Week - 2 - Explicit and Implicit Analysis

AIM

To solve this equation, F(u) = u3+9u2+4u using both Explicit and Implicit Methods ( have a tolerance of 0.001).

F = Force ; u = displacement

OBJECTIVE:

To solve the given equation using implicit and explicit methods with a residual tolerance of 0.01.

Equation: F(u)=u3+9u2+4u

F- Force in the bar, u- Displacement in the bar.

INTRODUCTION

1.  Preliminary comments regarding the incremental nature of Explicit and Implicit Analysis

A geometric and/or material nonlinear analysis requires incremental load (or displacement) steps.  At the end of each increment the structure geometry changes and possibly the material is nonlinear or the material has yielded.  Each of these things, geometry change or material change, may then need to be considered as you update your stiffness matrix for the next increment in the analysis.

2.  Explicit

An Explicit FEM analysis does the incremental procedure and at the end of each increment updates the stiffness matrix based on geometry changes (if applicable) and material changes (if applicable).  Then a new stiffness matrix is constructed and the next increment of load (or displacement) is applied to the system.  In this type of analysis the hope is that if the increments are small enough the results will be accurate.  One problem with this method is that you do need many small increments for good accuracy and it is time consuming.  If the number of increments are not sufficient the solution tends to drift from the correct solution.  Futhermore this type of analysis cannot solve some problems.  Unless it is quite sophisticated it will not successfully do cyclic loading and will not handle problems of snap through or snap back.  Perhaps most importantly, this method does not enforce equilibrium of the internal structure forces with the externally applied loads.

3.  Implicit

An Implicit FEM analysis is the same as Explicit with the addition that after each increment the analysis does Newton-Raphson iterations to enforce equilibrium of the internal structure forces with the externally applied loads.  The equilibirium is usually enforced to some user specified tolerance.  So this is the primary difference between the two types of anlysis, Implicit uses Newton-Raphson iterations to enforce equilibrium.  This type of analysis tends to be more accurate and can take somewhat bigger increment steps.  Also, this type of analysis can handle problems better such as cyclic loading, snap through, and snap back so long as sophisticated control methods such as arc length control or generalized displacement control are used.  One draw back of the method is that during the Newton-Raphson iterations one must update and reconstruct the stiffness matrix for each iteration.  This can be computationally costly.  (As a result there are other techniques that try to avoid this cost by using Modified Newton-Raphson methods.)  If done correctly the Newton-Raphson iterations will have a quadratic rate of convergence which is very desireable.

 A suggestion.  If you'd like to learn further about these two techniques it would be instructive for you to use both techniques and compare on the same problem.  Explicit can be done by simply turning off Newton-Raphson iterations in an Implicit routine, or by setting the equilibrium tolerance to a large number in an implicit routine.

As to the question of which method to use, the answer is that it depends.  The type of analysis that is sufficient for your needs will depend on the type of problem that you are trying to solve.  Often times since dynamic analyses are computationally intensive they are done with the explicit method.  However, for static problems now days it is becoming more common to do the full Implicit type of analysis.

Nonlinear analysis takes lots of experience and a careful understanding of what you want to accomplish and also a careful understanding of the anlaysis capabilities of the software you are trying to use.  As I mentioned I have worked with the above methods of analysis in graduate school and know a little about it, however, I would be happy for others here at iMechanica who have more experience than me to give their thoughts on this as well.

EXPLICIT/IMPLICIT METHOD

Explicit and implicit methods are mainly used to solve dynamic events with a governing equation [M]{x''}+[C]{x'}+[K]{x}={f}. Implicit and Explicit analysis differ in the approach to time incrementation. In Implicit analysis each time increment has to converge, but you can set pretty long time increments. Explicit on the other hand doesn’t have to converge each increment, but for the solution to be accurate time increments must be super small. This equation can also be used to solve static problems by considering the mass and damping matrix as null matrix. But solving this algorithm with this method are expensive as compared to direct static schemes.

IMPLICIT METHOD

• In implicit scheme the unknown factor (x) is found out by inversion of stiffness matrix.
• For each increment of load the stiffness matrix is updated (If needed mass matrix is also updated). Then an incremental load will be defined and when the governing equation is defined for the next increment , Newton Raphson method is used to solve the problem using iterative method.
• Thus this method gives more accurate results when compared to explicit method.
• This method is preferable when the strain rates are minimum.
• Implicit schemes are time consuming for large models and are unconditionally stable.
• Here there is no mathematical time limit for the solution and requires much more storage space in computer.

Applications:

 Good choice for structure dynamics problems for which load rate is slow. E.g.: Low frequency response, vibration, oscillation etc.

EXPLICIT METHOD

• In explicit scheme the unknown factor (x) is found out by inversion of mass matrix.
• Here also the matrix are updated and an incremental load is defined but like that of implicit method there is no need of iteration process, direct solution is possible.
• Less accurate when compared to implicit scheme.
• This method is preferable when the strain rates are high and can solve extremely discontinuous events or process.
• Time taken by the explicit analysis depends on wave propagation speed and characteristic length of the element and can predict the time taken for the analysis.

Applications:

For wave propagation types of problems such as crash analysis, drop, blast etc.

CALCULATION:

Using explicit method:

• Equation: F(u)=u3+9u2+4u
  
  
• If the nonlinear force versus displacement relationship is not known, but the stiffness is known, it is possible to construct the graph of force versus displacement numerically. 
• One method is to use an incremental explicit load control scheme. Here three equal load increments (or steps) of ΔFΔF unit each are used to load the tension bar.
• Differentiating the force with respect to displacement will give the stiffness. (F=kx)
• k(u)=dF/du=3u2+18u+4
• The relation ΔF=kΔu can be used to find the incremental displacement.
• F - External force.
• u - Displacement.
• k - Stiffness
• ΔF - Incremental force.
• Δu - Incremental displacement.

Step 1:

• At u0=0⇒k(u0)=4 , ΔF=1
• Δu1=ΔF/k(u0)=1/4=0.2514=0.25
• u1=u0+Δu1=0.25u1=u0+Δu1=0.25
• FExternal=ΔF=1
• FInternal=u3+9u2+4u=1.578
• uexcat=0.177(Solving, Fexternal=1=u3+9u2+4u)

Step 2:

•  u1=0.25⇒k(u1)=8.687,ΔF=1
• Δu2=ΔF/k(u1)=1/8.687=0.115
• u2=u1+Δu2=0.25+0.115=0.365
• FExternal=ΔF+ΔF=1+1=2
• FInternal=u3+9u2+4u=2.708
• uexcat=0.296 `(Solving, Fexternal=2=u3+9u2+4u)

Step 3:

•  u2=0.365⇒k(u2)=10.969,ΔF=1
• Δu3=ΔF/k(u2)=1/10.969=0.091
• u3=u2+Δu3=0.456
• FExternal=ΔF+ΔF+ΔF=1+1+1=3
• FInternal=u3=9u2+4u=3.790
• uexcat=0.391 (Solving, Fexternal=3=u3+9u2+4u)

Table 1: Summary of explicit analysis results

• It is evident from the results that external and internal forces are not in equilibrium because FExternal≠FInternal.

Using 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. This is accomplished by using Newton-Raphson iterations to minimize the residual, R(u)=FInternal−FExternal. Expanding the residual as a Taylor series about the current displacement uj gives,
• Rj+1=Rj(uj)+(dR(uj)/du).δuj+1+...=Rj+k(uj)δuj+1+...
• By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for δuj+1 the following correction is obtained.
• δuj+1=−[k(uj)]−1Rj
• Notice that the iteration variable is j and it may take several iterations for the norm of the residual, , to be reduced below the chosen tolerance criteria (for the example below a tolerance of 0.01 is chosen). The implicit analysis proceeds as follows (results are approximate, they are more precise if done in a computer to machine precision):

Step 1:

• At u0=0⇒k(u0)=4 , ΔF=1
• Δu1=ΔFk/(u0)=1/4=0.25
• u1=u0+Δu1=0.25
• Checking the residual,
• FExternal=ΔF=1
• FInternal=u^3+9u^2+4u=1.578
• R(0)=FInternal−FExternal=0.578
• R(0)>0.01⇒Newton Raphson iterations are necessary
• Calculating the correction to u1=u(0)1
• δu(1)=−[k(u(0)1)]^−1.R(0)=−(8.687)^−1∗(0.578)=0.066
• The updated value of u(1)1=u(0)1+δu(1)=0.25−0.066=0.184
• Checking the residual again,
• FExternal=ΔF=1
• FInternal=u^3+9u^2+4u, Putting (u=0.184) gives,
• FInternal=1.046
• R(1)=FInternal−FExternal=0.046
• R(1)>0.01⇒Newton Raphson iterations are necessary
• Calculating the new total correction to u1=u(0)
• δu(2)=δu(1)−[k(u(1)1)]−1R(1)=−0.066−(7.413)−1∗0.046=−0.072
• The updated value of u(2)1=u(0)1+δu(2)=0.25−0.072=0.178
• Checking for the residuals again,
• FExternal=ΔF=1
• FInternal=u^3+9u^2+4u Putting (u=0.178) gives,
• FInternal=1.002
• R(2)=FInternal−FExternal=0.002
• R2<0.01⇒Newton-Raphson iterations are not necessary
• ∴The final value is u1=u(2)1=0.178

Step 2:

• u1=0.178⇒k(u1)=7.299,ΔF=1
• Δu2=ΔF/k(u1)=1/7.299=0.137
• u2=u1+Δu2=0.178+0.137=0.315
• Checking the residuals,
• FExternal=ΔF+ΔF=1+1=2
• FInternal=u3+9u2+4u=2.184
• R(0)=FInternal−FExternal=0.184
• R(0)>0.01⇒Newton Raphson iterations are necessary
• Calculating the corrections to u1=u(0)0
• δu(1)=−[k(u(0)1)]−1R(0)=−(9.967)−1∗0.184=−0.018
• The updated value of u(1)2=u(0)1+δu(1)=0.315−0.018=0.297
• Checking the residuals again,
• FExternal=ΔF+ΔF=1+1=2
• FInternal=u^3+9u^2+4u=2.008
• R(0)=FInternal−FExternal=0.008
• R1<0.01⇒Newton-Raphson iterations are not necessary
• ∴The final value is u2=u(1)1=0.297

Step 3:

• u2=0.297⇒k(u2)=9.6106,ΔF=1
• Δu3=ΔF/k(u2)=19.6106=0.10405
• u3=u2+Δu3=0.297+0.10405=0.40105
• Checking the residuals,
• FExternal=ΔF+ΔF+ΔF=1+1+1=3
• FInternal=u3+9u2+4u=3.1162
• R(0)=FInternal−FExternal=0.1162
• R(0)>0.01⇒Newton Raphson iterations are necessary
• Calculating the corrections to u2=u(0)2
• δu(2)=−[k(u(0)3)]−1R(0)=−(11.701)−1∗0.1162=−0.00993
• u(1)3=u(0)3+δu(2)=0.40105−0.00993=0.39112
• Checking the residuals again,
• FExternal=ΔF+ΔF+ΔF=1+1+1=3
• FInternal=u^3+9u^2+4u=3.001
• R(0)=FInternal−FExternal=0.001
• R1<0.01⇒Newton-Raphson iterations are not necessary
• ∴The final value is u3=u(2)3=0.391

Table 2: Summary of implicit analysis results

It is evident from the table that the external forces and internal forces are coming into equilibrium.

RESULTS:

• The given equation have been solved using explicit and implicit methods to find out the variation in displacements and compared with the exact displacement. 
• The values have been tabulated and plotted graph between Force vs Displacement.

Table 3:

Graph: 

CONCLUSION:

• It is evident from the plot that the explicit analysis drifts from the exact solution. If 20 increments are used the explicit analysis more closely follows the exact solution. However, even with more increments the explicit analysis still drifts from the exact solution. The drift from the exact solution illustrates the lack of equilibrium between the internal and external forces. To correct for this problem an implicit analysis is required. The Newton-Raphson iterations correct the incremental steps so that they land on the exact solution according to the specified tolerance also an excellent agreement is achieved with just 3 increments. From the graph we can see that the implicit and exact results lie very close to each other.
• Here the problem is considered only for the load control scheme. For other schemes such as displacement control, arc length control etc. can also be carried out with this method.