Sample continuum mechanics for engineers third ( 3rd ) edition Thomas Mase solution manual pdf PDF

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Authors: G. Thomas Mase , Ronald E. Smelser , Jenn Stroud Rossmann
Published: CRC Press 2009
Edition: 3rd
Pages: 234
Type: pdf
Size: 9.53 MB
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FOLFNKHUHWRGRZQORDG

SOLUTIONS MANUAL FOR Continuum Mechanics for Engineers, Third Edition

by Thomas Mase Ronald E. Smelser

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FOLFNKHUHWRGRZQORDG

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FOLFNKHUHWRGRZQORDG

SOLUTIONS MANUAL FOR Continuum Mechanics for Engineers, Third Edition

by Thomas Mase Ronald E. Smelser

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2010 by Taylor and Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number: 978-1-4398-0898-6 (Paperback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

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Chapter 2 Solutions

Problem 2.1 Let v = a × b, or in indicial notation, ^k = εijk aj bk e ^i vie^i = aj ^ ej × bk e 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 ^p = εijk aj bk εpqs aq bs δip = εijk aj bk εiqs aq bs v · v = εijk aj bk ^ ei · εpqs aq bse = (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

^i ) · aq ^ a × b · a = v · a = (εijk aj bke eq = εijk aj bk aq δiq = εijk aj bk ai = 0 This is zero by symmetry in i and j. (c) This is ^i ) · bq e ^q = εijk aj bk bq δiq = εijk aj bk bi = 0 a × b · b = v · b = (εijk aj bke 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, u1 =

u2 × u3 ; [u1 , u2 , u3 ]

u2 =

u3 × u1 ; [u1 , u2 , u3 ]

u3 =

u1 × u2 [u1 , u2 , u3 ]

and determine the reciprocal basis for the specific base vectors u1 u2 u3

^2 , = 2^ e1 + e ^3 , = 2^ e2 − e = ^ e1 + e^2 + ^e3 .

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2 Continuum Mechanics for Engineers ttps://gioumeh.com/product/continuum-mechanics-for-engineers-solutions Answer u1 u2 u3

= = =

1 5 1 5 1 5

(3^ ^2 − 2^ e3 ) e1 − e ^3 ) e1 + 2^ e2 FOLFNKHUHWRGRZQORDG −e (−^ e1 + 2^ e2 + 4^ (−^ e3 )

Solution For the bases, we have u1 ·u1 = u1 ·

u2 × u3 = 1; [u1 , u2 , u3 ]

u2 ·u2 = u2 ·

u3 × u1 = 1; [u1 , u2 , u3 ]

u3 ·u3 = u3 ·

u1 × u2 =1 [u1 , u2 , u3 ]

since the triple scalar product is insensitive to the order of the operations. Now u2 · u1 = u2 ·

u2 × u3 =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

   e ^2 e ^3   ^1 e ^2 − 2^ e1 − e e3 ; 2 −1  = 3^ u2 × u3 =  0  1 1 1     ^ ^2 e^3   e1 e ^3 ; 1 1  = −^ u3 × u1 =  1 e1 + 2^e2 − e  2 1 0     e^1 ^ e2 ^ e3   e1 + 2^ e2 + 4^ e3 ; 1 0  = −^ u1 × u2 =  2  0 2 −1 

u1 =

1 ^2 − 2^ e1 − e (3^ e3 ) 5

u2 =

1 ^3 ) (−^ e1 + 2^ e2 − e 5

u3 =

1 e1 + 2^ e2 + 4^e3 ) (−^ 5

Problem 2.3 ^i be Let the position vector of an arbitrary point P (x1 x2 x3 ) be x = xi ^ ei , and let b = bi e 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 12 b. Solution For ^i − bie ^i ) · xj ^ej = (xi xj − bi xj ) δij = xi xi − bi xi = (x − b) · x = (xie = x12 + x22 + x32 − b1 x1 − b2 x2 − b3 x3 = 0 Now 2 2  2    1 1 2 1 1 1 = x1 − b 1 + x2 − b 2 b + b22 + b23 = b2 + x3 − b 3 4 4 1 2 2 2

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Chapter 2 Solutions 3 ttps://gioumeh.com/product/continuum-mechanics-for-engineers-solutions This is the equation of a sphere with the desired properties.

FOLFNKHUHWRGRZQORDG Problem 2.4 Using the notations A(ij) =

1 (Aij 2

+ Aji ) and A[ij] =

1 (Aij 2

− 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 decomposition, 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 + = A(ij) + A[ij] Aij = 2 2 (b) The trace of A is A(ii) =



Aii + Aii 2



= Aii

(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 1 Bij − Bji A(ij) B[ij] = = (Aij Bij + Aji Bij − Aij Bji − Aji Bji ) 4 2 2 1 = (Aij Bij + Aji Bij − Aji Bij − Aij Bij ) = 0 4 We have changed the dummy indices on the last two terms.

Problem 2.5 Expand the following expressions involving Kronecker deltas, and simplify where possible. (a) δij δij , (b) δij δjk δki , (c) δij δjk , (d) δij Aik Answer

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4 Continuum Mechanics for Engineers ttps://gioumeh.com/product/continuum-mechanics-for-engineers-solutions (a) 3, (b) 3, (c) δik ,

(d) Ajk

FOLFNKHUHWRGRZQORDG

Solution (a) Contracting on i or j, we have

δij δij = δjj = δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3 (b) Contracting on k and then j gives δij δjk δki = δij δji = δii = 3 (c) Contracting on j yields δij δjk = δik (d) Contracting on i gives δij Aik = Ajk Note: It may be helpful for beginning students to write out all terms.

Problem 2.6 If ai = εijk bj ck and bi = εijk gj hk , substitute bj into the expression for ai to show that ai = gk ck hi − hk ck gi , or in symbolic notation, a = (c · g)h − (c · h)g. Solution We begin by changing the dummy indices for bj = εjmn gm hn and ai = εijk bj ck = εijk εjmn gm hn ck = − (εjik εjmn gm hn ck ) = − (δim δkn − δin δkm ) gm hn ck = −gi hk ck + gk hi ck = gk ck hi − hk ck gi where we have used the anti-symmetry of εijk = −εjik and the ε−δ identity. Symbolically a = (c · g)h − (c · h)g

Problem 2.7 By summing on the repeated subscripts determine the simplest form of (a) ε3jk aj ak , (b) εijk δkj , (c) ε1jk a2 Tkj , (d) ε1jk δ3j vk . Answer (a) 0, (b) 0, (c) a2 (T32 − T23 ), (d) −v2 Solution (a) Summing gives ε3jk aj ak = ε31k a1 ak + ε32k a2 ak = ε312 a1 a2 + ε321 a2 a1 = a1 a2 − a2 a1 = 0

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Chapter 2 Solutions 5 ttps://gioumeh.com/product/continuum-mechanics-for-engineers-solutions (b) εijk δkj

= εij1 δ1j + εij2 δ2j + εij3 δ3j = εi21 δ12 +FOLFNKHUHWRGRZQORDG εi31 δ13 + εi12 δ21 + εi32 δ23 + εi13 δ31 + εi23 δ32 = 0

(c) ε1jk a2 Tkj

= ε12k a2 Tk2 + ε13k a2 Tk3 = ε123 a2 T32 + ε132 a2 T23 = a2 T32 − a2 T23 = a2 (T32 − T23 )

(d) ε1jk δ3j vk = ε12k δ32 vk + ε13k δ33 vk = 0 + ε132 δ33 v2 = −v2

Problem 2.8 Consider the tensor Bik = εijk vj . (a) Show that Bik is skew-symmetric. (b) Let Bij be skew-symmetric, and consider the vector defined by vi = εijk Bjk (often called the dual vector of the tensor B). Show that Bmq = 12 εmqi vi . Solution (a) For a tensor to be skew-symmetric, one has Aij = −Aji . For the given tensor Bik = εijk vj = −εkji vj = −Bki (b) For the dual vector of the tensor B, we have εmqi vi = εmqi εijk Bjk = (δmj δqk − δmk δqj ) Bjk = Bmq − Bqm = [Bmq − (−Bmq )] = 2Bmq since B is skew-symmetric.

Problem 2.9 Use indicial notation to show that Ami εmjk + Amj εimk + Amk εijm = Amm εijk where A is any tensor and εijk is the permutation symbol. Solution Multiply both sides by εijk and simplify Amm εijk εijk = 6Amm

= Ami εmjk εijk + Amj εimk εijk + Amk εijm εijk = Ami 2δmi + Amj 2δmj + Amk 2δmk = 6Amm

Problem 2.10 If Aij = δij Bkk + 3Bij , determine Bkk and using that solve for Bij in terms of Aij and its first invariant, Aii . Answer

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6 Continuum Mechanics for Engineers ttps://gioumeh.com/product/continuum-mechanics-for-engineers-solutions Bkk = 61 Akk ; Bij =

1 A 3 ij

Solution Taking the trace of Aij gives

−

1 δ A 18 ij kk

FOLFNKHUHWRGRZQORDG

Aii = δii Bkk + 3Bii = 3Bkk + 3Bii = 6Bkk since i and k are dummy indices. This gives Bkk =

1 Akk 6

Substituting for Bkk and solving for Bij gives 1 δij Akk 6

3Bij = Aij −

or

1 1 δij Akk Bij = Aij − 18 3

Problem 2.11 Show that the value of the quadratic form Tij xi xj is unchanged if Tij is replaced by its symmetric part, 12 (Tij + Tji ). Solution The quadratic form becomes Tij xi xj =

1 1 1 (Tij + Tji )xi xj = (Tij xi xj + Tji xi xj ) = (Tij xi xj + Tij xj xi ) = Tij xi xj 2 2 2

since i and j are dummy indices and multiplication commutes.

Problem 2.12 With the aid of Eq 2.7, show that any skew symmetric tensor W may be written in terms of an axial vector ωi given by 1 ωi = − εijk wjk 2 where wjk are the components of W . Solution Multiply by εimn εimn ωi

= − 21 εimn εijk wjk = −21 (δmj δnk − δmk δnj ) wjk = −21 (wmn − wnm ) = wnm ,

or, εmni ωi = wnm Problem 2.13 Show by direct expansion (or otherwise) that the determinant   a1 a2   b1 b2   c1 c2

the box product λ = εijk ai bj ck is equal to  a3  b3  . c3 

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