Application of Probability to Mechanical Design
65
If
Cvp ~P-q-O.O1 C,,b-- - -±0.01
.~
L
~h
= ~ = ~:0.01 Cvn --- ~0.01
C~ b
h
for 3 standard deviations
Pm~x = 1.03~ Lm~x = 1.03L bm~x = ~.03~ hmax = 1.03~
Pmin = 0.97~
Lmin = 0.97L bmin = 0.97~ hmin = 0.97~
_ 6(1.03~)(1.03L) _ 1.1624 [~L]
6pmaxZmax
ffmax -- bmin(hmin) 2 (0.97~)(0.97h) 2
Lbh2 j
for a card so~t, 4 terms selected from Fig. 2.4 and Table 2.2 the spread
amax--~ and ~ is
6.0737 ~ = am~x = 3
1.1624~ - a
~ =
= 0.02674 6
6.0737
using the partial derivative method Eq. (2.16)
~. = {[/o~ ~ 1/~
0o 6L oa 6~
Op bh ~ OL bh
2
~p = ~o.o1~ ~L = ~O.O1L ~ = ~0.01~ ~h = ~0.01~
substituting and collecting terms
~, = ~L [(0.01)2 + (0.01) 2 + (0.01) ~ + (2 x 0.01)2]
Uz
bh 2
~ = O.02646 6
0.026466 - 0.026746
%error -
x 100 = 1.06% to the high side
0.02646#
for the partial derivative
~a ~ ~[CTp d" Cv2L + Cv2b q- (2C~h)211/2