2
solutions for
Continuum Mechanics for Engineers
Fourth Edition
G. Thomas Mase
Ronald E. Smelser
Jenn Stroud Rossmann
** Immediate Download
** Swift Response
,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
solutions for
Continuum Mechanics for Engineers
Fourth Edition
G. Thomas Mase
Ronald E. Smelser
Jenn Stroud Rossmann
** Immediate Download
** Swift Response
,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
