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

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A B C D E F500 mmB y400 mm 600 mm 600 mm 400 mm200 N 350 N 400 N 200 NE yConstruct the Shear-Force andBending-Moment DiagramsThe shear-force and bending-moment diagrams can beconstructed in accordance with the six rules outlined inSection 7.3.The maximum internal bending moment occurs at Dand has a magnitude of M = 115 N · m.Moment of inertiaThe moment of inertia for the 40 mm diameter solid steelshaft isA B C D E FIzπ π= d = (40mm) = 125,664 mm64 644 4 4VM500 mm400 mm 600 mm 600 mm 400 mm200 N 350 N 400 N 200 N–200 N625 N–100 N·m425 N70 N·m75 N115 N·m–325 N525 N–80 N·m200 NFlexure FormulaThe maximum bending stress in the shaft occurs at D.Since the circular cross section is symmetric about the axisof bending, both the tensile and compressive bendingstresses have the same magnitude. In this situation, theflexure formula in the form of Equation (8.10) is convenientfor calculating bending stresses. The distance c usedin Equation (8.10) is simply the shaft radius. From thisform of the flexure formula, the magnitude of the maximumbending stress in the shaft isMc (115 Nm)(20 ⋅ mm)(1, 000 mm/m)σ max = =Iz125,664 mm4= 18.30 MPaAns.Section Modulus for a Solid Circular SectionAlternatively, the magnitude of the maximum bendingstress in the shaft can be computed from the section modulus.For a solid circular section, the following formulacan be derived for the section modulus:4Iz( π/64)d πS = = = dc d/2 323For the 40 mm diameter solid steel shaft considered here, the section modulus is,therefore,Sπ π= d = (40mm) = 6,283 mm32 323 3 3and the magnitude of the maximum bending stress in the shaft can be computed asM (115 Nm)(1, ⋅ 000 mm/m)σ max = =3S 6,283 mm= 18.30 MPaAns.260
mecmoviesExAmpLESm8.9 Determine the bending-moment diagram and the maximumtension and compression bending stresses for a tee shape.m8.13 Determine the bending-moment diagram, the centroidlocation, the moment of inertia, and the bending stressfor a simply supported beam with a U-shaped cross section.m8.10 Determine maximumbending moments, given allowabletension and compressionbending stresses.. . ....m8.11 Determine the bending-moment diagram, the momentof inertia, and the bending stress produced in a simplysupported wide-flange steel beam.m8.14 Determine the bending-moment diagram and thebending stress for a standard steel shape that is used as asimply supported beam with an overhang.m8.12 Determine the bending-moment diagram, the momentof inertia, and the bending stress produced in a cantileverbeam consisting of a tee shape.261
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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
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 237 and 238: ExAmpLE 7.7Draw the shear-force and
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 and 258: y5 kN20 kN.my12 kN.m18 kN/mAB3 m 3
Page 259 and 260: CHAPTER8bending8.1 IntroductionPerh
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: M = −6,000 lb · ft. For this neg
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
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Plan the SolutionBegin by calculati
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SuperpositionOften, curved bars are
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We now set the stress at point A (r
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p8.86 The curved tee shape shown in
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320SHEAR STRESS IN bEAMSy9 kN(2)xAB
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ExAmpLE 9.1M A = 11.0 kN·mAz(1)B(2
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84.2 MPa (C)y126.4 MPa (C)116.7 kN6
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326SHEAR STRESS IN bEAMSadA′MI zy
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328SHEAR STRESS IN bEAMSEquation (f
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SHEAR STRESS IN bEAMSyzzacbV(a) Box
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332SHEAR STRESS IN bEAMSThe shear s
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To determine the average horizontal
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p9.3 The beam segment shown in Figu
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9.6 Shear Stresses in Beams of circ
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36 kNVyMzxShear Stress FormulaThe m
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56 mmy13.5 mm 8.5 mm (3) K (4)z(5)7
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pRoBLEmSp9.11 A 0.375 in. diameter
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p9.23 A W14 × 34 standard steel se
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348SHEAR STRESS IN bEAMSNail ANail
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Plan the SolutionWhenever a cross s
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ExAmpLE 9.6An alternative cross sec
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ExERcISESm9.9 Five multiple-choice
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at intervals of s along each side o
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358SHEAR STRESS IN bEAMSCFdxtC′(2
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360SHEAR STRESS IN bEAMStb2stbdsd2y
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362SHEAR STRESS IN bEAMSIt is usefu
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PointSketchyy - ′(mm)A′(mm 2 )Q
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For this FBD, the calculation of Q
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ExAmpLE 9.971 mm89 mm280 mmV10 mm 1
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Point Sketch Calculation of Q = y
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be expressed as the product of the
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374SHEAR STRESS IN bEAMSThe exact l
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376SHEAR STRESS IN bEAMSd2d2FIGURE
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ExAmpLE 9.11Derive an expression fo
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Note that, since the shape is thin
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AByOzPED1.038 in.0.643 in.(a) Load
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τydAϕdϕyShear StressThe variatio
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p9.36 A plastic extrusion has the s
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p9.48 The beam cross section shown
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yePer3 in.z38 in. O38 in.10 in.5 in
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10.2 moment-curvature RelationshipW
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394bEAM dEFLECTIONS+ M + MPositive
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10.4 Determining Deflections by Int
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398bEAM dEFLECTIONS5. Boundary and
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Beam Deflection and Slope at AThe d
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Elastic Curve EquationSubstitute th
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Boundary ConditionsBoundary conditi
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integration over the interval a ≤
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mecmoviesExERcISEm10.1 Beam Boundar
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P10.11 For the beam and loading sho
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Elastic Curve EquationSubstitute th
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414beam deflectionsintegration give
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(a) Beam Deflection at AAt the tip
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Integrate again to obtain the beam
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The inclusion of the reaction force
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p10.24 The simply supported beam sh
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vwPvwvPALBv Bx=ALB( v B ) wx+ALB( v
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m10.5 Superposition Warm-up. A seri
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Beam deflection at C: The beam defl
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Avv30 kipsa= 13 ft b=7 ftBC D4 ft 6
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Next, consider the overhang span be
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By inspection, the rotation angle a
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v A70 kNAv210 kN.mBθ BCv Cθ D3 m
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Case 3—uniformly Distributed load
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m10.12 Use the superposition method
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v30 kNv12 kips2 kips/ftxxABHABC3 m
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p10.53 The simply supported beam sh
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446STATICALLy INdETERMINATEbEAMSM A
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— 1 w 0 Lv2M AA xAB2LA y 3L—3 B
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obtained from the continuity condit
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Solve for ReactionsSolve Equation (
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11.4 Use of Discontinuity Functions
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Integrate again to obtain the beam
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EI dvdx∫Integrate the beam load e
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v20 kips 30 kips 20 kipsA B C D E7
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462STATICALLy INdETERMINATEbEAMSThe
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corresponding beam deflection will
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Substitute these values into Equati
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By2 6 436(200,000 N/mm )(351 × 10
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The only unknown term in this equat
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The only unknown term in this equat
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p11.24-p11.26 For the beams and loa
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v90 kN/m30 kN/m(1)ABCx40 kN3 m5.5 m
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p11.49 Two steel beams support a co
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12.2 Stress at a General point in a
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12.3 Equilibrium of the Stress Elem
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484STRESS TRANSFORMATIONSsuch as th
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pRoBLEmSp12.1 A compound shaft cons
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p12.9 A steel pipe with an outside
Page 512 and 513:
The forces acting on the vertical a
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tynθxσn dAFigure 12.10b is a free
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ExAmpLE 12.3y842 MPa550 MPa16 MPaxA
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The angle θ from the redefined x a
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p12.19 Two steel plates of uniform
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500STRESS TRANSFORMATIONSprincipal
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502STRESS TRANSFORMATIONSIf the nor
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504STRESS TRANSFORMATIONSIn words,
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yFIGURE 12.15506There is nevershear
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ExAmpLE 12.5y9 ksi7 ksix11 ksiConsi
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planes whose normal is perpendicula
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ExERcISEm12.4 Sketching Stress tran
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514STRESS TRANSFORMATIONSmecmovies
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516STRESS TRANSFORMATIONSLabel the
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mecmoviesExAmpLEm12.10 Coach Mohr
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P 2(-7.22, 0)(-5, 6 ccw) yτ(2, 9.2
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in the x-y coordinate system is rot
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t73.7°τy (-16, 53)R = 55.22C (-31
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ExAmpLE 12.11y(-14, 0)y(-14, 0)14 k
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ExERcISESm12.10 Coach Mohr’s Circ
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p12.46 Figure P12.46 shows Mohr’s
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12.11 General State of Stress at a
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534STRESS TRANSFORMATIONSIn Equatio
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536STRESS TRANSFORMATIONSthird prin
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538STRESS TRANSFORMATIONSyσp2Arbit
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Strain TransformationsCHAPTER1313.1
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542STRAIN TRANSFORMATIONSyyyγ xy d
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544STRAIN TRANSFORMATIONSThus, diag
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tnFinally, to compute the shear str
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Table 13.2 Absolute maximum Shear S
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The in-plane principal directions c
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552STRAIN TRANSFORMATIONS13.6 mohr
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y24.2°279 μεπ2579 μεx- 858 μ
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556STRAIN TRANSFORMATIONSPlasticbac
Page 580 and 581:
Equation for gage b:2 2− 900 = e
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pRoBLEmSp13.25-p13.30 The strain ro
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yyyτ xyτ yzτ zxxxxzFIGURE 13.13a
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564STRAIN TRANSFORMATIONSHence, the
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SolutioN(a) Normal stresses: From E
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Then solve for γ xy :2 2380 = (
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(b) Principal and Maximum in-Plane
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(b) Principal and Maximum in-Plane
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y103.7 MPa40.2 MPa18.5°x63.5 MPa14
Page 598 and 599:
1εz = [ σ z − νσ ( x + σ y )
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σ xyFinally, suppose the material
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pRoBLEmSp13.31 An 8 mm thick brass
Page 604 and 605:
Problem e a e b e c E νycP13.47
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p13.65 A solid plastic [E = 45 MPa;
Page 608 and 609:
586PRESSuRE VESSELSThe wall compris
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588PRESSuRE VESSELSττ abs max = p
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590PRESSuRE VESSELSStresses on the
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592PRESSuRE VESSELSFor the outer su
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31.9 MPay63.8 MPat13.81 MPax30°39.
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p14.5 The normal strain measured on
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p14.20 The cylindrical pressure ves
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600PRESSuRE VESSELSSince the ends o
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602PRESSuRE VESSELSσ θτ maxσ r4
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604PRESSuRE VESSELSFor convenience,
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14.6 Deformations in Thick-Walled c
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Maximum Shear StressThe principal s
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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
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6.92 ksiz6.92 ksiHAHy11.54 ksi8.49
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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
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60 kNShear Force and Bending Moment
Page 650 and 651:
ExAmpLE 15.3A steel hollow structur
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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
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102,400 N · mm = 102.4 N · m. Sim
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yK12.02 MPax45°12.02 MPaStress tra
Page 666 and 667:
The equivalent moments at the secti
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Hy3,000 lb448 psiKThe 3,000 lb shea
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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
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z 2z 1dimensions of the assembly ar
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658COMbINEd LOAdSIf the naming conv
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660COMbINEd LOAdSfrom which it foll
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662COMbINEd LOAdSSimilarly, Equatio
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(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
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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
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16.3 The Effect of End conditionson
Page 704 and 705:
682COLuMNSwhere the first two terms
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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
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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
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Both tensile and compressive normal
Page 734 and 735:
ExAmpLE 16.940 kNexA 6061-T6 alumin
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pRoBLEmSp16.42 The structural steel
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716ENERgy METHOdSThe total energy o
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718ENERgy METHOdSydV = dx dy dzzxxS
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720ENERgy METHOdSSince the volume o
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The maximum force P that can be app
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17.5 Elastic Strain Energy for Flex
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segment AB of the beam. The second
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p17.8 A solid stepped shaft made of
Page 752 and 753:
730ENERgy METHOdSNote that the nega
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732ENERgy METHOdSSignificance: In t
Page 756 and 757:
From the positive root, the maximum
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Note: Throughout the previous chapt
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The right-hand side of this equatio
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(b) If the rod has a constant diame
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Note: Throughout the previous chapt
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assembly. The solid bronze post has
Page 768 and 769:
17.7 Work-Energy method for Single
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F 1(1)BSolutioNThe internal axial f
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17.8 method of Virtual WorkThe meth
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752ENERgy METHOdSIf the material be
Page 776 and 777:
754ENERgy METHOdSwhich is the mathe
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756ENERgy METHOdSThe virtual intern
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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
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764ENERgy METHOdSObtain the total v
Page 788 and 789:
766ENERgy METHOdSthe real external
Page 790 and 791:
ExAmpLE 17.141,400 N900 NCalculate
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1Ax 1 or x 2vmDraw a free-body diag
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p17.36 Determine the vertical displ
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p17.54 Determine the minimum moment
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776ENERgy METHOdSFor the general ca
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ExAmpLE 17.16Determine the vertical
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SolutioNWe seek the horizontal defl
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782ENERgy METHOdSprocedure for Anal
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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
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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
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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.
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YttffdXXt wYb fDesignationAreaADept
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YttffdXXt wYb fDesignationAreaADept
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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
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orσx + σ y σx − σ yσ n = + c
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Answers to Odd NumberedProblemsChap
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(c) φ E/C = 0.1408 rad(d) T A = 23
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P7.35 (a)wx ( ) =−5kN −x − 0
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(c)Chapter 8P8.1 σ = 1.979 ksiP8.3
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P10.7 v B = −8.27 mmP10.9 (a)wx 0
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P12.35 (a) 25.6 MPa(b) 23.1° clock
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P13.63 (a) Db = 1.272 mm,Dc = 0.254
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P17.41 (a) D F = 18.54 mm ←(b) D
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Biaxial stress, 566-568, 587, 590-5
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GGage length, 46, 47, 51, 54Gears,
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QQ section property, 327-332, 338,
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Torsionof circular shafts, 135-151e
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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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