Shearing stresses in Beams & Thin-walled Members .

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Shearing stresses in Beams & Thin-walled Members .

  1. 1. Shearing stresses in Beams & Thin-walled Members .<br />CH. 6<br />
  2. 2. 6.1 Transverse loading applied to abeam will result in normal and shearing stresses in any given transverse section of the beam .The normal stresses are created by the bending couple β€œM” and the shearing stresses by the shear β€œV” .<br />y<br />y<br />𝜏π‘₯𝑦𝑑𝐴<br />Β <br />M<br />V<br />𝜎π‘₯𝑑𝐴<br />Β <br />x<br />x<br />𝜏π‘₯𝑧𝑑𝐴<br />Β <br />Y components: 𝜏π‘₯𝑦𝑑𝐴 = -V<br />Z components: 𝜏π‘₯𝑧𝑑𝐴 = 0<br />Β <br />z<br />z<br />
  3. 3. The first of these equations shows that vertical shearing stresses must exit in a transverse section of a beam under transverse loading .<br />The second equation shows that the average horizontal shear stresses in the section = 0<br />So ,we conclude that shearing stresses in the horizontal plan = 0<br />πœΒ π‘‘π΄ = 0 (in the horizontal plan) .<br />Β <br />
  4. 4. A force P is applied .<br />So ,planks are observed to slide with respect to each other .<br />M is applied ,no shear happens ,no slide planks .<br />(a)<br />P<br />(b)<br />M<br />
  5. 5. We call shearing force Δ𝐻 in horizontal face in the direction shown before in x .<br />q is the shear per unit length β€œshear flow” .<br />Β <br />
  6. 6. 6.2 Shear in the horizontal face of a beam element .<br />If we took an element<br />w<br />y<br />P1<br />P2<br />z<br />x<br />y<br />Ξ”π‘₯<br />Β <br />c<br />y1<br />y1<br />z<br />
  7. 7. Vertical shearing forces 𝑉′𝐢 and 𝑉′𝐷 ,a horizontal shearing force Δ𝐻<br />∡𝐹π‘₯ = 0<br />βˆ΄Ξ”π» + (𝜎𝐷 - 𝜎𝐢 )dA = 0<br />, dA is the sheared area<br />∴𝜎=Β π‘€π‘ŒπΌ<br />Β <br />w<br />𝑉′𝐢<br />Β <br />𝑉′𝐷<br />Β <br />πœŽπΆΒ π‘‘π΄<br />Β <br />πœŽπ·Β π‘‘π΄<br />Β <br />Δ𝐻<br />Β <br />
  8. 8. , y = 𝑦1<br />From the former equation .<br />βˆ΄βˆ†π» = π‘€π·Β βˆ’Β π‘€πΆπΌπ‘¦Β π‘‘π΄<br />, 𝑦 𝑑𝐴 is the first moment with respect to the neutral axis of the portion located at β€œy” .<br />𝑦 𝑑𝐴 = Q<br />, π‘€π·Β βˆ’Β π‘€πΆ = βˆ†π‘€ = (𝑑𝑀𝑑π‘₯)βˆ†π‘₯ = Vβˆ†π‘₯<br />βˆ†π» = π‘‰π‘„βˆ†π‘₯𝐼 , q (shear flow) = βˆ†π»βˆ†π‘₯<br />I is the moment of inertia , Q = A y’<br />Β <br />
  9. 9. 6.3 Determination of the shearing stresses in a beam .<br />πœπ‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ = βˆ†π»βˆ†π΄ = 𝑉𝑄.βˆ†πΌπΌ.𝑑.βˆ†π‘₯<br />πœπ‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’= 𝑉𝑄𝐼𝑑<br />The average shearing stresses in the horizontal stress (πœπ‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’) .<br />Β <br />dA<br />βˆ†π»<br />Β <br />βˆ†π‘₯<br />Β <br />
  10. 10. 6.4 Shearing stresses 𝜏π‘₯𝑦 in common types of beams .<br />In common types at which b≀14h<br />Where b is the width of the beams , h is the depth .<br />𝜏π‘₯𝑦 = 0.8 % πœπ‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ .<br />∴𝜏π‘₯𝑦 = 𝑉𝑄𝐼𝑑<br /> t = L<br />Β <br />h<br />b or t<br />
  11. 11. For a rectangular cross section area ,shear stresses in x-y plane (horizontal plane) .<br />First we get Q (the 1st moment of the section) .<br />Q = Ay’ = b(c-y)(𝑐+𝑦2) = 12b(𝑐2βˆ’π‘¦2)<br />, 𝜏π‘₯𝑦 = 𝑉𝑄𝐼𝑑 = 𝑉𝑄𝐼𝑏= 𝑉 b(𝑐2βˆ’π‘¦2)2𝐼𝑏<br />𝐼 = 112𝑏(2𝑐)3 = 23𝑏𝑐3<br />∴𝜏π‘₯𝑦 = 34𝑉(𝑐2βˆ’π‘¦2)𝑏𝑐3<br />, A = 2bc (total area)<br />𝜏π‘₯𝑦 = 32𝑉𝐴 (1 - 𝑦2𝑐2)<br />When y = 0<br />πœπ‘šπ‘Žπ‘₯ = 32𝑉𝐴<br />Β <br />b<br />C=12h<br />Β <br />𝑦<br />Β <br />y<br />h<br />Z<br />
  12. 12. For 𝐼 beam .<br />π‘π‘šπ‘Žπ‘₯ = 𝑉𝐴𝑀𝑒𝑏<br />, 𝐴𝑀𝑒𝑏 = tb<br />Β <br />y<br />t<br />b<br />Z<br />

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