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

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YttffdXXt wYb fDesignationAreaADepthdWebthicknesst wFlangewidthb fFlangethicknesst f I x S x r x I y S y r yW12 × 5812 × 5012 × 4012 × 3012 × 2612 × 2212 × 14W10 × 5410 × 4510 × 3010 × 2610 × 2210 × 15W8 × 488 × 408 × 318 × 248 × 15W6 × 256 × 206 × 156 × 12in. 2 in. in. in. in. in. 4 in. 3 in. in. 4 in. 3 in.17.014.611.78.797.656.484.1615.813.38.847.616.494.4114.111.79.127.084.447.345.874.433.5512.212.211.912.312.212.311.910.110.110.510.310.210.08.508.258.007.938.116.386.205.996.030.3600 3700.2950.2600.2300.2600.2000.3700.3500.3000.2600.2400.2300.4000.3600.2850.2450.2450.3200.2600.2300.23010.08.088.016.526.494.033.9710.08.025.815.775.754.008.118.078.006.504.016.086.025.994.000.6400.6400.5150.4400.3800.4250.2250.6150.6200.5100.4400.3600.2700.6850.5600.4350.4000.3150.4550.3650.2600.28047539130723820415686.630324817014411868.918414611082.74853.441.429.122.178.064.251.538.633.425.414.960.049.132.427.923.213.843.235.527.520.911.816.713.49.727.315.285.185.135.215.174.914.624.374.324.384.354.273.953.613.533.473.423.292.702.662.562.4910756.344.120.317.34.662.3610353.416.714.111.42.8960.949.137.118.33.4117.113.39.322.9921.413.911.06.245.342.311.1920.613.35.754.893.971.4515.012.29.275.631.705.614.413.111.502.511.961.941.521.510.8480.7532.562.011.371.361.330.812.082.042.021.610.8761.521.501.450.918810
Wide-Flange Sections or W Shapes—SI UnitsDesignationAreaADepthdWebthicknesst wFlangewidthb fFlangethicknesst f I x S x r x I y S y r ymm 2 mm mm mm mm10 6mm 4 10 3mm 3mm10 6mm 4 10 3mm 3mmW610 × 140610 × 113610 × 101610 × 821790014500130001050061760760259913.111.210.510.023022822817822.217.314.912.81120874762562364028802520187025124624323145.434.329.312.139330225713650.348.847.534.0W530 × 101530 × 92530 × 74530 × 6612900118009480839053653352852610.910.29.658.8921020916616517.415.613.611.4616554410351229020801550134021821720820526.923.910.48.6225722912510445.745.033.032.0W460 × 82460 × 74460 × 60460 × 52105009480761066504604574554509.919.028.007.6219119115315216.014.513.310.837033325521216101460112094418818718317918.716.77.956.3719517510483.942.441.932.331.0W410 × 85410 × 75410 × 60410 × 46.11080094807610589041741440640410.99.657.756.9918118017814018.216.012.811.231627421615615101330106077317117016816317.915.512.05.1619817213573.640.640.439.929.7W360 × 101360 × 79360 × 72360 × 51360 × 44360 × 39360 × 32.912900101009100645057104960419035635335135635135334810.59.408.647.246.866.485.8425420520417117112812718.316.815.111.69.7810.78.5130122520114212110282.816901270115079668857847515315014914814614414150.424.021.49.708.163.712.9139723421011395.458.245.962.548.848.538.937.827.426.4811
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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
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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
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
Page 478 and 479:
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
Page 486 and 487:
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
Page 504 and 505:
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
Page 518 and 519:
The angle θ from the redefined x a
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p12.19 Two steel plates of uniform
Page 522 and 523:
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
Page 556 and 557:
534STRESS TRANSFORMATIONSIn Equatio
Page 558 and 559:
536STRESS TRANSFORMATIONSthird prin
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538STRESS TRANSFORMATIONSyσp2Arbit
Page 562 and 563:
Strain TransformationsCHAPTER1313.1
Page 564 and 565:
542STRAIN TRANSFORMATIONSyyyγ xy d
Page 566 and 567:
544STRAIN TRANSFORMATIONSThus, diag
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tnFinally, to compute the shear str
Page 570 and 571:
Table 13.2 Absolute maximum Shear S
Page 572 and 573:
The in-plane principal directions c
Page 574 and 575:
552STRAIN TRANSFORMATIONS13.6 mohr
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y24.2°279 μεπ2579 μεx- 858 μ
Page 578 and 579:
556STRAIN TRANSFORMATIONSPlasticbac
Page 580 and 581:
Equation for gage b:2 2− 900 = e
Page 582 and 583:
pRoBLEmSp13.25-p13.30 The strain ro
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yyyτ xyτ yzτ zxxxxzFIGURE 13.13a
Page 586 and 587:
564STRAIN TRANSFORMATIONSHence, the
Page 588 and 589:
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
Page 596 and 597:
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
Page 606 and 607:
p13.65 A solid plastic [E = 45 MPa;
Page 608 and 609:
586PRESSuRE VESSELSThe wall compris
Page 610 and 611:
588PRESSuRE VESSELSττ abs max = p
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590PRESSuRE VESSELSStresses on the
Page 614 and 615:
592PRESSuRE VESSELSFor the outer su
Page 616 and 617:
31.9 MPay63.8 MPat13.81 MPax30°39.
Page 618 and 619:
p14.5 The normal strain measured on
Page 620 and 621:
p14.20 The cylindrical pressure ves
Page 622 and 623:
600PRESSuRE VESSELSSince the ends o
Page 624 and 625:
602PRESSuRE VESSELSσ θτ maxσ r4
Page 626 and 627:
604PRESSuRE VESSELSFor convenience,
Page 628 and 629:
14.6 Deformations in Thick-Walled c
Page 630 and 631:
Maximum Shear StressThe principal s
Page 632 and 633:
610PRESSuRE VESSELScould be cooled
Page 634 and 635:
Maximum Circumferential Stress in t
Page 636 and 637:
212.5 MPa134.4 MPa145.2 MPa76.2 MPa
Page 638 and 639:
CHAPTER15Combined Loads61615.1 Intr
Page 640 and 641:
6.92 ksiz6.92 ksiHAHy11.54 ksi8.49
Page 642 and 643:
p15.5 A solid 1.50 in. diameter sha
Page 644 and 645:
622COMbINEd LOAdSPP2P2FIGURE 15.1 S
Page 646 and 647:
internal bending moment acting at t
Page 648 and 649:
60 kNShear Force and Bending Moment
Page 650 and 651:
ExAmpLE 15.3A steel hollow structur
Page 652 and 653:
1.216 ksiH1.216 ksiH1.373 ksi30.3°
Page 654 and 655:
Figure P15.13b/14b are d = 240 mm,
Page 656 and 657:
p15.25 A load P = 75 kN acting at a
Page 658 and 659:
636COMbINEd LOAdS b. Bending moment
Page 660 and 661:
20.05 MPacbyM z = 3.85 kN·m20.05 M
Page 662 and 663:
102,400 N · mm = 102.4 N · m. Sim
Page 664 and 665:
yK12.02 MPax45°12.02 MPaStress tra
Page 666 and 667:
The equivalent moments at the secti
Page 668 and 669:
Hy3,000 lb448 psiKThe 3,000 lb shea
Page 670 and 671:
z8.45 kN·m15.6 kN·mHK10.8 kN·mEq
Page 672 and 673:
Torsion shear13.438 MPaBeam shear3.
Page 674 and 675:
m15.5 Determine the stresses acting
Page 676 and 677:
p15.34 Forces P x = 580 lb and P y
Page 678 and 679:
z 2z 1dimensions of the assembly ar
Page 680 and 681:
658COMbINEd LOAdSIf the naming conv
Page 682 and 683:
660COMbINEd LOAdSfrom which it foll
Page 684 and 685:
662COMbINEd LOAdSSimilarly, Equatio
Page 686 and 687:
(b) What is the value of the Mises
Page 688 and 689:
p15.54 A thin-walled cylindrical pr
Page 690 and 691:
668COLuMNSStability of EquilibriumT
Page 692 and 693:
670COLuMNSSummaryBefore a compressi
Page 694 and 695:
672COLuMNSIf we let2Pk =(16.2)EIthe
Page 696 and 697:
674COLuMNS7060σY =50=σ cr4030σY
Page 698 and 699:
The Euler buckling load for this co
Page 700 and 701:
Spacer blockFIGURE p16.676 mm 76 mm
Page 702 and 703:
16.3 The Effect of End conditionson
Page 704 and 705:
682COLuMNSwhere the first two terms
Page 706 and 707:
684COLuMNSAnother way of expressing
Page 708 and 709:
LateralbracingxCBP17.5 ft17.5 ftzSt
Page 710 and 711:
xBPBuckling About the Strong AxisTh
Page 712 and 713:
p16.19 The aluminum column shown in
Page 714 and 715:
692COLuMNSIn this case, a relations
Page 716 and 717:
694COLuMNSwhich can be further simp
Page 718 and 719:
Pexp16.26 A steel [E = 200 GPa] pip
Page 720 and 721:
698COLuMNSFor short and intermediat
Page 722 and 723:
ExAmpLE 16.52 in.zSpacer blocky2 in
Page 724 and 725:
Since KL/ry > 133.7, the column is
Page 726 and 727:
The reduced Euler buckling stress t
Page 728 and 729:
Determine the allowable axial load
Page 730 and 731:
708COLuMNSwhere the compressive str
Page 732 and 733:
Both tensile and compressive normal
Page 734 and 735:
ExAmpLE 16.940 kNexA 6061-T6 alumin
Page 736 and 737:
pRoBLEmSp16.42 The structural steel
Page 738 and 739:
716ENERgy METHOdSThe total energy o
Page 740 and 741:
718ENERgy METHOdSydV = dx dy dzzxxS
Page 742 and 743:
720ENERgy METHOdSSince the volume o
Page 744 and 745:
The maximum force P that can be app
Page 746 and 747:
17.5 Elastic Strain Energy for Flex
Page 748 and 749:
segment AB of the beam. The second
Page 750 and 751:
p17.8 A solid stepped shaft made of
Page 752 and 753:
730ENERgy METHOdSNote that the nega
Page 754 and 755:
732ENERgy METHOdSSignificance: In t
Page 756 and 757:
From the positive root, the maximum
Page 758 and 759:
Note: Throughout the previous chapt
Page 760 and 761:
The right-hand side of this equatio
Page 762 and 763:
(b) If the rod has a constant diame
Page 764 and 765:
Note: Throughout the previous chapt
Page 766 and 767:
assembly. The solid bronze post has
Page 768 and 769:
17.7 Work-Energy method for Single
Page 770 and 771:
F 1(1)BSolutioNThe internal axial f
Page 772 and 773:
17.8 method of Virtual WorkThe meth
Page 774 and 775:
752ENERgy METHOdSIf the material be
Page 776 and 777:
754ENERgy METHOdSwhich is the mathe
Page 778 and 779:
756ENERgy METHOdSThe virtual intern
Page 780 and 781:
both P 1 and P 2 from the truss, ap
Page 782 and 783: Following is the table produced by
Page 784 and 785: SolutioNCalculate the member length
Page 786 and 787: 764ENERgy METHOdSObtain the total v
Page 788 and 789: 766ENERgy METHOdSthe real external
Page 790 and 791: ExAmpLE 17.141,400 N900 NCalculate
Page 792 and 793: 1Ax 1 or x 2vmDraw a free-body diag
Page 794 and 795: p17.36 Determine the vertical displ
Page 796 and 797: p17.54 Determine the minimum moment
Page 798 and 799: 776ENERgy METHOdSFor the general ca
Page 800 and 801: ExAmpLE 17.16Determine the vertical
Page 802 and 803: SolutioNWe seek the horizontal defl
Page 804 and 805: 782ENERgy METHOdSprocedure for Anal
Page 806 and 807: Castigliano’s second theorem appl
Page 808 and 809: Finally, derive the following equat
Page 810 and 811: p17.63 The truss shown in Figure P1
Page 812 and 813: APPENDIXAgeometric Propertiesof an
Page 814 and 815: 792gEOMETRIC PROPERTIESOF AN AREAyy
Page 816 and 817: mecmoviesExAmpLESA.2 Animated examp
Page 818 and 819: 796gEOMETRIC PROPERTIESOF AN AREAfo
Page 820 and 821: ExAmpLE A.340 mmDetermine the momen
Page 822 and 823: ExAmpLE A.440 mmDetermine the produ
Page 824 and 825: yy′yOxxdAθθcos θx′xy′sin
Page 826 and 827: 804gEOMETRIC PROPERTIESOF AN AREANo
Page 828 and 829: 806gEOMETRIC PROPERTIESOF AN AREAI
Page 830 and 831: I xy(38.8, 32.3) y1.654 in.yR = 38.
Page 834 and 835: YttffdXXt wYb fDesignationAreaADept
Page 836 and 837: x-XYttffXt wdYAmerican Standard cha
Page 838 and 839: b fdttffXYXy-t wYDesignationAreaADe
Page 840 and 841: dXYXtYbDesignationHSS304.8 × 203.2
Page 842 and 843: YxZXyXαYZDesignationMasspermeterAr
Page 844 and 845: Simply Supported BeamsBeam Slope De
Page 846 and 847: Average Properties ofSelected Mater
Page 848 and 849: Table D.1b Average properties of Se
Page 850 and 851: Fundamental Mechanicsof Materials E
Page 852 and 853: orσx + σ y σx − σ yσ n = + c
Page 854 and 855: Answers to Odd NumberedProblemsChap
Page 856 and 857: (c) φ E/C = 0.1408 rad(d) T A = 23
Page 858 and 859: P7.35 (a)wx ( ) =−5kN −x − 0
Page 860 and 861: (c)Chapter 8P8.1 σ = 1.979 ksiP8.3
Page 862 and 863: P10.7 v B = −8.27 mmP10.9 (a)wx 0
Page 864 and 865: P12.35 (a) 25.6 MPa(b) 23.1° clock
Page 866 and 867: P13.63 (a) Db = 1.272 mm,Dc = 0.254
Page 868 and 869: P17.41 (a) D F = 18.54 mm ←(b) D
Page 870 and 871: Biaxial stress, 566-568, 587, 590-5
Page 872 and 873: GGage length, 46, 47, 51, 54Gears,
Page 874 and 875: QQ section property, 327-332, 338,
Page 876: Torsionof circular shafts, 135-151e
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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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