CONTINUUM MECHANICS for ENGINEERS

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4 Kinematics of Deformation and Motion 4.1 Particles, Configurations, Deformation, and Motion In continuum mechanics we consider material bodies in the form of solids, liquids, and gases. Let us begin by describing the model we use to represent such bodies. For this purpose we define a material body B as the set of elements X, called particles or material points, which can be put into a oneto-one correspondence with the points of a regular region of physical space. Note that whereas a particle of classical mechanics has an assigned mass, a continuum particle is essentially a material point for which a density is defined. The specification of the position of all of the particles of B with respect to a fixed origin at some instant of time is said to define the configuration of the body at that instant. Mathematically, this is expressed by the mapping x = κ(X) (4.1-1) in which the vector function κ assigns the position x relative to some origin of each particle X of the body. Assume that this mapping is uniquely invertible and differentiable as many times as required; in general, two or three times will suffice. The inverse is written X = κ –1 (x) (4.1-2) and identifies the particle X located at position x. A change in configuration is the result of a displacement of the body. For example, a rigid-body displacement is one consisting of a simultaneous translation and rotation which produces a new configuration but causes no changes in the size or shape of the body, only changes in its position and/or orientation. On the other hand, an arbitrary displacement will usually include both a rigid-body displacement and a deformation which results in a change in size, or shape, or possibly both.
A motion of body B is a continuous time sequence of displacements that carries the set of particles X into various configurations in a stationary space. Such a motion may be expressed by the equation x = κ(X,t) (4.1-3) which gives the position x for each particle X for all times t, where t ranges from – ∞ to + ∞. As with configuration mappings, we assume the motion function in Eq 4.1-3 is uniquely invertible and differentiable, so that we may write the inverse X = κ –1 (x,t) (4.1-4) which identifies the particle X located at position x at time t. We give special meaning to certain configurations of the body. In particular, we single out a reference configuration from which all displacements are reckoned. For the purpose it serves, the reference configuration need not be one the body ever actually occupies. Often, however, the initial configuration, that is, the one which the body occupies at time t = 0, is chosen as the reference configuration, and the ensuing deformations and motions related to it. The current configuration is that one which the body occupies at the current time t. In developing the concepts of strain, we confine attention to two specific configurations without any regard for the sequence by which the second configuration is reached from the first. It is customary to call the first (reference) state the undeformed configuration, and the second state the deformed configuration. Additionally, time is not a factor in deriving the various strain tensors, so that both configurations are considered independent of time. In fluid mechanics, the idea of specific configurations has very little meaning since fluids do not possess a natural geometry, and because of this it is the velocity field of a fluid that assumes the fundamental kinematic role. 4.2 Material and Spatial Coordinates Consider now the reference configuration prescribed by some mapping function Φ such that the position vector X of particle X relative to the axes OX 1X 2X 3 of Figure 4.1 is given by X = Φ(X) (4.2-1) In this case we may express X in terms of the base vectors shown in the figure by the equation X = XAIˆ A (4.2-2)
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CONTINUUM MECHANICS for ENGINEERS S
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Library of Congress Cataloging-in-P
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University, for helpful comments on
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Reference Section for constitutive
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Nomenclature x 1, x 2, x 3 or x i o
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W Strain energy per unit volume, or
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4.5 The Material Derivative 4.6 Def
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1 Continuum Theory 1.1 The Continuu
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2 Essential Mathematics 2.1 Scalars
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FIGURE 2.1A Unit vectors in the coo
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1. Addition of vectors: 2. Multipli
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and, furthermore, that for repeated
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A sum of dyads such as is called a
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(b) Start with the first equation i
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Solution By definition of symmetric
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Addition of matrices is commutative
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A A A det A = Aij = A A A A A A (2.
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which is identical to Eq 2.4-9 and
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FIGURE 2.2A Rectangular coordinate
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Consider next an arbitrary vector v
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FIGURE E2.5-1 Vector νννν with
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or in expanded form ( Tij − λδi
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The transformation matrix here is o
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have a multiplicity of two, and det
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we write φ for ; vi,j for for ; an
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FIGURE 2.5B Bounding space curve C
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(b) the trace of A is expressed in
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2.15 Using the square matrices belo
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(a) Show that a multiplicity of two
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2.29 Transcribe the left-hand side
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3 Stress Principles 3.1 Body and Su
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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
Page 83 and 84: 1 Obviously, n from the second of t
Page 85 and 86: FIGURE 3.12 Normal and shear compon
Page 87 and 88: 1 1 1 n1 = 0 , n2 = ± , n3 = ± ;
Page 89 and 90: exterior to circle C1. Thus, combin
Page 91 and 92: FIGURE 3.15A Reference angles φ an
Page 93 and 94: FIGURE E3.8-1 Three-dimensional Moh
Page 95 and 96: FIGURE 3.16A Mohr’s circle for pl
Page 97 and 98: FIGURE 3.18A Representative rotatio
Page 99 and 100: Solution For this stress state, the
Page 101 and 102: which demonstrates that ( q) is als
Page 103 and 104: Example 3.11-1 Determine directly t
Page 105 and 106: FIGURE P3.6 Stress vectors represen
Page 107 and 108: FIGURE P3.9 Cylinder of radius r an
Page 109 and 110: (b) Project each of the stress vect
Page 111 and 112: Determine (a) the principal stress
Page 113 and 114: (b) Verify the result determined in
Page 115: ⎡ −12 − 12 + 3 2 −12 −32
Page 119 and 120: emphasize, however, that the materi
Page 121 and 122: Example 4.2-1 Let the motion of a b
Page 123 and 124: and v = v(X,t) = v[χ -1 (x,t),t] =
Page 125 and 126: Additionally, with regard to the ma
Page 127 and 128: In this equation, the first term on
Page 129 and 130: Therefore, substituting u i for P i
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Page 135 and 136: For diagonal BH, and for diagonal O
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Page 139 and 140: FIGURE 4.4 A rectangular parallelep
Page 141 and 142: FIGURE 4.6A Rotated axes for plane
Page 143 and 144: Consider once more the two neighbor
Page 145 and 146: where dx is the deformed magnitude
Page 147 and 148: In a similar fashion, from dX (1)
Page 149 and 150: Thus from Eq 4.8-4 the principal st
Page 151 and 152: and as is obvious, the inverse (U*)
Page 153 and 154: Finally, from Eq 4.9-11, [ RAB]= =
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Page 157 and 158: This rate of decrease in the angle
Page 159 and 160: FIGURE 4.8 Area dS° between vector
Page 161 and 162: FIGURE 4.9 Volume of parallelepiped
Page 163 and 164: (a) Show that the Jacobian determin
Page 165 and 166: 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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