Week - 2 - Explicit and Implicit Analysis
1. Write a detailed report about Implicit and explicit methods.
In most structural problems, the nonlinear equations fall into 3 categories:
- Material Nonlinearity: Where deformations and strains are large
- Geometric Nonlinearity: Where strains are small, but rotations are large
- Boundary Nonlinearity: Due to non-linearity of boundary conditions,
In linear problems, the PDE’s reduce to a matrix equation as:
[K]{x} = {f}
and for non-linear static problems as:
[K(x)]{x} = {f}
For dynamic problems, the matrix equations come down to:
`[M]{x´´} + [C]{x´} + [K]{x} = {f}`
IMPLICIT
One method of solving for the unknowns {x} is through matrix inversion (or equivalent processes). This is known as an implicit analysis. When the problem is nonlinear, the solution is obtained in a number of steps and the solution for the current step is based on the solution from the previous step. For large models, inverting the matrix is highly expensive and will require advanced iterative solvers (over standard direct solvers). Sometimes, this is also known as the backward Euler integration scheme These solutions are unconditionally stable and facilitate larger time steps. Despite this advantage, the implicit methods can be extremely time-consuming when solving dynamic and nonlinear problems.
EXPLICIT
Explicit analyses aim to solve for acceleration (or otherwise {x´´}). In most cases, the mass matrix is considered as “lumped” and thus a diagonal matrix. Inversion of a diagonal matrix is straightforward and includes inversion of the terms on the diagonal only. Once the accelerations are calculated at the nth step, the velocity at n+1/2 step and displacement at n+1 step are calculated accordingly. In these calculations, the scheme is not unconditionally stable and thus smaller time steps are required. To be more precise, the time step in an explicit finite element analysis must be less than the courant time step (i.e., the time taken by a sound wave to travel across an element) while implicit analyses have no such limitations.
WHAT TO DO IN CAE
implicit scheme
`{D}n+1 =f( {dotD}n+1, {ddot D}n+1, {D}n , {dot D}n, {ddot D}n,....)` -(1)
`[K ^(off)]``{D}n+1 = `{R ^(ext)}n+1`+[ M]`{1//β/_\t^2 { D }n + 1 // β/_\t {dot D} n +(1//2β -1){ddot D}n } +``[C] { r// β/_\t {D}n +(r//β -1) { dotD }n +/_\ t (r// 2β -1) {ddot D}n }` (-2)
equation ( 1) -thsi is the general governing differential equation implicit process .displacement of the n+1 increment the function of velocity and acceleration for the n+1 increment we cannot discriminate x and y parameter.
eqation (2) -is the implicit governing equation here in implicit scheme we need to make inverse of the stiffness to find the displacement.
explicit scheme .
{D}n+1 = `f( {D}n, {dot D}n , {ddotD}n ,{D}n-1,....)` -(3).
`[1///_\ t^2` M +`1//2/_\ t` C] {D}n+1 = `{R^(ext)}n -{R^(Int) }n`+ `2///_\t^2` [M] {D}n -`[1///_\t^2` M -`1//2/_\t`c ] {D} n-1. -( 4)
equation 3 -in explicit scheme n+1 increment displacement is not dependent on any parameter of n+1. RHS is the function n,n-1,n-2 ,...
equation 4 -this is the governing equation for explicit scheme we need to makeing mass and damping.
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2. Do the calculations in a spreadsheet, try to plot the graph and compare them (optional).
As demonstrated in the PDF, F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods ( have a tolerence of 10-2.
EXPLICIT CALCULATIONS
if the nonlinear force vs displacement relationship is not known,but the stiffness is known ,it is possible to construct the graph of force vs displacement numerically.
the following equation is solved using explicit method as follows,
`F(u)=u^3+9u^2+4u`.
differenciating force with respect to displacement gives the stiffness (k) of the bar ( since F=kx).
`k(u) =dF//du =3u^2+18u+4`
here three equal load increments of `/_\ F =1` unit each are used to load the tension bar.
the variables used are `u` for dispalcement , 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.
step -1
`u0 =0,0 , rArr k (u0) =4 , /_\ F=1 , /_\ u1 = /_\ F // K (u0) -0.25`
= `u1=u0+ /_\ u1=0.25.`
= `Fexternal = /_\ F =1`
=`Fi nternal =u^3 +9u^2 +4^u =2.708 ( :' u =u2).`
STEP -2.
u1 =0.25 k(u1) = 8.688.
=`/_\ F =1`
=`/_\ u2 = /_\F// k(u1) =0.1151.`
=u2 =u1 + `/_\ u2 =0.3651.`
=Fexternal =`/_\ F+V F =2.`
=Finternal = `u^3 +9u^2 +4u =2.708 ( :' u=u2) .`
STEP -3
u2 =0.3651 k(u2) =10.9717.
=`/_\ F =1`
=`u3 =/_\ F//k (u2) =0.0911.`
=`u3 = u2 +`/_\ u3 =0.4562.`
=Fexternal =`/_\ F+/_\F+/_\F =3.`
=Finternal = `U^3+9U^2+4U = 3.7928` `( :' u=u3)`
it is evident from the the result that external and internal forces are not in equilibrium because `Fexternal != Finternal.
Table for explicit method
it is evident from the above results that the residual is not equal to zero that is the internal and external forces are not in equilibrium. explicit method is generally used to solve fast processes where dynamic effects like mass acceleration damping are considered.the explicit method doesnot check for the static equilibrium (residual=internal force-external force ).thats why the time step size should be smaller than critical time step size (courant-Friedrich-Levt timestep).so basically explicit method is conditionally stable.
IMPLICIT METHOD CALCULATIONS
an implicit is the same as the explicit analysis ,except that at the end of each step Newton-Raphson iterations are used to enforce equillibrium before moving to the next step.basically an incremental force is applied to advance the solution forward at the begining of each step .internal force and external force will not be in equilibrium unless the stiffness is linear for the step.
in order to achieve equlibrium ,correction must be made to the displacement this is accomplished by using Newton-Raphson iterations to minimize the residual R=Finternal -Fexternal.Expanding the residual as a taylor series about the current dispalcement `u^j` gives
`R^(j+1`=`R^j (u^j) + dR^j (u^j)//du`. `deltau^(j+1` +......=`R^j` +K `(u^j)deltau^(j+1` +......
by neglecting higher order terms in the series ,setting `R^(j+1` =0 and solving `deltau^(j+1` the following correction is obtained .
`deltau^(j+1` =`-[k(u^j)]^-1 . R^j`. here the iteration is j and it may take several iterations for the norm of the residual,R,to be reduced below the chosen tolerance criteria of 0.001.
STEP 1.
`u0= 0,0` `k(u0) =4`
`/_\ F =1`
/_\ u1 =`/_\ F// k(u0) =0.25.`
u1=u0+/_\u1=0.25.
checking the residual
`Fexternal =/_\f=`1.
finternal =``u1^3+9u1^2+4u1=1.578.`
`R^(0) =`finternal -fexternal =0.578.
`R^(0) = `10^-2`= Newton-Raphson iteration are necessary`
calculate the correction to u1=`u1^(0).`
`delta u^(1)= [k (u0^10)]^-1` x `R ^(0)`
`(8.6875)^-1 x0.578 =-0.0665`
`u1^(1)``= u1^(0) +delta u^(1)`
0.25-0.0665= 0.1835
checking residual again
Fexternal = `/_\ F =1`
finternal =`(u1^(1))^3+9(u1^(1))^2+4u1^1.`=1.0432.
`R^(1)`= finternal -fexternal =0.0432.
`R^(1)`> `10^-2`.Newton-Raphson iteration are necessary`
calculate the new new total correction to `u1=u1^(0)=u1^(0).`
`deltau^(2)=deltau^(1)-[k(u1^(1))]^-1` X `R^(1)`.
-0.0665-`(7.404)^-1 X 0.0432`=-0.0723.
updating values `u1^2`
`u1^2`=`u1^0`+ `deltau^(2)`=0.25-0.723=0.1777.
checking the residual again
Fexternal =`/_\ F =1`
Finternal =`(u1^(2))^3+9(u1^(2))^2+4u1^2.`=1.0006`
`R^(2)= Finternal -Fexternal =0.0006.`u1 =0.1777`
`R^(2)<`10^-2`. newton
the final value is u1=`u1^(2)=0.1777`
STEP -2
`u1=0.1777` k(u1) =7.2933.
`/_\ F=1.`
/_\ u2 =`/_\F//k(u1)` =0.1371.
u2=u1+/_\u2=0.3148
checking residual
Fexternal = /_\F+/_\F=2
Finternal = `u2^3+9u2^2+4u2`
`R^(0)`=Finternal-Fexternal=0.1822.
`R^(0)`> `10^-2` newton -raphson iterations are necessary.
calculate the correction to u1= `u1^(0)`
`deltau ^1=[k(u2^0)]^-1` x `R^(0)`
`= -(9.9637)^-1 X 0.1822 =-0.0182`
the updated value
`u2^1=u2^0+deltau^1`=0.3148-0.0182=0.2966
checking the residual again
Fexternal =`/_\ F+`/_\ F=2
Finternal =`(u2^1)^3+9(u2^1)^2+4u2^1` =2.0042
`R^(1)` =Finternal -Fexternal =0.00042
`R^(1)` <`10^-2` Newton-Raphson iterations are not necessary
the final value is `u2=u2^1`=0.2966.
Step -3
u2=0.2966
k(u2)=9.6027
`/_\ F=1.`
`/_\ u3=``/_\ F//k(u2) =0.1041`u3=u2+/_\u3=0.4007
checking residual
`Fexternal=/_\F+/_\F+/_\F=`
Finternal =`u3^3+9u3^2+4u3=3.1121`
`R^(0)`=Finternal-Fexternal =0.1121
`R^(0) =`10^-2`
calculate the correction to u2=`u2^(0)`
`deltau^(2)`=`[k(u3^0)]^-1`x `R^0`
=`-(11.6942)-1 * 01121=-0.00958`
`u3^(1)=u3^(0)+deltau^(2)`
=0.4007 -0.00958=0.3911
checking the residual again
Fexternal = /_\F+/_\F+/_\F=3
Finternal = `(u3^(1))^3`+`9(u3^(1))^2`+`4u3^(1)`=3.0008
`R^(1)`=Finternal-Fexternal =0.0008
`R^(1)`< `10^-2.`
`u3^(2) = 0.3911.
Table results for implicit method
implicit method on the on the other hand ,is generally used for a slow process where inertia effects can be neglected iterations are done within time time steps if the stiffness is non-linear .the solution for this method is unconditionally stable but the time step has to be small enough to achieve convergence for non-linear problems.this method allows relatively large time steps but time consuming for dynamic and nonlinear problems.
PLOTS:
the force vs displacement behaviour for the given relation in the problem statement .for analytical soulution discrete displacement values considered and the force values are sought for those respective values by using the equaation 1 .numerical data for explicit and implicit is taken from the calculations done in the orevious section.
we can clearly see that for the explicit method the solution deifts from the actual solution .this is due to only incrementing the load without checking for the force equilibrium.for the implicit method the difference analytical and numerical values is less compared to the explicit method.the force equlibrium is checked after every load/time step and correction (Newton-Raphson iteration)is performed to reduce the residual .this reduces the residual and brings the solution close to the actualsolution
Conclusion
the explicit and implicit methods have been shown to calculate the force displacement response.from the analysis ,it is clear that explicit method drifts from the actual solution and one of the reasons are no equlibrium iterations are carried out within a load /time step.on the other hand implicit methods has less error and closely matched with the actual solution.this method utilizes the newton rapson method to enforce force equilibrium and brings down the residual within the tolerance values.