Application of Probability to Mechanical Design
61
The coupling, Eq. (2.42), is used
--3
with
6000
/~ = 156,000psi
8~2 11,700
ZF = 4300 psi z~ -Substituting and squaring both sides, two solutions for ? are found. They are
t=-3 is a structural solution and t= + 3 for a safety device which is
designed to be failed under these conditions.
Structural Member
R=0.999 Pu = O.O01
72 = 0.116" ± 0.00058"
~ = 156,000 ~F = 4,300 psi
~’2 ---- 141,000 psi ~,, = 2,559 psi
156,000
Safety factor
---- 1.106
141,000
The curves are shown in Fig. 2.8 and Fig. 2.9.
Safety Device
R=0.001 Pf=0.999
?l = 0.1055" ± 0.00053"
F = 156,000 psi ~F = 4,300 psi
~l = 171,500 psi ~ = 3,093 psi
156,000
Safety factor - -- - 0.909
171,500
EXAMPLE 2.15. Another application of the card sort may be used
to develop the standard deviation for the stress due to applied loads.
P P
A 7~r
2
Figure 2.8
156 ksi
Safety device t= +3 and R=0.001.
171.5 ksi