Week - 2 - Explicit and Implicit Analysis

1. Implicit and Explicit Analysis

Introduction

• Engineering problems can be solved using three approach, Analytical approach, Experimental approach, and Numerical Approach.
• Finite Element Method (FEM) is a numerical approach used to solve complex engineering problem with less time.
• FEM is basically a numerical technique used to find the approximate solution a partial differential equation of a particular region.
• FEM is categorized in the time-dependent and time-independent analysis where the concept of Implicit and Explicit analysis has introduced.

1.1 Implicit Analysis

The implicit analysis is a time-dependent approach in which the solution of every step requires the solution of previous steps. It solves the displacement matrix by matrix inversion which takes more time to get a stable solution. If we solve a big model, it will be highly expensive and require an advanced iterative solver for inverting the matrix. It is also known as the backward Euler Integration scheme. These solutions are unconditionally stable. Despite this advantage, the implicit methods can be extremely time-consuming when solving dynamic and nonlinear problems. We can use this method when the events are much slower and the effects of strain rate are minimal. This scheme uses the Newton- Raphson method to minimize the residual and get a stable solution in every step.

1.2 Explicit Analysis

The explicit analysis is a time-independent approach in which the solution of every step does not require a solution to previous steps. This approach solves the acceleration matrix instead of the displacement matrix to solve the problem and takes less time as compare to implicit analysis. It is not unconditionally stable and thus required smaller time steps. Time steps should be less than the courant time step for getting more precise solutions to the problem. Courant time is the time taken by a sound wave to travel across an element.

These two technique can be explain by the following example-

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

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

Where Force F is the function of u, which is displacement. We can see the equation 1 is nonlinear force. We will solve this problem using implicit and explicit analysis two analyze the difference between this two techniques. In this calculation, three equal loads incremental is used `∆F=1.`

1.2.1 Solution by Explicit Scheme

Step 1:- Initial condition

At `u_0 = 0,`

At this time stiffness `k (u_0) = 4;`

For first load increment `∆F=1;`

`(∆u)_(1 )= ∆F/(k(u_(0 )) = 1/4`

Then

`u_1= u_(0 )+(∆u)_(1 )= 0+0.25 = 0.25`

Step 2:- for `u_1=0.25,`

`k(u_(1 )) = 3*(0.25)^2+ 18*(0.25) +4 = 8.6875`

For second load increment `∆F=1;`

`(∆u)_(2 )= ∆F/(k(u_(1 )) = 1/8.6875 = 0.115107`

Then

`u_2= u_(1 )+(∆u)_(2 )=0.25+0.115107=0.365107`

Step 3 :- for `u_2=0.365107`

`k(u_(2 )) = 3*(0.365107)^2+ 18*(0.365107) +4 = 10.971835`

For third load increment `∆F=1;`

`(∆u)_(3 )= ∆F/(k(u_(2 )) = 1/10.971835 = 0.091142`

Then

`u_3= u_(2 )+(∆u)_(3 ) = 0.365107 +0.091142 = 0.456249`

From above calculation we have calculated the value of displacement at three load increment. Now we will check the difference between external applied force and the internal generated force to see whether these forces are in equilibrium or not. To satisfy this condition `F_(ext )` and `F_(Int )`should be same.

`F_(ext )= ∆F+∆F+∆F=3`

`F_(Int )= u_3^3+9u_3^2+4u_3 = 3.793438`

From the above calculation, we can find out that `F_(ext )≠ F_(Int )`, it means these forces, not in an equilibrium state. It shows that an explicit approach is conditionally stable.

Now `F_(Int )- F_(ext )= 3.793438-3 = 0.793438 = R`

Where R = Residual

Step	`(∆F)_i`	`(∆u)_(i )`	`u_i`	`F_(ext )`	`F_(Int )`	R
1	1	0.25	0.25	1	1.578	0.5781
2	1	0.115107	0.365107	2	2.7087	0.7087
3	1	0.091142	0.456249	3	3.821	0.821

1.2.2 Solution by Implicit Scheme

In an explicit scheme, it can be seen that at every step of estimation there is residual which means the solution is not stable. But in the case of the implicit scheme, the solution of the current step is to get stabilized by minimizing the residual. The Newton- Raphason method used to minimize the residual at every step until the solution gets stable.

Now we will solve the above problem with the implicit scheme-

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 δu^(j+1)+⋯=R^j+k(u^j )δu^(j+1)+⋯`

The higher-order in the above equation can be neglected due to very small value, setting `R^(j+1)` = 0 and solving for `δu^(j+1)` the following correction is obtained

`δu^(j+1)=-[k(u^j )]^(-1) R^j`

In the above relation j is the iteration variable which indicates that it may take several steps to reduce the residual below the tolerance criteria. In this analysis 0.01 is chosen.

Step 1:- At `u_0=0;`

`(k(u)_0)=3(u_0)^2+18u_0+4=4`

For first load increment `∆F=1;`

`(∆u)_(1 )= ∆F/(k(u_(0 ))= 1/4`

`Then; u_1= u_(0 )+(∆u)_(1 )= 0+0.25 = 0.25`

Now we will check the residual for this step

`R = F_(Int )- F_(ext )`

Where `F_(ext )= ∆F = 1`

and `F_(Int )= (u_(1 ))^3+9(u_(1 ))^2+4u_(1 )= (0.25)^3+9(0.25)^2+4*0.25 = 1.578125`

`R = F_(Int )- F_(ext )= 1.578125-1= 0.578125`

Here the value of calculated residual is greater than 0.01 so Newton Raphason method will be used to get stable solution.

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

`δu^1=-[k(u^0 )]^(-1) R^0 = -[8.6875]^(-1)*0.578125 = - 0.0665`

Now `u_1^((1))= u_1^((0))+δu^((1))= 0.25-0.0665 = 0.1835`

Check the residual R

`F_(Int )= (0.1835)^3+9(0.1835)^2+4*0.1835 = 1.04322`

`F_(ext )= 1`

R= 1.04322-1 = 0.04322

Here R`>0.01` , Newton-Raphason iterations are necessary.

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

`δu^((2))=-[k(u^1 )]^(-1) R^1= -[3*0.1835^2+18*0.1835+4]^(-1)*0.04322 = - 0.005837`

Now `u_1^((2))= u_1^((1))+δu^((2)) = 0.1835-0.005837= 0.1776`

Check the residual R,

`F_(Int )= (0.1776)^3+9(0.1776)^2+4*0.1776 = 0.9999`

`F_(ext )= 1`

|R|= 1-0.9999= 0.0001

Here R`<0.01` , Newton-Raphason iterations are not necessary.

Step 2:- stiffness at `u_1`

`k(u_1)=3u_1^2+18u_1+4= 7.2914`

For second load increment `∆F=1;`

`∆u_(2 )= ∆F/(k(u_(1 ))) =1/( 7.2914) = 0.1371`

`u_2=u_(1 )+∆u_(2 )= 0.1776+0.1371 = 0.3147`

Now

`F_(Int )= u_(2 )^3+9u_2^2+4u_2= (0.3147)^3+9*(0.3147)^2+4*(0.3147) = 2.1812`

R= 2.1812-2= 0.1812

Here R`>0.01` , Newton-Raphason iterations are necessary.

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

`δu^1=-[k(u_2^((0)) )]^(-1) R^0= -[3*0.3147^2+18*0.3147+4]^(-1)*0.1812=- 0.01818`

The new value of `u_2^((1) )= u_2^((0) )+δu^((1) )=0.3147-0.01818=0.2965`

Check the residual for this step

`F_(Int )= (0.2965)^3+9(0.2965)^2+4*0.2965= 2.00327`

`F_(ext )= ∆F = 1`

then R=2.00327-2= 0.00327

Here R`<0.01` , Newton-Raphason iterations are not necessary.

Therefore the final value of `u_2=u_2^((1))= 0.2965`

Step 3:- Stiffness at `u_2`

`k(u_2)=3u_2^2+18u_2+4= 9.6007`

For third load increment `∆F=1`

`∆u_(3 )= ∆F/(k(u_(2 ))) =1/( 9.6007) = 0.1041`

`u_3=u_(2 )+∆u_(3 )=0.2965+0.1041 = 0.4006`

Now

`F_(Int )= u_(3 )^3+9u_3^2+4u_3= (0.4006)^3+9*(0.4006)^2+4*(0.4006) = 3.1110`

`F_(ext )=3`

R= 3.1110-3= 0.1110

Here R`>0.01` , Newton-Raphason iterations are necessary.

Calculate the correction to `u_3=u_3^((0))`

`δu^1=-[k(u_3^((0)) )]^(-1) R^0= -[3*0.4006^2+18*0.4006+4]^(-1)*0.1110=- 0.00949`

The new value of `u_3^((1) )= u_3^((0) )+δu^((1) )=0.4006-0.00949=0.3911`

Check the residual for this step `F_(Int )=(0.3911)^3+9*(0.3911)^2+4*(0.3911) = 3.00085`

`F_(ext )=3`

R= 3.00085-3= 0.00085

Here R`<0.01` , Newton-Raphason iterations are not necessary.

Therefore the final value of `u_2=u_2^((1))= 0.3911`

Step	`∆F_i`	`∆u_(i )`	`u_i`	`F_(ext )`	`F_(Int )`	R
1	1	0.25	0.1776	1	1.578125	0.578125
2	1	0.1371	0.2965	2	2.00327	0.00327
3	1	0.1041	0.3911	3	3.00085	0.00085

From the graph, we can see that the solution of implicit analysis is more precise as compared to explicit analysis. So for this problem, Implicit analysis is best.