implicit and explicit
Force in bar f(u)=u3+9u2+4u --------(1)
Stiffness in bar from (1)
K(u)=dfdu=3u2+18u+4 -------(2)
Explicit Analysis
Cleary It is possible to plot a graph of force f, versus displacement by using equation (1)(2)
For this three equal load increments example ∆F=1
Step 1
u0=0, k(u)=1, ∆F=1, ∆u1=∆F/k(u0)=0.25 , u1=u0+∆u1=0+1/4=0.25
Step 2
k(u1)= 8.6875, ∆F=1, ∆u2=∆F/k(u1)=0.1151, u2=u1+∆u2=0.25+0.1151=0.3651
Step 3
k(u2)= 10.9716, ∆F=1, ∆u3=∆F/k(u1)=0.09115, u3=u2+∆u3=0.3651+0.09115=0.4562
Finally, from the here steps he following results are obtained
Fext=∆F+∆F+∆F=3.0
fint=u3+9u2+4u =3.7928
This evident from the results that External and internal forces are NOT In equilibrium because Fext≠fint. Table 1 summarizes the main results for the explicit analysis.
Table 1: Summary of explicit analysis results
|
Step i |
∆Fi |
ui |
(Fext)i |
(fint)i |
R= fint-Fext |
|
1 |
1 |
0.25 |
1 |
1.5781 |
0.5781 |
|
2 |
1 |
0.3651 |
2 |
2.7087 |
0.7087 |
|
3 |
1 |
0.4562 |
3 |
3.79280 |
0.7928 |
Implicit Analysis
Correction Factor formula From Tailor Seriesδj+1=-[k(uj)]-1Rj
Consider Residual∥R(0)∥>10-2
Step 1
u0=0, k(u)=1, ∆F=1, ∆u1=∆F/k(u0)=0.25 , u1=u0+∆u1=0+1/4=0.25
Check for Residual, Fext=∆F=1, fint=u13+9u12+4u1=1.578, R(0)= Fext - fint=0.578
∥R(0)∥>10-2 →Newton-Raphson iterations are necessary.
Calculate Correction to u1=u1(0)
u(1) = -[k(u1(0))]-1R(0)=-[8.6875]-1(0.578)= -0.0666
The updaed Value of u1(1)=u1(0) +u(1)=0.25+(-0.0666)=-0.1834
Check for Residual again Fext=1, fint=(u1(1))3+9(u1(1))2+4u1(1)=1.0424
R1)=fint-Fext=1.0424-1=0.424
∥R(1)∥>10-2 →Newton-Raphson iterations are necessary
Calculate the new total correction to u1=u1(0)
u(2) = u(1)-[k(u1(1))]-1R(1)= -0.0666-[7.4021-1]*0.424=-0.072338
The updaed Value of u1(2)=u1(0) +u(2)=0.25+(-0.072338)=0.1776616
Check for Residual again Fext=1, fint=(u1(2))3+9(u1(2))2+4u1(2)=1.000326
R(2)= Fext - fint=0.000326
∥R(1)∥>10-2 →no further iterations are necessary
Therefore, the final value is u1=u1(2)=0.1776616
Step 2
(k)u1=7.2925, ∆F=1, ∆u2=∆F/k(u1)=0.13712 ,u2= u1+∆u2=0.177661+0.13712 =0.3188
Check for Residual, Fext=∆F=1, fint=u23+9u22+4u2=2.18314, R(0)= Fext - fint=0.18314
∥R(0)∥>10-2 →Newton-Raphson iterations are necessary.
Calculate Correction to u2=u2(0)
u(1) = -[k(u2(0))]-1R(0)=-[9.95808]-1(0.18314)= -0.01839
The updated Value of u1(1)=u1(0) +u(1)=0.31488+(-0.0666)=0.2964
Check for Residual again Fext=1, fint=(u2(1))3+9(u2(1))2+4u2(1)=2.0026917
R(1)=fint-Fext=2.0026917-2=0.0026917
∥R(1)∥>10-2 →Newton-Raphson iterations are necessary
Calculate the new total correction to u2=u2(0)
u(2) = u(1)-[k(u1(1))]-1R(1)= -0.01839-[9.0065-1]* 0.0026917=-0.01867
The updated Value of u2(2)=u1(0) +u(2)=0.2964+(-0.01867)=0.296209
Check for Residual again Fext=1, fint=(u2(2))3+9(u2(2))2+4u2(2)=2.00039
R(2)= Fext - fint=0.00039
∥R(1)∥>10-2 →no further iterations are necessary
Therefore, the final value is u2=u2(2)=0.296209
Step 3
(k)u2=5.8424, ∆F=1, ∆u3=∆F/k(u1)=0.1711 ,u3= u2+∆u3=0.296204+0.1711=0.4673
Check for Residual, Fext=∆F=1, fint=u33+9u32+4u3=3.9365, R(1)= Fext - fint=0.9365
∥R(0)∥>10-2 →Newton-Raphson iterations are necessary.
Calculate Correction to u3=u2(0)
u(1) = -[k(u2(0))]-1R(1)=-[13.0665]-1(0.9365)= -0.07167
The updated Value of u3(1)=u3(0) +u(1)=0.4673+(-0.0666)=0.39562
Check for Residual again Fext=1, fint=(u2(1))3+9(u2(1))2+4u2(1)=3.05313
R(1=fint-Fext=3.05313-3=0.05313
∥R(1)∥>10-2 →Newton-Raphson iterations are necessary
Calculate the new total correction to u2=u2(0)
u(2) = u(1)-[k(u1(1))]-1R(1)= =--0.07167-[11.5907-1=0.05313=-0.07625
The updated Value of u3(2)=u3(0) +u(2)=0.4673+(-0.07625)=0.39105
Check for Residual again Fext=1, fint=(u3(2))3+9(u3(2))2+4u3(2)=3.002803
R(2)= Fext - fint=0.002803
∥R(1)∥>10-2 →no further iterations are necessary
Therefore, the final value is u3=u3(2)=0.39105
|
Step i |
∆Fi |
ui |
(Fext)i |
(fint)i |
R= fint-Fext |
|
1 |
1 |
0.1776616 |
1 |
1.000326 |
0.000326 |
|
2 |
1 |
0.296209 |
2 |
2.00039 |
0.00039 |
|
3 |
1 |
0.39105 |
3 |
3.002803 |
0.002803 |
graph