1. Shearing stresses in Beams & Thin-walled Members . CH. 6

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 “M” and the shearing stresses by the shear “V” . y y 𝜏𝑥𝑦𝑑𝐴 M V 𝜎𝑥𝑑𝐴 x x 𝜏𝑥𝑧𝑑𝐴 Y components: 𝜏𝑥𝑦𝑑𝐴 = -V Z components: 𝜏𝑥𝑧𝑑𝐴 = 0 z z

3. The first of these equations shows that vertical shearing stresses must exit in a transverse section of a beam under transverse loading . The second equation shows that the average horizontal shear stresses in the section = 0 So ,we conclude that shearing stresses in the horizontal plan = 0 𝜏 𝑑𝐴 = 0 (in the horizontal plan) .

4. A force P is applied . So ,planks are observed to slide with respect to each other . M is applied ,no shear happens ,no slide planks . (a) P (b) M

5. We call shearing force Δ𝐻 in horizontal face in the direction shown before in x . q is the shear per unit length “shear flow” .

6. 6.2 Shear in the horizontal face of a beam element . If we took an element w y P1 P2 z x y Δ𝑥 c y1 y1 z

7. Vertical shearing forces 𝑉′𝐶 and 𝑉′𝐷 ,a horizontal shearing force Δ𝐻 ∵𝐹𝑥 = 0 ∴Δ𝐻 + (𝜎𝐷 - 𝜎𝐶 )dA = 0 , dA is the sheared area ∴𝜎= 𝑀𝑌𝐼 w 𝑉′𝐶 𝑉′𝐷 𝜎𝐶 𝑑𝐴 𝜎𝐷 𝑑𝐴 Δ𝐻

8. , y = 𝑦1 From the former equation . ∴∆𝐻 = 𝑀𝐷 − 𝑀𝐶𝐼𝑦 𝑑𝐴 , 𝑦 𝑑𝐴 is the first moment with respect to the neutral axis of the portion located at “y” . 𝑦 𝑑𝐴 = Q , 𝑀𝐷 − 𝑀𝐶 = ∆𝑀 = (𝑑𝑀𝑑𝑥)∆𝑥 = V∆𝑥 ∆𝐻 = 𝑉𝑄∆𝑥𝐼 , q (shear flow) = ∆𝐻∆𝑥 I is the moment of inertia , Q = A y’

9. 6.3 Determination of the shearing stresses in a beam . 𝜏𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = ∆𝐻∆𝐴 = 𝑉𝑄.∆𝐼𝐼.𝑡.∆𝑥 𝜏𝑎𝑣𝑒𝑟𝑎𝑔𝑒= 𝑉𝑄𝐼𝑡 The average shearing stresses in the horizontal stress (𝜏𝑎𝑣𝑒𝑟𝑎𝑔𝑒) . dA ∆𝐻 ∆𝑥

10. 6.4 Shearing stresses 𝜏𝑥𝑦 in common types of beams . In common types at which b≤14h Where b is the width of the beams , h is the depth . 𝜏𝑥𝑦 = 0.8 % 𝜏𝑎𝑣𝑒𝑟𝑎𝑔𝑒 . ∴𝜏𝑥𝑦 = 𝑉𝑄𝐼𝑡 t = L h b or t

11. For a rectangular cross section area ,shear stresses in x-y plane (horizontal plane) . First we get Q (the 1st moment of the section) . Q = Ay’ = b(c-y)(𝑐+𝑦2) = 12b(𝑐2−𝑦2) , 𝜏𝑥𝑦 = 𝑉𝑄𝐼𝑡 = 𝑉𝑄𝐼𝑏= 𝑉 b(𝑐2−𝑦2)2𝐼𝑏 𝐼 = 112𝑏(2𝑐)3 = 23𝑏𝑐3 ∴𝜏𝑥𝑦 = 34𝑉(𝑐2−𝑦2)𝑏𝑐3 , A = 2bc (total area) 𝜏𝑥𝑦 = 32𝑉𝐴 (1 - 𝑦2𝑐2) When y = 0 𝜏𝑚𝑎𝑥 = 32𝑉𝐴 b C=12h 𝑦 y h Z

12. For 𝐼 beam . 𝑍𝑚𝑎𝑥 = 𝑉𝐴𝑤𝑒𝑏 , 𝐴𝑤𝑒𝑏 = tb y t b Z