CHAPTER 5: BEAM DESIGN

0 Steel Beam Design

0 Timber Beam Design

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Beam

The most frequently used and possibly the earliest used, structural element is beam.

Loading can be imposed on a beam from one or several of a number of sources , e.g. other secondary beams, columns, walls, floor systems or directly from installed plant or equipments.

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Steel Beam

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Steel BeamDesign Procedure (General)

The general steps in a beam design include: For ultimate limit state:

a) shear capacity – shear force due to the design loading must not exceed the shear capacity, the buckling due to shear also should be considered in some circumstance;

b) moment capacity – bending moments due to the design loading must not exceed the moment capacity. The reduction moment due to high shear and lateral torsional buckling due to insufficient restraint also required appropriate concern;

c) local buckling and bearing – when loads or reactions are applied through the flange to the web, the local resistance of the web should not be exceeded.

For serviceability limit state

a) deflection – the deflection due to design loading should not exceed the limits given in Table 8, BS 5950-1:2001

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0 The design path shown in Figure 5-1 below provides ageneral view of the necessary procedure in designinga beam.

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Figure 5-1 Flow chart of a beam designJan - May 2017 8

The shear force Fv, should not be greater than the shear capacity Pv

given by:

Pv = 0.6yAv

Where Av is

a) rolled I, H and channel sections, load parallel to web: tD

b) welded I-sections, load parallel to web: td

c) any other cases: 0.9A0

A0 is the area of that rectilinear element of the cross-section which has the largest dimension in the direction parallel to the shear force, see Figure 5-2.

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The d/t ratio also needs to be checked due to the strength of the web will be governed by its resistance to shear buckling if it is too high.

Provided in cl. 4.2.3. for a rolled section, the web should be checked for shear buckling when d/texceeds 70√(275/py) where the d is the web depth clear of the fillets.

Average shear stress on web = shear force/web area = V/ (D x t)

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Shear failure in concrete beam Failure in steel beam

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Moment Capacity

Mult = y x S

where Mult = ultimate moment

y = ultimate yield stress

S = plastic section modulus

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Table 1: Universal Steel Beams (Dimension and Properties)

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Example 1

0 A 254 x 146 x 31 kg/m universal beam is used as a 3 m longcantilever. What is the maximum ultimate load that can beapplied at the end of the cantilever if the steel yields at astress of 275 N/mm2?

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Solution

Mult = y x S

= 275 x 395/103

= 108.63kNm

Mult = 3 x F

3F = 108.63

F = 36.3 kN

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A single span beam is simply supported between two columns and carriesa reinforced concrete slab as shown in Figure below. Using the design loadindicated, determine the dimensions of suitable standard universal beam.Assume S275 steel.

Example 2

13.7 kN/m

6m

a) Universal Beam SectionJan - May 2017 16

Solution :

a) Design a suitable universal beam section

Solution

13.72kN/m

AB

6 m

Step 1: Determine reaction at support

Reaction RA = RB= (13.72 x 6)/2 = 41.2 kNMmax = wL2/8 = 13.72 x 62 /8 = 61.74 kNm

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Solution :

Step 2

Draw a shear force diagram

Solution

41.2

41.2

Step 3Draw a bending moment diagram

61.74

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Solution :

Step 4 : Calculation

A typical yield stress for mild steel is 275 N/mm2

S = Mult/ y = 61.74 x 106/ 275 x 103 = 224.5 cm3

Try 254 x 102 x 22 kg/m universal beam

( S = 262 cm3, D = 254 mm, t = 5.8 mm)

Average shear stress = 41.2 x 103/ 254 x 5.8 = 28 N/mm2 ( < 165N/mm2)

Use 254 x 102 x 22 kg/m universal beam

Example 2 : Full Lateral Restraint Beam

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A simply supported 406 x 178 x 74 UB S275 is required tospan 4.5 and to carry an ultimate design load of 40 kN/m.Check the suitability of the section with respect the shear.

Solution :

Section properties : 406 x 178 x 74 UB S275

t= 9.7 mm, D= 412.8 mm, d= 360.4 mm, d/t = 37.2

Design shear force at the end of beam Fv = (40 x 4.5)/2= 90kN

Clause 4.2.3: Pv = 0.6 yAv

For a rolled UB section Av = tD = 9.7 x 412.8 = 4 004 mm2

Example 3 : Shear check of a Simply Supported Beam

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Solution :

Clause 4.2.3 : Pv = 0.6yAv = 0.6x275x4 004/1000

= 661 kN > Fv = 90 kN

Shear is adequate

Example 3 : Shear check of a Simply Supported Beam

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A single span beam is simply supported between two columns andcarries a reinforced concrete slab in addition to the column and loadingshown below. Using the design load indicated, select a suitable sectionconsidering section classification, shear and bending only. Assume S275steel and that dead loads are inclusive of self-weights.

Exercise

41.2 kN

3.17 kN/m

3 m 3 m

Universal Beam Section

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Timber Beam

Glue Laminated Timber

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Timber Beam

Space Beam

• Blocked and securely nailed at frequent

intervals to enable individual member to

act as an integral unit.

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Timber Joists

Small timber beams are often used in the floor construction of houses, where they are known as joists.

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Timber Joists

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TIMBER JOISTS

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Timber Joists

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ExampleFigure below shows a house floor which uses 50 mm wide x 150 mm

deep timber joists spaced at 400 mm centres. If the maximum

permissible stress in the timber is limited to 7.5 N/mm2 , determine the

maximum allowable simply supported span of the floor if it supports the

following loads:

Dead load of floor = 0.45 kN/m2

Imposed load on floor = 1.5 kN/m2

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Solution:

Total load = 0.45 + 1.5 = 1.95 kN/m2

Load/m, w on each joist = 1.95 x 0.4 = 0.78 kN/m

For a simply supported floor joist with a uniformly distributed load:

M = wL2/8

= 0.78 L2/8

For a beam with a rectangular cross-section:

Elastic section modulus Z = BD2/6

= 50 x 1502 /6

= 187 500 mm3

From above M = max x Z = 7.5 x 187 500

= 1 406 000 Nmm

= 1.406 kNm

Equating the two values of M:

1.406 = 0.78 L2/8

L = 3.79 m

Allowable span = 3.79 mJan - May 2017 31

Exercise

32

Question 1

What is the origin of the word ‘Glulam’. Described how Glulam ismanufactured. State one structural advantages that Glulam has overconventional solid timber. State one example of the type of structural form thatcan be constructed from Glulam in buildings.