Week - 2 - Explicit and Implicit Analysis
week-2 - EXPLICIT AND IMPLICIT ANALYSIS
F(u) = u3+9u2+4u ---(1)
solve using both Explicit and Implicit Methods ( have a tolerence of 10-2).
Stiffness, k(U) = dF/du = 3u^2+18u+4 ---(2)
EXPLIIT ANALYSIS :
In explicit mostly dynamic analysis, nodal accelerations are solved directly (not iteratively) as the inverse of the diagonal mass matrix times the net nodal force vector where net nodal force includes contributions from exterior sources (body forces, applied pressure, contact, etc.), 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 implicit analysis. This would include treatment of contact and material nonlinearities. With this method is that you 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,
Step1 :
Take, u0 = 0.0, apply in (2)
(2) => k(u0)= 3x0^2+18x0+4 = 4
now, ∆u1 = ∆F/k(u0) = 1/4 = 0.25
u1 = u0+∆u1= 0.25
Step 2 :
Take u1= 0.25
(2) => k(u1)= 3x0.25^2+18x0.25+4 = 8.6875
now, ∆u1 = ∆F/k(u1) = 1/8.6875 = 0.1151
u2 = u1+∆u2= 0.3651
Step 3 :
Take u2= 0.3651
(2) => k(u2)= 3x0.3651^2+18x0.3651+4 = 10.9716
now, ∆u2 = ∆F/k(u2) = 1/10.9716 = 0.09114
u3 = u2+∆u3= 0.46
Finally, To determine wheather the analysis is in equlibrium,
F(ext) = ∆F+∆F+∆F = 3
f(int) = (1) => (0.46)^3+(9x0.46^2)+(4x0.46) = 3.841
F(ext) ≠ 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 |
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. Implicit uses Newton-Raphson iterations to enforce equilibrium. If done correctly the Newton-Raphson iterations will have a quadratic rate of convergence which is very desireable.
Implicit 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= 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+18x0+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)=> (0.25)^3+(9x(0.25)^2)+(4x0.25) =1.578
R0 = 1.578-1 = 0.578 > 10^-2
so, Newton-Raphson iterations are necessary.
calc. the correction to u1=u1(0)
wkt, ẟu(1) = -[k(u1(0)]^-1 x R0 = -[(3x0.25^2)+(18x0.25)+4]^-1 x 0.578 = -0.0665
updated, u1(1) = u1(0)+ẟu(1) = 0.25-0.0665 = 0.1835
Check residual again, R1,
F(ext)= 1, f(int) = (0.1835)^3+(9x0.1835)^2+(4x0.1835) = 1.0432
R1 = 1.0432-1 = 0.0432 > 10^-2, another Newton-Raphson iteration needed.
Calc. the correction to u1 = u1(1),
ẟu(2) = ẟu(1)-[k(u1(1)]^-1 x R1 = (-0.0665)-(8.6875)^-1 x(0.0432) = -0.0714
updated, u1(2) = u1(1)+ẟu(2) = 0.25-0.0714 = 0.1786
Checking Residual again,R2,
F(ext)=1, f(int)= (1) => (0.1786)^3+(9x0.1786^2)+(4x0.1786) = 1.0071
R2 = 1.0071-1 = 0.0071 < 10^-2
Therefore no iterations needed, final u1 = 0.1786.
Step 2&3:
similarly as step 1, we get u2=0.2966 , u3=0.3911.
| Step i | ∆Fi | ∆ui | 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 |
Exact values :
Take, F(u)= u3+9u2+4u = 1
using Scientific calculator, we get u=0.1776
take, F(u)= u3+9u2+4u = 2
using Scientific calculator, we get u= 0.2962
take, F(u)= u3+9u2+4u = 3
using Scientific calculator, we get u= 0.3910
Comparison Graph Implicit and Explicit Force vs Displacement :
The Implicit and Exact solutions are same line while explicit is away from exact solution.
Conclusion :
- The Implicit and Explicit numerical analysis is carried out for given equation.
- From the graph, Implicit method is more accurate but time consuming and solemnly depends on element size and timestep whereas explicit deviates slightly from exact result and needs more iterations to obtain good result.