Explicit and Implicit Analysis

AIM:

The aim of the challenge is to find solution for the function given below using both explicit and implicit methods.

`F(u) = u^3+9u^2+4u` with a tolerance of `10^-2`

EXPLANATION:

IMPLICIT METHOD:

`{x}=[K]^-1({f}-([M]{x''}+[C]{x'}))`

where, [K] is the stiffness matrix

[M] is the mass matrix

[C] is the damping matrix

In this method, the calculation of quantities in one time step is done using based on the quantities computed in the previous time step. This is known as Euler's time integration scheme. This method essentially uses inversion of stiffness matrix computation. Because of this, implicit method is extensively time consuming when compared to implicit method. This method is suitable for strucutrual dynamic problems in which load rate is low and this is more accurate than the explicit method.

EXPLICIT METHOD:

`{x''}=[M]^-1({f}-([C]{x'}+[K]{x}))`

where, [K] is the stiffness matrix

[M] is the mass matrix

[C] is the damping matrix

In this method, direct computation of dependent variable is done. Instead of verifying the global equilibrium, the solver assumes that equilibrium exists thereby calculating the local finite variables for that increment. This method uses inversion of mass matrix which is easier compared to stiffness matrix since all are diagonal elements in the former. The solver needs to take the reciprocal of the elements for inversion. So, the time taken is low compared to implicit method.

PROCEDURE:

EXPLICIT METHOD:

According to explicit analysis, if the non-linear force vs displacement is not known, but stifness is known, it is possible to construct the graph of force vs displacement numerically. One of the method used is incremental explicit load control scheme.

Given equation, `F(u) = u^3+9u^2+4u` ..... (1)

From equation (1), the stiffness is given by

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

Here, three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar.

The variables used are given below.

`u` represents displacement

`f` represents internal force in the bar

`F` represents external force applied on the bar

`k` represents stiffness

`Delta u` or `Delta F` represents incremental displacements or incremental externally applied forces.

The relationship `F=k*Deltau` is to used to find the incremental displacement.

Step-1:

`u_0 = 0.0 implies k(u_0) = 4` `DeltaF = 1, Deltau_1=(DeltaF)/(k(u_0)) = 1/4 = 0.25,` `u_1 = u_0+Deltau_1 =0.25`

Step-2:

`impliesk(u_1) = 3(0.25^2)+18(0.25)+4 = 8.6875` `DeltaF = 1, Deltau_2=(DeltaF)/(k(u_0)) = 1/8.675 = 0.11527,` `u_2 = u_1+Deltau_2 =0.36527`

Step-3:

`impliesk(u_2) = 3(0.36527^2)+18(0.36527)+4 = 10.97506 ` `DeltaF = 1, Deltau_3=(DeltaF)/(k(u_0)) = 1/10.97506 = 0.0911,` `u_3 = u_2+Deltau_3 =0.45637`

The results obtained are shown below:

`F_(ext) = DeltaF+DeltaF+DeltaF = 3`

`f_(Internal) = u_3^3+9u_3^2+4u_3 = 3.7949`

From the above calculations, it is clear that external and internal forces are not in equilibrium as `F_(ext) != f_(Internal)`

The table given below summarizes the main results for the explicit analysis.

Step i	`DeltaF_i`	`Deltau_i`	`u_i`	`(F_(ext))_i`	`(f_(Internal))_i`	`(f_(Internal))_i - (F_(ext))_i = R`
1	1	0.25	0.25	1.0	1.5781	0.5781
2	1	0.11527	0.36527	2.0	2.7106	0.7106
3	1	0.0911	0.45637	3.0	3.7949	0.7949

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, `(f_(Internal))_i - (F_(ext))_i = R`

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)]^(−1)*R^j`

The iteration variable is `j` and it may take several iterations for the norm of the residual, `||R||`, to be reduced below the chosen tolerance criteria of 0.01. The steps in implicit method is given below.

Step-1:

`u_0 = 0.0 implies k(u_0) = 4`, `DeltaF = 1, Deltau_1=(DeltaF)/(k(u_0)) = 1/4 = 0.25,` `u_1 = u_0+Deltau_1 =0.25`

Checking residual, `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) implies` Newton-Raphson Iterations are necessary

Calculating the new correction factor, `u_1 = u_1^((0))`

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

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

Checking the residual, `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) implies` Newton-Raphson Iterations are necessary

Calculating the new correction factor, `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`

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

Checking the residual, `F_(external)=DeltaF = 1`, `f_(Internal)=(u_1^((2)))^3+9(u_1^((2)))^2+4u_1^((2))=1.0006`, `R^((2))=f_(Internal)-F_(external)=0.0006`

`R^((1)) < 10^(−2) implies` Newton-Raphson Iterations are not necessary

Therefore, final value of residual is `u_1 = u_1^((2)) =0.1777`

Step-2:

`k(u_1) = 3(0.1777)^2 +18(0.1777)+4 = 7.2933`, `DeltaF = 1, Deltau_2 = (DeltaF)/(k(u_1)) = 0.1371, u_2 = u_1 + Deltau_2 = 0.3148`

Checking 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) implies` Newton-Raphson Iterations are necessary

Calculating the new correction factor, `u_1 = u_1^((0))`

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

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

Checking the residual, `F_(external)=DeltaF+DeltaF = 2`, `f_(Internal)=(u_2^((1)))^3+9(u_2^((1)))^2+4u_2^((1))=2.0042`, `R^((1))=f_(Internal)-F_(external)=0.0042`

`R^((1)) < 10^(−2) implies` Newton-Raphson Iterations are not necessary

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

Step-3:

`u_2 = 0.2966, ⇒ k(u_2) =9.6027 , DeltaF = 1, Deltau_3 = (DeltaF)/(k(u_2)) = 0.1041, u_3 = u_2 + Deltau_3 =0.4007`

Checking 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) implies` Newton-Raphson Iterations are necessary

Calculating the new correction factor, `u_2 = u_2^((0))`

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

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

Checking the residual, `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) implies` Newton-Raphson Iterations are not necessary

Therefore, final value of residual is `u_3 = u_3^((1)) =0.3911`

The table given below summarizes the main results for the explicit analysis.

Step i	`DeltaF_i`	`u_i`	`(F_(ext))_i`	`(f_(Internal))_i`	`(f_(Internal))_i - (F_(ext))_i = R`
1	1	0.1777	1.0	1.0006	0.0006
2	1	0.2966	2.0	2.0042	0.0042
3	1	0.3911	3.0	3.0008	0.0008

From the above plots, it is clear that deviation from the true solution is more for explicit analysis. This is because the explicit analysis does not consider global equilibrium. The time step is assumed for the explicit analysis. On the other hand, the implicit analysis considers the global equilibrium but there is a considerable increase in the computational time.

CONCLUSION:

It is evident from the plot that explicit analysis deviates more from the true solution when compared to implicit analysis but the explicit analysis is conditionally stable. To avoid this problem an implicit analysis is used, which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces. Therefore, implicit analysis gives more accurate results when compared to explicit analysis.