We have seen how contact stresses may be treated approximately under the heading of 'bearing' or 'crushing', using the concept of projected area. We now examine these stresses in more detail.
In real life however neither the base (b) below nor the cylinder (c) is completely rigid - rather both contacting elements must deform as indicated at (d). Elasticity requires gradual transition of both members from their free boundaries to the common interface area (e).
Shown are two spheres, or cylinders of length L in end view, with greatly exaggerated deflections. A Cartesian system is defined with the y-axis normal to the sketch plane. Both components deform; the generally non-plane contact area extends a distance 'a' from the z-axis of symmetry, and the contact pressure and loading severity are maxima on this axis.
The equivalent modulus E* and the equivalent diameter d* are defined by :
1/E* = 1/E'1 + 1/E'2 where E' = E/( 1-ν2 )
1/d* = 1/d1 + 1/d2 where |d2| >|d1| and is negative if internal.
The principal compressive stresses are then given by Shigley (op cit) :
| ( 5a) Spheres | ||||
| Contact zone size | a3 | = | 3 F d* / 8 E* | |
| Maximum contact pressure | pmax | = | 3 F / 2 π a2 | |
| Normalised principal stresses | sx | = | sy = ( 1 + ν )( 1 - φ arccot φ ) - 1/( 2 ( 1 + φ2 )) | |
| ( s = σ/pmax @ φ = z/a ) | sz | = | 1/( 1 +φ2 ) | |
| ( 5b) Cylinders | ||||
| Contact zone size | a2 | = | 2 F d* / π E* L ; L is the cylinders' contact length | |
| Maximum contact pressure | pmax | = | 2 F / π a L | |
| Normalised principal stresses | sx | = | 2 ν ( √( 1 + φ2 ) - φ ) ; sz = 1/ √( 1 + φ2 ) | |
| ( s = σ/pmax @ φ = z/a ) | sy | = | √( 1 + φ2 ) ( 2 - 1/( 1 + φ2 )) - 2 φ | |
Stresses are proportional to the peak contact pressure, which in turn depends on the relative size of the two bodies. If a reference body (cylinder or sphere) is contacted by an auxiliary body whose diameter varies (with other parameters constant) then the peak pressure will change as shown. The advantage of close conformity in reducing stress levels is apparent - ball bearings capitalise on this by using toroidal rather than cylindrical race surfaces.
E'1 = [ E/( 1 - ν2 ) ]1 = 207/( 1 - 0.292 ) = 226 GPa d1 = 100 mm
E'2 = [ E/( 1 - ν2 ) ]2 = 100/( 1 - 0.212 ) = 105 GPa d2 = ∞
E* = 1/( 1/E'1 + 1/E'2 ) = 1/( 1/226 + 1/105 ) = 71.7 GPa d* = 100 mm
Then, from the cylindrical equations ( 5b) with L = 5 mm :
a2 = 2 Fd* / π E* L = 2 x 500 x 100 / π x 71.7E3 x 5 whence a = 0.30 mm
pmax = 2F / π a L = 2 x 500 / π x 0.30 x 5 = 213 MPa
It will be recalled that a safety factor is essentially a load criterion - it is a stress criterion only when stress is proportional to load, which is NOT the case here. From the above pmax ∝ F /a where a ∝ √F, so that pmax ∝ √F.
For the steel, the 500 N load gives rise to a maximum equivalent stress of 0.6 pmax = 128 MPa, so the load necessary to produce yield will be 500 (350/128)2 = 3740 N. That is, the steel's safety factor is 3740/500 = 7.5.
Cast iron failure will occur when the maximum pressure, ie. the maximum surface compressive stress, reaches the compressive ultimate. Now the 500 N load above gives rise to pmax of 213 MPa, so the load necessary to induce fracture will be 500 (750/213)2 = 6200 N. That is, the cast iron's safety factor is 6200/500 = 12.4.
The overall factor of safety is therefore : minimum( 7.5, 12.4) = 7.5
The general case in which two contacting bodies 1 & 2 are each curved three-dimensionally is too complicated for closed-form evaluation of the stresses. Young (op cit) presents a numerical solution for the maximum contact pressure which may be approximated as a guide to safety. The minimum and maximum radii of curvature at the point of contact are r1 and R1 for body 1, and r2 and R2 for body 2. The reciprocals 1/r1 and 1/R1 are known as the principal curvatures of body 1, and 1/r2 and 1/R2 of body 2, and in each body the principal curvatures are mutually perpendicular. The plane containing curvature 1/r1 in body 1 makes an angle φ with the plane containing curvature 1/r2 in body 2.
The equivalent curvature radius r* and the parameter λ are defined by :
1/r* = 1/r1 + 1/R1 + 1/r2 + 1/R2
λ2 = (2/π) arccos { r* √ [ ( 1/r1 - 1/R1 )2 + ( 1/r2 - 1/R2 )2 + 2( 1/r1 - 1/R1 )( 1/r2 - 1/R2 ).cos 2φ ] }
Then, for λ > 0, the elliptical contact area Ac and the maximum contact pressure pmax are :
( 5c) Ac ≅ π ( 3 Fr*/2λE* )2/3 ; pmax = 3F /2Ac
In the case of ductiles, it is reasonable to take the maximum equivalent stress to be 0.6 pmax analogously to spheres and cylinders, since this figure does not appear to be affected significantly by the geometry. The approximation ( 5c) for the contact area can be 7-8% high, so that the predicted equivalent stress may be underestimated by this percentage.
Let the ball be body 1 and the inner race body 2. Then r1 = R1 = 7.5 mm; r2 = -8 mm (since concave) and R2 = 50 mm.
From the above r*= 6.19 mm, φ = 0 and so λ = 0.540. Since E*= 114 GPa for steel-on-steel, from ( 5c) Ac = 2.60 mm2, pmax = 2.88 GPa and σ* = 0.6 pmax = 1.73 GPa.
Repeating for the outer race, r1 = R1 = 7.5 mm; r2 = -8 mm and R2 = -65 mm (since concave). From the above r*= 7.92 mm, φ = 0 and so λ = 0.575; from ( 5c) Ac = 2.94 mm2, pmax = 2.55 GPa and σ* = 1.53 GPa.
For the bearing as a whole, the maximum equivalent stress is 1.73 GPa.
Copyright 1999-2005 Douglas Wright,
doug@mech.uwa.edu.au