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Solutions Manual for For Introduction to Continuum Mechanics (4th Edition) by Lai - Graded A+ Comprehensive Exam...

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### Complete Solutions Manual: For Introduction to Continuum Mechanics (4th Edition) by Lai | **Format:** Instant PDF Download | **Pages:** 396 Pages Master complex textbook exercises and exam problems with the complete, official **Solutions Manual** for **For Introduction to Continuum Mechanics** (4th Edition) by Lai. #### What is Included: - **100% Complete Worked Solutions:** Step-by-step mathematical derivations, conceptual reasoning, and formulas for all textbook exercises. - **All Chapter Coverage:** Detailed answers for all end-of-chapter problems, questions, and review sets. - **Homework & Exam Advantage:** Check your work, practice challenging problem sets, and prepare thoroughly for quizzes and exams. Essential resource for self-study and mastering course material. Instant download on Stuvia!

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STUDY NOTES & REFERENCE GUIDE




All Chapters Covered···




SOLUTION MANUAL ·




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STUDY NOTES & REFERENCE GUIDE


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


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 , (b) a = Q a' , (c) = a' Q .
' a a
p pm p qp m n mn
m q


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




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Lai et al, Introduction to Continuum Mechanics

b2 = B2 j a j = B21a1 + B22a2 + B23a3 = 2, b3 = B3 j a j = B31a1 + B32a2 + B33a3 = 2 .




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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 Ba = 1 0 20 5 1 0 = 1 0 22 = 2 + 4 = 6 .
T
·




     ·




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.
1 2 x2 3x2 x2 i i x x

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 in long form.
v i
t
j
x
j


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




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