CONTINUUM MECHANICS for ENGINEERS

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Solution In terms of the base vectors , the given vector v is expressed by the equation We note here that indices i, j, and k appear four times in this line; however, the summation convention has not been violated. Terms that are separated by a plus or a minus sign are considered different terms, each having summation convention rules applicable within them. Vectors joined by a dot or cross product are not distinct terms, and the summation convention must be adhered to in that case. Carrying out the indicated multiplications, we see that v = = = = = = ê i ( i i j j) + k k i i × ( j j × k k) v= aeˆ ⋅neˆ n eˆ neˆ aeˆ n eˆ ( an i jδij) neˆ + n an k k ieˆi ×( εjkq j eˆ k q) anneˆ + ε ε nan eˆ i i k k iqm jkq i j k m anneˆ + ε ε nan eˆ i i k k miq jkq i j k m ( ) anneˆ + δ δ −δ δ nan eˆ i i k k mj ik mk ij i j k m anneˆ + naneˆ −naneˆ i i k k i j i j i i k k nnaeˆ = aeˆ =a i i j j j j Since a must equal v, this example demonstrates that the vector v may be resolved into a component ( v⋅nˆ) nˆ in the direction of ˆn , and a component nˆ × v× nˆ perpendicular to ˆn . ( ) Example 2.2-3 Using Eq 2.2-13, show that (a) ε and that (b) . (Recall mkqεjkq = 2δmj ε ε = 6 jkq jkq that δ = 3 and δ δ = δ .) kk mk kj mj Solution (a) Write out Eq 2.2-13 with indice i replaced by k to get ε ε = δ δ −δ δ mkq jkq mj kk mk kj = 3δ − δ = 2δ mj mj mj
(b) Start with the first equation in Part (a) and replace the index m with j, giving Example 2.2-4 Double-dot products of dyads are defined by (a) (uv) · · (ws) = (v · w) (u · s) (b) (uv): (ws) = (u · w) (v · s) Expand these products and compare the component <strong>for</strong>ms. Solution (a) (uv) · · (ws) = (b) (uv): (ws) = 2.3 Indicial Notation ε ε = δ δ −δ δ jkq jkq jj kk jk jk = ( 3)( 3)− δ = 9− 3= 6. ( ) ⋅ By assigning special meaning to the subscripts, the indicial notation permits us to carry out the tensor operations of addition, multiplication, differentiation, etc. without the use, or even the appearance of the base vectors in the equations. We simply agree that the tensor rank (order) of a term is indicated by the number of “free,” that is, unrepeated, subscripts appearing in that term. Accordingly, a term with no free indices represents a scalar, a term with one free index a vector, a term having two free indices a secondorder tensor, and so on. Specifically, the symbol êi λ = scalar (zeroth-order tensor) λ vi = vector (first-order tensor) v, or equivalently, its 3 components ui vj = dyad (second-order tensor) uv, or its 9 components Tij = dyadic (second-order tensor) T, or its 9 components Qijk = triadic (third-order tensor) Q or its 27 components Cijkm = tetradic (fourth-order tensor) C, or its 81 components jj veˆ ⋅ w eˆ ( u eˆ s eˆ )= vwu s i i j j k k q q i i k k ( ) ⋅ ueˆ ⋅ w eˆ ( v eˆ s eˆ )= uwv s i i j j k k q q i i k k
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: A sum of dyads such as is called a
Page 29 and 30: Solution By definition of symmetric
Page 31 and 32: Addition of matrices is commutative
Page 33 and 34: A A A det A = Aij = A A A A A A (2.
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
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FIGURE 3.9A Traction vector at poin
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this system the shear stresses, σ
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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
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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
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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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