Week - 2 - Explicit and Implicit Analysis
AIM:
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).
SOLUTION:
f'(u)= k(u) = 3u2+18u+4
EXPLICIT METHOD
DELTA f= K DALTA u
STEP 1 u0= 0.0
k(u0)= 4
del f = 1
del u1= 1/4 = .25
u1 = u0+delu1= .25
fint =0.25^3+9.25(.25^2)+4*.25
=0.0156+.5625+1.00
=1.2781
step 2 k(u1)= (.25^2)*3+18*.25+4
= 0.1875+4.5+4
=8.6875
del f = 1
del u2= delf/ k(u1)= .115
u2= u1+ del u2= .365
fint= 0.0486+1.199+1.46
=2.7076
step 3 k(u2)= (0.365^2)*3+18*.365+0.365
=7.3346
del f= 1
del u3 1/7.3346=.126
u3= u2+del u3= .365+.126
=.491
fext= del f+ del f+ delf= 3.0
fint = .118+2.169+1.964=4.251
fext is not equal to fint
STEP i del fi del ui ui fexti finti fint-fext= R
1 1 0.25 .25 1.00 1.27 0.27
2 1 0.115 .365 2.00 2.70 0.70
3 1 0.126 .491 3.00 4.251 1.75
IMPLICIT ANALYSIS
STEP 1
u0=0.0 , k(u0)= 3u2+18u+4
=4
del F= 1
del u1= delF/ k(u0)= 1/4= .25
u1 = u0+ del u1 = 0.25
check residual fext= del f= 1.0
fint= u3+9u2+4u
= 1.278
R(0)= fint - f ext
= 0.278
r(0)> 10^-2
newton rapson iteration is neccsay
clculate the correction to u1= u1(0)
del u(1)= -[k(u1(0))]^-1 R(0)
k(u1(0)) = 8.6875
del u(1)= -0.032
u1(1)= u1+ del u(1) = 0.218
fint= 1.31
r(1)= fint -fext= 0.31
NR iteration is necassary
R(1)> 10^-2
calculate the correction to
u2=u1(1)
del u(1)= - [k(u1(1))]^-1* R(1)
k(u1(1)) = 8.066
del (u1(1))= -0.0384
u2= 0.179
fint = 1.009
R(2)= 0.009
R(2)< 10^-2 NR not needed final value
final value u1(2)= 0.179.
similarly For
step 2
del f= 1
u= .296
fext= 2
fint= 1.997
R= -.0021
similarly for step 3
del f= 1
u= .391
fext= 3
fint= 2.998
R= -.0.002
STEP DEL F u Fext f int R = fint - Fext
1 1 0.179 1 1.009 0.009
2 1 0.296 2 1.997 -0.0021
3 1 0.391 3 2.998 -0.002
CONCLUSION AND RESULT: