Please follow the pdf’s index, and why the index shown doesn’t match this 2-d function….42

CHAPTER 4. VECTORS AND TENSORS

where A, B are the projections of Ā onto the coordinate axes. The right

panel of figure 4.1 shows the same vector Ā and an orthogonal Cartestian

coordinate system, but now this system has been rotated relative to the

system in the left panel. In this new system, designated with a prime,

Ā= A’ + B’j’

Thus

Aſ + Bj = A’i’ + B’j’.

(4.2)

Let the primed axes be rotated by an angle & from the unprimed axes. Each

primed base vector can be expressed in terms of unprimed base vectors, as

in figure 4.2,

히 = cos di + sin 07,

j’= – sin 67 + cos 6j.

Using these in (4.2) produces

A = A’ cos 0 – B’ sin ,

B = A’ sin 6 + B’ cos 0.

Matrix notation is sometimes handy:

A

cose – sin

sin e

] ()

cos A

This can be inverted to produce

Α’

B’

() = [

cos sin

– sin cos

30] ()

While useful here, matrix notation becomes cumbersome for higher dimen-

sional cases, and will generally be avoided in favor of tensor notation.

4.2

More standard vector transformation

The notation of the previous section, where each component of A has its

own symbol, is cumbersome. Instead we choose tensor notation, also called

indicial notation, where the components of Ā are A1, A2, glibly refered to as

Aj. Correspondingly, the base vectors will now be called ēl, ēz instead of 7,j.

4.2. MORE STANDARD VECTOR TRANSFORMATION

43

ſ’

او

Figure 4.2: Base vector conversion.

AX2

AR

012)

022

X’i

021

011

11

Figure 4.3: Base vector conversion.

In order to generalize the transformation, it is convenient to refer to bij’

the angle between ē; and ēj, shown in figure 4.3. In two dimensions, all

these angles shown can be related to the angle 6 of the previous section, e.g.,

011′ = 6221 = 0, 021 = 1 – 0, etc. The cosine of these angles are the direction

cosines,

lij’

= cos Oija.

As above, the vector Ã may be expressed with components in either the

unprimed or primed system,

Ā= Ajēi + Azē2 = Alē + Alsē.

thus

Ajē; = A’e’,

remembering to use Einstein’s summation convention. Each primed base

vector can be expressed in terms of unprimed base vectors, as in figure 4.2,

7 = cos 011 ēj + cos 6211 ēm,

ē = cos 012, ēi + cos 622, 72,

44

CHAPTER 4. VECTORS AND TENSORS

012

o’11

011

012

021

022

Figure 4.4: Differential triangle.

Finally

A; = A’ cos bij!.

Again, the components A’

; are converted into A; with only the angles between

coordinate axes. This transformation rule may be expressed as

A; = lijı A’;;

which also applies to three dimensions.

4.3

Transformation of stress

As discussed in a previous chapter, the stress on an oblique face of a differ-

ential tetrahedron is found with a balance of forces on the tetrahedron. We

will consider the two-dimensional case first, which the tetrahedron is merely

a triangle. Choose the oblique face of the tetrahedron to be oriented with

one of the primed axes, e.g., choose the normal to be parallel to ē’, making

the normal and tangential stress components on this oblique face the same

as o’11 and 0-12, as shown in figure 4.4.

Stress is converted to the primed coordinate system by balancing the

forces in a convenient direction. Choose the x’ı direction, for example, and

remember that stress must be multiplied by area to get force:

011A1 – 011Aj cos 011 – 012A1 cos 621 021 A2 cos 0 12 – 022 A2 cos 6221 = 0,

4.4. EXERCISES

45

where A1, A2, Aſ are the three areas in figure 4.4. These areas are geometri-

cally related:

A1 = Aj cos 011′, A2 = Aſ cos 012′.

The result is

011 = 011 cos 011′ cos 011′ + 012 Cos 011cos 621

+021 Cos 012, cos 012+022 Cos 012 cos 622. (4.3)

This transformation is the same as

d’ab = lailB;’O ij.

This expression again is valid for three dimensions.

SI 5 wolus or less

Chapter 4 Vectors and Tensors

[1] Conaider a transformation that consists of a fototion of 180 degrees about the x-ars as shown in figure

(a) Find the direction Cosines.

Rotation of 180 degrees:

&

L

=

O

1 оо

Г1

1 o

O COSB Sino = O COBI80°Sinis

O Sin180° cos180°

Y’ –

Y

o-sino Cose

d 것

=

1 o

0

10=L

о

=LT

(b) Presume that a differential element in a state of plane stress, where Oxx = Oxy = Ox2=0, has value of

Oxy=10 0yz=2 8zz = 5

– 5

in some units. Find the stress components in the primed systein by performing a tensor transforma

1

Sirice Oxx = Sxy=0xz =O , we have

0

To ool

6=0

LO

og

1

Trio

o o=LTOL= 0 -1 0 0 2

0-111025 oo-1

2

5

2

La

2 5

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