Week - 2 - Explicit and Implicit Analysis
Aim :To solve the equation in both Explicit and implicit method
Procedure :
Explicit analysis :
Lsdyna is used to solve the explicit analysis mostly
Static analysis is done using an implicit solver in LS-DYNA. Dynamic analysis can be done via the explicit solver or the implicit solver.
In nonlinear implicit analysis, solution of each step requires a series of trial solutions (iterations) to establish equilibrium within a certain tolerance. In explicit analysis, no iteration is required as the nodal accelerations are solved directly.
Implicit analysis ;
An implicit analysis is the same as the explicit analysis, except that at the end of each step
Newton-Raphson iterations are used to enforce equilibrium before moving to the next step.
Basically, an incremental force is applied to advance the solution forward at the beginning
of a step. However, internal forces and external forces will not be in equilibrium unless the
stiffffness is linear for the given step. Hence, in order to achieve equilibrium, corrections must
be made to the displacement. This is accomplished by using Newton-Raphson iterations to
minimize the residual, R(u) = fint − Fext. Expanding the residual as a Taylor series about
the current displacement it is used
| 3(0)^2+18^(0)+4 | |||||||||
| 4 | |||||||||
| step I | Delta F | Delta U | Ui | F(ext)i | F(int)i | Fint-Fext=R | |||
| Delta U1=DeltaF/k(U0) | |||||||||
| (1/4) | 1 | 1 | 0.25 | 0.25 | 1 | 1.5 | 1 | ||
| 0.25 | 2 | 1 | 0.11 | 0.36 | 2 | 2.6 | 0.6 | ||
| 3 | 1 | 0.092 | 0.45 | 3 | 3.7 | 0.7 | |||
| U1=U0 +DeltaU1 | |||||||||
| (0+0.25) | |||||||||
| 0.25 | |||||||||
| k(U1)= 3u1^2+18^u1+4 | |||||||||
| 3(0.25)^2+18^(0.25)+4 | |||||||||
| 8.68 | |||||||||
| Delta U2 = DeltaF/k(U1) | |||||||||
| (1/8.68) | |||||||||
| 0.11 | |||||||||
| U2=U1+DeltaU2 | |||||||||
| 0.25+0.11 | |||||||||
| 0.36 | |||||||||
| k(U2)=3U2^2+18U2+4 | |||||||||
| 3(0.36)^2+18(0.36)+4 | |||||||||
| 10.868 | |||||||||
| Delta U3=DeltaF/K(u2) | |||||||||
| 1/10.868 | |||||||||
| 0.0920 | |||||||||
| U3 = U2+DeltaU3 | |||||||||
| 0.36+0.0920 | |||||||||
| 0.45 | |||||||||
| Fext=DeltaF1+DeltaF2+DeltaF3 | |||||||||
| 1+1+1 | |||||||||
| 3 | |||||||||
| Fint= | |||||||||
| F(U1)=U^3+9U^2+4U | |||||||||
| (0.25)^3+9(0.25)^2+4(0.25) | ans | ||||||||
| 1.578125 | 0.25 | ||||||||
| 2.653056 | 0.36 | ||||||||
| 3.713625 | 0.45 | ||||||||
| F(U2)=U^3+9U^2+4U | |||||||||
| (0.36)^3+9(0.36)^2+4(0.36) | |||||||||
| F(U3)=U^3+9U^2+4U | |||||||||
| (0.45)^3+9(0.45)^2+4(0.45) | |||||||||
| Implicit analysis | |||||||||
| F(U0) = U^3+9U^2+4U | |||||||||
| k(U0)= 3u^2+18^u+4 | |||||||||
| U0=(0) | |||||||||
| 3(0)^2+18(0)+4 | |||||||||
| K(U0)= 4 | |||||||||
| Delta U1=DeltaF/k(U0) | |||||||||
| (1/4) | |||||||||
| 0.25 | |||||||||
| U1=U0 +DeltaU1 | |||||||||
| 0+0.25 | |||||||||
| U1 =0.25 | |||||||||
| Fext=Delta F | |||||||||
| Fext=1 | |||||||||
| fint= U^3+9U^2+4U | |||||||||
| (0.25)^3+9(0.25)^2+4(0.25) | 0.01 | ||||||||
| Fint=1.5781 | |||||||||
| R^0=fint-fext | |||||||||
| 1.5781-1 | |||||||||
| R = 0.578 | |||||||||
| DeltaU^j+1= -(K(U^j))^-1R^j | |||||||||
| -0.06 | -0.06 | ||||||||
| U0= U0+CDelta U1 | |||||||||
| U0= 0.25-0.06 | |||||||||
| 0.19 | |||||||||
| Fext=Delta F | |||||||||
| Fext=1 | |||||||||
| Fint=U^3+9U^2+4U | |||||||||
| (0.19)^3+9(0.19)^2+4(-0.19) | |||||||||
| Fint= 1.08 | |||||||||
| R^1=Fint-Fext | |||||||||
| 0.08 | |||||||||
| DeltaU^j+1= CDelta U1-(K(U^j))^-1R^j | `∂ u ( 2 )^ 1 = ∂ u ( 1 )^ 1 − [ k ( u ( 1 )^ 1 ) ] ^(− 1) R^ ( 1 )` | ||||||||
| (-0.06-(10.868)^-1 (0.08) | |||||||||
| -0.067 | |||||||||
| U=U+CDelta | |||||||||
| 0.25-0.067 | |||||||||
| 0.183 | |||||||||
| Fext=1 | |||||||||
| Fint=U^3+9U^2+4U | |||||||||
| 1.017 | |||||||||
| R2=Fint-Fext | |||||||||
| 1.017-1 | |||||||||
| 0.017 | |||||||||
| u1=0.183*0.183=0.033 | |||||||||
| Step 2 | |||||||||
| k(U1)= 1+0.033 =1.033 | |||||||||
| Delta F=1 | |||||||||
| Delta U2=DeltaF/K(U1)= 0.966 | |||||||||
| U2=U1+DeltaU2 | |||||||||
| 1.033+0.966 =2 | |||||||||
| Fext=DeltaF=1 | |||||||||
| Fint=U2^3/3=2.66 | |||||||||
| R(0) = fint − Fext = 1.66 | |||||||||
| R(0) | |||||||||
| > 10−2 ⇒ Newton-Raphson iterations are necessary. | |||||||||
| Calculate the correction to u2 = u (0) 2 | |||||||||
| δu(1) = −[k(u (0) 2 )]−1R(0) = *10.868^-1 (1.66) = -0.1522 | |||||||||
| ⇒ The updated value of u (1) 2 = u (0) 2 + δu(1) = 2-0.1522=1.874 | |||||||||
| Check the residual again, Fext = 1, fint = u (1) 2 + (u (1) 2 )3/3 = 2.344 | |||||||||
| R(1) = fint − Fext = 2.344-1.41=0.933 | |||||||||
| R(1) | |||||||||
| > 10−2 ⇒ another Newton-Raphson iteration is necessary. | |||||||||
| Calculate the new total correction to u2 = u (0) 2 | |||||||||
| δu(2) = δu(1) − [k(u (1) 2 )]−1R(1) = 0.25*(8.68)^-1*0.933=-0.026 | |||||||||
| The updated value of u (2) 2 = u (0) 2 + δu(2) = 1.874 + (−0.026) = 1.847 | |||||||||
| Check the residual, Fext = 1, fint = u (2) 2 + (u (2) 2 )3/3 = 3.411 | |||||||||
| R(2) = fint − Fext = 1-3.411=-2.411 | |||||||||
| < 10−2 ⇒ no further iterations are necessary | |||||||||
| Step3 | |||||||||
| StepI | DELTAFi | Ui | F(ext)i | F(int)i | Fint-Fext=R | ||||
| 1 | 1 | 0.966 | 1 | 2.66 | 1.66 | ||||
| 2 | 1 | -0.1522 | 1.41 | 2.344 | 0.933 | ||||
| 3 | 1 | -0.026 | 1 | 3.411 | -2.411 | ||||
Conclusion : Thus the problem is solved In both Explicit and Implicit analysis method