124
Chapter 2
(2.131)
(~e)L = 7e/~ = 6"el1 - 2.576(0.1491)]
~eL = 25,984 psi
(2.171)
Now similar to Fig. 2.42 using ~’ti, e as a failure line in Eq. (2.134).
1
am aa
-
+
(2.172)
N Yl~ ~eL
with
then
N = 1
aa/a m = 2.10
O" a = 2.10 O’,~
substituting
O" m
2.10a m
1----t
141,000 25,984
am = 11,375 psi
aa ~ 2.10(11,375) = 23,888 psi
Now ~s
= IOa + a m]
Ys -- 26,458 psi
minimum from Eq. (2.159)
//= 4.25
minimum from Eq. (2.160)
(2.173)
(2.174)
~9 = 36.085 kpsi
(2.175)
The Weibull parameters Eqs. (2.172), (2.173_), (2.17_4) are used with
(2.165) in a Monte Carlo simulation to find b, bmax, h, hmax and the safety
factor. The next step is the Gaussian formulation for the material
Gaussian
The failure line is the low side of a Gaussian curve in the same fashion
as Example 2.21. On the am axis Eq. (2.138) using Eqs. (2.162) and (2.163).
SL ~- #min -- 2.576 ~max
= 176,684 psi - 2.576(7,494 psi)
(2.176)
Sz~ = 157,379 psi
The other end of the line is on ar axis and is (~3e)~ as per Eqs. (2.139)