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### fletched beam

1. 1. Lecture 6 Design of Fletched beam Structure II (AR-106 G) BY- AR. KHURRAM ALI Asst. Professor Gateway College of architecture Sonipat, Haryana
2. 2. Introduction • Composite beams obtained by bolting of steel plate or other steel sections to timber beams are known as fletched beam. • These beams have more load carrying capacity than pure timber beams, hence suitable for large spans or where there are heavy loads.
3. 3. Types of fletched beam
4. 4. Moment of Resistance of Fletched beams 1. Steel plate fixed to sides- in this case, we assume that steel and timber will strain together. or,strain in steel = strain in timber or,(fb/E)steel = (fb/E)timber or, fbtimber = (Et x fbsteel)/ Es hence, MR of fletched beam = (fb.Z)steel + (fb.Z)timber MR = [(fbsteel).(2).(txD2/6) + (fbtimber).(bxD2/6)]
5. 5. Contd. 2. Steel plate fixed to top and bottom- in this case, MR will be given by the relation, MR = fbsteel x [{b(D+2t)2/6} – {(b – t1)D2/6}] Where, t1 = [b(Et/Es)] = B.M= Equivalent width of timber in terms of steel m = modular ratio = Et/Es
6. 6. Contd. 3. Steel plate sand witched (symmetrically) between wooden sections- in that case, MR will be given by the relation, MR = [{(fbtimber).(m.b1.d1 3 + b.d3)}/6d] where, b1 = width of steel plate d1 = depth of steel plate
7. 7. Contd. 4. Steel plate fixed on the tension side only- MR will be given by the relation, MR = fbsteel x [{b(D+t)2/6} – {(b – t1)D2/6}]
8. 8. Numerical • Determine the superimposed UDL the fletched beam shown in figure can carry over an effective span of 5m. Assume inside location, unit weight of steel plate is 7.85kN/m3 and Es = 2x104kN/cm2 and permissible stress in bending in steel is 16.5kN/cm2. the timber used in beam is deodar wood.
9. 9. Solution For Deodar wood- fbt = 10.01N/mm2 Et = 9.32kN/mm2 = 9.32x102 kN/mm2 density of deodar wood = 5.346kN/m3 For Steel plate- Es = 2x104kN/cm2 fbs = 16.5x103/102 = 165N/mm2
10. 10. Contd. As steel and timber will stretch together, therefore at extreme fiber strain in steel is equal to strain in timber, (fb/E)steel = (fb/E)timber Max strain at extreme fiber in timber is say fab, fabtimber = Etxfbsteel/Es = 7.69N/mm2 As fab < fbt Safe.
11. 11. Contd. MR of beam = [(fbsteel).(2).(txD2/6) + (fbtimber).(bxD2/6)] = 46.39x106 N.mm = 46.39 kN.m to calculate UDL (w kN/m), Max BM = wl2/8 46.39 = wx25/8 w = 14.845kN/m Superimposed UDL the beam can carry = w – self weight of beam/m = 14.845 – [1x0.15x0.25x5.346 + 2(1x0.1x0.25x7.85)] = 14.25 kN/m