Week - 2 - Explicit and Implicit Analysis

#3 // Week _ 2 - Explicit and Implicit Analysis.

AIM:

Solve the above-given equation in both the Implicit and Explicit methods and compare the result with the exact result.

OBJECTIVE:

• To solve the given non-linear equation "F(u) = u3+9u2+4u" using both Explicit and Implicit Methods with a residual tolerance of '<0.01'.
• To perform 3 steps for both explicit and implicit Methods & calculate the various parameters such as displacement, stiffness, and internal & external forces. Then, to suggest the best method.
• To plot the obtained results for both explicit and implicit methods.

IMPLICIT Method ->

• This method is mostly used in static analysis.
• Acceleration will vary linearly within the given time step.
• Time step is more compared to the explicit method.
• It needs a numerical solver to invert the stiffness matrix, this costs more for large models.
• It Uses Newton-Raphson iterations to enforce equilibrium.

EXPLICIT Method ->

• This method of calculation is used in Crash analysis.
• Acceleration will be constant within the given time step.
• Because we consider the acceleration to be constant the time step should be very small.
• In this method we will inverse Mass diagonal matrix, this makes the solving timeless compared to the implicit method.

PROCEDURE:

Given non-linear equation: `F(u) = U^3 + (9*u^2) + 4u` ---->eq(1)

Where F = external force applied to the bar

u = Displacement.

f = Internal force in the bar.

k = Stiffness.

`Delta u` = Incremental displacements.

`Delta F` = Incremental externally applied forces.

Explicit Method:

For the nonlinear analysis, the force Vs displacement relationship is not known, but the stiffness is known, it is possible to construct the graph of force Vs displacement numerically. One method is, to use an incremental explicit load control scheme. For this example, three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar.

For finding the stiffness, the following relationship: `Delta F = k * Delta u` is used.

Then, `k = (dF) / (du)` ------->eq(2)

Explicit scheme:

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.

It can be done using an incremental explicit load control scheme.

The following equation is solved using explicit method as follows,

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

Differentiating force with respect to displacement gives the stiffness(k) of the bar (since F = kx).

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

Here,three equal load increments (or steps) of `Delta 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 `Delta u` or `Delta F`, respectively.

`Delta F = k * Delta u` This relation is used to find the incremental displacement.

Step 1: This step is performed by taking the initial conditions as follows:

u0 = 0

=> `k(u0) = (3 *(u0)^2) + (18*(u0)) +4` [From equations (1 &2)]

By substituting the value of u0 in the above equation, We get:

`k(u0) = (3 *(0)^2) + (18*(0)) + 4 = 4`

Therfore, k(u0) = 4

`Delta F = 1`

Then, `Delta u1 = (Delta F) / (k(u0)) = 1/4 = 0.25`

`u1 = u0 + Delta u1 = 0 + 0.25 = 0.25` ------->eq(3)

The internal force(f) can be calculated by substituting the "u1" value in Eqn(1):

`f1 = (0.25)^3 + (9*(0.25)^2) + (4* 0.25)`

= 1.578125

Therefore, the Internal force for the first increment is 1.578125.

Step 2:

In this step, The externally applied load is incremented by one unit (`Delta F = 1`).

From eq(3), The value for displacement for this iteration is u1= 0.25

Then, Stiffness for this step is calculated by: `k(u1) = (3 * u1^2) + (18*u1) + 4`

=> `k(u1) = (3* (0.25)^2) + (18*(0.25)) +4`

=> `k(u1) = 8.6875`

Now, The Incremental displacements for this step is: `Delta u2 = (Delta F) / (k(u1)) = 1 / 8.6875`

= 0.11512

`u2 = u1 + Delta u2 = 0.25 + 0.11512 = 0.36512`

Internal force, f2 = `(0.36512)^3 +(9 * (0.36512)^2) + (4 * 0.36512)`

= 2.708

Step 3:

In this step also, The externally applied load is incremented by one unit(`Delta F = 1`).

u2 = 0.36512

`k(u2) = (3 * u2^2) + (18 *u1) + 4`

= `(3 * (0.36512)^2) + (18 * (0.36512)) + 4`

k(u2) = 10.97

`Delta u3 = (Delta F) / (k(u2)) = 1 / 10.97`

`Delta u3 = 0.09115`

`u3 = u2 + Delta u3 = 0.36512 + 0.09115 = 0.45627`

Internal force, `f3 = (0.45627)^3 +(9 * (0.45627)^2) + (4 * 0.45627)`

= 3.794

Step No	Incremental external force (`Delta F`)	Incremental Displacement (`Delta u`)	Displacement(u)	External force(F)	Internal force(f)	Residual force= f-F
1	1	0.25	0.25	1	1.5781	0.5781
2	1	0.1151	0.3651	2	2.708	0.708
3	1	0.0911	0.4562	3	3.794	0.794

Result:

From the above table, The residual force(difference between internal & external forces are not equal) & is more in magnitude. This depicts that, external and internal forces are NOT in equilibrium. To reduce the residual forces, more iterations should be performed. The only advantage of choosing an explicit method for performing the analysis is, The processing time is very less when compared to the implicit method.

The obtained result in explicit methods is plotted & it appears as shown in below fig.

Implicit scheme:

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.

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, to achieve equilibrium, corrections must be made to the displacement.

This is accomplished by using Newton-Raphson iterations to minimize the residual, `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) * delta u ^(j+1) + ... = R ^j + k (u ^j) delta u ^(j+1) + ...`

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

`delta u ^(j+1) = - [k (u ^j)] ^-1 * R ^j`

Here, the iteration variable is j and it may take several iterations for the norm of the residual, R, to be reduced below the chosen tolerance criteria of 0.01.

The implicit analysis proceeds as follows,

Implicit Analysis:

Step 1:

In this step, The initial conditions are taken into consideration as follows:

u0 = 0

`Delta F = 1`

k(u0) = 4 ----> Find in explicit analysis step 1.

`Delta u1 = (Delta F) / (k(u0)) = 1/4 = 0.25`

`f1^0 = (0.25)^3 + (9 * (0.25)^2) + (4 * 0.25)`

= 1.5781

Residual force(`R^0`) = internal force(`f1^0`) - external force(F)

= 1.5781-1

= 0.5781, Which is Greater than 0.01. If the residual is `> 10^-2`, Then Newton-Raphson iterations are necessary to get the forces into equilibrium.

`u1 = u1^0`

`delta u1^1 = - [k (u1 ^0)] ^-1 * R ^0 = - [8.6875] ^(-1) * (0.57821)`

= - 0.066

Now the updated u1 value is

`u1 = u1^2 + delta u1^1 = 0.25 + (-0.066) = 0.184`

`f1^1 = (0.184)^3 + (9 * (0.184)^2) + (4 * 0.184) = 1.0469`

Now the residual is founded again, `R^1` = internal force(`f1^1`) - external force(F)

= 1.0469 - 1 = 0.0469 Which is > 0.01

So, The Newton-Raphson iterations are further performed.

Now, `u1 = u1^0`

`delta u1^2 = delta u1^1 - [k(u1)]^-1 * R^1 = (-0.066) - [8.6875]^(-1) * (0.0469) = (-0.066) - (0.0054) = - 0.0711`

Therefore, `delta u1^2 = -0.0711`

Now the updated u1 value is `u1 = u1^0 + delta u1^2 = 0.25 + (-0.0711) = 0.1789`

`f1^2 = (0.1789)^3 + (9 * (0.1789)^2) + (4 * 0.1789)`

= 1.0093

`R^2` = internal force(`f1^2`) - external force(F) = 1.0093 - 1 = 0.0093 Which is <0.01.

Therefore, The final value of u1 = 0.1789

Step 2:

u = u1 = 0.1789

`k(u1) = 3*(0.1789)^2+(18*(0.1789))+4`

= 7.316

`delta u2 = (Delta F) / (k(u1)) = 1 / 7.316 = 0.1366`

`u2^0 = u1 + delta u2 = 0.1789 + 0.1366 = 0.3115`

`f2^0 = (u2^0) ^3 + 9 * (u2^0) ^2 + 4 * (u2^0) = (0.3115) ^3 + (9 * (0.3155) ^2) + (4 * 0.3155) = 2.15`

Residual, `R^0 = f2^0 - F = 2.15 - 2 = 0.15` Which is `>0.01`. So, Newton's Raphson iterations are used to find u2.

`k(u2^1) = (3*(u2^0) ^2) + (18*(u2^0)) + 4 = (3*(0.3115) ^2) + (18*(0.3115)) + 4 = 9.898`

`delta u2 ^1 = - [k (u2 ^1)] ^-1 * R^0 = - 0.01515`

Therefore, `u2 ^1 = u2 ^0 + delta u2 ^1 = 0.3115 - 0.01515 = 0.2966`

Now, `f2 ^1 = (u2 ^1) ^3 + 9 * (u2 ^1) ^2 + 4 * (u2 ^1) = 2.0032`

Residual, `R ^1 = f2 ^1 - F = 2.0032 - 2 = 0.0032 <0.01`.

So, The updated u2 is `u2^1 =0.2966`

Step 3:

u= u2 = 0.2966

K(u2) =`k (u2) = 3 * (0.2966) ^2 + (18 * (0.2966)) + 4 = 9.6027`

`delta u3 ^0 = (Delta F) / (k (u2)) = 1 / 9.6027 = 0.1041`

`u3 ^0 = u2 + delta u3 = 0.2966 + 0.1041 = 0.4007`

`f3 ^0 = (u3 ^0) ^3 + (9 * (u3 ^0) ^2) + (4 * u3 ^0) = (0.4007) ^3 + (9 * (0.4007) ^2) + (4 * 0.4007) = 3.1121`

Residual, `R ^0 = f3 ^0 - F = 3.1121 - 3 = 0.1121`, Which is >0.01. So, The Newton-Raphson iteration is necessary.

`k (u3 ^1) = (3 * (u3 ^0) ^2) + (18 * (u3 ^0)) + 4 = (3 * (0.4007) ^2) + (18 * 0.4007) + 4 = 11.6942`

`delta u3 ^1 = - [k (u3 ^1)] ^-1 * R ^0 = - (11.6942) ^-1 * 0.1121 = -0.00958`

`u3 ^1 = u3 ^0 + delta u3 ^1 = 0.4007 + (- 0.00958) = 0.3911`

`f3 ^1 = (u3 ^1) ^3 + (9 * (u3 ^1) ^2) + (4 * u3 ^1) = (0.3911) ^3 + (9 * (0.3911) ^2) + (4 * 0.3911) = 3.0008`

Residual, `R^1 = f3^1 - F = 3.0008 - 3 = 0.0008 <0.01`

Therefore, The updated u3 is `u3 ^1 = 0.3911`

Step No.	Incremental external force (`Delta F`)	Displacement (u)	External force (F)	Internal force (f)	Residual force (Internal - External) = f- F
1	1	0.1789	1	1.0093	0.0093
2	1	0.2966	2	2.0032	0.0032
3	1	0.3911	3	3.0008	0.0008

Result:

From the above table, the residual force (difference between internal & external forces are not equal) & is less in magnitude & tries to attain the equilibrium m between the forces when compared to the results obtained in Explicit analysis. This depicts that, external and internal forces are almost brought to equilibrium by using the newton-Raphson iterations.

The obtained result in implicit methods is plotted & it appears as shown below in fig.

Inference:

A plot of the results is shown in the below figure.

It is evident from the plot that the explicit analysis drifts from the exact solution.

The drift from the exact solution illustrates the lack of equilibrium between the internal and external forces. To correct 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.

Discussion:

As discussed, we can use both implicit and explicit solvers to solve the same type of problems. The only difference is in the approach to time increment.

The implicit analysis allows you to select how big the time increment should be. This increment will take some time to compute since it has to iterate for global equilibrium first. But we can make it reasonably big for our analysis.

Explicit time increments calculate fast because they do not iterate for global equilibrium, but the time increment there is not the user’s choice. Solver simply assumes an acceptable time increment and goes with it.

The speed of computing is exactly the difference between implicit and explicit. Of course, the size of the explicit step depends on our model, and computing times depend on the system hardware.

The debate on what type of solver is better depends upon the type of problem and the accuracy we need.

Conclusion:

Final comparison graph distinguishing the implicit, explicit & exact calculations with force Vs Displacement:

From the above fig, the following aspects are concluded:

• The explicit solution drifts a bit from the exact solution. This is due to the residual force (the difference between internal & external forces are not equal) & more in magnitude. To attain equilibrium between forces, more iterations are needed.
• But the solution obtained by the implicit method with the newton-Raphson method iterations, Will help the solution to attain the equilibrium between the internal & external forces. Which results in getting the solution much closer to the exact solution.
• As the given equation F(u) = u3+9u2+4u is non-linear, the obtained result in terms of force Vs displacement will be in an exponential form. As shown in above fig.