| Author | Message | |||
     gqzhang (zgqfem) Member Username: zgqfem Post Number: 10 Registered: 07-2003 |
Dr, Nelson, I find you are really an expert on flexpde, thank you for your replies to me. And I am testing the Biot' theorem which is well-known in soil mechanics. For convience,if you are not in this realm,I would like to briefly describe it following: Soil is a porous material, which comprises 'skeleton' and voids, and voids are filled with water. The skeleton deforms elastically accordingly by the 'effective stress:the total stress minors the pore water pressure'. So this is acturally a two-field problem: elastic defomation will cause change in void, subsequently change the water pressure, and if the water pressure changes, the effective stress acting on the skeleton will change. I hope me explanation is roughly clear and helpful for your understanding my attached pde file. My testing is on a very simple rectangular domain: there is no force and pressure is fixed on the left side; the u and v are fixed along the bottom side, and there is no fluid flow across it; u is fixed along the right side, but free of force in y;the upper side is divivded in two segments(on the upper right the pressure is fixed,the force in u is zero, but the force in y is pointed;on the upper left the forces in x and y are zero,with the pressure is fixed) I verified the stress and displacements by ABAQUS, and the results are in good accuracy.The problem is Time takes no effect on the solution. I used some other software to solve the same problem, the pressure changes much with time. Can you give me some advice? thank you again. title "pressure-Elastic Stress" select errlim = 1.0e-2 nodelimit=800 ngrid=16 variables Pp(1.) Up Vp definitions E=4e6 { declare Young's Modulus - values come later } nu=0.25 { declare Poisson's Ratio - values come later } egma=1e3 {fluid density} mu=0.3e3 {viscocity coefficient, assumed to be 0.3} kx = 8.64e-4/mu/egma {declare the permeabilities in x and y directions} ky= 8.64e-4/mu/egma { define the constitutive relations } G = E/((1.+nu)*(1.-2.*nu)) C11 = G*(1-nu) C12 = G*nu C22 = G*(1-nu) C33 = G*(1-2*nu)/2 b=1. { define some utility functions } ex = dx(Up) ey = dy(Vp) gxy = dy(Up) + dx(Vp) Sx = C11*ex + C12*ey-Pp Sy = C12*ex + C22*ey-Pp Txy = C33*gxy initial values Pp = 0.0 { give FlexPDE an estimate of variable range } Up = 1.e-5 Vp = 1.e-5 equations dx(kx*dx(Pp)) + dy(ky*dy(Pp)) =-dt(ex+ey) {pressure equation} dx[Sx] + dy[txy] = 0. { the U-displacement equation } dx[txy] + dy[sy] = 0. { the V-displacement equation } boundaries region 1 { region one defines the problem domain as all copper and sets the boundary conditions for the problem } start(0,0) {top left} natural(Up)=0 {zero surface force in x direction} natural(Vp)=0 {zero surface force in y direction} value(Pp)=0 {pressure=0} line to (33,0) {top right} natural(Up)=0 natural(Vp)=-1000 {surface force in y direction=-1000} value(Pp)=0 line to (36,0) { right} value(Up)=0. natural(Vp)=0 natural(Pp)=0 {no fluid flow across this border} line to(36,-12) {bottom} value(Up)=0 value(Vp)=0. natural(Pp)=0. {no fluid flow across this border} line to(0,-12) {left} natural(Up)=0 natural(Vp)=0 value(Pp)=0 line to finish time from 0 to 2. by 0.02 plots for t=0 by 0.02 to 2. contour(Pp) as "pressure" contour(Up) as "X-displacement" contour(Vp) as "Y-displacement" contour(Sx) as "X-Stress" contour(Sy) as "Y-Stress" contour(Txy) as "Shear Stress" vector(Up,Vp) as "displacement" end | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 33 Registered: 06-2003 |
There are two difficulties here: 1) dt(ex+ey) = dt(dx(up))+dt(dy(vp)) The current version of FlexPDE does not handle cross-derivatives of time and space correctly. You should introduce an intermediate variable, such as esum = dx(up)+dy(vp), so that FlexPDE will see a clean time derivative of a variable. 2) Your displacement equations are not time dependent, so dt(ex+ey) is undefined. You can't use it as a source for the pressure, because it has no meaning. | |||
| gqzhang (zgqfem) Member Username: zgqfem Post Number: 11 Registered: 07-2003 |
I understood your first answer. For the second, as in my reply to your e-mail, some one treat this problem as the ramping load. But acturally, the displacement equation is time-dependent, because it contains Pp in Sx and Sy. When Pp changes, Sx and Sy will change at the same time. This group of equations are typically real-time two-field coupling ones. And there are many publications for the solution, as well as experimental verifications, but most of them used finite difference method. The physical process,also the experiment carried on,is: there is a rectangle area, the left and right sides are constrained in x directions, the bottom is constrained in y direction. The upper is fixed at zero pressure, and a force applied; the other three sides are impervious. The initial values are all zero. At time=zero , the force is applied suddenly, there must be deformation and pressure due to compaction of voids. the force does not change with time. And the fluid will excelled from the upper side gradually, thus pressure will decrease,then Sx and Sy will change for the reason of equalibrium at each moment. With Sx and Sy changing, defomation will develop further, causing the pressure equation changed. Hope this explanation clear enough. I just want to test it to explore the capacity of flexpde, for possible application to much complex situations-three field coupling. Thanks very much. I am upset for disturbing you so much. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 35 Registered: 06-2003 |
Frankly, I don't think this is a very meaningful test problem. Certainly, it isn't a test of FlexPDE's ability to solve coupled fields. When you present a "time dependent" problem, FlexPDE assumes you are trying to solve an initial value problem of the form dt(U,V,W)=F(U,V,W). That is, it assumes you have a differential equation in time that you want to integrate. In order to do this, FlexPDE approximates each variable as a polynomial function of time, and solves implicitly in each timestep for the values that produce the required time derivatives. It automatically scales the timestep to preserve an accurate polynomial representation of the solution functions in time. Some of the U,V,W variables can be steady-state equations, but at least one of the equations must be a differential equation in time. Your problem is not an initial value problem. It is a time series of steady-state problems, loosely tied together by a "time derivative" of strain. By removing all the time derivative terms in the equations describing the physics, you have effectively said that the system equilibrates instantaneously to any change in applied force. An instantaneously applied force implies an infinite time derivative, in order to equilibrate instantaneously. Applying instead a ramped application of force is not really a cure, because the resulting pressure is dictated solely by the arbitrarily chosen ramp time. So you can get any pressure you want by choosing an appropriate ramp. But more importantly for FlexPDE, there is no dt(U,V,W) which it must match on each timestep. This results in a singular coupling matrix and failure to converge. If you put some of the physics back into your equations, I think you will have more luck. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 37 Registered: 06-2003 |
I notice in the slides you sent that on page 7 Biot's Consolidation Model includes a term dt(P/M) which does not appear in your descriptor. This term should satisfy the needs of FlexPDE to have a true initial-value problem. Where did it go? | |||
| gqzhang (zgqfem) Member Username: zgqfem Post Number: 12 Registered: 07-2003 |
Ok,Thanks for your reply. Let me just put my question in the following way: Because you have read the slides I sent you, with the physics, the equations, and the results;and also I want to use FLEXPDE verify the resutls. Can I realize my goal with FLEXPDE of the current version or the future version? Thanks so much. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 39 Registered: 06-2003 |
Put the dt(P/M) back into the equations and try again. | |||
| gqzhang (zgqfem) Member Username: zgqfem Post Number: 13 Registered: 07-2003 |
Dr, I Put the dt(P/M) back into the equations and tried again with the following code, but as stated before stress takes no effect on pressure. And also you already have the slides describing the related theories. Would you please help me with this section of code? Also, the values of parameter appear in the code are in the practical use, so do make some sense. THANKS SO MUCH title "pressure-Elastic Stress" select errlim = 1.0e-2 nodelimit=800 ngrid=16 variables Pp Up Vp definitions E=4e7 { declare Young's Modulus } nu=0.25 { declare Poisson's Ratio } egma=1e3 {fluid density} mu=0.3e-9 {viscocity coefficient, assumed to be 0.3} kx =8.64e-18/mu {declare the permeabilities in x and y directions} ky= 8.64e-18/mu { define the constitutive relations } G = E/((1.+nu)*(1.-2.*nu)) C11 = G*(1-nu) C12 = G*nu C22 = G*(1-nu) C33 = G*(1-2*nu)/2 b=1. { define some utility functions } ex = dx(Up) ey = dy(Vp) gxy = dy(Up) + dx(Vp) Sx = C11*ex + C12*ey +Pp Sy = C12*ex + C22*ey +Pp Txy = C33*gxy initial values Pp = 0.0 { give FlexPDE an estimate of variable range } Up = 1.e-5 Vp = 1.e-5 equations dx(kx*dx(Pp)) + dy(ky*dy(Pp)) =0.07*1.5e-3*dt(Pp)+b*dt(dx(up)+dy(vp)) {pressure equation} dx[Sx] + dy[txy] = 0. { the U-displacement equation } dx[txy] + dy[sy] = 0. { the V-displacement equation } boundaries region 1 start(0,0) {top left} natural(Up)=0 natural(Vp)=10 natural(Pp)=0 line to (18,0) {top right} natural(Up)=0 natural(Vp)=0 natural(Pp)=0 line to (36,0) { right} value(Up)=0. natural(Vp)=0 natural(Pp)=0 line to(36,-12) {bottom} natural(Up)=0 value(Vp)=0. natural(Pp)=0. line to(0,-12) {left} value(Up)=0 natural(Vp)=0 natural(Pp)=0 line to finish time from 0 to 1e5 by 1e3 plots for t=0 by 1e3 to 1e5 contour(Pp) as "pressure" contour(Up) as "X-displacement" contour(Vp) as "Y-displacement" contour(Sx) as "X-Stress" contour(Sy) as "Y-Stress" contour(Txy) as "Shear Stress" vector(Up,Vp) as "displacement" end | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 48 Registered: 06-2003 |
You still have not addressed the issues I raised in previous postings: 1) FlexPDE does not handle cross-derivatives like dt(dx(u)) correctly. 2) FlexPDE thinks the time derivative of a steady-state quantity is meaningless. 3) The resultant "pressure" in your attempted model is purely a consequence of your chosen ramp time and has no physical meaning. The attached file is a modification of your earlier posting that I made at that time. I have introduced a new variable esum which is the sum of strains, and given it a relaxation equation. I have also make the displacement equations time dependent. It runs, but I have no idea whether the results are meaningful.
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| gqzhang (zgqfem) Member Username: zgqfem Post Number: 14 Registered: 07-2003 |
Thanks for your reply. 1 Because the displacement equations are balanced at any time, i.e. equal to zero, time dependent modification is not correct. 2 I have no idea about relaxation, can you explain some background about it? 3 Based on your issues you raised, you mean the mathematical model contained in the slide is not correct, right? With the best regards. |
soil1.pde