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# 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Mβ and the shearing stresses by the shear βVβ .<br />y<br />y<br />ππ₯π¦ππ΄<br />Β <br />M<br />V<br />ππ₯ππ΄<br />Β <br />x<br />x<br />ππ₯π§ππ΄<br />Β <br />Y components: ππ₯π¦ππ΄ = -V<br />Z components: ππ₯π§ππ΄ = 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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 Ξπ» in horizontal face in the direction shown before in x .<br />q is the shear per unit length βshear flowβ .<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 πβ²πΆ and πβ²π· ,a horizontal shearing force Ξπ»<br />β΅πΉπ₯ = 0<br />β΄Ξπ» + (ππ· - ππΆ )dA = 0<br />, dA is the sheared area<br />β΄π=Β πππΌ<br />Β <br />w<br />πβ²πΆ<br />Β <br />πβ²π·<br />Β <br />ππΆΒ ππ΄<br />Β <br />ππ·Β ππ΄<br />Β <br />Ξπ»<br />Β <br />
8. 8. , y = π¦1<br />From the former equation .<br />β΄βπ» = ππ·Β βΒ ππΆπΌπ¦Β ππ΄<br />, π¦Β ππ΄ is the first moment with respect to the neutral axis of the portion located at βyβ .<br />π¦Β ππ΄ = Q<br />, ππ·Β βΒ ππΆ = βπ = (ππππ₯)βπ₯ = Vβπ₯<br />βπ» = ππβπ₯πΌ , q (shear flow) = βπ»βπ₯<br />I is the moment of inertia , Q = A yβ<br />Β <br />
9. 9. 6.3 Determination of the shearing stresses in a beam .<br />πππ£πππππ = βπ»βπ΄ = ππ.βπΌπΌ.π‘.βπ₯<br />πππ£πππππ= πππΌπ‘<br />The average shearing stresses in the horizontal stress (πππ£πππππ) .<br />Β <br />dA<br />βπ»<br />Β <br />βπ₯<br />Β <br />
10. 10. 6.4 Shearing stresses ππ₯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 />ππ₯π¦ = 0.8 % πππ£πππππ .<br />β΄ππ₯π¦ = πππΌπ‘<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 = Ayβ = b(c-y)(π+π¦2) = 12b(π2βπ¦2)<br />, ππ₯π¦ = πππΌπ‘ = πππΌπ= πΒ b(π2βπ¦2)2πΌπ<br />πΌ = 112π(2π)3 = 23ππ3<br />β΄ππ₯π¦ = 34π(π2βπ¦2)ππ3<br />, A = 2bc (total area)<br />ππ₯π¦ = 32ππ΄ (1 -Β π¦2π2)<br />When y = 0<br />ππππ₯ = 32ππ΄<br />Β <br />b<br />C=12h<br />Β <br />π¦<br />Β <br />y<br />h<br />Z<br />
12. 12. For πΌ beam .<br />ππππ₯ = ππ΄π€ππ<br />, π΄π€ππ = tb<br />Β <br />y<br />t<br />b<br />Z<br />
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