Application of Probability to Mechanical Design
129
2 2
1 - t C~s = 0
1
t:
Cvs
Noting Eq. (2.42) and substituting Eq. (2.188)
1
t--- +7.168
4-0.1395
Table 2.3
Pf ~ 10 -12
Also note Eq.(2.188) inverted
3 51,250
t- ~ ----7.168
zs 7151
(2.199)
Now substitute C=0 into Eq. (2.197)
-B4-B
~-2~
(2.200)
for +B
~ = 0
This is a structural member which cannot be sized since member size
approaches infinity.
for - B
-(_~) _ (-~)
2A
4(3)
(2.201)
2(1 2 2
- t Cvs)
3 2 = 2.07573
}2 is a safety device not a structural stress like }1- It would appear that t
greater than Eq. (2.199) does not allow a structural design with the
Gaussian-Gaussian formulation for a structural
member. Note the
Gaussian-Weibull formulation may not be as severely limited as ~ is the
lowest stress in the Weibull formulation.
2. t and Pf in terms of the safety factor N
Now examine Eqs. (2.42) and (2.196)
t2 = (3 - ~)~
(Cyst) 2 q- (Cyst)
2