Week - 2 - Explicit and Implicit Analysis

Aim : Analytical solution for given equation using explicit and emplicit technique.

Equation : F(u) = U³+9u²+4u

Explicit method:

The equation for explicit method is as given below

u(n+1) = u(n)+ Δt. F(u(n))

equation to be solved as below

M (d²u / dt²) +k(u)= F(t)

where du /dt = F(u)

at t=0 we know thw value of u & v and then to compute M & k and then use them to compute the acceleration , from which velocity and displacement can be calculated at Δt.

a = [ M]^-1[F(t) - k(u)

v( Δt) = v(0) + a.(Δt)

u( Δt) = u(0) + v(0). (Δt) + 1/2. a.(Δt²)

in explicit mostly dynamic analysis, nodal accelerations are solved directly (not iteratively) as the inverse of the diagonal mass matrix times the net nodal force vector where net nodal force includes contributions from exterior sources (body forces, applied pressure, contact, etc.), element stress, damping, bulk viscosity, and hourglass control. Once accelerations are known at time n, velocities are calculated at time n+1/2, and displacements at time n+1. From displacements comes strain. From strain comes stress and the cycle is repeated.

Explicit analysis handles nonlinearities with relative ease as compared to implicit analysis. This would include treatment of contact and Material nonlinearities. With this method is that you do need many small increments for good accuracy and it is time consuming.

k(u)= (df)/(du) = d (U³+9u²+4u) / du

= 3u² + 18u +4

u= displacement

f= internal force

F=external force

k= stiffness

Δu= increamental displacement

ΔF= increamental external force applied

for 1st step :

ΔF = 1;

U₀=0.0

using stiffnerss equantion at u =u₀

k(u₀ )= 3u² + 18u +4

=4.0

now , we know that,

Δu₀ = ΔF / k(u₀ )

= 1 / 4

= 0.25

_now,

u₁ = u₀ + Δu₀

= 0 + 0.25

= 0.25

now calculate,

f_int = u³+9u²+4u

= 0.25³+9*0.25²+4*0.25

= 1.578125

F_ext = 1.

now find residue

R₁ = f_int - F_ext

_{= 1.578125 - 1}

= 0.578125

_{for 2nd step :
u1 = 0.25}

k(u₁ )= 3u₁² + 18u₁ +4

= 3* 0.25² + 18*0.25 +4

= 8.6875

ΔF = 1

Δu₁ = ΔF / k(u₁ )

= 1 / 8.6875

= 0.1151

NOw update the value for u.

u₂ = u₁ + Δu₁

= 0.25 + 0.1151

= 0.3651

now calculate

f_int = u₂³+9u₂²+4u₂

= 0.3651³+9*0.3651²+4*0.3651

= 2.7087

F_ext = 1+1 =2

now find residue.

R₂ = f_int - F_ext

_{= 2.7087 - 2}

= 0.7087

for 3rd step

u_{2 = 0.3651}

k(u₂ )= 3u₂² + 18u₂ +4

= 3* 0.3651² + 18*0.3651 +4

= 10.97169

ΔF = 1

Δu₃ = ΔF / k(u₂ )

=1 / 10.97169

= 0.0911436

u₃ = u₂ + Δu₃

=0.3651 + 0.0911436

= 0.45624

now calculate ,

f_int = u₂³+9u₂²+4u₂

= 0.45624³+9*0.45624²+4*0.45624

= 3.793325

F_ext = 2+1 =3

now find residue.

R₂ = f_int - F_ext

= 3.793325 - 3

= 0.793325

Table summrised the main result for explicit analysis method

Stepi	ΔFi	Δui	ui	fint	Fext	Ri = fint - Fext
1	1	0.25	0.25	1.578125	1	0.578125
2	1	0.1151	0.3651	2.7087	2	0.7087
3	1	0.09114	0.45624	3.793325	3	0.793325

it is evident from the result that the external and internal forces are not in equillibrilium because
f_int not equal to F_ext.

_{2. implicit method}

Equation for implicit method is as given below,

u(n+1) = u(n)+Δt . F(u(n+1))

= (I-A). Δt. u(n+1) = u(n).... where I is identity matrix

in implicit method we compute M and k not at Δt =0 , but at Δt and then proceed as above.

Required mpre calculation and cost of computing is large compare to explicit method.

forword and backword Euler mehtod are the example of explicit and implicit method to solve a differential equation .

each time step in explicit method is very fast, but the method is stable only when time step delta_t is small. courant condition dictates that the time step ,ust be smaller then the elastic wave to propogate from one side of element to the other

Courant fredrichs lavy criteria

Δt < sqrt( ρ(1-2nu) / 2μ ((1-nu)

here μ is shear modulus of material.

Implicit analysis requires a numerical solver to invert the stiffness matrix once or even several times over the course of a load/time step. This matrix inversion is an expensive operation, especially for large models. It is mostly used for static analysis. Implicit uses Newton-Raphson iterations to enforce equilibrium. If done correctly the Newton-Raphson iterations will have a quadratic rate of convergence which is very desireable.

Implicit transient analysis has no inherent limit on the size of the time step. As such, implicit time steps are generally several orders of magnitude larger than explicit time steps.

u= displacement ,

f= internal force in bar,

F= external force in bar

∆u = incremental displacements,

∆F = incremental external applied forces.

using relation,

∆F = k∆u,

Tolerance= 10^-2

we know that residual (R) = f_int - F_ext

Step 1:

ΔF = 1;

U₀=0.0

using stiffnerss equantion at u =u₀

k(u₀ )= 3u² + 18u +4

=4.0

now , we know that,

Δu₁ = ΔF / k(u₀ )

= 1 / 4

= 0.25

u₁ = u₀ + Δu₁

= 0 +0.25

= 0.25

now calculate,

f_int = u³+9u²+4u

= 0.25³+9*0.25²+4*0.25

= 1.578125

F_ext = 1.

now check for the resudual,

R⁰ = f_int - F_ext

= 1.578125 - 1

= 0.578125

| R⁰ | > 0.01 .... thus Newton Raphson iteration are neccessary

now, calculate the correction at u₁ = (u₁)⁰

δu¹= -[ k(u₀)⁰]^-1 x R

= -[ 3u² + 18u +4]^-1 x 0.578125 .... u = 0.25

= -[ 3 x 0.25² + 18x 0.25 +4]^-1 x 0.578125

= - 0.0665

The uodated value for u₁¹ is

u₁¹ = u₀¹ + δu¹

= 0.25 + (- 0.0665)

= 0.1835

now, check the residual again

F_ext = 1.

f_int = u³+9u²+4u ---- u = 0.1835

= 0.1835³+9*0.1835²+4*0.1835

= 1.04322

R¹ = f_int - F_ext

= 1.04322 - 1

= 0.04322

|R¹ | > 0.01... thus another newton raphson iteration are neccessary

calculate the new total correction u₁ = (u₁)¹

δu²= δu¹ -[ k(u₁)¹]^-1 x R

= - 0.0665 - [ 3u² + 18u +4]^-1 x 0.04322 .... u= 0.1835

= - 0.0665 - [ 3x 0.1835² + 18x 0.1835 +4]^-1 x 0.04322

= -0.072373

the updated value of (u₁)²

(u₁)² = u₀¹ + δu²

= 0.25 + ( -0.072373)

= 0.177662

check the residual again

F_ext = 1.

f_int = u³+9u²+4u ---- u = 0.177662

= 0.177662³+9*0.177662²+4*0.177662

= 1.00032

R² = f_int - F_ext

= 1.00032 - 1

= 0.00032

|R² | < 0.01 ... so no further iteration required

therefore the final value is

(u₁)= (u₁)² = 0.177662

Step 2:

ΔF = 1;

U₁=0.177662

k(u₁ )= 3u² + 18u +4 ... u= 0.177662

= 3x 0.177662² + 18x0.177662 +4

= 7.292607

Δu₂ = ΔF / k(u₁ )

= 1 / 7.292607

= 0.1371251

now ,

u₂ = u₁ + Δu₂

= 0.177662 + 0.1371251

= 0.314787

now calculate,

f_int = u³+9u²+4u ..... u= 0.314787

= 0.314787³+9*0.314787²+4*0.314787

= 2.182159

F_ext = 1+1 = 2.

check for the residual,

R⁰ = f_int - F_ext

= 2.182159 - 2

= 0.182159

|R⁰ | > 0.01... thus another newton raphson iteration are neccessary

Calculate the correction u₂ = (u₂)⁰

δu2¹= -[ k(u₂)⁰]^-1 x R

= -[ 3u² + 18u +4]^-1 x 0.182159 .... u = 0.314787

= -[ 3x 0.314787² + 18 x 0.314787 +4]^-1 x 0.182159

= -0.018326

The updated value for (u₂)¹

(u₂)¹= u₂ + δu2¹

= 0.314787 + (-0.018326)

= 0.29646

check for the residual again

f_int = u³+9u²+4u ..... u= 0.29646

= 0.29646³+9*0.29646²+4*0.29646

= 2.0028992

F_ext = 2.

R¹ = f_int - F_ext

= 2.0028992 - 2

= 0.0028992

|R¹ | < 0.01 ... so no further iteration required

therefore the final value for u₂ = (u₂)¹ = 0.2964

Step 3:

ΔF = 1;

U₂= 0.2964

k(u₂ )= 3u² + 18u +4 ... u=0.2964

= 3x 0.2964² + 18x0.2964 +4

= 6.93115888

Δu₃ = ΔF / k(u₂ )

= 1 / 6.93115888

= 0.144276

now ,

u₃ = u₂ + Δu₃

= 0.2964 + 0.144276

= 0.440676

check for the residual

f_int = u³+9u²+4u ..... u=0.440676

= 0.440676³+9*0.440676²+4*0.440676

= 3.59603922

F_ext = 2+1 =3.

R⁰ = f_int - F_ext

= 3.59603922 - 3

= 0.59603922

|R⁰ | > 0.01... thus another newton raphson iteration are neccessary

now Calculate the correction u₃ = (u₃)⁰

δu3¹= -[ k(u₃)⁰]^-1 x R

= - [ 3u² + 18u +4]^-1 x 0.182159 .... u = 0.440676

= - [ 3 x 0.440676² + 18x0.440676 +4]^-1 x 0.182159

= -0.04762378

updated value of (u₃)¹

(u₃)¹= u₃ + δu3¹

= 0.440676 -0.04762378

= 0.3930522

check for the residual

f_int = u³+9u²+4u ..... u=0.3930522

= 0.3930522³+9*0.3930522²+4*0.3930522

= 3.0233

F_ext = 3.

R⁰ = f_int - F_ext

= 3.0233 - 3

= 0.0233

|R⁰ | > 0.01... thus another newton raphson iteration are neccessary

now Calculate the correction u₃ = (u₃)⁰

δu3²= δu3¹ -[ k(u₃)¹]^-1 xR⁰

= -0.04762378 - [ 3u² + 18u +4]^-1 x 0.0233 .... u = 0.3930522

=-0.04762378 - [ 3x 0.3930522² + 18x 0.3930522 +4]^-1 x 0.0233

= -0.0067817

updated value of (u₃)²

(u₃)²= u₃ + δu3²

= 0.3930522 + (-0.0067817)

= 0.38627

now check for the residual

f_int = u³+9u²+4u ..... u=0.38627

= 0.38627³+9*0.38627²+4*0.38627

= 2.9465

F_ext = 3.

R² = f_int - F_ext

= 2.9465 - 3

= -0.005

|R²| < 0.01 ... thus no further newton raphson iteration are required.

therefore the final value for (u₃)² = (u₃)² = 0.38627

the summary of implicit analysis is given below

Stepi	ΔFi	Δui	ui	fint	Fext	Ri = fint - Fext
1	1	0.25	0.177662	1.00032	1	0.00032
2	1	0.1371251	0.2964	2.002899	2	0.00289
3	1	0.744276	0.38627	2.9465	3	0.005

conclusion : The Implicit and Explicit numerical analysis is carried out for given equation.

Implicit method is more accurate but time consuming and solemnly depends on element size and timestep whereas explicit deviates slightly from exact result and needs more iterations to obtain good result.