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

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PointSketchyy – ′(mm)A′(mm 2 )Q(mm 3 )τ(MPa)A′aay′118.0 400 47,200 19.46N.A.yA′b 93.0 800 74,400 30.67y′ bN.A.yA′c 71.5 1,144 81,796 33.72y′N.A.cyd 43.0 1,760 75,680 31.20N.A.y′dA′ye 43.0 880 37,840 15.60N.A.y′A′eyf 43.0 440 18,920 7.80N.A.y′A′f
The directions and intensities of the shear stress in the inverted-tee shape are shownin the accompanying sketch. Note that the shear stress in the tee stem is distributed parabolicallywhile the shear stress in the flange is distributed linearly. At the junction of thestem and the flange, the shear stress intensity is cut in half as shear flows outward in twoopposing directions.19.46 MPa30.67 MPa33.72 MPa7.80 MPa 15.60 MPa31.20 MPa15.60 MPa 7.80 MPaExAmpLE 9.8A beam with the cross section shown is subjected to a verticalshear force V = 65 kN. The location of the centroid appears on thesketch, and the moment of inertia of the cross section about the zcentroidal axis is I = 2,532,789 mm 4 . Determine the shear stress inthe cross section at point a.Plan the SolutionTo calculate the shear stress at a, we must cut an FBD through thecross section so that the shear stress at a is exposed. We will considerthree different FBDs that could be used to determine thequantity Q necessary to calculate the shear stress from the formulaτ = VQ/It.30.5 mm74.5 mmbcad6 mm50 mmz50 mm180 mmyVy6 mm6 mm99 mm105 mmSolutioNApproach 1The first FBD that we will consider is obtained by making twovertical cuts through the cross section. One vertical cut will bemade at point a, and a second vertical cut will be made on the ycentroidal axis, which is an axis of symmetry for the cross sectionconsidered here.The FBD exposed by these two cuts is shown in the accompanyingfigure. The pieces of the cross section have been separatedin the figure to help illustrate the shear stresses on the interiorsurfaces. The directions of the shear flow and the shear stresseson the various surfaces are also indicated. Notice that shear stresson the vertical plane of symmetry is zero.The shear stress at point a will be calculated from the formulaτ = VQ/It. the shear force V and the moment of inertia Ito be used in the formula are the same for any FBD that wemight consider.27.5 mm b acdVertical plane of symmetrybcdτ = 0azVτ = 0z6 mmyVx365
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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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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
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Construct the Bending-Moment Diagra
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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
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M (14,400 lb ⋅ ft)(12 in./ft)∴
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pRoBLEmSp8.26 A small aluminum allo
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Equivalent BeamsBefore considering
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Transformed-Section methodThe conce
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This equation reduces to∫A∫ydA
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ExAmpLE 8.7A cantilever beam 10 ft
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Bending Stresses at the intersectio
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(a) the normal stress in each mater
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283CM = PeF+ =bENdINg duE TO ANECCE
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and the combined normal stress on s
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Combined Stress at KThe combined st
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m8.23 Pipe AB (with outside diamete
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p8.54 The bracket shown in Figure P
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yFlangeyyz CM zWebM zM zCompressive
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These curvature expressions can now
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Coordinates of Points H and KThe (y
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Similarly, the moment of inertia I
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p8.63 A downward concentrated load
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the factor K depends only upon the
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SolutioNThe ultimate strength σ U
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MO307bENdINg OF CuRVEd bARSr or iθ
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Table 8.2 Area and Radial Distance
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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 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
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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
Page 442 and 443:
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
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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
Page 572 and 573:
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
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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
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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
Page 626 and 627:
604PRESSuRE VESSELSFor convenience,
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14.6 Deformations in Thick-Walled c
Page 630 and 631:
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
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
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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
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
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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
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
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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
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
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pRoBLEmSp16.42 The structural steel
Page 738 and 739:
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
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
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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
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
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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
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
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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.
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YttffdXXt wYb fDesignationAreaADept
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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
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Answers to Odd NumberedProblemsChap
Page 856 and 857:
(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,
Page 874 and 875:
QQ section property, 327-332, 338,
Page 876:
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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