CONTINUUM MECHANICS for ENGINEERS

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FIGURE 2.4 Principal axes Ox 1* x 2* x 3* relative to axes Ox 1x 2x 3. with [ ] = q Tij − λ δ n ( ) ( , (q = 1,2,3) (2.6-9) q) ij i 0 n ( q) n ( q) = 1, (q = 1,2,3) (2.6-10) i i If the ’s are distinct the principal directions are unique and mutually perpendicular. If, however, there is a pair of equal roots, say , then only the direction associated with will be unique. In this case any other two directions which are orthogonal to , and to one another so as to form a right-handed system, may be taken as principal directions. If , every set of right-handed orthogonal axes qualifies as principal axes, and every direction is said to be a principal direction. In order to reinforce the concept of principal directions, let the components of the tensor T be given initially with respect to arbitrary Cartesian axes , and let the principal axes of T be designated by , as shown in Figure 2.4. The transformation matrix A between these two sets of axes is established by taking the direction cosines as calculated from Eq 2.6-9 and Eq 2.6-10 as the elements of the qth λ( q) λ( λ 1) = ( 2) row of A. Therefore, by definition, as detailed in the table below. λ( 3) ( 3) ni λ( λ λ 1) = ( 2) = ( 3) * * * Ox1x2x3 Ox1x2x3 n ( q) i i aij ≡ nj () x or 1 1 * * 1 x orê a = n ( ) 1 1 11 1 * * 2 x orê a = n ( ) 2 2 x orê * * 3 3 21 1 3 a = n ( ) 31 1 ê x orê 2 2 1 a = n ( ) 12 2 2 a = n ( ) 22 2 3 a = n ( ) 32 2 x3 orê3 1 a = n ( ) 13 3 2 a = n ( ) 23 3 33 = 3 3 ( ) a n (2.6-11)
The transformation matrix here is orthogonal and in accordance with the transformation law for second-order tensors * * T T = a a T or T = ATA (2.6-12) where T * is a diagonal matrix whose elements are the principal values . λ ( q) Example 2.6-1 Determine the principal values and principal directions of the second-order tensor T whose matrix representation is Solution Here Eq 2.6-6 is given by which upon expansion by the third row becomes or ij iq jm qm [ Tij] = ⎡5 2 0⎤ ⎢ ⎥ ⎢ 2 2 0 ⎥ ⎣ ⎢0 0 3⎦ ⎥ 5−λ2 0 2 2−λ 0 = 0 0 0 3−λ 2 3−λ ( 10 7λ λ 4)= 0 ( ) − + − ( ) − 3−λ ( 6 λ) ( 1−λ)= 0 Hence, λ are the principal values of T. For , ( λ λ 1) = 3, ( 2) = 6, ( 1 3) = Eq 2.6-5 yields the equations λ( 3 1) = 2n + 2n = 0 1 2 2n − n = 0 1 2 which are satisfied only if n , and so from we have . 1 = n2 = 0 nn i i = 1 n3 =± 1 For 6 , Eq 2.6-5 yields λ( 2) =
Page 1 and 2: CONTINUUM MECHANICS for ENGINEERS S
Page 3 and 4: Library of Congress Cataloging-in-P
Page 5 and 6: University, for helpful comments on
Page 7 and 8: Reference Section for constitutive
Page 9 and 10: Nomenclature x 1, x 2, x 3 or x i o
Page 11 and 12: W Strain energy per unit volume, or
Page 13 and 14: 4.5 The Material Derivative 4.6 Def
Page 15 and 16: 1 Continuum Theory 1.1 The Continuu
Page 17 and 18: 2 Essential Mathematics 2.1 Scalars
Page 19 and 20: FIGURE 2.1A Unit vectors in the coo
Page 21 and 22: 1. Addition of vectors: 2. Multipli
Page 23 and 24: and, furthermore, that for repeated
Page 25 and 26: A sum of dyads such as is called a
Page 27 and 28: (b) Start with the first equation i
Page 29 and 30: Solution By definition of symmetric
Page 31 and 32: Addition of matrices is commutative
Page 33 and 34: 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: or in expanded form ( Tij − λδi
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 and 86: FIGURE 3.12 Normal and shear compon
Page 87 and 88: 1 1 1 n1 = 0 , n2 = ± , n3 = ± ;
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
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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.
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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
Page 292 and 293:
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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