3. Slabs
References
1 BRITISH STANDARDS INSTITUTION. BS EN 1992–1–1: Eurocode 2: Design of concrete structures – Part 1–1 General rules and rules for buildings. BSI, 2004.
2 BRITISH STANDARDS INSTITUTION. BS 8110–1: The structural use of concrete – Part 1, Code of practice for design and construction. BSI, 1997.
3 NARAYANAN, R S & BROOKER, O. How to design concrete structures using Eurocode 2: Introduction to Eurocodes. The Concrete Centre, 2005.
4 MOSS, R M & BROOKER, O. How to design concrete structures using Eurocode 2: Flat slabs. The Concrete Centre, 2006.
5 BROOKER, O. How to design concrete structures using Eurocode 2: Getting started. The Concrete Centre, 2005.
6 BRITISH STANDARDS INSTITUTION. BS EN 1992–1–2, Eurocode 2: Design of concrete structures. General rules – structural fire design, BSI 2004.
7 WEBSTER, R & BROOKER, O. How to design concrete structures using Eurocode 2: Deflection calculations. The Concrete Centre, 2006.
8 MOSS, R M & BROOKER, O. How to design concrete structures using Eurocode 2: Beams. The Concrete Centre, 2006.
9 THE INSTITUTION OF STRUCTURAL ENGINEERS/THE INSTITUTION OF CIVIL ENGINEERS. Manual for the design of concrete building structures to Eurocode 2. IStructE/ICE, 2006.
3. Slabs
Designing to Eurocode 2
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How to design concrete structures using Eurocode 2
4. Beams
R Moss
BSc, PhD, DIC, CEng, MICE, MIStructEO Brooker
BEng, CEng, MICE, MIStructEspecialist literature. Rather than giving a minimum cover, the tabular method chapter
in
e Chapter in this publication
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Table 1
Beam design procedure
Table 2
Minimum dimensions and axis distances for beams made with reinforced concrete for fire resistance
Figure 3
Simplified rectangular stress block for concrete up to class C50/60 from Eurocode 2
h d
Stress block and forces z
Fc
Fst
Figure 1
Section through structural member, showing nominal axis distances a and asd
b
a asd
h > b
Standard fire resistance Minimum dimensions (mm)
Possible combinations of a and bminwhere a is the average axis distance and bminis the width of the beam
Simply supported beams Continuous beams
A B C D E F G H
1 This table is taken from BS EN 1992–1–2 Tables 5.5 and 5.6.
2 The axis distance, asd, from the side of the beam to the corner bar should be a +10 mm except where bminis greater than the values in columns C and F.
3 The table is valid only if the detailing requirements (see note 4) are observed and, in normal temperature design, redistribution of bending moments does not exceed 15%.
4 For fire resistance of R90 and above, for a distance of 0.3lefffrom the centre line of each intermediate support, the area of top reinforcement should not be less than the following:
As,req(x) = As,req( 0 )( 1– 2.5 ( x/ leff) ) where:
x is the distance of the section being considered from the centre line of the support.
As,req( 0 ) is the area of reinforcement required for normal temperature design.
As,req(x) is the minimum area of reinforcement required at the section being considered but not less than that required for normal temperature design.
leff is the greater of the effective lengths of the two adjacent spans.
5 For fire resistances R120 – R240, the width of the beam at the first intermediate support should be at least that in column F, if both the following conditions exist:
athere is no fixity at the end support; and
bthe acting shear at normal temperature Vsd> 0.67 VRd,max. Key
a Normally the requirements of BS EN 1992–1–1 will determine the cover.
How to design concrete structures using Eurocode 2
2006
2 0 120 2
Outside scope of this publication Carry out analysis of beam to determine
design moments (M) (see Table 3)
Obtain lever arm z from Table 5 or use
Calculate tension reinforcement required from
Check minimum reinforcement requirements (see Table 6) No compression reinforcement required
Check maximum reinforcement requirements As,max= 0.04 Ac for tension or compression reinforcement outside lap locations Determine K’ from Table 4 or
K’ = 0.60d – 0.18d2– 0.21 where d ≤ 1.0
Compression reinforcement required
Calculate lever arm z from START
Procedure for determining flexural reinforcement
Table 3
Bending moment and shear coefficients for beams
member. It is a nominal (not minimum) dimension, so the designer should ensure that:
a ≥ cnom+ flink + fbar / 2 and asd = a + 10 mm
Table 2 gives the minimum dimensions for beams to meet the standard fire periods.
Flexure
The design procedure for flexural design is given in Figure 2; this includes derived formulae based on the simplified rectangular stress block from Eurocode 2. Table 3 may be used to determine bending moments and shear forces for beams, provided the notes to the table are observed.
Table 4 Values forK’
% redistribution d (redistribution ratio) K’
) *')) )'+)1Z
a Ebfbmbg`s mh)'2.] blghmZk^jnbk^f^gmh_>nkh\h]^+% [nmbl\hglb]^k^]mh[^`hh]ikZ\mb\^' for singly reinforced rectangular sections
Table 6
Minimum percentage of required reinforcement
fck fctm Minimum percentage (0.26 fctm/fyka)
Outer support 25% of span moment 0.45 (G + Q) G^Zkfb]]e^h_^g]liZg )')2)Gl + 0.100 Ql
At first interior support – 0.094 (G + Q) l 0.63 (G + Q)a At middle of interior spans 0.066 Gl + 0.086 Ql
At interior supports – 0.075 (G + Q) l 0.50 (G + Q) Key
a 0.55 (G + Q) may be used adjacent to the interior span.
Notes
1 Redistribution of support moments by 15% has been included.
2 Applicable to 3 or more spans only and where Qk≤ Gk. 3 Minimum span ≥ 0.85 longest span.
4 l is the span, G is the total of the ULS permanent actions, Q is the total of the ULS variable actions.
4. Beams
0.45 (Gd + Qd)
– 0.075 (Gd + Qd)l 0.50 (Gd + Qd) 0.55 (Gd + Qd) may be used adjacent to the interior span
l is the span, Gd is the design value of permanent actions (at ULS) and Qd is the design value of variable actions (at ULS).
0.066 Gdl + 0.086 Qdl – 0.094 (Gd + Qd)l
0.090 Gdl + 0.100 Qdl
0.63 (Gd + Qd)a
It is often recommended in the UK that K’ should be limited to 0.168 to ensure ductile failure.
Figure 5
Procedure for determining vertical shear reinforcement
Determine vEdwhere
vEd= design shear stress v[Ed= VEd/(bwz) = VEd/(0 .9 bwd)]
Yes (cot y = 2.5)
No No
START
Determine the concrete strut capacity vRd, max cot y = 2.5 from Table 7
Redesign section
Determine y from:
Calculate area of shear reinforcement:
Check maximum spacing for vertical shear reinforcement:
sl, max= 0.75 d s =
Asw
y = 0.5 sin-1
0.20 fck (1 – fck/250) vEd
T V
fywdcot y vEdbw Is
vEd< vRd,max coty = 2.5?
Is vEd< vRd, max cot y = 1.0?
(see Table 7)
Yes
Table 7
Minimum and maximum concrete strut capacity in terms of stress fck vRd,max cot y = 2.5 vRd,max cot y = 1.0
20 2.54 3.68
25 3.10 4.50
28 3.43 4.97
30 3.64 5.28
32 3.84 5.58
35 4.15 6.02
40 4.63 6.72
45 5.08 7.38
50 5.51 8.00
Figure 4
Strut inclination method
Longitudinal
reinforcement in tension