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- 1. Lecture 4 Design of timber beams Structure II (AR-106 G) BY- AR. KHURRAM ALI Asst. Professor Gateway College of architecture Sonipat, Haryana
- 2. Introduction • A timber beam may consist of a single member or may be built up from two or more members, called built up beams. • Timber beams are designed to resist- 1. Maximum bending moment 2. Maximum horizontal shear stress 3. Maximum stress at bearing Note: it is to be insured that maximum deflection in the beam does not exceed the permissible limits.
- 3. Main beams and secondary beams
- 4. Design for maximum bending moment • bending equation and the equation is modified to read as:- M= k˳ x fb x Z Where: • M is Maximum bending moment in beam in N/mm2 • 'k˳' is Form Factor depending on the shape of cross- section. • 'fb' is the permissible bending stress in the extreme fibre of beam in N/mm2
- 5. Contd. • 'Z' is section modulus of the beam in mm³ • But Z = I/y where I is moment of inertia of beam cross-section in mm⁴ • And 'y' is distance in mm from neutral axis to extreme fibre of beam • F or a rectangular cross - section beam, I = (bD³) /12 Where: • 'b' is breadth (width) of beam in mm. Width of beam shall not be less than 50 mm or 150 of the span, whichever is greater • Effective Span of a beam is taken as center to center of bearing. • D is depth of beam in mm and shall not be more than three times of its width without lateral stiffening.
- 6. Contd.
- 7. Contd.
- 8. Check for deflection • Deflection is the vertical displacement of neutral axis of beam. • Permissible deflection(δp)- Generally deflection in case of all beams supporting brittle materials like gypsum ceilings, slates, tiles and asbestos sheets shall not exceed 1/360 of the span.
- 9. Contd.
- 10. Contd.
- 11. Design examples Q.1 The roof of a room having clear dimensions 4.2 x 11.7m is supported on two timber beams equally spaced. Wall thickness is 30cm. Roof covering weighs 2.5 kN/m² and live load is 1.5 kN/m² Use Sal Wood. Solution: Beams are designed for inside location Step 1- Effective span(l) of beam, l = 4200+2x1/2x300 = 4500mm = 4.5m
- 12. Solution Step 2- Load per meter length of beam- a) Dead load Roof covering = 4x1x2.5 = 10kN Self wt. of beam/m = 0.5kN (assumed) b) Live load = 4x1x1.5 = 6kN Total load = 16.5kN/m Step 3- Max BM M = wl2/8 = 41.77kN.m = 41.77x106 N.mm
- 13. Contd. • Step 4- Estimation of beam size (bxD), • For a rectangular section beam- • M = k0.fbxZ • Assume k0 = 1 • fb is the permissible bending stress on extreme fiber • fb = 16.48N/mm2 for Sal wood.(from table) and Z = bD2/6
- 14. Contd. From, M = k˳x fb xZ 41.77x106 = 1 x 16.48 x bD2/6 bD2/6 = 2534321.91 we know that, min width of beam(b min) is greater of i) 50mm ii) span/50 b min = 4500/50 = 90mm assuming b = 150mm (least dimension) 150xD2/6 = 2534321.91 D = 318.39mm but D < 3b hence, choosing size 150mmX350mm
- 15. Contd. Step 5- Check for actual bending stress (fab), As beam is rectangular and has depth > 300mm hence, Form factor, fab = M/(k3xZ actual) = 14.06N/mm2 < 16.48N/mm2 hence, beam is safe from bending moment consideration.
- 16. Exercise Q 1. Design Teak wood timber joist of clear span 5m placed at center to center spacing of 3m in a roof. The bearing at each end is 20cm. The dead load of roof covering is 1.5kN/m² and live load is 3 kN/m² Q2. A Deodar wood beam has a size of 15cm x 40cm. Effective span is 4.3 meter. Calculate the safe central point load the beam can support when bearing at each end is 23cm and beam is used for inside location.

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