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

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ExERcISESm12.10 Coach Mohr’s Circle of Stress. Learn to constructand use Mohr’s circle to determine principal stresses, including theproper orientation of the principal stress planes.m12.12 Mohr’s Circle Game. Score a minimum of 1,800 points(out of a possible 2,000 points) in this game, which quizzes your abilityto recognize the principal stress element or maximum in-planestress element that corresponds to a given Mohr’s circle.FIGURE m12.10FIGURE m12.12m12.11 Mohr’s Circle Game. Score a minimum of 400 points(out of 450 points possible) in this game, which quizzes your abilityto recognize correctly constructed Mohr’s circles.m12.13 Determine the principal stress magnitudes, the maximumin-plane shear stress magnitude, and the absolute maximumshear stress for a given state of stress.m12.14 Sketching Stress transformation Results. Score atleast 100 points in this interactive activity.FIGURE m12.11 FIGURE m12.14528
pRoBLEmSp12.42 Figure P12.42 shows Mohr’s circle for a point in a physicalobject that is subjected to plane stress.(a) Determine the stresses σ x , σ y , and τ xy , and show them on astress element.(b) Determine the principal stresses and the maximum in-planeshear stress acting at the point, and show these stresses on anappropriate sketch (e.g., see Figure 12.15 or Figure 12.16).p12.44 Figure P12.44 shows Mohr’s circle for a point in a physicalobject that is subjected to plane stress.(a) Determine the stresses σ x , σ y , and τ xy , and show them on astress element.(b) Determine the principal stresses and the maximum in-planeshear stress acting at the point, and show these stresses on anappropriate sketch (e.g., see Figure 12.15 or Figure 12.16).τxyτCσCσyxFIGURE p12.42τ1 grid square = 500 psiτFIGURE p12.441 grid square = 10 MPap12.43 Figure P12.43 shows Mohr’s circle for a point in a physicalobject that is subjected to plane stress.(a) Determine the stresses σ x , σ y , and τ xy , and show them on astress element.(b) Determine the principal stresses and the maximum in-planeshear stress acting at the point, and show these stresses on anappropriate sketch (e.g., see Figure 12.15 or Figure 12.16).p12.45 Figure P12.45 shows Mohr’s circle for a point in a physicalobject that is subjected to plane stress.(a) Determine the stresses σ x , σ y , and τ xy , and show them on astress element.(b) Determine the principal stresses and the maximum in-planeshear stress acting at the point, and show these stresses on anappropriate sketch (e.g., see Figure 12.15 or Figure 12.16).ττyxCσCσyx1 grid square = 5 ksiττ1 grid square = 5 MPaFIGURE p12.43FIGURE p12.45529
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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
Page 500 and 501: p11.49 Two steel beams support a co
Page 502 and 503: 12.2 Stress at a General point in a
Page 504 and 505: 12.3 Equilibrium of the Stress Elem
Page 506 and 507: 484STRESS TRANSFORMATIONSsuch as th
Page 508 and 509: pRoBLEmSp12.1 A compound shaft cons
Page 510 and 511: p12.9 A steel pipe with an outside
Page 512 and 513: The forces acting on the vertical a
Page 514 and 515: tynθxσn dAFigure 12.10b is a free
Page 516 and 517: ExAmpLE 12.3y842 MPa550 MPa16 MPaxA
Page 518 and 519: The angle θ from the redefined x a
Page 520 and 521: p12.19 Two steel plates of uniform
Page 522 and 523: 500STRESS TRANSFORMATIONSprincipal
Page 524 and 525: 502STRESS TRANSFORMATIONSIf the nor
Page 526 and 527: 504STRESS TRANSFORMATIONSIn words,
Page 528 and 529: yFIGURE 12.15506There is nevershear
Page 530 and 531: ExAmpLE 12.5y9 ksi7 ksix11 ksiConsi
Page 532 and 533: planes whose normal is perpendicula
Page 534 and 535: ExERcISEm12.4 Sketching Stress tran
Page 536 and 537: 514STRESS TRANSFORMATIONSmecmovies
Page 538 and 539: 516STRESS TRANSFORMATIONSLabel the
Page 540 and 541: mecmoviesExAmpLEm12.10 Coach Mohr
Page 542 and 543: P 2(-7.22, 0)(-5, 6 ccw) yτ(2, 9.2
Page 544 and 545: in the x-y coordinate system is rot
Page 546 and 547: t73.7°τy (-16, 53)R = 55.22C (-31
Page 548 and 549: ExAmpLE 12.11y(-14, 0)y(-14, 0)14 k
Page 552 and 553: p12.46 Figure P12.46 shows Mohr’s
Page 554 and 555: 12.11 General State of Stress at a
Page 556 and 557: 534STRESS TRANSFORMATIONSIn Equatio
Page 558 and 559: 536STRESS TRANSFORMATIONSthird prin
Page 560 and 561: 538STRESS TRANSFORMATIONSyσp2Arbit
Page 562 and 563: Strain TransformationsCHAPTER1313.1
Page 564 and 565: 542STRAIN TRANSFORMATIONSyyyγ xy d
Page 566 and 567: 544STRAIN TRANSFORMATIONSThus, diag
Page 568 and 569: tnFinally, to compute the shear str
Page 570 and 571: Table 13.2 Absolute maximum Shear S
Page 572 and 573: The in-plane principal directions c
Page 574 and 575: 552STRAIN TRANSFORMATIONS13.6 mohr
Page 576 and 577: y24.2°279 μεπ2579 μεx- 858 μ
Page 578 and 579: 556STRAIN TRANSFORMATIONSPlasticbac
Page 580 and 581: Equation for gage b:2 2− 900 = e
Page 582 and 583: pRoBLEmSp13.25-p13.30 The strain ro
Page 584 and 585: yyyτ xyτ yzτ zxxxxzFIGURE 13.13a
Page 586 and 587: 564STRAIN TRANSFORMATIONSHence, the
Page 588 and 589: SolutioN(a) Normal stresses: From E
Page 590 and 591: Then solve for γ xy :2 2380 = (
Page 592 and 593: (b) Principal and Maximum in-Plane
Page 594 and 595: (b) Principal and Maximum in-Plane
Page 596 and 597: y103.7 MPa40.2 MPa18.5°x63.5 MPa14
Page 598 and 599: 1εz = [ σ z − νσ ( x + σ y )
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σ xyFinally, suppose the material
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pRoBLEmSp13.31 An 8 mm thick brass
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Problem e a e b e c E νycP13.47
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p13.65 A solid plastic [E = 45 MPa;
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586PRESSuRE VESSELSThe wall compris
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588PRESSuRE VESSELSττ abs max = p
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590PRESSuRE VESSELSStresses on the
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592PRESSuRE VESSELSFor the outer su
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31.9 MPay63.8 MPat13.81 MPax30°39.
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p14.5 The normal strain measured on
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p14.20 The cylindrical pressure ves
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600PRESSuRE VESSELSSince the ends o
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602PRESSuRE VESSELSσ θτ maxσ r4
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604PRESSuRE VESSELSFor convenience,
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14.6 Deformations in Thick-Walled c
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Maximum Shear StressThe principal s
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610PRESSuRE VESSELScould be cooled
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Maximum Circumferential Stress in t
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212.5 MPa134.4 MPa145.2 MPa76.2 MPa
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CHAPTER15Combined Loads61615.1 Intr
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6.92 ksiz6.92 ksiHAHy11.54 ksi8.49
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p15.5 A solid 1.50 in. diameter sha
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622COMbINEd LOAdSPP2P2FIGURE 15.1 S
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internal bending moment acting at t
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60 kNShear Force and Bending Moment
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ExAmpLE 15.3A steel hollow structur
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1.216 ksiH1.216 ksiH1.373 ksi30.3°
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Figure P15.13b/14b are d = 240 mm,
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p15.25 A load P = 75 kN acting at a
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636COMbINEd LOAdS b. Bending moment
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20.05 MPacbyM z = 3.85 kN·m20.05 M
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102,400 N · mm = 102.4 N · m. Sim
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yK12.02 MPax45°12.02 MPaStress tra
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The equivalent moments at the secti
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Hy3,000 lb448 psiKThe 3,000 lb shea
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z8.45 kN·m15.6 kN·mHK10.8 kN·mEq
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Torsion shear13.438 MPaBeam shear3.
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m15.5 Determine the stresses acting
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p15.34 Forces P x = 580 lb and P y
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z 2z 1dimensions of the assembly ar
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658COMbINEd LOAdSIf the naming conv
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660COMbINEd LOAdSfrom which it foll
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662COMbINEd LOAdSSimilarly, Equatio
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(b) What is the value of the Mises
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p15.54 A thin-walled cylindrical pr
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668COLuMNSStability of EquilibriumT
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670COLuMNSSummaryBefore a compressi
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672COLuMNSIf we let2Pk =(16.2)EIthe
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674COLuMNS7060σY =50=σ cr4030σY
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The Euler buckling load for this co
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Spacer blockFIGURE p16.676 mm 76 mm
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16.3 The Effect of End conditionson
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682COLuMNSwhere the first two terms
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684COLuMNSAnother way of expressing
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LateralbracingxCBP17.5 ft17.5 ftzSt
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xBPBuckling About the Strong AxisTh
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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
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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
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
Page 772 and 773:
17.8 method of Virtual WorkThe meth
Page 774 and 775:
752ENERgy METHOdSIf the material be
Page 776 and 777:
754ENERgy METHOdSwhich is the mathe
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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
Page 786 and 787:
764ENERgy METHOdSObtain the total v
Page 788 and 789:
766ENERgy METHOdSthe real external
Page 790 and 791:
ExAmpLE 17.141,400 N900 NCalculate
Page 792 and 793:
1Ax 1 or x 2vmDraw a free-body diag
Page 794 and 795:
p17.36 Determine the vertical displ
Page 796 and 797:
p17.54 Determine the minimum moment
Page 798 and 799:
776ENERgy METHOdSFor the general ca
Page 800 and 801:
ExAmpLE 17.16Determine the vertical
Page 802 and 803:
SolutioNWe seek the horizontal defl
Page 804 and 805:
782ENERgy METHOdSprocedure for Anal
Page 806 and 807:
Castigliano’s second theorem appl
Page 808 and 809:
Finally, derive the following equat
Page 810 and 811:
p17.63 The truss shown in Figure P1
Page 812 and 813:
APPENDIXAgeometric Propertiesof an
Page 814 and 815:
792gEOMETRIC PROPERTIESOF AN AREAyy
Page 816 and 817:
mecmoviesExAmpLESA.2 Animated examp
Page 818 and 819:
796gEOMETRIC PROPERTIESOF AN AREAfo
Page 820 and 821:
ExAmpLE A.340 mmDetermine the momen
Page 822 and 823:
ExAmpLE A.440 mmDetermine the produ
Page 824 and 825:
yy′yOxxdAθθcos θx′xy′sin
Page 826 and 827:
804gEOMETRIC PROPERTIESOF AN AREANo
Page 828 and 829:
806gEOMETRIC PROPERTIESOF AN AREAI
Page 830 and 831:
I xy(38.8, 32.3) y1.654 in.yR = 38.
Page 832 and 833:
YttffdXXt wYb fDesignationAreaADept
Page 834 and 835:
YttffdXXt wYb fDesignationAreaADept
Page 836 and 837:
x-XYttffXt wdYAmerican Standard cha
Page 838 and 839:
b fdttffXYXy-t wYDesignationAreaADe
Page 840 and 841:
dXYXtYbDesignationHSS304.8 × 203.2
Page 842 and 843:
YxZXyXαYZDesignationMasspermeterAr
Page 844 and 845:
Simply Supported BeamsBeam Slope De
Page 846 and 847:
Average Properties ofSelected Mater
Page 848 and 849:
Table D.1b Average properties of Se
Page 850 and 851:
Fundamental Mechanicsof Materials E
Page 852 and 853:
orσx + σ y σx − σ yσ n = + c
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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
Page 866 and 867:
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,
Page 874 and 875:
QQ section property, 327-332, 338,
Page 876:
Torsionof circular shafts, 135-151e
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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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