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- 1. Shearing stresses in Beams & Thin-walled Members .<br />CH. 6<br />
- 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 βMβ and the shearing stresses by the shear βVβ .<br />y<br />y<br />ππ₯π¦ππ΄<br />Β <br />M<br />V<br />ππ₯ππ΄<br />Β <br />x<br />x<br />ππ₯π§ππ΄<br />Β <br />Y components: ππ₯π¦ππ΄ = -V<br />Z components: ππ₯π§ππ΄ = 0<br />Β <br />z<br />z<br />
- 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 />πΒ ππ΄ = 0 (in the horizontal plan) .<br />Β <br />
- 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. We call shearing force Ξπ» in horizontal face in the direction shown before in x .<br />q is the shear per unit length βshear flowβ .<br />Β <br />
- 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 />Ξπ₯<br />Β <br />c<br />y1<br />y1<br />z<br />
- 7. Vertical shearing forces πβ²πΆ and πβ²π· ,a horizontal shearing force Ξπ»<br />β΅πΉπ₯ = 0<br />β΄Ξπ» + (ππ· - ππΆ )dA = 0<br />, dA is the sheared area<br />β΄π=Β πππΌ<br />Β <br />w<br />πβ²πΆ<br />Β <br />πβ²π·<br />Β <br />ππΆΒ ππ΄<br />Β <br />ππ·Β ππ΄<br />Β <br />Ξπ»<br />Β <br />
- 8. , y = π¦1<br />From the former equation .<br />β΄βπ» = ππ·Β βΒ ππΆπΌπ¦Β ππ΄<br />, π¦Β ππ΄ is the first moment with respect to the neutral axis of the portion located at βyβ .<br />π¦Β ππ΄ = Q<br />, ππ·Β βΒ ππΆ = βπ = (ππππ₯)βπ₯ = Vβπ₯<br />βπ» = ππβπ₯πΌ , q (shear flow) = βπ»βπ₯<br />I is the moment of inertia , Q = A yβ<br />Β <br />
- 9. 6.3 Determination of the shearing stresses in a beam .<br />πππ£πππππ = βπ»βπ΄ = ππ.βπΌπΌ.π‘.βπ₯<br />πππ£πππππ= πππΌπ‘<br />The average shearing stresses in the horizontal stress (πππ£πππππ) .<br />Β <br />dA<br />βπ»<br />Β <br />βπ₯<br />Β <br />
- 10. 6.4 Shearing stresses ππ₯π¦ in common types of beams .<br />In common types at which bβ€14h<br />Where b is the width of the beams , h is the depth .<br />ππ₯π¦ = 0.8 % πππ£πππππ .<br />β΄ππ₯π¦ = πππΌπ‘<br /> t = L<br />Β <br />h<br />b or t<br />
- 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 = Ayβ = b(c-y)(π+π¦2) = 12b(π2βπ¦2)<br />, ππ₯π¦ = πππΌπ‘ = πππΌπ= πΒ b(π2βπ¦2)2πΌπ<br />πΌ = 112π(2π)3 = 23ππ3<br />β΄ππ₯π¦ = 34π(π2βπ¦2)ππ3<br />, A = 2bc (total area)<br />ππ₯π¦ = 32ππ΄ (1 -Β π¦2π2)<br />When y = 0<br />ππππ₯ = 32ππ΄<br />Β <br />b<br />C=12h<br />Β <br />π¦<br />Β <br />y<br />h<br />Z<br />
- 12. For πΌ beam .<br />ππππ₯ = ππ΄π€ππ<br />, π΄π€ππ = tb<br />Β <br />y<br />t<br />b<br />Z<br />

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