Week - 2 - Explicit and Implicit Analysis
AIM: To study about implicit and explicit method to solve engineering problems.
Objectives: As demonstrated in the PDF, F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods with 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 nonlinear ordinary differential equation, this method is particularly suited for nonlinear 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 (tn-1. where tn is the current time step). The explicit method or algorithm works in time step increments i.e. the displacements are calculated as the time proceeds. Consider the simulation of a crash analysis. At time step 1 (t=O ms), there is no deformation since the impact is yet to occur. Gradual y as time would progresses the deformat ion also would change. Assume that at time step 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 especialy used for solving problems such as Crash, Impact, Penetration etc. Pamcrash 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 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 milions 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.
IMPLICIT METHOD VS EXPLICIT METHOD
IMPLICIT METHOD EXPLICIT METHOD
Deformation small Large
What cause large deformation Force and dynamic Velocity, acceleration with mass and inertia
K matrix inversion yes No
Time step size Large Small
What limits the time step Accuracy Stability
Cost per time step Large Small
What domiate the cost Matrix Inversion Element Process
Parallel process Possible Not possible
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 Liu or LiF respectively. Use is made of the relationship Lif = kLiu.
Explicit method
F(u) = u3+9u2+4u ........(1)
K(u) = dF/du = 3u2 +18u + 4 .........(2)
Step-1
Take u = 0 in (2)
(2) => k(uO) = 3 x 02 + 18 x 0 + 4 = 4
= /k(uO) = 1/4 = 0.25
u1= uO+ = 0.25
Fint = (0.25)3 + 9 x (0.25)2 +4 x 0.25 = 1.578
Step-2
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
Fint = (0.3651)3 + 9 x (0.3651)2 +4 x 0.3651 = 2.708
Step-3
Take u2 = 0.3651 in (2)
(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 equillibrium
Fext = 3
Fint = (0.46)3 + 9 x (0.46)2 +4 x 0.46 = 3.841
Hence Fext is not equal to Fint the analysis is not in equillibrium
Table 1: Summary of explicit analysis results
Step i δFi δui ui (Fext)i (fint)i Fint - Fext =R
1 1 0.25 0.25 1 1.578 0.578
2 1 0.1151 0.3654 2 2.708 0.708
3 1 0.09114 0.46 3 3.841 0.841
Implicit method
F(u) = u3+9u2+4u (1)
K(u) = dF/du = 3u2 +18u + 4 (2)
Step-1
Take u = 0 in (2)
(2) => k(uO) = 3 x 02 + 18 x 0 + 4 = 4
= /k(uO) = U. = 0.25
u1= uO+ = 0.25
Residual R
Fint = (0.25)3 + 9 x (0.25)2 + 4 x 0.25 = 1.578
RO = 1.578 - 1 = 0.578 > 10^2
So, Newton-Raphson iterations is necessary
The correction u1= u1(0)
u(1) = -[k(u1(0)^-1 x RO]
= -[(3 x 0.25)2 + (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
Fint = (0.1835)3 + 9 x (0.1835)2 +4 x 0.1835 = 1.0432
RO = 1.0432 -1 = 0.432 > 10^2
So, Newton-Raphson iterations is 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
Fint = (0.1786)3 + 9 x (0.1786)2 + 4 x 0.1786 = 1.0071
RO = 1.0071- 1= 0.0071 < 10^2
No residual needed U1 = 0.1786
Simiarly we derived u2 = .2966 and u3 = 0.3911
Table 2: Summary of explicit analysis results
Step i δFi δui ui (Fext)i (fint)i Fint - Fext =R
1 1 0.25 0.1786 1 1.0071 0.0071
2 1 0.1374 0.2966 2 2.0042 0.0042
3 1 0.1041 0.3911 3 3.008 0.008
CONCLUSION
Explicit and implicit analysis is done for the given equation, implicit method is more accurate than the explicit but the time consumption is more in implicit method.