Solutions Manual for Introduction to Continuum Mechanics 4th Edition by Lai, Rubin, and Krempl PDF | Complete Step-by-Step Solutions and Detailed Derivations | Covers Stress, Strain, Deformation, Tensor Analysis, Constitutive Equations, Elasticity, Fluid

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SOLUTION MANUAL

,CHAPTER 2, PART A

2.1 Given
1 0 2 1
Sij  0 1 2 and  ai 2
3 0 3 3
Evaluate (a) Sii , (b) Sij Sij , (c) S ji S ji , (d) S jk (e) amam , (f) Smn aman , (g) Snmaman
Skj

Ans. (a) Sii S11 S22 S33 1 1 3 5 .
(b) Sij Sij S 2 S 2 S 2 S 2 S2 S2 S2 S2 S2
11 12 13 21 22 23 31 32 33
1 0 4 0 1 4 9 0 9 28 .
(c) S ji S ji = Sij Sij =28.
(d) S jk Skj S1k Sk1 S2k Sk 2 S3k Sk 3
S11S11 S12 S21 S13S31 S21S12 S22 S22 S23S32 S31S13 S32 S23 S33S33
1 1 0 0 2 3 0 0 1 1 2 0 3 2 0 2
3 3 23 .
(e) amam a12 a22 a32 1 4 9 14 .
(f) Smn aman S1na1an S2na2an S3na3an
S11a1a1 S12a1a2 S13a1a3 S21a2a1 S22a2a2 S23a2a3 S31a3a1 S32a3a2 S33a3a3
1 1 1 0 1 2 2 1 3 0 2 1 1 2 2
2 2 3 3 3 1
0 3 2 3 3 3 1 0 6 0 4 12 9 0 27 59.
(g) Snmaman = Smn aman =59.

2.2 Determine which of these equations have an identical meaning with a Q a' .
i ij j
(a) a Q a' , (b) a Q a' , (c) a' Q .
a
p pm p qp q m n mn
m


Ans. (a) and (c)

2.3 Given the following matrices
1 2 3 0
ai   0 , Bij 0 5 1
2 0 21
Demonstrate the equivalence of the subscripted equations and corresponding matrix equations in
the following two problems. T
(a) b B a and b B a , (b) and s a B a
s B aa
i ij j ij i j

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Ans. (a)
bi Bija j b1 B1 ja j B11a1 B12a2 B13a3 2 1 3 0 0 2 2
b2 B2 j a j B21a1 B22a2 B23a3 2, b3 B3 j a j B31a1 B32a2 B33a3 2.




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, 2 3 0 1 2
b B a 0  5   1 0 2 . Thus, bi Bija j gives the same results as b
B a
0 2 2
(b) 1
2
s Bij aia j B11a1a1 B12a1a2 B13a1a3 B21a2a1 B22a2a2 B23a2a3
B31a3a1 B32a3a2 B33a3a3 2 (1)(1) 3 (1)(0) 0 (1)(2) 0 (0)(1)
5 (0)(0) 1 (0)(2) 0 (2)(1) 2 (2)(0) 1 (2)(2) 2 4 6.
2 3 0 1 2
T
and s a B a 1 0 2 0 5 1 0 1 0 2 2 2 4 6.
    
0 2 1 2 2

T
2.4 Write in indicial notation the matrix equation (a) A B C , (b) D B C
and (c)
T
E B C F .
T
Ans. (a) A B C A B C , (b) D B C A
B C .
ij im m j ij mi mj
T
(c) E B C F E B C F .
ij mi mk kj

2 2 2
2 2 2
2.5 Write in indicial notation the equation (a) s A1 A2 A3 and (b) 0.
1
x2 2
x2 3 x2

2 2 2 2
2 2 2

Ans. (a) s A1 A2 A3 Ai Ai . (b) 2
0 0.
1 2x 3 x2 xi2 i x x


2.6 Given that Si j =aiaj and Si j =ai a j, where ai =Qmi am and a j =Qn Qik Qjk ij .
jan , and

Show that Si i =Sii .

Ans. Si j =QmiamQn jan =QmiQn jaman Si i =QmiQniaman = a a =amam
mn m n Smm Sii .

vi
2.7 Write ai in long form.
v vi
t
j x
j




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Ans.
v1
i 1 a v1 v1 v1 v1 v1
v v v .
v
t 3
1 j xj t 1 2 x2 x3
x1
v2 v2 v2 v2 v2 v2
i 2 a v v v v .
t 3
2 j xj t 1 2 x2 x3
x1
v3 v3 v3 v3 v3 v3
i 3 a v v v v .
t 3
3 j xj t 1 2 x2 x3
x1




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,2.8 Given that Tij 2 Eij Ekk ij , show that
(a) T E 2 E E E 2 and (b) T T 4 2E E E 2 (4 3 2)
ij ij ij ij kk ij ij ij ij kk


Ans. (a)
Tij Eij (2 Eij Ekk ij )Eij 2 Eij Eij Ekk ij Eij 2 Eij Eij Ekk Eii 2 Eij Eij
(Ekk )2 (b)
2
TijTij (2 Eij Ekk ij )(2 Eij Ekk ij ) 4 Eij Eij 2 Eij Ekk ij 2 Ekk ij Eij
2 E 2 4 2E E 2 E E 2 E E 2 E 2
kk ij ij ij ij ii kk kk ii kk ii
4 2E E E 2 (4 3 2 ).
ij ij kk


2.9 Given that ai =Tijbj , and ai =Ti jb j, where ai =Qimam and Tij =QimQjnTm n.
(a) Show that QimTm nbn QimQjnTm nbj and (b) if Qik Qim = km , then Tk n (bn Qjnbj ) 0.

Ans. (a) Since ai =Qimam and Tij =QimQ jnTm n, therefore, ai =Tijbj .
Qimam QimQjnTm nbj (1), Now, ai =Tij b j am =Tm jb j Tm nbn , therefore, Eq. (1) becomes
Qi mTm nbn Qi mQj nTm nbj . (2)
(b) To remove Qim from Eq. (2), we make use of Qik Qim = km by multiplying the above equation,
Eq.(2) with Qik . That is,
Qik QimTm b
n n Qik QimQjnTm b
n j km m T b
n n km QjnTm b
n j Tk b
n n QjnTk b
n j

Tk n (bn Qjnbj ) 0.


1
0
Evaluate [di ] , if dk ijk aibj and show that this result is

2.10 Given ai 2 and b
   i
2
0 3
the same as dk a b
ek .


Ans. dk ijk aibj
d1 ij1aibj 231a2b3 321a3b2 a2b3 a3b2 (2)(3) (0)(2)
6 d2 ij2aibj 312a3b1 132a1b3 a3b1 a1b3 (0)(0)
(1)(3) 3 d3 ij3aibj 123a1b2 213a2b1 a1b2 a2b1
(1)(2) (2)(0) 2 Next, a b e1 + 2e2 2e2 + 3e3
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6e1 3e2 2e3 .
d1 a b e1 6, d2 a b e2 3, d3 a b e3 2.

2.11 (a) If ijkTij 0 , show that Tij Tji , and (b) show that ij ijk =0




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,Ans. (a) for k 1, ij1Tij 0 231T23 321T32 0 T23 T32 T23 T32 .
for k ij 2Tij 0 312T31 132T13 0 T31 T13 T31 T13 .
2,
ij3Tij 0 123T12 213T21 0 T12 T21 T12 T21 .
for k
3,
(b) ij ijk 11 11k 22 22k 33 33k 1 0 1 0 1 0 0.

2.12 Verify the following equation: ijm klm ik jl il jk .
(Hint): there are 6 cases to be considered (i) i j , (2) i k , (3) i l , (4) j k , (5) j l , and (6)
k l.


Ans. There are 4 free indices in the equation. Therefore, there are the following 6 cases to consider:
(i) i j , (2) i k , (3) i l , (4) j k , (5) j l , and (6) k l . We consider each case below
where we use LS for left side, RS for right side and repeated indices with parenthesis are not sum:
(1) For i j, LS= (i)(i)m klm 0, RS (i)k (i)l (i)l (i)k 0.
(2) For i k , LS= (i) j1 (i)l1 (i) j 2 (i)l 2 (i) j3 (i)l3 , RS (i)(i) jl (i)l j(i)
0 if j l
LS=RS = 0 if j l i .
 1 if j l i
(3) For i l , LS= (i) jm k (i)m , RS (i)k j(i) (i)(i) jk
if j k 0
LS=RS = 0 if j k i
 1 if j k i
(4) For j k , LS= i( j)m ( j)lm , RS i( j) ( j)l il ( j)( j)
if i l0
LS=RS = 0 if i l j
 1 if i l j
(5) For j l , LS= i( j)m k ( j)m , RS ik ( j)( j) i( j) ( j)k
if i k 0
LS=RS = 0 if i k j
 1 if i k j
(6) For k l , LS= ijm (k )(k )m =0, RS i(k ) j(k ) i(k ) j(k ) 0

2.13 Use the identity ijm klm ik jl il jk as a short cut to obtain the following results:
(a) ilm jlm 2 ij and (b) ijk ijk 6.

Ans. (a) ilm jlm ij ll il lj 3 ij ij 2 ij .
(b) ijk ijk ii jj ij ji (3)(3) ii 9 3 6.

2.14 Use the identity ijm klm ik jl il jk to a (b c) = (a c)b (a b)c .
show that
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Ans. a (b c) = amem ( ijkbjck ei ) = ijk ambjck (em ei )
= ijk ambjck ( nmien ) = ijk nmi ambjck en jki nmi ambjck en
( jn km jm kn )ambjck en jn kmambj ck en jm knambjck en
ak bnck en a jbjcnen (a c)b (a b)c .

2.15 (a) Show that if Tij Tji , Tij ai a j 0 and (b) if Tij Tji , and Sij S ji , then Tij Sij 0

Ans. Since Tij ai a j Tjia jai (switching the original dummy index i to j and the original index
j to i ), therefore Tij ai a j Tjia jai Tij a jai Tij ai a j 2Tij ai a j 0 Tijaia j 0.
(b) Tij Sij Tji S ji (switching the original dummy index i to j and the original index j to i ),
therefore, Tij Sij Tji S ji Tij S ji Tij Sij 2Tij Sij 0 Tij Sij 0.

2.16 Let Tij Sij S ji /2 and Rij Sij S ji / 2 , show that Tij Tji , Rij Rji ,
and Sij Tij Rij .


Ans. Tij Sij S ji /2 Tji S ji Sij /2 Tij .

Rij Sij S ji /2 Rji S ji Sij /2 Sij S ji /2 Rij .
Tij +Rij = Sij S ji /2+ Sij S ji /2 Sij .

2.17 Let f (x1, x2 , x3 ) be a function of x1, x2 , and x3 and vi (x1, x2 , x3 ) be three functions of
x1, x2 , and x3 . Express the total differential df and dvi in indicial notation.

Ans. df f
dx1 dx f dx3 f dxi .
f x 2
2
x
x 3 xi
vi 1 vi vi vi
dv dx dx dx dx .
i 1 2 3 m
x1 x2 x3 xm

2.18 Let Aij denote that determinant of the matrix Aij . Show that Aij ijk Ai1 Aj 2 Ak 3


Ans. ijk Ai1 Aj 2 Ak 3 1 jk A11 Aj 2 Ak 3 2 jk A21 Aj 2 Ak 3 3 jk A31 Aj 2 Ak 3

123 A11 A22 A33 132 A11 A32 A23 231 A21 A32 A13 213 A21 A12 A33 312 A31 A12 A23 321 A31 A22 A13
A11 A22 A33 A11 A32 A23 A21 A32 A13 A21 A12 A33 A31 A12 A23 A31 A22 A13
A11 A1 A1
2 3

A2 A2 A2
1 2 3

A3 A3 A3
1 2 3


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, CHAPTER 2, PART B

2.19 A transformation T operate on any vector a to give Ta = a / a , where a is the magnitude
of a . Show that T is not a linear transformation.

a a+ b a b
Ans. Since Ta = for any a , therefore T(a + b) = . Now Ta + Tb
a a+ b a b
therefore T(a + b) Ta + Tb and T is not a linear transformation.

2.20 (a) A tensor T transforms every vector a into a vector Ta = m a where m is a specified
vector. Show that T is a linear transformation and (b) If m = e1 e2 , find the matrix of the
tensor T .

Ans. (a) T( a + b) m ( a + b) m a +m b= m a+ m b= Ta + Tb.
Thus, the given T is a linear transformation.
(b) Te1 = m e1 (e1 e2 ) e1 e3 , Te2 = m e2 (e1 e2 ) e2 e3 ,
Te3 = m e3 (e1 e2 ) e3 e2 e1 . Thus,
0 0 1
T 0 0 1 .

1 10

2.21 A tensor T transforms the base vectors e1 and e2 such that Te1 = e1 + e2 and Te2 = e1 e2
. If a = 2e1 + 3e2 and b = 3e1 + 2e2 , use the linear property of T to find (a) Ta ,(b) Tb , and (c)
T(a + b) .

Ans.
(a)Ta = T(2e1 + 3e2 ) 2Te1 + 3Te2 2 e1 + e2 + 3 e1 e2 5e1 e2 .
(b)Tb = T(3e1 + 2e2 ) 3Te1 + 2Te2 =3 e1 + e2 + 2 e1 e2 =5e1 + e2 .
(c)T(a + b) = Ta + Tb = 5e1 e2 5e1 + e2 10e1.

2.22 Obtain the matrix for the tensor T which transforms the base vectors as follows:
Te1 = 2e1 + e3 , Te2 = e2 + 3e3 , Te3 = e1 + 3e2 .

2 0 1
Ans. T 0 1 3 .
 
1 3 0

2.23 Find the matrix of the tensor T which transforms any vector a into a vector b = m(a n)
where m = 2/2 e1 + e2 and n= 2/2 e1 + e3 .


  
 e1 + e 2    e1 + e 2 / 2 .
Ans. Te1 = m e1  n  n1m =   
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