Kiến trúc xây dựng - Chương 6: Mechanics of materials

MECHANICS OF MATERIALS Third Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER © 2002 The McGraw-Hill Companies, Inc. All rights reserved. Shearing Stresses in Beams and Thin- Walled Members © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 2 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 3 Introduction ( ) ( ) 00 0 00 =∫ −==∫= =∫=−=∫= =∫ −==∫= xzxzz xyxyy xyxzxxx yMdAF dAzMVdAF dAzyMdAF στ στ ττσ • Distribution of normal and shearing stresses satisfies • Transverse loading applied to a beam results in normal and shearing stresses in transverse sections. • 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 4 Shear on the Horizontal Face of a Beam Element • Consider prismatic beam • For equilibrium of beam element ( ) ∫−=∆ ∑ ∫ −+∆== A CD A DDx dAy I MMH dAHF σσ0 xVx dx dMMM dAyQ CD A ∆=∆=− ∫= • Note, flowshear I VQ x Hq x I VQH ==∆ ∆= ∆=∆ • Substituting, © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 5 Shear on the Horizontal Face of a Beam Element flowshear I VQ x Hq ==∆ ∆= • Shear flow, • where sectioncrossfullofmoment second above area ofmoment first ' 2 1 = ∫= = ∫= +AA A dAyI y dAyQ • Same result found for lower area HH QQ q I QV x Hq ∆−=′∆ = =′+ ′−=′=∆ ′∆=′ axis neutral to respect h moment witfirst 0 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 6 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 7 Example 6.01 ( )( ) ( )( ) ( )( ) ( )( ) 46 2 3 12 1 3 12 1 36 m1020.16 ]m060.0m100.0m020.0 m020.0m100.0[2 m100.0m020.0 m10120 m060.0m100.0m020.0 − − ×= ×+ + = ×= ×= = I yAQ SOLUTION: • Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank. m N3704 m1016.20 )m10120)(N500( 46- 36 = × ×== − I VQq • Calculate the corresponding shear force in each nail for a nail spacing of 25 mm. mNqF 3704)(m025.0()m025.0( == N6.92=F © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 8 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. It VQ xt x I VQ A xq A H ave = ∆ ∆=∆ ∆=∆ ∆=τ • On the upper and lower surfaces of the beam, τyx= 0. It follows that τxy= 0 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 9 Shearing Stresses τxy in Common Types of Beams • For a narrow rectangular beam, A V c y A V Ib VQ xy 2 3 1 2 3 max 2 2 = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −== τ τ • For American Standard (S-beam) and wide-flange (W-beam) beams web ave A V It VQ = = maxτ τ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 10 Further Discussion of the Distribution of Stresses in a Narrow Rectangular Beam ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −= 2 2 1 2 3 c y A P xyτ I Pxy x +=σ • Consider a narrow rectangular cantilever beam subjected to load P at its free end: • 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 11 Sample Problem 6.2 A timber beam is to support the three concentrated loads shown. Knowing that for the grade of timber used, psi120psi1800 == allall τσ determine the minimum required depth d of the beam. SOLUTION: • Develop shear and bending moment diagrams. Identify the maximums. • Determine the beam depth based on allowable normal stress. • Determine the beam depth based on allowable shear stress. • Required beam depth is equal to the larger of the two depths found. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 12 Sample Problem 6.2 SOLUTION: Develop shear and bending moment diagrams. Identify the maximums. inkip90ftkip5.7 kips3 max max ⋅=⋅= = M V © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 13 Sample Problem 6.2 ( ) ( ) 2 2 6 1 2 6 1 3 12 1 in.5833.0 in.5.3 d d db c IS dbI = = == = • Determine the beam depth based on allowable normal stress. ( ) in.26.9 in.5833.0 in.lb1090psi 1800 2 3 max = ⋅×= = d d S M allσ • Determine the beam depth based on allowable shear stress. ( ) in.71.10 in.3.5 lb3000 2 3psi120 2 3 max = = = d d A V allτ • Required beam depth is equal to the larger of the two. in.71.10=d © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 14 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’. ( )∑ ∫ −+∆== a dAHF CDx σσ0 • Except for the differences in integration areas, this is the same result obtained before which led to I VQ x Hqx I VQH =∆ ∆=∆=∆ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 15 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 16 Example 6.04 For the upper plank, ( )( )( ) 3in22.4 .in875.1.in3in.75.0 = =′= yAQ For the overall beam cross-section, ( ) ( ) 4 3 12 13 12 1 in42.27 in3in5.4 = −=I SOLUTION: • Determine the shear force per unit length along each edge of the upper plank. ( )( ) lengthunit per force edge in lb15.46 2 in lb3.92 in27.42 in22.4lb600 4 3 = == === qf I VQq • Based on the spacing between nails, determine the shear force in each nail. ( )in75.1 in lb15.46 ⎟⎠ ⎞⎜⎝ ⎛== AfF lb8.80=F © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 17 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 x I VQH ∆=∆ It VQ xt H xzzx =∆ ∆≈=ττ • The corresponding shear stress is • NOTE: 0≈xyτ 0≈xzτ in the flanges in the web • Previously found a similar expression for the shearing stress in the web It VQ xy =τ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 18 Shearing Stresses in Thin-Walled Members • The variation of shear flow across the section depends only on the variation of the first moment. I VQtq ==τ • 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 19 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 20 Plastic Deformations • The section becomes fully plastic (yY = 0) at the wall when pY MMPL == 2 3 • For PL > MY , yield is initiated at B and B’. For an elastoplastic material, the half-thickness of the elastic core is found from ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −= 2 2 3 11 2 3 c yMPx YY moment elastic maximum == YY c IM σ• Recall: • For M = PL < MY , the normal stress does not exceed the yield stress anywhere along the beam. • Maximum load which the beam can support is L M P p=max © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 21 Plastic Deformations • Preceding discussion was based on normal stresses only • Consider horizontal shear force on an element within the plastic zone, ( ) ( ) 0=−−=−−=∆ dAdAH YYDC σσσσ Therefore, the shear stress is zero in the plastic zone. • Shear load is carried by the elastic core, A P byA y y A P Y Y xy ′= =′⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −′= 2 3 2 where1 2 3 max 2 2 τ τ • As A’ decreases, τmax increases and may exceed τY © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 22 Sample Problem 6.3 SOLUTION: • For the shaded area, ( )( )( ) 3in98.15 in815.4in770.0in31.4 = =Q • The shear stress at a, ( )( )( )( )in770.0in394 in98.15kips50 4 3 == It VQτ ksi63.2=τ 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. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 23 Unsymmetric Loading of Thin-Walled Members • Beam loaded in a vertical plane of symmetry deforms in the symmetry plane without twisting. It VQ I My avex =−= τσ • Beam without a vertical plane of symmetry bends and twists under loading. It VQ I My avex ≠−= τσ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 24 • 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. 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 FdsqdsqFdsqV It VQ E D B A D B ave ′−=∫−=∫=∫==τ • F and F’ indicate a couple Fh and the need for the application of a torque as well as the shear load. VehF = © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 25 Example 6.05 • Determine the location for the shear center of the channel section with b = 4 in., h = 6 in., and t = 0.15 in. I hFe = • where I Vthb dshst I Vds I VQdsqF b bb 4 2 2 0 00 = ∫ ∫==∫= ( )hbth hbtbtthIII flangeweb +≅ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛++=+= 6 212 12 12 12 2 12 1 2 33 • Combining, ( ).in43 .in62 in.4 3 2 + = + = b h be .in6.1=e © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 6 - 26 Example 6.06 • Determine the shear stress distribution for V = 2.5 kips. It VQ t q ==τ • Shearing stresses in the flanges, ( ) ( )( ) ( ) ( )( ) ( )( )( ) ksi22.2in6in46in6in15.0 in4kips5.26 6 6 62 22 2 12 1 =+×= +=+= === hbth Vb hbth Vhb s I Vhhst It V It VQ Bτ τ • Shearing stress in the web,( )( ) ( ) ( ) ( ) ( )( ) ( )( )( ) ksi06.3in6in66in6in15.02 in6in44kips5.23 62 43 6 4 2 12 1 8 1 max =+× +×= + +=+ +== hbth hbV thbth hbhtV It VQτ