CONTINUUM MECHANICS for ENGINEERS

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2 2 2 Answer: (a) λ + λ + λ = 3 1 2 3 2( 1 3) 2 2 2 (b) λ λ + λ = 2 2 2 2 2 2 2 (c) λλ = 3 1 2 + λλ 2 3 + λλ 3 1 4.48 Let the unit cube shown in Problem 4.47 be given the motion 1 2 1 2 x 1 = X 1 + t 2 X 2, x 2 = X 2 + t 2 X 1, x 3 = X 3 Determine, at time t, (a) the rate-of-change of area ABFE (b) the volume of the body. Answer: (a) = /(1 – t4 ) – – (b) V = (1 — t4 • ( dS) t ) 4.49 For the homogeneous deformation 3 ê1 1 tê 4 2 1 4 x 1 = X 1 + αX 2 + αβX 3 x 2 = αβX 1 + X 2 + β 2 X 3 x 3 = X 1 + X 2 + X 3 where α and β are constants, determine the relationship between these constants if the deformation is isochoric. Answer: β = (α2 + α)/(α2 + α + 1) 4.50 Show that for any velocity field v derived from a vector potential ψ by v = curl ψ, the flow is isochoric. Also, for the velocity field v 1 = ax 1x 3 – 2x 3, v 2 = – bx 2x 3, v 3 = 2x 1x 2 determine the relationship between the constants a and b if the flow is isochoric. Answer: a = b t3 ê 3
5 Fundamental Laws and Equations 5.1 Balance Laws, Field Equations, Constitutive Equations A number of the fundamental laws of continuum mechanics are expressions of the conservation of some physical quantity. These balance laws, as they are often called, are applicable to all material continua and result in equations that must always be satisfied. In this introductory text, we consider only the conservation laws dealing with mass, linear and angular momentum, and energy. With respect to energy, we shall first develop a purely mechanical energy balance and follow that by an energy balance that includes both mechanical and thermal energies, that is, a statement of the first law of thermodynamics. In addition to that, the Clausius-Duhem form of the second law of thermodynamics is covered. The balance laws are usually formulated in the context of global (integral) relationships derived by a consideration of the conservation of some property of the body as a whole. As explained in Chapter One, the global equations may then be used to develop associated field equations which are valid at all points within the body and on its boundary. For example, we shall derive the local equations of motion from a global statement of the conservation of linear momentum. Constitutive equations, which reflect the internal constitution of a material, define specific types of material behavior. They are fundamental in the sense that they serve as the starting point for studies in the disciplines of elasticity, plasticity, and various idealized fluids. These equations are the topic of the final section of this chapter. Before we begin a discussion of the global conservation laws, it is useful to develop expressions for the material derivatives of certain integrals. This we do in the next section.
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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
Page 129 and 130: Therefore, substituting u i for P i
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Page 135 and 136: For diagonal BH, and for diagonal O
Page 137 and 138: which, upon factoring the left-hand
Page 139 and 140: FIGURE 4.4 A rectangular parallelep
Page 141 and 142: FIGURE 4.6A Rotated axes for plane
Page 143 and 144: Consider once more the two neighbor
Page 145 and 146: where dx is the deformed magnitude
Page 147 and 148: In a similar fashion, from dX (1)
Page 149 and 150: Thus from Eq 4.8-4 the principal st
Page 151 and 152: and as is obvious, the inverse (U*)
Page 153 and 154: Finally, from Eq 4.9-11, [ RAB]= =
Page 155 and 156: which becomes (after dividing both
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: FIGURE P4.47A Unit cube having diag
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
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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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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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