106
Chapter 2
Now to substitute into the coupling equation
O- S -- O" s
t - ,/~o ~ ~+ - -6 [six sigma criterion]
(2.115)
2.18
Scale from Fig. 2.38 as =~- 50,000= 78,056 psi. Then from the spread
1.45
6~td
~s --= 1--.~50,000
= 6,713 psi
substituting
78,056 - ~s
6=
[(6,713) 2 -4- (0.02411~s)2]
1/2
so both sides
2
36 = (78’056)2 - 2(78,056)a~ + s
(6,713) 2 + (0.0241 las) 2
Combining
0.979073 a 82 _ 2(78,056) as + [(78,056) 2
- 36(6713) 2] = 0
2
A a
s+ Bcrs+C=0
-B ± x/B 2 - 4Ac
~S ~
2A
[safety device pf high]
2(78,056) -4- 82,846
0"s2 =
= 122,033 psi
2(0.979073)
[Structural member no dynamic effects from loading]
2(78,056)- 82,846
68~ =
= 37,416
2(0.97073)
Now
ffs~ = 37,416 = ~3 (90801)
solve for d
3
d3 = 16 (90,801"]
-~ \37,016,/ 12.3596
d = 2.312
With no sudden loading taken into consideration