CONTINUUM MECHANICS for ENGINEERS

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FIGURE E6.7-5 Uniaxial loaded plate with a circular hole. Assume the width of the plate b is large compared to the radius of the hole. From consideration of the small triangular element at a distance b from the origin, the following boundary conditions must hold (1) σ r = S cos 2 θ at r = b S (2) τrθ=− sin 2θ at r = b 2 and the inner surface of the hole is stress free, which may be written as (3) σ r = 0 at r = a (4) τ rθ = 0 at r = a These conditions, when combined with the stress expressions, yield 6C4DS 2A + + = − 4 2 b b 2 2 6C2DS 2A+ 6Bb − − = − 4 2 b b 2 6C4D 2A + + = 0 4 2 a a 2 6C2D 2A+ 6Ba − − = 0 4 2 a a
Letting b → ∞, the above equations may be solved to determine So that now the stresses are given by Note that as r tends towards infinity and at θ = 0 the stresses are given by σ r = S; σ θ = τ rθ = 0 (simple tension). At r = a, σ r = τ rθ = 0 and σ θ = S – 2S cos 2θ, which indicates that when θ = π/2, or θ = 3π/2, the stresses become σ θ = S – 2S(–1) = 3S, a well-known stress concentration factor used in design. Also, when θ = 0, or θ = π, σ θ = –S a compression factor at the centerline of the hole. It should be pointed out that in this brief section only stresses have been determined. These, together with Hooke’s law, can be used to determine the displacements <strong>for</strong> the problems considered. 6.8 Torsion 4 2 S aS aS A =− ; B= 0; C=− ; D = 4 4 2 S⎛a ⎞ S⎛a a ⎞ σr= ⎜ − ⎟ + ⎜ + − ⎟ θ 2 ⎝ r ⎠ ⎝ r r ⎠ 1 2 4 2 3 4 1 cos2 2 4 2 2 S⎛a ⎞ S⎛a ⎞ σθ= ⎜ + ⎟ − ⎜ + ⎟ θ 2 ⎝ r ⎠ ⎝ r ⎠ 1 2 4 3 1 cos2 2 4 2 4 2 S⎛3a 2a⎞ τrθ=− ⎜1 − + ⎟ sin 2θ 4 2 2 ⎝ r r ⎠ We begin with a brief review of the solution to the simplest of torsion problems, the case of a shaft having a constant circular cross section when subjected to equilibrating end couples, M t as shown in Figure E6.7a. Let the end face at x 3 = 0 be fixed while the face at x 3 = L is allowed to rotate about the axis of the shaft. It is assumed that plane sections perpendicular to the axis remain plane under the twisting, and that each rotates through an angle proportional to its distance from the fixed end. Accordingly, a point in the cross section at coordinate x 3 will rotate an angle of θ x 3. Each point, say point P, in the cross section travels a distance θ x 3R which is proportional to the distance R from the x 3 axis as shown in Figure 6.7b. The distance squared to point P is the square of the x 1 and x 2 coordinates (Figure 6.7c). Using this distance, it is easy to define the cosine and sine of angle β. It is straight- <strong>for</strong>ward to write out displacements u1 and u2 in terms of θ since cos( 90 − β)= x and . Thus, 2 R sin 90 ( − β)= x1R u =− θx x ; u = θx x ; u = 1 2 3 2 1 3 3 0 (6.8-1)
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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
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
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 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 278 and 279: It should be pointed out that while
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
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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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