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

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NCR CR BNB= 42 teeth= 54 teethCφ Cφ BBTo calculate the maximum shear stress magnitudes, the absolute values of T 1 and T 2 willbe used. The maximum shear stress magnitude in the 35 mm diameter shaft (1) isTc 1 1(405 Nm)(35 ⋅ mm/2)(1, 000 mm/m)τ 1 = == 48.1 MPa Ans.4J147,324 mm1and the maximum shear stress magnitude in the 30 mm diameter shaft (2) isTc 2 2(315 Nm)(30 ⋅ mm/2)(1,000 mm/m)τ 2 = == 59.4 MPa Ans.4J79,552 mm2Angles of TwistThe angles of twist must be calculated with the signed values of T 1 and T 2 . Shaft (1) is 600 mmlong, and its shear modulus is G = 80 GPa = 80,000 MPa. The angle of twist in this shaft isTL 1 1( 405 Nm)(600 mm)(1, 000 mm/m)φ 1 = = − ⋅ =− 0.020618 rad =−0.0206 rad4 2JG 1 1 (147,324 mm )(80,000 N/mm )Ans.Shaft (2) is 850 mm long; therefore, its angle of twist isTL 2 2(315 Nm)(850 ⋅ mm)(1, 000 mm/m)φ 2 = == 0.042087 rad = 0.0421 rad Ans.4 2JG (79,522 mm )(80,000 N/mm )2 2Rotation Angles of Gears B and CThe rotation of gear B is equal to the angle of twist in shaft (1):φB = φ1= − 0.020618 rad = − 0.0206 radAns.Note: From the sign convention for rotation angles described in Section 6.6 and illustratedin Figure 6.13, a negative rotation angle for gear B indicates that gear B rotatesclockwise, as shown in the accompanying figure.The rotation angles of gears B and C are related because the arclengths associatedwith the respective rotations must be equal. Why? Because the gear teeth are interlocked.The gears turn in opposite directions, however. In this instance, gear B turns clockwise,which causes gear C to rotate in a counterclockwise direction. This change in the directionof rotation is accounted for in the calculations by a negative sign, so thatR φ=−RφC C B Bwhere R B and R C are the radii of gears B and C, respectively. Using this relationship, wecan express the rotation angle of gear C asRBφC=− φ BRCHowever, the ratio R B /R C is simply the gear ratio between gears B and C, and this ratiocan be equivalently expressed in terms of N B and N C , the number of teeth on gearsB and C, respectively. Thus,NBφC=− φ BNCTherefore, the rotation angle of gear C isNB54 teethφC=− φ B =− ( − 0.020618 rad) = 0.026509 rad = 0.0265 rad Ans.NC42 teethRotation Angle of Gear DThe rotation angle of gear D is equal to the rotation angle of gear C plus the twist thatoccurs in shaft (2):φD = φC + φ2= 0.026509 rad + 0.042087 rad = 0.068596 rad = 0.0686 rad Ans.158
mecmoviesExAmpLEm6.13 Two solid steel [G = 80 GPa] shafts are connectedby the gears shown. The diameter of each shaft is 35 mm. Atorque T = 685 N · m is applied to the system at D. Determine(a) the maximum shear stress in each shaft.(b) the angle of rotation at D.=== ′ExERcISESm6.9 Six multiple-choice questions concerning torque, rotationangle, and rotation speed of gears.m6.11 Six basic calculations involving three shafts connectedby gears.FIGURE m6.11m6.12 Five basic twist and rotation angle calculations involvingtwo shafts connected by gears.FIGURE m6.9m6.10 Six basic calculations involving two shafts connected bygears.FIGURE m6.12FIGURE m6.10159
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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
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 and 136: (1)Px Bx C113STATICALLy INdETERMINA
Page 137 and 138: ExERcISESm5.5 A composite axial str
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: will be dictated by the ratio of ge
Page 183 and 184: 6.8 power Transmission161POwER TRAN
Page 185 and 186: m6.17 A motor shaft is being design
Page 187 and 188: pulley is a belt having tensions F
Page 189 and 190: Step 2 — Geometry of Deformation:
Page 191 and 192: Polar moments of inertia for the al
Page 193 and 194: Successful application of the five-
Page 195 and 196: Plan the SolutionAA free-body diagr
Page 197 and 198: From Equation (e), the allowable in
Page 199 and 200: Next, consider a free-body diagram
Page 201 and 202: Polar moments of inertia for the sh
Page 203 and 204: m6.21 A composite torsion member co
Page 205 and 206: zD(3)Ay(1)L1,L3N BBL 2(a) Determine
Page 207 and 208: Stress concentrations also occur at
Page 209 and 210: For the case of the rectangular bar
Page 211 and 212: and the angle of twist for a 12 in.
Page 213 and 214: ExAmpLE 6.13A rectangular box secti
Page 215 and 216: CHAPTER7Equilibrium of beams7.1 Int
Page 217 and 218: beam (also called a simple beam). A
Page 219 and 220: ExAmpLE 7.1Draw the shear-force and
Page 221 and 222: The negative value for A y indicate
Page 223 and 224: Plot the FunctionsPlot the function
Page 225 and 226: ExAmpLE 7.5Draw the shear-force and
Page 227 and 228: yw aw byw 0xxABCABabLFIGURE p7.4FIG
Page 229 and 230: 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
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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
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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
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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
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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;
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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
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31.9 MPay63.8 MPat13.81 MPax30°39.
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p14.5 The normal strain measured on
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p14.20 The cylindrical pressure ves
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600PRESSuRE VESSELSSince the ends o
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602PRESSuRE VESSELSσ θτ maxσ r4
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604PRESSuRE VESSELSFor convenience,
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14.6 Deformations in Thick-Walled c
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Maximum Shear StressThe principal s
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610PRESSuRE VESSELScould be cooled
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Maximum Circumferential Stress in t
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212.5 MPa134.4 MPa145.2 MPa76.2 MPa
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CHAPTER15Combined Loads61615.1 Intr
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6.92 ksiz6.92 ksiHAHy11.54 ksi8.49
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p15.5 A solid 1.50 in. diameter sha
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622COMbINEd LOAdSPP2P2FIGURE 15.1 S
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internal bending moment acting at t
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60 kNShear Force and Bending Moment
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ExAmpLE 15.3A steel hollow structur
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1.216 ksiH1.216 ksiH1.373 ksi30.3°
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Figure P15.13b/14b are d = 240 mm,
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p15.25 A load P = 75 kN acting at a
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636COMbINEd LOAdS b. Bending moment
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20.05 MPacbyM z = 3.85 kN·m20.05 M
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102,400 N · mm = 102.4 N · m. Sim
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yK12.02 MPax45°12.02 MPaStress tra
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The equivalent moments at the secti
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Hy3,000 lb448 psiKThe 3,000 lb shea
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z8.45 kN·m15.6 kN·mHK10.8 kN·mEq
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Torsion shear13.438 MPaBeam shear3.
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m15.5 Determine the stresses acting
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p15.34 Forces P x = 580 lb and P y
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z 2z 1dimensions of the assembly ar
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658COMbINEd LOAdSIf the naming conv
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660COMbINEd LOAdSfrom which it foll
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662COMbINEd LOAdSSimilarly, Equatio
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(b) What is the value of the Mises
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p15.54 A thin-walled cylindrical pr
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668COLuMNSStability of EquilibriumT
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670COLuMNSSummaryBefore a compressi
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672COLuMNSIf we let2Pk =(16.2)EIthe
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674COLuMNS7060σY =50=σ cr4030σY
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The Euler buckling load for this co
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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
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684COLuMNSAnother way of expressing
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LateralbracingxCBP17.5 ft17.5 ftzSt
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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 832 and 833:
YttffdXXt wYb fDesignationAreaADept
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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