126
Chapter 2
Substituting
O’mL ~
1= 157,379
~J
~rmL = 14,114 psi
(2.183)
~r~L = 2.10(14,114 psi) = 29,640
,,2 11/2 32,829 psi
~s~ = [~.,~ + ~a~ =
Now to obtain ~s Eq. (2.141) using Eqs. (2.180) and (2.183)
2.576~s = 6s - ~sa
= 51,250 - 32,829
(2.184)
~s = 7,151 psi
Now
5s 7151 psi
Cvs - ~ -
- 4-0.1395
(2.185)
zs 51,250 psi
51,250 psi
(2.186)
~s -- 7151 psi
(2.187)
Equations (2.186) and (2.187_) are used with Eq. (2.165) in a Monte
simulation to find b, _bmax, h, hmax, and safety factor from Weibull Eqs.
(_2.144)-(2.151) with b=0.3279 and Gaussian Eqs. (2.152)-(2.158)
b=0.3813 in Table 2.13. All iterations started from the low side.
G. Bounds on Monte Carlo Fatigue Calculations
1. The minimum Pf for a structural member stress S 1.
There are some features of a Monte Carlo or probability solution which
should be discussed. They are, how small can the Pf be, and find a factor
of safety, N. How to select t and hence the Pf to produce N. First the
Cvs and Cvs for material and the stress due to loading should be developed.
The value for Cvs is from Eqs. (2.186) and (2.187)
~s
C~s = ~
7151
(2.188)
C~s - 51,250 4-0.1395