Question #234129 on Yade changed:
https://answers.launchpad.net/yade/+question/234129
Status: Answered => Open
lingran is still having a problem:
Hello Jan and other users,
The following is the new code I used to calculated plastic energy dissipation.
The result is quite close to yade's O.energy['plastDissip'] value, but huge
computation time is needed especially when used to a big sample of 200,000
particles.
Could you please have a look? there might be some mistakes inside, and
it would be great if someone could improve it.
E=O.energy['plastDissip']
e=0
energy=open('energy.txt','w')
for k in range (0,100):
a=open('state_1.txt','w')
b=open('state_2.txt','w')
for i in O.interactions:
Ft=i.phys.shearForce
a.write('%d %d %f %f %f\n' %(i.id1,i.id2,Ft[0],Ft[1],Ft[2]))
a.close()
O.step()
for i in O.interactions:
Ftt=i.phys.shearForce
du=i.geom.shearInc
kt=i.phys.ks
Fn=i.phys.normalForce
maxFs=Fn.norm()*i.phys.tangensOfFrictionAngle
b.write('%d %d %f %f %f %f %f %f %f %f\n'
%(i.id1,i.id2,Ftt[0],Ftt[1],Ftt[2],maxFs,du[0],du[1],du[2],kt))
b.close()
x = numpy.loadtxt('state_1.txt')
y = numpy.loadtxt('state_2.txt')
for i in range (0,len(y)):
Ftt=Vector3(y[i][2],y[i][3],y[i][4])
maxFs=y[i][5]
du=Vector3(y[i][6],y[i][7],y[i][8])
kt=y[i][9]
Ft=Vector3(0,0,0)
for j in range (0,len(x)):
if y[i][0]==x[j][0] and y[i][1]==x[j][1]:
Ft=Vector3(x[j][2],x[j][3],x[j][4])
Ftc=Ft-kt*du
if Ftc.norm()> maxFs:
#Ftt=Ftt/Ftt.norm()*maxFs
due=-(Ftt-Ft)/kt
#e+=Ftt.norm()*(du.norm()-due.norm())
e+=Ftt.norm()*(du-due).norm()
energy.write('%f %f %f\n'%(O.time,e,O.energy['plastDissip']-E))
energy.close()
Thanks and regards.
lingran
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