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
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Continuum Mechanics for Engineers
Answer u1 u2 u3
= = =
1 ^2 − 2^ (3^ e1 − e e3 ) 5 1 ^3 ) (−^ e + 2^ e − e 1 2 5 1 (−^ e + 2^ e + 4^ e3 ) 1 2 5
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 [u1 , u2 , u3 ] =
2 1 0 2 1 1
0 −1 1
=5
and u2 × u3 =
^1 e 0 1
^2 e 2 1
^3 e −1 1
^2 − 2^ = 3^ e1 − e e3 ;
u1 =
1 ^2 − 2^ (3^ e1 − e e3 ) 5
u3 × u1 =
^1 e 1 2
^2 e 1 1
^3 e 1 0
^3 ; = −^ e1 + 2^ e2 − e
u2 =
1 ^3 ) (−^ e1 + 2^ e2 − e 5
u1 × u2 =
^1 e 2 0
^2 e 1 2
^3 e 0 −1
= −^ e1 + 2^ e2 + 4^ e3 ;
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
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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 1 δij Akk Bkk = 61 Akk ; Bij = 13 Aij − 18
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 box product λ = εijk ai bj ck is equal to the determinant a1 a2 a3 b1 b2 b3 . c1 c2 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
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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 det A =
A11 A12 A13 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 Apq Apr Aps
= εmnp εqrs det A .
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 δnq δpq
δmr δnr δpr
δms δns 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
δ13 δ23 δ33
=
1 0 0
0 1 0
0 0 1
= 1 = ε123 ε123 det A
Setting p = q gives δmp δnp δpp
δmr δnr δpr
δms δns δps
= 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 with itself should be the identity matrix for it to be a square root. Thus 1 0 0 1 0 0 1 0 0 0 −1 0 0 −1 0 = 0 1 0 0 0 1 0 0 1 0 0 1 and
5 −12
2 −5
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 2.33 3 0 1 1 3 1 [Aij ] = 0 2 4 , [Bij ] = 2 2 5 . 5 1 2 4 0 3
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Continuum Mechanics for Engineers
Solution (a) For the matrix A, we have 3 [Aij ]2 = 0 5 and
0 2 1
1 3 4 0 2 5
0 2 1
1 14 4 = 20 2 25
1 8 4
5 3 1 0 2 1
0 2 4
5 14 1 = 1 2 5
20 25 8 4 16 13
T 2
Aij
3 = 0 1
0 2 4
T
= AT
2
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
This shows that A2
and
. Similarly for B, we have
(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. The resulting transformation is 0 t11 t012 t013 0 1 0 t11 t12 t021 t022 t023 = −1 0 0 t21 t22 t031 t032 t033 0 0 1 t31 t32 t22 −t21 t23 = −t12 t33 −t13 t32 −t31 t11
t13 0 t23 1 t33 0
−1 0 0
0 0 1
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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
x01 x02 = √1 √1 2 2 1 x03 2
x0 x0 x0 √1 + 2 − 3 2 2 2 x0 x0 √2 + √3 2 2 x0 x0 x0 √1 − 2 + 3 2 2 2
x0 x0 √2 + √3 2 2
Thus x1 + x2 + x3 =
x0 x0 x0 √1 + 2 − 3 2 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
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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 det (B − λI) =
17 − λ 0 0
0 −23 − λ 28
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
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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 0 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
(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
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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 0 √ 2 √ − 3 = (4 − λ) (11 − λ) (9 − λ) + 3 =0 9−λ = (4 − λ) λ2 − 20λ − 96 = 0 = (4 − λ) (λ − 8) (λ − 12) = 0
det [K − λI] =
4−λ 0 0
0 11 √ −λ − 3
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