Week - 2 - Explicit and Implicit Analysis
Explicit vs Implicit-
Explicit FEM is used to calculate the state of a given system at a different time from the current time. In contrast, an implicit analysis finds a solution by solving an equation that includes both the current and later states of the given system. This method requires additional computation and can be harder to implement. However, it will be used in lieu of explicit methodologies when problems are still, and using alternative analysis methods is impractical.
Explicit Analysis-
The explicit analysis offers a faster solution in events where there is a dynamic equilibrium or otherwise:
Sum of all forces = mass x acceleration
The explicit method should be used when the strain rates/velocity is over 10 units/second or 10 m/s respectively. These events can be best exampled by extreme scenarios such as an automotive crash, ballistic event, or even meteor impact. In these cases, the material models do not only need to account for the variation of stress with strain but also the strain rate. On this scale, the strain rates play a particularly important contribution.
Implicit Analysis-
The implicit method should be used when the events are much slower and the effects of strain rates are minimal. Once the growth of stress as a function of strain can be established, these can be analyzed using implicit methods. In this case, one can consider a static equilibrium such that:
Sum of all forces = 0
This covers many of the most common engineering problems.
Problem-
F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods ( have a tolerance of 10-2.
1- Solving with Explicite Method-
`F(u)=u^3+9u^2+4u`
Differentiate force with respect to displacemnt-
`k(u)=(dF)/(du)=3u^2+18u+4`
Here `DeltaF`=1 for used for every step
stiffness -force relation used to find incremental displacemt-
`DeltaF=k.Deltau`
Step-1-
`u0=0, k(u0)= 4, DeltaF=1, Delta u1= (DeltaF)/(k(u0))=0.25, u1= u0+Delta u1= 0.25`
`Fexternal (DeltaF)= 1`
`Fternal= u^3+9u^2+4u= 1.5781` at u=0.2
Step-2-
`u1=0.25, k(u1)= 8.688, DeltaF=1, Delta u2= (DeltaF)/(k(u1))=0.25, u2= u1+Delta u2= 0.3651`
`Fexternal (DeltaF)= DeltaF+DeltaF=1+1=2`
`Fternal= u^3+9u^2+4u= 2.708` at u= 0.3651
Step-3-
`u2=0.3651, k(u2)= 10.98, DeltaF=1, Delta u3= (DeltaF)/(k(u2))=0.0911, u3= u2+Delta u3= 0.4562`
`Fexternal (DeltaF)= DeltaF+DeltaF+DeltaF=1+1+1=3`
`Fternal= u^3+9u^2+4u= 3.7929` at u= 0.4562
1So from these results we can easily see that the Fexternal is not equal to Finternal
Table for the output-
| Step | ΔF | u | Δu | F ext | Fint | R= Fint-Fext |
| 1 | 1 | 0.25 | 0.25 | 1 | 1.578 | 0.578 |
| 2 | 1 | 0.3651 | 0.1151 | 1+1=2 | 2.708 | 0.708 |
| 3 | 1 | 0.4562 | 0.0911 | 1+1+1=3 | 3.792 | 0.792 |
1- Solving with Implicite Method-
So in this analysis, we will use the Newton-Raphson iteration for each step before moving to the next step for 10^-2 places. These changes made to reduce the error in solution and make the forces close to equilibrium, By minimization of Residue R= Finternal-Fexternal.
For this Residue will expand in the form of Taylors Series with neglecting the higher-order,
`deltau=-[k(u)]^-1.R`
Step-1-
`u0=0, k(u0)= 4, DeltaF=1, Delta u1= (DeltaF)/(k(u0))=0.25, u1= u0+Delta u1= 0.25`
`deltau^(1)=-[k(u^(0))]^-1.R^(0)= -(8.6874)^(-1).0.578= -0.0665`
the updated value= 0.25-0.0665= 0.1835
R1= 0.0432
`deltau^(2)=delta(u)^(1)-[k(u^(1))]^-1.R^(1)= -0.0665-(7.404)^(-1).0.0432= -0.0724`
`u1^(2)= u1^(0)+deltau^(2)=0.25-0.0723=0.1777`
Again Iteration with this we can get the value of minimum error= 0.00059
From this value, we can get the value of u1= 0.177
Step-2- and Step-3 can be calculated by the same procedure,
| Step | ΔF | u | Fext | Fint | R= Fint-Fext |
| 1 | 1 | 0.1778 | 1 | 1.0006 | 0.0006 |
| 2 | 1 | 0.2966 | 1+1=2 | 2.0041 | 0.0041 |
| 3 | 1 | 0.3912 | 1+1+1=3 | 3.0009 | 0.0009 |
Here we can see in this analysis the force is so close to equilibrium because we did 2-3 times iteration.
Discussion and Conclusion-
1- As we can see the explicit analysis is drifting from the exact solution.
2- Implicit analysis is the best analysis for minimum error as compared to explicit analysis.
3- To overcome the drift issue we come with the Implicit analysis.
4- We can use Implicit and Explicit analysis for the same problem, the on,y difference is the time increment.