Solving a simple non-linear problem with both Implicit and Explicit Methods
Solving a simple non-linear problem with both implicit and explicit:
IMPLICIT METHOD:
During the analysis when the deformation in the system is small then it is said to be a implicit analysis problem. In the Implicit time integration method the inversion of the Stiffness matrix [K] is expensive in terms of computation for multiple degree of freedom system.
In this method the slover has to slove a complex set of equations which gives the user the option of using comparatively bigger timestep than the Explicit analysis problem. Thus making this method to be "uncontionally stable".
Also when timestep is set to a larger value and there are possibilities of the solver jumping to the last timestep throwing a inaccurate results.
EXPLICIT METHOD:
Similarly during the analysis when the deformation is large, then it comes under the Explicit analysis. And in the Explicit time integration method for a multiple degree of freedom there is no inversion of the stiffness matrix but only the mass matrix [M], since the mass matrix here is a diagonal matrix less time is need for computation.
The solver here solves easier eqaution at each node when compared to implict method, since the solver takes less time for calculations this allows the user to use the minimal timestep.
In this case if the timestep set lower than the ciritiacal timestep the the system will run into instablity and the simulation will stop. This makes the Explicit method to "conditionally stable".
The critical timestep is minimum time required for the energy wave to propagate through the smallest side of the element.
PROBLEM:
Consider the case of a simple bar in tension as shown in Figure 1. Suppose that the forcein the bar is a nonlinear function of the displacement.
`f(u)=u^3+9u^2+4u` (1)
From (1) it follows that the stiffness of the bar is
`k(u)=3u^2+18u+4` (2)
EXPLICIT METHOD:
`f(u)=u^3+9u^2+4u`
`k(u)=3u^2+18u+4`
Step:1
`u0=0`;`triangleF=1`;`f"external"=1`
`triangleu=(triangleF)/(k(u))`
`k(u)=3(0)^2+18(0)+4`
`k(u)=4`
`triangleu=(1)/(4)` `triangleu=0.25`
`u1=u0+triangleu`
`u1=0.25`
`f"internal"=(0.25)^3+9(0.25)^2+4(0.25)`
`f"int"=1.5781`
`R=f"int"-f"ext"`
`Residue(R)=0.5781`
Step:2
`f"ext"=2`;`u1=0.25`
`k(u)=8.6875`
`u2=0.3651`
Similarly sloving for `f"int"`
`f"int"=2.7088"`
`R=0.7088`
Step:3
`f"ext"=3`;`u2=0.3651`
`k(u)=0.0911`
`u3=0.456`
Similarly sloving for `f"int"`
`f"int"=3.7902`
`R=0.7902`
IMPLICIT METHOD:
Step:1
`triangleF=1`;`u0=0`;`f"ext"=1`
`k(u)=4`
`triangleu=(1)/(4)=0.25`
`u1=u0+triangleu`
`u1=0.25`
`f"int"=(0.25)^3+(0.25)^2+4(0.25)`
`f"int"=1.5781`; `R^(0)=0.5781`
`||R^(0)||>10^-2, Newton-Raphson iterations are necessary.
ITERATION:1
`triangle^(j+1)=-[k(u)^j]^-1.R^j`
`u^1=u^0`
`k(u1)=3(0.25)^2+18(0.25)+4`
`k(u)=8.6875`
`triangleu^1=-(8.6875)^-1.0.5781`
`triangleu^1=-0.06656`
`u^(1)=u1+triangleu`
`u^(1)=0.18344`
`f"int"=(0.18344)^3+9(0.18344)^2+4(0.18344)`
`f"int"=1.04278`; `R^(1)=0.4278`
`||R^(0)||>10^-2`, Newton-Raphson iterations are necessary.
ITERATION:2
`triangle^(j+1)=triangleu^j-[k(u)^j]^-1.R^j`
`k(u)^(1)=7.40287`
`triangleu^2=-0.072338`
`u2^(2)=0.1777`
`f"int"=1.0006`; `R^(2)=0.0006`
`||R^(0)||<10^-2, Therefore `u1=0.1777`
Step:2
`u1=0.1777`;`f"ext"=2`
`k(u1)=7.2933`
`triangleu=0.1371`
`u2=0.3148`
`f"int"=2.1832`; `R^(0)=0.1832`
`||R^(0)||>10^-2`, Newton-Raphson iterations are necessary.
ITERATION:1
`u^(1)=u^(0)`
`k(u)=9.9637`
`triangleu=-0.0184`
`u2=0.2964`
`f"int"=2.00232`; `R^(1)=0.00232`
`||R^(0)||<10^-2`, Therefore `u2=0.2964`
Step:3
`u2=0.2964`;`f"ext"=3`
`k(u)=9.5988`
`triangleu=0.1042`
`u3=0.4006`
`f"int"=3.110`; `R^(0)=0.1110`
`||R^(0)||>10^-2`, Newton-Raphson iterations are necessary.
ITERATION:1
`u3=0.4006`
`k(u)=11.6922`
`triangleu=-0.0095`
`u3=0.3911`
`f"int"=3.0008`; `R^(1)=0.0008`
`||R^(0)||<10^-2`, Therefore `u3=0.3911`
Comparision of Explicit and Implicit Method :