Week - 2 - Explicit and Implicit Analysis

Explicit and Implicit Analysis

Objective: To use the given equation and solve it using Explicit and Implicit analysis methods.

• Explicit and Implicit analysis are the 2 types of approach that can be used to solve Finite elements problem.
• The implicit approach is useful in problems in which time dependency of the solution is not an important factor [ example: Static structures, harmonic, modal analysis etc.].
• Explicit Dynamic approach is most helpful in solving high deformation time dependent problem such as crash , blast , impact etc.

Calculations:

Given function: F(u) = u³ + 18u + 4

Sol:

F(u) = u³ + 9u² + 4u à eq 1

From equation 1

K(u) = df/du = 3u² + 18u + 4 à eq 2

We have F = Ku à eq 3

Here the force is F and displacement is u so we can rewrite eq 2 as

F = k * u à eq 4

Explicit analysis:

Step 1:

When there is no deformation that is in the initial condition

U₀ = 0 substitute in eq 2

F(u₀) = 0

K(u) = df/du = 3u² + 18u + 4

K(u₀) = 3(0)² + 18(0) + 4

= 4

Therefore K(u₀) = 4

For these 3 equal increments of ∆F = 1 is used to load the tension bar.

When ∆F =1 due to external force there will be deformation.

Now, from equation 4

∆u₁ = ∆F/K(u₀) = ¼ =0.25

U₁ = u₀ + ∆u₁

U₁ = 0 + 0.25

U₁ = 0.25

Step 2:

Consider u = u₁ = 0.25 and the load increments of ∆F = 1

K(u₁) = 3(0.25²) + 18(0.25) + 4

K(u₁) = 0.1875 + 4.5 + 4

Therefore K(u₁) = 8.6875

∆u₁ = ∆F/K(u₁) = 1/8.6875 = 0.115

U₂ = u₁ + ∆u₂

U₂ = 0.25 + 0.115

U₂ = 0.365

Step 3:

Consider u = u₂ = 0.365 and load increment ∆F = 1.

K(u₂) = 3(0.365²) + 18(0.365) +4

K(u₂) = 0.399 + 6.57 + 4

= 10.969

∆U₃ = u₂ + ∆u₂

U₃ = 0.36 + 0.911

Therefore u₃ = 0.4511

So, from the above, external force is given by 3 times in every instant.

F_external = ∆F ₊ ∆F + ∆F = 1 + 1 +1 = 3

And the force produced internally due to external force of every step is given by

F_internal = u₁³ + 9u₁² + 4u₁ = (0.25)³ + 9(0.25)² + 4(0.25) = 1.578

F_internal = u₂³ + 9u₂² + 4u₂ = (0.365)³ + 9(0.365)² + 4(0.365) = 2.707

F_internal = u₃³ + 9u₃³ + 4u₃ = (0.4511)³ + 9(0.4511)² + 4(0.4511) = 3.726.

Implicit analysis: An implicit analysis is the same as the explicit analysis, except that at the end of each step Newton-Raphson’s iterations are used to enforce equilibrium before moving to next step.

Δu^j+1 = -[k(u^j)]^-1 R^j

When there is deformation in the system, that is initial load is u₀ = 0.

Substitute in eq 2

F(u₀) = 0

K(u) = df/du = 3u² + 18u + 4

K(u₀) = 3(0) + 18(0) + 4 = 4

When ∆F =1 due to external force there will be deformation.

From eq 4

∆u₁ = ∆F/k(u₀) = ¼ =0.25

U₁ = u₀ + ∆ul = 0.25

F_external = ∆F = 1

F_internal = u₁³ + 9u₁² + 4u₁ = (0.25)³ + 9(0.25)² + 4(0.25) = 1.578

|R⁰| > 10^-2 This shows that Newton-Raphson’s Iterations are necessary.

Calculate the correction u₁ = u₁⁰

δu^j+1 = -[k(u^j)]]^-1 R^j where j = 0

δu¹ = δu¹ –[k(u₁⁰)]^-1 R⁰

k(u₁) = 3(0.25)² + 18(0.25) +4 =8.687

δu¹ = -[8.687]^-1 * 0.578 = -0.066

Hence, u₁¹ = u₁⁰ + δu₁⁰

= 0.25 – 0.066 = 0.184

Check the residual again,

F_external = ∆F = 1

F_internal = (u₁¹)³ + 9(u₁¹) + 4u₁¹

= (0.184)³ + 9(u₁¹) + 4u₁¹ = 1.046

R¹ = F_internal - F_external = 1.046 - 1

= 0.46>10^-2 Newton-Raphson’s iterations are necessary.

Calculate the correction to u₁ =u₁¹

δu^j+1 = -[k(u^j)]]^-1 R^j where j = 0

δu² = δu¹ –[k(u₁¹)]^-1 R¹

k(u₂) = 3(0.184)² + 18(0.184) + 4 = 7.41

δu¹ = -0.066 – (7.41)^-1 * 0.046

= -0.0722

Check the residual again

F_external = ∆F = 1

F_internal = (u₁²)³ + 9(u₁²)² + 4u₁

= 0.177³ + 9(0.177²) + 4(0.177)

= 0.995

R² = F_internal - F_external

R² = 0.995 – 1 = -0.005<10^-2

No further iterations are necessary.

Hence the u₁ = 0.177.

For u₂ and u₃ we carry the same steps

U₂ = 0.296

U₃ = 0.391

Observation: The given equation has solved using Explicit and Implicit methods.