Week - 2 - Explicit and Implicit Analysis

Week - 2 - Explicit and Implicit Analysis

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 0.001).

F(u) = u³+9u²+4u

Stiffness k(u) = df/du=3u²+18u+4

F=k(u)

∆F=k(u)* ∆u

Here,

∆F=1;U₀=0

U₁=u₀+ ∆u

∆u= ∆F/k(u)

U1=u0+ ∆u1 ∆u1= ∆F/k(u0) ∆u1=1/3u2+18u+4 ∆u1=1/3(0)2+18(0) +4 ∆u1=1/4=0.25 U1=u0+ ∆u1=0+0.25=0.25

U2=u1+ ∆u1 ∆u2= ∆F/k(u1) ∆u2=1/3u2+18u+4 ∆u2=1/3(0.25)2+18(0.25) +4 ∆u2=1/8.688=0.115 U2=u1+ ∆u2=0.25+0.115=0.365

U3=u2+ ∆u3 ∆u3= ∆F/k(u2) ∆u3=1/3u2+18u+4 ∆u3=1/3(0.365)2+18(0.365) +4 ∆u3=1/10.9696=0.0911 U3=u2+ ∆u3=0.365+0.0911=0.456

from the three steps the following results are obtained

External Forces F_ext = ∆F + ∆F + ∆F = 1+1+1=3.0

Internal Forces F_int = u³+9u²+4u = (0.456)³+9(0.456)²+4(0.456) =3.79

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

F_ext ≠F_int.

Step	∆Fi	∆ui	ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.25	0.25	1	1.578	0.578
2	1	0.115	0.365	2	2.708	0.708
3	1	0.0911	0.456	3	3.790	0.790

Implicit Analysis

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) = fint − Fext.

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 + ... = R ^j + k(u ^j )δu^j+1 + ...

By neglecting higher order terms in the series, setting R^j+1 = 0 and solving for δuj+1 the following correction is obtained

δu^j+1 = −[k(u ^j )]⁻¹R ^j .

Notice: that 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 (for the below a tolerance of 10−2 is chosen).

Here,

∆F=1;U₀=0

U₁=u₀+ ∆u

∆u= ∆F/k(u)

Find “U₁”

U₁=u₀+ ∆u_1,

∆u₁= ∆F/k(u₀)

∆u₁=1/3u²+18u+4

∆u₁=1/3(0)²+18(0) +4

∆u₁=1/4=0.25

U₁=u₀+ ∆u₁=0+0.25=0.25

After finding this u1, apply this F_ext.

check the residual, F_ext = ∆F = 1, f_int = u³+9u²+4u = (0.25)³+9(0.25)²+4(0.25) =1.578

R(u) = f_int − F_ext.

R(u₁) =1.578 -1=0.578

||R⁰|| > 10⁻² ⇒ another Newton-Raphson iteration is necessary

Apply Taylor equation.

Δu₁¹ = − [k(u₁⁰)]⁻¹ .R ^j .

consider u¹=u⁰

δu₁¹ =-[3u²+18u+4]^-1 *0.578=-[3(0.25)²+18(0.25)+4]^-1 *0.578=-0.0665

u₁¹ = u₁⁰ +δu₁¹=0.25+(-0.0665) =0.1835

use this u₁¹=0.1835 and find f _int,

f_int= u³+9u²+4u = (0.1835)³+9(0.1835)²+4(0.1835) = 1.0430

F_ext=1

R(u₁) = f_int − F_ext =1-1.0430=0.0430

(For the above more than a tolerance of 10⁻² repeats this up to below the value)----------------1

consider u₁²=u₁¹

δu₁²=-(k(u₁¹))^-1.R¹

δu₁² =-[3u²+18u+4]^-1 *0.0430=-[3(0.1835)²+18(0.1835)+4]^-1 *0.0430= -0.00581

u₁²₌ u₁¹+ δu₁²=0.1835+(-0. 0.00581) = 0.17769

use this u₁²=0.1777and find f _int,

f_int= u³+9u²+4u = (0.1777)³+9(0.1777)²+4(0.1777) = 1.0003

F_ext=1

R(u₁) = f_int − F_ext =1-1.0003=0.0003

R=0.0003<0.001

||R|| < 10−2 ⇒ no further iterations are necessary U₁=0.1777

Find “U₂”

U₂=u₁+ ∆u₂

∆u₂ = ∆F/k(u₁)

=1/(3u²+18u+4)=1/(3*0.1777²+18*0.1777+4)= 0.1371

U₂ =0.1777+0.1371=0.3148

Apply u₂ in f_ext F(u) = u³+9u²+4u

F(u₂) = u₂³+9u₂²+4u₂

= (0.3148)³+(9*0.3148)²+(4*0.3148)

F_ext(u₂) = 2.1824

R(u₂) = fint − Fext =2-2.1824=0.1824

Apply Taylor equation. Δu₂¹ = − [k(u⁰)]⁻¹ .R ^j .

consider u₂¹=u₂⁰

δu₂¹ =-[3u²+18u+4]^-1 *0.1824 =-[3(0.3148)²+18(0.3148)+4]^-1 *0.1824 =-0.0183

u₂¹₌ u₂⁰ +δu₂¹=0.3148+(-0.0183) =0.2965

use this u₁¹=0.2965and find f _int,

f_int= u³+9u²+4u = (0.2965)³+9(0.2965)²+4(0.2965) = 2.0032

F_ext=2

R(u¹₂) = f_int − F_ext =2-2.0032=0.0032

(For the above more than a tolerance of 10⁻² repeats this up to below the value)----------------1

consider u₂²=u₂¹

δu₂² = − [k(u₂¹)]⁻¹ .R ¹ .

δu₂² =-[3u²+18u+4]^-1 *0.0032=-[3(0.2965)²+18(0.2965)+4]^-1 *0.0032 = -0.00033

u₂²₌ u₂¹+ δu₂²=0.3148+(-0.00033) = 0.2962

use this u₂²=0.2962and find f _int,

f_int= u³+9u²+4u = (0.2962)³+9(0.2962)²+4(0.2962) = 2.000001

F_ext=2

R(u²₂) = f_int − F_ext =2-2.000001=0.000001

R=0.000001<0.001

||R|| < 10−2 ⇒ no further iterations are necessary U₂=0.2962

Find “U₃”

U₃=u₂+ ∆u₃

U₂=0.2962

∆u₃ = ∆F/k(u₂)

=1/(3u²+18u+4)=1/(3*0.2962+18*0.2962+4)= 0.10422

U₃ =0.2962+0.10422=0.4004

Apply u₂ in f_ext F(u) = u³+9u²+4u

F(u₃) = u₃³+9u₃²+4u₃

= (0.4004)³+(9*0.4004)²+(4*0.4004)

F_ext(u₂) = 3.1089

R(u₂) = fint − Fext =3-3.10894=-0.10894

Apply Taylor equation. Δu₃¹ = − [k(u⁰)]⁻¹ .R ^j .

consider u₃¹=u₂⁰

δu₃¹ =-[3u²+18u+4]^-1 *0.10894 =-[3(0.4004)²+18(0.4004)+4]^-1 *0.10894=-0.0093

u₃¹₌ u₃⁰ +δu₃¹=0.4004+(-0.0093) =0.3911

use this u₃¹=0.3911and find f _int,

f_int= u³+9u²+4u = (0.3911)³+9(0.3911)²+4(0.3911) = 3.000855237

F_ext=3

R(u₃) = f_int − F_ext =3-3.000855237=0.000855237

R=0.000855237<0.001

||R|| < 10−2 ⇒ no further iterations are necessary U₂=0.3911

Step	∆F	ui	Fext)i	(fint)i	fint − Fext = R
1	1	0.1777	1	1.001	0.001
2	1	0.2962	2	2.000	0.000
3	1	0.3911	3	3.001	0.001