In 4.4 the development of contact pressure between mating surfaces of joints, where loads are transmitted from one machine part to another, was briefly discussed. As noted in that dis- cussion, the Hertz contact stressdistributions at and below the surfaces within the contact re- gion of the mating parts are typically triaxial. Because of the triaxiality, a failure theory must be utilized if it is desired to assess the potential for failure by the governing failure mode.
The general case of contact stress occurs when each of the two contacting bodies loaded against each other has two mutually perpendicular principal curvatures at the con- tact site, measured by and for body 1, and and for body 2.19The two more common specific cases are two spheres in contact (including a sphere on a sphere, a sphere on a flat plane, and a sphere in a spherical cup) and two parallel cylinders in contact (including a cylinder on a cylinder, a cylinder on a flat plane, and a cylinder in a cylindrical groove). Examples of machine elements having characteristics of such con- tact geometries include ball bearings, roller bearings, cams, and gear teeth.
For the case of solid spheres with diameters and , pressed together by a force F, the “footprint” of the small contact area is circular, having radius aas shown in Figure 4.15.
The radius of the circular contact area is given by
(4-67)
The maximum contact pressure , at the center of the circular contact area, is (4-68) pmax = 3F
2pa2 pmax
v1,v2 = Poisson’s ratios for spheres 1 and 2, respectively whereE1,E2 = moduli of elasticity for spheres 1 and 2, respectively
a = 3 f
3Fc a1 - v21 E1
b + a1 - v22 E2
b d
8a1 d1
+ 1 d2
b d2
d1
R2min R2max
R1min R1max
Expressions (4-67) and (4-68) are equally valid for two spheres in contact ( and positive), a sphere on a plane for the plane), and a sphere in a spherical cup (dis negative for a cup).
The maximum stresses , and are generated on the z-axis, where they are prin- cipalstresses. On the z-axis then, the principal stresses are
(4-69)
and
(4-70)
Since , ⫽0, and
(4-71) Figure 4.16 depicts these stresses as a function of distance below the contact surfaces up to a depth of 3a. It may be noted that the maximum shearing stress, , reaches a peak value slightly below the contact surface, as alluded to in the discussion of surface fatigue failure in 2.3.
When two parallel solid cylinders of length L, with diameters and , are pressed together radially by a force F, the “footprint” of the narrow contact area is rectangular, having half-width bas shown in Figure 4.17.
The half-width of the narrow rectangular contact area may be calculated from
(4-72) b =
f
2Fc a1 - v21
E1 b + a1 - v22
E2 b d pLa1
d1 + 1 d2b
d2 d1
tmax
ƒt1ƒ = ƒt2ƒ = tmax = `s1 - s3
2 `
ƒt3ƒ s1 = s2
sz = s3 = -pmax J
1 az2
a2 + 1bK sx = s1 = sy = s2 = -pmax
J
11 + v 2a1 - z a tan-1a
zb - 1
2az2 a2 + 1bK sz
sx,sy
1d = q
d2
d1
Stresses Caused by Curved Surfaces in Contact 175
1 σ,τ
σz
σx,σy
τmax
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
00 0.5a a 1.5a 2a 2.5a 3a z
Distance below contact surface
Ratio of stress to pmax
Figure 4.16
Magnitudes of x, y, z, and maxas a function of maximum contact pressure between spheres for various distances z below the contact interface (for vⴝ0.3).
T S S S
z
F
F
Cylinder 2 Cylinder 1
d1
d2
E1,v1
E2,v2 pmax
x Rectangular
contact
footprint Semielliptic
pressure distribution
L 2b
Figure 4.17
Two cylinders in contact, loaded by force Funiformly distributed over their length L.
The maximum contact pressure , along the centerline of the narrow rectangular contact area, is
(4-73) Expressions (4-72) and (4-73) are equally valid for two parallel cylinders in contact and positive), a cylinder on a plane for the plane), or a cylinder in a cylin- drical groove (dis negative for the groove).
The maximum normal stresses (principal stresses) , and occur on the z-axis.
Thus the principal stresses are
(4-74)
(4-75)
and
(4-76)
The magnitudes of the principal shearing stresses20are
(4-77) ƒt1ƒ = `s2 - s3
2 `
s3 = sz = -pmax J
1 A
z2 b2 + 1K s2 = sy = -pmax
J P
2 - 1
az2
b2 + 1bQA z2
b2 + 1 - 2z bK s1 = sx = -2vpmaxc
A z2
b2 + 1 - z bd
sz
sx,sy
1d = q d2
1d1
pmax = 2F pbL pmax
20See, for example, ref. 8, pp. 100
Stresses Caused by Curved Surfaces in Contact 177
(4-78)
and
(4-79) Of these three principal shearing stresses, is a maximum at about z⫽0.75bbelow the surface, as shown in Figure 4.18, and is larger there than either or (although is not largest for allvalues of z /b).
In some instances the “normal approach” (displacement of the centers of two con- tacting spheres or cylinders toward each other) caused by the load-induced elastic contact deformations will be of interest. For example, the overall stiffnessof a machine assembly may be needed for certain design evaluations, requiring therefore that the stiffness be known for each componentin the assembly. If bearings, cams, or gear teeth are compo- nents of an assembly, the normal approach due to contact deformation may be a very sig- nificant part of the overall deformation.
For the case of two spheresin contact, the normal approach is given by21
(4-80) For the case of two parallel cylinders(of the same material) in contact, the normal ap-
proach may be calculated, defining and , as
(4-81a)
¢c = 2F11 - v22 pLE a2
3 + ln2d1
b + ln2d2
b b E1 = E2 = E v1 = v2 = v
¢c
¢s = 1.043 BF2a1
d1 + 1
d2b c a1 - v21
E1 b + a1 - v22
E2 b d2
¢s
t1
t3
t2
t1
ƒt3ƒ = `s1 - s2
2 `
ƒt2ƒ = `s1 - s3
2 `
1 σ,τ
σz
σy
σx
τmax
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
00 0.5b b 1.5b 2b 2.5b 3b
Distance below contact surface
Ratio of stress to pmax
z
Figure 4.18
Magnitudes of x, y, z, and maxas a function of maximum contact pressure between cylinders, for various distances z below the contact interface (for vⴝ0.3).
T S S S
21See, for example, ref. 4, Table 33.
2 in.
F = 4000 lb
2 in.
d1 = 1.000 in.
r2 = 0.531 in.
C = 1.469 in. (unloaded) L
Figure E4.14
Special-load transfer joint, made of AISI 4340 steel.
For a cylinder in a cylindrical groove
(4-81b) It should be noted that these displacements are highly nonlinear functions of the load;
hence the stiffnesscharacteristics of curved surfaces in contact are highly nonlinear.
¢c = 2F11 - v22
LE a1 - 2lnb 2b
Example 4.14 Cylindrical Surfaces in Direct Contact
Figure E4.14 shows the tentative concept for a special load transfer joint in a torque mea- suring fixture. It is important to accurately maintain distance Lat a constant value. The joint is to be constructed using a 2-inch-long cylinder of ultrahigh strength AISI 4340 steel (see Tables 3.3 and 3.9) with a diameter of 1.000 inch, residing in a 0.531-inch-radius semicir- cular groove in a 2-inch-wide block of the same material. It has been estimated that a to- tal force of 4000 lb will be distributed across the cylinder as shown. Experimentation has indicated that for satisfactory service the maximum surface contact pressure should not ex- ceed 250,000 psi. For the proposed design configuration:
a. Determine whether the surface contact pressure is below the specified limiting value.
b. Determine the magnitude of the maximum principal shearing stress in the contact re- gion, and its depth below the surface.
c. Estimate the change in dimension Cas the joint goes from the unloaded to the fully loaded condition.
Solution
a. For the parallel cylindrical surfaces in contact, the maximum contact pressure is given by (4-73) as
wherebis given by (4-72). To calculate b, we first find from Table 3.9 and the prob- lem statement that
F= 4000 lb L = 2.0 inches v1 = v2 = v = 0.3 E1 = E2 = 30 * 106 psi
pmax = 2F pbL
Load Sharing in Redundant Assemblies and Structures 179
hence
and
Since the limiting allowable contact pressure is given as 250,000 psi, the calcu- lated value of ⫽95,426 psi is acceptable.
b. To find the value of , the plot of Figure 4.18 may be utilized. Noting that peaks at about z⫽0.75b(forv⫽0.3), the corresponding ratio of may be read as
Thus
and it occurs at a depth
below the surface. The shearing stress and its depth below the surface are approxi- mately the same in both members since the material is the same for both.
c. The change in dimension C, when the joint is loaded as shown, may be calculated from (4-81b) as
¢c = 214000211 - 0.322
12.0230 * 106 a1 - 2ln0.013
2 b = 0.0013 inch z = 0.7510.0132 = 0.010 inch
t1max = 0.3195,4262 = 28,628 psi t1max
pmax = 0.3
t1max>pmax t1max
t1max
pmax
pmax = 2140002
p10.013212.02 = 95,426 psi b =
a
2140002c2a1 - 0.32 30 * 106b d p12.02c 1
1.000 + 1 -1.062d
= 0.013 inch