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

$\Delta$u = incremental displacements ,$\Delta$F=incremental external applied forces.

using relation,$\Delta F=k\Delta u,$

$Step1:$

Take,u0=0.0,apply in (2)

(2) $\Rightarrow$k(u0)=3(0)^2+18(0)+4=4

now,$\Delta$u1 =$\Delta F/k(u0)\; =1/4=0.25$

$u1=u0+\Delta u1=0.25$

$Step2:$

$take\; u2=0.3651$

$(2)\; \Rightarrow \; k(u1)=3(0.25)^2+18(0.25)+4=8.6875$

$now,\Delta u1\; =\; \Delta F/k(u1)\; =1/8.6875\; =0.1151$

$u2=u1+\Delta u2=0.3651$

$Step3:$

$Take\; u2=0.3651$

$(2)\; \Rightarrow \; k(u2)=3(0.3651)^2+18(0.3651)+4=10.9716$

$now,\Delta u2\; =\; \Delta F/k(u2)\; =1/10.9716\; =0.09114$

$u3=u2+\Delta u3=0.46$

$finally,to\; determine\; wheather\; the\; analysis\; is\; in\; equilibrium,$

$F(ext)=\; \Delta F+\; \Delta F+\; \Delta F=3$

$f(int)=(1)\; \Rightarrow \; (0.46)^3+(9x0.46^2)+(4x0.46)=3.841$

$F(ext)\ne \; f(int),so\; they\; are\; not\; in\; equilibrium.$

$step\; i$

$\Delta Fi$

$\Delta ui$

$ui$

$F(ext)i$

$f(int)i$

$f(int)i-F(ext)i\; =\; R$

$1$

$1$

$0.2500$

$0.25$

$1$

$1.578$

$0.578$

$2$

$1$

$0.1151$

$0.3651$

$2$

$2.708$

$0.708$

$3$

$1$

$0.09114$

$0.46$

$3$

$3.841$

$0.841$

$Implicit\; analysis\; :$

$implicit\; analysis\; requires\; a\; numerical\; solver\; to\; invert\; the\; stiffness\; matrix\; once\; or\; even\; several\; times\; over\; the\; course\; of\; a\; load/timestep.this\; matrix\; inversion\; is\; an\; expensive\; operation\; ,especially\; for\; large\; models.it\; is\; mostly\; used\; for\; static\; analysis\; .implicit\; uses\; Newton-Raphson\; iterations\; to\; enforce\; equilibrium.if\; done\; correctly\; the\; Newton\; Raphson\; iterations\; will\; have\; a\; quadratic\; rate\; of\; convergence\; which\; is\; very\; desireable\; .$

$Implicit\; transient\; analysis\; has\; no\; inherent\; limit\; on\; the\; size\; of\; the\; time\; step.as\; such,implicit\; time\; steps\; are\; generally\; several\; orders\; of\; magnitude\; larger\; than\; explicit\; time\; steps.$

$where\; u=displacement,f=internal\; force\; in\; bar,F=exteranal\; force\; in\; bar$

$\Delta u=incremental\; displacements,\Delta F=incremental\; external\; applied\; forces.$

$using\; relation,\Delta F=k\Delta u,Tolerance=10^-2$

$Step1:$

$Take\; u0=0,$

$(2)\; \Rightarrow \; k(u0)=3(0)^2+18(0)+4=4$

$now,$$\Delta$u1 =$\Delta F/k(u0)\; =1/4=0.25$

$u1=u0+\Delta u1=0.25$

$checking\; residual,R$

$F(ext)=\; \Delta F=1 \; f(int)=(1)\; \Rightarrow (0.25)^3+(9x(0.25)^2)+(4x0.25)=1.578$

$R0=1.578-1=0.578>10^-2$

$so,newton\; raphson\; iterations\; are\; necessary.$

$calculate\; the\; correction\; to\; u1=u1(0)$

$\delta u(1)=-[k(u1(0)]^-1\; x\; R0=-[(3x0.25^2)+(18x0.25)+4]^-1x0.578=-0.0665$

$updated\; u1(1)=u1(0)+\delta u(1)=0.25-0.0665=0.1835$

$check\; residual\; again,R1,$

$F(ext)=1 \; f(int)=(1)\; \Rightarrow (0.1835)^3+(9x(0.1835)^2)+(4x0.1835)=1.0432$

$R1=1.0432-1=0.0432>10^-2$

$so,another\; newton\; raphson\; iterations\; are\; necessary.$

$calculate\; the\; correction\; to\; u1=u1(1)$

$\delta u(2)=\delta u(1)-[k(u1(0)]^-1\; x\; R1=(-0.0665)-[(3x0.25^2)+(18x0.25)+4]^-1x0.04325=1.0071$

$R2=1.0071-1=0.0071<10^-2$

$there\; fore\; no\; iterations\; are\; needed,final\; u1=0.1786$

$Step2\&3:$

$similarly\; as\; step\; 1,we\; get\; u2=0.2966,u3=0.3911.$

$step\; i$

$\Delta Fi$

$\Delta ui$

$ui$

$F(ext)i$

$f(int)i$

$f(int)i-F(ext)i\; =\; R$

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)= 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.