CONTINUUM MECHANICS for ENGINEERS

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so that and by a direct comparison with Eq 6.1-10b we obtain (6.1-13) (6.1-14) The strain energy density, W (strain energy per unit volume) is defined by W = ρ 0u (6.1-15) and since ρ = ρ 0, a constant, under the assumptions we have made, it follows from Eq 6.1-14 that (6.1-16) It is worthwhile noting at this point that elastic behavior is sometimes defined on the basis of the existence of a strain energy function from which the stresses may be determined by the differentiation in Eq 6.1-16. A material defined in this way is called a hyperelastic material. The stress is still a unique function of strain so that this energy approach is compatible with our earlier definition of elastic behavior. Thus, in keeping with our basic restriction to infinitesimal deformations, we shall develop the linearized form of Eq 6.1-16. Expanding W about the origin, we have and, from Eq 6.1-16, u= ˙ ˙ u ∂ ij ∂ε ε (6.1-17) (6.1-18) It is customary to assume that there are no residual stresses in the unstrained state of the material so that σ ij = 0 when ε ij = 0. Thus, by retaining only the linear term of the above expansion, we may express the linear elastic constitutive equation as ij 1 ρ σ ∂ u ij = ∂ε σ ρ ∂ u ∂ W ij = = ∂ε ∂ε ( ) ij ij ij 2 ∂W 0 1 ∂ W 0 W( εij)= W( 0)+ εij + εε ij ∂ε 2 ∂ε ∂ε ij ( ) + ( ) ij km ( ) 2 ∂W ∂W 0 ∂ W 0 σ ij = = ε ∂ε ∂ε ∂ε ∂ε σ ij ij ij ij km ( ) 2 ∂ W 0 = ε = C ε ∂ε ∂ε ij km km km ijkm km km + L + L (6.1-19)
ased on the strain energy function. This equation appears to be identical to Eq 6.1-3, but there is one very important difference between the two — not only do we have the symmetries expressed by Eq 6.1-4, but now we also have due to the fact that C ijkm = C kmij ( ) = 2 ∂ W 0 2 ∂ W 0 ∂ε ∂ε ∂ε ∂ε ij km km ij (6.1-20) Thus, the existence of a strain energy function reduces the number of distinct components of C ijkm from 36 to 21. Further reductions for special types of elastic behavior are obtained from material symmetry properties in the next section. Note that by substituting Eq 6.1-19 into Eq 6.1-17 and assuming a linear stress-strain relation, we may now write which in the notation of Eq 6.1-8 becomes (6.1-21a) (6.1-21b) and by the symmetry condition C αβ = C βα we have only 21 distinct constants out of the 36 possible. 6.2 Hooke’s Law for Isotropic Media, Elastic Constants If the behavior of a material is elastic under a given set of circumstances, it is customarily spoken of as an elastic material when discussing that situation even though under a different set of circumstances its behavior may not be elastic. Furthermore, if a body’s elastic properties as described by the coefficients C ijkm are the same in every set of reference axes at any point for a given situation, we call it an isotropic elastic material. For such materials, the constitutive equation has only two elastic constants. A material that is not isotropic is called anisotropic; we shall define some of these based upon the degree of elastic symmetry each possesses. ( ) 1 1 W( εij)= Cijkmεijεkm = σijεij 2 2 1 1 W( ε )= C ε ε = σ ε 2 2 α αβ α β α α
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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.
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 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: where the factor of two on the shea
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
Page 256 and 257: FIGURE E6.7-1 (a) Rectangular regio
Page 258 and 259: Solution It is easily verified, by
Page 260 and 261: (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
Page 270 and 271: where ψ(x 1,x 2) is called the war
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
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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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