strain, reinforced concrete, confinement

8/20/2019 Strain, Reinforced concrete, confinement

1/125

Bachelor of Science in Civil Engineering

Construction Planning, Estimating Management,

Methods & Safety Engineering

1

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


2/125




2

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


3/125




3

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.


4/125




4

Short-term modulus of elasticity of concrete

28 day characteristic Static modulus E c,28

cube strength (kN/mm2)

f cu,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

E c,t = E c ,28 (0.4 + 0.6 f cu,t /f cu,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.


5/125




5

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/mm².

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


6/125




6

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/mm²)

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


7/125




7

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/m³) 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


8/125




8

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


9/125




9

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.


10/125




10

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.


11/125




11

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

f k = f m – 1.64s

where f k = characteristic strength, f m = 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.


12/125




12

Partial Factors of Safety

Partial Factors of Safety for Materials (γm)

Design Strength = Characteristic Strength ( f k ) / γm

e.g. 460kN/mm2 / 1.15 ≈ 0.87 f y

Factors to be considered in selecting γm

i) The strength of the material in an actual member.

e.g. The strength of concrete is easily affected by placing, compacting,

curing……=> higher γm

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

ii) The severity of the limit state being considered.

Higher for ULS

Small or even neglected for SLS

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

MaterialLimit 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 ( γ m )


13/125




13

Partial Factors of Safety for Loads (γf )

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 (γf ) on

loading, so that,

Design Load = Characteristic Load x (γf )

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

Ultimate

Concrete ImposedEarth &

WaterWind

Serviceability AllLoad Combination

(γG) (γQ) (γQ) (γW) (γG˙γQ˙γW)

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.


14/125




14

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.


15/125




15

Figure 3.1 Multi-span beam loading arrangements

Load Combination for SLS

A partial factor of safety γf = 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


16/125




16

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.


17/125




17

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


18/125




18

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 εo , from which point the strain increases while

the stress remains constant.

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


19/125




19

The ultimate strain = 0.0035 is typical for all grades of concrete. The ultimate

design stress is given by:-

0.67 f cu 0.67 f cu

γm =

1.5= 0.447 f cu ≈ 0.45 f cu

Where 0.67 allows the difference between the bending strength and cube crushing

strength of concrete and γm = 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

f y Design Yield Stress =

γm

Stress = Es × εs

f y ∴ Design Yield Strain εy = (

γm

) / Es

For high-tensile steel (T), f y = 460 N/mm².

εy = 460/(1.15 × 200 × 10³ ) = 0.002

For mild steel (R), f y = 250 N/mm²

εy = 250/(1.15 × 200 × 10³ ) = 0.00109


20/125




20

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 – xεst = εcc = (

x)

x – d’and εsc = εcc = (

x)

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


21/125




21

d – x

εst By rearranging (1), x =1 +

εcc

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

For steel with f y = 460 N/mm², the yield strain is 0.002.

d – x

εst ∴ x =1 +

εcc

= 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 ≤ (βb – 0.4)d

Where,

moment at the section after redistribution βb =

moment at the section before redistribution

Thus with moment redistribution not greater than 10%

βb ≤ 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.


22/125




22

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 f cu

= 0.45b · 2(d – z)z · f cu

= 0.9 f cu · b(d – z)z

Let K = M / (bd² f cu),

Eq.(3) becomes, (z/d)² - (z/d) + K/0.9 = 0

∴ z = d[0.5 + √(0.25-K/0.9)]

Also Fst = (f y / γm) ˙ As with γm = 1.15

M∴ As =

0.87 f y z


23/125




23

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/bd2f cu 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 0.775Table

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 f cu b(d – 0.775d) × 0.775d

=> M = 0.156 f cu bd²

The coefficient 0.156 is calculated from the more precise concrete stress.

MWhen

bd2 f cu

= K > 0.156, compression reinforcement is required.

Rectangular Section with Compression reinforcement at ULS


24/125




24

Figure 4.5 Section with compression reinforcement

When M > 0.156 f cu bd², 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 f y As = 0.45 f cu bs + 0.87 f y As’

and with s = 0.9 × d/2 = 0.45d

0.87 f y As = 0.201 f cu bd + 0.87 f y As’

Take moment at the centroid of the tension steel As ,

M = Fcc z + Fsc (d – d’)

= 0.201 f cu bd(0.775d) + 0.87 f y As’ (d – d’)

= 0.156 f cu bd² + 0.87 f y As’ (d – d’)

M – 0.156 f cu bd2

∴ As’ =0.87 f y (d – d’)

0.156 f cu bd2 and As =

0.87 f y z+ As’ with z = 0.775d

the value 0.156 is usually denoted by K’

Moment Redistribution


25/125




25

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 ≤ (βb - 0.4)d

Where d = effective depth

moment at the section after redistribution βb =

moment at the section before redistribution

∴ Depth of stress block, s = 0.9 ((βb - 0.4)d

and z = d- s/2

= d – 0.9(βb - 0.4)d/2

Moment of resistance of the concrete in compression,

Mc = Fcc z = 0.45f cu bsz

= 0.45 f cu b · 0.9(βb - 0.4)d[d – 0.9(βb - 0.4)d/2]

Mc

∴ bd2 f cu = 0.45 · 0.9(βb - 0.4) · [1-0.45(βb - 0.4)]

= 0.402(βb - 0.4)-0.18(βb - 0.4)²

Mc Let

bd2 f cu

= K’

∴ K’ = 0.402(βb - 0.4)-0.18(βb - 0.4)²

when M > K’bd²f cu , then compression steel is required such that

(K – K’) f cu bd2 As’ =

0.87 f y (d – d’)

K’ f cu bd2

and As =0.87 f y 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


26/125




26

steel will not be yielded and the compressive stress shall be taken as f sc = Es·εsc.

Redistribution

(per cent)

β 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


27/125




27

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 percentageSituation Definition of

percentage f y = 250 N/mm² f y = 450 N/mm²

% %

Tension reinforcement

Sections subjected mainly to pure tension

Sections subjected to flexure

100 As / Ac 0.8 0.45

(a) Flanged beams, web in tension:

(1) b W / b < 0.4 100 As /bwh 0.32 0.18

(2) b W / b ≥ 0.4 100 As /bwh 0.24 0.13

(b) Flanged beams, flange in tension

over a continuous support:

(1) T-beam 100 As /bwh 0.48 0.26

(2) L-beam 100 As /bwh 0.36 0.20

(c) Rectangular section (in solid slabs this minimum

should be provided in both directions)

100 As /Ac 0.24 0.13

Compression reinforcement (where such reinforcement is

required for the ultimate limit state)

General rule 100 Asc /Acc 0.4 0.4

Simplified rules for particular cases:

(a) rectangular column or wall 100 Asc /Ac 0.4 0.4

(b) flanged beam:

(1) flange in compression 100 Asc /bh t 0.4 0.4

(2) web in compression 100 Asc /bwh 0.2 0.2

(c) rectangular beam 100 Asc /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)

100 Asc /ht l 0.15 0.15


28/125




28

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.

ø = Bar size or nominal diameter.

f bu = Ultimate anchorage bond stress.

f s = Direct tensile or compressive stress in the bar.

Consider the forces on the bar,

πø2

Tensile pull-out force =4

•f s

Anchorage force = ( Lπø ) •f bu

πø2

∴ 4

= ( Lπø ) •f bu

f s 0.87 f y ∴ L =

4f bu •ø =

4 f bu •ø

The ultimate anchorage bond stress, f bu = β√(f cu)

Values of β are given in table 5.2

Table 5.2 Values of bond coefficient β

-Bar typeBars 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 ø


29/125




29

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

Grade 460Reinforcement 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


Compression lap length 33 61 43 34 27Concrete cube strength 40


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.


30/125




30

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 < 2ø (Multiply by 1.4)

b) At corners where min. cover to either face < 2ø or clear distance between

adjacent laps < 75 mm or 6ø. (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.


31/125




31

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.

f yv = Characteristic strength of the stirrup reinforcement

V = Shear force due to the ultimate load.

Consider section x – x

0.87 f yv Asv = V = vbd


32/125




32

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

If Sv is the actual spacing of stirrups, then

0.87 f yv Asv = vbd(sv /d)

Asv vb∴

sv =

0.87 f yv

Taken the shear resistance of concrete into account,

Asv b(v – vc)

sv =

0.87 f yv

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/mm²

0.45

0.53

0.67

0.77

0.84

0.97

1.06

1.22

N/mm²

0.43

0.51

0.64

0.73

0.81

0.92

1.02

1.16

N/mm²

0.41

0.49

0.62

0.71

0.78

0.89

0.98

1.12

N/mm²

0.40

0.47

0.60

0.68

0.75

0.86

0.95

1.08

N/mm²

0.39

0.46

0.58

0.66

0.73

0.83

0.92

1.05

N/mm²

0.38

0.45

0.56

0.65

0.71

0.81

0.89

1.02

N/mm²

0.36

0.43

0.54

0.62

0.68

0.78

0.86

0.98

N/mm²

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 γm 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

/ γm

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/mm², the values in the table may be

multiplied by (f cu /25) 1/3

. The value of f cu should not be taken as greater than 40.

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

tensile reinforcement.


33/125




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

vc·2d / 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√ (f cu ) or 5 N/mm² .


34/125




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/bd²f cu ≤ 0.156

With compression reinforcement,

M/bd²f cu ≤ 10 / f cu if the area of bending reinforcement

is no to be excessive.

ii) Shear stress v = V / bd and v should never exceed 0.8√f cu or 5 N/mm²

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/bd² and

the service stress in the tension reinforcement.


35/125




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 (βb ≤ 0.9)

Refer to the diagram below.


36/125




36

Figure 6.2 Singly reinforced section

The design procedures and can be summarized as follows:-

i) Calculate K = M/bd²f cu

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/bd2f cu 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 0.775Table

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:-

MAs =

0.87 f y 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)


37/125




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 f cu = 45

N/mm² for the concreter and f y = 460 N/mm ² 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/bd²f cu ⇒ 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,

MAs =

0.87 f y z

= 250 × 106 / (0.87 × 460 × 500.9) = 1247 mm2

Provide 2Y32 bars, area = 1608 mm² (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.


38/125




38

Moment Redistribution Factor βb ≥ 0.9 and d’/d ≤ 0.2

Compression reinforcement is required if M > 0.156 f cu bd²

that is K > 0.156

i) Area of compression steel

M – 0.156 f cu bd²As’ =

0.87 f sc (d-d’)

ii) Area of tension steel,

0.156 f cu bd²

As = 0.87 f y 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 βb < 0.9

The design procedures and can be summarized as follows:-

i) Calculate K = M/bd² f cu

ii)

Calculate K’ = 0.402 (βb – 0.4) – 0.18 (βb – 0.4)²

If K < K’, no compression steel required.

If K > K’, compression steel required.

iii) Calculate x = (βb – 0.4)d

If d’ /x < 0.43, the compression steel has yielded and f sc = 0.87 f y

If d’ /x < 0.43, calculate the steel compressive strain εsc and then


39/125




39

calculate f sc = Es ·εsc

iv) Calculate the area of compression steel by

(K - K’) f cu bd²

As’ = f sc (d-d’)

v) Calculate the area of tension steel by,

K’ f cu bd² f su As =

0.87 f y z+ As’

0.87 f y

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 ≤ 12φ 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, β b > 0.9

The beam section shown in figure 6.5 has characteristic material strengths of f cu = 30

N/mm² and f y = 460 N/mm². 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/bd²f cu ⇒ K = 160 × 106 / (230 × 3302× 30)

= 0.22 > 0.156 ∴ compression steel is required

d’/d = 50/330 = 0.15 < 0.2


40/125




40

∴ f cu = 0.87 f y

By formula, required compression steel

M – 0.156 f cu bd²As

’ =

0.87 f sc (d-d’)

165 × 106 – 0.156 × 30 × 230 × 3302=

0.87 × 460 (330 - 50)= 427 mm

2

and required tension steel

0.156 f cu bd²As =

0.87 f y z+ As

’

0.156 × 30 × 230 × 3302=

0.87 × 460 × 0.775 × 330+ 427 = 1572 mm

2

Provide 2Y20 bars for As, area = 628 mm² and 2Y32 bars for As, area = 1610 mm².

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, β b = 0.7

The beam section shown in figure 6.6 has the characteristic material strengths of f cu =

30 N/mm² and f y = 460 N/mm². The ultimate moment is 370 kNm causing hogging

of the beam.


41/125




41

Figure 6.6 Beam doubly reinforced to resist a hogging moment

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

x = (βb – 0.4)d

= (0.7 – 0.4) 540 = 162 mm

K = M/bd² f cu

= 370 × 106 /(300 × 540

2 × 30)

= 0.141

K = 0.402 (βb – 0.4) – 0.18 (βb – 0.4)²

= 0.104

K > K’ therefore compression steel is required

therefore f sc < 0.87 f y

0.0035 (x – d’)(1) Steel compressive strain εsc =

X

0.0035 (162 – 100)=

162= 0.00134

(2) Steel compressive stress = E s·εsc

= 200 000 × 0.00134

= 268 N/mm2

(K - K’) f cu bd²

(3) Compression steel As’ = f sc (d-d’)

(0.141 – 0.104) × 30 × 300 × 5402

=268 (540 – 100)

K’ f cu bd² f su (4) Tension Steel As =

0.87 f y z+ As’

0.87 f y

As = 0.104 × 30 × 300× 5402 + 823 × 268


42/125




42

0.87 × 460 (540 – 0.9 × 162/2) 0.87 × 460

Provide 2Y25 bars for As, area = 982 mm² and 2Y32 & 1Y25 bars for As area = 2101

mm², 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 b w 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


43/125




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/ bf d² f cu 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 mm² 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) 12φ

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


44/125




44

For condition (i) and (ii), AB is the greater of d or 12φ

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 0.8√f cu or 5 N/mm² )

Vertical Stirrups

Rules:

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


45/125




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.

f yv = 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 f yv = 250 N/mm².

(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

≥

0.87f yv

vc v <

2

Asv 0.4b

Sv =

0.87f yv


46/125




46

Shear V s = 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 < 0.8√f cu

(b) Shear linksDistance d from face of support

shear V d = V s - wu d

= 214 – 75.2 × 0.55 = 173 kN

shear stress v = 173 × 103 / ( 300 × 550)

= 1.05 N/mm²

Only 2 Nos. 25 mm bars extend a distanced d past the critical section.

Therefore for determining vc

100As 100 × 982bd

=300 × 550

= 0.59

From table 5.4. vc = 0.56 N/mm²

Asv b (v – vc) 300 (1.05 – 0.56)

Sv =

0.87f yv =

0.87 × 250= 0.68


47/125




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 =

0.87f yv =

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 f yv + 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 - 159s =

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


48/125




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(0.87f yv)


49/125




49

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

vi) Calculate the additional area of longitudinal steel.

Asv f yv As =

sv ( f y ) (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/mm²)


50/125




50

vt < vt min vt > vt min

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

shear reinforcement

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

Designed shear and torsionreinforcement

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 f cu

= 30N/mm² f y = 460 N/mm² and f

yv = 250 N/mm².

(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

2Tvt =

h2

min (hmax – hmin /3)

2 × 10 × 106

vt =300

2 (500 – 300/3)

= 0.56 N/mm2


51/125




51

So that vt < vtu y1 / 500 as required

(5)

Therefore

Provide R10 links at 100 centres

(6) Additional longitudinal steel

Asv f yv

As = sv ( f y ) (x1 + y1)

250= 0.55 ×

460(240 + 440) = 203 mm

2

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 TAdditional

sv =

0.8x1y1(0.87f yv)

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


52/125




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/bd² in table 6.4,

(b) The area of compression steel as in table 6.5

M/bdReinforcementservice

stress(N/mm²)

0.50 0.75 1.0 1.5 2.0 3.0 4.0 5.0 6.0

( f y = 250)

( f y = 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 – f s)Modification factor = 0.55 +

M≤ 2.0

120 ( 0.9 +bd 2

)

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 f y As, require 1 f s =

8 As, provided ×

βb

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 ⅝ f y.

Table 6.4 Tension reinforcement modification factors


53/125




53

100 As prov

bdFactor

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’, prov1 +

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 f y = 460 N/mm² 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 × 600

2= 3.7


54/125




54

thus, from table 6.4 for f y = 460 N/mm², 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/m³ 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 × 9.81/10³ = 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


55/125




55

12.42 × 7² / 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


56/125




56

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

moment diagram.


57/125




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 h f = 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 × 340

2 × 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 × 106As =

0.87 × 460 × 323= 589 mm

2

Provide 4Y16 mm, As = 804 mm². 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.


58/125




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 0.8√30 = 4.38 N/mm² or 5 N/mm². 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 mm²

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 = 100mm², 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 × 100sv =

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)


59/125




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 90º 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 × 2π /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 / 0.87f yv

(see note 2)

( vc + 0.4 ) < v < 0.8√f cu 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 ) / 0.87f yv

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


60/125




60

(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 × 340

2

= 0.82

The service stress is

5 × 460 × 589f s =

8 × 804= 210.6N/mm

2

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

is

477 – 210.60.55 +

120 × ( 0.9 + 0.82 )= 1.84 < 2.0

For the modification factor for compression reinforcement, with A’s.prov = 226

mm² 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


61/125




61

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 × 450² / 106 = 284 kNm. Compression

reinforcement is required.

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


62/125




62

moment diagram.

The stress in the compression reinforcement is 0.87 f y. The area of

compression steel is

( 364.8 – 274 ) × 106

As’ =( 450 – 55 ) × 0.87 × 460

= 508 mm2

Provide 2Y20, As’ = 628 mm²

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

0.87 × 460

Provide 6Y25, As = 2945 mm². 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 0.8√30 = 4.38 N/mm² or 5 N/mm². 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 1472mm².

100 As 100 × 1472

bd=

300 × 462.5= 1.06


63/125




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 mm². The spacing required is:

157 × 0.87 × 250sv =

300 × ( 1.16 – 0.68)= 237.1 mm

For minimum links the spacing is

157 × 0.87 × 250sv =

300 × 0.4= 284.6 mm

The spacing is not to exceed 0.75d = 346.8 mm. This distance x from thesupport 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 / 10³ = 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 × 450

2= 6.0


64/125




64

The service stress is

5 × 460 × 2631.2f s =

8 × 2945= 256.9 N/mm

2

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

is

477 – 256.90.55 +

120 × ( 0.9 + 6.0 )= 0.816 < 2.0

For the modification factor for compression reinforcement, with A’s.prov = 628

mm² 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.


65/125




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 f cu = 30 N/mm², f y = 460 N/mm² for the

longitudinal steel f yv and 250 N/mm² 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/met

strain, reinforced concrete, confinement

Documents