Week - 2 - Explicit and Implicit Analysis

To solve the below Force-Displacement non-linear response of a material using both explicit and implicit methods and to compare the results to find out which method gives better approximate results.

The Force-Displacement relation is given below :

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

where,

u - displacement

f(u) - internal force generated due to the application of external force, which is a function of displacement.

PROCEDURE:

Implicit Method:

An implicit method, in contrast, would evaluate some or all of the terms in S in terms of unknown quantities at the new time step n+1. Usually a matrix or iterative solution must be used to compute the new quantities. Numerical stability has to do with the behavior of the solution as the time-step dt is increased. Implicit analysis finds a solution by solving an equation that includes both the current and later states of the given system.

Explicit Method:

Explicit method is when the state of the analysis is advanced by assuming constant values for the velocities u̇ and the accelerations u¨ across half-time intervals. Explicit FEM is used to calculate the state of a given system at a different time from the current time. The best-known software analysis for the explicit method are: PAM-CRASH, LS-DYNA, and RADIOSS.

SOLUTION :

The given Force-Displacement response is given by,

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

We know that the stiffness of the material is given by the first derivative of the Force-Displacement relation.

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

Implicit Method

`u_i - displacement at 'i' th step`

`k(u) - stiffness of the material which is a function of displacement`

`DeltaF - Incemental change in externally applied force`

`Deltau_i - Change in displacement at 'i' th step`

`(F_(ext))_i - Externally applied force at 'i' th step`

`f_(i) - internal force generated at 'i' th step`

`R(u) - Residual = `abs(f_i-(F_(ext))_i)`

`Tolerance = 10^-2 (user-defined)`

`deltau^((j)) - Correction to `u_i` in the 'j' th Newton-Raphson iteration`

`u_i^((j)) - displacement at 'i' th step in the 'j' th Newton-Raphson iteration`

Step-1:

`Initial displacement u_0 = 0`

`Initial stiffness k(u_0) = 3*(0)^2+18*0+4 = 4 units`

`DeltaF = 1 unit`

`Deltau_1=(DeltaF)/(k(u_0))=1/4=0.25 units`

`u_1=u_0+Deltau_1=0+0.25=0.25 units`

Checking the residual,

`f_1=(u_1)^3+9(u_1)^2+4u_1=1.578 units`

`Tolerance = 10^-2`

`Residual R^((0))=f_1-(F_(ext))_1=1.578-1=0.578`

`R^((0))>10^-2`

By applying Newton-Raphson in iteration 1,

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

`k(u_1^((0)))=k(0.25)=8.6875 unit`

`deltau^((1))=-(8.6875)^-1*0.578=-0.0665 units`

`u_1=u_1^((1))=u_1^((0))+deltau^((1))=0.25-0.0665=0.1835 units`

Checking the residual again,

`f_1=(u_1^((1)))^3+9(u_1^((1)))^2+4u_1^((1))=1.0432 units`

`R^((1))=1.0432-1=0.0432`

`R^((1))>10^-2`

By applying Newton-Raphson in iteration 2,

`deltau^((2))=deltau^((1))-[k(u_1^((1)))]^-1*R^((1))`

`k(u_1^((1)))=k(0.1835)=7.404 units`

`deltau^((2))=-0.0665-(7.404)^-1*0.0432=-0.0723 units`

`u_1=u_1^((2))=u_1^((0))+deltau^((2))=0.25-0.0723=0.1777 units`

Checking the residual again,

`f_1=(u_1^((2)))^3+9(u_1^((2)))^2+4u_1^((2))=1.0006` units

`R^((2))=1.0006-1=0.0006`

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

Residual has come inside the tolerance value.

Therefore, no further iterations needed.

Final value of `u_1=u_1^((2))=0.1777` units

Therfore, steps 2 and 3 are carried out and the results are indicated in the mentioned below table.

Step `i`	`DeltaF_i`	`u_i`	`(F_(ext))_i`	`f_i`	`R=abs(f_i-(F_(ext))_i)`
1	1	0.1777	1	1.0006	0.0006
2	1	0.2966	2	2.0042	0.0042
3	1	0.3911	3	3.0008	0.0008

From the table external forces and internal forces are coming into equilibrium.

Explicit Method

`u_i` - displacement at 'i' th step

`k(u)` - stiffness of the material which is a function of displacement

`DeltaF` - Incemental change in externally applied force

`Deltau_i` - Change in displacement at 'i' th step

`(F_(ext))_i` - Externally applied force at 'i' th step

`f_(i)` - internal force generated at 'i' th step

Step-1:

Initial displacement, `u_0 = 0`

Initial stiffness,`k(u_0) = 3*(0)^2+18*0+4 = 4 units`

`DeltaF = 1 unit`

`Deltau_1=(DeltaF)/(k(u_0))=1/4=0.25 units`

`u_1=u_0+Deltau_1=0+0.25=0.25 units`

Step-2:

`u_1 =0.25 units`

`Updated stiffness k(u_1) = 3*(0.25)^2+(18*0.25)+4 = 8.6875 units`

`DeltaF = 1 unit`

`Deltau_2=(DeltaF)/(k(u_1))=1/8.6875=0.1151 units`

`u_2=u_1+Deltau_2=0.25+0.1151=0.3651 units`

Step-3:

`u_2 =0.3651 units`

`Updated stiffness k(u_2)=3*(0.3651)^2+(18*0.3651)+4 = 10.971 units`

`DeltaF = 1 unit`

`Deltau_3=(DeltaF)/(k(u_2))=1/10.971=0.0911 units`

`u_3=u_2+Deltau_3=0.3651+0.0911=0.4562 units`

`F_(ext)=3 units`

`Internal force f=(u_3)^3+9(u_3)^2+4u_3=3.79 units`

From the results of external and internal forces are NOT in equilibrium because `F_(ext)!=f`. Table summarizes the main results for the explicit analysis.

Step,`i`	`DeltaF_i`	`Deltau_i`	`u_i`	`(F_(ext))_i`	`f_i`	`R=f_i-(F_(ext))_i`
1	1	0.25	0.25	1	1.578	0.578
2	1	0.1151	0.3651	2	2.708	0.708
3	1	0.0911	0.4562	3	3.792	0.792

RESULTS AND OBSERVATIONS :

The following table shows the convergence in explicit and implicit method. We can see that the residual in implicit method in each step is much lower when compared to explicit method. This is because, in implicit method, convergence check is made in each step before proceding to the next step. Therefore, force balance is ensured in each step which is not in case of explicit method. From the residual values, we can clearly see that the implicit method gives better approximate results in the given problem.

Steps	Residual (R) in Explicit method	Residual (R) in Implicit method
1	0.578	0.0006
2	0.708	0.0042
3	0.792	0.0008