,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,
vi e ^j × bk e
^i = aj e ^k = εijk aj bk e
^i
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
v · v = εijk aj bk e
^i · εpqs aq bs e
^p = εijk aj bk εpqs aq bs δip = εijk aj bk εiqs aq bs
= (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
a × b · a = v · a = (εijk aj bk e
^i ) · aq e
^q = εijk aj bk aq δiq = εijk aj bk ai = 0
This is zero by symmetry in i and j.
(c) This is
a × b · b = v · b = (εijk aj bk e
^i ) · bq e
^q = εijk aj bk bq δiq = εijk aj bk bi = 0
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,
u2 × u3 u3 × u1 u1 × u2
u1 = ; u2 = ; u3 =
[u1 , u2 , u3 ] [u1 , u2 , u3 ] [u1 , u2 , u3 ]
and determine the reciprocal basis for the specific base vectors
u1 ^2 ,
e1 + e
= 2^
u2 ^3 ,
e2 − e
= 2^
u3 ^1 + e
= e ^2 + e^3 .
3
,4 Continuum Mechanics for Engineers
Answer
1
u1 = 5
(3^ ^2 − 2^
e1 − e e3 )
1
u2 = 5 e1 + 2^
(−^ ^3 )
e2 − e
1
u3 = 5 e1 + 2^
(−^ e2 + 4^
e3 )
Solution
For the bases, we have
u2 × u3 u3 × u1 u1 × u2
u1 ·u1 = u1 · = 1; u2 ·u2 = u2 · = 1; u3 ·u3 = u3 · =1
[u1 , u2 , u3 ] [u1 , u2 , u3 ] [u1 , u2 , u3 ]
since the triple scalar product is insensitive to the order of the operations. Now
u2 × u3
u2 · u1 = u2 · =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
^1
e ^2
e ^3
e
1
u2 × u3 = 0 2 −1 ^2 − 2^
e1 − e
= 3^ e3 ; u1 = ^2 − 2^
e1 − e
(3^ e3 )
1 1 1 5
^1
e ^2
e ^3
e
1
u3 × u1 = 1 1 1 e1 + 2^
= −^ ^3 ;
e2 − e u2 = e1 + 2^
(−^ ^3 )
e2 − e
2 1 0 5
^1
e ^2
e ^3
e
1
u1 × u2 = 2 1 0 e1 + 2^
= −^ e2 + 4^
e3 ; u3 = e1 + 2^
(−^ e2 + 4^
e3 )
0 2 −1 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
Let the position vector of an arbitrary point P (x1 x2 x3 ) be x = xi e ^i be
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.
,Chapter 2 Solutions 5
Solution
For
(x − b) · x = (xi e ^i ) · xj e
^i − bi e ^j = (xi xj − bi xj ) δij = xi xi − bi xi =
= x21 + x22 + x23 − b1 x1 − b2 x2 − b3 x3 = 0
Now
2 2 2
1 1 1 1 2 1
x1 − b1 + x2 − b2 + x3 − b3 = b + b22 + b23 = b2
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 decom-
position,
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
Aii + Aii
A(ii) = = Aii
2
(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 , (b) εijk δkj , (c) ε1jk a2 Tkj , (d) ε1jk δ3j vk .
Answer
,Chapter 2 Solutions 7
(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
1
Bkk = Akk
6
Substituting for Bkk and solving for Bij gives
1 1 1
3Bij = Aij − δij Akk or Bij = Aij − δij Akk
6 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
1 1 1
Tij xi xj = (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.
,Chapter 2 Solutions 9
Solution
Multiply by εimn
εimn ωi = − 21 εimn εijk wjk
= − 12 (δmj δnk − δmk δnj ) wjk
= − 12 (wmn − wnm ) = wnm ,
or,
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 a2 a3
b1 b2 b3 = a1 b2 c3 + a2 b3 c1 + a3 b1 c2 − a1 b3 c2 − a2 b1 c3 − a3 b2 c1 = λ
c1 c2 c3
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 A12 A13
A21 A22 A23 = A11 A22 A33 − A11 A23 A32 + A12 A23 A31 − A12 A21 A33
A31 A32 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 )
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,
vi e ^j × bk e
^i = aj e ^k = εijk aj bk e
^i
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
v · v = εijk aj bk e
^i · εpqs aq bs e
^p = εijk aj bk εpqs aq bs δip = εijk aj bk εiqs aq bs
= (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
a × b · a = v · a = (εijk aj bk e
^i ) · aq e
^q = εijk aj bk aq δiq = εijk aj bk ai = 0
This is zero by symmetry in i and j.
(c) This is
a × b · b = v · b = (εijk aj bk e
^i ) · bq e
^q = εijk aj bk bq δiq = εijk aj bk bi = 0
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,
u2 × u3 u3 × u1 u1 × u2
u1 = ; u2 = ; u3 =
[u1 , u2 , u3 ] [u1 , u2 , u3 ] [u1 , u2 , u3 ]
and determine the reciprocal basis for the specific base vectors
u1 ^2 ,
e1 + e
= 2^
u2 ^3 ,
e2 − e
= 2^
u3 ^1 + e
= e ^2 + e^3 .
3
,4 Continuum Mechanics for Engineers
Answer
1
u1 = 5
(3^ ^2 − 2^
e1 − e e3 )
1
u2 = 5 e1 + 2^
(−^ ^3 )
e2 − e
1
u3 = 5 e1 + 2^
(−^ e2 + 4^
e3 )
Solution
For the bases, we have
u2 × u3 u3 × u1 u1 × u2
u1 ·u1 = u1 · = 1; u2 ·u2 = u2 · = 1; u3 ·u3 = u3 · =1
[u1 , u2 , u3 ] [u1 , u2 , u3 ] [u1 , u2 , u3 ]
since the triple scalar product is insensitive to the order of the operations. Now
u2 × u3
u2 · u1 = u2 · =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
^1
e ^2
e ^3
e
1
u2 × u3 = 0 2 −1 ^2 − 2^
e1 − e
= 3^ e3 ; u1 = ^2 − 2^
e1 − e
(3^ e3 )
1 1 1 5
^1
e ^2
e ^3
e
1
u3 × u1 = 1 1 1 e1 + 2^
= −^ ^3 ;
e2 − e u2 = e1 + 2^
(−^ ^3 )
e2 − e
2 1 0 5
^1
e ^2
e ^3
e
1
u1 × u2 = 2 1 0 e1 + 2^
= −^ e2 + 4^
e3 ; u3 = e1 + 2^
(−^ e2 + 4^
e3 )
0 2 −1 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
Let the position vector of an arbitrary point P (x1 x2 x3 ) be x = xi e ^i be
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.
,Chapter 2 Solutions 5
Solution
For
(x − b) · x = (xi e ^i ) · xj e
^i − bi e ^j = (xi xj − bi xj ) δij = xi xi − bi xi =
= x21 + x22 + x23 − b1 x1 − b2 x2 − b3 x3 = 0
Now
2 2 2
1 1 1 1 2 1
x1 − b1 + x2 − b2 + x3 − b3 = b + b22 + b23 = b2
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 decom-
position,
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
Aii + Aii
A(ii) = = Aii
2
(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 , (b) εijk δkj , (c) ε1jk a2 Tkj , (d) ε1jk δ3j vk .
Answer
,Chapter 2 Solutions 7
(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
1
Bkk = Akk
6
Substituting for Bkk and solving for Bij gives
1 1 1
3Bij = Aij − δij Akk or Bij = Aij − δij Akk
6 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
1 1 1
Tij xi xj = (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.
,Chapter 2 Solutions 9
Solution
Multiply by εimn
εimn ωi = − 21 εimn εijk wjk
= − 12 (δmj δnk − δmk δnj ) wjk
= − 12 (wmn − wnm ) = wnm ,
or,
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 a2 a3
b1 b2 b3 = a1 b2 c3 + a2 b3 c1 + a3 b1 c2 − a1 b3 c2 − a2 b1 c3 − a3 b2 c1 = λ
c1 c2 c3
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 A12 A13
A21 A22 A23 = A11 A22 A33 − A11 A23 A32 + A12 A23 A31 − A12 A21 A33
A31 A32 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 )
