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Eccentric Loading In Welded Connections
- 1. NAME EMROLLMENT NO. PATELPRASHANT 141100106078 DESHMUKH BHAVIK 151103106002 KANSARAABHISHEK 151103106007 PATEL NIRMAL 151103106012 GUIDED BY :Asst. Prof. Sunil Jagania Sir
- 2. Welded Connections inEccentric loading
- 3. CONTENT Introduction Types of EccentricallyLoaded Connections Eccentric load causing twisting moment. Eccentric load causing bending moment. Examples
- 4. Eccentrically Loaded Connections Generally the structural members are subjected to the axial loading which is acting on the central vertical axis of the member. But sometimes it is possibility that the load acting on the members is not particularly on its axis but a far distance from its centre. That distance is considered as the Eccentric Distance and the load acting at that particular distance apart from its axis is defined as Eccentric Load.
- 6. • The weldedjoints subjected to eccentric load are two types:- Eccentric Load Causing Twisting Moment Eccentric Load Causing Bending Moment
- 7. (a) Eccentric LoadCausing Twisting Moment The centre of gravity of the weld lies in the plane of line of action of of the applied load.
- 8. Load Pwill cause direct shear and twisting moment in the weld. G = Centroid of weld x̄ and ȳ are the co-ordinates of centroid of the weld B is the critical point on the weld, x and y are the co-ordinates of the critical weld point, r = 𝑥2 + 𝑦2 Stress due to the direct shear, Pd = P/lw
- 9. Stress dueto the torsion, Pt = T * r / Ip Where, T = Twisting Moment r = distance of critical point from G Ip = Polar M.I. of weld Resultant Stress, R = 𝑃𝑑2 + 𝑃𝑡2 + 2 𝑃𝑑 𝑃𝑡 cos 𝜃 weld resistance = tt * fwd R = tt * fwd
- 10. (a) Eccentric LoadCausing Bending Moment The centre of gravity of the weld does not lies in the plane of line of action of the applied load. The load will cause direct shear and bending moment in the weld.
- 11. 1. Fillet Welds: •Direct Shear Stress = Load / Effective area of the weld Q = P / 2 * lw *tt • Bending Stress = Moment / Section Modulus fa = M / Z = M / I * y • Resultant Stress, 𝑓𝑎2 + 𝑞2
- 12. 2. Groove Welds: •Direct Shear Stress, Q = P / d * t Where, D = Depth of plate T = Thickness of Plate • Bending Stress = Moment / Section Modulus fa = M / Z = M / I * y • Resultant Stress, fe = 𝑓𝑎2 + 3𝑞2
- 13. Examples 1. A bracketplate is attached to the flange of column ISMB 250 * 5 mm fillet weld as shown in figure. Find the maximum factored load W carried by the bracket. Assume site welding and steel Fe 410.
- 14. The loadwill cause direct shear and tension due to bending. Assume throat thickness equal to unity. Direct shear stress = Load / Effective area of weld q = W / (200 * 2* 1) = 2.5 * 10−3 W N/mm Moment = M = W * 150 N/mm Izz = 2 * 1*(200³ / 12) = 1.33 * 106 𝑚𝑚4 y = 200 / 2 = 100 mm Bending Stress = M * y / I = W * 150 * 100 / 1.33 * 106 = 0.0113 W N/mm
- 15. Vector sumof stress, fe = 𝑓𝑎2 + 𝑞2 = (0.0113 𝑊 )2 + ( 2.5 ∗ 10−3 𝑊)2 = 0.0115 W N/mm Strength of 5 mm fillet weld = tt * fwd = 0.70 * 5 * 158 (fwd = 158 N/𝑚𝑚2 for the field weld) = 553 N/mm Now, 0.0115 W = 553 W = 48086 N W = 48.086 kN
- 16. 2. Determine thelength of a fillet weld required to carry 180 kN load as shown in figure. Use 8 mm shop weld.
- 17. Factored load= 180 kN For the shop weld fwd = 189 N/mm2 Size of weld = S = 8 mm tt = 0.7 * 8 = 5.6 mm lw = 6 𝑀/ 2 ∗ tt * fwd = 6 ∗ 18 ∗ 106 / 2 * 5.6 * 189 = 225.87 mm Adopt s length of weld = lw = 300 mm
- 18. Direct ShearStress, q = P / (2 * lw * tt ) = 180 * 103 / (2 * 300 * 5.6) = 53.53 N / 𝑚𝑚2 Bending Stress, fb = 6M / (2 * lw * tt ) = (6 * 18 * 106 ) / (2 * 5.6 * 3002 ) = 107.14 N / 𝑚𝑚2 Resultant Stress, fe = 𝑓𝑏2 + 𝑞2 = (114.28 )2 + (47.62)2 = 123.80 N / 𝑚𝑚2 < 189 N / 𝑚𝑚2 therefore O.K. Hence, 300 mm length of fillet weld is used 8 mm is sufficient.
- 19.