Page 162 - Marks Calculation for Machine Design

P1: Rakesh
14:16
January 4, 2005
Brown.cls
Brown˙C03
144
STRENGTH OF MACHINES
the distance (b) is given by Eq. (3.29),

2 2
1 − ν 1 − ν
1 2
2F E 1 + E 2
b = (3.29)
π L 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 cylinders contacts a flat surface, then set one of the diameters
to infinity (∞). In addition, if the cylinder and the flat surface are the same material, then
the distance (b) of the contact given in Eq. (3.29) becomes Eq. (3.30).

2
2
2F 2(1 − ν )d 4F (1 − ν )d
b = = (3.30)
π L E π L E
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. (3.31),
2F
p max = (3.31)
πbL
where the distance (b) is found from either Eq. (3.29) or Eq. (3.30).
Without providing the details of the development, the largest values of the stresses within
the two cylinders, which are all principal stresses, occur on three-dimensional stress ele-
ments along the (z) axis where (x = 0) and (y = 0). Using the axes notation from Fig. 3.11:
(x), (y), and (z), instead of the standard notation for principal stresses: (1), (2), and (3), the
three principal stresses, all of which are different, are given by the following equations.
  

1 z 2
  z 
σ x =−p max  2 −  1 + 2 − 2  (3.32)
 z 2  b b 
1 +
b 2


z 2 z
σ y =−p max (2ν)  1 + 2 −  (3.33)
b b
− p max
(3.34)
σ z =
z 2
1 +
b 2
There are three things to notice about Eqs. (3.32), (3.33), and (3.34). First, Poisson’s
ratio (ν) in Eq. (3.33) is for the cylinder of interest, either (ν 1 ) or (ν 2 ). Second, the maxi-
mum pressure (p max ) calculated from Eq. (3.31) is a positive number, so the minus sign in
Eqs. (3.32), (3.33), and (3.34) 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. (3.35) as
σ x − σ z σ y − σ z
τ max = or (3.35)
2 2
where the principal stresses (σ x ) and (σ y ) flip flop as to which is larger as the distance (z)
varies into the cylinder of interest.