Week - 2 - Explicit and Implicit Analysis

Aim: - To study about Implicit and explicit methods.

Objectives: - 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 0.01)

Explicit analysis: - 

Explicit analyses aim to solve for acceleration (or otherwise {x´´}). 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 the n+1/2 step and displacement at the 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.

 F(u) =  u^3+9u^2+4u   - (1) 

K(u) =  df/du =  3u^2+18u+4  -(2)

 

u - displacement

f - internal force in bar

F - external force in bar

△u - incremental displacement

△f - incremental external applied force

 

△F=k△u 

Step - (1)

u0 = 0 ---in (2)

k(u0) = 3u^2+18u+4

k(u0) =  3(0)+18(0)+4

k(u0) = 4

△F = 1

△u1 = △F/k(u0)

△u1 = 0.25

u1 = u0+△u1

u1 = 0 + 0.25

u1 = 0.25 

Step - (2) 

u1 = 0.25 ---put in equation (2)

k(u1) = 3u^2+18u+4

k(u1) =  3(0.25)+18(0.25)+4

k(u1) = 8.6875

△F = 1

△u2 = △F/k(u1)

△u2 = 0.1151

u2 = u1+△u2

u2 = 0.25 + 0.1151

u2 = 0.3651

Step -(3)

u2 = 0.3651 ---put in equation (2)

k(u2) = 3u^2+18u+4

k(u2) = 3(0.3651)^2 + 18(0.3651) + 4

k(u2) = 10.9716 

△F = 1

△u3 = △F/k(u2)

△u3 = 0.09114

u3 = u2+△u3

     = 0.3651 + 0.09114 

u3 = 0.4562

Finally, from the three steps, the following results are obtained.

Step (1)

Fext = ∆F = 1 

Fint = f(u1) = u^3+9u^2+4u 

                  = (0.25)^3 + 9(0.25)^2 + 4(0.25) 

  fint = f(u1)= 1.578

 External and internal forces are NOT in equilibrium because Fext ≠ fint

R = fint − Fext 

R = 1.578 - 1 = 0.578

Step (2)

Fext =  ∆F + ∆F  = 2 

Fint = f(u2) = u^3+9u^2+4u 

                  = (0.3651)^3 + 9(0.3651)^2 + 4(0.3651) 

  fint = f(u2)= 2.7087

 External and internal forces are NOT in equilibrium because Fext ≠ fint

R = fint − Fext 

R = 2.7087 - 2 = 0.7087

Step (3) 

Fext =  ∆F + ∆F + ∆F  = 3 

Fint = f(u3) = u^3+9u^2+4u 

                  = (0.4562)^3 + 9(0.4562)^2 + 4(0.4562) 

  fint = f(u3)= 3.7928

 External and internal forces are NOT in equilibrium because Fext ≠ fint

R = fint − Fext 

R = 3.7928 - 3 = 0.7928

 

Table 1: Summary of explicit analysis results.

Step i	∆Fi	∆ui	ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.25	0.25	1	1.578	0.578
2	1	0.1151	0.3651	2	2.7087	0.7087
3	1	0.09114	0.4562	3	3.7928	0.7928

Plot a graph of force(f) versus displacement(u)

 

Implicit analysis: - 

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). 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.

F(u) =  u^3+9u^2+4u   - (1) 

K(u) =  df/du =  3u^2+18u+4  -(2)

Step - (1)

u0 = 0 ---in (2)

k(u0) = 3u^2+18u+4

k(u0) = 4 

△F = 1

△u1 = △F/k(u0)

△u1 = 0.25

u1 = u0+△u1

u1 = 0 + 0.25

u1 = 0.25 

Check the residual, R

Fext = ∆F = 1 

fint = u1^3+9u1^2+4u1

     = (0.25)^3 + 9(0.25)^2 + 4(0.25)

     = 1.5781

R(0) = fint − Fext

       = 1.5781 - 1 

R(0) = 0.5781 > 0.01 

Newton-Raphson iterations are necessary.

Calculate the correction, u(1) = u1(0)

δu(1) = −[k(u1(0))^-1] R(0)

         = - (3(0.25)^2) + (18(0.25))+4 *(0.5781)

δu(1) = - 0.06653

The updated value of u1(1) = u1(0) + δu(1)

                                      = 0.25+(-0.06653)

                               u1(1) = 0.1835

 Check the residual again, 

Fext = 1

fint = u1(1)^3+9u1(1)^2+4u1(1)

     = (0.1835)^3 + 9(0.1835)^2 + 4(0.1835)

     = 1.0434

R(1) = fint − Fext

        = 1.0434 - 1 

R(1)  = 0.0434 > 0.01

Calculate the new total correction to u1 = u1(0)

δu(2) = δu(1) - [k(u1(1))^-1] R(1)

         = 0.06653 - ((8.6875)^-1) *(0.0432)

         = - 0.07147

The updated value of u1(2) = u1(1) + δu(2)

                                        = 0.25 +(-0.07147)

                                        = 0.1786

Check the residual again, 

Fext = 1

fint = (u1(2))^3 + 9(u1(2))^2 + 4(u1(2))

     = (0.1786)^3 + 9(0.1786)^2 + 4(0.1786)

     = 1.0071

R(2) = fint - fext 

       = 1.0071 - 1

R(2) = 0.0071 < 0.01

{no further iterations are necessary}

Therefore, the final value is u1 = u1(2) is 0.1786

Similarly, we calculated other values of u2 = 0.29654 and u3 = 0.3911 by same procedure.

Step i	∆Fi	∆ui	ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.25	0.1786	1	1.0071	0.0071
2	1	0.1367	0.29654	2	2.0036	0.0036
3	1	0.10415	0.3911	3	3.00085	0.00085

Plot a graph of force(f) versus displacement(u)

 

 

 

Difference between Implicit and Explicit.

Implicit	Explicit
1. Stain rate/velocity is lower, say less than 10 units/sec.	1. 1. Stain rate/velocity is high, say more than 10 units/sec.
2. Effect of strain rate is minimum.	2. Strain rate has a significant effect solve extremely discontinuous events or processes.
3. Easy or moderate contact condition.	3. Can solve most complex contact conditions.
4. More accurate.	4. Less accurate.
5. Extremely time-consuming for the large model.	5. Computationally efficient for a large model with relatively short dynamic response times.
6. No mathematical time limit for solution.	6. Time depends on wave propagation speed & characteristic length of the element.
7. Implicit method much more computer storage.	7. Less storage is required than an implicit method.
8. Good choice for the structural dynamic type of problems for which load rate is slow such as - low-frequency response, vibration, and oscillations. etc.	8. For wave propagation type of problems - car crash, Drop, Impact, etc.

Conclusion: - 

1. For a given equation the implicit, as well as explicit methods, are carried out.

2. Implicit method is more accurate than the explicit method because in the implicit we solve by the iterative method and in the explicit, we solve by the direct method.

3. Implicit is time-consuming and required more storage than the explicit method.