CONTINUUM MECHANICS for ENGINEERS

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where m and n are positive integers, or zero. Also, we note that and if BB = A then but the square root is not unique. Example 2.4-2 Show that for arbitrary matrices A and B: (a) A B A B , + T T T ( ) = + ( )T T T (b) AB = B A and (c) IB = BI= B where I is the identity matrix. (2.4-5) (2.4-6) Solution (a) Let , then in element form and therefore C T A + B = C C is ij = Aij + Bij given by C = C = A + B = A + B T T T ij ji ji ji ij ij (b) Let AB = C, then in element form Hence, (AB) T = B T A T . (c) Let IB = C, then in element form n T T ( A ) = ( A ) B A A by the substitution property. Thus, IB = BI = B. The determinant of a square matrix is formed from the square array of elements of the matrix, and this array evaluated according to established mathematical rules. The determinant of the matrix A is designated by either det A, or by , and for a 3 × 3 matrix A, A ij or n = =( )12 / C = A B = A B = B A = C T T T T T ij ik kj ki jk jk ki ji C = δ B = B = B δ ij ik kj ij ik kj ( ) = + C = A + B A B T T T T
A A A det A = Aij = A A A A A A (2.4-7) A minor of det A is another determinant formed by deleting the ith row and jth column of . The cofactor of the element Aij (sometimes referred to as the signed minor) is defined by Mij Aij ( ) ij ( ) c A = −1 M (2.4-8) where superscript (c) denotes cofactor of matrix A. Evaluation of a determinant may be carried out by a standard method called expansion by cofactors. In this method, any row (or column) of the determinant is chosen, and each element in that row (or column) is multiplied by its cofactor. The sum of these products gives the value of the determinant. For example, expansion of the determinant of Eq 2.4-7 by the first row becomes A22 A23 A21 A23 A21 A22 det A = A11 − A12 + A13 (2.4-9) A A A A A A 32 33 which upon complete expansion gives 11 12 13 21 22 23 31 32 33 (2.4-10) Several interesting properties of determinants are worth mentioning at this point. To begin with, the interchange of any two rows (or columns) of a determinant causes a sign change in its value. Because of this property and because of the sign-change property of the permutation symbol, the det A of Eq 2.4-7 may be expressed in the indicial notation by the alternative forms (see Prob. 2.11) det A = ε AA A = ε AA A (2.4-11) Furthermore, following an arbitrary number of column interchanges with the accompanying sign change of the determinant for each, it can be shown from the first form of Eq 2.4-11 that, (see Prob 2.12) i+ j ij 31 33 det A = A ( A A −A A )−A A A −A A 31 32 ( ) 11 22 33 23 32 12 21 33 23 31 ( ) + A A A −A A 13 21 32 22 31 ijk i1 j2 k3 ijk 1i 2j 3k ε det A = ε AA A qmn ijk iq jm kn (2.4-12)
Page 1 and 2: CONTINUUM MECHANICS for ENGINEERS S
Page 3 and 4: Library of Congress Cataloging-in-P
Page 5 and 6: University, for helpful comments on
Page 7 and 8: Reference Section for constitutive
Page 9 and 10: Nomenclature x 1, x 2, x 3 or x i o
Page 11 and 12: W Strain energy per unit volume, or
Page 13 and 14: 4.5 The Material Derivative 4.6 Def
Page 15 and 16: 1 Continuum Theory 1.1 The Continuu
Page 17 and 18: 2 Essential Mathematics 2.1 Scalars
Page 19 and 20: FIGURE 2.1A Unit vectors in the coo
Page 21 and 22: 1. Addition of vectors: 2. Multipli
Page 23 and 24: and, furthermore, that for repeated
Page 25 and 26: A sum of dyads such as is called a
Page 27 and 28: (b) Start with the first equation i
Page 29 and 30: Solution By definition of symmetric
Page 31: Addition of matrices is commutative
Page 35 and 36: which is identical to Eq 2.4-9 and
Page 37 and 38: FIGURE 2.2A Rectangular coordinate
Page 39 and 40: Consider next an arbitrary vector v
Page 41 and 42: FIGURE E2.5-1 Vector νννν with
Page 43 and 44: or in expanded form ( Tij − λδi
Page 45 and 46: The transformation matrix here is o
Page 47 and 48: have a multiplicity of two, and det
Page 49 and 50: we write φ for ; vi,j for for ; an
Page 51 and 52: FIGURE 2.5B Bounding space curve C
Page 53 and 54: (b) the trace of A is expressed in
Page 55 and 56: 2.15 Using the square matrices belo
Page 57 and 58: (a) Show that a multiplicity of two
Page 59 and 60: 2.29 Transcribe the left-hand side
Page 61 and 62: 3 Stress Principles 3.1 Body and Su
Page 63 and 64: FIGURE 3.2A Typical continuum volum
Page 65 and 66: where S I and S II are the bounding
Page 67 and 68: FIGURE 3.5 Free body diagram of tet
Page 69 and 70: FIGURE 3.6 Cartesian stress compone
Page 71 and 72: (b) The equation of the plane ABC i
Page 73 and 74: or But xj,q = δjq and by Eq 3.4-3,
Page 75 and 76: FIGURE E3.5-1 Rotation of axes x 1
Page 77 and 78: FIGURE 3.9A Traction vector at poin
Page 79 and 80: this system the shear stresses, σ
Page 81 and 82: FIGURE 3.10B Table displaying direc
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1 Obviously, n from the second of t
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FIGURE 3.12 Normal and shear compon
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1 1 1 n1 = 0 , n2 = ± , n3 = ± ;
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exterior to circle C1. Thus, combin
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FIGURE 3.15A Reference angles φ an
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FIGURE E3.8-1 Three-dimensional Moh
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FIGURE 3.16A Mohr’s circle for pl
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FIGURE 3.18A Representative rotatio
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Solution For this stress state, the
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which demonstrates that ( q) is als
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Example 3.11-1 Determine directly t
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FIGURE P3.6 Stress vectors represen
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FIGURE P3.9 Cylinder of radius r an
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(b) Project each of the stress vect
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Determine (a) the principal stress
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(b) Verify the result determined in
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⎡ −12 − 12 + 3 2 −12 −32
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A motion of body B is a continuous
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emphasize, however, that the materi
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Example 4.2-1 Let the motion of a b
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and v = v(X,t) = v[χ -1 (x,t),t] =
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Additionally, with regard to the ma
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In this equation, the first term on
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Therefore, substituting u i for P i
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upon X, in which case the deformati
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form consistent with deformation an
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For diagonal BH, and for diagonal O
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which, upon factoring the left-hand
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FIGURE 4.4 A rectangular parallelep
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FIGURE 4.6A Rotated axes for plane
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Consider once more the two neighbor
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where dx is the deformed magnitude
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In a similar fashion, from dX (1)
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Thus from Eq 4.8-4 the principal st
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and as is obvious, the inverse (U*)
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Finally, from Eq 4.9-11, [ RAB]= =
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which becomes (after dividing both
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This rate of decrease in the angle
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FIGURE 4.8 Area dS° between vector
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FIGURE 4.9 Volume of parallelepiped
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(a) Show that the Jacobian determin
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4.5 The Lagrangian description of a
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u 2 = - x 2 - (x 1 + x 2)e -t /2 +
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and from it the longitudinal (norma
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FIGURE P4.27 Unit square OBCD in th
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Answer: (a) (b) 2 2 (c) (1) a1a 2a
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FIGURE P4.35 Circular cylinder in t
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(a) the components of acceleration
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FIGURE P4.47A Unit cube having diag
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5 Fundamental Laws and Equations 5.
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which upon application of the diver
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which is known as the continuity eq
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FIGURE 5.1 Material body in motion
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where ∆f is the resultant force a
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which reduces to where we have used
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Ε ABδ iAδ jB = e ij = 0(ε) o Ex
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this text we consider only mechanic
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outward normal is n i (hence the mi
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θ = η ∂u ∂ (5.8-2) Furthermor
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The entropy production is always po
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One of the uses for the Clausius-Du
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Furthermore, assume the temperature
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In the first, a continuum body’s
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FIGURE 5.2 Reference frames Ox 1x 2
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Recall the definition t + = t + a f
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where the last substitution comes f
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With the use of Eqs 5.10-29 and 5.1
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This being the case, it is clear fr
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as the invariance requirement on th
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i ∂ {P} = p (5.12-7a) i i ∂ t i
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* 5.2 Let the property P in Eq 5.2-
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5.12 Determine the form which the e
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σ = - ij pδ + τ ij ij σ ij = -
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a. b. 5.30 Show that the Jacobian t
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FIGURE 6.1 Uniaxial loading-unloadi
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where the factor of two on the shea
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ased on the strain energy function.
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( q) ( q) σ = (λδijεkk + 2µε
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FIGURE 6.2 Simple stress states: (a
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FIGURE 6.3 For plane stress: (a) ro
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x 1x 2 plane to be one of elastic s
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FIGURE 6.4 (continued) From the def
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A most important feature of the fie
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FIGURE 6.5 (a) Plane stress problem
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For the plane strain situation (Fig
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By inserting Eq 6.6-2 into Eq 6.6-1
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FIGURE E6.7-1 (a) Rectangular regio
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Solution It is easily verified, by
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(6.7-8b) (6.7-8c) in which φ = φ
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These conditions result in the foll
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Solution From Eq 6.7-8 the stress c
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FIGURE E6.7-5 Uniaxial loaded plate
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FIGURE 6.7 (a) Cylinder with self-e
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where ψ(x 1,x 2) is called the war
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Thus, By eliminating ψ from this p
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where λ is a constant. Thus, Φ is
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Equilibrium equations, Eq 6.4-1 σi
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It should be pointed out that while
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This equation may be written in a m
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where A 1 and A 2 are constants of
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Upon setting the off-diagonal terms
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6.8 Show that the distortion energy
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6.16 For an elastic body whose x 3
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and B are constants, determine the
Page 292 and 293:
Thus, for the beam shown the stress
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Using the sketch on the facing page
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For a fluid in motion the shear str
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The first of this pair relates the
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posed in this section are relevant
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FIGURE E7.4-1A Flow down an incline
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and thus which describes the pressu
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where dx i is a differential tangen
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7.9 Show that for an incompressible
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8 Nonlinear Elasticity 8.1 Molecula
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strain. Highly cross-linked and fil
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FIGURE 8.3 A freely connected chain
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proportional to the probability per
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The deformation is volume preservin
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Note that is essentially the same a
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where dependence on I 3 has been in
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where P 11 is the 11-component of t
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This simplifies to where and ∂p
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σ yy ⎡⎛ ∂X ⎞ ⎛ ∂Y ⎞
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q( x)= Ae + Be kx −kx Eqs 8.4-19a
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FIGURE 8.6 Stresses on deformed rub
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8.2 Derive the following relationsh
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Answer: 8.9 Show F B iA = ij = g
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9 Linear Viscoelasticity 9.1 Introd
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˙ε ii ≈ Dii so that now, from E
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FIGURE 9.2 Mechanical analogy for s
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FIGURE 9.5 Three-parameter standard
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FIGURE 9.7 Graphic representation o
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Development of details relative to
Page 352 and 353:
FIGURE 9.9 Stress history with an i
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γ ()= t γ ( cos ωt+ isinωt)= γ
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FIGURE 9.10 Shear lag in viscoelast
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But by Eq 9.6-7, σ 12 (t) = γ o [
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Depending upon the particular state
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() (9.7-9a) (9.7-9b) (9.7-9c) From
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and so on for higher derivatives. N
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This expression may be inverted wit
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9.4 Develop the constitutive equati
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9.7 For the model shown the stress
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9.11 For the model shown determine
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9.16 Let the stress relaxation func
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9.24 For the rather complicated mod
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9.28 A viscoelastic body in the for
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9.32 The deflection at x = L for an
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CONTINUUM MECHANICS for ENGINEERS

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