for body 2. Before the
bodies are pressed together, the xi and x2 axes coincide, the yi and Y2
axes coincide, and the zi and z2 axes are directed toward the interiors
of bodies 1 and 2, respectively. When the bodies are pressed together,
the surface, zi = 0, is deformed into a new surface, zi = wi, where w,
is a function of x1y1. In general, a point xjylzj is displaced to xi + uI,
yi + vi, zi + wi, where uiviwl are zero when the point xiyizi is at an
infinite distance from the compressed area. For body 2, a point x2y2z2
is displaced to x2 + u2, y2 + v2, z2 + w2 in a similar fashion. The
original point of contact of bodies 1 and 2 is the origin (0, 0, 0) in both
systems of co6rdinates. When the bodies are pressed together, this
point is displaced to 0, 0, wi in the one system and 0, 0, w2 in the
*An English translation of his theory may be found in "Miscellaneous Papers" by H. Hertz
(Jones & Schott, London, 1896, p. 146).
tSince this was written there have come to the author's attention two bulletins written by Prof.
N. M. Belajef, Professor, Mechanical Laboratory, Institute of Ways of Communications, Leningrad,
Russia. In these bulletins Prof. Belajef solves this same problem, also making use of elliptic integrals
for this purpose. Although expressed in different form, his equations for X5, Yv, and Z, agree exactly
with equations (59). However, Prof. Belajef apparently considers that a tensile stress within the
body is numerically equal to the difference in stress components; in this bulletin the shearing stress,
equal to one half of the stress difference, is considered to be the critical stress. It is interesting to
note that he makes use of the mathematical analysis for computing the stresses in a railroad rail under
a wheel load.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 11
Pz a 1 P a
u 42r x \r/ 4=(X+ ) ax log (z + r)
Pz a 1\ P a
4y Ty -r 4i(W+ w1) y log (z + r)
rigdiie (a oul of elasicit in sh)r any
SPz a 1)i P(X + 21) a 2
4srf e rf 4b(d + ) log (z + r)
*Love, p. 189, Eq. 35.
other system and the displaced point is the center of the compressed
area. Then a = w1 + w2 is the distance by which the two bodies
approach each other when they are pressed together.
6. Definitions.-The following notation will be used:
X1, Ml; X2, p2 Lam6 elastic constants for bodies 1 and 2; I1, A2 are the
rigidities (moduli of elasticity in shear) and Xi + %21,
X2 + %Pi2 the moduli of compression.
E,, E2 Young's moduli (moduli of elasticity) for bodies 1 and 2.
*1, a2 Poisson's ratios for bodies 1 and 2.
pi, p'l Reciprocals of the principal radii of curvature of the
surface of body 1 at the origin.
p2, p'2 Ditto for body 2.
E1, 71i Distances from a point xiyizi to the planes of curvatures
p'l and pi respectively.
t2, 772 Distances from a point x2y2z2 to the planes of the
curvatures p'2 and P2 respectively.
e Angle which the planes of the curvatures pi and p2
make with each other.
Xx Y, Z, Normal components of the stress across planes perpen-
dicular to the x, y, and z directions through the point.
Tension is positive and compression negative.
P' Z, for points in the compressed area.
P Total force with which one body presses on the other.
4u, avI awl 4u2 4v2 a92
A 1 = + + A2= x+ y +
ax, 4yv azw Ox2 4Y2 4w2
a2 a2 a2
V2 + - + -
ax ay2 z2
7. Action of a Force Concentrated at a Point.-Consider first the
simplified but ideal case of a force P concentrated at the origin instead
of being distributed over a compressed area. The strain at a point
x, y, z in the solid is specified by the following equations written in
different form but equivalent to those of Love.*
(1)
ILLINOIS ENGINEERING EXPERIMENT STATION
where
r = x2 + y + z2 .
Then
A Pz - 1P P P /1\
4 4V2 r( -4 A V2log (z + r) --P +
4ir \ 4t() (X + /,) 47r az r
P(X + 3A) 02
47r(X + y) Oz2 log (z + r)
Since
V2 -
r (1) =
0 1
0, V2 log(z + r) = 0, and - log (z + r) -
az r
this becomes
_ P (1
22r(X + ) oz r
The stress may be determined from the strain.*
9u 0v Ow
X, = XA+2A ; Y, = XA+2A -; Z, = XA +2
=1-27
x-2a
Using (3) and (4) we have:
7r 9z r 2r 4x2 r
1P (1\ zPO (1)
' - 2 \r - 2 z2
2r ~ z a;02
(1 - 2a)P 02
- 2r Ox2 log (z + r)
(1 - 2u)P 02
- 2r y- log (z + r)
27r ay2
8. Solution of General Problem in Terms of P.-Let us return to
the actual case of a compressed area. The effect of P' distributed
over this area is the integrated effect of forces P'dx'dy' over the com-
pressed area. The components of displacement are then obtained
from (1) by letting r equal ý\(x - x')2 + (y - y')2 + z2 and taking
*Love, p. 124, Eq. 8; p. 101, Eq. 21.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 13
FIG. 1. VARIATION IN STRESS COMPONENTS IN DISTANCE FROM CONTACT SURFACE
the integral dx'dy' of each term over the compressed area. Introduc-
ing the symbols
S= J - dx'dy' and x = P' log(z + r)dx'dy' (6)
and interchanging the order of differentiation and integration
z O 1 ax
U ~ 47r4 ax 47r(X + u-) ax
v 1 (7)
4vy ay 47r(X + P) ay
z 06 X + 2_ u Ox
4 az+ 4irA(X + ju) az
By similar process from (2) and (5)
1 06
A 2 (1 J_ ) .3
7 A Z
ILLINOIS ENGINEERING EXPERIMENT STATION
or Q4 z a24 1 - 2l- a'x
,-r az 2r Ox2 27r x2
a ao z a8o 1 - 2o alx
y c = z - - 2 - - l (9)
S Oz 27r ay2 21r ay2
1 04 z 20)
Z'
S 2r 49z 2r O(2
Since a log (z + r) = (1/r) and since X. and Y, must evidently
vanish at infinity, we may write:
a2 a2 O2x f0a24
x -- x2 dz ; 2 f- 2 dz (10)
It is clear that Equations (9) contain the solution of the problem
when the curve of compression and the pressure P' within it are
known. The bodies have heretofore been regarded as having plane
surfaces. To determine the curve of compression and the value of P'
as a function of x', y' the surfaces must be regarded as curved, and
afterwards they will again be regarded as plane. In other words, the
curvature of the bodies is regarded as having no other effect on the
problem than the determination of the curve of compression and the
value of P' as a function x', y' within this curve.
9. Determination of P' and .-The equation of the surface of
body 1 in the neighborhood of the point of contact is zi = 2 pl12 +
12 p'1712 and similarly for body 2, z2 = Y p222 + ± 2 p2 22. Then by
the formula for rotation of axes through an angle e:
$2 = 1 cose + n71 sine, ,72 = - 1 sine + 71 cose
Hence z = Z1 + z2 = A'%i2 + 2H'%71 + B'712 (11)
where 2A' = pl + p2 cos2 + p'2 sin2e
2H' = (P2 - p'2) sine cose h (12)
2B' = p'l + p2 sin2 + p'2 cos J
By a rotation of the proper angle from the 1i, qi axes to the x, y axes
we get (Appendix A)
z zi1 + z2 = Ax2 + By2 (B > A) (13)
where A, B are the roots of the equation in x
(2x - pl - p2)(2x - p'l - p'2) + (pi - p'1)(p2 - p'2) sin2e = 0 (14)
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 15
If e = 0, then A = 1(PI + p2), B = 2(p'1 + p'2) (15)
When the bodies are pressed together, let the points (x, y, zi) of
body 1 and (x, y, z2) of body 2 come into contact within the com-
pressed area. Then zi + wi = - (z2 + w2) + a. The minus sign is
due to the opposite directions of the z1 and z2 axes. Hence by (13)
w + w2 = a- (1 + 2) = a - Ax2 - By2 (16)
For points outside the compressed area
wi + W2 > a - Ax2 - By2 (17)
For a point in the plane z = 0, we have from the third equation of (7)
Xi + 21 (Sx\
W i1 X,- -+±2, 0 x-) (18)
4TrAi(xi + I1A) az ,, o
Now
axz =f P' log (z + r)dx'dy' = P'(1/r)dx'dy' (19)
It is known that*
(3X1 + 21i) X(
E. - =; 'l e (20)
X, + A , 2(Xl + p1)
and therefore
1 - a12 X, + 2 (1
E, 4/12(Xi + (11)
Using (19) and (21) in (18) and letting 0 = fo when z = 0, we have:
(22)
1 - a 2
Similarly w2 = 020o where 02 -= rE2
Substituting wi and w2 from (22) into (16) and solving for o.
1
. - O +1 2 (a- Ax2 - By2) (23)
*Love, p. 124, Eq. 8; p. 101, Eq. 21.
ILLINOIS ENGINEERING EXPERIMENT STATION
Judging from (13) it seems probable that the curve of compression
is an ellipse. Now, as defined by (6), 0 is a Newtonian potential
function due to matter distributed over the compressed area witn
surface density P'. Hertz recalled that the potential at a point
x, y, z in the interior of a homogeneous ellipsoid is of the form*
a - Ax2 - By2 - Cz2 where the x, y, and z coordinate planes are the
planes of symmetry of the ellipsoid. For the plane z = 0 this yields
the form required by (23). Since k, defined by (6), is the potential
due to the attraction of a surface layer of mass, the potential due to
an infinitely flattened homogeneous ellipsoid must be taken for 0;
that is, the potential due to a mass density
3P = ./ x'2 y'2
P' - 1 - (24)
27rab a2 b2
x'2 y'2
distributed over an ellipse - + 2 = 1 in the plane z = 0.
From the known potential of a homogeneous ellipsoidt there is readily
found, for the elliptical layer just described,
0 3P "/ x2 y2 z2 d___(25)
S= , --a2 + - b2+ ) [(a2 + 4)(b + ) (25)
where y is the positive root of the equation
x2 y2 z2
f(x, y, z, y) 1 - - - y- = 0 (26)
a 2 + ly b2 2 + 5
Since y = z = 0 within the compressed area
3Pf - x2 y2 )d_,
40 = - - 2+ )[(a2+ )+ ) (27)
Since the values of fo in (27) and (23) are equal, the coefficients of 1,
x2, and y2 must be equal and we have:
3P t9) _( d±
a = + ( + 0 2) ) (28)
*Love, p. 194; Peirce, Newtonian Potential Function, 3rd ed., 1902, pp. 117-126.
tIbid.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 17
3P 0 d _
A = 4 (1 + 62) (a2 + k) (b2 + )
(29)
4B = (a + (bs +
Equations (29) may be solved for a and b. Hertz solves (29) for a in
terms of one transcendental function of B/A and for b in terms of
another such function. Tables of values are computed for these
functions and elliptic integrals are given a mere mention by Hertz as
the basis for computation. The problem of the solution of Equation
(29) for a and b will be taken up later.
10. Stress on Z Axis in Terms of Elliptic Integrals.-The value of k
from (25) can now be substituted in (9) and (10). Equations (7) and
(8) will only be applied for points (x = y = 0) on the z axis. In this
case we have from (26) 7' = z2. To find derivatives of 0 the rules
must be used
a a+ *aa+ a = a + 6 (30)
x x x y x ' Oy Oy - y ' az - z ay + z (30)
where 8 indicates differentiation when 7 is regarded as an independent
variable; that is, when no account is taken of (26). We thus find for
x= y = 0
94 3Pz f" db
z 2 , (a2 + #) (b2 + ) (31)
Changing the variable of integration to c where ^ = a2C2 and writing
k = b/a and ý = z/a
_ _ 3Pz d_______
Oz aJf ~w2(1 + a2)% (k2 + 202) (32)
Similarly
9024 3P f _ dw_____
Ox2 - a3f (1 + o2)% (k2 + 2) (33)
24, 3P _____d______
9y2 a3^ (1 r w2)li (k2 + w2)i (34)
02q4 3P f dw
az2 a3 ~ 2)2(1 + 2) (k2 + ,2)i
ILLINOIS ENGINEERING EXPERIMENT STATION
3P
+ a (1 + ~2) (k2 + 2) (35)
Differentiating (33) and (34) with respect to z = aý
a30 1 a30 3P
OzOx2 = a OQOx2 a4(1 + -2)% (k2 + -2) (36)
a30 1 a0 3P
OzOy2 a 4OOy2 a4(1 r2)+ (k2 + ( r) (37)
From (10) and integration by parts using (36) and (37)
a2x 2 '0 a2 a a23
Ox2 Ox2 dz = z + , z qxdz
x a 20 + 2fo " a 2
Ox2 + a2 d0
_3P d+ 3p + d
a2 J (+ w2)"(k2 + w2) a (1 + ) (k2 + -2)
3P_ dc 3P k2 2
a2 J (1 + w2) (k2 + ,2)" - a2(1 - k2) 1 + 2
3P
+ a2(1 - k2) (38)
y f ay2 dz = z- + z dz = at
(1J+ a z" dy
- 3P * d] 3P 1d) a
a2 r' (1 + w2)½ (kc2 + w2) a'(1 - _2) k2 + [2
3P
~ a2(1-k2) (39)
Substituting derivatives of 0 and x obtained in this paragraph
into (9) the normal components of stress X., Yy, Z, are obtained.
By consideration of symmetry for points on the z axis, the correspond-
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 19
ing tangential stress components must vanish; so that Xx, Y, and Z,
are the principal components of stress. The resulting formulas for
the principal components of stress are not written out here since the
integrals involved will be expressed in terms of elliptic integrals and
the equations for the stress will then reduce to Equations (59).
The author's innovation consists in expressing the integrals ap-
pearing in the last paragraph in terms of F(p, k') and E((p, k'), the
first and second elliptic integrals, for which tables have been com-
puted. These integrals are defined by
rf do __________
F(, k') - sin ; E(,') 1 - ksin-k 20 d (40)
and the corresponding definite integrals are obtained:
K (k') -F , k')- 1 - ksin 0
(41)
E(k') = E (,k -V) 1 -k' Vsin20 do
Changing the variable of integration to w = cot 0
)=f* dw (k2 +2)
F(ýo,k')- (1 ;E(Qp,k') mdr(1 )d(42)
f dw r(k + W2)%
K(k') - (1 + 2 (k2 + 2) E(k') -f (-2 d (43)
where k2 = 1 - k'2 and " = cot p (44)
For the integrals appearing in the preceding paragraph the following
results were obtained:
d, d _ .r E(p, k')
Jr (1 + 02)" (k2 + w2)= k k2 (45)
m de 1 r 1
S(1 + )(k + k 2 F(p, k') - E(,, k') (46)
(1 + 02)" (k2 + 2)" k'-= k) E(, k') - F(1, ')
r 1
k2k'2t k,2r
(47)
ILLINOIS ENGINEERING EXPERIMENT STATION
where k= +- 2 (48)
The direct derivation of these results is rather tedious but the inte-
gration may be verified by differentiation, since it is known from
(42) that
dF(, k') 1 dE( ,k') (k2 + ý2) (
dr (1+ +2) r (k2+ 2)+ d - (1 + 2) (49)
The problem of solving Equations (29) for a and b may now be
resumed. Changing the variable of integration in (29) to a where
,P = a2C2, the integrals become 2/a3 times the integrals of (46) and
(47), respectively, when the lower limit of these integrals is " = 0.
Hence
3P(O1 + 02)
A = 3P(2a1 2) [K(k') - E(k')] (50)
3P(O1 + 02) r1
B - 3P( 12) E(k') - K(k') (51)
Dividing (51) by (50)
B (1/k2)E(k') - K(k')
A K (k') - E(k')
From (52) a graph may be plotted connecting k and B/A, points on
this graph being found by assigning values to k and computing cor-
responding values of B/A. (See Fig. 2.) Since by (14) B and A are
functions of p1, p'l, p2, p'2 and e, this graph determines k in terms of
the geometrical configuration of the two bodies. Adding (50) and (51)
3P r(01i + 2)E(k')
A + B - 23a 4 k2 (53)
2ira3
Solving for p and using (22) we have:
27ra3 AE(k') AP 27rk2
3P - kc2 a3 3E(k') (54)
where
((,- + -,) 1 1 - ,12 1 - 2
A = A n A I? V +
_F - ±j \ Ld2 /
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 21
FIG. 2. VARIATION IN CERTAIN FACTORS WITH CHANGES IN B/A
By the theory of equations A + B must equal -Y4 times the coeffi-
cient of x in (14) when this equation is expanded, i.e.,
A + B = Y2(p + p' + p2 + p'2) (56)
Substituting this in (55)
(1 - T12 1 - 22
A=2 IE + ) - (P + P' +p2 +'2) (57)
E, E2/
r
I
M
ILLINOIS ENGINEERING EXPERIMENT STATION
When k has been found as just described, the substitution of this
value of k in (54) yields a; and thus the semi-axes, a and b ( = ka) of
the ellipse of pressure are determined.
With the value of A from (54) we have:
AXý 2k2 ra2 AYy, 2k 2 ra2 AZ, 2k2 7ra2
X =Y - Z (58)
a E(k') 3P ' a E(k') 3P "; a E(k') 3P (58)
Let X., Y,, Zz as given by (9) be expressed in terms of elliptic inte-
grals by means of the derivatives of 0 and x in the first paragraph of
this section and the integral evaluations (45), (46), and (47). Upon
substitution of the values of X., Y,, and Z, thus found in (58)
AXý
- = M(2, + ±t's)
a
AY,
- M(, + at'y)
AZ, M 1\
a 2 7
(59)
2k2
where M - k'2E(k') (60)
and 0, = - 1 + ([F(o, k') - E(p, k')]
'= = - k2 + 1 + ý[(1/k2)E(p, k') - F(p, k')]
1 1 -
S= 2r + 2 - + -[(1/k2)E(p, k') - F(p,k')]
, = -1 + + [F([p, k') - E(p, k')]
(61)
By these formulas, the principal components of stress can be found at
any depth z = at below the surface, since tables for F(p, k') and
E(p, k') are available.* When tables are computed for the stress
components at various depths (see Tables 1 and 2) it is found that for
a wide range in values of k, the maximum difference of stress com-
ponents, Z. - Y,, occurs at a point below the surface of contact and
*Hancock, Elliptic Integrals; Jahnke & Emde, Funktionentafeln.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 23
000 l
0000
I I I I
*00000000000000000
:1 eIco oI I lS l I I I I I I I I I
c00 00000000000000000 C
il : I ++++++++++.++++++
-00 N
0000
I I I I
:I I I++++++++++++
0
0
0
0
0
0
0
0
A
^
lill : I +++++++++++++++
0i0o0i *MioCi Mi0NQ
0cooo 00-000000000000
.... . ..... ...
o 0 -0ooo0oo0oooooooooo
000 0 *00000000000000000
I3 I 0 1 I I C+C++CO +(++
11i : ...........00000000000000000..
Noo o -coooooomo odddoooN
I : 1 1 : III +++++++++++++
So o o o o o .o o. o -o
11 111 10 1 0. . 0 00 ^ . . 0.
N00 -m a, m § 000 N 000
00000N ........... .0
ILLINOIS ENGINEERING EXPERIMENT STATION
0
CD
0
,4
C4
0
H'
'IV
co
Ad
OH
0
'H
N
N
H
N
H
t-MO1OO *HGOOlt-
O!G0H0! 0. .
0H'CHOGO- *CHGOGOH
0000 *0000
1 1 1 :+++1
0H'CH *ddoo'~
:0000
01000 G C 0000M
GomOGO-' O)O
HOOO *-tf~~l' 00G'
S1 1 +++
==m::W==m
Go CD) 01001 00ý
000000000
dod:ooooo
S~oo ;'ieooi
GO GO O~h0' G ' '
:::::::::
1 1 1 1 1') 0 0
0 O 0C t'- GO O 0
===000000
000000000
I III i Iooo
dddddddddG GO- O
O>le
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 25
not at the surface. This is discussed in more detail elsewhere. It
only remains to examine certain special cases.
11. Special Case, k = 1.-For k = 1, b = a and the compressed
area is a circle. The integral in (45) becomes, when [ = cot p,
SW(l w) =f tan2 v d = tan p - p (62)
The integrals in (46) and (47) become identical and equal to
(1 +d W12)2 J= sin2 dp = -ý sin ( cos ý + 2 (63)
Letting p = a2w2, Equations (29) become
3P (f dc 37rP
A = B 2a (1 (1++ ?')' 8a- (0+2) (64)
since by (63) (1 + w2)2 = [- sin 0cos P,+Y2 IP =
Since by (41) E(0) = 2 dO = - , we have on substitution of this
value in (58)
AX. 4a2 A Y = 4a2 AZ = 4a2
a - 3P X' a 3P Y; a -3P Z (65)
By the same process described for getting (59) except that the integral
evaluations no longer involve elliptic functions, we get
AX A Y, 2
aX - a - {(1 + o)(-1+ pcotp) + 2 sin2 p} (66)
AZ, 2
A - - - sin%2 (67)
M. T. Huber* has found the stress in this case (k = 1) for a point
x, y, z in a general position. His solution when specialized for the
z axis agrees with the above.
*Ann. d. Phys. 1904, Bd. 14, p. 153.
ILLINOIS ENGINEERING EXPERIMENT STATION
12. Special Case, k = 0.-Next consider the case of k = 0.
Equation (54) shows that a = oo since A and
fo '
E(1) = j -sin20 dO = 1
are both finite. A simple illustration of this case is that of a cylinder
z2 = By2 resting on a plane z1 = 0. Before the bodies are pressed
together, there is contact along the x axis. When pressed together,
the two bodies are in contact over the strip bounded by the lines
y = b and y = -b. Let P be the integrated pressure across the
minor axis of the ellipse before a approaches oo.
_ i 3P fb ,-- 3P vrb2 3P
P = b(P'),=ody'- J b f ', - y,2 dy' = - = (68)
-b 2rab b - 2irab2 2 4a
As a = co, let P = oc in such a way that P/a remains equal to a
finite constant. Then the value of P from (68) will in the limit be the
force per unit length of the strip of pressure. Then for a = co
3P y'2 27P y
2P' = ab 52 - b2 (69)
Since (Appendix B) at a distance r the potential of an infinitely long
straight line of constant linear density P' is -P' log r2
4 = - P' log r2dy' (70)
_-b
where r2 = (y - y')2 + z (71)
and where P' is given by (69).
By differentiation under the integral sign
.90 02,p x /'9 ,2
- 0; =0 - 0 dz = 0 (72)
b a6o' fb p' bb pi
az --2z dy'; -Z 2 -b dy' + 4z2 -b '4dy' (73)
Now for points on the z axis, i.e., for y = 0, we have by (69)
bp' 2p - b 1 y
dy' - 1 - - dy'
-b r2 7b -b r V2
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 27
and letting y' = b sin 0
b pi 2p bcos2 dO
b 2 dy 2 - b2 sin z2d
- - + -- ar tan btan
b6 2 z c- 2
S2 b2 + z2 - z)
b= 2z (74)
bpi 2F b / 12
Also dy' - 1-- dy
A-b Hrb f r dy'
bu
and letting y' = / u2
bpi' 2p du
- 4 [(b2 + z2)u2 + Z2 2
2P u 1 ub2 + z
7 L2z[(b2 + z2)u2 + z2] 2z3 + arc tan z -
= zi\/b ± 2 (75)
Substituting the integral evaluations from (74) and (75) in (73)
Q4 4P/ 2-
=z V-\2 + Z2 - z) (76)
2 4P/-- 4P
z2 - b2z b2+b 2 + z -b2 (77)
By substituting values of the derivatives of 4 from (72), (76) and (77)
into (9)
S= - - (y\2 + z2 - z) (78)
Xrb2
ILLINOIS ENGINEERING EXPERIMENT STATION
2P
z'= - ~ + z2 (79)
To get Y, it is noted that the stress at every point is a function only
of y and z and hence the strain is also a function only of y and z.
Thus Ou 9v 8w
S- =0 and A = - + -
ax ay Oz
Hence from (3) we have:
Y, + Z, = 2XA + 2p + z = 2(X + )A (80)
au
X, = XA + 2ju - = XA (81)
and hence by means of (4)
X. = o(Y, + Z,)
or XZ
Y - - Z' (82)
Substituting in (82) the value of X, from (78) and that of Z, from (79)
2Y (-Vb + z2- z)2
Y = - 7rb^^+z2 (83)
i.b2\/b 2 + Z2
By means of (54) and (68)
27rb2 a 27rb2 3 7rb2
A= 3 P- 3 4P 2P (84)
Hence with the values of A, Xý, Y,, Z. from (84), (78), (83), and (79)
respectively, we have
A X , z-- -
b -= -2 (Vl1 + (z/b)2 - z/b)
AY _ (\1 + (z/b)2 - z/b)2
b 1 + (z/b)2
AZ, 1
b 1 + (z/b)2
(85)
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 29
The stress components are thus algebraic functions of z/b and it is
readily found by the usual process that jZz - YJ will be a maximum
for z/b = - 0.7861. For z/b = 0.7861 we have from
(85)
AX.
- = - 0.9718a
b
AY,
S=- - 0.1856
AZ -
-- = - 0.7861
(86)
A
and - (Z, - Y,) = -0.6005 (87)
III. METHOD OF SOLUTION
13. Notation and Equations for Definite Problems.-For the con-
venience of those readers who find the mathematical derivation just
presented to be difficult reading, below are given the essential nota-
tion and equations to be used in the solution of definite problems.
The subscripts 1 and 2 are used to indicate one or the other of the
two bodies in contact. The following notation is used:
El, E2 Moduli of elasticity for the two bodies in contact.
Ca, a2 Poisson's ratios for the two bodies.
B Mean of reciprocals of the radii of curvature of the two
bodies in the y direction. B = 1 2 - +
A Mean of reciprocals of the radii of curvature of the two
bodies in the x direction. A = 2 (i R'2 . Note that
B/A is always equal to or greater than unity; this fact
determines the x and y directions. The ratio B/A is
related to the value of k as shown by Equation (52).
- 1 - 1 -22] 1
[ - E12 1 - 2 2 B (See Equation (55))
E, -E A + B
ILLINOIS ENGINEERING EXPERIMENT STATION
Note that this quantity A is a term which takes account of the
elastic properties of the two bodies in contact and also of the shape of
the two bodies at the point of contact. If the two bodies in contact
are similar, such that 0r = 02 = a and E1 = E2 = E, then the above
reduces to
2(1 - or2)
A-
E(A + B)
a is the semi-major axis of the ellipse of contact.
b is the semi-minor axis of the ellipse of contact.
k is the ratio b/a.
k' = 1 - k2
P is the total pressure between the two bodies.
3 APE (k')
2,k2
where E(k') is a definite elliptic integral. (See Equation (41)).
X , Y,, Z, Normal components of stress across planes perpendicular
to the x, y, and z directions through the point under con-
sideration. Attention is called to the fact that the stress
components considered in this bulletin are for points on
the z axis at various relative depths, z/a, from the contact
surface (see page 17).
A A A
The values of - Xx, - Yy,, and - Z, are given by Equation (59)
a a a
which will not be repeated here. The methods of solving those equa-
tions are discussed below.
14. Stress-Component Factors.-As suggested on a previous page,
the substitution of numerical values for the purpose of solving
Equations (59) for the stress components is practicable only by the
choice of certain values of k which will permit obtaining the values of
the elliptic integrals involved from published tables. The values of k
which can be substituted in this way probably will not be in agree-
ment with the numerical values for which a solution is desired, since
for convenience in the substitution it is desirable to take k = cos 0,
where 0 may vary from zero to 90 deg. If Hancock's table of Elliptic
Integrals is used, 0 should be some multiple of 5 deg. It will be found
much easier to substitute in these equations if p is taken in degrees
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 31
z
such that the values of - ( = cot p) will be approximately those de-
a
sired, instead of assuming values of z/a and solving for p. In order to
facilitate the solution of practical problems, the right hand members
of equations (59) have been evaluated for values of k ranging from
1 to 0.2588 and for z/a ranging from 0 to 4. (In some cases the values
of z/a are from zero to one). Poisson's ratio, a, is assumed as 1.
The computed values are given in Tables 1 and 2. In order to obtain
the stress components, the factors in these tables must be multiplied
a
by -, determined from Equations (54) and (57). In Fig. 1 have
been plotted for a value of k of 0.86603 the values of the factors for
the three stress components from Table 1, using the values of z/a as
abscissas. To avoid confusion the individual points are not indicated.
In this figure is also shown the curve for the maximum difference of
stress components, (Zz - Y,,). It is to be noted that the values of
the individual stress components are a maximum at the surface
(z/a = 0) and decrease with increase in distance from the surface of
contact. However, the values of (Z, - Y,) increase from a small
value at the surface to a maximum at a point below the surface where
z/a = 0.44 (for k = 0.86603). So far as tendency to cause damage
is concerned, it is evident that for k = 0.86603 the greatest value of
the difference of stress components, called (1Z. - 1Y,), is equal to
a
-0.384 -- and that it occurs at a point below the surface z2/a = 0.44,
as marked on the diagram. These curves are shown to illustrate the
method used in obtaining curves for the maximum differences of
stress components in Fig. 2, and to indicate how the stress compon-
ents vary with the distance from the contact surface.
In order to determine the greatest value of the difference of stress
components (1Z, - 1Y,) for the other values of k, curves similar to
those of Fig. 1 were plotted from quantities given in Tables 1 and 2
and the values thus determined for each value of k are given in the
A A
lower part of Table 3. The corresponding values of - 1Xx, - iY,,
a a
A Z\
- 1Zz, as well as of - , are also included. The stress component
a a
a
factors (- oXx, etc.) at the surface of contact, obtained from Tables 1
A
ILLINOIS ENGINEERING EXPERIMENT STATION
TABLE 3
SUMMARY OF FACTORS FOR CALCULATIONS
These factors are to be used for calculating stress-components, size of contact areas, and depth to
point of maximum stress difference. Poisson's ratio = Y4.
B/A ............
AP
a ............
A X ..........
A
a- z.........
A Z, ...........
a
o{z - o}..
21/ .Z. . ...Y .. .
£a 1/ ...........
Values of k
1
1.0000
1.3333
-0.4775
-0.4775
-0.6366
-0.1591
-0.1591
-0.1130
-0.1130
-0.5230
-0.410
0.467
0.86603
1.2409
1.0704
-0.4320
-0.4532
-0.5901
-0.1581
-0.1369
-0.105
-0.097
-0.481
-0.384
0.440
0.70711
1.6829
0.7753
-0.3702
-0.4151
-0.5235
-0.1533
-0.1084
-0.100
-0.084
-0.427
-0.343
0.390
0.57358
2.3083
0.5474
-0.3109
-0.3726
-0.4557
-0.1448
-0.0830
-0.090
-0.064
-0.367
-0.303
0.350
0.42262
3.6746
0.3214
-0.2355
-0.3092
-0.3631
-0.1276
-0.0539
-0.082
-0.052
-0.293
-0.241
0.280
0.34202
5.0831
0.2191
-0.1919
-0.2669
-0.3058
-0.1139
-0.0390
-0.070
-0.041
-0.243
-0.202
0.240
0.25882
7.8625
0.1303
-0.1451
-0.2160
-0.2407
-0.0956
-0.0247
-0.056
-0.028
-0.187
-0.159
0.200
and 2 for z/a = 0, have been included in this table, as have also the
AP
computed values of -3 (from Equation (54)) and of B/A (from
Equation (52)).
15. Method of Substitution.-Using B/A as abscissas, the values
AP zi A A
of k , -, - A (Zz- 1Y,) and - (oZ2 - oX.) from Table 3 have been
plotted in Fig. 2. This diagram (or a similar diagram plotted to a
larger scale) permits the determination of these factors for values
B/A other than those given in the table. In making use of this dia-
gram for the solution of a numerical problem, the first quantity to be
determined from the conditions of the problem is the value of B/A,
1 1
where A =- 2R , and B -= 2 (for crossed cylinders having radii
2R a 2R,,
of R1 and R2, respectively, and with axes perpendicular). Then for
AP
this B/A, the value of - is taken from the diagram, which permits
a3
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 33
solving for the value of a for given values of P and A. Having cal-
culated a and A, (1Zz - 1Y,) may be computed by multiplying the
A a
factor for - (1Z, - 1YY,), taken from the diagram, by - . Within the
a A
limits covered in this diagram, Z, - Y, is negative in sign. The depth
(in inches) to the point of maximum stress difference may be obtained
from the curve for zi/a, the value of a already having been deter-
mined. The semi-minor axis of the ellipse of pressure b is equal to ka
and if desired the pressure area may be calculated as irab.
16. Surface Stresses.-If the stress components at the surface of
contact are desired it will be necessary to construct a diagram similar
to that shown in Fig. 2, using for the purpose the values of the surface
stress-component factors in the upper part of Table 3. While the
stress components at the surface will be higher than those in the
interior, the stress differences will be less. To show the comparison,
a curve for the maximum difference of stress components at the sur-
A
face, - (oZz - 0X,), is included in Fig. 2. Attention is called to the
a
fact that at the surface the maximum stress difference is (Z, - Xý),
while in the interior it is (Z, - Yy).
17. Numerical Example.-As an illustration of the method to be
used in substituting definite values in the mathematical solution, a
numerical problem will be solved. The radii of curvature and shear-
ing stress used in this problem are those used in the check of the
mathematical analysis, as described in detail later. Assume crossed
cylinders of steel having axes perpendicular to each other, the radius
of one being 15.8 in. and of the other 12 in., Poisson's ratio ar = 14,
and the modulus of elasticity E = 30 000 000 lb. per sq. in.
For the problem under consideration
1 1
2RA 21 2 X 15.8 0.0316
1 1
B -
2R2 2 X 12 = 0.0417
0.0417
B/A = 0.0316 = 1.320
B + A = 0.0733
ILLINOIS ENGINEERING EXPERIMENT STATION
In Fig. 2 for B/A = 1.320, from the corresponding curves the fol-
lowing factors are found:
k = 0.838
AP
- = 1.010
a3
zl
- = 0.428
a
A
a (1Zý - lY,) = -0.375
From Equation (55), assuming al = a2 = o; E1 = E2 = E, we have:
A 2(1 - -2) 2(1 - 0.0625) = 8.52 X
E(A + B) 30 000 000 X 0.0733
S8.52 P -
a = .01 X 107-0.00946 P
a
A= 11 090 FP
The maximum difference of stress components is
a
Z - 1Y, = -0.375 A -4160 P
In order to calculate the load required to produce a shearing stress of
19 200 lb. per sq. in., corresponding to a stress difference of 38 400 lb.
per sq. in., we have 38 400 = 4160 *
from which
(38 400\
P 4160 =786 lb.
For calculating the area of contact,
a = 0.00946- P = 0.00946 /786 = 0.0873in.
b = ka = 0.838 X 0.0873 = 0.0731 in.
Area of contact = irab = 0.0200 sq. in.
Depth to point of maximum stress difference,
zi = 0.428a = 0.428 X 0.0873 = 0.037in.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 35
786
Average surface pressure =0.0200 - 39 300 lb. per sq. in.
The maximum surface stress must be calculated from the values
A
of - Zo in Table 3.
a
Additional numerical examples will be found in Appendix C.
18. Special Case.-It may be of value to indicate one further point
concerning the method to be used when one (or both) of the bodies
has more than one radius of curvature at the point of contact.
It will be remembered that in the definition of the quantities
A and B, it is assumed that the shape of the bodies in the neighbor-
hood of the surface of contact may, by a proper choice of axes, be
represented by a quadratic of the form
z = Ax2 + By2 (See page 14)
For crossed cylinders with axes perpendicular, as already indicated,
1 1
A = - B = --
2R, 2R2
where R1 and R2 represent the radii of curvature of the two cylinders
in the x and y directions, respectively. But if one of the bodies has
one radius of curvature in the x direction, R1, and another radius in
the y direction, R'l, while the second body has only, say, the radius
R2 in the x direction (in other words, R'2 = oo) then
A = i+ R) and
(1 1
B= - +- -2
That is, A is the mean of the reciprocals of the radii of curvature in
the x direction while B is the same function for the y direction. After
A and B have been calculated, the remainder of the computations
may be made as just described for crossed cylinders. It should be
remembered that the radius of curvature is to be considered as nega-
tive if the center of curvature lies outside of the body.
IV. CHECK OF MATHEMATICAL SOLUTION
19. Need for Checking.-In applying the mathematical calcula-
tions of stress components to certain practical problems, the existence
ILLINOIS ENGINEERING EXPERIMENT STATION
of such high shearing stresses was indicated that it was thought desir-
able to attempt to obtain a physical check of the correctness of the
mathematical assumptions and computations.
In the paper previously referred to, Hertz described the method
used by him for verifying his analysis. For this purpose he deter-
mined the areas of contact between a spherical glass lens and a glass
plate at various loads, and also between crossed cylindrical glass rods.
Since the measured areas showed good agreement with those calcu-
lated, he took this as a check also of the stress components.
Because of the difficulty of measuring accurately the areas of con-
tact, much more difficult for metals than for glass, and also in order
to obtain a more definite check of the calculated maximum shearing
stresses, it was decided to make use of the Fry strain-etch method*
for determining when a material has been stressed beyond the yield
point.
20. The Fry Strain-Etch Method.-By the Fry method of etching
(which is applicable only to certain low carbon steels), after the
material has been subjected to a stress exceeding the yield point, by
bending, direct loading, or other means, it is heated for half an hour at
a temperature of approximately 400 deg. F. After heating, the piece
may be cut at any section which it is desired to examine and on etch-
ing with a solution of cupric chloride in hydrochloric acid and water
it will be found that those regions which have been overstressed will
etch darker than the remainder of the section. According to Fry's
method, etching was accomplished by immersing the specimens in the
etching solution for as long as seemed necessary, followed by rubbing
with copper chloride crystals wet with the solution; but it was found
that etching could be accomplished very readily and more rapidly by
rubbing the surface with a swab made by twisting cotton around a
stick, the swab being dipped in the etching solution and in copper
chloride crystals from time to time. After a satisfactory etch has been
obtained the surface is cleaned with alcohol containing a small amount
of hydrochloric acid. The acid is used to prevent copper from being
deposited on the surface.
21. Special Materials and Equipment.-For this method of check-
ing the mathematical analysis there was obtained a bar of mild steel
21Y in. square and 6 feet long, which by preliminary tests it was found
*"Strain Figures in Mild Steel Evidenced by a New Etching Method," Ad. Fry, Stahl und Eisen,
Vol. 41, No. 32, Aug. 11, 1921, pp. 1093-97. A condensed translation of this paper may be found in the
Iron Age, Dec. 1, 1921, p. 1401. Additional work along these lines is given in a paper by J. Dudley
Jevons, "Strain Detection in Mild Steel by Special Etching," Journal of the Iron and Steel Institute,
Vol. 111.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 37
could be used. It is a peculiar fact that all mild steel will not give
strain-etch figures, but the reason for this is not known. Therefore,
in order to obtain a material which would strain-etch, six bars were
obtained from different sources and it was found that one of them was
satisfactory. The average physical properties of this material were:
Tensile strength ............. .47 400 lb. per sq. in.
Tensile yield point ........... .32 300 lb. per sq. in.
Compressive yield point ........ 28 800 lb. per sq. in.
Torsional yield point .......... 19 200 lb. per sq. in.
(solid specimen)
'o far as these tests are concerned only the yield points are of interest.
Two types of loading were used in checking the mathematics. The
first consisted in loading a plane specimen block through a cylindrical
loading block; the second, in loading a cylindrical specimen block
through the same loading block, the axes of the cylinders being at
right angles to each other.
The loading block consisted of a segment of a cylinder whose
radius was 15.8 in. The length of the block in the direction of the
axis of the cylinder was 21Y in., its radial thickness 34 in., and its
width 6 in. The block was machined to size, and then the cylindrical
surface was machined, the radius used being the largest which could
be obtained with the available equipment. After case-carburizing
and quenching, to prevent over-stress of this block during loading,
the cylindrical surface was ground to as accurate a finish as possible.
The specimen blocks for the tests of stresses between a cylinder
and a plane were machined from the steel already described. These
blocks were made 214 in. square and 12 in. long, one face being fin-
ished as accurately as possible to a true plane surface. For checking
the accuracy of the surface, the blocks were coated with prussian blue
and when tested with a straight edge showed practically perfect
contact. For the crossed-cylinder tests, a cross-radius of 12 in. was
machined on one of the square blocks just described. Considerable
difficulty was encountered in obtaining the degree of accuracy of
finish desired, both for the loading block and the specimen blocks,
but it is felt that the surfaces finally obtained were satisfactory.
22. Tests of Cylinder on a Plane.-In the case of a cylinder on a
plane, the calculation of stress components for a given load and a
given radius of a cylinder is relatively simple. By referring to
2PA
Equation (84), page 28, it is found that b2 = - where b is one-half
I
ILLINOIS ENGINEERING EXPERIMENT STATION
of the width of contact, P is the load in pounds per inch of length of
2(1 - o.2) 4R(1 - 2)
the cylinder, and A = - 1 - E . For steel, assuming
E -
2R
Poisson's ratio, a, as 0.25 and E as 30 000 000 lb. per sq. in., we have
4R(1 - 1/16)
30 000 000
R
-8 X 106
2 PR -
Then b = x , -2 0.000282 IPR
b \F]R
- = 0.000282 X 8-- X 10
A R
= 2256 -
IrR
The maximum difference of stress components, from Equation (87),
b
is, iZ, - =y = -0.6005-
= -1355 P
For the loading block used in these tests, R = 15.8 in., and hence
(1Z, - 1Y,) = 341 -p
Remembering that the shearing stress, S, is one-half of the stress
difference, we have
341
S = (Z, - Y,)/2 - P = 170.5 P
The load required, then, to produce an intensity of shearing stress
equal to the shearing yield point of the steel used will be
- 19 200 12700 lb. per in. of length
P = 175 = 12 700 lb. per in. of length
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 39
FIG. 3. STRAIN-ETCHED SPECIMEN BLOCK
For this load, the width of contact is
2b = 2 X 0.000282 \/12 700 X 15.8
= 0.000564 X 448
= 0. 252 in.
Depth to point of maximum shearing stress
zi = 0.786b = 0.786 X 0.126 = 0.099 in.
In order to check the load calculated as producing a shearing yield
point stress, the specimen block prepared from the strain-etch steel
was loaded at different points along its length with five loads as
follows:
Load
Load No. lb. per in. Remarks
1................................. 12 000 No strain lines
2 .............................. 12 450 No strain lines
3 ................................. 12 930 Strain lines
4................................. 15 100 Strain lines
5................................. 17 800 Strain lines
It will be noted that two of these loads are lower than the 12 700 lb.
per in. calculated above, while three are higher. The specimen block,
after heating as already described, was split longitudinally, the sec-
tion finished smooth, and then etched. It was found that for the load
of 12 930 lb. per in. and for all higher loads the sections showed
ILLINOIS ENGINEERING EXPERIMENT STATION
strain-etch patterns, while for the two loads lower than 12 700 lb.
per in. the sections showed no evidence of overstress.
Figure 3 is from a photograph of sections under loads 1, 4, and 5
and shows the kind of strain-etch patterns obtained. The magnitudes
and points of application of the loads are shown in the figure. It
should be noted that the strain lines practically disappear at the sur-
face and that the lines are at an angle of approximately 45 deg. at the
distance below the surface (0.1 in.) where the maximum shearing
stress is expected. Only small portions of the specimen block are
shown in this figure, since it was found desirable to cut the specimen
block into relatively small pieces for convenience in etching.
This same procedure was repeated with a second specimen block
and again the strain-etch figures were not obtained until the load was
such that the calculated shearing yield point stress of the material
was exceeded. The accuracy of the determination from the loads
appears to be on the order of plus or minus two per cent. This is
probably within the limits of accuracy of the determination of the
shearing yield point.
It may be worth while to compute the maximum intensity of the
direct compressive stress (o0Z) at the surface under the load of
12 700 lb. per in.
From Equation (85) we have
b
Ozý = - -
A
We have already found that
b =2256 P
A R
Hence:
/12 700
oZ, = -2256 1-- = -63 900 lb. per sq. in.
15.8
In general, for this type of loading (cylinder on a plane), the maxi-
4
mum stress at the surface is equal to - times the average stress or
4P
7 2b
While the compressive yield point of this material was only
28 800 lb. per sq. in., yielding in direct compression did not occur at
the surface because of the high stress components at right angles to
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 41
this. From Equations (85) it may be determined that at the surface
oY, is equal to oZý, and oXx is one-half as great, if Poisson's ratio is
taken as 1Y. That is, the maximum shearing stress at the surface is
63 900
only 4 , or approximately 16 000 lb. per sq. in., as compared to
19 200 lb. per sq. in. at a point in the interior.
23. Tests of Crossed Cylinders.-In checking the mathematical
analysis for stresses within the area of contact of crossed cylinders,
the same procedure was followed as has just been described for a
cylinder on a plane. After calculating the load required to produce a
shearing stress equal to the shearing yield point of the material a
series of loads extending below and above the calculated load was
applied, and after heating, the block was split, etched, and examined
for evidence of overstress.
In this case the mathematical computations are much more diffi-
cult than for the loading previously considered, unless the empirical
equation discussed in the following pages is used. Substituting the
conditions of this loading in that equation
= 23 500 P"
(Ri/R)0.271 R2
(158)0.271 3
or P = 2 X 19 200 ( 12 / (12)2
23 500
= 9.233 = 786 lb.
It is seen that this value of the load agrees with the value previously
obtained for these same conditions, using the exact method with
A
graphical interpolation for - (1Z, - Y,).
a
After loading the specimen block with a series of loads ranging
from below the calculated load to well above it, the block was split
through the centers of the loaded areas and on etching was examined
for evidence of overstress. It was found that for this condition of
loading the strain-etch lines were not clear and distinct as for the
loading first used. Although the lines were visible during the etching
on those sections where the load was such that the computed shearing
yield point was exceeded, it was found that on cleaning the surface to
ILLINOIS ENGINEERING EXPERIMENT STATION
prevent rusting or the deposition of copper, the lines became so
indistinct that they could not be photographed. Much effort was
spent in the attempt to produce etch-lines which could be photo-
graphed, but without success. However, it may be stated definitely
that the mathematical computation for crossed cylinders has been
verified within reasonable limits by means of the strain-etch method.
24. Approximate Equation-Crossed Cylinders.-As has been indi-
cated, the direct use of the mathematical analysis for the solution of a
definite problem is a long and tedious process. The most satisfactory
general method of substituting definite numerical values is by the use
of a diagram such as Fig. 2. However, for the case of crossed cylinders
it is possible to derive an approximate empirical equation which per-
mits an easy solution for values of the maximum difference of stress
components. Using the two equations
a = D or a -
where D is a constant (see second line, Table 3) for a given value of
B/A, and
2(1 - U2)
A--
E(A + B)
assuming that both bodies are steel and that a = 14, E = 30 000 000
lb. per sq. in., and remembering that R1 = (B/A)R2, it can readily
be determined that
a 40000 Pi
A B/A 1 V' R2"
_B/A + 1
Referring to Table 3 it will be noted that for a given B/A, the
value of (1Z, - 1Y,) is equal to a numerical factor multiplied by .
A
If then for each value of B/A in Table 3 we multiply the right-hand
side of the preceding equation by this factor, and substitute the cor-
responding numerical values of B/A and D, it will be found that the
maximum stress differences may be represented by the equation
phi
Z, - 1Y, = CR
where C is some function whose form is to be determined, but where
C has a definite numerical value for a given B/A. If the values of C
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 43
obtained as just described for values of B/A in Table 3 are plotted on
log paper using the corresponding values of B/A as ordinates, it will
be found that the points lie on a straight line giving the equation
23 500
C-
(B/A)0.271
Hence we may write:
23 500 P"
S- 1 = (R/R2)0.271 R (a)
For convenience in solving with a slide rule, this may be changed to a
slightly less accurate approximation:
23 000 P"
1Zz - (YV- = R2 (b)
Equation (b) gives values about two per cent low for = 1,
R2
and about 212 per cent high for R = 7.86. The error is less for
B
intermediate values. So long as -- is kept within the limits found in
Table 3 these equations may be used in solving for maximum differ-
ence of stress components. While the equations appear complicated
they may be solved with a slide rule, although (b) is the more easily
handled. In Fig. 4, Equation (a) has been plotted in the form of a
nomographic chart which may be used for solving for any one term
if the other three are given. In using it the intersection with the
"support" (ungraduated line) of a line joining known values on the
R1
lines for P and R2 must also be on the line joining corresponding
points on the 1Z, - 1Y, and R2 line. As an example, if P = 10 000
R1 R1
lb.' R = 1, R2 = 12 in., first connect R = 1 with P = 10 000, then
from R2 = 12 draw a line through the point where the first line cuts
the "support." This line, extended, intersects the 1Z, - 1Y, line in
the value desired, 96 500 lb. per sq. in.
As in the case of the empirical mathematical equation, the use of
this chart for values of R- outside of the limits indicated is not advis-
able, for the errors involved may be considerable.
ILLINOIS ENGINEERING EXPERIMENT STATION
Zl
Values of -, a, and k, may be more readily obtained from the
corresponding curves of Fig. 2 than by the use of any empirical equa-
tions which might be derived.
V. EFFECT OF VARIOUS FACTORS ON CALCULATED STRESSES
25. Significance of Calculated Stresses.-By the use of the method
of calculation presented in this bulletin, for many conditions it is
possible to compute the stress components and maximum shearing
stresses produced when one elastic body is pressed against another.
In general, the conditions of loading will probably be such that during
the useful life of the member it will be subjected to a large number of
applications of the load. Such loading produces a shearing stress
which varies from zero to a maximum for each application, and hence
it is evident that for materials used urider this kind of loading the
fatigue endurance limits of the materials when subjected to a shearing
(torsion) stress varying from zero to a maximum is one of great im-
portance. There is not a great amount of information available
concerning the endurance properties of materials subjected to this
kind of stress, and hence the interpretation of the significance or
seriousness of a particular calculated shearing stress is open to doubt.
An important question which cannot be settled at this time is
whether or not a repeated shearing stress having a maximum value
in the interior of. a body will cause a fatigue crack to form and spread
under this condition as readily as in the case of a torsion fatigue test
specimen where the maximum stress is at the surface. Some prelim-
inary tests made in the Fatigue of Metals Laboratory have a bearing
on this question, and the results seem to indicate that cracking and
rupture may not start in the interior of a body when the computed
shearing stress is somewhat higher than the torsion endurance limit
determined in the ordinary way. This needs considerable further
study, especially since it is difficult to make tests of this nature with-
out having a change occur in the curvature of the specimen due to
wear, and hence a quite appreciable change in calculated stresses.
Another question arising in this connection is the possibility of
work hardening of the material due to overstress, thus causing a
change in its physical properties. The residual internal stresses due
to cold work and rolling action may also have an effect both on the
amount of shearing stress produced at the point of contact of the two
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 45
ZOO 000
/00000
- 80000
VU 5000
- 0 L00
S40000
- 30000
- 25000
- 20000
FIG. 4.
(~)
\
- e3-- Soo0 _
,Z'--,>Y-
NOMOGRAPHIC CHART FOR APPROXIMATE EQUATION
bodies and possibly also on the endurance properties of the material.
In fact, it would seem that under some circumstances residual internal
compressive stresses due to cold work might so affect the shearing
stress as to cause the actual maximum shearing stress to be at right
angles to the direction which would be expected without taking ac-
count of the stress due to cold work. That is, the maximum stress
40 00
30000
/0000
9000
- 9 UUU
- /2
* 8000
7000
6000
_VJO,7 -
/--/-I /n^l/l
5000
ILLINOIS ENGINEERING EXPERIMENT STATION
difference might change to Z. - X, instead of Z, - Y,. This would
depend to some extent on the relative radii of curvature of the two
bodies, and for this to occur the internal stress due to cold work
would have to be greater than Xx - Y,.
26. Crossed Cylinders.-In order to study the effect of the various
factors entering into the equation for stress differences the empirical
equation
23 500 P_
1Zz - 1Y (R,/R2) 0.271 R2
can be changed to the form
23 500 P"
Z, - - R10.271 R20.396
From this it is seen that the maximum stress difference varies as the
cube root of the load, that it varies inversely as R1i0271 and inversely
as R20396. In this case R1 > R2, and hence it is seen that the effect
of changing the smaller radius a given amount is greater than that
for a corresponding change in R1. Again, attention is called to the
fact that these conclusions are valid only for values of n between
the limits of 1 and 8.
27. Cylinder on a Plane.-For the condition of loading of a cyl-
inder on a plane, the maximum stress difference varies directly with
the square root of the load per inch of length of cylinder and inversely
as the square root of the radius of curvature of the cylinder. That is,
the stress does not vary so long as the ratio of the load to the radius
of curvature is a constant.
28. Effect of Poisson's Ratio.--In all calculations in this bulletin,
Poisson's ratio has been assumed as Y1. The effect of changing
Poisson's ratio to some other value is to change the stress component
A A
factors - X, and - Y, in Tables 1 and 2. In order to determine the
a a
A
effect on - (1Z, - 1Y,) of assuming this ratio as Y instead of Y,
a
the values of the stress-difference factors have been calculated for
a- = Y for the various values of k. It is found that the values of
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 47
A
- (1Z, - 1Y,) are reduced, when o is changed from 14 to 13, in the
a
ratio shown in the second column below.
Stress-difference
Value of k Ratio, = ! 1 Net Effect
a = Y4
1.000..................... 0.94 0.97
0.866..................... 0.95 0.98
0.707..................... 0.95 0.98
0.573..................... 0.95 0.98
0.422..................... 0.96 0.99
0.342..................... 0.96 0.99
0.258..................... 0.96 0.99
a
However, we find that for crossed cylinders the quantity - is increased
by 3.5 per cent when a is changed from 14 to Y3. The net effect then
is to change the ratios to the values given in the last column. For
other than crossed cylinders it will be necessary to compute the effect
a
on - of a change in Poisson's ratio. If desired, calculated stress
A
difference may be corrected by the factors just given, but it should
be noted that in using the tables or diagrams for making calculations
a
the value of A must be based on a- = 1. The broken line in Fig. 1
shows the stress differences for k = 0.86603 and a = 13.
At the surface the change in Poisson's ratio has a greater effect.
A A
At this point the values of - (oZz - 0Y,) and of - (o0Z - oX.) for
a a
a = Y are two-thirds as large as for a = 1%, for all values of k.
For the case of a cylinder on a plane, there is no difficulty in direct
substitution of any value of Poisson's ratio desired, since the equations
are readily solved.
VI. SUMMARY AND CONCLUSIONS
29. Conclusions.-As the result of the investigation the following
conclusions were drawn:
(1) By using the methods of solution presented in this bulletin,
it is possible to compute the principal stress components within the
area of contact, both at the surface and at points below the surface,
ILLINOIS ENGINEERING EXPERIMENT STATION
when one elastic body is pressed against another. It is also possible
to compute the maximum shearing stress, due to the difference of
stress components, and to determine the distance from the contact
surface to the point of maximum shear, as well as the area of con-
tact.
(2) The accuracy of the calculations of maximum shearing stresses
in the interior of a body has been verified by a strain-etch method.
(3) For the case of crossed steel cylinders, if Poisson's ratio = 14,
the maximum stress difference is given by the following empirical
equation:
23 500 P"'
1zz - 1Y = - (R,/R2)0.271 R2
This maximum stress difference (which is equal to twice the shearing
stress) occurs at a point below the surface of contact. This equation
may only be used for values of R1/R2 between the limits of 1 and 8,
since calculations have not been carried beyond those limits. R1 is
the radius of curvature (expressed in inches) of the larger cylinder,
and R2 is the radius of curvature of the smaller. P is the total pressure
between the two bodies, in pounds.
The exact solution for this condition and also for more general
conditions is found in the text.
(4) For the case of a steel cylinder on a plane, the maximum
shearing stress equals 677 p , the width of contact is 2b =
0.000564 PR, and the depth to point of maximum shearing stress is
zi = 0.786b = 0.000222 PR. The maximum intensity of surface
4 P lp
pressure is equal to - - 2256 - . That is, the maximum
7r 2b R
surface pressure is 27 per cent higher than the average. In this case
P is the load per inch of length of the cylinder whose radius is R.
(5) The mathematical analysis is based on the assumption that
the bodies in contact are homogeneous and isotropic. While it is
well known that shapes and sections rolled from metals do not con-
form exactly to this assumption, it is probable that the errors in-
volved will not be large. The question as to the seriousness of a
given calculated shearing stress is one which can not be answered
definitely until more information is obtained on the endurance
limits of materials as commercially produced, when subjected to
repeated shear, and also the effect of cold work and the residual
stresses caused by cold work on the physical properties and behavior
of such materials.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 49
APPENDIX A
REDUCTION OF A QUADRATIC FORM TO Ax2 + By2
The expression A'1I2 + 2H'1,i1 + B'nv2 must be reduced to Ax2
+ By2 by a rotation of axes through angle 0. This rotation is
effected by making the substitutions
ýi = x cos 0 - y sin 0(
,i= x sin 0 + y cos O (1)
Thus we have after collecting terms
A'ý12 + 2H'1nr, + B',712 -. x2(A'cos'0 + 2H'sin0 cos 0
+ B'sin2 0)
+2xy[(B' - A')sin0 cos0 + H'(cos20 - sin20)] (2)
+y2(A'sin20 - 2H' sin 0 cos 0 + B'cos2 0)
Equating this to Ax2 + By2, it is found that
A = A'cos20 + 2H'sinO cos0 + B'sin20 0
0 = (B' - A')sin 0 cos 0 + H'(cos'0 - sin2 0) (3)
B = A'sin2 0 - 2H'sin 0 cos 0 + B'cos20 f
It may be verified that
A + B = A' + B' (4)
AB = A'B' - H'2 + [(B' - A')sin0 cos0
+ H'(cos20 - sin20)]2 = A'B' - H'2 (5)
Now the quadratic equation x2 - (A + B)x + AB = 0 has roots
A, B. With the values of the coefficients just found, the quadratic
equation with roots A, B becomes
x2 - (A' + B')x + A'B' - H'2 = 0 (6)
From (12) we have (see Chapter II):
2(A' + B') = p + P'l + 2 + p '2 (7)
4A'B' = [pI + P2 - (2 - p'2)sin2 e][p'1 + P'2 + (p2 - p'2) sin2 e
= (pi + P2)(P'i + p'2) + (P2 - P'2) (P - P'i + P2 - P2) sin2 e
- (P2 - p'2) sin4 E
4H'2 = (p2 - pl)2 sin2 e cos2 e = (P2 - p'2)2sin2 e - (p2 - p'2)2sin4D
4(A'B' - H'2) = (pI + P2)(p'i ± p'2) + (pI - p' )(p2 - P' )sin2 e (8)
ILLINOIS ENGINEERING EXPERIMENT STATION
Substituting the values of A' + B' from (7) and A'B' - H'2 from
(8) into (6) and multiplying the resulting equation by 4, we get
4x2 - 2x(pi + p', + p2 + p2) + (pI + P2)(P' + p'2)
+ (PI - P'i)(P2 - p'2)sin2 e = 0
or (2x - pi - p2)(2x - p', - p'2) + (PI - p')(P2 - 2)sin2 e = 0 (9)
The two roots of (9) are the values of A and B desired.
APPENDIX B
POTENTIAL OF AN INFINITELY LONG STRAIGHT FILAMENT
The potential of an infinitely long straight filament of constant
linear density P' may be found by means of the well-known theorem
of Gauss, viz., that the integrated value over any closed surface of
the component of attraction normal to this surface is 4-r times the
mass within. Take as the surface a cylinder of unit length with axis
coinciding with the attracting filament. The attraction, F, is evi-
dently everywhere in a direction perpendicular to the filament and
is a function only of r, the distance away from the filament. Hence
the surface integral of the attraction over the cylinder of unit length
is 2irrF and, by the theorem of Gauss,
2irrF = 47rP'
or F = 2 P'/r (1)
Then the potential is given by
0 = -fFdr = -P'logr2 (2)
where the arbitrary constant of integration is of no consequence in
our problem since only derivatives of 4 are involved.
APPENDIX C
ADDITIONAL NUMERICAL EXAMPLES
To show further the method of using the mathematical analysis
when solving certain practical problems, and also to indicate the mag-
nitude of the shearing stresses which may be expected, the following
examples will be solved. In all these cases assume Poisson's ratio as
0.25 and E = 30 000 000 lb. per sq. in.
Case I, Crossed Cylinders.-Assume R1 = 16.5 in., R2 = 14 in.,
P = 25 000 lb. This corresponds to the condition of a freight-car
wheel on a rail having a radius of head of 14 in.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 51
B/A = Ri/R2 = 16.5/14 = 1.18
B +A = 2 R1 + = 2 + =0. 066
From Fig. 2, for B/A = 1.18 we have
k = 0.90
AP
A - 1.13
a3
Zl
- = 0.44
a
A
- (1Z, - iY,) = 0.388
2(1 - ~r2) 1
A = = = 9.47 X 10-1
E(A+B) 16 X 106 (A + B)
a 9.47 X 25 000
a= = 1.13 X 10 0.276in.
a 0.388 X 0.276 X 107 = 113 000 lb.
(1Zz - 1YF) = 0.388- =
A 9.47 per sq. in.
Maximum shearing stress = 1Y(Z, - iY,) = 56 500 lb. per sq. in.
b = ka = 0.90 X 0.276 = 0.248 in.
Area of contact = 7rab = 3.1416 X 0.276 X 0.248 = 0.215 sq. in.
Depth to point of maximum shear = 0.44a= 0.44 X 0.276 = 0.121 in.
This stress difference may be checked by using the empirical
equation
23 500 P" 23 500 /25 000
1z 1 Y = (Ri/R2)0.271 R2 = (1.18)0o.271 '(14)2
23500 29.2
- 1.046 X 5.81 = 112 700 lb. per sq. in.
Case II, Crossed Cylinders.-Assume R, = 40 in., R2 = 14 in.,
P = 60 000 lb. This may be considered as corresponding to a loco-
motive driving wheel having a load of 30 000 lb., plus 100 per cent
allowance for impact (as frequently assumed for short spans), on a
rail having a radius of head of 14 in.
ILLINOIS ENGINEERING EXPERIMENT STATION
B/A = R1/R2 = 40/14 = 2.86
B + A = 2 - 0 4 0.0482
From Fig. 2,
k = 0.500
AP
-. = 0.430
a3
zi
- 0. 317
A
- (iZ, - 1Y,) = 0.274
a
1 1
A -- = = 12.95 X10-7
16(A + B)106 16 X 0.0482 X 106 12.95 X
./12.95 X 60 000
a = X 10 = 0.565 in.
b = 0.5a = 0.283 in.
Area of contact = irab = 0.502 sq. in.
a 0.274 X 0.565 X 107 119 600 lb.
Z, - Y = -0.274 - --
A 12.95 per sq. in.
Maximum shearing stress = 59 800 lb. per sq. in.
Depth to point of maximum shear zi = 0.317a = 0.179 in.
Note that this wheel and loading produce a shearing stress only six
per cent higher than the freight-car wheel, in spite of the fact that
the wheel load is 2.4 times as great.
Case III, Cylinder on Plane.-Assume, as in Case I, radius of
wheel of 16.5 in., load 25 000 lb., but assume the head of the rail is
flat with the wheel making contact on 2 in. of width of the head.
Width of contact 2b = 0.000564 ýPR
/25 000
= 0.000564 -- 2 X 16.5
= 0.256 in.
PRESSURE STRESSES OF ONE ELASTIC SOLID UPON ANOTHER 53
Maximum shearing stress
S2(Z - 1355 /P 1355 /12500
= Z2- RY) = 2 16.5
= 18 600 lb. per sq. in.
Depth to point of maximum shearing stress
0.256
zi = 0.786b = 0.786 X 2 = 0.100 in.
2
Case IV, Cylinder on Plane.-Assume, as in Case II, radius of
wheel of 40 in., load 60 000 lb., but a flat-head rail, with contact
over 2 in. of width of rail. Load per inch P = 30 000 lb.
Width of contact 2b = 0.000564 \130 000 X 40
= 0.618 in.
1355 iP
Shearing stress - 2 R = 18 600 lb. per sq. in.
Depth to maximum shearing stress
0.618
= 0.786 X - - 0.243 in.
2
Case V, Cylinder on Plane.-In University of Illinois Bulletin
No. 162,* page 54, are given the results of tests of a segment of a
roller for the Rall Type Bascule Bridge. This roller had a radius of
30 in., the Brinell hardness of the material being 160. Using the
average load per inch (39 800 lb.) given in Table 9 as causing yielding
in carbon steel, we have a calculated shearing stress of
Maximum shear = 12 X 1355 = 1355 3 800
= 24 700 lb. per sq. in.
No results of physical tests of the material are given in the bul-
letin, but it seems reasonable to expect the shearing yield point of the
material having a Brinell hardness of 160 to be in the neighborhood
of this calculated stress.
Case VI, Cylinder on Plane.-In Bulletin 162, just referred to,
on page 55 are given the loading conditions for the sole plates of the
Galveston Causeway Bascule Bridge. These plates had a radius of
*"Tests of the Bearing Value of Large Rollers," Univ. of Ill. Eng. Exp. Sta. Bul. 162, 1927.
ILLINOIS ENGINEERING EXPERIMENT STATION
309 in., were 2'2 in. thick and subjected to a design load of 112 400
lb. per inch of length. For this case:
Width of contact 2b = 0.000586 \V112 400 X 309 = 3.45 in.
Maximum shearing stress
1355 4/112 400
1 2 1 309 = 12 900 lb. per sq. in.
Depth to point of maximum shearing stress
zi = 0.786b = 0.786 X 1.725 = 1.355 in.
Based on the shearing stress calculated for this condition one
might judge that the plates would be satisfactory, and yet these
plates had to be removed,from the bridge because they did not stand
up. Professor Wilson reports that two pieces of these plates, finished
to a thickness of 2 in., having a length parallel to the axis of the cyl-
inder of 4.1 in., and a radius of 309 in. showed permanent set under
an average load of 53 000 lb. per in. For this load the calculated
shearing stress is 8 900 lb. per sq. in. The failure of the mathematical
methods in this case is not difficult to understand.
It will be remembered that in the derivation of the mathematical
equations for the condition of loading of a cylinder on a plane it was
assumed that the ratio of the length of contact (parallel to the axis
of the cylinder) to the width was infinite, and also that the thickness
of the bodies was infinite, or at least very large. It is evident, then,
that in using the mathematical methods for calculating the stresses
existing in this case they are being applied to conditions which are
entirely different from the assumptions on which the mathematical
analysis was based.
Professor Wilson found that, for a block 10 in. deep, length of
contact 4 in., radius 309 in., the load required (interpolated from
curve) to produce set in a material having a Brinell hardness of 117
would be about 113 000 lb. per in. In this case also the length of
contact is small in comparison with its width and hence the calculated
shearing stress of 13 000 lb. per sq. in. cannot be expected to be
correct.