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Concept of Shear centre with example

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- 1. SHEAR CENTRE PRESENTED BY: 1. Aiswarya Ray-B120100ME 2. Anirudh Ashok-B120325ME 3. Mahesh M.S.- 4. Harikumar 5. Ashish Ranjan 6. Deepesh
- 2. NEED FOR FINDING LOCATION OF SHEAR CENTRE In unsymmetrical sections, if the external applied forces act through the centroid of the section, then in addition to bending, twisting is also produced. To avoid twisting, and cause only bending, it is necessary for the forces to act through the particular point, which may not coincide with the centroid. The position of the this point is a function only of the geometry of the beam section. It is termed as shear center.
- 3. WHAT IS SHEAR CENTRE?? Shear center is defined as the point on the beam section where load is applied and no twisting is produced. - At shear center, resultant of internal forces passes. - On symmetrical sections, shear center is the center of gravity of that section. - Such sections in which there is a sliding problem, we place loads at the shear center.
- 4. 4 PROPERTIES OF SHEAR CENTRE 1) The shear center lays on the axis of symmetry. 2) Thus for twice symmetrical section the shear centre is the point of symmetry axes intersection. 3) If the cross section is composed of segments converging in a single point, this point is the shear centre. 4) The transverse force applied at the shear centre does not lead to the torsion in thin walled-beam. 5) The shear centre is the centre of rotation for a section of thin walled beam subjected to pure shear. 6) The shear center is a position of shear flows resultant force, if the thin-walled beam is subjected to pure shear.
- 5. DETERMINING LOCATION OF SHEAR CENTRE
- 6. SHEAR STRESSES IN THIN-WALLED OPEN SECTIONS Consider a beam having a thin-walled open section as shown above. Now consider an element of length Δx at section x as shown. Now for Force equilibrium in x-direction, 휏푠푥 푡푠Δx - 0∫s 휎푥tds + 0∫s (σx + 휕휎푥 휕x Δx) tds = 0
- 7. ie. 휏푠푥= – 1 푡푠 0∫s 휕σx 휕x tds ts → wall thickness at s Observing My = 0, normal stress, σx is given by σx = y퐼푦−푧퐼푦푧 퐼푦푧 2−퐼푦퐼푧 Mz Hence, 휕σx 휕x = y퐼푦−푧퐼푦푧 퐼푦푧 2−퐼푦퐼푧 휕Mz 휕푥 Recalling from strength of materials that 휕Mz 휕푥 = – Vy, and substituting We get, 휏푠푥 = Vy 푡푠 1 퐼푦푧 2−퐼푦퐼푧 푠 y퐼푦 − 푧퐼푦푧 푡푑푠 0 휏푠푥 = Vy 푡푠 1 퐼푦푧 2−퐼푦퐼푧 푠 푦푡 푑푠 − 퐼푦푧 0 [퐼푦 0 푠 푧푡 푑푠] 휏푠푥 = 휏푥푠 = Vy 푡푠 1 퐼푦푧 2−퐼푦퐼푧 [퐼푦푄푧 − 퐼푦푧푄푦]
- 8. 푄푧 , 푄푦 → 1st moments of area about the z and y-axis respectively Since for shear centre twisting caused is zero, The moment due to shear stress = The moment due to the load applied 푉푦푒푧 = Moment of 휏푠푥 about centroid
- 9. Shear Centre in Real Life Situations
- 10. Purlins Construction of Purlins A purlin is any longitudinal, horizontal, structural member in a roof.
- 11. The point of application of load is important , depending on the cross-section of purlin.
- 12. If an unsupported channel section is loaded closer to its shear centre, it'll take more load before buckling than if you put the load over the centre of the channel, the application being that you can get more load out of the same member. Useful in design of thin walled open steel sections as they are weak in resisting torsion.
- 13. SHEAR CENTRE PROBLEM A beam has the cross section composed of thin rectangles as shown in fig. The loads on the beam lie in a plane perpendicular to the axis of symmetry of cross section and so located that the beam does not twist .Bending load cause for any section a vertical shear V. determine the location of shear center.
- 14. Q. (a) Locate the shear center S of the hat section by determining the eccentricity, e. (b) If a vertical shear V =10 kips acts through the shear center of this hat what are the values of the shear stresses τA at the location and direction indicated in Figure.
- 15. Solution: The centroidal principal moment of inertia of the beam section is
- 16. The first moment areas of the locations shown in figure are
- 17. The shear stresses at these locations are τA = 2618 psi
- 18. The moment about point A is the resultant forces are

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