Week - 2 - Explicit and Implicit Analysis

>AIM

AS PER THE ARTICLE SHARED, A NON LINEAR STATIC PROBLEM IS SOLVED USING BOTH EXPLICITE AND IMPLICITE ANALYSIS. BOTH OF THE CASES USE INCREMENTAL LOAD CONTROL SCHEME AND HERE IT IS ADVISED TO USE THIS EQUATION "F(u) = u³+9u²+4u" AND SOLVE USING BOTH EXPLICITE AND IMPLICITE METHODS.

ALSO TO EXPLAIN IN DETAIL ABOUT IMPLICITE AND EXPLICITE METHODS.

>EXPLANATION

	IMPLICITE METHOD	EXPLICITE METHOD
DEFORMATION	SMALL	LARGE
CAUSE OF LARGE DEFORMATION	FORCE AND DYNAMIC EFFECT	VELOCITY, ACCELERATION
TIME STEP SIZE	LARGE	SMALL
COST PER TIME STEP	LARGE	SMALL
WHAT DOMINATE THE COST	MATRIX INVERSION	ELEMENT PROCESS
WHAT LIMITS THE TIME-STEP	ACCURACY	STABILITY
COST PER TIME-STEP	LARGE	SMALL

FEM EXPLICIT ANALYSIS IS A TIME-DEPENDENT PROCESS WHERE THE STIFFNESS MATRIX IS UPDATED AT EACH TIME STEP BASED ON GEOMETRY AND MATERIAL CHANGES. THIS METHOD USES SMALL INCREMENTS TO MAINTAIN ACCURACY, BUT IT REQUIRES MANY INCREMENTS, MAKING IT TIME-CONSUMING. UNLIKE IMPLICIT METHODS, EXPLICIT METHODS DON'T USE CONVERGENCE CRITERIA AT EACH STEP, LEADING TO POTENTIAL SOLUTION DRIFT IF INCREMENTS ARE TOO LARGE.

EXPLICIT METHODS ARE IDEAL FOR HIGH STRAIN RATES OR VELOCITIES, SUCH AS IN AUTOMOTIVE CRASHES OR BALLISTIC EVENTS. THEY USE CENTRAL DIFFERENCE TIME INTEGRATION (CDTI) TO CALCULATE FIELD VARIABLES AT EACH NODE, MAKING THEM SUITABLE FOR NON-LINEAR PROBLEMS.

IN EXPLICIT ANALYSIS, DISPLACEMENTS ARE CALCULATED INCREMENTALLY AS TIME PROGRESSES. FOR EXAMPLE; IN A CRASH SIMULATION, THE ALGORITHM CALCULATES DEFORMATIONS AT EACH TIME STEP GRADUALLY SHOWING THE IMPACT EFFECTS.

SOFTWARE LIKE LS-DYNA, PAM-CRASH, AND ABAQUS/EXPLICIT USE EXPLICIT DYNAMICS FOR SOLVING PROBLEMS INVOLVING CRASHES, IMPACTS, PENETRATION AND MUCH MORE.

FEM IMPLICIT ANALYSIS; DISPLACEMENTS AREN'T TIME-DEPENDENT, SO VELOCITIES AND ACCELERATIONS ARE ZERO AND MASS AND DAMPING ARE NEGLIGIBLE. THIS METHOD USES ALGORITHMS LIKE NEWTON-RAPHSON TO ENFORCE EQUILIBRIUM AFTER EACH TIME INCREMENT, WHICH MUST CONVERGE TO A SPECIFIED TOLERANCE. BECAUSE OF THIS, TIME INCREMENTS CAN BE LARGER, BUT EACH INCREMENT REQUIRES UPDATING AND INVERTING A LARGE STIFFNESS MATRIX, MAKING COMPUTATIONS SLOW.

IMPLICIT METHODS ARE BETTER FOR SLOW EVENTS WITH MINIMAL STRAIN RATE EFFECTS, LIKE CYCLIC LOADING, WHERE ACCURACY IS CRUCIAL. THEY ARE USED IN SOFTWARE LIKE ANSYS, NASTRAN, AND ABAQUS.

IN CONTRAST TO EXPLICIT METHODS, WHICH AVOID STIFFNESS MATRIX INVERSION AND WORK BEST FOR HIGH-STRAIN-RATE EVENTS, IMPLICIT METHODS HANDLE EQUILIBRIUM AND LARGE DEFORMATIONS MORE ACCURATELY BUT AT A HIGHER COMPUTATIONAL COST.

>PROCEDURE

AS PER THE SHARED FILE, WE ARE GIVEN WITH f(u) = u3+9u2+4u.----------(1)

THEREFORE k(u) = df/du = 3u²+18u+4.----------(2)

ALSO ΔF = k⋅Δu.----------(3)

WHERE F - EXTERNAL LOAD/FORCE APPLIED

f - INTERNAL FORCE/LOAD APPLIED

ΔF, Δu - LOAD/DISPLACEMENT INCREMENT

k - STIFFNESS

u - DISPLACEMENT

(EXPLICITE)

FIRST STEP ΔF = 1, u₀ = 0

k(u) =df/du =3u²+18u+4 i.e k(u₀) =3(u₀)²+18u₀+4 =3*0²+18*0+4 =4

Δu =ΔF/k =Δu₁ =ΔF/k(u₀) =1/4 =0.25

u₁ =u₀+Δu₁ =0+0.25 =0.25

f(u₁) =0.25³+9*0.25²+4*0.25 =1.578

.....(ΔF =1, u₁ =0.25, F =1, f(u₁) =1.578).....

SECOND STEP ΔF = 1, u₁ =0.25

k(u) =df/du =3u²+18u+4 i.e k(u₁) =3(u₁)²+18u₁+4 =3*0.25²+18*0.25+4 =8.687

Δu =ΔF/k =Δu₂ =ΔF/k(u₁) =1/8.687 =0.115

u₂ =u₁+Δu₂ =0.25+0.115 =0.365

f(u₂) =0.365³+9*0.365²+4*0.365 =2.707

.....(ΔF =1, u₂ =0.365, F =2, f(u₂) =2.707).....

THIRD STEP ΔF = 1, u₂ =0.365

K(u) =df/du =3u²+18u+4 i.e k(u₂) =3(u₂)²+18u₂+4 =3*0.365²+18*0.365+4 =10.969

Δu =ΔF/k =Δu₃ =ΔF/k(u₂) =1/10.969 =0.091

u₃ =u₂+Δu₃ =0.365+0.091 =0.456

f(u₃) =0.4563+9*0.4562+4*0.456 =3.790

.....(ΔF =1, u₃ =0.456, F =3, f(u₃) =3.790).....

TOTAL EXTERNAL FORCE, F =ΔF+ΔF+ΔF =1+1+1 =3 i.e F_ext =3.

F_ext ≠ f_int HENCE THE SYSTEM IS NOT IN EQUILIBRIUM.

STEP	ΔFi	Δui	ui	(Fext)i	(fint)i	Fext - fint
1	1	0.25	0.25	1	1.578	0.578
2	1	0.115	0.365	2	2.707	0.707
3	1	0.091	0.456	3	3.790	0.790

(IMPLICITE)

FIRST STEP ΔF = 1, u0 = 0

k(u) =df/du =3u2+18u+4 i.e k(u0) =3(u0)2+18u0+4 =3*02+18*0+4 =4

Δu =ΔF/k =Δu1 =ΔF/k(u0) =1/4 =0.25

u1 =u0+Δu1 =0+0.25 =0.25

f(u1) =0.253+9*0.252+4*0.25 =1.578

CHECKING RESIDUAL:

R₀ =f_int-F_ext =1.578-1 =0.578

THEREFORE R₀ =0.578 >10^-2, NEWTON RAPHSON ITERATIONS ARE NECESSARY.

CORRECTION 1, u₁=(u₁)⁰

Δu¹ =-(k(u1)0 )^-1*R₀ =-(3*((u1)0)2+18*((u1)0)+4)-1*R0

Δu1 =-(3*0.25²+18*0.25+4)^-1*0.578 =-0.0665

(u₁)¹ =(u₁)⁰+Δu1

(u1)1 =0.25-0.0665 =0.1835

f((u1)1) =0.1835³+9*0.1835²+4*0.1835 =1.0432

CHECKING RESIDUAL:

R0 =fint-Fext =1.0432-1 =0.0432

THEREFORE R0 =0.0432 >10-2, NEWTON RAPHSON ITERATIONS ARE NECESSARY.

CORRECTION 2, u1=(u₁)¹

Δu2 =Δu1-(k(u1)1)-1*R0 =Δu1-(3*((u1)1)2+18*((u1)1)+4)-1*R0

Δu2 =-0.0665-(3*(0.25)²+18*(0.25)+4)-1*0.0432 =-0.0714

(u1)² =(u1)1+Δu2

(u1)2 =0.25-0.0714 =0.1786

f((u1)2) =0.17863+9*0.17862+4*0.1786 =1.0071

CHECKING RESIDUAL:

R0 =fint-Fext =1.0071-1 =0.0071

THEREFORE R0 =0.0071<10-2, NO FURTHER ITERATIONS NEEDED (u₁ =0.1786)

SECOND STEP ΔF = 1, u1 =0.1786

k(u) =df/du =3u2+18u+4 i.e k(u1) =3(u₁)2+18u1+4 =3*0.17862+18*0.1786+4 =7.310

Δu =ΔF/k =Δu2 =ΔF/k(u1) =1/7.310 =0.1367

u2 =u1+Δu2 =0.1786+0.1367 =0.3153

f(u2) =0.31533+9*0.31532+4*0.3153 =2.1872

CHECKING RESIDUAL:

R0 =fint-Fext =2.1872-2 =0.1872

THEREFORE R0 =0.1872>10-2, NEWTON RAPHSON ITERATIONS ARE NECESSARY.

CORRECTION 1, u2=(u2)1

Δu2 =-(k(u2)1 )-1*R0 =-(3*((u2)1)2+18*((u2)1)+4)-1*R0

Δu2 =-(3*0.31532+18*0.3153+4)-1*0.1872 =-0.0187

(u2)2 =(u2)1+Δu2

(u2)2=0.3153-0.0187 =0.2966

f((u2)2) =0.29663+9*0.29662+4*0.2966 =2.0042

CHECKING RESIDUAL:

R0 =fint-Fext =2.0042-2 =0.0042

THEREFORE R0 =0.0042<10-2, NO FURTHER ITERATIONS NEEDED (u2 =0.2966)

THIRD STEP ΔF = 1, u2 =0.2966

k(u) =df/du =3u2+18u+4 i.e k(u2) =3(u2)2+18u2+4 =3*0.29662+18*0.2966+4 =9.6620

Δu =ΔF/k =Δu3 =ΔF/k(u2) =1/9.6620 =0.1034

u3 =u2+Δu3 =0.2996+0.1034 =0.403

f(u3) =0.4033+9*0.4032+4*0.403 =3.1391

CHECKING RESIDUAL:

R0 =fint-Fext =3.1391-3 =0.1391

THEREFORE R0 =0.1391>10-2, NEWTON RAPHSON ITERATIONS ARE NECESSARY.

CORRECTION 1, u3=(u3)2

Δu3 =-(k(u2)1 )-1*R0 =-(3*((u3)2)2+18*((u3)2)+4)-1*R0

Δu3 =-(3*0.4032+18*0.403+4)-1*0.1391 =-0.0118

(u3)3 =(u3)2+Δu3

(u3)3=0.403-0.0118 =0.3912

f((u3)3) =0.39123+9*0.39122+4*0.3912 =3.002

CHECKING RESIDUAL:

R0 =fint-Fext =3.002-3 =0.002

THEREFORE R0 =0.002<10-2, NO FURTHER ITERATIONS NEEDED (u3 =0.3912)

STEP	ΔFi	Δui	ui	(Fext)i	(fint)i	Fext - fint
1	1	0.25	0.1786	1	1.0071	0.0071
2	1	0.1367	0.2966	2	2.0042	0.0042
3	1	0.1034	0.3912	3	3.002	0.002

>RESULTS

X: EXTERNAL FORCE, Y: DISPLACEMENT

>CONCLUSION

THE DIFFERENCES BETWEEN EXPLICIT AND IMPLICIT METHODS FOR SOLVING A NON-LINEAR STATIC PROBLEM IS DEMONSTRATED. THE EXPLICIT METHOD, WHILE STRAIGHT FORWARD, TENDS TO DRIFT FROM THE EXACT SOLUTION DUE TO THE LACK OF EQUILIBRIUM CHECKS AT EACH STEP BUT FOR THE CASE OF IMPLICITE, EMPLOYS NEWTON-RAPHSON ITERATIONS TO ENSURE EQUILIBRIUM BETWEEN INTERNAL AND EXTERNAL FORCES AT EACH INCREMENT, RESULTING IN A MORE ACCURATE SOLUTION.

IMPLICIT METHOD IS MORE ACCURATE THAN THE EXPLICIT BUT THE TIME CONSUMPTION IS MORE IN IMPLICIT METHOD.