Week - 2 - Explicit and Implicit Analysis

AIM:-  Explicit and Implicit Analysis

OBJECTIVE:-  

As demonstrated in the PDF, F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods ( have a tolerance of 0.01).

As demonstrated in the PDF, F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods with the tolerance of 10^-2.

Explicit: As opposed to Implicit methods, explicit method 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 progresses the deformation also changes. Assume that at timestep 2, t is 5 ms, now The explicit algorithm will calculate the values of field variables at a time when t=5 ms.

This is the way in which the solution proceeds.

LS Dyna is one software that 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.

Implicit: - 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 on Newark’s method, Newton Raphson's Method, etc.

In order to solve a FEM problem using the 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 the stiffness matrix.

The larger the stiffness matrix longer the computational time required for its inversion.

Hence there is a need for an explicit method that would prevent the inversion of the stiffness matrix.

Implicit methods are mainly used in software such as Ansys, Nastran, Abaqus, etc. 

IMPLICIT METHOD VS. EXPLICIT METHOD

PROBLEM

            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 ∆u or ∆F, respectively.

Use is made of the relationship ∆F = k∆u.

Explicit method

F(u) = u3+9u2+4u                                   (1)

K(u) = dF/du = 3u² +18u + 4                (2)

Step1

Take u = 0 in (2)

(2) => k(u0) = 3 x 0² + 18 x 0 + 4 = 4

 = /k(u0) = ¼ = 0.25

u1= u0+ = 0.25

F_int  = (0.25)³ + 9 x (0.25)² +4 x 0.25 = 1.578

Step2

Take u1 = 0.25 in (2)

(2) => k(u1) = 3 x 0.25² + 18 x 0.25 + 4 = 8.6875

 = /k(u1) = 1/8.6875 = 0.1151

U2= u1+ = 0.3651

F_int  = (0.3651)³ + 9 x (0.3651)² +4 x 0.3651 = 2.708

Step3

Take u2 = 0.3651 in (2)

equation(2) = k(u2) = 3 x 0.3651² + 18 x 0.3651 + 4 = 10.9716

 = /k(u2) = 1/10.9716 = 0.09114

U3= u2+ = 0.46

To determine whether the analysis is equilibrium

F_ext =   = 3

F_int  = (0.46)³ + 9 x (0.46)² +4 x 0.46 = 3.841

Hence F_ext f_int the analysis is not equilibrium

Table 1: Summary of explicit analysis results

Implicit method

F(u) = u3+9u2+4u                                   (1)

K(u) = dF/du = 3u² +18u + 4                (2)

Step1

Take u = 0 in (2)

equation(2) => k(u0) = 3 x 0² + 18 x 0 + 4 = 4

 = /k(u0) = ¼ = 0.25

u1= u0+ = 0.25

Residual R

F_int  = (0.25)³ + 9 x (0.25)² +4 x 0.25 = 1.578

R0 = 1.578 – 1 = 0.578 > 10²

So, Newton-Raphson iterations are necessary

The correction u1 = u1(0)

 u(1) = -[k(u1(0)^-1 x R0

= -[(3 x 0.25² ) + (18 x 0.250) +4]^-1 x 0.578

=-0.0665

U1(1) = u1(0) + u(1)

= 0.25 – 0.0665 = 0.1835

Residual R1

F_int  = (0.1835)³ + 9 x (0.1835)² +4 x 0.1835 = 1.0432

R0 = 1.0432 – 1 = 0.432 > 10²

So, Newton-Raphson iterations are necessary

The correction u1 = u1(1)

 u(2) = u(2)  -[k(u1(1)^-1 x R1

=(-0.0665) -(8.6875 ^-1 )x 0.0432

=-0.0714

U1(2) = u1(1) + u(2)

= 0.25 – 0.0714 = 0.1786

Residual R1

F_int  = (0.1786)³ + 9 x (0.1786)² +4 x 0.1786 = 1.0071

R0 = 1.0071 – 1 = 0.0071 < 10>2

No residual needed

U1 = 0.1786

Similarly we derived u2 = 2966, u3 = 0.3911

Table 1: Summary of explicit analysis results