CONTINUUM MECHANICS for ENGINEERS

Recommendations

Info

Example 3.6-1 The components of the stress tensor at P are given in MPa with respect to axes Px 1x 2x 3 by the matrix Determine the principal stresses and the principal stress directions at P. Solution For the given stress tensor, Eq 3.6-5 takes the form of the determinant which, upon cofactor expansion about the first row, results in the equation or in its readily factored form [ σ ij]= ⎡57 0 24⎤ ⎢ ⎥ ⎢ 0 50 0 ⎥ ⎣ ⎢24 0 43⎦ ⎥ 57 − σ 0 24 0 50−σ0 = 0 24 0 43 − σ (57 )(50 )(43 ) – (24) 2 −σ −σ −σ (50 −σ ) = 0 (50 −σ )(σ – 25) (σ – 75) = 0 Hence, the principal stress values are σ (1) = 25, σ (2) = 50, and σ (3) = 75. Note that, in keeping with Eqs 3.6-7a and Eq 3.6-16a, we confirm that the first stress invariant, I = 57 + 50 + 43 = 25 + 50 + 75 = 150 To determine the principal directions we first consider σ (1) = 25, for which Eq 3.6-3 provides three equations for the direction cosines of the principal direction of σ (1), namely, ( ) ( ) 1 3 1 1 32n + 24n = 0 1 25n 0 ( ) = 1 1 24n + 18n = 0 2 ( ) ( ) 1 3
1 Obviously, n from the second of these equations, and from the other 2 0 two, ( 1) 4 ( 1) n so that, from the normalizing condition, nini = 1, we see 3 =− n1 3 2 ( 1) 9 ( 1) 3 ( 1) 4 that ( n1 ) = which gives n1 =± and n3 = m . The fact that the first 25 5 5 and third equations result in the same relationship is the reason the normalizing condition must be used. Next for σ (2) = 50, Eq 3.6-3 gives 2 2 which are satisfied only when n1= n3= 0 . Then from the normalizing ( 2) condition, nini = 1, n2 =± 1. Finally, for σ (3) = 75, Eq 3.6-3 gives as well as ( ) = ( ) ( ) 1 3 2 2 7n + 24n = 0 ( ) ( ) 1 3 2 2 24n − 7n = 0 ( ) ( ) ( ) ( ) 1 3 3 3 − 18n + 24n = 0 3 − 25n = 0 Here, from the second equation , and from either of the other two equations , so that from nini = 1 we have and . From these values of , we now construct the transformation matrix [aij] in accordance with the table of Figure 3.10b, keeping in mind that to assure a right-handed system of principal axes we must have . Thus, the transformation matrix has the general form n 3 2 0 ( 3) ( 3) ( 3) 4 ( 3) 3 4n3= 3n1 n1 =± n3 =± 5 5 n ( q) i ˆ( 3) 1 2 n nˆ( ) nˆ( ) = × [ aij] = ( ) 2 ( ) ( ) 1 3 3 3 24n − 32n = 0 ( ) = ⎡ 3 4 ⎤ ⎢± 0 m 5 5 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 ± 1 0⎥ ⎢ ⎥ ⎢ 4 3⎥ ⎢± 0 ± ⎥ ⎣ 5 5⎦
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 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: FIGURE 3.10B Table displaying direc
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
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
Page 137 and 138:
which, upon factoring the left-hand
Page 139 and 140:
FIGURE 4.4 A rectangular parallelep
Page 141 and 142:
FIGURE 4.6A Rotated axes for plane
Page 143 and 144:
Consider once more the two neighbor
Page 145 and 146:
where dx is the deformed magnitude
Page 147 and 148:
In a similar fashion, from dX (1)
Page 149 and 150:
Thus from Eq 4.8-4 the principal st
Page 151 and 152:
and as is obvious, the inverse (U*)
Page 153 and 154:
Finally, from Eq 4.9-11, [ RAB]= =
Page 155 and 156:
which becomes (after dividing both
Page 157 and 158:
This rate of decrease in the angle
Page 159 and 160:
FIGURE 4.8 Area dS° between vector
Page 161 and 162:
FIGURE 4.9 Volume of parallelepiped
Page 163 and 164:
(a) Show that the Jacobian determin
Page 165 and 166:
4.5 The Lagrangian description of a
Page 167 and 168:
u 2 = - x 2 - (x 1 + x 2)e -t /2 +
Page 169 and 170:
and from it the longitudinal (norma
Page 171 and 172:
FIGURE P4.27 Unit square OBCD in th
Page 173 and 174:
Answer: (a) (b) 2 2 (c) (1) a1a 2a
Page 175 and 176:
FIGURE P4.35 Circular cylinder in t
Page 177 and 178:
(a) the components of acceleration
Page 179 and 180:
FIGURE P4.47A Unit cube having diag
Page 181 and 182:
5 Fundamental Laws and Equations 5.
Page 183 and 184:
which upon application of the diver
Page 185 and 186:
which is known as the continuity eq
Page 187 and 188:
FIGURE 5.1 Material body in motion
Page 189 and 190:
where ∆f is the resultant force a
Page 191 and 192:
which reduces to where we have used
Page 193 and 194:
Ε ABδ iAδ jB = e ij = 0(ε) o Ex
Page 195 and 196:
this text we consider only mechanic
Page 197 and 198:
outward normal is n i (hence the mi
Page 199 and 200:
θ = η ∂u ∂ (5.8-2) Furthermor
Page 201 and 202:
The entropy production is always po
Page 203 and 204:
One of the uses for the Clausius-Du
Page 205 and 206:
Furthermore, assume the temperature
Page 207 and 208:
In the first, a continuum body’s
Page 209 and 210:
FIGURE 5.2 Reference frames Ox 1x 2
Page 211 and 212:
Recall the definition t + = t + a f
Page 213 and 214:
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 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
Page 304 and 305:
and thus which describes the pressu
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
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 [
Page 360 and 361:
Depending upon the particular state
Page 362 and 363:
() (9.7-9a) (9.7-9b) (9.7-9c) From
Page 364 and 365:
and so on for higher derivatives. N
Page 366 and 367:
This expression may be inverted wit
Page 368 and 369:
9.4 Develop the constitutive equati
Page 370 and 371:
9.7 For the model shown the stress
Page 372 and 373:
9.11 For the model shown determine
Page 374 and 375:
9.16 Let the stress relaxation func
Page 376 and 377:
9.24 For the rather complicated mod
Page 378 and 379:
9.28 A viscoelastic body in the for
Page 380:
9.32 The deflection at x = L for an
show all

CONTINUUM MECHANICS for ENGINEERS

Create successful ePaper yourself

Delete template?

Save as template?