chapter 2 mathematical background 2015 essentials of the finite element method (1)

CHAPTER

MATHEMATICAL BACKGROUND
2 2.1 VECTORS2.1.1 DEFINITION OF VECTORIt is well known from elementary physics that physical quantities can be represented by scalars or
vectors. For a physical quantity described by a scalar, we just need a single value for its specification.

Unlike scalar quantities, many other quantities (e.g., force, displacement, area, etc.), have direction in

addition to magnitude. These quantities are called vectors and are represented graphically by directed

line segments (Figure 2.1).

Any vector a!is correlated with a system of coordinates x, y, z through its components ax,ay,az and

can be expressed analytically by the following equation:

a!¼ ax i

!+ ay j

!+ az k

!(2.1)

where i!, j!, k!

are unit vectors directed along the x, y, z, axes, respectively.Between vectors, there are two types of products; the scalar product and the vector product.

2.1.2 SCALAR PRODUCTLet us consider the vectors a

!and b

!(Figure 2.2). If we denote by a,b their magnitudes, respectively, the

scalar product a! � b! is defined by the following equation:

a! � b!¼ a b cosϑð Þ¼ ab cosϑ (2.2)

Therefore, the scalar product of the vectors a!and b

!is the magnitude of a

!multiplied by the magnitude

of the projection of b!

on the direction of a!.

Taking into account Equation (2.2), it is obvious that

a! � b!¼b

! � a! (2.3)

According to the definition of the scalar product, the following properties can be obtained:

i! � i

!¼ 1ð Þ 1ð Þcos 0ð Þ¼ 1 (2.4)

i! � j

!¼ 1ð Þ 1ð Þcos π=2ð Þ¼ 0 (2.5)

Essentials of the Finite Element Method. http://dx.doi.org/10.1016/B978-0-12-802386-0.00002-5

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b cosja

b

j

FIGURE 2.2

Scalar product of vectors a!

and b!.

ik

jx

z

y

ax

az

a

ay

FIGURE 2.1

Graphical representation of a vector quantity a!.

20 CHAPTER 2 MATHEMATICAL BACKGROUND

i! � k!¼ 1ð Þ 1ð Þcos π=2ð Þ¼ 0 (2.6)

and so on.

Therefore, the scalar product

a! � b!¼ ax i

!+ ay j

!+ az k

!� �� bx i

!+ by j

!+ bz k

!� �(2.7)

yields

a! � b!¼ axbx i

! � i!

+ axby i! � j

!+ axbz i

! � k!

+ aybx j! � i

!+ ayby j

! � j!

+ aybz j! � k!

+ azbx k! � i

!+ azby k

! � j!

+ azbz k! � k!

(2.8)

or

a! � b!¼ axbx + ayby + azbz ¼ ax ay az½ � �

bxbybz

8<:

9=; (2.9)

212.1 VECTORS

2.1.3 VECTOR PRODUCTThe vector product of two vectors a

!and b

!(Figure 2.3) is defined by the following equation:

a!� b

!¼ deti!

j!

k!

ax ay azbx by bz

264

375 (2.10)

Taking into account the above definition, the following properties can be obtained:

i!� i

!¼ deti!

j!

k!

1 0 0

1 0 0

24

35¼ 0 (2.11)

i!� j

!¼ deti!

j!

k!

1 0 0

0 1 0

24

35¼k

!(2.12)

j!� i

!¼ deti!

j!

k!

0 1 0

1 0 0

24

35¼� k

!(2.13)

The resulting vector of the product a!� b

!is directed perpendicular to the plane formed by the vectors a

!

and b!, and its magnitude is

a!� b

!�� ¼ a b sinφð Þ (2.14)

Taking into account Figure 2.3, the area A of the triangle ABC is

A¼ 1

2a b sinφð Þ (2.15)

Therefore, combining Equations (2.14) and (2.15), the following result can be obtained:

A¼ 1

2a!� b

!�� (2.16)

b sinj

aA C

B

b

j

FIGURE 2.3

Vector product of vectors a!and b

!.


2.1.4 ROTATION OF COORDINATE SYSTEMA vector a

!can be expressed with respect to two different coordinate systems, namely x-y-z and η-ξ-ζ

(Figure 2.4). The vector’s components with respect to two coordinate systems can be correlated. If we

define the same vector a!

in the rotated coordinates η-ξ-ζ, the following expression can be written:

a!¼ aη i

0! + aξ j0! + aζ k

0! (2.17)

where i0!, j0!, k0!

are the unit vectors along the η-ξ-ζ directions.

If we denote by φx,φy,φz the angles of the vector a!with the axes x, y, z, respectively, the following

equations can be written:

i! � a!¼ 1ð Þ � a cosφxð Þ¼ ax (2.18)

j! � a!¼ 1ð Þ � a cosφy

� �¼ ay (2.19)

k! � a!¼ 1ð Þ � a cosφzð Þ¼ az (2.20)

Using Equation (2.17), Equations (2.18)–(2.20) can now be written in the following form:

ax ¼ i! � a!¼ an i

! � i0!

+ aξ i! � j0

!+ aζ i

! � k0!

(2.21)

ay ¼ j! � a!¼ an j

! � i0!

+ aξ j! � j0

!+ aζ j

! � k0!

(2.22)

az ¼k! � a!¼ an k

! � i0!

+ aξ k! � j0

!+ aζ k

! � k0!

(2.23)

k �

i�

a

j�

x

z

y

aζ

ζ

ξ

η

jx

jz

jy

FIGURE 2.4

Representation of a vector a!

with respect to two different systems of coordinates x-y-z and η-ξ-ζ.

232.1 VECTORS

Taking into account Equations (2.4)–(2.6), the above equations yield:

axayaz

8<:

9=;¼

cos x, nð Þ cos x, ξð Þ cos x, ζð Þcos y, nð Þ cos y, ξð Þ cos y, ζð Þcos z, nð Þ cos z, ξð Þ cos z, ζð Þ

24

35 �

anaξaζ

8<:

9=; (2.24)

where

cos x, nð Þ¼ i! � i0

!(2.25)

cos x, ξð Þ¼ i! � j0

!(2.26)

cos x, ζð Þ¼ i! � k0

!(2.27)

and so on are called directional cosines.

2.1.5 THE VECTOR DIFFERENTIAL OPERATOR (GRADIENT)The vector differential operator r! is defined by the following equation:

r! � @

@xi!

+@

@yj!

+@

@zk!

(2.28)

This operator is used for defining the following Green’s theorem.

2.1.6 GREEN’S THEOREMLet us consider a scalar field functionΦ¼Φ x, yð Þ. Its gradient in a two-dimensional (2D) domain is the

following vector:

r! Φ x, yð Þ�@Φ x, yð Þ@x

i!

+@Φ x, yð Þ

@yj!

(2.29)

If we denote by n!the unit vector normal to the boundary S of the 2D domain Ω (Figure 2.5), where

n!¼ nx i

!+ ny j

!(2.30)

Ω

S

h

x

y

FIGURE 2.5

A two-dimensional domain of a scalar function.


and

n!�� ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi

n2x + n2y

q¼ 1 (2.31)

the following theorem can be used:ððr! Φ x, yð Þdxdy¼

þS

Φ x, yð Þ n! dS (2.32)

The above equation is known as Green’s theorem and correlates an integral of a gradient of a function

Φ¼Φ x, yð Þ over an area Ω with a contour integral over the boundary S.

2.2 COORDINATE SYSTEMSIn applied mechanics, the three common coordinate systems are rectangular (or Cartesian), cylindrical,

and spherical. The aim of the present section is to describe these coordinate systems and to derive the

formulae for the transformation of the vector components of each coordinate system with respect to the

others.

2.2.1 RECTANGULAR (OR CARTESIAN) COORDINATE SYSTEMThe rectangular coordinate system has already been presented in Figure 2.1. The components of a vec-

tor a!are specified in terms of three mutually perpendicular unit vectors i

!, j!, k!. For convenience, we

denote them as x!, y!, z!, as shown in Figure 2.6.

yx

zy

x

z

az

ay

ax

a

FIGURE 2.6

Rectangular (or Cartesian) coordinate system.

zj

y

x

z

r

r

j

a

FIGURE 2.7

Cylindrical coordinate system.

252.2 COORDINATE SYSTEMS

2.2.2 CYLINDRICAL COORDINATE SYSTEMIn the cylindrical coordinate system, the components of a vector a

!are specified in terms of three mu-

tually perpendicular unit vectors r!, φ!, z!, as shown in Figure 2.7. Therefore, its coordinates are r,φ, z.

2.2.3 SPHERICAL COORDINATE SYSTEMIn the spherical coordinate system, the coordinates are r,ϑ,φ. Therefore, the components of any vector a

!

are specified in terms of the three mutually perpendicular unit vectors r!, ϑ!, φ!, as shown in Figure 2.8.

j

J

y

x

z

r

j

Ja

FIGURE 2.8

Spherical coordinate system.


2.2.4 COMPONENT TRANSFORMATIONThe components of a vector with respect to a certain coordinate system must often be transformed

to another coordinate system. The equations correlating the components of a vector in the

three coordinate systems can be derived using the properties of vector analysis. Let us consider

the vector a!

to be expressed in rectangular and spherical coordinate system as shown in

Figure 2.9.

The vector a!

can be expressed with respect to the system x,y, z by the following equation:

a!¼ ax x

!+ ay y

!+ az z

!(2.33)

Since the scalar product e! � u! of two arbitrary vectors e

!, u!expresses the product of the magnitude of

the vector u!with the magnitude of the projection of e

!on the direction of u

!, the following properties for

the products r! � x

!, r! � y

!, r! � z! can be used:

r! � x!¼ sinϑ cosφ (2.34)

r! � y!¼ sinϑ sinφ (2.35)

r! � z!¼ cosϑ (2.36)

and so on.

Therefore, if we multiply Equation (2.33) with r!, the following equation can be obtained:

r! � a!¼ ax r

! � x! + ay r! � y! + az r

! � z! (2.37)

y

x

z

r

yz

x

az

ax

ay

j

j

JJa

FIGURE 2.9

Transformation of a vector’s components from a rectangular coordinate system to a spherical one.

272.2 COORDINATE SYSTEMS

or

r! � a!¼ ax sinϑ cosφ + ay sinϑ sinφ+ az cosϑ (2.38)

Since

r! � a!¼ ar (2.39)

Equation (2.38) can now be written as:

ar ¼ sinϑ cosφ sinϑ sinφ cosϑ½ �axayaz

8<:

9=; (2.40)

Following similar procedure (i.e., to multiply Equation (2.33) with ϑ and φ), the following equations

can be derived:

aϑ ¼ cosϑ cosφ cosϑ sinφ �sinϑ½ �axayaz

8<:

9=; (2.41)

and

aφ ¼ �sinφ cosφ 0½ �axayaz

8<:

9=; (2.42)

Combination of Equations (2.40)–(2.42) yields the following transformation of a vector’s components

from a rectangular coordinate system to a spherical one:

araϑaφ

8<:

9=;¼

sinϑ cosφ sinϑ sinφ cosϑcosϑ cosφ cosϑ sinφ �sinϑ�sinφ cosφ 0

24

35 ax

ayaz

8<:

9=; (2.43)

Following the same concept, the transformation of a vector’s components from a rectangular coordi-

nate system to a cylindrical one can be performed by the following equation:

araφaz

8<:

9=;¼

cosφ sinφ 0

�sinφ cosφ 0

0 0 1

24

35 ax

ayaz

8<:

9=; (2.44)

Inversion of the above equations yields the transformation of the components from a spherical and

cylindrical coordinate system to a rectangular one, respectively:

axayaz

8<:

9=;¼

sinϑ cosφ cosϑ cosφ �sinφsinϑ sinφ cosϑ sinφ cosφ

cosϑ �sinϑ 0

24

35 ar

aϑaφ

8<:

9=; (2.45)

and

axayaz

8<:

9=;¼

cosφ �sinφ 0

sinφ cosφ 0

0 0 1

24

35 ar

aφaz

8<:

9=; (2.46)


Finally, transformation of the components from a cylindrical coordinate system to a spherical one, and

vice versa, can be performed from the following equations:

araϑaφ

8<:

9=;¼

sinϑ 0 cosϑcosϑ 0 �sinϑ0 1 0

24

35 ar

aφaz

8<:

9=; (2.47)

and

araφaz

8<:

9=;¼

sinϑ cosϑ 0

0 0 1

cosϑ �sinϑ 0

24

35 ar

aϑaφ

8<:

9=; (2.48)

2.2.5 THE VECTOR DIFFERENTIAL OPERATOR (GRADIENT) IN CYLINDRICALAND SPHERICAL COORDINATESTaking into account the above results, the vector differential operator r! presented in Equation (2.28)

for rectangular coordinates, can now be expressed in cylindrical and spherical coordinates:

r! � @

@rr!

+1

r

@

@φφ!

+@

@zz!

for cylindrical coordinates (2.49)

and

r! � @

@rr!

+1

r

@

@ϑϑ!

+1

rsinϑ

@

@φφ!

for spherical coordinates (2.50)

2.3 ELEMENTS OF MATRIX ALGEBRA2.3.1 BASIC DEFINITIONSA matrix [Amn] is an array of scalars consisting of m rows and n columns:

Amn½ � ¼a11 a12 … a1na21 a22 … a2n⋮ ⋮ ⋮am1 am2 … amn

2664

3775 (2.51)

The location of any element within the matrix is defined by the subscripts i, j. Therefore, the symbol aijmeans that this element is located at the i-th row and j-th column. A matrix {Bn} containing only one

column is called a vector:

Bnf g¼b1b2⋮bn

8>><>>:

9>>=>>; (2.52)

292.3 ELEMENTS OF MATRIX ALGEBRA

Inmechanical and structural engineering problems,we often encounter special cases ofmatrices such as:

1. Diagonal matrix, that is, square matrix containing non-zero elements only at the locations where

i 6¼ j, for example,

C4½ � ¼1 0 0 0

0 �12 0 0

0 0 7 0

0 0 0 3

2664

3775 (2.53)

Unit matrix, that is, a diagonal matrix containing unit elements at the locations where i¼ j, for
2.example,
I3½ � ¼1 0 0

0 1 0

0 0 1

24

35 (2.54)

Zero matrix, that is, a matrix containing elements with zero value for all locations i, j, for example,
3.
O3,2� ¼ 0 0

0 0

0 0

24

35 (2.55)

Symmetric matrix, that is, a matrix where aij ¼ aji for all i, j, for example,
4.
F½ � ¼1 2 �6

2 8 9

�6 9 11

24

35 (2.56)

2.3.2 BASIC OPERATIONSDeterminant of a square matrixThe determinant det[Ann] of a square matrix [Ann] is a scalar quantity given by the following equation:

det Ann½ � ¼X

φ1,φ2,…,φn

p φ1, φ2,…, φnð Þa1,φ1� a2,φ2

�⋯ � an,φn(2.57)

where p(φ1,φ2,…,φn) is a permutation equal to �1. For example, for a 3�3 matrix, we have the

following permutations:

p 1, 2, 3ð Þ¼ 1 (2.58)

p 1, 3, 2ð Þ¼�1 (2.59)

p 3, 1, 2ð Þ¼ 1 (2.60)

p 3, 2, 1ð Þ¼�1 (2.61)

p 2, 3, 1ð Þ¼ 1 (2.62)

p 2, 1, 3ð Þ¼�1 (2.63)


According to the above definition, the determinant of a 3�3 matrix is

det

a11 a12 a13a21 a22 a23a31 a32 a33

0@

1A¼ a11a22a33�a11a23a32 + a13a21a32�a13a22a31 + a12a23a31�a12a21a33 (2.64)

EXAMPLE 2.1Calculate the determinant of the following matrix:

A¼1 5 4

2 3 1

7 8 5

0@

1A

SolutionAccording to Equation (2.57), det[A] can be calculated as:

det A½ � ¼ 1�3�5ð Þ� 1�1�8ð Þ+ 4�2�8ð Þ� 4�3�7ð Þ+ 5�1�7ð Þ� 2�2�5ð Þ¼�28

Minor Dij of an element aijMinorDij of an element aij is the determinant of the submatrix obtained from the matrix [A] by deletingthe i-th row and j-th column, for example, for a 3�3 matrix, the minor D12 is:

11 12 1321 23

12 21 22 2331 33

31 32 33

det det

a a aa a

D a a aa a

a a a

ð2:65Þor

D12 ¼ a21a33�a31a23 (2.66)

Cofactor Aij of an element αij

The cofactor Aij of an element aij is defined as:

Aij ¼ �1ð Þi + jDij (2.67)

For the above example of the 3�3 matrix, the cofactor A12 is

A12 ¼ �1ð Þ1 + 2 a21a33�a31a23ð Þ¼�a21a33 + a31a23 (2.68)

The inverse of a matrixThe inverse of a square matrix [A], denoted by [A]�1 is defined by the following equation:

A½ ��1 ¼ 1

det A½ �

A11 A21 … An1

A12 A22 … An2

⋮ ⋮ ⋮A1n A2n … Ann

0BB@

1CCA (2.69)

where Aij are the cofactors of the elements aij.


EXAMPLE 2.2Calculate the inverse of the following matrix:

A¼1 5 4

2 3 1

7 8 5

0@

1A

SolutionThe minors of the above matrix are:

D11 ¼ det3 1

8 5

�¼ 7

D12 ¼ det2 1

7 5

�¼ 3

D13 ¼ det2 3

7 8

�¼�5

D21 ¼ det5 4

8 5

�¼�7

D22 ¼ det1 4

7 5

�¼�23

D23 ¼ det1 5

7 8

�¼�27

D31 ¼ det5 4

3 1

�¼�7

D32 ¼ det1 4

2 1

�¼�7

D33 ¼ det1 5

2 3

�¼�7

Then, the cofactors of the matrix [A] are

A11 ¼ �1ð Þ1 + 1D11 ¼ 7, A12 ¼ �1ð Þ1 + 2D12 ¼�3, A13 ¼ �1ð Þ1 + 3D13 ¼�5,

A21 ¼ �1ð Þ2 + 1D21 ¼ 7, A22 ¼ �1ð Þ2 + 2D22 ¼�23, A23 ¼ �1ð Þ2 + 3D23 ¼ 27,

A31 ¼ �1ð Þ3 + 1D31 ¼�7, A32 ¼ �1ð Þ3 + 2D32 ¼ 7, A33 ¼ �1ð Þ3 + 3D33 ¼�7


Taking into account (from the previous example) that

det A½ � ¼�28

then, according to Equation (2.69) the matrix A½ ��1is

A½ ��1 ¼ 1

�28

7 7 �7

�3 �23 7

�5 27 �7

24

35

Transpose of a matrixThe transpose of a matrix [A], denoted by [A]T, is given by the following equation:

A½ �T ¼a11 a12 … a1na21 a22 … a2n⋮ ⋮ ⋮an1 an2 … ann

0BB@

1CCA

T

¼a11 a21 … an1a12 a22 … an2⋮ ⋮ ⋮a1n a2n … ann

0BB@

1CCA (2.70)

Addition/subtraction of two matrices

Anm½ �� Bnm½ � ¼a11�b11 a12�b12 … a1m�b1ma21�b21 a22�b22 … a2m�b2m

⋮ ⋮ ⋮an1�bn1 an2�bn2 … anm�bnm

0BB@

1CCA (2.71)

Multiplication of two matricesIf a matrix [Anm] is multiplied by a scalar e, the matrix e�[A] is obtained by multiplying all elements of

[A] by the scalar e. For example,

e �a11 a12 a13a21 a22 a23a31 a32 a33

0@

1A¼

e � a11 e � a12 e � a13e � a21 e � a22 e � a23e � a31 e � a32 e � a33

0@

1A

If a matrix [Anm] is multiplied by another matrix [Bmk], then the obtained matrix

Cnk½ � ¼ Anm½ � � Bmk½ � (2.72)

is composed by elements given by the following equation:

cij ¼Xmφ¼1

aiφbφj (2.73)

Application of the above formula into multiplication of a matrix

R½ � ¼ r1 r2 r3½ � (2.74)

with a vector

Cf g¼c1c2c3

8<:

9=; (2.75)


yields

R½ � � Cf g¼ r1c1 + r2c2 + r3c3 (2.76)

An alternative procedure for multiplication of the following matrices

A23½ � ¼ a11 a12 a13a21 a22 a23

�(2.77)

B32½ � ¼b11 b12b21 b22b31 b32

0@

1A (2.78)

is to express the above matrices as

A23½ � ¼ R1½ �R2½ �

�(2.79)

B32½ � ¼ C1f g C2f g½ � (2.80)

where

R1½ � ¼ a11 a12 a13½ � (2.81)

R2½ � ¼ a21 a22 a23½ � (2.82)

C1f g¼b11b21b31

8<:

9=; (2.83)

C2f g¼b12b22b32

8<:

9=; (2.84)

Then,

[ ] [ ]

{ } { }

[ ] [ ]{ } [ ]{ }[ ] [ ]{ } [ ]{ }

A23 A32 R1 C1R2 C1

R1 C2R2 C2

R1R2

C1 C2↓ ↓

→→

. = ð2:85Þ

Following the same procedure, multiplication of a matrix with a vector can be performed as follows:

b1 b2 b3½ �a11 a12 a13a21 a22 a23a31 a32 a33

0@

1A �

b1b2b3

8<:

9=;¼

a11 a12 a13a21 a22 a23a31 a32 a33

0@

1A¼

a11b1 + a12b2 + a13b3a21b1 + a22b2 + a23b3a31b1 + a32b2 + a33b3

8<:

9=; (2.86)

Properties of matrix operations

A½ �+ B½ � ¼ B½ � + A½ � (2.87)

A½ � � B½ � 6¼ B½ � � A½ � (2.88)


A½ � � B½ �+ C½ �ð Þ ¼ A½ � � B½ � + A½ � � C½ � (2.89)

A½ � + B½ �ð Þ � C½ � ¼ A½ � � C½ � + B½ � � C½ � (2.90)

A½ �+ B½ �ð ÞT ¼ A½ �T + B½ �T (2.91)

A½ � � B½ �ð ÞT ¼ B½ �T � A½ �T (2.92)

A½ �+ B½ �ð Þ�1 ¼ A½ ��1+ B½ ��1

(2.93)

A½ � � B½ �ð Þ�1 ¼ B½ ��1 � A½ ��1(2.94)

2.4 VARIATIONAL FORMULATION OF ELASTICITY PROBLEMSAs it will be described in Chapters 7 and 9, the derivation of the finite element equation of elastic

solids can be based on the principle of minimum potential energy (MPE). This principle belongs to

a general category of energy methods that is known as the variational principle. The mathematical

background behind MPE assumes that some classes of boundary value problems can be solved by

the minimization of an integral functional (i.e., function of a function) whose necessary conditions

for a minimum implies that the differential equation and the associated boundary conditions

are satisfied. Therefore, the main task of the above procedure is the derivation of the integral

functional.

Since the derivation of an integral functional is based on concepts from calculus of variations, this

section explain the properties and the main concept of the variation of a function.

2.4.1 DEFINITION OF THE VARIATION OF A FUNCTIONLet us assume a continuous function u(x) (e.g., the axial displacement of a bar due to axial loads).

For simplicity, we have chosen a polynomial function

u xð Þ¼ a0 + a1x+ a2x2 +⋯+ anx

n (2.95)

For infinitesimal changes of the above coefficients, a0,a1,a2,…,an, namely, δa0,δa1,δa2,…,δan,a closely related function is written as follows:

�u xð Þ¼ a0 + δa0ð Þ+ a1 + δa1ð Þx+ a2 + δa2ð Þx2 +⋯+ an + δanð Þxn (2.96)

The variation δu(x) of u(x) is defined as the following difference

δu xð Þ¼ �u xð Þ�u xð Þ¼ δa0 + δa1x+ δa2x2 +⋯+ δanx

n (2.97)

Very often, apart from the variation of a function we need to calculate the variation of a functional

(i.e., function of a function) U(u(x)):

δU¼U �u xð Þð Þ�U u xð Þð Þ (2.98)

The calculation of the variation of a functional can be simplified by using the following properties

of variations.

352.4 VARIATIONAL FORMULATION OF ELASTICITY PROBLEMS

2.4.2 PROPERTIES OF VARIATIONS

δdU

dx

�¼ d δUð Þ

dx(2.99)

δ

ðUdx

�¼ðδUdx (2.100)

δ U +Vð Þ¼ δU + δV (2.101)

δ UVð Þ¼ δUð ÞV +U δVð Þ (2.102)

δ Unð Þ¼ nUn�1δU (2.103)

f xð ÞδU xð Þ¼ δ f xð ÞU xð Þð Þ (2.104)

U xð ÞδU xð Þ¼ 1

2δ U2 xð Þ� �

(2.105)

dU

dx� d δUð Þ

dx¼ 1

2δ

dU

dx

�2" #

(2.106)

F � minimum) δ F¼ 0 (2.107)

2.4.3 DERIVATION OF THE FUNCTIONAL FROM THE BOUNDARY VALUE PROBLEMFollowing mathematical manipulations, the functional to be minimized can be derived from the

differential equation and the boundary conditions. This procedure is illustrated through the following

example of a bar under axial loading.

Boundary value problem of a bar under axial loadingLet us consider a bar (Figure 2.10) with variable cross-sectional area A¼A(x). The bar is subjected to adistributed axial load q(x) and a concentrated force F at its right end. The equilibrium equation of an

infinitesimal part of the bar yields

N +@N

@xdx+ q xð Þdx�N¼ 0 (2.108)

or

@N

@x+ q xð Þ¼ 0 (2.109)

where N¼N(x) is the axial force acting on a cross-section located at x.According to the Hooke’s law, the force N(x) can be expressed as a function of strain ε(x)

N xð ÞA xð Þ ¼Eε xð Þ (2.110)

where E is the modulus of elasticity.

F

L

N dx∂N∂xN

( )q x dx

dx

FIGURE 2.10

Bar under axial loads.


Since

ε xð Þ¼ @u xð Þ@x

(2.111)

where u(x) is the axial displacement, Equation (2.110) can be written as:

N xð Þ¼EA xð Þ@u xð Þ@x

(2.112)

Using the above equation, Equation (2.109) yields

@

@xEA xð Þ@u xð Þ

@x

� + q xð Þ¼ 0 0� x� L (2.113)

Assuming that the bar is fixed at the left end and free at the right end, the following boundary conditions

may associate the governing Equation (2.113):

u 0ð Þ¼ 0 (2.114)

u0 0ð Þ¼ ε 0ð Þ¼ 0 (2.115)

u0 Lð Þ¼ ε Lð Þ¼ F

E �A Lð Þ (2.116)

Total potential energy of a bar under axial loadingTotal potential energy is defined by:

Π¼U +W (2.117)


whereU is the strain energy of stresses or internal forces, andW is the energy possessed by the external

loads (e.g., concentrated forces and surface tractions).

According to the above definition, the strain energy U and the work W are given by the following

equations:

U¼ 1

2

ðL0

σεAdx (2.118)

and

W¼�Fu Lð Þ�ðL0

u xð Þq xð Þdx (2.119)

It should be noted that the external loads F and q(x) are always acting at their full value (i.e., their workis independent of the elastic behavior of the bar). Their movement through the corresponding displace-

ments u(L) and u(x) is doing work in amount F � u Lð Þ and u xð Þ � q xð Þdx, losing potential of equal

amounts �F � u Lð Þ and �u xð Þ � q xð Þdx, respectively.Since

ε¼ @u

@x(2.120)

and

σ¼Eε¼E@u

@x(2.121)

Equation (2.118) can be written as:

U¼ 1

2

ðL0

EA xð Þ u0ð Þ2dx (2.122)

Therefore, using Equations (2.119) and (2.122), Equation (2.117) yields

Π¼�Fu Lð Þ+ðL0

1

2EA u0ð Þ2�q � u

� dx (2.123)

Derivation of the total potential of a bar from the boundary value problemThe aim of this section is to explain the derivation of Equation (2.123) from Equation (2.113). To

achieve this target, we have to manipulate Equation (2.113) in order to derive an expression of the form

δ ⋯ð Þ¼ 0. To achieve this, mathematical manipulation starts by multiplying the differential Equa-

tion (2.113) by the variation �δu and integrating over the solution domain (for simplicity, we assume

that the cross-sectional area A is constant):ðL0

�EAu00 �qð Þδudx¼ 0, 0� x� L (2.124)

Let us write the above equation as

I1 + I2 ¼ 0 (2.125)

where

I1 ¼�ðL0

EAu00δudx (2.126)


and

I2 ¼�ðL0

qδudx (2.127)

Taking into account the following well-known formulaðba

wdv¼wvjba�ðba

vdw (2.128)

and setting

a¼ 0 (2.129)

b¼ L (2.130)

w¼ δu (2.131)

dv¼ u00dx or v¼ u0 (2.132)

the integral I1 can be written as:

I1EA

¼�u0δujL0 +ðL0

u0d δuð Þ (2.133)

or

I1EA

¼�u0 Lð Þδu Lð Þ+ u0 0ð Þδu 0ð Þ+ðL0

u0d δuð Þ (2.134)

Taking into account Equation (2.99), that is,

d δuð Þdx

¼ δdu

dx

�(2.135)

the quantity d(δu) is

d δuð Þ¼ δu0dx (2.136)

Then, Equation (2.134) yields

I1EA

¼�u0 Lð Þδu Lð Þ+ u0 0ð Þδu 0ð Þ+ðL0

u0δu0dx (2.137)

We recall from the boundary conditions (Equation 2.115) that u0 0ð Þ¼ 0. In contrast, u0 Lð Þ¼ ε Lð Þ. Then,I1EA

¼�ε Lð Þδu Lð Þ +ðL0

u0δu0dx (2.138)

Taking into account Equation (2.105), the product u0δu0 can be written as:

u0δu0 ¼ 1

2δ u0ð Þ2 (2.139)

Therefore, Equation (2.138) can now take the form:

I1 ¼�AEε Lð Þδu Lð Þ+AEðL0

1

2δ u0ð Þ2dx (2.140)


Using the boundary condition given in Equation (2.116), the product, AEε(L), is AEε Lð Þ¼F.

Then, Equation (2.140) yields

I1 ¼�F � δu Lð Þ +AEðL0

1

2δ u0ð Þ2dx (2.141)

Using Equation (2.104), Equation (2.141) can be written as:

I1 ¼�δ F � u Lð Þð Þ+ðL0

δ AE1

2u0ð Þ2

�dx (2.142)

According to Equation (2.100), Equation (2.142) yields

I1 ¼�δ F � u Lð Þð Þ + δðL0

AE1

2u0ð Þ2dx (2.143)

Using Equation (2.101), Equation (2.143) can be formulated as:

I1 ¼ δ �F � u Lð Þ +ðL0

1

2AE u0ð Þ2dx

� (2.144)

Combining Equation (2.144) with Equations (2.125) and (2.127), the following formula can be

obtained:

δ �F � u Lð Þ+ðL0

1

2AE u0ð Þ2dx

� �ðL0

qδudx¼ 0 (2.145)

Using Equation (2.104), Equation (2.145) yields


1

2AE u0ð Þ2dx

� �ðL0

δ quð Þ � dx¼ 0 (2.146)

According to Equation (2.100), Equation (2.146) can now be written as:

δ �F � u Lð Þ +ðL0

1

2AE u0ð Þ2dx

� �δ

ðL0

q � u � dx¼ 0 (2.147)

or


1

2AE u0ð Þ2�q � u

� dx

� �¼ 0 (2.148)

Equation (2.148) expresses the principle of MPE:

δΠ¼ 0 (2.149)

Therefore,

Π¼�F � u Lð Þ +ðL0

1

2AE u0ð Þ2�q � u

� dx (2.150)

Equation (2.150) is same as Equation (2.123).


REFERENCES[1] Deif AS. Advanced matrix theory for scientists and engineers. 2nd ed. London: Abacus press; 1991.

[2] Wunderlich W, Pilkey WD. Mechanics of structures: variational and computational methods. Boca Raton:

CRC press; 2003.

[3] Hartmann F. Green’s functions and finite elements. Heidelberg: Springer; 2013.

[4] Bhatti MA. Fundamental finite element analysis and applications. Hoboken: John Wiley & Sons; 2005.

[5] Logan DL. A first course in the finite element method. Boston, MA: Gengage Learning; 2012.

[6] Fish J, Belytschko T. A first course in finite elements. New York: Wiley; 2007.

[7] Boresi AP, Schmidt RJ. Advanced mechanics of materials. New York: John Wiley & Sons; 2003.

http://refhub.elsevier.com/B978-0-12-802386-0.00002-5/rf0010








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