MECHANICS OF MATERIALS CHAPTER 6 SHEARING STRESSES IN BEAMS AND THIN-WALLED MEMBERS

Kinh Tế - Quản Lý - Công Nghệ Thông Tin, it, phầm mềm, website, web, mobile app, trí tuệ nhân tạo, blockchain, AI, machine learning - Cơ khí - Vật liệu MECHANICS OF MATERIALS CHAPTER 6 Shearing Stresses in Beams and Thin- Walled Members MECHANICS OF MATERIALS 4 - 26- 2 Introduction     MyMdAF dAzMVdAF dAzyMdAF xzxzz xyxyy xyxzxxx              0 0 00 Distribution of normal and shearing stresses satisfies ( from equilibrium) Transverse loading applied to beam results in normal and shearing stresses in transverse sections. Longitudinal shearing stresses must exist in any member subjected to transverse loading. When shearing stresses are exerted on vertical faces of an element, equal stresses exerted on horizontal faces MECHANICS OF MATERIALS Vertical and Horizontal Shear Stresses 4 - 3 MECHANICS OF MATERIALS Shear Stress in Beams Two beams glued together along horizontal surface When loaded, horizontal shear stress must develop along glued surface in order to prevent sliding between the beams. MECHANICS OF MATERIALS 6- 5 Shear on Horizontal Face of Beam Element Consider prismatic beam Equilibrium of element CDC’D’         A CD A CDx dAy I MM H dAHF    0 xVx dx dM MM dAyQ CD A   Let, flowshear     I VQ x H q x I VQ H MECHANICS OF MATERIALS 6- 6 Shear on Horizontal Face of Beam Element where sectioncrossfullofmomentsecond aboveareaofmomentfirst '''' 2 1      AA A dAyI y dAyQ Same result found for lower area HH q I QV I QV x H q dAyQQ AA            )( zero)isNAwrtareaofmomentfirst( 0 21  flowshear    I VQ x H q Shear flow, MECHANICS OF MATERIALS 6- 7 Example 6.1 Beam made of three planks, nailed together. Spacing between nails is 25 mm. Vertical shear in beam is V = 500 N. Find shear force in each nail. MECHANICS OF MATERIALS 6- 8 Example 6.1             46 2 3 12 1 3 12 1 36 m1020.16 m060.0m100.0m020.0 m020.0m100.02 m100.0m020.0 m10120 m060.0m100.0m020.0          I yAQ SOLUTION: Find horizontal force per unit length or shear flow q on lower surface of upper plank. m N3704 m1016.20 )m10120)(N500( 46- 36      I VQ q Calculate corresponding shear force in each nail for nail spacing of 25 mm. mNqF 3704)(m025.0()m025.0(  N6.92F MECHANICS OF MATERIALS 6- 9 Determination of Shearing Stress Average shearing stress on horizontal face of element is shearing force on horizontal face divided by area of horzontal face. xt x I VQ A xq A H ave           If width of beam is comparable or large relative to depth, the shearing stresses at D’1 and D’2 are significantly higher than at D, i.e., the above averaging is not good. Note averaging is across dimension t (width) which is assumed much less than the depth, so this averaging is allowed. On upper and lower surfaces of beam, tyx= 0. It follows that txy= 0 on upper and lower edges of transverse sections. qt It VQ ave     ; MECHANICS OF MATERIALS 6- 10 Shearing Stresses txy 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             For I beams web ave A V It VQ   max   MECHANICS OF MATERIALS 6- 11 Example 6.2 Timber beam supports three concentrated loads. MPa8.0MPa12  allall   Find minimum required depth d of beam. MECHANICS OF MATERIALS 6- 12 Example 6.2 kNm95.10 kN5.14 max max   M V MECHANICS OF MATERIALS 6- 13 Example 6.2     2 2 6 1 2 6 1 3 12 1 m015.0 m09.0 d d db c I S dbI     Determine depth based on allowable normal stress.   mm246m246.0 m015.0 Nm1095.10 Pa1012 2 3 6 max     d d S M all  Determine depth based on allowable shear stress.   mm322m322.0 m0.09 14500 2 3 Pa108.0 2 3 6 max    d d A V all  Required depth mm322d MECHANICS OF MATERIALS 6- 14 Longitudinal Shear Element of Arbitrary Shape Have examined distribution of vertical components txy on transverse section. Now consider horizontal components txz . So only the integration area is different, hence result same as before, i.e., Will use this for thin walled members also I VQ x H qx I VQ H     Consider element defined by curved surface CDD’C’.     A dAHF CDx  0 MECHANICS OF MATERIALS 6- 15 Example 6.3 Square box beam constructed from four planks. Spacing between nails is 44 mm. Vertical shear force V = 2.5 kN. Find shearing force in each nail. MECHANICS OF MATERIALS 6- 16 Example 6.3 SOLUTION: Determine the shear force per unit length along each edge of the upper plank.    lengthunitperforceedge mm N 8.7 2 mm N 6.15 mm10332 mm64296N2500 4 3    q f I VQ q Based on the spacing between nails, determine the shear force in each nail.  mm44 mm N 8.7        fF N2.343F For the upper plank,     3 mm64296 mm47mm768mm1   yAQ For the overall beam cross-section,     4 4 12 14 12 1 mm10332 mm76mm112  I MECHANICS OF MATERIALS 6- 17 Shearing Stresses in Thin-Walled Members Consider I-beam with vertical shear V. Longitudinal shear force on element is x I VQ H  It VQ xt H xzzx       Correspondi...