98
Chapter 2
Solutions may be solved for only one unknown, area or a cross sectional dimension. The factor of safety can be calculated after solving for
the unknown. Confidence levels can be found if the size of the input data
is known for ti’e, tiyt, and ti,tt.
The one piece of information needed for the tir--tim plot is the standard
deviation ~e of the endurance value ae from Eq. (2.55)
tie = kakbkc . . . kmtile
This is a calculation similar to Example 2.16 which is a well behaved product
~
[
2 "11/2
Zae : tie C~, c + Z Cvk, J
(2.101)
EXAMPLE 2.18. Use the cantilever
beam (Example 2.16) of
copper-based alloy
tie-’ = 0.35s,tt use s,~lt
: 80, 000 psi Fig. 2.28
tiu/t/5o _ ±0.0571
(2.102)
Cvo; - 0.35ti,~t
~a; = 0-0571~Ie = Cva’~r’e
km Eq.(2.89)10 ~° cycles
~m 0.0196
- -- -- 4-0.02305 = Cvkm
k,. 0.85
~,~ = 0.02305 km = Cvkm[Cm
k~ (Section A. 12) radiation k~ = Cv k~ = 0 .’ . ~t = 0
kk (Section A.11) dynamics (in stress calculation)/ok= 1, ~k, Cv k~ =0
0.01670.7~ -±0.0222---- C~k~
kj (Section A.10) fretting ~:j = 0.75,
~j
ki (Section A.9) surface treatment none k~ = 1 Cvk~ =
~h 0.035
kh Eq. (2.77) environment kh ---- 0.955,
-- -- 4-0.0367 ---- Cvkh
kh 0.955
kg (Section A.7) internal structure kg = 1 C% = 0
kf (Section A.8) residual stress, none kf = 1
C~k~ = 0
ke (Section A.5) stress concentration ke = 1 C~, e = 0
applied to stress