Week - 2 - Explicit and Implicit Analysis

F(u)= `u 3 + 9 u 2 + 4 u`

`k(u)= (df)/(du) = 3u2+18u+4`

Explicit Analysis

Three equal load increments (or steps) of ∆F = 1 is considered

u = Displacement, f= Internal Force, F0 External Force and k= Stiffness

Incremental displacements or incremental externally applied forces are represented as ∆u or ∆F,
respectively. Use is made of the relationship ∆F = k∆

Step 1:

u0 = 0

k(u0)= 4

∆F = 1

∆u1 = ∆F/k(u0) = 1/4

u1 = u0 + ∆u1 = 0+1/4 = 0.25

fint= (0.25)3+9(0.25)2+4(0.25) = 1.578

Step 2:

k(u1)= 3(0.25)2+18(0.25)+4= 8.68

∆F = 1

∆u2 = ∆F/k(u1) = 0.11

u2 = u1 + ∆u2 = 0.36

Step 3:

k(u2)= 3(0.36)2+18(0.36)+4 = 10.86

∆F = 1

∆u3 = ∆F/k(u2) = 0.092

u3 = u2 + ∆u3 = 0.45

Step i	∆Fi	∆ui	ui	Fext	fint	fint-Fext =R
1	1	0.25	0.25	1.0	1.57	0.57
2	1	0.11	0.36	2.0	2.65	0.65
3	1	0.092	0.45	3.0	3.71	0.71

Implicit Analysis

Step 1

u0 = 0.0, ⇒ k(u0) = 0.25, ∆F = 1, ∆u1 = ∆F/k(u0) = 0.25, u1 = u0 + ∆u1 = 0.25

Check the residual, Fext = ∆F = 1, fint = (u₁)³+ 9(u₁)²+ 4u₁= 1.57 R(0) = fint − Fext = 0.57

R(0)> 10⁻² ⇒ Newton-Raphson iterations are necessary.

Calculate the correction to u1 = u1(0)

δu(1) = −[k(u1(0)]-1 R(0) = -(8.68)-1 (0.57) = - 0.0656

⇒ The updated value of u1(1) = u0(1) + δu(1) = 0.25 + (−0.0656) = 0.1844

Check the residual again, Fext = 1, fint = (u1(1))3+ 9(u1(1))2+ 4u1(1) = 1.04

R(1) = fint − Fext = 0.04

R(1) > 10−2 ⇒ another Newton-Raphson iteration is necessary.

Calculate the new total correction to u1 = u0(1)

δu(2) = δu(1) − [k(u1)1)]−1R(1) =

−0.0656 − (7.42)−1 x 0.04 = - 0.070

The updated value of u1(2) = u1(0) + δu(2) = 0.25 + (−0.070) = 0.18

Check the residual, Fext = 1, fint = = (u1(2))3+ 9(u1(2))2+ 4u1(2) = 1.017

R(2) = fint − Fext = 0.017

R(2)> 10−2 ⇒ another Newton-Raphson iteration is necessary.

Calculate the new total correction to u1 = u0(1)

δu(3) = δu(2) − [k(u1)2)]−1R(2) =

−0.070 − (7.33)−1 x 0.017 = - 0.0723

The updated value of u1(3) = u1(0) + δu(3) = 0.25 + (−0.0723) = 0.17768

Check the residual, Fext = 1, fint = = (u1(3))3+ 9(u1(3))2+ 4u1(3) = 1.0004

R(2) = fint − Fext = 0.0004

R(2)> 10−2 ⇒ No Further iterations are necessary.

Final Value of u1 = 0.17768

Step 2

k(u1) = 3u2+18u+4 = 7.29, ∆F = 1, ∆u2 = ∆F/k(u1) = 1/7.29 = 0.137

u2 = u1 + ∆u2 = 0.31468

Check the residual, Fext = ∆F = 1, fint = (u2)3+ 9(u2)2+ 4u2 = 2.1810

R(0) = fint − Fext = 0.1810

R(0)> 10−2 ⇒ Newton-Raphson iterations are necessary.

Calculate the correction to u2 = u2(0)

δu(1) = −[k(u2(0))]−1R(0) = −9.9613−1 x (0.1810) = - 0.0181

⇒ The updated value of u2(1) = u2(0) + δu(1) = 0.314 + (−0.0181) = 0.2965

Check the residual again, Fext = 1, fint =(u2(1))3+ 9(u2(1))2+ 4u2(1) =2.0032

R(1) = fint − Fext = 0.0032

R(1)< 10−2 ⇒ No iteration required

Final Vlaue of u2 = u2(1)= 0.2965

Step 3

k(u2) = 3u2+18u+4 = 9.6007 , ∆F = 1, ∆u3 = ∆F/k(u2) = 1/9.6007 = 0.1041

u3 = u2 + ∆u3 = 0.4006

Check the residual, Fext = ∆F = 1, fint = (u3)3+ 9(u3)2+ 4u3 = 3.1110

R(0) = fint − Fext = 0.1110

R(0)> 10−2 ⇒ Newton-Raphson iterations are necessary.

Calculate the correction to u3 = u3(0)

δu(1) = −[k(u3(0))]−1R(0) = −11.6922−1 x (0.1110) = - 0.00949

⇒ The updated value of u3(1) = u3(0) + δu(1) = 0.4006 + (−0.00949) = 0.3911

Check the residual again, Fext = 1, fint =(u3(1))3+ 9(u3(1))2+ 4u3(1) = 3.0008

R(1) = fint − Fext = 0.0008

R(1)> 10−2 ⇒ No iteration required

Final Value of u3 = u3(1)= 0.3911

Step i	∆Fi	ui	Fext	fint	fint-Fext =R
1	1	0.17768	1.0	1.0004	0.0004
2	1	0.2965	2.0	2.0032	0.0032
3	1	0.3911	3.0	3.008	0.0080