Week - 2 - Explicit and Implicit Analysis

Basically problems are divided into two major categories as Static and Dynamic. While Static problems can be easily solved using Implicit methods Dynamic problems yeild much appropriate solution when solved by Explicit methods. But lets consider a detailed view on this, and decide which methods suits which problems.

Every Nonlinear problem is solved by breaking down the problem into smaller portions. It is also necessary to find out whether the problem is time dependant or Time independant. Further, this can be achieved by considering smaller time steps for computation.

Every Object in motion or static will have a will have some Total energy associated with it. Lets consider an time instant `T underset(i)()`to anlyse the effect of energies associated with that object. After some time at an instant T underset(f)()`the object might be in a different state depending on forces/energies acting on it. As the object moves around over time the energies associated with it (Gravitaional potential, Kinetic, Elastic Spring Potential, Thermal) change in value or form but the total energy of that isolated system will be conserved/remains same over time aaccording to Law of Conservation of Energy. This can be represented in the form of equation as:

`frac{1}{2}mv underset(i)() overset(2)() + mgh underset(i)() + frac{1}{2}kx underset( i)() overset(2)()` = `frac{1}{2}mv underset(f)() overset(2)() + mgh underset(f)() + frac{1}{2}kx underset(f)() overset(2)() +E underset(Hf)()`

where:

Kinetic energy = `frac{1}{2}mv overset(2)()`

Gravitaional Potential energy = `mgh`

Elastic Potential energy = `frac{1}{2}kx overset(2)()`

Thermal energy = `E underset(Hf)()`

Considering the instant where movement of object is so slow such that the difference in time instants for initial and final stages is huge. Lets say we analyse the instant which invloves objects undergoing bending/torsion/deformation over time due to some forces. The actual act of deformation is for very long duration such that the velocity of the object barely changes over that large time instances. Thus Kinetic energy is not considered in this equation. Also, notice that Gravitational Potential energy also barely changing over this time. Now if no Heat is addded or removed from the system the equation boils down to Elastic Potential energy changes which further involves formulation of Global Stifness matrix. Now, here every element within the object is considered as spring and a spring damper model is deployed to obatin the values for Stifness constant and further calculate the energy values. For a spring whose end points undergo the displacements x1 and x2, the elastic force developed in the spring is proportional to the spring deformation, which is the difference between the two end point displacements as:

`F(t) = k∆x underset(t)() = k (x1underset(t)() − x2 underset(t)())`

Also, from eqauation of motion,

`F underset(ext)()(t)=Mx¨ + cx˙ + kx`

Where, M is Mass Matrix, C=Damping matrix and K is stiffness matrix.

In static analysis, there is no effect of mass (inertia) or of damping. In dynamic analysis, nodal forces associated with mass/inertia and damping are included. As we dont need Mass matrix and Damping matrix in this case the equation becomes

`F underset(ext)()(t)= kx`.

Thus now we can calculate the values for global Stifness matrix and find the solution.

Now consider the case where object are colliding where difference between time instants is so small. Here actual act of collision can happen within from microseconds to a second. At such instances Kinetic energy is taken into consideration. Here too we can see that Gravitational potential energy is barely changing and if no heat is added or taken out of system, the equation reduces to Kinetic and Elastic potential energy.

Here, the equation becomes, `F underset(ext)()(t)=Mx¨ + cx˙ + kx`

Now we can clearly understand the differences in imlicit and Explicit solver as:

IMPLICIT	EXPLICIT
Always Stable	Conditional Stability
Used to solve Static and transient problems	Used to solve Dynamic and Shock problems
Global Resolution. Convergence is needed	Element by Element Calculation
Too Expensive	Low relative cost
Time step is too large	Time step is too small
Low Robustness	High Robustness

So factors that will decide which solver to select for a problems will depend on;

1. Time Step: The computational cost of a single explicit increment is small; all information needed is available and the calculation is straight-forward and fast. However, the time increment cannot be too large, because then method becomes unstable and the error on the solution increases exponentially. This makes sense if we think of it as an extrapolation: extrapolating far beyond the known range tends to give wrong results, especially if errors have a chance to add up. The largest time increment that can be used is calculated automatically by Abaqus and it is referred to as the stable time increment. The stable time increment reduces with smaller element size, lower density, and higher stiffness. The element with the smallest stable time increment determines the time increment for the entire analysis; a single badly shaped element can therefore hugely impact the simulation time. The stable time increment is usually approximately constant throughout the analysis, so that the time required to run an analysis can be estimated once the first increments have completed.
2. Linearity/Non-linearity: The computational cost of an implicit increment is much larger, because a system of equations needs to be solved. For non-linear analysis, this even needs to be done multiple times. It will thus take more time and memory to solve a single implicit increment. The time increment size, on the other hand, is not limited by stability issues for implicit algorithm. This typically allows much larger time increments than for explicit algorithms. The increment size in Abaqus/Standard (the implicit solver) is usually determined automatically by Abaqus, based on how easily the Newton-Raphson scheme converges. This can vary significantly throughout the analysis. The more non-linear the problem is, the more computationally expensive it is to find a converged solution.

IMPLICIT	EXPLICIT
BASED ON TIME INCREMENT
Large computational cost per increment	Small computational cost per increment
Large time increments	Small time increments
Strongly varying increment size for non-linear problems	Approximately constant increment size
BASED ON DYNAMIC EFFECTS
Static or dynamic	Dynamic
	Vibrations and oscillations
	Long simulation time for static analyses
	Mass scaling
Requirement to connect everything to the ground for static analyses
BASED ON CONTACT
Similar contact options available	Similar contact options available
Changes in contact extremely non-linear: computationally expensive	Small time increments beneficial to solve contact problem

We will be solving with help of similar taylor series expansion as shown in example.

For Implicit analysis:

Step 1

`u underset0()= 0, k(u underset0()) = F'(u underset0())= 2 (0) overset(2)()+18(0)+4 = 4 , Delta F=1 ,Deltau underset(1)() = (Delta F)/(k (u underset0 ()))= 0.25 , u underset1()= u underset0()+ Delta u underset1()= 0.25`

Step 2

`k(u underset1()) = F'(u underset1())= 2 (0.25) overset(2)()+18(0.25)+4 = 8.6875 , Delta F=1 ,Deltau underset(2)() = (Delta F)/(k (u underset1 ()))=0.1151 , u underset2()= u underset1()+ Delta u underset2()= 0.3651`

Step 3

`k(u underset2()) = F'(u underset2())= 2 (0.3651) overset(2)()+18(0.3651)+4 = 10.9718 , Delta F=1 ,Deltau underset(3)() = (Delta F)/(k (u underset2 ()))=0.0911 , u underset3()= u underset2()+ Delta u underset3()= 0.4562`

Finally, from the three steps the following results are obtained.

`F underset(EXT)() = Delta F + Delta F + Delta F = 3, f underset(INT)()= u overset(3)()+9u overset2() +4u = (0.3651) overset(3)()+9(0.3651) overset2() +4(0.3651) = 3.7934`

It is evident from the results that external and internal forces are NOT in equilibrium because
`F underset(EXT)() ne f underset(INT)()` . Table 1 summarizes the main results for the explicit analysis.

Table 1.

Explicit Analysis
	u0	`k(u)=F'(u)=2u overset(2)()+18u+4`	Step i	∆F=1	∆ui = ∆F/k(ui-1)	ui = ui-1 + ∆ui	(Fext)=∆F	(fint)	||R|| = fint − Fext
u0 = 0.0	0	4	1	1	0.25	0.25	1	1.5781	0.5781
u1 = 0.25	0.25	8.6875	2	1	0.1151	0.3651	1	2.7088	1.7088
u2 = 0.3651	0.36510	10.9718	3	1	0.0911	0.4562	1	3.7934	2.7934

An implicit analysis is the same as the explicit analysis, except that at the end of each step Newton-Raphson iterations are used to enforce equilibrium before moving to the next step. Basically, an incremental force is applied to advance the solution forward at the beginning of a step. However, internal forces and external forces will not be in equilibrium unless the stiffness is linear for the given step. Hence, in order to achieve equilibrium, corrections must be made to the displacement. This is accomplished by using Newton-Raphson iterations to minimize the residual, `R(u) = F underset(EXT)() - f underset(INT)()`. Expanding the residual as a Taylor series about the current displacement uj gives:

`R_(j+1) = R^j(u^j) + (dR(u^j))/ (du) δu^(j+1) + ...................... = R^j + k(u^j)δu^(j+1) + ... (3)`

By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for `δu^(j+1)` the following correction is obtained

`δu^(j+1) = −[k(uj)]^(−1).R^j`

Notice that the iteration variable is j and it may take several iterations for the norm of the residual, ||R||, to be reduced below the chosen tolerance criteria (for the example below a tolerance of 0.001 is chosen). The implicit analysis proceeds as follows (results are approximate, they are more precise if done in a computer to machine precision):

IMPLICIT ANALYSIS		F internal	F (ext)	`DeltaF`	K(u)	`Deltau`	R	Tolerance Check	Iteration #	u correction
`u underset(0)`	0	0	0	1	4	0.2500	0	R<0.001	None required	NA
`u underset(1)`	0.2500	1.5781	1	1	8.6875	0.1151	0.5781	R>0.001	1 st iteration	-0.0665
`u underset(1)overset(1)()`	0.1834	1.0428	1	1	7.4031	0.1350	0.0428	R>0.001	2 nd itereation	-0.0723
`u underset(1)overset(2)()`	0.1776	1.0003	1	1	7.2925	0.1371	0.0003	R<0.001	None required	NA
final value is u1 = u (2)1=0.1776606963
`u underset(2)()`	0.3147	2.1821	2	2	9.9634	0.1003	0.1821	R>0.001	1 st iteration	-0.0182
`u underset(2)overset(1)()`	0.2961	2.000001181	2	2	9.6008	0.1041	0.000001	R<0.001	None required	NA
final value is u2 = u (1)2=0.2961587551
`u underset(3)()`	0.4003	3.1076	3	3	11.6864	0.0855	0.1076	R>0.001	1 st iteration	-0.0092
`u underset(3)overset(1)()`	0.3911	3.000000058	3	3	11.4986	0.0869	0.000000058	R<0.001	None required	NA
final value is u3 = u (1)3=0.3910256231