Week - 2 - Explicit and Implicit Analysis
Explicit method
F(u) = u3+9u2+4u (1)
K(u) = dF/du = 3u2 +18u + 4 (2)
Step1:
Take u = 0 in (2)
(2) => k(u0) = 3 x 02 + 18 x 0 + 4 = 4
= /k(u0) = ¼ = 0.25
u1= u0+ = 0.25
Fint = (0.25)3 + 9 x (0.25)2 +4 x 0.25 = 1.578
Step2:
Take u1 = 0.25 in (2)
(2) => k(u1) = 3 x 0.252 + 18 x 0.25 + 4 = 8.6875
= /k(u1) = 1/8.6875 = 0.1151
U2= u1+ = 0.3651
Fint = (0.3651)3 + 9 x (0.3651)2 +4 x 0.3651 = 2.708
Step3:
Take u2 = 0.3651 in (2)
(2) => k(u2) = 3 x 0.36512 + 18 x 0.3651 + 4 = 10.9716
= /k(u2) = 1/10.9716 = 0.09114
U3= u2+ = 0.46
To determine whether the analysis is equilibrium
Fext = = 3
Fint = (0.46)3 + 9 x (0.46)2 +4 x 0.46 = 3.841
Hence Fext fint the analysis is not equilibrium
Step i | ∆Fi | ∆ui | ui | (Fext)i | (fint)i | fint − Fext = R |
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 |
Implicit method
F(u) = u3+9u2+4u (1)
K(u) = dF/du = 3u2 +18u + 4 (2)
Step1:
Take u = 0 in (2)
(2) => k(u0) = 3 x 02 + 18 x 0 + 4 = 4
= /k(u0) = ¼ = 0.25
u1= u0+ = 0.25
Residual R
Fint = (0.25)3 + 9 x (0.25)2 +4 x 0.25 = 1.578
R0 = 1.578 – 1 = 0.578 > 102
So, Newton-Raphson iterations is necessary
The correction u1 = u1(0)
u(1) = -[k(u1(0)-1 x R0
= -[(3 x 0.252 ) + (18 x 0.250) +4]-1 x 0.578
=-0.0665
U1(1) = u1(0) + u(1)
= 0.25 – 0.0665 = 0.1835
Residual R1
Fint = (0.1835)3 + 9 x (0.1835)2 +4 x 0.1835 = 1.0432
R0 = 1.0432 – 1 = 0.432 > 102
So, Newton-Raphson iterations is necessary
The correction u1 = u1(1)
u(2) = u(2) -[k(u1(1)-1 x R1
=(-0.0665) -(8.6875 -1 )x 0.0432
=-0.0714
U1(2) = u1(1) + u(2)
= 0.25 – 0.0714 = 0.1786
Residual R1
Fint = (0.1786)3 + 9 x (0.1786)2 +4 x 0.1786 = 1.0071
R0 = 1.0071 – 1 = 0.0071 < 102
No residual needed
U1 = 0.1786
Similarly we drived u2 = 2966, u3 = 0.3911
Step i | ∆Fi | ∆ui | ui | (Fext)i | (fint)i | fint − Fext = R |
1 | 1 | 0.25 | 0.1786 | 1 | 1.0071 | 0.0071 |
2 | 1 | 0.1374 | 0.2966 | 2 | 2.0042 | 0.0042 |
3 | 1 | 0.1041 | 0.3911 | 3 | 3.008 | 0.008 |