Week - 2 - Explicit and Implicit Analysis

Implicit and Explicit Methods :

Implicit and Explicit methods differ in the approach to time incrementation. In Implicit method 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.

Implicit :

• In each time increment the " global equilibrium " in the model is established, This mean that each increment has to converge (every iteration).
• After global equilibrium is ok solver calculates all the local finite element variables ( stress, strain, displacement etc) for this increment .
• Since global equilibrium is verified at each time increment those increments can be BIG.
• Each time increment computation is slow, since iteration are needed to get to the global equilibrium.

Explicit :

It works in way that you ddont have to converge each step there is no convergence criteria to check and no iteration, Instead of verifying the "Global Equilibrium" solver assumes that the equilibrium simply exists and solver goes straight to calculateion of local finite element variables.

• Calculates all of the local finite element variables for given increment and move to the next one.
• Each increment calculation is extremeley fast.
• The time step has to be extremely small or lese it is impossible to maintain this equilibrium that is assumed to simply exist.

Objective :

To solve F(u)= `u^3+9u^2+4u` using both Implicit and Explicit method with a tolerance of `10^2`.

Implicit :

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) = fint − Fext. Expanding the residual as a Taylor series about the current displacement `u^j` gives

`R^(j+1)`=`R^j(u^j)`+`dR(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 for `deltau^(j+1)` the following correction is obtained

`deltau^(j+1)` = `-[k(u^j)]^-1R^j`

Step 1 :

`u_1`=0.0 `=>``k(u_0)`=4, `DeltaF`=1, `Deltau_1`= `frac{DeltaF}{k(u_0)}`=0.25, `u_1=u_0+Deltau_1`=0.25

`F_external`=`DeltaF=1`

`F_Internal`=`u_1^3+9u_1^2+4u_1`=1.578

`R^0`=`F_Internal-F_external`= 0.578

`R^0>10^-2=>`Newton-Raphson iteration is necessary.

Calculate the correction to `u_1=u_1^((0))`

`deltau^((1))=-[k(u_1^((0)))]^-1*R^((0))` = `-(8.6875)^-1*0.578 =-0.0665`

The updated value of `u_1^((1)) = u_1^((0)) + deltau^((1))` =0.25-0.0665 =0.1835

Checking the residual again

`F_External =DeltaF=1`

`F_Internal =(u_1^((1)))^3 +9(u_1^((1)))^2+4u_1^((1)) =1.0432`

`R^((1))= F_Internal -F_External =0.0432`

`R^((1)) >10^-2 =>` Newton-Raphson itteration is neccessary

Calculate the correction to `u_1=u_1^((0))`

`deltau^((2))=deltau^((1))-[k(u_1^((1)))]^-1*R^((1))`= `-0.0665-(7.404)^-1*0.0432`=-0.0723

The updated value of `u_1^((2))=u_1^((0))+deltau^((2))=0.25-0.0723=0.1777`

Checking the residual again

`F_external=DeltaF=1``F_Internal =(u_1^((2)))^3+9(u_1^((2)))^2+4u_1^((2)) =1.006`

`R^((2))=F_Internal-F_External =0.0006`

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

Newton-Raphson itteration is not necessary.

The final value of `u_1=u_1^((2))=0.1777`

Step-2 :

`u_1= 0.1777, => k(u_1)=7.2933, DeltaF=1, Deltau_2=frac{DeltaF}{k(u_1)}=0.1371, u_2=u_1+Deltau_2=0.3148`

Checking the residual,

`F_External=DeltaF+DeltaF=2`

`F_Internal=u_2^3+9u_2^2+4u_2=2.1822`

`R^((0))=F_Internal-F_External=0.1822`

`R^((0))>10^-2=>`Newton-Raphson itteration is necessary.

Calculate the correction to `u_1=u_1^((0))`

`deltau^((1))=-[k(u_2^((0)))]^-1*R^((0))=-(9.9637)^-1*0.1822=-0.0182`

the updated value of `u_2^((1))=u_2^((0))+deltau^((1))=0.3148-0.0182=0.2966`

Checking the residual again.

`F_External=DeltaF+DeltaF=2`

`F_Internal=(u_1^((1)))^3+9(u_2^((1)))^2+4u_2^((1))`=2.0042

`R^((1))=F_Internal-F_External=0.0042`

`R^((1))<10^-2=>`Newton-Raphson iteration is not necessary

Therefore, the final value is `u_2=u_2^((1))=0.2966`

Step-3:

`u_2=0.2966, => k(u_2)=9.6027, DeltaF=1, Deltau_3=frac{DeltaF}{k(u_2)}=0.1041, u_3=u_2+Deltau_3=0.4007`

Checking the residual,

`F_external=DeltaF+DeltaF+DeltaF=3`

`F_Internal=u_3^3+9u_3^2+4u_3=3.1121`

`R^((0))=F_Internal-F_External=0.1121`

`R^((0))>10^-2=>`Newton-Raphson itteration is necessary.

Calculate the correction to `u_2=u_2^((0))`

`deltau^((2))=-[k(u_3^((0)))]^-1*R^((0))=-(11.6942)^-1*0.1121=-0.00958`

The updated value of `u_3^((1))=u_3^((0))+deltau^((2))=0.4007-0.00958=0.3911`

Checking the residual again.

`F_External=DeltaF+DeltaF+DeltaF=3`

`F_Internal=(u_3^((1)))^3+9(u_3^((1)))^2+4u_3^((1))=3.0008`

`R^((1))=F_Internal-F_External=0.0008`

`R^((1))<10^-2=>`Newton-Raphson itteration is not necessary

Therfore, the final value of `u_3=u_3^((2))=0.3911`

Implicit analysis results are summerized in the following table.

Step i	`DeltaF`	`u_i`	`(F_Ext)_i`	`(F_Int)_i`	`R=(F_Int)_i-(F_Ext)_i`
1	1	0.1777	1	1.006	0.006
2	1	0.2966	2	2.0042	0.0042
3	1	0.3911	3	3.0008	0.0008

From the Table we can see that external forces and internal forces are coming into equilibrium.

Explicit :

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.

The given equation is

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

Differentiating force with respect to displacement gives the stiffness(k)

`k(u)= frac{dF}{du}=3u^2+18u+4`

Here,three equal load increment of `DeltaF=1` unit each are used to load the tension bar.

`DeltaF=k*Deltau` this equation is used to find the incremental displacement.

Step-1:

`u_0=0.0, =>k(u_0)=4, DeltaF=1, Deltau_1=frac{DeltaF}{k(u_0)}=0.25, u_1=u_0+Deltau_1=0.25`

`F_External=DeltaF=1`

`F_Internal=u^3+9u^2+4u=1.5781 (therefore u=u_1)`

Step-2:

`u_1=0.25,=>k(u_1)=8.688,DeltaF=1,Deltau_2=frac{DeltaF}{k(u_1)}=0.1151, u_2=u_1+Deltau_2=0.3651`

`F_External=DeltaF+DeltaF=2`

`F_Internal=u^3+9u^2+4u=2.708 (therefore u=u_2)`

Step-3:

`u_2=0.3651,=>k(u_2)=10.9717, DeltaF=1, Deltau_3=frac{DeltaF}{k(u_2)}=0.0911, u_3=u_2+Deltau_3=0.4562`

`F_External=DeltaF+DeltaF+DelatF=3`

`F_Internal=u^3+9u^2+4u=3.7928 (therefore u=u_3)`

From the result we can see that external and internal forces are not equilibrium because `F_(external) ne F_Internal`

The table shows the results for explicit method :

Step i	`DeltaF_i`	`Deltau_i`	`u_i`	`(F_ext)_i`	`(F_Int)_i`	`R=(F_Int)_i-(F_ext)_i`
1	1	0.25	0.25	1	1.5781	0.5781
2	1	0.1151	0.3651	2	2.708	0.708
3	1	0.0911	0.4562	3	3.7928	0.7928

Conclusion:

Form the results we obtained, We can see that the results from explicit analysis drifts from the true solution.

Thus implicit analysis is used which uses Newton-Raphson method iteration to enforce equilibrium between internal and external forces