new .\ K0+b;
gen zone brick p0 0 0 0 p1 30 0 0 p2 0 1 0 p3 0 0 20 size 30 1 20 4+qoq$F</
gen zone brick p0 30 0 0 p1 100 0 0 p2 30 1 0 p3 30 0 20 size 70 1 20 bBA
#o\[
gen zone brick p0 30 0 20 p1 100 0 20 p2 30 1 20 p3 50 0 40 p4 100 1 20 & |giV<Sj
p5 50 1 40 p6 100 0 40 p7 100 1 40 size 70 1 20 ratio 1.03 1 1 FGY4 u4y
def SSR W9nmTz\8
ait1=0.01 9aky+
k11=0.0 2x%Xx3!
k12=3.0 [+<lm
5t
ks=(k11+k12)/2 D=uU:7m
loop while (k12-k11)>ait1 [(Ss^?AJW
E1=10.0e6 VX0q!Q
poi1=0.3 T<1*R>el
coh1=16000/ks ?*lpu
fri1=(atan((tan(20.8*pi/180))/ks))*180/pi wPdp!h7B~N
dila1=0.0 e=S51q_0
ten1=0.01e6 I/:M~ b
grav0=-10 :!H]gC
4
dens1=1914 5xKo(XNp
K1=E1/(3*(1-2*poi1)) 3m:[o`L
G1=E1/(2*(1+poi1)) w-9M{Es+j
command %2>ya>/M
model null r!A1Sfo4P
model elastic jI:5[. Y
pro bulk 1e10 she 3e9 dens dens1 P/uk]5H^
fix x y z ran z -.1 .1 L6S!?t.{Yv
fix x ran x 99.9 100.1 OIPJN8V
fix x ran x -.1 .1 vDl6TKXcu
fix y >Z@^R7_W
set grav 0 0 grav0 f'._{"
solve *\ZK(/V
ini xdisp 0 ydisp 0 zdisp 0 w ryjs!
ini xvel 0 yvel 0 zvel 0 xV@/z5Tq
model mohr WfYu-TK*
pro bulk K1 she G1 dens dens1 coh coh1 & R3=PV{`M
friction fri1 dil dila1 tens ten1 *F7ksLH|q
set mech ratio 9.8e-6 's#"~<L^e
solve step 10000 AG/?LPJ
endcommand y^pzqv
if mech_ratio<1.0e-5 yzJ
VU0s
k11=ks y
qDE|DIez
k12=k12 \1x<bx/1
else #Duz|F+%
k12=ks M_asf7|v
k11=k11 hZ6CiEJB
endif iv@ey-,<
ks=(k11+k12)/2 ,?s3%<\2
endloop ?[{_*qh
fosfile0='_fos3'+'.sav' E{+V_.tlu
command vZ3/t8$*
save fosfile0 Q v=F'
endcommand T v2d?y
end zQvp<IUq
SSR &cy@Be}|T
pr ks CJ0{>?
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。