a beam, viz. b and D (including d), and the area of tension steel Ast required to resist a known factored moment Mu. The material properties fck and fy are generally prescribed/selected on the basis of exposure conditions, availability and economy.
For normal applications, Fe 415 grade steel is used, and either M 20 or M 25 grade concrete is used (for exposures rated ‘severe’, ‘very severe’ and ‘extreme’, the minimum concrete grades specified are M 30, M 35 and M 40 respectively, as shown in Table 5.1). As explained earlier in Section 4.7.3, for under-reinforced sections, the influence of fck on the ultimate moment of resistance MuR is relatively small;
hence, the use of high strength concrete is not beneficial from the point of economy, although it is desirable from the point of durability.
The basic requirement for safety at the ‘ultimate limit state of flexure’ is that the factored moment Mu should not exceed the ultimate moment of resistance MuR, and that the failure at the limit state should be ductile. Accordingly, the design equation for flexure is given by:
uR
u M
M ≤ with xu≤xu,max
This implies that, for singly reinforced beam sections, Eq. 4.65 is applicable, with Mu = MuR: from Eq. 5.6 (or, alternatively from the analysis aids given in Table A.2). The limiting percentage tension reinforcement pt,lim and the corresponding
2 lim , lim bd R ≡Mu
are constants given by Eq. 4.62 and 4.61(a):
⎪⎩
5.5.1 Fixing Dimensions of Rectangular Section
Obviously, there are several combinations of pt, b and d (or D) which can satisfy Eq. 5.6. However, the problem is simplified if the values of b and D are either given (by architectural considerations) or arrived at on some logical basis.
In the case of slabs, b is taken as 1000 mm (as explained in Section 4.8) and d is governed by the limiting l/d ratios for deflection control (refer Section 5.3.2). As suggested in Section 5.4.2, a trial value of d may be assumed as approximately l/25 for simply supported spans, l/32 for continuous spans and l/8 for cantilevers. The overall depth D may be taken as d plus effective cover. The effective cover will be the sum of the clear cover, the diameter of the stirrup and half the bar diameter (in the case of a single layer of tension reinforcement). Assuming a stirrup diameter of 10mm and a bar diameter of 20mm, the effective cover will be in the range of 40 – 95 mm, depending on the exposure condition†.
In the case of beams, it is generally found economical to adopt under−reinforced sections with pt< pt,lim. The value of b may be suitably fixed as 200 mm, 250 mm, 300 mm, etc., and the value of d corresponding to any R ≤ Rlim is given by:
Rb
d = Mu (5.9)
where Mu is the factored moment‡ (in N mm) and R is given by Eq. 5.6 for the chosen value of pt. The minimum value of d corresponding to the limiting case pt = pt,lim is obtained by substituting R = Rlim (given by Eq. 5.8). It is desirable to adopt a value of d which is larger than dmin in order to obtain an under-reinforced section.
The overall depth of the beam may be taken as D > dmin + effective cover, and should be expressed in rounded figures (for ease in formwork construction). Multiples of 50 mm (or 25 mm) are generally adopted in practice. However, as explained earlier, the resulting D /b ratio should neither be excessive nor too small; ideally, it should be in the range 1.5 to 2.0. If the resulting D /b ratio is unacceptable and needs to be modified, this can be achieved by suitably modifying b, recalculating d (using Eq. 5.9) and fixing D.
Having fixed the rounded-off value of D, the correct value of the effective depth d can be obtained (assuming that the reinforcing bars can be accommodated in one layer) as follows:
cover) 2 clear
( tie φ
− φ
−
−
= D d
† effective cover (in mm) may be taken as 40, 50, 65, 70 and 95 respectively for mild, moderate, severe, very severe and extreme conditions of exposure.
‡ This will include the contribution of the self-weight of the flexural member. A conservative estimate of the size of the member may be made at the initial stage, for calculating self−weight. The unit weight of concrete should be taken as 25 kN/m3 [Cl. 19.2.1 of Code; see also Appendix B.1 of this book].
If the bars need to be accommodated in two or more layers, the values of D and d should be fixed accordingly [refer Fig. 5.1].
5.5.2 Determining Area of Tension Steel
At this stage of the design process, b and d are known, and it is desired to determine the required Ast so that the section has an ultimate moment of resistance MuR equal to the factored moment Mu.
From Eq. 4.60, considering Mu= MuR and xu < xu,max, it follows that:
which is a quadratic equation, whose solution gives:
[
ck]
u R f
d
x = 1.2021− 1−4.597 (5.11) where, as mentioned earlier, R≡Mu bd2.
It is possible to calculate (Ast)reqd directly, without having to determine xu/d. By re-arranging Eq 5.6,
which is a quadratic equation, whose solution gives:
[
ck]
The above formula provides a convenient and direct estimate of the area of tension reinforcement in singly reinforced rectangular sections.Alternative: Use of Design Aids
In practice, this is the most widely used method. Expressing the relationship between R≡Mu bd2and pt [Eq. 5.12] in the form of charts or tables for various combinations of fy and fck is relatively simple. These are available in design handbooks such as SP : 16 [Ref. 5.5]. The tabular format is generally more convenient to deal with than the Chart.
Accordingly, Tables A.3(a) and A.3(b) have been developed (based on Eq. 5.12) for M 20, M 25, M 30 and M 35 grades of concrete, each Table covering the three grades of steel [Fe 250, Fe 415 and Fe 500]; these Tables are placed in Appendix A of this book. For a given value of R, and specified values of fy and fck, the desired value of pt can be read off (using linear interpolation for intermediate values).
Converting Area of Steel to Bars
The calculated area of steel (Ast)reqd has to be expressed in terms of bars of specified nominal diameter φ and number (or spacing). Familiarity with the standard bar areas
(Ab =πφ2 4) [Table 5.4] renders this task easy.
Table 5.4 Standard bar areas (Ab=πφ2 4) and mass per metre (kg/m) φ
(mm) 6 8 10 12 14 16 18 20 22 25 28 32 36 Ab
(mm2) 28.3 50.3 78.5 113 154 201 254 314 380 491 616 804 1018
Mass kg/m
.222 .395 .616 .887 1.21 1.58 1.99 2.46 2.98 3.85 4.84 6.31 7.99
For a chosen bar diameter φ, the number of bars required to provide the area of tension steel Ast is given by Ast/Ab, taken as a whole number. Alternatively, for a chosen number of bars, the appropriate bar diameter can be worked out. In some cases, it may be economical to select a combination of two different bar diameters (close to each other) in order to arrive at an area of steel as close as possible to the Ast
calculated. As explained earlier, in the case of slab, the area of steel is expressed in terms of centre-to-centre spacing of bars, given by
reqd st b
reqd A A
s =1000 ( )
The actual spacing provided should be rounded off to the nearest lower multiple of 5 mm or 10 mm.
For convenience, Tables A.5 and A.6 (provided in Appendix A) may be referred to — for a quick selection of bar diameter and number/spacing of bars. The values of bar areas given in Table 5.4 are also obtainable from Table A.5. Table 5.4 also gives the mass per metre length of the bars which may be useful in cost estimation.
The arrangement of bars finally proposed must comply with the Code requirements for placement of flexural reinforcement described in Section 5.2.
5.5.3 Design Check for Strength and Deflection Control
The actual Ast and d provided should be worked out, and it should be ensured that the consequent pt is less than pt,lim (for ductile failure at the ultimate limit state). It is good practice to calculate the actual MuR of the section designed (using Eq. 4.65 or 4.66), and thereby ensure that the actual MuR
≥
Mu.A check on the adequacy of the depth provided for deflection control is also called for in flexural members. In the case of beams, the limiting (l/d) ratio given by Eq. 5.5 is generally more-than-adequately satisfied by singly reinforced sections.
However, in the case of slabs, the criteria for deflection control are generally critical.
In anticipation of this, it is necessary to adopt a suitable value of d at the initial stage of the design itself, as explained in Section 5.5.1.
The section should be suitably redesigned if it is found to be inadequate.
EXAMPLE 5.1
A rectangular reinforced concrete beam, located inside a building in a coastal town, is simply supported on two masonry walls 230 mm thick and 6m apart (centre-to-centre). The beam has to carry, in addition to its own weight, a distributed live load of 10 kN/m and a dead load of 5 kN/m. Design the beam section for maximum moment at midspan. Assume Fe 415 steel.
SOLUTION
The beam is located inside the building, although in a coastal area, and thereby protected against weather, and not directly exposed to ‘coastal environment’†. Hence, according to the Code (Table 3), the exposure condition may be taken as
‘moderate’. The corresponding grade of concrete may be taken as M 25 and the clear cover as 30 mm. This cover will be adequate for normal fire resistance requirement also.
Determining Mu for design
• Assume a trial cross-section b = 250 mm, and D = 600 mm (span/10).
Let d = D – 50 = 550 mm.
∴Effective span (Cl. 22.2 of Code)
⎩⎨
• Distributed load due to self-weight m
† Had the beam been located in the roof, the exposure condition would be ‘severe’. Further, if the structure is located at the seafront (subject to sea water spray), the exposure condition would be ‘very severe’, according to the Code.