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A research-based,online learning en
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MECHANICS OF MATERIALS:An Integrate
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About the AuthorTimothy A. Philpot
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xPREFACEAnimation also offers a new
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xiiPREFACE• Appendix E Fundamenta
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xivPREFACEWhat Do Instructors recei
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ContentsChapter 1 Stress 11.1 Intro
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Chapter 14 Pressure Vessels 58514.1
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2STRESSAddressing these concerns re
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4STRESSnumber begins with the digit
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ExAmpLE 1.3A(1)50 mmB80 kNA 50 mm w
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8STRESSFIGURE 1.2b Free-bodydiagram
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mecmoviesExAmpLEm1.5 A pin at C and
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12STRESSJeffery S. ThomasFIGURE 1.5
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14STRESSFIGURE 1.6 Bearing stress f
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m1.3 Use shear stress concepts for
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applied to beam ABC? Use dimensions
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p1.24 The two wooden boards shown i
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1.5 Stresses on Inclined Sectionsme
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24STRESSSignificanceAlthough one mi
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Thus, to provide the necessary weld
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p1.40 Two wooden members are glued
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30STRAINposition vector between H a
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32STRAINIn developing the concept o
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mecmoviesExAmpLESm2.1 A rigid steel
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p2.4 Bar (1) has a length of L 1 =
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38STRAINIn this expression, θ′ i
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pRoBLEmSp2.11 A thin rectangular po
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SOLUTIONThe thermal strain for a te
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CHAPTER3Mechanical Propertiesof Mat
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(3)(7)(4)(5)(6)Fracture47THE TENSIO
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8049THE STRESS-STRAIN dIAgRAM7060St
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The elastic limit is the largest st
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80Aluminum alloy××53THE STRESS-ST
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The second measure is the reduction
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Values vary for different materials
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before the strain measurement. From
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mecmoviesExERcISEm3.1 Figure M3.1 d
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material is initially 800 mm long a
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CHAPTER4design Concepts4.1 Introduc
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the distributed uniform area loadin
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In some instances, engineers may ne
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both members will be stressed to th
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MecMoviesExAMpLESM4.1 The structure
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bPtPFIGURE p4.3Splice plateBarBarPP
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4.5 Load and Resistance Factor Desi
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79FrequencyLarger γ factors combin
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Limit StatesLRFD is based on a limi
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CHAPTER5Axial deformation5.1 Introd
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The increased normal stress magnitu
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AAAA87dEFORMATIONS IN AxIALLyLOAdEd
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displacement in the horizontal dire
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mecmoviesExAmpLEm5.2 The roof and s
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t = 8 mm, and E = 72 GPa, determine
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d 0to 2d 0 at the other end. A conc
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How are the rigid-bar deflections v
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ExAmpLE 5.4A tie rod (1) and a pipe
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Since the sum of the four interior
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5.5 Statically Indeterminate Axiall
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Step 3 — Force-Deformation Relati
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Successful application of the five-
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Plan the SolutionConsider a free-bo
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Structures with a Rotating Rigid Ba
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(1)Px Bx C113STATICALLy INdETERMINA
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ExERcISESm5.5 A composite axial str
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p5.29 In Figure P5.29, a load P is
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tube has an outside diameter of 1.5
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ExAmpLE 5.7An aluminum rod (1) [E =
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mecmoviesExAmpLEm5.14 A rectangular
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Equations (h) and (i) can be solved
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pRoBLEmSp5.39 A circular aluminum a
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polymer [E = 370 ksi; α = 39.0 ×
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Stress-concentration factor K3.02.8
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of the hole has decayed to a value
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TorsionCHAPTER66.1 IntroductionTorq
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with respect to an adjacent cross s
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the longitudinal axis of the shaft.
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τStressxyτntσn141STRESSES ON ObL
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If a torsion member is subjected to
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ExAmpLE 6.1A hollow circular steel
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Plan the SolutionTo determine the l
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Plan the SolutionThe internal torqu
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mecmoviesExAmpLESm6.4 Determine the
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p6.10 The mechanism shown in Figure
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Furthermore, since gears have teeth
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will be dictated by the ratio of ge
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mecmoviesExAmpLEm6.13 Two solid ste
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6.8 power Transmission161POwER TRAN
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m6.17 A motor shaft is being design
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pulley is a belt having tensions F
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Step 2 — Geometry of Deformation:
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Polar moments of inertia for the al
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Successful application of the five-
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Plan the SolutionAA free-body diagr
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From Equation (e), the allowable in
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Next, consider a free-body diagram
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Polar moments of inertia for the sh
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m6.21 A composite torsion member co
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zD(3)Ay(1)L1,L3N BBL 2(a) Determine
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Stress concentrations also occur at
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For the case of the rectangular bar
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and the angle of twist for a 12 in.
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ExAmpLE 6.13A rectangular box secti
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CHAPTER7Equilibrium of beams7.1 Int
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beam (also called a simple beam). A
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ExAmpLE 7.1Draw the shear-force and
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The negative value for A y indicate
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Plot the FunctionsPlot the function
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ExAmpLE 7.5Draw the shear-force and
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yw aw byw 0xxABCABabLFIGURE p7.4FIG
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w(x) = w G at G. At A, where the di
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the left and right sides of a thin
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General procedure for constructing
- Page 235 and 236:
Construct the Bending-Moment Diagra
- Page 237 and 238:
ExAmpLE 7.7Draw the shear-force and
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j M(6) = 0 kN ⋅ m (Rule 4: ∆M =
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of M diagram = shear force V). The
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m7.3 Dynamically generated shear-fo
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p7.21-p7.22 Use the graphical metho
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x - a 0x - a 1x - a 2225dISCONTINuI
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conditions. The reactions for stati
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120 kN ⋅ m concentrated moment: F
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SolutioNWhen we refer to case 4 of
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Uniformly distributed load between
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y5 kN20 kN.my12 kN.m18 kN/mAB3 m 3
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CHAPTER8bending8.1 IntroductionPerh
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has a constant bending moment M, an
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The most common stress-strain relat
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All such moment increments that act
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The sense of σ x (either tension o
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A i(mm 2 )y i(mm)y i A i(mm 3 )(1)
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ExAmpLE 8.2The cross-sectional dime
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m8.7 Determine the centroid locatio
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zFIGURE p8.8p8.9 An aluminum alloy
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WidthWidthWidthDepthXYthicknessXWeb
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700 lb1,500 lb4 in.1 in.200 lb/ft1
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M = −6,000 lb · ft. For this neg
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mecmoviesExAmpLESm8.9 Determine the
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p8.16 A W18 × 40 standard steel sh
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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 343 and 344: a maximum intensity of (4.5 MPa −
- Page 345 and 346: I ci (mm 4 ) | d i | (mm) d i 2 A i
- Page 347 and 348: 9.3 The Shear Stress Formula325THE
- Page 349 and 350: What is the significance of Equatio
- Page 351 and 352: ad329THE FIRST MOMENT OF AREA, QVb
- Page 353 and 354: 9.5 Shear Stresses in Beams of Rect
- Page 355 and 356: mecmoviesExAmpLEm9.2 Derivation of
- Page 357 and 358: (b) Maximum Horizontal Shear Stress
- Page 359 and 360: ywaAaxLFIGURE p9.7a Simply supporte
- Page 361 and 362: the distribution of shear stress ma
- Page 363 and 364: (a) Shear Stress at HBefore proceed
- Page 365 and 366: m9.6 Determine the maximum horizont
- Page 367 and 368: p9.17 The extruded plastic shape sh
- Page 369 and 370: NailA(1)zy(2)Nail CNailB(3)Mz(2)(1)
- Page 371 and 372: Identifying the proper Area for QIn
- Page 373 and 374: Shear Flow FormulaThe shear flow fo
- Page 375 and 376: mecmoviesExAmpLESm9.9 Determine the
- Page 377 and 378: (a) If the screws are uniformly spa
- Page 379 and 380: F(1)dx357SHEAR STRESS ANd SHEARFLOw
- Page 381 and 382: τ xzCutFreesurfaceyFree surfaceCut
- Page 383 and 384: Next, consider the web of the thin-
- Page 385 and 386: Longitudinal planeof symmetryyB′
- Page 387 and 388: The directions and intensities of t
- Page 389 and 390: of Q. However, what is the negative
- Page 391 and 392:
used in Equation (b). that is, the
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(b) Distribution of shear stressThe
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Calculation of Shear Stress Distrib
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PA375SHEAR CENTERS OFTHIN-wALLEd OP
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SHEAR CENTERS OFTHIN-wALLEd OPENSEC
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the resultant force F w of the shea
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Distribution of Shear StressThe dis
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ExAmpLE 9.14Find the shear center O
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SHEAR CENTERS OFTHIN-wALLEd OPENSEC
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p9.42 The hat-shape extrusion shown
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btOet (typ.)cOh2beh2aFIGURE p9.54bF
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CHAPTER10beam deflections10.1 Intro
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When the beam is bent, points along
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P395THE dIFFERENTIAL EQuATIONOF THE
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continuity conditionsMany beams are
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integrationEquation (b) will be int
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1 ⎛ 2wx0 ⎞Σ M =⎛ ⎞⎝⎜
- Page 425 and 426:
order to complete this FBD, however
- Page 427 and 428:
ExAmpLE 10.4The simple beam shown s
- Page 429 and 430:
Since C 1 = C 3 ,C2 2Pb( L − b )=
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p10.5 For the beam and loading show
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ExAmpLE 10.5A beam is loaded and su
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p10.15 For the beam and loading sho
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to obtain the shear-force function
- Page 439 and 440:
Note that the second term in this e
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ExAmpLE 10.8For the beam shown, use
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pRoBLEmSp10.17 For the beam and loa
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p10.31 For the beam and loading sho
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calculation. Before proceeding, it
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The beam deflection at B can be cal
- Page 451 and 452:
mecmoviesExAmpLESm10.8 Determine th
- Page 453 and 454:
20 kipsb = 4 ft a=16 ftvxv Cx=10 ft
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MLθ =6EIBy the values defined prev
- Page 457 and 458:
The beam deflections at A, C, and E
- Page 459 and 460:
The beam deflection at E is compute
- Page 461 and 462:
Beam deflection at C: The beam defl
- Page 463 and 464:
m10.5 Superposition Warm-up. Exampl
- Page 465 and 466:
v4 kips/ft45 kipsvv180 kN.m70 kN80
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CHAPTERStatically Indeterminatebeam
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447P1P2P1P2P1P2THE INTEgRATION METH
- Page 471 and 472:
Evaluate ConstantsSubstitute the bo
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integrationIntegrate Equation (f) t
- Page 475 and 476:
p11.5 A beam is loaded and supporte
- Page 477 and 478:
SolutioN(a) Support ReactionsAn FBD
- Page 479 and 480:
ExAmpLE 11.4For the statically inde
- Page 481 and 482:
Equation (c) for the beam slope and
- Page 483 and 484:
11.5 The Superposition method461THE
- Page 485 and 486:
vPvPvAL—2L—2(a) Actual beamxBAS
- Page 487 and 488:
Notice that EI appears in both term
- Page 489 and 490:
that the roller support is displace
- Page 491 and 492:
Case 2—Cantilever Beam with Conce
- Page 493 and 494:
wooden beam. Determine an expressio
- Page 495 and 496:
ExERcISESm11.1 Propped Cantilevers.
- Page 497 and 498:
v60 kN/m125 kNv80 lb/in.140 lb/in.A
- Page 499 and 500:
p11.44 A timber [E = 12 GPa] beam i
- Page 501 and 502:
Stress TransformationsCHAPTER1212.1
- Page 503 and 504:
on the x, y, and z planes. Stresses
- Page 505 and 506:
The result of this simple equilibri
- Page 507 and 508:
Stresses at HThe forces and moments
- Page 509 and 510:
p12.5 An extruded polymer flexural
- Page 511 and 512:
also be added together to produce t
- Page 513 and 514:
pRoBLEmSp12.11-p12.14 The stresses
- Page 515 and 516:
Stress InvarianceThe normal stress
- Page 517 and 518:
The choice of either Equation (12.3
- Page 519 and 520:
ExERcISESm12.1 the Amazing Stress C
- Page 521 and 522:
p12.25-p12.26 The stresses shown in
- Page 523 and 524:
In this figure, we will assume that
- Page 525 and 526:
Method one. The first method is sim
- Page 527 and 528:
p2p2p2σ p2σp2σ p2σp1σp1σp1p1p
- Page 529 and 530:
• Substitute the value of θ s in
- Page 531 and 532:
Since θ p is negative, the angle i
- Page 533 and 534:
(b) The principal stresses and the
- Page 535 and 536:
p12.35 A shear wall in a reinforced
- Page 537 and 538:
Utility of mohr’s circleMohr’s
- Page 539 and 540:
8. Several points on Mohr’s circl
- Page 541 and 542:
Since points x and y are always the
- Page 543 and 544:
element counterclockwise; therefore
- Page 545 and 546:
and the normal stress acting on the
- Page 547 and 548:
ExAmpLE 12.10Stresses on an incline
- Page 549 and 550:
For element A, the absolute maximum
- Page 551 and 552:
pRoBLEmSp12.42 Figure P12.42 shows
- Page 553 and 554:
30.0 ksi2,250 psia5.5 ksi680 psiσ1
- Page 555 and 556:
The forces on the x, y, and z faces
- Page 557 and 558:
where a i , b i , and c i are the c
- Page 559 and 560:
Table 12.1 Direction cosines for pl
- Page 561 and 562:
p12.66 The stresses at point O in a
- Page 563 and 564:
ydx′541PLANE STRAINdxθ′ zxdy
- Page 565 and 566:
Next, the displacement vector AA′
- Page 567 and 568:
yyπ2γ xyxFIGURE 13.6a Positive sh
- Page 569 and 570:
13.4 principal Strains and maximumS
- Page 571 and 572:
• When θ p is positive, the elem
- Page 573 and 574:
p13.3 The thin rectangular plate sh
- Page 575 and 576:
SolutioNPlot point x as (ε x , −
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pRoBLEmSp13.16-p13.17 The principal
- Page 579 and 580:
Equations (13.9), (13.10), and (13.
- Page 581 and 582:
out-of-plane normal strain e z will
- Page 583 and 584:
Although stress is applied only in
- Page 585 and 586:
Unit Volume changeConsider an infin
- Page 587 and 588:
SolutioN(a) Determine the change in
- Page 589 and 590:
Also, Equation (13.17) becomes, sim
- Page 591 and 592:
e x and e y in terms of σ x and σ
- Page 593 and 594:
ExAmpLE 13.9On the free surface of
- Page 595 and 596:
Since the strain measurements were
- Page 597 and 598:
ExAmpLE 13.10A 2014-T6 aluminum all
- Page 599 and 600:
Whereas elongations and contraction
- Page 601 and 602:
ExAmpLE 13.11A unidirectional T-300
- Page 603 and 604:
p13.39 A thin brass [E = 100 GPa; G
- Page 605 and 606:
(a) Determine the normal strains e
- Page 607 and 608:
CHAPTER14Pressure Vessels14.1 Intro
- Page 609 and 610:
From this equilibrium equation, an
- Page 611 and 612:
yσ hoopzxσ longP = pπr 2σ longF
- Page 613 and 614:
τprτ abs max = +2tp2591STRAINS IN
- Page 615 and 616:
Stresses in the outlet PipeThe long
- Page 617 and 618:
m14.5 A cylindrical steel [E = 200
- Page 619 and 620:
diameter of 1,800 mm and a wall thi
- Page 621 and 622:
σ θ dr ∆xdθ599STRESSES IN THIC
- Page 623 and 624:
The radial stress in the thick-wall
- Page 625 and 626:
Similarly, the change in radial str
- Page 627 and 628:
External pressure on a Solid circul
- Page 629 and 630:
ExAmpLE 14.3An open-ended high-pres
- Page 631 and 632:
Then, set σ θ equal to the allowa
- Page 633 and 634:
Cylinder shrink fitted on a solid s
- Page 635 and 636:
The circumferential stress on the i
- Page 637 and 638:
p14.31 An open-ended compound thick
- Page 639 and 640:
normal stresses are identical at al
- Page 641 and 642:
mecmoviesExAmpLEm15.1 A tubular sha
- Page 643 and 644:
15.3 principal Stresses in a Flexur
- Page 645 and 646:
a. The normal stresses caused by F
- Page 647 and 648:
Shear Force and Bending Moment at H
- Page 649 and 650:
mecmoviesExAmpLESm15.2 A cantilever
- Page 651 and 652:
The following equilibrium equations
- Page 653 and 654:
m15.3 The rectangular tube is subje
- Page 655 and 656:
yPPQb ft fAx HHxB C DKx KEyL4FIGURE
- Page 657 and 658:
F yyM yy635gENERAL COMbINEdLOAdINgS
- Page 659 and 660:
SolutioNSection PropertiesThe cross
- Page 661 and 662:
m15.7 A rectangular post has cross-
- Page 663 and 664:
Combined Stresses at HThe normal an
- Page 665 and 666:
SolutioNEquivalent Force SystemA sy
- Page 667 and 668:
The 28,800 lb · ft bending moment
- Page 669 and 670:
Stress transformation Results at KT
- Page 671 and 672:
The 10.8 kN · m (i.e., 10.8 × 10
- Page 673 and 674:
The 8.45 kN · m (i.e., 8.45 × 10
- Page 675 and 676:
p15.28 Three loads are applied to t
- Page 677 and 678:
P y = 375 lb, and P z = 550 lb. Det
- Page 679 and 680:
σ Y(a) Stress element at yieldfor
- Page 681 and 682:
where P is some function of δ. The
- Page 683 and 684:
When the third term in the brackets
- Page 685 and 686:
CompressiontestτTensiontestσσUTp
- Page 687 and 688:
ExAmpLE 15.9The stresses on the fre
- Page 689 and 690:
CHAPTER16Columns16.1 IntroductionIn
- Page 691 and 692:
moment, then the system will tend t
- Page 693 and 694:
Buckled configurationIf the compres
- Page 695 and 696:
The critical load for an ideal colu
- Page 697 and 698:
• The Euler buckling load equatio
- Page 699 and 700:
Similarly, the moment of inertia of
- Page 701 and 702:
p16.11 An assembly consisting of ti
- Page 703 and 704:
P< PcrBxP=PcrBB yPBB yxPBL −x681T
- Page 705 and 706:
The equation of the buckled column
- Page 707 and 708:
where L is the actual length of the
- Page 709 and 710:
Critical StressThe critical load eq
- Page 711 and 712:
pRoBLEmSp16.15 An HSS152.4 × 101.6
- Page 713 and 714:
column is as shown in Figure 16.8b.
- Page 715 and 716:
1.15e = 0693THE SECANT FORMuLAPP cr
- Page 717 and 718:
several values of the eccentricity
- Page 719 and 720:
graph shows a scattered range of va
- Page 721 and 722:
For intermediate-length columns wit
- Page 723 and 724:
Controlling slenderness ratio: Sinc
- Page 725 and 726:
From this allowable stress, the all
- Page 727 and 728:
fixed at base A with respect to ben
- Page 729 and 730:
p16.41 A simple pin-connected wood
- Page 731 and 732:
ExAmpLE 16.8The W12 × 58 structura
- Page 733 and 734:
(b) Magnitude of the largest eccent
- Page 735 and 736:
where c is the outside radius of th
- Page 737 and 738:
CHAPTER17Energy Methods17.1 Introdu
- Page 739 and 740:
deformation vary. This dependency i
- Page 741 and 742:
σσModulus oftoughnessFracture719w
- Page 743 and 744:
or in terms of the deformation δ a
- Page 745 and 746:
strain energies in all the segments
- Page 747 and 748:
ExAmpLE 17.3A cantilever beam AB of
- Page 749 and 750:
or, since a + b = L,UPab 2 2 2= Ans
- Page 751 and 752:
In the loaded system, dynamic loadi
- Page 753 and 754:
Special cases. Two extreme situatio
- Page 755 and 756:
Plan the SolutionThe axial deformat
- Page 757 and 758:
so that the impact factor can be wr
- Page 759 and 760:
Multiplying the two factors of the
- Page 761 and 762:
SolutioN(a) The areas of the two ro
- Page 763 and 764:
This strain-energy density is repre
- Page 765 and 766:
pRoBLEmSp17.12 A 19 mm diameter ste
- Page 767 and 768:
p17.24 Figure P17.24 shows block D,
- Page 769 and 770:
The external work done by a force a
- Page 771 and 772:
pRoBLEmSp17.30 Determine the horizo
- Page 773 and 774:
PB751METHOd OF VIRTuAL wORkP 2ACδ
- Page 775 and 776:
internal forces f 1 and f 2 acting
- Page 777 and 778:
17.9 Deflections of Trussesby the V
- Page 779 and 780:
The right-hand sides of Equations (
- Page 781 and 782:
at B. The right-hand side represent
- Page 783 and 784:
MemberL(mm)A(mm 2 )F(kN)f(kN)⎛ FL
- Page 785 and 786:
vwPv1763dEFLECTIONS OF bEAMS byTHE
- Page 787 and 788:
Pw1765dEFLECTIONS OF bEAMS byTHE VI
- Page 789 and 790:
Real Moment M: Remove the virtual l
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BeamSegmentx Coordinateoriginlimits
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For segment CD, draw a free-body di
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P 2P 2P 2P PAC E G HB D F6 mp17.49
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Therefore, the total strain energy
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The terms L, A, and E are constant
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elastic modulus E = 200 GPa; theref
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17.13 Calculating Deflections of Be
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5. Integration: Perform the integra
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ExAmpLE 17.19Compute the deflection
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From Equation (17.40), the beam def
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120 N/mm40 kN/mAB C100 mm 65 mm 65
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Table A.1 properties of plane Figur
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ExAmpLE A.1Determine the location o
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Using the Pythagorean theorem, dist
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SolutioN(a) Moment of inertia About
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A.3 product of Inertia for an AreaT
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1.654 in.y1.654 in.y1.154 in.(1)1.8
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andcos2θ =p∓( I − I )/2x⎛ (
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ExAmpLE A.7Determine the principal
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mecmoviesExAmpLEA.6 The theory and
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geometric Properties ofStructural S
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Wide-Flange Sections or W Shapes—
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x-XYttffXt wdYb fAmerican Standard
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Shapes cut from Wide-Flange Section
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Hollow Structural Sections or HSS S
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Angle Shapes or L ShapesDesignation
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Table of beam Slopesand deflections
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cantilever BeamsBeam Slope Deflecti
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Table D.1a Average properties of Se
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Table D.2 Typical properties of Sel
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Gear relationships between gears A
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thin-walled pressure vesselsTangent
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P3.11 P = 42,000 lbP3.13 (a) E = 30
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P7.1310.0 kips40.00 kip·ftxP7.293.
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(c)(c)P7.45−(a) wx ( ) = 42.09 ki
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P8.83 (a) r n = 112.5 mm(b) σ A =
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P11.23 (a) P = 12.50 kips(b) M = 31
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P12.63 T max = 119.8 N · mP12.65 (
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P15.43 σ x = 1,545 psi, σ z = 0 p
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IndexAAbsolute maximum shear strain
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load, 207, 210, 211load-deflection,
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for indeterminate beam analysis,461
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flexural, 239-240in-plane shear, 54