Solutions for Fundamentals of Continuum Mechanics · Solutions for Fundamentals of Continuum...

Solutions for Fundamentals of Continuum

Mechanics

John W. Rudnicki

December 2, 2015

Contents

I Mathematical Preliminaries 1

1 Vectors 3

2 Tensors 7

3 Cartesian Coordinates 9

4 Vector (Cross) Product 13

5 Determinants 19

6 Change of Orthonormal Basis 25

7 Principal Values and Principal Directions 33

8 Gradient 37

II Stress 45

9 Traction and Stress Tensor 47

10 Principal Values of Stress 49

11 Stationary Values of Shear Traction 53

12 Mohr’s Circle 57

III Motion and Deformation 63

13 Current and Reference Configuration 65

14 Rate of Deformation 69

15 Geometric Measures of Deformation 73

iii

iv CONTENTS

16 Strain Tensors 81

17 Linearized Displacement Gradients 89

IV Balance of Mass, Momentum, and Energy 91

18 Transformation of Integrals 93

19 Conservation of Mass 95

20 Conservation of Momentum 97

21 Conservation of Energy 99

V Ideal Constitutive Relations 103

22 Fluids 105

23 Elasticity 109

Part I

Mathematical Preliminaries

1

Chapter 1

Vectors

1. According to the right hand rule the direction of u×v is given by puttingthe fingers of the right hand in the direction of u and curling in the

direction of v. Then the thumb of the right hand gives the direction of

u×v. Using the same procedure with v×u, the thumb of the right handis oriented in the opposite direction corresponding to the insertion of a

minus sign. Because w = u×v is orthogonal to both u and v, i.e., not inthe plane of u and v the scalar product of w with u and v is zero.

2. If u× v has a positive scalar product with w, then u, v and w are right

handed. In this case u × v ·w = v ×w · u = w × u · v are all equal tothe volume of the parallelopiped with u, v and w as edges. Interchanging

any two of the vectors in the vector product introduces a minus sign.

3. Letting w = u in (1.5) gives u × (v× u) = v + u, where and are

scalars. Because u× (v× u) is perpendicular to u

u · u× (v× u) = u · v+ u = 0= u · v+u · u = 0= cos + 2 = 0

Since 6= 0, cos + = 0.

4. If u · v = 0 for all v, then the scalar product must be zero for v = u

where is a nonzero scalar. Hence, u · u = 2 = 0. Therefore = 0

and u = 0.

5. (a)

(u+ v) · (u− v) = u · u+ v · u− u · v− v · v= 2+u · v − u · v− 2

= 2 − 2

3

4 CHAPTER 1. VECTORS

(b)

(u+ v)× (u− v) = u× u+ v× u− u× v− v × v= 0 + (−u× v)− u× v − 0= −2u× v

6. Define a, b, and c to coincide with the edges of the plane triangle with

c = a+ b.

a

b

c

(a)

2 = c · c =(a+ b) · (a+ b)= a · a+ a · b+ b · a+ b · b= 2 + 2a · b+ 2

= 2 + 2 cos( − ) + 2

= 2 − 2 cos() + 2

(b)

a× c = a× a+ a× b= a× b

Taking the magnitude of both sides gives sin ( − ) = sin or

sin=

sin

Similarly, b × a = b × c. Taking the magnitude of both sides gives sin ( − ) = sin or

sin=

sin =

sin

5

7. Two edges of the fourth face are b − a and c − a. The oriented area ofthis face is

n∆ =1

2(b− a)× (c− a)

=1

2(b× c)− 1

2(a× c)− 1

2(b× a)

Noting that

1

2(b× c) = −11, − 1

2(a× c) = −22, − 1

2(b× a) = −33

gives the result.

8. Forming the vector product of w with both sides of (1.5) gives

w × u× (v ×w) = w× vForming the scalar product of both sides with w × v and solving for gives

=(w × v) · [w× u× (v×w)]

|w × v|2Similarly

=(v×w) · [v × u× (v ×w)]

|v ×w|2

9. A line perpendicular to both of the given lines is w = c+ (l×m) . Thisline will intersect the line

u = a+ l

if

c+ (l×m) 13 = a+ l1for some values of 13 and 1. Forming the scalar product of both sides

with l×m and rearranging gives

13 =(l×m) · (a− c)

|l×m|Similarly, w = c+ (l×m) intersects v = b+m for

23 =(l×m) · (b− c)

|l×m|Because w = c+ (l×m) is perpendicular to both given lines, the mini-mum distance between them is

13 − 23 =(l×m) · (a− b)

|l×m|Consequently the two lines will intersect if

a · (l×m) = b · (l×m)

6 CHAPTER 1. VECTORS

10. Let P be a point on the line joining A and B and w be the vector joining

point O to point P. Then

v+AP = w

and

v+AB = u

But since AP and AB are collinear, AP = AB, where 0 ≤ ≤ 1.

Consequently,

w = u+ (1− )v

is the equation of the line joining A and B. By letting = (+ ) gives

w =u+v

+

11. Forming the scalar product of z = u+ v+ w with v ×w gives

=z · (v×w)u · (v ×w)

and similarly for and . If a, b, and c are coplanar, then u·(v ×w) = 0,and it is not possible to express an arbitrary vector z in the given form.

12. Let w be a vector that lies in both planes. Therefore w can be expressed

as

w = u+ v

and as

w = x+ y

Therefore (u× v) ·w = 0 and (x× y) ·w = 0. Hence, w is perpendicular

to the normal to the plane of u and v and to the normal to the plane of

x and y. Thus, w is proportional to (u× v) × (x× y). A unit vector inthe direction of w is

n =(u× v)× (x× y)|(u× v)| |(x× y)|

Chapter 2

Tensors

1. Let S = 12

¡F+F

¢. If S·u = u·S = u·S for any u then S is symmetric.

S · u =1

2

¡F+F

¢ · u=

1

2

nF · u+F · u

o=

1

2

nu · F + u · F

o=

1

2u · ©F +Fª = u · S

Note:

F · u = u · F = F · ufor all u implies the result F

= F. Let A = 12

¡F−F ¢. If A · u =

u ·A = −u ·A for any u then A is symmetric.

A · u =1

2

¡F−F ¢ · u

=1

2

nF · u−F · u

o=

1

2

nu · F − u · F

o=

1

2u · ©F − Fª = −u ·A

2. Substituting (2.10) into (2.5) and using (2.6) gives the result.

3.

(F ·G)−1 · u = vimplies

u = F ·G · v

7

8 CHAPTER 2. TENSORS

Multiplying from the left by F−1and then G−1 gives

G−1 · F−1 · u = vThe result follows by noting that the two expressions for v are equal for

all u.

4.

u = F · vimplies

v = F−1 · u = u · F−1

But u = F · v = v · F givesv = u · F−1

Noting that the two expressions for v are equal for all u gives the result.

5. Consider

v = (R · S) · uForming the scalar product of v with itself gives

v · v = u · (R · S) · (R · S) · uwhere the definition of the transpose has been used in the first expression

for v on the right hand side. Using the rule for the transpose of a product

(Example 2.5.2) gives

v · v = u · ¡S ·R¢ · (R · S) · u

Rearranging the parentheses gives

v · v = u · S · ¡R ·R¢ · S · uBut if R is orthogonal, R ·R = I yielding

v · v = u · S · S · uSimilarly, if S is orthogonal, S · S = I and v · v = u · u. Hence, R · S isorthogonal.

6. The angle between vectors u and v is given in terms of the scalar product

by (1.2). Consider the scalar product of A · u and A · v where A is

orthogonal:

(A · u) · (A · v) =³u ·A

´· (A · v)

where definition of the transpose has been used in the first term on the

right. Rearranging the parentheses gives

(A · u) · (A · v) = u· ¡A ·A¢ · v= u · v

and the second line follows because A is orthogonal. Hence, the angle

between u and v is the same as between A · u and A · v.

Chapter 3

Cartesian Coordinates

1. (a)

1 = 111 +212 +313

2 = 121 +222 +323

3 = 131 +232 +333

(b)

= 11 + 22 + 33

(c)

= 1111 + 1212 + 1313

+2121 + 2222 + 2323

+3131 + 3232 + 3333

(d)

11 = 11 12 = 12 13 = 1321 = 21 22 = 22 23 = 2331 = 31 32 = 32 33 = 33

2. (a)

= 11 + 22 + 33

= 1 + 1 + 1 = 3

9

10 CHAPTER 3. CARTESIAN COORDINATES

(b)

= 1111 + 1212 + 1313

+2121 + 2222 + 2323

+3131 + 3232 + 3333

= 1 + 0 + 0

+0 + 1 + 0

+0 + 0 + 1

= 3

3. ⎡⎣123

⎤⎦ =

⎡⎣11 21 31

12 22 32

13 23 33

⎤⎦⎡⎣123

⎤⎦=

£1 2 3

¤⎡⎣11 12 13

21 22 23

31 32 33

⎤⎦⎡⎣11 12 1321 22 2331 32 33

⎤⎦ =

⎡⎣11 12 1321 22 2331 32 33

⎤⎦4. (a)

v · I = (e) · (ee)= (e · e) e= e

= e = e = v

(b)

F · I = (ee) · (ee)= e (e · e) e= ee

= ee = ee = F

5. If u · v = = 0 for any v, then choose

1 = 1, 2 = 0, 3 = 0

⇒ 1 = 0. Choose

1 = 0, 2 = 1, 3 = 0

⇒ 2 = 0. Choose

1 = 0, 2 = 0, 3 = 1

⇒ 2 = 0. Hence, u = 0.

11

6.

F = F · ·I= (ee) · · (ee)= (e · e) (e · e)= =

= = 11 + 22 + 33

7.

F · ·G =

=

=

= (e · e) (e · e)= (ee) · · (ee)= G · ·F

8.

F · ·G =

=

= F : G

F · ·G =

=

= F : G

9.

F ·G · ·H = (ee) · (ee) · · (ee)

= (ee) · · (ee)

= (ee) · · (ee)

= =

H · F · ·G = (ee) · (ee) · · (ee)

= (ee) · · (ee)

= (ee) · · (ee)

= =

G ·H · ·F = (ee) · (ee) · · (ee)= (ee) · · (ee)= (ee) · · (ee)= =

12 CHAPTER 3. CARTESIAN COORDINATES

10.

F ·G :H = (ee) · (ee) : (ee)

= (ee) : (ee)

= (ee) : (ee)

= =

= (ee) : (ee)

= (ee) : (ee) · (ee)

= F : G ·H

11.

tr (F ·G) = tr (ee)

= I · · (ee)

= (e · e) (e · e) ()

= ()

= = F · ·G

Chapter 4

Vector (Cross) Product

1.

e · (e × e) = e · (e)= (e · e)= =

2. If , , and are the free indices, then the equality only needs to be

verified for the six cases: (i) = ; (ii) = ; (iii) = ; (iv) = ; (v)

= ; (vi) = .

3. (a)

= −

= 3 − = 2

(b)

= 2 = 6

4.

e × e = e

(e × e) = e

= 2e

⇒ =1

2e × e

13

14 CHAPTER 4. VECTOR (CROSS) PRODUCT

5. (a)

u× (v×w) = e × (e)

= e

= ( − )e

= e − e

= (u ·w)v− (u · v)w= u · (wv− vw)

(b)

u× (v×w) + v × (w × u) +w× (u× v)= (u ·w)v− (u · v)w+(v · u)w− (v ·w)u

+(w · v)u− (w · u)v= 0

6.

(u× v)×w = (e)× e

= e

= ( − ) e

= e − e

= (u ·w)v− (v ·w)uTherefore, = −v ·w and = u ·w

7. Let u = n and w = n in (4.14) to get

n× (v × n) = (n · n)v− (n · v)nSince n is a unit vector n · n = 1. Rearranging give

v =(n · v)n+ n× (v× n)

8. (a)

(a× b) · (c× d) =

= ( − ) ()

= () ()− () ()= (a · c) (b · d)− (a · d) (b · c)

(b)

(a× b)× (c× d) = e

= e ( − )

= (e) − (e) = [c · (d× a)]b− [c · (d× b)]a

15

9. Writing in index notation

=

Because this applies for all

=

Multiplying the preceding equation by gives

=

= −

= −2

= −2

Therefore,

= −12

IfW is not anti-symmetric, then w = 0.

10. (a)

n× n = n× (w× n)= n · (nw −wn)

where the second line follows from problem 5. Noting that n · n = 1because n is a unit vector and n ·w = 0 because n is orthogonal to

w gives the result.

(b)

=

= ()

= ( − )

= −

or

W = nn− nnin dyadic notation.

11.

u · (v×F) = e · (e × ee)

= e · (ee)= e

= e · ee

= (u× v) · F


12. Writing both sides in index form gives

∗e = e

Because this applies for all u and v,

∗ =

Multiplying both sides by and summing gives

∗ =

2∗ =

or

∗ =1

2

and re-labelling indices gives the result.

13.

= g1 · (g2 × g3)= (e2 + e3) · [(e1 + e3)× (e1 + e2)]= (e2 + e3) · [e3 + e2 − e1] = 2

Therefore

g1 =1

2(g2 × g3)

=1

2(e3 + e2 − e1)

g2 =1

2(g3 × g1)

=1

2(e3 − e2 + e1)

g3 =1

2(g1 × g2)

=1

2(−e3 + e2 + e1)

14. (a) Since g1 is perpendicular to g2, g1 · g2 = 0 and g1 also lies in the 12

plane. Therefore

g1 = g2 × e3= (3e2 + e1)

17

But g1 · g1 = 1 gives = 16. Therefore

g1 =1

6(e1 + 3e2)

Similarly

g2 =1

6(−e1 + 3e2)

e1

e2

g1

g2g

1

g2

Original and dual base vectors.

(b) v = e1 + 5e2 = g = g.

1 = g1 · v = 8

3

2 = g2 · v = 7

31 = g1 · v = 82 = g2 · v = 2

Therefore

v = 8g1 + 2g2

=8

3g1 +

7

3g2

g1

g2

v

v1

v2

Components of v relative to original base vectors.


v

g1g

2

v1v

2

Components of v relative to dual base vectors.

Chapter 5

Determinants

1.

det = 123

= 11 1232233 + 1323223+21 2133213 + 2131233+31 3121223 + 3212213

= 11 (+1)2233 + (−1)3223+21 (+1)3213 + (−1)1233+31 (+1)1223 + (−1)2213

= 11 2233 −3223−21 1233 −3213+31 1223 −2213

Therefore

det = 11

¯22 23

32 33

¯−21

¯12 13

32 33

¯+31

¯12 13

22 23

¯2.

det = 123

Interchanging row 1 and row 2 gives

213

= −213

= −123

= −det

19

20 CHAPTER 5. DETERMINANTS

3. Consider

=

Then 123 = 123 = det . If any two indices are the same,

the result is zero:

= , (no sum on )

= −, (no sum on )

= −, (no sum on )

= −, (no sum on )

= 0

The result is zero because the top and bottom lines are the negative of

each other. Cyclically rotating the indices does not change the value; e.g.,

=

=

=

=

=

But reversing any two indices introduces a minus sign. E.g.,

=

= −

= −

= −

= −

With these results it is possible to easy enumerate results for all numeric

values. Consequently,

= det

4.

1 = 23

=1

2(12323 + 13223)

=1

2 (12323 + 13232)

=1

21

Because the second index is arbitrary, it can be changed to “”. Rela-

belling indices then gives Equation (5.6).

21

5.

=1

2

21 =1

221

=1

22 12323 + 13232

=1

2223 − 232

=1

2223 − 232

=1

2223 − 232

= 223

= 3213 −1233

= (−1)2+1¯12 13

32 33

¯

13 =1

213

=1

23 (12323 + 31232)

=1

2(323 − 332)

= 323

= 3122132 + 3212231

= 2132 −2231

= (−1)3+1¯21 22

31 32

¯6. From Equation (4.10)

a · b× c =

¯¯1 2 31 2 31 2 3

¯¯

d · e× f =¯¯1 2 31 2 31 2 3

¯¯ =

¯¯1 1 12 2 23 3 3

¯¯

where the second equality results because the determinant of matrix is

equal to the determinant of its transpose. The product of the two matrices

is ⎡⎣1 2 31 2 31 2 3

⎤⎦⎡⎣1 1 12 2 23 3 3

⎤⎦ =⎡⎣a · d a · e a · fb · d b · e b · fc · d c · e c · f

⎤⎦


Noting that the determinant of a product is the product of the determi-

nants establishes the result.

7. Setting = gives

=

¯¯

¯¯

=

¯

¯−

¯

¯+

¯

¯= 3 ( − )− ( − )

+ ( − )

= 3 ( − )− ( − )

− ( − )

= ( − )

8.

det() =

Multiplying each side by and summing gives

det() =

6 det() =

or

det() =1

6

9.

(M · a×M · b) ·M = (e ×e) ·ee

= (e) ·ee

= e

= () e

= det()e

= det() (a× b)

10.

(M · a) · (M · b)× (M · c) = (e) · (e)× (e)

= (e) · (e)

= ()

= det()

= det() (a · b× c)

23

11.

A ·A = I

det¡A ·A¢ = det (I) = 1

det¡A¢det (A) = (det (A))

2= 1

Therefore

det (A) = ±1From Problem 10

(A · a) · (A · b)× (A · c) = det() (a · b× c)

If a, b, and c are right-handed, a · b × c 0. If A transforms a right-

handed coordinate system to a right-handed coordinate system, then the

left side is also positive and det() = +1. If A transforms a right-handed

coordinate system to a left-handed coordinate system, then the left side

is negative and det() = −1.

Chapter 6

Change of Orthonormal

Basis

1.

e0 = A · e = e ·A

Therefore, the components of a vector in the primed system are given by

0 = e0 · v =¡e ·A

¢ · v= e ·A · v

2.

= e · v= e · (0e0)= (e · e0) 0=

0

3.

= e · F · e= e · ( 0e0e0) · e= (e · e0) 0 (e0 · e)=

0

4. A = ee , A = ee and I = ee

e · (ee ·ee) · e = e · (ee) · ee · (ee) · e =

δ =

=

Beginning with A ·A = I establishes the first equality of (6.4).

25

26 CHAPTER 6. CHANGE OF ORTHONORMAL BASIS

5.

e03 = λ

Because e02 is perpendicular to e03 and e1

e02 = e03 × e1= (3e2 − 2e3)

where is a scalar. Choose so that e02 is a unit vector:

= ±q22 + 23

and take the + sign so that e02 has a positive projection on e2. Choose e01

so that the primed system is right-handed:

e01 = e02 × e03=

q22 + 23 e1 −

12q22 + 23

e2 − 13q22 + 23

e3

Noting that e0 = e and e0 · e = establishes the result.

6. Because = 0,

= 0

= 0 () =

for a scalar . Defining

0 =

establishes the as components of a vector because they transform like

one under a rotation of the coordinate system.

7. Because = 0,

0 =

0 () =

for a scalar . Defining

0 =

establishes the as components of a tensor because they transform like

one under a rotation of axes.

8.

=

= (0) (

0)

27

Multiplying both sides by and summing gives

= 00

= det()00

For transformation from a right (left) - handed system to a right (left) -

handed system det() = 1. In this case 0 = and the components

of w transform like components of a vector. If, however, the trans-

form from a right (left) - handed system to a left (right) - handed system

det() = −1 and the components of the vector product are not tensorcomponents as defined here.

9. Because

0 =

and

= 0

0 = 0

= 0

0

Therefore

0 =

10.

0 =

=

=

= 0

Hence the components are symmetric in any rectangular cartesian coordi-

nate system.

11.

0 =

= −

= −

= − 0Hence the components are anti-symmetric in any rectangular cartesian

coordinate system.

12.

0 =


trF = 0=

=

=

13.

=¡

0

¢(

0)

= () ()0

0

= 0

0

= 00

14.

e0 = A · e= (ee) · e= e

= e

From Figure 6.2

e01 = cos e1 + sin e2

e02 = − sin e1 + cos e2e03 = e3

Comparing yields the given result.

15. (a) ⎡⎣010203

⎤⎦ =⎡⎣ cos sin 0

− sin cos 0

0 0 1

⎤⎦⎡⎣123

⎤⎦01 = 1 cos + 2 sin

02 = −1 sin + 2 cos

03 = 3

(b)

[ 0] =

⎡⎣ cos sin 0

− sin cos 0

0 0 1

⎤⎦⎡⎣11 12 1321 22 2331 32 33

⎤⎦⎡⎣cos − sin 0

sin cos 0

0 0 1

⎤⎦

29

011 = 11 cos2 + (12 + 21) cos sin + 22 sin

2

012 = (22 − 11) cos sin + 12 cos2 − 21 sin

2

013 = 13 cos + 23 sin

021 = (22 − 11) cos sin + 21 cos2 − 12 sin

2

022 = 11 sin2 − (12 + 21) cos sin + 22 cos

2

023 = −13 sin + 23 cos

031 = 13 cos + 23 sin

032 = −13 sin + 23 cos

033 = 33

(c)

01 = 1 + 2

02 = −1 + 2

03 = 3

011 = 11 + (12 + 21)

012 = (22 − 11) + 12

013 = 13 + 23

021 = (22 − 11) + 21

022 = − (12 + 21) + 22

023 = −13 + 23

031 = 13 + 23

032 = −13 + 23

033 = 33

16.

A() = − sin() (e1e1 + e2e2) + cos () (e2e1 − e1e2)A(0) = (e2e1 − e1e2)A() · A(0) = A()

17.

e0 = A() · e

e0 = A() · e

but substituting e = A () · e0 gives the result. Because A ·A = I

A ·A +A · A = 0

A ·A = −³A ·A

´


18. (a) A rotation of base vectors 90 about the 2 is⎡⎣e01e02e03

⎤⎦ =⎡⎣0 0 −10 1 0

1 0 0

⎤⎦⎡⎣e1e2e3

⎤⎦or in dyadic form

A0 = e01e1 + e02e2 + e

03e3

A0 = −e3e1 + e2e2 + e1e3(b) A rotation of base vectors 90 about the 3 is⎡⎣e001e002

e003

⎤⎦ =⎡⎣ 0 1 0

−1 0 0

0 0 1

⎤⎦⎡⎣e01e02e03


A00 = e001e01 + e

002e02 + e

003e03

= e02e01 − e01e02 + e03e03

= −e2e3 + e3e2 + e1e1(c) Successive 90 about the 2 and 3 axes are given by⎡⎣e001e002

e003

⎤⎦ =

⎡⎣ 0 1 0

−1 0 0

0 0 1

⎤⎦⎡⎣0 0 −10 1 0

1 0 0

⎤⎦⎡⎣e1e2e3

⎤⎦=

⎡⎣0 1 0

0 0 1

1 0 0

⎤⎦⎡⎣e1e2e3

⎤⎦=

⎡⎣e2e3e1


A00 ·A0 = e2e1 + e3e2 + e1e3

19. (a) The matrix rotating the 3 direction into a direction λ is given by

the transpose of the matrix from Problem 5. Multiplying this matrix

by the matrix from Problem 14 rotates the axes through an angle .

Then multiplying by the matrix from Problem 5 rotates the back into

the original axes. Hence, the result is given by the matrix product⎡⎢⎢⎢⎣q22 + 23 0 112√22+

23

3√22+

23

2

− 13√22+

23

− 2√22+

23

3

⎤⎥⎥⎥⎦⎡⎣ cos sin 0

− sin cos 0

0 0 1

⎤⎦⎡⎢⎢⎣q22 + 23

12√22+

23

− 13√22+

23

0 3√22+

23

− 2√22+

23

1 2 3

⎤⎥⎥⎦

31

(b) The dyadic form follows from setting λ = e3, using I = e1e1 + e2e2+

e3e3 and noting that12 = −21 = −1 are components of the anti-symmetric matrix with axial vector e3.

20. Setting λ =(e1+e2 + e3) √3 and = 120 in the answer to Problem 19a

gives the matrix ⎡⎣0 1 0

0 0 1

1 0 0

⎤⎦which the same as the result of Problem 18c.

21.

0 =

= det()

=

Hence has the same components in any rectangular cartesian coor-

dinate system if det() = 1. This will be the case for transformations

from right (left) - handed systems to right (left) - handed systems. But

for reflections, i.e., transformations from a right (left) - handed systems

to a left (right) - handed system, det() = −1 and is not isotropic.

Chapter 7

Principal Values and

Principal Directions

1. Substituting = 5 back into equation (7.1) gives

2¡μ

¢1+ 0

¡μ

¢2− 2 ¡μ¢

3= 0

0 + (5− 5) ¡μ¢2− 0 = 0

−2 ¡μ¢1+ 0 + (−1) ¡μ¢

3= 0

The first and third equations are linearly independent and have only the

solution¡μ

¢1=¡μ

¢3= 0. Because the coefficient of

¡μ

¢2in the

second equation is zero, it can be arbitrary and is taken as¡μ

¢2= 1 to

make μ a unit vector.

2.

0 =

= =

00 = () ()

= () ()

= =

det ( 0) = 01

02

03

= (1) (2) (3)

= () (123)

= det() (123)

= det() () (123)

= det( ) (123)

= (det())2det( ) = det( )

33

34 CHAPTER 7. PRINCIPAL VALUES AND PRINCIPAL DIRECTIONS

because |det()| = 1.

3. Taking the trace of both sides of the Cayley Hamilton theorem (7.13)

yields

tr (F · F · F) = 1tr (F · F) + 2trF+ 3trI

But trF = 1, trI = 3 and

tr (F · F) = 22 + 21

Substituting and rearranging gives the result.

4. (a) Multiplying both sides of (7.13) by F−1 gives the result.

(b) Take the trace of both sides, use

tr (F · F) = 22 + 21

and rearrange.

5.

F · μ = μ

F · μ = μ

F · μ = μ

Form the scalar product of the first equation with μ and the second

with μ . Subtracting then gives

( − )μ · μ = 0

Because 6= , μ must be orthogonal to μ . By the same argument,

μ must be orthogonal to μ . Forming the scalar product of the second

equation with μ and the third with μ and subtracting yields

( − )μ · μ = 0

Because = , the equation is automatically satisfied and the only

requirement on μ and μ is that they be orthogonal to μ .

6. (a) Principal values are = 13, = 7 and = 3. Principal

directions are

μ = ± 2√5e1 ± 1√

5e2

μ = e3

μ = ± 1√5e1 ∓ 2√

5e2

35

(b) ⎡⎣± 2√5

± 1√5

0

0 0 1

± 1√5∓ 2√

50

⎤⎦⎡⎣11 4 0

4 5 0

0 0 7

⎤⎦⎡⎣± 2√5

0 ± 1√5

± 1√5

0 ∓ 2√5

0 1 0

⎤⎦=

⎡⎣13 0 0

0 7 0

0 0 3

⎤⎦7.

trF = a · b1

2( − ) = 0

det( ) = 0

8.

det

⎡⎣cos − sin 0

− sin cos − 0

0 0 1−

⎤⎦ = (1− )³(cos − )

2+ sin2

´= 0

Therefore = 1 is one solution and the remaining two satisfy

(cos − )2+ sin2 = 0

which has the solutions

= cos ± p1− cos2

Because cos2 ≤ 1, the only real solution is = 1 and corresponding

principal direction is e3.

9.

det ( − ) =1

6 ( − ) ( − ) ( − )

=1

6

−16 ( + + )

+1

62 ( + + )

−16

3

Using the − identities gives

det ( − ) = det ( ) +1

2 ( − ) +

2 − 3

36 CHAPTER 7. PRINCIPAL VALUES AND PRINCIPAL DIRECTIONS

Chapter 8

Gradient

1.

v = e

Substituting in ∇× v gives

∇× v = e

× e

= e2

Because is anti-symmetric with respect to interchange of and and

2 is symmetric, their summed product vanishes.

2. A vector in the direction of the normal to a surface is proportional to the

gradient:

y = e

()

= e ( + )

= e ( + )

= 2e ()

where the last line follows because = . Dividing by the magnitude

of y to get a unit vector gives

n = ep

3.

v · (∇u) = e · (ee)= e · e (e)= (v ·∇)u

37

38 CHAPTER 8. GRADIENT

4. (a)

∇ · (v) = e

· (e)

= e

· e + e ·

e

= v ·∇+ ∇ · v

(b)

∇× (v) = e

× (e)

= e

× (e) + e

× (e)

= ∇× v+ ∇× v

5. (a)

∇ · (u× v) = e

· (e × e)

= e

· (e)

=

µ

+

¶= () + ()

= () − ()

= (∇× u) · v− u · (∇× v)

(b)

∇ · (F · u) = e

· (ee · e)

= e

· (e)

=

+

= e ·µe

· ee

¶+ (e · e) (e · e)

= u · (∇ · F) +F · ·∇u

39

6. (a)

∇× (u× v) = e

× ( × )

= e

× (e)

= e ()

()

= e ( − )

()

= e

()− e

()

= e

+

(e)− e

−

(e)

= a∇ · b+ b ·∇a− b∇ · a− a·∇b

(b)

∇× (5× v) = e

×µe

× e

¶= e

×µe

¶= e

2

= ( − ) e2

= e

− 2

e

= ∇ (∇ · v)−∇2v

7. (a)

∇ = e

√

= e√

= e√

=e√

=x


(b)

∇x

= e

µe√

¶= ee

1√

+ ee

1√

=1√

+ ee

Ã−

()32

!=

ee

³ −

2

´(c)

∇2 = ∇ ·∇= e

·µ

e√

¶=

µ

−

3

¶=

2

8.

∇× u = e

= −e= −u×∇

9.

F×∇ =

µ

¶ee × e

=

µ

¶ee

=

µ

ee

¶= −

µ

ee

¶=

Ã

ee

!

= − ¡∇×F ¢

41

10.

M = ∇×E= e

×ee

= ee

Therefore,

=

11 = 11

= 312−213

22 = 22

= 123−321

12 = 12

= 322−223

21 = 21

= 113−311

11. To determine the Laplacian in cylindrical coordinates:

∇2 = ∇ · (∇)=

µe

+ e

1

+ e

¶·µe

+ e

1

+ e

¶=

2

2+1

22

2+e

· e

+

2

2

=2

2+1

22

2+1

+

2

2

=1

µ

¶+1

22

2+

2

2


12. To determine the curl in cylindrical coordinates

∇× v =

µe

+e

+ e

¶× (e + e + e)

= e × e

+ e × e

+ e × e

+ e × e

+e

× e

+e

× e

+

e

× e

= e

− e

+ e

− e

− e

+ e

+ e

= e

µ

−

¶+ e

µ

−

¶+ e

µ

−

+

¶

13. To determine ∇v in cylindrical coordinates

∇v =

µe

+e

+ e

¶(e + e + e)

= ee

+ ee

+ ee

+ee

+ e

e

+ ee

+ e

e

+ ee

+ee

+ ee

+ ee

= ee

+ ee

+ ee

+ee

µ

−

¶+ ee

µ

+

¶+ ee

+ee

+ ee

+ ee

43

14. To calculate the divergence of a tensor ∇ · F write

∇ · F =

µe

+e

+ e

¶· F

=

µe

+e

+ e

¶· (ee + ee + ee)

+

µe

+e

+ e

¶· (ee + ee + ee)

+

µe

+e

+ e

¶· (ee + ee + ee)

=

e +

e +

e +

e

· e

(e + e + e)

+1

e +

1

e +

1

e

+

µ

e

+

e

¶+

e +

e +

e

=

e +

e +

e +

1

(e + e + e)

+1

e +

1

e +

1

e

+

µ

e −

e

¶+

e +

e +

e

= e

µ

+1

+

+

−

¶+e

µ

+1

+

+

+

¶+e

µ

+1

+

+

¶

Part II

Stress

45

Chapter 9

Traction and Stress Tensor

1. On 1 = +, n = e1

t = e1 · σ= 11e1 + 12e2 + 13e3

= −µ1− 22

2

¶e1 + 2

2

e2 + 0e3

On 1 = −, n = −e1t = −e1 · σ= −11e1 − 12e2 − 13e3

=

µ1− 22

2

¶e1 + 2

2

e2 + 0e3

47

48 CHAPTER 9. TRACTION AND STRESS TENSOR

Chapter 10

Principal Values of Stress

1. (a)

det (σ − I) =

¯¯− (1 + ) −2 0

−2 − 2

0 2 1−

¯¯

= 3 − 9 = 0

Therefore the principal stresses are = +3, = 0 and = −3(b) To determine the principal direction corresponding to the principal

stress = +3:

−41 − 22 + 03 = 0

−21 − 32 + 23 = 0

01 + 22 − 23 = 0

The first and third equations give 1 = −22 and 3 = 2. Making

n a unit vector gives

n = ∓13e1 ± 2

3e2 ± 2

3e3

For the principal stress = −3:

21 − 22 + 03 = 0

−21 − 32 + 23 = 0

01 + 22 + 43 = 0

The first and third equations give 2 = −23 and 1 = 2. Making

n a unit vector gives

n = ∓23e1 ∓ 2

3e2 ± 1

3e3

49

50 CHAPTER 10. PRINCIPAL VALUES OF STRESS

Choosing n = n × n for a right-handed system gives

n = −23e1 +

1

3e2 − 2

3e3

2. (a)

det (σ − I) =

¯¯ − cos sin

cos − 0

sin 0 −

¯¯

= 3 − 2 = 0

Therefore the principal stresses are = + , = 0 and = − (b) To determine the principal direction corresponding to the principal

stress = + :

−1 + cos 2 + sin 3 = 0

cos 1 − 2 + 03 = 0

sin 1 + 02 − 3 = 0

The second and third equations give 2 = cos and 3 = sin .

Making n a unit vector gives

n =1√2e1 +

1√2cos e2 +

1√2sin e3

where, for definiteness, we have taken the positive sign. For the

principal stress = 0:

01 + cos 2 + sin 3 = 0

cos 1 + 02 + 03 = 0

sin 1 + 02 + 03 = 0

The second or third equations gives 1 = 0 and the first gives 2 =

− tan 3. Making n a unit vector givesn = 0e1 ∓ sin e2 ± cos e3

Choosing n = n × n for a right-handed system gives

n =1√2±e1 ∓ cos e2 ∓ sin e3

3.

2 =1

2

n(0)

2+ (0)

2+ (0)

2o

=1

2

n( − )

2+ ( − )

2+ ( − )

2o

51

where = ( + + ) 3.

2 =1

2

©2 + 2 + 2 − 32

ªSubstituting for , multiplying out and collecting terms gives the result.

4. Substituting (10.11) into (10.8) givesµ4

32

¶32sin3 − 2

r4

32 sin− 3 = 0

4 sin3 − 3 sin =

r27

4

3

322

Using the identity

4 sin3 − 3 sin = − sin 3gives (10.12).

5. (a)

0 = − 13

½ +

1

2( + ) +

¾= − 1

2( + ) = 0

Hence, 3 = 0, sin = 0 and = 0.

(b)

0 =2

3( − )

0 = 0 = −1

3( − )

Therefore

3 =2

27( − )

3

2 =1

3( − )

2

Hence sin 3 = −1 and = −30 .(c)

0 = 0 =1

3( − )

0 = −23( − )

52 CHAPTER 10. PRINCIPAL VALUES OF STRESS

Therefore

3 = − 227( − )

3

2 =1

3( − )

2

Hence sin 3 = 1 and = 30 .

6.

( − ) = 0

gives the principal directions, the , of the stress . Substituting the

deviatoric stress givesµ0 −

1

3 −

¶ = 0

Thus, the principal directions of 0 satisfy the same equation with analtered value of :

0 = +1

3

Chapter 11

Stationary Values of Shear

Traction

1. If none of the = 0, then the terms in braces must vanish. Elimi-nating leads to

21 − 21 = 22 − 22 = 23 − 23

From the first equality

21 − 22 = 2 (1 − 2)

Therefore, either 1 = 2 or 1 + 2 = 2. Similarly, the second and

third equality leads to either 2 = 3 or 2 + 3 = 2 and first and

third to either 3 = 1 or 1 + 3 = 2. Clearly, the first option leads

immediately to 1 = 2 = 3. If, for example, 1 = 2 but

2 + 3 = 2

In this case

2 + 3 = 2¡211 + 222 + 233

¢Because 1 = 2, this can be rewritten as

1

2(1 + 3) = 1

¡1− 23

¢+ 233

1

2(1 − 3) = 23 (1 − 3)

Hence either 1 = 3 or 3 = ±1√2. In the latter case the stress state is

axisymmetric (1 = 2) and the maximum shear traction occurs on any

plane with a normal that makes an angle of 4 with the distinguished

stress axis.

53

54 CHAPTER 11. STATIONARY VALUES OF SHEAR TRACTION

2. (a) Traction is given by⎡⎣−1 −2 0

−2 0 2

0 2 1

⎤⎦ 13

⎡⎣122

⎤⎦ = 1

3

⎡⎣−526

⎤⎦or

t =1

3(−5e1 + 2e2 + 6e3)

(b) Normal traction is

n · t =1

3(e1 + 2e2 + 2e3) · 1

3(−5e1 + 2e2 + 6e3)

=1

9(−5 + 4 + 12) = 11

9

Shear traction is

t = t−119n

= −2074e1 − 0148e2 + 1185e3

The magnitude of the shear traction is

=√t · t

=pt · t−2

= 2393

3. (a)

t = n · σ= (n · u)u= cos u

(b)

= n · t= cos2

(c)

= ±pt · t− 2

= ± cos p1− cos2

= ± cos sin

55

4. From Problem 10.2, the greatest and least principal stresses are and −and the corresponding principal directions are n and n . Thus, the

maximum shear stress is and it occurs on the plane with normal

n =1√2(n + n) = ±e1

and

n =cos e2 + sin e3

5. If the shear traction vanishes on every plane then

n · σ · s = 0for all unit vectors n and s such that n · s = 0. Substituting σ = σ0 − I

where σ0 is the deviatoric stress gives

n · σ0 · s− n · s = 0Because the second term vanishes, the first term will vanish only if the de-

viatoric stress σ0 vanishes. If σ = −I, then the first equation is satisfiedautomatically.

6. (a) The unit vector that makes equal angles with the principal axes is

given by

n =1√3(e1 + e2 + e3)

The traction on this plane is

t = n · σ=

1√3(1e1 + 2e2 + 3e3)

(b) The normal component of traction is

= n · t=

1

3(1 + 2 + 3)

(c) The magnitude of the shear traction is given by

=pt · t− 2

(d) The direction of the shear traction is

s =t− n

|t− n|=

1

(t− n)

=1√3

(01e1 + 02e2 + 03e3)

where the 0 are principal values of the deviatoric stress.

56 CHAPTER 11. STATIONARY VALUES OF SHEAR TRACTION

7. The normal traction is

= n · σ · n= 1

21 +

1

2(1 + 3)

22 + 3

23

Using 22 = 1−¡21 + 23

¢and setting equal to 1

2(1 + 3) gives¡

21 − 23¢(1 − 3) = 0

Because 2 0, 1 6= 3, and therefore 21 = 23 and

22 + 221 = 1

Calculating

t · t = 2121 +

∙1

2(1 + 3)

¸222 + 23

23

=¡21 + 23

¢21 +

1

4(1 + 3)

2 ¡1− 221

¢=

1

2(1 − 3)

21 +

1

4(1 + 3)

2

Computing the magnitude of the shear traction

=pt · t− 2

=1√21 (1 − 3)

Setting equal to (1 − 3) 4 gives

1 = 3 =1

2√2

and

2 =

q1− 221 =

√3

2

Therefore the normal to the plane is

n =1

2√2(e1 + e3) +

√3

2e2

Chapter 12

Mohr’s Circle

1. (a) Maximum shear stress is

1

2(0− (−)) =

2

and the normal stress on the plane of maximum shear is

1

2(0 + (−)) = −

2

tn

ts

( ) =ts max

Problem 12.1a

(b) Maximum shear stress is

1

2(− (−3)) = 2


1

2(+ (−3)) = −

57

58 CHAPTER 12. MOHR’S CIRCLE

tn

ts

aa

a

( ) = 2ats max

Problem 12.1b

(c) Maximum shear stress is 0 and the normal stress on the plane of

maximum shear is −.

tn

ts

(d) Principal stresses are −5, 0, +5. Therefore, maximum shear stressis

1

2(5− (−5)) = 5


1

2(5+ (−5)) = 0

59

tn

ts

5a 5a

( ) = 5ats max

(e) Maximum shear stress is

1

2(− (−5)) = 3


1

2(+ (−5)) = −2

tn

ts

aa5a

2. See graphical construction in the Figure.


tn

ts

3 1

ts

3. Failure occurs when the Mohr’s circle first becomes tangent to the failure

condition

|| = 0 −

Because 2 = (2)− , the magnitude of the shear traction at points A

and A’ is

|| =1

2(3 − 1) sin

h±³2−

´i=

1

2(3 − 1) cos

and the normal traction is

|| =1

2(3 + 1) +

1

2(3 − 1) cos

h±³2−

´i=

1

2(3 + 1) +

1

2(3 − 1) sin

1

ts

tn

A

A’

Part III

Motion and Deformation

63

Chapter 13

Current and Reference

Configuration

1. (a) Motion: 1 = 1 + ()2, 2 = 2, 3 = 3.

(b) Lagrangian description of velocity

1(X ) = ()2, 2(X ) = 3(X ) = 0

Eulerian description of velocity

1(x ) = ()2, 2(x ) = 3(x ) = 0

(c) 1 = 1 − ()2, 2 = 2, 3 = 3

2. (a)

=

+

= (1− )

(1 + )2

(b)

=

+

= 2 + 1 + 2

= 2 +1 + 2

1 +

3. (a) Lagrangian:

=

65

66 CHAPTER 13. CURRENT AND REFERENCE CONFIGURATION

To get Eulerian velocity, invert the motion:

=

(1 + )

Substituting into the Lagrangian velocity gives:

=

(1 + )

Note also that differentiating the inverted motion gives

=

(1 + )−

(1 + )2

The left hand side is zero because position is fixed in the reference

configuration and is the velocity. Hence solving for gives the

Eulerian description of the velocity; i.e.

= =

(1 + )

(b) = 0

(c) Substitution: = 0. Calculation from material derivative:

=

+

=−2(1 + )

2+

(1 + )

(1 + )

= 0

4. (a) Lagrangian description of velocity

1 = 2, 2 = −1

−, 3 = 0

Lagrangian description of acceleration:

1 = 2, 2 = 1

−, 3 = 0

(b) First invert the motion to get

1 =1 − 2 (

− 1) + − − 1

2 =2 + 1 (1− )

+ − − 1Then substitute into (a) to get

1 =2

+ 1 ( − 1)

+ − − 12 =

−1− + 2 (1− −) + − − 1

67

(c) Substitute the inverted motion into the answer from (a) to get

1 =2

+ 1 ( − 1)

+ − − 12 =

1− − 2 (1− −) + − − 1

Use the material derivative

=

+

to get the same result (after much algebra).

68 CHAPTER 13. CURRENT AND REFERENCE CONFIGURATION

Chapter 14

Rate of Deformation

1. Principal values are given by

det

⎡⎣ − 12 13

−12 − 23

−13 −23 −

⎤⎦ = 0

−©2 + ¡ 212 + 2

13 + 223

¢ª= 0

Thus, = 0 is one principal value and because ( ) 0, the remaining

two are purely imaginary.

2. The components of the axial vector are given by

= −12

= −12 ( − )

= −12(b× a) +

1

2(a× b)

Therefore

w = a× b3. (a) The axial vector of the anti-symmetric tensorW satisfies

W · u = w× ufor all u. Setting u = p gives w× p = 0. Because the cross productof w and p vanishes, they must be parallel and w = p.

(b) From (a)

W · u = p× uUsing p = q× r gives

W · u = (q× r)× u= −u× (q× r)

69

70 CHAPTER 14. RATE OF DEFORMATION

From exercise 4.5.a

W · u = −u · (rq− qr)= u · (qr− rq)

Writing the left side as u ·W and noting that the relation applies

for all u gives

W = (qr− rq)or

W = (rq− qr)Alternatively, using the result of (a) and Exercise 4.9 gives

=

Substituting p = q× r in index form =

gives

=

Using the − identity (4.13) gives

= ( − )

Converting to dyad form gives the result.

4. (a) From example 13.1, the Eulerian description of the velocity is

v(x ) = Q() ·Q−1 · x− c()+ c()Therefore the velocity gradient is L = Q() ·Q−1. The rate of defor-mation is

D =1

2

¡L+ L

¢=1

2

½Q() ·Q−1 +

³Q() ·Q−1

´¾and the spin tensor is

W =1

2

¡L− L ¢ = 1

2

½Q() ·Q−1 −

³Q() ·Q−1

´¾(b) If D = 0, then the distance between any two points is constant

|x − x| = |X −X|Substituting the motion yields Q ·Q = I and therefore Q−1 = Q .

Alternatively, D = 0 gives

Q ·Q−1 +³Q ·Q−1

´= 0

71

Multiplying from the left by Q and from the right by Q gives

Q · Q+Q ·³Q−1 · Q

´·Q = 0

©Q ·Qª = 0

Hence Q · Q is a constant. Setting the constant equal to I again

gives Q−1 = Q . If D = 0

Q ·Q−1 = −³Q ·Q−1

´andW = Q ·Q−1 is antisymmetric.

(c) The velocity can be written

v(x ) =W · x− c()+ c()From example 13.1, the acceleration is

a(x ) = Q ·Q−1 · x− c()+ c()Writing

Q =W ·Qand differentiating gives

Q = W ·Q+W · Q= W ·Q+W ·W ·Q

Substituting into the expression for the acceleration gives the result.

(d) Follows from definition of the axial vector of an anti-symmetric ten-

sor.

5. Position in a rigid body is given by

x() = a + c()

where the are fixed because the body is rigid. Solve for the by

forming the scalar product with a

a · x() = a · a + a · c()= + a · c()= + a · c()

or

= a · x()− c()The Lagrangian velocity is given by

v(X ) = a + c()

72 CHAPTER 14. RATE OF DEFORMATION

To obtain the Eulerian description, substitute for the to get

v(X ) = aa · x()− c()+ c()

Hence, the velocity gradient is L = aa and the anti-symmetric part is

=1

2aa − aa

The expression for the axial vector follows from exercise 14.2.

Chapter 15

Geometric Measures of

Deformation

1.

C =¡F · F¢

= F · F

= F · F = C

2.

detF = det (R ·U)= detRdetU

= detU

where the last line follows because R is an orthogonal tensor and, hence,

has a determinant equal to one, as long as the rotation is from right-handed

to right-handed.

3. (a)

11 =1

1

= cos

12 =1

2

= sin

21 =2

1

= − sin

22 =2

2

= cos

13 = 31 = 23 = 32 = 0

33 = 1

73

74 CHAPTER 15. GEOMETRIC MEASURES OF DEFORMATION

(b) C = F · F gives in matrix notation

[] = [ ][ ]

=

⎡⎣ cos − sin 0

sin cos 0

0 0 1

⎤⎦⎡⎣ cos sin 0

− sin cos 0

0 0 1

⎤⎦=

⎡⎣2 0 0

0 2 0

0 0 1

⎤⎦or, in dyadic notation,

C = 2e1e1 + 2e2e2 + e3e3

(c)

U =√C

= e1e1 + e2e2 + e3e3

(d)

R = F ·U−1

or, in matrix notation,

[] = [ ] [ ]−1

=

⎡⎣ cos sin 0

− sin cos 0

0 0 1

⎤⎦⎡⎣−1 0 0

0 −1 0

0 0 1

⎤⎦=

⎡⎣ cos sin 0

− sin cos 0

0 0 1

⎤⎦or

R = cos (e1e1 + e2e2) + sin (e2e1 − e1e2) + e3e3

(e)

n = R · e1= cos e1 + sin e2

n = R · e2= − sin e1 + cos e2

n = e3

75

4.

∇Xu =

µe

¶(e)

=

ee

=

ee

=

µ

ee

¶·µ

ee

¶= F ·∇xu

5. Forming the scalar product of each side of Nanson’s formula with itself

gives

()2= 2 ()

2 ¡N · F−1¢ · ¡N · F−1¢

= 2 ()2N ·

³F−1 · F−1

´·N

= 2 ()2 ¡N ·C−1 ·N¢

or

= √N ·C−1 ·N

6.

cosΘ = X · X

=¡F−1 · x

¢ · ¡F−1 · x¢= x ·

³F−1

· F−1´· x

= n ·B−1 · nRearranging gives

cosΘ =n ·B−1 · n

() ()

= −1 −1 n ·B−1 · n

7. Multiplying both sides of Nanson’s formula from the right by F gives

N = n · F

Forming the scalar product of each side with itself gives

22 = (n · F) · (n · F) 2= n · ¡F · F ¢ · n2


Rearranging gives µ

¶2= −2n ·B · n

8. See exercise 6.10.

9.

x = n = F · X= F · X

n

= F ·N

Λn = F ·N

10.

F = e1e2 + e1e1 + e2e2 + e3e3

C = e1e1 + (e1e2 + e2e1) +¡1 + 2

¢e2e2 + e3e3

11. Unit normals in the directions of the diagonals are given by

N+ =1√2(e1 + e2)

and

N− =1√2(−e1 + e2)

The stretches are

Λ+ =pN+ ·C ·N+

=

r1 + +

1

22

and

Λ− =pN− ·C ·N−

=

r1− +

1

22

12. (a)

[ ] =

⎡⎣1 1 0

0 1 0

0 0 1

⎤⎦

77

[] = [ ][ ] =

⎡⎣1 0 0

1 1 0

0 0 1

⎤⎦⎡⎣1 1 0

0 1 0

0 0 1

⎤⎦=

⎡⎣1 1 0

1 2 0

0 0 1

⎤⎦det( − ) = (1− ) (1− ) (2− )− 1 = 0

with solutions

=1

2

³3 +√5´= 2618

= 1

=1

2

³3−√5´= 0382

(b)

Λ =√ = 1618

Λ =√ = 1

Λ =√ = 0618

(c) For the principal direction N corresponding to

(1− ) ()1 + ()2 = 0

()1 + (2− ) ()2 = 0

(1− ) ()3 = 0

The third equation gives ()3 = 0. From the first or second equation

tan =()2()1

= 1618

or = 583 .

13. (a)

[ ] =

⎡⎣1 0

0 1 0

0 0 1

⎤⎦

[] = [ ][ ] =

⎡⎣1 0 0

1 0

0 0 1

⎤⎦⎡⎣1 0

0 1 0

0 0 1

⎤⎦=

⎡⎣1 0

1 + 2 0

0 0 1

⎤⎦


det( − ) = (1− )©(1− )

¡1− + 2

¢− 2ª= 0

with solutions = 1 and

± = 1 +1

22 ±

q1 + (2)

2

Note that since det() = 1, +− = 1. The principal stretches are

Λ =√+ =

Λ =√− = −1

Λ = 1

(b) The components of the principal direction corresponding to + satisfy⎡⎣1− + 0

1 + 2 − + 0

0 0 1− +

⎤⎦⎡⎣()1()2()3

⎤⎦ =⎡⎣000

⎤⎦or

(1− +) ()1 + ()2 = 0

()1 +¡1 + 2 − +

¢()2 = 0

(1− +) ()3 = 0

From the third equation 3 = 0. From the first

()2 = −1− +

()1 = ()1

Therefore the principal direction can be written as

N = cosΘ e1 + sinΘ e2

where tanΘ = . Because = 1 the corresponding principal

direction is given by N = e3. Then N = N ×N :

N = − sinΘe1 + cosΘ e2

Therefore N = R · e (Note that is not an index here and

and denote corresponding Roman and Arabic numbers) implies

R = cosΘ (e1e1 + e2e2) + sinΘ (e2e1 − e1e2)

(c) The deformation tensor is given in principal axis form by

U = NN + −1NN + e3e3

79

Substituting for N and N from (b) and multiplying out the dyads

gives

U = ©cos2Θe1e1 + cosΘ sinΘ (e2e1 + e1e2) + sin

2Θe2e2ª

+−1©sin2Θe1e1 − cosΘ sinΘ (e2e1 + e1e2) + cos2Θe2e2

ª+e3e3

Collecting the coefficients of the dyads gives

11 = cos2(Θ) + −1 sin2(Θ)

12 = 21 =¡− −1

¢cos(Θ) sin(Θ)

22 = −1 cos2(Θ) + sin2(Θ) =

33 = 1

and after some algebra

11 =2

2 + 1

12 = 21 =2 − 12 + 1

22 =3 + −1

2 + 1

33 = 1

14. First find the principal directions in the current state. From Exercise 15.9,

Λn = F ·N , (no sum on )

Therefore

Λn = F ·N

giving

n =1

√1 + 2

(1 + ) e1 + e2

Using 1 + = 2 gives

n =1√1 + 2

e1 + e2

Similarly

n =1√1 + 2

−e1 + e2

Noting that R = ne or comparing with the result of Exercise 6.14

give

R = (e1e1 + e2e2) cos + (e2e1 − e1e2) sin + e3e3


where

tan = −1

(see Figure below). Recall from Exercise 15.13 that

tanΘ =

and, hence, = 2−Θ. Because R rotates the N into the n ,

R = nN

Substituting for the n and N gives

R =2

1 + 2(e1e1 + e2e2)− 2 − 1

1 + 2(e2e1 − e1e2) + e3e3

(Alternatively, R can be found from F ·U−1.) Identifying

cos =2

1 + 2

sin =2 − 11 + 2

gives tan = 2. Note the minus sign preceding sin in R indicates that

is positive for a clockwise rotation about the e3. Using the trigonometric

identity

tan (± ) =tan± tan1∓ tan tan

and the expressions tanΘ = and tan = −1 gives = Θ − (See

figure below).

n

N

The figure shows the relation between the rotations. The principal axis in the

reference configuration N is rotated an angle clockwise about the e3 axis

into the principal axis in the current configuration n .

Chapter 16

Strain Tensors

1. Writing position in the reference configuration as = − gives

−1 =

= −

Substituting into (16.11)

=1

2

½ −

µ −

¶µ −

¶¾=

1

2

½ − +

+

+

¾=

1

2

½

+

+

¾

2. From (16.15), L = F · F−1. From F = R ·U

F = R ·U+R · U

and

F−1 = (R ·U)−1 = U−1 ·R−1= U−1 ·R

Substituting

L = R ·U ·U−1 ·RT +R · U ·U−1 ·R

= R ·RT +R · U ·U−1 ·R

81

82 CHAPTER 16. STRAIN TENSORS

3. (a)

E(−2) =1

2

X

¡1− Λ−2

¢NN

=1

2

(X

NN −X

1

ΛNN ·

X

1

ΛNN

)

=1

2

¡I−U−2¢

C−1 =¡U ·U¢−1 = U−1 ·U−1 = U−2

(b)

−1 =

=

−

= −

C−1 =¡F · F¢−1 = F−1 · F−1

−1 = −1−1 = −1

−1

=

µ −

¶µ −

¶= −

−

+

(−2) =

1

2

µ

+

−

¶(c)

e =1

2

³I−F−1 · F−1

´=

1

2

nI− (R ·U)−1 · (R ·U)−1

o=

1

2

nI− ¡U−1 ·R

¢ · ¡U−1 ·R¢o

=1

2

nI−R ·U−1 ·U−1 ·R

o= R ·

½1

2

³I−U−1 ·U−1

´¾·R

= R ·E(−2) ·R

83

4. (a)

e(−2) =1

2

X

¡−2 − 1

¢nn

=1

2

(X

1

nn ·

X

1

nn −

X

nn

)

=1

2

¡V2 − I¢

=1

2

¡F · F − I¢

=1

2(B− I)

(b)

(−2) =

1

2( − )

=1

2

½µ +

¶µ +

¶−

¾=

1

2

½

+

+

¾(c)

e(−2) =1

2

¡F · F − I¢

=1

2

n(R ·U) · (R ·U) − I

o=

1

2

©(R ·U) · ¡U ·R

¢− Iª= R ·

½1

2

¡U2 − I¢¾ ·R

= R ·E ·R

5.

lnΛ = Λ− 1− 12(Λ− 1)2 + 1

3(Λ− 1)3 +

Therefore

E(0) = lnU =X

ln(Λ)NN

=X

(Λ − 1)NN −X

1

2(Λ − 1)2NN +

= (U− I)−12(U− I)2 + 1

3(U− I)3 +


6. Principal values of the Green-Lagrange strain are

= N ·E ·N

= N ·½1

2(C− I)

¾·N

=1

2(N ·C ·N − 1)

=1

2

¡Λ2 − 1

¢Solving for Λ2 gives

Λ2 = 1 + 2

Prinipal values of the Almansi strain are

= n · e · n= n ·

½1

2

¡I−B−1¢¾ · n

=1

2

©1− n ·B−1 · n

ª=

1

2

½1− 1

Λ2

¾Solving for Λ2 gives

Λ2 =1

1− 2Equating the two expressions for Λ2 gives the result.

7. Differentiating

R ·R = I

gives

R ·R +R · R = 0

R ·R = −R · R

= −³R ·R

´8.

E(ln) =X

ln (Λ)NN

If the principal axes of U (i.e., the N) do not rotate

E(ln) =X

Λ

ΛNN

85

and the N can be replaced by n . But

Λ

Λ=

1

=1

Therefore ΛΛ are the principal values of D and

E(ln) = D

9.

= 123 + 123 + 123

Use

F = L · Fand rearrange the indices to get

= 123

where

= + +

10.

(n) =

¡N · F−1¢

= N³ · F−1 + · F−1

´Using = tr, from the preceding problem and F−1 = −F−1 · L give

(n) =

¡N · F−1¢ tr− ¡N · F−1¢ · L

= n tr− n · L

Forming the scalar product with n and noting that n · n = 1 gives

n · (n) = (tr− n · L · n)

The left side gives

(n) = n+ n

()

Forming the scalar product with n and noting that n ·n = 1 and n ·n = 0gives

() = (tr− n · L · n)


x = F · X

x

= F · X

but

= Λ

n =x

N =X

yielding the desired result.

11. (a) Differentiate Λn = F ·N (Exercise 15.9) to get

Λn+ Λn = F ·N (16.11.1)

Multiply both sides of Λn = F ·N by F−1 to get

N = ΛF−1 · n (16.11.2)

and substitute into the precding equation:

Λn+ Λn = ΛL · n

where L = F ·F−1. Dot both sides with n

Λn · n+ Λn · n = Λn ·L · n

But since n is a unit vector

n · n = 1 (16.11.4)

and.

n · n = 0 (16.11.5)

Since n ·W · n = 0 (because W = −W ). n · L · n = n ·D · nConverting to index notation and dividing through by Λ gives the

result.

(b) Differentiate x = F · X to get

v = F · X = L · x

where the second equality follows from the definition of L. Differen-

tiate again to get

a = F · X = L · x+L · v (16.11.5)

87

where a is the acceleration. Using X = F−1 ·x in the first equalityof (16.11.5) yields

a = F · F−1 · x = Q · x (16.11.6)

and gives

Q = F · F−1 (16.11.7)

Using v = L · x in the second equality of (16.11.5) yields an ex-pression for Q in terms of L

Q = L+L ·LDifferentiating (16.11.1) gives

Λn+ 2Λn+ Λn = F · (ΛF−1 · n) (16.11.8)

where (16.11.2) has been used forN . Dotting both sides of (16.11.8)

with n and using (16.11.7), (16.11.5) and (16.11.4) yield

Λ+ Λn · n = Λn ·Q · n (16.11.9)

Differentiating (16.11.4) yields

n · n+ n · n = 0Using n · n = −n · n in (16.11.9), dividing through by Λ and con-verting to index notation gives the result.

Another approach is to begin with

2

() = v · x+ x · v

Differentiating again gives

2

µ

()

¶2+ 2

2

2() = a · x+ x · a+ 2v · v

Using the last equality in (16.11.6), noting that Q is symmetric and dividing

through by 2()2 gives

1

2

2() = n ·Q · n+

µ1

¶2v · v −

µ1

()

¶2(16.11.10)

Multiplying the top and bottom of the left side by gives ΛΛ and the last

term on the right is (n ·D · n)2. Differentiate x = n to get

v = n+ n

()

Dividing through by and dotting the result with itself give

v

· v= n · n+

µ1

()

¶2Substituting back into (16.11.10) gives the desired result.

Chapter 17

Linearized Displacement

Gradients

1. Using sin ≈ , (17.8) and (17.11) in (17.13) gives

≈ N ·N + 2N · ε ·N

(1 +N · ε ·N) (1 +N · ε ·N)

Because N and N are orthogonal N ·N = 0. To first order in ε

1

(1 +N · ε ·N)≈ 1−N · ε ·N +

and similarly for the other term in the denominator. Retaining only terms

that are linear in ε gives (17.14).

2. Because

U =X

ΛNN

the inverse of U is given by

U−1 =X

1

ΛNN

Because

C ≈ I+ 2εC, U and ε have the same principal directions. Because Λ = 1+N · ε·N,the principal stretch ratios are

Λ = 1 +

where the are the principal values of ε. Therefore

1

Λ=

1

1 + ≈ 1−

89

90 CHAPTER 17. LINEARIZED DISPLACEMENT GRADIENTS

Substituting back into the expression for U−1 gives

U−1 =X

NN −X

NN

= I− ε

3. Because = +

−1 ≈ −

Also, ≈ 1 + trε. Substituting into Nanson’s formula gives

≈ (1 + trε)

½

µ −

¶¾

≈ (1 + trε)−

≈ (1 + trε)− ( +Ω)

4. The infinitesimal rotation vector

=1

2Ω

=1

2

∙1

2

µ

−

¶¸=

1

4

µ

+

¶=

1

2

Taking the divergence

=1

2

2

Because is skew symmetric and 2 is symmetric with re-

spect interchange of and , the product vanishes.

5. Setting the curl of the integrand in (17.32) equal to zero gives

+ = 0

+ = 0

+ = 0

+ − = 0

+ − = 0

Relabelling indices gives (17.33)

Part IV

Balance of Mass,

Momentum, and Energy

91

Chapter 18

Transformation of Integrals

1. Z Z

=

Z

Z 2()

1()

=

Z

[2() ]− [1() ]

=

Z

[2() ] +

Z

[1() ]

=

Z

From Figure 18.2

=

= cos

2. (a) Z

n · σ · v =

Z

=

Z

()

=

Z

½

+

¾

If is stress and is velocity then the second term can be re-

placed by , because = and is the symmetric part

of .

93

94 CHAPTER 18. TRANSFORMATION OF INTEGRALS

(b) Z

=

Z

()

=

Z

½

+

¾

=

Z

½ +

¾

=

Z

½ +

¾

If = , then the first term vanishes because = − .3. Connect L to the boundary by a surface AB as shown in the Figure.

Then apply the divergence theorem to the volume enclosed by the outer

surface S and a contour that wraps around ABC as shown. Because the

traction n ·σ continuous on AB, the contribution from this portion of the

surface cancels because the segment AB is traversed in opposite directions.

Because n ·σ is discontinuous on (BC) there are different contributionsfrom the + and − sides. The result isZ

∇ · σ =

Z

n · σ +Z

n(n · σ)+ − (n · σ)−

o

L-

L+ n V

S

A

B

C

Chapter 19

Conservation of Mass

1.

=

Z

A

=

Z

½

A +

A

+ A

¾

Use

= trL =∇ · v

and rearrange to get

=

Z

½∙

+ ∇ · v

¸A+

A

¾

The term [ ] is zero by mass conservation leaving

=

Z

A

2.

µ4

33

¶= 42

95

96 CHAPTER 19. CONSERVATION OF MASS

Chapter 20

Conservation of Momentum

1.

∇ · σ = e

· (x)

= e (x)

· nn

= ∇ (x) · nn

Because equilibrium (in the absence of body forces) requires that this

vanish and the unit vector n is not zero,

∇ (x) · n = 0

Hence, the gradient of (x) is orthogonal to n since their scalar product

vanishes.

2.

1

"−¡21 − 22

¢2

#+

2

h2

12

2

i=

−212+ 2

1

2= 0

2

"

¡21 − 22

¢2

#+

1

h2

12

2

i=

−222+ 2

2

2= 0

Therefore, equilibrium is satisfied.

97

98 CHAPTER 20. CONSERVATION OF MOMENTUM

3.

() +

() =

+

+

() + ()

=

½

+

()

¾+

½

+

¾=

where the first term vanishes because of mass conservation. The balance

of linear momentum for a control volume fixed in space expresses that the

work of the surface tractions on the boundary and the body forces in the

volume Z

t+

Z

b

plus the flux of momentum through the boundary into the volume

−Z

n · v (v)

(where the minus sign appears because the normal is outward) equals the

rate of change of momentum in the volume

Z

v

The time derivative can be moved inside the integral because the volume

is instantaneously fixed in space. Writing the traction as t = n · σ, usingthe divergence theorem on this term and the term for the momentum flux

and using the result from the first part of the problem gives the equation

of motion.

Chapter 21

Conservation of Energy

1.

∇ · (σ · v) = e

· [(ee) · e]

=

½

+

¾

=

+

=

½e

· (ee)

¾· e + (ee) · ·ee

= (∇ · σ) · v+ σ · · (∇v)

Because σ = σ , ∇v can be replaced by its symmetric part L.2. Substitute σ = −I into (21.6) to get

= − trL−∇ · q+

But

trL =∇ · v = −1

where the last equality follows from mass conservation (19.10). Substitut-

ing gives the result.

3. If the material is rigid, σ · ·L = 0 and

=

in (21.6). If the energy is only a function of the temperature

=

=

Substituting this and Fourier’s law and rearranging gives the result.

99

100 CHAPTER 21. CONSERVATION OF ENERGY

4. Energy conservation is given by (21.1). The power input express in terms

of the reference configuration is

=

Z

t0 · v+Z

0b0 · v

The first term can be rewritten asZ

t0 · v =Z

N ·T0 · v =Z

∇X ·¡T0 · v¢

Working out the integrand yields

∇X ·¡T0 · v¢ = ¡∇X ·T0

¢ · v +T0 · ·FThus the power input can be rewritten as

=

Z

©∇X ·T0 + 0b0ª · v + Z

T0 · ·F

The equation of motion (20.13) can be used in the first term to give

=

Z

0v

· v +

Z

T0 · ·F

and then the first term can be rearranged as follows:

=

Z

µ1

20v · v

¶ +

Z

T0 · ·F

The rate of heat input is

= −Z

N ·Q+Z

0

where the first term is the flux of heat through the surface area (and the

minus sign occurs because the normal points out of the body) and the

second term is the heat generated within the volume of the body. Using

the divergence theorem on the first term gives

= −Z

∇X ·Q +Z

0

The rate of change of the total energy is

=

Z

0 +

Z

µ1

20v · v

¶

where the first term is the rate of change of the internal energy and the

second is the rate of change of the kinetic energy. Because the volume

is fixed in the reference state the time derivatives can be taken inside

101

the integrals. Substituting into (21.1), cancelling the common term and

rearranging givesZ

½0

−T0 · ·F+∇X ·Q−0

¾

Because this must apply for any volume, the integrand must vanish which

gives (21.7). Because the heat flux must be the same whether referred to

the current configuration or the reference configuration

N ·Q = n · q

and using Nanson’s formula (15.12) gives the relation between Q and q.

5.

E(−2) =1

2

¡I−U−2¢

=1

2

¡I−C−1¢

=1

2

¡I−F−1 · F−1 ¢

Therefore

E(−2) = −12

³F−1 · F−1 +F−1 · F−1

´To compute F−1 differentiate F · F−1 = I to get

F · F−1 +F · F−1 = 0

Rearranging gives

F−1 = −F−1 · F · F−1= −F−1 · L

and the transpose is

F−1 = −L · F−1

Substituting into

E(−2) =1

2

©F−1 · L · F−1 +F−1 · L · F−1ª

= F−1 · 12

©L+ L

ª · F−1= F−1 ·D · F−1

or

D = F · E(−2) · F

102 CHAPTER 21. CONSERVATION OF ENERGY

Substituting into the stress working equality

S(−2) · ·E(−2) = σ · ·D= σ · ·

³F · E(−2) · F

´=

¡F · σ · F¢ · ·E(−2)

Hence

S(−2) = F · σ · F

6.

σ · ·D = S(1) · ·E(1)= S(1) · ·U

L = F · F−1

=³R ·U+R · U

´· (R ·U)−1

= R ·R +R · U ·U−1 ·R

Taking the transpose

L = R · R +R ·U−1 · U ·R

Noting that R · R is anti-symmetric and forming

D = R ·∙1

2

³U ·U−1 +U−1 · U

´¸·R

Substituting in the work-rate equality

S(1) · ·U = σ · ·D= σ · ·

½R ·

∙1

2

³U ·U−1 +U−1 · U

´¸·R

¾=

1

2

©U−1 · ¡R · σ ·R¢+ ¡R · σ ·R¢ ·U−1ª · ·U

Therefore

S(1) =1

2

©U−1 · ¡R · σ ·R¢+ ¡R · σ ·R¢ ·U−1ª

Part V

Ideal Constitutive Relations

103

Chapter 22

Fluids

1. For v = (2)e1, the flow is isochoric, ∇ · v = 0 and v ·∇v = 0. For

steady flow v = 0 and for no body force b = 0. The remaining terms

in equation (23.12) are

2

22+

1= 0

where 1 = constant 0 for flow in the positive 1 direction. Inte-

grating yields

(2) =1

2

122 +2 +

where and are constants. Because the velocity of the fluid must equal

the velocity of the boundary at 2 = ±,(±) = 0

Solving for the constants yields = 0 and

= −2

2

1

Substituting back into the velocity yields

(2) = − 1

2

1

¡2 − 22

¢The only non-zero component of shear stress is

=

2=

12

2. For velocity only in the direction and only a function of radial distance

v = ()e. From problem 8.13, the only nonzero component of D is

=1

2

³

´105

106 CHAPTER 22. FLUIDS

and the shear stress is

= 2 =

³

´From the answer to problem 8.14, the only non-trivial equilibrium equation

is

+

2

=

1

2

¡2

¢= 0

Substituting for gives

½3

³

´¾= 0

Integrating yields

() =

+

Because the fluid velocity must equal the velocity of the boundary

( = ) = Ω

and

( = ) = 0

Solving for the constants yields

= −Ω 2

2 − 2

and

= Ω22

2 − 2

The shear stress at =

( = ) = −2Ω 2

2 − 2

where this is the negative of the force per unit area on the cylinder. The

torque exerted on the cylinder is

= −22∆( = )

= 4∆Ω22

2 − 2

where∆ is the distance out of plane (along the axis of the cylinder).Alternate

method: the only non-zero term in equation (23.12) is

∇2v = 0

107

From the answer to problem 8.10

∇2 = 1

µ

¶+1

22

2+

2

2

Applying to the form of the velocity field gives

00() +1

0()− 1

2() = 0

where 0() = . Looking for a solution of the form gives = ±1and a velocity field of the same form as before.

3. The solution is again given by (23.15). The boundary conditions are now

( →∞ ) =

and

( → 0 ) = 0

The second corresponds to → 0 and gives = 0. The first corresponds

to →∞ and gives

=2√

Hence, the velocity is

( ) = erf

µ√4

¶At a distance of 1mm the velocity 1 s is 145%, 52% and 66% of for

air, water and SAE 30 oil.

4. Look for a solution of the form

( ) = 0 exp () exp()

and take the real part so that (0 ) satisfies

(0 ) = 0 cos()

Substituting into equation (23.13) and cancelling common terms gives

2 =

or

=√p

where √ = exp (4) , exp

µ

2

³2+ 2

´¶

108 CHAPTER 22. FLUIDS

Take the second solution so that ( )→ 0 as →∞Therefore,

( ) = 0 exp³−p2

´Rhexp

³−

p2

´iThe solution is plotted in the Figure below.

Velocity ( )0 vs. nondimensional distance 2 for four nondimensional

times = 00 20 50 100.

Chapter 23

Elasticity

1. If the strain energy is regarded as a function of F

=

where we have used (23.8). The components of the Green-Lagrange strain

are given by (16.4)

=1

2 −

Calculating the derivative

=1

2

½

+

¾=

1

2 +

=1

2 +

Substituting into the first equation above gives

=1

2 +

=1

2 +

= = 0

where the last line follows because = .

2. (a) Regarding C as a function of E gives

=

=

109

110 CHAPTER 23. ELASTICITY

Because = + 2 the second term is

= 2

This gives

= 2

Because the material is isotropic, is a function only of the invari-

ants of C.

= 2

½

1

1

+

2

2

+

3

3

¾= 2

½1

1

+2

2

+3

3

¾The forms of the first two invariants are given by (7.9) and (7.10).

For 3 = det(C) the form is given by problem 7.3. The derivatives of

the invariants can be computed as follows. For 1

1

=

= =

For 2

2

=

½1

2[ − ]

¾=

1

2 + − 2

=1

2 + − 21 = − 1

For 3

3

=

½1

3 − 12 − 1

331

¾=

1

3( + + )

− 1

2 − 12

− 211

=1

3( + + )

−2 − 1 ( − 1)− 21

= − 2 − 1

where the last line uses the symmetry of C. Substituting into =

2 gives

= 2 1 +2 ( − 1) +3 ( − 2 − 1)= 2 [1 − 12 − 23] + [2 − 13] +3

111

or

S = 2 I [1 − 12 − 23] +C [2 − 13] +3C ·C(b) The relation between the Cauchy stress σ and the second Piola-

Kirchhoff stress is given by (21.12) Substituting the result of (a)

gives

σ =2

F · I · F [1 − 12 − 23] +F ·C · F [2 − 13]

+3F ·C ·C · F Each of the terms can be expressed in terms of B as follows

F · I · F = F · F = BF ·C · F = F · ¡F · F¢ · F = ¡F · F ¢ · ¡F · F ¢ = B ·B

F ·C ·C · F =¡F · F ¢ · ¡F · F ¢ · ¡F · F ¢ = B ·B ·B

But because B and C have the same principal values, they have the

same invariants. Therefore

B3 = B ·B ·B = 1B ·B+2B+ 3

Also

2 = det(F · F ) = det(F · F ) = det (B)Substituting into the first equation of (b) gives the result.

(c) From problem 7.4.a

B ·B = 1B+2I+3B−1

Substituting this into the result of (b) gives the result.

3. With τ = σ, (21.12) gives

τ = F · S · F

or, in index notation

=

To linearize write the stresses as = + and = + where is the Cauchy stress in the reference state. The components of

the deformation gradient are

= +

where = . Substituting and retaining only terms of first order

gives

+ = ( + ) ( + ) ( + )

= + + +

= + + +


Cancelling yields

= + +

The result is symmetric with respect to interchange of and . Substitute

= +Ω

and rearrange to give

−Ω −Ω = + + Therefore

∗ =

where

= + +

and

∗ = −Ω −Ωis the Jaumann increment of the Kirchhoff stress (Prager, Introduction to

Mechanics of Continua, Dover, p. 155). This is the rate as computed by

an observer rotating with the material. Because = , can be

written symmetrically on and :

= +

1

2 + + +

The result is also symmetric on interchange of and . To linearize

=

note from Section 17.1.3 that

= det (F) ≈ 1 +

Substituting and writing the stresses as the sum of the Cauchy stress in

the reference state and an increment gives

+ = ( + ) (1 + )

= + +

Cancelling the reference stress on both sides gives

=

=¡ +

¢

or

=

−

Note that 6=

.

113

4. The Kirchhoff stress τ is related to the Cauchy stress σ by

τ = σ

Writing τ = σ + τ and σ = σ + σ gives

σ + τ = (σ + σ)

where σ is the Cauchy stress in the reference state. = det(F) can be

written as

= det

µ +

¶where is the component of displacement from the reference state. Lin-

earizing for small displacement gradients (Sec. 17.1.3), | | ¿ 1,

gives

' 1 + tr(ε)Substituting and cancelling σ on both sides gives

τ = σ + σ tr ε

For tr ε sufficiently small so that the product σ tr ε is small compared to

the stress increment, the Cauchy stress and Kirchhoff stress are approx-

imately equal. Alternatively, multiplying and summing (23.20) with the

infinitesimal strain gives

=

−

If the incremental volume strain (multiplied by the stress in the reference

state) is sufficiently small, the incremental moduli for the Cauchy and

Kirchhoff stress are the same.

5. The thermal conductivity tensor can be written as

=

where = .

(a) If the heat flux can be written

=

then

=2

=

2

=

because the derivatives can be taken in either order.


(b) If the 1 plane is a plane of symmetry, then

0 = =

for a reversal of the 2 axis corresponding to 11 = 33 = −22 = 1.Therefore

12 = 12 = 112212 = −12

and, hence, 12 = 0 and, similarly, 23 = 0.

(c) If the 12 plane is a plane of symmetry, the same argument as in

(b) with 11 = 22 = −33 = 1 gives 13 = 0.

(d) If the material is isotropic, then K must be an isotropic tensor, K =

I or = .

6. Consider a rotation through an angle around the 3 axis. The compo-

nents of a tensor in the primed (rotated) system are given in terms of the

components in the unprimed system by the answer to Problem 6.15b:

011 = 112 + 212 + 22

2

012 = (22 − 11) + 12¡2 − 2

¢ 013 = 13+ 23

022 = 112 − 212 + 22

2

023 = −13+ 23

033 = 33

where is symmetric and = cos and = sin . First show that

44 = 55. From above

013 = 13 + 23

= 25513 + 24423

Also

013 = 2055013

= 2055 13 + 23

Equating the coefficient of 23 in the two expressions for 013 gives 44 =

055 and equating the coefficients of 13 gives 055 = 55. Therefore 44 =

55. Now consider

033 = 33

with

33 = 1311 + 2322 + 3333

115

and

033 = 013011 + 023

022 + 33

033

= 013©11

2 + 212 + 222ª

+023©11

2 − 212 + 222ª

+03333

Equating the coefficients of 12 gives 013 = 023. Equating the coefficients

of 11 and 22 gives

2013 + 2 023 = 13

and

2013 + 2 023 = 23

respectively. With 013 = 023 we conclude that 013 = 13 = 23. Now

examine

012 = (22 − 11) + 12¡2 − 2

¢The left side can be written as

012 = 2 066012

= 2 066©(22 − 11) + 12

¡2 − 2

¢ªand the right side as

(22 − 11) + 12¡2 − 2

¢= (12 − 11) 11 + (22−12) 22

+266¡2 − 2

¢12

Again equating coefficients of the strain components gives

11 = 22

066 = 66 =1

2(11 − 12)

The results are independent of the particular value of and hence apply

for any rotation about the 3 axis.

7. Consider a rotation of the axis through an angle about the 3. Then

022 = 211 + 222 − 212

where = cos and = sin . But

11 = 1111 + 1222 + 1233

22 = 1211 + 1122 + 1233

12 = 24412


and

022 = 12011 + 11

022 + 033

= 12¡211 + 212 + 222

¢+11

¡211 − 212 + 222

¢+1233

Substituting and equating the coefficients of 12 give

44 =1

2(11 − 12)

8. From (23.32) with only 11 6= 0:

211 =2 (+ )

3+ 211

Therefore

11 = (3+ 2)

(+ )11 = 11

gives (23.33). Equation (23.32) with only 11 6= 0 also gives

222 = −

(3+ 2)11

Substituting for 11 from above and solving for 22 gives

22 = −

2 (+ )11

Comparing with the equation preceding (23.34) yields (23.34).

9. Equation (23.32) for only 12 6= 0 gives 212 = 12 The equation follow-

ing (23.34) gives

12 =(1 + )

12

Equating the expressions for 12 gives (23.35). Eliminating from (23.35)

and (23.33) gives (23.36).

10. From (23.31) with only 11 6= 0

11 = (+ 2) 11

22 = 11

Therefore = (+ 2) and

22

11=

+ 2

117

Using (23.36) gives

= 21−

1− 2and

22

11=

1− 2

11. (a) From the equation following (23.34)

33 = 0 =(1 + )

33 −

(11 + 22 + 33)

which gives

33 = (11 + 22)

(b) Substituting the result from (a) back into the equation following

(23.34) gives

=(1 + )

−

(11 + 22 + (11 + 22))

=(1 + )

− (11 + 22)

12. From (23.26) = (12) = (12) . Substituting (23.31)

gives

=1

2( + 2)

=1

2

n ()

2+ 2

oWriting = 0 + (13), where 0 is the deviatoric strain, andcarrying out the product gives (23.39).

13. From equation preceding (23.32) = + (23). Using (23.36) gives

= 2(1 + )

3 (1− 2)

Then 0 and 0 give (23.40). Using −1 and 0 in (23.35)

gives 0.

14. Setting = 3 and taking the stated limits gives = 0 and = 2.

Therefore = = 22, = −3, = 23 and

=

23

µ − 3

2

¶


15. (a) The initial and final volumes are = (43)3 and =

(43) (+ 0)3. For small (infinitesimal) strain

= −

= (1 + 0)3 − 1

≈ 30

where other terms are at least as small as (0)2. Therefore 0 is equal

to one-third of the volume strain.

(b) From (23.3.2), the radial displacement for the general spherically

symmetric solution has the form

= () =

2+

where and are constants. If the elastic constants are included

in calculating the form of the stresses

= (3+ 2) + 2

2

µ − 3

2

¶and

= (3+ 2) − 4 2

To calculate the pressure needed to restore the spherical region to its

original size, note that = 0, ( = ) = −0 and

( = ) = −0These conditions give = −0 and 0 = (3+ 2) 0. To calculate

the stress and displacement due to removing the force layer, note

that = 0 for and = 0 for . Therefore, for

+() =

2

+() = −4 3

and for

−() =

−() = (3+ 2)

The constants and are determined by the conditions of continuity

of displacements

+() = −()

119

and that the jump in equal minus the pressure due to the force

layer

+() + 0 = −()

Solving for and gives

= 0+ 23

+ 2

= 3

Since = −(), the is the actual strain of the inclusion is

= −() = = 0+ 23

+ 2

The pressure in the inclusion is the pressure due to restoring the

size of the inclusion due minus the pressure due to removing the force

layer −():0 − −() = −+() = 4

16. The form of the solutions for () and () are the same as in the

preceding problem. But now −() and +() are the total displacementsfrom the unstressed state. Hence they satisfy the following condition at

= :

+() = −() + 0

The tractions are now continuous at = :

+() = −()

Using these conditions to solve for and gives

= 03+ 23

+ 2

= − 4

3 (+ 2)

Recognizing that the net strain of the inclusion is +() and the pressure

is −−() gives the same results as the preceding question.

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