Solutions for Fundamentals of Continuum Mechanics. John W. Rudnicki

Solutions for Fundamentals of Continuum Mechanics John W. Rudnicki December, 015

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Contents I Mathematical Preliminaries 1 1 Vectors 3 Tensors 7 3 Cartesian Coordinates 9 4 Vector (Cross) Product 13 5 Determinants 19 6 Change of Orthonormal Basis 5 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 1 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 0 Conservation of Momentum 97 1 Conservation of Energy 99 V Ideal Constitutive Relations 103 Fluids 105 3 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 putting the 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. Usingthesameprocedurewithv u, the thumb of the right hand is 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 in the plane of u and v the scalar product of w with u and v is zero.. If u v has a positive scalar product with w, thenu, v and w are right handed. In this case u v w = v w u = w u v are all equal to the 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 + =0 Since 6= 0, cos + =0. 4. If u v =0for all v, then the scalar product must be zero for v = u where is a nonzero scalar. Hence, u u = =0. Therefore =0 and u =0. 5. (a) (u + v) (u v) = u u + v u u v v v = +u v u v = 3

4 CHAPTER 1. VECTORS (b) (u + v) (u v) = u u + v u u v v v = 0+( u v) u v 0 = u v 6. Define a, b, andc to coincide with the edges of the plane triangle with c = a + b. c b a (a) (b) = c c =(a + b) (a + b) = a a + a b + b a + b b = +a b + = + cos( )+ = cos()+ 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 of this face is n = 1 (b a) (c a) = 1 (b c) 1 (a c) 1 (b a) Noting that 1 (b c) = 1 1, 1 (a c) =, 1 (b a) = 3 3 gives the result. 8. Forming the vector product of w with both sides of (1.5) gives w {u (v w)} = w v Forming the scalar product of both sides with w v and solving for gives (w v) [w {u (v w)}] = w v Similarly = (v w) [v {u (v w)}] v w 9. A line perpendicular to both of the given lines is w = c +(l m). This line will intersect the line u = a + l if c +(l m) 13 = a + l 1 for 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 3 = (l m) (b c) l m Because w = c +(l m) is perpendicular to both given lines, the minimum distance between them is (l m) (a b) 13 3 = 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. Ifa, b, andc are coplanar, then u (v w) =0, and it is not possible to express an arbitrary vector z in the given form. 1. 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 in the direction of w is n = (u v) (x y) (u v) (x y)

Chapter Tensors 1. Let S = 1 F + F.IfS u = u S = u S for any u then S is symmetric. S u = 1 F + F u = 1 n o F u + F u = 1 nu o F + u F = 1 u F + F ª = u S Note: F u = u F = F u for all u implies the result F = F. Let A = 1 F F. If A u = u A = u A for any u then A is symmetric. A u = 1 F F u = 1 n o F u F u = 1 nu o F u F = 1 u F F ª = u A. Substituting (.10) into (.5) and using (.6) gives the result. 3. implies (F G) 1 u = v u = F G v 7

8 CHAPTER. TENSORS 4. Multiplying from the left by F 1 and then G 1 gives G 1 F 1 u = v The result follows by noting that the two expressions for v are equal for all u. implies But u = F v = v F gives u = F v v = F 1 u = u F 1 v = u F 1 Noting that the two expressions for v are equal for all u gives the result. 5. Consider v =(R S) u Forming the scalar product of v with itself gives v v = u (R S) (R S) u where 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.5.) gives v v = u S R (R S) u Rearranging the parentheses gives v v = u S R R S u But if R is orthogonal, R R = I yielding v v = u S S u Similarly, if S is orthogonal, S S = I and v v = u u. Hence, R S is orthogonal. 6. The angle between vectors u and v isgivenintermsofthescalarproduct by (1.). 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 = 11 1 + 1 + 31 3 = 1 1 + + 3 3 3 = 13 1 + 3 + 33 3 (b) = 11 + + 33 (c) = 11 11 + 1 1 + 13 13 + 1 1 + + 3 3 + 31 31 + 3 3 + 33 33 (d) 11 = 11 1 = 1 13 = 13 1 = 1 = 3 = 3 31 = 31 3 = 3 33 = 33. (a) = 11 + + 33 = 1+1+1=3 9

10 CHAPTER 3. CARTESIAN COORDINATES (b) = 11 11 + 1 1 + 13 13 + 1 1 + + 3 3 + 31 31 + 3 3 + 33 33 = 1+0+0 +0+1+0 +0+0+1 = 3 3. 1 = 11 1 31 1 3 1 3 13 3 33 3 = 1 3 11 1 13 1 3 31 3 33 11 1 13 1 3 = 11 1 13 1 3 31 3 33 31 3 33 4. (a) v I = ( e ) ( e e ) = (e e ) e = e = e = e = v (b) F I = ( e e ) ( e e ) = e (e e ) e 5. If u v = =0for any v, thenchoose = e e = e e = e e = F 1 =1, =0, 3 =0 1 =0.Choose 1 =0, =1, 3 =0 =0.Choose 1 =0, =0, 3 =1 =0. Hence, u =0.

11 6. F = F I = ( e e ) ( e e ) = (e e )(e e ) = = = = 11 + + 33 7. F G = = = = (e e )(e e ) = ( e e ) ( e e ) = G F 8. F G = = = F : G F G = = = F : G 9. F G H = ( e e ) ( e e ) ( e e ) = ( e e ) ( e e ) = ( e e ) ( e e ) = = H F G = ( e e ) ( e e ) ( e e ) = ( e e ) ( e e ) = ( e e ) ( e e ) = = G H F = ( e e ) ( e e ) ( e e ) = ( e e ) ( e e ) = ( e e ) ( e e ) = =

1 CHAPTER 3. CARTESIAN COORDINATES 10. 11. F G :H = ( e e ) ( e e ):( e e ) = ( e e ):( e e ) = ( e e ):( e e ) = = = ( e e ):( e e ) = ( e e ):( e e ) ( e e ) = F : G H tr (F G) = tr ( e e ) = I ( e e ) = (e e )(e e )( ) = ( ) = = F G

Chapter 4 Vector (Cross) Product 1. e (e e ) = e ( e ) = (e e ) = =. 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 = (b) = =6 4. e e = e (e e ) = e = e = 1 e e 13

14 CHAPTER 4. VECTOR (CROSS) PRODUCT 5. (a) 6. (b) u (v w) = e ( e ) = e = ( ) e = e e = (u w) v (u v) w = u (wv vw) 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 (u v) w = (e ) e Therefore, = v w and = u w 7. Let u = n and w = n in (4.14) to get = e = ( ) e = e e = (u w) v (v w) u n (v n) =(n n) v (n v) n Since n is a unit vector n n =1. Rearranging give 8. (a) (b) v =(n v) n + n (v n) (a b) (c d) = = ( )( ) = ( )( ) ( )( ) = (a c)(b d) (a d)(b c) (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 10. (a) Therefore, = = = = = 1 If W is not anti-symmetric, then w = 0. n ṅ = n (w n) = n (nw wn) where the second line follows from problem 5. Noting that n n =1 because n is a unit vector and n w =0because n is orthogonal to w gives the result. (b) = = ( ) = ( ) = or W = ṅn nṅ in dyadic notation. 11. u (v F) = e ( e e e ) = e ( e e ) = e = e e e = (u v) F

16 CHAPTER 4. VECTOR (CROSS) PRODUCT 1. Writing both sides in index form gives Because this applies for all u and v, e = e = Multiplying both sides by and summing gives = = or = 1 and re-labelling indices gives the result. 13. = g 1 (g g 3 ) = (e + e 3 ) [(e 1 + e 3 ) (e 1 + e )] = (e + e 3 ) [e 3 + e e 1 ]= Therefore g 1 = 1 (g g 3 ) = 1 (e 3 + e e 1 ) g = 1 (g 3 g 1 ) = 1 (e 3 e + e 1 ) g 3 = 1 (g 1 g ) = 1 ( e 3 + e + e 1 ) 14. (a) Since g 1 is perpendicular to g, g 1 g =0and g 1 also lies in the 1 plane. Therefore g 1 = g e 3 = (3e + e 1 )

17 But g 1 g 1 =1gives =16. Therefore g 1 = 1 6 (e 1 +3e ) Similarly g = 1 6 ( e 1 +3e ) e g g 1 g g 1 e 1 Original and dual base vectors. (b) v = e 1 +5e = g = g. 1 = g 1 v = 8 3 Therefore = g v = 7 3 1 = g 1 v =8 = g v = v = 8g 1 +g = 8 3 g 1 + 7 3 g v g g 1 v 1 v Components of v relative to original base vectors.

18 CHAPTER 4. VECTOR (CROSS) PRODUCT v v v 1 g g 1 Components of v relative to dual base vectors.

Chapter 5 Determinants 1. det = 1 3 = 11 { 13 33 + 13 3 3 } + 1 { 13 3 13 + 13 1 33 } + 31 { 31 1 3 + 31 13 } = 11 {(+1) 33 +( 1) 3 3 } + 1 {(+1) 3 13 +( 1) 1 33 } + 31 {(+1) 1 3 +( 1) 13 } = 11 { 33 3 3 } 1 { 1 33 3 13 } + 31 { 1 3 13 } Therefore det = 3 11 3 33 13 1 1 3 33 + 1 13 31 3. Interchangingrow1androwgives det = 1 3 1 3 = 1 3 = 1 3 = det 19

0 CHAPTER 5. DETERMINANTS 3. Consider = Then 13 = 1 3 =det. If any two indices are the same, the result is zero: =,(nosumon) =,(nosumon) =,(nosumon) =,(nosumon) = 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., 4. = = = = = With these results it is possible to easy enumerate results for all numeric values. Consequently, = det 1 = 3 = 1 ( 13 3 + 13 3 ) = 1 ( 13 3 + 13 3 ) = 1 1 Because the second index is arbitrary, it can be changed to. Relabelling indices then gives Equation (5.6).

1 5. = 1 1 = 1 1 = 1 { 13 3 + 13 3 } = 1 { 3 3 } = 1 { 3 3 } = 1 { 3 3 } = 3 = 3 13 1 33 = ( 1) +1 1 13 3 33 13 = 1 1 3 = 1 3 ( 13 3 + 31 3 ) = 1 ( 3 3 3 3 ) = 3 3 = 31 1 3 + 31 31 = 1 3 31 = ( 1) 3+1 1 31 3 6. From Equation (4.10) a b c = 1 3 1 3 1 3 1 3 d e f = 1 3 1 3 = 1 1 1 3 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 3 1 3 1 1 1 a d a e a f = b d b e b f 1 3 3 3 3 c d c e c f

CHAPTER 5. DETERMINANTS Noting that the determinant of a product is the product of the determinants establishes the result. 7. Setting = gives = = + = 3( ) ( ) + ( ) = 3( ) ( ) ( ) = ( ) 8. det() = Multiplying each side by and summing gives det() = 6det() = or det() = 1 6 9. (M a M b) M = ( e e ) e e = ( e ) e e = e = ( ) e = det() e = det()(a b) 10. (M a) (M b) (M c) = ( e ) ( e ) ( e ) = ( e ) ( e ) = ( ) = det() = det()(a b c)

3 11. A A = I det A A = det(i) =1 det A det (A) = (det(a)) =1 Therefore det (A) =±1 From Problem 10 (A a) (A b) (A c) =det()(a b c) If a, b, andc are right-handed, a b c 0. If A transforms a righthanded coordinate system to a right-handed coordinate system, then the left side is also positive and det() =+1.IfA transforms a right-handed coordinate system to a left-handed coordinate system, then the left side is negative and det() = 1.

4 CHAPTER 5. DETERMINANTS

Chapter 6 Change of Orthonormal Basis 1.. e 0 = A e = e A Therefore, the components of a vector in the primed system are given by 0 = e 0 v = e A v = e A v = e v = e (e 0 0 ) = (e e 0 ) 0 = 0 3. = e F e = e (e 0 0 e 0 ) e = (e e 0 ) 0 (e 0 e ) = 0 4. A = e e, A = e e and I = e e e ( e e e e ) e = e ( e e ) e e ( e e ) e = δ = = Beginning with A A = I establishes the first equality of (6.4). 5

6 CHAPTER 6. CHANGE OF ORTHONORMAL BASIS 5. e 0 3 = λ Because e 0 is perpendicular to e 0 3 and e 1 e 0 = e 0 3 e 1 = ( 3 e e 3 ) where is a scalar. Choose so that e 0 is a unit vector: q = ± + 3 and take the + sign so that e 0 has a positive projection on e.choosee 0 1 so that the primed system is right-handed: e 0 1 = e 0 e 0 3 q = + 3 e 1 q 1 e q 1 3 e 3 + 3 + 3 Noting that e 0 = e and e 0 e = establishes the result. 6. Because = 0, for a scalar. Defining = 0 = 0 ( )= 0 = establishes the as components of a vector because they transform like one under a rotation of the coordinate system. 7. Because = 0, 8. for a scalar. Defining 0 = 0 ( ) = 0 = establishes the as components of a tensor because they transform like one under a rotation of axes. = = ( 0 )( ) 0

7 Multiplying both sides by and summing gives = 0 0 = det() 0 0 For transformation from a right (left) - handed system to a right (left) - handed system det() =1.Inthiscase 0 = and the components of w transform like components of a vector. If, however, the transform from a right (left) - handed system to a left (right) - handed system det() = 1 and the components of the vector product are not tensor components as defined here. 9. Because and 0 = = 0 0 = 0 = 0 0 Therefore 0 = 10. 0 = = = = 0 Hence the components are symmetric in any rectangular cartesian coordinate system. 11. 0 = = = = 0 Hence the components are anti-symmetric in any rectangular cartesian coordinate system. 1. 0 =

8 CHAPTER 6. CHANGE OF ORTHONORMAL BASIS trf = 0 = = = 13. = 0 ( ) 0 = ( )( ) 0 0 = 0 0 = 0 0 14. From Figure 6. e 0 = A e = ( e e ) e = e = e e 0 1 = cos e 1 +sin e e 0 = sin e 1 +cos e e 0 3 = e 3 Comparing yields the given result. 15. (a) 0 1 cos sin 0 0 = sin cos 0 1 0 0 1 0 3 3 0 1 = 1 cos + sin 0 = 1 sin + cos 0 3 = 3 (b) cos sin 0 [ 0 ]= sin cos 0 11 1 13 cos sin 0 1 3 sin cos 0 0 0 1 31 3 33 0 0 1

9 11 0 = 11 cos +( 1 + 1 )cossin + sin 1 0 = ( 11 )cossin + 1 cos 1 sin 13 0 = 13 cos + 3 sin 1 0 = ( 11 )cossin + 1 cos 1 sin 0 = 11 sin ( 1 + 1 )cossin + cos 3 0 = 13 sin + 3 cos 31 0 = 13 cos + 3 sin 3 0 = 13 sin + 3 cos 33 0 = 33 (c) 0 1 = 1 + 0 = 1 + 0 3 = 3 11 0 = 11 +( 1 + 1 ) 1 0 = ( 11 ) + 1 13 0 = 13 + 3 1 0 = ( 11 ) + 1 0 = ( 1 + 1 ) + 3 0 = 13 + 3 31 0 = 13 + 3 3 0 = 13 + 3 0 33 = 33 16. 17. Ȧ() = { sin()(e 1 e 1 + e e )+cos()(e e 1 e 1 e )} Ȧ(0) = (e e 1 e 1 e ) A() Ȧ(0) = Ȧ() e 0 = A() e e0 = Ȧ() e but substituting e = A () e 0 gives the result. Because A A = I Ȧ A + A Ȧ = 0 Ȧ A = ³Ȧ A

30 CHAPTER 6. CHANGE OF ORTHONORMAL BASIS 18. (a) A rotation of base vectors 90 about the is e0 1 e 0 = 0 0 1 0 1 0 e 1 e 1 0 0 or in dyadic form e 0 3 e 3 A 0 = e 0 1e 1 + e 0 e + e 0 3e 3 A 0 = e 3 e 1 + e e + e 1 e 3 (b) A rotation of base vectors 90 about the 3 is e00 1 e 00 = 0 1 0 1 0 0 e0 1 e 0 0 0 1 or in dyadic form e 00 3 e 0 3 A 00 = e 00 1e 0 1 + e 00 e 0 + e 00 3e 0 3 = e 0 e 0 1 e 0 1e 0 + e 0 3e 0 3 = e e 3 + e 3 e + e 1 e 1 (c) Successive 90 about the and 3 axes are given by e00 1 e 00 = 0 1 0 1 0 0 0 0 1 0 1 0 e 1 e e 00 3 0 0 1 1 0 0 e 3 = 0 1 0 0 0 1 e 1 e 1 0 0 e 3 = e or in dyadic form e 3 e 1 A 00 A 0 = e e 1 + e 3 e + e 1 e 3 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 q + 3 0 1 1 3 + 3 + 3 1 3 + 3 3 + 3 cos sin 0 sin cos 0 0 0 1 q + 3 0 1 + 3 3 + 3 1 3 + 3 + 3 1 3

31 (b) The dyadic form follows from setting λ = e 3,usingI = e 1 e 1 + e e + e 3 e 3 and noting that 1 = 1 = 1 are components of the antisymmetric matrix with axial vector e 3. 0. Setting λ =(e 1 +e + e 3 ) 3 and =10 in the answer to Problem 19a gives the matrix 0 1 0 0 0 1 1 0 0 1. which the same as the result of Problem 18c. 0 = = det() = Hence has the same components in any rectangular cartesian coordinate 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.

3 CHAPTER 6. CHANGE OF ORTHONORMAL BASIS

Chapter 7 Principal Values and Principal Directions 1. Substituting =5back into equation (7.1) gives. μ 1 +0 μ μ = 0 3 0+(5 5) μ 0 = 0 μ 1 +0+( 1) μ = 0 3 The first and third equations are linearly independent and have only the solution μ 1 = μ 3 =0. Because the coefficient of μ in the second equation is zero, it can be arbitrary and is taken as μ =1to make μ a unit vector. 0 = = = 0 0 = ( )( ) = ( )( ) = = det ( 0 ) = 1 0 0 3 0 = ( 1 )( )( 3 ) = ( )( 1 3 ) = det()( 1 3 ) = det()( )( 1 3 ) = det( )( 1 3 ) = (det()) det( )=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) = 1 tr (F F)+ trf + 3 tri But trf = 1, tri =3and tr (F F) = + 1 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) = + 1 and rearrange. 5. F μ = μ F μ = μ F μ = μ Form the scalar product of the first equation with μ with μ. Subtracting then gives and the second ( ) 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 = ± e 1 ± 1 e 5 5 μ = e 3 μ = ± 1 e 1 e 5 5

35 (b) = ± 5 ± 1 5 0 0 0 1 11 4 0 ± 5 0 ± 1 5 4 5 0 ± 1 ± 1 5 5 0 5 5 0 0 0 7 0 1 0 13 0 0 0 7 0 0 0 3 7. 8. 9. trf = a b 1 ( ) = 0 det( ) = 0 cos sin 0 ³ det sin cos 0 = (1 ) (cos ) +sin 0 0 1 = 0 Therefore =1is one solution and the remaining two satisfy which has the solutions (cos ) +sin =0 =cos ± p 1 cos Because cos 1, the only real solution is =1and corresponding principal direction is e 3. det ( ) = 1 6 ( )( )( ) = 1 6 1 6 ( + + ) + 1 6 ( + + ) 1 6 3 Using the identities gives det ( )=det()+ 1 ( )+ 3

36 CHAPTER 7. PRINCIPAL VALUES AND PRINCIPAL DIRECTIONS

Chapter 8 Gradient 1. Substituting in v gives v = e v = e e = e Because is anti-symmetric with respect to interchange of and and is symmetric, their summed product vanishes.. A vector in the direction of the normal to a surface is proportional to the gradient: y = e ( ) = e ( + ) = e ( + ) = e ( ) where the last line follows because =. Dividing by the magnitude of y to get a unit vector gives n = e p 3. v ( u) = e ( e e ) = 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 ( e e e ) = e ( e ) = + µ = e e e e + (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) µ (5 v) = e e e µ = e e = e = ( ) e = e e = ( v) v 7. (a) = e = e = e = e = x

40 CHAPTER 8. GRADIENT (b) (c) x µ e = e 1 1 = e e + e e Ã! 1 = + e e = e e ³ ( ) 3 = µ e = e = µ = 3 8. u = e = e = u 9. F = = µ e e e µ e e µ = e e µ = e e à = e e = F!

41 10. M = E = e e e = e e Therefore, = 11 = 1 1 = 31 1 3 = = 1 3 3 1 1 = 1 = 3 3 1 = 1 = 11 3 31 1 11. To determine the Laplacian in cylindrical coordinates: = ( ) µ µ = e + e 1 + e e + e 1 + e = + 1 + e e + = + 1 + 1 + = 1 µ + 1 +

4 CHAPTER 8. GRADIENT 1. To determine the curl in cylindrical coordinates v = µ 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 = e e µ + e + 13. To determine v in cylindrical coordinates µ v = 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 e = e e + e e µ +e e + e e µ + e e + +e e + e e + e e + e e

43 14. To calculate the divergence of a tensor F write µ F = e + e + e F µ = e + e + e ( e e + e e + e e ) µ + e + e + e ( e e + e e + e e ) µ + e + e + e ( e e + e e + e e ) = 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 µ e e = e µ +e µ +e µ + 1 + 1 + 1 e e + 1 + e + e + + + + + + + + e

44 CHAPTER 8. GRADIENT

Part II Stress 45

Chapter 9 Traction and Stress Tensor 1. On 1 =+, n = e 1 On 1 =, n = e 1 t = e 1 σ = 11 e 1 + 1 e + 13 e µ 3 = 1 e 1 + e +0e 3 t = e 1 σ = 11 e 1 1 e 13 e µ 3 = 1 e 1 + e +0e 3 47

48 CHAPTER 9. TRACTION AND STRESS TENSOR

Chapter 10 Principal Values of Stress 1. (a) det (σ I) = (1 + ) 0 0 1 = 3 9 =0 Therefore the principal stresses are =+3, =0and = 3 (b) To determine the principal direction corresponding to the principal stress =+3: 4 1 +0 3 = 0 1 3 + 3 = 0 0 1 + 3 = 0 The first and third equations give 1 = and 3 =.Making n a unit vector gives n = 1 3 e 1 ± 3 e ± 3 e 3 For the principal stress = 3: 1 +0 3 = 0 1 3 + 3 = 0 0 1 + +4 3 = 0 The first and third equations give = 3 and 1 =.Making n a unit vector gives n = 3 e 1 3 e ± 1 3 e 3 49

50 CHAPTER 10. PRINCIPAL VALUES OF STRESS Choosing n = n n for a right-handed system gives n = 3 e 1 + 1 3 e 3 e 3. (a) det (σ I) = cos sin cos 0 sin 0 = 3 =0 Therefore the principal stresses are =+, =0and = (b) To determine the principal direction corresponding to the principal stress =+: 1 + cos + sin 3 = 0 cos 1 +0 3 = 0 sin 1 +0 3 = 0 The second and third equations give = cos and 3 = sin. Making n a unit vector gives n = 1 e 1 + 1 cos e + 1 sin e 3 where, for definiteness,wehavetakenthepositivesign. principal stress =0: For the 0 1 + cos + sin 3 = 0 cos 1 +0 +0 3 = 0 sin 1 +0 +0 3 = 0 The second or third equations gives 1 =0and the first gives = tan 3.Makingn a unit vector gives n =0e 1 sin e ± cos e 3 Choosing n = n n for a right-handed system gives n = 1 {±e 1 cos e sin e 3 } 3. = 1 n( 0 ) +( 0 ) +( 0 ) o = 1 n( ) +( ) +( ) o

51 where =( + + ) 3. = 1 + + 3ª Substituting for, multiplying out and collecting terms gives the result. 4. Substituting (10.11) into (10.8) gives µ 3 r 4 4 3 sin 3 3 sin 3 = 0 Using the identity gives (10.1). 4sin 3 3sin = 4sin 3 3sin = sin 3 r 7 4 3 3 5. (a) ½ + 1 ( + )+ ¾ 0 = 1 3 = 1 ( + )=0 (b) Hence, 3 =0, sin =0and =0. 0 = 3 ( ) 0 = 0 = 1 3 ( ) Therefore 3 = 7 ( ) 3 = 1 3 ( ) (c) Hence sin 3 = 1 and = 30. 0 = 0 = 1 3 ( ) 0 = 3 ( )

5 CHAPTER 10. PRINCIPAL VALUES OF STRESS Therefore Hence sin 3 =1and =30. 3 = 7 ( ) 3 = 1 3 ( ) 6. ( ) =0 gives the principal directions, the,ofthestress. Substituting the deviatoric stress gives µ 0 13 =0 Thus, the principal directions of 0 satisfy the same equation with an altered 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. Eliminating leads to From the first equality 1 1 = = 3 3 1 =( 1 ) Therefore, either 1 = or 1 + =. Similarly, the second and third equality leads to either = 3 or + 3 = and first and third to either 3 = 1 or 1 + 3 =. Clearly, the first option leads immediately to 1 = = 3. If, for example, 1 = but + 3 = In this case + 3 = 1 1 + + 3 3 Because 1 =,thiscanberewrittenas 1 ( 1 + 3 ) = 1 1 3 + 3 3 1 ( 1 3 ) = 3 ( 1 3 ) Hence either 1 = 3 or 3 = ±1. In the latter case the stress state is axisymmetric ( 1 = ) 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. (a) Traction is given by 1 0 0 1 1 = 1 5 3 3 0 1 6 or t = 1 3 ( 5e 1 +e +6e 3 ) (b) Normal traction is n t = 1 3 (e 1 +e +e 3 ) 1 3 ( 5e 1 +e +6e 3 ) Shear traction is = 1 9 ( 5+4+1)= 11 9 t = t 11 9 n = 074e 1 0148e +1185e 3 The magnitude of the shear traction is = t t = p t t = 393 3. (a) t = n σ = (n u) u = cos u (b) = n t = cos (c) = ± p t t = ± cos p 1 cos = ± cos sin

55 4. From Problem 10., 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 and n = 1 (n + n )=±e 1 n =cose +sine 3 5. If the shear traction vanishes on every plane then n σ s = 0 for all unit vectors n and s such that n s = 0. Substituting σ = σ 0 I where σ 0 is the deviatoric stress gives n σ 0 s n s =0 Because the second term vanishes, the first term will vanish only if the deviatoric stress σ 0 vanishes. If σ = I, then the first equation is satisfied automatically. 6. (a) The unit vector that makes equal angles with the principal axes is given by n = 1 (e 1 + e + e 3 ) 3 The traction on this plane is t = n σ = 1 ( 1 e 1 + e + 3 e 3 ) 3 (b) The normal component of traction is = n t = 1 3 ( 1 + + 3 ) (c) The magnitude of the shear traction is given by = p t t (d) The direction of the shear traction is s = t n t n = 1 (t n) = 1 3 ( 0 1e 1 + 0 e + 0 3e 3 ) 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 1 + 1 ( 1 + 3 ) + 3 3 Using =1 1 + 3 and setting equal to 1 ( 1 + 3 ) gives 1 3 (1 3 )=0 Because 0, 1 6= 3, and therefore 1 = 3 and Calculating + 1 =1 1 t t = 1 1 + ( 1 + 3 ) + 3 3 = 1 + 3 1 + 1 4 ( 1 + 3 ) 1 1 = 1 ( 1 3 ) 1 + 1 4 ( 1 + 3 ) Computing the magnitude of the shear traction = p t t = 1 1 ( 1 3 ) Setting equal to ( 1 3 ) 4 gives 1 = 3 = 1 and q 3 = 1 1 = Therefore the normal to the plane is n = 1 (e 1 + e 3 )+ 3 e

Chapter 1 Mohr s Circle 1. (a) Maximum shear stress is 1 (0 ( )) = and the normal stress on the plane of maximum shear is 1 (0 + ( )) = Problem 1.1a t s ( ) = t s max t n (b) Maximum shear stress is 1 ( ( 3)) = and the normal stress on the plane of maximum shear is 1 ( +( 3)) = 57

58 CHAPTER 1. MOHR S CIRCLE Problem 1.1b t s ( t s ) max = a a a a t n (c) Maximum shear stress is 0 and the normal stress on the plane of maximum shear is. t s t n (d) Principal stresses are 5, 0, +5. Therefore, maximum shear stress is 1 (5 ( 5)) = 5 and the normal stress on the plane of maximum shear is 1 (5 +( 5)) = 0

59 t s ( t s ) max = 5a 5a 5a t n (e) Maximum shear stress is 1 ( ( 5)) = 3 and the normal stress on the plane of maximum shear is 1 ( +( 5)) = t s 5a a a t n. See graphical construction in the Figure.

60 CHAPTER 1. MOHR S CIRCLE t s t s 3 1 t n 3. Failure occurs when the Mohr s circle first becomes tangent to the failure condition = 0 Because =(), the magnitude of the shear traction at points A and A is = 1 h ³ i ( 3 1 )sin ± and the normal traction is = 1 ( 3 1 )cos = 1 ( 3 + 1 )+ 1 ( 3 1 )cos = 1 ( 3 + 1 )+ 1 ( 3 1 )sin h ³ i ± t s A 1 t n A

61

6 CHAPTER 1. MOHR S CIRCLE

Part III Motion and Deformation 63

Chapter 13 Current and Reference Configuration 1. (a) Motion: 1 = 1 + (), =, 3 = 3. (b) Lagrangian description of velocity 1 (X)= (), (X)= 3 (X)=0 Eulerian description of velocity 1 (x)= (), (x)= 3 (x)=0. (a) (c) 1 = 1 (), =, 3 = 3 (b) = + (1 ) = (1 + ) = + = + 1 + = + 1 + 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 + ) The left hand side is zero because position is fixedinthereference 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: = + = (1 + ) + (1 + ) (1 + ) = 0 4. (a) Lagrangian description of velocity 1 =, = 1, 3 =0 Lagrangian description of acceleration: (b) First invert the motion to get 1 =, = 1, 3 =0 1 = 1 ( 1) + 1 = + 1 (1 ) + 1 Then substitute into (a) to get 1 = + 1 ( 1) + 1 = 1 + (1 ) + 1

67 (c) Substitute the inverted motion into the answer from (a) to get 1 = + 1 ( 1) + 1 = 1 (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 1 13 1 3 = 0 13 3 + 1 + 13 + 3 ª = 0 Thus, =0is one principal value and because () 0, the remaining two are purely imaginary.. The components of the axial vector are given by = 1 = 1 ( ) = 1 (b a) + 1 (a b) Therefore w = a b 3. (a) The axial vector of the anti-symmetric tensor W satisfies W u = w u for all u. Settingu = p gives w p = 0. Because the cross product of w and p vanishes, they must be parallel and w = p. (b) From (a) W u = p u Using 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()} + ċ() Therefore the velocity gradient is L = Q() Q 1. The rate of deformation is D = 1 L + L = 1 ½ ³ ¾ Q() Q 1 + Q() Q 1 and the spin tensor is W = 1 L L = 1 ½ ³ ¾ Q() Q 1 Q() Q 1 (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 =0gives Q Q 1 + ³ Q 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 Q ª = 0 Hence Q Q is a constant. Setting the constant equal to I again gives Q 1 = Q.IfD =0 ³ Q Q 1 = Q Q 1 and W = Q Q 1 is antisymmetric. (c) The velocity can be written v(x)=w {x c()} + ċ() From example 13.1, the acceleration is a(x)= Q Q 1 {x c()} + c() Writing Q = W Q and differentiating gives Q = Ẇ Q + W Q = Ẇ 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 tensor. 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)=ȧ + ċ()

7 CHAPTER 14. RATE OF DEFORMATION To obtain the Eulerian description, substitute for the to get v(x)=ȧ a {x() c()} + ċ() Hence, the velocity gradient is L = ȧ a and the anti-symmetric part is = 1 {ȧ a a ȧ } The expression for the axial vector follows from exercise 14..

Chapter 15 Geometric Measures of Deformation 1.. C = F F = F F = F F = C det F = det(r U) = detr det U = 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 = cos 1 1 = 1 = sin 1 = = sin 1 = = cos 13 = 31 = 3 = 3 =0 33 = 1 73

74 CHAPTER 15. GEOMETRIC MEASURES OF DEFORMATION (b) C = F F gives in matrix notation [] = [ ] [ ] cos sin 0 cos sin 0 = sin cos 0 sin cos 0 0 0 1 0 0 1 0 0 = 0 0 0 0 1 or, in dyadic notation, C = e 1 e 1 + e e + e 3 e 3 (c) U = C = e 1 e 1 + e e + e 3 e 3 (d) R = F U 1 or, in matrix notation, or [] = [ ][] 1 cos sin 0 1 0 0 = sin cos 0 0 1 0 0 0 1 0 0 1 cos sin 0 = sin cos 0 0 0 1 R =cos (e 1 e 1 + e e )+sin (e e 1 e 1 e )+e 3 e 3 (e) n = R e 1 = cose 1 +sine n = R e = sin e 1 +cose n = e 3

75 4. µ X u = e ( e ) = e e = e e µ µ = e e e e = F x u 5. Forming the scalar product of each side of Nanson s formula with itself gives () = () N F 1 N F 1 = () N ³F 1 F 1 N = () N C 1 N 6. or = N C 1 N cos Θ = X X = F 1 x F 1 x = x ³F 1 1 F x = n B 1 n Rearranging gives cos Θ = 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 = (n F) (n F) = n F F n

76 CHAPTER 15. GEOMETRIC MEASURES OF DEFORMATION Rearranging gives µ = n B n 8. See exercise 6.10. 9. x = n = F X = F X n = F N Λn = F N 10. F = e 1 e + e 1 e 1 + e e + e 3 e 3 C = e 1 e 1 + (e 1 e + e e 1 )+ 1+ e e + e 3 e 3 11. Unit normals in the directions of the diagonals are given by N + = 1 (e 1 + e ) and N = 1 ( e 1 + e ) The stretches are Λ + = p N + C N + r = 1+ + 1 and Λ = p N C N r = 1 + 1 1. (a) [ ]= 1 1 0 0 1 0 0 0 1

[] = [ ] [ ]= 1 0 0 1 1 0 1 1 0 0 1 0 0 0 1 0 0 1 = 1 1 0 1 0 0 0 1 77 det( ) =(1 ) {(1 )( ) 1} =0 with solutions = 1 ³ 3+ 5 =618 = 1 = 1 ³ 3 5 =038 (b) Λ = =1618 Λ = =1 Λ = =0618 (c) For the principal direction N corresponding to (1 )( ) 1 +( ) = 0 ( ) 1 +( )( ) = 0 (1 )( ) 3 = 0 The third equation gives ( ) 3 =0.Fromthefirst or second equation or =583. tan = ( ) ( ) 1 =1618 13. (a) [ ]= 1 0 0 1 0 0 0 1 [] = [ ] [ ]= 1 0 0 1 0 1 0 0 1 0 0 0 1 0 0 1 = 1 0 1+ 0 0 0 1

78 CHAPTER 15. GEOMETRIC MEASURES OF DEFORMATION det( ) =(1 ) (1 ) 1 + ª =0 with solutions =1and ± =1+ 1 q ± 1+() 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+ + 0 ( ) 1 ( ) = 0 0 0 0 1 + ( ) 3 0 or (1 + )( ) 1 + ( ) = 0 ( ) 1 + 1+ + ( ) = 0 (1 + )( ) 3 = 0 From the third equation 3 =0.Fromthefirst ( ) = 1 + ( ) 1 = ( ) 1 Therefore the principal direction can be written as N =cosθ e 1 +sinθ e where tan Θ =. Because = 1 the corresponding principal direction is given by N = e 3.ThenN = N N : N = sin Θ e 1 +cosθ e Therefore N = R e (Note that is not an index here and and denote corresponding Roman and Arabic numbers) implies R =cosθ (e 1 e 1 + e e )+sinθ (e e 1 e 1 e ) (c) The deformation tensor is given in principal axis form by U = N N + 1 N N + e 3 e 3

79 Substituting for N and N from (b) and multiplying out the dyads gives U = cos Θe 1 e 1 +cosθsin Θ (e e 1 + e 1 e )+sin ª Θe e + 1 sin Θe 1 e 1 cos Θ sin Θ (e e 1 + e 1 e )+cos ª Θe e +e 3 e 3 Collecting the coefficients of the dyads gives and after some algebra 11 = cos (Θ)+ 1 sin (Θ) 1 = 1 = 1 cos(θ)sin(θ) = 1 cos (Θ)+sin (Θ) = 33 = 1 11 = +1 1 = 1 = 1 +1 = 3 + 1 +1 33 = 1 14. First find the principal directions in the current state. From Exercise 15.9, Therefore Λ n = F N,(nosumon) Λ n = F N giving 1 n = 1+ {(1 + ) e 1 + e } Using 1+ = gives n = 1 1+ {e 1 + e } Similarly n = 1 1+ { e 1 + e } Noting that R = n e or comparing with the result of Exercise 6.14 give R =(e 1 e 1 + e e )cos +(e e 1 e 1 e )sin + e 3 e 3

80 CHAPTER 15. GEOMETRIC MEASURES OF DEFORMATION where tan = 1 (see Figure below). Recall from Exercise 15.13 that tan Θ = and, hence, = Θ. Because R rotates the N into the n, R = n N Substituting for the n and N gives R = 1+ (e 1e 1 + e e ) 1 1+ (e e 1 e 1 e )+e 3 e 3 (Alternatively, R can be found from F U 1.) Identifying cos = 1+ sin = 1 1+ gives tan =. Note the minus sign preceding sin in R indicates that is positive for a clockwise rotation about the e 3. Using the trigonometric identity tan ± tan tan ( ± ) = 1 tan tan and the expressions tan Θ = 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 e 3 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 = 1 = 1 ½ µ ½ + + ½ + + µ ¾ ¾ + ¾. From (16.15), L = Ḟ F 1.FromF = R U Ḟ = Ṙ U + R U and F 1 =(R U) 1 = U 1 R 1 = U 1 R Substituting L = Ṙ U U 1 R T + R U U 1 R = Ṙ R T + R U U 1 R 81

8 CHAPTER 16. STRAIN TENSORS 3. (a) E ( ) = 1 X 1 Λ N N ( = 1 X N N X = 1 I U 1 N N X Λ ) 1 N N Λ (b) C 1 = U U 1 = U 1 U 1 = U 1 = = { } = C 1 = F F 1 = F 1 F 1 1 = 1 = 1 µ = 1 1 µ = + ( ) = 1 µ + (c) e = 1 ³I 1 F 1 F = 1 ni (R U) 1 (R U) 1o = 1 ni U 1 R U 1 R o = 1 ni o R U 1 U 1 R ½ 1 = R ³I 1 ¾ U 1 U R = R E ( ) R

83 4. (a) e ( ) = 1 X ( = 1 X = 1 V I = 1 F F I 1 n n 1 n n X 1 n n X n n ) = 1 (B I) (b) ( ) = 1 ( ) = 1 ½µ + = 1 µ + ¾ ½ + + ¾ (c) e ( ) = 1 F F I = 1 n o (R U) (R U) I = 1 (R U) U R I ª ½ 1 = R U I ¾ R = R E R 5. ln Λ = Λ 1 1 (Λ 1) + 1 3 (Λ 1)3 + Therefore E (0) = lnu = X ln(λ )N N = X (Λ 1) N N X 1 (Λ 1) N N + = (U I) 1 (U I) + 1 3 (U I)3 +

84 CHAPTER 16. STRAIN TENSORS 6. Principal values of the Green-Lagrange strain are = N E N ½ ¾ 1 = N (C I) N = 1 (N C N 1) = 1 Λ 1 Solving for Λ gives Λ =1+ Prinipal values of the Almansi strain are = n e n ½ 1 = n I B 1 ¾ n = 1 1 n B 1 ª n = 1 ½ 1 1 ¾ Λ Solving for Λ gives Λ = 1 1 Equating the two expressions for Λ gives the result. 7. Differentiating gives R R = I Ṙ R + R Ṙ = 0 Ṙ R = R Ṙ = ³Ṙ R 8. E (ln) = X ln (Λ ) N N If the principal axes of U (i.e., the N )donotrotate Ė (ln) = X Λ Λ N N

85 and the N can be replaced by n.but Λ 1 = Λ = 1 Therefore Λ Λ are the principal values of D and Ė (ln) = D 9. Use = 1 3 + 1 3 + 1 3 and rearrange the indices to get Ḟ = L F = 1 3 where = + + 10. (n) = N F 1 ³ 1 = N F 1 + Ḟ Using = tr, from the preceding problem and Ḟ 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 =1gives n (n) = (tr n L n) Theleftsidegives (n) =ṅ + n () Forming the scalar product with n and noting that n n =1and ṅ n = 0 gives () = (tr n L n)

86 CHAPTER 16. STRAIN TENSORS but yielding the desired result. x = F X x = F X = Λ n = x N = X 11. (a) Differentiate Λn = F N (Exercise 15.9) to get Λn + Λṅ = F N (16.11.1) Multiply both sides of Λn = F N by F 1 to get and substitute into the precding equation: N = ΛF 1 n (16.11.) Λn + Λṅ = ΛL n where L = F F 1.Dotbothsideswithn But since n is a unit vector Λn n + Λn ṅ = Λn L n n n =1 (16.11.4) and. n ṅ =0 (16.11.5) Since n W n =0(because W = W ). n L n = n D n Converting 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. Differentiate 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 equality of (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 expression for Q in terms of L Q = L + L L Differentiating (16.11.1) gives Λn + Λṅ + Λ n = F (ΛF 1 n) (16.11.8) where (16.11.) has been used for N. Dotting both sides of (16.11.8) with n and using (16.11.7), (16.11.5) and (16.11.4) yield Λ + Λ n n = Λn Q n (16.11.9) Differentiating (16.11.4) yields n n + ṅ ṅ =0 Using n n = ṅ ṅ in (16.11.9), dividing through by Λ and converting to index notation gives the result. Another approach is to begin with () =v x + x v Differentiating again gives µ () + () =a x + x a +v v Using the last equality in (16.11.6), noting that Q is symmetric and dividing through by () gives 1 µ µ 1 1 () =n Q n + v v () (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).differentiate x = n to get v = ṅ + n () Dividing through by and dotting the result with itself give v v µ 1 = ṅ ṅ + () Substituting back into (16.11.10) gives the desired result.

88 CHAPTER 16. STRAIN TENSORS

Chapter 17 Linearized Displacement Gradients 1. Using sin, (17.8) and (17.11) in (17.13) gives N N +N ε N (1 + N ε N )(1+N ε N ) Because N and N are orthogonal N N =0.Tofirst 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).. Because U = X Λ N N the inverse of U is given by U 1 = X 1 Λ N N Because C I +ε C, 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 1 = 1 Λ 1+ 89

90 CHAPTER 17. LINEARIZED DISPLACEMENT GRADIENTS Substituting back into the expression for U 1 gives U 1 = X N N X N N = I ε 3. Because = + 1 Also, 1+trε. Substituting into Nanson s formula gives µ (1 + trε) ½ ¾ (1 + trε) (1 + trε) ( + Ω ) 4. The infinitesimal rotation vector Taking the divergence = 1 Ω = 1 1 = 1 4 µ µ + = 1 = 1 Because is skew symmetric and is symmetric with respect interchange of and, the product vanishes. 5. Setting the curl of the integrand in (17.3) equal to zero gives Relabelling indices gives (17.33) { + } = 0 + = 0 + = 0 + { } = 0 + = 0

Part IV Balance of Mass, Momentum, and Energy 91

Chapter 18 Transformation of Integrals 1. Z Z = = = = Z Z Z Z Z () 1 () { [ ()] [ 1 ()]} [ ()] + Z [ 1 ()] From Figure 18. = =cos. (a) Z n σ v = = = Z Z Z ( ) ½ ¾ + If is stress and is velocity then the second term can be replaced by,because = and isthesymmetricpart of. 93

94 CHAPTER 18. TRANSFORMATION OF INTEGRALS (b) Z Z = ( ) Z = = = Z Z ½ + ¾ ½ + ½ + ¾ 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 contributions from the + and sides. The result is Z Z Z n σ = n σ + (n σ) + (n σ) o B L + n V S A L - C

Chapter 19 Conservation of Mass 1. Use = Z = and rearrange to get Z = Z A ½ A + A + A = trl = v ¾ ½ + v A + A ¾ The term [] is zero by mass conservation leaving Z = A. µ 4 3 3 =4 95

96 CHAPTER 19. CONSERVATION OF MASS

Chapter 0 Conservation of Momentum 1. σ = e {(x) } (x) = e 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.. " # 1 1 + h i 1 = 1 + 1 = 0 " # 1 Therefore, equilibrium is satisfied. 97 + 1 h i 1 = + = 0

98 CHAPTER 0. 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 Z t + 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) equalsthe 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, using the 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 1 Conservation of Energy 1. (σ v) = e [( e e ) e ] = ½ + ¾ = + ½ ¾ = e ( e e ) e +( e e ) e e = ( σ) v + σ ( v) Because σ = σ, v can be replaced by its symmetric part L.. Substitute σ = I into (1.6) to get = trl q + But trl = v = 1 where the last equality follows from mass conservation (19.10). Substituting gives the result. 3. If the material is rigid, σ L =0and = in (1.6). If the energy is only a function of the temperature = = Substituting this and Fourier s law and rearranging gives the result. 99

100 CHAPTER 1. CONSERVATION OF ENERGY 4. Energy conservation is given by (1.1). The power input express in terms of the reference configuration is Z Z = t 0 v + 0 b 0 v The first term can be rewritten as Z Z Z t 0 v = N T 0 v = Working out the integrand yields X T 0 v X T 0 v = X T 0 v + T 0 Ḟ Thus the power input can be rewritten as Z Z = X T 0 ª + 0 b 0 v + T 0 Ḟ The equation of motion (0.13) can be used in the first term to give Z Z v = 0 v + T 0 Ḟ andthenthefirst term can be rearranged as follows: Z µ Z 1 = 0v v + T 0 Ḟ Therateofheatinputis Z = Z N Q + 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 Z = X Q + 0 The rate of change of the total energy is = Z 0 + Z µ 1 0v 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 fixedinthereferencestatethetimederivativescanbetakeninside

101 the integrals. Substituting into (1.1), cancelling the common term and rearranging gives Z ½ ¾ 0 T0 Ḟ + X Q 0 Because this must apply for any volume, the integrand must vanish which gives (1.7). 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 (15.1) gives the relation between Q and q. 5. E ( ) = 1 I U = 1 I C 1 = 1 I F 1 F 1 Therefore Ė ( ) = 1 ³Ḟ 1 F 1 + F 1 Ḟ 1 To compute Ḟ 1 differentiate F F 1 = I to get Ḟ F 1 + F Ḟ 1 =0 Rearranging gives Ḟ 1 = F 1 Ḟ F 1 = F 1 L and the transpose is Ḟ 1 = L F 1 Substituting into Ė ( ) = 1 F 1 L F 1 + F 1 L F 1 ª = F 1 1 L + L ª F 1 = F 1 D F 1 or D = F Ė ( ) F

10 CHAPTER 1. CONSERVATION OF ENERGY Substituting into the stress working equality S ( ) Ė ( ) = σ D = σ ³F Ė ( ) F = F σ F Ė ( ) Hence S ( ) = F σ F 6. σ D = S (1) Ė (1) = S (1) U Taking the transpose L = Ḟ F 1 = ³Ṙ U + R U (R U) 1 = Ṙ R + R U U 1 R L = R Ṙ + R U 1 U R Noting that R Ṙ is anti-symmetric and forming 1 ³ D = R U U 1 + U 1 U R Substituting in the work-rate equality S (1) U = σ D = σ ½ R 1 ³ ¾ U U 1 + U 1 U R = 1 U 1 R σ R + R σ R U 1 ª U Therefore S (1) = 1 U 1 R σ R + R σ R U 1 ª

Part V Ideal Constitutive Relations 103

Chapter Fluids 1. For v = ( )e 1,theflow is isochoric, v =0and v v =0. For steady flow v =0and for no body force b =0. The remaining terms in equation (3.1) are + =0 1 where 1 = constant 0 for flow in the positive 1 direction. Integrating yields ( )= 1 + + 1 where and are constants. Because the velocity of the fluid must equal the velocity of the boundary at = ±, (±) =0 Solving for the constants yields =0and = 1 Substituting back into the velocity yields ( )= 1 1 The only non-zero component of shear stress is = = 1. Forvelocityonlyinthe direction and only a function of radial distance v = ()e. From problem 8.13, the only nonzero component of D is = 1 ³ 105

106 CHAPTER. FLUIDS and the shear stress is = = ³ From the answer to problem 8.14, the only non-trivial equilibrium equation is + = 1 =0 Substituting for gives ½ 3 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 and The shear stress at = = Ω ( = ) = Ω where this is the negative of the force per unit area on the cylinder. The torque exerted on the cylinder is = ( = ) = 4 Ω where is the distance out of plane (along the axis of the cylinder).alternate method: the only non-zero term in equation (3.1) is v =0

107 From the answer to problem 8.10 = 1 µ + 1 + Applying to the form of the velocity field gives 00 ()+ 1 0 () 1 () =0 where 0 () =. Looking for a solution of the form gives = ±1 and a velocity field of the same form as before. 3. The solution is again given by (3.15). The boundary conditions are now and ( )= ( 0)=0 The second corresponds to 0 and gives =0.Thefirst corresponds to and gives = Hence, the velocity is µ ( ) = erf 4 At a distance of 1mm the velocity 1s is 145%, 5% 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 (3.13) and cancelling common terms gives = or where = p =exp(4), exp µ ³ +

108 CHAPTER. FLUIDS Take the second solution so that ( ) 0 as Therefore, ³ () = 0 exp p h ³ R exp p i The solution is plotted in the Figure below. Velocity ( ) 0 vs. nondimensional distance for four nondimensional times =00 0 50 100.

Chapter 3 Elasticity 1. If the strain energy is regarded as a function of F = where we have used (3.8). The components of the Green-Lagrange strain are given by (16.4) = 1 { } Calculating the derivative = 1 ½ ¾ + = 1 { + } = 1 { + } Substituting into the first equation above gives = 1 { + } = 1 { + } = = 0 where the last line follows because =.. (a) Regarding C as a function of E gives = = 109

110 CHAPTER 3. ELASTICITY Because = + the second term is = This gives = Because the material is isotropic, is a function only of the invariants of C. ½ 1 = + + ¾ 3 1 3 ½ ¾ 1 3 = 1 + + 3 The forms of the first two invariants are given by (7.9) and (7.10). For 3 =det(c) theformisgivenbyproblem7.3. Thederivativesof the invariants can be computed as follows. For 1 1 = = = For = ½ ¾ 1 [ ] = 1 { + } = 1 { + 1 } = 1 For 3 3 = ½ 1 3 1 1 ¾ 3 3 1 = 1 3 ( + + ) 1 1 1 1 = 1 3 ( + + ) 1 ( 1 ) 1 = 1 where the last line uses the symmetry of C. Substituting into = gives = { 1 + ( 1 )+ 3 ( 1 )} = { [ 1 1 3 ]+ [ 1 3 ]+ 3 }

111 or S ={I [ 1 1 3 ]+C [ 1 3 ]+ 3 C C} (b) The relation between the Cauchy stress σ and the second Piola- Kirchhoff stress is given by (1.1) Substituting the result of (a) gives σ = {F I F [ 1 1 3 ]+F C F [ 1 3 ] + 3 F C C F } Each of the terms can be expressed in terms of B as follows F I F = F F = B F 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 B 3 = B B B = 1 B B+ B + 3 Also =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 = 1 B+ I+ 3 B 1 Substituting this into the result of (b) gives the result. 3. With τ = σ, (1.1) gives or, in index notation τ = F S F = 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 + = ( + )( + )( + ) = + + + = + + +