Application of Probability to Mechanical Design
99
kd (Section A.4) temperature (R.T.) kd = 1 C,,k~
ko (Section A.3) Reliability ko = Cv k,, = 0
kb (Eq.(2.60)) size and shape kb = 0.85 Cvk~ =
~a 0.0406
ka (Fig. 2.14) Table 2.4 surface condition ka - 0.819
Now ~’e = kakbkc.., km(/e
= (0.819)(0.85)(0.955)(0.75)(0.85)(0.35180,
~e = 11, 867 psi
~ae = ~re C2vce q- E Cv2ki
i=a
-- - 0.0496 = C~,,
(2.103)
= 6-’e[(0.0571) 2 + (0.0496) 2 + (0.06) 2 + (0.0361)
2
+ (0.0222) 2 + (0.02305)2]
1/ 2
-- = ±0.1081
#e
~*e = +0.1081(11,867) = 1283 psi
(2.104)
Note:
Zae’ - 4-0.0571 for material alone
O" e
Now to construct a a~-~rm curve knowing
Zae
Zut
-- = 4-0.1081
-- = 4-0.05
Eq. (2.54)
~e
6~t
Construct the mean values on the line in Fig. 2.36
Gm
Gut
for a factor of safety 1.
The cantilever beam size is found from
~,=~+~
a~ is plotted on the at/am curve solutions obtained per Sect. C.2
5. Coupling Eq. (2.42)
~S ~ ~s
(2.105)