CONTINUUM MECHANICS for ENGINEERS

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We may demonstrate the tensor character of C by a consideration of the elastic constitutive equation when expressed in a rotated (primed) coordinate system in which it has the form (6.1-5) But by the transformation laws for second-order tensors, along with Eq 6.1-3, = which by a direct comparison with Eq 6.1-5 provides the result (6.1-6) that is, the transformation rule for a fourth-order Cartesian tensor. In general, the C ijkm coefficients may depend upon temperature, but here we assume adiabatic (no heat gain or loss) and isothermal (constant temperature) conditions. We also shall ignore strain-rate effects and consider the components C ijkm to be at most a function of position. If the elastic coefficients are constants, the material is said to be homogeneous. These constants are those describing the elastic properties of the material. The constitutive law given by Eq 6.1-3 is known as the generalized Hooke’s law. For certain purposes it is convenient to write Hooke’s law using a single subscript on the stress and strain components and double subscripts on the elastic constants. To this end, we define and σ 11 = σ 1 σ 23 = σ 32 = σ 4 σ 22 = σ 2 σ 31 = σ 13 = σ 5 (6.1-7a) σ 33 = σ 3 ε 11 = ε 1 σ 12 = σ 21 = σ 6 2ε 23 = 2ε 32 = ε 4 ε 22 = ε 2 2ε 31 = 2ε 13 = ε 5 (6.1-7b) ε 33 = ε 3 σ′ = C′ ε′ ij ijpn pn σ′ = a a σ =a a C ε ij iq js qs iq js qskm km a a C a a ε′ iq js qskm pk nm pn ′ = Cijpn aiqajsapkanmCqskm 2ε 12 = 2ε 21 = ε 6
where the factor of two on the shear strain components is introduced in keeping with Eq 4.7-14. From these definitions, Hooke’s law is now written σα = Cαβ εβ or = Cε (6.1-8) with Greek subscripts having a range of six. In matrix form Eq 6.1-8 appears as ⎡σ 1⎤ ⎡C C C C C C ⎢ σ ⎥ ⎢ ⎢ 2 C C C C C C ⎥ ⎢ ⎢σ ⎥ ⎢ 3 C C C C C C ⎢ ⎥ = ⎢ ⎢σ 4⎥ ⎢C C C C C C ⎢σ ⎥ ⎢ 5 C C C C C C ⎢ ⎥ ⎢ ⎣⎢ σ 6⎦⎥ ⎣⎢ C C C C C C 11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36 41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64 65 66 ⎤ ⎡ε1 ⎤ ⎥ ⎢ ε ⎥ ⎥ ⎢ 2⎥ ⎥ ⎢ε ⎥ 3 ⎥ ⎢ ⎥ ⎥ ⎢ε4 ⎥ ⎥ ⎢ε ⎥ 5 ⎥ ⎢ ⎥ ⎦⎥ ⎣⎢ ε6 ⎦⎥ (6.1-9) We point out that the array of the 36 constants C αβ does not constitute a tensor. In view of our assumption to neglect thermal effects at this point, the energy balance Eq 5.7-13 is reduced to the form ˙u= 1 ijDij ρ σ which for small-deformation theory, by Eq 4.10-18, becomes ˙ ˙ u= 1 ij ij ρ σε (6.1-10a) (6.1-10b) The internal energy u in these equations is purely mechanical and is called the strain energy (per unit mass). Recall now that, by the continuity equation in Lagrangian form, ρ o = ρJ and also that to the first order of approximation ⎛ ∂ u ⎞ i ∂ ui J = det F = det⎜δ iA + (6.1-11) ⎝ ∂ X ⎟ ≈ + ⎠ ∂ X 1 Therefore, from our assumption of small displacement gradients, namely ∂u i/∂X A
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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
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where S I and S II are the bounding
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FIGURE 3.5 Free body diagram of tet
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FIGURE 3.6 Cartesian stress compone
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(b) The equation of the plane ABC i
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or But xj,q = δjq and by Eq 3.4-3,
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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
Page 181 and 182: 5 Fundamental Laws and Equations 5.
Page 183 and 184: which upon application of the diver
Page 185 and 186: which is known as the continuity eq
Page 187 and 188: FIGURE 5.1 Material body in motion
Page 189 and 190: where ∆f is the resultant force a
Page 191 and 192: which reduces to where we have used
Page 193 and 194: Ε ABδ iAδ jB = e ij = 0(ε) o Ex
Page 195 and 196: this text we consider only mechanic
Page 197 and 198: outward normal is n i (hence the mi
Page 199 and 200: θ = η ∂u ∂ (5.8-2) Furthermor
Page 201 and 202: The entropy production is always po
Page 203 and 204: One of the uses for the Clausius-Du
Page 205 and 206: Furthermore, assume the temperature
Page 207 and 208: In the first, a continuum body’s
Page 209 and 210: FIGURE 5.2 Reference frames Ox 1x 2
Page 211 and 212: Recall the definition t + = t + a f
Page 213 and 214: where the last substitution comes f
Page 215 and 216: With the use of Eqs 5.10-29 and 5.1
Page 217 and 218: This being the case, it is clear fr
Page 219 and 220: as the invariance requirement on th
Page 221 and 222: i ∂ {P} = p (5.12-7a) i i ∂ t i
Page 223 and 224: * 5.2 Let the property P in Eq 5.2-
Page 225 and 226: 5.12 Determine the form which the e
Page 227 and 228: σ = - ij pδ + τ ij ij σ ij = -
Page 229 and 230: a. b. 5.30 Show that the Jacobian t
Page 231: FIGURE 6.1 Uniaxial loading-unloadi
Page 235 and 236: ased on the strain energy function.
Page 237 and 238: ( q) ( q) σ = (λδijεkk + 2µε
Page 240 and 241: FIGURE 6.2 Simple stress states: (a
Page 242 and 243: FIGURE 6.3 For plane stress: (a) ro
Page 244 and 245: x 1x 2 plane to be one of elastic s
Page 246 and 247: FIGURE 6.4 (continued) From the def
Page 248 and 249: A most important feature of the fie
Page 250 and 251: FIGURE 6.5 (a) Plane stress problem
Page 252 and 253: For the plane strain situation (Fig
Page 254 and 255: By inserting Eq 6.6-2 into Eq 6.6-1
Page 256 and 257: FIGURE E6.7-1 (a) Rectangular regio
Page 258 and 259: Solution It is easily verified, by
Page 260 and 261: (6.7-8b) (6.7-8c) in which φ = φ
Page 262 and 263: These conditions result in the foll
Page 264 and 265: Solution From Eq 6.7-8 the stress c
Page 266 and 267: FIGURE E6.7-5 Uniaxial loaded plate
Page 268 and 269: FIGURE 6.7 (a) Cylinder with self-e
Page 270 and 271: where ψ(x 1,x 2) is called the war
Page 272 and 273: Thus, By eliminating ψ from this p
Page 274 and 275: where λ is a constant. Thus, Φ is
Page 276 and 277: Equilibrium equations, Eq 6.4-1 σi
Page 278 and 279: It should be pointed out that while
Page 280 and 281: 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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