Solving the given Non-Linear Equation of Force using Explicit and Implicit Methods and Comparing the Results.

Objective:

• To solve the equation `F(u)=u^3+9u^2+4` using both Explicit and Implicit methods, setting a tolerance of `10^-2`.
• To write a detailed report about implicit and explicit methods.
• To do the calculations in a spreadsheet and to try and plot the graph and compare them.

Explicit Method:

The explicit method does not require convergence to occur at each step. There is no check for Residue and hence some errors can accumulate such that a less accurate solution is obtained.

The solver assumes that a global equilibrium exists while calculating the variables and does not verify the global equilibrium. It moves on to the next set of finite element variables after calculating the first set.

The time increment is very short in explicit analysis and this can speed up the solution process. Its the time taken for an elastic wave to travel through a single finite element in the model.

 

Implicit Method:

In this method, the global equilibrium is established after each time increment. This means that convergence happens to a certain degree at each increment.

Because a global equilibrium is verified at each time increment, the time increments are large and thus causes much higher computational time since the iterations need to reach the global equilibrium at a set tolerance value before proceeding to the next step.

 

 

Procedure:

Let us consider the given equation:

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

 

Explicit Method:

The Non-linear force vs Displacement relationship is not known but the stiffness is known by differentiating the force with respect to displacement. 

Step 1:

`k(u) = (dF)/(du) = 3u^2 +18u +4`

`Delta F = K Delta U`

Initial condition:

`u_0 = 0.0`

⇒ `k(u_0) = 1, DeltaF = 1`,

Substituting the values we get,

`k(0) = 4`

`Delta u_1 = DeltaF/(k(u_0)) = 1/4 = 0.25`,

 

`u_1 = u_0+ Deltau_1 = 0.25`

`F_(external) = 1`

`F_"internal" = u^3 + 9u^2 + 4u = 1.5781` `(because u = u_1)`

 

Step 2:

`k(u_1) = k(0.25) = 8.6875, DeltaF = 1`

`Delta (u_2) = DeltaF/(k(u_1)) = 1/8.6875 = 0.115`

 

`u_2 = u_1+ Deltau_2 = 0.25 + 0.115 = 0.3651`

`F_(external) = 1+1 = 2`

`F_"internal" = u^3 + 9u^2 + 4u = 2.708`;`(because u = u_2)`

 

Step 3:

`k(u_2) = k(0.3651) = 10.969, DeltaF = 1`

`Delta (u_3) = DeltaF/(k(u_2)) = 1/10.969 = 0.0911`

`u_3 = u_2+ Deltau_3 = 0.3651 + 0.0911 = 0.4562`

`F_(external) = 1+1+1 = 3`

`F_"internal" = u^3 + 9u^2 + 4u = 3.7928`;`(because u = u_3)`

 

The following table tabluates the values obtained along with the residual for each step.

From the table we can see that the external and internal forces are not in Equilibrium.

 

Implicit Method:

The Implicit method is similar to the Explicit one, but the different lies in the fact that at the end of each load step, Newton-Raphson iterations are used to create an equilibrium before moving on to the next step.

An incremental force is applied to advance the solution in the forward direction in the beginning of each step. and corrections are made to the displacement to achieve equilibrium by making sure the stiffness is linear for each step.

 

The Residual is minimized by using the Newton Raphson Method.

`R = F_"internal" - F_"external"`

 

Expanding the residual 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)]^(−1)R^j`

Here the iteration variable is j and tolerance value is `10^-2`

 

Step 1:

`u_0 = 0,`

`k(u_0) = 4,`

`DeltaF = 1`,

`Deltau_1 = DeltauF/k(u_0) = 0.25`

`u_1 = u_0 + Deltau_1 = 0.25`

 

Residual check:

`F_"internal" = u_1^3 + 9u_1^2 + 4u_1 = 1.578`

`F_"external" = DeltaF =1`

`R^(0) = F_"internal" - F_"external" = 0.578`

`R^(0) > 0.01` Newton Raphson iterations are necessary here

 Correction:

`deltau^(1) = −[k(u_1^(0))]^(−1)R^(0) = -(8.6875)^(-1) . 0.578 = -0.0665`

Updated value of `u_1^(1)= 0.25 - 0.0665 = 0.1835`

 

Checking residual again:

`F_"internal" = u_1^(1)^3 + 9u_1^(1)^2 + 4u_1^(1) = 1.0432`

`F_"external" = DeltaF =1`

`R^(1) = F_"internal" - F_"external" = 0.0432` this is greater than the tolerance and hence further iterations are necessary.

`deltau^(2) =deltau^(1) −[k(u_1^(1))]^(−1)R^(1) =-0.0665 -(7.404)^(-1) . 0.0432 = -0.0723`

Updated value of `u_1^(2)= 0.25 - 0.00723 = 0.1777`

Checking Residual:

`F_"internal" = u_1^(2)^3 + 9u_1^(2)^2 + 4u_1^(2) = 1.0006`

`F_"external" = DeltaF =1`

`R^(2) = F_"internal" - F_"external" = 0.0006` this is lesser than the tolerance and hence further iterations are not necessary.

Therefore final value of `u_1 = u_1^(2) = 0.1777`

Step 2:

`u_1 = 0.1777,`

`k(u_1) = 7.2933,`

`DeltaF = 1`,

`Deltau_2 = DeltauF/k(u_1) = 0.1371`

`u_2 = u_1 + Deltau_2 = 0.3148`

 

Residual check:

`F_"internal" = u_2^3 + 9u_2^2 + 4u_2 = 2.1822`

`F_"external" = DeltaF+DeltaF =2`

`R^(0) = F_"internal" - F_"external" = 0.1822`

`R^(0) > 0.01` Newton Raphson iterations are necessary here

Correction:

`deltau^(1) = −[k(u_2^(0))]^(−1)R^(0) = -(9.9637)^(-1) . 0.1822 = -0.0182`

Updated value of `u_2^(1)= 0.3148 - 0.0182 = 0.2966`

 

Checking residual again:

`F_"internal" = u_2^3+ 9u_2^2 + 4u_2^ = 2.0042`

`F_"external" = DeltaF+DeltaF =2`

`R^(1) = F_"internal" - F_"external" = 0.0042` this is lesser than the tolerance and hence further iterations are not necessary

 

Therefore final value of `u_2 = u_2^(1) = 0.2966`

 

Step 3:

`u_2 = 0.2966,`

`k(u_1) = 9.6027`

`DeltaF = 1`,

`Deltau_3 = DeltauF/k(u_2) = 0.1041`

`u_3 = u_2 + Deltau_3 = 0.4007`

 

Residual check:

`F_"internal" = u_3^3 + 9u_3^2 + 4u_3 = 3.1121`

`F_"external" = DeltaF+DeltaF+DeltaF =3`

`R^(0) = F_"internal" - F_"external" = 0.1121`

`R^(0) > 0.01` Newton Raphson iterations are necessary here

Correction:

`deltau^(2) = −[k(u_3^(0))]^(−1)R^(0) = -(11.6942)^(-1) . 0.1121 = -0.00958`

Updated value of `u_3^(1)= 0.4007 - 0.00958 = 0.3911`

 

Checking residual again:

`F_"internal" = u_3^3 + 9u_3^2 + 4u_3 = 3.0008

`F_"external" = DeltaF+DeltaF+DeltaF =3`

`R^(1) = F_"internal" - F_"external" = 0.0008` this is lesser than the tolerance and hence further iterations are not necessary

 

Therefore final value of `u_3 = u_3^(1) = 0.3911`

The Results of the implicit analysis is shown in the table below:

From the table we can see that the values are coming close to each other and hence equilibrium.

 

Results:

Inference from the Graph:

•  From the graph we notice that the Explicit force analysis does not come close to the exact solution and the Implicit solution is much more accurate.
• This difference can be accounted to the fact of inequilibrium in the Explicit solution which is fixed by using the Implicit solution.

Discussion and Conclusions:

• Both Implicit and Explicit solvers can be used to solve the given type of equations. The difference lies in the approach to the time increment.
• Implicit method gives freedom to the user to define how big the time increment should be whereas the same does not apply for the explicit solver.
• There is a compromise that needs to achieved with time and accuracy, where Implicit method takes more time to compute by arriving at equilibrium using the Newton Raphson method and the Explicit solution takes more time giving up on accuracy.
• Therefore the type of solver that needs to be used completely depends on the type of analyis that is being carried out, the time in hand and the accuracy that needs to be achieved.

Therefore, a non linear Analysis was illustrated using an incremental process by both Explicit and Implicit methods.

From the Results we can conclude that the explicit analysis is not accurate and drifts away from the solution and this is overcome by using the implicit analysis that uses Newton Raphson iterations to bring about equilibrium between internal and external forces.