Week - 2 - Explicit and Implicit Analysis
Objective:
1)Solve the given equation for the implicit and explicit analysis.
2)Explain details for implicit and explicit analysis.
Implicit analysis:
(a)In case of static analysis such as stress analysis in FEA, is solved using the simple linear equation {F}=[k]{U}.In this case time step is absent i.e the solution eqautions are not time dependent.
(b)In case of dynamic analysis soved by the the eqaution like
[M]{x''}+[C]{x'}+[K]{x}={F} are time dependent and defined by time step.
(c)Implicit analysis is the type of analysis in which the Eulers time integration is used i.e.the calculation of current quantities in one time step are based on the quantities calculated in the previous time step.
(d)The time steps defined are large and contains constant incremental intervals.Also it contains the inversion of stiffness matrix of the system of operation.
Explicit analysis:
(a)Each time step in explicit analysis is very fast and it is small.
(b)The time step in explicit analysis must be less than the courant time step (time it takes a sound wave to travel across an element).In explicit dynamic analysis, nodal accelerations are solved directly and not iteratively.
(c)It doesn't require inversion of stiffness matrix only mass matrix inversion is sufficient.Explicit analysis handles nonlinearities with relative ease as compared to implicit analysis.
[2]As demonstrated in the PDF, F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods
F(u) = u3+9u2+4u (1)
K(u)=df/du=3u^2+18u+4 (2)
Explicit analysis:
By using equation `delta`(F)=k*`delta`(u)
F=External force applied
k=Material stiffness
u=Displacement
(i) u(0)=0
`u(0) = 0.0, ⇒ k(u0) = 1, delta F = 1, delta u1 = delta F/k(u0) = 0.25, u1 = u0 + delta u1 =0.25`
(ii) u(1)=
`k(u1) = 3(1/4)^2+18(1/4)+4=8.6875, delta F = 1, delta u2 = delta F/k(u1) = 1/8.6875=0.1151, u2 = u1 + delta u2 = 0.3652`
`k(u2) = 3(0.3651)^2+18(0.3651)+4=10.9717, delta F = 1, delta u3 = delta F/k(u2) = 1/10.9717=0.0911, u3 = u2+delta u3=0.4562`
| Sr No | Delta(F) | Delta(u) | u(i) | F(ext) | F(int) | R=F(int)-F(ext) |
| 1 | 1 | 0.25 | 0.25 | 1 | 1.5781 | 0.5781 |
| 2 | 1 | 0.1151 | 0.3651 | 2 | 2.7087 | 0.7087 |
| 3 | 1 | 0.0911 | 0.4562 | 3 | 3.7928 | 0.7928 |
Implicit analysis:
`u(0) = 0.0, ⇒ k(u0) = 4, delta F = 1, delta u1 = delta F/k(u0) =1/4=0.25, u1 = u0 + delta u1 =0.25 Check the residual, Fext = delta F = 1, Fint = u1 + u31=1.5781, R(0) = Fint − Fext = 1.5781-1.0000=0.5781 ⇒ Newton-Raphson iterations are necessary.Calculate the correction to u1 = u(0) delta u(1) = −[k(u(0)^-1 )]−1R(0) =-(8.6875)^-1*(0.5781)= −0.0665.⇒ The updated value of u1(1) = u(0)1 + delta u(1) = 0.25 + (−0.16667) = 0.1835.Check the residual again,R(1) = Fint − Fext =0.25+(-0.0655)=0.0432 R(1)>10^-2⇒ Newton-Raphson iterations are necessaryCalculate the correction to u1 = u1^(0),delta u(2) = delta u(1) − [k(u(1) 1 )]−1R(1) = −0.0665 − (7.404)^(-1)*0.0432 = −0.0723,The updated value of u1(2)=u1(0)+delta u2-0.25+(-0.0723)=0.1777,Residual,R(2)=Fint-Fext=0.0006,Newton Raphson iterations are not necessary,Final value of u1=0.1777`
Similarly repeating above processu2=0.2132 and u3=0.3411
Conclusion:
The nonlinear problem solved for both implicit and explicit analysis.
The explicit analysis gives random values so implicit analysis is used with the help of Newton Raphson method of iterations.