Solution Manual for Introduction to Continuum Mechanics (4th Edition) by W. Michael Lai, David Rubin, and Erhard Krempl

All Chapters Covered




SOLUTION MANUAL

, www.konkur.in

Lai et al, Introduction to Continuum Mechanics



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 Skj (e) amam , (f) Smn aman , (g) Snmaman

Ans. (a) Sii = S11 + S22 + S33 = 1 + 1 + 3 = 5 .
(b) Sij Sij = S 2 + S 2 + S 2 + S 2 + S 2 + S 2 + S 2 + S 2 + S 2 =
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 + a23 = 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 = a' Q .
p pm m p qp q m n mn


Ans. (a) and (c)

2.3 Given the following matrices
1 2 3 0
ai  = 0 , Bij  = 0 5 1
  
2 0 2 1
Demonstrate the equivalence of the subscripted equations and corresponding matrix equations in
the following two problems.
(a) b = B a and b =  B  a  , (b) s = B a a and s = a Ba
T

i ij j ij i j


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 .



Copyright 2010, Elsevier Inc
2-1



forum.konkur.in

, www.konkur.in

Lai et al, Introduction to Continuum Mechanics


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 1 2 2
(b)
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
and s = a
T
Ba = 1 0 20 5 1  0  = 1 0 22 = 2 + 4 = 6 .
   
0 2 1 2 2


Write in indicial notation the matrix equation (a)  A = BC, (b) D = B C  and (c)
T
2.4
 E  = B C F  .
T



Ans. (a)  A = BC  → A = B C , (b) D = B C → A
T
=B C .
ij im m j ij mi mj
(c) E = B C F  → E
T
=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.
x12 x22 x23

2 2 2 2 2 2 2
Ans. (a) s = A1 + A2 + A3 = Ai Ai . (b) + + =0→ =0.
x12 x22 x23 x ixi

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

Ans. Sij =QmiamQn jan =QmiQn jaman → Sii =QmiQniaman =mnaman =amam = Smm = Sii .

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

Ans.
v
i = 1 → a = 1 + v v1 v1 v v1 v
= +v 1 +v + v3 1 .
1
t j
x j t 1
x1 2 x2 x3
v2 v2 v2 v2 v2 v2
i=2→a = +v = +v +v +v .
2
t j
x j t 1
x1 2
x2
3
x3
v3 v3 v3 v3 v3 v3
i = 3→ a = +v = +v +v +v .
3
t j
x j t 1
x1 2
x2
3
x3

__________________________________________________________________
Copyright 2010, Elsevier Inc
2-2



forum.konkur.in

, www.konkur.in

Lai et al, Introduction to Continuum Mechanics



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)
TijTij = (2 Eij +  Ekkij )(2 Eij +  Ekkij ) = 4 2 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nbn = Qik QimQjnTmnbj → kmTmnbn = kmQjnTmnbj → Tknbn = QjnTknbj
→ Tkn (bn − Qjnbj ) = 0 .


1 0
2.10 Given ai  = 2 and bi  = 2 Evaluate [di ] , if dk = ijk aibj and show that this result is
   
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 ) = 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



Copyright 2010, Elsevier Inc
2-3



forum.konkur.in