Explicit and Implicit Analysis

Description on explicit and implicit analysis

When we applied a load it was applied in steps. And we apply load in steps in implicit type of analysis .This phenomenon / approach is called as incremental approach . Now Implicit and explicit solver both apply this approach . Where they apply load in steps.

The only difference between this two solver is that implicit ,uses an iterative approach along with the incremental method.

The implicit solver try’s to achieve the equilibrium between the internal and external forces at each steps ,within that particular model, only then it will progressed to the next step.(force convergence graph)

But, for explicit solver it doesn’t try to achieve equilibrium at each step. If there are unbalanced forces these unbalanced forces are multiplied with step and we get wrong answer.

Here we have to specify a lowest time step

Also the research provides that data that the implicit solver is not suitable / less accurate for simulation stress wave propagation events,(drop event,car crash event).

Implicit solver only handle the static problem accurately.

Explicit calculation

Consider a case of simple bar in tension .suppose that the force in the bar is a non-linear function of displacement .

F(u) = u³ +u²+4u -------------1

From1 it follows that the stifness of the bar is

K(u)=df/dx = 3u²+2u+4---------------2

∆F=K *∆u

| u F

|------------------------------------------------------------------------------------->

Step1

u0 = 0.0 K(u0) =4 ∆F=1 ∆u1 =∆F/K(u0) = ¼

therefore, u1=u0+∆u1 = 0+1/4 =0.25

Step2

K(u1) =3*(1/4)²+2*1/2+4 =4.6875 ∆F=1 ∆u2 =∆F/K(u1) =0.2133

therefore, u2=u1+∆u2 = 0.25+0.2133=0.4633

Step3

K(u2) =3*(0.4633)²+2*0.4633+4 =5.5704 ∆F=1 ∆u3 =∆F/K(u2) =1/5.5704=0.1795

therefore, u3=u2+∆u3 =0.4633+0.1795=0.6383

Step i	∆Fi	∆ui	ui	(Fext)i	(Fint)i	Fint-Fext=R
1	1	0.25	0.25	1	1.0781	0.0781
2	1	0.2133	0.4633	2	2.1672	0.1672
3	1	0.1795	0.6383	3	3.2206	0.2206

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 th 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) = fint − Fext. Expanding the residual as a Taylor series about the current displacement uj gives
Rj+1 = Rj(uj) + dR(uj)du δuj+1 + ... = Rj + k(uj)δuj+1 + ... (3)3
By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for δuj+1 the
following correction is obtained
δuj+1 = −[k(uj)]−1Rj