Data Reduction
19
Weibull. A 1 l"x 17" plot similar to Fig. 1.2 is made and the/~ through the
data is 6.80 and the smallest /~ is 2.50 with the 50 percentile
of
43,500-45,500 psi. The Gaussian values Eq. (1.1) are plotted as the dotted
curve in Fig. 1.5
# = 46,507 psi
~ = 2158 psi
(1.64)
The Weibull values Eqs. (1.2), (1.14) are plotted as a solid line in Fig.
fl= 1.56090-1-0.4316
0=3766.922-1-739 ~ = 43,108 psi + 387 psi
(1.65)
The variation Eq. (2.13) on/~ is 0.1862641 and ~ (Eq. (2.16)) is 0.4310.
best goodness of fit found is the Weibull curve.
Again using Example 1.1 and following the format of Example 1.3 the
class A K= 3.15, Fig. E.1 and Class B K= 1.82, Fig. E.1 for N, 26, samples.
Then with #=Yc=46,507 psi and s=~=2158 psi using Eq. (1.32)
~A ---- 46,507 -- 3.15(2158) ---- 39,709 psi
(1.66)
46,507 -- 1.82(2158) = 42,579 psi
(1.67)
C
0
U
n
t
8
7
6
5
0
-43500
45000
46500
48000
49500
Tensile
Strength
Figure 1.5 Aluminium casting (A) Problem 1.1 tensile strength.
The solid Weibull line
/~ = 1.5609 + 0.4316
0 = 3766.922±739
# = 46,507 psi
~ = 2158 psi.
43,108 -t- 387 psi