| Author | Message | ||
     gqzhang (zgqfem) Junior Member Username: zgqfem Post Number: 3 Registered: 07-2003 |
Is flexpde capable of coping with one dimensional problems? I always use the two-dimensional model equavilent to one dimension,which is a little troubled. thanks | ||
| Robert G. Nelson (rgnelson) Member Username: rgnelson Post Number: 17 Registered: 06-2003 |
FlexPDE has no one-dimensional option, but you can deal with one-dimensional equations by creating a thin 2D domain with a fake y-dimension. The finite element equations are based on 2D cell integrals, so you MUST include a dyy() term and a natural()=0 boundary condition on the top and bottom in order to stabilize the fictitious dimension. See the example "Samples | Misc | One_Dim.pde". | ||
| Tan Hunseng (tan) New member Username: tan Post Number: 1 Registered: 08-2003 |
Dear all,, I have a 1D problem. My problem is my plots have the right shape but wrong value.all the constants are right. Can anybody suggest me what I should do? Thank you very much in advance. Here is the program. TITLE ' 1D EVP consolidation' SELECT contours = 5 VARIABLES Ez ue zigmap DEFINITIONS H=1 {m} Starttime= 0 {min} Endtime = 240000 {min} Gw=9.81 {kN/m3} incrstr=30 {increase vertical stress (kPa)} zigmat=(15/H+Gw)*H+incrstr {kPa} e0=5.98 k=2.5e-7 {m/min} e=e0-Ez*(1+e0) A=0.006 {K/V} C=0.0095 {phi/V} t0=0.5 {min} Ez0=0.03 B=0.138 {L/V} zigmap0=14.5 {kPa} V=1+e0 {Specific volume} INITIAL VALUES zigmap=14.5 {kPa} ue=30 {kPa} Ez=Ez0 EQUATIONS zigmap=zigmat-Gw*H-ue {effective stress concept} (1+e0)/Gw*dy(k/(1+e)*dy(ue))=-dt(Ez) {Continuity} dt(Ez)=A/zigmap*dt(zigmap)+C/t0*exp(-(Ez-Ez0)/C)*abs(zigmap/zigmap0)^(B/C) {EVP stress-strain} BOUNDARIES Region 1 start (0,0) Value(ue)=0 {drained boundary} line to (H/100,0) natural(ue)=0 natural(Ez)=0 line to (H/100,H) natural(ue)=0 natural(Ez)=0 line to (0,H) line to (0,0) TIME from Starttime to Endtime PLOTS for t = 0 by 1000 to endtime contour (ue) elevation(ue) from (H/100,0) to (H/100,H) range=(0,40) as "Ue" for t = 0 by 1000 to endtime contour (Ez) elevation(Ez) from (H/100,0) to (H/100,H) range=(0,0.5) as "average strain" HISTORIES HISTORY(ue) AT (0,0) (0,H*3/4) (0,H/2) (0,H/4) (0,H) HISTORY(Ez) AT (0,0) (0,H*3/4) (0,H/2) (0,H/4) (0,H) end | ||
| Robert G. Nelson (rgnelson) Member Username: rgnelson Post Number: 18 Registered: 06-2003 |
FlexPDE knows nothing about the numbers associated with any application. The numerical values returned by the solution follow directly from the values of coefficients, boundary conditions and initial values. If the scale of the solution is wrong, you must have a dimensional inconsistency somewhere in your coefficients. |