118
Chapter 2
where
-~t__ +0.01163 -~ = -4-0.01 }L = -4-0.01 ==~b +0.01
Kt
p
L
b
kt=2.10
~b=301bs
L=15" ~=solvefor
This is the Gaussian stress formulation to obtain S from the Weibull failure
line
1 ~m
O’a
_
-[__
N
~c (~e)L
where
N - one for calculations
O" a
Gm
~L - is 41,889 psi lOW side Eq. (1.74)
(?ae)L - is 2,482 psi Eq. (2.131)
substituting
1
41,889 2,482
~m = 2343 psi = ~
r 2 -- 2 11/2
"~a ~ [0" a ~- O’mJ
?~ = 4~O’m = 3314 psi
The parameters for the Weibull to obtain a conservative b.
7~ = 3314 low value for line intercept a~/am = 1
(2.134)
(2.135)
1.1299 Eq. (1.71) skews material property lower
(2.136)
0 -- 4878.47 Eq. (1.72) larger to make b larger
(2.137)
Equations (2.133) and (2.135), (2.137) are used in a Monte Carlo simulation
to obtain a solution for b. This is discussed in Appendix B.
The second solution is that of Gaussian stress versus Gaussian fatigue
strength.
Gaussian. The ar--~rm curve Fig. 2.43 is set up similar to Figs. 2.38 and 2.40
where the ultimate tensile strength/t and ~ are used on the am axis to obtain a