122
Chapter 2
where 3=5788 psi Eq. (2.143) and ~ is Eq. (2.133)
~ = 1.06066-/9
3
~ = 1,496
Substituting into Eq.(2.156) and Eq.(2.143)
5788
N1496
N = 3.86
(2.157)
(2.158)
EXAMPLE 2.22. The same tip loaded cantilever
beam used in
Example 2.21 is used but the material is ductile Ti-16V-2.5AI titanium with
the ultimate strength found in Example 1.5. The values for the conservative
sizing of/~ for the beam is
Weibull
Eq.
Eq.
Eq.
Gaussian
Eq.
Eq.
(1.81) low value for/~ --4.25
(2.159)
(1.82) high value for conservative ~ ~ = 36.0853
(2.160)
(1.83) low value for 7 ----- 141.000 kpsi
(2.161)
(1.84) low value for # = 176.684 kpsi
(2.162)
(1.85) larger value for ~ = 7.494 kpsi
(2.163)
The goodness of fit shows these curves close, hence the ~ solutions should be
closer than Example 2.21. Since a ductile material is being used, the stress
will be different than the casting in Example 2.21. In Eq. (2.132) the Kt
is used on the aa stress alone yielding for the titanium
~2 ± ~2 11/2
S ~ [t~ a
q- Um/
I/
0"
~2
/O"
\2-] 1/2
~
max
max
s= KtT) +~-~--)
(2.164)
0"max
s= 2 ~--t + 1]
1/2
Table 2.13 Cross sectional dimensions for Example 2.22 for Weibull and
Gaussian simulation
Solution
/, (in)
bmax (in)
~ (in)
hrnax (in)
N
Weibull
0.3279
0.3363
0.6558
0.6727
2.67
Gaussian
0.3813
0.3911
0.7626
0.7822
3.60
Chapter 2
where 3=5788 psi Eq. (2.143) and ~ is Eq. (2.133)
~ = 1.06066-/9
3
~ = 1,496
Substituting into Eq.(2.156) and Eq.(2.143)
5788
N1496
N = 3.86
(2.157)
(2.158)
EXAMPLE 2.22. The same tip loaded cantilever
beam used in
Example 2.21 is used but the material is ductile Ti-16V-2.5AI titanium with
the ultimate strength found in Example 1.5. The values for the conservative
sizing of/~ for the beam is
Weibull
Eq.
Eq.
Eq.
Gaussian
Eq.
Eq.
(1.81) low value for/~ --4.25
(2.159)
(1.82) high value for conservative ~ ~ = 36.0853
(2.160)
(1.83) low value for 7 ----- 141.000 kpsi
(2.161)
(1.84) low value for # = 176.684 kpsi
(2.162)
(1.85) larger value for ~ = 7.494 kpsi
(2.163)
The goodness of fit shows these curves close, hence the ~ solutions should be
closer than Example 2.21. Since a ductile material is being used, the stress
will be different than the casting in Example 2.21. In Eq. (2.132) the Kt
is used on the aa stress alone yielding for the titanium
~2 ± ~2 11/2
S ~ [t~ a
q- Um/
I/
0"
~2
/O"
\2-] 1/2
~
max
max
s= KtT) +~-~--)
(2.164)
0"max
s= 2 ~--t + 1]
1/2
Table 2.13 Cross sectional dimensions for Example 2.22 for Weibull and
Gaussian simulation
Solution
/, (in)
bmax (in)
~ (in)
hrnax (in)
N
Weibull
0.3279
0.3363
0.6558
0.6727
2.67
Gaussian
0.3813
0.3911
0.7626
0.7822
3.60