Week - 2 - Explicit and Implicit Analysis
AIM
To solve the given equation using explicit and impicit analysis and make a graph b/w F&D on each method.
Procedure
The given equation is F(u) = u^3+9u^2+4u
The force in the bar is a non linear function of displacement
F(u) = u^3+9u^2+4u (1)
stiffness of the bar is
k(u) =df/du=3u^2+18u+4 (2)
Explicit analysis calculation
u=dispalcement
f=internal force
F=external force
k=stiffness
∆u=incremental displacement
∆F=incremental external load
∆F=k.∆u (3)
step-1
u0=0.00 ⇒ k(u0)=4, ∆F=1, ∆u= ∆F/ k(u0), ∆u=1/4 ⇒ ∆u=0.25,u1=u0+∆u=0+0.25=0.25
Fint=(0.25)^3+9*(0.25)^2+4*0.25=1.5781
step-2
k(u1)=8.6875,∆F=1,∆u= ∆F/ k(u0),∆u=1/8.6875=0.1151,u2=u1+∆u2=0.25+0.1151=0.3651
Fint=(0.3651)^3+9*(0.3651)^2+4*0.3651=2.7086
step-3
k(u2)=10.9716,∆F=1,∆u= ∆F/ k(u0),∆u=1/10.9716=0.0911,u2=u1+∆u2=0.3651+0.0911=.0.4562
Fint=(0.4562)^3+9*(0.4562)^2+4*0.4562=3.7927
Fext = ∆F + ∆F + ∆F = 1+1+1=3.0
|
Step 1 |
∆Fi |
∆ui |
ui |
Fext |
fint |
fint − Fext = R |
|
1 |
1 |
0.25 |
0.25 |
1 |
1.5781 |
0.5781 |
|
2 |
1 |
0.1151 |
0.3651 |
2 |
2.7086 |
0.7086 |
|
3 |
1 |
0.0911 |
0.4562 |
3 |
3.7927 |
0.7927 |
IMPLICIT ANALYSIS
u0=0.00 ⇒ k(u0)=4, ∆F=1, ∆u= ∆F/ k(u0), ∆u=1/4 ⇒ ∆u=0.25,u1=u0+∆u=0+0.25=0.25
check the residual
Fext=∆F=1
Fint=(0.25)^3+9*(0.25)^2+4*0.25=1.5781
R(0)=1.5781-1=0.5781
Newton raphson iteration are necessary
put u^1=u1^(0)
δu ^(1) = −[k(u1^(0))]^ −1* R (0)=-8.6875^-1*0.5781=0.0665
u1^(1)=u1^(0)+δu^(1) = 0.25 -0.0665 = 0.1835
Check the residual again Fint=(0.1835)^3+9*(0.1835)^2+4*0.1835=1.0432
R^(1)=Fint-Fext=0.0432
R ^(1) > 10^(−2) ⇒ another Newton-Raphson iteration is necessary
put u^1=u1^(0)
δu ^(1) = −[k(u1^(1))]^ −1* R ^(1)=-0.0665-(7.404)^-1*0.0432=-0.0723
u1^(2) = u1^(0) + δu^(2)=0.25-0.0723=0.1777
Check the residual,Fext = 1,fint=(u1^(2))^3 + 9 (u1^(2))^2 + 4 (u1^(2)) = 1.0006
R (2) = fint − Fext = 1.0006-1= 0.0006
R (2) < 10−2 ⇒ no further iterations are necessary
u1 = u1^(2) = 0.1777
step-2
k(u1) = 3(0.1777)^2 + 18(0.1777) + 4=7.2933,∆F=1,∆u2=∆F/k(u^1)=1/7.2933=0.1371
u2 = u1 + ∆u2 =0.1777+0.1371=0.3148
Check the residual,Fext = ∆F = 1,fint=(u2)^3 + 9 (u2)^2 + 4 (u2)=2.1813
R^(0) = fint − Fext=2.1813-2=0.1813
R^(0) > 10−2 ⇒ Newton-Raphson iterations are necessary
u^2 = u2^(0)
δu^(1) = −[k(u2^(0))]^−1 *R^(0)=-9.9636^-1*0.1813=-0.01819
u2^(1) = u2^(0) + δu^(1) = 0.2966
Check the residual again,Fext =1,fint = (u2^(1))^3 + (9u2^(1))^2 + 4u2^(1)=2.0042
R^(1) = fint−Fext = 0.0042
R^(1) < 10^−2 ⇒ no further iterations are necessary
u2 = u2^(1) = 0.2966
step-3
k(u2) = 3( 0.2966)^2 + 18( 0.2966) + 4=9.6024,∆F=1,∆u3=∆F/k(u^1)=1/9.6024=0.1041
u3 = u2 + ∆u3 =0.2966+0.1041=0.4007
Check the residual,Fext = ∆F = 1,fint=(u3)^3 + 9 (u3)^2 + 4 (u3)=3.1121
R^(0) = fint−Fext = 0.1121
R^(0) > 10^−2 ⇒ Newton-Raphson iterations are necessary.
put u3 = u3^(0)
δu^(1) = −[k(u3^(0))]^−1 *R^(0)=-11.6942^-1*0.1121=-0.00958
u3^(1) = u3^(0) + δu^(1) =0.3911
Check the residual again,Fext =∆F =1,fint = (u3^(1))3 + (9u3^(1))2 + 4u3^(1)=3.00085
R^(1) = fint − Fext = 0.00085
R^(1) < 10^−2 ⇒ no further iterations are necessary
u3 = u3^(1) = 0.3911
Table-2 IMPLICIT ANALYSIS
|
Step |
∆Fi |
ui |
Fext |
Fint |
fint − Fext = R |
|
1 |
1 |
0.1777 |
1 |
1.0006 |
0.0006 |
|
2 |
1 |
0.2966 |
2 |
2.0042 |
0.0042 |
|
3 |
1 |
0.3911 |
3 |
3.00085 |
0.00085 |
EXPLICIT VS IMPLICIT
CONCLUSION
In week 2 challenge,after completed the week 2 challenge we can understand explicit and implicit method also their differentiation.Explicit is faster than implicit,which gives a smaller deformation, whereas implicit is slower and gives a large deformation.