CONTINUUM MECHANICS for ENGINEERS

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where P 11 is the 11-component of the Piola-Kirchhoff stress. The neo-Hookean material is the simplest form of the strain energy function and makes exact solutions much more tractable. A slightly more general model is a simple, or two-term, Mooney-Rivlin model. In this case, the strain energy function is assumed to be linear in the first and second invariants of the left deformation tensor. Again, assuming the material is isotropic and incompressible, the strain energy may be written as ⎛ ⎞ 2 2 2 1 1 1 W = C1( λ1+ λ2+ λ3−3)+ C2⎜ + + −3 2 2 2 ⎟ ⎝ λ λ λ ⎠ (8.3-4) 2 2 2 2 1 For a uniaxial test, the principal stretches are λ . Substi- 1 = λ λ2= λ3= λ tution of these stretches into Eq 8.3-4 and differentiating with respect to λ gives the uniaxial stress per unit undeformed area − , f P C C ⎛ 1 ⎞⎛ 2 ⎞ = 11 = 2 λ − ⎜ + 2 1 ⎟ ⎝ λ ⎠⎝ λ ⎠ (8.3-5) The Cauchy stress is easily formed by noting the area in the deformed configuration would be found by scaling dimensions in the x 2 and x 3 direction by λ 2 and λ 3, respectively. For the uniaxially case, this means that multiplying the force per undeformed area by stretch λ results in the uniaxial Cauchy stress ⎛ 2 1 ⎞⎛ 2 ⎞ σ11 = 2 λ − ⎜C1 + ⎟ ⎝ λ⎠⎝λ ⎠ C (8.3-6) Making use of Eq 4.8-8 while noting Λ in that equation is λ in Eq 8.3-6, a formula for stress in terms of strain is obtained ⎛ = ⎜ + + ⎝ σ 11 2C11 2E11 1 1+ 2E (8.3-7) Series expansion of the last three terms of Eq 8.3-7 followed by assuming E 11 is small results in (8.3-8) Hence, for small strain the modulus of elasticity may be written as E = 6(C 1 + C 2). Furthermore, since an incompressible material is assumed, Poisson’s ratio is equal to 0.5. This means, by virtue of Eq 6.2-8a, that the shear modulus is given by G = 3(C 1 + C 2) for small strain. ⎞ 1 ⎟ 2 11 11 ⎠ + + + ⎝ + ( ) σ 11 = 6 C1+ C2E11 2 ⎛ 2C ⎜ 1 2E 3 1 1 2E 11 ⎞ ⎟ ⎠
The simple, two-term Mooney-Rivlin model represents material response well for small to moderate stretch values, but for large stretch higher order terms are needed. Some of these different forms for the strain energy function are associated with specific names. For instance, an Ogden material (Ogden, 1972) assumes a strain energy in the form n n n n W = ∑ + + − µ α α α λ λ λ α (8.3-9) This form reduces to the Mooney-Rivlin material if n takes on values of 1 and 2 with α 1 = 2, α 2 = –2, µ 1 = 2C 1, and µ 2 = –2C 2. 8.4 Exact Solution for an Incompressible, Neo-Hookean Material n n [ 1 2 3 3] Exact solutions for nonlinear elasticity problems come from using the equations of motion, Eq 5.4-4, or for equilibrium, Eq 5.4-5, along with appropriate boundary conditions. A neo-Hookean material is one whose uniaxial stress response is proportional to the combination of stretch λ –1/λ 2 as was stated in Sec. 8.1. The proportionality constant is the shear modulus, G. The strain energy function in the neo-Hookean case is proportional to I 1 – 3 where I 1 is the first invariant of the left deformation tensor. A strain energy of this form leads to stress components given by σ =− pδ + GB ij ij ij (8.4-1) where p is the indeterminate pressure term, G is the shear modulus, and B ij is the left deformation tensor. The stress components must satisfy the equilibrium equations which, in the absence of body forces, are given as σ ij, j = 0 (8.4-2) Substitution of Eqs 4.6-16 and 8.4-1 into Eq 8.4-2 gives a differential equation governing equilibrium ∂ ∂x σ ij =− ∂p + ∂x j i ⎛ ∂ ∂ ⎞ ∂ ⎛ ∂ ∂ ⎞ ⎜ ⎟ + ⎜ ⎟ ⎝ ∂ ∂ ∂ ⎠ ∂ ⎝ ∂ ∂ ∂ ⎠ ∂ ⎡ ⎤ ⎢ ⎥ ∂ ⎣ ⎢ ⎦ ⎥ = 2 2 xi X x B j xj XB xi G 0 XA XB xj XA XA XB xj XA (8.4-3)
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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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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
Page 310 and 311: 8 Nonlinear Elasticity 8.1 Molecula
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
Page 318 and 319: The deformation is volume preservin
Page 320 and 321: Note that is essentially the same a
Page 322 and 323: where dependence on I 3 has been in
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
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
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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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