Explicit Analysis Vs Implicit Analysis - Hand Calculation Method

IMPLICIT METHOD :
Implicit methods find a solution by solving an equation involving both the current state of the system and the later one. Implicit analysis requires a numerical solver to invert the stiffness matrix once or even several times over the course of a load/time step. Implicit methods attempt to find a solution to the nonlinear system of equations iteratively by considering the current state of the system as well as its subsequent (or previous) time state.

EXPLICIT METHOD :
An explicit analysis aims to solve for acceleration. The accelerations are solved for the nth time step, and then the velocity is solved for the n+1/2 step, and displacement at the n+1 step. The solutions are not always stable and much smaller time steps are required. The time step in an explicit analysis must be less than the time taken by a sound wave to travel across an element and is hence dependent on the mesh size and the material properties. This is known as the Courant time step.

IMPLICIT METHOD VS EXPLICIT METHOD
CHARACTERISTICS	IMPLICIT METHOD	EXPLICIT METHOD
Time step	Large	Small
Strain Rate	small	Large
Strain Due to	Force	Velocity, acceleration with mass and inertia
K Matrix Inversion	yes	No
Advantages	Accuracy	Stability
Process Involved	Matrix inversion	Element process
Cost	large	Small

F(u) = u³+9u²+4u , Lets solve this Equation with both Explicit and Implicit Function,

EXPLICIT ANALYSIS:

k(u) = df/du = 3u²+18u+4

Initial cartesian Co ordinates , u₀ = (0,0)

ΔF = F1 = 1, (F= Split along the axis)

k(u₀) = K(0) = 3(0)²+18(0)+4 = 4

ΔU₁= ΔF/k(u₀) = 1/4 = 0.25

U₁= U₀+ΔU₁ = 0+0.25 = 0.25

Fₑₓₜ= F1 = 1

Fᵢₙₜ= u³+9u²+4u = (0.25)³+9(0.25)²+4(0.25) = 1.5781

u₁ = 0.25

ΔF = F2 = 1,

k(u₁) = K(0) = 3(0.25)²+18(0.25)+4 = 8.68

ΔU₂= ΔF/k(u₁) = 1/8.68 = 0.1152

U₂= U₁+ΔU₂ = 0.25+0.1152 = 0.3652

Fₑₓₜ = ΔF₁+ΔF₂= 1+1 = 2

Fᵢₙₜ= u₂³+9u₂²+4u₂ = (0.36)³+9(0.36)²+4(0.36) = 2.65

u₂ = 0.36

ΔF = F3 = 1,

k(u₂) = K(0) = 3(0.36)²+18(0.36)+4 = 10.86

ΔU₃= ΔF/k(u₂) = 1/10.86 = 0.0920

U₃= U₂+ΔU₃ = 0.36+0.0920 = 0.4520

Fₑₓₜ = ΔF₁+ΔF₂+ΔF₃= 1+1+1= 3

Fᵢₙₜ= u₃³+9u₃²+4u₃ = (0.45)³+9(0.45)²+4(0.45) = 3.71

Summary of explicit analysis results,

Steps	∆Fi	∆ui	ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.25	0.25	1	1.57	0.57
2	1	0.1152	0.3652	2	2.65	0.65
3	1	0.092	0.452	3	3.71	0.71

IMPLICIT ANALYSIS : (Newton Raphson Method)

F(u) = u³+9u²+4u

k(u) = df/du = 3u²+18u+4

Initial cartesian Co ordinates , u₀ = (0,0)

ΔF = F1 = 1, (F= Split along the axis)

k(u₀) = K(0) = 3(0)²+18(0)+4 = 4

ΔU₁= ΔF/k(u₀) = 1/4 = 0.25

U₁= U₀+ΔU₁ = 0+0.25 = 0.25

Fₑₓₜ= F1 = 1

Fᵢₙₜ= u³+9u²+4u = (0.25)³+9(0.25)²+4(0.25) = 1.5781

Fᵢₙₜ-Fₑₓₜ =0.5781

|R⁽⁰⁾|>10⁻²

Calculate the correction to u₁=u₁⁰

Δu⁽¹⁾=-[K(u₁⁽¹⁾)]⁻¹.R⁽⁰⁾

=-(8.68)⁻¹*0.5781

=-0.0666

U₁⁽¹⁾=U₁⁽⁰⁾+Δu⁽¹⁾=0.25-0.0666=0.1834

Residual

Fₑₓₜ = ΔF = 1

Fᵢₙₜ = (U₁⁽¹⁾)³+9(U₁⁽¹⁾)²+4U₁⁽¹⁾=

=(0.1834)³+9(0.1834)²+4(0.1834)

=1.042

R⁽¹⁾=Fᵢₙₜ -Fₑₓₜ =0.042

|R⁽¹⁾|>10⁻²

U₁=U₁⁽⁰⁾

ΔU⁽²⁾=ΔU⁽¹⁾-[K(u₁⁽¹⁾)]⁻¹.R⁽¹⁾

=-0.0666-(7.404)⁻¹*0.042

=0.072

U₁⁽²⁾=U₁⁽⁰⁾+Δu⁽²⁾=0.25-0.072=0.178

Residual

Fₑₓₜ = ΔF = 1

Fᵢₙₜ = (U₁⁽²⁾)³+9(U₁⁽²⁾)²+4U₁⁽²⁾=

=(0.178)³+9(0.178)²+4(0.178)

=1.0027

R⁽¹⁾=Fᵢₙₜ -Fₑₓₜ =0.0027

STEP 2

U₁⁽²⁾=0.178

K(U₁)=7.29

ΔF = 1

Δu₂=ΔF /K(U₁)=1/0.178=0.1371

u₂=u₁+Δu₂=0.315

Fₑₓₜ =ΔF+ΔF = 2

Fᵢₙₜ=(U₂)³+9(U₂)²+4U₂

=2.18

R⁽⁰⁾=Fᵢₙₜ -Fₑₓₜ =0.18

|R⁽¹⁾|>10⁻²

U₁=U₁⁽⁰⁾

ΔU⁽¹⁾=-[K(u₂⁽⁰⁾)]⁻¹.R⁽⁰⁾

=-0.0666- (7.404)⁻¹*0.042

=-0.018

U₂⁽¹⁾=U₂⁽⁰⁾+Δu⁽¹⁾=0.315-0.018=0.297

Residual

Fₑₓₜ = ΔF+ΔF= 1+1=2

Fᵢₙₜ = (U₂⁽¹⁾)³+9(U₂⁽¹⁾)²+4U₂⁽¹⁾=

=(0.178)³+9(0.178)²+4(0.178)

=2.00

R⁽¹⁾=Fᵢₙₜ -Fₑₓₜ =0.004

|R⁽⁰⁾|>10⁻² -> No further iterations are necessary.

Therefore the final value is u₂=u₂⁽¹⁾=0.297

STEP - 3

U₂⁽¹⁾=0.297,

K(U₂⁽¹⁾)=9.610,

ΔF= 1,

ΔU₃=ΔF /K(U₂)=0.1040,

U₃=U₂+ΔU₃=0.401

Checking the residual |R|

Residual

Fₑₓₜ = ΔF+ΔF+ΔF= 1+1+1=3

Fᵢₙₜ = (U₃⁽³⁾)³+9(U₃⁽³⁾)²+4U₃⁽³⁾

=(0.178)³+9(0.178)²+4(0.178)

=3.1156

R⁽⁰⁾=Fᵢₙₜ -Fₑₓₜ =0.1156

|R⁽⁰⁾|>10⁻² -> Newton Raphson iterations are necessary.

Therefore the final value is u₂=u₂⁽¹⁾=0.297

k(u) = df/du = 3u²+18u+4

sub U₃=0.401 in the above Eqn we get, K(u₃⁽⁰⁾) = 11.698

ΔU⁽²⁾=-[K(u₃⁽⁰⁾)]⁻¹.R⁽⁰⁾ = -(11.698)⁻¹*0.1156= -0.00988203

U₃⁽¹⁾=U₃⁽⁰⁾+Δu⁽²⁾=0.401-0.00988203=0.3911

Residual,

Fₑₓₜ = ΔF+ΔF+ΔF= 1+1+1=3

Fᵢₙₜ = (U₃⁽¹⁾)³+9(U₃⁽¹⁾)²+4U₃⁽¹⁾=3.0008

R⁽¹⁾=Fᵢₙₜ -Fₑₓₜ =3-3.0008=0.0008

|R⁽¹⁾|>10⁻² , No further iterations are necessary

U₃= U₃⁽²⁾=0.3911

Iterations	ΔF	Ui	Fₑₓₜ	(Fmt)i	R=Fᵢₙₜ -Fₑₓₜ
1	1	0.178	1	1.0027	0.0027
2	1	0.297	2	2.004	0.004
3	1	0.3911	3	3.0008	0.0008

Implicit Vs Explicit Analysis Graph:

RESULT :

CONCLUSION:

• 3rd Order Polynomial, for a simple Bar under a tensile force, is analytically solved both in the Implicit and Explicit analysis is done and e plotted.

• Implicit is detailed and Time-consuming and more accurate and depends on element size and time steps

• The explicit analysis is faster but less accurate and needs more iterations to obtain good results.