Week - 2 - Explicit and Implicit Analysis

TITLE: EXPLICIT AND IMPLICIT ANALYSIS

Implicit and Explicit are two types of approached that can be used to solve the finite element problem. The implicit approach is useful in problems in which time dependency of the solution is not an important factor [e.g. static structural, harmonic, modal analysis etc.] whereas Explicit Dynamics approach is most helpful in solving high deformation time dependent problems such as Crash, Blast, Impact etc.

The prime difference between the implicit and explicit scheme lies in the consideration of velocity or acceleration. You must be aware of the equation relating mass (m), damping (c), stiffness (k) and force (F). In equation 1, ‘x’ stands for displacement whereas ẋ and ẍ are resp. the first and second time derivatives of ‘x’. In other words they stand for velocity and acceleration resp.

mẍ + cẋ + kx = F (1)

Implicit Scheme: In an implicit scheme, the displacement is not a function of time (i.e. x = constant). Hence the velocities and accelerations which are time derivatives of displacement turn out to be zero and the mass and damping factors can be neglected. The implicit method can be based based on Newark’s method, Newton Raphson Method etc. In order to solve an FEM problem using implicit method, inversion of stiffness matrix (k) is required. Very Large deformation problems such as crash analysis can result in millions of degrees of freedom effectively increasing the size of stiffness matrix. Larger the stiffness matrix longer is the computational time required for its inversion. Hence there is a need for an explicit method which would prevent the inversion of stiffness matrix. Implicit methods are mainly used in softwares such as Ansys, Nastran, Abaqus etc.

Explicit Scheme: As opposed to Implicit methods, explicit scheme is a function of time. Being a function of time, the velocity and acceleration as well as the mass and damping need to be considered in this scheme. In an explicit method, Central Difference time integration (CDTI) is used to calculate field variables at respective nodal points. Since only a numerical solution is possible for a non linear ordinary differential equation, this method is particularly suited for non linear problems. It requires the inversion of the lumped mass matrix as opposed to that of the global stiffness matrix in the implicit methods. In the CDTI, the equation of motion is evaluated at the previous time step (t_n-1, where t_n is the current timestep).

The explicit method or algorithm works in timestep increments i.e. the displacements are calculated as the time proceeds. Consider the simulation of a crash analysis. At timestep 1 (t=0 ms), there is no deformation since the impact is yet to occur. Gradually as time would progresses the deformation also would change. Assume that at timestep 2, t is 5 ms, now the explicit algorithm will calculate the values of field variables at time when t=5 ms. This is the way in which the solution proceeds.

LS Dyna is one software which is based on explicit dynamics and is especially used for solving problems such as Crash, Impact, Penetration etc. Pam crash and Abaqus explicit are also based on the same.

The Problem

`F(u)=u^3+9u^2+4u` 1

From (1) it follows that the stiffness of the bar is

`k(u)=(dF)/(du)=2u^2+18u+4`

EXPLICIT ANALYSIS:

Clearly it is possible to plot a graph of force, f, versus displacement, u, by using equation(1) (see Figure 2). However, if the nonlinear force versus displacement relationship is
not known, but the stiffness is known, it is possible to construct the graph of force versus displacement numerically. One method is to use an incremental explicit load control scheme.
This is illustrated next. For this example three equal load increments (or steps) of `DeltaF` = 1 unit each are used to load the tension bar. The variables used are u for displacement, f for
internal force in the bar, F for external force applied to the bar and k for stiffness. Incremental displacements or incremental externally applied forces are represented as `Deltau` or `DeltaF`,
respectively. Use is made of the relationship `DeltaF=kDeltau`. The analysis proceeds as follows.

Step 1

`u0=0.0=k(u0)=4`

`DeltaF=1`

`Deltau1=(DeltaF)/(k(u0))=1/4=0.25`

`u1=u0+Deltau1=0.25`

Step 2

`k(u1)=2xx(0.25)^2+18(0.25)+4=8.687`

`Deltau2=(DeltaF)/(k(u1))=1/8.687=0.11511`

`u2=u1+Deltau2=0.25+0.11511=0.365`

Step 3

`k(u2)=2xx(0.365)^2+18(0.365)+4=10.96`

`Deltau3=(DeltaF)/(k(u2))=1/10.96=0.0911`

`u3=u2+Deltau3=0.365+0.0911=0.4561`

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

`Fext=1+1+1=3.0`

fint`=(u3)^3+9(u)^2+4u`

`(0.4561)^3+9(0.4561)^2+4(0.4561)cong3.7915`

It is evident from the results that external and internal forces are NOT in equilibrium because Fext is not equal to fint. Table 1 summarizes the main results for the explicit analysis.

Table 1: Summary of explicit analysis results

Step i	`DeltaFi`	`Deltaui`	ui	(Fext)i	(fint)i	fint − Fext = R
1	0.25	0.25	0.25	1	1.5781	0.5781
2	1.0	0.11511	0.365	2	2.7076	0.7076
3	1.0	0.0911	0.4561	3	3.7915	0.7915

IMPLICIT ANALYSIS:

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.

Step 1:

`u0=0.0=k(u0)=4`

`DeltaF=1,Deltau1=(DeltaF)/(k(u0))=1/4=0.25`

`u1=u0+Deltau1=0.25`

Check the residual,

`Fext=DeltaF=1`

fint=`u^3+9u^2+4u=0.25^3+9(0.25)^2+4(0.25)=1.5781`

`R^0=f`int`-Fext=1.5781-1=0.5781`

`||R^0||>10^-2,`,i.e Newton Rapson iterations are necessary.

Calculate the corrections to `u=u1^0`

`deltau^1=-[k(u1^0)]^-1xxR^0`

`=1/(8.6875)xx0.5781=-0.0665`

Hence,`u1^1=u1^0+deltau1^0=0.25-0.0665=0.1834`

Now check the residual again,

`Fext=DeltaF=1`

fint=`(u1^1)^3+9(u1^1)^2+4(u1^1)=1.04966`

`R1=1.0469-1=0.04>0.01`

`deltau^2=deltau^1-[ku^1]^-1.R^1=-0.0665-0.005=-0.0713`

Hence,`u1^2=u1^0+deltau^2=0.25-0.0713=0.1787`

fint=`(u1^2)^3+9(u1^2)^2+4(u1^2)=1.007`

`R^2=1.007-1=0.1787`

0.1787>0.01,Therefore no other iterations are needed.

Therefore the final value of u1 = 0.1787

Step 2 and Step 3 are completed similar to step 1,There fore the other values of u2 and u3 are,

u2 = 0.29658 and u3 = 0.3911

Step i	`DeltaFi`	ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.1787	1	1.007	0.007
2	1	0.2965	2	2.004	0.004
3	1	0.39114	3	3.001	0.001

Blue-Explicit

Orange-Implicit