Week - 2 - Explicit and Implicit Analysis

IMPLICIT METHOD :

In the Implicit method, the slope value at the unknown Time step is used. i.e, In the equation

`y(t_(n+1)) = y(t_(n)) + Deltat xx y^'` __________(1)

for the given curve as shown in the Figure below

The slope value at the Unknown time step is used for finding the value of 'y' at the time step `t_(n+1)` .i.e, `y^'` at `t_n+1` in equation (1)

thus, the equation (1) becomes

`y(t_((n+1))) = y(t_((n))) + Deltat xx y^'(t_((n+1)))`___________(2)

This equation is Non -Linear and it requires numerical methods such as Newton raphson's Iteration to solve it.

EXPLICIT METHOD :

The Explcit Method use the slope at the know time step `t_(n)`in equation (1) i.e `y^'` at `t_(n)` in equation(1)

i.e, the equation(1) becomes

`y(t_((n+1))) = y(t_((n))) +Deltat xx y^'(t_((n)))` _______(2)

No further effort is required unlike the Implicit method to determine the the value of 'y' at `t_(n+1)`

AIM : To use the equation `u^3 + 9u^2 + 4u` ` and solve using both explicit and Implicit methods . The Implicit Method should have a tolerance of 0.001.

ANSWER :

The Force is dependent on the non linear function of the displacement given as

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

The Load factor can be incremented in steps of 1.i.e, `Delta F=1`. Also, Stiffness `K(u)=(df//du)= 3u^2+18u^2+4

Explict method :

Step 1 : `u_0 =0 ; k(u_0)=0+0+4=4 ;DeltaF_1 =1 ; Delta u_1 = (DeltaF_1//k(u_o))=1/4_`

`u_1= Deltau_1 + u_0 = (1//4) + 0= 1/4=0.25`

Step 2 : `u_1 = 0.25 ; k(u_1) = 3(0.25)^2 + 18(0.25)+4= 8.6875 ; DeltaF_2=1 ; Deltau_2 = (DeltaF_2//k(u_1))=1/8.6875=0.1151`

`u_2=Deltau_2 + u_1 = 0.1151 + 0.25 = 0.3651`

Step 3 : `u_2= 0.3651 ; k(u_2)=3(0.3651)^2 + 18(0.3651) +4 = 10.971 ; DeltaF_3=1 ; (DeltaF_3//k(u_2))=1/10.971=0.09114`

`u_3=Deltau_3 + u_2 = 0.3651 + 0.09114=0.45921`

Explicit Method Chart

Step i	`DeltaF_i`	`Deltau_i`	ui	(Fext)i	(Fint)i	R= Fint- Fext
1	1.00	0.25	0.2500	1.00	1.5781	0.5781
2	1.00	0.1151	0.3651	2.00	2.7087	0.7087
3	1.00	0.0912	0.4593	3.00	3.8327	0.8327

Implcit Method :

Step 1 : `u_0 =0 ; k(u_0)=0+0+4=4 ;DeltaF_1 =1 ; Delta u_1 = (DeltaF_1//k(u_o))=1/4_`

`u_1= Deltau_1 + u_0 = (1//4) + 0= 1/4=0.25`

(f)int = `(0.25)^3 +9(0.25)^2+4(0.25) = 1.5781`

`R_0` =`F_(Internal)-F_(external) =`1.5781 - 1 =0.5781 > 0.001,

Hence using the Newton Raphson Method .

`del(u^1) = -[k(u_1)^0]^-1 xx R^0` = `-[4(8.675)]^-1xx0.5781` = -0.066543

Updated value of `(u_1)^1` = `(u_1)^0 + del(u^1)` = 0.25 + (-0.066543) = 0.183457

`F_(Internal)` =``(0.184)^3 + 9(0.184)^2 + 4(0.184)`

= 1.0469

`F_(Internal) - F_(external) = 1.0469 - 1`

= 0.0469 = `R^1 > 0.001`

Hence , Using the Newton Raphson Iteration Method again,

`k(u_1)^((1)) =` `3(u_1^1)^2 + 18(u_1^1) + 4`

= `3(0.1834)^2 + 18(0.1834) + 4`

= 7.414

`delu_2 = -k[(u_1)^((1))]^-1 xx R^1` = `-0.0469 / 7.414` = -0.06325

Update value of `u_1` = `(u_1)^((2)) = del u_2 + u_1^((1))` = 0.184 -0.06325 =0.1777

`F(Internal) = (u_1^((2)))^3 + 18 (u_1^((2)))^2 + 4(u_1^((2)))`

=`(0.1777)^3 + 9(0.1777)^2 + 4(0.1777)`

= 1.000607

`F_(Internal) - F_(external) =` 1.000607 - 1 = 0.000607 = `R_2 < 0.001`

Hence, the iterations are no longer required and the Final value of `u_1 = 0.1777`

Step 2 : `u_1 = 0.177` ; `k(u_1) = 3(0.1777)^2 + 18(0.1777) +4` = 7.2933

`DeltaF = 1` ; `Delta u_2 = DeltaF// k(u_1) =1/7.2933 = 0.1371`

`u_2 = u_1 + Delta u_2` = 0.1777 + 0.1371 =`= 0.2966 + (-4.41125xx10^-4)` 0.3149

`F_(Internal) = (0.3148)^3 + 9(0.3148)^2 + 4(0.3148)` = 2.1824

`-k[(u_2)^0] = 3(0.3149)^2 + 18(0.3149) +4` = 9.965

`u_2^((1)) = u_2^((0)) + delta u^((1)) =0.3149 + (-0.0183) = 0.2966`

`F_(Internal) = (0.2966)^3 + 9 (0.2966)^2 +4(0.2966)`= 2.004236

`F_((Internal)) - F_((external) = 2.004236 - 2 = 0.004236 = R^((1)) > 0.001`

Hence, using Newton Raphson Iterations again

`-k[u_2^((1))] = 3xx(0.2966)^2 + 18(0.2966) + 4 = 9.6027`

`deltau^((2)) = K[u_2^((1))]^-1R^((1)) = -0.004236/ 9.6027 = 4.41125xx10^-4`

`u_2^((2)) = u_2^((1)) + deltau^((2))`

`= 0.2966 + (-4.41125xx10^-4)`

= 0.29615

`F_(Internal) = (0.29615)^3 + 9xx(0.29615)^2 +4xx(0.29615) = 2.000002`

`F_((Intrnal)) - F_((external)) = 2.000002 - 2.000 = 0.000002 = R^(2)<0.001`

Hence, the iterations are no longer required and `u_2 = u_2^(2) = 0.29615`

Step 3 : `u_2 = 0.29615 ; k(u_2) = 3(0.29615)^2 + 18(0.29615) + 4 = 9.5938`

`DeltaF = 1 ; Delta u_3 = Delta F // k(u_2) = 1/9.5938 = 0.1042`

`u_3 = u_2 + Delta u_3 =0.29615 + 0.1042 = 0.40038`

`F_(Internal) = (0.40038)^3 + 9(0.40038)^2 + 4(0.40038) = 3.10848`

`F_(Internal) -F_(external) = 3.1084 - 3.00 = 0.10848 = R^0 > 0.001`

Hence, Using the Newton Raphson Iteration Method

`k[u_3^((0))] = 3(0.40038)^2 + 18(0.40038) + 4 = 11.687`

`deltau^((1)) = -k[u_3^((0))]^-1 R^((0)) = -0.10848/11.687 = 9.282xx10^-3`

`u_3^((1)) = u_3^((0)) +deltau^((1)) = 0.40038 + (-9.282 xx 10^-3) = 0.3910`

`F_(Internal) = (0.3910)^3 + 9 (0.3910)^2 + 4(0.3910) = 3.000832`

`F_(Internal) - F_(external) = 3.000832 - 3.000 = 0.00083224 =R_((1)) < 0.001`

Hence, The Newton Raphson Iterations are no longer required and `u_3 = u_3^((1)) = 0.3910`

Implicit Method Chart

Step i	`DeltaF_i`	`Deltau_i`	ui	(Fext)i	(fint)i	R= fint- Fext
1	1.00	0.25	0.1770	1.00	1.000607000	0.000607000
2	1.00	0.1371	0.2962	2.00	2.000002000	0.000002000
3	1.00	0.1042	0.3910	3.00	3.000832000	0.000832000

Force vs Displacement Comparison between Explicit and Implicit Methods

CONCLUSION :

It can be seen from the plot that the Value for the forces found are lot more closer to the Exact value of the forces as compared to the Force values determined using the Explicit methods. Thus the Implicit method is morte accurate as compared to the Explicit method but the Imlicit method more time consuming & cumbersome as compared to the Explicit method.