CONTINUUM MECHANICS for ENGINEERS

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FIGURE 9.10 Shear lag in viscoelastic material: (a) rotating so and go vectors, (b) stress-strain time curves. together with the resulting strain σ σ ω i t e 12 = o γ 12 γ = * e which when substituted into Eq 9.3-1 leads to * γ σ o (9.6-13) (9.6-14) (9.6-15) where the minus sign reflects the fact that the strain lags the stress by the loss angle δ, defined in this case by (9.6-16) In analogy with the complex modulus components, J′(ω) is called the storage compliance, J″(ω) the loss compliance, and iωt * = J ( iω)= J′( ω)− iJ′′( ω) ′( ) ω tan δ = ω ′′( ) J J 2 2 ˜J γ o ( ω)= [ J′( ω) ] + [ J′′( ω) ] = σ o (9.6-17)
the absolute compliance, in which γ o is the peak value of the strain as given by (9.6-18) Based upon the definitions of G * and J * , it is clear that these complex quantities are reciprocals of one another. Thus (9.6-19) A simple procedure for calculating G * (or J * ) of a specific model or material is to replace the partial differential operator ∂ t in the material’s constitutive equation by iω and solve the resulting algebraic equation for the ratio σ 12/γ 12 = G * , (or γ 12/σ 12 = J * if that is the quantity required). Accordingly, from Eq 9.3-6 for the Kelvin solid, the constitutive equation σ 12 = {G + η ∂ t }γ 12 becomes σ 12 = (G + iηω)γ 12 which yields σ 12/γ 12 = G(1 + iτω) = G * . Likewise, from Eq 9.3-7 for the Maxwell fluid, {∂ t + 1/τ}σ 12 = {G∂ t}γ 12 becomes (iω + 1/τ) σ 12 = (Giω)γ 12 from which σ 12/γ 12 = G(τ 2 ω 2 + iτω)/(1 + τ 2 ω 2 ) = G * . In developing the formulas for the complex modulus and the complex compliance we have used the differential operator form of the fundamental viscoelastic constitutive equations. Equivalent expressions for these complex quantities may also be derived using the hereditary integral form of constitutive equations. To this end we substitute Eq 9.6-2 into Eq 9.5-5a. However, before making this substitution it is necessary to decompose the stress relaxation function G(t) into two parts as follows, G(t) = G o [1 – φ(t)] (9.6-20) where G 0 = G(0), the value of G(t) at time t = 0. Following this decomposition and the indicated substitution, Eq 9.5-5a becomes (9.6-21) In this equation, let t – t′ = ξ so that dt′ = –dξ and such that when t′ = t, ξ = 0, and when t′ = –∞, ξ = ∞. Now which reduces to γ J˜ ω σ = ( ) o o ( ) ( )= * ∗ G iω J iω iωt′ σ ()= t iωγ G e G 1−φ t− t′ dt 12 o o ∫ t −∞ ω σ12 t iωtG ω e i t ⎧ o⎨ ⎩ i ∫0 o 1 [ ( ) ] ′ iωξ iωt ⎫ e e φξ dξ⎬ ⎭ ()= − ( ) ∞ ⎧ ⎫ σ12()= t γ o⎨Go − Go ( iωcosωξ+ ωsinωξ) φ( ξ) dξ⎬e ⎩ ∫0 ⎭ ∞ iωt (9.6-22) (9.6-23)
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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
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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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Page 308 and 309: 7.9 Show that for an incompressible
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Page 312 and 313: strain. Highly cross-linked and fil
Page 314 and 315: FIGURE 8.3 A freely connected chain
Page 316 and 317: proportional to the probability per
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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 358 and 359: But by Eq 9.6-7, σ 12 (t) = γ o [
Page 360 and 361: Depending upon the particular state
Page 362 and 363: () (9.7-9a) (9.7-9b) (9.7-9c) From
Page 364 and 365: and so on for higher derivatives. N
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Page 368 and 369: 9.4 Develop the constitutive equati
Page 370 and 371: 9.7 For the model shown the stress
Page 372 and 373: 9.11 For the model shown determine
Page 374 and 375: 9.16 Let the stress relaxation func
Page 376 and 377: 9.24 For the rather complicated mod
Page 378 and 379: 9.28 A viscoelastic body in the for
Page 380: 9.32 The deflection at x = L for an
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