CONTINUUM MECHANICS for ENGINEERS

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It should be pointed out that while Eq 6.9-12 is a solution of Eq 6.9-8, it is not the general solution of the Navier equations. If we choose , and ψ = 0 in Eq 6.9-12, the scalar function φ is known as the Lamé strain potential. By taking the divergence of Eq 6.9-10, and remembering that the divergence of a curl vanishes, we see that 2 φ = constant (6.9-13) is a solution of the resulting equation so that a bi-harmonic function as φ also yields a solution for u i. Similarly, by taking the curl of Eq 6.9-10 we find that (6.9-14) also provides for a solution ui. The second approach for solving the Navier equations is based on the premise of expressing the displacement field in terms of the second derivatives of a vector known as the Galerkin vector, and designated here by F= Feˆ. i i In this approach we assume the displacement ui is given in terms of the Galerkin vector specifically by the equation (6.9-15) which is substituted directly into Eq 6.9-8. Carrying out the indicated differentiation and reducing the resulting equations with the help of the identity λ = 2νµ 1−2ν we find that the Navier equations are satisfied if (6.9-16) Thus, any bi-harmonic vector is suitable as a Galerkin vector. As should be expected, because they are solutions to the same equation, there is a relationship between φ and with F. It can be been shown that and ( ) 4 4 φ = 0 ψ = 0 ( ) − u = − F F i 21 ν i, jj j, ji 4 F = 0 φ =−Fi, i ( ) ε ψ = 21−ν F ijk k, j ijj , (6.9-17a) (6.9-17b) If F i is not only bi-harmonic, but harmonic as well, Eq 6.9-17b reduces to εijkψk, j = 0 (6.9-18a)
and the relationship between φ and F i becomes φ , ii =−Fi, jji In this case φ is called the Lamé strain potential. (6.9-18b) Example 6.9-1 Consider a Galerkin vector of the form F= F eˆwhere F3 is a function of the 3 3 coordinates, that is F3 = F3( x1, x2, x3) . Apply this vector to obtain the solution to the problem of a concentrated force acting at the origin of coordinates in the direction of the positive x3 axis of a very large elastic body. This is called the Kelvin problem. Solution Let u = − F F as given by Eq 6.9-15. Accordingly, ( ) − i 21 ν i, jj j, ji u =−F 1 3, 31 u2 =−F3, 32 u = 21 ( −ν) ( F + F + F )−F 3 3, 11 3, 22 3, 33 3, 33 Take F3 to be proportional to the distance squared from the origin as defined 2 2 2 2 by F3 = BR where B is a constant and R = x x x . Thus, the displace- 1 + 2 + 3 ments are Bx3x1 u1 = 3 R Bx3x2 u2 = 3 R ⎡ 41−ν u3= B⎢ ⎣ R ( ) − 2 2 x1+ x ⎤ 2 3 ⎥ R ⎦ From these displacement components the stresses may be computed using Hooke’s law. In particular, it may be shown that 2 2 2 2 σ33 = 2 ν 2 2 2 2 3 1 2 3 3 ∂ ( − ) ∂ ∂ + ∂ ∂ + ∂ ∂ ⎡ ⎛ ⎞ ∂ ⎤ ⎢ ⎜ ⎝ ∂ ⎟ − ⎥ x ⎣⎢ x x x ⎠ ∂x ⎦⎥ BR which upon carrying out the indicated differentiation and combining terms becomes
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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.
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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
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 and 234: where the factor of two on the shea
Page 235 and 236: ased on the strain energy function.
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 280 and 281: This equation may be written in a m
Page 282 and 283: where A 1 and A 2 are constants of
Page 284 and 285: Upon setting the off-diagonal terms
Page 286 and 287: 6.8 Show that the distortion energy
Page 288 and 289: 6.16 For an elastic body whose x 3
Page 290 and 291: and B are constants, determine the
Page 292 and 293: Thus, for the beam shown the stress
Page 294 and 295: Using the sketch on the facing page
Page 296 and 297: For a fluid in motion the shear str
Page 298 and 299: The first of this pair relates the
Page 300 and 301: posed in this section are relevant
Page 302 and 303: FIGURE E7.4-1A Flow down an incline
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Page 306 and 307: where dx i is a differential tangen
Page 308 and 309: 7.9 Show that for an incompressible
Page 310 and 311: 8 Nonlinear Elasticity 8.1 Molecula
Page 312 and 313: strain. Highly cross-linked and fil
Page 314 and 315: FIGURE 8.3 A freely connected chain
Page 316 and 317: proportional to the probability per
Page 318 and 319: The deformation is volume preservin
Page 320 and 321: Note that is essentially the same a
Page 322 and 323: where dependence on I 3 has been in
Page 324 and 325: where P 11 is the 11-component of t
Page 326 and 327: 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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