Week - 2 - Explicit and Implicit Analysis

Explicit Analysis:

The explicit method solving allows problems element by elements. Compared with the implicit method it does not require the general matrix, since the velocity and displacement nodes can be counted directly, using the central difference integration.

Time increment in explicit analysis is very short. The time step is limited by the size of the numerical stability. If the time step is too large there will be a significant numerical error.

This method has no requirement for balance, so it is suitable for the higher dynamic analysis, such as high velocity impact.

Implicit Analysis:

In this method the general equilibrium is achieved by interaction at first elements, followed by evaluation of local element variables. If the balance is not achieved, this method becomes expensive for calculation, since the matrix coefficients should recalculated with different time steps.

Implicit method is unconditionally stable, allowing the use of large increment time steps. So it is suitable for problems that tend to be high linear, static and quasi-static.

As demonstrated in the PDF, F(u) = u³+9u²+4u use this equation and solve using both Explicit and Implicit Methods ( have a tolerence of 10^-2)

Explicit Method:

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 increment explicit load control scheme.

The following equation is solved using explicit method as follows,

F(u) = u³+9u²+4u

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

K(u) = = 3u² + 18u + 4

• Here, three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar.

Variables are as given below:

• U = Displacement
• F = External Force
• K = stiffness
• ∆u = Incremental Displacement
• ∆F = Incremental externally applied force

Step – 1:

u_{0 =} 0.0

k(u₀) = 3(0)2 + 18(0) + 4 = 4

∆F = 1,

∆ u₁ = = ¼ = 0.25,

u₁ = u₀ + ∆ u₁ = 0 + ¼ = 0.25

F_External = ∆F = 1,

F_Internal = u₁³ + 9u₁² + 4u₁ = (0.25)³ + 9(0.25)² + 4(0.25) = 1.5781,

R = F_I – F_E = 1.5781-1 = 0.5781.

Step – 2:

u₁ = 0.25,

k(u₁) = 3(0.25)2 + 18(0.25) + 4 = 8.6875,

∆F = 1,

∆ u₂ = = 1/ 8.6875 = 0.1151,

u₂ = u₁ + ∆ u₂ = 0.25 + 0.1151 = 0.3651,

F_External = ∆F + ∆F = 1+1 = 2

F_Internal = u₂³ + 9u₂² + 4u₂ = (0.3651)³ + 9(0.3651)² + 4(0.3651) = 2.7087,

R = F_I – F_E = 2.7087 – 2 = 0.7087.

Step – 3:

u₂ = 0.3651,

k(u₂) = 3(0.3651)2 + 18(0.3651) + 4 = 10.9713,

∆F = 1,

∆ u₃ = = 1/ 10.9713 = 0.0971,

u₃ = u₂ + ∆ u₃ = 0.3651 + 0.0971 = 0.4562,

F_External = ∆F + ∆F + ∆F = 1+1+1= 3,

F_Internal = u₃³ + 9u₃² + 4u₃ = (0.4562)³ + 9(0.4562)² + 4(0.4562) = 3.7928,

R = F_I – F_E = 3.7928 – 3 = 0.7928.

Summary of explicit analysis results:

Step i	∆Fi	∆ui	ui	(Fext)i	(fint)i	fint − Fext = R
1	1.0	0.25	0.25	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

It is evident from the results that external and internal forces are NOT in equilibrium because

F_ext /= f_int.

Implicit Method:

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) = f_int − F_ext.

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 (δu^j+1)+…= + k u^j (δu^j+1)+…

By neglecting higher order terms in the series, setting

R^j+1 = 0 and solving for δu^j+1

The correction is obtained

δu^j+1 = −[k(u^j)]⁻¹R^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.

Step – 1:

u_{0 =} 0.0

k(u₀) = 3(0)2 + 18(0) + 4 = 4

∆F = 1,

∆ u₁ = = ¼ = 0.25,

u₁ = u₀ + ∆ u₁ = 0 + ¼ = 0.25

F_External = ∆F = 1,

F_Internal = u₁³ + 9u₁² + 4u₁ = (0.25)³ + 9(0.25)² + 4(0.25) = 1.5781,

R = F_I – F_E = 1.5781-1 = 0.5781.

||R⁰|| > 10⁻²

Newton-Raphson iterations are necessary.

Calculate the correction to u₁ = u₁⁰

δu⁽¹⁾ = −[k(u⁽⁰⁾)]⁻¹R⁽⁰⁾

u₁ = 8.6875 & R⁰ = 0.578152

δu⁽¹⁾ = −[k(u₁⁰)]⁻¹R⁰ = -[8.6875]^-1 (0.578152)

δu¹ = -0.0665467,

Now, updating the value,

u₁⁽¹⁾= u₁⁽⁰⁾ + δu¹ = 0.25 – 0.0665 = 0.1835,

Now, checking Residuals,

F_External = ∆F = 1,

F_Internal = (u₁⁽¹⁾)³ + 9(u₁⁽¹⁾)² + 4u₁⁽¹⁾

= (0.18346)³ + 9(0.18346)² + 4(0.18346) = 1.0373

Now, R¹ = F_I – F_E = 1.03736- 1 = 0.03736,

||R⁰|| > 10⁻²

So, Newton-Raphson iterations are necessary

k(u₁) = 3 u₁² + 18u₁ + 4

= 3(0.18346)² + 18(0.18346) + 4

k(u₁) = 7.403252

So, calculating new total correction to u₁ = u₁⁰

δu⁽²⁾ = δu⁽¹⁾ −[k(u₁⁽¹⁾)]⁻¹R⁽¹⁾

= -0.06654 – [7.4032]^-1 (0.03736)

δu⁽²⁾ = -0.07159

Now, updating the value,

u₁⁽²⁾ = u₁⁰ + δu⁽²⁾

= 0.25 + (-0.071593)

u₁⁽²⁾ = 0.1784,

Now, Checking Residuals,

F_External = ∆F = 1,

F_Internal = (u₁⁽²⁾)³ + 9(u₁⁽²⁾)² + 4u₁⁽²⁾

= (0.1784)³ + 9(0.1784)² + 4(0.1784)

F_Internal = 1.0057

Now, R² = F_I – F_E = 1.0557-1 = 0.005767

||R⁰|| < 10⁻²

So, Newton-Raphson iterations are not necessary

So, Final value is u₁ = u₁⁽²⁾ = 0.1784

Step – 2:

u₁ = 0.1784,

k(u₁) = 3(0.1784)² + 18(0.1784) + 4 = 7.3068

∆F = 1,

∆ u₂ = = 1/ 7.306813 = 0.1368

u₂ = u₁ + ∆ u₂ = 0.1784 + 0.1368 = 0.3152

F_External = ∆F + ∆F = 1+1 = 2,

F_Internal = (u₂)³ + 9(u₂)² + 4u₂

= (0.3152)³ + 9(0.3152)² + 4(0.3152) = 2.1868

R = F_I – F_E = 2.1868-2 = 0.1868.

||R⁰|| > 10⁻²

Newton-Raphson iterations are necessary.

So, k(u₂) = 3 u₁² + 18u₁ + 4

= 3(0.3152)² + 18(0.3152) + 4 = 9.9727

So, calculating new total correction to u₁ = u₁⁰

δu⁽¹⁾ = −[k(u⁽⁰⁾)]⁻¹R⁽⁰⁾

= -[9.9727]^-1 (0.1868)

δu⁽¹⁾ = 0.01873.

Now, updating the value,

U₂⁽¹⁾ = u₂⁰ + δu⁽¹⁾

= 0.3152 + (-0.01873)

U₂⁽¹⁾ = 0.29653,

Now, checking Residuals,

F_External = ∆F + ∆F = 1+1 = 2,

F_Internal = (u₂⁽¹⁾)³ + 9(u₂⁽¹⁾)² + 4u₂⁽¹⁾

= (0.29653)³ + 9(0.29653)² + 4(0.29653) = 2.003563,

Now, R¹ = F_I – F_E = 2.003563-2 = 0.003563

||R⁰|| < 10⁻²

So, Newton-Raphson iterations are not necessary

So, Final value is u₂ = u₂⁽¹⁾ = 0.2965

Step – 3:

u₂ = 0.2965

k(u₂) = 3(0.2965)² + 18(0.2965) + 4 = 9.60133,

∆F = 1,

∆ u₃ = = 1/ 9.6013 = 0.1041

u₃ = u₂ + ∆ u₃ = 0.2965 + 0.1041 = 0.4006

F_External = ∆F + ∆F + ∆F = 1+1+1 = 3,

F_Internal = (u₃)³ + 9(u₃)² + 4u₃ = (0.4006)³ + 9(0.4006)² + 4(0.4006) = 3.1119

R = F_I – F_E = 3.1119-3 = 0.1119.

||R⁰|| > 10⁻²

Newton-Raphson iterations are necessary.

So, k(u₃) = 3 u₃² + 18u₃ + 4

= 3(0.4006)² + 18(0.4006) + 4 = 11.6938,

So, calculating new total correction to u₂ = u₂⁰

δu⁽²⁾ = −[k(u₃⁽⁰⁾)]⁻¹R⁽⁰⁾

= -[11.6938]^-1 (0.11194)

δu⁽²⁾ = -0.0095,

Now, updating the value,

u₃⁽¹⁾ = u₃⁰ + δu⁽²⁾

= 0.4006 + (-0.0095)

u₃⁽¹⁾ = 0.3911,

Now, checking Residuals,

F_External = ∆F + ∆F +∆F = 1+1+1= 3,

F_Internal = (u₃⁽¹⁾)³ + 9(u₃⁽¹⁾)² + 4u₃⁽¹⁾

= (0.3911)³ + 9(0.3911)² + 4(0.3911) = 3.0009,

Now, R¹ = F_I – F_E = 3.0009-3 = 0.003563

||R⁰|| < 10⁻²

So, Newton-Raphson iterations are not necessary

So, Final value is u₂ = u₃⁽¹⁾ = 0.3911,

Summary of implicit analysis results:

Step i	∆Fi	ui	(Fext)i	(fint)i	fint − Fext = R
1	1.0	0.1784	1.0	1.0057	0.0057
2	1.0	0.2966	2.0	2.0035	0.0035
3	1.0	0.3911	3.0	3.0009	0.0009

It is evident from the table s that the external forces and internal forces are coming into equilibrium because

F_External ≈ F_Internal

Plots:

Explicit Metod:

Implicit Method:

Exact solution:

Discussion:

As discussed, we can use both implicit and explicit solvers to solve the same type of problems. The only difference is in 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 really 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 explicit step depends on our model and computing times depends on the system hardware.

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