Application of Probability to Mechanical Design
133
The factor of safety
N = -
(2.213)
with ~ Eq. (2.190)
785.005
~ = /~3
(2.214)
and Eq. (2.207)
~ = 14,569 psi
(2.215)
and substituting into Eq. (2.213)
51,250
-14,569
(2.216)
N = 3.52
Gaussian- Weibull
1
The probability of failure Pf is that of P(smax >) =~ since Ys = 26,458 psi
the lowest value of the material representation
1
Pf = P(smax >)= 109
(2.217)
Now to find S(50 percentile) for the Weibull representation. From Eq.
(2.149) set
x - y _ 0.693147~//~
(2.218)
O
x is the 50 percentile value so substituting Eqs. (2.173)-(2.175)
x - 26,458
-- (0.693147)
1/425
36,085
,~ = 59,562 psi
Substitute ~ into Eq. (2.213) and Eq. (2.214) with b from Eq. (2.210) into
factor of safety.
59,562
-22,763
(2.219)
N = 2.617
These approximate values compare closely with values in Table 2.13 which
is a Monte-Carlo simulation with PT -- 1/10
6.