Application of Probability to Mechanical Design
127
Now to develop Cvs using a card sort for ~ Eq. (2.165)
~ = 0.75 ~[/f~ 2 + 1]
1/ 2
(2.189)
Kt will be used as [2.10].
~ = 0.75 30 lb__(15 in)[(2.1) 2 + 1]l/2 _ 785.005
b 3
D3
(2.190)
Now select maximum values for P, L, Kt and minimum values for b
Lmax = L[1 + 2.576(0.01)] = 15.3864
bmin = t~[1 - 0.02576] = 0.97424 D
Pmax = ~h[1.02576] = 30.7728 lbs
K~max = L[1.02576] = 2.1541
Note four cards or parameters are selected _2= 6.0737 Table 2.2 and Eq.
(2.25)
_~ ~s = Smax - ~
(2.191)
= 0 75 (30.7728)(15.3864)
Smax ¯
~ ttz.~541)
2 + 1]
1/ 2
912.037tu.y~’*z’~u)
(2.192)
Smax ~
Substituting Eq. (2.190) and (2.192) into Eq. (2.191)
912.037 785.005
6.0737 5s -- D3
/~3
(2.193)
20.9152
~s - ~3
Now
20.9152 D
3
(2.194)
Cvs = t~ 3 785.00~ - 4-0.0266433
Now using the coupling Eq. (2.42)
t -- /~s - #s
(2.195)