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)
stiffness k(u)=dF/du= 3u^2+18u+4 ---- (2)
Explicit Analysis:
In Explicit analysis mostly dynamic analysis and nodal accelerations are solved directly (not iteratively) as the inverse of the diagonal matrix times the net nodal force vector where net nodal force includes contributions from exterior sources(body forces,applied pressure,contact etc),elememt stress,damping,bulk viscosity,and hour glass control.once accelerations are known at time n,velocities are calculated at time (n+1)/2,and displacements at time n+1.strain can be calculated from displacements and stress will come from strain and the cycle repeated.
Explicit analysis handles nonliniarities with relative ease as compared to implicit analysis.This would include treatment of contact and material nonlinearities.with this method is that you 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
Δu = incremental displacements ,ΔF=incremental external applied forces.
using relation,ΔF=kΔu,
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 u2=0.3651
(2) ⇒ k(u1)=3(0.25)^2+18(0.25)+4=8.6875
now,Δu1 = ΔF/k(u1) =1/8.6875 =0.1151
u2=u1+Δu2=0.3651
Step3:
Take u2=0.3651
(2) ⇒ k(u2)=3(0.3651)^2+18(0.3651)+4=10.9716
now,Δu2 = ΔF/k(u2) =1/10.9716 =0.09114
u3=u2+Δu3=0.46
finally,to determine wheather the analysis is in equilibrium,
F(ext)= ΔF+ ΔF+ ΔF=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
ΔFi
Δui
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/timestep.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=exteranal force in bar
Δu=incremental displacements,ΔF=incremental external applied forces.
using relation,ΔF=kΔu,Tolerance=10^-2
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) ⇒(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.
calculate the correction to u1=u1(0)
δu(1)=-[k(u1(0)]^-1 x R0=-[(3x0.25^2)+(18x0.25)+4]^-1x0.578=-0.0665
updated u1(1)=u1(0)+δu(1)=0.25-0.0665=0.1835
check residual again,R1,
F(ext)=1 f(int)=(1) ⇒(0.1835)^3+(9x(0.1835)^2)+(4x0.1835)=1.0432
R1=1.0432-1=0.0432>10^-2
so,another newton raphson iterations are necessary.
calculate the correction to u1=u1(1)
δu(2)=δu(1)-[k(u1(0)]^-1 x R1=(-0.0665)-[(3x0.25^2)+(18x0.25)+4]^-1x0.04325=1.0071
R2=1.0071-1=0.0071<10^-2
there fore no iterations are needed,final u1=0.1786
Step2&3:
similarly as step 1,we get u2=0.2966,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
Exact values:
Take,F(u)= u^3+9u^2+4u =1
using scientific calculator ,we get u=0.1776
Take,F(u)= u^3+9u^2+4u =2
using scientific calculator ,we get u=0.2962
Take,F(u)= u^3+9u^2+4u =3
using scientific calculator ,we get u=0.3910
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 where as explicit deviates slightly from exact result and needs more iterations to obtain good result.