design of timber beam
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Lecture 4
Design of timber beams
Structure II (AR-106 G)
BY- AR. KHURRAM ALI Asst. Professor Gateway College of architecture Sonipat, Haryana
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.
Main beams and secondary beams
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
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.
Contd.
Contd.
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.
Contd.
Contd.
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
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
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
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
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.
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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