Week - 2 - Explicit and Implicit Analysis
EXPLICIT AND IMPLICIT ANALYSIS
F(u)=u^3+9u^2+4u ------[1]
solve using both explicit and implicit Methods (have a tolerance of 10^-2).
Stifness, K(U)=dF/dU=3u^2+18u+4 -------[2]
EXPLICIT 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 displacement at time n+1.From displacements comes strain .From strain comes stress and the cycle is repeated.
Explicit analysis handles nonlineraties with relative ease as compared to implicit analysis.This would include treatement of contact and material nonlineraties.with this method is that you do need many small increments for good accuracy and it is time consuming.
Given data:
where u-displacement ,f=internal force in bar,F=external force in bar
Δu=increment dispalcements,ΔF=increment external force applied forces
using relation,ΔF=kΔu, ΔF=1
step1:
Take,u0=0,0,apply in [2]
[2]==>k(u0)=3*0^2+18(0)+4=4
now,Δu1=ΔF/k(u0)=1/4=0.25
u1=u0+Δu1=0.25
step2:
Take,u1=0.25
[2]==>k(u1)=3*(0.25)^2+18(0.25)+4=8.6875
now,Δu2=ΔF/k(u1)=1/8.6875=0.1151
u2=u1+Δu2=0.25+0.1151=0.3651
step3:
Take,u2=0.3651
[2]==>k(u2)=3*(0.3651)^2+18(0.3651)+4=10.9716
now,Δu3=ΔF/k(u2)=1/10.9716=0.09114
u3=u2+Δu3=0.46
Finally,to determine wheather the analyis is in equalibrium ,
f(ext)=ΔF+ΔF+ΔF=1+1+1=3
f(int)=[1]==>(0.46)^3+9(0.46)^2+4(0.46)=3.841
f(ext)≠f(int),so they are not in equlibrium.
IMPLICIT ANALYSIS :
Implicit analysis requires a numerical solver to invert the stiffness matrix once or even several times over the course of load/time step.This matrix inversion is an expensive operation.especially for large models It is mostly used for static analysis .Implicit uses Newton-Rapshon iterations to enforce equlibrium .
Implicit transient analysis has no inherent limit on the size of the time step. As,such,implicit time steps are generallly several orders of magnitude larger than explicit time steps.
Where u=displacement,f=internal forces in bar, F=external force in bar
Δu=incremental displacements,ΔF=incremental external applied forces.
using relations,ΔF=kΔu,Tolerance=0.001
step1:
Take u0=0
[2]==>k(u0)=3(0)^2+18(0)+4=4
now,Δu1=ΔF/K(U0)=1/4=0.25
u1=u0+Δu1=0.25
checking residual.R
F(ext)=ΔF=1 f(int)=[1]==>(o.25)^3+9(0.25)^2+4(0.25)=1.578
R0=1.5781-1=0.578>0.001
So,Newton-Raphson iterations are necessary.
calc.the correction to u1=u1(0)
wkt,δu(1)=-k[(u1(0))]^-1*RO= -[(3*0.25^2)+18(0.25)+4]^-1*0.578= -0.0665
update, u1(1) =u1(0)+δu(1)=0.25-0.0665=0.1835
check residual again,R1
F(ext)=1.f(int)=(0.1835)^3+9(0.1835)^2+4(0.1835)=1.0432
R1=f(ext)-F(int)=1.0432-1=0.0432>0.001,another Newton-Raphson iteration needed.
calc.the correction to u1=u1(1),
[ku1(1)]=[3*(0.1835)^2+18(0.1835)+4=7.404
δu(2)=δu(1)-[k(u1(1)]^-1*R1=(-0.0065)-[3(0.1835)^2+18(0.1835)+4]^-1*(0.0432)=(-0.0665)-(7.404) -1*0.0432= -0.07233
updated,u1(2)=u1(0)+δu(2)=0.25-0.07233=0.1777
check the residual.F(ext)=1,f(int)=(0.1777)^3+9(0.177)^2+4(0.1777)=1.0006
R2=f(ext)-F(int)-1.0006-1=0.0006
R2<0.001 so,no furthur iterations are necessary
Therefore,the final value is u1=u1(2)=0.1777
similarly for step2 and step3 should be done .