为什么多荷载步求解后，提取弯矩剪力等结果为0呢？

jlm8521

在进行火车过桥分析是，加荷载步，可最后结果提取全为0？为什么？请哪位高手帮忙指点一下！
FINISH
/CLEAR
cj=1.27
/PREP7
!单元类型编号
ET,1,BEAM44
MP,EX,1,2.1E11
MP,PRXY,1,0.3
MP,DENS,1,10690
MP,ALPX,1,1E-5
sectype,1,beam,I
secdata,0.46,0.46,0.46,0.02,0.02,0.012
*DO,I,1,897,1
N,I,0.5*(I-1),0,0
*ENDDO
TYPE,1
MAT,1
SECNUM,1
*DO,I,1,896,1
E,I,i+1
*ENDDO

NSEL,S,,,1,1
D,ALL,,0,,,,,UY,UZ,,ROTY,
NSEL,S,,,253,253
D,ALL,ALL
NSEL,S,,,665,665
D,ALL,,0,,,,,UY,UZ,,ROTY,
NSEL,S,,,897,897
D,ALL,,0,,,,,UY,UZ,,ROTY,

/solu
antype,static
i=0
*do,i,0,896,1  
fdele,all,all
sfedele,all,all,pres
nsel,all
NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
NSEL,R,LOC,X,0.5*i
f,all,fy,-cj*110000
nsel,all
NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
NSEL,R,LOC,X,0.5*(i-3)
f,all,fy,-cj*110000
nsel,all
NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
NSEL,R,LOC,X,0.5*(i-6)
f,all,fy,-cj*110000

nsel,all
NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
NSEL,R,LOC,X,0.5*(i-9)
f,all,fy,-cj*110000

nsel,all
NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
NSEL,R,LOC,X,0.5*(i-12)
f,all,fy,-cj*110000
nsel,all
*if,i,le,14,then
*elseif,i,le,75
NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
nsel,r,loc,x,0,(i-15)*0.5,1
esln
esel,u,sec,,1
sfbeam,all,1,pres,cj*46000
*elseif,i,lt,129
NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
nsel,r,loc,x,(i-75)*0.5,(i-15)*0.5,1
esln
esel,u,sec,,1
sfbeam,all,1,pres,cj*46000
nsel,all

NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
nsel,r,loc,x,0,(i-77)*0.5,1
esln
esel,u,sec,,1
sfbeam,all,1,pres,cj*40000
nsel,all
*else

NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
nsel,r,loc,x,(i-75)*0.5,(i-15)*0.5,1
esln
esel,u,sec,,1
sfbeam,all,1,pres,cj*46000

nsel,all
NSEL,S,LOC,Z,0
NSEL,R,LOC,Y,0.61
nsel,r,loc,x,(i-128)*0.5,(i-77)*0.5,1
esln
esel,u,sec,,1
sfbeam,all,1,pres,cj*40000
*endif
allsel
outres,all,all
solve
*enddo
finish

/prep7
*creat,outt
*cfopen,max,txt
*vwrite,t0(1)
(f30.12)
*cfopen,min,txt
*vwrite,t2(1)
(f30.12)
*cfopen,maxx,txt
*vwrite,t3(1)
(f30.12)
*cfopen,minn,txt
*vwrite,t4(1)
(f30.12)
*cfclose
*end
/post1
*dim,t0,array,10 !定义t0数组存储第i个单元所有荷载步中弯矩最大的反应值
*dim,t1,array,897 !定义t1数组存储每个单元的I端弯距值
*dim,t5,array,897 !定义t1数组存储每个单元的I端剪力值
*dim,t2,array,10 !定义t2数组存储第i个单元所有荷载步中弯矩最小的反应值
*dim,t3,array,10 !定义t3数组存储第i个单元所有荷载步中剪力最大的反应值
*dim,t4,array,10!定义t4数组存储第i个单元所有荷载步中剪力最小的反应值
*do,i,1,10,1
*do,j,1,897,1
set,j
*GET,tt1,ELEM,i,smisc,5
*GET,tt5,ELEM,i,smisc,13
t1(j)=tt1
t5(j)=tt5
*enddo
*VSCFUN,t0(i),Max,t1 !t0(i)为t1数组的最大值
*VSCFUN,t2(i),Min,t1 !t0(i)为t1数组的最小值
*VSCFUN,t3(i),Max,t5 !t0(i)为t1数组的最大值
*VSCFUN,t4(i),Min,t5!t0(i)为t1数组的最小值
*enddo
*use,outt