CONTINUUM MECHANICS for ENGINEERS

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FIGURE E3.3-1 Plane P defined by points A, B, and C. Example 3.3-1 Let the components of the stress tensor at P be given in matrix form by [ σ ji]= in units of mega-Pascals. Determine (a) the stress vector on the plane at P having the unit normal (b) the stress vector on a plane at P parallel to the plane ABC shown in the sketch. Solution (a) From Eq 3.3-13 for the given data, ⎡ 21 −63 42⎤ ⎢ ⎥ ⎢ −63 0 84 ⎥ ⎣ ⎢ 42 84 −21⎦ ⎥ 1 ˆ ⎛ n= ⎝2eˆ − 3eˆ + 6eˆ 7 1 2 ⎞ 3⎠ ⎡ 21 −63 42⎤ 2 3 t ( nˆ) ˆ ˆ 6 1 , t ( n) 2 , t ( n) ⎡ ⎤ ⎢ ⎥ 3 = , − , −63 0 84 69 54 42 ⎣⎢ 7 7 7 ⎦⎥ ⎢ ⎥ = − ⎣ ⎢ 42 84 −21⎦ ⎥ [ ] [ ] or, in vector form, t ( nˆ ) = 69e ˆ + e ˆ ± eˆ . This vector represents the com- 1 54 2 42 3 ponents of the force per unit area (traction) on the plane defined by ⎡2 ⎣⎢ 7 3 − 7 6 ⎤ . 7 ⎦⎥
(b) The equation of the plane ABC in the sketch is easily verified to be 2x1 + 1 2x2 + x3 = 2, and the unit outward normal to this plane is ˆ ⎛ ˆ ˆ ˆ ⎞ n= ⎝2e1+ 2e2 + e3⎠ 3 so that, again from Eq 3.3-13, ⎡ 21 −63 42⎤ 2 2 t ( nˆ) ˆ ˆ 1 1 , t ( n) 2 , t ( n) ⎡ ⎤ ⎢ ⎥ 3 = , , −63 0 84 14 14 77 ⎣⎢ 3 3 3⎦⎥ ⎢ ⎥ = − − ⎣ ⎢ 42 84 −21⎦ ⎥ [ ] [ ] or, in vector form, t ( nˆ ) = −14eˆ eˆ eˆ . In this example, we clearly see 1− 14 2 + 77 3 the dependency of the cutting plane and the stress vector. Here, we have considered two different cutting planes at the same point and found that two distinct traction vectors arose from the given stress tensor components. 3.4 Force and Moment Equilibrium, Stress Tensor Symmetry In the previous section, we used a balance-of-forces condition for a tetrahedron element of a body in equilibrium to define the stress tensor and to develop the Cauchy stress formula. Here, we employ a force balance on the body as a whole to derive what are known as the local equilibrium equations. This set of three differential equations must hold for every point in a continuum body that is in equilibrium. As is well known, equilibrium also requires the sum of moments to be zero with respect to any fixed point, and we use this condition, together with the local equilibrium equations, to deduce the fact that the stress tensor is symmetric in the absence of concentrated body moments. Consider a material body having a volume V and a bounding surface S. Let the body be subjected to surface tractions t ( ˆn ) and body forces bi (force i per unit mass), as shown by Figure 3.7. As before, we exclude concentrated body moments from consideration. Equilibrium requires that the summation of all forces acting on the body be equal to zero. This condition is expressed by the global (integral) equation representing the sum of the total surface and body forces acting on the body, ∫ ∫ t ( ˆn ) dS+ ρbdV = 0 i i S V (3.4-1) where dS is the differential element of the surface S and dV that of volume V. Because t ( ˆn ) as a result of Eq 3.3-8, the divergence theorem Eq 2.8-1 i = σ jinj allows the first term of Eq 3.4-1 to be written as ∫ ji j S ∫V σ ndS= σ dV ji,j
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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
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 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: FIGURE 3.6 Cartesian stress compone
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 and 116: ⎡ −12 − 12 + 3 2 −12 −32
Page 117 and 118: A motion of body B is a continuous
Page 119 and 120: 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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