9 Contact Stresses

428
From Eqs. (9.5) and (9.6),
A + B = 1
1
+
4 R1
1
+
R2
1
1
+
4 R1
1
=
R2
1
1
2
A ∗ + B ∗ = 1
1
2 R∗ +
1
1
R∗
=
2
1
1
+
2λ R1
1
R2
and from (2),
Then
R1
τ 3 oct = F(cGcb) 3 / 2 , τ ∗3
oct = F(cGcb) 3 / ∗2
τ 3 oct
τ ∗3
=
oct
∗2
2 = [γ/(A∗ + B∗ )] 2
[γ/(A + B)]
Thus,
Since
it follows that
To check, we find
τ 3 ∗3
oct /τoct = λ2
2 3 τ 3 oct /τ 3 oct
CONTACT STRESSES
+ 1
R2
(A + B)2
=
2 (A∗ + B∗ 1
4
=
) 2 (1/R1 + 1/R2) 2
(1/4λ2 ) (1/R1 + 1/R2) 2
or τ ∗ oct
τ ∗ oct = 1 2 τoct
= λ2
= τoct/λ 2/3
(5)
(6)
or λ = √ 8 (7)
R ∗ 1 = √ 8(45) = 127.28 cm, R ∗ 2 = √ 8(35) = 98.995 cm
(A + B) ∗ = 1
1
2 1.2728 +
1
= 0.8979 m
0.98995
−1
∗ = 2 1 − (.29) 2
(2 × 10 11 )(0.8979) = 1.020 × 10−11 m 3 /N
τ ∗3 = (20 367)(0.2)3 (0.84) 3
(1.02 × 10 −11 ) 2
τ ∗ oct = 9.755 × 107 Pa or 97.55 MPa
= 9.282 × 1023 (N/m 2 ) 3
Since the yield value of maximum octahedral shear is 195.07 MPa and 97.55 is
one-half of the yield value, increasing R1 and R2 by a factor of √ 8 decreases the
maximum octahedral shear stress by one-half.