Application of Probability to Mechanical Design
117
The next step is to pick a 7 for the ar axis from the Gaussian represented ae. The following is used
(Ge)L = 7 = ae -- 2.576~e
= 6"e[1 -- 2.576(0.1735)]
(2.131)
(ae)Z~ = y = 2,482 psi
NOW the material failure line ends are defined. Next to define the
stress, "s" along the O-r, O’m slope of 1, then set the material value for S,
similar to Fig. 2.38 for interations using a Monte Carlo simulation can
be performed.
In Example 2.16 stress in the beam was found
6PL
bh 2
When the load P is varied from 0 to 30 lbs and back there are two
components formed on the Or --am curve
0"max -- ffmin
O- a -2
1
O" a = ~ O-max
The same can be said for am
0"ma x + O’mi
n
1
2
- 2 O-max
O" m -Now the stresses have Kt multiplied by each of them. This is because a
casting is sensitive to stress concentration in both aa and om. Now s is along
the line aa/a,,, equal to one. As in Fig. 2.38 s is required.
(2.132)
~2 11/2
S = [O’2a + t~m]
s = K 6PmaxL ,t 2-~-~/2
Now h = 2b so
6KtP,~axL ~/~
s- 2b(2b)
2
s = 1.06066 Kt_PL b 3
(2.133)