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

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ExAmpLE 9.6An alternative cross section is proposed for the simply supported beam of Example 9.5.In the alternative design, the beam is fabricated from two 1 in. by 10 in. wooden boardsnailed to a 2 in. by 6 in. flange board. The centroid location of the fabricated cross sectionis shown in the sketch, and the moment of inertia of the cross section about the z centroidalaxis is I z = 290.667 in. 4 . If the allowable shear resistance of each nail is80 lb, determine the maximum spacing interval s that is acceptable for the built-up beam.6 in.Nailsyys500 lbA B Cx3.50 in.6.5 in.z2 in.10 in.7 ft 7 ft1 in. 1 in.2.5 in.6 in.y2 in.z1 in. 1 in.10 in.Plan the SolutionThe shear flow formula [Equation (9.12)] and the fastener force–spacing relationship[Equation (9.13)] will be required to determine the maximum spacinginterval s. Since the 2 in. by 6 in. flange board is held in place by the nails, thefirst moment of area, Q, as well as the shear flow q, will be based on this regionof the cross section.SolutioNBeam internal Shear Force VThe shear-force and bending-moment diagrams for the simply supportedbeam are shown in Example 9.5. The shear force has a constant magnitude ofV = 500 lb throughout the entire beam span.First moment of area, Q: Q is calculated for the 2 in. by 6 in. flange board, which is theportion of the cross section held in place by the nails:Q = (6 in.)(2 in.)(2.5 in.) = 30 in. 3Shear flow q: The shear flow that must be transmitted through the pair of nails is3VQ (500 lb)(30 in. )q = = = 51.6 lb/in.4I 290.667 in.zMaximum nail-spacing interval s: The maximum spacing interval for the nails can becalculated from the fastener force–spacing relationship [Equation (9.13)]. The beam isfabricated with two nails installed in each interval; consequently, n f = 2, and we haveqs ≤nV f fnV f f (2 nails)(80 lb/nail)∴s≤ = = 3.10 in.Ans.q 51.6 lb/in.Thus, pairs of nails must be installed at intervals less than or equal to 3.10 in. In practice,nails would be driven at 3 in. intervals.352
mecmoviesExAmpLESm9.9 Determine the allowable shear force capacity oftwo wooden box beams fabricated with two different nailconfigurations.m9.12 Determine the shear stress developed in the boltsused to connect two channel shapes back to back.m9.10 Determine the maximum nail spacing that can beused to construct a U-shaped beam from wooden boards.m9.13 Determine the shear stress in the bolts used to fabricatea box beam.. . ....m9.11 Determine the maximum longitudinal bolt spacingrequired to support a 50 kip shear force.353
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A research-based,online learning en
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MECHANICS OF MATERIALS:An Integrate
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About the AuthorTimothy A. Philpot
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xPREFACEAnimation also offers a new
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xiiPREFACE• Appendix E Fundamenta
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xivPREFACEWhat Do Instructors recei
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ContentsChapter 1 Stress 11.1 Intro
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Chapter 14 Pressure Vessels 58514.1
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2STRESSAddressing these concerns re
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4STRESSnumber begins with the digit
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ExAmpLE 1.3A(1)50 mmB80 kNA 50 mm w
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8STRESSFIGURE 1.2b Free-bodydiagram
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mecmoviesExAmpLEm1.5 A pin at C and
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12STRESSJeffery S. ThomasFIGURE 1.5
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14STRESSFIGURE 1.6 Bearing stress f
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m1.3 Use shear stress concepts for
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applied to beam ABC? Use dimensions
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p1.24 The two wooden boards shown i
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1.5 Stresses on Inclined Sectionsme
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24STRESSSignificanceAlthough one mi
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Thus, to provide the necessary weld
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p1.40 Two wooden members are glued
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30STRAINposition vector between H a
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32STRAINIn developing the concept o
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mecmoviesExAmpLESm2.1 A rigid steel
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p2.4 Bar (1) has a length of L 1 =
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38STRAINIn this expression, θ′ i
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pRoBLEmSp2.11 A thin rectangular po
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SOLUTIONThe thermal strain for a te
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CHAPTER3Mechanical Propertiesof Mat
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(3)(7)(4)(5)(6)Fracture47THE TENSIO
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8049THE STRESS-STRAIN dIAgRAM7060St
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The elastic limit is the largest st
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80Aluminum alloy××53THE STRESS-ST
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The second measure is the reduction
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Values vary for different materials
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before the strain measurement. From
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mecmoviesExERcISEm3.1 Figure M3.1 d
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material is initially 800 mm long a
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CHAPTER4design Concepts4.1 Introduc
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the distributed uniform area loadin
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In some instances, engineers may ne
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both members will be stressed to th
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MecMoviesExAMpLESM4.1 The structure
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bPtPFIGURE p4.3Splice plateBarBarPP
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4.5 Load and Resistance Factor Desi
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79FrequencyLarger γ factors combin
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Limit StatesLRFD is based on a limi
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CHAPTER5Axial deformation5.1 Introd
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The increased normal stress magnitu
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AAAA87dEFORMATIONS IN AxIALLyLOAdEd
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displacement in the horizontal dire
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mecmoviesExAmpLEm5.2 The roof and s
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t = 8 mm, and E = 72 GPa, determine
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d 0to 2d 0 at the other end. A conc
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How are the rigid-bar deflections v
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ExAmpLE 5.4A tie rod (1) and a pipe
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Since the sum of the four interior
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5.5 Statically Indeterminate Axiall
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Step 3 — Force-Deformation Relati
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Successful application of the five-
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Plan the SolutionConsider a free-bo
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Structures with a Rotating Rigid Ba
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(1)Px Bx C113STATICALLy INdETERMINA
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ExERcISESm5.5 A composite axial str
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p5.29 In Figure P5.29, a load P is
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tube has an outside diameter of 1.5
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ExAmpLE 5.7An aluminum rod (1) [E =
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mecmoviesExAmpLEm5.14 A rectangular
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Equations (h) and (i) can be solved
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pRoBLEmSp5.39 A circular aluminum a
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polymer [E = 370 ksi; α = 39.0 ×
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Stress-concentration factor K3.02.8
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of the hole has decayed to a value
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TorsionCHAPTER66.1 IntroductionTorq
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with respect to an adjacent cross s
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the longitudinal axis of the shaft.
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τStressxyτntσn141STRESSES ON ObL
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If a torsion member is subjected to
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ExAmpLE 6.1A hollow circular steel
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Plan the SolutionTo determine the l
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Plan the SolutionThe internal torqu
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mecmoviesExAmpLESm6.4 Determine the
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p6.10 The mechanism shown in Figure
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Furthermore, since gears have teeth
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will be dictated by the ratio of ge
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mecmoviesExAmpLEm6.13 Two solid ste
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6.8 power Transmission161POwER TRAN
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m6.17 A motor shaft is being design
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pulley is a belt having tensions F
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Step 2 — Geometry of Deformation:
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Polar moments of inertia for the al
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Successful application of the five-
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Plan the SolutionAA free-body diagr
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From Equation (e), the allowable in
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Next, consider a free-body diagram
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Polar moments of inertia for the sh
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m6.21 A composite torsion member co
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zD(3)Ay(1)L1,L3N BBL 2(a) Determine
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Stress concentrations also occur at
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For the case of the rectangular bar
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and the angle of twist for a 12 in.
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ExAmpLE 6.13A rectangular box secti
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CHAPTER7Equilibrium of beams7.1 Int
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beam (also called a simple beam). A
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ExAmpLE 7.1Draw the shear-force and
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The negative value for A y indicate
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Plot the FunctionsPlot the function
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yw aw byw 0xxABCABabLFIGURE p7.4FIG
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w(x) = w G at G. At A, where the di
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the left and right sides of a thin
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General procedure for constructing
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Construct the Bending-Moment Diagra
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j M(6) = 0 kN ⋅ m (Rule 4: ∆M =
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of M diagram = shear force V). The
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m7.3 Dynamically generated shear-fo
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p7.21-p7.22 Use the graphical metho
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x - a 0x - a 1x - a 2225dISCONTINuI
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conditions. The reactions for stati
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120 kN ⋅ m concentrated moment: F
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SolutioNWhen we refer to case 4 of
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Uniformly distributed load between
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y5 kN20 kN.my12 kN.m18 kN/mAB3 m 3
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CHAPTER8bending8.1 IntroductionPerh
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has a constant bending moment M, an
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The most common stress-strain relat
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All such moment increments that act
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The sense of σ x (either tension o
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A i(mm 2 )y i(mm)y i A i(mm 3 )(1)
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ExAmpLE 8.2The cross-sectional dime
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m8.7 Determine the centroid locatio
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zFIGURE p8.8p8.9 An aluminum alloy
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WidthWidthWidthDepthXYthicknessXWeb
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700 lb1,500 lb4 in.1 in.200 lb/ft1
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M = −6,000 lb · ft. For this neg
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mecmoviesExAmpLESm8.9 Determine the
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p8.16 A W18 × 40 standard steel sh
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8.5 Introductory Beam Design for St
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M (14,400 lb ⋅ ft)(12 in./ft)∴
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pRoBLEmSp8.26 A small aluminum allo
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Equivalent BeamsBefore considering
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Transformed-Section methodThe conce
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This equation reduces to∫A∫ydA
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ExAmpLE 8.7A cantilever beam 10 ft
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Bending Stresses at the intersectio
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(a) the normal stress in each mater
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283CM = PeF+ =bENdINg duE TO ANECCE
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and the combined normal stress on s
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Combined Stress at KThe combined st
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m8.23 Pipe AB (with outside diamete
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p8.54 The bracket shown in Figure P
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yFlangeyyz CM zWebM zM zCompressive
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These curvature expressions can now
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Coordinates of Points H and KThe (y
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Similarly, the moment of inertia I
Page 323 and 324: p8.63 A downward concentrated load
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Page 327 and 328: SolutioNThe ultimate strength σ U
Page 329 and 330: MO307bENdINg OF CuRVEd bARSr or iθ
Page 331 and 332: Table 8.2 Area and Radial Distance
Page 333 and 334: Plan the SolutionBegin by calculati
Page 335 and 336: SuperpositionOften, curved bars are
Page 337 and 338: We now set the stress at point A (r
Page 339: p8.86 The curved tee shape shown in
Page 342 and 343: 320SHEAR STRESS IN bEAMSy9 kN(2)xAB
Page 344 and 345: ExAmpLE 9.1M A = 11.0 kN·mAz(1)B(2
Page 346 and 347: 84.2 MPa (C)y126.4 MPa (C)116.7 kN6
Page 348 and 349: 326SHEAR STRESS IN bEAMSadA′MI zy
Page 350 and 351: 328SHEAR STRESS IN bEAMSEquation (f
Page 352 and 353: SHEAR STRESS IN bEAMSyzzacbV(a) Box
Page 354 and 355: 332SHEAR STRESS IN bEAMSThe shear s
Page 356 and 357: To determine the average horizontal
Page 358 and 359: p9.3 The beam segment shown in Figu
Page 360 and 361: 9.6 Shear Stresses in Beams of circ
Page 362 and 363: 36 kNVyMzxShear Stress FormulaThe m
Page 364 and 365: 56 mmy13.5 mm 8.5 mm (3) K (4)z(5)7
Page 366 and 367: pRoBLEmSp9.11 A 0.375 in. diameter
Page 368 and 369: p9.23 A W14 × 34 standard steel se
Page 370 and 371: 348SHEAR STRESS IN bEAMSNail ANail
Page 372 and 373: Plan the SolutionWhenever a cross s
Page 376 and 377: ExERcISESm9.9 Five multiple-choice
Page 378 and 379: at intervals of s along each side o
Page 380 and 381: 358SHEAR STRESS IN bEAMSCFdxtC′(2
Page 382 and 383: 360SHEAR STRESS IN bEAMStb2stbdsd2y
Page 384 and 385: 362SHEAR STRESS IN bEAMSIt is usefu
Page 386 and 387: PointSketchyy - ′(mm)A′(mm 2 )Q
Page 388 and 389: For this FBD, the calculation of Q
Page 390 and 391: ExAmpLE 9.971 mm89 mm280 mmV10 mm 1
Page 392 and 393: Point Sketch Calculation of Q = y
Page 394 and 395: be expressed as the product of the
Page 396 and 397: 374SHEAR STRESS IN bEAMSThe exact l
Page 398 and 399: 376SHEAR STRESS IN bEAMSd2d2FIGURE
Page 400 and 401: ExAmpLE 9.11Derive an expression fo
Page 402 and 403: Note that, since the shape is thin
Page 404 and 405: AByOzPED1.038 in.0.643 in.(a) Load
Page 406 and 407: τydAϕdϕyShear StressThe variatio
Page 408 and 409: p9.36 A plastic extrusion has the s
Page 410 and 411: p9.48 The beam cross section shown
Page 412 and 413: yePer3 in.z38 in. O38 in.10 in.5 in
Page 414 and 415: 10.2 moment-curvature RelationshipW
Page 416 and 417: 394bEAM dEFLECTIONS+ M + MPositive
Page 418 and 419: 10.4 Determining Deflections by Int
Page 420 and 421: 398bEAM dEFLECTIONS5. Boundary and
Page 422 and 423: Beam Deflection and Slope at AThe d
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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°
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
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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
Page 688 and 689:
p15.54 A thin-walled cylindrical pr
Page 690 and 691:
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
Page 708 and 709:
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
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694COLuMNSwhich can be further simp
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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
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Determine the allowable axial load
Page 730 and 731:
708COLuMNSwhere the compressive str
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Both tensile and compressive normal
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ExAmpLE 16.940 kNexA 6061-T6 alumin
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pRoBLEmSp16.42 The structural steel
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716ENERgy METHOdSThe total energy o
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718ENERgy METHOdSydV = dx dy dzzxxS
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720ENERgy METHOdSSince the volume o
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The maximum force P that can be app
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17.5 Elastic Strain Energy for Flex
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segment AB of the beam. The second
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p17.8 A solid stepped shaft made of
Page 752 and 753:
730ENERgy METHOdSNote that the nega
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732ENERgy METHOdSSignificance: In t
Page 756 and 757:
From the positive root, the maximum
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Note: Throughout the previous chapt
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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
Page 776 and 777:
754ENERgy METHOdSwhich is the mathe
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756ENERgy METHOdSThe virtual intern
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both P 1 and P 2 from the truss, ap
Page 782 and 783:
Following is the table produced by
Page 784 and 785:
SolutioNCalculate the member length
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764ENERgy METHOdSObtain the total v
Page 788 and 789:
766ENERgy METHOdSthe real external
Page 790 and 791:
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
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
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APPENDIXAgeometric Propertiesof an
Page 814 and 815:
792gEOMETRIC PROPERTIESOF AN AREAyy
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mecmoviesExAmpLESA.2 Animated examp
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796gEOMETRIC PROPERTIESOF AN AREAfo
Page 820 and 821:
ExAmpLE A.340 mmDetermine the momen
Page 822 and 823:
ExAmpLE A.440 mmDetermine the produ
Page 824 and 825:
yy′yOxxdAθθcos θx′xy′sin
Page 826 and 827:
804gEOMETRIC PROPERTIESOF AN AREANo
Page 828 and 829:
806gEOMETRIC PROPERTIESOF AN AREAI
Page 830 and 831:
I xy(38.8, 32.3) y1.654 in.yR = 38.
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YttffdXXt wYb fDesignationAreaADept
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YttffdXXt wYb fDesignationAreaADept
Page 836 and 837:
x-XYttffXt wdYAmerican Standard cha
Page 838 and 839:
b fdttffXYXy-t wYDesignationAreaADe
Page 840 and 841:
dXYXtYbDesignationHSS304.8 × 203.2
Page 842 and 843:
YxZXyXαYZDesignationMasspermeterAr
Page 844 and 845:
Simply Supported BeamsBeam Slope De
Page 846 and 847:
Average Properties ofSelected Mater
Page 848 and 849:
Table D.1b Average properties of Se
Page 850 and 851:
Fundamental Mechanicsof Materials E
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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
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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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