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

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MECHANICS OF MATERIALS:An Integrated Learning System
Page 3: A research-based,online learning en
Page 8 and 9: VICE PRESIDENT AND DIRECTOREXECUTIV
Page 11 and 12: PrefaceAt the beginning of each sem
Page 13 and 14: topics for students. To help organi
Page 15 and 16: With each of these software feature
Page 17: Second Edition. John Baker, Univers
Page 20 and 21: 8.3 Normal Stresses in Beams 2408.4
Page 23 and 24: CHAPTER1Stress1.1 IntroductionThe t
Page 25 and 26: Greek letter σ (sigma). To determi
Page 27 and 28: ExAmpLE 1.2Rigid bar ABC is support
Page 29 and 30: of) the largest-magnitude internal
Page 31 and 32: 9dIRECT SHEAR STRESSJeffery S. Thom
Page 33 and 34: point O, which is the center of rot
Page 35 and 36: ExAmpLE 1.7A steel pipe column (6.5
Page 37 and 38: The projected area A b is equal to
Page 39 and 40: FIGURE p1.4/5p1.5 Three solid bars,
Page 41 and 42: Determine the shortest allowable le
Page 43 and 44: p1.29 A hollow box beam ABCD is sup
Page 45 and 46: To investigate the stresses acting
Page 47 and 48: ExAmpLE 1.9A 120 mm wide steel bar
Page 49 and 50: ExERcISESm1.12 The bar shown has a
Page 51 and 52: CHAPTER2Strain2.1 Displacement, Def
Page 53 and 54: Accordingly, a positive value of δ
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To distinguish clearly bet ween elo
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m2.2 A rigid steel bar AB is pinned
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Sanding sleeveDγ is the specific w
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In the undeformed plate, the angle
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2.4 Thermal StrainWhen unrestrained
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pRoBLEmSp2.16 An airplane has a hal
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46MECHANICAL PROPERTIESOF MATERIALS
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3.2 The Stress-Strain Diagrammecmov
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3.3 Hooke’s LawAs discussed previ
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(b) The lateral strain is0.004 in.
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ExAmpLE 3.340 mm20 mm80 mm(1)PTwo b
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p3.6 A nylon [E = 2,500 MPa; ν = 0
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p3.16 Compound axial member ABC sho
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66dESIgN CONCEPTS• Although their
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4.4 Allowable Stress DesignThe allo
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Note: The resultant force should al
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the design. Substitute the allowabl
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m4.2 The single-shear connection co
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p4.8 Rigid bar ABC is supported by
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78dESIgN CONCEPTSFrequencydistribut
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80dESIgN CONCEPTSFrequencyLoad fact
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The design strength is the product
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84AxIAL dEFORMATIONMore powerful co
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5.3 Deformations in Axially Loaded
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88AxIAL dEFORMATIONEquation (5.5) a
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Now compute the deformations in eac
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The total elongation of the bar is
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p5.9 The assembly shown in Figure P
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yA2.4 m 1.8 mRigid bar(1) (2) (3)de
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xBconfigurations due to a force app
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(b) Member DeformationsThe deformat
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y,va b cabx,uA B C Dp5.18 The truss
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F 1 F 1 F 2Since the axial members
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106AxIAL dEFORMATIONThe five-step p
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Equation Form Comments Typical Prob
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There are still two unknowns in thi
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112AxIAL dEFORMATION(1) (2)Px Bx CA
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114AxIAL dEFORMATIONAn equilibrium
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p5.24 A composite bar is fabricated
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p5.34 A pin-connected structure is
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m5.12 A rigid bar ABC is pinned at
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Step 2 — Geometry of Deformation:
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A(1)L 1+ v BL1100 mmCv BB(Note: Def
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and the deformation of member (2) i
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p5.46 At a temperature of 15°C, a
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130AxIAL dEFORMATIONA A BB CCA A BB
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132AxIAL dEFORMATIONUniform tensile
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pRoBLEmSp5.54 A 100 mm wide by 8 mm
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136TORSIONElectromagneticforcesA(1)
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138TORSIONIf the strain is small, t
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6.4 Stresses on oblique planesThe e
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6.5 Torsional DeformationsIf the sh
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144TORSIONA positive internal torqu
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SOLUTIONThe elastic torsion formula
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Rotation Angle at CThe angles of tw
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yA(1)0.85 m 1.00 m 0.70 m550 N.m900
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pRoBLEmSp6.1 A solid circular steel
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p6.15 Figure P6.15 shows a cutaway
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156TORSIONor in terms of the pitch
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NCR CR BNB= 42 teeth= 54 teethCφ C
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pRoBLEmSp6.18 The gear train system
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162TORSIONAnother common measure of
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Note: Only the magnitude of the tor
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respectively. The bearings shown al
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Step 2 — Geometry of Deformation:
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170TORSIONThe five-step procedure d
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mecmoviesExAmpLESm6.19 A composite
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For stainless steel core (2),τ1Tc
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m6.23 A composite shaft consists of
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R BNB= 40 teethφ BBwhere R B and R
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The angle of twist in shaft (2) isT
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zAyFIGURE p6.42/43L 1(1)T BBp6.43 T
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184TORSION3.02.8T2rT2.6DdStress-con
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p6.52 A stepped shaft has a major d
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188TORSIONabTbTaτ max3T=a 2 bFIGUR
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190TORSIONNote that the shear flow
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36 in. long, and its cross-sectiona
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194EQuILIbRIuM OF bEAMSyA xxAA y(a)
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196EQuILIbRIuM OF bEAMSyP1P2wABCxyA
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The sum of forces in the vertical d
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ExAmpLE 7.3ywDraw the shear-force a
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wx 2LA724 wL x—x3aaV2wL xMload th
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8 kN.m13 kNy8 kN.mAA2 m19 kN(3 kN/m
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206EQuILIbRIuM OF bEAMS from the ri
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208EQuILIbRIuM OF bEAMSw(x)w Aw Bw
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Table 7.1 construction Rules for Sh
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ExAmpLE 7.6AAAyA yy12 kips 10 kipsB
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214EQuILIbRIuM OF bEAMSfunction f (
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fBefore the V diagram is complete,
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e V(10) = +15 kips (Rule 2: ∆V =
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y120 kN.m130 kNV-130 kNM-120 kN.m60
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yA40 kN/m50 kNB60 kN.mxp7.17-p7.18
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y6 kips/ft4 kips/fty25 kN30 kN/mA B
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226EQuILIbRIuM OF bEAMSw(x)w(x)εP
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Table 7.2 Basic Loads Represented b
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Shear-force equation: Using the int
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ExAmpLE 7.129 kips/ftA9 kips/ftA5 k
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9 kips/ft5 kips/ft11 kips/ftBending
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y5 kips/ft9 kips/ft70 kN/m30 kN/m 5
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238bENdINgyLongitudinal plane of sy
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240bENdINgEquation (8.1) indicates
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242bENdINgproduce a resultant force
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244bENdINgyCompressivebending stres
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246bENdINgIn common practice, each
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The controlling section modulus is
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The moment of inertia of the cross
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pRoBLEmSp8.1 During the fabrication
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8.4 Analysis of Bending Stresses in
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256bENdINg“W12 by 50.” This sha
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11.5 in.6 in.Ref. axis0.5 in.1 in.4
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A B C D E F500 mmB y400 mm 600 mm 6
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m8.15 Moment-of-inertia calculation
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p8.22 A WT305 × 41 standard steel
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Table 8.1 Selecting Standard Steel
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55 kip·ft30 kips2.5 kips/ft30 kips
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Ptw2wAaFIGURE p8.32p8.33 The beam s
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272bENdINgSimilarly, the bending st
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274bENdINgas those in the actual cr
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276bENdINgNotice that the bending s
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Maximum Bending MomentThe maximum b
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ExERcISESm8.16 A composite-beam cro
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8.7 Bending Due to an Eccentric Axi
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HyF = 30 kipse = 8 in. 30 kipsM = 2
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Axial StressOn section a-a, the int
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m8.22 A steel inverted tee shape is
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a load P is applied at an eccentric
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outer faces of the flanges. Load P
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294bENdINgTo satisfy equilibrium, t
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296bENdINgEquation (8.24) is useful
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ExAmpLE 8.118 in.8 in.zHK(1)y4 in.(
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8 in.0.859 in.z53.6°H2.859 in.Comp
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(c) the principal moments of inerti
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304bENdINgStress-concentration fact
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p8.71 The notched bar shown in Figu
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308bENdINgThe circumferential norma
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310bENdINgTo remove θ from this ex
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At point A, r = 50 mm. Thus, the be
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SolutioNCentroid location and neutr
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DMr iOABp8.83 The curved tee shape
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CHAPTER9Shear Stress In beams9.1 In
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a maximum intensity of (4.5 MPa −
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I ci (mm 4 ) | d i | (mm) d i 2 A i
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9.3 The Shear Stress Formula325THE
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What is the significance of Equatio
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ad329THE FIRST MOMENT OF AREA, QVb
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9.5 Shear Stresses in Beams of Rect
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mecmoviesExAmpLEm9.2 Derivation of
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(b) Maximum Horizontal Shear Stress
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ywaAaxLFIGURE p9.7a Simply supporte
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the distribution of shear stress ma
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(a) Shear Stress at HBefore proceed
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m9.6 Determine the maximum horizont
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p9.17 The extruded plastic shape sh
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NailA(1)zy(2)Nail CNailB(3)Mz(2)(1)
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Identifying the proper Area for QIn
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Shear Flow FormulaThe shear flow fo
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mecmoviesExAmpLESm9.9 Determine the
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(a) If the screws are uniformly spa
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F(1)dx357SHEAR STRESS ANd SHEARFLOw
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τ xzCutFreesurfaceyFree surfaceCut
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Next, consider the web of the thin-
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Longitudinal planeof symmetryyB′
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The directions and intensities of t
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of Q. However, what is the negative
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used in Equation (b). that is, the
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(b) Distribution of shear stressThe
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Calculation of Shear Stress Distrib
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PA375SHEAR CENTERS OFTHIN-wALLEd OP
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SHEAR CENTERS OFTHIN-wALLEd OPENSEC
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the resultant force F w of the shea
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Distribution of Shear StressThe dis
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ExAmpLE 9.14Find the shear center O
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SHEAR CENTERS OFTHIN-wALLEd OPENSEC
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p9.42 The hat-shape extrusion shown
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btOet (typ.)cOh2beh2aFIGURE p9.54bF
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CHAPTER10beam deflections10.1 Intro
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When the beam is bent, points along
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P395THE dIFFERENTIAL EQuATIONOF THE
Page 419 and 420:
continuity conditionsMany beams are
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integrationEquation (b) will be int
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1 ⎛ 2wx0 ⎞Σ M =⎛ ⎞⎝⎜
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order to complete this FBD, however
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ExAmpLE 10.4The simple beam shown s
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Since C 1 = C 3 ,C2 2Pb( L − b )=
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p10.5 For the beam and loading show
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ExAmpLE 10.5A beam is loaded and su
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p10.15 For the beam and loading sho
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to obtain the shear-force function
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Note that the second term in this e
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ExAmpLE 10.8For the beam shown, use
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pRoBLEmSp10.17 For the beam and loa
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p10.31 For the beam and loading sho
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calculation. Before proceeding, it
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The beam deflection at B can be cal
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mecmoviesExAmpLESm10.8 Determine th
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20 kipsb = 4 ft a=16 ftvxv Cx=10 ft
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MLθ =6EIBy the values defined prev
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The beam deflections at A, C, and E
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The beam deflection at E is compute
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Beam deflection at C: The beam defl
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m10.5 Superposition Warm-up. Exampl
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v4 kips/ft45 kipsvv180 kN.m70 kN80
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CHAPTERStatically Indeterminatebeam
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447P1P2P1P2P1P2THE INTEgRATION METH
Page 471 and 472:
Evaluate ConstantsSubstitute the bo
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integrationIntegrate Equation (f) t
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p11.5 A beam is loaded and supporte
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SolutioN(a) Support ReactionsAn FBD
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ExAmpLE 11.4For the statically inde
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Equation (c) for the beam slope and
Page 483 and 484:
11.5 The Superposition method461THE
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vPvPvAL—2L—2(a) Actual beamxBAS
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Notice that EI appears in both term
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that the roller support is displace
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Case 2—Cantilever Beam with Conce
Page 493 and 494:
wooden beam. Determine an expressio
Page 495 and 496:
ExERcISESm11.1 Propped Cantilevers.
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v60 kN/m125 kNv80 lb/in.140 lb/in.A
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p11.44 A timber [E = 12 GPa] beam i
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Stress TransformationsCHAPTER1212.1
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on the x, y, and z planes. Stresses
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The result of this simple equilibri
Page 507 and 508:
Stresses at HThe forces and moments
Page 509 and 510:
p12.5 An extruded polymer flexural
Page 511 and 512:
also be added together to produce t
Page 513 and 514:
pRoBLEmSp12.11-p12.14 The stresses
Page 515 and 516:
Stress InvarianceThe normal stress
Page 517 and 518:
The choice of either Equation (12.3
Page 519 and 520:
ExERcISESm12.1 the Amazing Stress C
Page 521 and 522:
p12.25-p12.26 The stresses shown in
Page 523 and 524:
In this figure, we will assume that
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Method one. The first method is sim
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p2p2p2σ p2σp2σ p2σp1σp1σp1p1p
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• Substitute the value of θ s in
Page 531 and 532:
Since θ p is negative, the angle i
Page 533 and 534:
(b) The principal stresses and the
Page 535 and 536:
p12.35 A shear wall in a reinforced
Page 537 and 538:
Utility of mohr’s circleMohr’s
Page 539 and 540:
8. Several points on Mohr’s circl
Page 541 and 542:
Since points x and y are always the
Page 543 and 544:
element counterclockwise; therefore
Page 545 and 546:
and the normal stress acting on the
Page 547 and 548:
ExAmpLE 12.10Stresses on an incline
Page 549 and 550:
For element A, the absolute maximum
Page 551 and 552:
pRoBLEmSp12.42 Figure P12.42 shows
Page 553 and 554:
30.0 ksi2,250 psia5.5 ksi680 psiσ1
Page 555 and 556:
The forces on the x, y, and z faces
Page 557 and 558:
where a i , b i , and c i are the c
Page 559 and 560:
Table 12.1 Direction cosines for pl
Page 561 and 562:
p12.66 The stresses at point O in a
Page 563 and 564:
ydx′541PLANE STRAINdxθ′ zxdy
Page 565 and 566:
Next, the displacement vector AA′
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yyπ2γ xyxFIGURE 13.6a Positive sh
Page 569 and 570:
13.4 principal Strains and maximumS
Page 571 and 572:
• When θ p is positive, the elem
Page 573 and 574:
p13.3 The thin rectangular plate sh
Page 575 and 576:
SolutioNPlot point x as (ε x , −
Page 577 and 578:
pRoBLEmSp13.16-p13.17 The principal
Page 579 and 580:
Equations (13.9), (13.10), and (13.
Page 581 and 582:
out-of-plane normal strain e z will
Page 583 and 584:
Although stress is applied only in
Page 585 and 586:
Unit Volume changeConsider an infin
Page 587 and 588:
SolutioN(a) Determine the change in
Page 589 and 590:
Also, Equation (13.17) becomes, sim
Page 591 and 592:
e x and e y in terms of σ x and σ
Page 593 and 594:
ExAmpLE 13.9On the free surface of
Page 595 and 596:
Since the strain measurements were
Page 597 and 598:
ExAmpLE 13.10A 2014-T6 aluminum all
Page 599 and 600:
Whereas elongations and contraction
Page 601 and 602:
ExAmpLE 13.11A unidirectional T-300
Page 603 and 604:
p13.39 A thin brass [E = 100 GPa; G
Page 605 and 606:
(a) Determine the normal strains e
Page 607 and 608:
CHAPTER14Pressure Vessels14.1 Intro
Page 609 and 610:
From this equilibrium equation, an
Page 611 and 612:
yσ hoopzxσ longP = pπr 2σ longF
Page 613 and 614:
τprτ abs max = +2tp2591STRAINS IN
Page 615 and 616:
Stresses in the outlet PipeThe long
Page 617 and 618:
m14.5 A cylindrical steel [E = 200
Page 619 and 620:
diameter of 1,800 mm and a wall thi
Page 621 and 622:
σ θ dr ∆xdθ599STRESSES IN THIC
Page 623 and 624:
The radial stress in the thick-wall
Page 625 and 626:
Similarly, the change in radial str
Page 627 and 628:
External pressure on a Solid circul
Page 629 and 630:
ExAmpLE 14.3An open-ended high-pres
Page 631 and 632:
Then, set σ θ equal to the allowa
Page 633 and 634:
Cylinder shrink fitted on a solid s
Page 635 and 636:
The circumferential stress on the i
Page 637 and 638:
p14.31 An open-ended compound thick
Page 639 and 640:
normal stresses are identical at al
Page 641 and 642:
mecmoviesExAmpLEm15.1 A tubular sha
Page 643 and 644:
15.3 principal Stresses in a Flexur
Page 645 and 646:
a. The normal stresses caused by F
Page 647 and 648:
Shear Force and Bending Moment at H
Page 649 and 650:
mecmoviesExAmpLESm15.2 A cantilever
Page 651 and 652:
The following equilibrium equations
Page 653 and 654:
m15.3 The rectangular tube is subje
Page 655 and 656:
yPPQb ft fAx HHxB C DKx KEyL4FIGURE
Page 657 and 658:
F yyM yy635gENERAL COMbINEdLOAdINgS
Page 659 and 660:
SolutioNSection PropertiesThe cross
Page 661 and 662:
m15.7 A rectangular post has cross-
Page 663 and 664:
Combined Stresses at HThe normal an
Page 665 and 666:
SolutioNEquivalent Force SystemA sy
Page 667 and 668:
The 28,800 lb · ft bending moment
Page 669 and 670:
Stress transformation Results at KT
Page 671 and 672:
The 10.8 kN · m (i.e., 10.8 × 10
Page 673 and 674:
The 8.45 kN · m (i.e., 8.45 × 10
Page 675 and 676:
p15.28 Three loads are applied to t
Page 677 and 678:
P y = 375 lb, and P z = 550 lb. Det
Page 679 and 680:
σ Y(a) Stress element at yieldfor
Page 681 and 682:
where P is some function of δ. The
Page 683 and 684:
When the third term in the brackets
Page 685 and 686:
CompressiontestτTensiontestσσUTp
Page 687 and 688:
ExAmpLE 15.9The stresses on the fre
Page 689 and 690:
CHAPTER16Columns16.1 IntroductionIn
Page 691 and 692:
moment, then the system will tend t
Page 693 and 694:
Buckled configurationIf the compres
Page 695 and 696:
The critical load for an ideal colu
Page 697 and 698:
• The Euler buckling load equatio
Page 699 and 700:
Similarly, the moment of inertia of
Page 701 and 702:
p16.11 An assembly consisting of ti
Page 703 and 704:
P< PcrBxP=PcrBB yPBB yxPBL −x681T
Page 705 and 706:
The equation of the buckled column
Page 707 and 708:
where L is the actual length of the
Page 709 and 710:
Critical StressThe critical load eq
Page 711 and 712:
pRoBLEmSp16.15 An HSS152.4 × 101.6
Page 713 and 714:
column is as shown in Figure 16.8b.
Page 715 and 716:
1.15e = 0693THE SECANT FORMuLAPP cr
Page 717 and 718:
several values of the eccentricity
Page 719 and 720:
graph shows a scattered range of va
Page 721 and 722:
For intermediate-length columns wit
Page 723 and 724:
Controlling slenderness ratio: Sinc
Page 725 and 726:
From this allowable stress, the all
Page 727 and 728:
fixed at base A with respect to ben
Page 729 and 730:
p16.41 A simple pin-connected wood
Page 731 and 732:
ExAmpLE 16.8The W12 × 58 structura
Page 733 and 734:
(b) Magnitude of the largest eccent
Page 735 and 736:
where c is the outside radius of th
Page 737 and 738:
CHAPTER17Energy Methods17.1 Introdu
Page 739 and 740:
deformation vary. This dependency i
Page 741 and 742:
σσModulus oftoughnessFracture719w
Page 743 and 744:
or in terms of the deformation δ a
Page 745 and 746:
strain energies in all the segments
Page 747 and 748:
ExAmpLE 17.3A cantilever beam AB of
Page 749 and 750:
or, since a + b = L,UPab 2 2 2= Ans
Page 751 and 752:
In the loaded system, dynamic loadi
Page 753 and 754:
Special cases. Two extreme situatio
Page 755 and 756:
Plan the SolutionThe axial deformat
Page 757 and 758:
so that the impact factor can be wr
Page 759 and 760:
Multiplying the two factors of the
Page 761 and 762:
SolutioN(a) The areas of the two ro
Page 763 and 764:
This strain-energy density is repre
Page 765 and 766:
pRoBLEmSp17.12 A 19 mm diameter ste
Page 767 and 768:
p17.24 Figure P17.24 shows block D,
Page 769 and 770:
The external work done by a force a
Page 771 and 772:
pRoBLEmSp17.30 Determine the horizo
Page 773 and 774:
PB751METHOd OF VIRTuAL wORkP 2ACδ
Page 775 and 776:
internal forces f 1 and f 2 acting
Page 777 and 778:
17.9 Deflections of Trussesby the V
Page 779 and 780:
The right-hand sides of Equations (
Page 781 and 782:
at B. The right-hand side represent
Page 783 and 784:
MemberL(mm)A(mm 2 )F(kN)f(kN)⎛ FL
Page 785 and 786:
vwPv1763dEFLECTIONS OF bEAMS byTHE
Page 787 and 788:
Pw1765dEFLECTIONS OF bEAMS byTHE VI
Page 789 and 790:
Real Moment M: Remove the virtual l
Page 791 and 792:
BeamSegmentx Coordinateoriginlimits
Page 793 and 794:
For segment CD, draw a free-body di
Page 795 and 796:
P 2P 2P 2P PAC E G HB D F6 mp17.49
Page 797 and 798:
Therefore, the total strain energy
Page 799 and 800:
The terms L, A, and E are constant
Page 801 and 802:
elastic modulus E = 200 GPa; theref
Page 803 and 804:
17.13 Calculating Deflections of Be
Page 805 and 806:
5. Integration: Perform the integra
Page 807 and 808:
ExAmpLE 17.19Compute the deflection
Page 809 and 810:
From Equation (17.40), the beam def
Page 811 and 812:
120 N/mm40 kN/mAB C100 mm 65 mm 65
Page 813 and 814:
Table A.1 properties of plane Figur
Page 815 and 816:
ExAmpLE A.1Determine the location o
Page 817 and 818:
Using the Pythagorean theorem, dist
Page 819 and 820:
SolutioN(a) Moment of inertia About
Page 821 and 822:
A.3 product of Inertia for an AreaT
Page 823 and 824:
1.654 in.y1.654 in.y1.154 in.(1)1.8
Page 825 and 826:
andcos2θ =p∓( I − I )/2x⎛ (
Page 827 and 828:
ExAmpLE A.7Determine the principal
Page 829 and 830:
mecmoviesExAmpLEA.6 The theory and
Page 831 and 832:
geometric Properties ofStructural S
Page 833 and 834:
Wide-Flange Sections or W Shapes—
Page 835 and 836:
x-XYttffXt wdYb fAmerican Standard
Page 837 and 838:
Shapes cut from Wide-Flange Section
Page 839 and 840:
Hollow Structural Sections or HSS S
Page 841 and 842:
Angle Shapes or L ShapesDesignation
Page 843 and 844:
Table of beam Slopesand deflections
Page 845 and 846:
cantilever BeamsBeam Slope Deflecti
Page 847 and 848:
Table D.1a Average properties of Se
Page 849 and 850:
Table D.2 Typical properties of Sel
Page 851 and 852:
Gear relationships between gears A
Page 853 and 854:
thin-walled pressure vesselsTangent
Page 855 and 856:
P3.11 P = 42,000 lbP3.13 (a) E = 30
Page 857 and 858:
P7.1310.0 kips40.00 kip·ftxP7.293.
Page 859 and 860:
(c)(c)P7.45−(a) wx ( ) = 42.09 ki
Page 861 and 862:
P8.83 (a) r n = 112.5 mm(b) σ A =
Page 863 and 864:
P11.23 (a) P = 12.50 kips(b) M = 31
Page 865 and 866:
P12.63 T max = 119.8 N · mP12.65 (
Page 867 and 868:
P15.43 σ x = 1,545 psi, σ z = 0 p
Page 869 and 870:
IndexAAbsolute maximum shear strain
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load, 207, 210, 211load-deflection,
Page 873 and 874:
for indeterminate beam analysis,461
Page 875 and 876:
flexural, 239-240in-plane shear, 54
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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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