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.