CONTINUUM MECHANICS for ENGINEERS

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and similarly, σ N = t ( nˆ ) ⋅nˆby 2 2 σN= σIn1 + σIIn2+ σ n 2 III 3 Then, substituting Eqs 3.7-4 and 3.7-5 into Eq 3.7-2, we have ( ) 2 2 III 3 2 2 2 2 2 2 2 2 2 σS= σIn1 + σIIn2+ σIIIn3− σIn1 + σIIn2+ σ n (3.7-5) (3.7-6) 2 which expresses σ in terms of the direction cosines ni. But nini = 1, so that S 2 2 2 n and we are able to eliminate n3 from Eq 3.7-6, which then 3 = 1−n1−n2 becomes a function of n1 and n2 only, [ ] ( ) + ( − ) + − ( − ) + ( − ) + 2 2 2 2 2 2 2 2 2 2 σS= σI−σIII n1σII σIII n2σIII σIσIII n1σII σIII n2σ (3.7-7) 2 In order to obtain the stationary, that is, the extremal values of σ S , we must equate the derivatives of the right-hand side of this equation with respect to both n1 and n2 to zero, and solve simultaneously. After some algebraic manipulations, we obtain ( ) = − 2 ∂ σS ∂ n 1 2 ∂σS ∂ n 2 2 2 n1( σIσIII) { σI−σIII −2[ ( σI−σIII) n1 + ( σII −σIII) n2] }= 0 ( ) = − (3.7-8a) (3.7-8b) An obvious solution to Eq 3.7-8 is n1 = n2 = 0 for which n3 = ±1, and the corresponding value of is observed from Eq 3.7-7 to be zero. This is an expected result since n3 = ±1 designates a principal plane upon which the shear is zero. A similar calculation made with n1 and n3, or with n2 and n3 as the variables, would lead to the other two principal planes as minimum (zero) shear stress planes. It is easily verified that a second solution to Eq 3.7-8 is obtained by taking n1 = 0 and solving for n2. The result is and, from orthogonality, also. For this solution, Eq 3.7-7 yields the results σ 2 S n2 =± 1/ 2 n3 =± 1/ 2 2 III 2 2 n2( σII σIII) { σII −σIII −2[ ( σI−σIII) n1 + ( σII −σIII) n2] }= 0 σS σ σ = 2 1 − 4 ( ) 2 II III 1 or σ σ σ (3.7-9) S = ± ( II − III) 2 As before, if we consider in turn the formulation having n 1 and n 3, or n 2 and n 3 as the variable pairs, and assume n 3 = 0, and n 2 = 0, respectively, we obtain the complete solution which is presented here in the tabular form
1 1 1 n1 = 0 , n2 = ± , n3 = ± ; σS= σ −σ 2 2 2 (3.7-10a) (3.7-10b) (3.7-10c) where the vertical bars in the formulas for σ S indicate absolute values of the enclosed expressions. Because σ I ≥ σ II ≥ σ III, it is clear that the largest shear stress value is (3.7-11) It may be shown that, for distinct principal stresses, only the two solutions presented in this section satisfy Eq 3.7-8. 3.8 Mohr’s Circles For Stress Consider again the state of stress at P referenced to principal axes (Figure 3.12) and let the principal stresses be ordered according to σI > σII > σIII. As before, we may express σN and σS on any plane at P in terms of the components of the normal ˆn to that plane by the equations which, along with the condition ( ) II III 1 1 1 n1 =± , n2 = 0 , n3 =± ; σS= σIII −σI 2 2 2 ( ) 1 1 1 n1 =± , n2 =± , n3 = 0 ; σS= σ −σ 2 2 2 max 1 σS = σIII −σI 2 ( ) 2 2 σN= σIn1+ σIIn2+ σIIIn 2 2 2 2 2 2 2 σN + σS = σIn1+ σIIn2+ σIIIn 2 2 2 n + n + n = 1 1 2 ( ) I II (3.8-1a) (3.8-1b) (3.8-1c) provide us with three equations for the three direction cosines n 1, n 2, and n 3. Solving these equations, we obtain n 2 1 ( σ − σ ) ( σ − σ )+ σ = σ − σ σ σ 3 2 3 2 3 N II N III 2 S ( ) ( − ) I II I III (3.8-2a)
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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.
Page 35 and 36: which is identical to Eq 2.4-9 and
Page 37 and 38: FIGURE 2.2A Rectangular coordinate
Page 39 and 40: Consider next an arbitrary vector v
Page 41 and 42: FIGURE E2.5-1 Vector νννν with
Page 43 and 44: or in expanded form ( Tij − λδi
Page 45 and 46: The transformation matrix here is o
Page 47 and 48: have a multiplicity of two, and det
Page 49 and 50: we write φ for ; vi,j for for ; an
Page 51 and 52: FIGURE 2.5B Bounding space curve C
Page 53 and 54: (b) the trace of A is expressed in
Page 55 and 56: 2.15 Using the square matrices belo
Page 57 and 58: (a) Show that a multiplicity of two
Page 59 and 60: 2.29 Transcribe the left-hand side
Page 61 and 62: 3 Stress Principles 3.1 Body and Su
Page 63 and 64: FIGURE 3.2A Typical continuum volum
Page 65 and 66: where S I and S II are the bounding
Page 67 and 68: FIGURE 3.5 Free body diagram of tet
Page 69 and 70: FIGURE 3.6 Cartesian stress compone
Page 71 and 72: (b) The equation of the plane ABC i
Page 73 and 74: or But xj,q = δjq and by Eq 3.4-3,
Page 75 and 76: FIGURE E3.5-1 Rotation of axes x 1
Page 77 and 78: FIGURE 3.9A Traction vector at poin
Page 79 and 80: this system the shear stresses, σ
Page 81 and 82: FIGURE 3.10B Table displaying direc
Page 83 and 84: 1 Obviously, n from the second of t
Page 85: FIGURE 3.12 Normal and shear compon
Page 89 and 90: exterior to circle C1. Thus, combin
Page 91 and 92: FIGURE 3.15A Reference angles φ an
Page 93 and 94: FIGURE E3.8-1 Three-dimensional Moh
Page 95 and 96: FIGURE 3.16A Mohr’s circle for pl
Page 97 and 98: 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
Page 131 and 132: upon X, in which case the deformati
Page 133 and 134: form consistent with deformation an
Page 135 and 136: 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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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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