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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
- Page 85 and 86: material is initially 800 mm long a
- Page 87 and 88: CHAPTER4design Concepts4.1 Introduc
- Page 89 and 90: the distributed uniform area loadin
- Page 91 and 92: In some instances, engineers may ne
- Page 93 and 94: both members will be stressed to th
- Page 95 and 96: MecMoviesExAMpLESM4.1 The structure
- Page 97 and 98: bPtPFIGURE p4.3Splice plateBarBarPP
- Page 99 and 100: 4.5 Load and Resistance Factor Desi
- Page 101 and 102: 79FrequencyLarger γ factors combin
- Page 103 and 104: Limit StatesLRFD is based on a limi
- Page 105 and 106: CHAPTER5Axial deformation5.1 Introd
- Page 107 and 108: The increased normal stress magnitu
- Page 109 and 110: AAAA87dEFORMATIONS IN AxIALLyLOAdEd
- Page 111 and 112: displacement in the horizontal dire
- Page 113 and 114: mecmoviesExAmpLEm5.2 The roof and s
- Page 115 and 116: t = 8 mm, and E = 72 GPa, determine
- Page 117 and 118: d 0to 2d 0 at the other end. A conc
- Page 119 and 120: How are the rigid-bar deflections v
- Page 121 and 122: ExAmpLE 5.4A tie rod (1) and a pipe
- Page 123 and 124: Since the sum of the four interior
- Page 125 and 126: 5.5 Statically Indeterminate Axiall
- Page 127 and 128: Step 3 — Force-Deformation Relati
- Page 129 and 130: Successful application of the five-
- Page 131 and 132: Plan the SolutionConsider a free-bo
- Page 133 and 134: Structures with a Rotating Rigid Ba
- Page 135: (1)Px Bx C113STATICALLy INdETERMINA
- Page 139 and 140: p5.29 In Figure P5.29, a load P is
- Page 141 and 142: tube has an outside diameter of 1.5
- Page 143 and 144: ExAmpLE 5.7An aluminum rod (1) [E =
- Page 145 and 146: mecmoviesExAmpLEm5.14 A rectangular
- Page 147 and 148: Equations (h) and (i) can be solved
- Page 149 and 150: pRoBLEmSp5.39 A circular aluminum a
- Page 151 and 152: polymer [E = 370 ksi; α = 39.0 ×
- Page 153 and 154: Stress-concentration factor K3.02.8
- Page 155 and 156: of the hole has decayed to a value
- Page 157 and 158: TorsionCHAPTER66.1 IntroductionTorq
- Page 159 and 160: with respect to an adjacent cross s
- Page 161 and 162: the longitudinal axis of the shaft.
- Page 163 and 164: τStressxyτntσn141STRESSES ON ObL
- Page 165 and 166: If a torsion member is subjected to
- Page 167 and 168: ExAmpLE 6.1A hollow circular steel
- Page 169 and 170: Plan the SolutionTo determine the l
- Page 171 and 172: Plan the SolutionThe internal torqu
- Page 173 and 174: mecmoviesExAmpLESm6.4 Determine the
- Page 175 and 176: p6.10 The mechanism shown in Figure
- Page 177 and 178: Furthermore, since gears have teeth
- Page 179 and 180: will be dictated by the ratio of ge
- Page 181 and 182: mecmoviesExAmpLEm6.13 Two solid ste
- Page 183 and 184: 6.8 power Transmission161POwER TRAN
- Page 185 and 186: m6.17 A motor shaft is being design
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pulley is a belt having tensions F
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Step 2 — Geometry of Deformation:
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Polar moments of inertia for the al
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Successful application of the five-
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Plan the SolutionAA free-body diagr
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From Equation (e), the allowable in
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Next, consider a free-body diagram
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Polar moments of inertia for the sh
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m6.21 A composite torsion member co
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zD(3)Ay(1)L1,L3N BBL 2(a) Determine
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Stress concentrations also occur at
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For the case of the rectangular bar
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and the angle of twist for a 12 in.
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ExAmpLE 6.13A rectangular box secti
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CHAPTER7Equilibrium of beams7.1 Int
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beam (also called a simple beam). A
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ExAmpLE 7.1Draw the shear-force and
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The negative value for A y indicate
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Plot the FunctionsPlot the function
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ExAmpLE 7.5Draw the shear-force and
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yw aw byw 0xxABCABabLFIGURE p7.4FIG
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w(x) = w G at G. At A, where the di
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the left and right sides of a thin
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General procedure for constructing
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Construct the Bending-Moment Diagra
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ExAmpLE 7.7Draw the shear-force and
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j M(6) = 0 kN ⋅ m (Rule 4: ∆M =
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of M diagram = shear force V). The
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m7.3 Dynamically generated shear-fo
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p7.21-p7.22 Use the graphical metho
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x - a 0x - a 1x - a 2225dISCONTINuI
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conditions. The reactions for stati
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120 kN ⋅ m concentrated moment: F
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SolutioNWhen we refer to case 4 of
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Uniformly distributed load between
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y5 kN20 kN.my12 kN.m18 kN/mAB3 m 3
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CHAPTER8bending8.1 IntroductionPerh
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has a constant bending moment M, an
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The most common stress-strain relat
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All such moment increments that act
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The sense of σ x (either tension o
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A i(mm 2 )y i(mm)y i A i(mm 3 )(1)
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ExAmpLE 8.2The cross-sectional dime
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m8.7 Determine the centroid locatio
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zFIGURE p8.8p8.9 An aluminum alloy
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WidthWidthWidthDepthXYthicknessXWeb
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700 lb1,500 lb4 in.1 in.200 lb/ft1
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M = −6,000 lb · ft. For this neg
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mecmoviesExAmpLESm8.9 Determine the
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p8.16 A W18 × 40 standard steel sh
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8.5 Introductory Beam Design for St
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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
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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
- Page 500 and 501:
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
- Page 546 and 547:
t73.7°τy (-16, 53)R = 55.22C (-31
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ExAmpLE 12.11y(-14, 0)y(-14, 0)14 k
- Page 550 and 551:
ExERcISESm12.10 Coach Mohr’s Circ
- Page 552 and 553:
p12.46 Figure P12.46 shows Mohr’s
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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
- Page 586 and 587:
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
- Page 594 and 595:
(b) Principal and Maximum in-Plane
- Page 596 and 597:
y103.7 MPa40.2 MPa18.5°x63.5 MPa14
- Page 598 and 599:
1εz = [ σ z − νσ ( x + σ y )
- Page 600 and 601:
σ xyFinally, suppose the material
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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;
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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
- Page 616 and 617:
31.9 MPay63.8 MPat13.81 MPax30°39.
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p14.5 The normal strain measured on
- Page 620 and 621:
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,
- 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,
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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
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16.3 The Effect of End conditionson
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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
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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
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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
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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
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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
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Following is the table produced by
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SolutioNCalculate the member length
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764ENERgy METHOdSObtain the total v
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766ENERgy METHOdSthe real external
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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
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Finally, derive the following equat
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p17.63 The truss shown in Figure P1
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APPENDIXAgeometric Propertiesof an
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792gEOMETRIC PROPERTIESOF AN AREAyy
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mecmoviesExAmpLESA.2 Animated examp
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796gEOMETRIC PROPERTIESOF AN AREAfo
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ExAmpLE A.340 mmDetermine the momen
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ExAmpLE A.440 mmDetermine the produ
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yy′yOxxdAθθcos θx′xy′sin
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804gEOMETRIC PROPERTIESOF AN AREANo
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806gEOMETRIC PROPERTIESOF AN AREAI
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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
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b fdttffXYXy-t wYDesignationAreaADe
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dXYXtYbDesignationHSS304.8 × 203.2
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YxZXyXαYZDesignationMasspermeterAr
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Simply Supported BeamsBeam Slope De
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Average Properties ofSelected Mater
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Table D.1b Average properties of Se
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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