Explicit and Implicit Analysis
OBJECTIVE:
The main objective of this challenge is to solve the third-order polynomial equation with both the Implicit and Explicit analysis and compare it with the Exact solution. The given equation which is in tension is expressed as follows,
`F(u)=u^3+9u^2+4u`
Explicit Method :
Basically, the work done by the system is represented as force*displacement. If the non-linear force and the displacement are not known, but the stiffness of the system is a known parameter it is possible to construct a graph for the force (vs) displacement of the system.
To find these parameters a method called an Incremental Explicit Load control scheme is used. In this method, three equal load increments of `Deltat=1` unit each are used to load the tension bar.
The variables used are u for displacement, f for internal forces of the bar, F is the external force applied to the bar and k is the stiffness.
`F(u) = u^3 + 9u^2 + 4u`...........eqn(1)
From eqn(1) it follows the Stiffness of the bar,
`K(u) = (df)/(du)`
`K(u) = 3u^2+18u+4`
Step-1 :
`u_0 = 0.0`, `K(u_0)=4`, `DeltaF_1`,
`Deltau_1` = (`DeltaF_1`)/(K(u_0))=0.25`
`Deltau_1` = 0.25
`u_1=u_0+Deltau_1=0+0.25`
`u_1=0.25`
Step-2 :
`K(u_1) = 3(u_1)^2+18u_1+4`
`K(u_1) = 0.1875+4.5+4`
`K(u_1) = 8.6875`
`DeltaF_2=1`,
`DeltaU_2` = (`DeltaF_2`)/(K(u_1))=1/8.6875 = 0.1151`
`DeltaU_2`=0.1151
`u_2=u_1+`DeltaU_2` = 0.25+0.1151`
`u_2=0.3651`
Step-3 :
`K(u_2) = 3u^2+18u+4`
`K(u_2) = 0.3999+6.5718+4`
`K(u_2) = 10.9698`
`DeltaF_3=1`
`DeltaU_3` = (`DeltaF_3`)/(K(u_2))=1/10.9698 = 0.0911`
`u_3=u_2+`DeltaU_3` = 0.3651+0.0911`
`u_3=0.4562`
All the displacement values are
U_1=0.25; U_2=0.3651; U_3=0.4562.
External & Internal Force Calculation :
`F_(ext)=DeltaF_1+DeltaF_2+DeltaF_3`
`F_(ext)=3`
`f_(i_1)` = u_1^3+9u_1^2+4u_1`
= 0.0156 + 0.5625 + 1
`f_(i_1)` = 1.5781`
`f_(i_2)` = u_2^3+9u_2^2+4u_2`
= 0.0486+1.199+1.46
`f_(i_2)` = 2.7076`
`f_(i_3)`= u_3^3+9u_3^2+4u_3`
= 0.0948 + 1.871 + 1.824
`f_(i_3)`= 3.7898`
Table of results for Explicit analysis:
| Step i | `DeltaF_i` | `DeltaU_i` | `u_i` | `(F_(ext))_i` | `(f_i)_i` | `f_i-`(F_(ext))_i` = R` |
| 1 | 1 | 0.25 | 0.25 | 1 | 1.5781 | 0.5781 |
| 2 | 1 | 0.1151 | 0.3651 | 2 | 2.7076 | 0.7076 |
| 3 | 1 | 0.0911 | 0.4562 | 3 | 3.7898 | 0.7898 |
Implicit Method :
The Implicit analysis is also the same as 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 the step.
However, internal forces and external forces will not be in equilibrium unless the stiffness is linear for the given timestep. Hence in order to achieve equilibrium, corrections must be made to the displacement.
`R(u)=f_(Int)-F_(ext)`
By expanding the Newton-Raphson iterations to minimize the residual to a simpler form.
`deltau^(j+1)=-[K(u^j)]^(-1)*R^j`
From that, the iterations variable is (j) and it may take several iterations for the norm of the residual below the chosen tolerance criteria(tolerance value is about below 10^(-2) is chosen).
`F(u) = u^3 + 9u^2 + 4u`...........eqn(2)
From eqn(2) it follows the Stiffness of the bar,
`K(u) = (df)/(du)`
`K(u) = 3u^2+18u+4`
Step-1 :
`u_0 = 0.0`, `K(u_0)=4`,`DeltaF_1=1`,
`Deltau_1` = (`DeltaF_1`)/(K(u_0))=0.25`
`u_1=u_0+`Deltau_1` =0+0.25`
`u_1=0.25`
Check the residual
`F_(ext)=DeltaF=1``
`f_i = u_1^3+9u_1^2+4u_1`
= 0.015625+0.5625+1
`f_i = 1.5781`
`R^(0)=f_i-F_(ext)`=0.5781
`||R^(0)||>10^-2` - Newton-Raphson iterations are necessary.
Calculate the correction to `u_1 = u_1^(0)`
`deltau^1` =-[K(u_1^(0))]^-1*R^0`
`deltau^1` =-(8.6875)^-1*0.5781`
`deltau^1`= -0.0665
The updated value of`u_1^(1) = u_1^(0) + `deltau^1`
`u_1^(1) = 0.25-0.0665`
`u_1^(1) = 0.1834`
Check the residual again, `F_(ext)=DeltaF=1`,
`f_i = (u_1^(1))^3+9(u_1^(1))^2+4(u_
= 0.0062+0.303+0.73384
`f_i = 1.04304`
`R^(1)=f_i-F_ext`
`R^(1)= 1.04304-1`
`R^(1)= 0.043`
`||R^(1)||>10^-2` - Another Newton-Raphson iterations are necessary.
Calculate the new total correction to `u_1 = u_1^(0)`
`deltau^2`= `deltau^1` - [K(u_1^(1))]^-1*R^1`
`deltau^2`= -0.0665 - (7.40318)^-1*0.043`
`deltau^2` = -0.0724
The updated value of`u_1^(2) = u_1^(1) + `deltau^2`
`u_1^(2) = 0.25-0.0724`
`u_1^(2) = 0.1776`
Check the residual again, `F_(ext)=DeltaF=1`,
`f_i = (u_1^(2))^3+9(u_1^(2))^2+4(u_
= 0.0056+0.282+0.7108
`f_i = 0.9984`
`R^(2)=f_i-F_ext`
`R^(2)= 0.9984-1`
`R^(2)= -0.0016`
`||R^(1)||<10^-2` - No further iterations are necessary.
Therefore the final value of `u_1=u_1^(2) = 0.1777`
Step-2 :
`K(u_1) = 3u^2+18u+4`,
`K(u_1) = 0.0947+3.1986+4`,
`K(u_1) = 7.2933`,
`DeltaF_2=1`,
`Deltau_2`= (`DeltaF_2`)/(K(u_1))=0.1371`
`u_2=u_ 1+ `Deltau_2`=0.1777+0.1371`
`u_2=0.3148`
Check the residual
`F_(ext)`=`DeltaF`=2,`
`f_i = u_2^3+9u_2^2+4u_2`
= 0.0312+0.8918+1.2592
`f_i = 2.1822`
`R^0=f_i-F_ext= 0.1822`
`||R^(0)||>10^-2` - Newton-Raphson iterations are necessary.
Calculate the correction to `u_2 = u_2^(0)`
`K(u_2) = 3u^2+18u+4`
= 0.2972+5.6664+4
`K(u_2) = 9.9636`
`deltau^(1) =-[K(u_2^(0))]^-1*R^0`
`deltau^(1) =-(9.9636)^-1*0.1822`
`deltau^(1) = -0.0183`
The updated value of`u_2^(1) = u_2^(0) + deltau^(1)`
`u_2^(1) = 0.3148-0.0183`
`u_2^(1) = 0.2965`
Check the residual again, `F_(ext)=DeltaF=2`,
`f_i = (u_2^(1))^3+9(u_2^(1))^2+4(u_
= 0.0260+0.7912+1.186
`f_i = 2.0032`
`R^(1)=f_i-F_ext`
`R^(1)= 2.0032-2`
`R^(1)= 0.0032`
`||R^(1)||<10^-2` - No further iterations are necessary.
Therefore the final value of `u_2=u_2^(1) = 0.2965`
Step-3 :
`K(u_2) = 3u^2+18u+4`,
`K(u_2) = 0.2637+5.337+4`,
`K(u_2) = 9.6007`,
`DeltaF_2=1`,
`Deltau_3` = (`DeltaF_2`)/(K(u_2))=0.1041`
`u_3=u_ 2+triangleu_3=0.2965+0.1041`
`u_3=0.4006`
Check the residual
`F_(ext)=DeltaF=3`
`f_i = u_3^3+9u_3^2+4u_3`
= 0.0643+1.4443+1.6024
`f_i = 3.111`
`R^0=f_i-F_ext= 0.111`
`||R^(0)||>10^-2` - Newton-Raphson iterations are necessary.
Calculate the correction to `u_3 = u_3^(0)`
`K(u_3) = 3u^2+18u+4`
= 0.4814+7.2108+4
`K(u_3) = 11.6922`
`deltau^(1) =-[K(u_2^(0))]^-1*R^0`
`deltau^(1) =-(11.6922)^-1*0.111`
`deltau^(1) = -0.0095`
The updated value of`u_3^(1) = u_3^(0) + deltau^(1)`
`u_3^(1) = 0.4006-0.0095`
`u_3^(1) = 0.3911`
Check the residual again, `F_(ext)=DeltaF=3`,
`f_i = (u_3^(1))^3+9(u_3^(1))^2+4(u_
= 0.0598+1.3766+1.5644
`f_i = 3.0008`
`R^(1)=f_i-F_e`
`R^(1)= 3.0008-3`
`R^(1)= 0.0008`
`||R^(1)||<10^-2` - No further iterations are necessary.
Therefore the final value of `u_3=u_3^(2) = 0.3911`
All the displacement values are
u_1=0.1777; u_2=0.2965; u_3=0.3911
Summary of Implicit analysis results :
| Step i | `DeltaF_i` | `Deltau_i` | `u_i` | `(F_(ext))_i` | `(f_i)_i` | `f_i-`(F_(ext))_i` = R` |
| 1 | 1 | 0.25 | 0.1777 | 1 | 0.9984 | -0.0016 |
| 2 | 1 | 0.1371 | 0.2965 | 2 | 2.0032 | 0.0032 |
| 3 | 1 | 0.1041 | 0.3911 | 3 | 3.0008 | 0.0008 |
Exact Solution by solving third order Polynomial :
`F(u) = u^3 + 9u^2 + 4u`
`u^3 + 9u^2 + 4u - F(u) = 0`
| Step i | `(F_(ext))_i` | `u_i` |
| 1 | 1 | 0.17762 |
| 2 | 2 | 0.29616 |
| 3 | 3 | 0.39103 |
Thus the third-order polynomial equation under a tensile force is analytically solved both in the Implicit and Explicit method. The results are compared with the exact solutions and the graph is plotted for Implicit, Explicit, and exact solution.