9 Contact Stresses

426
Alternatively, use the formulas of case 1e of Table 9-2:
CONTACT STRESSES
1 − ν2 1 − 0.32
γ = 2 = 2
E 2 × 1011 = 0.91 × 10−11 m 2 /N
1
1
K =
=
= 0.01532
2/R1 − 1/R2 + 1/R3 2/0.019 − 1/0.02 − 1/0.10
A = 1
1
−
2 R1
1
=
R2
1
1 1
− = 1.316 m
2 0.019 0.020
−1
B = 1
1
+
2 R1
1
=
R3
1
1 1
+ = 31.32 m
2 0.019 0.1
−1
A/B = 0.04202
From Table 9-3,
na = 3.385, nb = 0.4390, nc = 0.6729, nd = 0.6469 (7)
The semimajor axis of the contact ellipse is given by
a = 1.145na(FKγ) 1/3
= 1.145 × 3.385 × 4500(0.91 × 10 −11 1/3 )0.01532
and the semiminor axis is
(6)
= 3.31 × 10 −3 m = 3.31 mm (8)
b = 1.145nb(FKγ) 1/3 = 0.4303 mm (9)
Furthermore, the maximum compressive stress is
σc = 0.365nc
and the rigid approach of the two bodies becomes
F/(K 2 γ 2 1/3 ) = 1508 MPa (10)
d = 0.655nd(F 2 γ 2 /K ) 1/3 = 0.02027 mm (11)
Example 9.3 Wheel–Rail Analyses Consider again the wheel and rail shown in
Fig. 9-5. In Example 9.1, the maximum octahedral shear stress was found to be 256.8
MPa and to be located 0.22 cm below the initial contact point. Suppose now that the
rail steel has a tensile yield strength of 413.8 MPa.
1. Determine whether yielding occurs in the rail according to the maximum octahedral
shear stress yield theory (equivalent to the von Mises–Hencky theory).