Week - 2 - Explicit and Implicit Analysis

Objective: Solve a Simple bar in Tension with Explicit and Implicit Method

Equation of Force in the bar as non linear function of displacement is given as:

`f(u)=u^3+9u^2+4u`

From force equation , Stiffness of bar is obtained as:

`k(u)=(df)/(du)= 3u^2+9u+4`

Explicit Analysis

Clearly it is possible to plot a graph of force, f, versus displacement, u, by using force equation . However, if the nonlinear force versus displacement relationship is
not known, but the stiffness is known, it is possible to construct the graph of force versus displacement numerically. One method is to use an incremental explicit load control scheme.
This is illustrated next. For this example three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar. The variables used are u for displacement, f for
internal force in the bar, F for external force applied to the bar and k for stiffness. Incremental displacements or incremental externally applied forces are represented as ∆u or ∆F,
respectively. Use is made of the relationship ∆F = k∆u. The analysis proceeds as follows:

Step 1:

`u_0=0.0,rArrk(u_0)=4` where `DeltaF=1`,

`Deltau_1=(DeltaF)/(k(u_0))=1/4=0.25`,

`u_1=u_0+Deltau_1=0.25` since `u_0=0`

Step 2:

`k(u_1)=3(0.25)^2+18xx0.25+4=8.6875`where `DeltaF=1`,

`Deltau_2=(DeltaF)/(k(u_1))=1/8.6875=0.1151`,

`u_2=u_1+Deltau_2=0.25+0.1151 =0.3651`since `u_1=0.25`

Step 3:

`k(u_2)=3(0.3651)^2+18xx0.3651+4=10.9717` where `DeltaF=1`,

`Deltau_3=(DeltaF)/(k(u_2))=1/10.9717=0.0911`,

`u_3=u_2+Deltau_3=0.3651+0.0911 =4562` since `u_2=0.3651`

Now External force,

`F_ext=f1+f2+f3`

For step 1 , `F_ext_1=1` ,

`Step 2 ,`F_ext_2=2`

`Step 3 ,`F_ext_3=3`

Internal forces

`f_int(1)=u_1^3+9u_1^2+4u_1=1.5781`

`f_int(2)=u_2^3+9u_2^2+4u_2=2.7087`

`f_int(3)=u_3^3+9u_3^2+4u_3=3.7928`

It is evident from the results that the external and internal forces are NOT in equilibrium because `F_ext!=f_int`

The following table summarizes the main results for Explicit method:

Step i	`DeltaF_i`	`Deltau_i`	`u_i`	`(F_ext)_i`	`(f_int)_i`	Residual `R= (f_int)_i-(F_ext)_i`
1	1.0	0.2500	0.2500	1.0	1.5781	0.5781
2	1.0	0.1151	0.3651	2.0	2.7087	0.7087
3	1.0	0.0911	0.4562	3.0	3.7928	0.7928

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
stiffness 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 uj gives
`R^(j+1 )= R^j(u^j) + dR(u^j)/(du)δu^(j+1)+ ... = R^j + k(u^j)δu^(j+1) + ...`

By neglecting higher order terms in the series, setting `R^(j+1)=0` and solving for `delta^(j+1)` the
following correction is obtained

`δu^(j+1) = −[k(u^j)]^-1R^j`

Notice that the iteration variable is j and it may take several iterations for the norm of the residual, `absabsR`, 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 (results are approximate, they are more precise if done in a computer to machine precision):

Step 1:

`u_0=0.0, →k(u_0 )=4, ∆F=1, ∆u_1=∆F/k(u_0 ) =0.25, u_1=u_0+∆u_1=0.25`

`check the residual, F_ext=∆F=1, f_int=〖u_1〗^3+9〖u_1〗^2+4u_1=1.5781, 〖 R〗^((0) )=f_int-F_ext=0.578`

`R^((0) ) ‖>〖10〗^(-2)rarr`Newton-Raphson iterations are necessary.

Calculate the correction to `u_1=〖u_1〗^((0) )`

`δu^((1) )=-[k(〖u_1〗^((0) ) )]^(-1) R^((0) )=-〖8.6875〗^(-1)×0.5781≅-0.0665.`

`rarr` The updated value of `〖u_1〗^((1) )=〖u_1〗^((0) )+〖δu〗^((1) )=0.25-0.0665=0.1835.`

Check the residual again , `F_ext=1, f_int=(〖u_1〗^((1) ) )^3+9(〖u_1〗^((1) ) )^2+4〖u_1〗^((1) )≅1.0432`

`〖 R〗^((1) )=f_int-F_ext=0.0432`

`‖R^((1) ) ‖>〖10〗^(-2)→` another Newton-Raphson is necessary.

Calculate the new total total correction to `u_1=〖u_1〗^((0) )`

`δu^((2) )=δu^((1) )-[k(〖u_1〗^((1) ) )]^(-1) R^((1) )=-0.0665-(〖7.404〗^(-1) )×0.0432≅-0.0723.`

The updated value of `〖u_1〗^((2) )=〖u_1〗^((0) )+〖δu〗^((2) )=0.25-0.0723=0.1777.`

Check the residual, `F_ext=1, f_int=(〖u_1〗^((2) ) )^3+9(〖u_1〗^((2) ) )^2+4〖u_1〗^((2) )≅1.0006`

`〖 R〗^((2) )=f_int-F_ext=0.0006`

`‖R^((2) ) ‖<〖10〗^(-2)→`no further iterations are necessary.

Therefore, the final value is `u_1=〖u_1〗^((2) )=0.1777.`

Step 2:

`k(u_1 )=3〖×0.1777〗^2+18×0.1777+4≅7.2933, ∆F=1, ∆u_2=∆F/k(u_1 ) ≅0.1371,`

`u_2=u_1+∆u_2=0.3148`

Check the residual, `F_ext=2×∆F=2, f_int=〖u_2〗^3+9〖u_2〗^2+4u_2≅2.1823`

`〖 R〗^((0) )=f_int-F_ext=0.1823`

`‖R^((0) ) ‖>〖10〗^(-2)→` Newton-Raphson iterations are necessary.

Calculate the correction to `u_2=〖u_2〗^((0) )`

`δu^((1) )=-[k(〖u_2〗^((0) ) )]^(-1) R^((0) )=-〖9.9637〗^(-1)×0.1823≅-0.0183.`

`→`The updated value of `〖u_2〗^((1) )=〖u_2〗^((0) )+〖δu〗^((1) )=0.3148-0.0183=0.2965.`

Check the residual again , `F_ext=2, f_int=〖〖u_2〗^((1) )〗^3+9〖〖u_2〗^((1) )〗^2+4〖u_2〗^((1) )≅2.0033`

`〖 R〗^((1) )=f_int-F_ext=0.0033`

`‖R^((1) ) ‖<〖10〗^(-2)→`no further iterations are necessary.

Therefore, the final value is `u_2=〖u_2〗^((1) )=0.2965.`

Step 3:

`k(u_2 )=3〖×0.2965〗^2+18×0.2965+4≅9.6007, ∆F=1, ∆u_3=∆F/k(u_2 ) ≅0.1042,`

`u_3=u_2+∆u_3=0.4007`

Check the residual, `F_ext=3×∆F=3, f_int=〖u_3〗^3+9〖u_3〗^2+4u_3≅3.1122`

`〖 R〗^((0) )=f_int-F_ext=0.1122`

`‖R^((0) ) ‖>〖10〗^(-2)→` Newton-Raphson iterations are necessary.

Calculation the correction to `u_3=〖u_3〗^((0) )`

`δu^((1) )=-[k(〖u_3〗^((0) ) )]^(-1) R^((0) )=-〖11.6943〗^(-1)×0.1122≅-0.0096.`

`→` The updated value of `〖u_3〗^((1) )=〖u_3〗^((0) )+〖δu〗^((1) )=0.4007-0.0096=0.3911.`

Check the residual again, `F_ext=3, f_int=〖〖u_3〗^((1) )〗^3+9〖〖u_3〗^((1) )〗^2+4〖u_3〗^((1) )≅3.0009`

`〖 R〗^((1) )=f_int-F_ext=0.0009`

`‖R^((1) ) ‖<〖10〗^(-2)→`no further iterations are necessary.

Therefore, the final value is `u_2=〖u_2〗^((1) )=0.3911.`

The following table summarizes the main results for Implicit method:

Step i	`DeltaF_i`	`u_i`	`(F_ext)_i`	`(f_int)_i`	Residual `R= (f_int)_i-(F_ext)_i`
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	3.0009	0.0009

It is evident from the results that the external and internal forces are in equilibrium

Comparision of Explicit and Implicit Results:

The exact solution curve plot lies closer to Implicit results curve plot but the exact solution plot is not shown here. This is because the Implicit method uses Newton-Raphson iterations to enforce equilibrium in each step while the Explicit method does not take into account the equilibrium solutions.

Comparing with the results in the reference, using the new function of force vs displacement `f(u)= u^3+9u^2+4u`has helped the Implicit results reach equilibrium faster for each step.