CONTINUUM MECHANICS for ENGINEERS

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8 Nonlinear Elasticity 8.1 Molecular Approach to Rubber Elasticity Many of today’s challenging design problems involve materials such as butadiene rubber (BR), natural rubber (NR), or elastomers. Rubber materials might be most easily characterized by the stretching and relaxing of a rubber band. The resilience of rubber, the ability to recover intial dimensions after large strain, was not possible with natural latex until Charles Goodyear discovered vulcanization in 1939. Vulcanization is a chemical reaction known as cross-linking which turned liquid latex into a non-meltable solid (thermoset). Cross-linked rubber would also allow considerable stretching with low damping; strong and stiff at full extension, it would then retract rapidly (rebound). One of the first applications was rubber-impregnated cloth, which was used to make the sailor’s “mackintosh.” Tires continue to be the largest single product of rubber although there are many, many other applications. These applications exhibit some or all of rubber’s four characteristics, viz. damping in motor mounts, rebound/resilience in golf ball cores, or simple stretching in a glove or bladder. While thermoset rubber remains dominant in rubber production, processing difficulties have led to the development and application of thermoplastic elastomers (TPEs). These materials are easier to process and are directly recyclable. While TPEs are not as rubberlike as the thermosets, they have found wide application in automotive fascia and as energy-absorbing materials. There are several reasons why designing with plastic and rubber materials is more difficult than with metals. For starters, the stress-strain response, that is, the constitutive response, is quite different. Figure 8.1A shows the stress-strain curves for a mild steel specimen along with the response of a natural rubber used in an engine mount. Note that the rubber specimen strain achieves a much higher stretch value than the steel. The dashed vertical line in Figure 8.1B represents the strain value of the mild steel at failure. This value is much less than the 200% strain the rubber underwent without failing. In fact, many rubber and elastomer materials can obtain 300 to 500%
FIGURE 8.1A Nominal stress-stretch curve for mild steel. FIGURE 8.1B Nominal stress-stretch curve for natural rubber. Note the large stretch compared to the mild steel curve.
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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
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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
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 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 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
Page 328 and 329: σ yy ⎡⎛ ∂X ⎞ ⎛ ∂Y ⎞
Page 330 and 331: q( x)= Ae + Be kx −kx Eqs 8.4-19a
Page 332 and 333: FIGURE 8.6 Stresses on deformed rub
Page 334 and 335: 8.2 Derive the following relationsh
Page 336 and 337: Answer: 8.9 Show F B iA = ij = g
Page 338 and 339: 9 Linear Viscoelasticity 9.1 Introd
Page 340 and 341: ˙ε ii ≈ Dii so that now, from E
Page 342 and 343: FIGURE 9.2 Mechanical analogy for s
Page 344 and 345: FIGURE 9.5 Three-parameter standard
Page 346 and 347: FIGURE 9.7 Graphic representation o
Page 348: Development of details relative to
Page 352 and 353: FIGURE 9.9 Stress history with an i
Page 354 and 355: γ ()= t γ ( cos ωt+ isinωt)= γ
Page 356 and 357: FIGURE 9.10 Shear lag in viscoelast
Page 358 and 359: 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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