Week - 2 - Explicit and Implicit Analysis

Aim:-

• `f(u)=u^3+9*u^2+4*u`, The aim is to analyse the given equation using the equal load increment method with both explicit and implicit load control schemes.

Procedure:-

Explicit Analysis

`f(u)=u^3+9*u^2+4*u`

`k(u)=(df)/(du)=3*u^2+18*u+4`

`DeltaF=k*Deltau`

`DeltaF =1` , Incremental step for external force

• f = Internal force
• k= Stiffness
• F = External force
• u = Displacement

Step-1:

Initially,

`u underset(0)()=0.0`

`k(u underset(0)())=4`

`DeltaF=1`

`Deltau underset(1)()=(DeltaF)/(k(u underset(0)()))=1/4=0.25`

`u underset(1)()=u underset(0)()+Deltau underset(0)()=0.25`

`(f underset(Internal)()) underset(1)()=u underset(1)()^3+9*u underset(1)()^2+4*u underset(1)()=1.5781`

Step-2:

`k(u underset(1)())=8.6875`

`DeltaF=1`

`Deltau underset(2)()=(DeltaF)/(k(u underset(1)()))=1/8.6875=0.1151`

`u underset(2)()=u underset(1)()+Deltau underset(2)()=0.3651`

`(f underset(Internal)()) underset(2)()=u underset(2)()^3+9*u underset(2)()^2+4*u underset(2)()=2.7087`

Step-3:

`k(u underset(2)())=10.971`

`DeltaF=1`

`Deltau underset(3)()=(DeltaF)/(k(u underset(2)()))=1/10.971=0.0911`

`u underset(3)()=u underset(2)()+Deltau underset(3)()=0.4562`

`(f underset(Internal)()) underset(3)()=u underset(3)()^3+9*u underset(3)()^2+4*u underset(3)()=3.7928`

• A Summary of the above analysis is given below

Step (i)	`DeltaF underset(i)()`	`Deltau underset(i)()`	`u underset(i)()`	`(F underset(ext)()) underset(i)()`	`(f underset(Internal)()) underset(i)()`	`(f underset(Internal)()) underset(i)()-(F underset(ext)()) underset(i)()=R`
1	1	0.25	0.25	1	1.5781	0.5781
2	1	0.1151	0.3651	2	2.7087	0.7087
3	1	0.0911	0.4562	3	3.7928	0.7928

Implicit Analysis:

• In the case of explicit analysis, we haven't considered the equilibrium of forces in order to get the value of displacement but for Implicit Analysis, we have to get as close to the equilibrium(i.e as low residual as possible) according to the tolerance value(for this problem which is `10^-3`).
• To get as close to the tolerance value we use the Newton-Raphson iterations to minimize the residual (R).

`R(u)=f underset(Internal)()-F underset(ext)()`

• For reducing the residual we follow the below-given correction method

`deltau^(j+1)=-[k(u^j)]^-1*R^j`

Step-1:

`u underset(0)()=0.0`

`k(u underset(0)())=4`

`DeltaF=1`

`Deltau underset(1)()=(DeltaF)/(k(u underset(0)()))=1/4=0.25`

`u underset(1)()=u underset(0)()+Deltau underset(0)()=0.25`

`(f underset(Internal)()) underset(1)()=u underset(1)()^3+9*u underset(1)()^2+4*u underset(1)()=1.5781`

`R^0=0.5781`

`R^0>10^-3`

• Iteration-1:

`u underset(1)()=u underset(1)()^0`

`deltau^(1)=-[k(u underset(1)()^0)]^-1*R^0=-8.6875^-1*0.5781=-0.6654`

`u underset(1)()^1=u underset(1)()^0+deltau^1 =0.25-0.06654=0.18345`

`f underset(Internal)()^1=1.04285`

`R^1=0.04285`

`R^1>10^-3`, So we will reiterate until we get the value within toleration.

• Iteration-2:

`deltau^(2)=deltau^(1)-[k(u underset(1)()^1)]^-1*R^1=-0.06654-7.403^-1*0.04285=-0.07232`

`u underset(1)()^2=u underset(1)()^0+deltau^2 =0.25-0.07232=0.177671`

`f underset(Internal)()^1=1.00039`

`R^2=0.00039`

`R^2<10^-3`, So we will consider the 2nd iteration as final value of step 1.

`u underset(1)()=u underset(1)()^2=0.177671`

Step-2:

`k(u underset(1)())=7.29277`

`DeltaF=1`

`Deltau underset(2)()=(DeltaF)/(k(u underset(1)()))=1/7.29277=0.137121`

`u underset(2)()=u underset(1)()+Deltau underset(2)()=0.177671+0.137121=0.314792`

`(f underset(Internal)()) underset(2)()=u underset(2)()^3+9*u underset(2)()^2+4*u underset(2)()=2.1822`

`R^0=0.1822`

`R^0>10^-3`

• Iteration-1:

`u underset(1)()=u underset(1)()^0`

`deltau^(1)=-[k(u underset(2)()^0)]^-1*R^0=-9.9635^-1*0.1822=-0.018286`

`u underset(2)()^1=u underset(2)()^0+deltau^1 =0.314792-0.018286=0.29650`

`f underset(Internal)()^1=2.00327`

`R^1=0.00327`

`R^1>10^-3`, So we will reiterate until we get the value within toleration.

• Iteration-2:

`deltau underset(2)()^(2)=deltau underset(2)()^(1)-[k(u underset(2)()^1)]^-1*R^1=-0.018286-9.6^-1*0.00327=-0.0186265`

`u underset(2)()^2=u underset(2)()^0+deltau underset(2)()^2 =0.314792-0.0186265=0.296165`

`f underset(Internal)()^1=2.00061`

`R^2=0.00061`

`R^2<10^-3`, So we will consider the 2nd iteration as final value of step 1.

`u underset(2)()=u underset(2)()^2=0.296165`

Step-3

`k(u underset(2)())=9.5941`

`DeltaF=1`

`Deltau underset(3)()=(DeltaF)/(k(u underset(2)()))=1/9.5941=0.10423`

`u underset(2)()=u underset(2)()+Deltau underset(3)()=0.400395`

`(f underset(Internal)()) underset(3)()=u underset(3)()^3+9*u underset(3)()^2+4*u underset(3)()=3.108615`

`R^0=0.108615`

`R^0>10^-3`, So we will reiterate until we get the value within toleration.

• Iteration-1:

`u underset(3)()=u underset(3)()^0`

`deltau underset(3)()^(1)=-[k(u underset(3)()^0)]^-1*R^0=-11.688058^-1*0.108615=-0.0092928`

`u underset(3)()^1=u underset(3)()^0+deltau underset(3)()^1 =0.400395-0.0092928=0.391102`

`f underset(Internal)()^1=3.000878`

`R^1=0.000878`

`R^1<10^-3`, So we will consider the 2nd iteration as final value of step 1.

`u underset(3)()=u underset(3)()^1=0.400395`

• A summary of the above analysis is given below

Step (i)	`DeltaF underset(i)()`	`Deltau underset(i)()`	`u underset(i)()`	`(F underset(ext)()) underset(i)()`	`(f underset(Internal)()) underset(i)()`	`(f underset(Internal)()) underset(i)()-(F underset(ext)()) underset(i)()=R`
1	1	-0.07232	0.177671	1	1.00039	0.00039
2	1	-0.0186265	0.296165	2	2.000061	0.000061
3	1	-0.0092928	0.391102	3	3.000878	0.000878