new FZ?eX`,
gen zone brick p0 0 0 0 p1 30 0 0 p2 0 1 0 p3 0 0 20 size 30 1 20 "mQcc}8
gen zone brick p0 30 0 0 p1 100 0 0 p2 30 1 0 p3 30 0 20 size 70 1 20 "n`z`{<n
gen zone brick p0 30 0 20 p1 100 0 20 p2 30 1 20 p3 50 0 40 p4 100 1 20 & @$n
$f
p5 50 1 40 p6 100 0 40 p7 100 1 40 size 70 1 20 ratio 1.03 1 1 ;Tp9)UP)
def SSR !cYID \}S,
ait1=0.01 +:-xV
k11=0.0 Ec}%!p_$
k12=3.0 _1qR1<V
ks=(k11+k12)/2 iiMS3ueF
loop while (k12-k11)>ait1 Ao$k[#px
E1=10.0e6 _<FUS'"
poi1=0.3 h=gtuaR4
coh1=16000/ks L;u 5
fri1=(atan((tan(20.8*pi/180))/ks))*180/pi zMu9A|
dila1=0.0 GRbbU#/=G
ten1=0.01e6 ~{tO8
]
grav0=-10 ]6[+tpx
dens1=1914 l^nvwm`f#:
K1=E1/(3*(1-2*poi1)) #jrlNg4(
G1=E1/(2*(1+poi1)) $zp|()_
command k|?[EWIi^
model null H.4ISmXU
model elastic i79$D:PcLa
pro bulk 1e10 she 3e9 dens dens1 wCg7JW#
fix x y z ran z -.1 .1 W/xPVmnV
fix x ran x 99.9 100.1 [B+F}Q^;
fix x ran x -.1 .1 4S~kNp$
fix y un_NBv}
set grav 0 0 grav0 |
U"fhG=g
solve ZxSFElDD]E
ini xdisp 0 ydisp 0 zdisp 0 Se'SDJl=
ini xvel 0 yvel 0 zvel 0 ~M{/cv
model mohr 1:](=%oM&k
pro bulk K1 she G1 dens dens1 coh coh1 & /go[}X5QR[
friction fri1 dil dila1 tens ten1 qe{;EH*
set mech ratio 9.8e-6 g^(gT
solve step 10000 9QXsbd6
endcommand gM>?w{!LBx
if mech_ratio<1.0e-5 f^B'BioW(
k11=ks bUm%#a
k12=k12 \f?
K74
else
^ExuIe
k12=ks |<h}'
k11=k11 ^AaE$G&:
endif >pLJ ,Z
ks=(k11+k12)/2 FEu"b@v
endloop ,o%by5j"^N
fosfile0='_fos3'+'.sav' .,xyE--;d
command B
os`+Y
save fosfile0 CU\gx*=E
endcommand bRy(`
end ;9mRumLG"
SSR 7uL.=th'
pr ks h`\$8oV
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。