Week - 2 - Explicit and Implicit Analysis
Aim- To sove for the equation using explicit and implicit method.
Implicit Method
The displacement is independent function of time, which implies the velocity and accleration which are time derivaties of displacement are zero. The mass and the damping factors are also negelected. Implicit function can be represented with the newton raphson method. In this method the stiffness matrix is inverted. Implicit matrix is genrally used in the softwares such as Ansys, Nastran, Abaqus.
Explicit Method
Explicit Method is a time dependent function. Since displacement is a time dependent function the velocit and acceleration as well as the mass and damping are considered in the calculation. central difference time integration is used to calculate field variables at the respective nodal points. Explicit function is generally used for non linear problems because a numerical solution is possible for a non linear ordinary differential equation.Explicit method work on time step increments. Ls Dyna is one of the software used to solve explicit dynamics problems such as Crash, Impact, Penetration etc.
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
Fint = (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
Fint = (0.3651)3 + 9 x (0.3651)2 +4 x 0.3651 = 2.708
Step3
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 equilibrium
Fext = = 3
Fint = (0.46)3 + 9 x (0.46)2 +4 x 0.46 = 3.841
Hence Fext fint the analysis is not equilibrium
Table 1: Summary of explicit analysis results
|
Step i |
∆Fi |
∆ui |
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)
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
Residual R
Fint = (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 is 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
Fint = (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 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
R0 = 1.0071 – 1 = 0.0071 < 102
No residual needed
U1 = 0.1786
Similarly we drived u2 = 2966, u3 = 0.3911
Table 1: Summary of explicit analysis results
|
Step i |
∆Fi |
∆ui |
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 |