Week - 2 - Explicit and Implicit Analysis

Aim:

The aim of the analysis is to calculate the given derivation in both implicit analysis and in the explicit analysis method

Explaination:

Implicit analysis:

One method of solving for the unknowns {x} is through matrix inversion is known as an implicit analysis. When the problem is nonlinear, the solution is obtained in several 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). Sometimes, this is also known as the backward Euler integration scheme. 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.

Explicit analysis:

Explicit analyses aim to solve for acceleration {x´´}. In most cases, the mass matrix is considered as “lumped” and thus a diagonal matrix. 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 analysis has no such limitations.

Procedure:

F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods ( have a tolerence of 0.01).

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

For this example three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar. The variables used are u for displacement, f for internal force in the bar, F for external force applied to the bar and k for stiffness. Incremental displacements or incremental externally applied forces are represented as ∆u or ∆F, respectively. Use is made of the relationship ∆F = k∆u

fint - internal force

Fext - external force

u -displacement

k -stiffness

Explicit calculation:

Step 1

u0 = 0.0, ⇒ k(u0) =3*(0)^2+18*0+4= 4, ∆F = 1, ∆u1 = ∆F/k(u0) = 0.25, u1 = u0 + ∆u1 = 0.25

Step 2

k(u1) = 3*(0.25)^2+18*0.25+4=8.6875, ∆F = 1, ∆u2 = ∆F/k(u1) = 0.115, u2 = u1 + ∆u2 = 0.365

Step 3

k(u2) = 3*(0.365)^2+18*0.365+4=10.969, ∆F = 1, ∆u3 = ∆F/k(u2) = 0.0911, u3 = u2+∆u3 = 0.4561

Finally, from the three steps the following results are obtained.
Fext = ∆F + ∆F + ∆F = 3.0

fint = u1^3 + 9u1^2+ 4u1 = 1.5781

fint = u2^3 + 9u2^2+ 4u2 = 2.708

fint = u3^3 + 9u3^2+ 4u3 = 3.7908

It is evident from the results that external and internal forces are NOT in equilibrium because Fext ≠ fint.
Table 1 summarizes the main results for the explicit analysis.

Step	ΔFi	Δui	ui	(Fext)i	(fint)i	fint- Fext = R
1	1.0	0.25	0.25	1.0	1.578	0.578
2	1.0	0.115	0.365	2.0	2.708	0.708
3	1.0	0.0911	0.456	3.0	3.7908	0.7908

Implicit calculation:

Step 1

u0 = 0.0, ⇒ k(u0) = 4, ∆F = 1, ∆u1 = ∆F/k(u0) = 0.25, u1 = u0 + ∆u1 = 0.25

Check the residual, Fext = ∆F = 1, fint = u13 + 9 u12 + 4 u1= 1.578, R (0) = fint − Fext = 0.578

R (0) > 10−2 ⇒ Newton-Raphson iterations are necessary.

Calculate the correction to u1 = u (0)

δu (1) = −[k(u1(0))] −1 R (0) = −8.6875 −1 * 0.578 = −0.0665

The updated value of u1 (1) = u1(0) + δu(1) = 0.25 + (−0.0665) = 0.1835

Check the residual again, Fext = 1, fint = (u1(1))3 + 9 (u1(1))2 + 4 (u1(1)) = 1.0432

R (1) = fint − Fext = 0.0432

R (1) > 10−2 ⇒ another Newton-Raphson iteration is necessary.

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

δu(2) = δu(1) − [k(u1(1))] −1 R(1) = -0.0665 – (7.404)-1*0.0432= -0.07233

The updated value of u1 (2) = u1 (0) + δu(2) = 0.25 + (-0.0723) = 0.1777

Check the residual, Fext = 1, fint = (u1(2))3 + 9 (u1(2))2 + 4 (u1(2)) = 1.0006

R (2) = fint − Fext = 0.0006

R (2) < 10−2 ⇒ no iterations needed.

Therefore, the final value is u1 = u1 (2) = 0.1777

Step 2:

k(u1) = 3(0.1777) + 18(0.1777) + 4 = 7.293, ∆F = 1, ∆u2 = ∆F/k(u1) = 0.1371

u2 = u1 + ∆u2 = 0.3148

Check the residual, Fext = ∆F = 1, fint = u23 + 9 u22 + 4 u2 = 2.1822

R (0) = fint − Fext = 0.1822

R (0) > 10−2 ⇒ Newton-Raphson iterations are necessary.

Calculate the correction to u2 = u2(0)

δu (1) = −[k(u2(0))] −1 R (0) = -(9.9637)-1 * 0.1822 = -0.0183

⇒ The updated value of u2 (1) = u2 (0) + δu (1) = 0.2966

Check the residual again, Fext = ∆F = 1, fint = (u2 (1))3 + (9 u2 (1))2 + 4 u2 (1) = 2.0042

R (1) = fint − Fext = 0.0042

R (1) < 10−2 ⇒ Newton-Raphson iterations are not necessary.

Therefore, the final value is u2 = u2(1) = 0.2966

It is repeated for the step 3

u3 = u3(1) = 0.39112

Step	ΔFi	ui	(Fext)i	(fint)i	fint - Fext = R
1	1.0	0.1777	1.0	1.0006	0.0006
2	1.0	0.2966	2.0	2.00042	0.0042
3	1.0	0.39112	3.0	3.00085	0.00085

Things learned:

1. The calculation method of the implicit and explicit analysis

Conclusion:

The given equation is calculation in both implicit and explicit analysis method