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...

CHAPTER MECHANICS OF 6 MATERIALS Shearing Stresses in Beams and Thin- Walled Members MECHANICS OF MATERIALS Introduction Transverse loading applied to beam results in normal and shearing stresses in transverse sections Distribution of normal and shearing stresses satisfies ( from equilibrium) Fx   xdA  0 M x   y xz  z xy dA  0 Fy   xydA  V Fz   xzdA  0 M y   z xdA  0 M z    y x   M When shearing stresses are exerted on vertical faces of an element, equal stresses exerted on horizontal faces Longitudinal shearing stresses must exist in any member subjected to transverse loading 46 22 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 Shear on Horizontal Face of Beam Element Consider prismatic beam Equilibrium of element CDC’D’  Fx  0  H    D   C dA A H  M D  MC  y dA IA Let, Q   y dA A M D  M C  dM x  V x dx H  VQ x I q  H  VQ  shear flow x I 6- 5 MECHANICS OF MATERIALS Shear on Horizontal Face of Beam Element Shear flow, q  H  VQ  shear flow x I where Q   y dA A  first moment of area above y1 I   y2dA A A'  second moment of full cross section Same result found for lower area Q  Q   y dA  0 A1  A2 (first moment of area wrt NA is zero) q  H   VQ  V (Q)  q x I I H   H 6- 6 MECHANICS OF MATERIALS 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 6- 7 MECHANICS OF MATERIALS Example 6.1 SOLUTION: Find horizontal force per unit length or shear flow q on lower surface of upper plank VQ (500N)(120 106 m3) q  I 16.20 10-6 m4 Q  Ay  3704 N m  0.020m  0.100m0.060m Calculate corresponding shear force in each nail for nail spacing of 25 mm  120 106 m3 F  (0.025m)q  (0.025m)(3704 N m I  112 0.020m0.100m3 F  92.6 N  2[ 112 0.100m0.020m3  0.020m  0.100m0.060m2]  16.20 106 m4 6- 8 MECHANICS OF MATERIALS Determination of Shearing Stress Average shearing stress on horizontal face of element is shearing force on horizontal face divided by area of horzontal face  ave  H  q x  VQ x A A I t x    ave  VQ ; tq It 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 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 6- 9 MECHANICS OF MATERIALS Shearing Stresses txy in Common Types of Beams For a narrow rectangular beam, VQ 3 V  y2   xy   1  2  Ib 2 A  c     max  3 V 2A For I beams  ave  VQ It  max  V Aweb 6- 10 MECHANICS OF MATERIALS Example 6.3 SOLUTION: Determine the shear force per unit length along each edge of the upper plank VQ 2500 N64296 mm3  N q   15.6 I 10332 mm 4 mm f  q  7.8 N 2 mm For the upper plank,  edge force per unit length Q  Ay  18mm76 mm47 mm Based on the spacing between nails, determine the shear force in each  64296 mm3 nail For the overall beam cross-section,  N I  1 112 mm4  1 76 mm4 F  f   7.8 44 mm 12 12  mm   10332 mm4 F  343.2 N 6- 16 MECHANICS OF MATERIALS Shearing Stresses in Thin-Walled Members Shear stress assumed constant through thickness t, i.e., due to thinnness our averaging is now accurate/exact Consider I-beam with vertical shear V Longitudinal shear force on element is H  VQ x I Corresponding shear stress is  zx   xz  H  VQ t x It Previously had similar expression for shearing stress web  xy  VQ It NOTE:  xy  0 in the flanges  xz  0 in the web 6- 17 MECHANICS OF MATERIALS Shearing Stresses in Thin-Walled Members The variation of shear flow across the section depends only on the variation of the first moment q   t  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 6- 18 MECHANICS OF MATERIALS Shearing Stresses in Thin-Walled Members For wide-flange beam, shear flow q increases symmetrically from zero at A and A’, reaches a maximum at C and then 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 6- 19 MECHANICS OF MATERIALS Example 6.4 Q  108 mm19.6 mm122.2 mm  258700 mm3 Shear stress at a, Vertical shear is 200 kN in a VQ 200103 N258.7 106 m3  W250x101 rolled-steel beam Find   horizontal shearing stress at a It 16410 m 0.0196 m 6 4   16.1MPa 6- 20