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Unformatted text preview: corresponding radii, respectively. If the cylinders
have the same elastic modulus E and Poisson’s ratio = 0.3, these expressions reduce to
c = 0.418 PE r1 + r2
,
r1r2
AL b = 1.52 r1r2
P
A EL r1 + r2 (3.61) Figure 3.20b shows the special case of contact between a circular cylinder of radius
r and a flat surface, both bodies of the same material. After rearranging the terms
and taking r1 = r and r2 = q in Eqs. (3.61), we have
c = 0.418 PE
,
A Lr b = 1.52 Pr
A EL (3.62) Two Curved Surfaces of Different Radii
Consider now two rigid bodies of equal elastic moduli E, compressed by force P
(Fig. 3.21). The load lies along the axis passing through the centers of the bodies
and through the point of contact and is perpendicular to the plane tangent to both
132 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 133 FIGURE 3.21. Curved surfaces of different radii of two bodies compressed by forces P. bodies at the point of contact. The minimum and maximum radii of curvature of the
œ
œ
surface of the upper body are r1 and r1; those of the lower body are r2 and r2 at the
œ
œ
point of contact. Thus, 1/r1, 1/r1, 1/r2, and 1/r2 are the principal curvatures. The sign
convention of the curvature is such that it is positive if the corresponding center of
curvature is inside the body. If the center of the curvature is outside the body, the
œ
œ
curvature is negative. (For example, in Fig. 3.22a, r1, r1 are positive, while r2, r2 are
negative.)
Let be the angle between the normal planes in which radii r1 and r2 lie. Subsequent to loading, the area of contact will be an ellipse with semiaxes a and b
(Table C.1). The maximum contact pressure is
c = 1.5 3 Pm
,
An P
ab In this expression the semiaxes are given by
a = ca b = cb (3.63) 3 Pm
An (3.64) Here
m= 4
,
1
1
1
1
+ œ+
+œ
r1
r2
r1
r2 n= 4E
311  2 2 (3.65) The constants ca and cb are read in Table 3.3. The first column of the table lists values of , calculated from
cos 3.13 Contact Stresses = B
A (3.66) 133 ch03.qxd 12/20/02 7:21 AM Page 134 TABLE 3.3
ca cb 3.778
2.731
2.397
2.136
1.926
1.754
1.611
1.486
1.378
1.284
1.202
1.128
1.061
1.000 0.408
0.493
0.530
0.567
0.604
0.641
0.678
0.717
0.759
0.802
0.846
0.893
0.944
1.000 (degrees) 20
30
35
40
45
50
55
60
65
70
75
80
85
90
where
2
A= ,
m 1
1
1
1
1
B = B¢ ≤ +¢ ≤
r2
2 r1
r¿1
r¿2
2 + 2¢ 2 1
1
1
1
≤¢ ≤ cos 2 R
r1
r ¿ 1 r2
r¿2 1/2 (3.67) By applying Eq. (3.63), many problems of practical importance may be treated,
for example, contact stresses in ball bearings (Fig. 3.22a), contact stresses between a FIGURE 3.22. Contact load: (a) in a singlerow ball bearing; (b) in a cylindrical wheel and rail. 134 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:21 AM Page 135 cylindrical wheel and a rail (Fig. 3.22b), and contact stresses in cam and pushrod
mechanisms.
EXAMPLE 3.6
A railway car wheel rolls on a rail. Both rail and wheel are made of steel
= 0.3. The wheel has a radius of
for which E = 210 GPa and
r1 = 0.4 m, and the cross radius of the rail top surface is r2 = 0.3 m (Fig.
3.22b). Determine the size of the contact area and the maximum contact
pressure, given a compression load of P = 90 kN.
Solution For the situation described, 1/r ¿ 1 = 1/r ¿ 2 = 0, and, because
the axes of the members are mutually perpendicular, = /2. The first
of Eqs. (3.65) and Eqs. (3.67) reduce to
m= 4
,
1/r1 + 1/r2 A= 11
1
¢ + ≤,
r2
2 r1 B= ; 11
1
¢  ≤ (3.68)
r2
2 r1 The proper sign in B must be chosen so that its values are positive. Now
Eq. (3.66) has the form
cos =; 1/r1  1/r2
1/r1 + 1/r2 (3.69) Substituting the given numerical values into Eqs. (3.68), (3.69), and
the second of (3.65), we obtain
m=
cos 4
= 0.6857,
1/0.4 + 1/0.3 =; 1/0.4  1/0.3
= 0.1428
1/0.4 + 1/0.3 n= 41210 * 1092
= 3.07692 * 1011
310.912 or = 81.79° Corresponding to this value of , interpolating in Table 3.3, we have
ca = 1.1040, cb = 0.9113 The semiaxes of the elliptical contact are found by applying Eqs. (3.64):
a = 1.1040 B b = 0.9113 B 90,000 * 0.6857 1/3
R = 0.00646 m
3.07692 * 1011 90,000 * 0.6857 1/3
R = 0.00533 m
3.07692 * 1011 The maximum contact pressure, or maximum principal stress, is thus
c 3.13 Contact Stresses = 1.5 90,000
= 1248 MPa
10.00646 * 0.005332 135 ch03.qxd 12/20/02 7:21 AM Page 136 A hardened steel material is capable of resisting this or somewhat
higher stress levels for the body geometries and loading conditions described in this section.
PROBLEMS
Secs. 3.1 through 3.7
3.1. A stress distribution is given by
x = pyx3  2c1xy + c2y y = pxy 3  2px3y xy 3
22
2 px y = (a) + c1y2 + 1
4
2 px + c3 where the p and c’s are constants. (a) Verify that this field represents a solution for a thin plate of thickness t (Fig. P3.1); (b) obtain the corresponding stress function; (c) find the resultant normal and shearing boundary
forces ( Py and Vx ) along edges y = 0 and y = b of the plate.
3.2. If the stress field given by Eq. (a) of Prob. 3.1 acts in the thin plate shown
in Fig. P3.1 and p is a known constant, determine the c’s so that edges
x = ;...
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This note was uploaded on 01/09/2011 for the course ME 201 taught by Professor Yok during the Spring '08 term at Boğaziçi University.
 Spring '08
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