Week - 2 - Explicit and Implicit Analysis

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,

1.Step 1

u=0, k(u)=3x0^2+18(0)+4

k(u)=4

∆F=1

∆U1=∆F/K(U)=1/4=0.25

U1=0+0.25=0.25

Step 2

K(U1)=3X(0.25)^2+18(0.25)+4

0.125+4.5+4

=8.625

∆U2=∆F/K(U1)=1/8.625=0.115

U2=U1+∆U2=0.25+0.115=0.365

Step 3

K(u2)=3x(0.3651)^2+18x(0.3651)+4=10.97

∆U3=∆F/K(U2)=1/10.97=0.9114

U3=U2+∆U3=0.365+0.9114=0.46 *(please explain how to find the (fint)i value for explicit analysis)*

Step i	∆F	∆u	ui	(Fext)	(fint)i	fint − Fext = R
1	1	0.25	0.25	1
2	1	0.115	0.365	1
3	1	0.9114	0.46	1

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