Bearing Stresses (Contact Stresses)
This page provides the chapter on bearing stress (also referred to as contact stress) from the "Stress Analysis Manual," Air Force Flight Dynamics Laboratory, October 1986.
11.1 Introduction to Bearing Stresses
The stresses developed when two elastic bodies are forced together are termed bearing stresses. They are localized on the surface of the material and may be very high due to the small areas in contact. The design information given in this section assumes either static loading or low velocity loading. The application to ball and roller antifriction bearings is not covered.
A brief discussion of bearing stresses in riveted joints is followed by a presentation of theoretically derived equations for the bearing stresses between various shapes in contact. An empirical treatment for the determination of allowable loads is also presented.
It is noted here that the design of ball and roller bearings is a very specialized area; however, their selections for various applications can be made based on data published by the various manufacturers.
11.2 Nomenclature for Bearing Stresses
a  =  1/2 the major diameter of an ellipse 
c  =  1/2 the minor diameter of an ellipse 
D  =  diameter 
E  =  modulus of elasticity 
F_{br}  =  allowable bearing load 
F_{cp}  =  proportional limit in compression 
F_{cy}  =  compressive yield stress 
f_{br}  =  calculated bearing stress 
f_{brc}  =  calculated compressive bearing stress 
f_{brs}  =  calculated shear bearing stress 
f_{brt}  =  calculated tensile bearing stress 
K_{1}, K_{2}, K_{3}  =  coefficients in Table 111 
P  =  axial load 
P_{a}  =  allowable axial load 
R  =  minimum radius of curvature 
R'  =  maximum radius of curvature 
r  =  radius 
r  =  cylindrical coordinate 
t  =  thickness 
t  =  width of rectangular area 
w  =  load per length (lb/in) 
w_{a}  =  allowable load per length (lb/in) 
x, y  =  rectangular coordinates 
δ  =  deflection 
θ  =  cylindrical coordinate 
μ  =  Poisson's ratio 
ϕ  =  angle shown in diagram for general case of two bodies in contact in Table 111 
11.2 Bearing Stresses in Riveted Connections
Figure 111 shows a riveted connection between two plates.
Excessive bearing stresses result in yielding of the plate, the rivet, or both.
The intensity of the bearing stress between the rivet and the hole is not constant but varies from zero at the edges to a maximum value directly in back of the rivet. The difficulty caused by considering a variable stress distribution may be avoided by the common practice of assuming the bearing stress to be uniformly distributed over the projected area of the rivet hole. The bearing stress is thus,
The allowable load is
where F_{br} is the allowable bearing stress.
11.4 Sample Problem  Bearing Stresses in Riveted Connections
Given: The riveted plate in Figure 112
Find: The bearing stress between the rivets and the plate.
Solution: The load per rivet is 20,000/4 = 5,000 lb. From Equation (111),
$$ f_{br} = {P \over Dt} = {5,000 \over (0.5)(0.25)} = 40,000 ~\text{psi} $$We have a number of structural calculators to choose from. Here are just a few:
11.5 Elastic Stresses and Deformation of Various Shapes in Contact
Table 111 treats the elastic stress and deformations produced by pressure between bodies of various forms. The first column of this table gives the form of the bodies and the dimensions that describe them. The second column indicates the shape and size of the surface of contact between the two bodies as well as the combined deformation of the bodies, δ. The maximum compressive, tensile, and shear bending stresses (f_{brc}, f_{brt} and f_{brs}) are given in the third column of Table 111. The maximum compressive and tensile bearing stresses occur at the center of the surface of contact and at the edge of the surface of contact, respectively, and the maximum shear bearing stress occurs in the interiors of the compressed parts. The equations in Table 111 are based on the assumption that the length of the cylinder and the dimensions of the plate are infinite. For a very short cylinder or for a plate having a width of less than five or six times that of the contact area or a thickness of less than five or six times the depth to the point of maximum shear stress, the actual stresses may vary considerably from those given by the equation in Table 111.
Because of the very small area involved in what initially approximates a point or line contact, the stresses obtained from the equations in Table 111 are high even for light loads. However, since the stress is highly localized and triaxial, the stress intensity may be very high (above the yield point) without producing apparent damage. Since this is the case and the formulas in Table 111 hold only in the elastic range, the empirical formulas for allowable loads given in Section 11.7 are most useful for practical design. However, the formulas in Table 111 are useful as a guide to design, especially when empirical formulas are not available for a given case.
Form of Contacting Bodies  Shape of Contact Surface and Combined Deformation  Maximum Bearing Stress  

Sphere on Sphere 
$$ r = 0.721 \sqrt[\uproot{1} \scriptstyle 3]{ P \left({D_1 D_2 \over D_1 + D_2}\right) \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$
$$ \delta = 1.04 \sqrt[\uproot{1} \scriptstyle 3]{ { P^2 (D_1 + D_2) \over D_1 D_2} \left( {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right)^2 } $$

$$ \text{Max}~ f_{brc} = 0.918 \sqrt[\uproot{1} \scriptstyle 3]{ P \left({ D_1 + D_2 \over D_1 D_2 }\right)^2 \over \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right]^2 } $$
If E_{1} = E_{2} = E and μ_{1} = μ_{2} = 0.3,
$$ \text{Max}~ f_{brt} = 0.0820 \sqrt[\uproot{1} \scriptstyle 3]{ P E^2 \left({ D_1 + D_2 \over D_1 D_2 }\right)^2 } $$
and
$$ \text{Max}~ f_{brs} = 0.205 \sqrt[\uproot{1} \scriptstyle 3]{ P E^2 \left({ D_1 + D_2 \over D_1 D_2 }\right)^2 } $$


Sphere in Spherical Socket 
$$ r = 0.721 \sqrt[\uproot{1} \scriptstyle 3]{ P \left({D_1 D_2 \over D_1  D_2}\right) \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$
$$ \delta = 1.04 \sqrt[\uproot{1} \scriptstyle 3]{ { P^2 (D_1  D_2) \over D_1 D_2} \left( {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right)^2 } $$

$$ \text{Max}~ f_{brc} = 0.918 \sqrt[\uproot{1} \scriptstyle 3]{ P \left({ D_1  D_2 \over D_1 D_2 }\right)^2 \over \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right]^2 } $$
If E_{1} = E_{2} = E and μ_{1} = μ_{2} = 0.3,
$$ \text{Max}~ f_{brt} = 0.0820 \sqrt[\uproot{1} \scriptstyle 3]{ P E^2 \left({ D_1  D_2 \over D_1 D_2 }\right)^2 } $$
and
$$ \text{Max}~ f_{brs} = 0.205 \sqrt[\uproot{1} \scriptstyle 3]{ P E^2 \left({ D_1  D_2 \over D_1 D_2 }\right)^2 } $$


Sphere on Flat Plate 
$$ r = 0.721 \sqrt[\uproot{1} \scriptstyle 3]{ P D \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$

$$ \text{Max}~ f_{brc} = 0.918 \sqrt[\uproot{1} \scriptstyle 3]{ P \over D^2 \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right]^2 } $$
If E_{1} = E_{2} = E and μ_{1} = μ_{2} = 0.3,
$$ \text{Max}~ f_{brt} = 0.0820 \sqrt[\uproot{1} \scriptstyle 3]{ P E^2 \over D^2 } $$
and
$$ \text{Max}~ f_{brs} = 0.205 \sqrt[\uproot{1} \scriptstyle 3]{ P E^2 \over D^2 } $$
at r/4 beneath the surface of the plate 

Cylinder on Cylinder 
$$ b = 1.6 \sqrt{ {w D_1 D_2 \over D_1 + D_2} \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$
If E_{1} = E_{2} = E and μ_{1} = μ_{2} = 0.3,
$$ \delta = {2 (1  \mu^2) w \over \pi E} \left( {2 \over 3} + \ln{2 D_1 \over b} + \ln{2 D_2 \over b} \right) $$

$$ \text{Max}~ f_{brc} = 0.798 \sqrt{ {w (D_1 + D_2) \over D_1 D_2} \over \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$


Cylinder in Cylindrical Groove 
$$ b = 1.6 \sqrt{ {w D_1 D_2 \over D_1  D_2} \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$
If E_{1} = E_{2} = E and μ_{1} = μ_{2} = 0.3,
$$ \delta = {2 (1  \mu^2) w \over E} \left( 1  2 ~\ln{b \over 2} \right) $$

$$ \text{Max}~ f_{brc} = 0.798 \sqrt{ {w (D_1  D_2) \over D_1 D_2} \over \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$


Cylinder on Flat Plate 
$$ b = 1.6 \sqrt{ w D \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$
If E_{1} = E_{2} = E and μ_{1} = μ_{2} = 0.3, the total compression of the cylinder between two plates is
$$ \Delta D = 4w \left({1  \mu^2 \over \pi E}\right) \left( {1 \over 3} + \ln{2 D \over b} \right) $$

$$ \text{Max}~ f_{brc} = 0.798 \sqrt{ w \over D \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$
If E_{1} = E_{2} = 30×10^{6} and μ_{1} = μ_{2} = 0.25,
$$ f_{brs} = 0.958 \sqrt{ w \over D } $$
at a depth of 0.393b below the surface of the plate. 

Cylinder on Cylinder 
$$ a = K_1 \sqrt[\uproot{1} \scriptstyle 3]{ P \left({D_1 D_2 \over D_1 + D_2}\right) \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$
$$ b = K_2 c $$
$$ \delta = K_3 ~\sqrt[\uproot{1} \scriptstyle 3]{ { P^2 \over \left( {E_1 \over 1  \mu_1^2} + {E_2 \over 1  \mu_2^2} \right)^2 } { (D_1 + D_2) \over D_1 D_2 } } $$
where K_{1}, K_{2} and K_{3} have values as follows:

$$ \text{Max}~ f_{brc} = {1.5 P \over \pi a b} $$
If E_{1} = E_{2} = 30×10^{6}, μ_{1} = μ_{2} = 0.25, and 1 < D_{1}/D_{2} < 8,
$$ \text{Max}~ f_{brs} = {1175 ~O \over \left({D_1 \over D_2}\right)^{0.271} } ~\sqrt[\uproot{1} \scriptstyle 3]{ 4 P \over D_2^2 } $$


General Case of Two Bodies in Contact At point of contact, minimum and maximum radii of curvature are R_{1} and R'_{1} for Body 1 and R_{2} and R'_{2} for Body 2.
$$ \gamma = { 4 \over {1 \over R_1} + {1 \over R_2} + {1 \over R'_1} + {1 \over R'_2} } $$
$$ K = {8 \over 3} { E_1 E_2 \over E_2 (1  \mu_1^2) + E_1 (1  \mu_2^2) } $$

$$ \delta = K_3 ~\sqrt[\uproot{1} \scriptstyle 3]{ P^2 \over K^2 \gamma } $$ 
$$ \text{Max}~ f_{brc} = {1.5 P \over \pi a b} $$
where
$$ a = K_1 ~\sqrt[\uproot{1} \scriptstyle 3]{ P \gamma \over K } $$
and
$$ b = K_2 ~\sqrt[\uproot{1} \scriptstyle 3]{ P \gamma \over K } $$


K_{1}, K_{2} and K_{3} are given in the following table, where:
$$ \theta = \arccos{1 \over 4} ~\gamma ~\left[ \left({1 \over R_1}  {1 \over R'_1}\right)^2 + \left({1 \over R_2}  {1 \over R'_2}\right)^2 + 2 \left({1 \over R_1}  {1 \over R'_1}\right) \left({1 \over R_2}  {1 \over R'_2}\right) \cos{2 \phi} \right]^{1/2} $$


Knife Edge on Plate 
At any point Q,
$$ f_{brc} = { 2 w \cos{\theta} \over \pi r } $$


Rigid Block on Plate 
At any point Q on surface of contact,
$$ f_{brc} = { w \over \pi \sqrt{ {t^2 \over 4}  x^2 } } $$


Rigid Cylindrical Die on Plate 
$$ \delta = { P (1  \mu^2) \over D E } $$ 
At any point on surface of contact,
$$ f_{brc} = { P \over \pi D \sqrt{ {D^2 \over 4}  r^2 } } $$
\( \text{Max}~ f_{brc} = \infty \) at edge

11.6 Sample Problem  Elastic Stress and Deformation of Cylinder on a Cylinder
Given: The cylinders shown in Figure 113.
Find: The contact surface, total deflection, and maximum compressive stress.
Solution: From Table 111, the surface of contact between two cylinders with their axes parallel is a rectangular strip of width, b, given by
$$ b = 1.6 \sqrt{ {w D_1 D_2 \over D_1 + D_2} \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$In this case,
The combined deflection of the cylinders is given by
$$ \delta = {2 (1  \mu^2) w \over \pi E} \left( {2 \over 3} + \ln{2 D_1 \over b} + \ln{2 D_2 \over b} \right) $$if E_{1} = E_{2} = E and μ_{1} = μ_{2} = 0.3, which is true for the given cylinders. Thus, in this case,
From the third column of Table 111, the maximum compressive bearing stress between two parallel cylinders is
$$ \text{Max}~ f_{brc} = 0.798 \sqrt{ {w (D_1 + D_2) \over D_1 D_2} \over \left[ {1  \mu_1^2 \over E_1} + {1  \mu_2^2 \over E_2} \right] } $$In this case,
Thus, the cylinders will not yield and since Max f_{brc} < F_{cy}, the equations in Table 111 are valid.
We have a number of structural calculators to choose from. Here are just a few:
11.7 Empirical Treatment of Allowable Bearing Loads
Many tests have been made to determine the bearing strength of spheres and cylinders. However, it is difficult to interpret the results due to the lack of any satisfactory criterion for failure. Some permanent deformation is shown to be produced even for very small loads. This deformation increases progressively with increasing load, but there is no sharp break in the loadset curve. Thus, it is necessary to select some arbitrary criterion for the amount of plastic yielding that may be considered to represent failure. The circumstances of use determine the degree of permanent deflection necessary to make a part unfit for service.
The following sections present empirical formulas for the maximum allowable bearing loads for various shapes in contact.
11.7.1 Empirical Formulas for Allowable Bearing Loads of a Cylinder on a Flat Plate
Figure 114 shows a cylinder on a flat plate under a loading of w lb. per linear inch.
Table 112 gives empirical formulas for the allowable load (w_{a}) for various diameters of steel cylinders on flat steel plates. It should be noted that there is little difference between failure under static conditions and that under slow rolling conditions if slipping does not occur. If slipping occurs, tests are necessary to obtain reliable information.
Although the allowable load (w_{a}) is dependent upon length for short cylinders, it is independent of length if the cylinders are longer than 6 inches. The last equation in Table 112 is based upon an elongation of 0.001 in./in. in the bearing plate.
Diameter  Loading Condition  Allowable Load (lb/in) 

D < 25 in.  static 
$$ w_a = \left({ F_{cy}  13000 \over 20000 }\right) 600 D $$

25 < D < 125 in.  static 
$$ w_a = \left({ F_{cy}  13000 \over 20000 }\right) 3000 \sqrt{D} $$

116 < D < 476 in.  slow rolling 
$$ w_a = (18000 + 120 D) \left({ F_{cy}  13000 \over 23000 }\right) $$

11.7.2 Empirical Formula for Allowable Bearing Load of Steel Spheres in Contact
Figure 115 shows two similar spheres in contact.
The crushing load P is given by
$$ P = 1960 (8D)^{1.75} $$The test sphere used to derive this formula was steel of hardness 64 to 66 Rockwell C.