Week - 2 - Explicit and Implicit Analysis

EXPLICIT AND IMPLICIT ANALYSIS

AIM= To solve the given problem by both an Explicit and Implicit analysis.

Time Step = We divide the total response time into much smaller time intervals called Time Step or Increment. The equilibrium equations are solved and the value of unknown are determined at (t+Δt) based on the knowledge of their values at time t.

Explicit Method:

• Explicit methods are those in which the information at time step n+1 can be obtained in terms of previous time steps and there is no dependence on the current time step.
• Time steps are smaller of the order of 1e-06 sec for a mesh size of 5mm.
• Conditional stability governed by CFL condition.
• In explicit method computational speed is very fast as there is no matrix inversion. Mass matrix is lumped or diagonal.
• The maximum size of computational problem is bounded by number of CPU's.
• Central difference numerical scheme is used in explicit method.
• Linear to highly nonlinear problems can be efficiently handled. This is the very reason that we see explicit codes being used for crash analysis.
• Explicit codes are best suitable for wave propagation or hyperbolic problem.
• In explicit analysis while filtering of frequencies no significant effect observed.

Implicit Method:

• Implicit methods are those where the information at time step n+1 is dependent on previous time levels (n, n-1,.....) and the current i.e. n+1.
• Usually large time steps of the order of 1e-03 sec are used.
• Large time steps often lead to lot of numerical oscillations. Hence Unconditionally stable.
• In implicit method computational speed is equally fast as one can take larger time steps.
• The maximum size of computational problem is bounded by the input and output.
• Newmark beta integration and Hilbert-Hughes-Taylor numerical scheme used in implicit method.
• Linear- moderate nonlinear problems can be efficiently handled.
• The most suitable application are static to vibration problems such as frequency response and quasi- static simulations.
• In implicit analysis while filtering of frequencies removal of high frequency component observed.

Given Problem:

F(u) = u³+9u²+4u

Explanation:

Therefore,

K(u) = 3u²+18u+4

The Solution for above problem by both Explicit and Implicit methods are as follows:

Explicit Method:

For explicit method three equal load steps of ΔF=1 unit each are used to load the tension bar.

The variables used are u=Displacement, f=Internal force, F=External force applied to the bar and K=Stiffness. Incremental displacement and incremental externally applied forces are represented as Δu and ΔF respectively.

Hence, ΔF=kΔu. The analysis proceeds as follows:

Step-1:

u0=0.0, k(u0)=1, ΔF=1, Δu= ΔF/k(u0)=0.25, u1=u0+Δu1=0.25

F_external = ΔF=1

F_internal = u³+9u²+4u = 1.5781 (u=u1)

Step-2:

u1=0.25, k(u1)=8.675, ΔF=1, Δu= ΔF/k(u1)=0.1151, u2=u1+Δu2=0.3611

F_external = ΔF+ ΔF=2

F_internal = u³+9u²+4u = 2.7088 (u=u2)

Step-3:

u2=0.3651, k(u2)=10.9718, ΔF=1, Δu= ΔF/k(u2)=0.0911, u3=u2+Δu3=0.4562

F_external = ΔF+ ΔF+ ΔF =3

F_internal = u³+9u²+4u = 3.7934 (u=u3)

From above results it is clear that external and internal forces are not in equilibrium because

F_{external ≠} F_internal

The following table shows the results for Explicit analysis method:

Step	ΔF	Δu	u	Fexternal	Finternal	R=Fint-Fext
1	1	0.25	0.25	1	1.578125	0.578125
2	1	0.115108	0.365108	2	2.708836	0.708836
3	1	0.091142	0.45625	3	3.793454	0.793454

Implicit method:

An implicit analysis is the same as the explicit analysis, except that at the end of each step Newton-Raphson iterations are used to enforce equilibrium before moving to the next step.

An incremental force is applied to advance the solution forward at the beginning of a step. However, internal forces and external forces will not be in equilibrium unless the stiffness is linear for the given step.

Hence, in order to achieve equilibrium, corrections must be made to the displacement.

This is accomplished by using Newton-Raphson iterations to minimize the residual

R = F_internal - F_external,

Expanding the residual as a Taylor series ab=out the current displacement u^j gives

R^j+1 = R^j (u^j) + `dR(uj)/du` . δu^j+1 +………. = R^j + k(u^j). δu^j+1 + ……..

By neglecting higher order terms in the series, setting R^j+1 = 0 and solving for δu^j+1 the following correction is obtained.

δu^j+1 = - [k(u^j)]^-1 . R^j

Here, the iteration variable is j and it may take several iterations for the norm of the residual, R, to be reduced below the chosen tolerance criteria of 0.01.

The implicit analysis proceeds as follows:

Step-1:

u0=0.0, k(u0)=4, ΔF=1, Δu1= ΔF/k(u0)= 0.25, u1=u0+Δu1=0.25

F_external = ΔF=1

F_internal = u1³+9u1²+4u1 = 1.5781

Checking the residual

Residual R1⁽⁰⁾ = F_internal - F_external = 0.578125

R1⁽⁰⁾ > 10^-2 → Newton Raphson iteration are necessary.

Calculation of correction u1 = u1⁽⁰⁾

δu⁽¹⁾ = - [k(u1⁽⁰⁾)]^-1 . R1⁽⁰⁾ = - (8.6875)^-1 . 0.578125 = - 0.06654

→ u1 ⁽¹⁾ = u1 ⁽⁰⁾ + δu⁽¹⁾ = 0.25 - 0.06554 = 0.1834

Checking the residual again,

F_external = ΔF=1

F_internal = (u1⁽¹⁾)³+9(u1⁽¹⁾)²+4u1⁽¹⁾ = 1.04288

Residual R1⁽¹⁾ = F_internal - F_external = 0.04288

R1⁽¹⁾ > 10^-2 → Newton Raphson iteration are necessary.

Calculation of correction u1 = u1⁽⁰⁾

δu⁽²⁾ = δu⁽¹⁾ - [k(u1⁽¹⁾)]^-1 . R1⁽¹⁾ = -0.06654 - (7.4031)^-1 . 0.04288 = - 0.0723

→ u1 ⁽²⁾ = u1 ⁽⁰⁾ + δu⁽²⁾ = 0.25 - 0.0723 = 0.1776

Checking the residual again,

F_external = ΔF=1

F_internal = (u1⁽²⁾)³+9(u1⁽²⁾)²+4u1⁽²⁾ = 1.0003

Residual R1⁽²⁾ = F_internal - F_external = 0.0003

R1⁽²⁾ < 10^-2 → Newton Raphson iteration are not necessary.

The final value of u1 = u1⁽²⁾ = 0.1776

Step-2:

u1=0.1776, k(u1)=7.2925, ΔF=2, Δu2=ΔF/k(u1)= 0.1371, u2=u1+Δu2=0.3147

F_external = ΔF + ΔF =2

F_internal = u2³+9u2²+4u2 = 2.1821

Checking the residual

Residual R2⁽⁰⁾ = F_internal - F_external = 0.1822

R2⁽⁰⁾ > 10^-2 → Newton Raphson iteration are necessary.

Calculation of correction u1 = u1⁽⁰⁾

δu⁽¹⁾ = - [k(u2⁽⁰⁾)]^-1 . R2⁽⁰⁾ = - (9.9634)^-1 . 0.1821 = - 0.01828

→ u2 ⁽¹⁾ = u2 ⁽⁰⁾ + δu⁽¹⁾ = 0.3147 - 0.01828 = 0.2965

Checking the residual again,

F_external = ΔF + ΔF =2

F_internal = (u2⁽¹⁾)³+9(u2⁽¹⁾)²+4u2⁽¹⁾ = 2.003

Residual R2⁽¹⁾ = F_internal - F_external = 0.0033

R2⁽¹⁾ < 10^-2 → Newton Raphson iteration are not necessary.

The final value of u2 = u2⁽¹⁾ = 0.2965

Step-3:

u1=0.2965, k(u1)=9.6008, ΔF=1, Δu3=ΔF/k(u2)= 0.1041, u3=u2+Δu3=0.4006

F_external = ΔF + ΔF + ΔF =3

F_internal = u3³+9u3²+4u3 = 3.1117

Checking the residual,

Residual R3⁽⁰⁾ = F_internal - F_external = 0.1117

R3⁽⁰⁾ > 10^-2 → Newton Raphson iteration are necessary.

Calculation of correction u1 = u1⁽⁰⁾

δu⁽¹⁾ = - [k(u3⁽⁰⁾)]^-1 . R3⁽⁰⁾ = - (11.6942)^-1 . 0.1117 = - 0.00956

→ u3 ⁽¹⁾ = u3 ⁽⁰⁾ + δu⁽¹⁾ = 0.4006 - 0.00956 = 0.3911

Checking the residual again,

F_external = ΔF + ΔF + ΔF =3

F_internal = (u3⁽¹⁾)³+9(u3⁽¹⁾)²+4u3⁽¹⁾ = 3.0009

Residual R3⁽¹⁾ = F_internal - F_external = 0.00093

R3⁽¹⁾ < 10^-2 → Newton Raphson iteration are not necessary.

The final value of u3 = u3⁽¹⁾ = 0.3911

The results of the implicit analysis are summarized below in the table

Step	ΔF	u	Fexternal	Finternal	R=Fint-Fext
1	1	0.1776607	1	1.0003202	0.00032
2	1	0.2965043	2	2.0033317	0.00331
3	1	0.3911065	3	3.0009326	0.00093

Conclusion:

A nonlinear analysis problem is solved using an incremental explicit and implicit procedure. It is evident from the results that the explicit analysis drifts from the true solution. To overcome this problem an implicit analysis is used which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces.