Solutions Manual for Introduction to Continuum Mechanics, 4th Edition by William Lai

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SOLUTION MANUAL

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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 = S11
2
+ S12
2
+ S13
2
+ S 221 + S 222 + S 223 + S 231 + S 232 + S 233 =
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 ,  Bij  = 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 i = B ija jand b =  B  a  , (b) s = B aij ai and s = a   B a
T
j
-------------------------------------------------------------------------------
Ans. (a)
bi = Bija j → b1 = B1j a j = B11a1 + B12 a2 + 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 
1   2
(b)  2

s = Bij aia j = B11a1a1 + B12a1a2 + B13a1a3 + +B21a2a1 + B22a2a2 + B23a2a3
+B31a3a1 + B32 a3a2 + B33 a3 a3 = ( 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  → Aij = Bi mC m j , (b)  D  =  B C  → Aij
T
= BmiC mj .
(c)  E  =  B C  F  → E = B C F .
T

ij mi mk kj


2 2  2 
2.5 Write in indicial notation the equation (a) s = A21 + A22 + A23 and (b) 2
+ 2
+ 2
=0.
x1 x2 x3
-------------------------------------------------------------------------------
2 2 2 2
Ans. (a) s = A2 + A2 + A2 = A A . (b)       =0.
1 2 3 i i
2
+ 2
+ 2
=0→
x1 x2 x3 xixi

2.6 Given that Si j =ai a j and Sij =aiaj , where ai=Qmi am and aj =Qn j an , and Qik Q jk = ij .
Show that Sii =Sii .
-------------------------------------------------------------------------------
Ans. Sij =Qmi am Qn j an =Qmi Qn j am an → Sii =Qmi Qni am an = mn am an =am am = Smm = Sii .


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

-------------------------------------------------------------------------------
Ans.
v1
i =1→ a = + v v1 v1
=
v
+v 1 +v
v1
+
v1
.
1 j 1 2 v 3
t x j t x x2 x3
1
v2
i = 2 → a2 = + v j v2
=
v2
+ v1
v2
+ v2
v2
+ v3
v2
.

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Lai et al, Introduction to Continuum Mechanics

t x j t x x2 x3
1
v3
i = 3 → a3 = + v j v3
=
v3
+ v1
v3
+ v2
v3
+ v3
v3
.
t x j t x x2 x3
1




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