Application of Probability to Mechanical Design
(2.140). First using Eq. (2.128) and (2.129).
Suit __ ,ttmi n 176,684 psi
0-re-2
2 -
2
125
(2.177)
8)(085)(088) 17_~__846_6
0-el : (0.
.
~e = 52,864 psi
(2.178)
Now using the mean lines for # = 176,684 and #e = 52,864 psi in Eq. (2.139)
~-- 176,68~ ~O" a
--=2.10
and N=I
O" m
1 = [ am
2.10am’]
176,68~
~ ~J
(2.179)
Now
Now
O" m = 22,034 psi
0-a = 2.10(22,034 psi) = 46,271 psi
~’S = [ 0-2 a + O’2m] 1/2 = 51,250 psi
Eq. (2.128) and Eq. (2.168)
~ae : ~’e[(0-0667) 2 + (0.11) 2 + (0.06) 2 + (0.0455)2])
1/2
-~,~e = ~0.1491 6e
using Eq. (2.140)
1
[0-mL
0-aLl
~ = L SL "~- (O’e)LJ
~L = 157,379 psi Eq. 2.176
(0-e)2 = ~’e -- 2.576(0.1491 ~e)
= 52,86411 - 2.576(0.1491)]
(0-e)L
32,560 psi
0-aL - 2.10 N = 1
0-mL
(2.180)
(2.181)
(2.182)