Download - CHAPTER 4- Beams Design(5)
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CHAPTER 4: 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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Figure 5-1 Flow chart of a beam designJan - May 2016 8
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The d/t ratio also needs to be checked due to thestrength of the web will be governed by itsresistance to shear buckling if it is too high.
Provided in cl. 4.2.3. for a rolled section, the webshould be checked for shear buckling when d/t exceeds 70√(275/py) where the d is the web depthclear of the fillets.
Average shear stress on web = shear force/webarea = 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 long
cantilever. What is the maximum ultimate load that can beapplied at the end of the cantilever if the steel yields at a
stress of 275 N/mm2?
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Solution
Mult = y x S
= 275 x 396/103
= 108.9 kNm
Mult = 3 x F 3F = 108.9
F = 36.3 kN
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A single span beam is simply supported between two columns and carries
a reinforced concrete slab as shown in Figure below. Using the design load indicated, determine the dimensions of suitable standard universal beam.
Assume S275 steel.
Example 2
13.7 kN/m
6m
a) Universal Beam SectionJan - May 2016 16
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Solution :
a) Design a suitable universal beam section
Solution
13.72kN/m
A B
6 m
Step 1: Determine reaction at support
Reaction R A = R B= (13.72 x 6)/2 = 41.2 kNMmax = wL
2 /8 = 13.72 x 62 /8 = 61.74 kNm
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Solution :
Step 4 : CalculationA 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 = 2 6 2 c m3, D = 254 mm, t = 5.8 mm)
Average shear stress = 41.2 x 103/ 254 x 5.8 = 28 N/mm2 ( < 165
N/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 yAvFor a rolled UB section Av = tD = 9.7 x 412.8 = 4 004 mm
2
Example 3 : Shear check of a Simply SupportedBeam
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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.
Glue Laminated Timber
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Timber Joists
Small timber beams are often used in the floor construction ofhouses, 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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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 kNmEquating the two values of M:
1.406 = 0.78 L2/8
L = 3.79 m
Allowable span = 3.79 mJan - May 2016 30
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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 loading
shown below. Using the design load indicated, select a suitable section
considering section classification, shear and bending only. Assume S275
steel 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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