new r|l?2 eO~
gen zone brick p0 0 0 0 p1 30 0 0 p2 0 1 0 p3 0 0 20 size 30 1 20 TU6s~
gen zone brick p0 30 0 0 p1 100 0 0 p2 30 1 0 p3 30 0 20 size 70 1 20 >5t!
Xt
gen zone brick p0 30 0 20 p1 100 0 20 p2 30 1 20 p3 50 0 40 p4 100 1 20 & eWFkUjz
p5 50 1 40 p6 100 0 40 p7 100 1 40 size 70 1 20 ratio 1.03 1 1 3@" :&
def SSR AUD)=a>
ait1=0.01 ,P9F*;Dj
k11=0.0 $IQPB_:
k12=3.0 *6yY>LW
ks=(k11+k12)/2
uF<34
loop while (k12-k11)>ait1 [)V~U?
E1=10.0e6 nT?+^Ruc
poi1=0.3 H~yHSm 3
coh1=16000/ks /xUF@%rT
fri1=(atan((tan(20.8*pi/180))/ks))*180/pi Q\4tzb]
dila1=0.0 {}s/p9F4
ten1=0.01e6 }.o.*N
grav0=-10 AE:(:U\
dens1=1914 L;0
NR(b!
K1=E1/(3*(1-2*poi1)) yBy7d!@2
G1=E1/(2*(1+poi1)) tU?BR<q
command {m*lt3$k
model null [;wJM|Z J0
model elastic kTH""h{
pro bulk 1e10 she 3e9 dens dens1 jSpj6:@B
fix x y z ran z -.1 .1 S${%T$>
fix x ran x 99.9 100.1 :fj>JF\[
fix x ran x -.1 .1 xT 06*wQ
fix y ;+DEU0|pe
set grav 0 0 grav0 ^`!+7!
solve (9`dLw5
ini xdisp 0 ydisp 0 zdisp 0 *IOrv)
ini xvel 0 yvel 0 zvel 0 |?V7E\S
model mohr :;_}Gxx
pro bulk K1 she G1 dens dens1 coh coh1 & B& @ pZYl
friction fri1 dil dila1 tens ten1 @RPQ1da
set mech ratio 9.8e-6 AZ(zM.y!#_
solve step 10000 BI%^7\HZ
endcommand {#kCqjWG
if mech_ratio<1.0e-5 QKjn/%l"@
k11=ks 68j1svz9
k12=k12 ,<
g%}P/
else `<g]p-=":
k12=ks :m`D
k11=k11 t*= nI $
endif 2OUx@Vj
ks=(k11+k12)/2 !-)!UQ~|8
endloop U@q5`4-!8
fosfile0='_fos3'+'.sav' {>
,M
command sl-wNIQ
save fosfile0 ]r#b:W\
endcommand $ ,K@xq5
end DY#195H
SSR w4P;Z-Cd
pr ks }Hb0@
b_
上面为我的计算程序,可是结果不太对劲,与不同数据的其他结果一样,烦请哪位高手给我指点指点,看有没有不合适的地方。该程序是用强度折减法计算安全系数的。