Simplified engineering for architecture students

1. CHAPTER 5: BEAM DESIGN
0 Steel Beam Design
0 Timber Beam Design
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2. 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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3. Steel Beam
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6. Steel Beam
Design 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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7. 0 The design path shown in Figure 5-1 below provides a
general view of the necessary procedure in designing
a beam.
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8. Figure 5-1 Flow chart of a beam design
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9.  The shear force Fv, should not be greater than the shear capacity Pv
given by:
Pv = 0.6yAv
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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10.  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/t
exceeds 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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11. Shear failure in concrete beam Failure in steel beam
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12. Moment Capacity
 Mult = y x S
where Mult = ultimate moment
y = ultimate yield stress
S = plastic section modulus
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13. Table 1: Universal Steel Beams (Dimension and Properties)
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14. 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 be
applied at the end of the cantilever if the steel yields at a
stress of 275 N/mm2?
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15. 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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16. 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 Section
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17. Solution :
a) Design a suitable universal beam section
Solution
13.72kN/m
A
B
6 m
Step 1: Determine reaction at support
Reaction RA = RB= (13.72 x 6)/2 = 41.2 kN
Mmax = wL2/8 = 13.72 x 62 /8 = 61.74 kNm
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18. Solution :
Step 2
Draw a shear force diagram
Solution
41.2
41.2
Step 3
Draw a bending moment diagram
61.74
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19. 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 ( < 165
N/mm2)
Use 254 x 102 x 22 kg/m universal beam
Example 2 : Full Lateral Restraint Beam
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20. A simply supported 406 x 178 x 74 UB S275 is required to
span 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= 90
kN
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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21. Solution :
Clause 4.2.3 : Pv = 0.6yAv = 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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22. A single span beam is simply supported between two columns and
carries 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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24. Timber Beam
Glue Laminated Timber
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25. 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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26. Timber Joists
Small timber beams are often used in the floor construction of
houses, where they are known as joists.
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27. Timber Joists
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28. TIMBER JOISTS
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29. Timber Joists
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30. Example
Figure 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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31. 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 m
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32. Exercise
32
Question 1
What is the origin of the word ‘Glulam’. Described how Glulam is
manufactured. State one structural advantages that Glulam has over
conventional solid timber. State one example of the type of structural form that
can be constructed from Glulam in buildings.