Application of Probability to Mechanical Design
59
Table 2.3 Values of minus t and P(t) for Eqs.
(2.42) and (2.43) with P(t) = 10
-D
-t
D
-t
D
zero
infinity
7.3488
13
1.2816
1
7.6506
14
2.3263
2
7.9413
15
3.0912
3
8.2221
16
3.7190
4
8.4938
17
4.2649
5
8.7573
18
4.7534
6
9.0133
19
5.1993
7
9.2623
20
5.6120
8
9.5050
21
5.9478
9
9.7418
22
6.3613
10
9.9730
23
6.7060
11
10.1992
24
7.0345
12
10.4205
25
EXAMPLE 2.13. A material part has a yield coefficient of variation
CA = 4-0.07 and a yield strength mean/~n of 35,000 psi with an applied
mean stress of 20,000 psi, #a, and a coefficient of variation of C, = + 0.10.
Find t for Eq. (2.42) and the reliability and failure.
11 A -- 12
a
t=-[(~A) 2 --1- (~a)2]
1/2
~ = CA#~ = 4-0.07(35,000 psi) ~a = Ca#a = 4-0.10(20,000 psi)
~A = 4-2450 psi
~, = 4-2000 psi
35,000 -- 20,000
t
[(2450)2 + (2000)2]t/2 --4.7428
from Table 2.3
1
t = -4.7534 is P(t) ,-~ ma king R(t)
A value 0.999999. Also note the factor of safety is
F.S.
#~ _ 35 1.75
#. 20
Now both P(t) and factor of safety defines the parts safety.
EXAMPLE 2.14. A simple example to give a feel for what can be
done with these concepts [2.19]. A tension sample Fig. 2.7 has the following