CONTINUUM MECHANICS for ENGINEERS

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strain. Highly cross-linked and filled rubbers can result in materials not intended for such large strains. Two-piece golf ball cores are much stiffer than a rubber band, for instance. Each of these products is designed for different strain regimes. A golf ball’s maximum strain would be on the order of 40 to 50%. Its highly crosslined constitution is made for resilience, not for large strain. The rubber stress-strain curve exhibits nonlinear behavior from the very beginning of its deformation, whereas steel has a linear regime below the yield stress. The reason rubber materials exhibit drastically different behavior than metals results from their sub-microscopic characteristics. Metals are crystalline lattices of atoms all being, more or less, well ordered: in contrast, rubber material molecules are made up of carbon atoms bonded into a long chain resembling a tangled collection of yarn scraps. Since the carbon-carbon (C- C) bond can rotate, it is possible for these entangled long chain polymers to rearrange themselves into an infinite number of different conformations. While the random coil can be treated as a spring, true resilience requires a cross-link to stop viscous flow. In a thermoset rubber, a chemical bond, often with sulfur, affords the tie while physical entanglements effect the same function in a TPE material. The degree of cross-linking is used to control the rubber’s stiffness. In the context of this book’s coverage of continuum mechanics, material make-up on the micro-scale is inconsistent with the continuum assumption discussed in Chapter One. However, a rubber elasticity model can be derived from the molecular level which somewhat represents material behavior at the macroscopic level. In this chapter, rubber elasticity will be developed from a first-principle basis. Following that, the traditional continuum approach is developed by assuming a form of the strain energy density and using restrictions on the constitutive response imposed by the second law of thermodynamics to obtain stress-stretch response. One of the major differences between a crystalline metal and an amorphous polymer is that the polymer chains have the freedom to rearrange themselves. The term conformation is used to describe the different spatial orientations of the chain. Physically, the ease at which different conformations are achieved results from the bonding between carbon atoms. As the carbon atoms join to form the polymer chain, the bonding angle is 109.5°, but there is also a rotational degree of freedom around the bond axis. For most macromolecules, the number of carbon atoms can range from 1,000 to 100,000. With each bond having a certain degree of freedom to orient itself, the number of conformations becomes quite large. Because of this substantial amount, the use of statistical thermodynamics may be used to arrive at rubber elasticity equations from first principles. In addition to the large number of conformations for a single chain, there is another reason the statistical approach is appropriate: the actual polymer has a large number of different individual chains making up the bulk of the material. For
FIGURE 8.2 A schematic comparison of molecular conformations as the distance between molecule’s ends varies. Dashed lines indicated other possible conformations. instance, a cubic meter of an amorphous polymer having 10,000 carbon atoms per molecule would have on the order of 1024 molecules (McCrum et al., 1997). Clearly, the sample is large enough to justify a statistical approach. At this point, consider one particular molecule, or polymer chain, and its conformations. The number of different conformations the chain can obtain depends on the distance separating the chain’s ends. If a molecule is formed of n segments each having length l , the total length would be L = n l . Separating the molecule’s ends, the length L would mean there is only one possible conformation keeping the chain intact. As the molecule’s ends get closer together, there are more possible conformations that can be obtained. Thus, a Gaussian distribution of conformations as a function of the distance between chain ends is appropriate. Figure 8.2 demonstrates how more conformations are possible as the distance between the molecule’s ends is reduced. Molecule end-to-end distance, r, is found by adding up all the segment lengths, l as is shown in Figure 8.3. Adding the segment lengths algebra- ically gives distance r from end-to-end, but it does not give an indication of the length of the chain. If the ends are relatively close and the molecule is long there will be the possibility of many conformations. Forming the magnitude squared of the end-to-end vector, r, in terms of the vector addition of individual segments r 2 = r⋅ r = ⎛l + l + L+ l ⎞ ⋅ ⎛l + l + L+ l ⎞ ⎝ ~ 1 ~ 2 ~ n⎠⎝~ 1 ~ 2 ~ n⎠ (8.1-1) where is the vector defining the ith l segment of the molecule chain. Mul- ~i tiplying out the right-hand side of Eq 8.1-1 leads to 2 2 r = nl + ⎡l ⋅ l + l ⋅ l + L+ l ⋅l ⎤ ~ 1 ~ 2 ~ 1 ~ 3 ~ n−1 ~ ⎣⎢ n⎦⎥ (8.1-2) This would be the square of the end-to-end vectors for one molecule of the polymer. In a representative volume of the material there would be many chains from which we may form the mean square end-to-end distance
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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
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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
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
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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
Page 282 and 283: where A 1 and A 2 are constants of
Page 284 and 285: Upon setting the off-diagonal terms
Page 286 and 287: 6.8 Show that the distortion energy
Page 288 and 289: 6.16 For an elastic body whose x 3
Page 290 and 291: and B are constants, determine the
Page 292 and 293: Thus, for the beam shown the stress
Page 294 and 295: Using the sketch on the facing page
Page 296 and 297: For a fluid in motion the shear str
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Page 300 and 301: posed in this section are relevant
Page 302 and 303: FIGURE E7.4-1A Flow down an incline
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Page 308 and 309: 7.9 Show that for an incompressible
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Page 314 and 315: FIGURE 8.3 A freely connected chain
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Page 320 and 321: Note that is essentially the same a
Page 322 and 323: where dependence on I 3 has been in
Page 324 and 325: where P 11 is the 11-component of t
Page 326 and 327: This simplifies to where and ∂p
Page 328 and 329: σ yy ⎡⎛ ∂X ⎞ ⎛ ∂Y ⎞
Page 330 and 331: q( x)= Ae + Be kx −kx Eqs 8.4-19a
Page 332 and 333: FIGURE 8.6 Stresses on deformed rub
Page 334 and 335: 8.2 Derive the following relationsh
Page 336 and 337: Answer: 8.9 Show F B iA = ij = g
Page 338 and 339: 9 Linear Viscoelasticity 9.1 Introd
Page 340 and 341: ˙ε ii ≈ Dii so that now, from E
Page 342 and 343: FIGURE 9.2 Mechanical analogy for s
Page 344 and 345: FIGURE 9.5 Three-parameter standard
Page 346 and 347: FIGURE 9.7 Graphic representation o
Page 348: Development of details relative to
Page 352 and 353: FIGURE 9.9 Stress history with an i
Page 354 and 355: γ ()= t γ ( cos ωt+ isinωt)= γ
Page 356 and 357: FIGURE 9.10 Shear lag in viscoelast
Page 358 and 359: But by Eq 9.6-7, σ 12 (t) = γ o [
Page 360 and 361: 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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