new E) >~0jv
gen zone brick p0 0 0 0 p1 30 0 0 p2 0 1 0 p3 0 0 20 size 30 1 20 W?G4\ubM3<
gen zone brick p0 30 0 0 p1 100 0 0 p2 30 1 0 p3 30 0 20 size 70 1 20 _$By c(.c
gen zone brick p0 30 0 20 p1 100 0 20 p2 30 1 20 p3 50 0 40 p4 100 1 20 & }.7!@!q.
p5 50 1 40 p6 100 0 40 p7 100 1 40 size 70 1 20 ratio 1.03 1 1 (j+C&*u
def SSR -Xkdu?6Eh
ait1=0.01 rB}UFS)
k11=0.0 28-6(oG
k12=3.0 [syuoJ
ks=(k11+k12)/2 tq?lF$mM:
loop while (k12-k11)>ait1 0b=OK0n!%
E1=10.0e6 BSG_),AH
poi1=0.3 ~-Rr[O=E
coh1=16000/ks \0Zm3[
fri1=(atan((tan(20.8*pi/180))/ks))*180/pi V#|#%
8
dila1=0.0 O: sjf?z
ten1=0.01e6 jcN84AaRFI
grav0=-10 KGkzE
dens1=1914 MwL'
H<
K1=E1/(3*(1-2*poi1)) kqSCKY1
G1=E1/(2*(1+poi1)) `pN"T?Pk
command {!xPq%
model null mUzNrkG(G
model elastic &~U8S^os
pro bulk 1e10 she 3e9 dens dens1 7[QU
*1bk
fix x y z ran z -.1 .1 BG"~yyKA
fix x ran x 99.9 100.1 __$IbF5
fix x ran x -.1 .1 <FMW%4
fix y 3~BL!e,
set grav 0 0 grav0 B} gi /
solve }#q9>gx
ini xdisp 0 ydisp 0 zdisp 0 z?h\7
R
ini xvel 0 yvel 0 zvel 0 tuK"}HepB
model mohr J}TS-j0
pro bulk K1 she G1 dens dens1 coh coh1 & =R!=uml(
friction fri1 dil dila1 tens ten1 7CABM
set mech ratio 9.8e-6 +M
(\R?@gr
solve step 10000 )__vPPko i
endcommand /H@k;o
if mech_ratio<1.0e-5 F$ x@]
k11=ks WKqNJN C
k12=k12 X(1nAeQ
else cg<10KT
k12=ks s'ntf
k11=k11 +GgWd=X.Y
endif T.!GEUQ
ks=(k11+k12)/2 Oe^3YOR#j{
endloop }+QgRGQ
fosfile0='_fos3'+'.sav' Vy{=Y(cpF2
command /]T#@>('
save fosfile0 yEtSyb~GK
endcommand Xcicqywe?
end J& +s
SSR }.4`zK&SB
pr ks ()K%Rn
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。