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Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi

Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi. Shear Forces-Shear stress Shear flow-Shear center. Introduction. Transverse loading applied to a beam results in normal and shearing stresses in transverse sections. Distribution of normal and shearing stresses satisfies.

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Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi

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  1. Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi ShearForces-Shearstress Shearflow-Shearcenter

  2. Introduction • Transverse loading applied to a beam results in normal and shearing stresses in transverse sections. • Distribution of normal and shearing stresses satisfies • 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.

  3. Substituting, Shearflowon the Horizontal Face of a Beam Element • Consider prismatic beam • For equilibrium of beam element • Note,

  4. where Shearflowon the Horizontal Face of a Beam Element Shear flow,

  5. Shear on the Horizontal Face of a Beam Element • Same result isfound for lower area

  6. ShearStresson the Horizontal Face of a Beam Element • Shear flow, • Shear stress is found by dividing the shear flowqwithbz. • Shear stress

  7. Örnek: Şekildeki yükleme durumu ve kesiti görülen kiriş için; C noktasındaki asal gerilmeleri ve doğrultularını bulunuz. Kesitteki kayma gerilmesi dağılımını gösteriniz. C G 20 20 6 kN 6 kN 40 A B E D 20 0.5 m 0.5 m 1 m 60

  8. Çözüm: KKD - EMD 6 kN 6 kN A B E D 6 kN 6 kN 0.5 m 0.5 m 1 m (+) 6 kN 6 kN (-) 3 kNm (+) (+) (+)

  9. Ağırlık merkezi ve Atalet momenti 20 20 C 40 G 20 60

  10. İç kuvvetler: Birinci moment: D Kayma gerilmesi: Normal gerilme:

  11. İnce Cidarlı Kirişlerde Simetrik Olmayan Yüklemeler: Kesme Kuvveti-Kayma Akımı Kayma Merkezi-Kayma Gerilmesi

  12. İNCE CİDARLI AÇIK KESİTLERDE KAYMA GERİLMELERİ VE KAYMA MERKEZİ Kesitlerdeki iç kuvvetler (normal kuvvetleri):

  13. Denge denklemi: Kesme kuvveti:

  14. b – b kesitindeki ortalama gerilme:

  15. Kayma Gerilmesi: Kayma Akımı:

  16. «U» şeklindeki kesitin kayma merkezi Kesitteki Kayma Gerilmesi Değişimi:

  17. Kanattaki kesme kuvveti: Kayma merkezi:

  18. Kanattaki kayma akımı ve kesme kuvveti:

  19. ÖRNEK: Şekilde görülen profilin boyutları b=100 mm, h=150 mm ve t=3 mm olup, profil P=800 N’luk bir kesme kuvvetine maruz kaldığına göre: Kayma merkezinin yerini bulunuz. Kesit çevresi boyunca kayma gerilmesi dağılımını gösteriniz. P b A B e h O’ O t D E

  20. Çözüm: b=100 mm, h=150 mm, t=3 mm İhmal edilebilir b A B e h O’ O t Veya kısaca D E

  21. AB kolundaki kayma akımını bulmak için s uzunluğundaki bir eleman dikkate alınır. ve s A B Statik momenti: A1 Kayma akımı: D E

  22. AB kolundaki kesme kuvvetini hesaplamak için A’dan B’ye kadar integral almak gerekir. s A B A1 D E

  23. O’ noktasına göre moment alınırsa kayma merkezi b A B şeklinde bulunur. e h V O’ O t D E

  24. Kesit çevresi boyunca kayma gerilmesi dağılımı A-B ve E-D kesitindeki kayma gerilmesi değişimi Statik momenti: ve Kayma gerilmesi denkleminden: şeklinde bulunur. A1 s

  25. Kesit çevresi boyunca kayma gerilmesi dağılımı B-D kesitindeki kayma gerilmesi (maksimum kayma gerilmesi) Maksimum kayma gerilmesi T.E. üzerinde meydana gelir. T.Ü. deki alanın statik momenti Statik momenti: Kayma gerilmesi denkleminden: bulunur.

  26. Örnek: Şekilde kesiti görülen kirişin • Kanatlarda oluşan iç kuvvetleri hesaplayınız. • Kayma merkezinin yerini bulunuz. t1 t2 t =6 mm t1 =4 mm t2 =5 mm h1 =60 mm h2 =40 mm b=50 mm P=800 N P h1 h2 t b

  27. t1 t2 V1 V2 P Çözüm: Denge denklemleri x A O h2 h1 e f b veya

  28. Atalet momentleri Tüm kesitin Atalet momenti t1 t2 V1 V2 P x A O h2 h1 Başlıkların atalet momentleri e f b

  29. Sağ başlıktaki maksimum kayma gerilmesi

  30. Sağ başlıktaki kesme kuvveti Sol başlıktaki kesme kuvveti Kayma merkezinin yeri:

  31. Example:For the channel section, and neglecting stress concentrations, determine the maximum shearing stress caused by a V=800-N vertical shear applied at centroid C of the section, which is located to the right of the center line of the web BD. A B b=100 mm h=150 mm t=3 mm V h C x t D E b

  32. Solution: V V B B B B A A A A V T = = + T C C C C D D D D E E E E

  33. V B A C D E

  34. O B A T C D E The maximum shearing stress

  35. EXAMPLE 6.16 (fromCraig): A pipe conveying fluid over a narrow stream crossing must act as a beamas well as a conduit, as indicated in Fig. 1. Assuming that the ratio of mean diameter to wall thickness satisfiesthe requirement d/t>>1,determine the shear flow and shear stress distributionat a section due to the transverse shear force, V, at that section,Also, determine the maximum shear stress on the cross section, andexpress τmaxin terms of V/A.

  36. Solution:Because this is a circular cross section, it is convenient to selecta FBDthat is defined by radial cutting planes at angleθeither side of the xy-plane, as indicated in Fig. 2. In Fig. 2b there are twosurfaces that have equal shear forces ∆Hwhich serve to balance the netforce (F2 — F1) due to the flexural stresses.Therefore, we get where, from the definition of shear flow

  37. The geometric properties for a thinring and for a sector of a thin ring are shown in Fig. 3. Using the formulasin Fig. 3b, we have, fort<< r,

  38. Review the Solution:As a check on the expression that we obtainedfor q, Eq. (6). we can see if the resultant of this distribution is V. as itis supposed to be. The force on the elemental area highlighted in Fig.4 is The vertical component, as shown on the insert in Fig. 4, is, therefore, So, due to symmetry, we have

  39. Since the distribution looks “reasonable” (the shear flow has its maximumat the neutral axis and is zero on the plane of symmetry), and sinceit gives the correct resultant, we can assume that our answer is correct.

  40. ÖRNEK: Şekilde görülen profil P=12 kN’luk bir kesme kuvvetine maruz kaldığına göre: Kayma merkezinin yerini bulunuz. Kesit çevresi boyunca kayma gerilmesi değişimini gösteriniz.

  41. 10 18 MPa

  42. The longitudinal shear force on the element is • The corresponding shear stress is • Previously found a similar expression for the shearing stress in the web • NOTE: in the flanges in the web Shearing Stresses in Thin-Walled Members • Consider a segment of a wide-flange beam subjected to the vertical shear V.

  43. The variation of shear flow across the section depends only on the variation of the first moment. Shearing Stresses in Thin-Walled Members • 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.

  44. Sample Problem 6.3 Knowing that the vertical shear is 50 kips in a W10x68 rolled-steel beam, determine the horizontal shearing stressesin the top flange at the pointsaandC.

  45. SOLUTION: • First moment for the shaded area, • The shear stress at a,

  46. The shear stress at C, • First moment for the areaover point C,

  47. Beam loaded in a vertical plane of symmetry deforms in the symmetry plane without twisting. • Beam without a vertical plane of symmetry bends and twists under loading. Unsymmetric Loading of Thin-Walled Members

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