CONTINUUM MECHANICS for ENGINEERS

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known as the right principal directions of stretch. By the transformation Eq 4.9-2a, line elements along the directions , (i = 1,2,3) are stretched by an amount U (i), (i = 1,2,3), respectively, with no change in direction. This is followed by a rigid body rotation given by Eq 4.9-2b. The second decomposition of Eq 4.9-1 reverses the sequence: first a rotation by R, then the stretching by V. In a general deformation, a rigid body translation may also be involved as well as the rotation and stretching described here. As a preliminary to determining the rotation and stretch tensors, we note that an arbitrary tensor T is positive definite if v ⋅ T ⋅ v > 0 for all vectors v ≠ 0. A necessary and sufficient condition for T to be positive definite is for all its eigenvalues to be positive. In this regard, consider the tensor C = F T ⋅ F. Inasmuch as F is non-singular (det F ≠ 0) and F ⋅ v ≠ 0 if v ≠ 0, so that (F ⋅ v) ⋅ (F ⋅ v) is a sum of squares and hence greater than zero. Thus ˆN i (F ⋅ v) ⋅ (F ⋅ v) = v ⋅ F T ⋅ F ⋅ v = v ⋅ C ⋅ v > 0 (4.9-3) and C is positive definite. Furthermore, (F T ⋅ F) T = F T ⋅ (F T ) T = F T ⋅ F (4.9-4) which proves that C is also symmetric. By the same arguments we may show that c = (F –1 ) T ⋅ (F –1 ) is also symmetric and positive definite. Now let C be given in principal axes form by the matrix ⎡C ⎤ ( 1) 0 0 * ⎢ ⎥ [ CAB]= ⎢ 0 C( 2) 0 ⎥ ⎢ ⎣ 0 0 C ⎥ ( 3) ⎦ (4.9-5) and let [a MN] be the orthogonal transformation that relates the components of C* to the components of C in any other set of axes through the equation expressed here in both indicial and matrix form * C = a a C AB AQ BP QP or C* = ACA T (4.9-6) We define U as the square root of C — that is, U = C or U ⋅ U = C — and since the principal values C (i), (i = 1,2,3) are all positive we may write C( ) * [ C ] C( ) C = ⎡ 1 0 0 ⎢ ⎢ 0 2 0 ⎢ ⎣ 0 0 ⎤ ⎥ * ⎥ = [ U ] ⎥ ⎦ AB AB ( 3) (4.9-7)
and as is obvious, the inverse (U*) –1 by (4.9-8) Note that both U and U –1 are symmetric positive definite tensors given by and UAB = or U = AT * a a U U*A (4.9-9) or U –1 = A T (U*) –1 A (4.9-10) respectively. Therefore, now, from the first decomposition in Eq 4.9-1, so that C ⎡ * ( U AB) ⎤ C ⎣⎢ ⎦⎥ C = ⎡1/ ( 1) 0 0 −1 ⎢ ⎢ 0 1/ ( 2) 0 ⎢ ⎣ 0 0 1/ -1 U =a a U AB QA PB QP ( ) −1 QA PB * QP R = F ⋅ U –1 (4.9-11) R T ⋅ R = (F ⋅ U –1 ) T ⋅ (F ⋅ U –1 ) = (U –1 ) T ⋅ F T ⋅ F ⋅ U –1 = U –1 ⋅ C ⋅ U –1 = U –1 ⋅ U ⋅ U ⋅ U –1 = I (4.9-12) which shows that R is proper orthogonal. The second decomposition in Eq 4.9-1 may be confirmed by a similar development using C –1 = F ⋅ F T = V 2 . Example 4.9-1 A homogeneous deformation is given by the equations x 1 = 2X 1 – 2X 2, x 2 = X 1 + X 2 and x 3 = X 3 . Determine the polar decomposition F = R ⋅ U for this deformation. Solution The matrix form of the tensor F iA ≡ x i,A is easily determined to be [ Fi,A]= ⎡2 −2 0⎤ ⎢ ⎥ ⎢ 1 1 0 ⎥ ⎣ ⎢0 0 1⎦ ⎥ ( 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
Page 99 and 100: Solution For this stress state, the
Page 101 and 102: which demonstrates that ( q) is als
Page 103 and 104: Example 3.11-1 Determine directly t
Page 105 and 106: FIGURE P3.6 Stress vectors represen
Page 107 and 108: FIGURE P3.9 Cylinder of radius r an
Page 109 and 110: (b) Project each of the stress vect
Page 111 and 112: Determine (a) the principal stress
Page 113 and 114: (b) Verify the result determined in
Page 115 and 116: ⎡ −12 − 12 + 3 2 −12 −32
Page 117 and 118: A motion of body B is a continuous
Page 119 and 120: emphasize, however, that the materi
Page 121 and 122: Example 4.2-1 Let the motion of a b
Page 123 and 124: and v = v(X,t) = v[χ -1 (x,t),t] =
Page 125 and 126: Additionally, with regard to the ma
Page 127 and 128: In this equation, the first term on
Page 129 and 130: Therefore, substituting u i for P i
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Page 133 and 134: form consistent with deformation an
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: Thus from Eq 4.8-4 the principal st
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 and 200: 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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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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