Week - 2 - Explicit and Implicit Analysis

AIM

To study about implicit and explicit method to solve a given polynomial equation

F(u) = u3+9u2+4u

EXPLICIT and IMPLICIT Method

Implicit and Explicit analysis differ in the approach to time incrementation.

Explicit methods calculate the state of a system at the next time step from the state of the system at the current time, Explicit analysis handles non linearities with relative ease compared to implicit analysis. In nonlinear implicit analysis, solution of each step requires a series of trial solutions (iterations) to establish equilibrium within a certain tolerance.

In explicit analysis, no iteration is required as the nodal accelerations are solved directly. However, each time increment has to converge, but you can set quite large time increments. Explicit analysis doesn’t need to have a convergence at each increment, but time increments must be super small for the accuracy of solution since the time step in explicit analysis must be less than the Courant time step for it to be valid. Implicit method checks for convergence after every time-step this is achieved by incorporating a correction scheme using newton Raphson iteration to minimize the residual This method is suitable for highly linear problems.

1.Explicit Method:

F(u) = u³+9u³+4u, to solve this equation an incremental explicit load control scheme is used.

Three equal load increment of ΔF = 1 unit are used,

u - displacement

f - internal force

F - external force applied

K – stiffness

Δu - Incremental displacement and

ΔF - Externally applied load

Force(F) = Stiffness (K)* displacement (u)

F(u) = u³+9u³+4u

K(u) = dF/du = 3u²+18u+4

Step 1,

Take u = 0

k(u₀) = 3 x 0 + 18 x 0 + 4 = 4

Δu₁ = ΔF/k(u₀) = ¼ = 0.25

u_{1 =} u₀ + Δu₁ = 0 + 0.25 = 0.25

Step 2,

Take u₁= 0.25

k(u₁)= 3 x 0.25² + 18 x 0.25 + 4 = 8.6875

Δu₂ = ΔF/k(u₁) = 1/8.6875 = 0.1151

u₂ = u₁+Δu₂= 0.3651

Step 3,

Take u₂= 0.3651

k(u₂)= 3 x 0.3651² + 18 x 0.3651 + 4 = 10.9716

Δu₂ = ΔF/k(u₂) = 1/10.9716 = 0.09114

u₃ = u₂+Δu₃ = 0.46

Finally,

F(ext) = ΔF+ΔF+ΔF = 3

f(int) = (0.46³) + (9 x 0.46²) + (4 x 0.46) = 3.841

Thus F(ext) and f(int) are not in equilibrium

Step	ΔF	Δui	ui	Fext	Fint	R = Fext - fint
1	1	0.25	0.25	1	1.578	0.578
2	1	0.1151	0.3654	2	2.708	0.708
3	1	0.09114	0.46	3	3.841	0.841

2.Implicit Method:

F(u) = u³+9u³+4u

K(u) = dF/du = 3u²+18u+4

Step1,

Take u₀ = 0

k(u₀) = 3 x 0 + 18 x 0 + 4 = 4

Δu₁ = ΔF/k(u₀) = ¼ = 0.25

u_{1 =} u₀ + Δu₁ = 0 + 0.25 = 0.25

Residual R⁰

F = (0.25) + 9 x (0.25) + 4 x 0.25 = 1.578

R⁰ = 1.578 – 1 = 0.578 > 10^-2

Thus, Newton-Raphson iterations is necessary

u₁⁽⁰⁾ = u₁⁽⁰⁾ = 0.25

du⁽¹⁾ = -[k(u₁⁽⁰⁾]^-1 x R⁰

= - [(3 x 0.25 ) + (18 x 0.250) +4]^-1 x 0.578

= - 0.0665

u₁⁽¹⁾ = u₁⁽⁰⁾ + du⁽¹⁾

= 0.25 – 0.0665 = 0.1835

Residual R¹

F = (0.1835) + 9 x (0.1835) +4 x 0.1835 = 1.0432

R¹ = 1.0432 – 1 = 0.0432 > 10^-2

Thus, Newton-Raphson iterations is necessary

u₁⁽¹⁾ = 0.1835

du⁽²⁾= du⁽¹⁾ - [k(u₁⁽¹⁾)]^-1 x R1

= (-0.0665) - (8.6875 )^-1 x 0.0432

= -0.0714

u₁⁽²⁾ = u₁⁽¹⁾ + du⁽²⁾

= 0.25 – 0.0714 = 0.1786

Residual R²

F = (0.1786) + 9 x (0.1786) +4 x 0.1786 = 1.0071

R0 = 1.0071 – 1 = 0.0071 < 10^-2

Thus u1 = 0.1786

Step 2&3:

we get u2=0.2968 , u3=0.3913 following a similar approach

Step	ΔF	ui	Fext	Fint	R = Fext - fint
1	1	0.1786	1	1.004	0.004
2	1	0.2968	2	2.0038	0.0038
3	1	0.3911	3	3.0007	0.0007

CONCLUSION:

The above equation was solved using both implicit and explicit approaches. The difference being the approach to time increment between the cases. The explicit method gives faster result where as the implicit is more computationally intensive. The results obtained from implicit method match the analytical results where as the explicit method results tends to deviate.