CONTINUUM MECHANICS for ENGINEERS

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One more application of the Clausius-Duhem inequality for a special process will lead to an expression for the Cauchy stress in terms of the free energy. For this process, select the temperature gradient to be an arbitrary constant and the time rate-of-change of the temperature gradient to be identically zero. Eq 5.9-5 becomes (5.9-14) which must hold for all velocity gradients L ij. Picking a L ij that would violate the inequality unless the coefficient of L ij vanishes implies (5.9-15) In arriving at Eq 5.9-14 it was noted that the free energy had already been shown to be independent of the temperature gradient by virtue of Eq 5.9-11. Thus, g k was dropped from the independent variable list in Eq 5.9-14. Finally, substituting the results of Eqs 5.9-15, 5.9-13, and 5.9-11 into 5.9-5, a last restriction for an elastic material is found to be 5.10 Invariance ˜ σ˜ , θ, ρ ˜ , , ψ ⎛ o ∂ ⎞ 1 o o ⎜ ij( F g θ kB k )−FjA Lij qiF g g kB k i ⎝ ∂F ⎠ θ iA ⎟ − ( ) ≥ σ σ θ ρ ψ ij ij FkB F F = ∂ ˜ ˜ ( , )= ∂ ˜qg i i ≤ 0 (5.9-16) The concept of invariance has been discussed in Section 2.5 with respect to tensors. A tensor quantity is one which remains invariant under admissible coordinate transformations. In other words, all the different stress components represented by Mohr’s circle refer to a single stress state. This invariance is crucial for consolidating different stress components into a yield criterion such as the maximum shearing stress (or Tresca criterion). Invariance plays another important role in continuum mechanics. Requiring a continuum to be invariant with regards to reference frame, or to have unchanged response when a superposed rigid body motion is applied to all material points, produces some significant results. The most important of these consequences might be restrictions placed on constitutive models, as discussed in the next section. There are two basic methods for examining invariance of constitutive response functions: material frame indifference and superposed rigid body motion. iA jA 0
In the first, a continuum body’s response to applied forces or prescribed motion must be the same as observed from two different reference frames. The body and the applied forces remain the same; only the observer’s reference frame changes. In superposing a rigid body motion to the body, the observer maintains the same reference frame. Here, each material point has a superposed motion added to it. The forces applied to the body are rotated with the superposed motion. Both of these methods produce the same restrictions on constitutive responses within the context of this book. In this text, the method of superposed rigid body motion will be presented as the means for enforcing invariance. The superposition of a rigid motion along with a time shift can be applied to the basic definition of the motion x i = χ i(X A, t) (5.10-1) From this state a superposed rigid body motion is applied, maintaining all relative distances between material points. Also, since the motion is a function of time, a time shift is imposed on the motion. After the application of the superposed motion, the position of the material point becomes x i + at time t + = t + a where a is a constant. A superscript “+” is used for quantities having the superposed motion. Some literature uses a superscript “*” to denote this, but since we use this to represent principal stress quantities, the “+” is used. The motion in terms of the superposed motion is written as + + x χ X , t = ( ) i i A (5.10-2) Assuming sufficient continuity, the motion can be written in terms of the current configuration since X A = χ –1 (x i, t). That is to say ˜χ i + + + x χ X , t ˜ χ x , t = ( )= ( ) i i A i j (5.10-3) where is written because the substitution of XA = χ –1 + (xi, t) results in a + different function than χ . i To represent the relative distance between two particles, a second material point is selected. In an analogous manner, it is straightforward to determine + + + y χ Y , t ˜ χ y , t = ( )= ( ) i i A i j Relative distance between material particles X A and Y A is written as ( ) − [ ] [ ( )− ( ) ] ( )= ( )− ( ) + + + + x − y x y ˜ χ x , t ˜ χ y , t ˜ χ x , t ˜ χ y , t i i i i i j i j i j i j (5.10-4) (5.10-5)
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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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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
Page 167 and 168: u 2 = - x 2 - (x 1 + x 2)e -t /2 +
Page 169 and 170: and from it the longitudinal (norma
Page 171 and 172: FIGURE P4.27 Unit square OBCD in th
Page 173 and 174: Answer: (a) (b) 2 2 (c) (1) a1a 2a
Page 175 and 176: FIGURE P4.35 Circular cylinder in t
Page 177 and 178: (a) the components of acceleration
Page 179 and 180: 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: Furthermore, assume the temperature
Page 209 and 210: FIGURE 5.2 Reference frames Ox 1x 2
Page 211 and 212: Recall the definition t + = t + a f
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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 and 232: FIGURE 6.1 Uniaxial loading-unloadi
Page 233 and 234: where the factor of two on the shea
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
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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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CONTINUUM MECHANICS for ENGINEERS

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