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# Shear stresses on beam (MECHANICS OF SOLIDS)

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### Shear stresses on beam (MECHANICS OF SOLIDS)

1. 1. SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING
2. 2. Subject:-Mechanics Of Solids Topic:-Shear stresses on
3. 3. Presented by:- Name  Arvindsai  Dhaval Chavda  Fahim Patel  Navazhushen Patel Enrollment no. 130454106002 130454106001 140453106005 140453106008
4. 4. Topics To Be Covered 1. Shear Force 2. Shear Stresses In Beams 3. Horizontal Shear Stress 4. Derivation Of Formula 5. Shear Stress Distribution Diagram 6. Numericals
5. 5. Shear force Any force which tries to shear-off the member, is termed as shear force. Shear force is an unbalanced force, parallel to the cross-section, mostly vertical, but not always, either the right or left of the section.
6. 6. Shear Stresses To resist the shear force, the element will develop the resisting stresses, Which is known as Shear Stresses(). = = Shear force Cross sectional area S A
7. 7. Example:-  For the given figure if we want to calculate the ..  Then it will be Let shear force be S =S/(bxd) d b S
8. 8. Shear Stresses In Beams  Shear stresses are usually maximum at the neutral axis of a beam (always if the thickness is constant or if thickness at neutral axis is minimum for the cross section, such as for I-beam or T-beam ), but zero at the top and bottom of the cross section as normal stresses are max/min. NA NA NA
9. 9.  When a beam is subjected to a loading, both bending moments, M, and shear forces, V, act on the cross section. Let us consider a beam of rectangular cross section. We can reasonably assume that the shear stresses τ act parallel to the shear force V. v n V z m O b h
10. 10.  Shear stresses on one side of an element are accompanied by shear stresses of equal magnitude acting on perpendicular faces of an element. Thus, there will be horizontal shear stresses between horizontal layers of the beam, as well as, Vertical shear stresses on the vertical cross section. m n
11. 11. Horizontal Shear Stress  Horizontal shear stress occurs due to the variation in bending moment along the length of beam.  Let us assume two sections PP' and QQ', which are 'dx' distance apart, carrying bending moment and shear forces 'M and S' and 'M+ ∆M and S+ ∆S‘ respectively as shown in Fig.
12. 12.  Let us consider an elemental cylinder P"Q" of area 'dA' between section PP' and QQ' . This cylinder is at distance 'y' from neutral axis.
13. 13.   F)(FQ' F be,shallcylinderinforce horizontalunbalancedHence, dAδσσ Q' F dAxσ p' F be,shallQ'and'P'at stressesthesetodueforcesThe xy I dMM δσσ be,will'Q'at stressbendingSimilarly, xy I M σ bewillP"atstressbendingHence,      
14. 14.  This unbalanced horizontal force is resisted by the cylinder along its length in form of shear force. This shear force which acts along the surface of cylinder, parallel to the main axis of beam induces horizontal shear stress in beam. Aaxy I dM dA)(σd(σpF Q FHF  
15. 15. DERIVATION OF FORMULA: SHEAR STRESS DISTRIBUTION ACROSS BEAM SECTION  Let us consider section PP' and QQ' as previous.  Let us determine magnitude of horizontal shear stress at level 'AB' which is at distance YI form neutral axis.  The section above AA' can be assumed to be made up of numbers of elemental cylinder of area 'dA'. Then total unbalance horizontal force at level of' AS' shall be the summation of unbalanced horizontal forces
16. 16. a _ yx I dM FM dA.y 1yy 1yy .X I dM dA.xy 1yy 1yy I dM HF        
17. 17.  Here, y = distance of centroid of area above AB from neutral axis, And a= area of section above AB.  This horizontal shear shall be resisted by shear area ABA'B‘ parallel to the Neutral plane. The horizontal resisting area here distance of centroid of area above AB from neutral axis and a=area of section above AB.  Ah = AB x AA’=b x dx where ‘b’is width of section at AB.
18. 18.  We know that shear force is defined as S=dM/dx  Therefore, horizontal shear stress acting at any level across the cross sections. Ib ya x dx dM dx.b ay I dM x A F forceshearhorizontal resistingshear stressshearHorizontal _ H _ H M H   
19. 19. SHEAR STRESS DISTRIBUTION DIAGRAM 1.Rectangular section 2.Circular section maxNA maxNA
20. 20. 3.Triangular section 4.Hollow circular section h/2 max avg  NA maxNA
21. 21. 5.Hollow Rectangular section 6. “I” section maxNA maxNA
22. 22. 7. “C” section 8. “+” section maxNA maxNA
23. 23. 9. “H” section 10. “T” section maxNA maxNA
24. 24. Numericals
25. 25. Rectangular section sum Example-1: Two wooden pieces of a section 100mm X100mm glued to gather to for m a beam cross section 100mm wide and 200mm deep. If the allowable shear stress at glued joint is 0.3 N/mm2 what is the shear force the section can carry ?
26. 26. KN4FeShear forc KN4N4000F 10061067.66 50810000F 3.0 I.b yFA τ Νοω. mm50 2 100 Υ 2mm000,10100100A joint.gluedabovebeamofpartconsider 2N/mm3.0τ 4mm61067.66 12 3200100 I :Solution             100mm 100mm 100mm wooden piece
27. 27. Circular section sum Example-2: A circular a beam of 100mm Diameter is subjected to a Shear force of 12kN, calculate The value of maximum shear Stress and draw the variation of shear stress along the Depth of the beam.
28. 28. 22.03N/mm 1.531.33 1.33 Now. 21.53N/mm 7853.98 31012 A F 27853.98mm2100 4 π A N3101212kNF τmax τavgτmax τavg :Solution        
29. 29. 22.03N/mmτmax D =100mm NA
30. 30. I section sum Example-3: A rolled steel joist of I section overall 300 mm deep X 100mm wide has flange and web of 10 mm thickness. If permissible shear stress is limited to 100N/mm2, find the value of uniformly distributed load the section can carry over a simply supported span of 6m.  Sketch the shear stress distribution across the section giving value at the point of maximum shear force.
31. 31. 100mm 300mm 10mm 150mm __ Y  10 NA 25.63N/mminm  261.63N/mmavg  2100N/mmmax 
32. 32. mm10b (givan)3N/mm100τ 3mm257500 )7010140()14510110(yA N.A.atmaximumbewillstressshear 4mm61055.64Ixx I 3xxI 2xxI 1xxIxx 4mm61029.1802800 12 328010 I 2xx 4mm61013.2321451100 12 310110 2ahIgI 1xx Solution:             
33. 33. kN/m56.83w kN/mm64.83 6 36.501 l W wu.d.l. )loadtotalkN(36.501W 2 W 68.250 2 W Now, F kN68.250F N6.250679F 1061055.66 257500F 100 I.b yFa τ           
34. 34. Triangular section sum Example-4: A beam of triangular section having base width 150mm and height 200mm is subjected to a shear force of 20kN the value of maximum shear stress and draw shear stress distribution diagram.
35. 35. 2N/mm2 33.15.1 τave5.1τmax 2N/mm33.1 15000 31020 A F τave 2mm15000200150 2 1 bh 2 1 A 20kNFforceshear Solution:         
36. 36. max=2N/mm2 avg=1.33N/mm2 200mm 150mm h/2 2/3.h NA
37. 37. Cross section sum Example-5: FIG Shows a beam cross section subjected to shearing force of 200kN. Determine the shearing stress at neutral axis and at a-a level. Sketch the shear stress distribution across the section. 50mm 100mm100mm 100mm 100mm 100mm 50mm 50mm X
38. 38. 215.48N/mm 50610129.16 1005000310200 I.b yFA τ 50mmb 200kNF 100mm5050y 25000mm10050A :lineb_batstressshear 4mm610129.6 ] 12 3100100 [2] 12 330050 [xxI :Solution             
39. 39. 2N/mm03.5 250610129.16 812500310200 I.b yFA τ 3mm812500 25)50(250100)100(50yA :axisx_xatstressShear 2 mm/N09.3 250 50 15.48X        
40. 40. 5.03N/mm2 15.48N/mm2 3.09N/mm2 100mm100mm 100m m 100mm 100mm 50mm 50mm X 50mm NA
41. 41. Inverted T section sum Example-6: Shows the cross section of a beam which is subjected to a vertical shearing force of 12kN.find the ratio the maximum shear stress to the mean shear stress. 60mm 20mm 60mm 20mm
42. 42. 46 2 3 2 3 xx2xx1 21 2211 2 2 2 1 2 1 mean max 3 mm101.36 30)1200(5012 12 6020 10)(301200 12 2060 III inertiaofMoment 30mm 12001200 501200101200 AA yAyA y 50mm20 2 60 y1200mm6020A 10mmy1200mm2060A AxisNeutralofPosition τ τ :findTo N101212kNs:Given :Solution                 
43. 43. 2.21 τ τ 5 11.029 τ τ stressshearmeantostressshearmaximumofRatio 5MPa 2400 1012 areac/s forceShear τ stressshearMean 11.029MPa 101.3620 2510001050 τ axisneutralaboveareaConsider bI SAy τ axis,neutralatproducedstressshearMaximum mean max mean max 3 mean 6 3 max           
44. 44. max=11.29MPA avg=5 MPA min =2.21 MPA 20mm 60mm 20mm NA
45. 45. L section sum Example-7: An L section 10mm X 2mm show in the fig. is subjected to a shear force F. Find the value Of shear force F if max. shear stress developed is 5N/mm2.
46. 46. 4208.09mm 24)(6.7716 12 382 2ahgI xx2 I 4106.12mm26.77)(920 12 3210 2ahgIxxI 6.77mm 1620 416920 2 a 1 a 2 y 2 a 1 y 1 a y 4mm 2 y 216mm28 2 a 9mm 1 y 220mm210 1 a :Solution                   
47. 47. 2mm 2mm 10mm 10mm 6.77mm __ Y  3.23mm
48. 48. 68.14NF 2314.21 46.11F 5 I.b yFA τ 46.11mm 2 1.23 2)(1.232.232)(10yA N.A.atoccurwillstressshearMaximum 314.21mm 208.09106.12 III 3 4 xx2xx1xx          
49. 49. Tee section sum Example-8: A beam is having and subjected to load as shown in fig. Draw shear stress distribution diagram across the section at point of maximum shear force, indication value at all important points. 100kN A B 3m 3m
50. 50. 100200 200 300 100 25 275mm __ Y
51. 51. K.N502 100F 4mm3750000275)-(350(500x100)xyA :web&flangeofjunctiontheatstressShear 4mm91.02x10 6693.57x106322.91x10 12 2150)-100)x(275x(5003100x(300) 12 275)-(350x(500x100)3500x(100)I mm275 80000 612x10 100)x(300100)x(500 100)x15x(300100)x350x(500 Y XX            2
52. 52. 2 9max 3 2 avg 2 9min N/mm85.1 500x10x02.1 3781250x3)10x(50 I.b yFA τ mm3781250 )5.12x25x100()75x100x500(yA axis,neutralatstressShear mmN/84.1 100 500x367.0 τ to,increasesuddenlywillstressShear mmN/367.0 500x10x02.1 3750000x3)10x(50 I.b yFA τ mm500b      
53. 53. max=1.85N/mm2 avg=1.84N/mm2 min=0.367N/mm2 100 200 200 300 100 25 275mm __ Y  NA
54. 54. I section sum Example-9: Find the shear stress at the junction of the flange and web of an I section shown in fig. If it is subjected to a shear force of 20 kN. 100mm 200mm 20mm 100mm 20
55. 55. 2 2 6 3 3 _ 46 3 xx mm/N57.4915.0x 2 100 = to,increasedsuddenlywillstressShear mm915.0 100x10x36.39 180000x10x20 b.I _ yAF mm18000090x)20x100(yA :webandflangofjunetionatstreesShear mm10x36.39 12 200x100 I N.k20F :Solution     
56. 56. 100mm 200mm 20mm 100mm 20NA 2N/mm0.915τmin  2N/mm915.0avg  2 N/mm75.4max 
57. 57. Rectangular section sum Example-10: A 50mm x l00mm in depth rectangular section of a beam is s/s at the ends with 2m span the beam is loaded with 20 kN point load at o.5m from R.H.S. Calculate the maximum shearing stress in the beam. 20kN R B B 0.5m 2.0m RA A
58. 58. 2N/mm5.4 00.35.1 τave5.1τmax 2N/mm00.3 100x50 31015 A F τave kN51520B B B R kN15R 5.1x202xR at AmomentTaking Solution:         
59. 59. max=4.5N/mm2 50mm 100mm NA