Week - 2 - Explicit and Implicit Analysis
OBJECTIVE:-
The objective of this challenge is to give a detailed report about implicit and explicit methods and solve the give non-linear equation (`f(u) = u^3+9u^2+4u`) in both implicit and explicit methods.
PROCEDURE:-
Implicit & Explicit Methods:
Implicit Method :
Implicit Method is a time indepndent method. It is used to solve linear equation problems. Since implicit method is a time independent one so it dosen't have a condition to be stable so it is called as unconditionally stabe method. Mostly in Implicit Method we will mainly concentrate on the final output and don't mind about the things running in between the process. the disadvantage of this type is in calculation we have to calculate the inverse k matrix which is a very complex one.
Explicit Method :
Explicit Method is a time depndent method. It is used to solve Non-linear equation problems. Since there is a condition for the method to be stable, it is called as conditionally stable method. In this method the force and acceleration of the methods various rapidly within the short period of time. As compared to implicit method here we dont need to solve inverse k matrix instead we we solve inverse mass(m) matrix which will be a diagonal matrix.
Problem:
`f(u)=u^3+9u^2+4u`. (equ-1)
From equ-1 stiffness {k} is
`k(u)=(df)/(du)=3u^2+18u+4`. (equ-2)
Explicit Analysis
The non-linear force vs diaplacement relationship is not known but the stiffness matrix is known, so it is Clearly possible to plot a graph with froce (f) vs displacement (u) numerically by using equation-1. We are using incremental explicit load control scheme method to solve, Here load increment (or step) ΔF=1 unit each to load. The variables used are displacement (u), internal force (f), external froce (F), stiffness (K), incremental displacement (Δu) and incremental extenal force (ΔF). Thus with incremental displacement and incremental external force a relationship is formed ΔF = kΔu.
step-1:
`u_0=0.0`.
`k(u_0)=3(u_0)^2+18(u_0)+4`.
`=3(0)^2+18(0)+4`.
`k(u_0)=4`.
`ΔF_1=1`, `=>` `Δu_1=(ΔF_1)/(k(u_0))`.
`=1/4`.
`Δu_1=0.25`.
`u_1=u_0+Δu_1`.
`=0+0.25`.
`u_1=0.25`.
step-2:
`u_1=0.25`.
`k(u_1)=3(u_1)^2+18(u_1)+4`.
`=3(0.25)^2+18(0.25)+4`.
`=0.19+4.5+4`.
`k(u_1)=8.69`.
`ΔF_2=1`, `=>` `Δu_2=(ΔF_2)/(k(u_1))`.
`=1/8.69`.
`Δu_2=0.12`.
`u_2=u_1+Δu_2`.
`=0.25+0.12`.
`u_2=0.37`.
step-3:
`u_2=0.37`.
`k(u_2)=3(u_2)^2+18(u_2)+4`.
`=3(0.37)^2+18(0.37)+4`.
`=0.41+6.66+4`.
`k(u_1)=11.07`.
`ΔF_3=1`, `=>` `Δu_3=(ΔF_3)/(k(u_1))`.
`=1/11.07`.
`Δu_3=0.09`.
`u_3=u_2+Δu_3`.
`=0.37+0.09`.
`u_3=0.46`.
Finally, From the above three step the following results are obtained.
`F_(ext)=ΔF_1+ΔF_2+ΔF_3`.
`=1+1+1`.
`F_(ext)=3`.
`F_(Int)=(u_3)^3+9(u_3)^2+4(u_3).`
`=(0.46)^3+9(0.46)^2+4(0.46).`
`=0.10+1.91+1.84.`
`F_(Int)~=3.85.`
It is evident from the result that external and internal force are not in equilibrium because `F_(ext)≠F_(Int)`.
Table-1 summarizes the main results for the explicit analysis.
| Step | `Δ F_i` | `Δ u_i` | `u_i` | `(F_(ext))_i` | `(F_(Int))_i` | `R=F_(Int)-F_(ext)` |
| 1 | 1.0 | 0.25 | 0.25 | 1 | 1.58 | 0.58 |
| 2 | 1.0 | 0.115 | 0.37 | 2 | 2.76 | 0.76 |
| 3 | 1.0 | 0.09 | 0.46 | 3 | 3.85 | 0.85 |
Table-1: Summary of Explicit Analysis Result
Implicit Analysis
An implicit analysis is 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 begining of a step. However, internal force and external force will not be in equilibrium unless the stiffness is linear for the given step. Hence, in order to achive equilibrium, correction must be made to the displacement. This is accomplished by using Newton-Raphos iteration to minimize the resudial, `R(u)=F_(Int)-F_(ext).` Expanding the resudial as a taylor series about the current displacement `u^j` gives
`R^(j+1)=R^j(u^j)+(dR(u^j))/(du)deltau^(j+1)+...=R^j+k(u^j)deltau^(j+1)+...`
By neglecting higher order terms in the series, setting `R^(j+1)=0` and solving for `deltau^(j+1)` the following correction is obtained
`deltau^(j+1)=-[k(u^j)]^-1R^j`
Notice that the iteration variable is j and it may take several iterationfor the norm of the resudial, `||R||`, to be reduced below the chosen tolerance criteria (for the example below a tolerance of `10^-2` is chosen). the Implicit Analysis proceeds as follows
Step-1:
`u_0=0.0`.
`k(u_0)=3(u_0)^2+18(u_0)+4`.
`=3(0)^2+18(0)+4`.
`k(u_0)=4`.
`ΔF_1=1`, `=>` `Δu_1=(ΔF_1)/(k(u_0))`.
`=1/4`.
`Δu_1=0.25`.
`u_1=u_0+Δu_1`.
`=0+0.25`.
`u_1=0.25`.
Check the residual, `F_(ext)=ΔF=1,` `F_(Int)=u_(1)^3+9u_(1)^2+4u_(1)`
`=(0.25)^3+9(0.25)^2+4(0.25)`
`F_(Int)=1.5751.`
`||R^(0)||=F_(Int)-F_(ext)`
`=1.5781-1`
`||R^(0)||=0.5781.`
`||R^((0))||>10^-2 =>` Newton-Raphson iteration are necessary.
calculate the correction of `u_1=u_1^(0)`
`deltau^(1)=-[k(u_1^(0))]^(-1)*R^(0)`
`=-8.6875^(-1)(0.5781)`
`deltau^(1)=0.0665.`
`=>` the updated value of `u_1^(1)=u_1^(0)+deltau^(1)`
`=0.25-0.0665`
`u_1^(1)=0.1835`
check the residual again `F_(ext)=1, F_(Int)=[u_1^(1)]^3+9[u_1^(1)]^2+4[u_1^(1)].`
`=(0.1835)^3+9(0.1835)^2+4(0.1835)`
`F_(Int)=1.0433`
`R^(1)=F_(Int)-F_(ext)`
`R^(1)=0.0433`
`||R^(1)||>10^(-2) =>` another Newton-Raphson iteration is necessary.
calculate the new total correction of `u_1^(0)=u_1^(1)`
`deltau^(2)=deltau^(1)-k[u_1^(1)]^-1(R^(1)).`
`=0.0665-(0.0433/7.404)`
`deltau^(2)=-0.0723`
the updated value of `u_1^(2)=u_1^(0)+deltau^(2)`
`=0.25-0.0723`
`u_1^(2)=0.1777.`
check the residual again `F_(ext)=1, F_(Int)=[u_1^(2)]^3+9[u_1^(2)]^2+4[u_1^(2)]`
`=(0.1777)^3+9(0.1777)^2+4(0.1777)`
`F_(Int)=1.0006.`
`||R^(2)||=F_(Int)-F_(ext)`
`=1.0006-1`
`||R^(2)||=0.0006`
`||R^(2)||<10^(-2)=>` no further iterations are necessary.
therefore, the final value is `u_1=0.1777.`
Step-2:
`u_1=0.1777`.
`k(u_1)=3(u_1)^2+18(u_1)+4`.
`=3(0.1777)^2+18(0.1777)+4`.
`k(u_1)=7.2933`.
`ΔF_2=1`, `=>` `Δu_2=(ΔF_2)/(k(u_1))`.
`=1/7.2933`.
`Δu_2=0.1371`.
`u_2=u_1+Δu_2`.
`=0.1777+0.1371`.
`u_2=0.3148`.
Check the residual, `F_(ext)=ΔF_1+ΔF_2=2,` `F_(Int)=u_(2)^3+9u_(2)^2+4u_(2)`
`=(0.3148)^3+9(0.3148)^2+4(0.3148)`
`F_(Int)=2.1823.`
`||R^(0)||=F_(Int)-F_(ext)`
`=2.1823-2`
`||R^(0)||=0.1823.`
`||R^((0))||>10^-2 =>` Newton-Raphson iteration are necessary.
calculate the correction of `u_2=u_2^(0)`
`deltau_2^(1)=-[k(u_2^(0))]^(-1)*R^(0)`
`=-9.9637^(-1)(0.1823)`
`deltau_2^(1)=-0.0183.`
`=>` the updated value of `u_2^(1)=u_2^(0)+deltau_2^(1)`
`=0.3148-0.0183`
`u_2^(1)=0.2965`
check the residual again `F_(ext)=2, F_(Int)=[u_2^(1)]^3+9[u_2^(1)]^2+4[u_2^(1)].`
`=(0.2965)^3+9(0.2965)^2+4(0.2965)`
`F_(Int)=2.0033`
`R^(1)=F_(Int)-F_(ext)`
`R^(1)=0.0033`
`||R^(1)||<10^(-2) =>` no further iterations are necessary.
therefore, the final value is `u_2=0.2965.`
Step-3:
`u_2=0.2965`.
`k(u_2)=3(u_2)^2+18(u_2)+4`.
`=3(0.2965)^2+18(0.2965)+4`.
`k(u_2)=9.6007`.
`ΔF_3=1`, `=>` `Δu_3=(ΔF_3)/(k(u_2))`.
`=1/9.6007`.
`Δu_1=0.1042`.
`u_3=u_2+Δu_3`.
`=0.2965+0.1.42`.
`u_1=0.4007`.
Check the residual, `F_(ext)=ΔF_1+ΔF_2+ΔF_3=3,` `F_(Int)=u_(3)^3+9u_(3)^2+4u_(3)`
`=(0.4007)^3+9(0.4007)^2+4(0.4007)`
`F_(Int)=3.1121.`
`||R^(0)||=F_(Int)-F_(ext)`
`=3.1121-3`
`||R^(0)||=0.1121.`
`||R^((0))||>10^-2 =>` Newton-Raphson iteration are necessary.
calculate the correction of `u_3=u_3^(0)`
`deltau_3^(1)=-[k(u_3^(0))]^(-1)*R^(0)`
`=-11.68^(-1)(0.1121)`
`deltau^(1)=-0.0096.`
`=>` the updated value of `u_3^(1)=u_3^(0)+deltau_3^(1)`
`=0.4007-0.0096`
`u_3^(1)=0.3911`
check the residual again `F_(ext)=3, F_(Int)=[u_3^(1)]^3+9[u_3^(1)]^2+4[u_3^(1)].`
`=(0.3911)^3+9(0.3911)^2+4(0.3911)`
`F_(Int)=3.0012`
`R^(1)=F_(Int)-F_(ext)`
`R^(1)=0.0012`
`||R^(1)||<10^(-2) =>`no further iterations are necessary.
therefore, the final value is `u_3=0.3911.`
The result of the implicit analysis are summarized in Table 2. It is evident from the table that the external force and internal force are coming into equilibrium.
| Step | `Δ F_i` | `u_i` | `(F_(ext))_i` | `(F_(Int))_i` | `R=F_(Int)-F_(ext)` |
| 1 | 1.0 | 0.1777 | 1.0 | 1.0006 | 0.0006 |
| 2 | 1.0 | 0.2965 | 2.0 | 2.0033 | 0.0033 |
| 3 | 1.0 | 0.3911 | 3.0 | 3.0012 | 0.0012 |
Table-2: Summary of Implicit Analysis Result
Plot of Results
It is evident from the results that the explicit analysis drifts from the exact solution. If 20 increment are used the explicit analysis more closely follows the exact solution. However, even with more increments the explicit analysis still drifts from the exact solution. The drift from the exact solution illustrates the lack of equilibrium between the internal and external forces. To correct for this problem an implicit analysis is required. The Newton-Raphson iterations correct the incremental steps so that they land on the exact solution according to the specified tolerance. Excellent agreement is achieved with just 3 increments and in each step only two Newton-Raphson iterations are required.
RESULT:-
A nonlinear analysis is illustrated using an incremental explicit and implicit procedure. It is evident from the results that the explicit analysis drifts from the true solution. To overcome this problem an implicit analysis is used, which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces.