CE116 Notes

Bachelor of Science in Civil Engineering Construction Planning, Estimating Management,

Methods & Safety Engineering

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1. INTRODUCTION

Reinforced Concrete Structures

Concrete is the most important building material, due to its ability to be moulded to take up the shapes required for the various structural forms. It is also very durable and fire resistant.

Structural Elements and Frames

Its utility and versatility is achieved by combining the best features of concrete and steel. Consider some of the widely differing properties of these two materials that are listed below.

Concrete Steel strength in tension poor good strength in compression good good, but slender bars will buckle strength in shear fair good durability good corrodes if unprotected fire resistance good poor suffers rapid loss of strength a high

temperatures

It can be see from this list that the materials are more or less complementary.

The complete building structure can be broken down into the following elements:

Beams horizontal members carrying vertical and/or lateral loads

Slab horizontal plate elements carrying vertical loads Columns vertical members carrying primarily axial load, but

generally also subjected to lateral load and moment Wall vertical plate elements resisting vertical, lateral or

in-plane loads Bases and foundations pads or strips supported directly on the ground that

spread the loads from columns or walls so that they can be supported by the ground without excessive settlement



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Composite Action

The tensile strength of concrete is only about 10 per cent of the compressive strength; therefore, nearly all reinforced concrete structures are not designed to resist any tensile forces, while all tensile forces are designed to be carried by reinforcement, which are transferred by bond between the interface of the two materials. If the bond is not adequate, the reinforcing bars will just slip within the concrete and there will not be a composite action.

In the analysis and design of the composite reinforced concrete section, it is assumed that there is perfect bond, so that the strain in the reinforcement is identical to the strain in the adjacent concrete.

Figure 1.1 Composite action

Figure 1.1 illustrates the behaviour of a simply supported beam subjected to bending and shows the position of steel reinforcement to resist the tensile forces, while the compression forces in the top of the beam are carried by the concrete.

1.4 Stress-Strain Relations

1.4.1 Concrete

Figure 1.2 Stress-strain curve for concrete in compression



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A typical curve for concrete in compression is shown in above fig.1.2. As the load is applied, the ratio between the stresses and strains is approximately linear at first and the concrete behaves almost as an elastic material. Eventually, the curve is no longer linear and the concrete behaves more and more as a plastic material. The ultimate strain for most structural concrete tends to be a constant value of approximately 0.0035, irrespective of the strength of the concrete.

1.4.2 Modulus of Elasticity of Concrete

This is measured for a particular concrete by means of a static test in which a cylinder is loaded to just above one-third of the corresponding control cube stress and then cycled back to zero stress. This removes the effect of initial bedding in and minor stress redistribution in the concrete under load. Load is then reapplied and the behaviour will then be almost linear; the average slope of the line up to the specified stress is taken as the value for EC.

Figure 1.3 Moduli of elasticity of concrete

The magnitude of the modulus of elasticity is required when investigating the deflection and cracking of a structure. When considering short-term effects, member stiffness will be based on the static modulus EC. If long-term effects are being considered such as creep, modified value of EC should be used.



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Short-term modulus of elasticity of concrete 28 day characteristic Static modulus Ec,28 cube strength (kN/mm2) fcu,28 (N/mm2) Typical Mean 25 19 31 25 30 20 32 26 40 22 34 28 50 24 36 30 60 26 38 32 Elastic modulus at an age other than 28 days may be estimated from

Ec,t = Ec,28 (0.4 + 0.6 f cu,t /fcu,28)

1.4.3 Steel

Figure 1.4 Stress-strain curves for reinforcing bars.

Fig.1.4 shows typical stress-strain curves for (a) hot rolled mild steel & high yield steel, and (b) cold worked high yield steel. The hot rolled bars have definite yield point, but the cold worked bar does not have a definite yield point. The specified strength used in design is based on the yield stress for hot rolled mild steel and high yield steel, whereas for cold worked high yield steel, the strength is based on a specified proof stress. A 0.2% proof stress is defined in fig.1.4.



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1.5 Durability

Concrete structures, properly designed and constructed, are long lasting and should required little maintenance. The durability of the concrete is influenced by:

i) the exposure conditions ii) the concrete quality iii) the cover to the reinforcement iv) the width of any cracks

1.6 Specification of Materials

1.6.1 Concrete

The selection of the type of concrete is usually governed by the strength and durability requirements. The concrete strength is assessed by measuring the crushing strength of cubes or cylinders of concrete made from the mix. These are usually cured and tested at 28 days according to standard procedures. A grade 25 concrete has a characteristic cube crushing strength of 25 N/mm.

The table below shows the lowest grade of concrete for use as specified.

GRADE Lowest Grade for Use as Specified

C7 C10

Plain concrete

C15 C20

Reinforced concrete with lightweight aggregate

C25

Reinforced concrete with dense aggregate

C30

Concrete with post-tensioned tendons

C40

Concrete with pre-tensioned tendons

Table 1.1 Lowest grade of concrete for use



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1.6.2 Reinforcing Steel

The characteristic strength of the more common types of reinforcement are shown below.

Designation Nominal Sizes (mm) Specified Characteristic Strength (N/mm)

Hot-rolled mild steel (CS2:1995)

(6), 8, 10, 12, 16, 25, 32, 40 & (50) 250

Hot-rolled high yield (CS2:1995)

Cold-worked High Yield (BS4461)

(6), 8, 10, 12, 16, 25, 32, 40 & (50) 460

Hard drawn steel wire (BS4482) (6), 8, 10 & 12 485

Table 1.2 Nominal Steel with strength

1.7 Concrete Cover

1.7.1 Nominal cover against corrosion

The nominal cover should protect steel against corrosion and fire. The cover to a main bar should not be less than the bar size or in the case of pairs or bundles the size of a single bar of the same cross-sectional area.

The cover depends on the exposure conditions shall be as follows.

Mild concrete is protected against weather Moderate concrete is sheltered from severe rain concrete under water concrete in non-aggressive soil Severe concrete exposed to severe rain or to alternate wetting and drying Very severe concrete exposed to sea water, de-icing salts or corrosive fumes Extreme concrete exposed to abrasive action



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Limiting values for nominal cover are given in the following Table 1.3. Note that the water-to-cement ratio and minimum cement content are specified. Good workmanship is required to ensure that the steel is property placed and that the specified cover is obtained.

Conditions of exposure Nominal cover (mm) Mild 25 20 20 Moderate 35 30 Severe 40 Very severe 50 Extreme

-- -- --

Maximum free water-to-cement ratio 0.65 0.6 0.55 Minimum cement content (kg/m) 275 300 325 Lowest grade of concrete C30 C35 C40 Table 1.3 Nominal cover to all reinforcement including links to meet durability requirements

1.7.2 Cover as fire protection

Nominal cover to all reinforcement shall meet a given fire resistance period for various elements in a building is given in Table 1.4 and in the design code. Minimum dimensions of members from the Code of Practice are shown in following Fig. 1.5. Reference should be made to the complete table and figure in the Hong Kong Code of Practice for Fire Resisting Construction.

Nominal Cover mm Fire Resistance

Beams

Floors

Ribs Hour SS C SS C SS C

Column

1.0 20 20 20 20 20 20 20 1.5 20 20 25 20 35 20 20 2.0 40 30 35 25 45 35 25 SS simply supported, and C continuous Table 1.4 Nominal cover to all reinforcement including links to meet specified FRP



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Fire resistance

(hour)

Min. beam

b

Rib

b

Min. floor thickness

h

Column width full exposed

b

Min. wall thickness

0.4% < p < 1% 1.0 1.5 2.0

200 200 200

125 125 125

95 110 125

200 250 300

120 140 160

Dimensions mm p = area of steel relative to concrete

Figure 1.5 Minimum dimensions for fire resistance



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2. LIMIT STATE DESIGN

Introduction The design of an engineering structure must ensure that i) Under the worst loadings the structure is safe. ii) During the normal working conditions the deformation of the members does not

detract from the appearance, durability or performance of the structure.

Design Methods Three basic methods using factors of safety to achieve safe, workable structures have been developed and they are:

i) The permissible stress method in which ultimate strengths of the materials are divided by a factor of safety to provide design stresses, which are usually within the elastic range. e.g. actual strength = 460N/mm2, applying F.O.S. = 2 using 230N/mm2 for design.

The permissible stress method has proved to be a simple and useful method but it does have some serious inconsistencies. Because it is based on an elastic stress distribution, it is not really applicable to a semi-plastic material such as concrete.

ii) The load factor method in which the working loads are multiplied by a factor of safety. e.g. actual load = 20kN, applying F.O.S. = 2 using 40kN for design.

In load factor method the ultimate strength of the materials should be used in the calculations. As this method does not apply factors of safety to the material stresses, it cannot directly take account of the variability of the material.

iii) The limit state method which multiplies the working loads by partial factors of safety and also divides the materials ultimate strength by further partial factors of safety.

The limit state of design overcomes many of the disadvantages of the above two methods.



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Limit States

The purpose of design is to achieve acceptable probabilities that a structure will not become unfit for its intended use that is, that it will not reach a limit state.

The two principal types of limit state are the Ultimate Limit State and the Serviceability Limit State.

i) Ultimate Limit State

The most important serviceability limit states are:

a) Deflection the appearance or efficiency of any part of the structure must not be adversely affected by deflection.

b) Cracking local damage due to cracking and spalling must not affect the appearance, efficiency or durability of the structure.

c) Durability this must be considered in terms of the proposed life of the structure and its conditions of exposure.

Other limit states that may be reached include:

d) Excessive vibration which may cause discomfort or alarm as well as damage.

e) Fatigue must be considered if cyclic loading is likely.

f) Fire resistance this must be considered in terms of resistance to collapse, flame penetration and heat transfer.

g) Special circumstances any special requirements of the structures which are not covered by any of the more common limit states.



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Characteristic Strength

It is assumed that for a given material, the distribution of strength will be approximately normal, so that a frequency distribution curve of a large number of sample results would be of the form shown in the Fig.2.1 The characteristic strength is taken as that value below which it is unlikely that more than 5% of the results will fall. This is given by

fk = fm 1.64s

where fk = characteristic strength, fm = mean strength, s = standard deviation.

Figure 2.1 Normal frequency distribution of strengths

Characteristic Loads

Ideally, it should be possible to assess loads statistically, that

Characteristic Load = mean load 1.64 Standard Deviations

These characteristic values represent the limits within which at least 90% of values will lie in practice. It is to be expected that not more than 5% will exceed the upper limit and not more than 5% will fall below the lower limit.

* In Hong Kong, design loads should be obtained from the Building (Construction) Regulations.



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Partial Factors of Safety

Partial Factors of Safety for Materials (m)

Design Strength = Characteristic Strength ( fk ) / m

e.g. 460kN/mm2 / 1.15 0.87 fy

Factors to be considered in selecting m

i) The strength of the material in an actual member. e.g. The strength of concrete is easily affected by placing, compacting, curing=> higher m

Steel is a relative consistent material, therefore, using small m

ii) The severity of the limit state being considered.

Higher for ULS Small or even neglected for SLS

Recommended values for m, see table below. (to BS8110)

Material Limit state Concrete Steel

Ultimate Flexure 1.5 1.15 Shear 1.25 1.15 Bond 1.4

Serviceability 1.0 1.0

Table 2.1 Partial factors of safety applied to materials (m)



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Partial Factors of Safety for Loads (f )

Errors may arise due to: -

i) Design assumptions and inaccuracy of calculation ii) Possible unusual load increases iii) Unforeseen stress redistributions iv) Constructional inaccuracies

These are taken into account in design by applying a partial factor safety (f ) on loading, so that,

Design Load = Characteristic Load x (f )

The partial factor of safety (to BS8110) is as shown below.

Ultimate

Concrete Imposed Earth & Water

Wind Serviceability All

Load Combination

(G) (Q) (Q) (W) (GQW) Dead & Imposed (+ Earth & Water)

1.4 (or 1.0)

1.6 (or 0.0) 1.4 - 1.0

Dead & Wind (+ Earth & Water)

1.4 (or 1.0) - 1.4 1.4 1.0

Dead & Imposed & Wind (+ Earth & Water)

1.2 1.2 1.2 1.2 1.0

Table 2.2 Partial factors of safety for loadings

*It should be noted that the Partial Factors of Safety for Loadings in local code ( Code of Practice for Structural Use of Concrete ) uses a different for LL and DL.



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3. ANALYSIS OF THE STRUCTURE

Loading

Dead Load

Dead load includes the weight of the structure, and all architectural components such as exterior cladding, partitions and ceilings. Equipment and static machinery, when permanent fixtures, are considered as part of the dead load.

* The typical value for the self-weight of reinforced concrete is 24 kN/m3

Imposed Load

Imposed loads on buildings are: the weight of its occupants, furniture, or machinery; the pressure of wind, the weight of snow, and of retained earth and water.

Load Combinations

For ULS the loading combination to be considered are as follows:

i) Dead and imposed load 1.4 Gk + 1.6 Qk

ii) Dead and wind load 1.0 Gk + 1.4 Wk

iii) Dead, imposed and wind load 1.2 Gk + 1.2 Qk + 1.2 Wk

Fig. 3.1 shows the arrangement of vertical loading for multi-span continuous beam to cause (i) maximum sagging moment in alternate spans and (ii) maximum moment at support A.

*BS8110 & HK Code of Practice allow the ultimate design moments at the supports to be calculated from one loading condition with all spans fully loaded.



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Figure 3.1 Multi-span beam loading arrangements

Load Combination for SLS

A partial factor of safety f = 1.0 is applied to all load combinations at SLS.

Frame Analysis

Braced Frame

A braced frame is where the sway deflection is reduced substantially by the presence of cross bracing, shear wall or core wall.

Sub-frame analysis

All columns and beam spanning between joints which allow rotation take the full stiffness. Beams having one end fixed take a reduced stiffness of 50% to allow for the remote end of the beam is fixed against rotation. (see fig.3.2)

The structure will be subjected to the following loading cases:-

i) All spans with full loading (1.4Gk + 1.6 Qk) ii) Alternative span full loading and the remainder with 1.0 Gk



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Unbraced Frame

An unbraced frame is where the sway due to the imposition of horizontal loading is not limited (except the inherent stiffness of the columns and beams) and the beam-column connection is required to resist the moment induced by sway.

Sub-frame analysis

Only the two frames shown in fig. 3.2b can be considered, i.e. the sub-frame (viii) and the sub-frame (ix). The sub-frame (ix) is analysed by using 1.4Gk + 1.6 Qk and 1.2Gk + 1.2 Qk. The sub-frame (viii) is analysed by using 1.2 Wk. The most critical loading case for the structural members is obtained by comparing the results of the two loading cases.

i) 1.4Gk + 1.6 Qk ii) 1.2Gk + 1.2 Qk + 1.2 Wk

The sub-frame (viii) is analysed under the loading of and is carried out by assuming point of contraflexure at the mid-spans of all beams and mid-storey heights for all columns.



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Fig. 3.2 Basic frame with typical subframes for both the braced and unbraced case.



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4. ANAYLSIS OF THE SECTION

The most important principles are:-

i) The stresses and strains are related by the material properties, including the stress-strain curves for concrete and steel.

ii) The distribution of strains must be compatible with the distorted shape of the cross-section.

iii) The resultant forces developed by the section must balance the applied loads for static equilibrium.

Stress-Strain Relations

Concrete

The behavior of structural concrete (Fig. 4.1) is represented by a parabolic stress-strain relationship, up to strain o , from which point the strain increases while the stress remains constant.

Figure 4.1 Short term design stress-strain curve for normal-weight concrete



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The ultimate strain = 0.0035 is typical for all grades of concrete. The ultimate design stress is given by:-

0.67 fcu 0.67 fcu m

=

1.5 = 0.447 fcu 0.45 fcu

Where 0.67 allows the difference between the bending strength and cube crushing strength of concrete and m = 1.5 is the usual partial safety factor.

Steel Reinforcement

Fig. 4.2 shows a typical short-term design stress-strain curve for reinforcement. The behaviour of steel is identical in tension and compression.

Figure 4.2 Short term design stress-strain curve for reinforcement

fy Design Yield Stress = m

Stress = Es s

fy Design Yield Strain y = ( m

) / Es

For high-tensile steel (T), fy = 460 N/mm. y = 460/(1.15 200 10 ) = 0.002

For mild steel (R), fy = 250 N/mm y = 250/(1.15 200 10 ) = 0.00109



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Distribution of Strains and Stresses Across a Section

Assumptions:-

i) Concrete cracks in the region of tensile strain ii) After cracking, all tension is carried by the reinforcement. iii) Plane sections remain plane after straining.

The following Fig. 4.3 shown the cross-section of a member subjected to bending, and the resultant stain diagram, together with three types of stress distribution in the concrete.

Figure 4.3 Section with strain diagram and stress blocks

i) The triangular stress distribution applies to the serviceability limit state. (SLS)

ii) The rectangular parabolic stress block represents the distribution at failure when compressive strains are within the plastic range and if is associated with the design for ultimate limit state. (ULS).

iii) The equivalent rectangular stress block is a simplified alternative to the rectangular-parabolic distribution. It is adopted in BS8110.

From the compatibility of stains, d x

st = cc = ( x

)

x d and sc = cc = (

x )

Where d and d are the effective depth & the depth of comp. reinf. Respectively



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d x st By rearranging (1), x = 1 + cc

At ULS, the maximum compressive strain in concrete is 0.0035.

For steel with fy = 460 N/mm, the yield strain is 0.002.

d x st x = 1 + cc

= 0.636 d

Hence to ensure yielding of the tension steel at ULS, x 0.636d

To be very certain of the tension steel yielding, BS8110 Limits the depth of neutral axis so that x (b 0.4)d

Where, moment at the section after redistribution

b = moment at the section before redistribution

Thus with moment redistribution not greater than 10% b 0.9, x 0.5d

Equivalent Rectangular Stress Block

The rectangular stress block shown in Fig. 4.4 may be used in preference to the more rigorous rectangular-parabolic stress block. This simplified stress distribution will facilitate the analysis and provide more manageable design equations.

Note that the stress block does not extend to the neutral axis of the section but has a depth s = 0.9x. This will result in the centroid of the stress block being s/2 = 0.45x from the top edge of the section, which is very nearly the same location as for the more precise rectangular-parabolic stress block; also the areas of the two types of the stress block are approximately equal.



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Figure 4.4 Singly reinforced section with rectangular stress block

Singly Reinforced Rectangular Section in Bending

Fst tensile force in the reinforcing steel Fcc resultant compressive force in concrete

For equilibrium, the ultimate design moment M must be balanced by the moment of resistance of the section.

M = Fcc z = Fst z, z lever arm

Fcc = Stress Area of action = 0.45 Fcu bs and z = d- s/2 => s = 2(d-z)

M = 0.45bs z fcu = 0.45b 2(d z)z fcu = 0.9 fcu b(d z)z

Let K = M / (bd fcu), Eq.(3) becomes, (z/d) - (z/d) + K/0.9 = 0

z = d[0.5 + (0.25-K/0.9)]

Also Fst = (fy / m) As with m = 1.15

M As = 0.87 fy z



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The lever arm z can be found by using formula, table or design chart as shown below.

z = d[0.5 + (0.25-K/0.9)] Formula

K = M/bd2fcu 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.156

la = z/d 0.941 0.928 0.915 0.901 0.887 0.873 0.857 0.842 0.825 0.807 0.789 0.775 Table

Design Chart

The % values on the K axis mark the limits for singly reinforced sections with moment redistribution applied.

The upper limit of the lever-arm curve, z = 0.95d, is specified by BS8110. The lower limit of z = 0.775d is when the depth of neutral axis x = d/2, which is the maximum value allowed by the code for a singly reinforced section in order to provide a ductile section which will have a gradual tension type failure.

With z = 0.775d, M = 0.9 fcu b(d 0.775d) 0.775d

=> M = 0.156 fcu bd The coefficient 0.156 is calculated from the more precise concrete stress.

M When

bd2 fcu = K > 0.156, compression reinforcement is required.

Rectangular Section with Compression reinforcement at ULS



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Figure 4.5 Section with compression reinforcement

When M > 0.156 fcu bd, the design ultimate moment exceeds the moment of resistance of the concrete. compression reinforcement is required.

Z = d s/2 = d 0.9 x/2 = d 0.9 0.5d/2 = 0.775d

For equilibrium, Fst = Fcc + Fsc 0.87 fy As = 0.45 fcu bs + 0.87 fy As and with s = 0.9 d/2 = 0.45d 0.87 fy As = 0.201 fcu bd + 0.87 fy As

Take moment at the centroid of the tension steel As ,

M = Fcc z + Fsc (d d) = 0.201 fcu bd(0.775d) + 0.87 fy As (d d) = 0.156 fcu bd + 0.87 fy As (d d)

M 0.156 fcu bd2 As = 0.87 fy (d d) 0.156 fcu bd2

and As = 0.87 fy z + As with z = 0.775d

the value 0.156 is usually denoted by K

Moment Redistribution



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The moment derived from an elastic analysis may be redistributed based on the assumption that plastic hinges have formed at the sections with largest moments. The formulation of plastic hinges requires relatively large rotations with yielding of the tension reinforcement. To ensure large strains in the tension steel, the code of practice restricts the depth of neutral axis.

X (b - 0.4)d

Where d = effective depth

moment at the section after redistribution b =

moment at the section before redistribution

Depth of stress block, s = 0.9 ((b - 0.4)d and z = d- s/2 = d 0.9(b - 0.4)d/2

Moment of resistance of the concrete in compression, Mc = Fcc z = 0.45fcu bsz

= 0.45 fcu b 0.9(b - 0.4)d[d 0.9(b - 0.4)d/2]

Mc bd2 fcu

= 0.45 0.9(b - 0.4) [1-0.45(b - 0.4)]

= 0.402(b - 0.4)-0.18(b - 0.4)

Mc Let bd2 fcu

= K

K = 0.402(b - 0.4)-0.18(b - 0.4)

when M > Kbdfcu , then compression steel is required such that

(K K) fcu bd2 As = 0.87 fy (d d)

K fcu bd2 and As = 0.87 fy z

+ As

Table 4.1 shows the various design factors associated with the moment redistribution. If the value of d/d for the section exceeds that shown in the table, the compression



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steel will not be yielded and the compressive stress shall be taken as fsc = Essc.

Redistribution (per cent)

b x/d z/d K d/d

10 0.9 0.5 0.775 0.156 0.215 15 0.85 0.45 0.797 0.144 0.193 20 0.8 0.4 0.82 0.132 0.172 25 0.75 0.35 0.842 0.119 0.150 30 0.7 0.3 0.865 0.104 0.129

Table 4.1 Moment redistribution design factors



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5. BOND & ANCHORAGE

Minimum Distance Between Bars i) The recommended distance between bars are given in C1.3.12.11 of BS8110

Part 1, hagg (the maximum size of coarse aggregate). ii) Horizontal Distance between bars hagg + 5mm. iii) Vertical Distance between bars 2 hagg /3 iv) And if the bar size exceeds hagg + 5mm, a spacing of bar size is required.

Minimum Percentages of Reinforcement in Beams, Slabs and Columns Minimum percentage Situation Definition of

percentage fy = 250 N/mm fy = 450 N/mm % %

Tension reinforcement Sections subjected mainly to pure tension Sections subjected to flexure

100As / Ac

0.8

0.45

(a) Flanged beams, web in tension: (1) b W / b < 0.4 100As /bwh 0.32 0.18 (2) b W / b 0.4 100As /bwh 0.24 0.13 (b) Flanged beams, flange in tension over a continuous support:

(1) T-beam 100As /bwh 0.48 0.26 (2) L-beam 100As /bwh 0.36 0.20 (c) Rectangular section (in solid slabs this minimum

should be provided in both directions) 100As /Ac 0.24 0.13

Compression reinforcement (where such reinforcement is required for the ultimate limit state)

General rule 100Asc /Acc 0.4 0.4

Simplified rules for particular cases:

(a) rectangular column or wall 100Asc /Ac 0.4 0.4 (b) flanged beam: (1) flange in compression 100Asc /bht 0.4 0.4 (2) web in compression 100Asc /bwh 0.2 0.2 (c) rectangular beam 100Asc /Ac 0.2 0.2 Transverse reinforcement in flanges of flanges of flanged beams (provided over full effective flange width near top surface to resist horizontal shear)

100Asc /htl

0.15

0.15



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Anchorage Bond

The reinforcing bar subject to tension shown in Fig. 5.1 must be firmly anchored if it not be pulled out of the concrete The anchorage depends on the bond between the bar and the concrete, and the contract area.

Figure 5.1 Anchorage bond

L = Min. anchorage length to prevent pull out. = Bar size or nominal diameter. fbu = Ultimate anchorage bond stress. fs = Direct tensile or compressive stress in the bar.

Consider the forces on the bar, 2

Tensile pull-out force = 4

fs

Anchorage force = ( L ) fbu 2

4 = ( L ) fbu

fs 0.87 fy L = 4fbu

= 4 fbu

The ultimate anchorage bond stress, fbu = (fcu) Values of are given in table 5.2

Table 5.2 Values of bond coefficient - Bar type

Bars in tension Bars in compression Plain bars Type 1 : deformed bars Type 2 : deformed bars Fabric

0.28 0.40 0.50 0.65

0.35 0.50 0.53 0.81

anchorage length L can be written as: L = KA



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Table 5.3 Ultimate anchorage bond lengths and lap lengths as multiples of bar size

Grade 460 Reinforcement Grade 250

plain Plain Deformed

type 1

Deformed

type 2

Fabric

Concrete cube strength 25

Tension anchorage and lap length 39 72 51 41 31

1.4 x tension lap 55 101 71 57 44

2.0 x tension lap 78 143 101 81 62

Compression anchorage 32 58 41 32 25

Compression lap length 39 72 51 40 31



1.4 x tension lap 50 92 64 52 40

2.0 x tension lap 71 131 92 74 57





1.4 x tension lap 46 85 60 48 37

2.0 x tension lap 66 121 85 68 53





1.4 x tension lap 43 80 56 45 35

2.0 x tension lap 62 113 80 64 49



NOTE, The values are rounded up to the whole number and the length derived from these values may

differ slightly from those calculated directly for each bar or wire size.

Furthermore, anchorage length may be provided by hooks and bends in the reinforcement as shown in below figure 5.2.



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Figure 5.2 Anchorage provided by hooks & bends

Lapping of Reinforcement

Rules for lapping are:-

i) The laps should be staggered and be away from sections with high stresses. ii) Tension laps should be equal to at least the design tension anchorage length,

but in certain conditions, it should be increased. a) At top section and with min. cover < 2 (Multiply by 1.4) b) At corners where min. cover to either face < 2 or clear distance between

adjacent laps < 75 mm or 6. (Multiply by 1.4) c) Where both (a) and (b) apply. (Multiply by 2.0)

Figure 5.3 Lapping of reinforcing bars iii) Compression laps should be at least 25% greater than the compression

anchorage length. iv) Lap lengths for unequal size bars may be based on the smaller bar.



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Shear

Shear reinforcement can be in the form of stirrups and inclined bars. However, inclined bars are less frequently used in construction today, due to difficult in actual construction work.

Stirrups

Fig. 5.4 shows an analogous truss in which the longitudinal reinforcement forms the bottom chord, the stirrups are the vertical members and the concrete acts as the diagonal and top chord compression member.

Figure 5.4 Stirrups and the analogous truss

The spacings of the stirrups are equal to the effective depth d.

Let Asv = Cross-sectional area of the two legs of the stirrup. fyv = Characteristic strength of the stirrup reinforcement V = Shear force due to the ultimate load.

Consider section x x 0.87 fyv Asv = V = vbd



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Where v = V/bd is the average shear stress.

If Sv is the actual spacing of stirrups, then 0.87 fyv Asv = vbd(sv/d)

Asv vb sv

= 0.87 fyv

Taken the shear resistance of concrete into account, Asv b(v vc)

sv =

0.87 fyv

where vc is the ultimate shear stress that can be resisted by the concrete. The values of vc is given in table 5.3

Table 5.4 Values of Vc, design concrete shear stress 100 As Effective depth (in mm)

bvd

125 150 175 200 225 250 300 >400

0.15 0.25

0.50 0.75 1.00 1.50 2.00

3.00

N/mm 0.45 0.53 0.67 0.77 0.84 0.97 1.06 1.22

N/mm 0.43 0.51 0.64 0.73 0.81 0.92 1.02 1.16

N/mm 0.41 0.49 0.62 0.71 0.78 0.89 0.98 1.12

N/mm 0.40 0.47 0.60 0.68 0.75 0.86 0.95 1.08

N/mm 0.39 0.46 0.58 0.66 0.73 0.83 0.92 1.05

N/mm 0.38 0.45 0.56 0.65 0.71 0.81 0.89 1.02

N/mm 0.36 0.43 0.54 0.62 0.68 0.78 0.86 0.98

N/mm 0.34 0.40 0.50 0.57 0.63 0.72 0.80 0.91

NOTE 1. Allowance has been made in these figures for a m of 1.25 NOTE 2. The values in the table are derived from the expression: 0.79 (100 As /(bvd)1/3(400/d)1/4/m where 100A Vs /bvd should not be taken as greater than 3: 400/d should not be taken as less than 1. For characteristic concrete strengths greater than 25 N/mm, the values in the table may be multiplied by (fcu/25) 1/3. The value of fcu should not be taken as greater than 40.

The values of Vc increases for shallow members and those with larger percentages of tensile reinforcement.



33

Enhanced Shear Resistance

Within a distance of 2d from a support or a concentrated load, the design concrete shear stress vc may be increased to

vc2d / av

The distance av is measured from the support or concentrate load to the section being designed.

* Average shear stress should never exceed the lesser of 0.8(fcu) or 5 N/mm.



34

6 DESIGN OF REINFORCED CONCRETER BEAM

Preliminary Analysis and Member Sizing

The preliminary analysis need only provide the max. moments and shears in order to ascertain reasonable dimensions. Beam dimension required are:-

i) Cover to reinforcement (c) ii) Breath (b) iii) Effective depth (d) iv) Overall depth (h)

The strength of a beam is affected more by its depth than its breath. A suitable breath may be 1/3 1/2 of the depth.

Suitable dimensions for b and d can be decided by a few trial calculations as follows:-

i) For no compression reinforcement, M/bdfcu 0.156

With compression reinforcement, M/bdfcu 10 / fcu if the area of bending reinforcement

is no to be excessive.

ii) Shear stress v = V / bd and v should never exceed 0.8fcu or 5 N/mm whichever is the lesser. To avoid congested shear reinforcement, v, shall be in about 1/2 of the max. allowable value.

iii) The span-effective depth ratio for span 10m should be within the basic values given below.

Cantilever Beam Simply Supported Beam Continuous Beam

7 20 26 } Basic span- effective depth ratio.

* The basic span-effective depth ratio shall be modified according to M/bd and the service stress in the tension reinforcement.



35

* For span greater than 10m, the basic ratios shall be multiplied by 10/span. iv) The overall depth of the beam is given by

h = d + cover + t where t = estimated distance from the outside of the link to the center of the tension bar.

Figure 6.1 Beam dimensions

Effective Span of Beam

i) Simply supported beam:- the smaller of the distance between the centre of bearings, or the clear distance between support plus the effective depth.

ii) Continuous beam:- the distance between the centre of supports. iii) Cantilever beam:- the length to the face of the support plus 1/2 effective depth,

or the distance to the center of the support if the beam is continuous.

Design for Bending

* An excessive amount of reinforcement indicates that a member is undersized and it may also cause difficulty in fixing the bars and pouring of concrete.

Singly Reinforced Rectangular Section (b 0.9)

Refer to the diagram below.



36

Figure 6.2 Singly reinforced section The design procedures and can be summarized as follows:-

i) Calculate K = M/bdfcu ii) Determine the lever-arm, z, by from formula, table or the design chart below.

z = d[0.5 + (0.25-K/0.9)] Formula

K = M/bd2fcu 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.156

la = z/d 0.941 0.928 0.915 0.901 0.887 0.873 0.857 0.842 0.825 0.807 0.789 0.775 Table

Design Chart

The % values on the K axis mark the limits for singly reinforced sections with moment redistribution

applied.

iii) The area of tension steel is:- M

As = 0.87 fy z iv) Select suitable bar sizes v) Check that the area of steel provided is within the limits required by the code,

i.e.: As /bh 4.0% And As /bh 0.24% (for mild steel) 0.13% (for high yield steel)



37

Example 6.1. Design of Tension Reinforcement for a Rectangular Section

The beam section shown in figure 6.3 has characteristic material strengths of fcu = 45 N/mm for the concreter and fy = 460 N/mm for the steel with R12 stirrups and 30mm concrete cover. The design moment at the ultimate limit state is 250 kN m which causes sagging of the beam

Figure 6.3

Solution: K = M/bdfcu assuming Y32 steel to be used, d = 600 30 12 32/2 = 542 mm K = 250 106 / (300 5422 45) = 0.063 < 0.156 compression steel is not required

by formula, lever arm, la = 0.5 +(0.25 - K/0.9) = 0.5 +(0.25 -0.063/0.9) = 0.942 < 0.95 use 0.942 la = z/d z = 0.942 542 mm = 500.9 mm

Calculating the required tension reinforcement, M

As = 0.87 fy z = 250 106 / (0.87 460 500.9) = 1247 mm2

Provide 2Y32 bars, area = 1608 mm (As% = 0.99%) > 0.13% & < 4% the section is satisfy for resisting the applied moment

Doubly Reinforced Concrete Beam

* Compression steel is required whenever the concrete in compression is unable to develop the necessary moment of resistance.



38

Moment Redistribution Factor b 0.9 and d/d 0.2

Compression reinforcement is required if M > 0.156 fcu bd that is K > 0.156

i) Area of compression steel M 0.156 fcu bd

As = 0.87 fsc (d-d)

ii) Area of tension steel, 0.156 fcu bd

As = 0.87 fy z + As and z = 0.775d

* If d/d > 0.2, the stress in compression steel should be determined by the stress-strain relationship.

Figure 6.4 Beam doubly reinforced to resist a sagging moment

Moment Redistribution Factor b < 0.9

The design procedures and can be summarized as follows:- i) Calculate K = M/bd fcu ii) Calculate K = 0.402 (b 0.4) 0.18 (b 0.4)

If K < K, no compression steel required. If K > K, compression steel required.

iii) Calculate x = (b 0.4)d If d /x < 0.43, the compression steel has yielded and fsc = 0.87 fy If d /x < 0.43, calculate the steel compressive strain sc and then



39

calculate fsc = Es sc iv) Calculate the area of compression steel by

(K - K) fcu bd As = fsc (d-d)

v) Calculate the area of tension steel by,

K fcu bd fsu As = 0.87 fy z

+ As 0.87 fy

where z = d 0.9 x/2

* Links should be provided to give lateral restraint to the outer layer of compression steel according to the following rules.

A) The links should pass round the corner bars and each alternative bar. B) The link size 1/4 of the size of largest compression bar. C) The spacing of links 12 of the smallest compression bars. D) No compression bar should be more than 150 mm from a restrained bar

Example 6.2 Design of Tension and Compression Reinforcement, b > 0.9

The beam section shown in figure 6.5 has characteristic material strengths of fcu = 30 N/mm and fy = 460 N/mm. The ultimate moment is 165 kNm, causing hogging of the beam.

Figure 6.5 Beam doubly reinforced to resist a hogging moment Solution:

K = M/bdfcu K = 160 106 / (230 3302 30) = 0.22 > 0.156 compression steel is required d/d = 50/330 = 0.15 < 0.2



40

fcu = 0.87 fy By formula, required compression steel

M 0.156 fcu bd As

=

0.87 fsc (d-d) 165 106 0.156 30 230 3302

=

0.87 460 (330 - 50) = 427 mm2

and required tension steel 0.156 fcu bd

As = 0.87 fy z + As

0.156 30 230 3302 =

0.87 460 0.775 330 + 427 = 1572 mm2

Provide 2Y20 bars for As, area = 628 mm and 2Y32 bars for As, area = 1610 mm.

Checking of As%, 100As 100 628

bh =

230 390 = 0.70

100As 100 1610 bh

=

230 390 = 1.79

> 0.13% & < 4% the section is satisfy for resisting the applied moment

Example 3 Design of Tension and Compression Reinforcement, b = 0.7

The beam section shown in figure 6.6 has the characteristic material strengths of fcu = 30 N/mm and fy = 460 N/mm. The ultimate moment is 370 kNm causing hogging of the beam.



41

Figure 6.6 Beam doubly reinforced to resist a hogging moment

As the moment reduction factor b = 0.7, the limiting depth of the neutral axis is

x = (b 0.4)d = (0.7 0.4) 540 = 162 mm K = M/bd fcu

= 370 106/(300 5402 30) = 0.141 K = 0.402 (b 0.4) 0.18 (b 0.4) = 0.104 K > K therefore compression steel is required therefore fsc < 0.87 fy

0.0035 (x d) (1) Steel compressive strain sc = X 0.0035 (162 100)

=

162 = 0.00134

(2) Steel compressive stress = Essc = 200 000 0.00134 = 268 N/mm2

(K - K) fcu bd (3) Compression steel As = fsc (d-d) (0.141 0.104) 30 300 5402

=

268 (540 100) K fcu bd fsu (4) Tension Steel As = 0.87 fy z + As 0.87 fy

As = 0.104 30 300 5402 + 823 268



42

0.87 460 (540 0.9 162/2) 0.87 460

Provide 2Y25 bars for As, area = 982 mm and 2Y32 & 1Y25 bars for As area = 2101 mm, which also meet the requirements of the code for steel areas..

T-Beam and L-Beam

Fig. 6.7 shows a T-beam and an L-beam. Where the beams are resisting sagging moment, part of the slab acts as a compression flange and the members may be designed as a T-beam or L-beam.

* With hogging moment, the slab will be in tension and assumed to be crack, the beam must be designed as a rectangular section of width bw and overall depth h.

Figure 6.7 T-beam and L-beam

BS8110 defines effective width as follows:-

i) T-section, the lesser of the actual flange width, or the width of the web + 1/5 of the distance between point of zero moment.

ii) L-section, the lesser of the actual flange width or the width of the web + 1/10 of the distance between point of zero moment.

* As a simple rule, the distance between the points of zero moment may be taken as 0.7 times the effective span for a continuous beam.

* When the N.A. falls within the flange, design the T-beam or L-beam as an equivalent rectangular section of breath bf.

* Transverse reinforcement should be placed across the top flange with an area



43

of not less than 0.15% of the longitudinal cross-section of the flange.

Design Procedures for T or L Beam

i) Calculate K = M/ bfd fcu and determine the lever-arm. ii) If d z < hf /2, the stress block falls within the flanged depth, and the design is

proceeded as for a rectangular section. iii) Provide transverse steel in the top of the flange.

Area = 0.15hf 1000/100 = 1.5hf mm per meter length of the slab.

Curtailment of Bars

In every flexural member every bar should be extend beyond the theoretical cut-off point for a distance equal to greater of :-

i) The effective depth of the member or ii) 12

This applies to compression and tension reinforcement, but reinforcement in the tension zone should satisfy the following additional requirements.

iii) The bars extend a full anchorage bond length beyond the theoretical cut-off point.

iv) At the physical cut-off point, the shear capacity is at least twice the actual shear force.

v) At the physical cut-off point, the actual bending moment is not more than half the moment at theoretical cut-off point.

For example:-

Figure 6.8 Locations of Cut-off Points along a beam



44

For condition (i) and (ii), AB is the greater of d or 12 For condition (iii), AB equals the full anchorage bond length. For condition (iv), At B, actual shear < half the shear capacity. For condition (v), At b, moment < half moment at A.

* Where the loads on a beam are substantially uniformly distributed, simplified rules for curtailment may be used. These rules only apply to continuous beam if the characteristic imposed load does not exceed the characteristic dead load and the spans are equal. Fig. 6.9 shows the rules in a diagrammatic form.

Figure 6.9 Simplified rules for curtailment of bars in beams

Design for Shear

If V is the shear force at a section, then the shear stress v is given by v = V/bd. ( * v must never exceed the lesser of 0.8fcu or 5 N/mm )

Vertical Stirrups

Rules: i) Min. spacing Sv of stirrup should not be less than 80 mm.



45

ii) Max. spacing Sv of stirrup should not exceed 0.75d longitudinally along the span.

iii) At right angles to the span, the spacing of the vertical legs should not exceed d, and all tension bars should be within 150 mm of a vertical leg.

* The choice of steel type is often governed by the fact that mild steel may be bent to a smaller radius than high-yield steel. This is important in narrow members to allow correct positioning of the tension reinforcement.

The size and spacing of the stirrups is given by the following equation.

Where Asv = Area of the legs of a stirrup Sv = Spacing of the stirrup B = Breath of the beam V = V/bd = shear stress vc = The ultimate shear stress of conc. fyv = Characteristic strength of the link reinforcement.

If v is less than vc nominal links must still be provided unless the beam is a very minor one and

The nominal links should be provided such that

Example 6.4 Design of Shear Reinforcement for a Beam

Shear reinforcement is to be designed for the one span beam of example as shown in figures 6.10. The characteristic strength of the mild steel links in fyv = 250 N/mm.

(a) Check maximum shear stress Total load on span F = wu span = 75.2 x 6.0 = 451 kN At face of support

Asv b(v-vc) Sv

0.87fyv

vc v <

2

Asv 0.4b Sv

=

0.87fyv



46

Shear Vs = F/2 - wu support width /2 = 451/2 758.2 0.15= 214 kN

Figure 6.10 Non-continuous beam-shear reinforcement

shear stress vs 214 103

v = bd

=

300 550

= 1.3 N/mm2 < 0.8fcu

(b) Shear links Distance d from face of support shear Vd = Vs - wu d = 214 75.2 0.55 = 173 kN shear stress v = 173 103 / ( 300 550) = 1.05 N/mm

Only 2 Nos. 25 mm bars extend a distanced d past the critical section. Therefore for determining vc

100As 100 982 bd

=

300 550 = 0.59

From table 5.4. vc = 0.56 N/mm Asv b (v vc) 300 (1.05 0.56) Sv

=

0.87fyv =

0.87 250 = 0.68



47

Provide R10 links at 220 mm centers Asv 2 78.5 Sv

=

220 = 0.71

(c) Nominal links for mild steel links

Asv 0.4b 0.4 300 Sv

=

0.87fyv =

0.87 250 = 0.55

Provide R10 links at 280 mm centers

Asv 2 78.5 Sv

=

280 = 0.56

(d) Extent of shear links shear resistance of nominal links + concrete is

Asv Vn = ( Sv 0.87 fyv + bvc ) d

= (0.56 0.87 250 + 300 0.56) 550 = 159 kN

shear reinforcement is required over a distance s given by

Vs - Vn 214 - 159 s =

Wu =

75.2 = 0.73 metres from the face of the support

Number of R10 links at 220 mm required at each end of the beam is l + (s/220) = l + (730/220) = 5

* There is a section at which the shear resistance of the concrete plus the nominal stirrups equals the shear force form the envelope diagram. At this section, the stirrups necessary to resist shear can stop and replaced by the nominal stirrups. The shear resistance Vn of the concrete plus the nominal stirrups is given by



48

Vn = (0.4 + Vc )bd

Design for Torsion

Design procedures consist of determining an additional area of longitudinal and link reinforcement required to resist the torsional shear forces.

i) Determine As and Asv to resist the bending moments and shear forces.

ii) Calculate the torsional shear stress.

T Torsional moment hmin The smaller dimension of the beam section. hmax The larger dimension of the beam section.

iii) If vt > vtmin , in Table 6.1, then torsional reinforcement required. Refer to Table 6.2 for the reinforcement requirements with a combination of torsion and shear stress v.

iv) v + vt vtu (refer to Table 6.1), where v is the shear stress due to shear force. Also, for sections with yl < 550mm

where yl is the larger c/c dimension of link

v) Calculate additional shear reinforcement required from

where xl is the smaller c/c dimension of link.

This value of Asv /sv is added to (i), and a suitable link size and spacing is chosen.

2T vt = h2min (hmax hmin/3)

vtu y1 vt ( 550

Asv T sv

=

0.8x1y1(0.87fyv)



49

* Sv < 200 mm or xl and the link should be of the closed type.

vi) Calculate the additional area of longitudinal steel.

Asv fyv As = sv

( fy ) (x1 + y1)

Asv Where sv

is the value from step (v).

* As should be evenly distributed around the inside perimeter of the links. At least four corner bars should be used and the clear distance between bars should not exceed 300mm.

Figure 6.11 Torsion example

Concrete grade 25 30 40

or more

yl min 0.33 0.37 0.40 vtu 4.00 4.38 5.00 Table 6.1 Ultimate torsion shear stresses (N/mm)



50

vt < vt min vt > vt min

v vc + 0.4 Nominal shear reinforcement. Designed torsion reinforcement only, but not less than nominal shear reinforcement

v > vc + 0.4 Designed shear reinforcement, not torsion reinforcement

Designed shear and torsion reinforcement

Table 6.2 Reinforcement for shear and torsion

Example 5 Design of Torsional Reinforcement

The rectangular section of figure 6.11 resists a bending moment of 170 kN m, a shear of 160 kN and a torsional moment of 10kN m. The characteristic material strengths are fcu = 30N/mm fy = 460 N/mm and fyv = 250 N/mm.

(1) Calculations for bending and shear would give As = 1100mm2

and

Asv sv

= 0.79

(2) Torsional shear stress

(3) 0.56> 0.37 from table 6.1. Therefore torsional reinforcement is required. (4)

V 160 103 v =

bd =

300 450 = 1.19 N/mm2

Therefore v + vs = 1.19 + 0.56 = 1.75 N/mm2

vut from table 6.1 = 4.38 N/mm2, therefore

2T vt = h2min (hmax hmin/3)

2 10 106 vt = 3002 (500 300/3) = 0.56 N/mm

2



51

So that vt < vtu y1 / 500 as required (5)

Therefore

Provide R10 links at 100 centres

(6) Additional longitudinal steel

Asv fyv As = sv

( fy ) (x1 + y1) 250

= 0.55 460

(240 + 440) = 203 mm2

Therefore Total steel area 1100 + 203 = 1303 mm2 Provide the longitudinal steel shown in figure 6.11.

(7) The torsional reinforcement should extend at least hmax beyond where it is required to resist the torsion.

vtu y1 4.38 400 550

=

550 = 3.5

Asv T Additional sv

=

0.8x1y1(0.87fyv) 10.0 106

=

0.8 240 440 0.87 250 = 0.55

Asv Total sv

= 0.79 + 0.55 = 1.34

Asv sv

= 1.57



52

6.10 Checking of Deflection

BS 8110 specifies a set of basic span-effective depth ratios to control deflections which are given in table 6.3 for rectangular sections and for flanged beams with spans less than 10 m. Where the ratios for spans > 10 m are factored by 10/Span.

Rectangular section

Flanged (bw > 0.3b)

Cantilever Simply supported

Continuous

7 20 26

5.6 16.0 20.8

Table 6.3 Basic span-effective depth ratios

The basic ratios given in table 6.3 are modified in particular cases according to (a) The service stress in the tension steel and the value of M/bd in table 6.4, (b) The area of compression steel as in table 6.5

M/bd Reinforcement service stress

(N/mm) 0.50 0.75 1.0 1.5 2.0 3.0 4.0 5.0 6.0

(fy = 250)

(fy = 460)

100 150 156 200 250 288 300

2.0 2.0 2.0 2.0

1.90 1.68 1.60

2.0 2.0 2.0

1.95 1.70 1.50 1.44

2.0 1.98 1.96 1.76 1.55 1.38 1.33

1.86 1.69 1.66 1.51 1.34 1.21 1.16

1.63 1.49 1.47 1.35 1.20 1.09 1.06

1.36 1.25 1.24 1.14 1.04 0.95 0.93

1.19 1.11 1.10 1.02 0.94 0.87 0.85

1.08 1.01 1.00 0.94 0.87 0.82 0.80

1.01 0.94 0.94 0.88 0.82 0.78 0.76

NOTE 1. The values in the table derive from the equation:

(477 fs) Modification factor = 0.55 +

M 2.0

120 ( 0.9 + bd2 ) where M is design ultimate moment at the center of the span or, for a cantilever, at the support

NOTE 2. The design service stress in the tension reinforcement in a member may be estimated

from the equation:

5 fy As, require 1 fs = 8 As, provided

b

NOTE 3. For a continuous beam, if the percentage of redistribution is not known but the design

ultimate moment at mid-span is obviously the same as or greater than the elastic

ultimate moment, the stress, fs, in this table may be taken as fy. Table 6.4 Tension reinforcement modification factors



53

100 As prov bd

Factor

0.00 0.15 0.25 0.35 0.50 0.75 1.0 1.5 2.0 2.5 3.0

1.00 1.05 1.08 1.10 1.14 1.20 1.25 1.33 1.40 1.45 1.50

NOTE 1. The values in this table are derived from the following

equation:

Modification factor for compression reinforcement =

100 As, prov 100 As, prov 1 +

bd /( 3 + bd ) 1.5 NOTE 2. The area of compression reinforcement As, prov used in this

table may include all bars in the compression zone, even

those not effectively tied with links.

Table 6.5 Compression reinforcement modification factors

1 Example 6.6 Span-Effective Depth Ratio Check

A rectangular continuous beam spans 12 m with a mid-span ultimate moment of 400 kN m. If the breadth is 300 mm, check the acceptability of an effective depth of 600 mm when high yield reinforcement fy = 460 N/mm is used. Two 16 mm bars are located within the compressive zone.

Basic span-effective depth ratio from Table 6.3 = 26

To avoid damage to finishes modified ratio = 26 10/12 = 21.7. Tensile reinforcement modification factor:

M 400 106

bd2 =

300 6002 = 3.7



54

thus, from table 6.4 for fy = 460 N/mm, modification factor = 0.89.

Compression reinforcement modification factor: 100As 100 402

bd =

300 600 = 2.2

thus from table 6.5, modification factor = 1.07

Hence, modified span-effective depth ratio is equal to 21.7 0.89 1.07 = 20.7

12 103 Span-effective depth ratio provided =

600 = 20

Which is less than the allowable upper limit, thus deflection requirements are likely to be satisfied.

6.11 Worked Examples

Example 6.6 Design of a simply supported L-beam in footbridge

(a) Specification The section through a simply supported reinforced concrete footbridge of 7 m span shown in Fig.6.12. The imposed load is 5 kN/m2 and the materials to be used are grade 30 concrete and grade 460 reinforcement. Design the L-beams that support the bridge. Concrete weights 2400 kg/m and the weight of the handrails are 16 kg/m per side.

(b) Loads, shear force and bending moment diagram The dead load carried by one L-beam is

[(0.15 0.8) + (0.2 0.28)] 23.5 + 16 9.81/10 = 4.3 kN/m

The imposed load carried by one L-beam is 0.8 5 = 4 kN/m. The design load

(1.4 4.3) + (1.6 4) = 12.42 kN/m The ultimate moment at the center of the beam is



55

12.42 7 / 8 = 76.1 kNm

The load, shear force and bending moment diagrams are shown in Fig 6.12(b), 6.12(c) and 6.12(d) respectively.

(c) Design of moment reinforcement

The effective width of the flange of the L-beam is given by the lesser of

1. the actual width, 800mm, or 2. b = 200 + 7000/10 = 900 mm

From Table 1.3 (BS 8110: Part 1. Table 3.4) the cover for moderate exposure is 35 mm. The effective depth, d = 400 35 8 12.5 = 344.5 mm say 340 mm



56

Fig.6.12 Section through footbridge; (b) design load; (c) shear force diagram; (d) bending moment diagram.



57

Figure 6.13 (a) Beam section; (b) beam support; (c) beam elevation.

The L-beam is shown in Fig. 6.13(a)

The moment of resistance of the section when 0.9 of the depth to the neutral axis is equal to the slap depth hf = 120 mm is

MR = 0.45 30 800 120 (340 0.5 120)/106 = 362.9 kNm

The neutral axis lies in the flange.

76.1 106 K =

800 3402 30 = 0.027

z = 340 [0.5 + (0.25 0.027/0.9)] = 329.5 mm > 0.95d using 0.95d = 0.95 340 = 323 mm

76.1 106 As = 0.87 460 323

= 589 mm2

Provide 4Y16 mm, As = 804 mm. Using the simplified rules for curtailment of bars, 2 bars are cut off as shown in Fig. 6.13(c) at 0.08L of the span from each end.



58

(d) Design of shear reinforcement

The enhancement of shear strength near the support using the simplified approach is taken into account in the design of shear. The maximum shear stress at the support is

43.5 103 v =

300 340 = 0.64 N/mm2

This is less than 0.830 = 4.38 N/mm or 5 N/mm. The shear at d = 340mm from the support is

V = 43.5 0.34 12.43 = 38.8 kN 38.8 103

v = 200 340

= 0.57 N/mm2

The effective area of steel at d from the support is 2Y16 of area 402 mm 100 As 100 402

bd =

200 340 = 0.59

The design concrete shear strength from the formula in Table 5.4 (BS 8110: Part 1, Table 3.9) is

vc = 0.79 (0.59)1/3(400/340)1/4(30/20)1/3/1.25 = 0.586 N/mm2

Provide R8 vertical links (two legs), Asv = 100mm, in grade 250 reinforcement. The spacing required is determined using Table 6.6 (BS 8110: Part 1, Table 3.8). v < vc

The spacing for minimum links is 0.87 250 100

sv = 0.4 200

= 271.9

0.75 340 = 255 mm

The links will be spaced at 250 mm throughout the beam. 2Y20 are to be provided to carry the links at the top of the beam. The shear reinforcement is shown in Figs 6.13(b) and 6.13(c)



59

(e) End anchorage

The anchorage of the bars at the supports must comply with BS8110: Part 1, clause 3.12.9.4. The bars are to be anchored 12 bar diameters past the center of the support. This will be provided by a 90 bend with an internal radius of three bar diameters. From clause 3.12.8.23, the anchorage length is the greater of

1. 4 internal radius = 4 48 = 192 mm but not greater than 12 16 = 192 mm

2. the actual length of the bar (4 16) + 56 2pi/4 = 151.9 mm

The anchorage is 12 bar diameters.

Value of v

(N/mm2) Form of shear reinforcement

to be provided

Area of shear reinforcement to

be provided

Less than 0.5 vc throughout

the beam

See note 1 -

0.5 vc < v < ( vc + 0.4 ) Minimum links for whole length of beam

Asv 0.4 bvsv / 0.87fyv

(see note 2) ( vc + 0.4 ) < v < 0.8fcu or 5 N/mm2

Links or links combined with

bent-up bars. Not more than

50% of the shear resistance

provided by the steel may be

in the form of bent-up bars

(see note 3)

Where links only provided:

Asv bvsv ( v - vc ) / 0.87fyv

NOTE 1. While minimum links should be provided in all beams of structural importance,

it will be satisfactory to omit them in members of minor structural importance

such as lintels or where the maximum design shear stress is less than half vc.

NOTE 2. Minimum links provide a design shear resistance of 0.4 N/mm2.

NOTE 3. Spacing of links. The spacing of links in the direction of the span should not exceed 0.75d. At right-angles to the span, the horizontal spacing should be such

that no longitudinal tension bar is more than 150mm from a vertical leg; this

spacing should in any case not exceed d.

Table 6.6 Form and area of shear reinforcement in beams



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(f) Deflection check The deflection of the beam is checked using the rules given in table 6.4 & 6.5 (BS8110: Part 1, clause 3.4.6).

The basic span-to-effective depth ration is 16 M 76.1 106 bd2

=

800 3402 = 0.82

The service stress is 5 460 589

fs = 8 804

= 210.6N/mm2

The modification factor for tension reinforcement using the formula in Table 6.4 is

477 210.6 0.55 +

120 ( 0.9 + 0.82 ) = 1.84 < 2.0

For the modification factor for compression reinforcement, with As.prov = 226 mm according to Table 6.5 is

100 As, prov 100 226 bd

=

800 340 = 0.083

Therefore, the modification factor by the formula is 1 + [0.083 / (3 + 0.083)] = 1.027 allowable span / d = 16 1.84 1.027 = 30.23 actual span / d = 7000/340 = 20.5 < 30.23

The beam is satisfactory with respect to deflection.

web width bw 200 effective flange width

=

b =

800 = 0.25 < 0.3



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Example 6.7 Design of simply supported doubly reinforced rectangular beam

(a) Specification

A rectangular beam is 300 mm wide by 450 mm effective depth with inset to the compression steel of 55 mm. The beam is simply supported and spans 8 m. The dead load including an allowance for self-weight is 20 kN/m and the imposed load is 11 kN/m. The materials to be used are grade 30 concrete and grade 460 reinforcement. Design the beam.

(b) Loads and shear force and bending moment diagrams are shown in Fig. 6.14. design load = ( 1.4 20 ) + (1.6 11 ) = 45.6 kN/m ultimate moment = 45.6 82 / 8 = 364.8 kN

(c) Design of the moment reinforcement (section 4.6)

When the depth x to the neutral is 0.5d, the moment of resistance of the concrete only is MRC = 0.156 30 300 450 / 106 = 284 kNm. Compression reinforcement is required.

Figure 6.14 (a) Design Loading; (b) ultimate shear force diagram; (c) ultimate bending



62

moment diagram.

The stress in the compression reinforcement is 0.87 fy. The area of compression steel is

( 364.8 274 ) 106 As = ( 450 55 ) 0.87 460

= 508 mm2

Provide 2Y20, As = 628 mm

( 0.203 30 300 450 ) + ( 0.87 460 508 ) As =

0.87 460

Provide 6Y25, As = 2945 mm. The reinforcement is shown in Fig. 6.15(a). In accordance with the simplified rules for curtailment, 3Y25 tension bars will be cut off at 0.08 of the span from each support. The compression bars will be carried through to the ends of the beam to anchor the links. The end section and the side elevation of the beam are shown in Figure 6.15(b) and 6.15(c) respectively. The cover to the reinforcement is taken as 35 mm for moderate exposure.

(d) Design of shear reinforcement

The maximum shear stress at the support is

182.4 103 v =

300 462.5 = 1.31 N/mm2

This is less than 0.830 = 4.38 N/mm or 5 N/mm. The shear at d = 462.5 mm from the support is

V = 182.4 0.46 45.6 = 161.4 kN 161.4 103

v = 300 462.5

= 1.16 N/mm2

The area of steel at the section is 1472mm. 100 As 100 1472

bd =

300 462.5 = 1.06



63

The design shear strength from the formula in BS8110: Part 1, Table 3.9. is vc = 0.79 ( 1.06 )1/3 ( 30/25 )1/3 / 1.25 = 0.68 N/mm2

Provide R10 (two-leg vertical) links, Asv = 157 mm. The spacing required is: 157 0.87 250

sv = 300 ( 1.16 0.68)

= 237.1 mm

For minimum links the spacing is 157 0.87 250

sv = 300 0.4

= 284.6 mm

The spacing is not to exceed 0.75d = 346.8 mm. This distance x from the support where minimum links only are required is determined. In this case v = vc. The design shear strength where As = 2945 mm2 is

100 As / bd = 2.18 vc = 0.87 N/mm2

For the distance x, it is found by solving the equation 0.87 300 450 / 10 = 182.4 45.6x x = 1.42m

Space links at 200 mm centres for 2 m from each support and then at 250 mm centres over the center 4 m. Note that the top layer of three 25 mm diameter bars continues for 780 mm greater than d past the section when v = vc

(e) Deflection check

The basic span-effective depth ratio from Table 6.3 is 20 for a simply supported rectangular beam

Tensile reinforcement modification factor: M 364.8 106

bd2 =

300 4502 = 6.0



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The service stress is 5 460 2631.2

fs = 8 2945

= 256.9 N/mm2

The modification factor for tension reinforcement using the formula in Table 6.4 is

477 256.9 0.55 +

120 ( 0.9 + 6.0 ) = 0.816 < 2.0

For the modification factor for compression reinforcement, with As.prov = 628 mm according to Table 6.5 is

100 As, prov 100 628 bd

=

300 450 = 0.46

Therefore, the modification factor by the formula is 1 + [0.083 / (3 + 0.46)] = 1.13 allowable span / d = 20 0.816 1.13 = 18.4 actual span / d = 8000/450 = 17.8 < 18.4

The beam is satisfactory with respect to deflection.



65

Figure 6.15 (a) Section at centre; (b) end section; (c) part side elevation.

Example 6.8 Design of a Continuous Beam

The beam is 300 mm wide by 660 mm deep with three equal 5.0 m spans. In the transverse direction, the beams are at 4.0 m centres with a 180 mm thick slab, as shown in figure 6.17.

The live load qk on the beam is 50 kN/m and the dead load gk, including self weight is 85 kN/m.

Characteristic material strengths are fcu = 30 N/mm, fy = 460 N/mm for the longitudinal steel fyv and 250 N/mm for the links. For a mild exposure the minimum concrete cover is to be 25 mm.

For each span Ultimate load wu = (1.4gk + 1.6qk) kN/metre (1.4 85 + 1.6 50) = 199 kN/metre



66

Figure 6.16 Continuous beam with ultimate bending moment and shear force coefficients

Total ultimate load on a span is F = 199 5.0 = 995 kN

As the loading is uniformly distributed, qk ( gk and the spans are equal, the coefficients shown in figure 6.16 have been used to calculate the design moment and shears.

Bending

(a) Mid-span of 1st and 3rd Spans Design as a T section Moment M = 0.09 FL = 0.09 x 995 5 = 448 kNm Effective width of flange = bw + 0.7 L/5

0.7 5000 = 300 +

5 = 1000 mm

therefore M 448 106

bd2 fcu =

1000 6002 30 = 0.041

From the lever-arm curve, la = 0.95 therefore z = 0.95 600 = 570 mm

and d z = 600 570 = 30 < hf / 2

so that the stress block must lie within the 180 mm thick flange. Therefore

M 448 106 As = 0.87 fy z

=

0.87 460 570 = 1964 mm2



67

Provide 2Y32 plus 1Y25, As = 2101 mm (bottom steel)

(b) Interior Supports Design as a Rectangular Section

M = 0.11FL = 0.11 995 5 = 547 kNm hogging

Thus. Compression steel is required,

This area of steel will be provided by extending the span reinforcement beyond the supports

Provide 2Y32 plus 3Y25, As = 3080 mm (top steel).

(c) Mid-span of 2nd Span Design as a T-section

M = 0.07 FL = 0.07 995 5 = 348 kNm

Using the lever-arm curve, it is found that la = 0.95 M 348 106

As = 0.87 fy z =

0.87 460 (0.95 600) = 1525 mm2

Provide 1Y32 plus 2Y25, As = 1786 mm (bottom steel)

M 547 106

bd2 fcu =

300 5802 30 = 0.18 > 0.156

M 0.156 fcu bd2 As =

0.87 fy ( d d )

547 106 0.156 30 300 5802 =

0.87 460 ( 580 50 ) = 352 mm2

0.156 fcu bd2 As =

0.87 fy z + As

0.156 30 300 5802 =

0.87 460 0.775 580 + 352 = 2977 mm2



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Shear

(a) Check maximum shear stress Maximum shear at face of support is

Vs = 0.6F wu support width / 2 = 0.6 995 199 0.15 = 567 kN

= 3.26 N/mm2 < 0.8fcu

(b) Nominal links Asv 0.4 b 0.4 300 sv

=

0.87 fyv =

0.87 250 = 0.55

Provide R10 links at 280 mm centres, Asv / sv = 0.56

(c) End supports Shear distance, d from support face is

Vd = 0.45F - wu ( d + support width / 2 ) = 0.45 995 199 (0.6 + 0.15) = 299 kN

Vd 299 103 v =

bd =

300 600 = 1.66 N/mm2

Therefore from table 5.4 vc = 0.70 N/mm

Asv b ( v vc ) sv

=

0.87 fyv = = 1.32

Provide R10 links at 110 mm centres Asv / Sv = 1.41 Shear resistance of nominal links + concrete is

Asv Vn = ( sv 0.87 fyv + bvc ) d = (0.56 0.87 250 + 300 0.7) 600= 199 kN

Vs 567 103 v =

bd =

300 580



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Shear reinforcement other than the nominal is required over a distance

Vd - Vn 299 199 s =

wu + d =

199 + 0.6 = 1.1 m

from the face of the support.

Figure 6.17 End-span reinforcement details



70

(d) First and third spans interior supports Distance d from support face Vd = 0.6 995 199 (0.58 + 0.15)

= 452 kN

452 103 v =

300 580 = 2.60 N/mm2

100 Asv 100 3080 bd

=

300 580 = 1.77

therefore from table 5.4 vc = 0.81 Asv 300 ( 2.6 0.81) sv

=

0.87 250 = 2.47

Provide R12 links in pairs at 180 mm centres, Asv / sv = 2.51. Using Vn from part (c) as a conservative value, shear links are required over a distance

Vd - Vn 452 - 199 s =

wu + d =

199 + 0.58 = 1.85 m

A similar calculation would show that single R12 links at 120 mm centres would be adequate 1.0 m from the support face.

(e) Second span

Distance d from support face

Vd = 0.55 995 199 (0.58 + 0.15) = 402 kN Calculation would show that R10 links in pairs at 150 mm centres would be adequate.



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7. SLAB

Types of Slab

Slabs are plate elements forming floors and roofs in buildings which normally carry uniformly distributed loads. Slabs may be simply supported or continuous over supports. Slabs can be classified according to the method of support as follows:

i) Spanning one way between beams or walls

ii) Spanning two ways between the supports

iii) Flat slabs carried on columns and edge beams or walls with no interior beams.

One Way Spanning Solid Slabs

Idealization for Design

a) Uniformly Loaded Slabs

One way slabs carrying uniform load are designed on the assumption that they consists of a series of rectangular beam 1 m wide spanning between supporting beams or walls. The sections through a simply supported slab and a continuous slab are shown in Fig. 7.1.

Figure 7.1 (a) Simply Supported Slab; (b) Continuous One-way Slab



72

Effective Span, Loading and Analysis

a) Effective Span

The effective spans for one-way slabs are:- i) Simply supported slabs:- the smaller of the centres of bearings or the

clear span + d. ii) Continuous slabs:- centres of supports.

b) Arrangement of Loads

The code states that the slab should be designed to resist the most unfavourable arrangement of loads. However, it is possible to design for a single-load case of max. design load on all spans. This is permitted subject to the following conditions:

i) The area of each bay, i.e. the building width column spacing > 30m. ii) The ratio of Qk to Gk 1.25 iii) Qk 5 kN/m (excluding partition)

c) Analysis and Redistribution of Moments

The code states that if the analysis is carried out for a single-load case of all spans loaded, the support moment except at the supports of cantilevers should be reduced by 20%. This gives an increase in span moment. (No further redistribution is to be carried out).

d) Analysis using Moment Coefficients

BS8110 states that where the spans of the slab are approx. equal, the moment and shears for design may be taken from Table 7.1. This table allows for 20% redistribution.

Table 7.1 Ultimate bending moment and shear force coefficients in one way spanning slabs

Outer support

Middle of end span

First interior support

Middle of interior span

Interior supports

Moment 0 0.086 FL 0.086 FL 0.063 FL - 0.063 FL Shear 0.4F - 0.6F - 0.5F

Note: F is the total design ultimate load on the span and L is the effective span



73

Section Design and Slab Reinforcement

a) Main Reinforcement

The min. area of main reinf. is:- For rectangular sections and solid slabs:-

b) Distribution Steel

The distribution steel runs at right angles to the main moment steel and serves the purpose of tying the slab together. The area of distribution steel is the same as the min. area for main reinforcement in (a).

c) Curtailment of Bars in Slabs

BS8110 sets out simplified rules for curtailment of bars for slabs. These rules may be used subject to the following provisions:-

i) The slabs are designed for predominately uniformly distributed loads; ii) In continuous slabs the design has been made for the single load case of

max. design load on all spans. The simplified rules are shown in Fig. 7.2

d) Cover

The amount of cover required for durability and fire protection is taken from Table 1.3 and 1.4 (Table 3.4 and 3.5 of BS8110).

100 As Mild Steel, fy = 250 N/mm2,

Ac = 0.24

100 As High Yield Steel, fy = 250 N/mm2,

Ac = 0.13



74

Figure 7.2 (a) Simply Supported Span; (b) Cantilever; (c) Continuous Span

Shear

Under normal loads shear stresses are not critical and shear reinforcement is not required. Shear reinforcement is provided in heavily loaded thick slabs but should not be used in slabs less than 200 mm thick.

In design, the average shear stress v is given by



75

v = V/bd (where b is normally taken as 1m wide)

The design procedures for slab is essentially the same as that for beams. However, in the design of slab in shear, the following points should be noted.

i) v should not exceed 0.8fcu or 5 N/mm whichever is less. ii) if v < vc , then no shear reinforcement is required. iii) if vc < v (vc + 0.4), then provide min. links as :-

iv) if v > (vc + 0.4), provide shear links as

Deflection

In slab design, deflections are usually controlled by limiting the span/effective depth ratio.

iii) Basic span/effective depth ratio:- Cantilever slab 7 Simply supported slabs 20 Continuous slabs 26

iv) The basic span/effective depth ratio is multiplied by a modification factor for tension reinforcement. Only the tension steel at the centre of span is taken into account.

Crack Control

To control cracking in slabs, max. values for clear spacing between bars are set out in BS8110, Cl.3.12.11.2.7. The clause states that in no case should the clear spacing exceed the lesser of three times the effective depth or 750 mm.

Asv 0.4b Sv

0.87fyv

Asv b(v-vc) Sv

0.87fyv



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Example 7.1 Design of a Simply Supported Slab

Figure 7.3 Simply Supported Reinforced Concrete Slab

The slab shown in figure 7.3 is to be designed to carry a live load of 3.0 kN/m plus floor finishes and ceiling loads of 1.0 kN/m. The characteristic material strengths are fcu = 30 N/mm and fy = 460 N/mm. Basic span-effective depth ratio = 20.

Span therefore minimum effective depth d =

20 modification factor, m.f. 4500 225

=

20 m.f. =

m.f.

(1) Estimating the modification factor to be of the order of 1.3 for a lightly reinforced slab. Try effective depth d = 170 mm. For a mild exposure the cover = 25 mm. Allowing say 5 mm as half the diameter of the reinforcing bar

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