Design of RCC Columns as per IS 456:2000 || CIVIL Design

R.C.C. Design (As Per IS : 456-2000)

* Concrete is good in resisting compressive stress but is very weak in resisting tensile stress. Hence,

concrete structure is reinforced with steel wherever tension develops. Steel is the best

reinforcement since its tensile strength is quite high and the bond between steel and concrete is

excellent.

* Characteristic strength of concrete is defined as the compressive strength of 150 mm concrete cubes

at 28 days in N/mm2

, below which not more than 5 per cent cubes give the result. Table 7.1 gives

grades of concrete.

Nowadays ultra high strength concrete of M 500 are also produced in the laboratories and M 250

concrete has been used in bridge construction.

* Minimum grade of concrete for different exposure with normal weight aggregates of 20 mm nominal

maximum size are as shown below:

Mild – M : 20 Moderate M 25

Severe – M : 30 Very severe M 35

Extreme – M : 40

* Tensile strength: A designer may use the following expression for finding flexural tensile strength of

concrete.

Ft = 0.7 N/mm2

* As per IS 456–2000, modulus of elasticity for concrete

Ec = 5000

* Poisson’s ratio: m = 0.1 for high strength concrete

= 0.2 for weak concrete

Usually, m = 0.15 for strength calculation

and = 0.20 for serviceability calculation

* Shrinkage: Approximate value of the total shrinkage strain may be taken as 0.0003.

* Creep: The creep coefficient is defined as the ratio of ultimate creep strain to elastic strain at the

age of loading and it may be taken as given below:

7 days 2.2

28 days 1.6

1 year 1.1

* Grades of steel:

yield/0.2 % proof stress

Mild steel 255 N/mm2

HYSD bars 415 N/mm2 and 500 N/mm2

Hard drawn steel 480 N/mm2

* Characteristic strength fy

in tension and compression is taken equal.

* Young’s modulus: It is taken as 2 × 10

5 N/mm2

for all grades of steel.

* Available diameter of bars in mm are:

Mild steel: 6, 10, 12, 16, 20, 25, and 32

HYSD bars: 8, 10, 12, 16, 20, 22, 25, 28 and 32

* Arrangement of bars: They may be arranged singly or in pairs. Use of 3 to 4 bundled bars is also

permitted.

* Loads to be considered and their combinations are as per the IS 875–1987.

* Structural analysis may be by linear elastic theory. Classical methods or finite element method may

be used. Standard packages may be used.

7.1 METHODS OFR.C.C. DESIGN

* With an appropriate safety the structure should

1. sustain all loads expected in its life cycle

2. sustain deformations during and after construction

3. have adequate durability

4. have adequate resistance to misuse and fire.

Methods used for design are given below.

1. Working Stress Method The designer aims at keeping stresses at working loads as close to

permissible stresses as possible but without exceeding them. Permissible stress is defined as the

ultimate stress divided by a factor of safety in case of concrete. Factor of safety in concrete is taken

as 3. In case of steel permissible stress is yield stress or 0.2% proof stress divided by factor of

safety. Factor of safety is taken as 1.75 to 1.85 only, since steel is a more reliable material.

Modular ratio is taken as

m = .

where sebc = permissible compressive stress in bending.

Working stress method is conservative and hence being given up. However, for checking

serviceability conditions this method is also used now.

2. Load Factor Method (LFM) In this method ultimate load is used as design load and the collapse

criteria for the design. Ultimate load is defined as load factor times the working load. A load factor

of 2 was used.

3. Limit State Method (LSM) This is a comprehensive method which takes care of the structure not

only for its safety but its fitness throughout the period of service. Various limit states to be

considered are:

(a) Limit states of collapse: Tension, compression, flexure, tension and shear.

(b) Limit states of serviceability: Limit states of deflection and cracking.

(c) Other limit states: Limit states of vibration, fire resistance, chemical and environmental actions,

accidental or catastrophic collapse.

Characteristic Load

Characteristic load means the value of the load above which not more than 5 per cent results are

expected to fall. It may be taken as

Characteristic load = Mean load + kS

when for normal distribution k = 1.64 and S is standard deviation.

IS code 456–2000 permits use of the values given in IS 875 as characteristic loads.

Characteristic Strength

The strength that one can safely assume for a material is known as characteristic strength. It may be

taken as

Characteristic strength = Mean strength – kS

when k = 1.64 and S is standard deviation

Partial Safety Factor for Loads

gf =

where Fd = Design load and F = characteristic load.

Table 18 of IS 456–2000 gives the values for various loads and load combinations for limit state of

collapse and the limit state of serviceability.

Partial Safety Factor for Materials

gm =

where f = characteristic strength of the material

fd = design strength of material

Partial safety factor for material are as given in Table 7.2.

Table 7.2 Partial safety factor for material strength.

Idealized Stress-Strain Curves

Irrespective of grade, stress strain curve for concrete is taken as:

1. Parabolic for 0 to 0.002 strain

2. Constant for 0.002 to 0.0035 strain set

3. Ultimate strain 0.0035.

* Strength achieved in the structure in the field may be taken as 0.67 for (i.e., 2/3 fck

).

* In addition to the above, code recommend use of partial safety factor of 1.5 for concrete.

Hence, design stress in concrete = = 0.446 fck

.

= 0.45 fck

\ Stress block for a beam is as shown in Fig. 7.1.

For steel, partial safety factor used is 1.15:

\ Max stress in steel is restricted to

The maximum strain in the tensile reinforcement of failure

, i.e., 0.87 fy + 0.002

* For balanced section esu = + 0.002.

* If strain in steel reach 0.002 + earlier to compressive strain in concrete reaching 0.0035 the

section is called under reinforced section. The excess strain in steel beyond critical strain amounts

to considerable cracks in concrete. The deflection increases. They serve as warning. This type of

failure in under reinforced section is called primary tensile failure. Code recommends design

should be to such failures only.

Stress block parameters:

Area of stress block = 0.36 fck xu

where xu depth of neutral axis.

\ Compressive force in section

C = 0.36 fck bxu

Distance of centroid of compressive force from extreme compression flange

= 0.42 xu

.

Limiting value of xu

is given by

=

\ For mild steel = 0.93

For Fe 415 = 0.48

For Fe 500 = 0.46.

For under reinforced sections

=

Strength of rectangular section in flexure

Mu = 0.36 fck

= k fck bd

2

for balanced section

where k = 0.36

For mild steel Mu lim = 0.148 fck bd

2

For Fe 415, Mu lim = 0.138 fck bd

2

For Fe 500, Mu lim = 0.133 fck bd

2

.

Approximate expression for moment of resistance

Mu = 0.87 fy Ast

However, to avoid compression failures, the strength of such sections should be considered as that of

balanced section, i.e.,

Mu = Mu lim = 0.36 fck xu lim (d – 0.42 xu lim).

Flanged Section in Flexure

* Effective width of flanges:

For T-beams bf = + bw + 6Df

For L-beams bf = + bw + 3 Df

where lo = distance between points of zero moments in the beam.

Note: For continuous beams and frames lo may be taken as 0.7 × the effective span.

* Strength of flanged section in flexure: Based on the value of xu

, the following three cases arise:

1. xu £ Df

.

Mu = 0.36 fck bf xu

(d – 0.42 xu

).

2. xu

> Df

.

IS 456 says, if £ 0.2, treat that it belongs to this case. In this case

C = Cw + Cf

where Cw = 0.36 fck bw xu

.

and Cf = 0.446 (bf – bw) fck Df

To find depth of neutral axis

0.36 fck bw xu + 0.446 (bf – bw) fck Df = 0.87 fy Ast

.

\ Mu = 0.36 fck bw xu

(d – 0.42 xu

) + 0.446 fck

(bf – bw) × Df

(d – Df

/2)

3. xu

< Df or < 0.2 and xu

> Df

.

0.36 fck bw xu + 0.45 fck yf

(bf – bw) = 0.87 fy Ast

where yf = 0.15 xu + 0.65 Df

From it xu may be found.

Mu = 0.36 fck bw xu

(d – 0.42 xu

) + 0.45 fck yf

(bf – bw) × (d – 0.5 y6

)

Mulim = 0.36 fck bw xulim (1 – 0.42 xulim) + 0.45 fck yf

(bf – bw) (d – 0.5 yf

).

where yf = 0.15 xu lim + 0.65 Df but not more than Df

.