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[lingran <question234129@xxxxxxxxxxxxxxxxxxxxx>]

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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