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

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60 kNShear Force and Bending Moment at KAt the location of point K, the internal shear force is V =−45 kN and the internal bending moment is M = 180 kN · m.These internal forces act in the directions shown.M = 180 kN·mKBxV = 45 kN0.5 m1.5 m105 kN200 mm16 mmNormal and Shear Stresses at KPoint K is located 75 mm above the z centroidal axis; therefore, y = 75 mm. Thebending stress at K can be calculated from the flexure formula:75 mmKyMy (180 kN⋅m)(75 mm)(1,000 N/kN)(1, 000 mm/m)s x =− =−6 4I202.74 × 10 mmzz350 mm=− 66.6 MPa = 66.6 MPa(C)9 mmNote that this compressive normal stress acts parallel to the longitudinal axis ofthe beam—that is, in the x direction.To compute the shear stress at K, Q must be calculated for the highlightedarea. The first moment of the highlighted area about the z centroidal axis is Q =622,852 mm 3 . The shear stress at K due to beam flexure is thenVQτ = =Itz3(45 kN)(622,852 mm )(1, 000 N/kN)6 4(202.74 × 10 mm )(9 mm)= 15.36 MPaGenerally, the magnitude of V is used in this calculation and the direction of the shearstress is determined by inspection. The shear stress acts in the same direction as the internalshear force V. Therefore, on the left face of the stress element, the shear stress τ actsdownward.15.36 MPaK66.59 MPaxStress Element for Point KThe compressive bending stress acts on the x faces of the stress element, and theshear stress acts downward on the −x face of the stress element. After the propershear stress direction has been established on one face, the shear stress directionson the other three faces are known.15.36 MPaK66.59 MPa12.4°33.3 MPa36.7 MPaxStress transformation Results at KThe principal stresses and the maximum shear stress at K areshown in the accompanying figure.70.0 MPa3.37 MPa626
mecmoviesExAmpLESm15.2 A cantilever beam has a uniformly distributed loadof 2 kips/ft. The beam cross section is a tee shape. Determinethe principal stresses and the maximum shearing stress atpoint D, located 4 in. above the bottom of the tee stem and ata distance of 1 ft from the fixed support.m15.3 A steel rectangular tube shape is used as a beam tosupport the loads shown. Determine the principal stressesand the maximum shear stress at point H, which is located 1 mto the right of pin support A.m15.4 A steel wide-flange beam carries loads that createan internal shear force of V = 60 kips and an internal bendingmoment of M = 150 kip · ft at a particular point along thespan. Determine the normal and shear stresses that act atpoint B, located on the surface of the steel shape, 3 in. abovethe centroid.627
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A research-based,online learning en
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MECHANICS OF MATERIALS:An Integrate
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About the AuthorTimothy A. Philpot
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xPREFACEAnimation also offers a new
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xiiPREFACE• Appendix E Fundamenta
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xivPREFACEWhat Do Instructors recei
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ContentsChapter 1 Stress 11.1 Intro
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Chapter 14 Pressure Vessels 58514.1
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2STRESSAddressing these concerns re
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4STRESSnumber begins with the digit
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ExAmpLE 1.3A(1)50 mmB80 kNA 50 mm w
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8STRESSFIGURE 1.2b Free-bodydiagram
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mecmoviesExAmpLEm1.5 A pin at C and
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12STRESSJeffery S. ThomasFIGURE 1.5
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14STRESSFIGURE 1.6 Bearing stress f
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m1.3 Use shear stress concepts for
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applied to beam ABC? Use dimensions
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p1.24 The two wooden boards shown i
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1.5 Stresses on Inclined Sectionsme
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24STRESSSignificanceAlthough one mi
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Thus, to provide the necessary weld
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p1.40 Two wooden members are glued
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30STRAINposition vector between H a
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32STRAINIn developing the concept o
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mecmoviesExAmpLESm2.1 A rigid steel
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p2.4 Bar (1) has a length of L 1 =
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38STRAINIn this expression, θ′ i
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pRoBLEmSp2.11 A thin rectangular po
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SOLUTIONThe thermal strain for a te
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CHAPTER3Mechanical Propertiesof Mat
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(3)(7)(4)(5)(6)Fracture47THE TENSIO
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8049THE STRESS-STRAIN dIAgRAM7060St
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The elastic limit is the largest st
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80Aluminum alloy××53THE STRESS-ST
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The second measure is the reduction
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Values vary for different materials
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before the strain measurement. From
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mecmoviesExERcISEm3.1 Figure M3.1 d
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material is initially 800 mm long a
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CHAPTER4design Concepts4.1 Introduc
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the distributed uniform area loadin
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In some instances, engineers may ne
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both members will be stressed to th
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MecMoviesExAMpLESM4.1 The structure
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bPtPFIGURE p4.3Splice plateBarBarPP
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4.5 Load and Resistance Factor Desi
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79FrequencyLarger γ factors combin
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Limit StatesLRFD is based on a limi
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CHAPTER5Axial deformation5.1 Introd
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The increased normal stress magnitu
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AAAA87dEFORMATIONS IN AxIALLyLOAdEd
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displacement in the horizontal dire
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mecmoviesExAmpLEm5.2 The roof and s
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t = 8 mm, and E = 72 GPa, determine
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d 0to 2d 0 at the other end. A conc
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How are the rigid-bar deflections v
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ExAmpLE 5.4A tie rod (1) and a pipe
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Since the sum of the four interior
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5.5 Statically Indeterminate Axiall
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Step 3 — Force-Deformation Relati
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Successful application of the five-
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Plan the SolutionConsider a free-bo
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Structures with a Rotating Rigid Ba
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(1)Px Bx C113STATICALLy INdETERMINA
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ExERcISESm5.5 A composite axial str
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p5.29 In Figure P5.29, a load P is
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tube has an outside diameter of 1.5
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ExAmpLE 5.7An aluminum rod (1) [E =
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mecmoviesExAmpLEm5.14 A rectangular
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Equations (h) and (i) can be solved
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pRoBLEmSp5.39 A circular aluminum a
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polymer [E = 370 ksi; α = 39.0 ×
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Stress-concentration factor K3.02.8
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of the hole has decayed to a value
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TorsionCHAPTER66.1 IntroductionTorq
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with respect to an adjacent cross s
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the longitudinal axis of the shaft.
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τStressxyτntσn141STRESSES ON ObL
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If a torsion member is subjected to
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ExAmpLE 6.1A hollow circular steel
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Plan the SolutionTo determine the l
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Plan the SolutionThe internal torqu
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mecmoviesExAmpLESm6.4 Determine the
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p6.10 The mechanism shown in Figure
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Furthermore, since gears have teeth
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will be dictated by the ratio of ge
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mecmoviesExAmpLEm6.13 Two solid ste
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6.8 power Transmission161POwER TRAN
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m6.17 A motor shaft is being design
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pulley is a belt having tensions F
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Step 2 — Geometry of Deformation:
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Polar moments of inertia for the al
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Successful application of the five-
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Plan the SolutionAA free-body diagr
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From Equation (e), the allowable in
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Next, consider a free-body diagram
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Polar moments of inertia for the sh
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m6.21 A composite torsion member co
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zD(3)Ay(1)L1,L3N BBL 2(a) Determine
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Stress concentrations also occur at
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For the case of the rectangular bar
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and the angle of twist for a 12 in.
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ExAmpLE 6.13A rectangular box secti
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CHAPTER7Equilibrium of beams7.1 Int
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beam (also called a simple beam). A
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ExAmpLE 7.1Draw the shear-force and
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The negative value for A y indicate
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Plot the FunctionsPlot the function
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yw aw byw 0xxABCABabLFIGURE p7.4FIG
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w(x) = w G at G. At A, where the di
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the left and right sides of a thin
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General procedure for constructing
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Construct the Bending-Moment Diagra
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j M(6) = 0 kN ⋅ m (Rule 4: ∆M =
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of M diagram = shear force V). The
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m7.3 Dynamically generated shear-fo
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p7.21-p7.22 Use the graphical metho
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x - a 0x - a 1x - a 2225dISCONTINuI
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conditions. The reactions for stati
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120 kN ⋅ m concentrated moment: F
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SolutioNWhen we refer to case 4 of
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Uniformly distributed load between
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y5 kN20 kN.my12 kN.m18 kN/mAB3 m 3
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CHAPTER8bending8.1 IntroductionPerh
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has a constant bending moment M, an
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The most common stress-strain relat
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All such moment increments that act
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The sense of σ x (either tension o
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A i(mm 2 )y i(mm)y i A i(mm 3 )(1)
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ExAmpLE 8.2The cross-sectional dime
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m8.7 Determine the centroid locatio
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zFIGURE p8.8p8.9 An aluminum alloy
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WidthWidthWidthDepthXYthicknessXWeb
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700 lb1,500 lb4 in.1 in.200 lb/ft1
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M = −6,000 lb · ft. For this neg
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mecmoviesExAmpLESm8.9 Determine the
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p8.16 A W18 × 40 standard steel sh
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8.5 Introductory Beam Design for St
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M (14,400 lb ⋅ ft)(12 in./ft)∴
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pRoBLEmSp8.26 A small aluminum allo
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Equivalent BeamsBefore considering
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Transformed-Section methodThe conce
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This equation reduces to∫A∫ydA
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ExAmpLE 8.7A cantilever beam 10 ft
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Bending Stresses at the intersectio
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(a) the normal stress in each mater
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283CM = PeF+ =bENdINg duE TO ANECCE
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and the combined normal stress on s
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Combined Stress at KThe combined st
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m8.23 Pipe AB (with outside diamete
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p8.54 The bracket shown in Figure P
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yFlangeyyz CM zWebM zM zCompressive
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These curvature expressions can now
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Coordinates of Points H and KThe (y
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Similarly, the moment of inertia I
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p8.63 A downward concentrated load
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the factor K depends only upon the
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SolutioNThe ultimate strength σ U
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MO307bENdINg OF CuRVEd bARSr or iθ
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Table 8.2 Area and Radial Distance
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Plan the SolutionBegin by calculati
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SuperpositionOften, curved bars are
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We now set the stress at point A (r
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p8.86 The curved tee shape shown in
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320SHEAR STRESS IN bEAMSy9 kN(2)xAB
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ExAmpLE 9.1M A = 11.0 kN·mAz(1)B(2
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84.2 MPa (C)y126.4 MPa (C)116.7 kN6
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326SHEAR STRESS IN bEAMSadA′MI zy
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328SHEAR STRESS IN bEAMSEquation (f
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SHEAR STRESS IN bEAMSyzzacbV(a) Box
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332SHEAR STRESS IN bEAMSThe shear s
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To determine the average horizontal
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p9.3 The beam segment shown in Figu
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9.6 Shear Stresses in Beams of circ
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36 kNVyMzxShear Stress FormulaThe m
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56 mmy13.5 mm 8.5 mm (3) K (4)z(5)7
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pRoBLEmSp9.11 A 0.375 in. diameter
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p9.23 A W14 × 34 standard steel se
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348SHEAR STRESS IN bEAMSNail ANail
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Plan the SolutionWhenever a cross s
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ExAmpLE 9.6An alternative cross sec
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ExERcISESm9.9 Five multiple-choice
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at intervals of s along each side o
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358SHEAR STRESS IN bEAMSCFdxtC′(2
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360SHEAR STRESS IN bEAMStb2stbdsd2y
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362SHEAR STRESS IN bEAMSIt is usefu
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PointSketchyy - ′(mm)A′(mm 2 )Q
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For this FBD, the calculation of Q
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ExAmpLE 9.971 mm89 mm280 mmV10 mm 1
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Point Sketch Calculation of Q = y
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be expressed as the product of the
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374SHEAR STRESS IN bEAMSThe exact l
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376SHEAR STRESS IN bEAMSd2d2FIGURE
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ExAmpLE 9.11Derive an expression fo
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Note that, since the shape is thin
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AByOzPED1.038 in.0.643 in.(a) Load
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τydAϕdϕyShear StressThe variatio
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p9.36 A plastic extrusion has the s
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p9.48 The beam cross section shown
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yePer3 in.z38 in. O38 in.10 in.5 in
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10.2 moment-curvature RelationshipW
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394bEAM dEFLECTIONS+ M + MPositive
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10.4 Determining Deflections by Int
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398bEAM dEFLECTIONS5. Boundary and
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Beam Deflection and Slope at AThe d
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Elastic Curve EquationSubstitute th
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Boundary ConditionsBoundary conditi
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integration over the interval a ≤
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mecmoviesExERcISEm10.1 Beam Boundar
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P10.11 For the beam and loading sho
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Elastic Curve EquationSubstitute th
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414beam deflectionsintegration give
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(a) Beam Deflection at AAt the tip
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Integrate again to obtain the beam
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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
Page 598 and 599: 1εz = [ σ z − νσ ( x + σ y )
Page 600 and 601: σ xyFinally, suppose the material
Page 602 and 603: pRoBLEmSp13.31 An 8 mm thick brass
Page 604 and 605: Problem e a e b e c E νycP13.47
Page 606 and 607: p13.65 A solid plastic [E = 45 MPa;
Page 608 and 609: 586PRESSuRE VESSELSThe wall compris
Page 610 and 611: 588PRESSuRE VESSELSττ abs max = p
Page 612 and 613: 590PRESSuRE VESSELSStresses on the
Page 614 and 615: 592PRESSuRE VESSELSFor the outer su
Page 616 and 617: 31.9 MPay63.8 MPat13.81 MPax30°39.
Page 618 and 619: p14.5 The normal strain measured on
Page 620 and 621: p14.20 The cylindrical pressure ves
Page 622 and 623: 600PRESSuRE VESSELSSince the ends o
Page 624 and 625: 602PRESSuRE VESSELSσ θτ maxσ r4
Page 626 and 627: 604PRESSuRE VESSELSFor convenience,
Page 628 and 629: 14.6 Deformations in Thick-Walled c
Page 630 and 631: Maximum Shear StressThe principal s
Page 632 and 633: 610PRESSuRE VESSELScould be cooled
Page 634 and 635: Maximum Circumferential Stress in t
Page 636 and 637: 212.5 MPa134.4 MPa145.2 MPa76.2 MPa
Page 638 and 639: CHAPTER15Combined Loads61615.1 Intr
Page 640 and 641: 6.92 ksiz6.92 ksiHAHy11.54 ksi8.49
Page 642 and 643: p15.5 A solid 1.50 in. diameter sha
Page 644 and 645: 622COMbINEd LOAdSPP2P2FIGURE 15.1 S
Page 646 and 647: internal bending moment acting at t
Page 650 and 651: ExAmpLE 15.3A steel hollow structur
Page 652 and 653: 1.216 ksiH1.216 ksiH1.373 ksi30.3°
Page 654 and 655: Figure P15.13b/14b are d = 240 mm,
Page 656 and 657: p15.25 A load P = 75 kN acting at a
Page 658 and 659: 636COMbINEd LOAdS b. Bending moment
Page 660 and 661: 20.05 MPacbyM z = 3.85 kN·m20.05 M
Page 662 and 663: 102,400 N · mm = 102.4 N · m. Sim
Page 664 and 665: yK12.02 MPax45°12.02 MPaStress tra
Page 666 and 667: The equivalent moments at the secti
Page 668 and 669: Hy3,000 lb448 psiKThe 3,000 lb shea
Page 670 and 671: z8.45 kN·m15.6 kN·mHK10.8 kN·mEq
Page 672 and 673: Torsion shear13.438 MPaBeam shear3.
Page 674 and 675: m15.5 Determine the stresses acting
Page 676 and 677: p15.34 Forces P x = 580 lb and P y
Page 678 and 679: z 2z 1dimensions of the assembly ar
Page 680 and 681: 658COMbINEd LOAdSIf the naming conv
Page 682 and 683: 660COMbINEd LOAdSfrom which it foll
Page 684 and 685: 662COMbINEd LOAdSSimilarly, Equatio
Page 686 and 687: (b) What is the value of the Mises
Page 688 and 689: p15.54 A thin-walled cylindrical pr
Page 690 and 691: 668COLuMNSStability of EquilibriumT
Page 692 and 693: 670COLuMNSSummaryBefore a compressi
Page 694 and 695: 672COLuMNSIf we let2Pk =(16.2)EIthe
Page 696 and 697: 674COLuMNS7060σY =50=σ cr4030σY
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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
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p16.19 The aluminum column shown in
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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
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698COLuMNSFor short and intermediat
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ExAmpLE 16.52 in.zSpacer blocky2 in
Page 724 and 725:
Since KL/ry > 133.7, the column is
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The reduced Euler buckling stress t
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Determine the allowable axial load
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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
Page 740 and 741:
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
Page 746 and 747:
17.5 Elastic Strain Energy for Flex
Page 748 and 749:
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
Page 758 and 759:
Note: Throughout the previous chapt
Page 760 and 761:
The right-hand side of this equatio
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(b) If the rod has a constant diame
Page 764 and 765:
Note: Throughout the previous chapt
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assembly. The solid bronze post has
Page 768 and 769:
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
Page 774 and 775:
752ENERgy METHOdSIf the material be
Page 776 and 777:
754ENERgy METHOdSwhich is the mathe
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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
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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
Page 792 and 793:
1Ax 1 or x 2vmDraw a free-body diag
Page 794 and 795:
p17.36 Determine the vertical displ
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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.
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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
Page 856 and 857:
(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,
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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