Page 158 - Marks Calculation for Machine Design

P1: Rakesh
January 4, 2005
14:16
Brown.cls
Brown˙C03
140
If the two spheres are made of two different materials, then the radius (a) of the area of
contact is given by Eq. (3.19), STRENGTH OF MACHINES

1 − ν 2 1 − ν 2
1 2
3F +

E 1 E 2

a = 3 (3.19)
8 1 + 1
d 1 d 2
where (ν) is Poisson’s ratio and (E) is the modulus of elasticity.
As stated earlier, if one of the spheres contacts a flat surface, then set one of the diameters
to infinity (∞). In addition, if the sphere and the flat surface are the same material, then the
radius (a) of the contact given in Eq. (3.19) becomes Eq. (3.20).

2
2
3F 2(1 − ν )d 3F(1 − ν )d
3 3
a = = (3.20)
8 E 4E
The pressure distribution over the area of contact is elliptical with the maximum pressure
(p max ), which is a negative stress, occurring at the center of the contact area and given by
Eq. (3.21),
3F
p max = 2 (3.21)
2πa
where the radius (a) is found either from Eq. (3.19) or Eq. (3.20).
Without providing the details of the development, the largest values of the stresses within
the two spheres, which are all principal stresses, occur on three-dimensional stress elements
along the (z) axis where (x = 0) and (y = 0). Using the axes notation from Fig. 3.9: (x),
(y), and (z), instead of the standard notation for principal stresses: (1), (2), and (3), the
three principal stresses, two of which are equal, are given by the following equations.
  

z 1 1
 −1  
σ x = σ y =−p max (1 + ν) 1 − tan z −  (3.22)
a z 2 
a 2 1 +
a 2
−p max
σ z = (3.23)
z 2
1 +
a 2
There are three things to notice about Eqs. (3.22) and (3.23). First, Poisson’s ratio (ν) in
Eq. (3.22) is for the sphere of interest, either (ν 1 ) or (ν 2 ). Second, the maximum pressure
(p max ) calculated from Eq. (3.21) is a positive number, so the minus sign in Eqs. (3.22) and
(3.23) makes all three principal stresses negative, or compressive. Third, as the principal
stress (σ z ) has the largest magnitude, but negative, and as the three principal stresses form
a triaxial stress element, the maximum shear stress (τ max ) is given by Eq. (3.24) as
σ x − σ z σ y − σ z
τ max = = (3.24)
2 2
To determine the maximum shear stress (τ max ) at the plane of contact between the two
spheres, subsitute z = 0 in Eqs. (3.22) and (3.23) to give
  

0 −1 1 1
σ x = σ y =−p max (1 + ν) 1 − tan −  

a 0 0 2
a 2 1 +
a 2
(3.25)
!
1 1
=−p max (1 + ν)(1) − =−p max + ν
2(1) 2
−p max
σ z = =−p max (3.26)
0 2
1 +
a 2