Application of Probability to Mechanical Design
105
now
o- s _~ [(O’~n) 2 --1- (O-’r)2]
1/2
16
~,/~Tm 2+I(Km2Mm)2 3(K, rT r)
2}
~zd 3
evaluating for average values
~@3 [(~/-J[45,000 in lb]) ~ + {(2.025[2 x 4500 in lb])
~
+ 3(1.65115,000 in lb)~}]
1/~
16
if, = ~ (90,802)
2. Card Sort
We have six variables in Eq. (2.113) to generate a maximum value
for a card sort and find ~ from Example 2.14, for 3~ measuring
errors 0.01, Eq. (2.69), and with
~a=O.OlO ~.
yielding
dmind = [1 -- 3(0.01)]~
= 0.970~
(Tm)max = 1.01Tin
(Tr)max -- 1.01Zr
(Mr)max = 1.01Mr
Then from Fig. 2.4, and Table 2.2 with
X ~s = 7.593 ~s = as max -- ~’s
Substituting into Eq. (2.113)
16
O’s max - gd 3 107,425
Evaluating
16
~ = nd ~ (107,425 - 90,801)
7.593
16
~s nd 3 2189.40
--- 16
= 0.02411(2.411%)
~s red3 90,801
So
16
2,189.40
= ~d
3
~, = +0.02411~s
(gtB)max 1.3498[(~B
(KtT)max = 1.3498~,r
(2.114)