CONTINUUM MECHANICS for ENGINEERS

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internal energy u (5.8-9a) free energy ψ = −ηθ (5.8-9b) enthalpy (5.8-9c) u χ = u −∑τa a v free enthalpy ζ = χ − ηθ = u −ηθ −∑τ (5.8-9d) a a v These potentials are related through the relationship (5.8-10) All of the energy potentials may be written in terms of any one of the following independent variable sets η, v a; θ, v a; η, τ a; θ, τ a (5.8.11) In order to describe the motion of a purely mechanical continuum the function x i = x i(X A, t) is needed. Adding the thermodynamic response requires the addition of temperature, θ, or, equivalently, entropy, η, both being a function of position and time θ = θ(X A, t) or η = η(X A, t) (5.8-12) When considered for a portion P of the body, the total entropy is given as and the entropy production in the portion P is given by a u − ψ + ζ − χ = 0 Η= ∫ ρηdV P Γ= ∫ ργ dV P (5.8-13) (5.8-14) where the scalar γ is the specific entropy production. The second law can be stated as follows: the time rate-of-change in the entropy equals the change in entropy due to heat supply, heat flux entering the portion, plus the internal entropy production. For a portion P of the body, this is written as d ρr qi⋅ni ρηdV = dV − dS + ργ dV dt∫ ∫ θ ∫ θ ∫ P P a ∂P P (5.8-15)
The entropy production is always positive, which leads to a statement of the second law in the form of the Clausius-Duhem inequality d r qi⋅ni ρηdV ≥ dV − dS dt∫ ∫θ ∫ θ P P (5.8-16) This global form can easily be posed locally by the now-familiar procedures. Applying the divergence theorem to the heat flux term yields ∫ ∫ ∂P Furthermore, the differentiation of the entropy term is simplified by the fact that it is a specific quantity (see Section 5.3, Eq 5.3-11). Thus, we write ∫ P (5.8-17) and since this must hold for all arbitrary portions of the body, and the integrand is continuous, then Thus, the local form of the Clausius-Duhem equation is (5.8-18a) Often, the gradient of the temperature is written as g i = θ ,i in which case Eq 5.8-18a becomes (5.8-18b) Combining this result with Eq 5.7-13 brings the stress power and internal energy into the expression, giving a reduced form of the Clausius-Duhem equation ∂P qi⋅ni qi dS = ⎛ ⎞ ⎜ ⎟ θ ⎝ θ ⎠ , P ρη˙ − ρ + θ θ ⎛ ⎡ r qj ⎞ ⎢ ⎜ ⎟ ⎢ ⎝ ⎠ ⎣ , j i dV ⎤ ⎥dV ≥ 0 ⎥ ⎦ , ρη˙ − ρ + ρη˙ ρ θ, θ θ θ θ θ ⎛ r qj ⎞ r qjj 1 ⎜ ⎟ = − + − qj j ≥0 2 ⎝ ⎠ , j ρθη ρ θ θ ˙ 1 − r+ qii , − qi, i≥0 ρθη˙ 1 − ρr+ qii , − qigi≥0 θ ρθη˙ 1 − ρu˙+ Dijσ ij − qigi≥0 θ (5.8-19)
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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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Page 153 and 154: 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: θ = η ∂u ∂ (5.8-2) Furthermor
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 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
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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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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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