Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Recommendations

Info

y5 kips/ft9 kips/ft70 kN/m30 kN/m 50 kN/mA B CDFIGURE p7.456 ft 10 ft11 ftxAB1.5 mFIGURE p7.475.5 mC2.0 mDxy25 kN/m70 kN/m60 kN90 kN/mA B C D2.5 m4.0 m1.5 mxA1.5 mB4.5 mC2.0 mDxFIGURE p7.46FIGURE p7.48236
CHAPTER8bending8.1 IntroductionPerhaps the most common type of structural member is the beam. In actual structures and machines,beams can be found in a wide variety of sizes, shapes, and orientations. The elementarystress analysis of the beam constitutes one of the more interesting facets of mechanics of materials.Beams are usually long (compared with their cross-sectional dimensions), straight,prismatic members that support transverse loads, which are loads that act perpendicular tothe longitudinal axis of the member (Figure 8.1a). Loads on a beam cause it to bend (orflex) as opposed to stretching, compressing, or twisting. The applied loads cause theinitially straight member to deform into a curved shape (Figure 8.1b), which is called thedeflection curve or the elastic curve.In this chapter, we will consider beams that are initially straight and that have alongitudinal plane of symmetry (Figure 8.2a). The member cross section, the supportThe term transverse refers toloads and sections that areperpendicular to the longitudinalaxis of the member.yyPw(x)Pw(x)zxzxFIGURE 8.1a Transverse loads applied to a beam.FIGURE 8.1b Deflection caused by bending.237
Page 3:
A research-based,online learning en
Page 7 and 8:
MECHANICS OF MATERIALS:An Integrate
Page 9:
About the AuthorTimothy A. Philpot
Page 12 and 13:
xPREFACEAnimation also offers a new
Page 14 and 15:
xiiPREFACE• Appendix E Fundamenta
Page 16 and 17:
xivPREFACEWhat Do Instructors recei
Page 19 and 20:
ContentsChapter 1 Stress 11.1 Intro
Page 21:
Chapter 14 Pressure Vessels 58514.1
Page 24 and 25:
2STRESSAddressing these concerns re
Page 26 and 27:
4STRESSnumber begins with the digit
Page 28 and 29:
ExAmpLE 1.3A(1)50 mmB80 kNA 50 mm w
Page 30 and 31:
8STRESSFIGURE 1.2b Free-bodydiagram
Page 32 and 33:
mecmoviesExAmpLEm1.5 A pin at C and
Page 34 and 35:
12STRESSJeffery S. ThomasFIGURE 1.5
Page 36 and 37:
14STRESSFIGURE 1.6 Bearing stress f
Page 38 and 39:
m1.3 Use shear stress concepts for
Page 40 and 41:
applied to beam ABC? Use dimensions
Page 42 and 43:
p1.24 The two wooden boards shown i
Page 44 and 45:
1.5 Stresses on Inclined Sectionsme
Page 46 and 47:
24STRESSSignificanceAlthough one mi
Page 48 and 49:
Thus, to provide the necessary weld
Page 50 and 51:
p1.40 Two wooden members are glued
Page 52 and 53:
30STRAINposition vector between H a
Page 54 and 55:
32STRAINIn developing the concept o
Page 56 and 57:
mecmoviesExAmpLESm2.1 A rigid steel
Page 58 and 59:
p2.4 Bar (1) has a length of L 1 =
Page 60 and 61:
38STRAINIn this expression, θ′ i
Page 62 and 63:
pRoBLEmSp2.11 A thin rectangular po
Page 64 and 65:
SOLUTIONThe thermal strain for a te
Page 67 and 68:
CHAPTER3Mechanical Propertiesof Mat
Page 69 and 70:
(3)(7)(4)(5)(6)Fracture47THE TENSIO
Page 71 and 72:
8049THE STRESS-STRAIN dIAgRAM7060St
Page 73 and 74:
The elastic limit is the largest st
Page 75 and 76:
80Aluminum alloy××53THE STRESS-ST
Page 77 and 78:
The second measure is the reduction
Page 79 and 80:
Values vary for different materials
Page 81 and 82:
before the strain measurement. From
Page 83 and 84:
mecmoviesExERcISEm3.1 Figure M3.1 d
Page 85 and 86:
material is initially 800 mm long a
Page 87 and 88:
CHAPTER4design Concepts4.1 Introduc
Page 89 and 90:
the distributed uniform area loadin
Page 91 and 92:
In some instances, engineers may ne
Page 93 and 94:
both members will be stressed to th
Page 95 and 96:
MecMoviesExAMpLESM4.1 The structure
Page 97 and 98:
bPtPFIGURE p4.3Splice plateBarBarPP
Page 99 and 100:
4.5 Load and Resistance Factor Desi
Page 101 and 102:
79FrequencyLarger γ factors combin
Page 103 and 104:
Limit StatesLRFD is based on a limi
Page 105 and 106:
CHAPTER5Axial deformation5.1 Introd
Page 107 and 108:
The increased normal stress magnitu
Page 109 and 110:
AAAA87dEFORMATIONS IN AxIALLyLOAdEd
Page 111 and 112:
displacement in the horizontal dire
Page 113 and 114:
mecmoviesExAmpLEm5.2 The roof and s
Page 115 and 116:
t = 8 mm, and E = 72 GPa, determine
Page 117 and 118:
d 0to 2d 0 at the other end. A conc
Page 119 and 120:
How are the rigid-bar deflections v
Page 121 and 122:
ExAmpLE 5.4A tie rod (1) and a pipe
Page 123 and 124:
Since the sum of the four interior
Page 125 and 126:
5.5 Statically Indeterminate Axiall
Page 127 and 128:
Step 3 — Force-Deformation Relati
Page 129 and 130:
Successful application of the five-
Page 131 and 132:
Plan the SolutionConsider a free-bo
Page 133 and 134:
Structures with a Rotating Rigid Ba
Page 135 and 136:
(1)Px Bx C113STATICALLy INdETERMINA
Page 137 and 138:
ExERcISESm5.5 A composite axial str
Page 139 and 140:
p5.29 In Figure P5.29, a load P is
Page 141 and 142:
tube has an outside diameter of 1.5
Page 143 and 144:
ExAmpLE 5.7An aluminum rod (1) [E =
Page 145 and 146:
mecmoviesExAmpLEm5.14 A rectangular
Page 147 and 148:
Equations (h) and (i) can be solved
Page 149 and 150:
pRoBLEmSp5.39 A circular aluminum a
Page 151 and 152:
polymer [E = 370 ksi; α = 39.0 ×
Page 153 and 154:
Stress-concentration factor K3.02.8
Page 155 and 156:
of the hole has decayed to a value
Page 157 and 158:
TorsionCHAPTER66.1 IntroductionTorq
Page 159 and 160:
with respect to an adjacent cross s
Page 161 and 162:
the longitudinal axis of the shaft.
Page 163 and 164:
τStressxyτntσn141STRESSES ON ObL
Page 165 and 166:
If a torsion member is subjected to
Page 167 and 168:
ExAmpLE 6.1A hollow circular steel
Page 169 and 170:
Plan the SolutionTo determine the l
Page 171 and 172:
Plan the SolutionThe internal torqu
Page 173 and 174:
mecmoviesExAmpLESm6.4 Determine the
Page 175 and 176:
p6.10 The mechanism shown in Figure
Page 177 and 178:
Furthermore, since gears have teeth
Page 179 and 180:
will be dictated by the ratio of ge
Page 181 and 182:
mecmoviesExAmpLEm6.13 Two solid ste
Page 183 and 184:
6.8 power Transmission161POwER TRAN
Page 185 and 186:
m6.17 A motor shaft is being design
Page 187 and 188:
pulley is a belt having tensions F
Page 189 and 190:
Step 2 — Geometry of Deformation:
Page 191 and 192:
Polar moments of inertia for the al
Page 193 and 194:
Successful application of the five-
Page 195 and 196:
Plan the SolutionAA free-body diagr
Page 197 and 198:
From Equation (e), the allowable in
Page 199 and 200:
Next, consider a free-body diagram
Page 201 and 202:
Polar moments of inertia for the sh
Page 203 and 204:
m6.21 A composite torsion member co
Page 205 and 206:
zD(3)Ay(1)L1,L3N BBL 2(a) Determine
Page 207 and 208: Stress concentrations also occur at
Page 209 and 210: For the case of the rectangular bar
Page 211 and 212: and the angle of twist for a 12 in.
Page 213 and 214: ExAmpLE 6.13A rectangular box secti
Page 215 and 216: CHAPTER7Equilibrium of beams7.1 Int
Page 217 and 218: beam (also called a simple beam). A
Page 219 and 220: ExAmpLE 7.1Draw the shear-force and
Page 221 and 222: The negative value for A y indicate
Page 223 and 224: Plot the FunctionsPlot the function
Page 227 and 228: yw aw byw 0xxABCABabLFIGURE p7.4FIG
Page 229 and 230: w(x) = w G at G. At A, where the di
Page 231 and 232: the left and right sides of a thin
Page 233 and 234: General procedure for constructing
Page 235 and 236: Construct the Bending-Moment Diagra
Page 239 and 240: j M(6) = 0 kN ⋅ m (Rule 4: ∆M =
Page 241 and 242: of M diagram = shear force V). The
Page 243 and 244: m7.3 Dynamically generated shear-fo
Page 245 and 246: p7.21-p7.22 Use the graphical metho
Page 247 and 248: x - a 0x - a 1x - a 2225dISCONTINuI
Page 249 and 250: conditions. The reactions for stati
Page 251 and 252: 120 kN ⋅ m concentrated moment: F
Page 253 and 254: SolutioNWhen we refer to case 4 of
Page 255 and 256: Uniformly distributed load between
Page 257: y5 kN20 kN.my12 kN.m18 kN/mAB3 m 3
Page 261 and 262: has a constant bending moment M, an
Page 263 and 264: The most common stress-strain relat
Page 265 and 266: All such moment increments that act
Page 267 and 268: The sense of σ x (either tension o
Page 269 and 270: A i(mm 2 )y i(mm)y i A i(mm 3 )(1)
Page 271 and 272: ExAmpLE 8.2The cross-sectional dime
Page 273 and 274: m8.7 Determine the centroid locatio
Page 275 and 276: zFIGURE p8.8p8.9 An aluminum alloy
Page 277 and 278: WidthWidthWidthDepthXYthicknessXWeb
Page 279 and 280: 700 lb1,500 lb4 in.1 in.200 lb/ft1
Page 281 and 282: M = −6,000 lb · ft. For this neg
Page 283 and 284: mecmoviesExAmpLESm8.9 Determine the
Page 285 and 286: p8.16 A W18 × 40 standard steel sh
Page 287 and 288: 8.5 Introductory Beam Design for St
Page 289 and 290: M (14,400 lb ⋅ ft)(12 in./ft)∴
Page 291 and 292: pRoBLEmSp8.26 A small aluminum allo
Page 293 and 294: Equivalent BeamsBefore considering
Page 295 and 296: Transformed-Section methodThe conce
Page 297 and 298: This equation reduces to∫A∫ydA
Page 299 and 300: ExAmpLE 8.7A cantilever beam 10 ft
Page 301 and 302: Bending Stresses at the intersectio
Page 303 and 304: (a) the normal stress in each mater
Page 305 and 306: 283CM = PeF+ =bENdINg duE TO ANECCE
Page 307 and 308: and the combined normal stress on s
Page 309 and 310:
Combined Stress at KThe combined st
Page 311 and 312:
m8.23 Pipe AB (with outside diamete
Page 313 and 314:
p8.54 The bracket shown in Figure P
Page 315 and 316:
yFlangeyyz CM zWebM zM zCompressive
Page 317 and 318:
These curvature expressions can now
Page 319 and 320:
Coordinates of Points H and KThe (y
Page 321 and 322:
Similarly, the moment of inertia I
Page 323 and 324:
p8.63 A downward concentrated load
Page 325 and 326:
the factor K depends only upon the
Page 327 and 328:
SolutioNThe ultimate strength σ U
Page 329 and 330:
MO307bENdINg OF CuRVEd bARSr or iθ
Page 331 and 332:
Table 8.2 Area and Radial Distance
Page 333 and 334:
Plan the SolutionBegin by calculati
Page 335 and 336:
SuperpositionOften, curved bars are
Page 337 and 338:
We now set the stress at point A (r
Page 339:
p8.86 The curved tee shape shown in
Page 342 and 343:
320SHEAR STRESS IN bEAMSy9 kN(2)xAB
Page 344 and 345:
ExAmpLE 9.1M A = 11.0 kN·mAz(1)B(2
Page 346 and 347:
84.2 MPa (C)y126.4 MPa (C)116.7 kN6
Page 348 and 349:
326SHEAR STRESS IN bEAMSadA′MI zy
Page 350 and 351:
328SHEAR STRESS IN bEAMSEquation (f
Page 352 and 353:
SHEAR STRESS IN bEAMSyzzacbV(a) Box
Page 354 and 355:
332SHEAR STRESS IN bEAMSThe shear s
Page 356 and 357:
To determine the average horizontal
Page 358 and 359:
p9.3 The beam segment shown in Figu
Page 360 and 361:
9.6 Shear Stresses in Beams of circ
Page 362 and 363:
36 kNVyMzxShear Stress FormulaThe m
Page 364 and 365:
56 mmy13.5 mm 8.5 mm (3) K (4)z(5)7
Page 366 and 367:
pRoBLEmSp9.11 A 0.375 in. diameter
Page 368 and 369:
p9.23 A W14 × 34 standard steel se
Page 370 and 371:
348SHEAR STRESS IN bEAMSNail ANail
Page 372 and 373:
Plan the SolutionWhenever a cross s
Page 374 and 375:
ExAmpLE 9.6An alternative cross sec
Page 376 and 377:
ExERcISESm9.9 Five multiple-choice
Page 378 and 379:
at intervals of s along each side o
Page 380 and 381:
358SHEAR STRESS IN bEAMSCFdxtC′(2
Page 382 and 383:
360SHEAR STRESS IN bEAMStb2stbdsd2y
Page 384 and 385:
362SHEAR STRESS IN bEAMSIt is usefu
Page 386 and 387:
PointSketchyy - ′(mm)A′(mm 2 )Q
Page 388 and 389:
For this FBD, the calculation of Q
Page 390 and 391:
ExAmpLE 9.971 mm89 mm280 mmV10 mm 1
Page 392 and 393:
Point Sketch Calculation of Q = y
Page 394 and 395:
be expressed as the product of the
Page 396 and 397:
374SHEAR STRESS IN bEAMSThe exact l
Page 398 and 399:
376SHEAR STRESS IN bEAMSd2d2FIGURE
Page 400 and 401:
ExAmpLE 9.11Derive an expression fo
Page 402 and 403:
Note that, since the shape is thin
Page 404 and 405:
AByOzPED1.038 in.0.643 in.(a) Load
Page 406 and 407:
τydAϕdϕyShear StressThe variatio
Page 408 and 409:
p9.36 A plastic extrusion has the s
Page 410 and 411:
p9.48 The beam cross section shown
Page 412 and 413:
yePer3 in.z38 in. O38 in.10 in.5 in
Page 414 and 415:
10.2 moment-curvature RelationshipW
Page 416 and 417:
394bEAM dEFLECTIONS+ M + MPositive
Page 418 and 419:
10.4 Determining Deflections by Int
Page 420 and 421:
398bEAM dEFLECTIONS5. Boundary and
Page 422 and 423:
Beam Deflection and Slope at AThe d
Page 424 and 425:
Elastic Curve EquationSubstitute th
Page 426 and 427:
Boundary ConditionsBoundary conditi
Page 428 and 429:
integration over the interval a ≤
Page 430 and 431:
mecmoviesExERcISEm10.1 Beam Boundar
Page 432 and 433:
P10.11 For the beam and loading sho
Page 434 and 435:
Elastic Curve EquationSubstitute th
Page 436 and 437:
414beam deflectionsintegration give
Page 438 and 439:
(a) Beam Deflection at AAt the tip
Page 440 and 441:
Integrate again to obtain the beam
Page 442 and 443:
The inclusion of the reaction force
Page 444 and 445:
p10.24 The simply supported beam sh
Page 446 and 447:
vwPvwvPALBv Bx=ALB( v B ) wx+ALB( v
Page 448 and 449:
m10.5 Superposition Warm-up. A seri
Page 450 and 451:
Beam deflection at C: The beam defl
Page 452 and 453:
Avv30 kipsa= 13 ft b=7 ftBC D4 ft 6
Page 454 and 455:
Next, consider the overhang span be
Page 456 and 457:
By inspection, the rotation angle a
Page 458 and 459:
v A70 kNAv210 kN.mBθ BCv Cθ D3 m
Page 460 and 461:
Case 3—uniformly Distributed load
Page 462 and 463:
m10.12 Use the superposition method
Page 464 and 465:
v30 kNv12 kips2 kips/ftxxABHABC3 m
Page 466 and 467:
p10.53 The simply supported beam sh
Page 468 and 469:
446STATICALLy INdETERMINATEbEAMSM A
Page 470 and 471:
— 1 w 0 Lv2M AA xAB2LA y 3L—3 B
Page 472 and 473:
obtained from the continuity condit
Page 474 and 475:
Solve for ReactionsSolve Equation (
Page 476 and 477:
11.4 Use of Discontinuity Functions
Page 478 and 479:
Integrate again to obtain the beam
Page 480 and 481:
EI dvdx∫Integrate the beam load e
Page 482 and 483:
v20 kips 30 kips 20 kipsA B C D E7
Page 484 and 485:
462STATICALLy INdETERMINATEbEAMSThe
Page 486 and 487:
corresponding beam deflection will
Page 488 and 489:
Substitute these values into Equati
Page 490 and 491:
By2 6 436(200,000 N/mm )(351 × 10
Page 492 and 493:
The only unknown term in this equat
Page 494 and 495:
The only unknown term in this equat
Page 496 and 497:
p11.24-p11.26 For the beams and loa
Page 498 and 499:
v90 kN/m30 kN/m(1)ABCx40 kN3 m5.5 m
Page 500 and 501:
p11.49 Two steel beams support a co
Page 502 and 503:
12.2 Stress at a General point in a
Page 504 and 505:
12.3 Equilibrium of the Stress Elem
Page 506 and 507:
484STRESS TRANSFORMATIONSsuch as th
Page 508 and 509:
pRoBLEmSp12.1 A compound shaft cons
Page 510 and 511:
p12.9 A steel pipe with an outside
Page 512 and 513:
The forces acting on the vertical a
Page 514 and 515:
tynθxσn dAFigure 12.10b is a free
Page 516 and 517:
ExAmpLE 12.3y842 MPa550 MPa16 MPaxA
Page 518 and 519:
The angle θ from the redefined x a
Page 520 and 521:
p12.19 Two steel plates of uniform
Page 522 and 523:
500STRESS TRANSFORMATIONSprincipal
Page 524 and 525:
502STRESS TRANSFORMATIONSIf the nor
Page 526 and 527:
504STRESS TRANSFORMATIONSIn words,
Page 528 and 529:
yFIGURE 12.15506There is nevershear
Page 530 and 531:
ExAmpLE 12.5y9 ksi7 ksix11 ksiConsi
Page 532 and 533:
planes whose normal is perpendicula
Page 534 and 535:
ExERcISEm12.4 Sketching Stress tran
Page 536 and 537:
514STRESS TRANSFORMATIONSmecmovies
Page 538 and 539:
516STRESS TRANSFORMATIONSLabel the
Page 540 and 541:
mecmoviesExAmpLEm12.10 Coach Mohr
Page 542 and 543:
P 2(-7.22, 0)(-5, 6 ccw) yτ(2, 9.2
Page 544 and 545:
in the x-y coordinate system is rot
Page 546 and 547:
t73.7°τy (-16, 53)R = 55.22C (-31
Page 548 and 549:
ExAmpLE 12.11y(-14, 0)y(-14, 0)14 k
Page 550 and 551:
ExERcISESm12.10 Coach Mohr’s Circ
Page 552 and 553:
p12.46 Figure P12.46 shows Mohr’s
Page 554 and 555:
12.11 General State of Stress at a
Page 556 and 557:
534STRESS TRANSFORMATIONSIn Equatio
Page 558 and 559:
536STRESS TRANSFORMATIONSthird prin
Page 560 and 561:
538STRESS TRANSFORMATIONSyσp2Arbit
Page 562 and 563:
Strain TransformationsCHAPTER1313.1
Page 564 and 565:
542STRAIN TRANSFORMATIONSyyyγ xy d
Page 566 and 567:
544STRAIN TRANSFORMATIONSThus, diag
Page 568 and 569:
tnFinally, to compute the shear str
Page 570 and 571:
Table 13.2 Absolute maximum Shear S
Page 572 and 573:
The in-plane principal directions c
Page 574 and 575:
552STRAIN TRANSFORMATIONS13.6 mohr
Page 576 and 577:
y24.2°279 μεπ2579 μεx- 858 μ
Page 578 and 579:
556STRAIN TRANSFORMATIONSPlasticbac
Page 580 and 581:
Equation for gage b:2 2− 900 = e
Page 582 and 583:
pRoBLEmSp13.25-p13.30 The strain ro
Page 584 and 585:
yyyτ xyτ yzτ zxxxxzFIGURE 13.13a
Page 586 and 587:
564STRAIN TRANSFORMATIONSHence, the
Page 588 and 589:
SolutioN(a) Normal stresses: From E
Page 590 and 591:
Then solve for γ xy :2 2380 = (
Page 592 and 593:
(b) Principal and Maximum in-Plane
Page 594 and 595:
(b) Principal and Maximum in-Plane
Page 596 and 597:
y103.7 MPa40.2 MPa18.5°x63.5 MPa14
Page 598 and 599:
1εz = [ σ z − νσ ( x + σ y )
Page 600 and 601:
σ xyFinally, suppose the material
Page 602 and 603:
pRoBLEmSp13.31 An 8 mm thick brass
Page 604 and 605:
Problem e a e b e c E νycP13.47
Page 606 and 607:
p13.65 A solid plastic [E = 45 MPa;
Page 608 and 609:
586PRESSuRE VESSELSThe wall compris
Page 610 and 611:
588PRESSuRE VESSELSττ abs max = p
Page 612 and 613:
590PRESSuRE VESSELSStresses on the
Page 614 and 615:
592PRESSuRE VESSELSFor the outer su
Page 616 and 617:
31.9 MPay63.8 MPat13.81 MPax30°39.
Page 618 and 619:
p14.5 The normal strain measured on
Page 620 and 621:
p14.20 The cylindrical pressure ves
Page 622 and 623:
600PRESSuRE VESSELSSince the ends o
Page 624 and 625:
602PRESSuRE VESSELSσ θτ maxσ r4
Page 626 and 627:
604PRESSuRE VESSELSFor convenience,
Page 628 and 629:
14.6 Deformations in Thick-Walled c
Page 630 and 631:
Maximum Shear StressThe principal s
Page 632 and 633:
610PRESSuRE VESSELScould be cooled
Page 634 and 635:
Maximum Circumferential Stress in t
Page 636 and 637:
212.5 MPa134.4 MPa145.2 MPa76.2 MPa
Page 638 and 639:
CHAPTER15Combined Loads61615.1 Intr
Page 640 and 641:
6.92 ksiz6.92 ksiHAHy11.54 ksi8.49
Page 642 and 643:
p15.5 A solid 1.50 in. diameter sha
Page 644 and 645:
622COMbINEd LOAdSPP2P2FIGURE 15.1 S
Page 646 and 647:
internal bending moment acting at t
Page 648 and 649:
60 kNShear Force and Bending Moment
Page 650 and 651:
ExAmpLE 15.3A steel hollow structur
Page 652 and 653:
1.216 ksiH1.216 ksiH1.373 ksi30.3°
Page 654 and 655:
Figure P15.13b/14b are d = 240 mm,
Page 656 and 657:
p15.25 A load P = 75 kN acting at a
Page 658 and 659:
636COMbINEd LOAdS b. Bending moment
Page 660 and 661:
20.05 MPacbyM z = 3.85 kN·m20.05 M
Page 662 and 663:
102,400 N · mm = 102.4 N · m. Sim
Page 664 and 665:
yK12.02 MPax45°12.02 MPaStress tra
Page 666 and 667:
The equivalent moments at the secti
Page 668 and 669:
Hy3,000 lb448 psiKThe 3,000 lb shea
Page 670 and 671:
z8.45 kN·m15.6 kN·mHK10.8 kN·mEq
Page 672 and 673:
Torsion shear13.438 MPaBeam shear3.
Page 674 and 675:
m15.5 Determine the stresses acting
Page 676 and 677:
p15.34 Forces P x = 580 lb and P y
Page 678 and 679:
z 2z 1dimensions of the assembly ar
Page 680 and 681:
658COMbINEd LOAdSIf the naming conv
Page 682 and 683:
660COMbINEd LOAdSfrom which it foll
Page 684 and 685:
662COMbINEd LOAdSSimilarly, Equatio
Page 686 and 687:
(b) What is the value of the Mises
Page 688 and 689:
p15.54 A thin-walled cylindrical pr
Page 690 and 691:
668COLuMNSStability of EquilibriumT
Page 692 and 693:
670COLuMNSSummaryBefore a compressi
Page 694 and 695:
672COLuMNSIf we let2Pk =(16.2)EIthe
Page 696 and 697:
674COLuMNS7060σY =50=σ cr4030σY
Page 698 and 699:
The Euler buckling load for this co
Page 700 and 701:
Spacer blockFIGURE p16.676 mm 76 mm
Page 702 and 703:
16.3 The Effect of End conditionson
Page 704 and 705:
682COLuMNSwhere the first two terms
Page 706 and 707:
684COLuMNSAnother way of expressing
Page 708 and 709:
LateralbracingxCBP17.5 ft17.5 ftzSt
Page 710 and 711:
xBPBuckling About the Strong AxisTh
Page 712 and 713:
p16.19 The aluminum column shown in
Page 714 and 715:
692COLuMNSIn this case, a relations
Page 716 and 717:
694COLuMNSwhich can be further simp
Page 718 and 719:
Pexp16.26 A steel [E = 200 GPa] pip
Page 720 and 721:
698COLuMNSFor short and intermediat
Page 722 and 723:
ExAmpLE 16.52 in.zSpacer blocky2 in
Page 724 and 725:
Since KL/ry > 133.7, the column is
Page 726 and 727:
The reduced Euler buckling stress t
Page 728 and 729:
Determine the allowable axial load
Page 730 and 731:
708COLuMNSwhere the compressive str
Page 732 and 733:
Both tensile and compressive normal
Page 734 and 735:
ExAmpLE 16.940 kNexA 6061-T6 alumin
Page 736 and 737:
pRoBLEmSp16.42 The structural steel
Page 738 and 739:
716ENERgy METHOdSThe total energy o
Page 740 and 741:
718ENERgy METHOdSydV = dx dy dzzxxS
Page 742 and 743:
720ENERgy METHOdSSince the volume o
Page 744 and 745:
The maximum force P that can be app
Page 746 and 747:
17.5 Elastic Strain Energy for Flex
Page 748 and 749:
segment AB of the beam. The second
Page 750 and 751:
p17.8 A solid stepped shaft made of
Page 752 and 753:
730ENERgy METHOdSNote that the nega
Page 754 and 755:
732ENERgy METHOdSSignificance: In t
Page 756 and 757:
From the positive root, the maximum
Page 758 and 759:
Note: Throughout the previous chapt
Page 760 and 761:
The right-hand side of this equatio
Page 762 and 763:
(b) If the rod has a constant diame
Page 764 and 765:
Note: Throughout the previous chapt
Page 766 and 767:
assembly. The solid bronze post has
Page 768 and 769:
17.7 Work-Energy method for Single
Page 770 and 771:
F 1(1)BSolutioNThe internal axial f
Page 772 and 773:
17.8 method of Virtual WorkThe meth
Page 774 and 775:
752ENERgy METHOdSIf the material be
Page 776 and 777:
754ENERgy METHOdSwhich is the mathe
Page 778 and 779:
756ENERgy METHOdSThe virtual intern
Page 780 and 781:
both P 1 and P 2 from the truss, ap
Page 782 and 783:
Following is the table produced by
Page 784 and 785:
SolutioNCalculate the member length
Page 786 and 787:
764ENERgy METHOdSObtain the total v
Page 788 and 789:
766ENERgy METHOdSthe real external
Page 790 and 791:
ExAmpLE 17.141,400 N900 NCalculate
Page 792 and 793:
1Ax 1 or x 2vmDraw a free-body diag
Page 794 and 795:
p17.36 Determine the vertical displ
Page 796 and 797:
p17.54 Determine the minimum moment
Page 798 and 799:
776ENERgy METHOdSFor the general ca
Page 800 and 801:
ExAmpLE 17.16Determine the vertical
Page 802 and 803:
SolutioNWe seek the horizontal defl
Page 804 and 805:
782ENERgy METHOdSprocedure for Anal
Page 806 and 807:
Castigliano’s second theorem appl
Page 808 and 809:
Finally, derive the following equat
Page 810 and 811:
p17.63 The truss shown in Figure P1
Page 812 and 813:
APPENDIXAgeometric Propertiesof an
Page 814 and 815:
792gEOMETRIC PROPERTIESOF AN AREAyy
Page 816 and 817:
mecmoviesExAmpLESA.2 Animated examp
Page 818 and 819:
796gEOMETRIC PROPERTIESOF AN AREAfo
Page 820 and 821:
ExAmpLE A.340 mmDetermine the momen
Page 822 and 823:
ExAmpLE A.440 mmDetermine the produ
Page 824 and 825:
yy′yOxxdAθθcos θx′xy′sin
Page 826 and 827:
804gEOMETRIC PROPERTIESOF AN AREANo
Page 828 and 829:
806gEOMETRIC PROPERTIESOF AN AREAI
Page 830 and 831:
I xy(38.8, 32.3) y1.654 in.yR = 38.
Page 832 and 833:
YttffdXXt wYb fDesignationAreaADept
Page 834 and 835:
YttffdXXt wYb fDesignationAreaADept
Page 836 and 837:
x-XYttffXt wdYAmerican Standard cha
Page 838 and 839:
b fdttffXYXy-t wYDesignationAreaADe
Page 840 and 841:
dXYXtYbDesignationHSS304.8 × 203.2
Page 842 and 843:
YxZXyXαYZDesignationMasspermeterAr
Page 844 and 845:
Simply Supported BeamsBeam Slope De
Page 846 and 847:
Average Properties ofSelected Mater
Page 848 and 849:
Table D.1b Average properties of Se
Page 850 and 851:
Fundamental Mechanicsof Materials E
Page 852 and 853:
orσx + σ y σx − σ yσ n = + c
Page 854 and 855:
Answers to Odd NumberedProblemsChap
Page 856 and 857:
(c) φ E/C = 0.1408 rad(d) T A = 23
Page 858 and 859:
P7.35 (a)wx ( ) =−5kN −x − 0
Page 860 and 861:
(c)Chapter 8P8.1 σ = 1.979 ksiP8.3
Page 862 and 863:
P10.7 v B = −8.27 mmP10.9 (a)wx 0
Page 864 and 865:
P12.35 (a) 25.6 MPa(b) 23.1° clock
Page 866 and 867:
P13.63 (a) Db = 1.272 mm,Dc = 0.254
Page 868 and 869:
P17.41 (a) D F = 18.54 mm ←(b) D
Page 870 and 871:
Biaxial stress, 566-568, 587, 590-5
Page 872 and 873:
GGage length, 46, 47, 51, 54Gears,
Page 874 and 875:
QQ section property, 327-332, 338,
Page 876:
Torsionof circular shafts, 135-151e
show all

Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Create successful ePaper yourself

Delete template?

Save as template?