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- G. Thomas Mase
- Ronald E. Smelser
- Jenn Stroud Rossmann

2

solutions for

Continuum Mechanics for Engineers

Fourth Edition

G. Thomas Mase Ronald E. Smelser Jenn Stroud Rossmann

Chapter 2 Solutions

Problem 2.1 Let v = a × b, or in indicial notation, ^i = aj e ^j × bk e ^k = εijk aj bk e ^i vi e Using indicial notation, show that, (a) v · v = a2 b2 sin2 θ , (b) a × b · a = 0 , (c) a × b · b = 0 . Solution (a) For the given vector, we have ^i · εpqs aq bs e ^p = εijk aj bk εpqs aq bs δip = εijk aj bk εiqs aq bs v · v = εijk aj bk e = (δjq δks − δjs δkq ) aj bk aq bs = aj aj bk bk − aj bk ak bj = (a · a) (b · b) − (a · b) (a · b) = a2 b2 − (ab cos θ)2 = a2 b2 1 − cos2 θ = a2 b2 sin2 θ (b) Again, we find ^i ) · aq e ^q = εijk aj bk aq δiq = εijk aj bk ai = 0 a × b · a = v · a = (εijk aj bk e This is zero by symmetry in i and j. (c) This is ^i ) · bq e ^q = εijk aj bk bq δiq = εijk aj bk bi = 0 a × b · b = v · b = (εijk aj bk e Again, this is zero by symmetry in k and and i.

Problem 2.2 With respect to the triad of base vectors u1 , u2 , and u3 (not necessarily unit vectors), the triad u1 ,u2 , and u3 is said to be a reciprocal basis if ui · uj = δij (i, j = 1, 2, 3). Show that to satisfy these conditions, u1 =

u2 × u3 ; [u1 , u2 , u3 ]

u2 =

u3 × u1 ; [u1 , u2 , u3 ]

u3 =

u1 × u2 [u1 , u2 , u3 ]

and determine the reciprocal basis for the specific base vectors u1 u2 u3

^2 , = 2^ e1 + e ^3 , = 2^ e2 − e ^1 + e ^2 + e ^3 . = e 3

4

Continuum Mechanics for Engineers

Answer u1 u2 u3

= = =

1 5 1 5 1 5

^2 − 2^ (3^ e1 − e e3 ) ^3 ) (−^ e1 + 2^ e2 − e (−^ e1 + 2^ e2 + 4^ e3 )

Solution For the bases, we have u1 ·u1 = u1 ·

u2 × u3 = 1; [u1 , u2 , u3 ]

u2 ·u2 = u2 ·

u3 × u1 = 1; [u1 , u2 , u3 ]

u3 ·u3 = u3 ·

u1 × u2 =1 [u1 , u2 , u3 ]

since the triple scalar product is insensitive to the order of the operations. Now u2 · u1 = u2 ·

u2 × u3 =0 [u1 , u2 , u3 ]

since u2 ·u2 ×u3 = 0 from Pb 2.1. Similarly, u3 ·u1 = u1 ·u2 = u3 ·u2 = u1 ·u3 = u2 ·u3 = 0. For the given vectors, we have 2 1 0 [u1 , u2 , u3 ] = 0 2 −1 = 5 1 1 1 and e ^2 e ^3 ^1 e ^2 − 2^ 2 −1 = 3^ u2 × u3 = 0 e1 − e e3 ; 1 1 1 e ^2 e ^3 ^1 e ^3 ; 1 1 = −^ u3 × u1 = 1 e1 + 2^ e2 − e 2 1 0 e ^2 e ^3 ^1 e 1 0 = −^ u1 × u2 = 2 e1 + 2^ e2 + 4^ e3 ; 0 2 −1

u1 =

1 ^2 − 2^ (3^ e1 − e e3 ) 5

u2 =

1 ^3 ) (−^ e1 + 2^ e2 − e 5

u3 =

1 (−^ e1 + 2^ e2 + 4^ e3 ) 5

Problem 2.3 If the base vectors u1 , u2 , and u3 are eigenvectors of a tensor A , prove that the reciprocal basis vectors u1 , u2 , and u3 are eigenvectors of the tensor’s transpose, AT .

Problem 2.4 If the base vectors u1 , u2 , and u3 form an orthonormal triad, prove that nk nk = I where I is the identity matrix.

Problem 2.5 ^i , and let b = bi e ^i be Let the position vector of an arbitrary point P (x1 x2 x3 ) be x = xi e a constant vector. Show that (x − b) · x = 0 is the vector equation of a spherical surface having its center at x = 21 b with a radius of 21 b.

5

Chapter 2 Solutions Solution For ^i − bi e ^i ) · xj e ^j = (xi xj − bi xj ) δij = xi xi − bi xi = (x − b) · x = (xi e = x21 + x22 + x23 − b1 x1 − b2 x2 − b3 x3 = 0 Now 2 2 2 1 1 1 1 1 2 b + b22 + b23 = b2 x1 − b1 + x2 − b2 + x3 − b3 = 2 2 2 4 1 4 This is the equation of a sphere with the desired properties.

Problem 2.6 Using the notations A(ij) = 12 (Aij + Aji ) and A[ij] = 21 (Aij − Aji ) show that (a) the tensor A having components Aij can always be decomposed into a sum of its symmetric A(ij) and skew-symmetric A[ij] parts, respectively, by the decomposition, Aij = A(ij) + A[ij] , (b) the trace of A is expressed in terms of A(ij) by Aii = A(ii) , (c) for arbitrary tensors A and B, Aij Bij = A(ij) B(ij) + A[ij] B[ij] . Solution (a) The components can be written as Aij + Aji Aij − Aji Aij = + = A(ij) + A[ij] 2 2 (b) The trace of A is A(ii) =

Aii + Aii 2

= Aii

(c) For two arbitrary tensors, we have Aij Bij = A(ij) + A[ij] B(ij) + B[ij] = A(ij) B(ij) + A[ij] B(ij) + A(ij) B[ij] + A[ij] B[ij] = A(ij) B(ij) + A[ij] B[ij] since the product of a symmetric and skew-symmetric tensor is zero Aij + Aji Bij − Bji 1 A(ij) B[ij] = = (Aij Bij + Aji Bij − Aij Bji − Aji Bji ) 2 2 4 1 = (Aij Bij + Aji Bij − Aji Bij − Aij Bij ) = 0 4

6

Continuum Mechanics for Engineers

We have changed the dummy indices on the last two terms.

Problem 2.7 Expand the following expressions involving Kronecker deltas, and simplify where possible. (a) δij δij ,

(b) δij δjk δki ,

(c) δij δjk ,

(d) δij Aik

Answer (a) 3,

(b) 3,

(c) δik ,

(d) Ajk

Solution (a) Contracting on i or j, we have δij δij = δjj = δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3 (b) Contracting on k and then j gives δij δjk δki = δij δji = δii = 3 (c) Contracting on j yields δij δjk = δik (d) Contracting on i gives δij Aik = Ajk Note: It may be helpful for beginning students to write out all terms.

Problem 2.8 If ai = εijk bj ck and bi = εijk gj hk , substitute bj into the expression for ai to show that a i = g k ck h i − h k ck g i , or in symbolic notation, a = (c · g)h − (c · h)g. Solution We begin by changing the dummy indices for bj = εjmn gm hn and ai = εijk bj ck = εijk εjmn gm hn ck = − (εjik εjmn gm hn ck ) = − (δim δkn − δin δkm ) gm hn ck = −gi hk ck + gk hi ck = gk ck hi − hk ck gi where we have used the anti-symmetry of εijk = −εjik and the ε−δ identity. Symbolically a = (c · g)h − (c · h)g

Problem 2.9 By summing on the repeated subscripts determine the simplest form of (a) ε3jk aj ak , Answer

(b) εijk δkj ,

(c) ε1jk a2 Tkj ,

(d) ε1jk δ3j vk .

7

Chapter 2 Solutions (a) 0,

(b) 0,

(c) a2 (T32 − T23 ),

(d) −v2

Solution (a) Summing gives ε3jk aj ak = ε31k a1 ak + ε32k a2 ak = ε312 a1 a2 + ε321 a2 a1 = a1 a2 − a2 a1 = 0 (b) εijk δkj

= εij1 δ1j + εij2 δ2j + εij3 δ3j = εi21 δ12 + εi31 δ13 + εi12 δ21 + εi32 δ23 + εi13 δ31 + εi23 δ32 = 0

(c) ε1jk a2 Tkj

= ε12k a2 Tk2 + ε13k a2 Tk3 = ε123 a2 T32 + ε132 a2 T23 = a2 T32 − a2 T23 = a2 (T32 − T23 )

(d) ε1jk δ3j vk = ε12k δ32 vk + ε13k δ33 vk = 0 + ε132 δ33 v2 = −v2

Problem 2.10 Consider the tensor Bik = εijk vj . (a) Show that Bik is skew-symmetric. (b) Let Bij be skew-symmetric, and consider the vector defined by vi = εijk Bjk (often called the dual vector of the tensor B). Show that Bmq = 12 εmqi vi . Solution (a) For a tensor to be skew-symmetric, one has Aij = −Aji . For the given tensor Bik = εijk vj = −εkji vj = −Bki (b) For the dual vector of the tensor B, we have εmqi vi = εmqi εijk Bjk = (δmj δqk − δmk δqj ) Bjk = Bmq − Bqm = [Bmq − (−Bmq )] = 2Bmq since B is skew-symmetric.

Problem 2.11 Use indicial notation to show that Ami εmjk + Amj εimk + Amk εijm = Amm εijk where A is any tensor and εijk is the permutation symbol. Solution Multiply both sides by εijk and simplify

8

Continuum Mechanics for Engineers Amm εijk εijk = 6Amm

= Ami εmjk εijk + Amj εimk εijk + Amk εijm εijk = Ami 2δmi + Amj 2δmj + Amk 2δmk = 6Amm

Problem 2.12 If Aij = δij Bkk + 3Bij , determine Bkk and using that solve for Bij in terms of Aij and its first invariant, Aii . Answer Bkk = 61 Akk ; Bij = 13 Aij −

1 18 δij Akk

Solution Taking the trace of Aij gives Aii = δii Bkk + 3Bii = 3Bkk + 3Bii = 6Bkk since i and k are dummy indices. This gives Bkk =

1 Akk 6

Substituting for Bkk and solving for Bij gives 1 3Bij = Aij − δij Akk 6

or

Bij =

1 1 Aij − δij Akk 3 18

Problem 2.13 Show that the value of the quadratic form Tij xi xj is unchanged if Tij is replaced by its symmetric part, 21 (Tij + Tji ). Solution The quadratic form becomes Tij xi xj =

1 1 1 (Tij + Tji )xi xj = (Tij xi xj + Tji xi xj ) = (Tij xi xj + Tij xj xi ) = Tij xi xj 2 2 2

since i and j are dummy indices and multiplication commutes.

Problem 2.14 With the aid of Eq 2.7, show that any skew symmetric tensor W may be written in terms of an axial vector ωi given by 1 ωi = − εijk wjk 2 where wjk are the components of W.

9

Chapter 2 Solutions Solution Multiply by εimn εimn ωi

or,

= − 21 εimn εijk wjk = − 12 (δmj δnk − δmk δnj ) wjk = − 12 (wmn − wnm ) = wnm , εmni ωi = wnm

Problem 2.15 Show by direct expansion (or otherwise) that the determinant a1 a2 b1 b2 c1 c2

the box product λ = εijk ai bj ck is equal to a3 b3 . c3

Thus, by substituting A1i for ai , A2j for bj and A3k for ck , derive Eq 2.42 in the form det A = εijk A1i A2j A3k where Aij are the elements of A. Solution Direct expansion gives λ = εijk ai bj ck = ε1jk a1 bj ck + ε2jk a2 bj ck + ε3jk a3 bj ck = ε12k a1 b2 ck + ε13k a1 b3 ck + ε21k a2 b1 ck + ε23k a2 b3 ck + ε31k a3 b1 ck + ε32k a3 b2 ck = ε123 a1 b2 c3 + ε132 a1 b3 c2 + ε213 a2 b1 c3 + ε231 a2 b3 c1 + ε312 a3 b1 c2 + ε321 a3 b2 c1 = a1 b2 c3 − a1 b3 c2 − a2 b1 c3 + a2 b3 c1 + a3 b1 c2 − a3 b2 c1 and a1 b1 c1

a2 b2 c2

a3 b3 c3

= a1 b2 c3 + a2 b3 c1 + a3 b1 c2 − a1 b3 c2 − a2 b1 c3 − a3 b2 c1 = λ

Using the suggested substitutions for ai , bi , ci , we have A3 λ

= εijk A1i A2j A3k = ε1jk A11 A2j A3k + ε2jk A12 A2j A3k + ε3jk A13 A2j A3k = ε12k A11 A22 A3k + ε13k A11 A23 A3k + ε21k A12 A21 A3k + ε23k A12 A23 A3k +ε31k A13 A21 A3k + ε32k A13 A22 A3k = ε123 A11 A22 A33 + ε132 A11 A23 A32 + ε213 A12 A21 A33 + ε231 A12 A23 A31 +ε312 A13 A21 A32 + ε321 A13 A22 A31 = A11 A22 A33 − A11 A23 A32 − A12 A21 A33 + A12 A23 A31 + A13 A21 A32 − A13 A22 A31

and A11 A21 A31

A12 A22 A32

A13 A23 A33

= A11 A22 A33 − A11 A23 A32 + A12 A23 A31 − A12 A21 A33 +A13 A21 A32 − A13 A22 A31 = λ

Problem 2.16 Starting with Eq 2.42 of the text in the form det A = εijk Ai1 Aj2 Ak3

10

Continuum Mechanics for Engineers

show that by an arbitrary number of interchanges of columns of Aij we obtain εqmn det A = εijk Aiq Ajm Akn which is Eq 2.43. Further, multiply this equation by the appropriate permutation symbol to derive the formula 6 det A = εqmn εijk Aiq Ajm Akn . Solution Each row or column change introduces a minus sign. After an arbitrary number of row and column changes, we have εqmn det A = εijk Aiq Ajm Akn Multiplying by εqmn gives εqmn εqmn det A

= (δmm δnn − δmn δnm ) det A = (3 · 3 − δnn ) det A = (9 − 3) det A = εqmn εijk Aiq Ajm Akn

from the ε − δ identity.

Problem 2.17 Let the determinant of the tensor Aij be given by A11 A12 A13 det A = A21 A22 A23 A31 A32 A33

.

Since the interchange of any two rows or any two columns causes a sign change in the value of the determinant, show that after an arbitrary number of row and column interchanges Amq Amr Ams Anq Anr Ans = εmnp εqrs det A . Apq Apr Aps Now let Aij = δij in the above determinant which results in det A = 1 and, upon expansion, yields εmnp εqrs = δmq (δnr δps − δns δpr ) − δmr (δnq δps − δns δpq ) + δms (δnq δpr − δnr δpq ) . Thus, by setting p = q, establish Eq 2.7 in the form εmnq εqrs = δmr δns − δms δnr . Solution Letting Aij = δij in the determinant gives δmq δmr δms δnq δnr δns δpq δpr δps = δmq (δnr δps − δns δpr ) − δmr (δnq δps − δns δpq ) + δms (δnq δpr − δnr δpq )

11

Chapter 2 Solutions and εmnp εqrs = δmq (δnr δps − δns δpr ) − δmr (δnq δps − δns δpq ) + δms (δnq δpr − δnr δpq ) since

δ11 δ21 δ31

δ12 δ22 δ32

Setting p = q gives δmp δmr δms δnp δnr δns δpp δpr δps

δ13 δ23 δ33

1 = 0 0

0 1 0

0 0 1

= 1 = ε123 ε123 det A

= δmp (δnr δps − δns δpr ) − δmr (δnp δps − δns δpp ) + δms (δnp δpr − δnr δpp ) = δnr δms − δns δmr − δmr (δns − 3δns ) + δms (δnr − 3δnr ) = δnr δms − δns δmr + 2δmr δns − 2δms δnr = δmr δns − δms δnr = εpmn εprs Problem 2.18 Show that the square matrices 1 0 [Bij ] = 0 −1 0 0

0 0 1

and

[Cij ] =

5 −12

2 −5

are both square roots of the identity matrix. Solution The product of the matrix root. Thus 1 0 0 and

5 −12

2 −5

with itself should be the identity matrix for it to be a square 0 0 1 0 0 1 0 0 −1 0 0 −1 0 = 0 1 0 0 1 0 0 1 0 0 1 5 −12

2 −5

=

25 − 24 −60 + 60

10 − 10 −24 + 25

=

1 0

0 1

Problem 2.19 Using the square matrices below, demonstrate (a) that the transpose of the square of a matrix is equal to the square of its transpose (Eq 2.36 with n = 2), (b) that (AB)T = BT AT as was proven in Example 3 0 1 1 [Aij ] = 0 2 4 , [Bij ] = 2 5 1 2 4

2.33 3 2 0

1 5 . 3

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Continuum Mechanics for Engineers

Solution (a) For the matrix A, we have 3 [Aij ]2 = 0 5 and

T 2

3 = 0 1

T

= AT

Aij

This shows that A2

0 2 1

1 3 4 0 2 5

0 2 1

1 14 4 = 20 2 25

1 8 4

0 2 4

5 3 1 0 2 1

0 2 4

5 14 1 = 1 2 5

20 25 8 4 16 13

2

. Similarly for B, we have

1 [Bij ]2 = 2 4

3 2 0

1 1 5 2 3 4

3 2 0

1 11 5 = 26 3 16

9 10 12

1 2 BTij = 3 1

2 2 5

4 1 0 3 3 1

2 2 5

4 11 0 = 9 3 19

26 16 10 12 27 13

and

(b) For (AB)T = BT AT , we have 3 0 [Aij ] [Bij ] = 0 2 5 1 and

5 16 13

1 BTij ATij = 3 1

1 1 4 2 4 2 2 2 5

4 3 0 0 3 1

3 2 0

19 27 13

7 1 5 = 20 15 3

9 4 17

6 22 16

5 7 1 = 9 2 6

20 4 22

15 17 16

0 2 4

The result is demonstrated.

Problem 2.20 Let A be any orthogonal matrix, i.e., AAT = AA−1 = I, where I is the identity matrix. Thus, by using the results in Examples 2.9 and 2.10, show that det A = ±1. Solution From Example 2.9 det AAT = det A det AT and from Example 2.10, det A = det AT Then det AAT = det A det AT = det A det A = (det A)2 = det I = 1 and (det A) = ±1

13

Chapter 2 Solutions

Problem 2.21 A tensor is called isotropic if its components have the same set of values in every Cartesian coordinate system at a point. Assume that T is an isotropic tensor of rank two with 0 0 0 components tij relative to axes Ox1 x2 x3 . Let axes Ox1 x2 x3 be obtained with respect to √ ^ = (^ ^+e ^) / 3. Show Ox1 x2 x3 by a righthand rotation of 120◦ about the axis along n e+e by the transformation between these axes that t11 = t22 = t33 , as well as other relationships. 00 00 00 Further, let axes Ox1 x2 x3 be obtained with respect to Ox1 x2 x3 by a right-hand rotation of 90◦ about x3 . Thus, show by the additional considerations of this transformation that if T is any isotropic tensor of second order, it can be written as λI where λ is a scalar and I is the identity tensor. Solution √ ^ = (^ ^2 + e ^3 ) / 3, the transformation matrix is For a 120◦ rotation about the axis n e1 + e 0 1 0 [aij ] = 0 0 1 1 0 0 0 with det A = 1. The transformation is Tij = aiq ajm Tqm or

t011 t021 t031

t012 t022 t032

t013 t11 0 1 0 t023 = 0 0 1 t21 t033 1 0 0 t31 t22 t23 t21 = t32 t33 t31 t12 t13 t11

t12 t22 t32

t13 0 t23 1 0 t33

0 0 1

1 0 0

Thus if T is isotropic, the transformation will not distinguish between the primed and unprimed coordinates. This gives t11 = t22 ;

t22 = t33 ;

t33 = t11

or

t11 = t22 = t33

and t12 = t23 ;

t13 = t21 ;

t21 = t32 ;

t23 = t31 ;

t31 = t12 ;

t32 = t13

This results in t12 = t23 = t31

and t13 = t21 = t32

For a 90◦ rotation about the x3 axis, the transformation matrix is 0 1 0 [aij ] = −1 0 0 0 0 1 with det A = 1. 0 t11 t021 t031

The resulting transformation is t012 t013 0 1 0 t11 t022 t023 = −1 0 0 t21 t032 t033 0 0 1 t31 t22 −t21 t23 = −t12 t33 −t13 t32 −t31 t11

t12 t22 t32

t13 0 t23 1 t33 0

−1 0 0

0 0 1

14

Continuum Mechanics for Engineers

Again t11 = t22 = t33 = λ and t12 = −t21 ;

t13 = t23 ;

t21 = −t12 ;

t23 = −t13 ;

t31 = t32 ;

t32 = −t31

This results in t12 = −t12 = −t21 = 0;

t13 = −t13 = t23 = 0;

t32 = −t31 = t31 = 0

For a tensor to be isotropic, T = λI

Problem 2.22 0 0 0 For a proper orthogonal transformation between axes Ox1 x2 x3 and Ox1 x2 x3 show the invariance of δij and εijk . That is, show that 0

(a) δij = δij , 0

(b) εijk = εijk . 0

Hint: For part (b) let εijk = aiq ajm akn εqmn and make use of Eq 2.43. Solution (a) The transformation is δ0ij = aiq ajm δqm = aim ajm = δij (b) The transformation is ε0ijk = aiq ajm akn εqmn From 2.43, we have ε0ijk = εijk det A =εijk (+1) since the transformation is proper orthogonal.

Problem 2.23 0 0 0 The angles between the respective axes of the Ox1 x2 x3 and the Ox1 x2 x3 Cartesian systems are given by the table below x1 x2 x3 x10 45◦ 90◦ 45◦ x20 60◦ 45◦ 120◦ x30 120◦ 45◦ 60◦ Determine (a) the transformation matrix between the two sets of axes, and show that it is a proper orthogonal transformation, √ (b) the equation of the plane x1 + x2 + x3 = 1/ 2 in its primed axes form, that is, 0 0 0 in the form b1 x1 + b2 x2 + b3 x3 = b.

15

Chapter 2 Solutions Answer (a)

[aij ] = 0

√1 2 √1 2 1 √ − 2

0

0 √1 2 √1 2

√1 2 − √12 √1 2

0

(b) 2x1 + x2 + x3 = 1 Solution (a) The transformation matrix is cos 45◦ cos 60◦ [aij ] = cos 120◦

√1 cos 45◦ 2 cos 120◦ = 12 cos 60◦ − 12

cos 90◦ cos 45◦ cos 45◦

0 √1 2 √1 2

√1 2 − 12 1 2

and for a proper orthogonal matrix 1 1 1 1 1 1 √ + √ √ + √ det A = √ +√ =1 2 2 2 2 2 2 2 2 2 2 (b) The transformation is x0 = Ax or x = AT x0 This gives √1 x1 2 x2 = 0 x3 √1

2

and

1 2 √1 2 − 21

x1 x2 = x3

− 12 √1 2 1 2

x0 √1 + 2 x0 √2 2 x0 √1 − 2

x01 1 0 x2 = √ 2 x03

x02 2

+ x02 2

− x0 √3 2

+

x03 2 x03 2

Thus x1 + x2 + x3 =

x0 x0 x0 √1 + 2 − 3 2 2 2

+

x0 x0 √2 + √3 2 2

+

x0 x0 x0 √1 − 2 + 3 2 2 2

1 =√ 2

and 2x01 + x02 + x03 = 1 Problem 2.24 Making use of Eq 2.42 of the text in the form det A = εijk A1i A2j A3k write Eq 2.72 as |tij − λδij | = εijk (t1i − δ1i ) (t2j − δ2j ) (t3k − δ3j ) = 0

16

Continuum Mechanics for Engineers

and show by expansion of this equation that 1 3 2 (tii tjj − tij tji ) λ − εijk t1i t2j t3k = 0 λ − tii λ + 2 to verify Eq 2.74 of the text. Solution Expansion of the equation gives |tij − λδij | = εijk (t1i − λδ1i ) (t2j − λδ2j ) (t3k − λδ3k ) = εijk t1i t2j − λ (t2j δ1i + t1i δ2j ) + λ2 δ1i δ2j (t3k − λδ3k ) = εijk [t1i t2j t3k − λ (t3k t2j δ1i + t3k t1i δ2j ) − λδ3k t1i t2j +λ2 δ3k (t2j δ1i + t1i δ2j ) + λ2 δ1i δ2j t3k − λ3 δ1i δ2j δ3j = εijk t1i t2j t3k − λεijk (t3k t2j δ1i + t3k t1i δ2j + δ3k t1i t2j ) +λ2 εijk (t2j δ1i δ3k + t1j δ2j δ3k + δ1i δ2j t3k ) − λ3 εijk δ1i δ2j δ3j = εijk t1i t2j t3k − λ (ε1jk t3k t2j + εi2k t3k t1j + εij3 t1i t2j ) +λ2 (ε1j3 t2j + εi23 t1i + ε12k t3k ) − λ3 ε123 = εijk t1i t2j t3k − λ (t33 t22 − t32 t23 + t33 t11 − t31 t13 + t11 t22 − t12 t21 ) +λ2 (t22 + t11 + t33 ) − λ3 ε123 This is the desired result since tii tjj − tij tji = (t11 + t22 + t33 ) (t11 + t22 + t33 ) − (t1 jtj1 + t2j tj2 + t1j tj3 ) and (t11 + t22 + t33 ) (t11 + t22 + t33 ) = t211 + t222 + t233 + 2 (t11 t22 + t22 t33 + t11 t33 ) t1 jtj1 + t2j tj2 + t1j tj3 = t211 + t12 t21 + t13 t31 + t21 t12 + t222 + t23 t32 + t31 t13 + t32 t23 + t233 = t211 + t222 + t233 + 2 (t21 t12 + t13 t31 + t23 t32 ) or tii tjj − tij tji = 2 (t11 t22 + t22 t33 + t11 t33 ) − 2 (t21 t12 + t13 t31 + t23 t32 )

Problem 2.25 For the matrix representation of tensor B shown below, 17 0 0 [bij ] = 0 −23 28 0 28 10 determine the principal values (eigenvalues) and the principal directions (eigenvectors) of the tensor. Answer

17

Chapter 2 Solutions λ1 = 17, λ2 = 26, λ3 = −39

√ √ ^ (1) = e ^1 , n ^ (2) = (4^ ^ (3) = (−7^ n e2 + 7^ e3 ) / 65, n e2 + 4^ e3 ) / 65 Solution The eigenvalues are given by 17 − λ 0 0 0 −23 − λ 28 det (B − λI) = 0 28 10 − λ = (17 − λ) (−23 − λ) (10 − λ) − 282 = 0 = (17 − λ) λ2 + 13λ − 230 − 784 = (17 − λ) (λ + 39) (λ − 26) = 0 For λ = 17, the eigenvector is found from 17 − 17 0 0 n1 0 n2 = 0 0 −23 − 17 28 0 28 10 − 17 n3 0 or −40n2 + 28n3 = 0 28n2 − 7n3 = 0 The result is n2 = n3 = 0 together with the unit vector relation n21 + n22 + n23 = 1 yields ^ (1) = e ^1 n For λ = 26, the eigenvector is found from 17 − 26 0 0 n1 0 n2 = 0 0 −23 − 26 28 0 28 10 − 26 0 n3 or −9n1 = 0 −49n2 + 28n3 = 0 28n2 − 16n3 = 0 The result is n1 = 0 7n2 = 4n3 together with the unit vector relation n21 + n22 + n23 = 1

18

Continuum Mechanics for Engineers

yields 7 4^ e2 + 7^ e3 4 ^ (2) = √ or n n2 = √ ; n3 = √ 65 65 65 Similarly, for λ = −39, the eigenvector is found from 17 + 39 0 0 n1 0 n2 = 0 0 −23 + 39 28 0 28 10 + 39 n3 0 or 56n1 = 0 16n2 + 28n3 = 0 28n2 + 49n3 = 0 The result is n1 = 0 4n2 = −7n3 together with the unit vector relation n21 + n22 + n23 = 1 yields 7 n2 = − √ ; 65

4 n3 = √ 65

^ (3) = or n

−7^ e2 + 4^ e3 √ 65

Problem 2.26 Consider the symmetrical matrix

5 4

0

3 2

[Bij ] = 0

4

0 .

3 2

0

5 2

(a) Show that a multiplicity of two occurs among the principal values of this matrix. (b) Let λ1 be the unique principal value and show that the transformation matrix √1 √1 0 − 2 2 [aij ] = 0 1 0 √1 √1 0 2 2 gives B∗ according to B∗ij = aiq ajm Bqm . h i (c) Taking the square root of B∗ij and transforming back to Ox1 x2 x3 axes show that 3 1 0 2 hp i 2 Bij = 0 2 0 . 1 3 0 2 2

19

Chapter 2 Solutions (d) Verify that the matrix [Cij ] =

− 21

0

0

2

− 32

0

− 32

0

− 12

is also a square root of [Bij ]. Solution (a) The principal values are found from 5

3 0 2 4−λ 0 =0 det[Bij − λδij ] = det 3 5 0 2 −λ "2 2 2 # 3 5 −λ − = (4 − λ) 2 2 = (4 − λ) λ2 − 5λ + 4 = (4 − λ)2 (λ − 1) = 0 2

−λ 0

(b) For λ = 1, the principal vector is given by 5 3 n1 0 0 2 −1 2 0 4−1 0 n2 = 0 3 5 n3 0 0 2 2 −1 or n1 = −n3 ;

n2 = 0

together with the unit vector relation yields 1 1 ^ (1) = √ e ^1 − √ e ^3 n 2 2 ^ (2) to be perpendicular to n ^ (1) for the λ2 = λ3 = 4 roots. Select n ^ (2) = e ^2 n ^ (3) = n ^ (1) × n ^ (2) and n

1 1 ^1 + √ e ^3 ^ (3) = √ e n 2 2

The transformation matrix is

√1 2

[aij ] = 0

√1 2

and B∗ = ABAT .

√1 2

B∗ij = 0

√1 2

5 0 − √12 2 1 0 0 3 0 √12 2

0 4 0

0 − √12 1 0 1 0 √2

0 3 2

√1 2

5 2

0 − √12

0 1 0

√1 2

1 0 = 0 √1 0 2

0 4 0

0 0 4

20

Continuum Mechanics for Engineers

(c) Taking the square root yields

hq

B∗ij

i

1 = 0 0

0 2 0

0 0 2

√ √ and transforming back gives B = AT B∗ A √1 √1 √1 0 1 0 0 hp i 2 2 2 1 0 0 2 0 0 Bij = 0 √1 0 0 2 − √1 0 √1 2

2

2

0 1 0

3 − √12 2 0 = 0 √1 2

(d) For [Cij ] to be a square root of [Bij ], we must have 1 1 5 − 2 0 − 32 − 2 0 − 32 2 0 0 2 0 = 0 [Cij ] [Cij ] = 0 2 3 − 32 0 − 12 − 32 0 − 12 2

1 2

0 4 0

3 2

0 2 0

1 2

0 3 2

0 = [Bij ]

5 2

Problem 2.27 Determine the principal values of the matrix 4 0 √0 , [Kij ] = 0 √11 − 3 0 − 3 9 and show that the principal axes Ox∗1 x∗2 x∗3 are obtained from Ox1 x2 x3 by a rotation of 60◦ about the x1 axis. Answer λ1 = 4,

λ2 = 8, λ3 = 12.

Solution The principal values are found from 4−λ 0 0 √ 2 √ 11 √ − λ − 3 = (4 − λ) (11 − λ) (9 − λ) + 3 =0 det [K − λI] = 0 0 − 3 9−λ = (4 − λ) λ2 − 20λ − 96 = 0 = (4 − λ) (λ − 8) (λ − 12) = 0 The eigenvalues are λ1 = 4;

λ2 = 8;

λ3 = 12

The associated principal vectors are given by Eq 2.58. For λ1 = 4 we have √ 7n2 − 3n3 = 0 √ − 3n2 + 5n3 = 0

21

Chapter 2 Solutions or n2 = n3 = 0 so that ^ (1) = e ^1 n For λ2 = 8, we have −4n1 = 0 √ 3n2 − 3n3 = 0 √ − 3n2 + n3 = 0 or n3 =

√ 3n2

and n21 + n22 + n33 = 1

so that ^ (2) = n

√ 1 3 ^2 + ^3 e e 2 2

For λ3 = 12, we have −8n1 = 0 √ −n2 − 3n3 = 0 √ − 3n2 − 3n3 = 0 or

√ n2 = − 3n3

so that (3)

^ n The transformation matrix is 1 0 1 [aij ] = 0 2 √ 0 − 23

0

√ 3 2 1 2

and n21 + n22 + n33 = 1 √ 3 1 ^2 + e ^3 =− e 2 2

cos 0◦ = cos 90◦ cos 90◦

cos 90◦ cos 60◦ cos 120◦

cos 90◦ cos 30◦ cos 60◦

This is the desired result. Problem 2.28 ^ (q) (q = 1, 2, 3) Determine the principal values λ(q) (q = 1, 2, 3) and principal directions n for the symmetric matrix 3 − √12 √12 1 9 3 [Tij ] = − √1 2 2 2 2 √1 2

Answer

3 2

9 2

22

Continuum Mechanics for Engineers λ(1) ^ (1) n ^ (2) n ^ (3) n

= 1, λ(2) = 2 λ(3) = 3 √ ^2 − e ^3 2^ e1 + e = 12 √ ^2 + e ^3 2^ e1 − e = 12 √ ^3 ) / 2 = − (^ e2 + e

Solution The principal values are found from 3 −λ 2 1 det [T − λI] = − 2√ 2 1 √

1 − 2√ 2 9 4 −λ

2 2

3 4

1 √ 2 2 3 4 9 − λ 4

=0

= (λ − 3) (λ − 2) (λ − 1) = 0 The eigenvalues are λ1 = 1;

λ2 = 2;

λ3 = 3

The associated principal vectors are determined from Eq 2.71. For λ1 = 1 we have 1 1 1 n1 − √ n2 + √ n3 = 0 2 2 2 2 2 5 3 1 − √ n1 + n2 + n3 = 0 4 4 2 2 1 3 5 √ n1 + n2 + n3 = 0 4 4 2 2 Only two of these are independent. This gives √ n1 = 2n2 and n2 = −n3 and together with n21 + n22 + n33 = 1 we have ^ (1) = n

1 √ ^2 − e ^3 2^ e1 + e 2

For λ2 = 2, we have 1 1 1 − n1 − √ n2 + √ n3 = 0 2 2 2 2 2 1 1 3 − √ n1 + n2 + n3 = 0 4 4 2 2 1 3 1 √ n1 + n2 + n3 = 0 4 4 2 2 Only two of these are independent. This gives √ n1 = − 2n2 and n2 = −n3 and together with n21 + n22 + n33 = 1

23

Chapter 2 Solutions we have ^ (2) = n

1 √ ^2 + e ^3 2^ e1 − e 2

For λ3 = 3, we have 1 1 3 − n1 − √ n2 + √ n3 = 0 2 2 2 2 2 1 3 3 − √ n1 − n2 + n3 = 0 4 4 2 2 1 3 3 √ n1 + n2 − n3 = 0 4 4 2 2 Only two of these are independent. This gives and n2 = n3

n1 = 0 and together with

n21 + n22 + n33 = 1 we have

1 ^3 ) ^ (3) = √ (^ e2 + e n 2

Problem 2.29 For the second-order tensor Cij = ui vj : (a) Calculate the principal invariants IC , IIC , and IIIC . Reduce your answer to its simplest form, but leave it in index notation. State the reasons for simplification. (b) If u = 1 2 3 and v = 5 10 4 write out the matrix form for Cij . (c) Use the numbers of (b) to validate the answer of (a).

Solution (a) The invariants are IC = ui vi IIC

= =

1 2 1 2

[(ui vi )(uj vj ) − (ui vj )(uj vi )] [ui uj vi vj − ui uj vj vi ] = 0

IIIC = det ui vj = εijk u1 u2 u3 vi vj vk = 0 |{z} | {z } sym

skew

(b)

5 [Cij ] = 10 15

10 4 20 8 30 12

(c) ? 5 + 20 + 12 = 1

2

Here is a Matlabr check of the numbers:

3 · 5

10

4

24

Continuum Mechanics for Engineers >> C = [5 10 4; 10 20 8; 15 30 12]; >> I_C = trace(C) I_C =

37

>> II_C = 0.5*((trace(C))^2 - trace(C*C)) II_C =

0

>> III_C = det(C) III_C =

0

>> Problem 2.30 Let D be a constant tensor whose components do not depend upon the coordinates. Show that ∇ (x · D) = D ^i is the position vector. where x = x i e Solution Using indicial notation, we have (xi dij ) ,k = xi,k dij + xi dij,k = δik dij + 0 = dkj = dT jk Problem 2.31 ^i having a magnitude squared x2 = x21 + x22 + x33 . Determine Consider the vector x = x i e (a) grad x , (b) grad (x−n ) , (c) ∇2 (1/x) , (d) div (xn x) , (e) curl (xn x), where n is a positive integer. Answer (a) xi /x,

(b) −nxi /x(n+2) ,

(c) 0,

(d) xn (n + 3),

Solution (a) Using indicial notation, we have (grad x)i = x,i Now

∂ x2 ∂x = 2x = 2xx,i ∂xi ∂xi

(e) 0.

25

Chapter 2 Solutions and

∂ (xj xj ) = 2δij xj = 2xi ∂xi

Equating these expressions gives 2xx,i = 2xi

or

x,i =

xi x

(b) We have grad x−n

i

= x−n

,i

= −nx−n−1 x,i = −n

−n xi −nxi x,i = = n+2 xn+1 xn+1 x x

(c) For this calculation ∇2 (1/x) = x−1 Now

∂ ∂xi

∂x−1 ∂xi

∂ = ∂xi

,ii

−xi x3

from (b). Repeating this process gives ∂ −xi xi,i −3xi δii xi xi 3 x2 = − − x =− 3 +3 5 =− 3 +3 5 =0 i 3 3 5 ∂xi x x x x x x x (d) The expression div(xn x) is ∂ nxn−1 xi xi (xn xi ) = nxn−1 x,i xi + xn xi,i = + δii xn ∂xi x 2 x = nxn + 3xn = (n + 3) xn x2

div(xn x) =

(e) For curl(xn x), where n is a positive integer, we have xj xk εijk (xn xk ),j = εijk nxn−1 x,j xk + xn xk,j = εijk nxn−1 + xn δkj x n−2 n = εijk nx xj xk + x δkj = 0 by symmetry and skew symmetry

Problem 2.32 If λ and ϕ are scalar functions of the coordinates xi , verify the following vector identities. Transcribe the left-hand side of the equations into indicial notation and, following the indicated operations, show that the result is the right-hand side. (a) v × (∇ × v) = 12 ∇(v · v) − (v · ∇)v (b) v · u × w = v × u · w (c) ∇ × (∇ × v) = ∇(∇ · v) − ∇2 v (d) ∇ · (λ∇ϕ) = λ∇2 ϕ + ∇λ · ∇ϕ (e) ∇2 (λϕ) = λ∇2 ϕ + 2(∇λ) · (∇ϕ) + ϕ∇2 λ (f) ∇ · (u × v) = (∇ × u) · v − u · (∇ × v)

26

Continuum Mechanics for Engineers

Solution We note that ∇(v · v) is (vi vi ),j = 2vi,j vi in indicial notation. (a) Now v × (∇ × v) is εpqi vq εijk vk,j = (δpj δqk − δpk δqj ) vq vk,j = vk vk,p − vq vp,q or

1 ∇(v · v) − (v · ∇)v 2 (b) For v · u × w = v × u · w, we have for this scalar quantity v × (∇ × v) =

vi εijk uj wk = wk εkij uj vi or v·u×w=w·v×u (c) For ∇ × (∇ × v) = ∇(∇ · v) − ∇2 v, we have for this vector operation εpqi (εijk vk,j ),q = (δpj δqk − δpk δqj ) vk,jq = vk,pk − vp,qq or ∇ × (∇ × v) = ∇(∇ · v) − ∇2 v (d) For this scalar operation ∇ · (λ∇ϕ) = λ∇2 ϕ + ∇λ · ∇ϕ, we have (λϕ,i ),i = λ,i ϕ,i + λϕ,ii or ∇ · (λ∇ϕ) = λ∇2 ϕ + ∇λ · ∇ϕ (e) For this scalar operation ∇2 (λϕ) = λ∇2 ϕ + 2(∇λ) · (∇ϕ) + ϕ∇2 λ, we have (λϕ),ii = (λ,i ϕ + λϕ,i ),i = (λ,ii ϕ + λ,i ϕ,i + λ,i ϕ,i + λϕ,ii ) = (λϕ,ii + 2λ,i ϕ,i + λ,ii ϕ) or ∇2 (λϕ) = λ∇2 ϕ + 2(∇λ) · (∇ϕ) + ϕ∇2 λ (f) For the scalar ∇ · (u × v) = (∇ × u) · v − u · (∇ × v), we have (εijk uj vk ),i = εijk (uj,i vk + uj vk,i ) = εkij uj,i vk − εjik vk,i uj or ∇ · (u × v) = (∇ × u) · v − u · (∇ × v)

Problem 2.33 Let the vector v = b × x be one for which b does not depend upon the coordinates. Use indicial notation to show that (a) curl v = 2b , (b) div v = 0 .

27

Chapter 2 Solutions Solution (a) The expression in indicial notation is (curl v)i = εijk vk,j = εijk (εkpq bp xq ),j = (δip δjq − δiq δjp ) bp xq,j = (δip δjq − δiq δjp ) bp δqj = bi δjj − δij bj = (3 − 1) bi = 2bi (b) The divergence is div v = vi,i = (εijk bk xj ),i = εijk bk xj,i = εijk bk δji = 0 since two arguments for the permutation symbol are equal.

Problem 2.34 Transcribe the left-hand side of the following equations into indicial notation and verify that the indicated operations result in the expressions on the right-hand side of the equations for the scalar ϕ, and vectors u and v. (a) div (ϕv) = ϕ div v + v · grad ϕ (b) u × curl v + v × curl u = −(u · grad)v − (v · grad)u + grad(u · v) (c) div (u × v) = v · curl u − u · curl v (d) curl(u × v) = (v · grad)u − (u · grad)v + u div v − v div u (e) curl(curl u) = grad(div u) − ∇2 u

Solution (a) In indicial notation div(ϕv) = (ϕvi ),i = ϕ,i vi + ϕvi,i = v · grad ϕ + ϕ div v (b) Using indicial notation (u × curl v + v × curl u)i = εijk uj (εkpq vq,p ) + εijk vj (εkpq uq,p ) = (δip δjq − δiq δjp ) (uj vq,p + vj uq,p ) = uq vq,i − up vi,p + vq uq,i − vp ui,p = [−(u · grad)v − (v · grad)u + grad(u · v)]i since grad(u · v)i = up vp,i + vp up,i . (c) We have div(u × v) = (εijk uj vk ),i = εijk uj,i vk + εijk uj vk,i = εkij uj,i vk − εjik uj vk,i = v · curl u − u · curl v (d) For curl(u × v), we have [curl(u × v)]p = εpqi (εijk uj vk ),q = εipq εijk (uj,q vk + uj vk,q ) = (δpj δqk − δpk δqj ) (uj,q vk + uj vk,q ) = up,q vq − uj,j vp + up vk,k − uq vp,q = (v · grad)u − v div u + u div v − (u · grad)v

28

Continuum Mechanics for Engineers

(e) We have [curl(curl u)]p = εpqi (εijk uk,j ),q = εipq εijk uk,jq = (δpj δqk − δpk δqj ) uk,jq = uq,pq − up,qq = uq,qp − up,qq = grad(div u)p − ∇2 u p

Problem 2.35 Let the volume V have a bounding surface S with an outward unit normal ni . Let xi be the position vector to any point in the volume or on its surface. Show that Z (a) xi nj dV = δij S , S Z ^ dS = 6V , ∇ (x · x) · n (b) S Z Z ^ dS = (c) λw · n w · grad λdV, where w = curl v and λ = λ(x) , S V Z ^j , n ^ ] dS = 2Vδij where e ^i and e ^j are coordinate base vectors. [^ (d) ei × x, e S

Hint: Write the box product ^j , n ^ ] = (^ ^) [^ ei × x, e ei × x) · (^ ej × n and transcribe into indicial notation. Solution (a) Applying the divergence theorem gives Z Z Z xi nj dS = xi,j dV = δij dV = δij V S

V

(b) Using indicial notation and the divergence theorem, we have Z Z Z Z Z (xj xj ),i ni dS = (xj xj ),ii dV = (2xj,i xj ),i dV = 2δij xj,i dV = 2δij δij dV S V V V V Z = 2δii dV = (2) (3) V V

(c) Again using indicial notation and the divergence theorem, we have Z Z Z Z ^ dS = λwi ni dS = (λwi ),i dV = (λεijk vk,j ),i dV λw · n S V V ZS (λ,i εijk vk,j + λεijk vk,ji ) dV = V

The second term in parentheses is zero by symmetry arguments, and Z Z Z ^ dS = (λ,i εijk vk,j ) dV = λw · n w · grad λ dV S

V

V

29

Chapter 2 Solutions (d) Using the hint, we find ^j , n ^ ] = (^ ^ ) = εpik xk e ^p · εqjr nr e ^q = εpik xk εqjr nr δpq [^ ei × x, e ei × x) · (^ ej × n = εpik xk εpjr nr and Z

Z

S

Z Z (εpik εpjr xk ),r dV = εpik xk εpjr nr dS = εpik εpjr xk,r dV V ZS Z V Z (δij δrr − δir δjr ) dV = εpik εpjr δkr dV = εpir εpjr δkr dV = V V ZV (3δij − δij ) dV = 2Vδij =

^j , n ^ ] dS = [^ ei × x, e

V

Problem 2.36 Use Stokes’ theorem to show that upon integrating around the space curve C having a differential tangential vector dxi that for ϕ(x). I ϕ,i dxi = 0 C

Solution Applying Stokes’ theorem gives I Z ϕ,i dxi = εpqi np ϕ,iq dS = 0 C

S

by symmetry of the indices.

Problem 2.37 For the position vector xi having a magnitude x, show that x,j = xj /x and therefore, δij xi xj − 3 , x x 3x δij i xj (b) x−1 ,ij = − 3 , x5 x 2 (c) x,ii = . x (a) x,ij =

Solution xj From Prob 2.31, we have x,j = where x = xj xj . x (a) We have x xi,j xj δij xi xj i x,ij = − xi x−2 = − 3 = x ,j x x x x (b) For the given expression, we have xi,j xi xj δij xi x−1 ,ij = −x−2 x,i ,j = −x−2 = 3x−4 x,j xi − x−2 =3 5 − 3 x ,j x x x

30

Continuum Mechanics for Engineers

(c) The expression is x,ii = (x,i ),i =

x i

x

,i

3 x 3 2 xi,i xi xi = −x−2 x,i xi + =− 2 + =− 2 + = x x x x x x

Problem 2.38 Show that for arbitrary tensors A and B, and arbitrary vectors a and b, (a) (A · a) · (B · b) = a · (AT · B) · b , (b) b × a = 21 (B − BT ) · a, if 2bi = εijk Bkj , (c) a · A · b = b · AT · a . Solution (a) Using indicial notation, we have Aij aj Bik bk = aj Aij Bik bk or

(A · a) · (B · b) = a · (AT · B) · b

(b) We can write the cross product of two vectors as 1 1 1 (b × a)i = εijk bj ak = εijk εjpq Bqp ak = (δkp δqi − δkq δpi ) Bqp ak = (Bik ak − Bki ak ) 2 2 2 or b×a=

1 (B − BT ) · a 2

(c) We have ai Aij bj = bj Aij ai or a · A · b = b · AT · a

Problem 2.39 Use Eqs 2.42 and 2.43 as necessary to prove the identities (a) [Aa, Ab, Ac] = (det A)[a, b, c] , (b) AT · (Aa × Ab) = (det A)(a × b) , for arbitrary vectors a, b, c, and tensor A. Solution (a) The triple scalar product is [Aa, Ab, Ac] = εijk Aip ap Ajq bq Akr cr

31

Chapter 2 Solutions From the definition in Chapter 2, εpqr det A =εijk Aip Ajq Akr and [Aa, Ab, Ac] = εijk Aip ap Ajq bq Akr cr = εpqr det Aap bq cr = (det A) [a, b, c] (b) The r component of the vector quantity is T A · (Aa × Ab) r = Air εijk Ajp ap Akq bq = εijk Air Ajp Akq ap bq = εpqr det Aap bq = det A (a × b)r

Problem 2.40 Let ϕ = ϕ(xi ) and ψ = ψ(xi ) be scalar functions of the coordinates. Recall that in the indicial notation ϕ,i represents ∇ϕ and ϕ,ii represents ∇2 ϕ. Now apply the divergence theorem, Eq 2.103, to the field ϕψ,i to obtain Z Z (ϕ,i ψ,i + ϕψ,ii ) dV . ϕψ,i ni dS = S

V

Transcribe this result into symbolic notation as Z Z Z ∂ψ ^ dS = ϕ dS = ∇ϕ · ∇ψ + ϕ∇2 ψ dV ϕ∇ψ · n ∂n V S S which is known as Green’s first identity. Show also by the divergence theorem that Z Z (ϕψ,i − ψϕ,i ) ni dS = (ϕψ,ii − ψϕ,ii ) dV , S

V

and transcribe into symbolic notation as Z Z ∂ϕ ∂ψ ϕ −ψ dS = ϕ∇2 ψ − ψ∇2 ϕ dV ∂n ∂n S V which is known as Green’s second identity. Solution The divergence theorem applied to the field is Z Z Z (ϕψ,i ),i dV = (ϕ,i ψ,i + ϕψ,ii ) dV ϕψ,i ni dS = S

V

V

Symbolically, this is Z

Z S

and

S

^ dS = ϕ∇ψ · n

Z

S

ϕ

∂ψ dS ∂n

Z V

The result is

ϕψ,i ni dS =

Z

(ϕ,i ψ,i + ϕψ,ii ) dV = Z S

ϕ

∂ψ dS = ∂n

V

∇ϕ · ∇ψ + ϕ∇2 ψ dV

Z V

∇ϕ · ∇ψ + ϕ∇2 ψ dV

32

Continuum Mechanics for Engineers Again applying the divergence theorem gives Z Z (ϕψ,i − ψϕ,i ) ni dS = (ϕψ,i − ψϕ,i ),i dV S V Z (ϕ,i ψ,i + ϕψ,ii − ψ,i ϕ,i − ψϕ,ii ) dV = ZV Z (ϕψ,ii − ψϕ,ii ) dV = = ϕ∇2 ψ − ψ∇2 ϕ dV V

and

Z

Z (ϕψ,i ni − ψϕ,i ni ) dS = S Z ∂ψ ∂ϕ ϕ = dS −ψ ∂n ∂n S

V

Z ^ − ψ∇ϕ · n ^ ) dS (ϕ∇ψ · n

(ϕψ,i − ψϕ,i ) ni dS = S

This gives

S

Z Z ∂ψ ∂ϕ ϕ −ψ dS = ϕ∇2 ψ − ψ∇2 ϕ dV ∂n ∂n S V

Chapter 3 Solutions

Problem 3.1 At a point P, the stress tensor relative to axes Px1 x2 x3 has components tij . On the area n1 ) ^ 1 , the stress vector is t(^ element dS(1) having the unit normal n , and on area element (^ ) (2) n n1 ) ^ 2 the stress vector is t 2 . Show that the component of t(^ dS with normal n in the (^ ) n ^ 2 is equal to the component of t 2 in the direction of n ^1. direction of n Solution ^ 2 in the direction n ^1 For the traction on the plane with normal n (^ n2 )

ti

(1)

ni

(2)

(1)

= tji nj ni

(2)

(1)

= tij ni nj

(2)

(1)

= tji ni nj

(^ n1 )

= ti

(2)

ni

where the symmetry of the stress and dummy indices substitution was used.

Problem 3.2 Verify the result established in Problem 3.1 for the area elements having normals 1 (2^ e1 + 3^ e2 + 6^ e3 ) 7 1 ^ 2 = (3^ n e1 − 6^ e2 + 2^ e3 ) 7

^1 = n

if the stress matrix at P is given with respect to axes Px1 x2 x3 by 35 0 21 [tij ] = 0 49 0 . 21 0 14 Solution n1 ) For t(^ , we have

(^ n ) t1 1 35 (^ n ) t2 1 = 0 (^ n ) 21 t3 1

0 21 2 28 1 49 0 3 = 21 18 0 14 7 6

and n1 ) ^ 2 = (28^ t(^ ·n e1 + 21^ e2 + 18^ e3 ) · n2 ) For t(^ , we have

(^ n ) t1 2 35 (^ n ) t2 2 = 0 (^ n ) 21 t3 2

6 1 (3^ e1 − 6^ e2 + 2^ e3 ) = − 7 7

3 21 0 21 1 49 0 −6 = −42 0 14 7 2 13

33

34

Continuum Mechanics for Engineers

and n2 ) ^ 1 = (21^ t(^ ·n e1 − 42^ e2 + 13^ e3 ) ·

6 1 (2^ e1 + 3^ e2 + 6e3 ) = − 7 7

Problem 3.3 The stress tensor at P relative to axes Px1 x2 x3 has components in MPa given by the matrix representation t11 2 1 [tij ] = 2 0 2 1 2 0 ^ at P for which the plane perpendicular to where t11 is unspecified. Determine a direction n n) ^ will be stress-free, that is, for which t(^ n = 0 on that plane. What is the required value of t11 for this condition? Answer ^= n

1 ^2 − 2^ (2^ e1 − e e3 ), t11 = 2 MPa 3

Solution n) For t(^ = 0, we have t11 2 1 n1 t11 n1 + 2n2 + n3 2 0 2 n2 = 2n1 + 2n3 1 2 0 n3 n1 + 2n2

0

= 0 0

This gives n1 = −n3 and since

and n1 = −2n2

2 1 ni ni = n21 + − n1 + (−n1 )2 = 1 2

we have n1 = ±

2 3

n2 = ∓

1 3

n3 = ∓

2 3

Finally, we find t11

2 1 2 −2 − =0 3 3 3

and t11 = 2 Problem 3.4 The stress tensor has components at point P in ksi units as specified by the matrix −9 3 −6 9 . [tij ] = 3 6 −6 9 −6

35

Chapter 3 Solutions Determine: (a) the stress vector on the plane at P whose normal vector is ^= n

1 (^ e1 + 4^ e2 + 8^ e3 ) , 9

(b) the magnitude of this stress vector, (c) the component of the stress vector in the direction of the normal, (d) the angle in degrees between the stress vector and the normal. Answer n) (a) t(^ = −5^ e + 11^ e2 − 2^ e3 √ 1 (^ n) (b) t = 150 23 (c) 9 (d) 77.96◦

Solution (a) The traction vector is

−9 3 −6

3 6 9

−6 1 −5 1 9 4 = 11 9 −6 8 −2

(b) The magnitude of the stress vector is 2 (^ n) n) · t(^ = (−5)2 + (11)2 + (−2)2 = 150 t n) = t(^ and

(^ t n) = 12.25

(c) The component of the stress vector in the direction of the normal is n) ^ = (−5^ t(^ ·n e1 + 11^ e2 − 2^ e3 ) ·

23 1 (^ e1 + 4^ e2 + 8^ e3 ) = 9 9

(d) The angle between the stress vector and the normal is 23 n) n) ^ = t(^ t(^ ·n n| cos θ = (12.25) (1) cos θ = |^ 9 and cos θ = 0.209

or

θ = 77.96◦

Problem 3.5 Let the stress tensor components at a point be given by tij = ±σ0 ni nj where σ0 is a positive constant. Show that this represents a uniaxial state of stress having a magnitude ±σ0 and acting in the direction of ni .

36

Continuum Mechanics for Engineers

Solution The stress vector is

(^ n)

ti

= tij nj = ±σ0 ni nj nj = ±σ0 ni

since nj nj = 1

Problem 3.6 Show that the sum of squares of the magnitudes of the stress vectors on the coordinate planes is independent of the orientation of the coordinate axes, that is, show that the sum (^ e1 ) (^ e ) ti 1

ti

(^ e2 ) (^ e ) ti 2

(^ e3 ) (^ e ) ti 3

+ ti

+ ti

x3

ti

is an invariant. ^3 ) (e

^2 ) (e

ti

^ ) (e ti 1

x2 x1 Solution The sum is (^ e1 ) (^ e ) ti 1

ti

(^ e2 ) (^ e ) ti 2

+ ti

(^ e3 ) (^ e ) ti 3

+ ti

= ti1 n1 ti1 n1 + ti2 n2 ti2 n2 + ti3 n3 ti3 n3 = ti1 ti1 + ti2 ti2 + ti3 ti3 = tij tij

since n1 n1 = n2 n2 = n3 n3 = 1

Problem 3.7 With respect to axes Ox1 x2 x3 the stress state is given in terms of the coordinates by the matrix x1 x2 x22 0 x2 x3 x23 [tij ] = x22 2 0 x3 x3 x1 Determine (a) the body force components as functions of the coordinates if the equilibrium equations are to be satisfied everywhere (b) the stress vector at point P(1, 2, 3) on the plane whose outward unit normal makes equal angles with the positive coordinate axes. Answer:

37

Chapter 3 Solutions −3x2 −3x3 −x1 , b2 = , b3 = ρ ρ ρ (6^ e1 + 19^ e2 + 12^ e3 ) √ = 3

(a) b1 = n) (b) t(^

Solution (a) The equilibrium equations are tij,j + ρbi = 0 For i = 1 t11,1 + t12,2 + t13,3 + ρb1 = 0 x2 + 2x2 + 0 + ρb1 = 0 and b1 = −

3x2 ρ

For i = 2 t21,1 + t22,2 + t23,3 + ρb2 = 0 0 + x3 + 2x3 + ρb2 = 0 and b2 = −

3x3 ρ

For i = 3 t31,1 + t32,2 + t33,3 + ρb3 = 0 0 + 0 + x1 + ρb3 = 0 and b3 = −

x1 ρ

(b) The stress at the point P (1, 2, 3) is

2 [tij ] = 4 0

4 6 9

0 9 3

^ = √1 (^ ^2 + e ^3 ). The stress vector is e1 + e The direction is n 3 (^ n) t1 2 4 0 1 6 (^ t n) = 4 6 9 √1 1 = √1 19 2 3 3 (^ n) t3 0 9 3 1 12 Problem 3.8 Relative to the Cartesian axes Ox1 x2 x3 a stress field is given by the matrix 2 3 2 2 − 4 − x2 x1 0 1 − x1 x2 + 3 x2 1 . 2 3 [tij ] = − 4 − x2 x1 − x2 − 12x2 0 3 0 0 3 − x21 x2

38

Continuum Mechanics for Engineers (a) Show that the equilibrium equations are satisfied everywhere for zero body forces. (b) Determine the stress vector at the point P(2, −1, 6) of the plane whose equation is 3x1 + 6x2 + 2x3 = 12.

Answer n) (b) t(^ =

1 (−29^ e1 − 40^ e2 + 2^ e3 ) 7

Solution (a) For zero body force bi = 0, the equilibrium equations are tij,j = 0 For i = 1 t11,1 + t12,2 + t13,3 + ρb1 = 0 −2x1 x2 + 2x1 x2 + 0 + 0 = 0 For i = 2 t21,1 + t22,2 + t23,3 + ρb2 = 0 − 4 − x22 + −x22 + 4 + 0 + 0 = 0 For i = 3 t31,1 + t32,2 + t33,3 + ρb3 = 0 0+0+0+0=0 (b) At the point P (2, −1, 6), the stress is 7 −6 0 3 11 [tij ] = 0 −6 − 3 0 0 1

^ = 17 (3^ The normal to the plane is n e1 + 6^ e2 + 2^ e3 ) and the stress vector is (^ n)

t1 n) t(^ 2 (^ n) t3

7 3 −6 0 3 −29 1 1 = −6 − 11 0 6 = −40 7 7 3 0 0 1 2 2

Problem 3.9 The stress components in a circular cylinder of length L and radius r are given by Ax2 + Bx3 Cx3 −Cx2 . Cx3 0 0 [tij ] = −C2 0 0

39

Chapter 3 Solutions

(a) Verify that in the absence of body forces the equilibrium equations are satisfied. (b) Show that the stress vector vanishes at all points on the curved surface of the cylinder.

x3

L x2 r x1 Solution (a) The equilibrium equations are tij,j = 0 For i = 1 t11,1 + t12,2 + t13,3 = 0 0+0+0=0 For i = 2 t21,1 + t22,2 + t23,3 = 0 0+0+0=0 For i = 3 t31,1 + t32,2 + t33,3 = 0 0+0+0=0 x x 2 ^= ^2 + 3 e ^3 , the stress vector is (b) The normal to the cylindrical surface is n e r r 1 (^ n) 0 (Cx3 x2 − Cx2 x3 ) t1 Ax2 + Bx3 Cx3 −Cx2 x r (^ 2 t2n) = = Cx3 0 0 0 r x3 (^ n) t3 −C2 0 0 0 r Problem 3.10 Rotated axes Px01 x02 x03 are obtained from axes Px1 x2 x3 by a right-handed rotation about the line PQ that makes equal angles with respect to the Px1 x2 x3 axes (see sketch). Determine the primed stress components for the stress tensor in (MPa) 3 0 6 0 [tij ] = 0 0 6 0 −3

40

Continuum Mechanics for Engineers

if the angle of rotation is (a) 120◦ , or (b) 60◦ .

x3 Q

β

β P β

x1

x2

Answer (a)

0 h i t0ij = 0 0

0 0 −3 6 MPa, 6 3

Solution (a) The stress is

(b)

h i 1 −5 t0ij = 10 3 10

10 −11 −2

10 −2 MPa 16

t0ij = aiq ajm tqm

^2 + e ^3 ) is ^ = √1 (^ e1 + e The transformation matrix for 120◦ about n 3 0 1 0 [aij ] = 0 0 1 1 0 0 This gives

t0ij

0 = 0 1

1 0 0

3 0 1 0 6 0

0 0 0

0 6 0 1 0 −3

(b) For a rotation of 60◦ , the rotation matrix is 2 2 1 [aij ] = −1 2 3 2 −1

0 0 1

0 1 0 = 0 0 0

0 −3 6

0 6 3

−1 2 2

This gives

t0ij

1 = 3 1 = 9

2 −1 2 −15 30 30

−1 3 2 0 2 6 30 30 −33 −6 = −6 48

2 2 −1

0 0 0 1 3

2 6 1 0 2 −3 3 −1 −5 10 10

10 −11 −2

−1 2 2

2 −1 2

10 −2 16

Problem 3.11 At the point P, rotated axes Px01 x02 x03 are related to the axes Px1 x2 x3 by the transformation matrix √ √ a√ 1 − 3 1 + √3 1 [aij ] = 1 + √3 b√ 1− 3 3 1− 3 1+ 3 c

41 h i where a, b, and c are to be determined. Determine t0ij if the stress matrix relative to axes Px1 x2 x3 is given in MPa by 1 0 1 [tij ] = 0 1 0 . 1 0 1

Chapter 3 Solutions

Answer √ 11 + 2√ 3 h i 1 t0ij = 5 + 3 9 −1

√ 5+ 3 5√ 5− 3

−1√ 5− √ 3 MPa 11 − 2 3

Solution From the orthogonality condition, or aji aki = δjk

aij aik = δjk This results in a√ 1 1 + √3 9 1− 3

√ 1− 3 b√ 1+ 3

√ 1 + √3 a√ 1 − 3 1 − √3 c 1+ 3

√ 1+ 3 b√ 1− 3

√ 1 − √3 1 1+ 3 = 0 0 c

Thus √ 2 √ 2 a2 + 1 − 3 + 1 + 3 = 9 √ 2 √ 2 1 + 3 + b2 + 1 − 3 = 9 √ 2 √ 2 1 − 3 + 1 + 3 + c2 = 9 and a2 = b2 = c2 = 1

or a = b = c = ±1

Also √ √ √ √ a 1+ 3 +b 1− 3 + 1+ 3 1− 3 =0 √ √ √ √ a 1− 3 +b 1+ 3 + 1− 3 1+ 3 =0 or a+b=2 Thus a=b=1 Finally

√ √ √ √ 1− 3 1+ 3 +b 1+ 3 +c 1− 3 =0

and c=1

0 1 0

0 0 1

42

Continuum Mechanics for Engineers

The stress is t0ij = aiq ajm tqm √ √ 1√ 1 − 3 1 + √3 1 = 1 + √3 1√ 1 − 3 3 1− 3 1+ 3 1 √ √ 11 + 2√ 3 5 + 3 −1√ 1 = 5+ 3 5√ 5− √ 3 9 −1 5 − 3 11 − 2 3

t0ij

1 0 1

0 1 0

1√ 1 1 0 1 − √3 3 1 1+ 3

√ 1+ 3 1√ 1− 3

√ 1 − √3 1+ 3 1

Problem 3.12 The stress matrix referred to axes Px1 x2 x3 is given in ksi by

14 [tij ] = 0 21

0 21 0

21 0 7

Let rotated axes Px01 x02 x03 be defined with respect to axes Px1 x2 x3 by the table of base vectors ^1 e

^2 e

^3 e

^01 e

2/7

3/7

6/7

^02 e

3/7 −6/7

2/7

^03 e

6/7

2/7 −3/7

(a) Determine the stress vectors on planes at P perpendicular to the primed axes; e01 ) e02 ) e03 ) ^1 , e ^2 , and e ^3 . determine t(^ , t(^ , and t(^ in terms of base vectors e (b) Project each of the stress vectors h iobtained in (a) onto the primed axes to determine the nine components of t0ij . (c) Verify the result obtained in (b) by a direct application of Eq 3.33 of the text.

Answer h i 1 143 t0ij = 36 7 114

36 166 3

114 3 ksi −15

Solution (a) The stress vectors are

(e^0 ) t1 1 (e^0 ) t2 1 (e^0 ) t3 1

=

14

0

0

21

21

0

21

2

22

1 0 3 = 9 7 7 6 12

43

Chapter 3 Solutions 0 or t(e^1 ) = 22^ e1 + 9^ e2 + 12^ e3 ; ^0 (e ) t 2 14 1 0 (e^2 ) = 0 t2 0 (e^ ) 21 t3 2 0 or t(e^2 ) = 12^ e1 − 18^ e2 + 11^ e3 ; ^0 (e ) t 3 14 1 0 (e^3 ) = 0 t2 0 ^ e ( ) 21 t3 3

0 21 0

0

21

3

12

1 = 0 −18 −6 7 7 11 2

21

21 0

6

1 0 2 7 7 −3

=

3

6 15

0 or t(e^3 ) = 3^ e1 + 6^ e2 + 15^ e3 ; (b) Projecting these stress vectors onto the e0i axes gives 0 t(e^1 ) = 22^ e1 + 9^ e2 + 12^ e3 9 12 22 = 2^ e01 + 3^ e02 + 6^ e03 + 3^ e01 − 6^ e02 + 2^ e03 + 6^ e01 + 2^ e02 − 3v03 7 7 7 1 0 0 0 143^ e1 + 36^ e2 + 114^ e3 = 7

0 t(e^2 ) = 12^ e1 − 18^ e2 + 11^ e3 18 11 12 = 2^ e01 + 3^ e02 + 6^ e03 − 3^ e01 − 6^ e02 + 2^ e03 + 6^ e01 + 2^ e02 − 3^ e03 7 7 7 1 0 0 0 36^ e1 + 166^ e2 + 3^ e3 = 7

0 t(e^3 ) = 3^ e1 + 6^ e2 + 15^ e3 6 15 3 = 2^ e01 + 3^ e02 + 6^ e03 + 3^ e01 − 6^ e02 + 2^ e03 + 6^ e01 + 2^ e02 − 3^ e03 7 7 7 1 0 0 0 114^ e1 + 3^ e2 − 15^ e3 = 7

Thus

t0ij

143 1 36 = 7 114

36 166 3

114 3 −15

(c) Transforming the stresses by t0ij = aiq ajm tqm , we obtain

t0ij

2 3 6 14 1 = 3 −6 2 0 7 6 2 −3 21 143 36 114 1 3 = 36 166 7 114 3 −15

0 21 2 1 21 0 3 0 7 7 6

3 −6 2

6 2 −3

44

Continuum Mechanics for Engineers

The results are identical. Problem 3.13 At point P, the stress matrix is given in MPa with respect to axes 6 4 0 2 0 Case 2: [tij ] = 1 Case 1: [tij ] = 4 6 0 0 −2 1

Px1 x2 x3 by 1 1 2 1 1 2

Determine for each case (a) the principal stress values, (b) the principal stress directions. Answer (a) Case 1: σ(1) = 10 MPa, σ(2) = 2 MPa, σ(3) = −2 MPa Case 2: σ(1) = 4 MPa, σ(2) = σ(3) = 1 MPa ^ −e ^ ^ +e ^ e e ^ (2) = ± 1 √ 2 , n ^ (3) = ∓^ ^ (1) = ± 1 √ 2 , n e3 (b) Case 1: n 2 2 ^ ^ + 2^ ^ +e ^ +e ^3 (2) −^ e1 + e −^ e1 − e e3 e ^ = ^ (3) = ^ (1) = 1 √2 √ 2, n √2 ,n Case 2: n 3 2 6 Solution Case 1: The principal values are found from 6−σ 6−σ 4 0 4 4 6−σ 0 = det 0 0 0 −2 − σ

0 =0 0 −2 − σ = (−2 − σ) σ2 − 12σ + 20 = 0 = (−2 − σ) (σ − 10) (σ − 2) = 0 4 6−σ 0

The results are σ(1) = 10,

σ(2) = 2,

σ(3) = −2

For σ(1) = 10, the principal direction is found from

6 − 10 4 0

4 6 − 10 0

(1) n1 0 0 (1) 0 n2 = 0 (1) −2 − 10 0 n3

^ (1) · n ^ (1) = 1. The result is and n (1)

(1)

(1)

(1)

−4n1 + 4n2 = 0 +4n1 − 4n2 = 0 (1)

−12n3 = 0 or

(1)

(1)

n1 = n2

(1)

and n3 = 0

45

Chapter 3 Solutions Using the unit vector relation 2 2 2 2 2 (1) (1) (1) (1) (1) n1 + n2 = n1 + n1 = 2 n1 =1 and

1 1 ^ (1) = ± √ e ^1 ± √ e ^2 n 2 2

1 (1) (1) n1 = n2 = ± √ 2

For σ(2) = 2, the principal direction is found from (2) n 6−2 4 0 1 (2) n2 4 6−2 0 (2) n3 0 0 −2 − 2

0

= 0 0

^ (2) · n ^ (2) = 1. The result is and n (2)

(2)

(2)

(2)

4n1 + 4n2 = 0 4n1 + 4n2 = 0 (2)

−4n3 = 0 or

(2)

(2)

n1 = −n2

(2)

and n3 = 0

Using the unit vector relation 2 2 2 2 2 (2) (2) (2) (2) (2) n1 + n2 = n1 + −n1 = 2 n1 =1 and

1 (2) (2) n1 = −n2 = ± √ 2

1 1 ^ (2) = ± √ e ^1 ∓ √ e ^2 n 2 2

For σ(3) = −2, the principal direction is found from (3) n 6+2 4 0 1 (3) n2 4 6+2 0 (3) n3 0 0 −2 + 2 ^ (3) · n ^ (3) = 1. The result is and n (3)

(3)

(3)

(3)

8n1 + 4n2 = 0 4n1 + 8n2 = 0 (3)

0n3 = 0 or

(3)

(3)

n1 = n2 = 0 Using the unit vector relation 2 (3) n3 = ±1

0

= 0 0

46

Continuum Mechanics for Engineers

and

(3)

^ (3) = ±^ n e3

n3 = ±1

Case 2: The principal values are found from 2−σ 2−σ 1 1 1 1 2−σ 1 det = 1 1 1 2−σ

1 =0 1 2−σ = (σ − 1) −σ2 + 5σ − 4 = 0 = (σ − 1) (σ − 1) (σ − 4) = 0 1 2−σ 1

The results are σ(1) = 4,

σ(2) = σ(3) = 1

For σ(1) = 4, the principal direction is found from (1) 2−4 1 1 n 1 (1) 1 2−4 1 n2 (1) 1 1 2−4 n3

0

= 0 0

^ (1) · n ^ (1) = 1. The result is and n (1)

(1)

(3)

(1)

(1)

(1)

−2n1 + 1n2 + n3 = 0 n1 − 2n2 + n3 = 0 (1)

(1)

(1)

n1 + n2 − 2n3 = 0 or

(1)

(1)

(1)

n1 = n2 = n3 Using the unit vector relation

2 2 2 2 2 2 2 (1) (1) (1) (1) (1) (1) (1) n1 + n2 + n3 = n1 + n1 + n1 = 3 n1 =1 and

1 (1) (1) (1) n1 = n2 = n3 = ± √ 3

1 1 1 ^ (1) = ± √ e ^1 ± √ e ^2 ± √ e ^3 n 3 3 3

For σ(2) = 1, the principal direction is found from (2) 2−1 1 1 n 1 (2) 1 2−1 1 n2 (2) 1 1 2−1 n3 ^ (2) · n ^ (2) = 1. The result is and n (2)

(2)

(2)

(2)

(2)

(2)

(2)

(2)

(2)

n1 + n2 + n3 = 0 n1 + n2 + n3 = 0 n1 + n2 + n3 = 0

0

= 0 0

47

Chapter 3 Solutions

^ (2) to be perpendicular to n(1) Using the This is one equation in three unknowns. Select n unit vector relation and ^ (2) · n ^ (1) = 0 n we can choose

(2)

(2)

(2)

and n3 = 0

n1 = −n2 and

2 2 2 2 2 (2) (2) (2) (2) (2) n1 + n2 = n1 + −n1 = 2 n1 =1

This gives 1 1 ^ (2) = ∓ √ e ^1 ± √ e ^2 n 2 2

1 (2) (2) n1 = −n2 = ± √ 2

^ (3) that is perpendicular to For σ(3) = σ(2) = 1, the multiple root requires that we find n (1) (2) ^ and n ^ . This is found from the principal directions n e ^1 ^2 e ^3 e 1 ^2 + 2^ ^ (3) = n ^ (1) × n ^ (2) = √1 √1 √1 = √ (−^ e1 − e e3 ) n 3 3 3 6 √1 − √1 0 2

2

Problem 3.14 When referred to principal axes at P, the stress matrix in ksi units is 2 0 0 ∗ tij = 0 7 0 0 0 12 If the transformation matrix between the 1 [aij ] = √ 2

principal axes and axes Px1 x2 x3 is − 53 1 − 54 a21 a22 a23 −1 − 54 − 35

where a21 , a22 , and a23 are to be determined, calculate [tij ]. Answer

7 [tij ] = 3 0

3 7 4

0 4 ksi 7

Solution The orthogonality condition requires aij aik = δjk or

− 35

1 a 2 21 − 53

1 a22 −1

− 45

or aji aki = δjk

− 35

a23 1 − 45 − 45

a21 a22 a23

− 53

1

−1 = 0 − 54 0

0 1 0

0

0 1

48

Continuum Mechanics for Engineers

This gives 4 3 − a21 + a22 − a23 = 0 5 5 3 4 − a21 − a22 − a23 = 0 5 5 2 2 a21 + a22 + a223 = 2 The first two equations give a22 = 0,

3a21 = −4a22

This together with the final equation yields a221

2 3 + − a21 =2 4

√ 4 2 =± 5

or a21

and

a23

√ 3 2 =∓ 5

The transformation matrix is 1 [aij ] = √ 2

− 53

1

8 5

0

− 53

−1

− 45

− 65 − 45

The stress is [tij ] = [aij ]T t∗ij [aij ] 8 − 35 5 1 =√ 1 0 2 − 45 − 65

− 35

2 −1 0 4 −5 0

0 7 0

0

− 35

1 0 √ 85 2 12 − 35

Problem 3.15 The stress matrix in MPa when referred to axes Px1 x2 x3 is 3 −10 0 0 30 . [tij ] = −10 0 30 −27 Determine (a) the principal stresses, IT , IIT , IIIT , (b) the principal stress directions. Answer (a) IT = 23 MPa, IIT = 0 MPa, IIIT = −47 MPa ^ (1) = −0.394e ^1 + 0.788e ^2 + 0.473e ^3 (b) n ^ (2) = 0.931e ^1 + 0.274e ^2 + 0.304e ^3 n ^ (3) = 0.110e ^1 + 0.551e ^2 − 0.827e ^3 n

1 0 −1

− 54

− 56 4 −5

49

Chapter 3 Solutions Solution (a) The principal values are found from 3−σ 3 − σ −10 0 −10 0 − σ 30 det = −10 0 0 30 −27 − σ

0 30 −27 − σ

−10 0−σ 30

=0

= σ3 + 24σ2 − 1081σ = 0 = σ (σ − 23) (σ + 47) = 0 The results are σ(1) = 23,

σ(2) = 0,

σ(3) = −47

(b) For σ(1) = 23, the principal direction is found from (1) n 3 − 23 −10 0 1 (1) n2 −10 0 − 23 30 (1) n3 0 30 −27 − 23

0 = 0 0

^ (1) · n ^ (1) = 1. The result is and n (1)

(1)

(1)

(1)

(1)

(1)

−20n1 − 10n2 = 0 (1)

−10n1 − 23n2 + 30n3 = 0 30n2 − 50n3 = 0 or

1 (1) (1) n1 = − n2 2 Using the unit vector relation

and

(1)

n3 =

3 (1) n 5 2

2 2 2 2 2 1 2 3 (1) (1) (1) (1) (1) (1) n2 + n2 + =1 n1 + n2 + n3 = − n2 2 5 and (1)

n2 = ±0.788

(1)

n1 = ∓0.394

(1)

n3 = ±0.473

^ (1) = ∓0.394^ n e1 ± 0.788^ e2 ± 0.473^ e3 For σ(2) = 0, the principal direction is found from (2) n 3 − 0 −10 0 1 (2) −10 0 − 0 n2 30 (2) 0 30 −27 − 0 n3 ^ (2) · n ^ (2) = 1. The result is and n (2)

(2)

(2)

(2)

(2)

(2)

3n1 − 10n2 = 0 −10n1 + 30n3 = 0 30n2 − 27n3 = 0

0

= 0 0

50

Continuum Mechanics for Engineers

or

(2)

(2)

10n2 = 3n1

(2)

(2)

and 3n3 = n1

Using the unit vector relation 2 2 2 2 2 2 (2) (2) (2) (2) (2) (2) n1 + n2 + n3 = n1 + 0.3n1 + 3n1 =1 and (2)

n1 = ±0.913

(2)

n2 = ±0.274

(2)

n3 = ±0.304

^ (2) = ±0.913^ n e1 ± 0.274^ e2 ± 0.304^ e3 For σ(3) = −47, the principal direction is found from 3 − (−47) −10 0 −10 0 − (−47) 30 0 30 −27 − (−47)

(3)

n1

(3) n2 (3) n3

0

= 0 0

^ (3) · n ^ (3) = 1. The result is and n (3)

(3)

(3)

(3)

(3)

(3)

50n1 − 10n2 = 0 (3)

−10n1 + 47n2 + 30n3 = 0 30n2 + 20n3 = 0 or (3)

n1 =

1 (3) n 5 2

and

3 (3) (3) n3 = − n2 2

Using the unit vector relation 2 2 2 2 2 1 2 3 (3) (3) (3) (3) (3) (3) n2 n1 + n2 + n3 = + n2 + − n2 =1 5 2 and (3)

n2 = ±0.110

(3)

n1 = ±0.551

(3)

n3 = ∓0.827

^ (3) = ±0.110^ n e1 ± 0.551^ e2 ∓ 0.827^ e3 Problem 3.16 At point P, the stress matrix relative to axes Px1 x2 x3 is given in MPa by 5 a −a [tij ] = a 0 b −a b 0 where a and b are unspecified. At the same point relative to axes Px∗1 x∗2 x∗3 the matrix is σ(1) 0 0 ∗ 2 0 . tij = 0 0 0 σ(2) If the magnitude of the maximum shear stress at P is 5.5 MPa, determine σI and σIII . Answer

51

Chapter 3 Solutions σ(1) = 7 MPa, σ(3) = −4 MPa Solution The invariance of the trace gives tr T = tii = 5 + 0 + 0 = σ(1) + 2 + σ(3) or σ(1) + σ(3) = 3 The maximum shear stress is or

σ(1) − σ(3) = 5.5 2 σ(1) − σ(3) = 11

The result is σ(1) = 7

and

σ(3) = −4

Problem 3.17 The state of stress at point P is given in ksi with 1 0 [tij ] = 0 1 2 0

respect to axes Px1 x2 x3 by the matrix 2 0 . −2

Determine (a) the principal stress values and principal stress directions at P, (b) the maximum shear stress value at P, ^ = ni e ^i to the plane at P on which the maximum shear stress acts. (c) the normal n Answer (a) σ(1) = 2 ksi, σ(2) = 1 ksi, σ(3) = –3 ksi ^ 2^ e1 + e −^ e1 + 2^ e3 ^ (1) = ^ (2) = e ^2 , n ^ (3) = √ 3, n √ n 5 5 (b) (σS )max = ± 2.5 ksi ^ + 3^ e e3 ^ = 1√ (c) n 10 Solution (a) The principal values are found from 1−σ 1−σ 0 2 = 0 1−σ 0 0 det 2 0 −2 − σ 2

2 =0 0 −2 − σ = (1 − σ) σ2 + σ − 6 = 0 = (1 − σ) (σ + 3) (σ − 2) = 0 0 1−σ 0

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Continuum Mechanics for Engineers

The results are σ(1) = 2,

σ(2) = 1,

σ(3) = −3

For σ(1) = 2, the principal direction is found from (1) n 1−2 0 2 1 (1) n2 0 1−2 0 (1) n3 2 0 −2 − 2

0

= 0 0

^ (1) · n ^ (1) = 1. The result is and n (1)

(1)

−n1 + 2n3 = 0 (1)

−n2 = 0 (1)

(1)

2n1 − 4n3 = 0 or

(1)

(1)

n1 = 2n3

and

(1)

n2 = 0

Using the unit vector relation 2 2 2 2 2 (1) (1) (1) (1) (1) n1 + n3 = 2n3 + n3 = 5n3 =1 and

1 (1) n3 = ± √ 5

or

2 (1) n1 = ± √ 5

(1)

n2 = 0

2 1 ^ (1) = ± √ e ^1 ± √ e ^3 n 5 5

For σ(2) = 1, the principal direction is found from (2) n 1−1 0 2 1 (2) n2 0 1−1 0 (2) n3 2 0 −2 − 1

0

= 0 0

^ (2) · n ^ (2) = 1. The result is and n (2)

2n3 = 0 (2)

(2)

2n1 − 3n3 = 0 or

(2)

(2)

n3 = n1 = 0 Using the unit vector relation 2 2 2 2 (2) (2) (2) (2) n1 + n2 + n3 = n2 =1 and

(2)

n1 = 0

(2)

n2 = ±1

(2)

n3 = 0

53

Chapter 3 Solutions or

^ (2) = ±^ n e2

For σ(3) = −3, the principal direction is found from (3) n 1 − (−3) 0 2 1 (3) n2 0 1 − (−3) 0 (3) n3 2 0 −2 − (−3)

0

= 0 0

^ (3) · n ^ (3) = 1. The result is and n (3)

(3)

4n1 + 2n3 = 0 (3)

4n2 = 0 (3)

(3)

2n1 + n3 = 0 or

(3)

n2 = 0

and

(3)

(3)

n3 = −2n1

Using the unit vector relation 2 2 2 2 2 (3) (3) (3) (3) (3) n1 + n2 + n3 = n1 + −2n1 =1 and

1 (3) n1 = ± √ 5

or

(3)

n2 = 0

2 (3) n3 = ∓ √ , 5

1 2 ^ (3) = ± √ e1 ∓ √ e3 n 5 5

^ (1) and n ^ (2) . This gives the negative The result can be found through the cross product of n of the result for a right-hand coordinate system. (b) The maximum shear stress is (σS )max =

σ(1) − σ(3) 2 − (−3) = = 2.5 2 2

(c) The plane of maximum shear stress is ^= n

^ (1) + n ^ (3) ^3 −^ ^1 + 2^ n 2^ e1 + e e1 + 2^ e3 e e3 √ √ = √ + = √ 2 10 10 10

^ (3) for a right-hand coordinate system was used. Here n Problem 3.18 The stress tensor at P is given with respect to by 4 [tij ] = b b

Ox1 x2 x3 in matrix form with units of MPa b 7 2

b 2 4

where b is unspecified. If σ(3) = 3 MPa and σ(1) = 2σII , determine

54

Continuum Mechanics for Engineers (a) the principal stress values, (b) the value of b, (c) the principal stress direction of σ(2) .

Answer (a) σ(1) = 8 MPa, σ(2) = 4 MPa, σ(3) = 3 MPa ^ (2) = e ^1 (b) b = 0, (c) n Solution (a) The principal stress values are found from the invariance of the trace of the stress tensor. tr T = t11 + t22 + t33 = 4 + 7 + 4 = 15 = σ(1) + σ(2) + σ(3) = 2σ(2) + σ(2) + 3 and σ(1) = 8

σ(2) = 4

σ(3) = 3

(b) To determine the value of b, we can use the invariance of the determinate. 4 b b det T = b 7 2 = σ(1) σ(2) σ(3) = 8 · 4 · 3 = 96 b 2 4 or 4 (24) − b (4b − 2b) + b (2b − 7b) = 96 − 2b2 − 5b2 = 96 This gives b = 0. (c) To find the principal direction for σ(2) = 4, we have (2) n 4−4 0 0 1 (2) 0 7−4 2 n2 (2) n3 0 2 4−4

0

= 0 0

This gives (2)

(2)

3n2 + 2n3 = 0 (2)

2n2 = 0 and

(2)

(2)

n2 = n3 = 0 The unit vector condition gives 2 (2) n1 =1 and

or

(2)

n1 = ±1

^ (2) = ±^ n e1

Problem 3.19 The state of stress at P, when referred to axes 9 [tij ] = 3 0

Px1 x2 x3 is given in ksi units by the matrix 3 0 9 0 0 18

55

Chapter 3 Solutions Determine (a) the principal stress values at P,

^ ∗ = ni e ^∗i of the plane on which σN = 12 ksi and σS = 3 ksi. (b) the unit normal n Answer (a) σ(1) = 18 ksi, σ(2) = 12 ksi, σ(3) = 6 ksi √ ∗ ^∗1 + 6^ ^∗3 e e +e ∗ ^ = √2 (b) n 2 2 Solution (a) The principal stresses are found by 9−σ 9−σ 3 0 3 9−σ 0 3 = det 0 0 18 − σ 0

0 =0 0 18 − σ = (18 − σ) σ2 − 18σ + 72 = 0 = (18 − σ) (σ − 12) (σ − 6) = 0 3 9−σ 0

The principal values are σ(1) = 18

σ(2) = 12

σ(3) = 6

(b) With respect to the principal axes ∗

n t(^

)

=T ·n

or

18n1

0

0

0 0

12

0 n2 = 12n2 6 n3 6n3

0

n1

18

This gives ∗

n t(^

)

and

^∗1 + 12n2 e ^∗2 + 6n3 e ^∗3 = 18n1 e ∗

n σN = t(^

)

^ = 18n21 + 12n22 + 6n23 = 12 ·n

Thus 3n21 + 2n22 + n23 = 2 Now

∗

n σ2S = t(^

)

∗

n · t(^

)

− σ2N = 324n21 + 144n22 + 36n23 − 144 = 9

or 9n21 + 4n22 + n23 = 4.25 Finally n21 + n22 + n23 = 1

56

Continuum Mechanics for Engineers

These three equations can be solved to give √ 3 n2 = , 2

1 n1 = √ , 2 2

1 n3 = √ 2 2

and 1 ∗ √ ∗ ^∗ = √ e ^1 + 6^ ^∗3 n e2 + e 2 2 Problem 3.20 Verify the result listed for Problem 3.19b above by use of Eq 3.63.

Solution From Eq 3.63, we have n21 =

(σN − σII ) (σN − σIII ) + σ2S (12 − 12) (12 − 6) + 32 9 1 = = = (σI − σII ) (σI − σIII ) (18 − 12) (18 − 6) (6) (12) 8

n22 =

(σN − σIII ) (σN − σI ) + σ2S (12 − 6) (12 − 18) + 32 −36 + 9 3 = = = (σII − σIII ) (σII − σI ) (12 − 6) (12 − 18) −36 4

n22 =

(σN − σI ) (σN − σII ) + σ2S (12 − 18) (12 − 12) + 32 9 1 = = = (σIII − σI ) (σIII − σII ) (6 − 18) (6 − 12) (−12) (−6) 8

Problem 3.21 Sketch the Mohr’s circles for the various stress states shown on the cube which is oriented along the coordinate axes.

σ0

σ0 σ0

σ0

σ0 σ0 (a)

(b)

57

Chapter 3 Solutions

σ0

σ0

2σ0

σ0 σ0

2σ0 σ0

σ0 (c)

(d)

Problem 3.22 The state of stress referred to axes Px1 x2 x3 is 9 [tij ] = 12 0

given in MPa by the matrix 12 0 −9 0 . 0 5

Determine (a) the normal and shear components, σN and σS , respectively, on the plane at P whose unit normal is 1 ^ = (4^ n e1 + 3^ e2 ) , 5 (b) verify the result determined in (a) by a Mohr’s circle construction similar to that shown in the figure from Example 3.5. Answer σN = 14.04 MPa, σS = 5.28 MPa Solution (a) To find σN , we have n) ^ t(^ =T ·n

or

(^ n)

t1

(^ t2n) (^ n) t3

9

= 12 0

The stress vector is n) t(^ =

12

0

4 5

0 35 0 5 0

−9

=

72 5 21 5

0

21 72 ^1 + e ^2 e 5 5

The normal stress component is 72 21 4 3 288 + 63 (^ n) ^= ^1 + e ^2 · ^1 + 3^ σN = t ·n e e e2 = = 14.04 5 5 5 5 25

58

Continuum Mechanics for Engineers

The shear stress component is n) n) σ2S = t(^ · t(^ − σ2N =

(72)2 + (21)2 − (14.04)2 = (5.28)2 25

(b) Verify this by construction of the Mohr’s circle. Problem 3.23 Sketch the Mohr’s circles for σ0 0 (a) [tij ] = 0 σ0 σ0 0 σ0 0 (b) [tij ] = 0 2σ0 0 0

the simple states of stress given by σ0 0 , σ0 0 0 , −σ0

and determine the maximum shear stress in each case. Answer (a) (σS )max = σ0 (b) (σS )max = 32 σ0 Solution (a) The principal values are σ0 − σ 0 0 σ0 − σ det σ0 0

σ0 − σ σ0 0 0 = σ0 σ0 − σ

0 σ0 − σ 0

σ0 0 σ0 − σ = (σ0 − σ) σ2 − 2σ0 σ = (σ0 − σ) σ (σ − 2σ0 ) = 0

=0

and

2σ0 − 0 σI − σIII = = σ0 2 2 (b) The principal values are σI = 2σ0 , σII = σ0 , σIII = −σ0 , and (σS )max =

(σS )max =

2σ0 − (−σ0 ) 3 σI − σIII = = σ0 2 2 2

Problem 3.24 Relative to axes Ox1 x2 x3 , the state of stress at O is represented by the matrix 6 −3 0 6 0 [ksi] . [tij ] = −3 0 0 0 Show that, relative to principal axes Ox∗1 x∗2 x∗3 , 3 0 [tij ] = 0 9 0 0

the stress matrix is 0 0 [ksi] , 0

59

Chapter 3 Solutions

and that these axes result from a rotation of 45◦ about the x3 axis. Verify these results by Eq 3.72. Solution The principal stresses are found from 6−σ 6−σ −3 0 = −3 6−σ 0 det −3 0 0 0 0−σ

0 0 0−σ = −σ σ2 − 12σ + 27 = 0 = −σ (σ − 9) (σ − 3) = 0 −3 6−σ 0

=0

and σ(1) = 9

σ(2) = 3

σ(3) = 0

For a counterclockwise rotation about the x3 axis, we have the transformation matrix √1 √1 0 2 2 √1 [aij ] = − √1 0 2 2 0 0 1 t0ij = aiq ajm tqm gives

t0ij =

√1 2 √ − 12

√1 2 √1 2

0

0

0

6

0 −3 0 1

−3 6 0

0

√1 2 √1 2

0 0 0

− √12 √1 2

0

0

t11 + t22 t11 − t22 + cos 2θ + t12 sin 2θ 2 2 (6 + 6) (6 − 6) π π = + cos + (−3) sin = 3 2 2 2 2

t011 =

t11 + t22 t11 − t22 − cos 2θ − t12 sin 2θ 2 2 (6 + 6) (6 − 6) π π = − cos − (−3) sin = 9 2 2 2 2 t11 − t22 sin 2θ + t12 cos 2θ 2 (6 − 6) π π =− sin + (−3) cos = 0 2 2 2

t012 = −

Problem 3.25 The stress matrix representation at P is given by 29 0 0 [tij ] = 0 −26 6 [ksi] . 0 6 9

3

0 = 0 0 1

From Eq 3.72, we have

t022 =

0 9 0

0

0 0

60

Continuum Mechanics for Engineers

Decompose this matrix into its spherical and deviator parts, and determine the principal deviator stress values. Answer SI = 25 ksi, SII = 6 ksi, SIII = −31 ksi

Solution The spherical stress is σM =

1 1 1 tr T = tii = (29 − 26 + 9) = 4 3 3 3

The deviatoric stress is

1 Sij = tij − δij tkk 3

Thus the deviatoric stress is 4 29 0 0 [Sij ] = 0 −26 6 − 0 0 0 6 9

0 4 0

25 0 0 = 0 0 4

0 0 −30 6 6 5

The principal deviatoric stress is found from

25 − S 0 det 0

0 −30 − S 6

0 6 = 0 5−S = (25 − S) S2 + 25S − 186 = 0 = (25 − S) (S − 6) (S + 31) = 0

25 − S 0 0 6 = 0 5−S

0 −30 − S 6

The principal deviatoric stresses are SI = 25,

SII = 6,

SIII = −31

Problem 3.26 Let the second invariant of the stress deviator be expressed in terms of its principal values, that is, by IIS = SI SII + SII SIII + SIII SI . Show that this sum is the negative of two-thirds the sum of squares of the principal shear stresses, as given by Eq 3.60. Solution The second invariant of the stress deviator is IIS = SI SII + SII SIII + SIII SI where

SI = σI −

σI + σII + σIII 3

= σI − σM

61

Chapter 3 Solutions similarly for the other values. Then IIS = (σI − σM ) (σII − σM ) + (σII − σM ) (σIII − σM ) + (σIII − σM ) (σI − σM ) = σI σII + σII σIII + σIII σI − 2σM (σI + σII + σIII ) + 3σ2M = σI σII + σII σIII + σIII σI − 3σ2M

σI + σII + σIII 3 σ2I + σ2II + σ2III − σI σII − σII σIII − σIII σI =− 3

(σI σII + σII σIII + σIII σI ) −3 =9 9

2

From Eq 3.60, we have σS1 =

σI − σIII 2

σS2 =

σII − σIII 2

σS3 =

σIII − σI 2

Therefore, σ2S1

+

σ2S2

+

σ2S3

σ2 − 2σI σIII + σ2III + σ2II − 2σII σIII + σ2III + σ2III − 2σIII σI + σ2I = I 4 σ2I + σ2II + σ2III − σI σII − σII σIII − σIII σI = 2

Finally, we find −

2 2 σS1 + σ2S2 + σ2S3 = SI SII + SII SIII + SIII SI 3

Problem 3.27 Verify the results presented in Eqs 3.86 and 3.87 for the octahedral shear stress. Solution Equation 3.85 is 1 1 2 σI + σ2II + σ2III − (σI + σII + σIII )2 3 9 1 2 3 2 2 2 σI + σII + σIII − σI + σ2II + σ2III + 2 (σI σII + σII σIII + σIII σI ) = 9 9 2 2 2 2 = σI + σII + σIII − 2 (σI σII + σII σIII + σIII σI ) 9 i 1h (σI − σII )2 + (σII − σIII )2 + (σIII − σI )2 = 9

σ2oct =

and σoct =

i1/2 1h (σI − σII )2 + (σII − σIII )2 + (σIII − σI )2 3

This is Eq 3.86. Now σI = SI + σM for each principal stress. This results in i 1h (SI + σM − SII − σM )2 + (SII + σM − SIII − σM )2 + (SIII + σM − SI − σM )2 σ2oct = 9 i 1h (SI − SII )2 + (SII − SIII )2 + (SIII − SI )2 = 9 1 = 2 S2I + S2II + S2III − 2 (SI SII + SII SIII + SIII SI ) 9

62

Continuum Mechanics for Engineers

Also (SI + SII + SIII )2 = 0 and S2I + S2II + S2III = −2 (SI SII + SII SIII + SIII SI ). This gives 1 2 2 1 2 2 2 2 2 2 3 SI + SII + SIII = S + SII + SIII = − (SI SII + SII SIII + SIII SI ) σoct = 9 3 I 3 and from Problem 3.26

1/2 2 σoct = − IIS 3

This is Eq 3.87. Problem 3.28 At point P in a continuum body, the stress tensor components are given in MPa with respect to axes Px1 x2 x3 by the matrix √ 1 −3 √2 [tij ] = √ −3 1 − 2 . √ 2 − 2 4 Determine (a) the principal stress values σI , σII , and σIII , together with the corresponding principal stress directions, (b) the stress invariants IT , IIT , and IIIT , (c) the maximum shear stress value and the normal to the plane on which it acts, (d) the principal deviator stress values, (e) the stress vector on the octahedral plane together with its normal and shear components, (f) the stress matrix for axes rotated 60◦ counterclockwise with respect to the axis PQ, which makes equal angles relative to the coordinate axes Px1 x2 x3 . Answer (a) σI = 6 MPa, σII = 2 MPa, σIII = −2 MPa √ 1 (1) ^ =2 e ^1 − e ^2 + 2^ n e3 √ ^1 − e ^2 − 2^ ^ (2) = 21 e e3 n ^ +e ^ e ^ (3) = 1√ 2 n 2 (b) IT = 6 MPa, IIT = −4 MPa, IIIT = −24 MPa √ √ √ ^1 − 1 − 2 e ^2 + 2^ 1+ 2 e e3 ^ max = √ (c) (σS )max = 4 MPa n 2 2 (d) SI = 4 MPa, SII = 0 MPa, SIII = −4 MPa q 6^ e∗1 + 2^ e∗ − 2^ e∗3 n) √2 (e) ^ t(^ = , σN = 2, σoct = 32 3 3 √ √ −12 √ −12 + 3√ 2 −12 − 3 2 h i 1 (f) t0ij = −12 + 3√2 33 − 12 2 −3 √ MPa 9 −12 − 3 2 −3 33 + 12 2

63

Chapter 3 Solutions Solution (a) The principal values are found from √ 1−σ 1−σ −3 √2 = −3 det √ −3 1− σ − 2 √ √ 2 − 2 4−σ 2

√ √2 − 2 = 0 4−σ = (1 − σ) σ2 − 5σ + 2 − 3 (10 − 3σ) + 2 (2 + σ) = 0 −3 1− √σ − 2

= σ3 − 6σ2 − 4σ + 24 = (σ − 6) (σ − 2) (σ + 2) = 0 The results are σ(1) = 6,

σ(2) = 2,

σ(3) = −2

For σ(1) = 6, the principal direction is found from √ (1) 1−6 −3 2 n1 √ (1) −3 1 − 6 − 2 n2 √ √ (1) 2 − 2 4−6 n3

0

= 0 0

^ (1) · n ^ (1) = 1. The result is and n √ (1) 2n3 = 0 √ (1) (1) (1) −3n1 − 5n2 − 2n3 = 0 √ (1) √ (1) (1) 2n1 − 2n2 − 2n3 = 0 (1)

(1)

−5n1 − 3n2 +

From the first two equations, we have (1)

(1)

n1 = −n2

and this result together with the third equation gives √ (1) (1) n3 = 2n1 Using the unit vector relation 2 2 2 2 2 √ 2 (1) (1) (1) (1) (1) (1) n1 + n2 + n3 = n1 + −n1 + 2n2 =1 and (1) n1

^ (1) n

√ 1 1 2 (1) (1) n2 = ∓ n3 = ± =± 2 2√ 2 1 1 2 ^1 ∓ e ^2 ± ^3 =± e e 2 2 2

For σ(2) = 2, the principal direction is found from √ (2) 2 n 1−2 −3 1 √ (2) −3 1 − 2 − 2 n2 √ √ (2) 2 − 2 4−2 n3

0

= 0 0

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Continuum Mechanics for Engineers

^ (2) · n ^ (2) = 1. The result is and n √ (2) 2n3 = 0 √ (2) (2) (2) −3n1 − n2 − 2n3 = 0 √ (2) √ (2) (2) 2n1 − 2n2 + 2n3 = 0 (2)

(2)

−1n1 − 3n2 +

or

(2)

(2)

and

n2 = −n1

√ (2) (2) n3 = − 2n1

Using the unit vector relation 2 2 2 2 2 √ 2 (2) (2) (2) (2) (2) (2) n1 + n2 + n3 = n1 + −n1 + − 2n1 =1 and (2) n1

^ (2) n

√ 1 2 1 (2) (2) n2 = ∓ n =∓ =± 2 2√ 3 2 1 1 2 ^1 ∓ e ^2 ∓ ^3 =± e e 2 2 2

For σ(3) = −2, the principal direction is found from √ (3) n1 1 − (−2) −3 2 √ (3) −3 1 − (−2) − 2 n2 √ √ (3) n3 2 − 2 4 − (−2)

0

= 0 0

^ (3) · n ^ (3) = 1. The result is and n √ (3) 2n3 = 0 √ (3) (3) (3) −3n1 + 3n2 − 2n3 = 0 √ (3) √ (3) (3) 2n1 − 2n2 + 6n3 = 0 (3)

(3)

3n1 − 3n2 +

or from the second and third equation (3)

(3)

n1 = n2

(3)

and n3 = 0

Using the unit vector relation 2 2 2 2 2 (3) (3) (3) (3) (3) n1 + n2 + n3 = n2 + n2 =1 and 1 1 (3) (3) n1 = ± √ n2 = ± √ 2 2 1 1 n(3) = ± √ e1 ± √ e2 2 2

(3)

n3 = 0

(b) The invariants of T are IT = σI + σII + σIII = 6 + 2 − 2 = 6 IIT = σI σII + σII σIII + σIII σI = 6 (2) + 2 (−2) − 2 (6) = −4 IIIT = σI σII σIII = 6 (2) (−2) = −24

65

Chapter 3 Solutions (c) The maximum shear stress is (σS )max =

σI − σIII 6 − (−2) = =4 2 2

The normal to the plane of maximum shear stress is ! " # √ ^∗1 + e ^∗3 1 1 1 1 e 2 ^ max = √ e1 − e2 + e3 + √ (e1 + e2 ) =√ n 2 2 2 2 2 2 i h √ √ √ 1 = √ 1 + 2 e1 − 1 − 2 e2 + 2e∗3 2 2 (d) The principal stress deviators are given that the mean stress is σM = 1 (6 + 2 − 2) = 2 3

1 3

(σI + σII + σIII ) =

SI = σI − σM = 6 − 2 = 4 SII = σII − σM = 2 − 2 = 0 SIII = σIII − σM = −2 − 2 = −4 ^ = √1 (^ ^2 + e ^3 )is (e) The stress vector on the octahedral plane with normal n e1 + e 3 ^ t(n) = T · n or

6 0 0

0 2 0

0 1 6 1 1 0 √ 1 = √ 2 3 3 −2 1 −2

This gives 1 n) t(^ = √ (6^ e1 + 2^ e2 − 2^ e3 ) 3 The normal stress on this plane is 1 1 ^2 + e ^3 ) = 2 ^ = √ (6^ σN = t(n) · n e1 + 2^ e2 − 2^ e3 ) · √ (^ e1 + e 3 3 and the octahedral stress is n) n) · t(^ − σ2N = σ2oct = t(^

36 + 4 + 4 32 −4= 3 3

This gives r σoct =

32 3

(f) The stress on the plane with a normal making equal angles to the coordinate axes and a 60◦ counterclockwise rotation is given by t0ij = aiq ajm tqm . The transformation matrix is (Problem 3.10) 2 2 −1 1 2 2 [aij ] = −1 3 2 −1 2

66

Continuum Mechanics for Engineers

and √ 1 −3 2 2 −1 2 √2 1 0 1 2 −1 2 2 tij = −3 1 − 2 √ √ 3 3 −1 2 −1 2 2 − 2 4 √ √ −12 √ −12 + 3√ 2 −12 − 3 2 1 = −12 + 3√2 33 − 12 2 −3 √ 9 −3 33 + 12 2 −12 − 3 2

−1 2 2

2 −1 2

Problem 3.29 In a continuum, the stress field relative to axes Ox1 x2 x3 is given by x21 x2 x1 1 − x22 0 1 2 [tij ] = x1 1 − x22 . x − 3x 0 2 2 3 0 0 2x23 Determine (a) the body force distribution if the equilibrium equations are to be satisfied throughout the field, √ (b) the principal stresses at P(a, 0, 2 a), (c) the maximum shear stress at P, (d) the principal deviator stresses at P. Answer 4x3 ρ (b) σI = 8a, σII = a, σIII = −a (c) (σS )max = ±4.5a 16 5 11 (d) SI = a, SII = − a, SIII = − a 3 3 3 (a) b1 = b2 = 0, b3 = −

Solution (a) The body force distribution is found from the equilibrium equations tij,j + ρbi = 0 For i = 1 t11,1 + t12,2 + t13,3 + ρb1 = 2x1 x2 − 2x1 x2 + 0 + ρb1 = 0 or b1 = 0 For i = 2 t21,1 + t22,2 + t23,3 + ρb2 = 1 − x22 + and b2 = 0

1 3x22 − 1 + 0 + ρb2 = 0 3

67

Chapter 3 Solutions For i = 3 t31,1 + t32,2 + t33,3 + ρb3 = 4x3 + ρb3 = 0 and b3 = −

4x3 ρ

√ (b) At the point P a, 0, 2 a the stress is

0 [tij ] = a 0

a 0 0 0 0 8a

The principal stress is found from the eigen-problem 0−σ 0−σ a 0 = a 0−σ 0 det a 0 0 0 8a − σ

a 0−σ 0

0 0 8a − σ 2 2 = (8a − σ) σ − a = 0

=0

and σI = 8a

σII = a

σIII = −a

(c) The maximum shear stress at P is (σS )max =

| σI − σIII | | 8a − (−a) | = = ±4.5a 2 2

(d) The principal deviatoric stresses are 8a 16a = 3 3 8a 5a SII = σII − σM = a − =− 3 3 8a 11a SIII = σIII − σM = −a − =− 3 3 SI = σI − σM = 8a −

given σM =

1 3

(σI + σII + σIII ) =

1 3

(8a + a − a) = 8a/3.

Problem 3.30 In describing the yield surface in plasticity the second invariant of the deviator stress, often denoted by J2 , plays an important role. Starting with the second invariant of the deviator stress J2 = IIS = − S(1) S(2) + S(2) S(3) + S(1) S(3) derive the formula J2 =

3 2 σ . 2 oct

Solution Substitute S(i) = σ(i) − σM and carryout some algebra.

68

J2 = IIS

Continuum Mechanics for Engineers

=

−(S(1) S(2) + S(2) S(3) + S(1) S(3) )

=

−((σ(1) − σM )(σ(2) − σM ) + (σ(2) − σM )(σ(3) − σM ) + (σ(1) − σM )(σ(3) − σM ))

=

−(σ(1) σ(2) + σ(2) σ(3) + σ(1) σ(3) − 2(σ(1) + σ(2) + σ(3) )σM + 3σ2M )

=

−(σ(1) σ(2) + σ(2) σ(3) + σ(1) σ(3) − 3σ2M )

=

−

σ(1) + σ(2) + σ(3) 2(σ(1) σ(2) + σ(2) σ(3) + σ(1) σ(3) ) −3 2 9

(σ(1) + σ(2) + σ(3) )2 − (σ2(1) + σ2(2) + σ2(3) )

σ(1) + σ(2) + σ(3) − − 2 3 ! 3(σ2(1) + σ2(2) + σ2(3) ) 3(σ(1) + σ(2) + σ(3) )2 + − − 2·3 2·3·3 ! (σ2(1) + σ2(2) + σ2(3) ) σ(1) + σ(2) + σ(3) 2 3 − 2 3 3

=

=

=

!

3 2 2 σOCT

= Problem 3.31 Show that

J2 = IIS = −(S(1) S(2) + S(2) S(3) + S(1) S(3) ) =

1 2 S(1) + S2(2) + S2(3) 2

where J2 is the second invariant of the deviator stress and S(1) , S(2) , S(3) are its principal values. Solution Use the fact that IS = 0 and substitute S(1) = −S(2) − S(3) S(2) = −S(1) − S(3) S(3) = −S(1) − S(2) J2

= S(1) (S(2) + S(3) ) + S(2) (S(1) + S(2) ) + S(3) (S(2) + S(3) ) = S2(1) + S2(2) + S2(3) − J2

Problem 3.32 Let the stress tensor components tij be derivable from the symmetric tensor field ϕij by the equation tij = εiqk εjpm ϕkm,qp . Show that, in the absence of body forces, the equilibrium equations are satisfied. Recall from Problem 2.17 that δji δjq δjk εiqk εjpm = δpi δpq δpk . δmi δmq δmk

69

Chapter 3 Solutions Solution The equilibrium equations without body forces are tij,j = 0 = εiqk εjpm ϕkm,qpj For εiqk εjpm

δji = δpi δmi

δjq δpq δmq

δjk δpk δmk

= δji (δpq δmk − δpk δmq ) − δjq (δpi δmk − δpk δmi ) + δjk (δpi δmq − δpq δmi ) then for i the free index tij,j = εiqk εjpm ϕkm,qpj = [δji (δpq δmk − δpk δmq ) − δjq (δpi δmk − δpk δmi ) + δjk (δpi δmq − δpq δmi )] ϕkm,qpj = ϕkk,ppi − ϕkq,qki − ϕkk,qiq + ϕki,jkj + ϕkq,qik − ϕki,ppk = 0 Problem 3.33 Verify that ∂tij /∂tmn = δim δjn and use this result (or otherwise) to show that ∂IIS = −Sij . ∂tij that is, the derivative of the second invariant of the deviatoric stress with respect to the stress components is equal to the negative of the corresponding component of the deviatoric stress.

Chapter 4 Solutions

Problem 4.1 The motion of a continuous medium is specified by the component equations x1

=

1 2 (X1

+ X2 )et + 12 (X1 − X2 )e−t ,

x2

=

1 2 (X1

+ X2 )et − 12 (X1 − X2 )e−t ,

x3

= X3 .

(a) Show that the Jacobian determinant J does not vanish, and solve for the inverse equations X = X(x, t). (b) Calculate the velocity and acceleration components in terms of the material coordinates. (c) Using the inverse equations developed in part (a), express the velocity and acceleration components in terms of spatial coordinates. Answer (a) J = cosh2 t − sinh2 t = 1 X1 = 12 (x1 + x2 )e−t + 21 (x1 − x2 )et X2 = 12 (x1 + x2 )e−t − 21 (x1 − x2 )et X3 = x3 (b) v1 = 12 (X1 + X2 )et − 21 (X1 − X2 )e−t v2 = 12 (X1 + X2 )et + 21 (X1 − X2 )e−t v3 = 0 a1 = 12 (X1 + X2 )et + 12 (X1 − X2 )e−t a2 = 12 (X1 + X2 )et − 12 (X1 − X2 )e−t a3 = 0 (c) v1 = x2 , v2 = x1 , v3 = 0 a1 = x1 , a2 = x2 , a3 = 0 Solution (a) For the motion to be invertible, the Jacobian of the deformation gradient is to be non-zero. −t t −t t ∂xi (et + e−t ) /2 (et − e−t ) /2 0 J = det F = |FiA | = = (e − e ) /2 (e + e ) /2 0 ∂XA 0 0 1 t 2 2 et − e−t e + e−t − = cosh2 t − sinh2 t = 1 6= 0 = 2 2

71

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Continuum Mechanics for Engineers

Now x1 + x2 = (X1 + X2 ) et x1 − x2 = (X1 − X2 ) e−t and X1 + X2 = (x1 + x2 ) e−t X1 − X2 = (x1 − x2 ) et This yields (x1 + x2 ) e−t + (x1 − x2 ) et 2 (x1 + x2 ) e−t − (x1 − x2 ) et X2 = 2 X3 = x3

X1 =

(b) The velocity is v =

dx or dt X

1 (X1 + X2 )et − 2 1 v2 = (X1 + X2 )et + 2 v3 = 0 dv or The acceleration is a = dt X

1 (X1 − X2 )e−t 2 1 (X1 − X2 )e−t 2

1 (X1 + X2 )et + 2 1 a2 = (X1 + X2 )et − 2 a3 = 0

1 (X1 − X2 )e−t 2 1 (X1 − X2 )e−t 2

v1 =

a1 =

(c) From part (a), we have the sum and differences X1 + X2 and X1 − X2 to find the spatial forms 1 (x1 + x2 ) e−t et − 2 1 (x1 + x2 ) e−t et + v2 = 2 v3 = 0 v1 =

1 (x1 − x2 ) et e−t = x2 2 1 (x1 − x2 ) et e−t = x1 2

The acceleration in spatial form is 1 (x1 + x2 ) e−t et + 2 1 (x1 + x2 ) e−t et − a2 = 2 a3 = 0 a1 =

1 (x1 − x2 ) et e−t = x1 2 1 (x1 − x2 ) et e−t = x2 2

73

Chapter 4 Solutions Problem 4.2 Let the motion of a continuum be given in component form by the equations x1 = X1 + X2 t + X3 t2 , x2 = X2 + X3 t + X1 t2 , x3 = X3 + X1 t + X2 t2 . (a) Show that J 6= 0, and solve for the inverse equations. (b) Determine the velocity and acceleration

(1) at time t = 1 s for the particle which was at point (2.75, 3.75, 4.00) when t = 0.5 s. (2) at time t = 2 s for the particle which was at point (1, 2, −1) when t = 0. Answer (a) J = (1 − t3 )2 X1 = (x1 − x2 t)/(1 − t3 ) X2 = (x2 − x3 t)/(1 − t3 ) X3 = (x3 − x1 t)/(1 − t3 ) (b) (1) v = 8^ e1 + 5^ e2 + 5^ e3 , a = 6^ e1 + 2^ e2 + 4^ e3 (2) v = −2^ e1 + 3^ e2 + 9^ e3 , a = −2^ e1 + 2^ e2 + 4^ e3 Solution (a) The Jacobian is 1 J = det F = |FiA | = t2 t

t 1 t2

t2 t 1

= 1 − t3 2

The inverse is using Cramer’s rule x1 t t2 x2 1 t x3 t2 1 x1 − x2 t x1 (1 − t3 ) − x2 (t − t4 ) = = X1 = 2 3 (1 − t3 ) J (1 − t ) 1 x1 t2 2 t x2 t t x3 1 x2 (1 − t3 ) − x3 (t − t4 ) x2 − x3 t = = X2 = 2 3 (1 − t3 ) J (1 − t ) 1 t x1 2 t 1 x2 t t2 x3 x3 (1 − t3 ) − x1 (t − t4 ) x3 − x1 t X3 = = = (1 − t3 ) J (1 − t3 )2 dx (b) The velocity is v = or dt X v1 = X2 + 2X3 t;

v2 = X3 + 2X1 t;

v3 = X1 + 2X2 t

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Continuum Mechanics for Engineers

Using the inverse equations from part (a), we have the spatial forms v1 =

x2 + x3 t − 2x1 t2 (1 − t3 )2

v2 =

;

x3 + x1 t − 2x2 t2 (1 − t3 )2

v3 =

;

x1 + x2 t − 2x3 t2 (1 − t3 )2

dv or The acceleration is a = dt X a1 = 2X3 ;

a2 = 2X1 ;

a3 = 2X2

and the spatial form is a1 =

2 (x3 − x1 t) (1 −

2 t3 )

a2 =

;

2 (x1 − x2 t) (1 −

2 t3 )

;

a3 =

2 (x2 − x3 t) (1 − t3 )2

Problem 4.3 A continuum body has a motion defined by the equations x1 = X1 + 2X2 t2 , x2 = X2 + 2X1 t2 , x3 = X3 . (a) Determine the velocity components at t = 1.5 s of the particle which occupied the point (2, 3, 4) when t = 1.0 s. (b) Determine the equation of the path along which the particle designated in part (a) moves. (c) Calculate the acceleration components of the same particle at time t = 2 s. Answer (a) v1 = 2, v2 = 8, v3 = 0 (b) 4x1 − x2 = 5 in the plane x3 = 4 (c) a1 = 4/3, a2 = 16/3, a3 = 0. Solution

dx (a) The velocity is v = or dt X v1 = 4X2 t;

v2 = 4X1 t;

v3 = 0

When t = 1.0 s, the initial location of the particle is found from 2 = X1 + 2X2 3 = X2 + 2X1 4 = X3 The solution is X1 =

4 ; 3

X2 =

1 ; 3

X3 = 4

75

Chapter 4 Solutions The velocity components at t = 1.5 s are 3 4 3 1 = 2; v2 = 4X1 t = 4 = 8; v1 = 4X2 t = 4 3 2 3 2

v3 = 0

(b) The path for the particle at the position in (a) is given by 4 1 2 2 x1 = X1 + 2X2 t = +2 t 3 3 1 4 2 x2 = X2 + 2X1 t2 = +2 t 3 3 x3 = X3 = 4 Eliminating t gives 12x1 − 16 = 3x2 − 1 or

4x1 − x2 = 5 in the plane x3 = 4 dv (c) The acceleration is a = or dt X a1 = 4X2 ; or a1 = 4X2 = 4

4 1 = ; 3 3

a2 = 4X1 ;

a3 = 0

a2 = 4X1 = 4

4 16 = ; 3 3

a3 = 0

Problem 4.4 If the motion x = x(X, t) is given in component form by the equations x1 = X1 (1 + t),

x2 = X2 (1 + t)2 ,

x3 = X3 (1 + t2 ) ,

determine expressions for the velocity and acceleration components in terms of both Lagrangian and Eulerian coordinates. Answer v1 = X1 = x1 /(1 + t) v2 = 2X2 (1 + t) = 2x2 /(1 + t) v3 = 2X3 t = 2x3 t/(1 + t2 ) a1 = 0 a2 = 2X2 = 2x2 /(1 + t)2 a3 = 2X3 = 2x3 /(1 + t2 ) Solution The velocity is v =

dx or dt X v1 = X1 ;

v2 = 2X2 (1 + t) ;

v3 = 2X3 t

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Continuum Mechanics for Engineers

To establish the Eulerian form we must invert the position equations to get X1 =

x1 ; 1+t

X2 =

x2 (1 + t)

2

;

X3 =

x3 (1 + t2 )

and

2x2 x1 ; v2 = 2X2 (1 + t) = ; 1+t 1+t dv or The acceleration is a = dt X

v3 = 2X3 t =

v1 = X1 =

a1 = 0;

a2 = 2X2 ;

or a1 = 0;

a2 =

2x2 (1 + t)2

;

2x2 t (1 + t2 )

a3 = 2X3

a3 =

2x3 (1 + t2 )

Problem 4.5 The Lagrangian description of a continuum motion is given by x1 = X1 e−t + X3 (e−t − 1) , x2 = X2 et − X3 (1 − e−t ) , x3 = X3 et . Show that these equations are invertible and determine the Eulerian description of the motion. Answer X1 = x1 et − x3 (et − 1) X2 = x2 e−t + x3 (e−2t − e−3t ) X3 = x3 e−t Solution The Jacobian must be non-zero for the equations be invertible. −t e 0 e−t − 1 et − (1 − e−t ) = e−t e2t = et 6= 0 J = det F = [FiA ] = 0 0 0 et Inverting the equation for x3 gives X3 = x3 e−t Then x1 = X1 e−t + X3 (e−t − 1) = X1 e−t + x3 e−t (e−t − 1) x2 = X2 et − X3 (1 − e−t ) = X2 et − x3 e−t (1 − e−t ) X3 = x3 e−t

77

Chapter 4 Solutions The Eulerian form of the motion is X1 = x1 et x3 (e−t − 1) X2 = x2 e−t + x3 e−2t (1 − e−t ) X3 = x3 e−t Also by using Cramer’s rule, we have x1 0 e−t − 1 −t t − (1 − e−t ) X1 = e x2 e x3 0 et

;

−t X3 = e

−t e X2 = e−t 0 0 e−t 0 x1 0 et x2 0 0 x3

e−t − 1 − (1 − e−t ) et

x1 x2 x3

Problem 4.6 A velocity field is given in Lagrangian form by v2 = X2 et ,

v1 = 2t + X1 ,

v3 = X3 − t .

Integrate these equations to obtain x = x(X, t) with x = X at t = 0, and using that result compute the velocity and acceleration components in the Eulerian (spatial) form. Answer v1 = (x1 + 2t + t2 )/(1 + t) v2 = x2 v3 = (2x3 − 2t − t2 )/2(1 + t) a1 = 2, a2 = x2 , a3 = −1 Solution Integrating the velocity equations gives x1 = t2 + X1 t + C1 ;

x2 = X2 et + C2 ;

x3 = X3 t −

t2 + C3 2

The initial conditions are x = X at t = 0. This gives x1 = X1 = C1 ;

x2 = X2 = X2 + C2 and C2 = 0;

x3 = X3 = C3

This results in x1 = t2 + X1 t + X1 x2 = X2 et x3 = X3 t −

t2 + X3 2

Inverting directly gives X1 =

x1 − t2 ; 1+t

X2 = x2 e−t ;

X3 =

x3 + 12 t2 1+t

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Continuum Mechanics for Engineers

Substituting these positions gives the Eulerian form 2t (1 + t) + x1 − t2 , 1+t dv or The acceleration is a = dt X v1 =

a1 = 2;

v2 = x2 ,

a2 = X2 et = x2 ,

v3 =

2x3 − 2t − t2 2 (1 + t)

a3 = −1

Problem 4.7 If the motion of a continuous medium is given by x1 = X1 et − X3 (et − 1) , x2 = X2 e−t + X3 (1 − e−t ) , x3 = X3 , determine the displacement field in both material and spatial descriptions. Answer u1 = (X1 − X3 )(et − 1) = (x1 − x3 )(1 − e−t ) u2 = (X2 − X3 )(e−t − 1) = (x2 − x3 )(1 − et ) u3 = 0 Solution The displacement field is u = x − X. For the given motion, we have u1 = x1 − X1 = X1 et − X3 (et − 1) − X1 = (X1 − X3 ) et − 1

u2 = x2 − X2 = X2 e−t + X3 (1 − e−t ) − X2 = (X3 − X2 ) 1 − e−t

u3 = x3 − X3 = X3 − X3 = 0 This is the material form for the displacement field. To find the spatial form, the motion must be inverted. This can be accomplished directly without the calculation of the Jacobian. x1 = X1 et − X3 (et − 1) = X1 et − x3 (et − 1) x2 = X2 e−t + X3 (1 − e−t ) = X2 e−t + x3 (1 − e−t ) x3 = X3 and X1 = x1 e−t + x3 1 − e−t X2 = x2 et − x3 et − 1

X3 = x3 The spatial form of the displacement field is u1 = x1 − X1 = (X1 − X3 ) et − 1 = x1 e−t + x3 1 − e−t − x3 et − 1 = (x1 − x3 ) 1 − e−t

79

Chapter 4 Solutions u2 = x2 − X2 = (X3 − X2 ) 1 − e−t = x3 − x2 et − x3 et − 1 1 − e−t = (x3 − x2 ) et − 1 = (x2 − x3 ) 1 − et and u3 = x3 − X3 = x3 − x3 = 0 Problem 4.8 The temperature field in a continuum is given by the expression θ = e−3t /x2

where

x2 = x21 + x22 + x23 .

The velocity field of the medium has components v1 = x2 + 2x3 ,

v2 = x3 − x1 ,

v3 = x1 + 3x2 .

Determine the material derivative dθ/dt of the temperature field. Answer dθ/dt = −e−3t (3x2 + 6x1 x3 + 8x2 x3 )/x4 Solution The material time derivative of the temperature field is dθ ∂θ e−3t ∂θ ∂ = + vi = −3 2 + vi dt ∂t ∂xi x ∂xi

e−3t x2

Now, using the chain rule ∂ ∂xi

e−3t x2

=e

−3t

−3 ∂x −2x ∂xi

The last term in the parentheses is ∂ (xj xj )1/2 ∂xj ∂x 1 1 xi −1/2 = = (xk xk ) 2xj = xj δij = ∂xi ∂xi 2 ∂xi x x So that ∂ ∂xi

e−3t x2

=e

−3t

x xi i −3 ∂x −2x = e−3t −2x−3 = e−3t −2 4 ∂xi x x

This gives x dθ e−3t e−3t i = −3 2 − 2e−3t vi 4 = − 4 3x2 + 2vi xi dt x x x e−3t = − 4 3x2 + 2 [x1 (x2 + 2x3 ) + x2 (x3 − x1 ) + x3 (x1 + 3x2 )] x e−3t e−3t = − 4 3x2 + 2 [3x1 x3 + 4x2 x3 ] = − 4 3x2 + 6x1 x3 + 8x2 x3 x x

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Continuum Mechanics for Engineers

Problem 4.9 In a certain region of a fluid the flow velocity has components v1 = A(x31 + x1 x22 )e−kt ,

v2 = A(x21 x2 + x33 )e−kt ,

v3 = 0

where A and k are constants. Use the (spatial) material derivative operator to determine the acceleration components at the point (1, 1, 0) when t = 0. Answer a1 = −2A(k − 5A), a2 = −A(k − 5A), a3 = 0 Solution The acceleration is a=

∂v dv = + v · grad v dt ∂t

or ai =

∂vi ∂vi + vk ∂t ∂xk

So that a1 = −kA(x31 + x1 x22 )e−kt + A(x31 + x1 x22 )e−kt 3Ax21 e−kt + A(x21 x2 + x33 )e−kt 2Ax1 x2 e−kt

a2 = −kA(x21 x2 + x33 )e−kt + A(x31 + x1 x22 )e−kt (2Ax1 x2 ) e−kt + A(x21 x2 + x33 )e−kt Ax21 e−kt and a3 = 0 At the point (1, 1, 0) and t = 0, we have a1 |(1,1,0),t=0 = −kA (2) + A (2) (3A) + A (1) (2A) = −2A (k − 4A) a2 |(1,1,0),t=0 = −kA (2) + A (2) (2A) + A (1) (A) = −A (k − 5A) a3 |(1,1,0),t=0 = 0 Problem 4.10 A displacement field is given in terms of the spatial variables and time by the equations u1 = x2 t2 ,

u2 = x3 t,

u3 = x1 t .

Using the (spatial) material derivative operator, determine the velocity components. Answer v1 = (2x2 t + x3 t2 + x1 t3 )/(1 − t4 ) v2 = (x3 + x1 t + 2x2 t3 )/(1 − t4 ) v3 = (x1 + 2x2 t2 + x3 t3 )/(1 − t4 )

81

Chapter 4 Solutions Solution The velocity is v=

∂u du = + v · grad u dt ∂t

or vi =

dui ∂ui ∂ui = + vk dt ∂t ∂xk

This is an implicit equation for the velocity field. For the given displacement field v1 = 2x2 t + v1 (0) + v2 t2 + v3 (0) = 2x2 t + v2 t2 v2 = x3 + v1 (0) + v2 (0) + v3 (t) = x3 + v3 (t) v3 = x1 + v1 (t) + v2 (0) + v3 (0) = x1 + v1 (t) Now, we have v1 = 2x2 t + t2 (x3 + tv3 ) = 2x2 t + t2 (x3 + t [x1 + tv1 ]) and v1 =

2x2 t + x3 t2 + x1 t3 1 − t4

Similarly v2 = x3 + t (x1 + tv1 ) = x3 + t x1 + t 2x2 t + t2 v2 and v2 =

x3 + x1 t + 2x2 t3 1 − t4

Finally v3 = x1 + t 2x2 t + t2 v2 = x1 + t 2x2 t + t2 [x3 + tv3 ] and v3 =

x1 + 2x2 t2 + x3 t3 1 − t4

Alternatively, the problem can be solved using Cramer’s rule. We write the system of equations as v1 −t2 v2 = 2x2 t v2 −tv3 = x3 −tv1 +v3 = x1 The determinant is

and 1 v1 = 1 − t4

1 0 −t 2x2 t x3 x1

−t2 1 0

−t2 1 0

0 −t 1 1 v3 = 1 − t4

0 −t 1 ; 1 0 −t

= 1 − t4 1 0 −t 2x2 t x3 x1

1 v2 = 1 − t4 −t2 1 0

2x2 t x3 x1

0 −t 1

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Continuum Mechanics for Engineers

Problem 4.11 For the motion given by the equations x1 = X1 cos ωt + X2 sin ωt , x2 = −X1 sin ωt + X2 cos ωt , x3 = (1 + kt)X3 , where ω and k are constants, determine the displacement field in Eulerian form. Answer u1 = x1 (1 − cos ωt) + x2 sin ωt u2 = −x1 sin ωt + x2 (1 − cos ωt) u3 = x3 kt/(1 + kt) Solution The Jacobian is cos ωt J = det F = |FiA | = − sin ωt 0

sin ωt cos ωt 0

0 0 1 + kt

= cos2 ωt + sin2 ωt (1 + kt)

which is not zero. Multiplying the first equation in the motion by sin ωt and the second equation by cos ωt yields x1 sin ωt = X1 sin ωt cos ωt + X2 sin2 ωt x2 cos ωt = −X1 cos ωt sin ωt + X2 cos2 ωt and adding gives

x1 sin ωt + x2 cos ωt = X2

Also x1 cos ωt = X1 cos2 ωt + X2 sin ωt cos ωt x2 sin ωt = −X1 sin2 ωt + X2 cos ωt sin ωt and subtracting yields

x1 cos ωt − x2 sin ωt = X1

The displacements are u1 = x1 − X1 = x1 − x1 cos ωt + x2 sin ωt = x1 (1 − cos ωt) + x2 sin ωt u2 = x2 − X2 = x2 − x1 sin ωt − x2 cos ωt = −x1 sin ωt + x2 (1 − cos ωt) 1 x3 kt = u3 = x3 − X3 = x3 − x3 1 + kt 1 + kt Problem 4.12 Show that the displacement field for the motion analyzed in Problem 4.1 has the Eulerian form u1 = x1 − (x1 + x2 )e−t /2 − (x1 − x2 )et /2 , u2 = −x2 − (x1 + x2 )e−t /2 + (x1 − x2 )et /2 ,

83

Chapter 4 Solutions

and by using the material derivative operator (dui /dt = ∂ui /∂t + vj ∂ui /∂xj ), verify the velocity and acceleration components calculated in Problem 4.1. Solution From Problem 4.1 the inverted motion is (x1 + x2 ) e−t + (x1 − x2 ) et 2 (x1 + x2 ) e−t − (x1 − x2 ) et X2 = 2 X3 = x3 X1 =

The displacement field is (x1 + x2 ) e−t + (x1 − x2 ) et 2 (x1 + x2 ) e−t − (x1 − x2 ) et u2 = x2 − X2 = x2 − 2 u3 = x3 − X3 = x3 − x3 = 0 u1 = x1 − X1 = x1 −

The velocity field is du1 ∂u1 ∂u1 = + vk dt ∂t ∂xk −t (x1 + x2 ) e−t (x1 − x2 ) et et e−t et e = − + v1 1 − − − v2 − 2 2 2 2 2 2

v1 =

and ∂u2 du2 ∂u2 = + vk dt ∂t ∂xk −t (x1 + x2 ) e−t (x1 − x2 ) et e et e−t et = + − v1 − + v2 1 − − 2 2 2 2 2 2

v2 =

Adding and subtracting these equations, yields v1 + v2 = x1 + x2 v1 − v2 = −x1 + x2 and v1 = x2 ;

v2 = x1 ;

v3 = 0

The material description is v1 = x2 =

(X1 + X2 ) et (X1 − X2 ) e−t − 2 2

v2 = x1 =

(X1 + X2 ) et (X1 − X2 ) e−t + 2 2

and

The accelerations are a1 = x1 ;

a2 = x1 ;

a3 = 0

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Continuum Mechanics for Engineers

Problem 4.13 The Lagrangian description of a deformation is given by x1 = X1 + X3 (e2 − e−2 ) , x2 = X2 − X3 (e2 − 1) , x3 = X3 e2 . Determine the components of the deformation matrix FiA and from it show that the Jacobian J does not vanish. Invert the mapping equations to obtain the Eulerian description of the deformation. Answer J = e2 X1 = x1 − x3 (1 − e−4 ) X2 = x2 + x3 (1 − e−2 ) X3 = x3 e−2 Solution The deformation gradient is 1 0 ∂xi [FiA ] = = 0 1 ∂XA 0 0

The Jacobian is

1 0 J = det F = 0 1 0 0

e2 − e−2 e2 − 1 e2

e2 − e−2 e2 − 1 e2

= e2

and is non-vanishing. From the third equation of the motion, we have X3 = x3 e−2 Then we have x1 = X1 + x3 e−2 (e2 − e−2 );

x2 = X2 − x3 e−2 (e2 − 1)

and X1 = x1 − x3 (1 − e−4 );

X2 = x2 + x3 (1 − e−2 );

X3 = x3 e−2

Problem 4.14 A homogeneous deformation has been described as one for which all of the deformation and strain tensors are independent of the coordinates, and may therefore be expressed in general by the displacement field ui = Aij Xj where the Aij are constants (or in the case of a motion, functions of time). Show that for a homogeneous deformation with the Aij constant: (a) plane material surfaces remain plane, (b) straight line particle elements remain straight, (c) material surfaces which are spherical in the reference configuration become ellipsoidal surfaces in the deformed configuration.

85

Chapter 4 Solutions Solution (a) The displacement field is ui = xi − Xi = Aij Xj The spatial points are given by xi = Xi + Aij Xj = (δij + Aij ) Xj Let a material plane be given by βi Xi + α = 0 and let Xi = (δij + Bij ) xj be the inverse of the spatial points. Then βi (δij + Bij ) xj + α = 0 is the spatial description of the deformed plane or λj xj + α = 0

where λj = βj + βi Bij is a constant. Thus the deformed surface is a plane. (b) Since planes deform into planes, the intersection of two planes is a straight line, and straight lines deform into straight lines. (c) Spherical material surfaces are given by Xi Xi = R2 Then in the spatial or deformed configuration this becomes (δij + Bij ) xj (δik + Bik ) xk = R2 and (δjk + Bkj + Bjk + Bij Bik ) xj xk = R2 This gives βjk xj xk = R2 an ellipsoid. Problem 4.15 An infinitesimal homogeneous deformation ui = Aij Xj is one for which the constants Aij are so small that their products may be neglected. Show that for two sequential infinitesimal deformations the total displacement is the sum of the individual displacements regardless of the order in which the deformations are applied. Solution Let the two displacements be (1)

ui

(1)

(2)

= Aij Xj

(2)

(2) = δij + Aij Xj

= Aij Xj ;

ui

Then the spatial positions are (1) (1) xi = δij + Aij Xj ;

xi

(2)

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Continuum Mechanics for Engineers

Successive motions give (2)

xi

(2) (1) = δij + Aij δjk + Ajk Xk (1) (2) (2) (1) = δik + Aik + Aik + Aij Ajk Xk

Neglecting the product terms, this gives (2) (1) (2) xi = δik + Aik + Aik Xk and

(2)

xi

(1) (2) (1) (2) − Xi = Aik + Aik Xk = ui + ui

Problem 4.16 For the homogeneous deformation defined by x1 = αX1 + βX2 , x2 = −βX1 + αX2 , x3 = µX3 , where α, β and µ are constants, calculate the Lagrangian finite strain tensor E. Show that if α = cos θ, β = sin θ and µ = 1 the strain is zero and the mapping corresponds to a rigid body rotation of magnitude θ about the X3 axis. Answer α2 + β2 − 1 1 0 =2 0

EAB

0 0 α2 + β2 − 1 0 0 µ2 − 1

Solution The deformation gradient is

α [FiA ] = −β 0

β α 0

0 0 µ

The Green’s deformation tensor is C = FT · F and 2 α + β2 α −β 0 α β 0 0 [CAB ] = β α 0 −β α 0 = 0 0 µ 0 0 µ 0

0 α2 + β2 0

0 0 µ2

The Lagrangian finite strain tensor is E = 12 (C − I) or 2 0 0 α + β2 − 1 1 0 α2 + β2 − 1 0 [EAB ] = 2 0 0 µ2 − 1 For α = cos θ, β = sin θ, and µ = 1, the strain is cos2 θ + sin2 θ − 1 0 0 0 1 = 0 [EAB ] = 0 cos2 θ + sin2 θ − 1 0 2 0 0 0 12 − 1

0 0 0

0 0 0

87

Chapter 4 Solutions The mapping is x1 = cos θX1 + sin θX2 x2 = − sin θX1 + cos θX2 x3 = X3 This is a rotation in the plane. Problem 4.17 Given the deformation defined by x1 = X1 ,

1 x2 = X2 + X23 , 2

x3 = X3

(a) Sketch the deformed shape of the unit square OABC in the plane X1 = 0. (b) Determine the differential vectors dx(2) and dx(3) which are the deformed vectors I2 and dX(3) = dX(3)^ I3 , respectively, that were resulting from dX(2) = dX(2)^ originally at corner C. (c) Calculate the dot product dx(2) · dx(3) , and from it determine the change in the original right angle between dX(2) and dX(3) at C due to the deformation. (d) Compute the stretch Λ at B in the direction of the unit normal √ ^ = ^ N I2 + ^ I3 / 2 .

X3 C

B

O A

X2

X1

Unit square OABC in the reference configuration. Answer ^2 , dx(3) = dX(3) (^ ^3 ) (b) dx(2) = dX(2) e e2 + e ◦ (c) ∆θ = −45 √ (d) Λ(N 2.5 ^) = Solution (a) The points move to Point O (0, 0, 0) → O (0, 0, 0) Point B (0, 1, 1) → b (0, 01.5, 1)

Point A (0, 1, 0) → a (0, 1, 0) Point C (0, 0, 1) → c (0, 0.5, 1)

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Continuum Mechanics for Engineers

(b) The differential vectors dx(2) and dx(3) result from dX(2) = dX(2)^ I2 and dX(3) = (3)^ dX I3 . This is determined from the deformation gradient through dx = F · dX. The deformation gradient is 1 0 0 [FiA ] = 0 1 1 0 0 1 Now

1 0 0

0 1 0

0 0 0 1 dX(2) = dX(2) 1 0 0

and ^2 dx(2) = dX(2) e Also

1 0 0

0 1 0

0 0 0 1 0 = dX(3) 1 dX(3) dX(3)

and ^3 ) dx(3) = dX(3) (^ e2 + e (c) The dot product is (2)

dx

(3)

· dx

q 2 (2) (3) (2) 2 dX(3) cos θ = dx dx cos θ = dX √ = 2dX(2) dX(3) cos θ

and cos θ =

dX(2) · dX(3) dX(2) dX(3) 1 √ = =√ (2) (3) dx(2) · dx(3) 2dX dX 2

This gives

θ = 45◦

(d) The stretch is found from C = FT · F and 1 1 0 0 [CAB ] = 0 1 0 0 0 0 1 1

0 1 0

1 0 1 = 0 0 1

Now Λ2(0,1,1) =

h

0

√1 2

√1 2

i

1 0 0

0 1 1

0 1 1

0 1 2

0 0 1 1 √2 = 2.5 √1 2 2

Problem 4.18 Given the deformation expressed by x1 = X1 + AX22 ,

x2 = X2 ,

x3 = X3 − AX22

where A is a constant (not necessarily small), determine the finite strain tensors E and e, and show that if the displacements are small so that x ≈ X, and if squares of A may be neglected, both tensors reduce to the infinitesimal strain tensor . Answer

Chapter 4 Solutions 0 [ij ] = Ax2 0

89

Ax2 0 −Ax2

0 −Ax2 0

Solution The finite strain tensors E is E =

(C − I) = 21 FT · F − I . The deformation gradient is 1 2AX2 0 1 0 [FiA ] = 0 0 −2AX2 1 1 2

and

1 [CAB ] = 2AX2 0 1 = 2AX2 0

0 1 0

0 1 −2AX2 0 1 0

2AX2 1 + 8A2 X22 −2AX2

2AX2 1 −2AX2

0 0 1

0 −2AX2 1

The finite strain tensor E is 0 1 2AX2 [EAB ] = 2 0

0 −2AX2 0

2AX2 8A2 X22 −2AX2

Neglecting squares of A we have AX2 0 0 −AX2 −AX2 0 T The finite strain tensors e is e = 12 (I − c) = 12 I − F−1 · F−1 . Inverting the deformation X1 = x1 − Ax22 , X2 = x2 , X3 = x3 + Ax22

0 [ij ] = AX2 0

Then, the inverse deformation gradient is 1 −1 FiA = 0 0

−2Ax2 1 2Ax2

0 0 1

and

1 [cAB ] = −2Ax2 0 1 = −2Ax2 0

0 0 1 1 2Ax2 0 0 1 0 −2Ax2 1 + 8A2 x22 2Ax2

−2Ax2 1 2Ax2

0 2Ax2 1

The finite strain tensor e is

0 1 [eAB ] = −2Ax2 2 0

−2Ax2 8A2 x22 2Ax2

0 2Ax2 0

0 0 1

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Continuum Mechanics for Engineers

Neglecting squares of A we have

0 [ij ] = −Ax2 0

−Ax2 0 Ax2

0 2Ax2 0

Now if x ≈ X the infinitesimal strain measures are the same. Problem 4.19 For the infinitesimal homogeneous deformation xi = Xi + Aij Xj where the constants Aij are very small, determine the small strain tensor it the longitudinal (normal) , and from √ ^ ^ ^ strain in the direction of the unit vector N = I1 − I3 / 2. Answer 2e(N ^ ) = A11 − A13 − A31 + A33 Solution The infinitesimal strain tensor is ij = 21 (ui,j + uj,i ). The displacement is ui = xi − Xi = Aij Xj , and since xi ≈ Xi , we have 2A11 A12 + A21 A13 + A31 1 A21 + A12 2A22 A23 + A32 [ij ] = 2 A +A A +A 2A 31

13

32

23

33

Then e(N ^ ) in the direct given is 2e(N ^) =

h

√1 2

0

− √12

i

2A11 A21 + A12 A31 + A13

A12 + A21 2A22 A32 + A23

√1 A13 + A31 2 A23 + A32 0 2A33 − √12

= A11 − A13 − A31 + A33 Problem 4.20 A deformation is defined by x1 = X1 / X21 + X22 ,

x2 = X2 / X21 + X22 ,

x3 = X3 .

Determine the deformation tensor C together with its principal values. Answer C(1) = C(2) = X21 + X22

−2

,

C(3) = 1

Solution The deformation gradient is −X21 + X22 ∂xi 1 [FiA ] = = 2 −2X1 X2 ∂XA X21 + X22 0

−2X1 X2 X21 − X22 0

0 0 2 X21 + X22

91

Chapter 4 Solutions The deformation tensor C = FT · F or CAB = FAi FiB −X21 + X22 −2X1 X2 0 1 X21 − X22 0 [CAB ] = 4 −2X1 X2 2 2 2 X21 + X22 0 0 X1 + X2 2 X21 + X22 0 0 1 2 2 2 = 0 X + X 0 4 1 2 4 X21 + X22 2 0 0 X1 + X22

−X21 + X22 −2X1 X2 0

−2X1 X2 X21 − X22 0

0 0 2 2 2 X1 + X2

Since the deformation tensor is diagonal, the principal values are the diagonal values. C(1) = C(2) = X21 + X22

−2

,

C(3) = 1

Problem 4.21 For the deformation field given by x1 = X1 + αX2 ,

x2 = X2 − αX1 ,

x3 = X3

where α is a constant, determine the matrix form of the tensors E and e, and show that the circle of particles X21 + X22 = 1 deforms into the circle x21 + x22 = 1 + α2 . Answer α2 1 0 2 0

0 0 α2 0 0 0 −α2 1 0 2 2 (1 + α2 ) 0

[Eij ]

[eij ]

=

=

0 −α2 0

0 0 0

Solution The finite strain tensor E requires the deformation gradient F since E = 1 T 2 F ·F−I 1 α 0 [FiA ] = −a 1 0 0 0 1 and 1 −α 1 a 1 [EAB ] = 2 0 0 2 α 0 1 = 0 α2 2 0 0

0 1 0 −a 1 0

α 0 1 1 0 − 0 0 1 0

0 0 1

To compute e, we need to invert the deformation. The Jacobian is 1 α 0 J = det F = −a 1 0 = 1 + α2 0 0 1

0 1 0

0 0 1

1 2

(C − I) =

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Continuum Mechanics for Engineers

This is non-zero, and X1 =

x1 − αx2 ; 1 + α2

X2 =

αx1 + x2 ; 1 + α2

X3 = x3

This gives 1 −α 0 1 a 1 0 1 + α2 0 0 1 + α2 T and e = 12 (I − c) = 12 I − F−1 · F−1 1 α 0 1 0 0 1 1 −a 1 0 [eAB ] = 0 1 0 − 2 (1 + α2 )2 0 0 1 0 0 1 + α2 2 α 0 0 1 0 α2 0 = 2 (1 + α2 ) 0 0 0

F−1 iA =

1 −α 0 a 1 0 0 0 1 + α2

The circle X21 + X22 = 1 gives 2 2 αx1 + x2 x1 − αx2 + =1 1 + α2 1 + α2 and

(x1 − αx2 )2 + (αx1 + x2 )2 = 1 + α2

2

Simplifying, we have x21 + x22 = 1 + α2 This is a circle. Problem 4.22 Let the deformation of a continuum be given by the equations x1 = X1 + kX22 ,

x2 = X2 − kX21 ,

x3 = X3

where k is a constant. Determine the Lagrangian finite strain tensor E, and from it, assuming k is very small, deduce the infinitesimal strain tensor . Verify this by calculating the displacement field and using the definition 2ij = ui,j + uj,i for the infinitesimal theory. Solution The finite strain tensor is E =

FT · F − I . The deformation gradient F is 1 2kX2 0 1 0 [FiA ] = −2kX1 0 0 1 1 2

(C − I) =

1 2

and 1 −2kX1 0 1 2kX2 1 2kX2 1 0 −2kX1 1 [EAB ] = 2 0 0 1 0 0 2 2 4k X1 2kX2 − 2kX1 0 1 4k2 X22 0 = 2kX2 − 2kX1 2 0 0 0

0 1 0 − 0 1 0

0 1 0

0 0 1

93

Chapter 4 Solutions Neglecting squares of k, gives the linear strain tensor 0 kX2 − kX1 0 [ij ] = kX2 − kX1 0 0

0 0 0

The displacement field is u1 = x1 − X1 = kX22 ,

u2 = x2 = X2 = −kX21 ,

u3 = x3 − X3 = 0

The infinitesimal strain tensor 0 1 [ij ] = [ui,j + uj,i ] = kX2 − kX1 2 0

kX2 − kX1 0 0

0 0 0

The results are identical. Problem 4.23 Given the displacement field u1 = AX2 X3 ,

u2 = AX23 ,

u3 = AX21

where A is a very small constant, determine (a) the components of the infinitesimal strain tensor , and the infinitesimal rotation tensor ω. (b) the principal values of , at the point (1, 1, 0). Answer (a) 11 = 22 = 33 = 0, 12 = AX3 , 13 = 12 A(X2 + 2X1 ) and 23 = AX3 ω11 = ω22 = ω33 = 0, ω12 = −ω21 = 21 AX3 , ω13 = −ω31 = 12 AX1 − AX1 and ω23 = −ω32 = AX3 (b) (I) = 32 A, (II) = 0, (III) = − 23 A Solution (a) The infinitesimal strain tensor is ij = 12 (ui,j + uj,i ) and 0 AX3 AX2 + 2AX1 1 AX3 0 2AX3 [ij ] = 2 AX + 2AX 2AX 0 2

1

3

The infinitesimal rotation is ωij = 12 (ui,j − uj,i ) and 0 AX3 1 −AX3 0 [ωij ] = 2 AX − 2AX −2AX 2

1

(b) At (1, 1, 0), the infinitesimal strain tensor is 0 1 [ij ](1,1,0) = 0 2 3A

3

0 0 0

AX2 − 2AX1 2AX3 0

3A 0 0

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Continuum Mechanics for Engineers

The principal values are given by − det ( − I) = 0 3A 2

0 − 0

" 2 # 3 2 0 = − − A =0 2 − 3 2A

and (I) =

3 A; 2

3 (III) = − A 2

(II) = 0;

Problem 4.24 Suppose that a triangular element in a Finite Element Modeling (FEM) code deforms as shown. 1 3

2 INITIAL

2

DEFORMED

1

3

Using a ruler and protractor to measure lengths and angles in these images, estimate the components of the 2x2 deformation gradient matrix. Problem 4.25 A 45◦ strain rosette measures longitudinal strains along the X1 , X2 , and X01 axes shown below. At point O the strains recorded are 11 = 6 × 10−4 ,

22 = 4 × 10−4

and

011 = 8 × 10−4

Determine the shear strain γ12 at O, together with ε022 , and verify that ε11 +ε22 = ε011 +ε022 . See Eq 4.88. X2 45

o

X1‛

45

o

X2‛

O

X1

95

Chapter 4 Solutions Solution The transformation equation is 011 =

γ12 11 + 22 11 − 22 + cos 2θ + sin 2θ 2 2 2

This gives 8=

6+4 6−2 γ12 + cos 90◦ + sin 90◦ 2 2 2

and solving γ12 = 6 × 10−4 The second principal strain is 022 =

11 + 22 11 − 22 γ12 + cos 2θ + sin 2θ 2 2 2

and 022 =

6+4 6−4 3 + cos 270◦ + sin 270◦ 2 2 2

Solving gives We have

022 = 2 11 + 22 = 6 + 4 = 011 + 022 = 8 + 2

Problem 4.26 By a direct expansion of Eq 4.93, 2ωi = εijk ωkj , show that ω1 = ω32 = −ω23 , etc. Also, show that only if A is a very small constant does the mapping x1 = X1 − AX2 + AX3 , x2 = X2 − AX3 + AX1 , x3 = X3 − AX1 + AX2 , represent a rigid body rotation (E ≡ 0). Additionally, determine the infinitesimal rotation tensor ωij in this case; from it, using the result proven above, deduce the rotation vector ωi . Answer ^1 + e ^2 + e ^3 ) ω = A(e Solution Expanding 2ωi = ijk ωkj , gives 2ω1 = 12k ωk2 + 13k ωk3 = 123 ω32 + 132 ω23 = ω32 − ω23 = 2ω32 = −2ω23 2ω2 = 23k ωk3 + 21k ωk1 = 231 ω13 + 213 ω31 = ω13 − ω31 = 2ω13 = −2ω31 2ω3 = 31k ωk1 + 32k ωk2 = 312 ω21 + 321 ω12 = ω21 − ω12 = 2ω21 = −2ω12 To compute the finite strain tensor E = 12 FT · F − I . the deformation gradient is 1 −A A 1 −A [FiA ] = A −A A 1

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Continuum Mechanics for Engineers

and

1 A −A 1 1 1 A A [EAB ] = −A 2 A −A 1 −A 2 2 2 2A −A −A 1 = −A2 2A2 −A2 2 −A2 −A2 2A2

A 1 −A − 0 1 0

−A 1 A

0 0 1

0 1 0

This is only zero when A is small. The displacement field is u1 = x1 − X1 = −AX2 + AX3 u2 = x2 − X2 = −AX3 + AX1 u3 = x3 − X3 = −AX1 + AX2 h i The infinitesimal rotation is ω = 21 (grad u) − (grad u)T and

0 [grad u] = [ui,j ] = A −A The infinitesimal rotation is 0 −A 1 0 [ωij ] = A 2 −A A

0 A −A − −A A 0

−A 0 A

A 0 −A

A −A 0

−A 0 A = A 0 −A

−A 0 A

A −A 0

This gives ω1 = ω23 = A;

ω2 = ω13 = A;

ω3 = ω21 = A

The rotation vector is ^2 + e ^3 ) ω = A (^ e1 + e Problem 4.27 For the displacement field u1 = kX1 X2 ,

u2 = kX1 X2 ,

u3 = 2k(X1 + X2 )X3

where k is a very small constant, determine the rotation tensor ω, and show that it has only one real principal value at the point (0, 0, 1). Answer √ √ ω(1) = 0, ω(2) = −ω(3) = ik 2, where i = −1 Solution The rotation tensor ω requires the displacement gradient grad u. This is kX2 kX1 0 kX1 0 [ui,j ] = kX2 2kX3 2kX3 2k (X1 + X2 )

97

Chapter 4 Solutions and kX2 kX1 0 kX2 1 kX kX 0 [ωij ] = − kX1 2 1 2 2kX3 2kX3 2k (X1 + X2 ) 0 k (X1 − X2 ) −kX3 0 2 k (X1 − X2 ) = − 0 −kX3 2 kX3 kX3 0

kX2 kX1 0

2kX3 2kX3 2k (X1 + X2 )

The rotation tensor at (0, 0, 1) is

[ωij ](0,0,1)

0 = 0 k

0 0 k

−k −k 0

The principal value is found from −λ 0 −k det (ω − λI) = 0 −λ −k = −λ λ2 + k2 − k2 λ = λ λ2 + 2k2 = 0 k k −λ and λI = 0;

√ λII , λIII = ±ik 2

√ ω(II) = −ω(III) = ik 2

or ω(I) = 0;

Problem 4.28 Let the deformation x1 = X1 + AX2 X3 , x2 = X2 + AX23 , x3 = X3 + AX21 , where A is a constant be applied to a continuum body. For the unit square of material line elements OBCD as shown, calculate at point C (a) the stretch and unit elongation for the element in the direction of diagonal OC, (b) the change in the right angle at C if A = 1; if A = 0.1.

D

X3

C

1

O 1

B

X2

X1

Unit square OBCD in the reference configuration.

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Continuum Mechanics for Engineers

Answer (a) Λ2(OC) = 1 + 2A + 4A2 , e(OC) =

√ 1 + 2A + 4A2 − 1

(b) ∆θ(A=1) = 60◦ , ∆θ(A=0.1) = 11.77◦ Solution (a) The deformation gradient is [FiA ] =

∂xi ∂XA

1 = 0 2AX1

AX3 1 0

AX2 2AX3 1

The Green’s deformation tensor is 1 0 2AX1 1 AX3 AX2 1 0 0 1 2AX3 [CAB ] = [FAi ] [FiB ] = AX3 AX2 2AX3 1 2AX1 0 1 2 2 1 + 4A X1 AX3 AX2 + 2AX1 AX3 1 + A2 X23 A2 X2 X3 + 2AX3 = 2 AX2 + 2AX1 A X2 X3 + 2AX3 1 + A2 X22 + 4A2 X23 At the point C, we have (0, 1, 1) and CAB we have 1 A [CAB ](0,1,1) = A 1 + A2 A A2 + 2A ^ · C · N. ^ Along the diagonal N ^ = Now Λ2 = N Λ2(0,1,1) =

h

0

√1 2

√1 2

1

i A A

√1 2

A A2 + 2A 2 2 1 + A + 4A ^ I2 + ^ I3 and

A

A

1 + A2

A2 + 2A

A2 + 2A

1 + A2 + 4A2

This gives e( N ^) = Λ − 1 =

p

1 + A2 + 4A2 − 1

^ 1 and N ^ 2 is (b) The angle between N cos θ = p

^1 · C · N ^ N p 2 ^1 · C · N ^1 N ^2 · C · N ^2 N

For A = 1, we have

[CAB ](0,1,1)

1 = 1 1

and

h i ^1 · C · N ^2 = 0 N

1

0

1 1 1

1 3 6

1 2 3 1 2 3

1 0 3 0 = 3 6 1

0 √1 2 √1 2

99

Chapter 4 Solutions

h i ^1 · C · N ^1 = 0 N

1

h i ^2 · C · N ^2 = 0 N

0

So that

1 2 3 1 2 3

1 3 6 1 3 6

0 1 =2 0 0 0 =6 1

√ 3 3 cos θ = √ √ = 2 2 6

and

θ = 30◦

For A = 0.1, we have

and ∆θA=1 = 60◦

[CAB ](0,1,1) and

1 0 1 1 1 1 1 1

^1 · C · N ^ 2 = 0.21; N

1 = 0.1 0.1

0.1 1.01 0.21

h i ^1 · C · N ^ 1 = 1.01; N

0.1 0.21 1.05 h i ^2 · C · N ^ 2 = 1.05 N

This gives 0.21 0.21 √ = cos θ = √ 1.03 1.01 1.05 and

θ = 78.24◦

and

∆θA=0.1 = 11.76◦

Problem 4.29 For the homogeneous deformation expressed by the equations √ √ 3 2 x1 = 2X1 + X2 , 4 √ 3 2 x2 = −X1 + X2 + X3 , 4 √4 3 2 x3 = X1 − X2 + X3 , 4 4 determine ^ for the line element originally in the direction of (a) the unit normal n √ ^ = ^I1 − ^I2 + ^I3 / 3. N (b) the stretch Λ(N ^ ) of this element. (c) the maximum and minimum stretches at the point X1 = 1, X2 = 0, X3 = −2 in the reference configuration. Answer (a)

√ √ √ ^1 + ^2 + ^3 2e 2−7 e 2+7 e ^= √ n 104

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Continuum Mechanics for Engineers (b)

Λ( N ^ ) = 1.472

(c)

Λ(max) = 2; Λ(min) = 0.5

Solution (a) The deformation gradient is √ 2 [FiA ] = −1 1

^ = B. This gives Now F · N √ 2 [Bij ] = −1 1

√ 3 2 4 3 4 3 − 4

0 √ 2 √4 2 4

√ 1 2 √ √ 0 3 4 3√ √ 1 −7 + 2 2 −√ = √ 3 4 √ 3 √4 2 1 7 + 2 √ √ 4 3 4 3

√ 3 2 4 3 4 3 − 4

^ = B/B and n 104 B · B = |B| = 48 2

√ 104 and |B| = √ 4 3

Then

√ √ √ ^2 + 7 + 2 e ^3 2^ e1 + −7 + 2 e ^= √ n 104 2 ^ ^ (b) Square of the stretch is Λ = N · C · N and √ √ √ 3 2 2 −1 1 2 4 √ 3 2 3 3 3 [CAB ] = [FAi ] [FiB ] = − −1 4 4 √4 √4 2 2 3 0 1 − 4 4 4 4 = 0 0

0 9 4 0

4

Λ2 =

1 √ 3

0 √ 2 √4 2 4

0 0 1

Then

1 −√ 3

4 1 0 √ 3 0

0 9 4 0

1 √ 0 3 13 1 = √ − 0 6 3 1 1 √ 4 3

101

Chapter 4 Solutions and (c) Since C in (b) is C∗ then Λ(max)

Λ = 1.47 p √ = 4 = 2 and Λ(min) = 1/4 = 12 .

Problem 4.30 Let the deformation x1 = a1 (X1 + 2X2 ),

x2 = a2 X2 ,

x3 = a3 X3

where a1 , a2 , and a3 are constants be applied to the unit cube of material shown in the sketch. Determine (a) the deformed length l of diagonal OC, (b) the angle between edges OA and OG after deformation, (c) the conditions which the constants must satisfy for the deformation to be possible if (1) the material is incompressible, (2) the angle between elements OC and OB is to remain unchanged.

X3 , x3 E F D C

O

G

A

X2 , x2 B

X1 , x1

Unit cube having diagonal OC. Answer (a) l2 = 9a21 + a22 + a23 2a1 (b) cos θ = q 4a21 + a22 (c) (1) a1 a2 a3 = 1, (2) 9a21 + a22 = 2a23

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Solution (a) The deformation gradient is

a1 [FiA ] = 0 0

0 0 a3

2a1 a2 0

^ =^ ^ ^ =F·N The diagonal is N I1 + ^ I2 + ^ I3 . The deformed diagonal is n a1 2a1 0 1 3a1 a2 0 1 = a2 [^ n] = 0 0 0 a3 1 a3 The length is l2 = (3a1 )2 + a22 + a23 = 9a21 + a22 + a23 (b) The angle between OA and OG ^ OA · C · N ^ N p OG cos θ = p ^ OA · C · N ^ OA N ^ OG · C · N ^ OG N where C = FT · F a1 a1 0 0 [CAB ] = [FAi ] [FiB ] = 2a1 a2 0 0 0 0 0 a3 2 2 a1 2a1 0 0 = 2a21 4a21 + a22 0 0 a23

2a1 a2 0

0 0 a3

Now h i ^ OA · C · N ^ OG = 1 N

0

a21 2a21 0

2a21 2 4a1 + a22 0

a21 2a21 0

2a21 2 4a1 + a22 0

0

0

h i ^ OA · C · N ^ OA = 1 N

0

a21 2a21 0

h i ^ OG · C · N ^ OG = 0 N

1

0

a1

q

2a21 + a22 0

4a21

0 0 0 1 = 2a21 0 a23 0 1 0 0 = a21 0 a23 0 0 0 1 = 4a21 + a22 0 a23

This gives 2a21

cos θ =

2a1 =q 4a21 + a22 4a21 + a22

(c) (1) For the material to be incompressible J = 1. This gives a1 2a1 0 a2 0 = a1 a2 a3 = 1 J = det [FiA ] = 0 0 0 a3

103

Chapter 4 Solutions (2) The angle between OB and OC is cos β =

^ I +^ I +^ I3 2 I1 + ^ I ^ √ 2 · 1 √2 =√ 2 3 6

The angle between OB and OC after deformation is ^ OB · C · N ^ N p OC cos θ = p ^ OB · C · N ^ OB N ^ OC · C · N ^ OC N where

^ +^ I2 ^ OB = I1√ N 2

Now

^ ^ I2 + ^ I3 ^ OC = I1 + √ and N 3

2 2 ^ OB · C · N ^ OC = 9a1√+ a2 N 6 2 2 ^ OB · C · N ^ OB = 9a1 + a2 N 2

2 2 2 ^ OC · C · N ^ OC = 9a1 + a2 + a3 N 3 Equating the cosine of the two angles gives

9a21 + a22 √ 2 r6 = cos β = √ cos θ = r 2 2 2 2 2 6 9a1 + a2 9a1 + a2 + a3 2 3 and simplifying 9a21 + a22 = 2a23 Problem 4.31 A homogeneous deformation is defined by x1 = αX1 + βX2 ,

x2 = −αX1 + βX2 ,

x3 = µX3

where α, β and µ are constants. Determine (a) the magnitudes and directions of the principal stretches, (b) the matrix representation of the rotation tensor R, (c) the direction of the axis of the rotation vector, and the magnitude of the angle of rotation. Answer (a) Λ2(1) = Λ2(e^1 ) = 2a2 , Λ2(2) = Λ2(e^2 ) = 2β2 , Λ2(3) = Λ2(e^3 ) = µ2 √1 √1 0 2 2 √1 (b) [Rij ] = − √1 0 2 2 0 0 1

104

Continuum Mechanics for Engineers ^ = ^I3 ; Φ = 45◦ (c) n

Solution (a) The deformation gradient is α β ∂ui [FiA ] = = −α β ∂XA 0 0

0 0 µ

The Green’s deformation tensor is C = FT · F or α −α 0 α β [CAB ] = [FAi ] [FiB ] = β β 0 −α β 0 0 µ 0 0

2α2 0 0 = 0 µ 0

0 2β2 0

0 0 µ2

Since this is diagonal, we have 2 Λ2(1) = Λ2(^ e1 ) = 2a ;

2 Λ2(2) = Λ2(^ e2 ) = 2β ;

(b) The right stretch tensor is U = C1/2 so that √ 2α √0 [UAB ] = 0 2β 0 0 and R = F · U−1 . The inverse of the stretch tensor is √1 0 2α −1 √1 UAB = 0 2β 0 0 and

α

β

[RAB ] = −α 0

0

0 µ

β 0

√1 2α

0

0

√1 2β

0

0

0

2 Λ2(3) = Λ2(^ e3 ) = µ

0 0 µ

0

0

1 µ

0 =

√1 2 √ − 12

√1 2 √1 2

0

0

1 µ

(c) The angle of rotation is 1 1 + 2 cos Φ = Rii = 1 + 2 √ 2 and

1 cos Φ = √ 2

The direction is found from

√1 2 − √12

√1 2 √1 2

0

0

0

or Φ = 45◦

n1

n1

0 n2 = n2 1 n3 n3

0

0 1

105

Chapter 4 Solutions and √ 2n1 √ −n1 + n2 = 2n2 n1 + n2 =

n3 = n3 The first two equations give n1 = n2 = 0 and since n21 + n22 + n23 = 1 we have n3 = ±1 The direction of the rotation is

^=e ^3 = ^ n I3

Problem 4.32 Consider the deformation field x1 = X1 − AX2 + AX3 , x2 = X2 − AX3 + AX1 , x3 = X3 − AX1 + AX2 , where A is a constant. Show that the principal values of the right stretch tensor have a √ ^ ^ ^ ^ multiplicity of two, and that the axis of the rotation tensor is along N = I1 + I2 + I3 / 3. Determine the matrix of the rotation vector together with the angle of rotation φ. Answer √ Λ(1) = 1, Λ(2) = Λ(3) = 1 + 3A2 = β β+2 β − 1 − 3A 1 β − 1 + 3A β+2 [Rij ] = 3β β − 1 − 3A β − 1 + 3A

β − 1 + 3A β − 1 − 3A β+2

φ = cos−1 (1/β) Solution The deformation gradient is [FiA ] =

∂xi ∂XA

1 = A −A

−A 1 A

A −A 1

The right stretch tensor is

1 A −A 1 1 A A [CAB ] = [FAi ] [FiB ] = −A A −A 1 −A 2 2 2 1 + 2A −A −A 1 + 2A2 −A2 = −A2 −A2 −A2 1 + 2A2

−A 1 A

A −A 1

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Continuum Mechanics for Engineers

The principal values are found from 1 + 2A2 − λ −A2 −A2 2 2 −A 1 + 2A − λ −A2 det (C − λI) = 2 2 −A −A 1 + 2A2 − λ = 1 + 3A2 − λ (1 − λ) 1 + 3A2 − λ = 0

=0

The roots are Λ(2) = Λ(3) = 1 + 3A2

Λ(1) = 1; The principal stretch tensor is

[U∗AB ] = [C∗AB ]1/2 and

[U∗AB ]−1

1 = 0 0

1 = 0 0

√ 0 1 + 3A2 0

0 −1/2 1 + 3A2 0

0 1 = 0 0 √ 0 1 + 3A2

0 1 0 = 0 −1/2 0 1 + 3A2

0 β 0

0 1/β 0

0 0 β 0 0 1/β

The rotation tensor is R = F · U−1 . The inverse stretch tensor can be found from the eigenvectors associated with the principal stretches and inverse principal stretch tensor. For Λ(1) = 1. We have (1) 0 n1 2A2 −A2 −A2 −A2 2A2 −A2 n2(1) = 0 (1) 0 n3 −A2 −A2 2A2 and (1)

(1)

(1)

(1)

(1)

2n1 − n2 − n3 = 0 (1)

−n1 + 2n2 − n3 = 0 (1)

(1)

(1)

−n1 − n2 + 2n3 = 0 This gives 1 (1) (1) (1) n1 = n2 = n3 = √ 3 1 ^ =√ ^ This is the axis of rotation N I1 + ^ I2 + ^ I3 since the other two roots are equal. 3 For Λ(2) = Λ(3) = 1 + 3A2 , we have to select two vectors that are perpendicular to the above vector. One choice is 1 ^2 ) n(2) = √ (^ e1 − e 2 and 1 1 (3) (1) (2) ^2 + e ^3 ) × √ (^ ^2 ) n = n × n = √ (^ e1 + e e1 − e 3 2 1 ^2 − 2^ = √ (^ e1 + e e3 ) 6

107

Chapter 4 Solutions The transformation matrix is [aij ] =

√1 3 √1 2 √1 6

√1 3 − √12 √1 6

√1 3

0 − √26

The inverse stretch tensor is U−1 = AT · (U∗ )−1 · A or √1 √1 √1 1 0 0 3 2 6 −1 1 1 1 √ UAB = √ 0 1/β 0 − √2 6 3 √1 0 0 1/β 0 − √26 3 =

√1 3 √1 3 √1 3

√1 2 − √12

0

√1 6 √1 6 − √26

√1 3 − √12β √1 6β

√1 3 √1 2β √1 6β

√1 3 √1 2 √1 6

√1 3

0 − √26β

√1 3 − √12 √1 6

√1 3

0 − √26

β+2 β−1 β−1 1 = β−1 β+2 β−1 3β β−1 β−1 β+2 The rotation tensor is 1 −A A A 1 −A [RiB ] = [FiA ] U−1 = AB −A A 1 β+2 β − 1 − 3A 1 β − 1 + 3A β+2 = 3β β − 1 − 3A β − 1 + 3A

β+2 β−1 β−1 1 β−1 β+2 β−1 3β β − 1 β − 1 β + 2 β − 1 + 3A β − 1 − 3A β+2

The angle of rotation is 1 + 2 cos Φ = Rii =

2 3 (β + 2) =1+ 3β β

or cos Φ =

1 β

Problem 4.33 For the deformation field x1 =

√ 3X1 + X2 ,

x2 = 2X2 , x3 = X3 . determine (a) the matrix representation of the rotation tensor R,

108

Continuum Mechanics for Engineers (b) the right stretch tensor U and the left stretch tensor V, then show that the principal values of U and V are equal, (c) the direction of the axis of rotation and the magnitude of the angle of rotation.

Answer √ √ 3+1 3−1 √ 1 √ (a) [Rij ] = √ − 3 + 1 3+1 2 2 0 0 √ √ (b) Λ(1) = 6, Λ(2) = 2, Λ(3) = 1 ^ = ^I3 ; φ = 15◦ (c) N

0 0 √ 2 2

Solution (a) The deformation gradient is √ 3 ∂xi [FiA ] = = 0 ∂XA 0

1 2 0

0 0 1

√ 0 3 0 0 1 0

1 2 0

0 √3 0 = 3 1 0

The right stretch tensor is √ 3 [CAB ] = [FAi ] [FiB ] = 1 0

0 2 0

The principal stretches are 3−λ √ det (C − λI) = 3 0

√ 3 5−λ 0

0 0 1−λ

=0

= (1 − λ) (λ − 6) (λ − 2) = 0 The principal stretch tensors are

6 [C∗AB ] = 0 0 and

√ 6 [U∗AB ] = 0 0

0 0 1

0 2 0

√0 2 0

0 0 1

The inverse principal stretch is √ 1/ 6 = 0 0

[U∗AB ]−1

0√ 1/ 2 0

0 0 1

√ 3 5 0

0 0 1

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Chapter 4 Solutions

The inverse stretch is determined from U−1 = AT · (U∗ )−1 · A where the A is found from the eigenvectors for the principal stretches. √ 1 3 0 2 2√ 1 [aij ] = − 23 0 2 0 0 1 This gives

1 2 √ 3 2

−

√ 3 2

0

√ 1/ 6

0 √ 1/ 2

0 0 0 0 1 0 0 √ √ 3− 0 1+3 3 √3 1 √ = √ 3−3 3+ 3 0 √ 4 6 0 0 4 6

U−1 AB =

1 2

Since F = R · U, we have R = F · U−1 , we find √ √ 1√+ 3 3 3 1 0 1 [RiA ] = 0 2 0 √ 3−3 4 6 0 0 1 0 √ √ 1 + √3 −1 +√ 3 0 1 = √ 1− 3 1+ 3 0 √ 2 2 0 0 2 2

0

0 1

1 2 √ − 23

√ 3− √3 3+ 3 0

1 2

0

0

0 1

0 0 √ 4 6

(b) The square of the left stretch tensor is V 2 = F · FT or √ √ 4 3 1 0 3 0 0 2 V = 0 2 0 1 2 0 = 2 0 0 0 1 0 0 1 The principal values are 4−λ 2 0 2 4 − λ 0 0 0 1−λ

0

√ 3 2

2 4 0

0 0 1

h i = 0 = (1 − λ) (4 − λ)2 − 4 = (1 − λ) (λ − 6) (λ − 2)

These are the same as the principal values of the square of the right stretch tensor. ^ = ni^ (c) Let the axis of rotation be N Ii then √ √ 1 + √3 −1 +√ 3 0 n1 n1 1 √ 1− 3 1+ 3 0 n2 = n2 √ 2 2 n3 n3 0 0 2 2 and √ √ √ 1 + 3 − 2 2 n1 + 3 − 1 n2 = 0 √ √ √ 1 − 3 n1 + 1 + 3 − 2 2 n2 = 0

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Continuum Mechanics for Engineers

^ =^ Solving gives n1 = n2 = 0 and for a unit vector n3 = 1. Thus N I3 . The angle of 1 ^ ^ ^2 ) and ^ = R · N where N = √ (^ e + e rotation can be found from n 1 2 √ 3 √ 1 √ 1− 3 2 2 0

1+

√ −1 + 3 √ 1+ 3

0 0 √ 2 2

0

√1 2 √1 2

0

=

√ 3 2

n1

= n2 n3 0 1 2

This gives

√ 3 1 ^= ^1 + e ^2 n e 2 2 The unit vector is at Φ = 60◦ and the angle change is ∆Φ = 15◦ clockwise. Problem 4.34 Let a displacement field be given by u1 =

1 (X3 − X2 ), 4

u2 =

1 (X1 − X3 ), 4

u3 =

1 (X2 − X1 ) . 4

Determine (a) the volume ratio dV/dV0 , (b) the change inthe right angle between line elements originally along the unit √ √ ^ 1 = 3^I1 − 2^I2 − ^I3 / 14 and N ^ 2 = ^I1 + 4^I2 − 5^I3 / 42. Explain vectors N your answer. Answer (a) dV/dV0 = 1.1875 (b) ∆θ = 0◦ Solution (a) The Jacobian of the deformation gradient gives the volume ratio. The deformations are 1 1 1 x1 = X1 + (X3 − X2 ), x2 = X2 + (X1 − X3 ), x3 = X3 + (X2 − X1 ) 4 4 4 This gives 4 −1 1 ∂x1 1 1 4 −1 [FiA ] = = ∂XA 4 −1 1 4 and

4 1 J = det F = 1 4 −1

−1 4 1

1 −1 4

19 = 16 ;

dV 19 = = 1.1875 dV 0 16

(b) The angle is by definition cos θ = p

^1 · C · N ^ N p 2 ^1 · C · N ^1 N ^2 · C · N ^2 N

111

Chapter 4 Solutions The right deformation tensor is C = FT · F and 4 1 1 −1 4 [CAB ] = [FAi ] [FiB ] = 16 1 −1 18 −1 −1 1 −1 18 −1 = 16 −1 −1 18

−1 4 1 1 4 −1

−1 4 1

1 −1 4

Now ^1 · C · N ^2 = N

h

√3 14

− √214

− √114

18

−1

i 1 −1 16 −1

18 −1

−1

−1 18

57 − 152 + 95 = √ √ =0 14 42(16) and

cos θ = 0;

and

θ = 90◦

^ 1 and N ^ 2 are perpendicular initially, ∆θ = 0. Since N Problem 4.35 Consider again the deformation given in Example 4.16, namely x1 = 2(X1 − X2 ), x2 = X1 + X2 , x3 = X3 . Determine (a) the left stretch tensor V, (b) the direction normals of the principal stretches of V. Answer √ 2 2 √0 0 (a) [VAB ] = 0 2 0 0 0 1 ^ 2 = ^I2 , N ^ 3 = ^I3 ^ 1 = ^I1 , N (b) N

Solution (a) The deformation gradient is 2 ∂xi [FiA ] = = 1 ∂XA 0

The square of the left deformation tensor is V 2 2 −2 0 2 2 V = 1 1 0 −2 0 0 1 0

−2 1 0

0 0 1

= F · FT 1 0 8 1 0 = 0 0 1 0

0 2 0

0 0 1

√1 42 √4 42 − √542

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Continuum Mechanics for Engineers

Since this is diagonal,

√ 2 2 [VAB ] = 0 0

√0 2 0

0 0 1

(b) The stretch tensor is diagonal, V = V ∗ and ^1 =^ N I1 ,

^2 =^ N I2 ,

^3 =^ N I3

Problem 4.36 A deformation field is expressed by x1 = µ(X1 cos βX3 + X2 sin βX3 ) , x2 = µ(−X1 sin βX3 + X2 cos βX3 ) , x3 = νX3 , where µ, β, and ν are constants. (a) Determine the relationship between these constants if the deformation is to be a possible one for an incompressible medium. (b) If the above deformation is applied to the circular cylinder shown by the sketch, determine (1) the deformed length l in terms of L, the dimension a, and the constants µ, β, and ν of an element of the lateral surface which has unit length and is parallel to the cylinder axis in the reference configuration, and (2) the initial length L of a line element on the lateral surface which has unit length and is parallel to the cylinder axis after deformation.

X3 a

L

X2 X1

113

Chapter 4 Solutions Answer (a) µ2 ν = 1 p (b) (1) l = µ2 β2 a2 + ν2 1p 2 2 β a +1 (2) L = ν

Solution (a) The deformation gradient is [FiA ] =

∂xi ∂XA

µ cos βX3 = −µ sin βX3 0

µ sin βX3 µ cos βX3 0

−µβX1 sin βX3 + µβX2 cos βX3 −µβX1 cos βX3 − µβX2 sin βX3 ν

For an incompressible material, the Jacobian of the deformation gradient must be one. Thus µ cos βX3 µ sin βX3 −µβX1 sin βX3 + µβX2 cos βX3 J = det F = −µ sin βX3 µ cos βX3 −µβX1 cos βX3 − µβX2 sin βX3 0 0 ν 2 2 2 2 2 = ν µ cos βX3 + µ sin βX3 = νµ = 1 (b) (1) The deformed length is

l1 0 l2 = [FiA ] 0 l3 1 µ cos βX3 µ sin βX3 −µβX1 sin βX3 + µβX2 cos βX3 0 = −µ sin βX3 µ cos βX3 −µβX1 cos βX3 − µβX2 sin βX3 0 1 0 0 ν −µβX1 sin βX3 + µβX2 cos βX3 = −µβX1 cos βX3 − µβX2 sin βX3 ν and l2 = l21 + l22 + l23 = µ2 β2 X21 + X22 + ν2 Since X21 + X22 = a2 , the result is l=

p

µ2 β2 a2 + ν2

(2) To calculate the initial length, we need the inverse of the deformation gradient. This is µν cos βX3 1 µν sin βX3 νµ2 0

F−1 iA =

−µν sin βX3 µν cos βX3 0

−µ2 βX2 −µ2 βX1 µ2

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Continuum Mechanics for Engineers

The initial length is L1 −1 0 L2 = F 0 iA L3 1 µν cos βX3 1 µν sin βX3 = νµ2 0 2 −µ βX2 1 2 µ βX1 = νµ2 µ2

−µν sin βX3 µν cos βX3 0

−µ2 βX2 0 µ2 βX1 0 1 µ2

The initial length squared is L2 = L21 + L22 + L23 = and L=

1 2 2 β X1 + X22 + 1 ν2

1p 2 2 β a +1 ν

Problem 4.37 A velocity field is defined in terms of the spatial coordinates and time by the equations, v1 = 2tx1 sin x3 ,

v2 = 2tx2 cos x3 ,

v3 = 0 .

At the point (1, −1, 0) at time t = 1, determine (a) the rate of deformation tensor and the vorticity tensor,

√ ^ = (^ ^2 + e ^) / 3, (b) the stretch rate per unit length in the direction of the normal n e1 + e (c) the maximum stretch rate per unit length and the direction in which it occurs, (d) the maximum shear strain rate. Answer

(a) (b) (c) (d)

0 0 [dij ] = 0 2 1 0 ˙ Λ/Λ = 4/3 ˙ Λ/Λ = 2, max γ˙ max = 1.5

0 0 1 1 0 , [wij ] = 0 0 0 −1 0 0 0 ^ =e ^2 n

Solution (a) The velocity gradient is

∂vi ∂xj

2t sin x3 0 = 0

0 2t cos x3 0

and

∂vi ∂xj

(1,−1,0),t=1

0 = 0 0

2tx1 cos x3 −2tx2 sin x3 0 0 2 0

2 0 0

115

Chapter 4 Solutions The rate of deformation tensor is D = 12 grad v + (grad v)T or

0 1 [dij ] = 0 2 0 The vorticity tensor is W =

1 2

2 0 0 + 0 0 2

0 2 0

0 0 0 = 0 0 1

0 2 0

1 0 0

0 2 0

grad v − (grad v)T

0 1 0 [wij ] = 2 0

0 2 0

2 0 0 − 0 0 2

0 0 0 0 = 0 0 0 −1 0

0 2 0

1 0 0

√ ^ = (^ ^2 + e ^) / 3 is (b) The rate of stretching along n e1 + e h Λ˙ = dij ni nj = Λ

√1 3

√1 3

√1 3

0

i 0 1

0 2 0

1

0 0

√1 3 √1 3 √1 3

4 = 3

(c) The principal rate of stretching is −λ 0 0 2−λ 1 0

1 0 −λ

= (−λ) (2 − λ) (−λ) − (2 − λ) = 0 = (2 − λ) λ2 − 1 = 0

and λ(1) = 2;

λ(2) = 1;

λ(3) = −1

The maximum rate of stretching is

Λ˙ Λ

=2 max

The direction is the eigenvector associated with the eigenvalue 2. This is

−2 0 0 0 1 0

1 n1 0 0 n2 = 0 −2 n3 0

or −2n1 + n3 = 0 n1 − 2n3 = 0 Solving gives n1 = n3 = 0 or ^ (1) = e ^2 n

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Continuum Mechanics for Engineers

(d) The maximum shear strain is found from γ˙ max ^ 2 are perpendicular. Thus n 2 0 i h γ˙ max = √1 0 √1 0 1 2 2 0 0

^1 · D · n ^ 2 where the vectors n ^ 1 and =n

0

√1 2

0 0 −1 − √12

3 = 2

Problem 4.38 Let NA and ni denote direction cosines of a material line element in the reference and current configurations, respectively. Beginning with Eq 4.147, ni Λ = xi,A NA , and using the indicial notation throughout, show that ˙ (a) Λ/Λ = dij ni nj , ¨ (b) Λ/Λ = Qij ni nj + n˙ i nj where Qij = 21 (ai,j + aj,i ) with ai being the components of acceleration. Solution (a) Taking the time derivative gives n˙ i Λ + ni Λ˙ = x˙ i,A NA = vi,A NA since NA is fixed. Taking the dot product with ni gives n˙ i ni Λ + ni ni Λ˙ = vi,A ni NA Now, ni ni = 1 and 2n˙ i ni = 0. This gives ∂vi ∂xj ∂vi ∂vi ni NA = ni NA = ni nj Λ Λ˙ = ∂XA ∂xj ∂XA ∂xj where Eq 4.147 was used. Thus ∂vi Λ˙ = ni nj = (dij + wij ) ni nj = dij ni nj Λ ∂xj since wij ni nj = 0. (b) We begin by taking the time derivative of n˙ i Λ + ni Λ˙ = x˙ i,A NA = vi,A NA to obtain

¨ = v˙ i,A NA = ai,A NA = ∂ai ∂xj NA n¨ i Λ + 2 n˙ i Λ˙ + ni Λ ∂xj ∂XA

Taking the dot product with ni gives ¨ = ∂ai ∂xj ni NA n¨ i ni Λ + 2 n˙ i ni Λ˙ + ni ni Λ ∂xj ∂XA or

¨ = ∂ai ∂xj NA n¨ i ni Λ + Λ ∂xj ∂XA

117

Chapter 4 Solutions Now

d (ni n˙ i ) = n˙ i n˙ i + ni n¨ i = 0 dt

or

ni n¨ i = −n˙ i n˙ i

This gives ¨ = −n˙ i n˙ i Λ + Λ

∂xj ∂ai ∂ai ni NA = ni nj Λ ∂xj ∂XA ∂xj

and ¨ ai,j − aj,i ai,j + aj,i ∂ai Λ + ni nj + n˙ i n˙ i = ni nj + n˙ i n˙ i = Λ ∂xj 2 2 ai,j + aj,i = ni nj + n˙ i n˙ i = Qij ni nj + n˙ i n˙ i 2 The second term in parentheses with ni nj is zero since it is antisymmetric. Problem 4.39 In a certain region of flow the velocity components are v1 = x31 + x1 x22 e−kt , v2 = x32 − x21 x2 e−kt ,

v3 = 0

where k is a constant, and t is time in s. Determine at the point (1, 1, 1) when t = 0, (a) the components of acceleration, (b) the principal values of the rate of deformation tensor, (c) the maximum shear rate of deformation. Answer (a) a1 = 2(4 − k), a2 = −4, a3 = 0 (b) D(1) = 4, D(2) = 2, D(3) = 0 (c) γ˙ max = ±2 Solution (a) The acceleration is ai =

∂vi ∂vi dvi = + vj dt ∂t ∂xj

Now a1 = −k x31 + x1 x22 e−kt + 3x21 + x22 e−kt x31 + x1 x22 e−kt + (2x1 x2 ) e−kt x31 − x1 x22 e−kt and

a1 |(1,1,1),t=0 = 2 (4 − k)

Also a2 = −k x32 − x21 x2 e−kt + (−2x1 x2 ) e−kt x32 + x21 x2 e−kt + 3x22 − x21 e−kt x32 − x21 x2 e−kt

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Continuum Mechanics for Engineers

and

a2 |(1,1,1),t=0 = −4

Finally, a3 = 0 (b) The rate of deformation tensor is D = 12 grad v + (grad v)T

∂vi ∂xj

3x21 + x22 e−kt = (−2x1 x2 ) e−kt 0

(2x1 x2 ) e−kt 3x22 − x21 e−kt 0

0 0 0

At the given point and time, we have

∂vi ∂xj

(1,1,1),t=0

0 0 0

4 2 = −2 2 0 0

The rate of deformation tensor is

4 [dij ] = 0 0 and

γ˙ max =

h

√1 2

0

√1 2

4

i 0 0

0 2 0

0 0 0

0

0

2 0

√1 2

0 0 0 − √12

= ±2

where the ± results from the selection of the orthogonal vectors.

Problem 4.40 For the motion x1 = X1 , x2 = X2 et + X1 (et − 1) , x3 = X1 (et − e−t ) + X3 , determine the velocity field vi = vi (x), and show that for this motion (a) L = F˙ · F−1 , (b) D = ˙ at t = 0. Solution (a) Inverting the equations of motion gives X1 = x1 X2 = x2 e−t − x1 1 − e−t X3 = x3 − x1 et − e−t

119

Chapter 4 Solutions The deformation gradient is 1 ∂xi [FiA ] = = et − 1 ∂XA et − e−t and

1 ∂X A −t e −1 = F−1 = Ai ∂xi − (et − e−t )

Also

F˙ iA

0 et = et + e−t

0 0 1

0 et 0

0 0 1

0 e−t 0 0 0 0

0 et 0

The velocity is dx1 =0 dt X dx2 v2 = = x1 + x2 dt X dx3 = x1 et + e−t v3 = dt X v1 =

and

0 1 [Lij ] = [vi,j ] = et + e−t

0 0 0

0 1 0

Also, we have h −1 i [Lij ] = F˙ iA FAj =

=

0 1 et + e−t

0 1 0

0 et et + e−t 0 0 0

0 1 0 e−t − 1 0 − (et − e−t )

0 et 0

The results are identical. (b) The rate of deformation tensor is D =

1 2

e−t 0

0 0 1

grad v + (grad v)T . This is at t = 0

0 0 0 0 1 1 1 0 + 0 [dij ] = 2 et + e−t 0 0 0 t −t 0 1 e +e 1 1 2 0 = 2 t −t e +e 0 0 At t = 0

0

0 1 [dij ] = 1 2 2

1 2 0

2 0 0

1 1 0

et + e−t 0 0

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Continuum Mechanics for Engineers

The displacement field is u1 = x1 − X1 = 0 u2 = x2 − X2 = (X1 + X2 ) et − 1 = (x1 + x2 ) 1 − e−t u3 = x3 − X3 = X1 et − e−t = x1 et − e−t and = 12 grad u + (grad u)T = 21 (ui,j + uj,i ) 0 0 0 0 1 − e−t et − e−t 1 0 [ij ] = 1 − e−t 1 − e−t 0 + 0 1 − e−t 2 et − e−t 0 0 0 0 0 0 1 − e−t et − e−t 1 0 = 1 − e−t 2 (1 − e−t ) 2 et − e−t 0 0 The time derivative is 0 1 e−t [˙ ij ] = 2 et + e−t At t = 0 [˙ ij ]t=0

e−t 2e−t 0

0 1 1 = 2 2

1 2 0

et + e−t 0 0 2 0 0

This is identical to the rate of deformation D. Problem 4.41 Determine an expression for the material derivative d (ln dx) /dt in terms of the rate of ^ = dx/dx. deformation tensor D and the unit normal n Answer ^ ·D·n ^ d (ln dx) /dt = n Solution The derivative is

.

d (ln dx) dx = dt dx and

.

d Λ˙ = dt So that

dx dX

dx = dX

.

d (ln dx) dx dX Λ˙ ^·D·n ^ = = Λ˙ = =n dt dx dx Λ

from Problem 4.37. Problem 4.42 A velocity field is given in spatial form by v1 = x1 x3 ,

v2 = x22 t,

v3 = x2 x3 t .

121

Chapter 4 Solutions (a) Determine the vorticity tensor W and the vorticity vector w. (b) Verify the equation εpqi wi = wqp for the results of part (a).

(c) Show that at the point (1, 0, 1) when t = 1, the vorticity tensor has only one real root. Answer (a) w1 = 12 x3 t, w2 = 21 x1 , w3 = 0 √ √ (c) W(1) = 0, W(2) = −W(3) = ı/ 2, where ı = −1 Solution (a) The vorticity tensor is W =

1 2

grad v − (grad v)T where (grad v)ij =

x3 [vi,j ] = 0 0

0 2x2 t x3 t

∂vi ∂xj .

This gives

x1 0 x2 t

This gives

x3 1 [wij ] = 0 2 0 0 1 = 0 2 −x

0 2x2 t x3 t 0 0 x3 t

1

x3 x1 0 − 0 x1 x2 t x1 −x3 t 0

0 2x2 t 0

0 x3 t x2 t

The vorticity vector is w=

1 curl v or wi = εijk vk,j 2

So that 1 1 (v3,2 − v2,3 ) = (x3 t) 2 2 1 1 w2 = (v1,3 − v3,1 ) = (x1 ) 2 2 1 w3 = (v2,1 − v1,2 ) = 0 2

w1 =

(b) For εpqi wi = wqp , we have ε123 w3 = w21 = 0;

ε231 w1 = w32 =

1 (x3 t) ; 2

ε312 w2 = w13 =

1 x1 2

and ε213 w3 = −w3 = w12 = 0;

1 ε132 w2 = −w2 = w31 = − x1 ; 2

1 ε321 w1 = −w1 = w23 = − (x3 t) 2

(c) At the point (1, 0, 1) and t = 1, the vorticity tensor is 0 0 1 1 [wij ](1,0,1),t=1 = 0 0 −1 2 −1 1 0

122

Continuum Mechanics for Engineers

The eigenvalues are 1 −λ 0 2 1 1 2 1 −λ =0 det (wij − λδij ) = 0 −λ − 2 = −λ λ + 4 4 1 1 −2 −λ 2 1 =0 = −λ λ2 + 2 or λ(1) = 0;

1 λ(2) , λ(3) = ±i √ 2

Problem 4.43 In materials science, a single crystal deforms by shearing on a single active slip system, as shown. Here, γ denotes shear. The loading on the crystal is such that the slip direction s and normal to the slip plane m each maintain a constant direction during the deformation. m

m

-1

tan γ

s

s

(a) Show that the deformation gradient can be expressed in terms of the components of the slip direction s and the normal to the slip plane m as Fij = δij + γsi mj . (b) Suppose shearing occurs at a shear rate γ. ˙ At the moment when γ = 0, calculate the velocity gradient tensor, the stretch rate tensor, and the vorticity tensor associated with the deformation. (c) For a material fiber parallel to a unit vector n in the deformed solid, find an expression for the stretch rate and angular velocity in terms of γ, ˙ s, and m.

Problem 4.44 Show that the velocity field v1 = 1.5x3 − 3x2 ,

v2 = 3x1 − x3 ,

v3 = x2 − 1.5x1

corresponds to a rigid body rotation, and determine the axis of spin (the vorticity vector). Answer ^1 + 1.5e ^ 2 + 3e ^3 w=e Solution The velocity gradient is [Lij ] =

∂vi ∂xj

0 = 3 −1.5

−3 0 1

1.5 −1 0

123

Chapter 4 Solutions

The rate of deformation tensor is identically zero, D = 0. This is a rigid motion. Since the velocity gradient is skew symmetric, W=L The vorticity vector is w1 = w32 = 1;

w2 = w13 = 1.5;

w3 = w21 = 3

or ^1 + 1.5^ w=e e2 + 3^ e3 Problem 4.45 For the steady velocity field v1 = x21 x2 ,

v2 = 2x22 x3 ,

v3 = 3x1 x2 x3

^3 ) /5. determine the rate of extension at (2, 0, 1) in the direction of the unit vector (4^ e1 − 3 e Answer 48 ˙ Λ/Λ =− 25 Solution The velocity gradient is [Lij ] =

∂vi ∂xj

x21 4x2 x3 3x1 x3

2x1 x2 0 = 3x2 x3

0 2x22 3x1 x2

and 1 [dij ] = 2

x21 8x2 x3 3x1 x3 + 2x22

4x1 x2 ∂vj ∂vi 1 x21 + = ∂xj ∂xi 2 3x2 x3

3x2 x3 3x1 x3 + 2x22 6x1 x2

At the point (2, 0, 1), we have

[dij ](2,0,1) and Λ˙ = Λ

4 5

− 35

0

0 = 2 0

0 2 0

2 0 3

2 0 3

0 3 0

4 0 5 48 3 − 53 = − 25 0 0

Problem 4.46 Prove that d(ln J) /dt = div v and, in particular, verify that this relationship is satisfied for the motion x1 = X1 + ktX3 , where k is a constant. Answer

x2 = X2 + ktX3 ,

x3 = X3 − kt(X1 + X2 )

124

Continuum Mechanics for Engineers ˙ = div v = 4k2 t/(1 + 2k2 t2 ) J/J

Solution The material time derivative is

.

det F J˙ = J J and from Eq 4.167

. det F = det F tr F˙ · F−1 = J tr L = J div v

This gives

.

J˙ det F 1 = = (J div v) = div v J J J For the motion, the velocity field is v1 = kX3 ;

v2 = kX3 ;

v3 = −k (X1 + X2 )

The deformation gradient is 1 ∂xi [FiA ] = = 0 ∂XA −kt

and

1 J = det F = 0 −kt

This gives

0 1 −kt

kt kt 1

0 1 −kt

kt kt 1

= 1 + 2k2 t2

J˙ = 4k2 t

Inverting the motion, we have 1 + k2 t2 x1 − k2 t2 x2 − ktx3 X1 = 1 + 2k2 t2 2 2 −k t x1 + 1 + k2 t2 x2 − ktx3 X2 = 1 + 2k2 t2 ktx1 + ktx2 + ktx3 X3 = 1 + 2k2 t2 Substituting into the velocity field gives v1 =

k (ktx1 + ktx2 + ktx3 ) ; 1 + 2k2 t2

v2 =

k (ktx1 + ktx2 + ktx3 ) ; 1 + 2k2 t2

v3 =

−k (x1 + x2 − 2ktx3 ) 1 + 2k2 t2

Then vi,i = v1,1 + v2,2 + v3,3 = =

J˙ 4k2 t = 2 2 1 + 2k t J

k2 t k2 t 2k2 t + + 2 2 2 2 1 + 2k t 1 + 2k t 1 + 2k2 t2

125

Chapter 4 Solutions Problem 4.47 Prove the identity

xi,A J

=0,

J = det(xi,A )

,i

Solution Computing the derivative gives

xi,A J

= ,i

(xi,A ),i = Use

(xi,A ),i xi,A J,i − J J2

∂xi,A ∂XC = xi,AC XC,i ∂XC ∂xi ∂J = XC,j J ∂xj,C

to get J,i =

∂xj,C ∂J ∂xj,C = XC,j J = XC,j Jxj,CB XB,i ∂xj,C ∂xi ∂xi

Substitution gives

xi,A J

=

xi,AC XC,i xi,A XC,j Jxj,CB XB,i − J J2

=

xi,AC XC,i − δAB XC,j xj,CB J

=

xi,AC XC,i − XC,j xj,CA J

,i

Assume xi are smooth functions of XA such that order of partial differentiation does not matter. xi,AC XC,i − XC,j xj,AC xi,A = =0 J J ,i Problem 4.48 Equation 4.158 gives the material derivative of dx2 in terms of dij . Using that equation as the starting point, show that d2 (dx2 )/dt2 is given in terms of dij and its time derivative by d2 (dx2 )/dt2 = 2(d˙ ij + vk,i dkj + vk,j dik )dxi dxj . Solution Equation 4.158 is d dx2 = dx · 2D · dx = 2dij dxi dxj dt and d2 dx2 d = 2 dt dt

d dx2 dt

!

.

.

= 2d˙ ij dxi dxj + 2dij dxi dxj + 2dij dxi dxj

126

Continuum Mechanics for Engineers .

Now dxi = vi,j dxj . This gives d2 dx2 = 2d˙ ij dxi dxj + 2dij vi,k dxk dxj + 2dij vj,k dxk dxi dt2 = 2 d˙ ij + dkj vk,i + dik vk,j dxi dxj where we have changed the dummy indices. Problem 4.49 A continuum body in the form of the unit cube shown by the sketch undergoes the homogeneous deformation x1 = λ1 X1 , x2 = λ2 X2 , x3 = λ3 X3 where λ1 , λ2 , and λ3 are constants. Determine the relationships among λ1 , λ2 , and λ3 if (a) the length of diagonal OC remains unchanged, (b) the rectangular area ABFE remains unchanged, (c) the triangular area ACE remains unchanged. X3 , x3

X3 , x3

E

E F

F

D

D C

C

O

O

G

A B

X1 , x1

G

A

X2 , x2

X2 , x2 B

X1 , x1

(a)

(b)

X3 , x3 E F D C

O

G

A

X2 , x2 B

X1 , x1

(c)

127

Chapter 4 Solutions Answer (a) λ21 + λ22 + λ23 = 3 (b) λ22 λ21 + λ23 = 2 (c) λ21 λ22 + λ22 λ23 + λ23 λ21 = 3 Solution (a) The deformation gradient, Jacobian and right deformation tensor are 2 λ1 λ1 0 0 [FiA ] = 0 λ2 0 , J = det F = λ1 λ2 λ3 , C = FT · F, [CAB ] = 0 0 0 λ3 0

0 λ22 0

The length OC is Λ2 = 1 =

h

√1 3

√1 3

√1 3

λ21

0

0

i 0 0

λ22

0 λ23

0

√1 3 √1 3 √1 3

1 2 λ + λ22 + λ33 = 1 3 1 (b) Equation 4.166 gives dSq = XA,q JdS0A =

Now

[XA,q ] =

1 λ1

∂XA = 0 ∂xq 0

0

0

1 λ2

h i −1 0 = FAq

1 λ3

0

This gives h

√1 2

0

√1 2

i

dS = λ1 λ2 λ3

h

= λ1 λ2 λ3

h

and dS2 =

√1 2

(λ1 λ2 λ3 )2 2

0

i 0 λ1 2 0 0 i √1 dS0

0

0

0 0 dS

√1 2

0

√1 2λ1

1 λ1

1 λ3

2λ3

1 1 + λ21 λ22

dS0

2

2 Since dS2 = dS0 , we have 2 = λ22 λ23 + λ21 λ23 = λ23 λ22 + λ21

(c) For the triangular area, h

√1 3

− √13

√1 3

i

dS = λ1 λ2 λ3

=

h

λ√ 2 λ3 3

h

√1 3

− √13

1 λ3 − λ√ 3

√1 3

λ√ 1 λ2 3

i

1 λ1

i 0 0 dS0

0

0

1 λ2

0 0 dS

0

1 λ3

0 0 λ23

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Continuum Mechanics for Engineers

and dS2 =

2 1 2 2 λ2 λ3 + λ21 λ23 + λ21 λ22 dS0 3

Now dS/dS0 = 1, and λ22 λ23 + λ21 λ23 + λ21 λ22 = 3 Problem 4.50 Let the unit cube shown in Problem 4.47 be given the motion 1 x1 = X1 + t2 X2 , 2

1 x2 = X2 + t2 X1 , 2

x3 = X3

Determine, at time t, (a) the rate-of-change of area ABFE, (b) the volume of the body. Answer .

^1 /(1 − 14 t4 ) − te ^ 2 − t3 e ^3 (a) dS = t3 e (b) V = (1 − 14 t4 ) Solution (a) The time rate of change of the area is , Eq 4.169, .

dSi = vk,k dSi − vi,j dSj The velocity field is v1 = X2 t;

v2 = X1 t;

v3 = 0

Inverting the motion gives X1 =

x1 − 12 t2 x2 1 − 14 t4

X2 =

;

x2 − 12 t2 x1 1 − 14 t4

X3 = x3

;

The velocity field is v1 =

t x2 − 12 t2 x1 1 − 14 t4

;

v2 =

t x1 − 21 t2 x2 1 − 14 t4

;

The velocity gradient is 3

− t2 1 t [vi,j ] = 1 − 41 t4 0

and vk,k = −

t 3 − t2 0

t3 1 − 14 t4

0 0 0

v3 = 0

129

Chapter 4 Solutions This gives .

dSi = vk,k dSi − vi,j dSj h t3 =− 1 − 14 t4 =

√1 2

h 1 t3 −√ 1 4 2 1 − 4t

0 √t 2

√1 2

h 1 dS − 1 − 41 t4

3

t −√ 2

i

√1 2

0

√1 2

i

3

− t2 t 0

i

t 3 − t2 0

0 0 dS 0

dS

(b) The current volume is given by the Jacobian, t2 2

1 2 J = det F = t2 0

1 0

0 0 1

= 1 − 1 t2 4

Problem 4.51 For the homogeneous deformation x1 = X1 + αX2 + αβX3 , x2 = αβX1 + X2 + β2 X3 , x3 = X1 + X2 + X3 , where α and β are constants, determine the relationship between these constants if the deformation is isochoric. Answer β = (α2 + α)/(α2 + α + 1) Solution For an isochoric motion J = det F = 1. The deformation gradient is 1 α αβ ∂xi [FiA ] = = αβ 1 β2 ∂Xa 1 1 1 and

1 J = αβ 1

α αβ 1 β2 = 1 + αβ2 + α2 β2 − αβ − β2 − α2 β = 1 1 1

This gives β αβ + α2 β − α − β − α2 = 0 or β α + α2 − 1 − α − α2 = 0 The relation is β=

α2 + α +α−1

α2

130

Continuum Mechanics for Engineers

Problem 4.52 Show that for any velocity field v derived from a vector potential ψ by v = curl ψ, the flow is isochoric. Also, for the velocity field v1 = ax1 x3 − 2x3 ,

v2 = −bx2 x3 ,

v3 = 2x1 x2

determine the relationship between the constants a and b if the flow is isochoric. Answer a=b Solution The velocity field is

v = curl ψ

or vi = εijk ψk,j

For an isochoric motion, div v = 0. This gives vi,i = εijk ψk,ji = 0 since the permutation symbol is skew symmetric in ji. For the velocity field, vi,i = ax3 − bx3 + 0 = 0 and a=b

Chapter 5 Solutions

Problem 5.1 Determine the material derivative of the flux of any vector property Q∗i through the spatial area S. Specifically, show that Z Z d ˙ ∗i + Q∗i vk,k − Q∗k vi,k ni dS Q∗i ni dS = Q dt S S in agreement with Eq 5.5. Solution Transcribing to symbolic notation Z Z Z d d d Q∗i ni dS = Q∗i dSi = Q∗ · dS dt S dt S dt S Transforming to the reference configuration Z Z Z d d d Q∗ · dS = Q∗ · JdS0 · F−1 = Q∗ · JdS0 · F−1 Eq 4.166 dt S dt 0 S0 dt Z S . ˙ ∗ · JdS0 · F−1 + Q∗ · JdS ˙ 0 · F−1 + Q∗ · JdS0 · (F−1 ) = Q S0

Now

.

.

.

(F · F−1 ) = (I) = 0 = F˙ · F−1 + F · (F−1 ) or

.

(F−1 ) = −F−1 · F˙ · F−1 = −L · F−1 so that d dt

In indicial form

R S

Q∗ · dS

d dt

=

R

J˙ = J div v

∗ 0 ∗ ˙ v − Q∗ · LT · JdS · F−1 RQ +∗Q · div ∗ ∗ T ˙ + Q · div v − Q · L · dS = S Q

S0

Z

Z S

and

Q∗ · dS =

S

˙ ∗i + Q∗i vk,k − Q∗k vi,k ni dS Q

Problem 5.2 ∗ Let the property Pij... in Eq 5.1 be the scalar 1 so that the integral in that equation represents the instantaneous volume V. Show that in this case Z Z d P˙ ij... = dV = vi,i dV . dt V V 131

132

Continuum Mechanics for Engineers

Solution From d dt

Z V

dV =

d dt Z

= V0

Z JdV0 (Eq 4.170) Z Z Z d 0 0 0 ˙ JdV = JdV = Jvi,i dV = vi,i dV dt V0 V0 V V0

Problem 5.3 Verify the identity εijk ak,j = 2 (w ˙ i + wi wj,j − wj vi,j ) , and by using this identity as well as the result of Problem 5.1, prove that the material derivative of the vorticity flux equals one half the flux of the curl of the acceleration; that is, show that Z Z 1 d wi ni dS = εijk ak,j ni dS . dt S 2 S Solution The acceleration is dvi ∂vi ∂vi = + vk dt ∂t ∂xk ∂vi ∂vi ∂vk ∂vk ∂vi ∂vk + vk − vk + vk = + vk (2Wik ) + vk = ∂t ∂xk ∂xi ∂xi ∂t ∂xi

ai =

But Wik = εkip wp (Eq 4.161) This gives ai =

∂vi ∂vk ∂vi + vk (2εkip wp ) + vk = + 2vk εipk wp + vk vk,i ∂t ∂xi ∂t

Taking the curl of a εpqi ai,q = εpqi =

∂vi ∂t

+ 2εpqi εkij (vk wj ),q + εpqi (vk vk,i ),q ,q

∂ (εpqi vi,q ) + 2 (δpj δqk − δpk δqj ) (vk,q wj + vk wj,q ) ∂t

+εpqi (vk,q vk,i + vk vk,iq ) The last term is zero because of the skew symmetric - symmetric product. Now εpqi ai,q = 2

∂wp + vk wp,k +2vk,k wp −2wq vp,q −2vp wj,j = 2 (w ˙ p + wp vk,k − wk vp,k ) ∂t

From Problem 5.1 Z Z Z 1 wi ni dS = (w ˙ i + wi vk,k − wk vi,k ) ni dS = εijk ak,j ni dS S S S 2

133

Chapter 5 Solutions Problem 5.4 Making use of the divergence theorem of Gauss together with the identity 1 ∂wi = εijk ak,j − εijk εkmq (wm vq ),j ∂t 2 show that

∂ ∂t

Z

Z V

wi dV =

S

Solution From Problem 5.3

εijk ak,j = 2

∂ ∂t

Z

1 εijk ak + wj vi − wi vj nj dS . 2

∂wi ∂t

+ 2εijk εkpq (wp vq ),j

Z

Z ∂wi 1 wi dV = dV = εijk ak,j − εijk εkpq (wp vq ),j dV V V ∂t V 2 Z 1 = εijk ak − (δip δjq − δiq δjp ) (wp vq ) nj dS S 2 Z 1 εijk ak − wi vj + wj vi nj dS = S 2

Problem 5.5 Show that the material derivative of the vorticity of the material contained in a volume V is given by Z Z 1 d wi dV = εijk ak + wj vi nj dS . dt V 2 S Solution From Eq 5.3 Z Z Z d ∂wi wi dV = dV + vk wi nk dS dt V ∂t S ZV Z 1 = εijk ak − wi vj + wj vi nj dS + vk wi nk dS (Prob. 5.4) S 2 S Z 1 = εijk ak + wj vi nj dS S 2 Problem 5.6 Given the velocity field q v1 = ax1 − bx2 , v2 = bx1 + ax2 , v3 = c x21 + x22 where a, b, and c are constants, determine (a) whether or not the continuity equation is satisfied, (b) whether the motion is isochoric. Answer

134

Continuum Mechanics for Engineers (a) only when ρ = ρ0 e−2at

(b) only if a = 0

Solution (a) Continuity requires ρ˙ + ρ div v = ρ˙ + ρvi,i = 0 Now div v = v1,1 + v2,2 + v3,3 = a + a + 0 and ρ˙ + ρ (2a) = 0

or

ρ˙ = −2a ρ

Integrating gives ln ρ = −2at + C If at t = 0, ρ = ρ0 then C = ln ρ0 and ρ = ρ0 e−2at (b) For an isochoric motion div v = vi,i = 2a = 0. The motion is isochoric only if a = 0. Problem 5.7 For a certain continuum at rest, the stress is given by tij = −p0 δij where p0 is a constant. Use the continuity equation to show that for this case the stress power may be expressed as p0 ρ˙ tij dij = . ρ Solution The stress power is tij dij = −p0 δij dij = −p0 dii = −p0 vi,i From the continuity equation ρ˙ + ρvi,i = 0

or

and tij dij =

vi,i = −

ρ˙ ρ

p0 ρ˙ ρ

Problem 5.8 Consider the motion xi = (1 + t/k)Xi where k is a constant. From the conservation of mass and the initial condition ρ = ρ0 at t = 0, determine ρ as a function of ρ0 , t, and k. Answer ρ=

ρ0 k 3 (k + t)3

135

Chapter 5 Solutions Solution The motion xi = (1 + t/k) Xi inverts to xi Xi = t 1+ k The velocity is vi =

Xi dxi = = dt Xi k

xi x i = t (k + t) k 1+ k

The divergence of the velocity field is vi,i =

xi,i δii 3 = = (k + t) (k + t) (k + t)

Continuity requires ρ˙ + ρvi,i = ρ˙ + ρ

3 =0 (k + t)

dρ 3 =− dt (k + t) ρ

or

Integration gives ln ρ = − ln (k + t)3 + ln C and ρ=

where C = ρ0 k3

ρ0 k3 (k + t)3

Problem 5.9 By combining Eqs 5.17b and 5.13, verify the result presented in Eq 4.168. Solution Eqs 5.17b and 5.13 are

.

(ρJ) = ρJ ˙ + ρJ˙ = 0 and ρ˙ + ρvi,i = 0 Combining

−ρvi,i J + ρJ˙ = 0

or

J˙ = vi,i J

the desired result, Eq 4.168. Problem 5.10 Using the identity εijk ak,j = 2 (w ˙ i + wi vj,j − wj vi,j ) as well as the continuity equation, show that εijk ak,j + 2wj vi,j d wi = . dt ρ 2ρ

136

Continuum Mechanics for Engineers

Solution Differentiating d dt

wi ρ

=

w ˙ i wi w ˙ i wi − 2 ρ˙ = + vk,k ρ ρ ρ ρ

where the continuity equation ρ˙ + ρvk,k = 0 was used. Now 1 1 d wi = εijk ak,j − wi vj,j + wj vi,j + wi vj,j dt ρ ρ 2 1 (εijk ak,j + 2wj vi,j ) = 2ρ Problem 5.11 State the equations of motion and from them show by the use of the material derivative v˙ i =

∂vi + vj vi,j , ∂t

and the continuity equation that ∂ (ρvi ) = (tij − ρvi vj ),j + ρbi . ∂t Solution The equations of motion are tij,j + ρbi = ρv˙ i = ρ

∂vi + ρvj vi,j ∂t

Rearranging tij,j + ρbi − ρvj vi,j − ρvi vj,j − ρ,j vi vj = ρ

∂vi − ρvi vj,j − ρ,j vi vj ∂t

or ∂ (ρvi ) ∂ρ − vi − ρvi vj,j − ρ,j vi vj ∂t ∂t ∂ (ρvi ) ∂ρ − vi + ρ,j vj − ρvi vj,j = ∂t ∂t ∂ (ρvi ) = − vi (ρ˙ + ρvj,j ) ∂t

(tij − ρvi vj ),j + ρbi =

The last term is zero from the continuity equation, ρ˙ + ρvj,j = 0. Problem 5.12 Determine the form which the equations of motion take if the stress components are given by tij = −pδij where p = p(x, t). Answer ρai = −p,i + ρbi

137

Chapter 5 Solutions Solution The equations of motion are So that

ρv˙ i = tij,j + ρbi

ρv˙ i = (−pδij ),j + ρbi = −p,j δij + ρbi

or ρai = −p,i + ρbi Problem 5.13 Let a material continuum have the constitutive equation tij = αδij dkk + 2βdij where α and β are constants. Determine the form which the equations of motion take in terms of the velocity gradients for this material. Answer ρv˙ i = ρbi + (α + β)vj,ij + βvi,jj Solution The equations of motion are

ρv˙ i = tij,j + ρbi

Now tij,j = (αδij dkk + 2βdij ),j = αδij dkk,j + 2βdij,j = αdkk,i + 2βdij,j = αvk,ki + β (vi,jj + vj,ij ) = (α + β) vj,ji + βvi,jj and

ρv˙ i = (α + β) vj,ji + βvi,jj + ρbi

Problem 5.14 Assume that distributed body moments mi act throughout a continuum in motion. Show that the equations of motion are still valid in the form of Eq 5.22, but that the angular momentum principle now requires εijk tjk + mi = 0 implying that the stress tensor can no longer be taken as symmetric. Solution The balance of linear momentum, Eq 5.22, is unaffected by the distributed body moments. The balance of angular momentum, Eq 5.52 is modified by adding an additional term Z Z Z . εijk (xj vk )ρ dV = εijk xj tqk nq dS + (εijk xj ρbk + mi ) dV V V Z ZS h i εijk (vj vk + xj ak ) ρ dV = εijk (xj tqk ),q + (εijk xj ρbk + mi ) dV V

S

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Since εijk vj vk = 0, Z V

[εijk xj ρak − εijk (xj,q tqk + xj tqk,q ) − (εijk xj ρbk + mi )] dV = 0

Recognizing xj,q = δjq and rearranging Z V

[εijk xj (ρak − ρbk − tqk,q ) − (εijk tjk + mi )] dV = 0

The first term is zero from the equations of motion leaving εijk tjk + mi = 0 Problem 5.15 For a rigid body rotation about the origin, the velocity field may be expressed by vi = εijk ωj xk where ωj is the angular velocity vector. Show that for this situation the angular momentum principle is given by .

Mi = (ωj Iij ) where Mi is the total moment about the origin of all surface and body forces, and Iij is the moment of inertia of the body defined by the tensor Z Iij = ρ (δij xk xk − xi xj ) dV . V

Solution The angular momentum principal is Mi = = = =

Z Z d d εijk xj ρvk dV = εijk xj ρεkmn ωm xn dV dt V dt V Z d (δim δjn − δin δjm ) xj ρωm xn dV dt V Z d (xj ρωi xj + xj ρωj xi ) dV dt V Z . d ωj ρ (δij xk xk + xj xi ) dV = (ωj Iij ) dt V

Problem 5.16 Determine expressions for the stress power tij dij in terms of (a) the first Piola-Kirchoff stress tensor, (b) the second Piola-Kirchoff stress tensor.

Answer (a) tij dij = ρF˙ iA PiA /ρ0

(b) tij dij = ρsAB C˙ AB /2ρ0

139

Chapter 5 Solutions Solution (a) From Eq 5.46 Jtji = PAi xj,A or tij =

ρxi,A PAj ρ0

since J = ρ0 /ρ. Now tij dij = tij vj,i =

ρvj,i xi,A PAj ρ0

but L · F = F˙ so tij dij = tij vi,j =

ρF˙ jA PAj ρ0

The summation makes this equivalent to the answer in the text. (b) The second Piola-Kirchoff stress tensor gives tij dij =

ρxi,A xj,B dij sAB ρ0

From Eq 4.156 xi,A xj,B dij = E˙ AB and 2EAB = CAB + δAB from Eq 4.49, so that C˙ AB E˙ AB = 2 and tij dij =

ρsAB C˙ AB 2ρ0

Problem 5.17 Show that, for a rigid body rotation about the origin, the kinetic energy integral Eq 5.55 reduces to the form given in rigid body dynamics, that is, K=

1 ωi ωj Iij 2

where Iij is the inertia tensor defined in Problem 5.15. Solution The kinetic energy is 1 K (t) = 2

Z V

ρvi vi dV

The rigid velocity field is vi = εijk ωj xk and Z Z 1 1 K (t) = ρεijk ωj xk εipq ωp xq dV = ρ (δjp δkq − δjq δkp ) ωj xk ωp xq dV 2 V 2 V Z Z 1 1 = ρ (ωj ωk δjk xq xq − ωj ωk xj xk ) dV = ωj ωk ρ (δjk xq xq − xj xk ) dV 2 V 2 V 1 = ωj ωk Ijk 2

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Problem 5.18 Show that one way to express the rate of change of kinetic energy of the material currently occupying the volume V is by the equation Z Z Z (^ n) ˙K = ρbi vi dV − tij vi,j dV + vi ti dS V

V

S

and give an interpretation of each of the above integrals. Solution The rate of change of the kinetic energy is Z Z Z 1 d ρvi vi dV = ρv˙ i vi dV = vi (Tij,j + ρbi ) dV K˙ (t) = dt 2 V V Z hV Z Z i (vi Tij ),j − vi,j Tij + ρvi bi dV = (ρvi bi − vi,j Tij ) dV + vi Tij nj dS = V S ZV Z (^ n) (ρvi bi − vi,j Tij ) dV + vi ti dS = V

S

The first term is the rate of work of the body forces. The second term is the stress power, and the final term is the rate of surface force working.

Problem 5.19 Consider a continuum for which the stress is tij = −p0 δij and which obeys the heat conduction law qi = −κθ,i . Show that for this medium the energy equation takes the form ρu˙ = −p0 vi,i + ρr + κθ,ii . Solution From Eq 5.67, the energy equation is ρu˙ − tij dij − ρr + qi,i = 0 The stress power is tij dij = −p0 δij vi,j = −p0 vi,i This gives

ρu˙ = −p0 vi,i + ρr + κθ,ii

Problem 5.20 If mechanical energy only is considered, the energy balance can be derived from the equations of motion. Thus, by forming the scalar product of each term of Eq 5.22 with the velocity vi and integrating the resulting equation term-by-term over the volume V, we obtain the energy equation. Verify that one form of the result is . 1 ρ(v · v) + tr (T · D) − ρb · v − div (T · v) = 0 . 2

Solution The equation of motion multiplied by the velocity and integrated over the volume is Z Z (tij,j vi + ρbi vi ) dV = (ρv˙ i vi ) dV V

V

141

Chapter 5 Solutions .

.

.

but (vi vi ) = 2v˙ i vi and v˙ i vi = 21 (vi vi ) = 12 (v · v). Now or

(tij vi ),j = tij,j vi + tij dij Therefore

Z V

tij,j vi = div (T · v) − tr (T · D)

. 1 div (T · v) − tr (T · D) + ρb · v − ρ(v · v) dV = 0 2

The result follows from localization. Problem 5.21 Show that for a continuum body experiencing volumetric growth Z Z d ρf˙ + ρcf dV . ρf dV = dt P P Solution Begin by transferring to the reference configuration Z Z Z d d ˙ dV0 (ρfJ) [ρ (X, t) f (X, t)] JdV0 = ρf dV = dt P dt P0 P0 Z Z ˙ + ρfJ˙ dV0 = = ρfJ ˙ + ρfJ ρf ˙ + ρf˙ + ρfvk,k JdV0 P0

P0

Transferring back to the current configuration gives Z Z d ρf dV = ρf˙ + (ρ˙ + ρvk,k ) f dV dt P P0 Using the mass conservation equation ρ˙ + ρvk,k = ρc, the result is obtained. Problem 5.22 Using the local form of the balance of mass, Eq 5.91, show that the balance of linear momentum for a material with volumetric growth is Eq 5.94 ρ

dvi = tji,j + ρbi . dt

Solution The balance of linear momentum for a growing body is Z Z Z Z d (^ n) ρvi dV = ti dS + ρbi dV + ρcvi dV dt P ∂P P P Using the result for the derivative of an integral with a growing mass, we have Z Z Z Z Z d (^ n) ρvi dV = ρv˙ i + ρcvi dV = ti dS + ρbi dV + ρcvi dV dt P P P P ∂P This leaves

Z

Z P

Z (^ n)

ρv˙ i dV = ∂P

ti

dS +

P

ρbi dV

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Continuum Mechanics for Engineers (^ n)

and using ti

= tji nj together with the divergence theorem yields the desired result.

Problem 5.23 Using the local form of the balance of mass, Eq 5.91, and linear momentum, Eq 5.94, show that the stress in a volumetrically growing body is symmetric. Solution The balance of moment of momentum is Z Z Z d (^ n) εijk xj ρvk dV = εijk xj tk dS + εijk xj ρ (bk + cvk ) dV dt P ∂P P The derivative of the first integral is Z Z Z d εijk xj ρvk dV = εijk ρ (x˙ j vk + xj v˙ k + cxj vk ) dV = εijk ρ (xj v˙ k + cxj vk ) dV dt P P P The first term is the cross product of a vector with itself. The remaining terms are Z Z Z (^ n) εijk ρ (xj v˙ k + cxj vk ) dV = εijk xj tk dS + εijk xj ρ (bk + cvk ) dV P

P

∂P

The remaining integral is Z Z εijk ρxj v˙ k dV = P

Z ∂P

(^ n) εijk xj tk

dS +

P

εijk xj ρbk dV

This is the balance of angular momentum and symmetry follows directly. Problem 5.24 If a continuum has the constitutive equation tij = −pδij + αdij + βdik dkj where p, α and β are constants, and if the material is incompressible (dii = 0), show that tii = −3p − 2βIID where IID is the second invariant of the rate of deformation tensor. Solution Computing the trace of the constitutive equation gives tii = −pδii + αdii + βdik dki Now IID =

1 2

(dii djj − dik dki ) =

1 2

and

tii = −3p + βdik dki

(−dik dki ) for an incompressible material. Thus tii = −3p − 2βIID

Problem 5.25 Starting with Eq 5.174 for isotropic elastic behavior, show that tii = (3λ + 2µ)ii ,

143

Chapter 5 Solutions and using this result, deduce that ij =

tij −

1 2µ

λ δij tkk 3λ + 2µ

.

Solution Equation 5.174 gives tij = λδij kk + 2µij for a linear isotropic material. Taking the trace gives tii = λδii kk + 2µii = (3λ + 2µ) ii Solving and substituting for ii yields tij = or

λ δij tkk + 2µij 3λ + 2µ

1 ij = 2µ

tij −

λ δij tkk 3λ + 2µ

Problem 5.26 For a Newtonian fluid, the constitutive equation is given by tij = −pδij + τij = −pδij + λ∗ δij dkk + 2µ∗ dij

(see Eq 5.176) .

By substituting this constitutive equation into the equations of motion, derive the equation ρv˙ i = ρbi − p,i + (λ∗ + µ∗ )vj,ji + µ∗ vi,jj . Solution The equation of motion is

tij,j + ρbi = ρv˙ i

Substituting the constitutive equation yields ρv˙ i = (−pδij + λ∗ δij dkk + 2µ∗ dij ) ,j +ρbi = −p,j δij + λ∗ δij dkk,j + 2µ∗ dij,j + ρbi = −p,i + λ∗ dkk,i + 2µ∗ dij,j + ρbi = −p,i + (λ∗ + µ∗ ) vj,ji + µ∗ vi,jj Problem 5.27 Combine Eqs 5.178a and Eq 5.178b into a single viscoelastic constitutive equation having the form tij = δij {R}kk + {S}ij where the linear time operators {R} and {S} are given, respectively, {R} =

3K {P} − 2 {Q} 3 {P}

and

{S} =

2 {Q} . {P}

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Continuum Mechanics for Engineers

Solution Equations 5.178a and 5.178b are {P}Sij = 2{Q}ηij tii = 3Kii Substituting for the stress and strain deviators gives 1 1 {P} tij − δij tkk = 2 {Q} ij − δij kk 3 3 This gives 1 {P} (tij − δij Kkk ) = 2 {Q} ij − δij kk 3 or

1 {P} (tij ) = 2 {Q} ij − δij kk + {P} δij Kkk 3

Rearranging tij =

2 {Q} 2 {Q} ij + δij K − kk {P} 3 {P}

Problem 5.28 Assume a viscoelastic medium is governed by the constitutive equations Eq 5.178. Let a slender bar of such material be subjected to the axial tension stress t11 = σ0 f(t) where σ0 is a constant and f(t) some function of time. Assuming that t22 = t33 = t13 = t23 = t12 = 0, determine 11 , 22 , and 33 as functions of {P}, {Q}, and f(t). Answer 3K {P} + {Q} σ0 f (t) 9K {Q}

11

=

22

= 33 =

2 {Q} − 3K {P} σ0 f (t) 18K {Q}

Solution From the constitutive equations, Eq 5.178, and the given data tii = t11 + t22 + t33 = t11 and

and so

t11 = 3Kkk

1 1 1 {P} tij − δij t11 = 2 {Q} ij − δij kk = 2 {Q} ij − δij t11 3 3 9K

For i = j = 1, 1 1 {P} t11 − δ11 t11 = 2 {Q} 11 − δ11 t11 3 9K or 2 {Q} 11 =

2 {P} 2 {Q} + 3 9K

t11

or

11 =

3K {P} + {Q} t11 9K {Q}

145

Chapter 5 Solutions For i = j = 2, 1 1 {P} − δ22 t11 = 2 {Q} 22 − δ22 t11 3 9K or 2 {Q} 22 =

2 {Q} {P} − 9K 3

t11

or

22 =

2 {Q} − 3K {P} t11 18K {Q}

For i = j = 3 1 1 {P} − δ33 t11 = 2 {Q} 33 − δ33 t11 3 9K or 2 {Q} 33 =

2 {Q} {P} − 9K 3

t11

or

33 =

2 {Q} − 3K {P} t11 18K {Q}

where t11 = σ0 f (t) Problem 5.29 Use the definition of the free energy along with the reduced form of the Clausius-Duhem equation to derive the local dissipation inequality. Solution The free energy is ψ = u − ηθ and the reduced form of the Clausius-Duhem equation is, Eq 5.85 1 ρθη˙ − ρu˙ + dij tij − qi gi > 0 θ ˙ Taking the time derivative of the free energy gives, ψ˙ = u˙ − ηθ ˙ − ηθ˙ or u˙ = ψ˙ + ηθ ˙ + ηθ. Substituting for u˙ gives 1 ρθη˙ − ρ ψ˙ + ηθ ˙ + ηθ˙ + dij tij − qi gi > 0 θ or

1 −ρ ψ˙ + ηθ˙ + dij tij − qi gi > 0 θ (See A. E. Green and N. Laws, On the Formulation of Constitutive Equations in Thermodynamical Theories of Continua, Quarterly Journal of Mechanics and Applied Mathematics, Vol 20, 1967, Eq 2.11) Problem 5.30 The constitutive model for a compressible, viscous, and heat-conducting material is defined by ˜ θ, gk , FiA , F˙ iA , ψ=ψ η = η˜ θ, gk , FiA , F˙ iA , tij = t˜ ij θ, gk , FiA , F˙ iA , qi = q˜ i θ, gk , FiA , F˙ iA . Deduce the following restrictions on these constitutive response functions: ˜ θ, gk , FkB , F˙ kB ∂ψ = 0, (a) ∂gi

146

Continuum Mechanics for Engineers ˜ θ, gk , FkB , F˙ kB ∂ψ = 0, ∂F˙ iA ˜ (θ, FkB ) ∂ψ (c) η˜ (θ, FiA ) = − , ∂θ ˜ (θ, FkB ) ∂ψ (d) t˜ ij = ρFjA , ∂FiA 1 (e) − q˜ i (θ, gk , FiA , 0) gi > 0. θ

(b)

Solution Beginning with the free energy ˜ ˜ ˜ ˜ ∂ψ ∂ψ ∂ψ ∂ψ ψ˙ = θ˙ + g˙ k + F˙ iA + F¨ iA ˙ ∂θ ∂gk ∂FiA ∂FiA Substituting into the reduced Clausius-Duhem inequality, Eq 5.86 and collecting terms, gives ˜ ˜ ˜ ˜ ∂ψ ∂ψ ∂ψ ∂ψ 1 + η˜ θ˙ + tij − ρ g˙ k − −ρ F˙ iA − ρ F¨ iA − qi gi > 0 ∂θ ∂FiA ∂gk θ ∂F˙ iA The quantities θ, gi ,FiA , F˙ iA , and F¨ iA can be chosen arbitrarily by Eq 5.104. Select a process with FiA and θ = θ0 , a constant. Also, assume a uniform temperature field so that gi = 0. Then ˜ ∂ψ F¨ iA > 0 − ∂F˙ iA Since F¨ iA can be chosen arbitrarily, this inequality can be violated unless −

˜ ∂ψ ˜ (θ, FiA , gi ) = 0 which implies ψ = ψ ∂F˙ iA

Now proceed as in the text from Eq 5.100. (See Coleman and Mizel, 1964.) Problem 5.31 Assume the constitutive relationships u = u˜ (CAB , η) , θ = θ˜ (CAB , η) , tij = t˜ ij (CAB , η) , qi = q˜ i (CAB , η, gk ) , for an elastic material. Use the Clausius-Duhem inequality to show ∂u˜ , ∂η u = u˜ (CAB , η) , 1 ∂u˜ ∂u˜ tij = ρFiA + FjB , 2 ∂CAB ∂CBA q˜ i gi > 0 . θ=

147

Chapter 5 Solutions Solution Begin with the Clausius-Duhem inequality, Eq 5.85 ρθη˙ − ρu˙ + dij tij − Now u˙ = Combining

1 q i gi > 0 θ

∂u˜ ˙ ∂u˜ CAB + η˙ ∂CAB ∂η

∂u˜ ˙ ∂u˜ 1 η˙ − ρ ρ θ− CAB + dij tij − qi gi > 0 ∂η ∂CAB θ

Now C˙ AB = F˙ iA FiB + FiA F˙ iB = Lij FjA FiB + FiA Lij FjB = 2FiA FjB Lij = 2FiA FjB dij where dummy indices were switched and dij replaces Lij since only the symmetric part is retained in the product. ∂u˜ ∂u˜ 1 η˙ + tij − 2ρFiA FjB ρ θ− Lij − q˜ i gi > 0 ∂η ∂CAB θ We know η, ˙ Lij , and gi are independent and can be chosen independently. Choose a thermodynamic process in which η˙ = 0 and Lij = 0 to conclude 1 q˜ i gi > 0 θ

or q˜ i gi > 0

since the temperature is always positive. Now choose a process where η˙ = 0 and gi = 0. Then ∂u˜ tij − 2ρFiA FjB Lij > 0 ∂CAB Since this must hold for all process and Lij may be chosen independently, for this inequality to always hold ∂u˜ tij = 2ρFiA FjB ∂CAB where it is assumed that ∂u˜ 1 ∂u˜ ∂u˜ = + ∂CAB 2 ∂CAB ∂CBA Finally choose Lij = 0 and gi = 0 resulting in ∂u˜ ρ θ− η˙ > 0 ∂η which must hold for all processes. Since η˙ is arbitrary, it coefficient must be zero for the inequality to hold. This gives ∂u˜ θ= ∂η Problem 5.32 Use the basic kinematic result of superposed rigid body motion given in Eq 5.122b to show the following:

148

Continuum Mechanics for Engineers ∂x+ i = Qik , ∂xk (b) F+ iA = Qij FjA . (a)

Solution a. From Eq 5.122b x+ i = ai + Qim xm differentiating gives ∂x+ ∂ai ∂xm i = + Qim = Qim δmj = Qij ∂xj ∂xj ∂xj b. The deformation gradient is + F+ iA = xi,A = (ai + Qim xm ),A = Qim xm,A = Qij FjA

where dummy indices have been exchanged. Problem 5.33 Show that the Jacobian transforms as follows under a superposed rigid body motion: J+ = J . Solution By definition J = det FiA and J+ = det F+ iA = det (Qij FjA ) = det Qij det FjA = +1 det FjA = J where the results form Problem 5.32b. was used. Problem 5.34 Utilize the result of Problem 5.33 along with the law of conservation of mass to show that ρ+ = ρ. Solution The continuity equation gives ρ+ = ρ0 det J+ = ρ0 det J = ρ Problem 5.35 Show that the gradient of the stress components transforms under superposed rigid body motion as follows: ∂t+ ∂tmn ij . + = Qim ∂xn ∂xj Solution The divergence of t+ ij is ∂t+ ij ∂x+ j

=

∂t+ ∂xn ∂ ij ∂xn (Qim Qjq tmq ) = + ∂xn ∂x+ ∂x j j ∂xn

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Chapter 5 Solutions

where we have used the chain rule and the transformation equation for stress. Now + from x+ k = ak + Qkq xq , Qkn Qkq xq = δnq xq = xn = Qkn xk − ak since Q is a proper orthogonal tensor. Then ∂x+ ∂xn k = Q = Qkn δkj = Qjn kn ∂x+ ∂x+ j j and

∂t+ ij ∂x+ j

∂tmq ∂tmq ∂tmn = Qjn Qim Qjq = δnq Qim = Qim ∂xn ∂xn ∂xn

Problem 5.36 Use the superposed rigid body motion definitions to show the following relationships: (a) C+ AB = CAB , (b) U+ AB = UAB , (c) R+ iA = Qij RjA , (d) B+ ij = Qim Bmk Qkj .

Solution From Problem 5.32 + + (a) C+ AB = FiA FiB = Qij FjA Qik FkB = δjk FjA FkB = FjA FjB = CAB + + (b) U+ AB = FiA RiB = Qij FjA Qik RkB = δjk FjA RkB = FkB RkB = UAB + + + + + (c) F+ iA = RiB UAB = RiB UAB from result b. Now FiA = Qij FjA = RiB UAB or −1 + RiB = Qij FjA UAB = Qij RjB + + (d) B+ ij = FiA FjA = Qim FmA Qjn FnA = Qim Qjn Bmn

Problem 5.37 In the context of rigid body dynamics, consider the motion defined by x (X, t) = X along with x = QT (t) x+ − a (t) show that v+ = a˙ (t) + ω (t) × x+ − a (t) where ω is the angular velocity of the body. Solution The velocity is ˙ + Qx˙ v+ = x˙ + = a˙ + Qx but for a rigid body x˙ = X˙ = 0

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and ˙ = a˙ + ΩQx = a˙ + ΩQ QT x+ − a v+ = a˙ + Qx = a˙ + Ω x+ − a = a˙ + ω × x+ − a where ω is the axial vector of the skew tensor Ω. Eq 5.134 was used in the final step ^i = −εijk ωk bj e ^i = εikj ωk bk e ^i Ωij bj e or Ωb = ω × b

Chapter 6 Solutions

Problem 6.1 In general, the strain energy density W may be expressed in the form W = C∗αβ α β

(α, β = 1, . . . , 6)

where C∗αβ is not necessarily symmetric. Show that this equation may be rearranged to appear in the form 1 W = Cαβ α β 2 where Cαβ is symmetric, so that now ∂W = Cαβ β = tβ ∂β in agreement with Eq 6.8. Solution Let C∗αβ =

C∗αβ + C∗βα 2

+

C∗αβ − C∗βα 2

Now W= = and

C∗αβ + C∗βα 2 ∗ Cαβ + C∗βα 2

α β + α β =

C∗αβ − C∗βα 2

α β

1 Cαβ α β 2

1 1 1 ∂W = Cαβ δγα β + Cαβ α δγβ = (Cγβ β + Cαγ α ) ∂γ 2 2 2

But Caβ = Cβα and

∂W 1 = (Cβγ β + Cαγ α ) = Cαγ α ∂γ 2

Problem 6.2 Let the stress and strain tensors be decomposed into their respective spherical and deviator components. Determine an expression for the strain energy density W as the sum of a dilatation energy density W(1) and a distortion energy density W(2) . 151

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Continuum Mechanics for Engineers

Answer W = W(1) + W(2) =

1 1 tii jj + Sij ηij 6 2

Solution The strain energy density is 1 1 W = tij ij = 2 2 1 Sij ηij + = 2

1 1 Sij + δij tkk ηij + δij kk 3 3 1 1 1 ηii tkk + Sii kk + δii tkk jj . 3 3 9

But Sii = ηii = 0 and W=

1 1 tkk jj + Sij ηij 6 2

since δii = 3.

Problem 6.3 If the strain energy density W is generalized in the sense that it is assumed to be a function of the deformation gradient components instead of the small strain components, that is, if W = W(FiA ), make use of the energy equation and the continuity equation to show that in this case Eq 6.16 is replaced by ∂W FjA . Jtij = ∂FiA

Solution The strain energy density is W = ρ0 u and

˙ = ρ0 u˙ W

Now from the energy equation u˙ = and

1 1 tij dij = tij vi,j ρ ρ ˙ = ρ0 tij vi,j W ρ

Now we also have for W (FiA ) ˙ (FiA ) = ∂W (FiA ) F˙ iK = ∂W (FiA ) vj,q FqK W ∂FjK ∂FjK ˙ we have Then from the expressions for W, ˙ = ρ0 tij vi,j = ∂W (FqA ) vi,j FjK W ρ ∂FiK

153

Chapter 6 Solutions and Jtij =

∂W (FqK ) FjA ∂FiA

Problem 6.4 For an isotropic elastic medium as defined by Eq 6.23, express the strain energy density in terms of (a) the components of ij , (b) the components of tij , (c) the invariants of ij . Answer 1 (λii jj + 2µij ij ), 2 (3λ + 2µ) tij tij − λtii tjj , (b) W = 4µ (3λ + 2µ) (c) W = 12 λ + µ (I )2 − 2µII (a) W =

Solution (a) For the isotropic linear material, tij = 2µij + λδij kk and 1 1 1 tij ij = ij (2µij + λδij kk ) = (2µij ij + λii kk ) 2 2 2 λ 1 tij − δij tkk and (b) From Eq 6.23, tii = (3λ + 2µ) ii and ij = (3λ + 2µ) 2µ 1 1 λ W = tij ij = tij tij − tjj tkk (3λ + 2µ) 2 4µ W=

(c) I = tr = ii and II =

(ii jj − ij ij ) or ij ij = I2 − 2II . From (a) 1 2 λ W= λI + 2µ I2 − 2II = + µ I − 2µII 2 2 1 2

Problem 6.5 Let tij be any second-order isotropic tensor such that t0ij = aim ajn tmn = tij for any proper orthogonal transformation aij . Show that by successive applications of the transformations 0 0 −1 0 0 1 0 and [aij ] = −1 0 0 [aij ] = −1 0 0 1 0 0 −1 0 every second-order isotropic tensor is a scalar multiple of the Kronecker delta, δij .

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Solution For the first transformation t11 t12 t13 0 0 −1 t11 t12 t21 t22 t23 = −1 0 0 t21 t22 t31 t32 t33 0 1 0 t31 t32 t33 t31 −t32 t11 −t12 = t13 −t23 −t21 t22 Thus t11 = t22 = t33 , t12 = t31 , t13 = −t32 , t21 = t13 , t23 or we can write t12 = t31 = −t23 ; t21 = t13 = −t32 (a) For the second transformation t11 t12 t13 0 0 1 t11 t12 t21 t22 t23 = −1 0 0 t21 t22 0 −1 0 t31 t32 t33 t31 t32 t33 −t31 −t32 t12 = −t13 t11 −t23 t21 t22

t13 0 t23 0 t33 −1

−1 0 0 1 0 0

= −t12 , t31 = −t23 , t32 = −t21 ,

t13 0 t23 0 1 t33

−1 0 0 −1 0 0

Here, we have t11 = t22 = t33 , t12 = −t31 , t13 = −t32 , t21 = −t13 , t23 = t12 , t31 = −t23 , t32 = t21 , or we can write t12 = −t31 = t23 ; t21 = −t13 = −t32 (b) Combining (a) and (b) gives t12 = t31 = t23 = t21 = t13 = t32 = 0. Now let t11 = t22 = t33 = λ, then tij = λδij .

Problem 6.6 Verify that Eqs 6.33a and 6.33b when combined result in Eq 6.31 when Eqs 6.32 are used.

Solution From Eqs 6.33a and 6.33b 1 1 1 δij tkk tij − δij tkk = 2G ij − δij kk = 2G ij − 3 3 9K or 1 2G − 3K 1+ν −3ν tij + δij tkk = tij + δij tkk 2G 9K E 3E ν 1+ν = tij − δij tkk E E

ij =

Problem 6.7 For an elastic medium, use Eq 6.33 to express the result obtained in Problem 6.2 in terms of the engineering elastic constants K and G. Answer W = 12 Kii jj + G ij ij − 13 ii jj

155

Chapter 6 Solutions Solution From Problem 6.2 3Kii jj 2Gηij ηij tii jj Sij ηij + = + 6 2 6 2 K 1 1 = ii jj + G ij − δij kk ij − δij mm 2 3 3 K 1 = ii jj + G ij ij − ii kk 2 3

W=

Problem 6.8 Show that the distortion energy density W(2) (see Problem 6.2) for a linear elastic medium may be expressed in terms of (a) the principal stresses, σ(1) , σ(2) , σ(3) and (b) the principal strains, (1) , (2) , (3) in the form σ(1) − σ(2)

(a) W(2) = (b) W(2) =

2

+ σ(2) − σ(3) 12G

2

+ σ(3) − σ(1)

2 ,

2 2 2 i 1h (1) − (2) + (2) − (3) + (3) − (1) G. 3

Solution (a) For the stresses, 1 1 1 1 Sij Sij = tij − δij tkk tij − δij tkk 4G 4G 3 3 1 1 = tij tij − tii tjj 4G 3

W(2) =

Note that tii = σ(1) + σ(2) + σ(3) in terms of the principal stresses, σ(i) , and tij tij = σ(1) + σ(2) + σ(3)

σ(1) + σ(2) + σ(3) − 2 σ(1) σ(2) + σ(2) σ(3) + σ(3) σ(1)

Then W(2)

2 2 1 = −2 σ(1) σ(2) + σ(2) σ(3) + σ(3) σ(1) + σ(1) + σ(2) + σ(3) 4G 3 2 i 1 h −6 σ(1) σ(2) + σ(2) σ(3) + σ(3) σ(1) + 2 σ(1) + σ(2) + σ(3) = 12G 1 = −6 σ(1) σ(2) + σ(2) σ(3) + σ(3) σ(1) 12G i +2 σ2(1) + σ2(2) + σ2(3) + 2σ(1) σ(2) + 2σ(2) σ(3) + 2σ(3) σ(1) 2 2 2 i 1 h = σ(1) − σ(2) + σ(2) − σ(3) + σ(3) − σ(1) 12G

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(b) Similarly for the strains, 1 1 1 W(2) = Gηij ηij = G ij − δij kk ij − δij kk = G ij ij − ii jj 3 3 3 2 2 = G −2 (1) (2) + (2)(3) + (3)(1) + (1) + (2) + (3) 3 G −6 (1) (2) + (2) (3) + (3)(1) = 3 i +2 2(1) + 2(2) + 2(3) + 2(1) (2) + 2(2) (3) + 2(3)(1) 2 2 2 i Gh = (1) − (2) + (2) − (3) + (3) − (1) 3 Problem 6.9 Beginning with the definition for W (Eq 6.21a), show for a linear elastic material represented by Eq 6.23 or by Eq 6.31 that ∂W/∂ij = tij and ∂W/∂tij = ij (Note that ∂ij /∂mn = δim δjn .) Solution From Eq 6.21a, we have using Eq 6.23 W=

1 1 tij ij = (2µij + λδij kk ) ij 2 2

Now ∂W 1 = [4µδim δjn ij + (λδij δkm δkn ij + λδij kk δim δjn )] ∂mn 2 1 = [4µmn + (λδmn ii + λδmn kk )] = 2µmn + λδmn kk 2 = tmn From Eq 6.31, we have W=

1 1 [(1 + ν) tij − νδij tkk ] tij tij ij = 2 2E

and ∂W 1 [2 (1 + ν) δim δjn tij − ν (δij δkm δkn tij + δij δim δjn tkk )] = ∂tmn 2E 1 [2 (1 + ν) tmn − ν (δmn tjj + δmn tkk )] = 2E 1 = [(1 + ν) tmn − νδmn tkk ] = mn E Problem 6.10 For an isotropic, linear elastic solid, the principal axes of stress and strain coincide, as was shown in Example 6.1. Show that, in terms of engineering constants E and ν this result is given by (1 + ν) σ((q) − ν σ((1) + σ((2) + σ((3) (q) = , (q = 1, 2, 3). E

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Chapter 6 Solutions

Thus, let E = 106 psi and ν = 0.25, and determine the principal strains for a body subjected to the stress field (in ksi) 12 0 4 [tij ] = 0 0 0 . 4 0 6 Answer (1) = −4.5 × 10−6 , (2) = 0.5 × 10−6 , (3) = 13 × 10−6

Solution Consider

1 [(1 + ν) tij − νδij tkk ] E ^ (q) , (q = 1, 2, 3) be the eigenvectors corresponding to σ(1) , σ(2) , σ(3) . Then Let n ij =

(q)

ij nj

=

but

1 (q) [(1 + ν) tij − νδij tkk ] nj E

(q) tij − δij σ(q) nj = 0

This gives (q)

ij nj

i 1h (q) (q) (1 + ν) δij σ(q) nj − νδij tkk nj E (q) 1 (1 + ν) σ(q) − νtkk ni = E =

For principal strains (q) ij − δij (q) nj = 0 or (q)

ij nj and

(q) =

(q)

= (q) ni

=

(q) 1 (1 + ν) σ(q) − νtkk ni E

1 (1 + ν) σ(q) − ν σ(1) + σ(2) + σ(3) E

since tkk is aninvariant. 12 0 4 0 For [tij ] = 0 0 0 the matrix of principal stresses is σ(q) = 0 4 0 6 0 Now 1 ij = 6 [(1.25) tij − (0.25) δij (18)] 10 and 10.5 0 5 −4.5 0 10−6 [ij ] = 0 5 0 3 This gives

(q)

−4.5 0 0.5 = 0 0 0

0 0 10−6 13

0 4 0

0 0 14

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From (q) =

1 (1 + ν) σ(q) − ν σ(1) + σ(2) + σ(3) E

and (1) = −4.5 × 10−6

(2) = 0.5 × 10−6

(3) = 13 × 10−6

Problem 6.11 Show for an isotropic elastic medium that 1 2 (λ + µ) ν λ = , (b) = , 1+ν 3λ + 2µ 1+ν λ + 2µ (d) 2µ(1 + ν) = 3K(1 − 2ν). (a)

(c)

2µν 3Kν = , 1 − 2ν 1+ν

Solution (a) 1 = 1+ν

1 λ 1+ 2 (λ + µ)

=

2 (λ + µ) 3λ + 2µ

(b) ν = 1−ν

λ λ 2 (λ + µ) = λ λ + 2µ 1− 2 (λ + µ)

(c)

3κ (1 − 2ν) 2ν 2µν 2 (1 + ν) = 1 − 2ν 1 − 2ν

=

3Kν 1+ν

(d)

3κ (1 − 2ν) (1 + ν) = 3K(1 − 2ν) 2µ(1 + ν) = 2 2 (1 + ν) Problem 6.12 Let the x1 x3 plane be a plane of elastic symmetry such that the transformation matrix between Ox1 x2 x3 and Ox01 x02 x03 is 1 0 0 [aij ] = 0 −1 0 . 0 0 1 Show that, as the text asserts, this additional symmetry does not result in a further reduction in the elastic constant matrix, Eq 6.40. Solution The transformation is 1 0 0 t1 0 −1 0 t6 0 0 1 t5

t6 t2 t4

t5 1 t4 0 t3 0

0 −1 0

0 t1 0 = −t6 1 t5

−t6 t2 −t4

t5 −t4 t3

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Chapter 6 Solutions

Thus t0α = tα for (α = 1, 2, 3, 5) and t0a = −ta for (α = 4, 6). Similarly 0α = α for (α = 1, 2, 3, 5) and 0a = −a for (α = 4, 6). These conditions are identical to those satisfied in deriving Eq 6.40. Problem 6.13 Let the x1 axis be an axis of elastic symmetry of order N = 2. Determine the form of the elastic constant matrix Cαβ , assuming Cαβ = Cβα . Answer [Cαβ ] =

C11 C12 C13 C14 0 0

C12 C22 C23 C24 0 0

C13 C23 C33 C34 0 0

C14 C34 C34 C44 0 0

0 0 0 0 C55 C56

0 0 0 0 C56 C66

Solution 2π For N = 2, we have θ = π since N = . Thus θ 1 0 0 0 [aij ] = 0 −1 0 0 −1 and

1 0 0

0 0 t1 −1 0 t6 0 −1 t5

t6 t2 t4

t5 1 t4 0 t3 0

0 −1 0

0 t1 0 = −t6 −1 −t5

−t6 t2 t4

−t5 t4 t3

Thus t0α = tα for (α = 1, 2, 3, 4) and t0a = −ta for (α = 5, 6). Similarly 0α = α for (α = 1, 2, 3, 4) and 0a = −a for (α = 5, 6). This gives t01 = C11 01 + C12 02 + C13 03 + C14 04 + C15 05 + C16 060 and t1 = C11 1 + C12 2 + C13 3 + C14 4 + C15 5 + C16 6 or t1 = C11 01 + C12 02 + C13 03 + C14 04 − C15 05 − C16 06 The result is C16 = C61 = C15 = C51 = 0. Likewise from t02 = t2 , C25 = C52 = C26 = C62 = 0 from t03 = t3 , C35 = C53 = C36 = C63 = 0 from t04 = t4 , C45 = C54 = C46 = C64 = 0, also t05 = C55 05 + C56 06 = −C55 5 − C56 6 = −t5 t06 = C56 05 + C66 06 = −C56 5 − C66 6 = −t6

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Continuum Mechanics for Engineers

Problem 6.14 Assume that, by the arguments of elastic symmetry, the elastic isotropic body has been reduced to the form C11 C12 C12 0 0 0 C12 C11 C12 0 0 0 C12 C12 C11 0 0 0 [Cαβ ] = 0 0 0 C 0 0 44 0 0 0 0 C44 0 0 0 0 0 0 C44

constant matrix for an .

Show that, if the x1 axis is taken as an axis of elastic symmetry of any order (θ is arbitrary), C11 = C12 + 2C44 . (Hint: Expand t023 = a2q a3m tqm and 023 = a2q a3m qm .) Solution The transformation matrix is

1 [aij ] = 0 0

0 0 cos θ sin θ − sin θ cos θ

Now from t023 = a2q a3m tqm = − sin θ cos θt22 + cos2 θ − sin2 θ t23 + sin θ cos θt33 or in single subscript notation t04 = − sin θ cos θt2 + cos2 θ − sin2 θ t4 + sin θ cos θt3 Similarly, 023 = a2q a3m qm = − sin θ cos θ22 + cos2 θ − sin2 θ 23 + sin θ cos θ33 or 04 = −2 sin θ cos θ2 + cos2 θ − sin2 θ 4 + 2 sin θ cos θ3 Now t04 = C44 044 = 2C44 sin θ cos θ (3 − 2 ) + C44 cos2 θ − sin2 θ 4 Also from the stress transformation and the constitutive equation, we have t04 = − sin θ cos θ (C12 1 + C11 2 + C12 3 ) + cos2 θ − sin2 θ C44 4 + sin θ cos θ (C12 1 + C12 2 + C11 3 ) = sin θ cos θ (C11 − C22 ) (3 − 2 ) + cos2 θ − sin2 θ C44 4 Comparing the expressions for t04 , we find 2C44 = C11 − C22 Problem 6.15 If the axis which makes equal angles with the coordinate axes is an axis of elastic symmetry of order N = 3, show that there are twelve independent elastic constants and that the elastic

161

Chapter 6 Solutions matrix has the form [Cαβ ] =

C11 C13 C12 C41 C43 C42

C12 C11 C13 C42 C41 C43

C13 C12 C11 C43 C42 C41

Solution 2π For N = 3, we have N = = 3 and θ = 120◦ . θ 0 [aij ] = 0 1

C14 C16 C15 C44 C46 C45

C15 C14 C16 C45 C44 C46

C16 C15 C14 C46 C45 C44

.

The transformation matrix is 0 1 0

1 0 0

This gives

0 0 1

1 0 0

t1 0 1 t6 t5 0

t6 t2 t4

t5 0 t4 1 t3 0

0 0 1

1 t2 0 = t4 0 t6

t4 t3 t5

t6 t5 t1

Thus we have t01 = t2 , and

01 = 2 ,

Now

t02 = t3 ,

t03 = t1 ,

t04 = t5 ,

t05 = t6 ,

02 = 3 ,

03 = 1 ,

04 = 5 ,

05 = 6 ,

t06 = t4 06 = 4

t01 = C11 01 + C12 02 + C13 03 + C14 04 + C15 05 + C16 06

and t2 = C21 1 + C22 2 + C23 3 + C24 4 + C25 5 + C26 6 Using the strain transformations, we have t2 = C21 03 + C22 01 + C23 02 + C24 06 + C25 04 + C26 05 Equating t01 = t2 yields C11 = C22 ,

C12 = C23 ,

C13 = C21 ,

C14 = C25 ,

C15 = C26 ,

C16 = C24

C22 = C33 ,

C23 = C31 ,

C24 = C35 ,

C25 = C36 ,

C26 = C34

C32 = C13 ,

C33 = C11 ,

C34 = C15 ,

C35 = C16 ,

C36 = C14

C42 = C53 ,

C43 = C51 ,

C44 = C55 ,

C45 = C56 ,

C46 = C54

Likewise from t02 = t3 C21 = C32 , from t03 = t1 C31 = C12 , from t04 = t5 C41 = C52 ,

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Continuum Mechanics for Engineers

from t05 = t6 C51 = C62 ,

C52 = C63 ,

C53 = C61 ,

C54 = C65 ,

C55 = C66 ,

C56 = C64

C62 = C43 ,

C63 = C41 ,

C64 = C45 ,

C65 = C46 ,

C66 = C44

from t06 = t4 C61 = C42 ,

Problem 6.16 For an elastic body whose x3 axis is an axis of elastic symmetry of order N = 6, show that the nonzero elastic constants are C11 = C22 , C33 , C55 = C44 , C66 = 12 (C11 − C12 ), and C13 = C23 . Solution 2π For N = 6, we have N = = 6 and θ = 60◦ . The transformation matrix for the x3 as a θ symmetry axis is √ 1 2 0 2 2√ 1 [aij ] = − 2 0 2 2 0 0 1 and T 0 = AT AT . This yields t01 t06 t05

t06 t02 t04

√ t1 + 3t2√ + 2 3t6 t05 √ 1 t04 = − 3t√ 3t2 − 2t6 1+ 4 t03 2 3t4 + 2t5

√ √ − 3t1 + 3t2√− 2t6 3t1 + t2 −√2 3t6 2t4 − 2 3t5

√ 2 3t4 + √2t5 2t4 − 2 3t5 4t3

In a similar manner 0 = AAT . This gives 01 06 /2 05 /2

√ 2 + 2 36 06 /2 05 /2 1 + 62 √ √ 1 02 04 /2 = −2 3√ 1 + 2 32 − 26 8 04 /2 03 2 34 + 25

√ √ −2 31 + 2 3√ 2 − 26 61 + 22 − √ 2 36 24 − 2 35

√ 2 34 + √25 24 − 2 35 83

Now t01 = C11 01 + C12 02 + C13 03 + C14 04 + C15 05 + C16 06 1 √ √ 1 = C11 21 + 62 + 2 36 + C12 61 + 22 − 2 36 + C13 3 8 8 1 √ 1 √ √ √ 1 + C14 24 − 2 35 + C15 2 34 + 25 + C16 −2 31 + 2 32 − 26 4 4 4 Also √ 1 t1 + 3t2 + 2 3t6 4 1 = (C11 1 + C12 2 + C13 3 + C14 4 + C15 5 + C16 6 ) 4 3 + (C21 1 + C22 2 + C23 3 + C24 4 + C25 5 + C26 6 ) 4 √ 3 (C61 1 + C62 2 + C63 3 + C64 4 + C65 5 + C66 6 ) + 2

t01 =

163

Chapter 6 Solutions Equating coefficients on i with Cij = Cji , we find √ √ for 1 2√3C16 + 3C12 = 3C12 − 2√3C16 for 2 2 3C26 + 3C22√= 3C11 + 2 3C16 for 3 √ −3C13 + 2 3C36 = 3C√23 for 4 2√ 3C46 + 3C24 = C14 √ + 2 3C15 for 5 2 3C√56 + 3C25 = −2 √3C14 + C√15 for 6 3C16 + 2 3C66 + 3C26 = 3C11 − 3C12

Continue in the same way for t02 = t2 , t03 = t3 , etc. to obtain 36 equations (not all independent) in the 21 unknown Cij s. Solve to verify that all coefficients are zero except C11 = C22 ,

C13 = C21 ,

C33 ,

C44 = C55 ,

C66 =

1 (C11 − C12 ) 2

Problem 6.17 Develop a formula in terms of the strain components for the strain energy density W for the case of an orthotropic elastic medium. Answer W

1 (C11 1 + 2C12 2 2 1 + 2 C33 23 + C44 24

=

1 (C22 2 2 2 C66 6

+ 2C13 3 ) 1 + +

C55 25

+

+ 2C23 3 ) 2

Solution The strain energy is W=

tα α 1 = [(C11 1 + C12 2 + C13 3 ) 1 + (C21 1 + C22 2 + C23 3 ) 2 2 2 + (C31 1 + C32 2 + C33 3 ) 3 + C44 24 + C55 25 + C66 26

where the constitutive relation for an orthotropic material was used. Collecting terms gives 1 {[C11 1 + (C12 + C21 ) 2 + (C13 + C31 ) 3 ] 1 2 + [C22 2 + (C23 + C32 ) 3 ] 2 + C33 23 + C44 24 + C55 25 + C66 26

W=

Since Cij = Cji , we have 1 1 (C11 1 + 2C12 2 + 2C13 3 ) 1 + (C22 2 + 2C23 3 ) 2 2 2 1 2 2 2 2 + C33 3 + C44 4 + C55 5 + C66 6 2

W=

Problem 6.18 Show that, for an elastic continuum having x1 as an axis of elastic symmetry of order N = 2, the strain energy density has the same form as for a continuum which has an x2 x3 plane of elastic symmetry.

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Continuum Mechanics for Engineers

Answer 2W

= (C11 1 + 2C12 2 + 2C13 3 + 2C14 4 ) 1 + (C22 2 + 2C23 3 + 2C24 4 ) 2 + (C33 3 + 2C34 4 ) 3 + C44 24 + (C55 5 + 2C56 6 ) 5 + C66 26

Solution 2π For N = = 2, we have θ = π and θ

0 0 −1 0 0 −1

1 [aij ] = 0 0 From Problem 6.13,

and

t0α = tα

for

(α = 1, 2, 3, 4)

and t0α = −tα

for

(α = 5, 6)

0α = α

for

(α = 1, 2, 3, 4)

and 0α = −α

for

(α = 5, 6)

For the x2 x3 plane of symmetry, we have −1 [aij ] = 0 0

0 1 0

0 0 −1

Thus we can show

and

t0α = tα

for

(α = 1, 2, 3, 4)

and t0α = −tα

for

(α = 5, 6)

0α = α

for

(α = 1, 2, 3, 4)

and 0α = −α

for

(α = 5, 6)

In each case, we have C15 = C16 = C25 = C26 = C35 = C45 = C46 = 0 with Cαβ = Cβα (See Problem 6.13) This gives 1 {[C11 1 + 2C12 2 + 2C13 3 + 2C14 4 ] 1 + [C22 2 + 2C23 3 + 2C24 4 ] 2 2 + [C33 3 + 2C34 4 ] 3 + C44 24 + [C55 5 + 2C56 6 ] 5 + C66 26

W=

Problem 6.19 Let the stress field for a continuum be given by x1 + x2 t12 x1 − x2 [tij ] = t12 0 0

0 0 x2

where t12 is a function of x1 and x2 . If the equilibrium equations are satisfied in the absence of body forces and if the stress vector on the plane x1 = 1 is given by t(e^1 ) = ^2 , determine t12 as a function of x1 and x2 . ^1 + (6 − x2 ) e (1 + x2 ) e Answer

165

Chapter 6 Solutions t12 = x1 − x2 + 5

Solution The equilibrium equations are tij,j = 0 t11,1 + t12,2 + t13,3 = 0

for

i=1

t12,1 + t22,2 + t23,3 = 0

for

i=2

t13,1 + t23,2 + t33,3 = 0

for

i=3

The resulting equations are ∂t12 +0=0 ∂x2

and

t12 = −x2 + f1 (x1 ) + C1

∂t12 −1+0=0 ∂x1 0+0+0=0

and

t12 = x1 + f2 (x2 ) + C2

1+

Now we find df2 ∂t12 =1+0+ + 0 = 0 and f2 = −x2 ∂x2 dx2 ∂t12 df1 − 1 = −1 + 0 + + 0 = 0 and f1 = x1 ∂x1 dx1

1+

This gives t12 = x1 − x2 + C On the plane x1 = 1, we have 1 + x2 1 − x2 + C 1 − x2 + C 1 − x2 0 0

0 1 1 + x2 1 + x2 0 0 = 1 − x2 + C = 6 − x2 x2 0 0 0

Equating, we have 1 − x2 + C = 6 − x2 and C = 5. Problem 6.20 Invert Eq 6.45b to obtain Hooke’s law in the form ν kk δij tij = 2G ij + 1 − 2ν which, upon combination with Eqs 6.44 and 6.43, leads to the Navier equation 1 G ui,jj + uj,ij + ρbi = 0 . 1 − 2ν This equation is clearly indeterminate for ν = 0.5. However, show that in this case Hooke’s law and the equilibrium equations yield the result 1 Gui,jj + tjj,i + ρbi = 0 . 3

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Continuum Mechanics for Engineers

Solution Eq 6.45b is Eij = (1 + ν) tij − νδij tkk and Eii = (1 − 2ν) tkk This gives E tij = 1+ν Now

ij +

ν δij kk 1 − 2ν

tij,j + ρbi = 2G ij,j +

We have 2ij,j = ui,jj + uj,ij

= 2G ij +

ν δij kk,j 1 − 2ν and

ν δij kk 1 − 2ν

+ ρbi = 0

kk,j = uk,kj

and the Navier equations are 2ν 1 uj,ji + ρbi = G ui,jj + uj,ji + ρbi = 0 G ui,jj + uj,ij + 1 − 2ν 1 − 2ν For ν = 0.5, this result is indeterminate. However, from Eii = (1 − 2ν) tkk we find ii = ui,i = 0. Hooke’s law gives Eij,j =

E (ui,jj + uj,ij ) = (1 + ν) tij,j − νδij tkk,j 2

From the equilibrium equations and ui,i = 0, we have ν E ui,jj = −ρbi − tjj,i 2 (1 + ν) 1+ν and Gui,jj + since

ν 1 tjj,i + ρbi = Gui,jj + tjj,i + ρbi = 0 1+ν 3

v 0.5 1 = = for ν = 0.5. 1+ν 1.5 3

Problem 6.21 Let the displacement field be given in terms of some vector qi by the equation ui =

2 (1 − ν) qi,jj − qj,ji . G

Show that the Navier equation (Eq 6.49) is satisfied providing bi ≡ 0 and qi is bi-harmonic so that qi,jjkk = 0. If q1 = x2 /r and q2 = −x1 /r where r2 = xi xi , determine the resulting stress field. Answer t11 = −t22 = 6QGx1 x2 /r5 ; t33 = 0 t12 = t21 = 3QG x22 − x21 /r5 t13 = t31 = 3QGx2 x3 /r5 ; t23 = t32 = −3QGx1 x2 /r5 ; where Q = 4(1 − ν)/G

167

Chapter 6 Solutions Solution The Navier equation is (Eq 6.49) µui,kk + (λ + µ) uk,ki + ρbi = 0 For the given displacement field, the Navier equation is 2 (1 − ν) qi,jjkk − qj,jikk + (λ + µ) since G = µ. Since 2 (1 − ν) =

[2 (1 − ν) qj,kkji − qk,kjji ] + ρbi = 0 µ

(λ + 2µ) and if qi,jjkk = 0 for a bi-harmonic function (λ + µ)

qi,jikk

(−µ + λ + 2µ − λ − µ) + ρbi = 0 µ

for zero body forces. Thus the Navier equation is satisfied. 2 (1 − ν) qi,jj − qj,ji x2 x1 If q1 = and q2 = − where r2 = xi xi = x21 +x22 +x23 and ui = , r r G then 2 (1 − ν) qi,kkj − qk,kij ui,j = G To compute the stresses, we use the strains and Hooke’s law. Now 2 (1 − ν) −2r−3 + 6x22 r−5 2 (1 − ν) 6x1 x2 −2 (1 − ν) 6x2 x3 u1,1 = u1,2 = u1,3 = Gr5 −3 G Gr5 2 (1 − ν) 2r − 6x21 r−5 −2 (1 − ν) 6x1 x2 2 (1 − ν) 6x1 x3 u2,2 = u2,3 = u2,1 = G Gr5 Gr5 u3,1 = 0 u3,2 = 0 u3,3 = 0 Hooke’s law is tij = 2µij + λδij kk = G (ui,j + uj,i ) + λδij tkk . The stresses are 24 (1 − ν) x1 x2 6GQx1 x2 = r5 r5 −6GQx1 x2 −4 (1 − ν) 6x1 x2 = = 2Gu2,2 + λuk,k = 2Gu2,2 = Gr5 r5 = 2Gu3,3 + λuk,k = 2Gu3,3 = 0 12G (1 − ν) x22 − x21 3GQ x22 − x21 = G (u1,2 + u2,1 ) = t21 = = Gr5 r5 −12G (1 − ν) x2 x3 −3GQx2 x3 = = G (u1,3 + u3,1 ) = t31 = Gr5 r5 12G (1 − ν) x1 x3 3GQx1 x3 = G (u2,3 + u3,2 ) = t32 = u2,3 = = 5 Gr Gr5

t11 = 2Gu1,1 + λuk,k = 2Gu1,1 = t22 t33 t12 t13 t23 where Q =

4 (1 − ν) . G

Problem 6.22 If body forces are zero, show that the elastodynamic Navier equation (Eq 6.55) will be satisfied by the displacement field ui = φ,i + εijk ψk,j provided the potential functions φ and ψk satisfy the three-dimensional wave equation.

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Solution The elastodynamic Navier equation (Eq 6.55) is µui,kk + (λ + µ) uk,ki = ρu¨ i with zero body forces. Then ui,jj = φ,ijj + εiqk ψk,qjj uj,ji = φ,jji + εiqk ψk,qji and

¨ ,i + εiqk ψ ¨ k,q u¨ i = φ

Combining, we obtain ¨ ,i + εiqk ψ ¨ k,q µ (φ,ijj + εiqk ψk,qjj ) + (λ + µ) φ,jji = ρ φ

since εiqk ψk,qji = 0. Now ¨ + εiqk µψk,jj − ρψ ¨k (λ + 2µ) φ,jj − ρφ =0 ,i ,q This is true if each term is zero individually. This is the three-dimensional wave equation. Problem 6.23 Show that, for plane stress, Hooke’s law Eq 6.120a and Eq 6.121 may be expressed in terms of the Lam´e constants λ and µ by λ 1 tij − δij tkk (i, j, k = 1, 2) , ij = 2µ 3λ + 2µ λ tii (i = 1, 2) . kk = − 2µ (3λ + 2µ)

Solution Equation 6.120a can be written as ij = Now

and

1 [(1 + ν) tij − νδij tkk ] E

(i, j, k = 1, 2)

(1 + ν) 1 ν λ = and = , then E 2µ E 2µ (3λ + 2µ) 1 λ ij = tij − δij tkk 2µ 3λ + 2µ

(i, j, k = 1, 2)

ν λ 33 = − tkk = − tkk E 2µ (3λ + 2µ)

(k = 1, 2)

Problem 6.24 For the case of plane stress, let the stress components be defined in terms of the function φ = φ(x1 , x2 ), known as the Airy stress function, by the relationships, t11 = φ,22 ,

t22 = φ,11 ,

t12 = −φ,12 .

169

Chapter 6 Solutions

Show that φ must satisfy the biharmonic equation ∇4 φ = 0 and that, in the absence of body forces, the equilibrium equations are satisfied identically by these stress components. If φ = Ax31 x22 − Bx51 where A and B are constants, determine the relationship between A and B for this to be a valid stress function. Answer A = 5B

Solution The compatibility equation (Eq 6.119) is 11,22 + 22,11 = 212,12 Using Eq 6.120a, this can be written 1 [t11,22 − νt22,22 + t22,11 − νt11,11 − 2 (1 + ν) t12,12 ] = 0 E Substituting the stress function gives φ,2222 − νφ,1122 + φ,1111 − νφ,2211 − 2 (1 + ν) (−φ,1212 ) = 0 or φ,1111 + φ,2222 + 2φ,1212 − ν (φ,1122 + φ,2211 − 2φ,1212 ) = 0 This gives φ,1111 + φ,2222 + 2φ,1212 = ∇4 φ = 0 the bi-harmonic equation since the second term in parentheses is zero. The equilibrium equations are tij,j = 0

(i, j = 1, 2)

and t11,1 + t12,2 = φ,221 − φ,122 = 0 t21,1 + t22,2 = −φ,211 + φ,112 = 0 For φ = Ax31 x22 − Bx51 , we find φ,1111 = −120Bx1 φ,2222 = 0 φ,1212 = 12Ax1 and φ,1111 + φ,2222 + 2φ,1212 = −120Bx1 + 0 + 24Ax1 = 0 Thus A = 5B if the bi-harmonic equation is satisfied.

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Problem 6.25 Develop an expression for the strain energy density, W, for an elastic medium in (a) plane stress and (b) plane strain. Answer (a) W = t211 + t222 − 2νt11 t22 + 2 (1 + ν) t212 /2E (b) W = µ + 12 λ 211 + 222 + λ11 22 + 2µ212

Solution (a) For plane stress (Eq 6.120a), the strain energy is 1 1 tij ij = (t11 11 + t22 22 + 2t12 12 ) 2 2 1 2 = t − νt11 t22 + t222 − νt22 t11 + 2 (1 + ν) t212 2E 11

W=

(b) For plane strain (Eq 6.23), the strain energy is 1 2µ211 + λ (11 + 22 ) 11 + 2µ222 + λ (11 + 22 ) 22 + 4µ212 2 1 (2µ + λ) 211 + 222 + 2λ11 22 + 4µ212 = 2

W=

Problem 6.26 Show that φ = x41 x2 + 4x21 x32 − x52 is a valid Airy stress function, that is, that ∇4 φ = 0, and compute the stress tensor for this case assuming a state of plane strain with ν = 0.25. Answer 24x21 x2 − 20x32 [tij ] = −4x31 − 24x1 x22 0

−4x31 − 24x1 x22 12x21 x2 + 8x32 0

1 4

0 0 2 3 9x1 x2 − 3x2

Solution The Airy stress function must satisfy the bi-harmonic equation ∇4 φ = φ,1111 + φ,2222 + 2φ,1212 = 0 Now

φ,1111 = 24x2 φ,2222 = −120x2 φ,1212 = 48x2

and ∇4 φ = 24x2 − 120x2 + 2 (48x2 ) = 0 The stresses are

t11 t22 t12 t33

= φ,22 = 24x21 x2 − 20x32 = φ,11 = 12x21 x2 + 8x32 = −φ,12 = −4x31 − 24x1 x22 = ν (t11 + t22 ) = 14 9x21 x2 − 3x32

171

Chapter 6 Solutions

Problem 6.27 Verify the inversion of Eq 6.147 into Eq 6.148. Also, show that the two equations of Eq 6.149 may be combined to produce Eq 6.148. Solution Equation 6.147 is ij =

1+ν ν tij − δij tkk + α (θ − θ0 ) δij E E

and ii =

1+ν ν tii − 3 tkk + 3α (θ − θ0 ) E E

This gives tkk = Then tij =

E [ii − 3α (θ − θ0 )] 1 − 2ν

E [(1 − 2ν) ij + νδij kk − (1 + ν) δij α (θ − θ0 )] (1 + ν) (1 − 2ν)

This is the desired result. Equation 6.149a is Sij = or

E ηij (1 + ν)

E 1 E tij − δij tkk = ηij = (1 + ν) (1 + ν) 3

1 ij − δij kk 3

This gives E tij = (1 + ν)

1 1 ij − δij kk + δij tkk 3 3

From Eq 6.149b, we have tkk = or

E [kk − 3α (θ − θ0 )] (1 − 2ν)

1 E 1 [kk − 3α (θ − θ0 )] ij − δij kk + δij (1 − 2ν) 3 3 E [(1 − 2ν) ij + νδij kk − (1 + ν) δij α (θ − θ0 )] = (1 + ν) (1 − 2ν)

tij =

E (1 + ν)

Problem 6.28 Develop appropriate constitutive equations for thermoelasticity in the case of (a) plane stress and (b) plane strain. Answer (a) ij = [(1 + ν)tij − νδij tkk ]/E + δij (θ − θ0 )α (i, j, k = 1, 2) 33 = νtii /E + α(θ − θ0 ) (i = 1, 2) (b) tij = λδij kk + 2µij − δij (3λ + 2µ)α(θ − θ0 ) (i, j, k = 1, 2) t33 = νtii − αE(θ − θ0 ) = λii − (3λ + 2µ)α(θ − θ0 ) (i = 1, 2)

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Solution (a) From Eq 6.147, we have ij =

ν 1+ν tij − δij tkk + α (θ − θ0 ) δij E E

(i, j, k = 1, 2)

and

ν (t11 + t22 ) + α (θ − θ0 ) E (b) From Eq 6.23 and Eq 6.147, we have 33 = −

tij = 2µij + λδij kk − (3λ + 2µ) δij α (θ − θ0 )

(i, j, k = 1, 2)

and t33 = ν (t11 + t22 ) − α (θ − θ0 ) E = λ (11 + 22 ) − (3λ + 2µ) α (θ − θ0 ) Problem 6.29 Consider the Airy stress function φ5 = D5 x21 x32 + F5 x52 . (a) Show that for this to be valid stress function, F5 = −D5 /5. (b) Construct the composite stress function φ = φ5 + φ3 + φ2 where

1 1 1 φ = D5 x21 x32 − x52 + B3 x21 x2 + A2 x21 . 5 2 2

For this stress function show that the stress components are t11 = D5 6x21 x2 − 4x32 , t22 = 2D5 x32 + B3 x2 + A2 , t12 = −6D5 x1 x22 − B3 x1 .

Solution (a) Take derivatives and substitute into the biharmonic φ,1111 = 0 φ,1122 = 120F5 x2 φ,1122 = 12D5 x2 φ,1111 + 2φ,1122 + φ,1122 = 120F5 + 24D5 = 0 (b) Simply take partial derivatives t11 = φ,22 , t22 = φ,11 and t12 = −φ,12 . Problem 6.30 A rectangular beam of width unity and length 2L carries a uniformly distributed load of q lb/ft as shown. Shear forces V support the beam at both ends. List the six boundary conditions for this beam the stresses must satisfy using stresses determined in Problem 6.30.

173

Chapter 6 Solutions q [lb/ft]

x2

V

V

c c

x1

O

L

L

Answer 1. t22 = −q at x2 = +c 2. t22 = 0 at x2 = −c 3. t12 = 0 at x2 = ±c R+c 4. −c t12 dx2 = qL at x1 = ±L R+c 5. −c t11 dx2 = 0 at x1 = ±L R+c at x1 = ±L 6. −c t11 x2 dx2 = 0

Solution See Answer. Problem 6.31 Using boundary conditions 1, 2, and 3 listed in Problem 6.31, show that the stresses in Problem 6.30 require that A2 = −

q , 2

B3 = −

3q , 4c

D5 =

q . 8c3

Thus, for the beam shown the stresses are q 2 3 2 t11 = x1 x2 − x2 , 2I 3 2 3 q 1 3 2 x − c x2 − c , t22 = 2I 3 2 3 q t12 = − x1 x22 − c2 x1 , 2I where I = 23 c3 is the plane moment of inertia of the beam cross section. Solution Take partial derivatives t11 = φ,22 , t22 = φ,11 and t12 = −φ,12 to get 2D3 c3 + B3 c + A2 = −q −2D3 c3 − B3 c + A2 = 0 6D3 c2 + B3 = 0

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Continuum Mechanics for Engineers

Solve to get solution. Problem 6.32 Show that, using the stresses calculated in Problem 6.32, the boundary conditions 4 and 5 are satisfied, but boundary condition 6 is not satisfied. Solution

Problem 6.33 Continuing Problems 6.32 and 6.33, in order for boundary condition 6 to be satisfied an additional term is added to the stress function, namely φ3 = D3 x32 . Show that, from boundary condition 6, D3 =

3q 4c

1 L2 − 2 15 6c

,

so that finally t11 =

q 2 6 x21 − x22 + c2 − L2 x2 . 21 3 15

Solution

Problem 6.34 Show that for the shaft having a cross section in the form of an equilateral triangle the warping function is ψ (x1 , x2 ) = λ x32 − 3x21 x2 . Determine (a) the constant λ in terms of the shaft dimension, (b) the torsional rigidity K, (c) the maximum shearing stress.

x2 a

2a

O

x1 Mt

175

Chapter 6 Solutions Answer 1 6a√ 9G 3a4 (b) K = 5 3Mt a (c) t23 |max = at x1 = a, x2 = 0 2K (a) λ = −

Solution (a) The warping function must satisfy the harmonic equation ∇2 ψ = ψ,11 + ψ,22 = 0 The function ψ = λ x2 3 − 3x1 2 x2 satisfies this equation. The constant λ is found from the boundary condition dψ = ψ,1 n1 + ψ,2 n2 = x2 n1 + x1 n2 dn ^=^ On x1 = a, the normal is n e1 . This gives ψ,1 = x2 or on x1 = a

λ (−6x1 x2 ) = x2 Then λ=−

1 6a

It can be verified that this warping function also satisfies the boundary condition on the inclined boundaries. (b) The torsional rigidity is, from Eq. 6.79, K

= 2G = 2G

Ra

R(x1 +2a)√3

−2a

Ra

0

R(x1 +2a)√3

−2a

Ra

0

"

= 2G −2a √ 9 3Ga4 = 5

x21

x21 + x22 + x1 ψ,2 − x2 ψ,1 dx2 dx1 x21 + x22 + 3λx1 x22 − x21 + 6λx2 (x1 x2 ) dx2 dx1

x1 + 2a √ 3

1 + 3

x1 + 2a √ 3

3

x1 − 6a

x1 + 2a √ 3

x1 + 2a √ 3

!

2 −

3x21

(c) The shear stresses are t13 t23

= Gθ (ψ,1 − x2 ) = Gθ (−6λx1 x2 − x2 ) = Gθ (ψ,2 − x1 ) = Gθ 3λ x22 − x21 − x1

The stress is maximized at the locations closest to the centroid. The easiest location for calculation is at x2 = 0 and x1 = a. The result follows.

x1 − 3a

x1 + 2a √ 3

3 # dx1

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Continuum Mechanics for Engineers

Alternatively, we can write out Φ and differentiate √ √ Φ = k(x1 − a)(x1 − x2 3 + 2a)(x1 + x2 3 + 2a) h i = k(x1 − a) (x1 + 2a)2 − 3x22 Differentiate to find k from ∇2 Φ = −2Gθ. h i Φ,1 = k (x1 + 2a)2 − 3x22 + k(x1 − a)2(x1 + 2a) Φ,11

= 2k(x1 + 2a) + 2k(x1 + 2a) + 2k(x1 − a)

Φ,11

= 6kx1 + 6ka

Φ,2 Φ,22

= k(x1 − a)(−6x2 ) = −6kx2 (x1 − a) = −6kx1 + 6ka

So ∇2 Φ = Φ,11 + Φ,22 = 6kx1 + 6ka − 6kx1 + 6ka = 12ka = −2Gθ giving k=− The moment is found from Z Mt = 2 ΦdA A

Za

Z (x1 +2a)/√3

1 6a

Gθ 6a

h i (x1 − a) (x1 + 2a)2 − 3x22 dx2 dx1 −2a 0 " 3 # Za 1 x1 + 2a 2 x1 + 2a √ √ = 4Gθ − − (x1 − a) (x1 + 2a) dx1 6a 3 3 −2a = 2Gθ · 2

−

This form is easily evaluated using Matlab below. >> syms a x >> f = -(x-a)*((x+2*a)^2*((x+2*a)/sqrt(3))-(x+2*a)^3/(3*sqrt(3)))/(6*a) f = 1/27*(-x+a)*(x+2*a)^3*3^(1/2)/a >> m = int(f,-2*a,a) m = 9/20*3^(1/2)*a^4 So and Compute the shear stresses t31 t32

√ 9 3Gθa4 Mt = 5 √ 9 3Ga4 K = Mt /θ = 5

x2 (x1 − a) a i Gθ h (x1 + 2a)2 − 3x22 + (x1 − a)2(x1 + 2a) = −Φ,1 = − 6a Gθ 2 =− 3x1 + 6ax1 − 3x22 6a = Φ,2 = −Gθ

177

Chapter 6 Solutions

Maximum shear stress occurs at three locations nearest the centroid. The easiest to compute is at x1 = 0 , x2 = a. Here, t31 = 0 and t32 = 9Gθa/6. So √ 5 3Mt 3Gθa 3GMt a 3GMt a 5 √ |t23 |max = = = = 2 2K 2 18a3 9 3Ga4 Problem 6.35 Consider the Galerkin vector that is the sum of three double forces, that is, let

x3

^r e ^θ e ^ψ e

ψ r O

x2 θ x1 x2 x ^2 + 3 e ^3 e r r r where B is a constant and r2 = xi xi . Show that the displacement components are given by F=B

x

1

^1 + e

2B (1 − 2ν) xi . r3 Using the sketch above, (spherical coordinates) observe that the radial displacement ur (subscript r not a summed indice, rather indicating the radial component of displacement) ui xi ur = , r ui = −

and show that ur = −2B (1 − 2ν) r−2 . Also, show that uψ = uθ = 0. Thus ∂ur 4B (1 − 2ν) 2B (1 − 2ν) = and ψ = θ = − , 3 ∂r r r3 so that the cubical dilatation is r + ψ + θ = 0. From Hooke’s law r =

tij = λδij uk,k + 2µ (ui,j + uj,i ) which reduces here to tij = 2µij so that trr =

8BG (1 − 2ν) , r3

and tψψ = tθθ = −

4BG (1 − 2ν) . r3

Chapter 7 Solutions

Problem 7.1 Introduce the stress deviator Sij and the viscous stress deviator τ∗ij = τij − 31 δij τkk into Eq 7.5 to prove that Sij = τ∗ij . Solution Eq 7.5 is tij = −pδij + τij and tkk = −3p + τkk The stress deviator is Now

1 Sij = tij − δij tkk 3

δij τkk 1 Sij + δij tkk = tij = −pδij + τij = −pδij + τ∗ij + 3 3 τkk tkk Sij + δij +p− = τ∗ij 3 3

or

The term in parentheses is zero, and the result is shown Problem 7.2 Determine an expression for the stress power (a) tij dij and (b) τij dij for a Newtonian fluid. First, show that 2 ∗ ∗ τij = κ − µ δij dkk + 2µ∗ dij . 3 Answer (a) tij dij = −pdii + κ∗ dii djj + 2µ∗ βij βij (b) τij dij = κ∗ (tr D)2 + 2µ∗ βij βij

Solution (a) The stress is from Eq 7.10 tij = −pδij + λ∗ δij dkk + 2µ∗ dij with (Eq 7.12a) κ∗ =

1 (3λ∗ + 2µ∗ ) 3 179

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Continuum Mechanics for Engineers

gives 2 tij dij = −pδij dij + κ∗ − µ∗ δij dkk dij + 2µ∗ dij dij 3 1 = −pdjj + κ∗ dkk djj + 2µ∗ dij dij − dkk djj 3 ∗ ∗ = −pdjj + κ dkk djj + 2µ βij βij where βij = dij − 13 δij dkk (b) From Eq 7.5 tij = −pδij + τij and Eq 7.10 and Eq 7.12a yields 2 ∗ ∗ τij = κ − µ δij dkk + 2µ∗ dij 3 Now 2 κ∗ − µ∗ δij dkk dij + 2µ∗ dij dij 3 1 ∗ ∗ = κ dkk dii + 2µ dij dij − dkk djj 3

τij dij =

= κ∗ dkk dii + 2µ∗ βij βij = κ∗ tr (D)2 + 2µ∗ βij βij Problem 7.3 Determine the constitutive equation for a Newtonian fluid for which Stokes condition holds, that is, for κ∗ = 0. Answer tij = −pδij + 2µ∗ βij

Solution From Eq 7.10

tij = −pδij + λ∗ δij dkk + 2µ∗ dij

with (Eq 7.12a) κ∗ =

1 (3λ∗ + 2µ∗ ) 3

Now 2 tij = −pδij + κ∗ − µ∗ δij dkk + 2µ∗ dij 3 1 = −pδij + κ∗ δij dkk + 2µ∗ dij − δij dkk 3 ∗ = −pδij + 2µ βij if κ∗ = 0 Problem 7.4 Develop an expression of the energy equation for a Newtonian fluid assuming the heat conduction follows Fourier’s law.

181

Chapter 7 Solutions Answer ρu˙ = −pvi,i + λ∗ vi,i vj,j + 21 µ∗ (vi,j + vj,i ) + κ∗ θ,ii + ρr

Solution From Eq 5.67

ρu˙ = tij dij + ρr − qi,i

From Eq 7.10 ρu˙ = (−pδij + λ∗ δij dkk + 2µ∗ dij ) dij + ρr + κθ,ii = −pdii + λ∗ dkk dii + 2µ∗ dij dij + ρr + κθ, ii but dij = (vi,j + vj,i ) /2 and so ρu˙ = −pdii + λ∗ vk,k vi,i +

µ∗ (vi,j + vj,i )2 + κθ,ii 2

Problem 7.5 The dissipation potential Ψ for a Newtonian fluid is defined as a function of D and β by Ψ=

1 ∗ κ djj dii + µ∗ βij βij , 2

2 where κ∗ = λ∗ + µ∗ . 3

Show that ∂Ψ/∂dij = τij . Solution The derivative of Ψ is ∂Ψ κ∗ ∂ ∂ (dii djj ) + µ∗ (βij βij ) = ∂dpq 2 ∂dpq ∂dpq ∂βij κ∗ (δip δiq djj + δjp δjq dii ) + 2µ∗ βij = 2 ∂dpq Now βij

∂βij ∂ 1 = βij dij − δij dkk ∂dpq ∂dpq 3 δij δpq = βij δip δjq − δpk δqk = βpq − βii 3 3 = βpq

since βii = 0. This gives ∂Ψ = κ∗ δpq djj + 2µ∗ βpq ∂dpq δpq = κ∗ δpq djj + 2µ∗ dpq − djj 3 2 = κ∗ − µ∗ δpq djj + 2µ∗ dpq = τpq 3 This last result follows from Problem 7.2

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Problem 7.6 Verify the derivation of the Navier-Stokes equations for a Newtonian fluid as given by Eq 7.25. Solution Eq 7.25 is ρv˙ i = ρbi − p,i + (λ∗ + µ∗ ) vj,ji + µ∗ vi,jj The constitutive equation is, Eq 7.10 tij = −pδij + λ∗ δij dkk + 2µ∗ dij and the equation of motion is Eq 7.17 tij,j + ρbi = ρv˙ i Computing the divergence of the stress tij,j = −p,j δij + λ∗ δij dkk,j + 2µ∗ dij,j = −p,i + λ∗ vk,ki + µ∗ (vi,ji + vj,ii ) Changing dummy indices and collecting terms yield the desired result −p,i + (λ∗ + µ∗ ) vk,ki + µ∗ vj,ii + ρbi = ρv˙ i Problem 7.7 Consider a two-dimensional flow parallel to the x2 x3 plane so that v1 = 0 throughout the fluid. Assuming that an incompressible, Newtonian fluid undergoes this flow, develop a Navier-Stokes equation and a continuity equation for the fluid. Answer (Navier-Stokes) (Continuity)

ρv˙ i = ρbi − p,i + µ∗ vi,jj (i, j = 2, 3)

vi,i = 0 (i = 2, 3)

Solution From Eq 7.28 the Navier-Stokes equation is ρv˙ i = ρbi − p,i + µ∗ vi,jj Then p,1 = ρb1

for i = 1

ρv˙ i = ρbi − p,i + µ∗ vi,jj

for i, j = 2, 3

The continuity equation is, Eq 7.16 ρ˙ + ρvi,i = 0 where ρ˙ = 0 for an incompresible material, and vi,i = 0

for i = 2, 3

183

Chapter 7 Solutions

Problem 7.8 Consider a barotropic, inviscid fluid under the action of conservative body forces. Show that the material derivative of the vorticity of the fluid in the current volume V is Z Z d wi dV = vi wj nj dS . dt V S

Solution From Problem 5.5

d dt

Z

Z

V

wi dV =

S

1 εijk ak + wj vi nj dS 2

The equation of motion for an inviscid fluid is,Eq 7.29 ρv˙ i = ρbi − p,i where from Eq 7.34,bi = −Ω,i 7.35, P,i =

1 p,i and the equation of motion is ρ

ρak = −ρ (Ω + P),k Now

Z

Z Z 1 1 ρ εijk ak nj dS = εijk ak,j dV = − εijk (Ω + P),kj dV = 0 2 2 S V V 2 from the skew symmetry of εijk . Then Z Z d wi dV = vi wj nj dS dt V S Problem 7.9 Show that for an incompressible, inviscid fluid the stress power vanishes identically as one would expect. Solution The constitutive equation for an incompressible, inviscid fluid is tij = −pδij . The stress power is 1 tij dij = − pδij (vi,j + vj,i ) = −pvi,i = 0 2 since vi,i = 0. An alternate solution exploits the symmetry of the stress tensor tij dij =

1 tij (vi,j + vj,i ) = tij vi,j = −pδij vi,j = −pvi,i = 0 2

Problem 7.10 Show that the vorticity and velocity of a barotropic fluid of constant density moving under conservative body forces are related through the equation w ˙ i = wj vi,j . Deduce that for a steady flow of this fluid vj wi,j = wj vi,j . Solution From Problem 5.10

d dt

wi ρ

=

εijk ak,j + 2wj vi,j 2ρ

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Continuum Mechanics for Engineers

For a constant density 1 εijk ak,j + wj vi,j 2 For a barotropic fluid with conservative body forces, Eq 7.43 gives w ˙i=

ρv˙ i = (Ω + P),i and hence εijk ak,j = and

1 εijk (Ω + P),kj = 0 ρ

w ˙ i = wj vi,j

Expanding w ˙ j gives ∂wi + vj wi,j = wj vi,j ∂t and for steady flow vj wi,j = wj vi,j Problem 7.11 In terms of the vorticity vector w, the Navier-Stokes equations for an incompressible fluid may be written as ρv˙ i = ρbi − p,i − 2µ∗ εijk wk,j . Show that, for an irrotational motion, this equation reduces to the Euler equation ρv˙ i = ρbi − p,i .

Solution The equation of motion is, Eq 7.28 ρv˙ i = ρbi − p,i + µ∗ vi,jj From Eq 4.160 vi,jj = vj,ji − 2εijk wk,j and ρv˙ i = ρbi − p,i + µ∗ (vj,ji − 2εijk wk,j ) = ρbi − p,i − 2µ∗ εijk wk,j since vj,ji = 0 from incompressibility. Now for irrotational flow wi = 0 and ρv˙ i = ρbi − p,i This is Eq 7.29. Problem 7.12 Carry out the derivation of Eq 7.41 by combining the Euler equation with the continuity equation, as suggested in the text.

185

Chapter 7 Solutions Solution The Euler equation is ρv˙ i = ρ

∂vi + vj vi,j ∂t

= ρbi − p,i

and the continuity equation is ρ˙ + ρvi,i = For steady flow

∂ρ + vi p,i + ρvi,i = 0 ∂t

∂vi ∂ρ = = 0 and bi = 0. This gives ∂t ∂t

ρvj vi,j = −p,i (Euler)

and

vi p,i + ρvi,i = 0 (continuity)

For a barotropic flow p,i =

dp ρ,i = c2 ρ,i dρ

Substituting into the Euler equation gives ρvj vi,j + c2 ρ,i = 0 Multiplying by vi gives ρvi vj vi,j + c2 vi ρ,i = ρvi vj vi,j − c2 ρvi,i = 0 where the continuity equation was used. This gives ρ vi vj − c2 ρδij vi,j = 0 This is Eq 7.41. Problem 7.13 Consider the velocity potential φ = x2 x3 /r2 where r2 = x21 + x22 . Show that this satisfies the Laplace equation φ,ii = 0. Derive the velocity field and show that this flow is both incompressible and irrotational. Solution The individual terms of the Laplacian are ∂ x2 x3 ∂ x2 x3 ∂r ∂ = −2 3 φ,11 = ∂x1 ∂x1 r2 ∂x1 r ∂x1 Now

h i x i 1 ∂r ∂ h i (xj xj )1/2 = (2δij xj ) (xj xj )−1/2 = = ∂xi ∂xi 2 r

Thus ∂ h x2 x3 x1 i h x2 x3 x2 x3 x1 x1 i −2 3 = −2 4 + 8 ∂x r r r r5 r 12 −6 2 2 2 = 8x1 x2 x3 − 2x2 x3 x1 + x2 r = 6x1 − 2x32 x3 r−6

φ,11 =

Similarly φ,22 = −6x21 + 2x32 x3 r−6

and φ,33 = 0

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Continuum Mechanics for Engineers

so that φ,11 + φ,22 + φ33 = 0 Now the velocity is v1 = φ,1 =

−2x1 x2 x3 ; r4

v2 = φ,2 =

x3 x21 − x22 ; r4

v3 = φ,3 =

x2 r2

and the continuity equation is vi,i = v1,1 + v2,2 = v3,3 = φ,11 + φ,22 + φ,33 = φ,ii = 0 This is an incompressible flow. For irrotational motion ^i = εijk φ,kj e ^i = 0 curl v = εijk vk,j e Problem 7.14 If the equation of state of a barotropic fluid has the form p = λρk where k and λ are constants, the flow is termed isentropic. Show that the Bernoulli equation for a steady motion in this case becomes Ω+

1 kp + v2 = constant . (k + 1) ρ 2

Also, show that for isothermal flow the Bernoulli equation takes the form Ω+

p ln ρ 1 2 + v = constant . ρ 2

Solution The Bernoulli equation, Eq 7.44, is Z x2 x1

∂vi v2 dxi + + Ω + P = G (t) ∂t 2

and for steady flow becomes v2 + Ω + P = G0 2 Now

dp = λkρk−1 dρ

and

Zp P= p0

Thus

dp k = ρ k−1

p p0 − ρ ρ0

=

kp + C1 ρ (k − 1)

v2 kp +Ω+ = constant 2 ρ (k − 1)

For an isothermal flow dp = λdρ and Z Z dp λ p P= = dρ = λ ln ρ + C2 = ln ρ + C2 ρ ρ ρ

187

Chapter 7 Solutions and

v2 p ln ρ +Ω+ = constant 2 ρ

Problem 7.15 Derive Eq 7.44 by taking the scalar product of dxi (the differential displacement along a streamline) with Eq 7.43 and integrating along the streamline, that is, by the integration of Z x2 (v˙ i + Ω,i + P,i ) dxi . x1

Solution Eq 7.43 is v˙ i + Ω,i + P,i =

∂vi + vj vi,j ∂t

+ Ω,i + P,i = 0

Taking the scalar product with dxi and integrating gives Z x2 ∂vi + vj vi,j + Ω,i + P,i dxi = G (t) ∂t x1 or

Z x2

∂vi + vj vi,j dxi + Ω + P = G (t) ∂t x1 v v i i Along a streamline, dxi = ds and so vj vi,j dxi = vj vi,j ds = vi vi,j dxj = vi dvi v v and Z x2 v2 ∂vi dxi + + Ω + P = G (t) 2 x1 ∂t Problem 7.16 Verify that Eq 7.47 is the material derivative of Eq 7.46. Also, show that for a barotropic, inviscid fluid subjected to conservative body forces the rate of change of the circulation is zero (See Eq 7.47). Solution The material derivative of Eq 7.46 is I I ˙ΓC = d vi dxi = (v˙ i dxi + vi,q vi dxq ) dt from Eq 5.7. Now vi,q dxq = dvi and v˙ i = − (Ω + P),i for a barotropic, inviscid fluid. So that I I I 2 x1 ˙ΓC = (v˙ i dxi + vi dxi ) = − (Ω + P),i dxi + vi dvi = − (Ω + P) + v =0 2 x1 Problem 7.17 Determine the circulation ΓC around the square in the x2 x3 plane shown in the figure if the velocity field is given by ^2 + (x3 + x2 ) e ^3 . v = x3 − x22 e

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Continuum Mechanics for Engineers

x3 (1,1)

(−1,1)

x2

O

(−1,−1)

(1,−1)

Answer ΓC = 0

Solution Beginning at the lower right-hand corner, the circulation, ΓC , is I Z1 Z −1 Z −1 Z1 ΓC = vi dxi = v3 (1, x3 ) dx3 + v2 (x2 , 1) dx2 + v3 (−1, x3 ) dx3 + v2 (x2, − 1) dx2 Z1

−1

Z −1

−1

Z −1

1 − x22 , dx2 +

(x3 + 1) dx3 +

=

1

1

1

−1 − x22, dx2

(x3 − 1) dx3 + 1

−1

Z1 −1

8 4 =2− +2− =0 3 3 Alternatively, from Eq 7.46 with ni = n1 Z ZZ (ε123 v3,2 + ε132 v2,3 ) dx2 dx3 ΓC = ε1jk vk,j n1 dS = S ZZ (1 − 1) dx2 dx3 =

Chapter 8 Solutions

Problem 8.1 Referred to principal axes, the invariants of the Green deformation tensor CAB are I1 = λ21 + λ22 + λ23 , I2 = λ21 λ22 + λ21 λ23 + λ22 λ23 , I3 = λ21 λ22 λ23 . For an isotropic, incompressible material, show that I1 = λ21 + λ22 + I2 =

1 λ21 λ22

,

1 1 + 2 + λ21 λ22 . 2 λ1 λ2

Solution For an incompressible material, I3 = 1 = λ21 λ22 λ23

or

Then I1 = λ21 + λ22 + and I2 = λ21 λ22 + λ21 + λ22

1 λ21 λ22

λ23 =

1 λ21 λ22

1 λ21 λ22 =

1 1 + 2 + λ21 λ22 2 λ1 λ2

Problem 8.2 Derive the following relationships between invariants I1 , I2 , and I3 , and the deformation gradient, CAB : ∂I1 = δAB , ∂CAB ∂I2 (b) = CAB − I1 δAB , ∂CAB ∂I3 (c) = I3 C−1 AB . ∂CAB (a)

Solution

189

190

Continuum Mechanics for Engineers (a) For I1 = CMM and ∂I1 ∂CMM = = δMA δMB = δAB ∂CAB ∂CAB (b) For I2 =

(CMM CNN − CMN CMN ) and 1 ∂ ∂I2 (CMM CNN − CMN CMN ) = ∂CAB ∂CAB 2 1 = CNN δMA δMB − [CMM δNA δNB + CMN δAM δBN ] 2 1 = (I1 δAB + I1 δAB ) − CAB = I1 δAB − CAB 2

1 2

(c) For I3 = 16 εLMN εPQR CLP CMQ CNR and differentiating ∂I3 1 1 = εLMN εPQR δAL δBP CMQ CNR + εLMN εPQR CLP δAM δBQ CNR ∂CAB 6 6 1 + εLMN εPQR CLP CMQ δAN δBR 6 1 1 1 = εAMN εBQR CMQ CNR + εLAN εPBR CLP CNR + εLMA εPQB CLP CMQ 6 6 6 1 = εAMN εBQR CMQ CNR 2 where the property of εPOR and dummy indices have been used in the last step. This is the cofactor of CAB and, so from Eq 2.47, we have ∂I3 −1 = (det CAB ) C−1 AB = I3 CAB ∂CAB Problem 8.3 Use the definitions of I1 and I2 in terms of the principal stretches λ1 , λ2 , and λ3 to show ∂W ∂W 2 2 2 ∂W 2 = λ − λ3 + λ2 , (a) ∂λ1 λ1 1 ∂I1 ∂I2 ∂W ∂W 2 ∂W (b) = λ2 − λ23 + λ21 . ∂λ2 λ2 2 ∂I1 ∂I2

Solution (a) For an incompressible material W (I1 , I2 ), differentiating gives ∂W ∂I1 ∂W ∂I2 ∂W = + ∂λ1 ∂I1 ∂λ1 ∂I2 ∂λ1 2 ∂W 2 ∂W = 2λ1 − 3 2 + − 3 + 2λ1 λ22 ∂I1 ∂I2 λ1 λ2 λ 1 2 ∂W ∂W = λ2 − λ33 + λ22 λ1 1 ∂I1 ∂I2 where results from Problem 8.1 were used noting that λ21 λ22 λ23 = 1

191

Chapter 8 Solutions (b) Now ∂W ∂W ∂I1 ∂W ∂I2 = + ∂λ2 ∂I1 ∂λ2 ∂I2 ∂λ2 ∂W 2 2 ∂W = 2λ2 − 2 3 + − 3 + 2λ21 λ2 ∂I1 ∂I2 λ1 λ2 λ 2 ∂W ∂W 2 λ2 − λ33 + λ21 = λ2 2 ∂I1 ∂I2

Problem 8.4 Let W(I1 , I2 , I3 ) be the strain energy per unit volume for a homogeneous, isotropic material. Show that the first Piola-Kirchhoff stress components may be written as follows: PiA ≡

∂W ∂W ∂W ∂W −1 (Bij − I1 δij ) FjA + 2I3 =2 FiA + 2 F . ∂FiA ∂I1 ∂I2 ∂I3 iA

Solution Using the chain rule gives PiA ≡

∂W ∂W ∂I1 ∂W ∂I2 ∂W ∂I3 = + + ∂FiA ∂I1 ∂FiA ∂I2 ∂FiA ∂I3 ∂FiA

Each term must be evaluated in order. terms of CAB . We first calculate

From Problem 8.2, the invariants are given in

∂ ∂CMN ∂ ∂ ∂ (FjM FjN ) = = ∂FiA ∂CMN ∂FiA ∂CMN ∂FiA ∂ (δij δAM FjN + FjM δij δAN ) = ∂CMN ∂ (δAM FiN + FiM δAN ) = ∂CMN From Problem 8.2, we have ∂I1 ∂CMN ∂I1 = = δMN (δAM FiN + FiM δAN ) ∂FiA ∂CMN ∂FiA = FiA + FiA = 2FiA ∂I2 ∂CMN ∂I2 = = (I1 δMN − CMN ) (δAM FiN + FiM δAN ) ∂FiA ∂CMN ∂FiA = I1 FiA + I1 FiA − CAN FiN + CMA FiM = 2I1 FiA − FjA FjN FiN − FjM FjA FiM = 2 (I1 δij − Bij ) FjA and ∂I3 ∂I3 ∂CMN = = I3 C−1 MN (δAM FiN + FiM δAN ) ∂FiA ∂CMN ∂FiA −1 −1 = I3 C−1 AN FiN + I3 CMA FiM = 2I3 CAM FiM

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Continuum Mechanics for Engineers

−1 −1 −1 −1 −1 −1 since C−1 FiM = F−1 AM = CMA . Now CAM FiM = (FiA FiM ) iM FiA FiM = FiM δAM = FiA , so ∂I3 = 2I3 F−1 iA ∂FiA

This gives ∂W ∂W ∂I1 ∂W ∂I2 ∂W ∂I3 = + + ∂FiA ∂I1 ∂FiA ∂I2 ∂FiA ∂I3 ∂FiA ∂W ∂W ∂W −1 (I1 δij − Bij ) FjA + 2I3 =2 FiA + 2 F ∂I1 ∂I2 ∂I3 iA

PiA =

Problem 8.5 The Cauchy stress is given by tij =

1 FjA PiA . J

Start with the result of Problem 8.4 to show that 2 ∂W ∂W ∂W (Bik − I1 δik ) Bjk + I3 tij = Bij + δij . J ∂I1 ∂I2 ∂I3

Solution Forming the indicated product with the result of Problem 8.4 gives 2 ∂W 1 ∂W ∂W (Bik − I1 δik ) FjA FkA + I3 FjA PiA = FjA FiA + FjA F−1 iA J J ∂I1 ∂I2 ∂I3 ∂W 2 ∂W ∂W (Bik − I1 δik ) Bjk + I3 Bij + δij = J ∂I1 ∂I2 ∂I3

tij =

Problem 8.6 Assuming a strain energy of the form W = w (λ1 ) + w (λ2 ) + w

1 λ 1 λ2

for an isotropic, incompressible material, show that λ1

∂3 W ∂3 W = λ . 2 ∂λ21 ∂λ2 ∂λ22 ∂λ1

Solution Let g (λ1 , λ2 ) = λ21 λ22

−1

for an incompressible material, then

W = w (λ1 ) + w (λ2 ) + w (g (λ1, λ2 ))

193

Chapter 8 Solutions Now differentiating W gives ∂2 ∂3 W = ∂λ21 ∂λ2 ∂λ21

∂ [w (λ1 ) + w (λ2 ) + w (g (λ1, λ2 ))] ∂λ2 dw (λ2 ) ∂2 ∂ = 2 + w (g (λ1, λ2 )) dλ2 ∂λ2 ∂λ1 ∂ dw (λ2 ) ∂ ∂ = + w (g (λ1, λ2 )) ∂λ1 ∂λ1 dλ2 ∂λ2 ∂3 w (g (λ1 , λ2 )) = ∂λ21 ∂λ2

Using the chain rule gives ∂3 w (g (λ1 , λ2 )) ∂2 ∂w ∂g ∂ ∂2 w ∂g ∂g ∂w ∂2 g = = + ∂λ1 ∂g2 ∂λ1 ∂λ2 ∂g ∂λ1 ∂λ2 ∂λ21 ∂λ2 ∂λ21 ∂g ∂λ2 2 ∂3 w ∂g ∂g ∂2 w ∂g ∂2 g ∂2 w ∂2 g ∂g ∂w ∂3 g = +2 2 + + 2 3 2 ∂g ∂λ1 ∂λ2 ∂g ∂λ1 ∂λ1 ∂λ2 ∂g ∂λ1 ∂λ2 ∂g ∂λ21 ∂λ2 Similarly ∂3 W ∂3 w (g (λ1 , λ2 )) = 2 ∂λ1 ∂λ2 ∂λ1 ∂λ22 and ∂3 w (g (λ1 , λ2 )) ∂3 w = 2 ∂g3 ∂λ1 ∂λ2

∂g ∂λ2

2

∂g ∂2 w ∂g ∂2 g ∂2 w ∂2 g ∂g ∂w ∂3 g +2 2 + + 2 2 ∂λ1 ∂g ∂λ2 ∂λ2 ∂λ1 ∂g ∂λ2 ∂λ1 ∂g ∂λ22 ∂λ1

Now ∂g ∂ −2 −2 −2 = −2λ−3 λ λ = 1 λ2 ∂λ1 ∂λ1 1 2

and

∂g ∂ −2 −2 −3 = −2λ−2 λ λ = 1 λ2 ∂λ2 ∂λ2 1 2

∂2 g ∂2 g −3 = = 4λ−3 1 λ2 ∂λ2 ∂λ1 ∂λ1 ∂λ2 and finally ∂3 g −3 = −12λ−4 1 λ2 ∂λ21 ∂λ2

and

∂3 g −4 = −12λ−3 1 λ2 ∂λ1 ∂λ22

Substituting yields ∂2 w ∂w ∂3 w ∂3 W −7 −5 −3 = −8λ−8 + −28λ−6 + −12λ−4 1 λ2 1 λ2 1 λ2 2 3 2 ∂g ∂g ∂g ∂λ1 ∂λ2 and

∂2 w ∂w ∂3 W ∂3 w −7 −8 −6 −4 = −8λ λ + −28λ−5 + −12λ−3 1 2 1 λ2 1 λ2 2 3 2 ∂g ∂g ∂g ∂λ1 ∂λ2

Then λ1

∂3 W ∂3 W = λ 2 ∂λ21 ∂λ2 ∂λ1 ∂λ22

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Continuum Mechanics for Engineers

Problem 8.7 For biaxial loading of a thin vulcanized rubber sheet the strain energy may be written as W = C1 (I1 − 3) + C2 (I2 − 3) + C3 (I2 − 3)2 (Rivlin and Saunders, 1951). (a) Use the definitions of invariants I1 , I2 , and I3 in terms of stretches λ1 , λ2 , and λ3 to show 1 1 1 (I2 − 3)2 = 4 + 4 + 4 + 2I1 − 6I2 + 9 . λ1 λ2 λ3 (b) Substitute the results from (a) into the strain energy above to obtain −4 w (λ1 ) = (C1 + 2C3 ) λ21 + (C2 − 6C3 ) λ−2 1 + C3 λ3 − (C1 + C2 − 3C3 )

where W = w (λ1 ) + w (λ2 ) + w

1 λ1 λ2

.

Solution (a) We note that for an incompressible material, λ21 λ22 λ23 = 1 and (I2 − 3)2 = I22 − 6I2 + 9 2 1 1 1 1 1 1 + + −6 + + +9 = λ21 λ22 λ23 λ21 λ22 λ23 1 1 1 1 1 1 + + = 4 + 4 + 4 +2 λ1 λ2 λ3 λ21 λ22 λ22 λ23 λ23 λ21 1 1 1 −6 + + +9 λ21 λ22 λ23 1 1 1 = 4 + 4 + 4 + 2 λ23 + λ21 + λ22 λ1 λ2 λ3 1 1 1 −6 + + +9 λ21 λ22 λ23 This gives (I3 − 3)2 =

1 1 1 + 4 + 4 + 2I1 − 6I2 + 9 4 λ1 λ2 λ3

(b) Now 1 1 1 W = C1 λ21 + λ22 + λ23 − 3 + C2 + + − 3 λ21 λ22 λ23 1 1 1 + C3 + + + 2I1 − 6I2 + 9 λ41 λ42 λ43

195

Chapter 8 Solutions Collecting terms yields 1 1 C2 2 + C + 2λ − 6 + 3 − C1 − C2 3 1 λ21 λ41 λ21 C2 1 1 2 + C1 λ22 + 2 + C3 + 2λ − 6 + 3 − C1 − C2 2 λ2 λ42 λ22 C2 1 1 2 + C1 λ23 + 2 + C3 + 2λ − 6 + 3 − C1 − C2 3 λ3 λ43 λ23

W = C1 λ21 +

where the constant terms were distributed equally across the three terms. Now let w (λ) = (C1 + 2C3 ) λ2 + (C2 − 6C3 ) λ−2 − (C1 + C2 − 3C3 ) Then W = w (λ1 ) + w (λ2 ) + w (λ3 ) = w (λ1 ) + w (λ2 ) + w

1 λ1 λ 2

where incompressibility was used to eliminate λ3 . Problem 8.8 Consider a material having reference configuration coordinates (R, Θ, Z) and current configuration coordinates (r, θ, z). Assume a motion defined by r=

R , θ = f (Θ) , z = Z . g (Θ)

Determine FiA and Bij in terms of g, g0 , and f0 . Answer

FiA

1 g = 0 0

g0 g2 f0 g 0

−

0 , 0 1

0 2 1 g + g2 2 g Bij = f0 g0 − g3 0

f0 g0 − 3 g 0 2 f g 0

0 0 1

Solution The physical components of the deformation gradient are (Malvern, Appendix II). [FiA ] =

∂r ∂R ∂θ r ∂R ∂z ∂R

1 ∂r R ∂Θ r ∂θ R ∂Θ 1 ∂z R ∂Θ

∂r ∂Z ∂θ r ∂Z ∂z ∂Z

1 g = 0 0

g0 g2 f0 g 0

−

0 0 1

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Continuum Mechanics for Engineers

Now B = F · FT or Bij = FiA FjA in indicial notation. In matrix form 1 1 g0 g − g2 0 g g0 f0 [Bij ] = [FiA ] [FiA ]T = 0 − 2 0 g g 0 0 0 1 0 2 1 g f0 g 0 − 3 0 2+ 2 g g g 0 2 0 0 = fg f − 3 0 g g 0 0 1

0

0 0 1

f0 g 0

Note: For further discussion or student projects see Rajagopal and Tao, “On an inhomogeneous deformation of a generalized Neo-Hookean material,” Journal of Elasticity, 28, 165-184, 1992. Problem 8.9 Show

t22

∂W

2 2 ∂W − 2 λ−2 , 1 − λ 1 λ2 ∂I1 ∂I2 −2 ∂W 2 2 ∂W = 2 λ22 − λ−2 − 2 λ−2 , 1 λ2 2 − λ 1 λ2 ∂I1 ∂I2

−2 t11 = 2 λ21 − λ−2 1 λ2

are the nonzero Cauchy stress components for biaxial tensile loading of a homogeneous, isotropic, incompressible rubber-like material by solving for the indeterminant pressure p from the fact that t33 = 0 condition.

Solution Consider the principal axes where

λ1 [FiA ] = 0 0

0 λ2 0

0 0 λ3

λ21 [Bij ] = 0 0

and

0 λ22 0

0 0 λ23

From Eq 8.47

1 [tij ] = p 0 0

0 1 0

2 λ1 0 ∂W 0 0 + ∂I1 1 0

Now t33 = p + and p=−

0 λ22 0

−2 λ1 0 ∂W 0 0 + ∂I2 λ23 0

∂W 2 ∂W −2 λ + λ =0 ∂I1 3 ∂I2 3

∂W 2 ∂W −2 ∂W 1 ∂W 2 2 λ − λ =− − λ λ ∂I1 3 ∂I2 3 ∂I1 λ21 λ22 ∂I2 1 2

0 λ−2 2 0

0 0 λ−2 3

197

Chapter 8 Solutions Substituting for p, The stresses are ∂W −2 −2 ∂W 2 2 ∂W 2 ∂W −2 λ λ − λ λ + λ + λ ∂I1 1 2 ∂I2 1 2 ∂I1 1 ∂I2 1 ∂W −2 ∂W 2 −2 = λ1 − λ−2 + λ1 − λ21 λ22 1 λ2 ∂I1 ∂I2

t11 = −

and ∂W −2 −2 ∂W 2 2 ∂W 2 ∂W −2 λ λ − λ λ + λ + λ ∂I1 1 2 ∂I2 1 2 ∂I1 2 ∂I2 2 ∂W −2 ∂W 2 −2 = λ2 − λ−2 + λ2 − λ21 λ22 1 λ2 ∂I1 ∂I2

t22 = −

Chapter 9 Solutions

Problem 9.1 By substituting Sij = tij − 31 δij tkk and ηij = ij − 13 δij kk into Eq 9.7 and combining those two equations, determine expressions in operator form for (a) the Lam´e constant, λ, (b) Young’s modulus, E, (c) Poisson’s ratio, ν. Answer (a) {λ} = K − 2{Q}/3{P} (b) {E} = 9K{Q}/(3K{P} + {Q}) (c) {ν} = (3K{P} − 2{Q})/(6K{P} + 2{Q})

Solution From Eqs 9.7 {P} Sij = 2 {Q} ηij tii = 3Kii (a) Now 1 1 {P} tij − δij tkk = 2 {Q} ij − δij kk 3 3 and tij =

2 {Q} 2 {Q} ij + δij K − kk {P} 3 {P}

From Eq 6.23, we have {λ} = K −

2 {Q} 3 {P}

(b) and (c) Inverting we have {P} ij = tij − δij 2 {Q}

3K {P} − 2 {Q} 18K {Q}

tkk

From Eq6.31, we have {P} 1 + {ν} = {E} 2 {Q}

{ν} 3K {P} − 2 {Q} = {E} 18K {Q} 199

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Continuum Mechanics for Engineers

Solving simultaneously gives {E} =

9K {Q} 3K {P} + {Q}

{ν} =

3K {P} − 2 {Q} 6K {P} + 2 {Q}

Problem 9.2 Compliances are reciprocals of moduli. Thus, in elasticity theory D = 1/E, J = 1/G, and B = 1/K. Show from the stress-strain equations of a simple one-dimensional tension that D=

1 1 J+ B . 3 9

Solution From Eq 6.30a, D=

1 λ+µ = E µ (3λ + 2µ)

By partial fractions C λ+µ A + = µ 3λ + 2µ µ (3λ + 2µ) and and 2A + C = 1

3A = 1 or A=

1 3

and

C=

1 3

This gives D=

1/µ 3 J B + = + 3 9 (2λ + 3µ) 3 9

See Table 6.1 Problem 9.3 The four-parameter model shown consists of a Kelvin unit in series with a Maxwell unit. Knowing that γMODEL = γKELVIN +γMAXWELL , together with the operator equations Eqs 9.13 and 9.14, determine the constitutive equation for this model.

G1

G2

T

η2

T

η1 Answer G2 η1 γ¨ + G1 G2 γ˙ = η1 T¨ + (G1 + G2 + η1 /τ2 ) T˙ + (G1 /τ2 ) T

Solution

201

Chapter 9 Solutions The model and Eqs 9.13 and 9.14, give γ=

(∂t + 1/τ2 ) T + T (G1 + η1 ∂t ) (G2 ∂t )

or [(G1 + η1 ∂t ) (G2 ∂t )] γ = (G2 ∂t ) T + [(G1 + η1 ∂t ) (∂t + 1/τ2 )] T Expanding, we find G1 η1 ˙ G1 G2 γ˙ + G2 η1 γ¨ = G2 T˙ + G1 T˙ + η1 T¨ + T+ T τ2 τ2 Rearranging gives G1 η1 ˙ ¨ T ++ T G2 η1 γ¨ + G1 G2 γ˙ = η1 T + G1 + G2 + τ2 τ2 Problem 9.4 Develop the constitutive equations for the three-parameter models shown.

T

G1

η2

T

T

T

η1

G1 (b)

(a)

η2

G2

T

T η1 (c) Answer (a) (b) (c)

γ¨ + γ/τ ˙ 1 = [(η1 + η2 ) /η1 η2 ] T˙ + (1/τ1 η2 ) T T˙ + T/τ2 = (G2 + G1 ) γ˙ + (G1 /τ2 ) γ T˙ + T/τ2 = η1 γ¨ + (G2 + η1 /τ2 ) γ˙

Solution (a) For this model γ˙ = This gives

η2

G2

T˙ T + (G1 + η1 ∂t ) η2

G1 η2 γ˙ + η1 η2 γ¨ = η2 T˙ + G1 T + η1 T˙

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Continuum Mechanics for Engineers

and

γ˙ γ¨ + = τ1

η1 + η2 η1 η2

T˙ +

1 T η2 τ1

(b) For this model T=

G2 γ˙ + G1 γ (∂t + 1/τ2 )

This gives G1 G1 1 γ = (G1 + G2 ) γ˙ + γ T˙ + T = G2 γ˙ + G1 γ˙ + τ2 τ2 τ2 (c) For this model T= Then

G2 γ˙ + η1 γ˙ (∂t + 1/τ2 )

η1 1 γ˙ = T˙ + T = G2 γ˙ + η1 γ¨ + τ2 τ2

η1 G2 + γ˙ + η1 γ¨ τ2

Problem 9.5 A proposed model consists of a Kelvin unit in parallel with a Maxwell unit. Determine the constitutive equation for this model.

G2 T

η2 T

G1 η1

Answer T˙ + T/τ2 = η1 γ¨ + (G1 + G2 + η1 /τ2 ) γ˙ + (G1 /τ2 ) γ

Solution This model gives T = (G1 + η1 ∂t ) γ +

G2 γ˙ (∂t + 1/τ2 )

Expanding gives 1 G1 η1 T˙ + T = G1 γ˙ + η1 γ¨ + γ+ γ˙ + G2 γ˙ τ2 τ2 τ2 and collecting terms 1 η1 G1 T˙ + T = η1 γ¨ + G1 + G2 + γ˙ + γ τ2 τ2 τ2 Problem 9.6 For the four-parameter model shown, determine

203

Chapter 9 Solutions

(a) the constitutive equation, (b) the relaxation function, G(t). [Note that G(t) is the sum of the G(t)’s of the parallel joined units.]

G1 η2

T

T

G2 η3 Answer (a) T˙ + T/τ2 = η3 γ¨ + (G1 + G2 + η3 /τ2 ) γ˙ + (G1 /τ2 ) γ (b) G(t) = G1 + G2 e−t/τ2 + η3 δ (t) Solution (a) The model yields T = G1 γ +

G2 γ˙ + η3 γ˙ (∂t + 1/τ2 )

Expanding gives G1 1 γ + G2 γ˙ + η3 γ¨ + T˙ + T = G1 γ˙ + τ2 τ2 Collecting terms, we find ˙T + 1 T = η3 γ¨ + G1 + G2 + η3 γ˙ + τ2 τ2

η3 γ˙ τ2 G1 γ τ2

(b) To compute the relaxation function, we apply a step change in strain, γ = γ0 U (t). Now γ˙ = γ0 δ (t), and γ¨ = γ0 δ˙ (t). This gives ˙T + 1 T = η3 γ0 δ˙ (t) + G1 + G2 + η3 γ0 δ (t) + G1 γ0 U (t) τ2 τ2 τ2 Integrating gives Te

t/τ2

Zt η3 0 0 0 ˙ = η3 γ0 e γ0 et /τ2 δ (t0 ) dt0 δ (t ) dt + G1 + G2 + τ2 0 0 Zt G1 0 γ0 et /τ2 U (t0 ) dt0 + τ2 0 Zt

t0 /τ2

or Te

and

t/τ2

U (t) η3 t/τ2 = η3 γ0 e δ (t) − + G1 + G2 + γ0 U (t) τ2 τ2 + G1 γ0 et/τ2 − 1 U (t) i h T = γ0 η3 δ (t) + G2 e−t/τ2 U (t) + G1 U (t)

Problem 9.7 For the model shown the stress history is given by the accompanying diagram. Determine the strain γ(t) for this loading during the intervals

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Continuum Mechanics for Engineers (a) 0 6 t/τ 6 2, (b) 0 6 t/τ 6 4.

Use superposition to obtain answer (b).

t

G G

T

T

η

T0 0

1

2

3

4

5 t/τ

Answer (a) γ(t) = T0 J 2 − e−t/τ U (t) (b) γ(t) = T0 J 2 − e−t/τ U (t) − T0 J 2 − e−(t−2τ)/τ U (t − 2τ)

Solution The constitutive equation is 1 1 2 γ˙ + γ = T˙ + T τ G η (a) For T = T0 U (t) as shown in the diagram, we have 1 2 1 γ˙ + γ = T0 δ (t) + T0 U (t) τ G η Integrating gives Z Z 2T0 t t0 /τ T0 t t0 /τ e δ (t0 ) dt0 + e U (t0 ) dt0 G 0 η 0 T0 2T0 t/τ e − 1 U (t) = U (t) + G G

γet/τ =

and for 0 6 t 6 2τ γ (t) = T0

e−t/τ 2 + 1 − e−t/τ U (t) = T0 J 2 − e−tτ U (t) G G

(b) By superposition for 0 6 t 6 4τ, we have γ (t) = T0 J 2 − e−tτ U (t) + T0 J 2 − e−(t−2τ)/τ U (t − 2τ) Problem 9.8 For the model shown determine (a) the constitutive equation, (b) the relaxation function, G(t),

205

Chapter 9 Solutions

(c) the stress, T (t) for 0 6 t 6 t1 , when the strain is given by the accompanying graph.

γ

G η

G

T

T

λ 1

η

0

t

Answer (a) T˙ + T/τ = ηγ¨ + 3Gγ˙ + (G/τ) γ (b) G(t) = ηδ (t) + G 1 + e−t/τ2 (c) T (t) = λ 2η − 2e−t/τ Gt U (t)

Solution (a) The model’s constitutive equation gives Gγ˙ + ηγ˙ T = Gγ + 1 ∂t + τ Expanding 1 G T˙ + T = ηγ¨ + (3G) γ˙ + γ τ τ (b) See the solution to Problem 9.6 with G1 = G2 = G and η2 = η3 = η, then G (t) = ηδ (t) + G 1 + e−t/τ (c) From the graph, γ (t) = tλU (t) ,

γ˙ (t) = λU (t)

γ¨ (t) = λδ (t)

and from (a) 1 λG T˙ + T = ηλδ (t) + (3G) λU (t) + tU (t) τ τ Integrating, we have Zt Te

t/τ

= ηλ

e 0

and or

t0 /τ

Zt 0

0

δ (t ) dt + (3G) λ

e 0

t0 /τ

λG U (t ) dt + τ 0

Zt

0

0

t0 et /τ U (t0 ) dt0 0

h i T et/τ = ηλ + 3ηλ et/τ − 1 + λG et/τ (t − τ) + τ U (t) T (t) = λ 2η − ηe−t/τ + Gt U (t)

Problem 9.9 For the model shown in Problem 9.8, determine T (t) when γ(t) is given by the diagram shown here.

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Continuum Mechanics for Engineers

γ λ 1

γ0 0

t

Answer T (t) = γ0 ηδ (t) + G 1 + e−t/τ U (t) + λ η 2 − e−t/τ + Gt U (t)

Solution For the γ0 portion from answer (b), Problem 9.8, we have h i T (t) = γ0 ηδ (t) + G 1 + e−t/τ U (t) For the ramp portion from answer (c), Problem 9.8, we have h i T (t) = λ η 2 − e−t/τ + Gt U (t) Combining we have the final result h i h i T (t) = γ0 ηδ (t) + G 1 + e−t/τ U (t) + λ η 2 − e−t/τ + Gt U (t) Problem 9.10 The three-parameter model shown is subjected to the strain history pictured in the graph. Use superposition to obtain T (t) for the t > t1 , from T (t) for t 6 t1 . Let γ0 /t1 = λ.

γ

G2 T

T G1

η1

γ0 0

t1

Answer For t 6 t1 ; T (t) = λ η1 1 − e−t/τ1 + G2 t U(t) For t > t1 ; T (t) = λ η1 1 − e−t/τ1 + G2 t U(t) −λ η1 1 − e−(t−t1 )/τ1 + G2 (t − t1 ) U(t − t1 )

Solution The constitutive equation for the model is G2 1 T˙ + T = (G1 + G2 ) γ˙ + γ τ τ1 For t 6 t1 , γ = λtU (t), γ˙ = λU (t), and 1 G2 T˙ + T = (G1 + G2 ) λU (t) + λtU (t) τ τ1

t

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Chapter 9 Solutions Integrating, we find Zt Te

t/τ

= (G1 + G2 ) λ

e

t0 /τ

0

and

G2 λ U (t ) dt + τ1 0

Zt

0

et /τ t0 U (t0 ) dt0

0

0

h i T (t) = λ η1 1 − e−t/τ1 + G2 t U (t)

For t > t1 by superposition of adding a segment with a negative slope at t1 to give a constant strain history beyond t1 , we have h i T (t) = λ η1 1 − e−t/τ1 + G2 t U (t) h i − λ η1 1 − e−(t−t1 )/τ1 + G2 (t − t1 ) U (t − t1 ) Problem 9.11 For the model shown determine the stress, T (t) at (a) t = t1 ; (b) t = 2t1 ; and (c) t = 3t1 , if the applied strain is given by the diagram. Use superposition for (b) and (c).

γ

η

G

T

T η

γ0 t1

0

2t1

3t1

Answer (a) T (t1 ) = (γ0 η/t1 ) 2 − e−t1 /τ (b) T (2t1 ) = (γ0 η/t1 ) −2 + 2e−t1 /τ − e−2t1 /τ (c) T (3t1 ) = (γ0 η/t1 ) −e−t1 /τ + 2e−2t1 /τ − 3e−3t1 /τ

Solution (a) The constitutive equation is 1 T˙ + T = ηγ¨ + 2Gγ˙ τ γ0 γ0 γ0 tU (t); γ˙ (t) = U (t); and γ˙ (t) = δ (t). Substituting t1 t1 t1 into the constitutive model gives For t 6 t1 , we have γ (t) =

1 γ0 γ0 T˙ + T = η δ (t) + 2G U (t) τ t1 t1 Integrating we obtain, T et/τ = η

γ0 t1

Zt

0

et /τ δ (t0 ) dt0 + 2G 0

and T (t) =

γ0 t1

Zt

0

et /τ U (t0 ) dt0 0

ηγ0 2 − e−t/τ U (t) t1

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Continuum Mechanics for Engineers

For t = t1 , we have T (t1 ) =

ηγ0 2 − e−t1 /τ t1

(b) Next for t > t1 , we have by superposing a ramp with a slope with −2 γt10 tU (t). This gives i h i ηγ0 h T (t) = 2 − e−t/τ U (t) − 2 2 − e−(t−t1 )/τ U (t − t1 ) t1 For t = 2t1 , we have T (2t1 ) =

i ηγ0 h −2 + 2e−t1 /τ − e−2t1 /τ t1

(c) Now for 2t1 6 t 6 3t1 , we superpose a term with the original positive slope to have a zero strain history i h i ηγ0 h 2 − e−t/τ U (t) − 2 2 − e−(t−t1 )/τ U (t − t1 ) T (t) = t h 1 i + 2 − e−(t−2t1 )/τ U (t − 2t1 ) and for t = 3t1 , we have T (3t1 ) =

i ηγ0 h −3t1 /τ −e + e−2t1 /τ − e−t1 /τ t1

Problem 9.12 If the model shown in the sketch is subjected to the strain history γ (t) = (σ0 /2G) 2 − e−t/2τ U (t), as pictured in the time diagram, determine the stress, T (t).

γ

η

G

T

T

γ (t) = T0 2G

G

T0 t 2 − e− 2τ 2G

0

t

Answer T (t) = T0 e−t/τ + U(t)

Solution The constitutive equation for the model is 1 G T˙ + T = 2Gγ˙ + γ τ τ For the given strain history, i h T0 1 −t/2τ γ˙ (t) = e U (t) + 2 − e−t/2τ δ (t) 2G 2τ and substituting into the constitutive equation gives h i i 1 −t/2τ T0 h −t/2τ ˙T + 1 T = T0 e U (t) + 2 − e δ (t) + 2 − e−t/2τ U (t) τ 2τ 2τ

209

Chapter 9 Solutions Integrating, we find Zt Te

t/τ

= T0

e

t0 /τ

0

Z h i T0 t t0 /τ −t0 /2τ 0 0 e U (t0 ) dt0 2−e δ (t ) dt + τ 0

This gives T et/τ = T0 (2 − 1) + T0 et/τ U (t) , or T et/τ = T0 e−t/τ + T0 U (t) , Problem 9.13 For the hereditary integral, Eq 9.38 Zt γ (t) =

J (t − t0 ) (dT (t0 ) /dt0 ) dt0

−∞

assume T (t) = est where s is a constant. Let τ = t − t0 be the “elapsed time” of the load application and show that γ(t) = sest J¯ (s) where J¯ (s) is the Laplace transform of J(t). Solution For the hereditary integral Zt J (t − t0 )

γ (t) = −∞

dT (t0 ) 0 dt dt0

0

With T (t) = est , we have T (t0 ) = est , and dT (t0 ) 0 = sest 0 dt For τ = t − t0 , dτ = −dt0 , and when t0 = t, τ = 0 and when t0 = −∞, τ = ∞. Then Z0 Z∞ Z∞ γ (t) = − J (τ) ses(t−τ) dτ = J (τ) sest e−sτ dτ = sest J (τ) e−sτ dτ ∞

0

and

0

γ (t) = sest J¯ (s)

Problem 9.14 Using T (t) = est as in Problem 9.13, together with the hereditary integral Eq 9.41a Zt G (t − t0 ) (dγ (t0 ) /dt0 ) dt0 ,

T (t) = −∞

¯ (s) J¯ (s) = 1/s2 where G ¯ (s) is the Laplace and the result of Problem 9.13 show that G transform of G(t). Assume s is real. Solution The stress is Zt G (t − t0 ) (dγ (t0 ) /dt0 ) dt0

T (t) = −∞

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Continuum Mechanics for Engineers

0 Now from Problem 9.13, γ (t0 ) = sest J¯ (s), then

dγ (t0 ) 0 = s2 est J¯ (s) = s2 es(t−τ) J¯ (s) 0 dt Thus

Z0 T (t) = −

∞

G (τ) s2 es(t−τ) J¯ (s) dτ = s2 est

Z∞

G (τ) e−sτ J¯ (s) dτ = est

0

from Problem 9.13. Then

¯ (s) J¯ (s) = 1 s2 G

or ¯ (s) J¯ (s) = 1 G s2 Problem 9.15 Taking the hereditary integrals for viscoelastic behavior in the form Eq 9.40 Zt T (t0 ) [dJ (t − t0 ) /d (t − t0 )] dt0

γ (t) = J0 T (t) + 0

and Eq 9.41c

Zt γ (t0 ) [dG (t − t0 ) /d (t − t0 )] dt0 ,

T (t) = G0 γ (t) + 0

show that for the stress loading T (t) = est and with τ = t − t0 , the expression ¯ (s) + J0 B¯ (s) + A ¯ (s) B¯ (s) = 0 G0 A results, where here ¯ (s) = A

Z∞

e−sτ (dJ/dτ) dτ

0

and B¯ (s) =

Z∞

e−sτ (dG/dτ) dτ .

0

Solution For T (t) = est , consider the integral Zt T (t0 ) 0

dJ (t − t0 ) 0 dt d (t − t0 )

Let τ = t − t0 , then when t0 = 0, τ = t and dt0 = −dτ. The above integral becomes Z0

0

est

− t

and from Eq 9.40

dJ (τ) dτ = dτ

Zt es(t−τ) 0

dJ (τ) dτ = est dτ

Zt e−sτ 0

¯ (s) γ (t) = J0 est + est A

dJ (τ) ¯ (s) dτ = est A dτ

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Chapter 9 Solutions From Eq 9.41c e

st

Zt

st0 dG (t − t0 ) ¯ dt0 = G0 J0 + A (s) e d (t − t0 ) 0 Z t s(t−τ) dG (τ) ¯ (s) est + J0 + A ¯ (s) = G0 J0 + A e dτ dτ 0 Zt ¯ (s) est + J0 + A ¯ (s) est e−sτ dG (τ) dτ = G0 J0 + A dτ 0

Now

¯ (s) est + J0 + A

¯ (s) est + J0 + A ¯ (s) est B¯ (s) est = G0 J0 + A

and ¯ (s) + J0 + A ¯ (s) B¯ (s) 1 = G0 J0 + A ¯ (s) + J0 B¯ (s) + A ¯ (s) B¯ (s) = 1 + G0 A This

¯ (s) + J0 B¯ (s) + A ¯ (s) B¯ (s) = 0 G0 A

Problem 9.16 Let the stress relaxation function be given as G(t) = a(b/t)m where a, b, and m are conm 1 stants and t is time. Show that the creep function for this material is J (t) = amπ sin mx bt ¯ (s) J¯ (s) = 1/s2 where barred quantities are Laplace transwith m < 1. Use the identity G forms. Solution Let G (t) = abm t−m = abm tk−1 Since m < 1, k > 0. Then

m ¯ (s) = ab Γ (k) G sk where Γ (k) is the gamma function. Then

1 sk sk−2 J¯ (s) = ¯ = = m 2 2 ab Γ (k) s abm Γ (k) G (s) s But Γ (k) Γ (1 − k) =

π , and so sin πk (sin πk) Γ (1 − k) J¯ (s) = abm πs2−k

By definition Γ (k + 1) = kΓ (k) and (1 − k) Γ (1 − k) = Γ (2 − k), this gives (sin πk) (1 − k) Γ (1 − k) (sin πk) Γ (2 − k) = J¯ (s) = m 2−k (1 − k) ab πs abm πs2−k (1 − k) and since m = 1 − k, we have [sin (1 − m) π] Γ (2 − k) J¯ (s) = abm mπs2−k

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Continuum Mechanics for Engineers

Inverting gives J (t) =

sin (mπ) t2−k−1 sin (mπ) tm sin (mπ) = = m m ab mπ ab mπ amπ

m t b

Problem 9.17 A three-parameter solid has the model shown. Derive the constitutive equation for this model and from it determine (a) the relaxation function, and (b) the creep function for the model.

η1

G1

T

T G2 Answer

(a) G(t) = G2 + G1 e−t/τ1 ∗ ∗ (b) J (t) = (1/ (G1 + G2 )) e−t/τ1 + (1/G2 ) 1 − e−t/τ1 where τ∗1 = (G1 + G2 ) τ1 /G2

Solution The constitutive equation is T=

G1 γ˙ + G2 γ 1 ∂t + τ1

and expanding gives 1 G2 T˙ + T = (G1 + G2 ) γ˙ + γ τ1 τ1 (a) The relaxation function is given by applying a step function in strain, γ = γ0 U (t); and γ˙ = γ0 δ (t). This gives a differential equation for the stress as 1 G2 T˙ + T = (G1 + G2 ) γ0 δ (t) + γ0 U (t) τ1 τ1 Integrating, we find Zt

Te

or

t/τ1

Zt G2 0 δ (t ) dt + = (G1 + G2 ) γ0 e γ0 et /τ1 U (t0 ) dt0 τ1 0 0 G2 t/τ1 = (G1 + G2 ) γ0 U (t) + γ0 e − 1 U (t) τ1 t0 /τ1

0

0

h i T (t) = γ0 (G1 + G2 ) e−t/τ1 + G2 1 − e−t/τ1 U (t)

This gives G (t) = G2 + G1 e−t/τ1

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Chapter 9 Solutions

(b) The relaxation function results from applying a step function in stress, T = T0 U (t), and T˙ = T0 δ (t). The constitutive equation can be rewritten as 1 G2 1 γ= T T˙ + (G1 + G2 ) τ1 (G1 + G2 ) τ1 G1 + G2

γ˙ + or

1 T0 T0 δ (t) + γ= U (t) τ∗1 G1 + G2 G2 τ∗1

γ˙ + where τ∗1 =

(G1 + G2 ) τ1 . Integrating, we have G2 Zt Zt T0 T0 ∗ 0 ∗ 0 ∗ γet/τ1 = et /τ1 δ (t0 ) dt0 + et /τ1 U (t0 ) dt0 G1 + G2 0 G2 τ∗1 0

and t/τ∗ 1

γe

= T0

1 t/τ∗1 1 + e − 1 U (t) G1 + G2 G2

This gives γ (t) = T0

1 1 ∗ ∗ e−t/τ1 + 1 − e−t/τ1 U (t) G1 + G2 G2

and the relaxation function is J (t) =

1 1 ∗ ∗ e−t/τ1 + 1 − e−t/τ1 G1 + G2 G2

Problem 9.18 A material is modeled as shown by the sketch. (a) For this model determine the relaxation function, G(t). (b) If a ramp function strain as shown by the diagram is imposed on the model, determine the stress, using the appropriate hereditary integral involving G(t).

G T

2G

γ(t) η

T

λ 1

G

2η

0

t

Answer (a) G(t) = G + 2Ge−2t/τ + Ge−t/2τ (b) T (t) = Gλ t + 3τ − τe−2t/τ − 2τe−t/2τ U(t)

Solution (a) From Eq 9.35, G (t) = Ge−t/τ for a Maxwell unit, and G (t) = G for a spring. Elements in parallel are additive for relaxation functions G (t) = G + 2Ge−2t/τ + Ge−t/2τ

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Continuum Mechanics for Engineers

(b) Using the hereditary integral, Eq 9.41a in the form Zt dγ (t0 ) G (t − t0 ) dt0 T (t) = dt0 0 With γ (t) = λtU (t); γ˙ (t) = λU (t) + λtδ (t) this gives Zt i h 0 0 T (t) = [λU (t0 ) + λt0 δ (t0 )] G + 2Ge−2(t−t )/τ + Ge−(t−t )/2τ dt0 0

and

h i T (t) = Gλ t + 3τ − τe−2t/τ − 2τe−t/2τ U (t)

Problem 9.19 Determine the complex modulus, G∗ (iω) for the model shown using the substitution iω for ∂t in the constitutive equation.

η1

G1

T

T G2 Answer

G∗ (iω) = G2 + (G1 + G2 ) τ21 ω2 + iG1 τ1 ω / 1 + τ21 ω2

Solution The constitutive equation for the model is T=

G1 γ˙ + G2 γ 1 ∂t + τ1

and expanding 1 G2 T˙ + T = (G1 + G2 ) γ˙ + γ τ1 τ1 Now, we can substitute iω for ∂t in the above or rewrite the constitutive equation as iω γ + G2 γ T = G1 1 iω + τ1 and T=

(G1 + G2 ) iω + (G2 /τ1 ) τ1 (G1 + G2 ) iω + G2 γ= γ 1 (1 + iωτ1 ) iω + τ1

Rationalizing the denominator gives G2 + (G1 + G2 ) ω2 τ21 + iG1 τ1 ω T τ1 (G1 + G2 ) iω + G2 1 − iωτ1 = = (1 + iωτ1 ) γ 1 − iωτ1 1 + ω2 τ21

215

Chapter 9 Solutions and

G2 + (G1 + G2 ) ω2 τ21 + iG1 τ1 ω G (iω) = 1 + ω2 τ21 ∗

Problem 9.20 Show that, in general, J0 = 1/G0 (1 + tan 2δ) and verify that G0 and J0 for the Kelvin model satisfies this identity. (Hint: Begin with G∗ J∗ = 1.) Solution Now we have G∗ J∗ = 1 But

1 G0 + iG00 J0 − iJ00 = 0 0 0 0 G J GJ

or

G00 J00 = = tan δ G0 J0

and so (1 + i tan δ) (1 − i tan δ) =

1 G0 J0

This gives J0 =

G0

1 1 + tan2 δ

For the Kelvin model, T = [G + η∂t ] γ or replacing ∂t with iω, we have γ 1 G − iωη G − iωη 1 + iωτ = J∗ = = 2 = T G + iωη G − iωη G + ω2 η2 G (1 + ω2 τ2 ) This gives J0 =

1 1 = 0 2 2 G (1 + ω τ ) G 1 + tan2 δ

since G0 = G for the Kelvin model and J00 /J0 = tan δ = ωτ Problem 9.21 Let the complex viscosity (denoted here by η∗ (iω)) be defined through the equation T0 eiωt = η∗ iωγ0 eiωt . Determine η∗ (iω) in terms of G∗ (iω) (see Eq 9.49) and calculate η∗ (iω) for the model shown below.

η

G

T

T η Answer η∗ (iω) = η 2 + ω2 τ2 − iτηω / 1 + ω2 τ2

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Continuum Mechanics for Engineers

Solution From the definition of the complex viscosity T0 eiωt = η∗ iωγ0 eiωt and Eq 9.49 G∗ = T0 /γ0 , we find T0 = G∗ = η∗ iω γ0 Thus η∗ =

G∗ iω

For the model, the constitutive equation is 1 T˙ + T = ηγ¨ + 2Gγ˙ τ or

h i 1 T = η (iω)2 + 2Giω γ iω + τ

The complex relaxation modulus is h i 2 (iω) η + 2Giω T G∗ = = γ iω + τ1 and the complex viscosity is h i 2 (iω) η + 2Giω G −τηω2 + 2τGiω η∗ = = = iω iω − ω2 τ iω + τ1 iω ∗

This simplifies to 2 + ω2 τ2 η − iηωτ η (iω) = 1 + ω2 τ2 ∗

Problem 9.22 From Eq 9.55a in which G0 = T0 cos δ/γ0 and 9.55b in which G00 = (T0 sin δ)/γ0 show that J0 = (γ0 cos δ)/T0 and that J00 = (γ0 sin δ)/T0 . Use G∗ J∗ = 1. Solution From the identity G∗ J∗ = (G0 + iG00 ) (J0 − iJ00 ) = (G0 J0 + G00 J00 ) + i (G00 J0 − G0 J00 ) = 1 Collecting the real and imaginary parts gives G0 J0 + G00 J00 = 1

(a)

G00 J0 − G0 J00 = 0

(b)

From (b) and Eq 9.55b, we have T0 sin δ 0 T0 cos δ 00 J = J γ0 γ0

217

Chapter 9 Solutions and so from (a) T0 cos δ 0 J + γ0

T0 sin δ tan δ J0 = 1 γ0

Solving J0 =

γ0 cos δ T0

and J00 =

γ0 sin δ T0

Problem 9.23 Show that the energy dissipated per cycle is related directly to the loss compliance, J00 R by evaluating the integral T dγ over one complete cycle assuming T (t) = T0 sin ωt. (See Fig. 9.11.) Answer Z T dγ = T02 πJ00

Solution For a complete cycle, Z 2π/ω T 0

dγ dt = dt

Z 2π/ω T0 (sin ωt) γ0 ω cos (ωt − δ) dt 0

and expanding Z 2π/ω T 0

dγ dt = T0 γ0 ω dt

Z 2π/ω sin ωt (cos ωt cos δ + sin ωt sin δ) dt 0

From the result of Problem 9.22 " Z # Z 2π/ω Z 2π/ω 2π/ω dγ sin 2ωt 2 0 00 2 T dt = T0 ω J dt + J sin ωtdt = T02 πJ00 dt 2 0 0 0 Problem 9.24 For the rather complicated model shown here, determine the constitutive equation and from it G∗ (iω). Sketch a few points on the curve G00 vs. ln(ωτ).

G T

2G G

Answer

η 2η

T

218

Continuum Mechanics for Engineers T¨ + (5/2τ) T˙ + 1/τ2 T = 4Gγ¨ + (11G/2τ) γ˙ + G/τ2 γ G∗ (iω) = G0 + iG00 where G0 = G 1 + (35/4) τ2 ω2 + 4τ4 ω4 / 1 + (17/4) τ2 ω2 + τ4 ω4 G00 = G 3τω + (9/2) τ3 ω3 / 1 + (17/4) τ2 ω2 + τ4 ω4 For ln(ωτ) = 0, G00 = 1.2G For ln(ωτ) = 1, G00 = 1.13G For ln(ωτ) = 2, G00 = 0.572G For ln(ωτ) = ∞, G00 = 0

Solution The constitutive equation for this model is T=

G 2G γ˙ + γ˙ + Gγ 1 ∂t + 2τ ∂t + τ2

or upon expansion 1 5 1 G ∂2t + ∂t + 2 T = 4G∂2t + ∂t + 2 γ 2τ τ 2Gτ τ Let ∂kt = (iω)k and 5 1 G 1 2 2 (iω) + (iω) + 2 T = 4G (iω) + (iω) + 2 γ 2τ τ 2Gτ τ Now T/γ = G∗ and ∗

G =

G

1+

35 2 2 4 τ ω

+ 4τ4 ω4 + i 3τω + 92 τ3 ω3

1+

17 2 2 4 τ ω

+ τ4 ω4

For ln(ωτ) = 0, G00 = 1.2G For ln(ωτ) = 1, G00 = 1.13G For ln(ωτ) = 2, G00 = 0.572G For ln(ωτ) = ∞, G00 = 0 Problem 9.25 A block of viscoelastic material in the shape of a cube fits snugly into a rigid container. A uniformly distributed load p = −p0 U(t) is applied to the top surface of the cube. If the material is Maxwell in shear and elastic in dilatation determine the stress component t11 (t) using Eq 9.71. Evaluate t11 (0) and t11 (∞).

x2

p = −p0 U (t)

x1

219

Chapter 9 Solutions Answer t11 (t) = −p0 [1 − (6G/(3K + 4G))]e−(3K/(3K+4G)τ)t ]U(t) t11 (0) = −p0 [(3K − 2G)/(3K + 4G)] t11 (∞) = −p0

Solution In compact operator notation {P} Sij = 2 {Q} ηij

and tkk = 3Kkk and {Q} = (G∂t ). Now from symmetry, for a Maxwell shear behavior {P} = ∂t + t11 = t33 , and from the loading condition = −p0 U (t). Also, 11 = 33 = 0. From Eq 9.71a, we have 1 1 1 ∂t + t11 − (2t11 + t22 ) = 2 (G∂t ) − 22 τ 3 3 1 τ t22

or t˙ 11 +

1 3K − 2G 1 t11 = t˙ 22 + t22 τ1 3K + 4G τ

3K + 4G τ. Inserting t22 = −p0 U (t), and t˙ 22 = −p0 δ (t) integrating and 3K simplifying, gives 6G t11 (t) = −p0 1 − e−t/τ U (t) 3K + 4G 3K − 2G , and when t → ∞, t11 = −p0. When t = 0, t11 = −p0 3K + 4G where τ1 =

Problem 9.26 A slender viscoelastic bar is loaded in simple tension with the stress T (t) = t11 (t) = T0 U(t). The material may be modeled as a standard linear solid in shear having the model shown, and as elastic in dilatation. Using the hereditary integrals, Eq 9.42, determine the axial strain 11 (t) and the lateral strain 22 (t).

G T

G

T

t11

t11

η Answer 11 (t) = T0 {[(6K + G)/3K − e−t/τ ]/3G}U(t) 22 (t) = T0 [(1/9K) − (2 − e−t/τ )/6G)]U(t)

Solution The creep function for the model is the sum of the creep functions of the individual components. 2 − e−t/τ J (t) = G

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From Eq 9.42a and tkk = 3Kkk where tkk = T0 U (t) and Sij = 23 T0 U (t) Zt 0 1 2 − e−(t−t )/τ 0 2 2 − e−t/τ 2 2 11 − kk = U (t) T0 δ (t0 ) dt = T0 3 G 3 G 0 3 Thus 11

T0 6K + G −t/τ U (t) = −e G 3K

and since 22 = 33 by symmetry, from tkk = 3Kkk , T0 U (t) = 3K (11 + 222 ), we find 22 (t) = T0

2 − e−t/τ 1 − U (t) 9K 6G

Problem 9.27 A cylinder of viscoelastic material fits snugly into a rigid container so that 11 = 22 = rr = 0 (no radial strain). The body is elastic in dilatation and has a creep compliance JS = J0 (1 + t) with J0 a constant. Determine t33 (t) if ˙ 33 = A (a constant).

x3

x2

x1

r

Answer t33 (t) = {A[Kt + 4(1 − e−t )/3J0 ]}U(t)

Solution For JS = J0 (1 + t), the Laplace transform is J¯ S = J0

1 1 + 2 s s

¯ S = 1 , we find and from J¯ S G s2 ¯ S (s) = G

= J0

J0 1+s

Inverting, we have G (t) =

e−t J0

1+s s2

221

Chapter 9 Solutions From Eq 9.43a 1 t33 − tii = 3

Zt 0

2 3

0

2Ae−(t−t ) J0

dt0 =

4A 1 − e−t U (t) 3J0

and since tii = 3Kii = 3K33 , then t˙ ii = 3KA and 13 tii = KAt. So 4 (1 − e−t ) t33 (t) = A Kt + U (t) 3J0 Problem 9.28 A viscoelastic body in the form of a block is elastic in dilatation and obeys the Maxwell law in distortion. The block is subjected to a pressure impulse t11 = −p0 δ(t) distributed uniformly over the x1 face. If the block is constrained so that 22 = 33 = 0, determine 11 (t) and t22 (t).

t33

t22 t11= −poδ(t) Answer t22 (t) = p0 [(2G − 3K)δ(t)/(3K + 4G) − (6G/(3K + 4G))e[−3K/(3K+4G)τ]t ]U(t) 11 (t) = p0 [3δ(t)/(3K + 4G) − [4G/(3K + 4G)K]e[−3K/(3K+4G)τ]t ]U(t)

Solution Given t11 = −p0 δ (t); 11 = 33 = 0; this yields tii = −p0 δ (t) + t22 + t33

and

ii = 11 (t)

From the elastic response (a)

− p0 δ (t) + t22 + t33 = 3K11 (t)

and Eq 9.71

1 1 (b) {P} −p0 δ (t) + (p0 δ (t) − t22 − t33 ) = 2 {Q} 11 − 11 3 3 1 1 (c) {P} t22 + (p0 δ (t) − t22 − t33 ) = 2 {Q} − 11 3 3 1 1 (d) {P} t33 + (p0 δ (t) − t22 − t33 ) = 2 {Q} − 11 3 3

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From (c) and (d), t22 = t33 , and from (a) and (c) p0 δ (t) − 2t22 1 {P} t22 + (p0 δ (t) − 2t22 ) = 2 {Q} 3 9K Then (3 {P} K + 4 {Q}) t22 = (2 {Q} − 3K) p0 δ (t) For a Maxwell material {P} = ∂t + τ1 and {Q} = (G∂t ) (3K + 4G) t˙ 22 +

3K 3K t22 = (2G − 3K) p0 δ˙ (t) − p0 δ (t) τ τ

Taking the Laplace transform gives (3K + 4G) st¯ 22 +

3K 3K t¯ 22 = (2G − 3K) p0 s − p0 τ τ

or (s + λ) t¯ 22 = where λ =

2G − 3K p0 s − λp0 3K + 4G

3K 3K + 4G t¯ 22 =

2G − 3K s λ p0 − p0 3K + 4G s + λ s + λ

Inverting gives t22 =

6K 2G − 3K p0 δ (t) − p0 e−λt U (t) 3K + 4G 3K + 4G

To find 11 (t) use −p0 δ (t) + 2t22 = 3K11 (t) 4G 3δ (t) −λt 11 (t) = p0 − e U (t) 3K + 4G (3K + 4G) K Problem 9.29 A viscoelastic cylinder is inserted into a snug fitting cavity of a rigid container. A flat, smooth plunger is applied to the surface x1 = 0 of the cylinder and forced downward at a constant strain rate ˙ 11 = 0 . If the material is modeled as the three-parameter solid shown in shear and as elastic in dilatation, determine t11 (t) and t22 (t) during the downward motion of the plunger.

x2

G T

T G

η

x3

Answer t11 (t) = −0 [(4Gτ/3)(1 − e−t/τ ) + (K + 4G/3)t]U(t) t22 (t) = 0 [(2Gτ/3)(1 − e−t/τ ) + (−K + 2G/3)t]U(t)

x1

223

Chapter 9 Solutions Solution The constitutive equation is 1 G T˙ + T = 2Gγ˙ + γ τ τ G 1 Thus {P} = ∂t + τ and {Q} = 2G∂t + τ . Here ˙ 11 = 0 and 11 = t22 = t33 , so that tii = t11 + 2t22 . From Eq 9.71 1 G 1 t11 − (t11 + 2t22 ) = 2 2G∂t + −0 t + ∂t + τ 3 τ

−0 t, ii = −0 t, 1 0 t 3

and tii = t11 + 2t22 = 3K11 = −3K0 t Thus

1 8G 4G t˙ 11 + t11 = −K0 − K0 t − 0 − 0 t τ 3 3τ

Integrating and simplifying t11 (t) = −0

4G 4Gτ −t/τ 1−e + K+ t U (t) 3 3

From t11 + 2t22 = −3K0 t, we have t22 = − 12 t11 − 32 K0 t and t22 (t) = 0

2 2 −t/τ Gτ 1 − e + −K + G t U (t) 3 3

Problem 9.30 For a thick-walled elastic cylinder under internal pressure p0 , the stresses are trr = A−B/r2 ; tθθ = A + B/r2 and the radial displacement is given by ur = 1+ν E [A(1 − 2ν)r + B/r] where A and B are constants involving p0 , E is Young’s modulus, and ν is Poisson’s ratio. Determine trr , tθθ , and ur for a viscoelastic cylinder of the same dimensions that is Kelvin in shear and elastic dilatation if p = p0 U(t).

trr

tθθ trr

r

tθθ p0

Answer trr = same as elastic solution but with p0 now p0 U(t) tθθ = same as elastic solution but with p0 now p0 U(t) ur (t) = (3Ar/(6K + 2G))(1 − e−(3K+G)t/Gτ) )U(t) + (B/2Gr)(1 − e−t/τ )U(t)

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Continuum Mechanics for Engineers

Solution Because there are no elastic constants in the expressions for the stresses, the stresses will be the same for the viscoelastic tube but the pressure will be replaced by p0 U (t). For the displacement u, the operator for (1 + ν) /E is {P} /2 {Q} and for (1 + ν) (1 − 2ν) /E is [3P/ (6KP + 2Q)]. For a Kelvin material, {P} = 1 and {Q} = (G + η∂t ). Taking A = A1 p0 U (t) and B = B1 p0 U (t) 1 B1 p0 3 A1 rp0 U (t) + U (t) u (t) = 6K + 2G + 2η∂t 2G + 2η∂t r Taking the Laplace transform, we have u¯ (s) = A1 rp0

3 B1 p0 1 + (6K + 2G + 2ηs) s r (2G + 2ηs) s

Inverting, we find u (t) =

3Ar B 1 − e−(3K+G)t/Gτ U (t) + 1 − e−t/τ U (t) 6K + 2G 2Gr

Problem 9.31 A viscoelastic half-space is modeled as Kelvin in shear, elastic in dilatation. If the point force P = P0 e−t is applied at the origin of a stress free material at time t = 0, determine trr (t) knowing that for an elastic half-space the radial stress is trr = P0 [(1 − 2ν)A − B]/2π where A and B are functions of the coordinates only.

P = P0e−t x2 θ r x1 x3 Answer trr = (P0 /2π)[(3(G − η)e−t + 9Ke−(3K+G)t/η ]A/(3K + G − η) − e−t B

Solution Given P = P0 e−t , we have trr =

P0 e−t [(1 − 2ν) A − B] π

225

Chapter 9 Solutions

The operator form of (1 − 2ν) = 3 {Q} / (3K {P} + {Q}). for a Kelvin material {P} = 1 and {Q} = (G + η∂t ). Therefore, the transform is 3 (1/τ + s) B P ¯trr (s) = 0 A− 2π (1 + s) (3K + G) s/η 1+s Inverting using partial fractions trr (t) =

P0 3(G − η)e−t + 9Ke−(3K+G)t/η A − e−t B U (t) 2π 3K + G − η

Problem 9.32 The deflection at x = L for an end-loaded cantilever elastic beam is w = P0 L3 /3EI. Determine the deflection w(L, t) for a viscoelastic beam of the same dimensions if P = P0 U(t) assuming (a) one-dimensional analysis based on Kelvin material, and (b) three-dimensional analysis with the beam material Kelvin in shear, elastic in dilatation. Check w(L, ∞) in each case.

P0

L Answer (a) w(L, t) = (P0 L3 /3EI)(1 − e−t/τE )U(t), τE = η/E w(L, ∞) = (P0 L3 /3EI), the elastic deflection. (b) w(L, t) = (P0 L3 /3EI)[(1 − e−t/τ ) + (1/9K)e−t/τ ]U(t) w(L, ∞) = (P0 L3 /3EI), the elastic deflection.

Solution For an elastic beam w (L) = PL3 /3EI. For a viscoelastic beam, w (L, t) = P (t) L3 /3 {E} I (a) in the one-dimensional theory, the Kelvin constitutive model is T = (E + η∂t ) . Thus w (L, t) = or w ˙ +

P0 U (t) L3 3 (E + η∂t ) I

1 P0 L3 w= U (t) τE 3ηI

Integrating, we have we1/τE =

P0 L3 3ηI

Zt

0

et /τe U (t0 ) dt0 = 0

and w (L, t) =

P0 L3 t/τe e − 1 U (t) 3ηI

P0 L3 1 − e−t/τe U (t) 3EI

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Continuum Mechanics for Engineers

where τe = η/E. Thus w (L, ∞) =

P0 L3 3EI

the elastic response. (b) In 3-D theory, for a Kelvin material {P} = 1 and {Q} = (G + ∂t ) so that{E} = 9K (G + η∂t ) / (3K + G + ∂t ). Thus P0 (3K + G + ∂t ) U (t) L3 w (L, t) = 27K (G + η∂t ) I and

1 P0 L3 [(3K + G) U (t) + ηδ (t)] w ˙ + w= τ 27KηI

Integrating we

1/τ

P0 L3 = 27KηI

Z t (3K + G) e

t0 /τ

Zt 0

0

U (t ) dt +

0

ηe

t0 /τ

0

δ (t ) dt

0

0

continuing and simplifying P0 L3 1 − e−t/τ e−t/τ w (L, t) = + 3I E 9K Thus w (L, ∞) =

P0 L3 3EI

the elastic response. Problem 9.33 A simply-supported viscoelastic beam is subjected to the time-dependent loading f(x, t) = q0 t where q0 is a constant and t is time. Determine the beam deflection w(x, t) in terms of the elastic beam shape X(x) if the beam material is assumed to be (a) one-dimensional Kelvin, and (b) three-dimensional Kelvin in shear, and elastic in dilatation. Compare the results.

q0t

L Answer (a) w(x, t) = X(x) t − τE (1 − e−t/τE ) U(t) (b) w(x, t) = X(x) t − (3τE / (3K + G)) (1 − e−t/τE ) U(t) where τE = η/E

Solution (a) For a Kelvin material, T = (E + ∂t ) . The beam deflection differential equation is {EI}

d4 w = f (x, t) dx4

227

Chapter 9 Solutions Assume a separable solution w (x, t) = X (x) T (t). Then

and

{E} T = t

and

EIXIV = q0

1 T˙ + T = Et/η τE

where τE = η/E. Integrating T et/τe = and

E η

Zt

0

et /τE t0 dt0 0

h i T (t) = t − τE 1 − e−t/τE U (t)

9K{Q}/{P} and {P} = 1 and {Q} = (E + ∂t ) for a Kelvin material. From (b) With {E} = 3K+{Q}/{P} {E} T = Et, we have 1 E [(3K + G) t + η] T˙ + T = τ 9Kη

Integrating Te and

t/τ

E = 9Kη

Z t (3K + G) e 0

t0 /τ 0

Zt 0

t dt +

ηe

t0 /τ

0

U (t ) dt

0

0

3KτE T (t) = t − 1 − e−t/τ U (t) 3K + G

If the material is incompressible, K → ∞ and 3/ (3 + GK) → 1 so that (a) and (b) are equal if τE = τ.

Chapter 10 Solutions

Problem 10.1 A uniaxial tensile specimen is loaded past initial yield to point B under load control. (Servo-hydraulic machines can operate in stroke (displacement) or load control.) The load is then removed completely. Will the cross-head displacement be in its starting position? Explain your answer.

Y

Problem 10.2 Refer to Fig 10.1 and explain how both points A and B on the stress-strain curve can be considered the yield point. Can only one be the initial yield point? Problem 10.3 Figure 10.2(a) shows a reversed load pattern for a material that is not purely isotropic or kinematic in hardening. Make accurate sketches similar to Fig 10.2(a) for an isotropic and kinematic hardening material. Problem 10.4 Show that the hydrostatic line is normal to the Π-plane. Solution 229

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Continuum Mechanics for Engineers

The unit normal to a plane is given by n=

∇f , |∇f|

where f is the equation of the plane. For the Π-plane, the equation is or

σ(1) + σ(2) + σ(3) = 0

f (x) = x1 + x2 + x3 = 0

in terms of the xi coordinates. Then, ^1 + e ^2 + e ^3 ∇f = e

|∇f| =

and

√ √ 1 + 1 + 1 = 3.

This gives n=

1 ∇f ^2 + e ^3 ) . = √ (^ e1 + e |∇f| 3

Problem 10.5 Verify that the σ∗(i) in Eq 10.13 are orthogonal. Solution The σ∗(i) axes transformation is orthogonal if the transformation matrix is orthogonal. That is Q · QT = QT · Q = I and from the transformation √ √ √ √ 1/ √3 1/√3 1/ 3 1/√3 −1/ 2 1/ 2 0√ 1/√3 √ √ −1/ 6 −1/ 6 2/ 6 1/ 3

√ −1/√ 2 1/ 2 0

√ −1/√6 1 −1/√ 6 = 0 0 2/ 6

0 1 0

0 0 1

Alternatively, the dot product of the σ∗(i) stresses can be taken as the base vectors to show that σ∗(i) · σ∗(j) = δij . These are the orthogonality conditions of the base vectors. Problem 10.6 Verify that the Tresca-Coulomb and the von Mises yield criteria agree at the vertices in Figure 10.8. Solution At the point 0, σ∗(3) , The Tresca-Coulomb criterion is r

3 ∗ σ + 2 (3)

r

1 ∗ σ = 2 (2)

or

r

3 ∗ σ + 0 = 2k = Py 2 (3)

r σ∗(3)

=

2 Py . 3

The Hencky-Huber-von Mises criterion is ∗2 σ∗2 (2) + σ(3) =

2 2 P 3 y

or ∗2 ∗2 σ∗2 (2) + σ(3) = 0 + σ(3) =

2 2 P . 3 y

231

Chapter 10 Solutions This gives r

2 Py 3 which is the same as the Tresca-Coulomb criterion. At the vertex in the first quadrant, the intersection is given by r r √ ∗ 3 ∗ 1 ∗ 2σ(2) = σ + σ 2 (3) 2 (2) or r r 3 ∗ 1 √ 2k = σ(3) + 2k . 2 2 This gives r 2 ∗ k. σ(3) = 3 q √ The vertex is at 2k, 23 k , the Tresca-Coulomb criterion is 2k. The Hencky-Huber-von Mises criterion is r !2 √ 2 2 2 8 2 ∗2 ∗2 σ(2) + σ(3) = 2k + k = 2k2 + k2 = k2 = Py 3 3 3 3 σ∗(3) =

since 2k = Py . Problem 10.7 Show that the result of the derivative of the yield function with respect to the stress for the von Mises yield function is 1 ∂f 3 =√ Sij . ∂tij −3J2 2

Solution The derivative is ∂f ∂f ∂Skl ∂ = = ∂tij ∂Skl ∂tij ∂Skl

r

3 ∂Skl 1 Smn Smn = 2 ∂tij 2

3 Smn Smn 2

−1/2

3 ∂Skl (2) Smn δmk δnl . 2 ∂tij

The derivative of the stress deviator with respect to the stress is ∂Skl ∂ 1 1 1 = tkl − δkl (tpp ) = δik δjl − δkl δip δjp = δik δjl − δkl δij . ∂tij ∂tij 3 3 3 Combining the two expressions gives ∂f 1 1 1 1 3 3 3 =√ Skl δik δjl − δkl δij = √ Sij − Skk δij = √ Sij ∂tij 3 −3J2 2 −3J2 2 −3J2 2 since Skk = 0.

Problem 10.8 The plastic working is

˙ p = T : ˙ P . W

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Show that for uniaxial tension and the von Mises yield condition, this gives r ∂f σ , W p 2 eq p ˙ = Λ˙ σeq W . 3 ∂T

Solution For uniaxial tension, the plastic working is ˙ p = t11 ˙ 11 , W and from Eqs 10.30 and 10.31, the plastic working is ˙ p = σeq ˙ eff . W Equation 10.32 gives ˙ eff

r ∂f σeq , W p 2 = Λ˙ . 3 ∂T

The result is ˙ = Λ˙ W p

r

∂f σ , W p 2 eq σeq . 3 ∂T

Problem 10.9 Ziegler’s proposed modification to Prager’s hardening model assumes that the back stress evolves as ˙ = µ˙ (T − a) α or α˙ ij = (tij − αij ) . Give a geometric interpretation of this evolution law.

Problem 10.10 The hardening of many metals exhibits both kinematic and isotropic hardening. (a) Using the yield function in the form r f

T , α, p eff

=

3 (S − α) : (S − α) = 2

r

3 (Sij − αij ) (Sij − αij ) = K p eff , 2

determine consistency parameter Λ˙ and the hardening modulus Hp . (b) Show that for a uniaxial deformation, this reproduces the Bauschinger effect for the Prager evloution equation ˙ = c˙ p . α

Problem 10.11 Determine the torque-twist curve for a solid round shaft assuming that it is made of an elastic linearly hardening material.

233

Chapter 10 Solutions

Problem 10.12 p An elastic-plastic thin-walled tube is loaded in such a way that M3 / (T3 r0 ) = l0/ /l. The materail has a hardening curve specified by σeq = 105 εp eff

1/4

.

Determine expressions for the strains 33 and zθ when the elastic strains are negligible. (a) Compare this result to a simple tension test for the material. (b) Compare this result to a torsion test for the material.

Problem 10.13 Show that the result of the derivative of the yield function with respect to the stress for the Huber-Henky-von Mises yield function is ∂f 1 3 =√ Sij . ∂tij −3J2 2

Solution The derivative is ∂f ∂Skl ∂ ∂f = = ∂tij ∂Skl ∂tij ∂Skl

r

3 ∂Skl 1 Smn Smn = 2 ∂tij 2

3 Smn Smn 2

−1/2

3 ∂Skl (2) Smn δmk δnl . 2 ∂tij

The derivative of the stress deviator with respect to the stress is ∂Skl ∂ 1 1 1 = tkl − δkl (tpp ) = δik δjl − δkl δip δjp = δik δjl − δkl δij . ∂tij ∂tij 3 3 3 Combining the two expressions gives ∂f 1 3 3 3 1 1 1 =√ Skl Sij − Skk δij = √ Sij δik δjl − δkl δij = √ ∂tij 3 −3J2 2 −3J2 2 −3J2 2 since Skk = 0.