Application of Probability to Mechanical Design
63
Again as in the resistor Example 2.10 two variables are selected to obtain
0-max and 0-min and each is separated from the respective mean by 4.056 standard deviations, 4.056 ~
2(4.056 ~,~) = 0-max - °’min
/’2057.05 1770.41.’~
~1
~ = \ ~5
72 ,1 2(4.056)
35.34
The previous calculated value, Example 2.14
~2 11,700
Za -- ~2~4
34.43
~ - ?2
The percent error is the difference of ~ in Example 2.14 and Example 2.15
divided by ~ in Example 2.14
% error = (34.43 - 35.34) 100
34.43
?2__
~2
% error = 2.64% on the high side.
35.34
Now compare ~ = ?~ solution in Eq. (2.42) for Example 2.14
t=-3=
[~2 F q- ~2a]1/2
Noting
~ 35.34
1
6 - 72 6000 = +0.0185
Substituting
-3 = [156,000 - 0-][(4300) 2 + (0.0185a)21
-l /2
Squaring and transposing
9[(4300) 2 + 3.4240 x 10-4o "2] = (156,000) 2 - 2(156,000)0- + 0-2
[1 - 3.08158 x 10-310 -2 - 2(156,000)0- - 9(4300) 2 + (156,000) 2 = 0
A0- 2 + Ba + C = 0