ContinuumMechanics forEngineers

2 solutions for ContinuumMechanics forEngineers FourthEdition

• ThomasMase

RonaldE. Smelser JennStroudRossmann 1 / 4

Chapter 2 Solutions Problem 2.1 Letv=ab, or in indicial notation, vi^ei=aj^ejbk^ek="ijkajbk^ei Using indicial notation, show that, (a)vv=a 2 b 2 sin 2 , (b)aba=0, (c)abb=0.Solution (a) For the given vector, we have vv="ijkajbk^ei"pqsaqbs^ep="ijkajbk"pqsaqbsip="ijkajbk"iqsaqbs =(jqks-jskq)ajbkaqbs=ajajbkbk-ajbkakbj =(aa) (bb)-(ab) (ab)=a 2 b 2 -(abcos) 2 =a 2 b 2

1-cos 2

=a 2 b 2 sin 2

(b) Again, we nd aba=va=("ijkajbk^ei)aq^eq="ijkajbkaqiq="ijkajbkai=0 This is zero by symmetry iniandj.(c) This is abb=vb=("ijkajbk^ei)bq^eq="ijkajbkbqiq="ijkajbkbi=0 Again, this is zero by symmetry inkand andi.Problem 2.2 With respect to the triad of base vectorsu1,u2, andu3(not necessarily unit vectors), the triadu 1 ,u 2 , andu 3 is said to be areciprocal basisifuiu j =ij(i; j=1; 2; 3). Show that to satisfy these conditions, u 1 = u2u3 [u1;u2;u3] ;u 2 = u3u1 [u1;u2;u3] ;u 3 = u1u2 [u1;u2;u3] and determine the reciprocal basis for the specic base vectors u1=2^e1+^e2; u2=2^e2-^e3;

u3=^e1+^e2+^e3:

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4Continuum Mechanics for Engineers Answer u 1 = 1 5 (3^e1-^e2-2^e3) u 2 = 1 5 (-^e1+2^e2-^e3) u 3 = 1 5 (-^e1+2^e2+4^e3) Solution For the bases, we have u1u 1 =u1 u2u3 [u1;u2;u3] =1;u2u 2 =u2 u3u1 [u1;u2;u3] =1;u3u 3 =u3 u1u2 [u1;u2;u3] =1 since the triple scalar product is insensitive to the order of the operations. Now u2u 1 =u2 u2u3 [u1;u2;u3] =0 sinceu2u2u3=0from Pb 2.1. Similarly,u3u 1 =u1u 2 =u3u 2 =u1u 3 =u2u 3 =0.For the given vectors, we have [u1;u2;u3]=

• 1 0
• 2-1
• 1 1

=5 and u2u3=

^e1^e2^e3

0 2 -1

• 1 1

=3^e1-^e2-2^e3;u 1 = 1 5 (3^e1-^e2-2^e3) u3u1=

^e1^e2^e3

• 1 1
• 1 0

= -^e1+2^e2-^e3;u 2 = 1 5 (-^e1+2^e2-^e3) u1u2=

^e1^e2^e3

• 1 0

0 2 -1

= -^e1+2^e2+4^e3;u 3 = 1 5 (-^e1+2^e2+4^e3) Problem 2.3 If the base vectorsu1,u2, andu3are eigenvectors of a tensorA, prove that the reciprocal basis vectorsu 1 ,u 2 , andu 3 are eigenvectors of the tensor's transpose,A T .Problem 2.4 If the base vectorsu1,u2, andu3form an orthonormal triad, prove thatnknk=Iwhere Iis the identity matrix.Problem 2.5 Let the position vector of an arbitrary pointP(x1x2x3)bex=xi^ei, and letb=bi^eibe aconstant vector. Show that(x-b)x=0is the vector equation of a spherical surface having its center atx= 1 2 bwith a radius of 1 2

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Chapter 2 Solutions5 Solution For (x-b)x=(xi^ei-bi^ei)xj^ej=(xixj-bixj)ij=xixi-bixi= =x 2 1+x 2 2+x 2 3-b1x1-b2x2-b3x3=0 Now

x1- 1 2 b1

2 +

x2- 1 2 b2

2 +

x3- 1 2 b3

2 = 1 4

b 2 1+b 2 2+b 2 3

= 1 4 b 2 This is the equation of a sphere with the desired properties.Problem 2.6 Using the notationsA (ij)= 1 2 (Aij+Aji)andA [ij]= 1 2 (Aij-Aji)show that (a)<> the tensorAhaving componentsAijcan always be decomposed into a sum of its symmetricA (ij)and skew-symmetricA [ij]parts, respectively, by the decom- position, Aij=A (ij)+A [ij]; (b)<> the trace ofAis expressed in terms ofA (ij)by Aii=A (ii); (c)<> for arbitrary tensorsAandB, AijBij=A (ij)B (ij)+A [ij]B

[ij]:

Solution (a) The components can be written as Aij=

Aij+Aji 2

+

Aij-Aji 2

=A (ij)+A [ij] (b) The trace ofAis A (ii)=

Aii+Aii 2

=Aii (c) For two arbitrary tensors, we have AijBij=

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 A (ij)B [ij]=

Aij+Aji 2

Bij-Bji 2

= 1 4 (AijBij+AjiBij-AijBji-AjiBji) = 1 4 (AijBij+AjiBij-AjiBij-AijBij)=0

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