Design of singly reinforced rectangular sections There are two design approaches for singly reinforced rectangular sections: free design

and restricted design.

In a free design the applied moment is given together with the material properties of the section. The designer selects a value for the steel ratio (pt), based on which the

dimensions b and d can be determined.

In a restricted design, b and D are specified, as well as the material properties. The design process leads to the required steel ratio.

In both approaches, the requirements of the Standard for reinforcement spacing and concrete cover for durability and fire resistance, if applicable, must be complied with. Details of such requirements may be found in Section 1.4.

3.4.1 Free design

In a free design there is no restriction on the dimensions of the concrete section. Equation 3.3(10) may be rewritten as,

Mu = ptfs ybd2(1− ξo) Equation 3.4(1)

where ξo = ptfs y 2α2fc

Equation 3.4(1)a Using Equation 1.1(1) gives,

where M∗is the action effect that results from the most critical load combination as discussed in Section 1.3.3.

Substituting Equation 3.4(1) into 3.4(2) leads to

bd2= M ∗ φpt fs y(1− ξo) Equation 3.4(3) Let R= d/b, then d =3 R M∗ φptfs y(1− ξo) Equation 3.4(4) The upper limit for ptin design (i.e. pall) is given in Equation 3.3(6) and according to

Clause 8.1.6.1 of the Standard, Mu ≥ 1.2Mcr, which is deemed to be the case if the

minimum steel that is provided for rectangular sections is

pt,min ≥ 0.20(D/d)2f ’ct/fs y Equation 3.4(5)

For economy, ptshould be about 2/3pall(Darval and Brown 1976). A new survey

may be conducted to check current practice.

With a value for ptchosen and the desired R value set, the effective depth of the

section can be determined using Equation 3.4(4). Note that the design moment M∗is a function of the self-weight, amongst other variables. This leads to an iterative process in computing d.

As there is no restriction on D, Astcan be readily accommodated. Load

combination requirements may lead to both the maximum (positive) and minimum (negative) values of M∗to be designed at a given section of the beam. The absolute maximum value should be used to determine d. Then b= d/R. These values of b and d must be adopted for a prismatic beam. Thus, for the other sections of the same beam, where M∗(positive or negative) is smaller than the absolute maximum, the design becomes a restricted one. There are other situations where a restricted design is necessary or specified.

3.4.2 Restricted design

In a restricted design, M∗, fc, fsy, b and D are given, while ptis to be determined either

by solving Equation 3.4(3) or by pt = ξ − ξ2₋ 2ξ M∗ φbd2_{fs y} Equation 3.4(6) where ξ = α2fc fs y Equation 3.4(7)

Chapter 3 Ultimate strength analysis and design for bending 33

The design procedure may be summarised in the following steps:

(i) Assume the number of layers of steel bars with adequate cover and spacing. This yields the value for d.

(ii) With M∗, fc, fsy, b and d in hand, compute ptusing Equation 3.4(6). This

equation will give an imaginary value if b and d are inadequate to resist M∗. If this should occur, a doubly reinforced section will be required (see Section 3.5). At this stage,φ as per Equation 3.3(20)a is an unknown, thereby requiring a trial and error process for determining pt.

(iii) Ensure that

pt,min ≤ pt ≤ pall Equation 3.4(8)

where pt,minand pallare defined in Equations 3.4(5) and 3.3(6), respectively.

(iv) Compute Ast= ptbd and select the bar group (using Table 2.2(1)), which gives an

area greater than but closest to Ast.

(v) Check that b is adequate for accommodating the number of bars in every layer. Choose a different bar group or arrange the bars in more layers as necessary. Also ensure that kuo≤ 0.36 as stipulated in Clause 8.1.5 of the Standard.

(vi) A final check should be carried out to ensure that

φMu≥ M∗ _{Equation 3.4(9)}

where Muis computed using Equation 3.4(1) for the section designed.

3.4.3 Design example

Problem

Using the relevant clauses of AS 3600-2009, design a simply supported beam of 6 m span to carry a live load of 3 kN/m and a superimposed dead load of 2 kN/m plus self-weight. Given that fc= 32 MPa, fsy= 500 MPa for 500N bars, the maximum

aggregate size a= 20 mm, the stirrups are made up of R10 bars, and exposure classification A2 applies.

Solution

Live load moment

Mq=wl

2

8 =

3× 62

8 = 13.5 kNm

Superimposed dead load moment

MSG= 2× 6

2

8 = 9 kNm

Take b× D = 150 × 300 mm and assume pt= 1.4% (by volume). Then

Equation 2.3(1):ρw= 24 + 0.6 × 1.4 = 24.84 kN/m3 Thus, self-weight= 0.15 × 0.30 × 24.84 = 1.118 kN/m The moment due to self-weight is

MSW=1.118 × 6

2

8 = 5.031 kNm

then

Equation 1.3(2): M∗= 1.2 Mg+ 1.5 Mqor M∗= 1.2 × 14.031 + 1.5 × 13.5 =

37.09 kNm

Equation 3.2(2)a:α2= 1.0 − 0.003 × 32 = 0.904 but 0.67 ≤ α2≤ 0.85, thereforeα2= 0.85

Equation 3.2(2)b:γ =1.05 − 0.007 × 32 = 0.826

Equation 2.1(2): f_{ct. f} = 0.6 ×√32= 3.394 MPa

Adopting N20 bars as the main reinforcement in one layer with 25 mm cover gives,

d= D – cover – diameter of stirrup – db/2= 300 – 25 – 10 – 20/2 = 255 mm

Equation 3.4(5): pt,min = 0.20  300 255 2 3.394 500 = 0.00191 Equation 3.3(6): pall= 0.4 × 0.85 × 0.826 × 32 500 = 0.01797 Say use pt = 2

3pall = 0.01198 > pt,minthis is acceptable. Then Equation 3.4(3): bd2₌ M∗ φptfs y  1− 1 2α2 × pt× fs y f_c  Equation 3.3(8): ku = 0.01198 × 500 0.85 × 0.826 × 32 = 0.267

For a single layer of bars, we have d= do. Thus, kuo= ku= 0.267. Then from

Equation 3.3(20)a with b,φ = 0.8. Thus Equation 3.4(3): 150 d2= 37.09 × 10 6 0.8 × 0.01198 × 500 ×  1− 1 2× 0.85× 0.01198 × 500 32  from which d= 240.80 mm Finally, Ast= ptbd= 0.01198 × 150 × 240.80 = 432.72 mm2

From Table 2.2(1), there are three options: (i) 2 N20: Ast= 620 mm2

(ii) 3 N16: Ast= 600 mm2

(iii) 4 N12: Ast= 440 mm2

Taking option (i), we have 2 N20 bars (see Figure 3.4(1)) and Table 1.4(2) gives a cover c= 25 mm to stirrups at top and bottom. Hence the cover to the main bars

35 2N20 10 35 20 35 20 40 25

Chapter 3 Ultimate strength analysis and design for bending 35

= c + diameter of stirrup = 25 + 10 = 35 mm, and d = D – cover to main bars –

db/2= 300 – 35 – 20/2 = 255 mm > 240.80 mm; therefore this is acceptable (see Figure 3.4(1)).

Item (a) in Table 1.4(4) specifies a minimum spacing sminof [25, db, 1.5a]max.Thus

smin= [25, 20, 30]max = 30 mm.

The available spacing= 150 − 2 × 35 − 2 × 20 = 40 mm > smin= 30 mm;

therefore, this is acceptable (see Figure 3.4(1)).

Note also that since kuo= 0.267 < 0.36, the design is acceptable without providing

any compression reinforcement (see Section 3.3.1).

Taking option (ii), we have 3 N16 bars as shown in Figure 3.4(2).

35 3N16 10 35 16 35 16 25 16 16

Figure 3.4(2) Checking accommodation for 3 N16 bars

The available spacing= (150 − 2 × 35 − 3 × 16)/2 = 16 mm < smin= 30 mm.

To provide a spacing of 30 mm would require b> 150 mm (Figure 3.4(2)); therefore, option (ii) is not acceptable.

Option (iii) is similarly unacceptable. Thus, option (i) should be adopted, but noting the following qualifications:

(a) Option (i) is slightly over-designed (i.e. d= 255 mm is about 5.9% higher than required and Ast= 620 mm2is 43.3% higher than necessary).

(b) The percentage of steel by volume for the section is [620/(150 × 300)] × 100 = 1.38%≈ 1.4% as assumed in self-weight calculation; hence this is acceptable. (c) If the beam is to be used repeatedly or frequently, a closer and more economical

design could be obtained by having a second or third trial, assuming different

b× D.

(d) If design for fire resistance is specified, ensure that concrete cover of 25 mm is adequate by checking Section 5 of AS 3600-2009.

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