120
Chapter 2
Now to obtain ~s
2.576 ~s = 5s - aSL
= 5788 -- 3298 psi
(2.141)
~S = 967 psi standard deviation
(2.142)
and
~ = 5788 psi Gaussian mean
(2.143)
The stress due to the load is Eq. (2.133).
Again a Monte Carlo simulation yields a value for/~ using Eqs (2.133),
(2.142) and (2.143). Pf i s 10 -6. The Mont e Carlo simulation is d isc ussed
in Appendix B. The computer results for iterations starting from the low
side, are
Weibull
/~ = 0.7015 in
(2.144)
bmax =/,[1 ÷ 2.576(0.01)]
(2.145)
bmax = 1.02576/~
bmax = 0.7196
~ = 2/~ = 1.4030
(2.146)
hmax --= 2bmax = 1.43914 in
(2.147)
The factor of safety is
N- S(50%)
(2.148)
S(50%) is derived from Eq. (1.18) where x is S(50%)
pf=0.5=exp
-
The natural log of both sides is
-0.693147
: -(-~)#
The 1//~ root of both sides yields with/~= 1.1299
(~-~) = 0.722978
Chapter 2
Now to obtain ~s
2.576 ~s = 5s - aSL
= 5788 -- 3298 psi
(2.141)
~S = 967 psi standard deviation
(2.142)
and
~ = 5788 psi Gaussian mean
(2.143)
The stress due to the load is Eq. (2.133).
Again a Monte Carlo simulation yields a value for/~ using Eqs (2.133),
(2.142) and (2.143). Pf i s 10 -6. The Mont e Carlo simulation is d isc ussed
in Appendix B. The computer results for iterations starting from the low
side, are
Weibull
/~ = 0.7015 in
(2.144)
bmax =/,[1 ÷ 2.576(0.01)]
(2.145)
bmax = 1.02576/~
bmax = 0.7196
~ = 2/~ = 1.4030
(2.146)
hmax --= 2bmax = 1.43914 in
(2.147)
The factor of safety is
N- S(50%)
(2.148)
S(50%) is derived from Eq. (1.18) where x is S(50%)
pf=0.5=exp
-
The natural log of both sides is
-0.693147
: -(-~)#
The 1//~ root of both sides yields with/~= 1.1299
(~-~) = 0.722978