1

FRONTAL SOLUTION TECHNIQUE

Introduction:

We have noted that, irrespective of the type of problem, the solution

involves formulation and solution of a set of simultaneous equations governing

the phenomenon being analyzed. For example, in case of a single stage linear

deformation analysis, the ultimate need was to formulate & solve the equations

governing the equilibrium of the idealized system, as defined in equation 1.

[[ ]] [[ ]] [[ ]] )1(Fk KK==δδ

Wherein, [δδ] is the nodal displacement vector of the idealized system,

[F] is the nodal load vector of the idealized system,

[k] is the structural stiffness matrix of the idealized system.

With ‘N’ as total nodal degree of freedom, eq.1 represents ‘N’

simultaneous equations, with ‘N’ unknowns. It thus has unique solution.

Computational exercise with respect to eq.1 involves following features:

a) For each element of the idealized system the element stiffness

matrix [ke] and the element nodal vector [Fe] are derived.

b) [k] is developed by superpositioning [ke] of the elements meeting at

the nodes of the idealized system.

c) [F] = [F]1 + [F]2 ; wherein, [F]1 is derived by superpositioning [Fe] of

the element meeting at the nodes of the idealized system; whereas,

[F]2 represents the load components directly applied at the nodes of

the idealized system.

d) The set of simultaneous equations are solved, subject to the

prescribed nodal restraints on solution, the displacements at the free

nodes & reactive components at the restrained nodes are

established.

Most straight forward approach to the problem would be to implement

the algorithm. It solves the problem in two sequential steps. In the first step the

complete set of equations is formulated; & in the second step the step is

solved. In the early stage of Finite Element development, this two step

approach was invariably adopted, in the formulation of the Finite Element

analysis softwares. As the computational technology evolved, however, more

efficient solution strategies emerged. Of these the most widely accepted

2

algorithm was the Frontal Solution technique, promoted by Bruce Trons in

1970. As of today the Frontal algorithm is an integral component of majority of

the finite element softwares, marketed in the world. In Frontal algorithm, the

formulation and solution of the set of equations progress simultaneously;

consequently it provides a most natural solution strategy.

In this chapter the Frontal algorithm & its computer implementation is

thoroughly discussed. By the way of illustration details pertaining to the single

stage embankment analysis presented in chapter 3 of the text are utilized.

Principles of Elimination:

In the frontal solution technique, equations are solved by the Gauss’s

method, which involves two stages of operations. In the first stage the

equations are reduced through forward elimination of variables; & in the

second stage the values of the variables are established through the process

of back substitution. In this section we shall discuss the principles of forward

elimination. For the sake of illustration we consider eq.1, with N=5; such that,

details are as shown in eq.2

)2(

F

F

F

F

F

kkkkk

kkkkk

kkkkk

kkkkk

kkkkk

5

4

3

2

1

5

4

3

2

1

5554535251

4544434241

3534333231

2524232221

1514131211

KK

==

δδδδδδδδδδ

In a conventional Gauss’s solution scheme, the solution of eq.2 will be

undertaken by eliminating sequentially the variables ),,,,( 54321 δδδδδδδδδδ & by

deriving the values ),,,,( 12345 δδδδδδδδδδ through back substitution. We shall

recognize at a later stage that, in the frontal solution technique, elimination of

the variables would follow a random order. For example, the variables may be

required to be eliminated in the order ),,,,( 34152 δδδδδδδδδδ . Consequently,

through the back substitution the values of ),,,,( 25143 δδδδδδδδδδ would get

established, i.e. in the reverse order.

Let us assume that we need to eliminate variable δδ3 from the set of

equations in eq.2. In theory, we could achieve this by considering any of the

3

equation from the set from the view point of requirements in the frontal solution

technique. We shall consider the third equation, wherein, the diagonal

coefficient k33 happens to be the multiplier to the variable δδ3 being eliminated.

It may be noted that the sequential elimination process in the conventional

application of the Gauss’s method, employs invariable such diagonal

coefficients referred to as pivot, while undertaking the process of elimination.

We shall thus begin by extracting the third equation from the set of

equation. That is:

)3(Fkkkkk 3535434333232131 KK==δδ++δδ++δδ++δδ++δδ

The same is preserved to facilitate the operations associated with the

back substitution process.

On elimination of the variable δδ3 the original set of five equations gets

reduced to a set of four equations, with the coefficients of the remaining four

equations in the set, i.e. the equations other than the one shown in eq.3 getting

appropriately modified. While undertaking the modifications, it is however

necessary to identify the nature of variable being eliminated. In context with the

finite element solution, a variable which happens to be a nodal degree of

freedom having freedom to develop is a free variable. On the other hand if it is

subjected to boundary conditions, it will have a prescribed value. The

modifications referred to above depend upon, whether the variable being

eliminated is a free variable or a variable with the prescribed value.

a) Free Variable:

From eq.3 it follows that;

)4(k

k

k

k

k

k

k

k

k

F5

33

354

33

342

33

321

33

31

33

33 KKδδ−−δδ−−δδ−−δδ−−==δδ

On substituting eq.4 into remaining four equations of eq.2, the set of

equations gets modified to the on shown in eq.5

[[ ]][[ ]] [[ ]] )5(Fk *** KK==δδ

Wherein, * denotes the modifications due to elimination of δδ3. The

details are thus as shown in eq.6

4

)6(

F

F

0

F

F

0

kk0kk

kk0kk

00000

kk0kk

kk0kk

*5

*4

*2

*1

5

4

2

1

*55

*54

*52

*51

*45

*44

*42

*41

*25

*24

*22

*21

*15

*14

*12

*11

KK

==

δδδδ

δδδδ

NOTE: ‘0’ denotes absence.

The coefficients *ijk & *

iF are as defined in eq.7

)7(Fk

kFF

k

kkkk

333

3ii

*i

33

j33iij

*ij

KK−−==

−−==

We could generalize the above process in the following manner:

In a set containing ‘n’ equations defined by eq.1, δδm the (m+n) variable

could be eliminated through the (m+n) equation of the set. This involves

extraction of the (m + n) equation & its preservation. It also leads to the

absence of this equation from the set. The remaining coefficients are modified

to *ijk & *

iF as shown in eq.8.

)8(Fk

kFF

k

kkkk

mmm

imi

*i

mm

mjimij

*ij

KK−−==

−−==

Having undertaken these modifications, kim is set to zero. The set thus

gets transformed to (n-1) no. of equations.

b) Prescribed Variable:

In case of a prescribed variable, the modifications get considerably

simplified, because the process of elimination involves direct substitution of the

prescribed value of the variable. We should however recollect that a prescribed

nodal parameter does not reduce the no. of unknowns, because a prescribed

nodal displacement introduces unknown nodal reactive component.

Let δδ3 = ∆∆ . Now, leaving the third expression that has been extracted &

preserved (for calculating reactive component at the back substitution stage)

the condition δδ3 = ∆∆ is substituted in the remaining four equations. Assuming

eq.6 to be symbolic representation of the modified set of equations, we should

5

note that in the modified set ij*ij kk == , i.e. the original coefficients kij remain

unaffected. On the other hand the coefficients Fi get modified to Fi* as defined

in eq.9

)9(k.FF 3ii*i KK∆∆−−==

We could generalize the process in the following manner:

In a set containing ‘n’ equations defined by eq.1, δδm the (m+n) variable

with a prescribed value of ∆∆ is eliminated by extracting & preserving

expressions unaffected.

But, Fi should be modified to Fi* as defined in eq.10

)10(k.FF imi*i KK∆∆−−==

Having undertaken these modifications, kim is set to zero. The set thus

transformed to (n - 1) no. of equations. In the subsequent developments, we

shall be continuously referring to eq.8 & eq.10.

Essence of Frontal Solution Technique:

We shall now present the basic concepts of Frontal Solution Technique.

Let us consider a finite element idealization comprising of three triangular

elements & five nodes as shown in Fig.1.

Assuming a single nodal degree of freedom denoted through ‘φ’; the

governing set of equations are as presented in eq.11.

)11(]B[][.]A[ KK==φφ

The details of the equations are given below:

1

2

3

1 2 3

4 5

Fig. No. 1 Illustrative Idealization for Essence of Frontal Solution Technique

6

)12(

B

B

B

B

B

AAAA0

AA0AA

A0AA0

AAAAA

0A0AA

5

4

3

2

1

5

4

3

2

1

55545352

45444241

353332

2524232221

141211

KK

==

φφφφφφφφφφ

Wherein, the coefficients of [A] & [B] are generated by the

superpositioning technique. We should note that being a normal finite element

formulation [A] is symmetrical, i.e. Aij = Aji.

Before the advent of the Frontal Solution Technique the entire set of

simultaneous equations were formulated & the same was then solved through

Gauss’s method. In this the variables (φφ1, φφ2, φφ3, φφ4, φφ5) were eliminated

sequentially through the forward elimination process, & the values (φφ5, φφ4, φφ3,

φφ2, φφ1), were established through the back substitution process.

Frontal Solution Technique also employs Gauss’s method, but it differs

from the above mentioned sequential technique in an important manner. In the

Frontal Solution Technique the entire set of equations is not required to be

formulated before hand. A closer examination of eq.9 & eq.10 would reveal

that once the equation employed in carrying out the elimination is completely

formed, the modifications to the coefficients of the remaining equations depend

only on a summation process. And the order in which such summations are

carried out is immaterial to the ultimate solution. The Frontal Solution

Technique takes advantages of this fact, by eliminating a variable as soon as

its governing equation is completely formed.

In case of the problem of Fig. No. 1, the solution by the Frontal Solution

Technique progresses by calling the elements , & respectively.

Element :

The analysis begins by activating element 1. Due to this a set of three

equations as shown below gets generated:

Wherein, the superfix 1 attached to various coefficients denotes element

1. From Fig. No. 1 it is clear that only one element i.e. the element 1

1 2 3

1

)13(

B

B

B

AAA

AAA

AAA

14

12

11

3

2

1

144

142

131

124

122

121

114

112

111

KK

==

φφφφφφ

7

meets at the node -1. This means the equation with respect to φφ, is completely

formed on activating the element 1. Consequently, φφ1 be eliminated at this

stage of analysis. Therefore, equation corresponding to φφ1 is extracted &

preserved as shown in eq.14 (a), whereas the other two equations get

modified due to the elimination of φφ1 as shown in eq.14 (b):

)14(

B

B

xx

AAx

AAx

xxx

)b(

BAAA)a(

*4

*2

4

2*44

*42

*24

*22

'14

'142

'121

'11

KKK

==

φφφφ

==φφ++φφ++φφ

Note:

1) ‘X’ denotes the vacancy created by the extraction of equation related to

φφ1. In this process the row & column of [A] get vacated.

2) ‘∗’ denotes the modifications due to elimination of ‘φφ1’. Due to

symmetrical character of [A], *24

*42 AA == .

Element :

We should note that the next step of the analysis begins with residual

characteristics, as per eq.14 (b), & the residual portion of the idealization as

shown in Fig. No. 2(a). This is because in the previous step element 1 was

considered, therefore the same goes out of the analysis.

2

Fig. No. 2 Illustration for the Frontal Solution Process

2 3

4 5

2

3

( a )

3

2 3

5

( b )

8

Now element 2 is activated, wherein its characteristics are

superpositioned over the residual characteristics of the previous step. The

resulting set of equations is as shown in eq.15.

)15(

BB

BB

B

AAAAA

AAAAA

AAA

24

*4

22

*2

25

4

2

5

244

*44

242

*42

245

224

*24

222

*22

225

254

252

255

KK

++++==

φφφφφφ

++++++++

Note:

1) Superfix ‘2’ denotes element 2.

2) Vacant space of the residual block is utilized to accommodate the

characteristics of node 5. Economy of required core space for managing

the equation is thus obvious.

It follows from Fig. No. 2 (a), that at this stage of the analysis only one

element meets at the node 4. This means on activating the element 2, the

equation with respect to ‘φφ4’ is completely formed, hence ‘φφ4’ can be eliminated.

Towards this the equation corresponding to ‘φφ4’ is extracted & preserved as

shown in eq.16 (a), whereas the remaining equations get modified due to

elimination of ‘φφ4’ as shown in eq.16 (b).

)16(

x

B

B

xxxx

xAA

xAA

)b(

)BB()AA()AA(A)a(

**2

**5

2

5

22**

25

**52

**55

24

*44

244

*442

242

*425

245

KKK

==

φφφφ

++==φφ++++φφ++++φφ

Note:

1) ‘X’ denotes the vacancy created by extracted of equation related to ‘φφ4’.

In this both 3rd row and 3rd column are vacated.

2) ‘∗∗’ denotes the modifications due to elimination of ‘φφ4’. As [A] is

symmetrical **25

**52 AA == .

Element :

In the final step, we are left with the residual characteristics as defined

in eq.16 (b) & the residual portion of the idealization as shown in Fig. No. 2(b).

Now the element 3 is activated, wherein, its characteristics are superpositioned

3

9

over the residual characteristics of the previous step. The resulting set of

equations is as shown in eq.17.

)17(

B

BB

BB

AAA

AAAAA

AAAAA

33

32

**2

35

**5

3

2

5

333

332

335

323

322

**22

325

**25

353

352

**52

355

**55

KK

++++

==

φφφφφφ

++++++++

Wherein, superfix 3 denotes element 3.

We should note that at this stage of analysis at each node of the

element only one element, viz. element 3 meets. Therefore, equations

governing each of them are completely formed. The remaining variable could

therefore be eliminated one after the other. With this the elimination stage gets

completed & values of all the variables could be established through back

substitution process.

Illustrative Computations

With a view to provide in depth prescription of the Frontal Solution

Technique, we shall present complete set of computational details related to

the frontal solution for the single stage embankment analysis considered. The

idealization details are as shown in Fig. No. 3.

The details regarding [ke] & [Fe] of the four element are reproduced

here below for the sake of convenience as shown in Table 1.

Fig. No. 3 Illustrative Problem

2 3

4 5

2

3 1

4

1

6

10

Table No. 1(a) : [ke] of elements 1, 3 and 4

448 448 -88 268 -356 -716

448 1118 -268 1026 -180 -2144

-88 -268 448 -448 -356 716

268 1026 -448 1118 180 -2144

-356 -180 -356 180 716 0

-716 -2144 716 -2144 0 4288

Table No. 1(b) : [ke] of elements 2

716 0 -356 -180 -356 180

0 4288 -716 -2144 716 -2144

-356 -716 448 448 -88 268

-180 -2144 448 1118 -268 1026

-356 716 -88 -268 448 -448

180 -2144 268 1026 -448 1118

Table No. 1(c) : [Fe] of elements 1, 2, 3 & 4

0

3400−−

0

3400−−

0

3400−−

The base nodes were assumed to be completely restrained in both x &

y direction. For better appreciation of the Frontal Solution Technique we shall

however assume that the base nodes (1, 2, 3) rather than completely

11

restrained, suffer vertical displacement of -0.1m. Thus the boundary conditions

are:

u1 = 0.0; u2 = 0.0; u3 = 0.0;

v1 = - 0.1m; v2 = - 0.1m; v3 = - 0.1m

These nodal displacements represent uniform settlement of the

embankment; hence in the final result all the six nodes would display additional

vertical displacement of -0.1m. The strains and stresses induced in the dam

would be same as the ones estimated. Thus, emphasizing a basic structural

concept that rigid body motion of a structure does not induce additional

stresses in the structure.

Elimination Phase: In the elimination phase of the frontal solution technique,

the elements are activated sequentially to formulate the equations; & to

eliminate the variables in the appropriate manner, with a view to present the

computational details in comprehensive manner. We shall adopt the notations

having following character.

a) Algorithm

Let us assume that the analysis is completed up to ith element, & we are

considering the operations in respect of activating of (i+1)st element. In this

connection, we should note the following points:

1. Let [kR]I & [FR]I represent respectively the residual stiffness matrix &

residual nodal load vector, at the completion of the analysis up to ith element.

2. Let [kei+1] & [Fe

I+1] be the element stiffness matrix and element load

vector respectively, of the active element, i.e. (i+1)st element.

3. We superposition [kei+1] & [Fe

I+1] respectively over [kR]I & [FR]I, and

define the condition of equilibrium at that stage through eq.18.

)18(]F[][]k[ 1iA1iA1iA KK++++++ ==δδ

Wherein,

[δδA]i+1 represents vector of active nodal variables,

[FA]i+1 represents active load vector,

[kA]i+1 represents active stiffness matrix.

4. We eliminate from eq.18 all the variables for which the governing

equations are completely formulated. This involves extraction & preservation

of such completely formulated equations for making them available at the

12

back substitution stage. And modifications of the coefficients of remaining

equations. With this, we shall be left with the residual stiffness matrix [kR]i+1

& residual load vector [FR]i+1. The process is continued till all the elements

are considered.

b) Computational Details

We shall now demonstrate the application of the algorithm presented in

(A) above, by presenting the computational details in respect of the illustrative

problem under consideration.

1. Activate Element 1

We begin the analysis by activating element 1. As the element is the

first one to be handled, the residual stiffness matrix [kR]0 = 0 & the residual

nodal load vector [FR]0 = 0. (The outer suffix ‘0’ denotes zero or initial

condition) Consequently the assembled stiffness matrix [kA] & the assembled

nodal load vector [FA]1 are respectively equal to the element stiffness matrix

[k1e] & the element load vector [F1

e].

Nodal connections for the element 1 are (1, 2, 4), which defines

[[ ]] [[ ]] [[ ]] [[ ]] [[ ]] [[ ]].FF,kk, 1e1A

1e1A

1e1A ====δδ==δδ These are available from Table 1. Thus the

equations of equilibrium for the active assemblage are as presented in Table

no. 2 (a).

Table no. 2 (a)

−−

−−

−−

==

−−−−−−−−−−−−

−−−−−−−−−−−−

−−−−−−−−−−−−

3400

0

3400

0

3400

0

v

u

v

u

v

u

4288021447162144716

0716180356180356

214418011184481026268

71635644844826888

214418010262681118448

71635626888448448

4

4

2

2

1

1

[[ ]] [[ ]] [[ ]]1A1A1A Fk ==δδ

We should note that node 1 is appearing in the analysis for the last time;

hence the governing equations in respect of (u1, v1) are completely formulated.

In view of this, we shall eliminate (u1, v1) from the set of equations presented in

Table no. 2 (a).

Elimination of u1

13

u1 has a prescribed value of u1 = 0, hence following operations are

performed. The expression against u1 form Table no. 2 (a) is extracted &

preserved as shown in eq.19.

)19(FxFx0v716u356v268u88v448u448 R1

R1442211 K==++==−−−−++−−++

Wherein, Fx1R denotes the reaction due to the restraint over u1. This is

added to formulate the extracted equation properly. It should be emphasized

that in eq.19 u1 is known & Fx1R

is unknown.

As u1 has a prescribed value kij the coefficients of the remaining

equations of [kA]1 are unaffected by elimination of u1. Further as the prescribed

value of u1 is equal to zero, it follows from eq.10 that, the coefficients of the

remaining portion of [FA]1 are also unaffected. With all these, the residual

matrices governing the equations of equilibrium acquire the details as shown in

Table no. 2 (b).

Table no. 2 (b)

−−

−−

−−

==

−−−−−−−−

−−−−−−−−−−

−−−−−−

3400

0

3400

0

3400

X

v

u

v

u

v

X

4288021447162144X

0716180356180X

214418011184481026X

716356448448268X

214418010262681118X

XXXXXX

4

4

2

2

1

‘X’ in the table indicates the vacancies created by the extracted

equation.

Elimination of v1

v1 has a prescribed value of v1 = - 0.1, hence following operations are

performed. The expression against v1 form Table no. 2 (b) is extracted &

preserved as shown in eq.20.

)20(Fy3400

v2144u180v1026u268v1118 R144221 K++

−−==−−−−−−−−

Wherein, Fy1R denotes the reaction due to the restraint over v1. It

should be noted that v1 is known, whereas Fy1R is unknown.

As v1 has a prescribed value kij the coefficients of the remaining

equations in Table no 2 (b) are unaffected by the elimination of v1. On the

other hand the remaining coefficients Fi of the nodal load vector are modified

14

as per eq.10. With this the residual set of equations are as shown in Table no

2 (c).

Table no. 2 (c)

−−−−

−−−−

==

−−−−

−−−−−−−−

73.347

18

73.30

8.26

X

X

v

u

v

u

X

X

428802144716XX

0716180356XX

21441801118448XX

716356448448XX

XXXXXX

XXXXXX

4

4

2

2

Active Front Width

Before undertaking the operations with respect to the subsequent

elements, let us define an important parameter of Frontal Solution technique

called Front Width. The number of rows (or column) of [kA] in Table no 2 (a) is

called its active front width. As [kA], is of the size (6 x 6) the active front width =

6. The implication of this parameter would get clarified with the progress of

analysis.

2. Activate Element 2

We continue the analysis by activating the element 2. Now, the details

in Table no 2 (c) represents the residual characteristics [kR]1, [ Rδδ ]1, [FR]1 & the

residual idealization as shown in Fig no. 4.

Over the residual characteristics, we superposition the characteristics of

the active element, i.e. element 2, as shown in Table no. 3(a). Thus [ke2] &

Fig. No. 4 Active Element 2 followed by element 3

2 3

4 5

2

3

4

6

15

[Fe2] are appropriately superpositioned. Keeping in view that [ Rδδ 2] represents

the nodal displacement vector, with respect to the element nodes (2, 5, 4).

Table no. 3 (a)

−−−−−−

−−−−−−

−−

==

++++−−++−−++++−−++++−−−−−−−−

−−−−++++−−++−−−−++−−−−−−++−−−−

−−−−−−−−−−−−

73.347)3400(

180

73.30)3400(

8.260

3400

0

v

u

v

u

v

u

428811180448214421447161801026268

044871644818071635635626888

21442144180716111842884482144716

7161803563564480448716180356

102626821441801118448

26888716356448448

5

4

2

2

5

5

The coefficients of [ke2] & [Fe

2] are derived from Table no.1. We should note

the following points:

a) In Table no. 3(a), the free terms are picked up from Table no. 2(c); & the

same belong to the residual characteristics, whereas the terms within

the parenthesis represent the contribution from the active element 2.

b) From Table no. 2(c), it follows that [kA]2, [δδA]2, [FA]2 will have the

accommodations in respect of the nodes 2 & 4, which are present

already. As the node 5 gets introduced into the analysis for the first

time, provision will have to be made for accommodating the related

characteristics. Fortunately, we could utilize the vacancies created by

elimination of (u1, v1) for the purpose. Thus, without exceeding the

active front width of 6, the node 5 gets accommodate into the analysis.

c) Carrying out the algebraic sum of the terms in Table no. 3(a), the

equations of equilibrium for the active assemblage gets established as

shown in Table no. 3(b).

Table no. 3 (b)

−−−−

−−−−

−−

==

−−−−−−−−−−−−

−−−−−−−−−−−−−−−−−−−−−−−−−−

06.481

18

06.164

8.26

3400

0

v

u

v

u

v

u

540644842888961026268

448116489671226888

428889654064482144716

8967124481164180356

102626821441801118448

26888716356448448

4

4

2

2

5

5

From Fig. No. 4, it follows that none of the active nodes viz, nodes (2, 4, 5) are

appearing for the last time; hence there is no scope for the elimination process.

16

The details in Table no. 3(b), thus automatically represents the residual

characteristics [kR]2, [ Rδδ ]2, [FR]2.

3. Activate Element 3

We continue further the analysis by activating the element 3. As in the

previous step, none of the nodes could be eliminated, Fig. No. 4 continues to

represent the residual idealization. Now, [ke3] & [Fe

3] are available in Table no.

1, & the same are superpositioned over [kR]2 & [FR]2 respectively. The result

represents [kA]3 & [FA]3 the characteristics for the active assemblage. Towards

this process [δδ3e] is defined by the element nodes (2, 3, 5), hence [δδA]3 gets

accordingly modified. The entire process is similar to the one considered

earlier while activating the element 2. The results are presented in Table no.

4(a).

Table no. 4 (a)

−−

−−−−

−−−−

−−

==

−−−−−−−−−−−−

−−−−−−−−−−−−

−−−−−−−−−−−−−−−−

−−−−−−−−−−−−−−−−

3400

0

481

18

39.297

8.26

3800

0

v

u

v

u

v

u

v

u

11184480010262682144180

4484480026888716356

00540644842888961026268

00448116489671226888

10262684288896652404288896

2688889671201612896712

2144716102626842888965406448

180356268888967124481164

3

3

4

4

2

2

5

5

We should note the following points:

a) Locations for the degree of freedom related to the nodes (2, 4, 5) are

already available as per Table no. 3(b).

b) As node no 3 is appearing in the analysis for the first time, it would be

necessary to make provision for accommodating its characteristics.

Now, within the frame work of the residual blocks, vacant spaces do not

exist, hence the sizes of [kA]3, [FA]3 & [δδA]3 would get increased. In this

process the active front width also gets increased to 8.

Degree of freedom (u2, v2, u3, v3) related to nodes 2 & 3 are eliminated one

after the other because these nodes are appearing in the analysis for the last

time. It may be noted that each of the eliminatable variables has prescribed

values; hence as demonstrated earlier the equations could be modified. In

17

Table no. 4(b), (c), (d) & (e), the set of equations resulting from the elimination

of u2, v2, u3 & v3 are respectively presented. The corresponding expression

extracted prior to each stage of elimination are shown in eq.21 (a), (b), (c) &

(d) respectively.

)21(

Fy73.30v118v2144u180)d(

Fx8.26v448u448v716u365)c(

Fy39.297v1026u268v4288u896v6524v4288u896)b(

Fx8.26v1026u88v896u712u1612v896u712)a(

R3355

R33355

R23344255

R23344255

KKK

++−−==++−−

++−−==−−++++−−

++−−==++−−−−++++−−−−

++−−==++−−++−−++−−−−

Wherein, )Fy,Fx,Fy,Fx( R3

R3

R2

R2 denoting reactions induced due to the

restraints over (u2, v2, u3, v3), are added to formulate the extracted equations

completely. We should note that (u2, v2, u3, v3) are known, whereas

)Fy,Fx,Fy,Fx( R3

R3

R2

R2 are unknowns.

Table no. 4 (b)

−−

−−−−

−−

−−

==

−−−−−−−−−−

−−−−−−−−−−

−−−−−−−−

−−−−−−−−−−−−

3400

0

481

18

39.297

x

3800

0

v

u

v

u

v

x

v

u

1118448001026x2144180

44844800268x716356

0054064484288x1026268

004481164896x26888

102626842888966524x4288896

xxxxxxxx

214471610262684288x5406448

18035626888896x4481164

3

3

4

4

2

5

5

Table no. 4 (c)

−−−−

−−

−−−−

==

−−−−−−−−

−−−−−−−−

−−−−−−−−

73.30

8.26

66.909

6.71

x

x

46.695

6.89

v

u

v

u

x

x

v

u

111844800xx2144180

44844800xx716356

005406448xx1026268

004481164xx26888

xxxxxxxx

xxxxxxxx

21447161026268xx5406448

18035626888xx4481164

3

3

4

4

5

5

18

Table no. 4 (d)

−−

−−

−−−−

==

−−

−−−−−−−−

−−−−−−

73.30

x

66.909

6.71

x

x

46.695

6.89

v

x

v

u

x

x

v

u

1118x00xx2144180

xxxxxxxx

0x5406448xx1026268

0x4481164xx26888

xxxxxxxx

xxxxxxxx

2144x1026268xx5406448

180x26888xx4481164

3

4

4

5

5

Table no. 4 (e)

−−

−−−−

==

−−−−−−−−

−−−−

x

x

86.909

6.71

x

x

86.909

6.71

x

x

v

u

x

x

v

u

xxxxxxxx

xxxxxxxx

xx5406448xx1026268

xx4481164xx26888

xxxxxxxx

xxxxxxxx

xx1026268xx5406448

xx26888xx4481164

4

4

5

5

4. Activate Element 4

In the final step we activate element 4. Its element characteristics [ke4] & [Fe

4]

are available in Table no. 1. These are superpositioned over the residual

characteristic [kR]3 & [FR]3 represented in Table no. 4(e), to derive [kA]4 &

[FA]4. Towards this [δδ4e] & [δδA]3 are defined by the nodes (4, 5, 6). The details

of the resulting set of equations are presented in Table no. 5(a).

Table no. 5 (a)

−−

−−

−−−−

==

−−−−−−−−

−−−−−−−−−−−−

−−−−−−

19.1043

6.71

3400

0

19.1043

6.71

v

u

v

u

v

u

65240214418020520

016127163560176

2144716428802144716

1803560716181356

20520214418065240

017671635601612

4

4

6

6

5

5

19

We should note the following points in this context:

a) Locations for the degree of freedom related with nodes 4 & 5 are

already available as per Table no. 4(e).

b) Node 6 is appearing in the analysis for the first time. Vacant spaces are

however available to accommodate its characteristics. As the available

vacancies are more than the ones required for the purpose, we have an

option. For example, the vacancies in between the characteristics of

nodes 5 & 4 may be chosen or the ones available at the end portion of

[kR]3, [FR]3 could be utilized. As the former option involves active front

width of 6, which happens to be lesser than the active front width of 8

required in the later option, we choose the former alternative.

Degree of freedom (u4, v4, u5, v5, u6, v6) related to the nodes (4, 5, 6) are

eliminated one by one; because in this final stage all the nodes appear for the

last time. As these are free nodal degree of freedom, at each stage of

elimination the process is governed by eq.8 in Table no. 5(b), (c), (d), (e) & (f),

the set of equations resulting from the elimination of u4, v4, u5, v5 & u6

respectively are presented. Table no. 5(f) thus contains only one equation, with

one variable v6. To complete the process of elimination, at various stages the

equations are extracted & preserved to keep them ready for the back

substitution process. The details are shown in eq.22:

)22(

59.597v62.2642)f(

0u02.525)e(

07.715v71.1469u62.236v65.5878)d(

78.63v82.637u87.394u78.1592)c(

19.1043v6524v2144u180v2052)b(

6.71u1612v716u356u176)a(

6

6

665

665

4665

4665

KKKK

−−==

==−−==−−++

−−==++−−−−==++−−−−

==++−−−−−−

Table no. 5 (b)

−−

−−

−−−−

==

−−−−

−−−−−−−−−−−−

−−−−

19.1043

x

53.101

81.15

19.1043

78.63

v

x

v

u

v

u

6524x214418020520

xxxxxx

2144x95.396912.158214482.637

180x12.1584.63718087.394

2052x214418065240

0x82.63787.394078.1592

4

6

6

5

5

20

Table no. 5 (c)

−−−−

−−−−

==

−−−−−−−−−−

−−

x

x

36.444

97.12

07.715

78.63

x

x

v

u

v

u

xxxxxx

xxxxxx

xx43.326528.21771.146982.637

xx28.21743.63262.23687.394

xx71.146962.23665.58780

xx82.63787.394078.1592

6

6

5

5

Table no. 5 (d)

−−−−

−−==

−−−−−−

−−

82.418

78.28

07.715

x

v

u

v

x

05.301017.5971.1469x

17.5954.53462.236x

71.146962.23665.5878x

xxxx

6

6

5

Table no. 5 (e)

−−

==

59.597

0

x

x

v

u

x

x

62.26420xx

002.525xx

xxxx

xxxx

6

6

Table no. 5 (f)

−−

==

59.597

x

x

x

v

x

x

x

62.2642xxx

xxxx

xxxx

xxxx

6

Back Substitution phase:

Equations that have been extracted at various stages of elimination,

constitute the basic record through which the values of the unknown nodal

parameters are established. This means, at the free nodes the displacements,

whereas at the restrained nodes the reactions are computed. All these are

derived through the second basic stage of Gauss’s solution scheme viz. the

back substitution phase.

We begin the process of back substitution from eq.22 (f) & proceed

backward to the eq.22 (e), (d), (c), (b) & (a); thereby establish the values of the

nodal displacements v6, u6, v5, u5, v4 & u4 respectively. These nodal

21

displacements, along with the prescribed values of displacements at the

restrained nodes, are utilized to carry out the back substitution process further;

by operating through eq.21, eq.20 & eq.19. The complete set of results are

presented in Table no.6

Table no. 6 (a)

Nodal Displacements

Node 1 2 3 4 5 6

u (m) 0.00 0.00 0.00 -0.0505 0.0505 0.00

v (m) -0.10 -0.10 -0.10 -0.1782 -0.1782 -0.2261

Table no. 6 (b)

Reactions

R1Fx R

2Fx R3Fx R

1Fy R2Fy R

3Fy

73.87t 0.0 -73.87t 309.91t 980.18t 309.91t

NOTE: In the above problem, vertical displacement of -0.1m was prescribed at

the restrained nodes, i.e. v1 = v2 = v3 = -0.1m, for demonstrating certain

inherent aspect of elimination of prescribed variables. These displacements at

the restrained nodes being of identical nature & magnitude, constituted

prescribed rigid body movements of the embankment section. In view of this

solution with v1 = v2 = v3 = 0, will differ from the present results in the following

manner:

a) ‘u’ at all the nodes would be identical to the ones as shown in Table no.

6(a)

b) ‘v’ at all the nodes would be less by -0.1m, to the ones shown in Table

no. 6(a)

c) Reactions at the restrained nodes would be identical to the ones shown

in Table no. 6(b).

22

Resolution:

Gauss’s elimination technique for the solution of simultaneous

equations has inherent advantage in offering resolution facilities, for the types

of problems under consideration, need for the resolution arise due to two basic

possibilities. These are:

a) The amount of prescribed displacements may be different from the one

already considered.

b) The load vector may change from the one considered earlier.

In the set of equations governed by [k] & [F], the resolution does not affect

the modifications in [k], while undertaking the elimination phase. It changes

only the details in the [F] vector. We should note that in the elimination phase

the modification to [k] is labor intensive; hence, in the resolution, due to saving

in this computational effort, the solution takes.