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The following will be discussed in this chapter: Introduction, shear on the horizontal face of a beam element, determination of the shearing stress, longitudinal shear on a beam element of arbitrary shape, shearing stresses in thin-walled members, unsymmetric loading of thin-walled members.
Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P Beer E Russell Johnston, Jr John T DeWolf Lecture Notes: J Walt Oler Texas Tech University Shearing Stresses in Beams and ThinWalled Members © 2002 The McGraw-Hill Companies, Inc All rights reserved Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shearing Stresses in Beams and Thin-Walled Members Introduction Shear on the Horizontal Face of a Beam Element Example 6.01 Determination of the Shearing Stress in a Beam Shearing Stresses τxy in Common Types of Beams Further Discussion of the Distribution of Stresses in a Sample Problem 6.2 Longitudinal Shear on a Beam Element of Arbitrary Shape Example 6.04 Shearing Stresses in Thin-Walled Members Plastic Deformations Sample Problem 6.3 Unsymmetric Loading of Thin-Walled Members Example 6.05 Example 6.06 © 2002 The McGraw-Hill Companies, Inc All rights reserved 6-2 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Introduction • Transverse loading applied to a beam results in normal and shearing stresses in transverse sections • Distribution of normal and shearing stresses satisfies Fx = ∫ σ x dA = Fy = ∫ τ xy dA = −V Fz = ∫ τ xz dA = ( ) M x = ∫ y τ xz − z τ xy dA = M y = ∫ z σ x dA = M z = ∫ (− y σ x ) = • When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces • Longitudinal shearing stresses must exist in any member subjected to transverse loading © 2002 The McGraw-Hill Companies, Inc All rights reserved 6-3 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shear on the Horizontal Face of a Beam Element • Consider prismatic beam • For equilibrium of beam element ∑ Fx = = ∆H + ∫ (σ D − σ D )dA A ∆H = M D − MC ∫ y dA I A • Note, Q = ∫ y dA A M D − MC = dM ∆x = V ∆x dx • Substituting, VQ ∆x I ∆H VQ q= = = shear flow ∆x I ∆H = © 2002 The McGraw-Hill Companies, Inc All rights reserved 6-4 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shear on the Horizontal Face of a Beam Element • Shear flow, q= ∆H VQ = = shear flow ∆x I • where Q = ∫ y dA A = first moment of area above y1 I= ∫ y dA A + A' = second moment of full cross section • Same result found for lower area ∆H ′ VQ′ = = − q′ ∆x I Q + Q′ = = first moment with respect to neutral axis ∆H ′ = − ∆H q′ = © 2002 The McGraw-Hill Companies, Inc All rights reserved 6-5 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 6.01 SOLUTION: • Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank • Calculate the corresponding shear force in each nail A beam is made of three planks, nailed together Knowing that the spacing between nails is 25 mm and that the vertical shear in the beam is V = 500 N, determine the shear force in each nail © 2002 The McGraw-Hill Companies, Inc All rights reserved 6-6 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 6.01 SOLUTION: • Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank VQ (500 N )(120 × 10− m3 ) q= = I 16.20 × 10-6 m = 3704 N m Q = Ay = (0.020 m × 0.100 m )(0.060 m ) = 120 × 10− m3 I (0.020 m )(0.100 m )3 = 12 (0.100 m )(0.020 m )3 + 2[12 • Calculate the corresponding shear force in each nail for a nail spacing of 25 mm + (0.020 m × 0.100 m )(0.060 m )2 ] = 16.20 ì 10 m â 2002 The McGraw-Hill Companies, Inc All rights reserved F = (0.025 m)q = (0.025 m)(3704 N m F = 92.6 N 6-7 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Determination of the Shearing Stress in a Beam • The average shearing stress on the horizontal face of the element is obtained by dividing the shearing force on the element by the area of the face ∆H q ∆x VQ ∆x = = ∆A ∆A I t ∆x VQ = It τ ave = • On the upper and lower surfaces of the beam, τyx= It follows that τxy= on the upper and lower edges of the transverse sections • If the width of the beam is comparable or large relative to its depth, the shearing stresses at D1 and D2 are significantly higher than at D © 2002 The McGraw-Hill Companies, Inc All rights reserved 6-8 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shearing Stresses τxy in Common Types of Beams • For a narrow rectangular beam, VQ V ⎛⎜ 1− τ xy = = ⎜ Ib A ⎝ τ max = y ⎞⎟ c ⎟⎠ 3V 2A • For American Standard (S-beam) and wide-flange (W-beam) beams VQ It V τ max = Aweb τ ave = © 2002 The McGraw-Hill Companies, Inc All rights reserved 6-9 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Further Discussion of the Distribution of Stresses in a Narrow Rectangular Beam • Consider a narrow rectangular cantilever beam subjected to load P at its free end: P ⎛⎜ τ xy = 1− A ⎝⎜ y ⎞⎟ c ⎟⎠ σx = + Pxy I • Shearing stresses are independent of the distance from the point of application of the load • Normal strains and normal stresses are unaffected by the shearing stresses • From Saint-Venant’s principle, effects of the load application mode are negligible except in immediate vicinity of load application points • Stress/strain deviations for distributed loads are negligible for typical beam sections of interest © 2002 The McGraw-Hill Companies, Inc All rights reserved - 10 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 6.2 SOLUTION: Develop shear and bending moment diagrams Identify the maximums Vmax = kips M max = 7.5 kip ⋅ ft = 90 kip ⋅ in © 2002 The McGraw-Hill Companies, Inc All rights reserved - 12 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 6.2 • Determine the beam depth based on allowable normal stress σ all = M max S 1800 psi = 90 × 103 lb ⋅ in (0.5833 in.) d d = 9.26 in bd3 I = 12 I S = = 16 b d c = 16 (3.5 in.)d = (0.5833 in.)d • Determine the beam depth based on allowable shear stress Vmax A 3000 lb 120 psi = (3.5 in.) d d = 10.71in τ all = • Required beam depth is equal to the larger of the two d = 10.71in © 2002 The McGraw-Hill Companies, Inc All rights reserved - 13 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Longitudinal Shear on a Beam Element of Arbitrary Shape • We have examined the distribution of the vertical components τxy on a transverse section of a beam We now wish to consider the horizontal components τxz of the stresses • Consider prismatic beam with an element defined by the curved surface CDD’C’ ∑ Fx = = ∆H + ∫ (σ D − σ C )dA a • Except for the differences in integration areas, this is the same result obtained before which led to ∆H = © 2002 The McGraw-Hill Companies, Inc All rights reserved VQ ∆x I q= ∆H VQ = ∆x I - 14 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 6.04 SOLUTION: • Determine the shear force per unit length along each edge of the upper plank • Based on the spacing between nails, determine the shear force in each nail A square box beam is constructed from four planks as shown Knowing that the spacing between nails is 1.5 in and the beam is subjected to a vertical shear of magnitude V = 600 lb, determine the shearing force in each nail © 2002 The McGraw-Hill Companies, Inc All rights reserved - 15 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 6.04 SOLUTION: • Determine the shear force per unit length along each edge of the upper plank ( ) VQ (600 lb ) 4.22 in lb q= = = 92 I in 27.42 in q lb = 46.15 in = edge force per unit length f = For the upper plank, Q = A′y = (0.75in.)(3 in.)(1.875 in.) = 4.22 in For the overall beam cross-section, (4.5 in ) − (3 in ) I = 12 12 3 = 27.42 in © 2002 The McGraw-Hill Companies, Inc All rights reserved • Based on the spacing between nails, determine the shear force in each nail lb ⎞ ⎛ F = f A = ⎜ 46.15 ⎟(1.75 in ) in ⎠ ⎝ F = 80.8 lb - 16 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shearing Stresses in Thin-Walled Members • Consider a segment of a wide-flange beam subjected to the vertical shear V • The longitudinal shear force on the element is ∆H = VQ ∆x I • The corresponding shear stress is τ zx = τ xz ≈ ∆H VQ = t ∆x It • Previously found a similar expression for the shearing stress in the web τ xy = VQ It • NOTE: τ xy ≈ τ xz ≈ © 2002 The McGraw-Hill Companies, Inc All rights reserved in the flanges in the web - 17 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shearing Stresses in Thin-Walled Members • The variation of shear flow across the section depends only on the variation of the first moment q =τt = VQ I • For a box beam, q grows smoothly from zero at A to a maximum at C and C’ and then decreases back to zero at E • The sense of q in the horizontal portions of the section may be deduced from the sense in the vertical portions or the sense of the shear V © 2002 The McGraw-Hill Companies, Inc All rights reserved - 18 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shearing Stresses in Thin-Walled Members • For a wide-flange beam, the shear flow increases symmetrically from zero at A and A’, reaches a maximum at C and the decreases to zero at E and E’ • The continuity of the variation in q and the merging of q from section branches suggests an analogy to fluid flow © 2002 The McGraw-Hill Companies, Inc All rights reserved - 19 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Plastic Deformations I c • Recall: M Y = σ Y = maximum elastic moment • For M = PL < MY , the normal stress does not exceed the yield stress anywhere along the beam • For PL > MY , yield is initiated at B and B’ For an elastoplastic material, the half-thickness of the elastic core is found from ⎛ yY2 ⎞ Px = M Y ⎜1 − ⎟ ⎜ 3c ⎟ ⎝ ⎠ • The section becomes fully plastic (yY = 0) at the wall when PL = M Y = M p • Maximum load which the beam can support is Pmax = © 2002 The McGraw-Hill Companies, Inc All rights reserved Mp L - 20 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Plastic Deformations • Preceding discussion was based on normal stresses only • Consider horizontal shear force on an element within the plastic zone, ∆H = −(σ C − σ D )dA = −(σ Y − σ Y )dA = Therefore, the shear stress is zero in the plastic zone • Shear load is carried by the elastic core, P ⎛⎜ τ xy = 1− A′ ⎜⎝ τ max = y ⎞⎟ where A′ = 2byY 2⎟ yY ⎠ 3P A′ • As A’ decreases, τmax increases and may exceed τY © 2002 The McGraw-Hill Companies, Inc All rights reserved - 21 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 6.3 SOLUTION: • For the shaded area, Q = (4.31in )(0.770 in )(4.815 in ) = 15.98 in • The shear stress at a, Knowing that the vertical shear is 50 kips in a W10x68 rolled-steel beam, determine the horizontal shearing stress in the top flange at the point a © 2002 The McGraw-Hill Companies, Inc All rights reserved ( ) ) VQ (50 kips ) 15.98 in τ= = It 394 in (0.770 in ) ( τ = 2.63 ksi - 22 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Unsymmetric Loading of Thin-Walled Members • Beam loaded in a vertical plane of symmetry deforms in the symmetry plane without twisting σx = − My I τ ave = VQ It • Beam without a vertical plane of symmetry bends and twists under loading σx = − © 2002 The McGraw-Hill Companies, Inc All rights reserved My I τ ave ≠ VQ It - 23 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Unsymmetric Loading of Thin-Walled Members • If the shear load is applied such that the beam does not twist, then the shear stress distribution satisfies D VQ τ ave = V = ∫ q ds It B B E A D F = ∫ q ds = − ∫ q ds = − F ′ • F and F’ indicate a couple Fh and the need for the application of a torque as well as the shear load F h = Ve • When the force P is applied at a distance e to the left of the web centerline, the member bends in a vertical plane without twisting © 2002 The McGraw-Hill Companies, Inc All rights reserved - 24 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 6.05 • Determine the location for the shear center of the channel section with b = in., h = in., and t = 0.15 in e= Fh I • where b b VQ Vb h F = ∫ q ds = ∫ ds = ∫ st ds I0 0 I Vthb = 4I ⎡1 3 ⎛ h⎞ ⎤ I = I web + I flange = th + ⎢ bt + bt ⎜ ⎟ ⎥ 12 ⎝ ⎠ ⎥⎦ ⎢⎣12 th (6b + h ) ≅ 12 • Combining, e= b h 2+ 3b = in in 2+ 3(4 in.) © 2002 The McGraw-Hill Companies, Inc All rights reserved e = 1.6 in - 25 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 6.06 • Determine the shear stress distribution for V = 2.5 kips τ= q VQ = t It • Shearing stresses in the flanges, VQ V h Vh = (st ) = s It It 2I Vhb 6Vb = τB = th (6b + h ) th(6b + h ) τ= (12 ) = 6(2.5 kips )(4 in ) = 2.22 ksi (0.15 in )(6 in )(6 × in + in ) • Shearing stress in the web, ( ) VQ V ht (4b + h ) 3V (4b + h ) = = τ max = It th (6b + h )t 2th(6b + h ) 12 = © 2002 The McGraw-Hill Companies, Inc All rights reserved 3(2.5 kips )(4 × in + in ) = 3.06 ksi 2(0.15 in )(6 in )(6 × in + in ) - 26 ... Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shearing Stresses in Beams and Thin-Walled Members Introduction Shear on the Horizontal Face of a Beam Element Example 6.01 Determination of. .. Pxy I • Shearing stresses are independent of the distance from the point of application of the load • Normal strains and normal stresses are unaffected by the shearing stresses • From Saint-Venant’s... the Shearing Stress in a Beam Shearing Stresses τxy in Common Types of Beams Further Discussion of the Distribution of Stresses in a Sample Problem 6.2 Longitudinal Shear on a Beam Element of