Week - 2 - Explicit and Implicit Analysis

EXPLICIT AND IMPLICIT METHODS IN FINITE ELEMENT ANALYSIS

OBJECTIVE:

1. To compare the implicit and explicit methods in the finite element analysis.
2. To solve the equation `u^3+9*u^2+4*u` using impicit and explicit methods.
3. To plot and compare the results of implicit, explicit and exact solutions.

INTRODUCTION:

For all nonlinear and non-static analyses, incremental load (also known as displacement steps) are needed. This means we need to break down the physics/time relationship to solve a mathematical problem. To do this, we form two groups: either time-dependent or time-independent problems. To solve these problems, we commonly use ‘implicit’ and/or ‘explicit’ methods.

EXPLICIT METHOD:

1. Explicit methods calculate the state of a system at a later time from the state of the system at the current time.
2. Acceleration is assumed to be constant over timestep. Therefore, the timestep has to be very small.
3. Requires the inversion of mass matrix.

IMPLICIT METHOD:

1. Implicit methods find a solution by solving an equation involving both the current state of the system and the later one.
2. Acceleration is assumed to vary linearly over timestep.
3. Large timestep size can applied.
4. Requires the inversion of stiffness matrix.

COMPARISON BETWEEN IMPLICIT AND EXPLICIT METHODS IN FINITE ELEMENT ANALYSIS:

IMPLICIT VS EXPLICIT METHODS IN FEA

	IMPLICIT METHOD	EXPLICIT METHOD
Time step size	Large	Small
What limits the time step	Accuracy	Stability
Stiffness Matrix Inversion	Yes	No
Deformation	Small	Large
Main cost factor	Matrix Inversion	Elemental processing
Cost for doubling Mesh density	x 128	x 16
Cost per time step	Large	Small
Acceleration	Vary Linearly over timestep	Assumed to be constant over timestep

EXAMPLE:

Consider the case of a simple bar in tension as shown in Figure. Suppose that the force
in the bar is a nonlinear function of the displacement

`F(u)=u^3+9u^2+4u`

The stiffness of the bar is

`k(u)=(df)/(du)=3u^2+18u+4`

1)Explicit method:

`k(u)=(df)/(du)=3u^2+18u+4`

Three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar. 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 ∆u or ∆F, respectively. Use is made of the relationship ∆F = k∆u. The analysis proceeds as follows:

Step 1:

`∆F = k∆u`

`u0=1 , k(u)=3(0)+18(0)+4=4`

`∆F=1`

`∆u1=`∆F/k(u0)=0.25`

`u1=u0+∆u1=0+0.25=0.25`

Step 2:

`k(u1)=3*0.25^2+18*(0.25)+4=8.6875`

`∆F=1`

`∆u2=(∆F)/(k(u1))=1/(8.68750)=0.11510`

`u2=u1+∆u2=0.25+0.11510=0.3651`

Step 3:

`k(u2)=3*0.3651^2+18*(0.3651)+4=10.9716`

`∆F=1`

`∆u3=(∆F)/(k(u2))=1/(10.9716)=0.09114`

`u3=u2+∆u3=0.3651+0.09114=0.4562`

Finally, from the three steps the following results are obtained.

`underset(ext)(∆F)=∆F+∆F+∆F=3`

`f_1=u_1^3+9*u_1^2+4*u_1=1.5781`

`f_2=u_2^3+9*u_2^2+4*u_2=2.7087`

`f_3=u_3^3+9*u_3^2+4*u_3=3.7928`

It is evident from the results that external and internal forces are NOT in equilibrium because `F_(ext) != f_i`

Summary of explicit analysis results

Step i	`triangleF`	`triangleu_i`	`u_i`	`(F_(ext))_i`	`f_i`	`R=f_i-(F_(ext))_i`
1	1	0.25	0.25	1	1.5781	0.5781
2	1	0.1151	0.3651	2	2.7087	0.7087
3	1	0.09114	0.4562	3	3.7928	0.7928

2)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.
Basically, 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. The correction obtained is:

`deltau_(j+1) = −[k(u_ j )]^(−1)R _j`

Step 1:

`∆F = k∆u`

`u0=1 , k(u)=3(0)+18(0)+4=4`

`∆F=1`

`∆u1=`∆F/k(u0)=0.25`

`u1=u0+∆u1=0+0.25=0.25`

Check the residual, `R^0=f_(Int) − Fext rightarrow 1.5781-1=0.5781`

`||R||>10^(-2)``

⇒ Newton-Raphson iterations are necessary.

Calculate the correction to `u_1=u_1^0`

`deltau_(j+1) = −[k(u_ j )]^(−1)R _j`

`deltau^1=-8.6875^-1*0.5781=-0.06654`

⇒ The updated value of `u_1^1=0.25-0.06654=0.18346`

Check the residual again, `R^1=1.0424-1=0.0424`

`||R||>10^(-2)`

⇒ another Newton-Raphson iteration is necessary.

`deltau_(j+1) = deltau_j−[k(u_ j )]^(−1)R _j`

`deltau^2=-0.0665-(3*0.1834^2+18*0.1834+4)^-1*0.0424=-0.0722`

⇒ The updated value of

`u_1^2=0.25-0.0722=0.1778`

Check the residual again,

`R^2=(0.1778^3+9*0.1778^2+4*0.1778)-1=0.00133`

`||R||<10^(-2)``

Therefore, the value of `u_1=0.1778`

Step2:

`u_1=0.1778`

`k(u_1)=7.295` `triangleF=1`

`triangleu_2=0.137`

`u_2=u_1+triangleu_2=0.1778+0.137=0.3148`

Check the residual,

`f_(Int)=2.182`

`R^0=f_(Int) − Fext rightarrow 2.182-2=0.182`

`||R||>10^(-2)`

Therefore, Newton-Raphson iterations are necessary.

Calculate the correction to

`u_2=u_2^0`

`deltau^(j+1) = −[k(u_ j )]^(−1)R _j`

`deltau^1=-0.0182`

`u_2^1=0.296`

`R^0=f_(Int) − Fext rightarrow (0.296^3+9*0.269^2+4*0.296)-2=0.00522`

`||R||<10^(-2)``

Therefore, the value of

`u_2=0.0.296`

Step 3:

`u_2=0.296, k(u_2)=9.6, triangle F=1,`

`triangle u_3=0.104 rightarrowu_3=u_2+triangleu_3=0.296+0.104=0.4004`

Check the residual,

`R^0=f_(Int) − Fext rightarrow (0.4004^3+9*0.4004^2+4*0.4004)-3=0.108`

`||R||>10^(-2)`

Therefore, Newton-Raphson iterations are necessary.

Calculate the correction to

`u_3=u_3^0`

`deltau^(j+1) = −[k(u_ j )]^(−1)R _j`

`deltau^1=-(3*0.4004^2+18*0.4004+4)^-1*0.108=-0.0092`

`u_3^1=0.4004-0.0092=0.3912`

`R^1=f_(Int) − Fext rightarrow (0.3912^3+9*0.3912^2+4*0.3912)-3=0.0025`

`||R||<10^(-2)``

Therefore, the value of

`u_3=0.3912`

Summary of implicit analysis results

Step i	`triangleF`	`u_i`	`(F_(ext))_i`	`f_i`	`R=f_i-(F_(ext))_i`
1	1	0.1778	1	1.0013	0.0013
2	1	0.296	2	2.0052	0.0052
3	1	0.3912	3	3.0025	0.0025

Plots of Results:

CONCLUSION:

Implicit and Explicit refer to two types of time integration methods used to perform dynamic simulations. Explicit time integration is more accurate and efficient for simulations involving – Shock wave propagation – Large deformations and strains – Non-linear material behaviour – Complex contact – Fragmentation – Non-linear buckling. Typical applications are drop tests and impact and penetration.