Appendix A A Short Introduction to Tensor Algebra

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Appendix A

A Short Introduction to Tensor Algebra

A.1 Notational Conventions A.1.1 Some Vector Operations

Let a and b be two non-zero vectors. The scalar product s of the two vectors is denoted in the usual way:

s= a · b . (A.1.1)

The operation sign is the reason for calling it dot product.

The scalar product is a commutative vector operation: a· b = b · a.

If a· b = 0, then a and b are perpendicular to each other. This condition is known as condition of perpendicularity.

The zero vector is perpendicular to any other vector.

The cross product c of the vectors a and b is also denoted in the usual way:

c= a × b . (A.1.2)

The cross product is not commutative: a× b = −b × a.

If c= a × b = 0, then a and b are parallel to each other. This condition is that of the parallelism for a and b.

The zero vector is parallel to any other vector.

The dyadic product of two vectors is denoted by

W = a ◦ b . (A.1.3)

The left side of this equation is the name we have given to the product. We remark that any name can be given to a dyadic product. These names are, in general, typeset by boldface and italic letters.

G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8

379

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380 Appendix A: A Short Introduction to Tensor Algebra

Properties:

1. If we dot multiply it from left or right by c, we get

W · c = (a ◦ b) · c = a (b · c) , (A.1.4a) c· W = c · (a ◦ b) = (c · a) b . (A.1.4b) Let d be a non-zero vector. With regard to (A.1.4a) and (A.1.4b), it is clear that the product

d· W · c = d · (a ◦ b) · c = (d · a) (b · c) (A.1.4c) is a scalar.

2. Equations (A.1.4) show that the dyadic product is not a commutative operation.

3. The dot product of the two dyadic products a◦ b and c ◦ d is also a dyadic product defined by the relation

(a ◦ b) · (c ◦ d) = (a ◦ d) b · c . (A.1.5) 4. The inner product a◦ b and c ◦ d is a scalar:

(a ◦ b) · · (c ◦ d) = (a · c) (b · d) . (A.1.6) 5. If we cross multiply W from left or right by c, we get two dyadic products:

W× c = (a ◦ b) × c = a ◦ (b × c) , (A.1.7a) c× W = c × (a ◦ b) = (c × a) ◦ b . (A.1.7b) The box product of the vectors a, b, and c is defined by the following equation:

[abc]= (a × b) · c. (A.1.8)

The box product has the following properties:

1. The placement of the operation signs is interchangeable:

(a × b) · c = a · (b × c) . (A.1.9a) 2. Cyclic interchangeability:

[abc]= [bca] = [cab] = − [bac] = − [cba] = − [acb] . (A.1.9b) If the three vectors are measured from the same point, the box product results in the volume of the parallelepiped determined by the three vectors.

The box product is zero if the three vectors lie in the same plane.

The triple cross products of the vectors a, b, and c can be expended as a× (b × c) = (a · c) b − (a · b) c , (a × b) × c = (a · c) b − (b · c) a .

(A.1.10)

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Appendix A: A Short Introduction to Tensor Algebra 381

A.1.2 Base Vectors

Let g1, g2, and g3be three non-zero vectors. It will be assumed that

g₁g₂g₃

= γ◦= 0, γ^∗_◦ = 1/γ◦. (A.1.11) Then, the three vectors (if measured from the same point) have no common plane. It is obvious that any vector v can be given in the form

v= v¹g¹+ v²g2+ v³g3, (A.1.12) where v ( = 1, 2, 3) is the component of v in the direction g. This statement follows form the fact that the above equation system, which is detailed here

⎡

⎣vx

vy

vz

⎤

⎦ =

⎡

⎣g_x1g_x2g_x3 g_y1 g_y2 g_y3 g_z1 g_z2 g_z3

⎤

⎦

⎡

⎣v¹ v² v³

⎤

⎦ , (A.1.13)

always has a unique solution forv. The triplet gis called basis in the 3D space since any vector v can be given in terms of g. The dual basis (The dual base vectors) is (are) defined by the following equations:

g^∗₁= g₂× g3

γ_◦ , g^∗₂ =g₃× g1

γ_◦ , g₃^∗=g₁× g2

γ_◦ . (A.1.14a)

It can be shown that g1= g^∗₂× g₃^∗

γ_◦^∗ , g2=g^∗₃× g^∗₁

γ_◦^∗ , g3= g^∗₁× g^∗₂

γ_◦^∗ . (A.1.14b) The base vectors gkand g^∗have the following properties:

gk· g^∗=

1 if k= ,

0 if k= . k, = 1, 2, 3 (A.1.15) For example

g₁· g^∗1 = g1· g2× g3

γ_◦ =

g1g2g3

γ_◦ =γ_◦

γ_◦ = 1 , or

g2· g^∗₃= g2·g1× g2

γ_◦ = g1× g2

γ_◦ · g2=

g₁g₂a₂

γ_◦ = g1· (g2× g2) γ_◦ = 0 .

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Since

ix× iy= iz, iy× iz= ix, iz× ix= iy, and ixiyiz

= 1, (A.1.16a)

it follows that

ix = i^∗x, iy= i^∗y, iz = i^∗z. (A.1.16b) This means that the basis and the dual basis is the same in the Cartesian coordinate system. If we utilize relation (A.1.15), we get from Eq. (A.1.12) that

v= v · g^∗. (A.1.17)

In the Cartesian coordinate system(xyz), it also holds that

vm= v · im, m= x, y, z . (A.1.18)

A.2 Tensors

A.2.1 Classification

A scalar field, which describes a physical state of a body, is said to be a tensor field of order zero. (For example, the temperature field in a solid body.) A vector field, which describes a physical state of a body, is said to be a tensor field of order one.

(For example, the displacement field of a solid body.)

A common property of tensors of order zero and one is that they are independent of the coordinate system, i.e., they do not change if we translate and rotate the coordinate system (if the coordinate system performs a rigid body motion).

In the following subsection, we shall deal with tensors of order two.

A.2.2 Homogeneous Linear Vector-Vector Functions:

Tensors of Order Two

Consider the vector-vector function

w= f(v), (A.2.19)

in which w is the dependent variable and f denotes the prescription made on the independent variable v. Function (A.2.19) is a mapping of the vectors v (measured from the origin O_v) onto the vectors w (measured from the origin O_w—see Fig.A.1):

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v

O_v O_w

w y x

z z

x

y

object vector

image vector

Fig. A.1 Mapping of v onto w

The vector-vector function w= f(v) is a homogeneous linear function if the following functional equation is satisfied:

w= f(v) = f(vxix+ vyiy+ vyiz) = vxf(i x)

w_x

+ vyf(iy)

w_y

+ vzf (iz)

wz

=

= wxvx+ wyvy+ wzvz, (A.2.20) where

wx= wx xix+ wyxiy+ wzxiz, wy= wx yix+ wyyiy+ wzyiz, wz= wx zix+ wyziy+ wzziz

(A.2.21)

are the images of the base vectors ix, iy, and iz.

Mapping (A.2.20) is uniquely determined if we know the nine scalars wx x, wyx, . . . , wzz.

Mapping (A.2.20) is degenerated if the 3D space (the vectors v) is (are) mapped onto a plane, or a straight line, or the origin O_w.

Since

vxwx = wx(i x· v)

vx

= (wx◦ ix) · v ,

vywy= wy(i y· v)

vy

= (wy◦ iy) · v ,

vxwz = wz(i z· v)

vz

= (wz◦ iz) · v,

(A.2.22)

we can rewrite Eq. (A.2.20) in the following form:

w=

wx◦ ix+ wy◦ iy+ wz◦ iz

· v, (A.2.23)

where the sum

W = wx◦ ix+ wy◦ iy+ wz◦ iz (A.2.24) is called tensor of order two (or simply tensor). With the knowledge of W mapping (A.2.20) takes the form

w= W · v . (A.2.25)

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Because of its definition mapping (A.2.20) is independent of the coordinate sys- tem. Hence, so is the tensor W . The sum (A.2.24) is, however, the presentation of the tensor W in the Cartesian coordinate system(xyz).

It is obvious that the image vectors wx, wy, and wzcan be given in terms of W as wx = W · ix, wy= W · iy, wz = W · iz. (A.2.26) If we dot multiply these equations by imand take relations (A.1.4c) and (A.2.21) into account, we get

wmn = im· W · in, m, n = x, y, z . (A.2.27)

We remark that the simple dyad (A.1.3) is also a tensor:

A= a ◦ b = a ◦

bxix+byiy+bziz

= ab x image of ix

◦ ix+ aby

image of iy

◦ iy+ ab z image of iz

◦ iz.

(A.2.28) Observe that tensor a◦ b is degenerated since the image vectors A · c = (a ◦ b) · c= a (b · c) are all parallel to a independently of the value c has.

A.2.3 Special Tensors

The zero tensor 0, for which the images of the unit vectors ix, iy, and izare all zero vectors, maps any vector onto a zero vector.

The unit tensor 1 maps each vector onto itself. It follows from the transformation v= vxix+ vyiy+ vziz = ix(ix· v) +i^y

iy· v

+ iz(iz· v) =

=

ix◦ ix+ iy◦ iy+ iz◦ iz

· v

that

1= ix◦ ix+ iy◦ iy+ iz◦ iz. (A.2.29) Consider the vector triplet

n, |n| = 1 , nk· n=

1 if k= ,

0 if k= k, = 1, 2, 3 . (A.2.30) Since the unit vectors n1, n2, and n3are mutually perpendicular to each other, they are said to [constitute] (determine) [an orthonormal basis] (a Cartesian coordinate system) in the 3D space. It is obvious that the unit tensor assumes the form

1= n1◦ n1+ n2◦ n2+ n3◦ n3 (A.2.31) in this basis.

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The inverse of the tensor

W = wx◦ ix+ wy◦ iy+ wz◦ iz, wo= wxwywz

= 0

is denoted by W⁻¹and is defined by the following equation:

W· W⁻¹= W⁻¹· W = 1 . (A.2.32) Sincewo=

wxwywz

= 0, there is a dual base for the image vectors wx, wy, and wz:

w^∗_x =wy× wz

wo

, w^∗y= wz× wx

wo

, w^∗z = wx× wy

wo

. (A.2.33)

The inverse W⁻¹can be given in terms of the dual base:

W⁻¹ = ix◦ w^∗x+ iy◦ w^∗y+ iz◦ w^∗z. (A.2.34) Proof

W⁻¹·W =

ix◦ w^∗x+ iy◦ w^∗y+ iz◦ w^∗z

·

wx◦ ix+ wy◦ iy+ wz◦ iz

=

= ix◦ ix

w^∗_x· wx

=1

+ ix◦ iy

w^∗_x· wy

=0

+ ix◦ iz

w^∗_x· wz

=0

+

+ iy◦ ix

w^∗_y· wx

=0

+ iy◦ iy

w^∗_y· wy

=1

+ iy◦ iz

w^∗_y· wz

=0

+

+ iz◦ ix

w^∗_z · wx

=0

+ iz◦ iy

w_z^∗· wy

=0

+ iz◦ iz

w^∗_z· wz

=1

=

= ix◦ ix+ iy◦ iy+ iz◦ iz= 1 .

The transpose of the tensor W is defined by the following equation:

W^T = ix◦ wx+ iy◦ wy+ iz◦ wz. (A.2.35) Note that

W· v = wx(ix· v) + w^y iy· v

+ wy

iy· v

=

= (v · ix) wx+ v· iy

wy+ (v · iz) wz = v · W^T.

If we dot multiply the above equation by the vector u, we get the following relation:

u· W · v = v · W^T · u, (A.2.36) which should hold for any v and u.

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Let p and q be two non-zero vectors. Further, let S= u ◦ v and T = p ◦ q be two tensors. Some properties of the transpose are detailed below:

W^TT

= W , (A.2.37a)

(S · T)^T = T^T· S^T. (A.2.37b) Property (A.2.37a) is obvious. As regards property (A.2.37a), compare the following two equations:

(S · T)^T =

(u ◦ v) · (p ◦ q)T

=

(v · p) u ◦ qT

= (v · p) q ◦ u ,

T^T · S^T = (q ◦ p) · (v ◦ u) = (v · p) q ◦ u . The tensor W is symmetric if

W = W^T. (A.2.38)

For a symmetric W , Eq. (A.2.36) yields

v· W · u = u · W · v . (A.2.39) If we write imfor v, infor u(m, n = x, y, z) in the above equation and take (A.2.27) into account, we get for the symmetric tensor W that

wmn= im· W · in = in· W · im = wnm. (A.2.40) The tensor W is skew if

W = −W^T. (A.2.41)

For the skew tensor W , it follows from Eq. (A.2.36) that

v· W · u = −u · W · v . (A.2.42) If we now write imfor v, infor u(m, n = x, y, z) in (A.2.42), and take (A.2.27) into account, we obtain for the skew tensor W that

wmn = im· W · in= −in· W · im= −wnm hence wmm= 0 . (A.2.43) The tensor W is said to be

positive definite positive semidefinite negative semidefinite negative definite

⎫⎪

⎪⎬

⎪⎪

⎭ if, for any u,

⎧⎪

⎪⎨

⎪⎪

⎩

u· W · u > 0 u· W · u ≥ 0 u· W · u ≤ 0 u· W · u < 0

.

and it is indefinite if none of the above relations is satisfied.

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It follows from the identity

W =1 2

W+ W^T +1

2

W− W^T

(A.2.44) that any tensor can be resolved into symmetric and skew parts:

W = Wsym+ Wskew

Wsym= 1 2

W+ W^T

, Wskew=1 2

W− W^T

. (A.2.45)

This theorem is that of the additive resolution.

A.2.4 Axial Vector

Consider the mapping that belongs to W_skew. We can write

W_skew· v = 1 2

W− W^T

· v =

=1 2

wx◦ ix+ wy◦ iy+ wz◦ iz− ix◦ wx+ iy◦ wy+ iz◦ wz

· v =

= −1 2

ix(wx·v)−w^x(ix·v)+i^y wy·v

−wy

iy·v

+iz(wz·v)−w^z(iz·v)

=

= ↑

(A.1.10)

= −1 2

wx× ix+ wy× iy+ wz× iz

× v . (A.2.46)

Here the vector

wa= −1 2

wx× ix+ wy× iy+ wz× iz

(A.2.47)

is called axial vector. With wait holds that

W_skew· v = wa× v . (A.2.48)

It is obvious that

(a) the axial vector wais independent of the coordinate system,

(b) mapping (A.2.48) is degenerated since the image vector is always perpendicular to the axial vector wa.

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By utilizing the cross products

wx× ix= wzxiy− wyxiz, wy× iy= wx yiz− wzyix, wz× iz= wyzix− wx ziy

we get

wa = waxix+ wayiy+ waziz, wax=−1

2

wyz−wzy

, way=−1

2(wzx−wx z) , waz=−1 2

wx y−wyx

.

(A.2.49) If the tensor W is symmetricwmn = wnm(m, n = x, y, z)—see (A.2.40). Hence, it follows from (A.2.49) that the axial vector of a symmetric tensor is zero vector.

A.2.5 Tensors and Matrices

Recalling Eqs. (A.2.25) and (A.2.20), we can write

W· v = wxvx+ wyvy+ wzvz. (A.2.50) By introducing the column matrices

w

(3×1)=

⎡

⎣wx

wy

wz

⎤

⎦ , wx (3×1)=

⎡

⎣wx x

wyx

wzx

⎤

⎦ , wy (3×1)

=

⎡

⎣wx y

wyy

wzy

⎤

⎦ , wz (3×1)=

⎡

⎣wx z

wyz

wzz

⎤

⎦ , (A.2.51) we can manipulate Eq. (A.2.50) into the following form:

w

(3×1)=

⎡

⎣wx

wy

wz

⎤

⎦ =

⎡

⎣wx x

wyx

wzx

⎤

⎦ vx+

⎡

⎣wx y

wyy

wzy

⎤

⎦ vy+

⎡

⎣wx z

wyz

wzz

⎤

⎦ vz =

=

⎡

⎣wx x wx y wx z

wyx wyy wyz

wzx wzy wzz

⎤

⎦

W

(3×3)

⎡

⎣vx

vy

vz

⎤

⎦

v

(3×1)

(A.2.52)

or shortly

w

(3×1)= w_x

(3×1)| wy (3×1)

| wz (3×1)

W

(3×3)

v

(3×1)= W

(3×3) v

(3×1), (A.2.53)

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where the matrix W is that of the tensor W in the Cartesian coordinate system(xyz).

It follows from (A.2.28) that the matrix of the tensor A= a ◦ b is of the form

A

(3×3)=

⎡

⎣axbx axbyaxbz

aybx aybyaybz

azbx azby azbz

⎤

⎦ =

⎡

⎣ax

ay

az

⎤

⎦ bx| by| bz

= a

(3×1) b

(1×3)

T = a b^T. (A.2.54) This means that the matrix product

a

(3×1) b

(1×3)

T is equivalent to the dyadic product a◦ b (A.2.55)

if we use matrix notation in the Cartesian coordinate system(xyz).

Consider now the dot product v= u · W = ↑

(A.2.24)= u · w x vx

ix+ u · wy

vy

iy+ u · w z vz

iz. (A.2.56)

It can be checked with ease that the matrix product

v

(1×3) T =

vx| vy|vz

=

ux| uy|uz

⎡

⎣wx x wx y wx z

wyx wyy wyz

wzx wzy wzz

⎤

⎦ = u

(1×3)

T W

(3×3) (A.2.57) yields the very same result as a row matrix.

We shall proceed with the cross product u× W = ↑

(A.1.7b)= u × w x image of ix

◦ ix+ u × wy

image of iy

◦ iy+ u × w z image of iz

◦ iz (A.2.58)

which results in a tensor. If we recall Eqs. (1.14) and introduce the matrix

u_×

(3×3)=

⎡

⎣ 0 −uz uy

uz 0 −ux

−uy ux 0

⎤

⎦ , (A.2.59)

we arrive at the result that the matrix of the tensor

u× W (A.2.60a)

can be obtained by performing the following matrix product:

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u× W

(3×3) = u_×

(3×3)

W

(3×3)=

⎡

⎣ 0 −uz uy

uz 0 −ux

−uy ux 0

⎤

⎦

⎡

⎣wx x wx y wx z

wyx wyy wyz

wzx wzy wzz

⎤

⎦ . (A.2.60b)

Let T and S be two tensors:

T = tx◦ ix+ ty◦ iy+ ty◦ iy, S= sx◦ ix+ sy◦ iy+ sy◦ iy. (A.2.61) Making use of (A.1.5), we can determine the dot product of these tensors:

T· S =

txsx x + tysyx+ tzszx

the first column in the matrix T·S

(3×3)

◦ ix+

txsx y+ tysyy+ tzszy

the second column in the matrix T·S

(3×3)

◦ iy+

+

txsx z+ tysyz+ tzszz

the third column in the matrix T·S

(3×3)

◦ iz. (A.2.62)

We can read off from this equation that the matrix of the dot product T · S is given by the following matrix product:

T· S

(3×3) = T

(3×3) S

(3×3)=

⎡

⎣tx x tx y tx z

tyx tyy tyz

tzx tzy tzz

⎤

⎦

⎡

⎣sx x sx y sx z

syx syy syz

szx szy szz

⎤

⎦ . (A.2.63)

It also holds that

(T · S)

(3×3) T =

(3×3)T S

(3×3)

T

= S(1×3)

T T

(1×3)

T. (A.2.64)

The determinant of a tensor is that of its matrix.

A.2.6 Eigenvalue Problem of Symmetric Tensors

Let W be a symmetric tensor. Find the direction n for which it holds that the unit vector (Fig.A.2)

n= nxix+ nyiy+ nziz; n²_x+ n²y+ n²z = 1, (A.2.65) which belongs to the direction n satisfies the relation

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wn = W · n = λn. (A.2.66)

which means that the object vector and the image vector are parallel to each other.

Note that here the scalar λ—which is referred to as eigenvalue—and the vector n (nx, nyand nz)—which is called eigenvector—are the unknowns.

Fig. A.2 Parallel object and image vectors

O_w=O_v y z

wn

x

n

Since the unit tensor 1 maps each vector onto itself, we can rewrite Eq. (A.2.66) in the following form:

(W − λ1) · n = 0 . (A.2.67)

If we now utilize the matrices of W , 1, and n, we get a homogeneous linear equation

system: ⎡

⎣wx x − λ wx y wx z

wyx wyy− λ wyz

wzx wzy wzz− λ

⎤

⎦

W−λ1

⎡

⎣nx

ny

nz

⎤

⎦

n

= 0. (A.2.68)

Let P₃(λ) = − det W−λ1

. This function is a cubic polynomial of λ. Solution for n different from the trivial one exists if and only if

P3(λ) = −

wx x − λ wx y wx z

wyx wyy− λ wyz

wzx wzy wzz− λ

= λ³− WIλ²+ WI Iλ − WI I I = 0, (A.2.69) where

WI = wx x + wyy+ wzz , (A.2.70a) WI I =

wx x wx y

wyx wyy

+

wx x wx z

wzx wzz

+

wyy wyz

wzy wzz

(A.2.70b)

and

WI I I =

wx x wx y wx z

wyx wyywyz

wzx wzy wzz

 . (A.2.70c)

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Since mapping (A.2.66) is independent of the coordinate system, the solutions for λ should also be coordinate system independent quantities. This requirement is satisfied if and only if the coefficients WI, WI I, and WI I Iare also independent of the coordinate system. For this reason, they are called scalar invariants of the tensor W . The most important results for the eigenvalue problem considered are as follows [1]:

Assume that the three eigenvalues are ordered: λ₁≥ λ2≥ λ3. Assume further the eigenvectors are attached to the same point, i.e., to the point where the tensor W is defined.

(a) If the three eigenvalues are different, there exist three eigenvectors n1, n2, and n3which are mutually perpendicular to each other. Therefore, they constitute an orthonormal basis in the 3D space in which

W =

3

=1

λn◦ n (A.2.71)

is the form of the tensor.

(b) If λ1is a single root and λ2is a double root, then the eigenvector n1is unique while any direction in the plane perpendicular to n1is a principal direction. Select the vectors n2and n3perpendicularly to each other and n1. Then the vectors n1, n2, and n3constitute a basis in the 3D space and

W = λ1n1◦ n1+ λ2(n2◦ n2+ n3◦ n3) = ↑

(A.2.31)=

= (λ1− λ2) n1◦ n1+ λ2 (n1◦ n1+ n2◦ n2+ n3◦ n3)

1

=

= (λ1− λ2) n1◦ n1+ λ21 (A.2.72) is the form of the tensor W in the basis mentioned. Here, we have not assumed that the roots are ordered.

(c) If the three roots are the same denoted, say, by λ1, then every direction is a principal direction. This means that we can select again an orthonormal basis n1, n2, and n3in which

W = λ1(n1◦ n1+ n2◦ n2+ n3◦ n3) = λ11 (A.2.73) is the form of the tensor W .

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Appendix B

Some Useful Relationships

B.1 The Symbolic Dirac Function and Its Most Important Properties

B.1.1 The Dirac Function in One Variable

Let f(x) be a continuous function in the neighborhood of the point P1with coordinate x= 1. Further let a < b. The Dirac function of the variable x is defined by the following equations [2,3] :

δ(x − 1) =

0 if x= 1,

∞ if x− 1= 0 (B.1.1)

and

b a

f(x) δ(x − 1) dx =

⎧⎪

⎪⎨

⎪⎪

⎩

0 if 1 /∈ [a, b] , 1

2f(1) if 1= a or 1= b , f(1) if 1∈ (a, b) ,

(B.1.2)

which shows that the Dirac¹functions tend to infinity for x= 1in such a way that relations (B.1.2) are also satisfied.

The Dirac function is not a real function, and its integrals do not exist in the traditional sense of the integral concept of the classical analysis. Instead, it is such a generalized function, or symbol, which can be used to calculate the value f(1) of the function f(x) at x = 1by performing a formal integral transformation. The latter property of the Dirac function is very useful since it makes possible to model a concentrated load as if it were a distributed one.

1Paul Adrien Maurice Dirac (1902–1984).

393

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394 Appendix B: Some Useful Relationships

Since the independent variable x belongs to a point on the x axis, denoted, say, by P—the point P is naturally not fixed—and the coordinate1= xQbelongs to a fixed point denoted, say, by Q we can rewrite Eq. (B.1.2) in the following form:

b a

f(P) δ(P − Q) dx =

⎧⎪

⎪⎨

⎪⎪

⎩

0 if Q /∈ [a, b]

1

2 f(Q) if Q coincides with a or b, f(Q) if ξ1Q∈ (a, b)

(B.1.3)

where the arc element dsP taken at the point P corresponds to dx.

B.1.2 The Dirac Function in Two Variables

Let P and Q be two points in the coordinate plane(x, y) not necessarily different from each other. The corresponding position vectors are denoted by rP = xix+ yiy

and rQ= xQix+ yQiy. The Dirac delta function in the coordinate plane(x, y) is defined by the following equation:

δ(P − Q) = δ(x − xQ)δ(y − yQ) . (B.1.4) Let S be a plane region. Further let f(x, y) = f (P) be a continuous function in two variables in the coordinate plane(x, y). Then it follows from (B.1.2) that

S

f(P) δ(P − Q) sAP =

0 if Q /∈ S ,

f(Q) if Q∈ S . (B.1.5)

B.1.3 The Dirac Function in the 3D Space

Definition of the Dirac function in the 3D space is basically the same as its definition in the 2D space. Let P and Q two points in the 3D space with coordinates x, y, z and xQ, yQ, zQ. Further let f(Q) be a continuous function in the 3D space. Then it holds that

V

f(P) δ(P − Q) dVP =

0 if Q /∈ V ,

f(Q) if Q∈ V (B.1.6)

where VPis a volume region of the 3D space.

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Appendix B: Some Useful Relationships 395

B.2 Bessel Functions—Fundamental Relations

Let us denote the Bessel²functions Jn(x) and Yn(x) uniformly by Hn(x). The following relations hold:

d H₀(x)

d x = −H1(x) , (B.2.7a)

d H1(x) d x = 1

x(H0(x) x − H1(x)) = H0(x) − 1

xH₁(x) = 1

2(H0(x) − H2(x)) , (B.2.7b) d Hn(x)

d x =1

2(Hn−1(x) − Hn+1(x)) , (B.2.7c)

2n

x Hn(x) = Hn−1(x) + Hn+1(x) (B.2.7d)

and

Yn(x)Jn+1(x) − Yn+1(x)Jn(x) = 2

πx . (B.2.7e)

See [4, 1977a] a for more details.

As regards the Bessel functions In(x) and Kn(x) use has been made of the following relations:

d I₀(x)

d x = I1(x) , d K₀(x)

d x = −K1(x) , (B.2.8a)

d I₁(x) d x = 1

x(x I0(x) − I1(x)) = I0(x) −1

xI₁(x) =1

2(I0(x) + I2(x)) , (B.2.8b) d I_n(x)

d x = 1 2

I_n₋₁(x) + In+1(x)

, (B.2.8c)

2n

x In(x) = In−1(x) − In+1(x) , (B.2.8d)

d K₁(x) d x = −1

x (x K0(x) + K1(x)) = −K0(x) −1

xK₁(x) = −1

2(K0(x) + K2(x)) , (B.2.8e) d K_n(x)

d x = −1 2

K_n₋₁(x) + Kn+1(x)

, (B.2.8f)

−2n

x In(x) = Kn−1(x) − Kn+1(x) (B.2.8g)

and

In(x)Kn+1(x) + In+1(x)Kn(x) = 1

x . (B.2.8h)

See [4, 1977b] for more details.

2Friedrich Wilhelm Bessel (1784–1846).

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B.3 Gaussian Quadrature Abscissae and Weights

This section contains the tabulations for calculating integral approximations in the following form³:

J(φ) =

_η=1

η=−1φ(η) dη =

n j=1

wjφ(ηj) , (B.3.9)

in whichwjis the weight and ηjis the abscissa where the function to be integrated should be taken. The datawj and ηj in TablesB.1,B.2,B.3,B.4,B.5andB.6are presented for n= 4, 5, 6, 7, 8, 10.

Table B.1 Weights and abscissae of the Gauss points if n= 4 j Abscissa ηi Weightwi

1 0.6521451548625461 -0.3399810435848563 2 0.6521451548625461 0.3399810435848563 3 0.3478548451374538 -0.8611363115940526 4 0.3478548451374538 0.8611363115940526 Table B.2 Weights and abscissae of the Gauss points if n= 5

j Abscissa ηi Weightwi

1 -0.9061798459386640 0.2369268850561891 2 -0.5384693101056831 0.4786286704993665 3 0.0000000000000000 0.5688888888888889 4 0.5384693101056831 0.4786286704993665 5 0.9061798459386640 0.2369268850561891 Table B.3 Weights and abscissae of the Gauss points if n= 6

1 -0.9324695142031521 0.1713244923791704 2 -0.6612093864662645 0.3607615730481386 3 -0.2386191860831969 0.4679139345726910 4 0.2386191860831969 0.4679139345726910 5 0.6612093864662645 0.3607615730481386 6 0.9324695142031521 0.1713244923791704

3These data are taken from the homepagehttps://pomax.github.io/bezierinfo/legendre-gauss.html.

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Appendix B: Some Useful Relationships 397

Table B.4 Weights and abscissae of the Gauss points if n= 7

1 -0.9491079123427585 0.129484966168869 2 -0.7415311855993945 0.2797053914892766 3 -0.4058451513773972 0.3818300505051189 4 0.000000000000000 0.4179591836734694 5 0.4058451513773972 0.3818300505051189 6 0.7415311855993945 0.2797053914892766 7 0.9491079123427585 0.129484966168869

1 -0.9602898564975363 0.1012285362903763 2 -0.7966664774136267 0.2223810344533745 3 -0.5255324099163290 0.3137066458778873 4 -0.1834346424956498 0.3626837833783620 5 0.1834346424956498 0.3626837833783620 6 0.5255324099163290 0.3137066458778873 7 0.7966664774136267 0.2223810344533745 8 0.9602898564975363 0.1012285362903763

1 -0.9739065285171717 0.0666713443086881 2 -0.8650633666889845 0.1494513491505806 3 -0.6794095682990244 0.2190863625159820 4 -0.4333953941292472 0.2692667193099963 5 -0.1488743389816312 0.2955242247147529 6 0.1488743389816312 0.2955242247147529 7 0.4333953941292472 0.2692667193099963 8 0.6794095682990244 0.2190863625159820 9 0.8650633666889845 0.1494513491505806 10 0.9739065285171717 0.0666713443086881

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B.4 Shear Correction Factor

A few shear correction factors are presented on the basis of [5,6] in TableB.7where ν is the Poisson number:

Table B.7 Shear correction factors

Cross section Name κy

y z

Rectangle 10(1 + ν)

12+ 11ν

y z

Thin-walled 20(1 + ν)

48+ 39ν Square Tube

y z

Circle 6(1 + ν)

7+ 6ν

R z y r

Hollow Cylinder 6(1 + ν)(1 + r/R)²

(7 + 6ν)(1 + r/R)²+ 4(5 + 3ν)(r/R)²

z

y Thin-walled 2(1 + ν)

4+ ν Circular Tube

y z

Semi-circle 1+ ν

1.305 + 1.273ν

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Appendix C

Solutions to Selected Problems

C.1 Problems in Chap. 1

Problem1.1The proof is presented by using tensorial notations. FigureC.1shows the mass center G, the position vector ρ of the mass element with respect to the mass center G, the direction vectors n and m for which it holds that|n| = |m| = 1, n· m = 0, ρ · n = ρnandρ · m = ρm.

Fig. C.1 Resolution ofρ into two components

Recalling (1.22) for the product (1.17d)1we can write n· JG· n=

V

n·

ρ²1−ρ ◦ ρ

·n ρdV dm

=

V

ρ²n· 1 · n n

− n · ρ ρn

◦ ρ · n ρn

dm= ↑ n·n=1=

=

V

ρ²− ρ²n

dm= ↑

ρ²−ρ²n=ρ²m

=

V

ρ²_mdm= Jn,

which is really the moment of inertia with respect to the axis n.

For the product (1.17d)₂, a similar line of thought yields

−m· JG· n = −

V

m·

ρ²1− ρ ◦ ρ

· n ρdV

dm

=

= −

V

ρ²m· 1 · n

n

− m · ρ

ρm

◦ ρ · n

ρn

dm= ↑

m·n=0=

V

ρmρndm= Jmn,

399

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400 Appendix C: Solutions to Selected Problems

which is the product of inertia with respect to the axes n and m.

Problem1.2The moment of momentum about the point A is

HA=

V

ˆρ × v ρdV in which

v= vA+ ω × ˆρ . Hence

HA =

V

ˆρ ×

vA+ ω × ˆρ ρdV =

V

ˆρ ρdV × vA+

V

ˆρ × ω × ˆρ

ρdV, (C.1.1)

where

V

ˆρ ρdV × vA= ↑

(1.7)= Q A mr_AG

× vA= mrAG× vA (C.1.2a)

and

V

ˆρ × ω × ˆρ

ρdV =

V

ˆρ²ω − ˆρ ˆρ · ω

ρdV =

=

V

ˆρ²1− ˆρ ◦ ˆρ

ρdV · ω . (C.1.2b)

In this equation, 1 is the unit tensor and◦ stands for the operation sign of a dyadic product. Integral

JA =

V

ˆρ²1− ˆρ ◦ ˆρ ρdV

defines the tensor of inertia at A. Using the tensor of inertia JA, integral (C.1.2b) can be given in the form

V

ˆρ × ω × ˆρ

ρdV = JA · ω . (C.1.2c)

Comparison of Eqs. (C.1.1), (C.1.2a) and (C.1.2c) yields the equation to be proved in tensorial notions:

HA= mrAG× vA+ JA · ω. (C.1.3) Problem 1.3To prove the parallel axis theorem rewrite Eq. (C.1.2c) by taking in account that ˆρ = rAG + ρ. We obtain