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The following will be discussed in this chapter: Pure bending, other loading types, symmetric member in pure bending, bending deformations, strain due to bending, beam section properties, properties of american standard shapes, deformations in a trans verse cross section, bending of members made of several materials,...
Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P Beer E Russell Johnston, Jr John T DeWolf Pure Bending Lecture Notes: J Walt Oler Texas Tech University © 2002 The McGraw-Hill Companies, Inc All rights reserved Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Pure Bending Pure Bending Other Loading Types Symmetric Member in Pure Bending Bending Deformations Strain Due to Bending Beam Section Properties Properties of American Standard Shapes Deformations in a Transverse Cross Section Sample Problem 4.2 Bending of Members Made of Several Materials Example 4.03 Reinforced Concrete Beams Sample Problem 4.4 Stress Concentrations Plastic Deformations Members Made of an Elastoplastic Material © 2002 The McGraw-Hill Companies, Inc All rights reserved Example 4.03 Reinforced Concrete Beams Sample Problem 4.4 Stress Concentrations Plastic Deformations Members Made of an Elastoplastic Material Plastic Deformations of Members With a Single Plane of S Residual Stresses Example 4.05, 4.06 Eccentric Axial Loading in a Plane of Symmetry Example 4.07 Sample Problem 4.8 Unsymmetric Bending Example 4.08 General Case of Eccentric Axial Loading 4-2 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Pure Bending Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane © 2002 The McGraw-Hill Companies, Inc All rights reserved 4-3 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Other Loading Types • Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple • Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple • Principle of Superposition: The normal stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress © 2002 The McGraw-Hill Companies, Inc All rights reserved 4-4 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Symmetric Member in Pure Bending • Internal forces in any cross section are equivalent to a couple The moment of the couple is the section bending moment • From statics, a couple M consists of two equal and opposite forces • The sum of the components of the forces in any direction is zero • The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane • These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forces Fx = ∫ σ x dA = M y = ∫ zσ x dA = M z = ∫ − yσ x dA = M © 2002 The McGraw-Hill Companies, Inc All rights reserved 4-5 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Bending Deformations Beam with a plane of symmetry in pure bending: • member remains symmetric • bends uniformly to form a circular arc • cross-sectional plane passes through arc center and remains planar • length of top decreases and length of bottom increases • a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change • stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it © 2002 The McGraw-Hill Companies, Inc All rights reserved 4-6 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Strain Due to Bending Consider a beam segment of length L After deformation, the length of the neutral surface remains L At other sections, L′ = ( ρ − y )θ δ = L − L′ = ( ρ − y )θ − ρθ = − yθ εx = εm = δ =− L c ρ yθ ρθ or =− ρ= y ρ (strain varies linearly) c εm y c ε x = − εm © 2002 The McGraw-Hill Companies, Inc All rights reserved 4-7 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stress Due to Bending • For a linearly elastic material, y c σ x = Eε x = − Eε m y = − σ m (stress varies linearly) c • For static equilibrium, y Fx = = ∫ σ x dA = ∫ − σ m dA c σ = − m ∫ y dA c First moment with respect to neutral plane is zero Therefore, the neutral surface must pass through the section centroid • For static equilibrium, ⎛ y ⎞ M = ∫ − yσ x dA = ∫ − y⎜ − σ m ⎟ dA ⎝ c ⎠ σ σ I M = m ∫ y dA = m c c σm = Mc M = I S y Substituting σ x = − σ m c σx = − © 2002 The McGraw-Hill Companies, Inc All rights reserved My I 4-8 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Beam Section Properties • The maximum normal stress due to bending, Mc M = I S I = section moment of inertia I S = = section modulus c σm = A beam section with a larger section modulus will have a lower maximum stress • Consider a rectangular beam cross section, I 12 bh S= = = 16 bh3 = 16 Ah c h2 Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending • Structural steel beams are designed to have a large section modulus © 2002 The McGraw-Hill Companies, Inc All rights reserved 4-9 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Properties of American Standard Shapes © 2002 The McGraw-Hill Companies, Inc All rights reserved - 10 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 4.05, 4.06 • Thickness of elastic core: M = ⎛ M ⎜1 − yY Y⎜ ⎞ ⎟ c ⎟⎠ ⎝ 36.8 kN ⋅ m = ⎞ ⎟ c ⎟⎠ ⎛ (28.8 kN ⋅ m )⎜1 − yY ⎜ yY yY = = 0.666 c 60 mm ⎝ yY = 80 mm • Radius of curvature: • Maximum elastic moment: ( )( ) I 2 −3 −3 = bc = 50 × 10 m 60 × 10 m c = 120 × 10− m3 ( ) I M Y = σ Y = 120 × 10− m3 (240 MPa ) c = 28.8 kN ⋅ m © 2002 The McGraw-Hill Companies, Inc All rights reserved εY = σY E = 240 × 106 Pa 200 × 109 Pa = 1.2 × 10−3 εY = yY ρ= yY ρ εY = 40 × 10−3 m 1.2 × 10−3 ρ = 33.3 m - 28 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 4.05, 4.06 • M = 36.8 kN-m yY = 40 mm σ Y = 240 MPa • M = -36.8 kN-m Mc 36.8 kN ⋅ m = I 120 × 106 m3 = 306.7 MPa < 2σ Y ′ = σm • M=0 At the edge of the elastic core, εx = σx E = − 35.5 × 106 Pa 200 × 109 Pa = −177.5 × 10− ρ =− yY εx = 40 × 10−3 m 177.5 × 10 − ρ = 225 m © 2002 The McGraw-Hill Companies, Inc All rights reserved - 29 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Eccentric Axial Loading in a Plane of Symmetry • Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment σ x = (σ x )centric + (σ x )bending = • Eccentric loading F=P M = Pd P My − A I • Validity requires stresses below proportional limit, deformations have negligible effect on geometry, and stresses not evaluated near points of load application © 2002 The McGraw-Hill Companies, Inc All rights reserved - 30 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 4.07 SOLUTION: • Find the equivalent centric load and bending moment • Superpose the uniform stress due to the centric load and the linear stress due to the bending moment • Evaluate the maximum tensile and compressive stresses at the inner and outer edges, respectively, of the superposed stress distribution An open-link chain is obtained by bending low-carbon steel rods into the shape shown For 160 lb load, determine • Find the neutral axis by determining the location where the normal stress (a) maximum tensile and compressive is zero stresses, (b) distance between section centroid and neutral axis © 2002 The McGraw-Hill Companies, Inc All rights reserved - 31 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 4.07 • Normal stress due to a centric load A = πc = π (0.25 in )2 = 0.1963 in P 160 lb σ0 = = A 0.1963 in = 815 psi • Equivalent centric load and bending moment P = 160 lb M = Pd = (160 lb )(0.6 in ) = 104 lb ⋅ in • Normal stress due to bending moment I = 14 πc = 14 π (0.25)4 = 3.068 × 10−3 in Mc (104 lb ⋅ in )(0.25 in ) σm = = I 068 × 10−3 in = 8475 psi © 2002 The McGraw-Hill Companies, Inc All rights reserved - 32 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 4.07 • Maximum tensile and compressive stresses σt = σ0 +σm = 815 + 8475 σc = σ −σ m = 815 − 8475 σ t = 9260 psi σ c = −7660 psi © 2002 The McGraw-Hill Companies, Inc All rights reserved • Neutral axis location 0= P My0 − A I P I 3.068 × 10−3 in y0 = = (815 psi ) AM 105 lb ⋅ in y0 = 0.0240 in - 33 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 4.8 The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MPa in compression Determine the largest force P which can be applied to the link SOLUTION: • Determine an equivalent centric load and bending moment • Superpose the stress due to a centric load and the stress due to bending From Sample Problem 2.4, A = × 10−3 m Y = 0.038 m I = 868 ì 10 m Evaluate the critical loads for the allowable tensile and compressive stresses • The largest allowable load is the smallest of the two critical loads © 2002 The McGraw-Hill Companies, Inc All rights reserved - 34 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 4.8 • Determine an equivalent centric and bending loads d = 0.038 − 0.010 = 0.028 m P = centric load M = Pd = 0.028 P = bending moment • Superpose stresses due to centric and bending loads (0.028 P )(0.022) = +377 P P Mc A P + =− + A I × 10−3 868 × 10−9 (0.028 P )(0.022) = −1559 P P Mc P σB = − − A = − − A I × 10−3 868 × 10 −9 σA = − • Evaluate critical loads for allowable stresses σ A = +377 P = 30 MPa P = 79.6 kN σ B = −1559 P = −120 MPa P = 79.6 kN • The largest allowable load © 2002 The McGraw-Hill Companies, Inc All rights reserved P = 77.0 kN - 35 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Unsymmetric Bending • Analysis of pure bending has been limited to members subjected to bending couples acting in a plane of symmetry • Members remain symmetric and bend in the plane of symmetry • The neutral axis of the cross section coincides with the axis of the couple • Will now consider situations in which the bending couples not act in a plane of symmetry • Cannot assume that the member will bend in the plane of the couples • In general, the neutral axis of the section will not coincide with the axis of the couple © 2002 The McGraw-Hill Companies, Inc All rights reserved - 36 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Unsymmetric Bending • = Fx = ∫ σ x dA = ∫ ⎛⎜ − σ m ⎞⎟dA y ⎝ c ⎠ or = ∫ y dA neutral axis passes through centroid Wish to determine the conditions under which the neutral axis of a cross section of arbitrary shape coincides with the axis of the couple as shown • The resultant force and moment from the distribution of elementary forces in the section must satisfy Fx = = M y M z = M = applied couple © 2002 The McGraw-Hill Companies, Inc All rights reserved ⎞ ⎛ y M M y = = − ⎜ − σ m ⎟dA ∫ z • ⎠ ⎝ c σ I I = I z = moment of inertia or M = m c defines stress distribution • = M y = ∫ zσ x dA = ∫ z⎛⎜ − σ m ⎞⎟dA y ⎝ c ⎠ or = ∫ yz dA = I yz = product of inertia couple vector must be directed along a principal centroidal axis - 37 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Unsymmetric Bending Superposition is applied to determine stresses in the most general case of unsymmetric bending • Resolve the couple vector into components along the principle centroidal axes M z = M cosθ M y = M sin θ • Superpose the component stress distributions σx = − Mzy Myy + Iz Iy • Along the neutral axis, σx = = − tan φ = (M cosθ ) y + (M sin θ ) y Mzy Myy + =− Iy Iz Iy Iz y Iz tan θ = z Iy © 2002 The McGraw-Hill Companies, Inc All rights reserved - 38 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 4.08 SOLUTION: • Resolve the couple vector into components along the principle centroidal axes and calculate the corresponding maximum stresses M z = M cosθ M y = M sin θ • Combine the stresses from the component stress distributions Mzy Myy σx = − + Iz Iy A 1600 lb-in couple is applied to a rectangular wooden beam in a plane • Determine the angle of the neutral forming an angle of 30 deg with the axis vertical Determine (a) the maximum y Iz tan φ tan θ = = stress in the beam, (b) the angle that the z Iy neutral axis forms with the horizontal plane © 2002 The McGraw-Hill Companies, Inc All rights reserved - 39 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 4.08 • Resolve the couple vector into components and calculate the corresponding maximum stresses M z = (1600 lb ⋅ in ) cos 30 = 1386 lb ⋅ in M y = (1600 lb ⋅ in )sin 30 = 800 lb ⋅ in (1.5 in )(3.5 in )3 = 5.359 in I z = 12 (3.5 in )(1.5 in )3 = 0.9844 in I y = 12 The largest tensile stress due to M z occurs along AB σ1 = M z y (1386 lb ⋅ in )(1.75 in ) = = 452.6 psi Iz 5.359 in The largest tensile stress due to M z occurs along AD σ2 = M yz Iy = (800 lb ⋅ in )(0.75 in ) = 609.5 psi 0.9844 in • The largest tensile stress due to the combined loading occurs at A σ max = σ + σ = 452.6 + 609.5 © 2002 The McGraw-Hill Companies, Inc All rights reserved σ max = 1062 psi - 40 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 4.08 • Determine the angle of the neutral axis Iz 5.359 in tan φ = tan θ = tan 30 Iy 0.9844 in = 3.143 φ = 72.4o © 2002 The McGraw-Hill Companies, Inc All rights reserved - 41 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf General Case of Eccentric Axial Loading • Consider a straight member subject to equal and opposite eccentric forces • The eccentric force is equivalent to the system of a centric force and two couples P = centric force M y = Pa M z = Pb • By the principle of superposition, the combined stress distribution is σx = P Mz y M yz − + A Iz Iy • If the neutral axis lies on the section, it may be found from My Mz P y− z= Iz Iy A © 2002 The McGraw-Hill Companies, Inc All rights reserved - 42 ...Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Pure Bending Pure Bending Other Loading Types Symmetric Member in Pure Bending Bending Deformations Strain Due to Bending Beam Section... Problem 4.8 Unsymmetric Bending Example 4.08 General Case of Eccentric Axial Loading 4-2 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Pure Bending Pure Bending: Prismatic members... ×1 0-9 m ) © 2002 The McGraw-Hill Companies, Inc All rights reserved = 20.95 × 10−3 m -1 ρ ρ = 47.7 m - 14 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Bending of Members Made of