Week - 2 - Explicit and Implicit Analysis
AIM:
To do Explicit and Implicit Analysis.
PROCEDURE:
EXPLICT AND IMPLICIT ANALYSIS:
F(u) = u3+9u2+4u ----(1) use this equation and solve using both Explicit and Implicit Methods ( have a tolerence of 0.001).
Stiffness, k(U) = dF/du
= 3u^2+18u+4 ----(2)
EXPLICIT ANALYSIS:
In explicit mostly dynamic analysis nodal accelerations are solved directly as the inverse of the diagonal mass matrix times the net nodal force vector where not nodal forces includes contributions from exterior sources element stress, damping, bulk viscosity and hourglass control.
Once accelerations are known at time n, velocities are calculated at time n+1/2 and displacements at time n+1. From displacements comes strain. From strain comes stress and the cycle is repeated.
Explicit analysis handles nonlinearities with relative ease as compared to implict analysis. This would include treatment of contact and material nonlinearties. With this method is that do need many small increments for good accuracy and it is time consuming.
where,
u = displacement
f = internal force in bar
F = external force in bar
△u = incremental displacements
△F = incremental external applied forces.
using relation △F = k△u
Step-1:
Take u0 = apply in (2)
we get,
k(u0) = 3*0^2 + 18*0 + 4 = 4
now △u1 = △F/k(u0) = 1/4 = 0.25
u1 = u0+△u1 = 0.25.
Step-2:
Take u1 = 0.25 in (2)
we get
k(u1) = 3*0.25^2 + 18*0.25 + 4 = 8.6875
now △u1 = △F/k(u0) = 1/4 = 0.1151
u1 = u0+△u1 = 0.3651.
Step-3:
Take u2 = 0.3651 in (2)
k(u2) = 3*0.3651^2 + 18*0.3651 + 4 = 10.9716
now △u1 = △F/k(u0) = 1/4 = 0.09114
u1 = u0+△u1 = 0.46.
To determine wheather the analysis is in equlibrium,
F(ext) = △F+△F△F = 3
f(int) = (1) implies (0.46)^3 + (9*0.462^2) + (4*0.46) = 3.841
Fext ≠ f(int) so they are not in equilibrium.
| Step i | △Fi | △ui | ui | F(ext) i | f(int) i | f(int) i - F(ext) i = R |
| 1 | 1 | 0.2500 | 0.25 | 1 | 1.578 | 0.578 |
| 2 | 1 | 0.1151 | 0.3651 | 2 | 2.708 | 0.708 |
| 3 | 1 | 0.09114 | 0.46 | 3 | 3.841 | 0.841 |
FORCE VS DISPLACEMENT GRPAH:
IMPLICIT ANALYSIS:
Implicit analysis requires a numerical solver to invert the stiffness matrix once or even several times over the course of a load / time step.This matrix inversion is an expensive operation especially for large models.
It is mostly used for static analysis. Implict uses Newton-Raphson iterations to enforce equilibrium. If done correctly the iterations will have a quadratic rate of convergence which is very desireable.
Implict transient analysis has no inherent limit on the size of the time step. As such implicit time steps are generally several orders of magnitude larger than explicit time steps.
where,
u = incremental displacement
f = internal force in bar
F = external force in bar
△u = incremental displacements
△F = incremental external applied forces.
using relation
△F = k△u, Tolerance = 10^-2
Step-1:
take u0 = 0
(2) = k(u0) = 3(0)^2 +18*0 +4 = 4
now
△u1 = △F/k(u0) = 1/4 = 0.25
u1 = u0+△u1 = 0.25
Checking residual R
F(ext) = △F =1
f(int) = (1) we get 1.578
R0 = 1.578-1 = 0.578 > 10^-2
so, Newton -Raphson iterations are necessary.
calculation the correction to u1 = u1(0)
we know that
δu(1) = -[k(u1(0)]^-1 * R0
= -0.6665
updated u1(1) = u1(0) + δu(1) = 0.25 - 0.0665 = 0.1835.
Checking residual again R1,
F(ext) = 1, f(int) = 1.0432
R1 = 1.0432-1 = 0.0432 > 10^-2 another Newton Raphson iteration needed.
calculate the correction to u1 = u1(1),
δu(2) = δu(1) - -[k(u1(0)]^-1 * R1
= -0.0714
updated u1(2) = u1(1) + δu(2) = 0.25-0.0714 = 0.1786
Checking Residual again, R2
F(ext) = 1
f(int) = (1) = 0.0071
R2 = 1.0071 - 1
= 0.0071 < 10^-2
Therefore no iteraitons needed, final u1 = 0.1786.
Step 2:
Similarly as same as step 1 we get u2 = 0.2966
Step-3:
Similarly as same as step 1 we get u3 = 0.3911
| Step i | ΔFi | Δui | ui | F(ext)i | f(int)i | f(int)i - F(ext)i = R |
| 1 | 1 | 0.25 | 0.1786 | 1 | 1.0071 | 0.0071 |
| 2 | 1 | 0.1374 | 0.2966 | 2 | 2.0042 | 0.0042 |
| 3 | 1 | 0.1041 | 0.3911 | 3 | 3.0008 | 0.0008 |
FORCE VS DISPLACEMENT GRAPH:
Excat Values:
Take, F(u) = u3+9u2+4u = 1
u = 0.1776
Take, F(u) = u3+9u2+4u = 2
u = 0.2962
Take, F(u) = u3+9u2+4u = 3
u = 0.3910
Comparision Graph Implicit and Explicit Force VS Displacement:
The implicit and exact solutions are same line while explicit is away from excat solution.
OUTPUT:
From the graph implicit method is more accurate but time consuming depends on element size and time step whereas explicit deviates slightly from exact result and needs more iterations to obtain good result.
RESULT:
Thus the Implicit and Explicit numerical analysis is carried out for the given equation.