Week - 2 - Explicit and Implicit Analysis

Aim:

F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods ( have a tolerance of 0.01).

Theory required:

Explicit Analysis:

The explicit analysis is preferred for the following reasons:

• In Explicit analysis no iterations are required.
• It is conditionally stable, Courant-Friedrich-Levy(CFL) criteria.
• It requires low memory.
• It has relatively smaller time steps.
• Cost per timestep is low.
• Explicit analysis is mainly used where large strain deformation problems.
• Acceleration is assumed to be constant over the time step.
• It will calculate the state of system at different time by considering the current time, Which results in finding out the solution faster.
• Each increment is founded fast, This results in less computation.
• Explicit method is used for large strain deformation problems, Which includes the applications such as: Automotive Crash analysis, Drop test & few dynamic analysis.
• One of the con in explicit analysis is: More increments are needed to match up with the exact solution.
• Less accuracy in the obtained results when compared to the solution obtained by implicit method.
• This method is used where the events to be performed are more.

Implicit Analysis:

The implicit analysis is preferred for the following reasons:

• In Implicit analysis iterations are required.
• Newton Raphson method is used for the iterations.
• It requires high memory.
• Time is high in implicit analysis.
• Cost per timestep is high.
• Implicit analysis is mainly used for small strain deformation problems.
• In implicit analysis, Acceleration is assumed to vary linearly over the time step(Which could be either positive(or)Negative).
• It will calculate the solution by considering the current & further states of the system, Which helps in attaining the equilibrium between the external & internal forces before moving in to next step with the help of newton-raphson iterations.
• Implicit method is used for small strain deformation problems.
• More accuracy in the obtained results when compared to the solution obtained by explicit method.
• This method is used where the events to be performed are less.

Procedure:

Given non-linear equation: F(u) = u3">u^3 + 9u^2 + 4u -------->eqn (1)

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: ∆F =k⋅∆u">k⋅Δu
is used.

Then, k = dF/du

k = 3u^2 + 18u + 4 -------->eqn (2)

Therefore, The analysis proceeds as follows:

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

Put u0 = 0 value in equation 2

⇒ k(u0) = 0+0+4 = 4, ΔF = 1, Δu1 = ΔF/k(u0) = 1/4 = 0.25 , u1 = u0 + Δu1 = 0 + 0.25 = 0.25

f(u1) = (0.25)^3 + 9x(0.25)^2 + 4x0.25 = 1.5781

R = f - F = 1.5781- 1 = 0.5781

Step 2: Put u1 = 0.25 value in equation 2

⇒ k(u1) = 3x(0.25)^2 + 18x0.25 + 4 = 8.6875, ΔF = 1, Δu2 = ΔF/k(u1) = 1/8.6875 = 0.1151, u2 = u1 + Δu2 = 0.25 + 0.1151 = 0.3651

f(u2) = (0.3651)^3 + 9x(0.3651)^2 + 4x0.3651 = 2.7087

R = f - F = 2.7087 - 2 = 0.7087

Step 3: Put u2 = 0.3657 value in equation 2

⇒ k(u2) = 3x(0.3657)^2 + 18x0.3657 + 4 = 10.7293, ΔF = 1, Δu3 = ΔF/k(u2) = 1/10.7293 = 0.0932, u3 = u2+Δu3 = 0.3657 + 0.0932 = 0.4589

f(u3) = (0.4589)^3 + 9x(0.4589)^2 + 4x0.4589 = 3.8275

R = f - F = 3.8275 - 3 = 0.8275

Step No.	Incremental External Force (ΔF)	Incremental Displacement (Δu)	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	2.7087	0.7087
3	1	0.0932	0.4589	3	3.8275	0.8275

From the above table, The residual force(difference between internal & external forces are not equal) & more in magnitude. This depicts that, external and internal forces are NOT in equilibrium.

Implicit Method:

F = k.Δu -------->eqn (3)

Step 1:

u0 = 0.0, ⇒ k(u0) = 0 + 0 + 4, ΔF = 1, Δu1 = ΔF/k(u0) = 1/4 = 0.25, u1 = u0 + Δu1 = 0 + 0.25 = 0.25

f(u1) = (0.25)^3 + 9x(0.25)^2 + 4x0.25 = 1.5781

R(0) = f(u1) - F = 1.5781- 1 = 0.5781 > 0.01 ---------> Newton-Raphson iterations are necessary.

Calculate the correction to u1 = u0

k(uo)=3(uo)^2 + 18(uo) + 4 = 3(0.25)^2 + 18(0.25) + 4 = 8.6875

Δu1 = R(0) / -k(u0) = -0.5781 / 8.6875 = -0.0665

u1(1) = u1 + Δu1 = 0.25 + (-0.0665) = 0.1835

check the residual again

f(u1(1)) = (0.1835)^3 + 9x(0.1835)^2 + 4x0.1835 = 1.0432

R(1) = f(u1(1)) - F = 1.0432 - 1 = 0.0432 > 0.01 ---------> Newton-Raphson iterations are necessary.

Calculate the correction to u1(1) = u0

k(u1(1))=3(u1(1))^2 + 18(u1(1)) + 4 = 3(0.1835)^2 + 18(0.1835) + 4 = 7.4040

Δu2 = Δu1 - R(1) x k(u1(1))^-1 = -0.0665 -0.0432 x (7.4040)-1 = -0.0723

u1(2) = u1 + Δu2 = 0.25 + (-0.0723) = 0.1777

check the residual again

f(u1(2)) = (0.1777)^3 + 9x(0.1777)^2 + 4x0.1777 = 1.0006

R(2) = f(u1(2)) - F = 1.0006 -1 = 0.0006 < 0.01 ---------> No further iterations are necessary

Therefore for step 1 Final values are u1 = 0.1777, f = 1.0006, R = 0.0006

Step 2:

k(u1) = 3(u1) + 18(u1) + 4 = 3(0.1777)^2 + 18 (0.1777) + 4 = 7.2933

Δu2 = ΔF / k(u1) = 1 / 7.2933 = 0.1371

u2 = u1 + Δu2 = 0.1777 + 0.1371 = 0.3148

check the residual

F = ΔF + ΔF = 2

f(u2) = (u2)^3 + 9(u2)^2 + 4(u2) = (0.3148)^3 + 9(0.3148)^2 + 4(0.3148) = 2.1822

R(0) = f(u2) - F = 2.1822 - 2 = 0.1822 > 0.01 ---------> Newton-Raphson iterations are necessary.

Calculate the correction to u1 = u2^(0)

k(u2) = 3(u2)^2 + 18(u2) + 4 = 3(0.3148)^2 + 18(0.3148) + 4 = 9.9636

Δu2 = R(0) / -k(u2) = -0.1822 / 9.9636 = -0.01828

u2(1) = u2 + Δu2 = 0.3148 + (-0.01828)=0.2965

check the residual again

f(u2(1)) = (u2(1))^3 + 9(u2(1))^2 + 4(u2(1)) = (0.2965)^3 + 9(0.2965)^2 + 4(0.2965) = 2.0032

R(1) = f(u2(1)) - F = 2.0032 - 2 = 0.0032 < 0.01 ---------> No further iterations are necessary

Therefore for step 2 Final values are u2 = 0.2965, f = 2.0032, R = 0.0032

Step 3:

k(u2) = 3(u2)^2 + 18(u2) + 4 = 3(0.2965)^2 + 18(0.2965) + 4 = 9.6007

Δu3 = ΔF / k(u2) = 1 / 9.6007=0.1041

u3 = u2 + Δu3 = 0.2965 + 0.1041 = 0.4006

check the residual

F = ΔF + ΔF + ΔF = 3

f(u3) = (u3)^3 + 9(u3)^2 + 4(u3) = (0.4006)^3 + 9(0.4006)^2 + 4(0.4006) = 3.1110

R(0) = f(u3) - F = 3.1110 - 3 = 0.1110 > 0.01 ---------> Newton-Raphson iterations are necessary.

calculate the correction to u3 = u3^0

k(u3^0) = 3(u3^0)^2 + 18(u3^0) + 4 = 3(0.4006)^2 + 18(0.4006) + 4 = 11.6922

Δu3^0 = R(0) / -k(u2^0) = -0.1110 / 11.6922 = 0.009493

u3^1 = u3 + Δu3^0 = 0.4006 - 0.00949 = 0.3911

check the residual again

f(u3(1)) = (u3^1)^3 + 9(u3^1)^2 + 4(u3^1) = (0.3911)^3 + 9(0.3911)^2 + 4(0.3911) = 3.00085

R(1) = f(u3(1)) - F = 3.00085 - 3 = 0.00085 < 0.01 ---------> No further iterations are necessary

Therefore for step 3 Final values are u3 = 0.3911, f = 3.00085, R = 0.00085

Step No.	Incremental External Force (ΔF)	Displacement (u)	External Force (F)	Internal Force (f)	Residual Force = f - F
1	1	0.1777	1	1.0006	0.0006
2	1	0.2965	2	2.0032	0.0032
3	1	0.3911	3	3.00085	0.00085

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

Result:

Implicit Method

Step No.	Incremental External Force (ΔF)	Displacement (u)	External Force (F)	Internal Force (f)	Residual Force = f - F
1	1	0.1777	1	1.0006	0.0006
2	1	0.2965	2	2.0032	0.0032
3	1	0.3911	3	3.00085	0.00085

Explicit Method

Step No.	Incremental External Force (ΔF)	Incremental Displacement (Δu)	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	2.7087	0.7087
3	1	0.0932	0.4589	3	3.8275	0.8275

From the above table it seen that the explicit analysis deviates from the precise solution. Since the residual value is more in explicit anaysis. Implicit analysis is more accurate.

The graphic clearly shows that the explicit analysis deviates from the precise solution. The deviation from the exact solution demonstrates the absence of internal and external force equilibrium. An implicit analysis is necessary to address this issue. The incremental stages are corrected using the Newton-Raphson iterations, leading to the precise solution within the predetermined tolerance. Only three increments are needed to reach excellent agreement, and only two Newton-Raphson iterations are needed for step.

Conclusion:

The equation F(u) = u3+9u2+4u is used to solve using both Explicit and Implicit Methods ( have a tolerance of 0.01). It as per the results to get the accurate result Implicit method is best suited to solve this.