ce6505 design of reinforced concrete elements l t p c …

CE6505 DESIGN OF REINFORCED CONCRETE ELEMENTS L T P C 3 0 0 3

OBJECTIVES:

To introduce the different types of philosophies related to design of basic structural elements such as slab, beam, column and footing which form part of any structural system with reference to Indian standard code of practice.

UNIT I METHODS OF DESIGN OF CONCRETE STRUCTURES 9 Concept of Elastic method, ultimate load method and limit state method – Advantages of Limit State Method over other methods – Design codes and specification – Limit State philosophy as detailed in IS code – Design of beams and slabs by working stress method.

UNIT II LIMIT STATE DESIGN FOR FLEXURE 9 Analysis and design of singly and doubly reinforced rectangular and flanged beams - Analysis and design of one way, two way and continuous slabs subjected to uniformly distributed load for various boundary conditions.

UNIT III LIMIT STATE DESIGN FOR BOND, ANCHORAGE SHEAR & TORSION 9 Behaviour of RC members in bond and Anchorage - Design requirements as per current code - Behaviour of RC beams in shear and torsion - Design of RC members for combined bending shear and torsion.

UNIT IV LIMIT STATE DESIGN OF COLUMNS 9 Types of columns – Braced and unbraced columns – Design of short Rectangular and circular columns for axial, uniaxial and biaxial bending.

UNIT V LIMIT STATE DESIGN OF FOOTING 9 Design of wall footing – Design of axially and eccentrically loaded rectangular pad and sloped footings – Design of combined rectangular footing for two columns only.

TOTAL: 45 PERIODS

OUTCOMES:

The student shall be in a position to design the basic elements of reinforced concrete structures.

TEXT BOOKS: 1. Varghese, P.C., “Limit State Design of Reinforced Concrete”, Prentice Hall of India, Pvt.

Ltd., New Delhi, 2002.2. Gambhir.M.L., "Fundamentals of Reinforced Concrete Design", Prentice Hall of India

Private Limited, New Delhi, 2006.3. Subramanian,N.,”Design of Reinforced Concrete Structures”,Oxford University Press, New

Delhi, 2013.

REFERENCES: 1. Jain, A.K., “Limit State Design of RC Structures”, Nemchand Publications, Roorkee, 19982. Sinha, S.N., “Reinforced Concrete Design”, Tata McGraw Hill Publishing Company Ltd.,

New Delhi, 20023. Unnikrishna Pillai, S., Devdas Menon, “Reinforced Concrete Design”, Tata McGraw Hill

Publishing Company Ltd., 20094. Punmia.B.C., Ashok Kumar Jain, Arun Kumar Jain, “Limit State Design of Reinforced

Concrete”,Laxmi Publication Pvt. Ltd., New Delhi, 2007.5. Bandyopadhyay. J.N., "Design of Concrete Structures"., Prentice Hall of India Pvt.

Ltd., New Delhi, 2008.6. IS456:2000, Code of practice for Plain and Reinforced Concrete, Bureau of Indian

Standards, New Delhi, 20007. SP16, IS456:1978 “Design Aids for Reinforced Concrete to Bureau of Indian Standards,

New Delhi, 19998. Shah V L Karve S R., "Limit State Theory and Design of Reinforced Concrete", Structures

Publilcations, Pune, 2013

Sl.No Contents Page No.

UNIT I METHODS OF DESIGN OF CONCRETE STRUCTURES

1.1 Introduction 1

1.2 Design loads 2

1.3 Materials for Reinforced Concrete 4

1.4 Design codes and Hand books 6

1.5 Design Philosophies 7

1.6 Partial safety factor 8

1.7 Characteristic strength and loads 10

1.8 Working stress metnod design 11

1.9 Design Problems 141.9 Design Problems 14

UNIT II WATER TANKS

2.1 Pre-requisite Discussion 21

2.2 General specification for flexure design of beams 23

2.3 General Aspects of Serviceability 24

2.4 Limit state of serviceability for flexural members 25

2.5 Water Tanks 26

2.6 Cracking in structural members 26

2.7 Design Problems 28

UNIT III LIMIT STATE DESIGN FOR BOND, ANCHORAGE SHEAR& TORSION

3.1 Design of torsion 36

3.2 Equilibrium torsion and compatibility torsion 36

3.3 Estimation of torsional stiffness 39

Sl.No Contents Page No.


UNIT IV LIMIT STATE DESIGN OF COLUMNS


4.2 Introduction 48

4.3 Classification of columns 49

4.4 Behaviour of tied and bspiral columns 50

4.5 Specifications for covers and reinforcement in column 50

4.6 Effective length of compression member 51

4.7 Specifications as per IS: 456-2000 52

4.8 Transverse reinforcement 534.8 Transverse reinforcement 53

4.9 Limit state design of collapse 55

4.10Procedure for using of Non-dimensional Interaction Diagrams as DesignAids to find steel

58


UNIT V LIMIT STATE DESIGN OF FOOTING


5.2 Types of foundations 70

5.2.1 Shallow foundations 70

5.2.2 Deep foundations 73

5.3 Bearing Capacity of Soil 74

5.4 Design of Isolated Column Footing 74

5.5 Specifications for Design of footings as per IS 456 : 2000 75


1.1 Introduction:

A structure refers to a system of connected parts used to support forces (loads). Buildings,

bridges and towers are examples for structures in civil engineering. In buildings, structure

consists of walls floors, roofs and foundation. In bridges, the structure consists of deck,

supporting systems and foundations. In towers the structure consists of vertical, horizontal and

diagonal members along with foundation.

A structure can be broadly classified as (i) sub structure and (ii) super structure. The portion of

building below ground level is known as sub-structure and portion above the ground is called as

super structure. Foundation is sub structure and plinth, walls, columns, floor slabs with or

without beams, stairs, roof slabs with or without beams etc are super structure.

Many naturally occurring substances, such as clay, sand, wood, rocks natural fibers are used to

construct buildings. Apart from this many manmade products are in use for building

construction. Bricks, tiles, cement concrete, concrete blocks, plastic, steel & glass etc are

manmade building materials.

Cement concrete is a composites building material made from combination of aggregates (coarse

and fine) and a binder such as cement. The most common form of concrete consists of mineral

aggregate (gravel & sand), Portland cement and water. After mixing, the cement hydrates and

eventually hardens into a stone like material. Recently a large number of additives known as

concrete additives are also added to enhance the quality of concrete. Plasticizers, super

plasticizers, accelerators, retarders, pazolonic materials, air entertaining agents, fibers, polymers

and silica furies are the additives used in concrete. Hardened concrete has high compressive

strength and low tensile strength. Concrete is generally strengthened using steel bars or rods

known as rebars in tension zone. Such elements are “reinforced concrete” concrete can be

moulded to any complex shape using suitable form work and it has high durability, better

appearance, fire resistance and economical. For a strong, ductile and durable construction the

reinforcement shall have high strength, high tensile strain and good bond to concrete and thermal

compatibility. Building components like slab walls, beams, columns foundation & frames are

constructed with reinforced concrete. Reinforced concreted can be in-situ concreted or precast

concrete.

For understanding behavior of reinforced concrete, we shall consider a plain concrete beam

subjected to external load as shown in Fig. 1.1. Tensile strength of concrete is approximately

one-tenth of its compressive strength.

UNIT I METHODS OF DESIGN OF CONCRETE STRUCTURES

CE6505-Design of RC Elements III Year/ V Semester Civil Engineering

K.ChandraBose 1 Academic Year 2015-2016

Hence use of plain concrete as a structural material is limited to situations where significant

tensile stresses and strains do not develop as in solid or hollow concrete blocks , pedestal and in

mass concrete dams. The steel bars are used in tension zone of the element to resist tension as

shown in Fig 1.2 The tension caused by bending moment is chiefly resisted by the steel

reinforcements, while concrete resist the compression. Such joint action is possible if relative

slip between concrete and steel is prevented. This phenomena is called “bond”. This can be

achieved by using deformed bass which has high bond strength at the steel-concrete interface.

Rebars imparts “ductility” to the structural element, i.e RC elements has large deflection before

it fails due to yielding of steel, thus it gives ample warning before its collapse.

1.2 Design Loads

For the analysis and design of structure, the forces are considered as the “Loads” on the

structure. In a structure all components which are stationary, like wall, slab etc., exert forces

due to gravity, which are called as “Dead Loads”. Moving bodies like furniture, humans etc

exert forces due to gravity which are called as “Live Loads”. Dead loads and live loads are

gravity forces which act vertically down ward. Wind load is basically a horizontal force due to

wind pressure exerted on the structure. Earthquake load is primarily a horizontal pressure

exerted due to movement of the soil on the foundation of a structure. Vertical earthquake

force is about 5% to 10% of horizontal earthquake force. Fig. 1.3 illustrates the loads that are

considered in analysis and design.



Bead loads Live loads Load due Vibrator Fatigue

(Fixes) (movable) (to settlement)

Self due to occupancy environment Harmonic Impact shock blast

Weigth fixed (ex: water flow) or

furniture random

or

equipment

Internal Wind forces

Force

Machinery Ground

Induced vibration

Ex: Earthquake

IS875 -1987 part 1 gives unit weight of different materials, Part – 2 of this code describes live

load on floors and roof. Wind load to be considered is given in part 3 of the code. Details of

earthquake load to be considered is described in 1893 – 2002 code and combination of loads is

given in part 5 of IS875 – 1987.

Static Dynamic

Loads

Fig 1.3 Types of loads on Structure



Concrete

Concrete is a composite material consists essentially of

a) A binding medium cement and water called cement paste

b) Particles of a relatively inert filler called aggregate

The selection of the relative proportions of cement, water and aggregate is called “mix design”

Basic requirement of a good concrete are workability, strength, durability and economy.

Depending upon the intended use the cement may be OPC (33,43 & 53 Grade), Rapid hardening

cements Portland slag, Portland pozzolona etc. High cement content give rise to increased

shrinkage, creep and cracking. Minimum cement content is 300Kg/m3 and maximum being

450Kg/m3 as per Indian code. Mineral additives like fly ash , silica fume, rice husk ash,

metakoline and ground granulated blast furnace slag may be used to reduce micro cracks . The

aggregate used is primarily for the purpose of providing bulk to the concrete and constitutes 60

to 80 percent of finished product. Fine aggregates are used to increase the workability and

uniformly of concrete mixture. Water used for mixing and curing shall be clean and free from

oil, acids, alkalis, salts, sugar etc. The diverse requirements of mixability, stability,

transportability place ability, mobility, compatibility of fresh concrete are collectively referred

to as workability.

Compressive strength of concrete on 28th

day after casting is considered as one of the measure

of quality. At least 4 specimens of cubes should be tested for acceptance criteria.

Grade of concrete

Based on the compressive strength of concrete, they are designated with letter H followed by

an integer number represented characteristic strength of concrete, measured using 150mm size

cube. Characteristic strength is defined us the strength of material below which not more than

5% of test results are expected to full. The concrete grade M10, M15 and M20 are termed as

ordinary concrete and those of M25 to M55 are termed as standard concrete and the concrete

of grade 60 and above are termed as high strength concrete. The selection of minimum grade

of concrete is dictated by durability considerations which are based on kind of environment to

which the structure is exposed, though the minimum grade of concrete for reinforced concrete

is specified as M20 under mild exposure conditions, it is advisable to adopt a higher grade. For

moderate, severe, very severe and extreme exposure conditions, M25, M30, M35 & M40

grades respectively are recommended. Typical stress-strain curves of concrete is shown in

Fig.1.4

1.3. Materials for Reinforced Concrete



Reinforcing steel

Steel bars are often used in concrete to take case of tensile stresses. Often they are called as

rebars, steel bar induces ductility to composite material i.e reinforced concrete steel is stronger

than concrete in compression also. Plain mild steel bars or deformed bars are generally used.

Due to poor bond strength plain bars are not used. High strength deformed bars generally cold

twisted (CTD) are used in reinforced concrete. During beginning of 21st century, Thermo-

mechanical tream (TMT) bars which has ribs on surface are used in reinforced concrete. Yield

strength of steel bars are denoted as characteristic strength. Yield strength of mild steel is

250MPa, yield strength of CTD &TMT bars available in market has 415 MPa or 500 MPa or

550MPa. TMT bars have better elongation than CTD bars. Stress-strain curve of CTD bars or

TMT bars do not have definite yield point, hence 0.2% proof stress is used as yield strength. Fig

1.5 shows stress strain curve of different steel grades. Steel grades are indicated by Fe followed

by yield strength. In the drawings of RCC, � denotes MS bar and # denotes CTD or TMT bars



A code is a set of technical specifications intended to control the design and construction. The

code can be legally adopted to see that sound structure are designed and constructed code

specifies acceptable methods of design and construction to produce safe and sound structures.

National building code have been formulated in different countries to lay down guidelines for

the design and construction of structures. International building code has been published by

international code council located in USA. National building code (NBC – 2005) published in

India describes the specification and design procedure for buildings.

For designing reinforced concrete following codes of different countries are available

India – IS456 – 2000 – Plain and reinforced concrete code practice.

USA - ACI 318-2011 – Building code requirements for Structural concrete (American concrete

institute)

UK - BS8110 –part1 – structural use of concrete –code of practice for design and

construction. (British standard Institute)

Europe – EN 1992(Euro code 2) - Design of concrete structures

Canada – CAN/CSA – A23.3-04 - Design of concrete structures (Reaffirmed in 2010),

Australia – As 3600 -2001 – concrete structures.

Germany – Din 1045 – Design of concrete structures

Russia – SNIP

China - GB 50010 -2002 code for design of concrete structures to help the designers, each

country has produced “handbook”. In India following hand books called special publication are

available.

Fig 1-5 stress –strain curve

1.4 Design codes and Hand books



SP – 16-1980- Design Aid for Reforced concrete to IS456-1978

SP – 23-1982- Hand book on concrete mixes

Sp – 24 -1983 – Explanatary hand book on IS456 – 1978

SP – 34-1987 - Hand book on concrete reinforcement and detailing.

Structural design is process of determining the configuration (form and proportion) of a

structure subject to a load carrying performance requirement. Form of a structure describes

the shape and relative arrangements of its components. The determination of an efficient

form is basically a trial and error procedure.

In the beginning of 20th

century (1900 to 1960) to late 50’s of this century, members were

proportioned so that stresses in concrete and steel resulting from service load were within the

allowable stresses. Allowable stresses were specified by codes. This method of design is called

“working stress method” (WSM). This method of design resulted in conservative sections and

was not economical. This design principle satisfies the relation �

��> �.

Where R is resistance of structural element, RS is factor of safety and L is applied external load.

In 1950’s ultimate load method or load factor method was developed. In this method, using

non linear stress – strain curve of concrete and steel, the resistance of the element is

computed. The safety measure in the design is introduced by an appropriate choice of the load

factor (ultimate load/working load). Different load factors are assigned for different loads.

Following equations are used for finding ultimate load as per IS456 – 1964

U = 1.5 DL + 2.2 LL

U = 1.5 DL +2.2LL to 5WL or 1.5 DL +0.5LL +2.2 WL

Here DL = Dead load, LL = Live load WL= wind load or earthquake load. The design principle

should satisfy R≥LF etc or R ≥ U, Where, R= Resistance, LF= Load factor, L= load. Ultimate load

method generally results in more slender section, but leads to larger deformation. Due to the

disadvantage of larger deflection, this method was discontinued. To over come the

disadvantages of working stress method and ultimate method, a probabilistic design concept

called as “Limit state method, was developed during 1970’s. IS456 -1978 recommended this

method and is continued in 2000 version also. This method safe guards the risk of both

collapse and unserviceability. Limits state method uses multiple safety factory format, which

attempt to provide adequate safety at ultimate loads or well as a denote serviceability at

1.5 Design Philosophies�



service loads by considering all limit states, The acceptable limit for safety and serviceability

requirements before failure or collapse is termed as “Limit state” Two principal limit states are

considered i.e 1. Limit state of collapse 2. Limit state of serviceability. The limit state of

collapse include one or more of i) flexure, II) shear, III) torsion and IV) compression the limit

state of collapse is expressed as µR>ΣXiLi Where, µ and λ are partial safety factors, Here µ<1 &

λ>1. The most important limit state considered in design are of deflection, other limit state of

serviceability are crack and vibration. For deflection δmax ≤ �

∝ where δmax is maximum

deflection, l=span 4 ∝ is an integer numbers. For over all deflection ∝ is 250 and for short term

deflection ∝ =350.

Design strength = ��

γm

Design load = γf x characteristic load

As per clause 36.4.2 page 68 of IS 456, γm= 1.5 for concrete and γm =1.15 for steel. Similarly

clause 36.4.1 page 68 of code gives γf in table 18 for different values for different load

combinations and different limit states.

IS 456 – 2000 Recomendations

(i) Partial safety factors for materials to be multiplied with characteristic

strength is given below.

Values of partial safety factor rm

Material Limit state

Collapse Deflection Cracking

Concrete 1.5 1.0 1.3

Steel 1.15 1.0 1.0

Design strength �� =��

γ�

1.6 Partial safety factor

To account for the different conditions like for material strength, load etc. Different partial

factors are used for material and load. M indicate safety factor for material & for load



(ii) Partial safety factors for loads to be multiplied with characteristic load is

given below.

Value of partial safety factors γγγγf

Load combination Ultimate limit

state

Serviceability limit

state

1) Dead load & live load 1.5(DL+LL) DL+LL

2) Dead seismic/wind load

a) Dead load contributes to

Stability

b) Dead load assists overturning

0.9DL+1.5(E2/WL)

1.5(DL+E2/WL)

DL + EQ/WL

DL+EQ/wL

3) Dead, live load and Seismic/wind load 1.2(DL+LL+EQ/WL) Dl+0.LL+0.8EQ/WL

DL-Dead load, LL- Live load WL- Wind load EQ- Earthquake load

(iii) The code has suggested effective span to effective depth ratios as given

below

Basic effective span to effective depth ratio (l/l) basic

Type of beam one /slab Span≤10m Span>10m

1)Cantilever 7 Deflection should be

Be calculated

2) Simply supported 20 (20X10)/span

3)continuous beam 26 (26X10)/span

The above values are to be modified for (i) the type and amount of tension steel (Fig 4 page 38

of T5456-2000)

(ii) The amount of compression steel (Fig 5 page 39 of I5456-2000)

(iv) The type of beam ie flanged beams etc (Fig 6 page 39 of I5456 – 2000).



For slabs spanning in two directions, the l/d ratio is given below.

For slabs spanning in two directions, the l/d ratio is given below

Type of slab l/d for grade of steel

Fe250 Fe415

1)Simply supported 35 28

2) Continuous 40 32

∴ fk = fm -1.646 � (2.6) here fm = mean strength.

Similarly “characteristic load” is that value of load which has an accepted probability of not

being exceeded during the life span of structure. In practice the load specified by IS875 – 1987

is considered as characteristic load. Equation for characteristic load is

Lk = Lm + 1.64�.

1.7 Characteristic strength and loads

Limit state method is based on statistical concepts. Strength of materials and loads are highly

variable in a range of values. The test in laboratory on compressive strength of concrete has

indicated coefficient of variation of ±10%. Hence in reinforced concrete construction, If is not

practicable to specify a precise cube strength. Hence in limit state design uses the concept of

“characteristic strength” fck indicates characteristics strength of concrete & by characteristic

strength of steel. In general fk indicates the characteristic strength of material.



(a) General features

method extensively during the 20th

century and presently the method is incorporated as an alternative to

the limit state method in Annexure – B of the recently revised Indian Standard Code Is : 456 – 2000 for

specific applications.

The permissible stresses in concrete under service loads for the various stress states of compressive,

flexure and bond is compiled in Table 2.1 (Table 21 of IS ; 456 – 2000)

The permissible stress in different types of steel reinforcement is shown in table 2.2 (Table 22 of IS 456

– 2000)

The permissible shear stress for various grades of concrete in beams is shown in Table 12.1 (Table 23 of

IS: 456 – 2000)

The maximum shear stress permissible in concrete for different grades is shown in Table 12.2 Table

12.2 (Table 24 of IS: 456 – 2000)

In the case of reinforced concrete slabs, the permissible shear stress in concrete is obtained by

multiplying the values given in Table 2.1 by factor „k‟ whose values depend upon the thickness of slab

as shown in Table 12.3 (Section 40.2.1.1. of IS; 456 – 2000)

Table 12.1Permissible Shear Stresses in Concrete (τc N/mm2) (Table 23 of IS:456 – 2000)

100 As / bd Permissible shear stresses in concrete τc

N/mm2

M15 M20 M25 M30 M35 M40 &

During the early part of 20th

century, elastic theory of reinforced concrete sections outlined was

developed which formed the basis of the working stress or permissible stress method of design of

reinforced concrete members. In this method, the working or permissible stress in concrete and steel are

obtained applying appropriate partial safety factors to the characteristics strength of the materials. The

permissible stresses in concrete and steel are well within the linear elastic range of the materials.

The design based on the working stress method although ensures safety of the structures at working or

services loads, it does not provide a realistic estimate of the ultimate or collapse load of the structure in

contrast to the limit state method of design. The working stress method of design results in

comparatively larger and conservative sections of the structural elements with higher quantities of steel

reinforcement which results in conservative and costly design. Structural engineers have used this

1.8 WORKING STRESS METHOD DESIGN

GENERAL PRINCIPLES OF WORKING STRESS DESIGN



ABOVE

≤ 0.15 0.18 0.18 0.19 0.20 0.20 0.20

0.25 0.22 0.22 0.23 0.23 0.23 0.23

0.50 0.29 0.30 0.31 0.31 0.31 0.32

0.75 0.34 0.35 0.36 0.37 0.37 0.38

1.00 0.37 0.39 0.40 0.41 0.42 0.42

1.25 0.40 0.42 0.44 0.45 0.45 0.46

1.50 0.42 0.45 0.46 0.48 0.49 0.49

1.75 0.44 0.47 0.49 0.50 0.52 0.52

2.00 0.44 0.49 0.51 0.53 0.54 0.55

2.25 0.44 0.51 0.53 0.55 0.56 0.57

2.50 0.44 0.51 0.55 0.57 0.58 0.60

2.75 0.44 0.51 0.56 0.58 0.60 0.62

3.00 & above 0.44 0.51 0.57 0.60 0.62 0.63

Note: As is that area of longitudinal tension reinforcement which continues at least one effective depth

beyond the section being considered except at supports where the full area of tension reinforcement may

be used provided the detailing conforms to 26.2.3.

Table 12.2 Maximum Shear Stress (τc, max N/mm2) (Table 24 of IS: 456 – 2000)

Concrete grade

(τc max N/mm2)

M – 15

1.6

M – 25

1.8

M – 30

1.9

M – 35

2.3

M – 40 & above

2.5

The maximum shear stress permissible in concrete for different grades is shown in Table 12.2 (Table 24

of IS 456 – 2000)

In the case of reinforced concrete slabs, the permissible shear stress in concrete is obtained by

multiplying the3 values in Table 2.1 by a factor „k‟ whose values depend upon the thickness of slab as

shown in Table 12.3 (Section 40.2.1.1. of IS 456 – 2000)

(b) General design procedure

In the working stress design, the cross – sectional dimensions are assumed based on the basic span /

depth ratios outlined in Chapter 5 (Table 5.1 and 5.2) (Section 23.2.1. of IS: 456 – 2000)

The working load moments and shear forces are evaluated at critical sections and the required effective

depth is checked by using the relation:

d = √ M / Q.b

Where d = effective depth of section

M = working load moment

b = width of section

Q = a constant depending upon the working stresses in concrete and steel, neutral axis depth

factor (k) and lever arm coefficient (f).

For different grades of concrete and steel the value of constant „Q‟ is compiled in Table 2.3. The depth

provided should be equal to or greater than the depth computed by the relation and the area of

reinforcement required in the section to resist the moment „M‟ is computed using the relation:

Ast = ( M/ σst . j. d



The number of steel bars required is selected with due regard to the spacing of bars and cover

requirements.

After complying with flexure, the section is generally checked for resistance against shear forces by

calculating the nominal shear stress τc given by τv = (V / bd)

Where V = Working shear force at critical section.

The permissible shear stress in concrete (τc) depends upon the percentage reinforcements in the cross –

section and grade of concrete as shown in Table 12.1

If τc < τv suitable shear reinforcements are designed in beams at a spacing sv given by the relation;

Sv = [ 0.87 fy Asv d / Vus]

Where sv = spacing of stirrups

Asv = cross – sectional area of stirrups legs

fy = Characteristics strength of stirrup reinforcement

d = effective depth

Vs = [ V – τc .b .d]

If τv < τc, nominal shear reinforcements are provided in beams are provided in beams at a spacing given

by

Sv [ 0.87 fy Ast / 0.4 b]

In case of slabs, the permissible shear stress if „k‟ is a constant depending upon the thickness of the slab.

Also in the case of slabs the nominal shear stress (τv) should not exceed half the value of τc max shown in

Table 12.2. In such cases the thickness of the slab is increased and the slab is redesigned.

In the case of compression members, the axial load permissible on a short column reinforced with

longitudinal bars and lateral ties is given by

P = (σcc Ac + σsc Asc)

Where scc = permissible stress in concrete in direct compression (Refer Table 2.1)

Ac = cross – sectional area of concrete excluding the area of reinforcements.

Ssc = permissible compressive stress in reinforcement

Asc = cross – sectional area of longitudinal steel bars.



1. Design a R.C beam to carry a load of 6 kN/m inclusive of its own weight on an

effect span of 6m keep the breath to be 2/3 rd of the effective depth .the permissible

stressed in the concrete and steel are not to exceed 5N/mm2 and 140 N/mm2.take

m=18. (NOV-DEC 2012)

Step 1: Design constants.

Modular ratio, m =18.

A Coefficient n σbc.m/(σbc.m + σst) 0.39

Lever arm Coefficient, j=1-(n/3) = 0.87

Moment of resistance Coefficient Q σbc/2. n. j 0.84

Step 2: Moment on the beam.

M = (w.l2)/8 = (6x62)/8 = 27kNm

M = Qbd2

d2 = M/Qb = (27x106)/ (0.84x2/3xd)

d = 245mm.

Step 3: Balanced Moment.

Mbal = Qbd2 = 0.84x245x3652 = 27.41kNm. > M. it can be designed as singly

1.9 Design Problems:



reinforced section.

Step 4: Area of steel.

Ast = Mbal / (σst.j.d) 616.72mm2

Use 20mm dia bars ast π/4 (202) = 314.15mm2

No. of bars = Ast/ast = 616.72/314.15 = 1.96 say 2nos.

Provide 2#20mm dia bars at the tension side.

2. Design a doubly reinforced beam of section 240X500mm to carry a bending

moment of 80kNm.Assume clear cover at top a bottom as 30mm and take

m=18.adopt working stress method. Assume the permissible stressed in the

concrete and steel are not to exceed 5N/mm2 and 140 N/mm2.







M = 80kNm

M = Qbd2

D = 500mm, b = 240mm

d = 500-30mm = 470mm

.Step 3: Balanced Moment.

Mbal = Qbd2 = 0.84x240x4702 = 44.53kNm. < M. it can be designed as doubly



reinforced section.

Step 4: Area of Tension steel.

Ast = Ast1 + Ast2

Ast1 = Mbal / (σst.j.d) (44.53x106)/(140x0.87x470) = 777.87mm2



Ast2 = (M-Mbal) / (σst.(d-d1)) = (80x106-44.53x106)/(140x(470-30)) = 575.8mm2



Step 5: Area of Compression steel:

Asc = (M-Mbal) / (σsc.(d-d1)) = (80x106-44.53x106)/(51.8x(470-30))=1580.65 mm2



Provide 6#20mm dia bars as compression reinforcement.

3. Design a beam subjected to a bending moment of 40kNm by working stress

design. Adopt width of beam equal to half the effective depth. (NOV-DEC 2010)

Assume the permissible stressed in the concrete and steel are not to exceed

5N/mm2 and 140 N/mm2.take m=18.









M = 40kNm

M = Qbd2

d2 = M/Qb = (40x106)/ (0.84x1/2xd)

d = 456.2 say 460 mm.

b = ½ d = 0.5x460 = 230mm



reinforced section.


Ast = Mbal / (σst.j.d) (40.88x106)/(140x0.87x460) = 729.64mm2




4. Determine the moment of resistance of a singly reinforced beam 160X300mm

effective section, if the stress in steel and concrete are not to exceed 140N/mm2

and 5N/mm2.effectve span of the beam is 5m and the beam carries 4 nos of 16mm

dia bars. Take m=18.find also the minimum load the bam can carry. Use WSD

method. (NOV-DEC 2009)



Step 1: Actual NA.

b xa2/2 = m.Ast.(d- xa)

160. xa2/2 = 18 X 804.24(300 –xa)

Xa = 159.42mm

Step 2: Critical NA.

xc σbc.d/(σst/.m + σcbc) 117.39mm < Xa 159.42mm

it is Over reinforced Section.

Step 3: Moment of Resistance

M (b. xa/2 .σcbc )(d- xa/3) = (160x159.42/2x5)(300-159.42/3) = 15.74kNm

Step 4: Safe load.

M = (w.l2)/8

W = (8 x 15.74)/52 = 5.03 kN/m

5. Design an interior panel of RC slab 3mX6m size, supported by wall of 300mm

thick. Live load on the slab is 2.5kN/m2.the slab carries 100mm thick lime

concrete (density 19kN/m3).Use M15 concrete and Fe 415 steel. (NOV-DEC

2009)

Step 1: Type of Slab.

ly/lx = 6/3 = 2 = 2.it has to be designed as two way slab.

Step 2:Effective depth calculation.

For Economic consideration adopt shorter span to design the slab.

d = span/(basic value x modification factor) = 3000/(20x0.95) = 270mm

D = 270 + 20 + 10/2 = 295mm

Step 3: Effective Span.

For shorter span:

Le = clear span + effective depth = 3000 + 270 = 3.27m (or)

Le =c/c distance b/w supports = 3000 + 2(230/2) =3.23m



Adopt effective span = 3.23m least value.

For longer span:


Le =c/c distance b/w supports = 6000 + 2(230/2) = 6.23m


Step 4: load calculation

Live load = 2.5kN/m2

Dead load = 1x1x0.27x25 = 6.75kN/m2


Floor Finish = 1kN/m2

Total load = 12.15kN/m2

Factored load = 12.15 x 1.5 = 18.225kN/m2

Step 5: Moment calculation.

Mx αx . w . lx 0.103x18.225x3.23 = 9.49kNm

My αy . w . lx 0.048 x18.225x3.23 = 4.425kNm

Step 6: Check for effective depth.

M = Qbd2

d2 = M/Qb = 9.49/2.76x1 = 149.39mm say 150mm.

For design consideration adopt d = 150mm.

Step 7: Area of Steel.

For longer span:

Mu = 0.87 fy Ast d (1- (fy ast)/(fck b d))

4.425x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150))

Ast = 180mm2

Use 10mm dia bars

Spacing ,S = ast/Astx1000 = (78.53/300)1000 = 261mm say 260mmc/c

Provide 10mm dia @260mm c/c.

For shorter span:




9.49x106 = 087x415xAstx150(1-(415

Ast)/(20x1000x150)) Ast = 200mm2

Use 10mm dia bars



6. A reinforced concrete rectangular section 300 mm wide and 600 mm overall depth

is reinforced with 4 bars of 25 mm diameter at an effective cover of 50 mm on the

tension side. The beam is designed with M 20 grade concrete and Fe 415 grade

steel. Determine the allowable bending moment and the stresses developed in steel

and concrete under this moment. Use working stress method. (MAY JUNE 2009)

Step 1: Actual NA.


300. xa2/2 = 18 X 1963.50(550 –xa) Xa = 117.81mm


xc σbc.d/(σst/.m + σcbc) = 194.66mm > Xa =

117.81mm it is Under reinforced Section.


For steel:

M = (Ast.σst )(d- xa/3) = (1963.5x230)(550-117.81/3) = 230.64kNm

For concrete:




A structural member that support transverse (Perpendicular to the axis of the member)

load is called a beam. Beams are subjected to bending moment and shear force. Beams are

also known as flexural or bending members. In a beam one of the dimensions is very large

compared to the other two dimensions. Beams may be of the following types:

a. Singly or doubly reinforced rectangular beams

Fig 1: Singly reinforced rectangular beam

Fig 2: Doubly reinforced rectangular beam

UNIT – 2

LIMIT STATE DESIGN FOR FLEXURE

2.1 Pre-requisites

Beam



b. Singly or doubly reinforced T-beams

Fig 3: Singly reinforced T beam

Fig 4: Doubly reinforced T beam

c. Singly or doubly reinforced L-beams



Fig 5: Singly reinforced L beam

Fig 6: Doubly reinforced L beam

Beams are designed on the basis of limit state of collapse in flexure and checked for

other limit states of shear, torsion and serviceability. To ensure safety the resistance to

bending, shear, torsion and axial loads at every section should be greater than the appropriate

values at that produced by the probable most unfavourable combination of loads on the

structure using the appropriate safety factors. The following general specifications and

practical requirements are necessary for designing the reinforced cement concrete beams.

a. Selection of grade of concrete

Apart from strength and deflection, durability shall also be considered to

select the grade of concrete to be used. Table 5 of IS 456:2000 shall be referred for

the grade of concrete to be used. In this table the grade of concrete to be used is

recommended based on the different environmental exposure conditions.

b. Selection of grade of steel

Normally Fe 250, Fe 415 and Fe 500 are used. In earthquake zones and other

places where there are possibilities of vibration, impact, blast etc, Fe 250 (mild steel)

is preferred as it is more ductile.

c. Size of the beam

The size of the beam shall be fixed based on the architectural requirements,

placing of reinforcement, economy of the formwork, deflection, design moments and

shear. In addition, the depth of the beam depends on the clear height below the beam

and the width depends on the thickness of the wall to be constructed below the beam.

The width of the beam is usually equal to the width of the wall so that there is no

projection or offset at the common surface of contact between the beam and the wall.

The commonly used widths of the beam are 115 mm, 150 mm, 200 mm, 230

mm, 250 mm, 300 mm.

d. Cover to the reinforcement

2.2 General specification for flexure design of beams



Cover is the certain thickness of concrete provided all round the steel bars to

give adequate protection to steel against fire, corrosion and other harmful elements

present in the atmosphere. It is measured as distance from the outer concrete surface

to the nearest surface of steel. The amount of cover to be provided depends on the

condition of exposure and shall be as given in the Table 16 of IS 456:2000. The cover

shall not be less than the diameter of the bar.

e. Spacing of the bars

The details of spacing of bars to be provided in beams are given in clause

26.3.2 of IS 456. As per this clause the following shall be considered for spacing of

bars.

The horizontal distance between two parallel main bars shall usually be not less than

the greatest of the following

i. Diameter of the bar if the diameters are equal

ii. The diameter of the larger bar if the diameters are unequal

iii. 5mm more than the nominal maximum size of coarse aggregate

Greater horizontal spacing than the minimum specified above should be

provided wherever possible. However when needle vibrators are used, the horizontal

distance between bars of a group may be reduced to two thirds the nominal maximum

size of the coarse aggregate, provided that sufficient space is left between groups of

bars to enable the vibrator to be immersed.

Where there are 2 or more rows of bars, the bars shall be vertically in line and

the minimum vertical distance between the bars shall be of the greatest of the

following

i. 15 mm

ii. Maximum size of aggregate

iii. Maximum size of bars

Maximum distance between bars in tension in beams:

The maximum distance between parallel reinforcement bars shall not be greater than

the values given in table 15 of IS 456:2000.

The members are designed to withstand safely all loads liable to act on it throughout

its life using the limit state of collapse. These members designed should also satisfy the

serviceability limit states. To satisfy the serviceability requirements the deflections and

cracking in the member should not be excessive and shall be less than the permissible values.

Apart from this the other limit states are that of the durability and vibrations. Excessive

values beyond this limit state spoil the appearance of the structure and affect the partition

walls, flooring etc. This will cause the user discomfort and the structure is said to be unfit for

use.

2.3 General Aspects of Serviceability:



The different load combinations and the

used for the limit state of serviceability are given in Table 18 of IS 456:2000.

Deflection

The check for deflection is done through the following two

456:2000 (Refer clause 42.1)

1 Empirical Method

In this method, the deflection criteria of the member is said to be satisfied

when the actual value of span to depth ratio of the member is less than the

permissible values. The IS code

are as given below

a. Choosing the basic values of span to effective depth ratios (l/d) from the

following, depending on the type of beam

1. Cantilever = 8

2. Simply supported = 20

3. Continuous

b. Modify the value of basic span to depth ratio to get the allowable span to depth

ratio.

Allowable l/d = Basic l/d x M

Where, Mt = Modification factor obtained from fig 4 IS 456:2000. It depends

on the area of tension

Mc = Modification factor obtained from fig 5 IS 456:2000. This depends on

the area of compression steel used.

Mf = Reduction factor got from fig 6 of IS 456:2000

The different load combinations and the corresponding partial safety factors to be

used for the limit state of serviceability are given in Table 18 of IS 456:2000.

The check for deflection is done through the following two methods specified by IS



permissible values. The IS code procedure for calculating the permissible values

Choosing the basic values of span to effective depth ratios (l/d) from the

following, depending on the type of beam

Cantilever = 8

Simply supported = 20

= 26

the value of basic span to depth ratio to get the allowable span to depth

Allowable l/d = Basic l/d x Mt x Mc x Mf

= Modification factor obtained from fig 4 IS 456:2000. It depends

on the area of tension reinforcement provided and the type of steel.

= Modification factor obtained from fig 5 IS 456:2000. This depends on

the area of compression steel used.

= Reduction factor got from fig 6 of IS 456:2000

corresponding partial safety factors to be

methods specified by IS



the permissible values

Choosing the basic values of span to effective depth ratios (l/d) from the

the value of basic span to depth ratio to get the allowable span to depth

= Modification factor obtained from fig 4 IS 456:2000. It depends

= Modification factor obtained from fig 5 IS 456:2000. This depends on

2.4 Limit state of serviceability for flexural members:



Note: The basic values of l/d mentioned above is valid upto spans of 10m. The basic values

are multiplied by 10 / span in meters except for cantilever. For cantilevers whose span

exceeds 10 m the theoretical method shall be used.

2 Theoretical method of checking deflection

The actual deflections of the members are calculated as per procedure given in

annexure ‘C’ of IS 456:2000. This deflection value shall be limited to the

following

i. The final deflection due to all loads including the effects of temperature, creep and

shrinkage shall not exceed span / 250.

ii. The deflection including the effects of temperature, creep and shrinkage occurring

after erection of partitions and the application of finishes shall not exceed

span/350 or 20 mm whichever is less.

Cracking of concrete occurs whenever the tensile stress developed is greater than the

tensile strength of concrete. This happens due to large values of the following:

1. Flexural tensile stress because of excessive bending under the applied load

2. Diagonal tension due to shear and torsion

3. Direct tensile stress under applied loads (for example hoop tension in a circular

tank)

4. Lateral tensile strains accompanying high axis compressive strains due to

Poisson’s effect (as in a compression test)

5. Settlement of supports

In addition to the above reasons, cracking also occurs because of

1. Restraint against volume changes due to shrinkage, temperature creep and

chemical effects.

2. Bond and anchorage failures

Cracking spoils the aesthetics of the structure and also adversely affect the durability

of the structure. Presence of wide cracks exposes the reinforcement to the atmosphere due to

which the reinforcements get corroded causing the deterioration of concrete. In some cases,

such as liquid retaining structures and pressure vessels cracks affects the basic functional

requirement itself (such as water tightness in water tank).

Permissible crack width

The permissible crack width in structural concrete members depends on the type of

structure and the exposure conditions. The permissible values are prescribed in clause 35.3.2

IS 456:2000 and are shown in table below

2.6 Cracking in structural members



Table: Permissible values of crack width as per IS 456:2000

No. Types of Exposure Permissible widths of crack

at surface (mm)

1 Protected and not exposed to aggressive

environmental conditions 0.3

2 Moderate environmental conditions 0.2

Control of cracking

The check for cracking in beams are done through the following 2 methods specified in

IS 456:2000 clause 43.1

1. By empirical method:

In this method, the cracking is said to be in control if proper detailing (i.e. spacing) of

reinforcements as specified in clause 26.3.2 of IS 456:2000 is followed. These specifications

regarding the spacing have been already discussed under heading general specifications. In

addition, the following specifications shall also be considered

i. In the beams where the depth of the web exceeds 750 mm, side face reinforcement

shall be provided along the two faces. The total area of such reinforcement shall

not be less than 0.1% of the web area and shall be distributed equally on two faces

at a spacing not exceeding 300 mm or web thickness whichever is less. (Refer

clause 25.5.1.3 IS456:2000)

ii. The minimum tension reinforcement in beams to prevent failure in the tension

zone by cracking of concrete is given by the following

As = 0.85 fy / 0.87 fy (Refer clause 26.5.1.1 IS 456:2000)

iii. Provide large number of smaller diameter bars rather than large diameter bars of

the same area. This will make the bars well distributed in the tension zone and will

reduce the width of the cracks.

2. By crack width computations

In the case of special structures and in aggressive environmental conditions, it is

preferred to compute the width of cracks and compare them with the permissible

crack width to ensure the safety of the structure at the limit state of serviceability. The

IS 456-2000 has specified an analytical method for the estimation of surface crack

width in Annexure-F which is based on the British Code (BS : 8110) specifications

where the surface crack width is less than the permissible width, the crack control is

said to be satisfied.



1. Given the following data of a simply supported T beam, check the deflection criteria by

empirical method

Width of the beam (b) = 230 mm

Effective depth (d) = 425 mm

Effective span = 8.0 m

Area of tension steel required = 977.5 mm2

Area of tension steel provided = 1256 mm2

Area of compression steel provided = 628 mm2

Type of steel = Fe 415

Width of flange (bf) = 0.9 m

Width of web (bw) = 0.3 m

Solution:

Basic �� = 20 for simply supported beam from clause 23.2.1

Allowable �� = Basic

�� x Mt x Mc x Mf …………. (1)

�� = 1265�100230�425 = 1.30%

�� = 0.58�� × �� !��"��# !�!

�� = 0.58 × 415 × 977.51256 = 187.3

From fig 4, for Pt = 1.3%, fs = 187.5 N/mm2

Mt = 1.1 ………. (a)

�& = 628 × 100230 × 425 = 0.65%

From fig 5, for Pc = 0.65%, Mc = 1.15 ………………(b)

From fig 6, for '(') =

*.+**.,* = 0.33, Mf = 0.80 ………(c)

Substituting a, b and c in equation (1)

We get allowable �� = 20 x 1.1 x 1.15 x 0.80 = 20.2

2.7 Design Problems:



Actual �� =

-*../0 = 18.82 < allowable

��

Hence OK

2. A rectangular beam continuous over several supports has a width of 300 mm and overall

depth of 600 mm. The effective length of each of the spans of the beam is 12.0 m. The

effective cover is 25 mm. Area of compression steel provided is 942 mm2 and area of

tension steel provided is 1560 mm2. Adopting Fe 500 steel estimate the safety of the

beam for deflection control using the empirical method

Solution:


�� x Mt x Mc x Mf …………. (1)

Basic �� = 26 as the beam is continuous

�� = 0.58�� × �� !��"��# !�!

�� = 0.58 × 500 × 15601560 = 290

From fig 4, for fs = 290, Pt = 0.90, Mt = 0.9……….(a)

From fig 5, for Pc = 0.54%, Mc = 1.15 ……………….(b)

From fig 6, for '(') = 1.0, Mf = 1 ……………………(c)

The equation (1) shall be multiplied by 2*

�345 . � 2*2/ as the span of the beam is greater

than 10.0 m

Allowable �� =

2*2/ x 26 x 0.9 x 1.15 x 1 = 22.4

Actual �� =

2/*.060 = 20.86 < allowable

��

Hence deflection control is satisfied.



3. Find the effective depth based on the deflection criteria of a cantilever beam of 6m span.

Take fy = 415 N/mm2, Pt = 1%, Pc = 1%.

Solution:


�� x Mt x Mc x Mf

Basic �� = 7 for cantilever beam

Assume 789:;<=>:;�7893:?@>�;� = 1.0

fs = 0.58 x 415 x 1= 240.7

From fig 4, for fs = 240, Pt = 1%, Mt = 1.0

From fig 5, for Pc = 1%, Mc = 1.25

From fig 6, for '(') = 1.0, Mf = 1

Allowable �� = 7 x 1.0 x 1.25 x 1.0 = 8.75

! = �-.60 =

A***-.60 = 685 mm

4. A simply supported beam of rectangular cross section 250mm wide and 450mm overall

depth is used over an effective span of 4.0m. The beam is reinforced with 3 bars of

20mm diameter Fe 415 HYSD bars at an effective depth of 400mm. Two anchor bars of

10mm diameter are provided. The self weight of the beam together with the dead load on

the beam is 4 kN/m. Service load acting on the beam is 10 kN/m. Using M20 grade

concrete, compute

a. Short term deflection

b. Long term deflection

Solution:

Data b = 250 mm, D = 450 mm, d = 400 mm, fy = 415 N/mm2

Ast = 3 x B.x 20

2 = 942 mm

2, l = 4.0 m, D.L = 4 kN/m, Service load = 10 kN/m,

Total load = 14 kN/m, fck = 20, Asc = 2 x B.x 10

2 = 158 mm

2

Es = 2.1 x 105 , Ec = 5000 C�&D = 22360 N/mm

2



m = /-*

+EFGF = /-*

+H6= 13.3

fcr = 0.7 C�&D = 0.7 √20 = 3.13 N/mm2

a. Short term deflection

To determine the depth of N.A

Equating the moment of compression area to that of the tension area, we get

b * x * H/ = m * Ast * (d-x)

‘m’ is used to convert the steel into equivalent concrete area

250 * HJ/ = 13 * 942 * (400-x)

Solving, x = 155 mm from the top

Cracked MOI Ir = /0*×200K

2/ + L250 × 155M x (155/2)2 + 13 x 942 (400 – 155)

= 10.45 x 108 mm

4

(2) Igr = Gross MOI = /0*H.0*K

2/ = 18.98 x 108 mm

4

(3) M = Maximum BM under service load

M= N�J

- = 2.×.J

- = 28 kN = 28 x 106 N-mm

(4) Cracked moment of inertia

Mr = OFPQRP�9 =

+.2+×2-.,-×2*S*.0×.0* = 26 x 10

6 N-mm

Lever arm = z = T! − H+V

= T400 − 200+ V = 348.34 mm

(5) Ieff = W XP2./YTZPZ VT[\VT2Y]\VTG(G V^

_ 2*..0×2*S2./Y`Ja×bca

JS×bcadTKeS.Keecc VT2YbffeccVL2Mg



Ieff = 14.93 x 108 mm

4

Further Ir < Ieff < Igr

(6) Maximum short term deflection

ai(perm) = h(N�eiFXj)) =

0+-. 2.×L.***MJ

//+A*×2..,+×2*S = 1.39 mm

Kw = 0

+-. for SSB with UDL

b. Long term deflection

(1) Shrinkage deflection (acs):

acs = K3 ψcs L2

K3 = 0.125 for simply supported beam from Annexure C-3.1

ψcs = Shrinkage curvature = k. TlF8m V

n&� = Ultimate shrinkage strain of concrete (refer 6.2.4) = 0.0003

�� = 2**×,.//0*×.** = 0.942

�& = 2**×20-/0*×.** = 0.158

Pt - Pc = (0.942 – 0.158) = 0.784

This is greater than 0.25 and less than 1.0 Hence ok.

Therefore k. = 0.72 ×o9YoFCo9 = 0.72 ×*.,./Y*.20-√*.,./

K4 = 0.58

ψcs = *.0-×*.***+

.0* = 3.866 x 10-7

acs = K3 ψcs L2

= 0.125 x 3.866 x 10-7

x (4000)2

= 0.773 mm

(2) Creep deflection [acc(perm)]

Creep deflection acc(perm) = aicc(perm) – ai(perm)

Where, acc(perm) = creep deflection due to permanent loads



aicc(perm) = short term deflection + creep deflection

ai(perm) = short term deflection

aicc (perm) = kN ` N�KiFjXj))d

p&; = iFL2qrM = iFL2q2.AM

s = Creep coefficient = 1.6 for 28 days loading

aicc(perm) = 2.6 x short term deflection

= 2.6 x ai(perm)

= 2.6 x 1.39 = 3.614 mm

Creep deflection acc(perm) = 3.614 – 1.39 = 2.224 mm

Total long term deflection = shrinkage deflection + Creep deflection

= 0.773 + 2.224 = 3.013 mm

Total deflection = Short term deflection + Long term deflection

= 1.39 + 3.013 = 4.402 mm

5. A simply supported beam of rectangular section spanning over 6 m has a width of

300mm and overall depth of 600 mm. The beam is reinforced with 4 bars of 25 mm

diameter on the tension side at an effective depth of 550 mm spaced 50 mm centres. The

beam is subjected to a working load moment of 160 kN.m at the centre of the span

section. Using M-25 grade concrete and Fe-415 HYSD bars, check the beam for

serviceability limit state of cracking according to IS:456-2000 method. The beam is

protected and not exposed to aggressive environmental conditions.

a. Data:

b = 300 mm fck = 25 N/mm2

h = 600 mm fy = 415 N/mm2

d = 550 mm Es = 2 x 105 N/mm

2

M = 160 kN.m spacing between bars s = 5 mm

Ast = 1963 mm2

cover = 50mm

For fck = 25 N/mm2, from table 21 IS 456

t&'& = 8.5u/ww/

w = /-*+EFGF

= 11



b. Neutral axis depth

Let x = depth of neutral axis

Then we have 0.5 b�/ = m Ast (d-x)

0.5 x 300�/ = 11 x 1963 (550-x)

Solving, x = 220 mm

c. Cracked moment of area (Ir):

Ir = (bx3/3) + m Ast r

2

Where r = (d-x) = (550-220) = 330 mm

Ir = T+**×//*K+ V +L11 × 1963 ×330/M

= 34.1 x 108 mm

4

d. Maximum width of cracks

Cover = Cmin = (50 - 12.5) = 37.5

acr = [(0.5 S)2 + C

2min]

1/2

= [(0.5 x 50)2 + 37.5

2]1/2

= 45

Crack width will be maximum at the soffit of the beam

Distance of the centroid of steel from neutral axis

= r = (d-x) = (50-220) = 330 mm

Therefore n2 = TO8i8V yzYH�YH{

Where, �� = w y|�XP { = 11 y2A*×2*a×++*

+..2×2*S { = 170 N/mm2

Therefore, n2 = T 26*/×2*fV yA**Y//*

00*Y//*{ = 9.78 x 10-4

n} = n2 − y'9LzYHMLzYHM+i878L�YHM {

n} = L9.78 × 10Y.M − y +**LA**Y//*MLA**Y//*M+×/×2*f×2,A+L00*Y//*M{

= 8.67 x 10-4



Maximum width of crack is expressed as:

~&: = W +��єZ2q/��FP�FZ��] �^

~&: = W+×.0×-.A6×2*�e2q/� ef�K�.facc�JJc� ^

= 0.113 mm< Permissible crack width of 0.3mm from clause 35.3.2

page 67. Hence ok.



UNIT III LIMIT STATE DESIGN FOR BOND, ANCHORAGE SHEAR & TORSION

INTRODUCTION

Torsion when encountered in reinforced concrete members usually occurs in combination with flexure shear.

Torsion in its „pure‟ form (generally associated with metal shafts) is rarely encountered in reinforced concrete.

The interactive behavior of torsion with bending moment and flexural shear in reinforced concrete beams is

fairly complex, owing to the no homogeneous, nonlinear and composite nature of the material and the presence

of cracks. For convenience in design, codes prescribe highly simplified design procedures, which reflect a

judicious blend of theoretical considerations and experimental results.

These design procedures and their bases are described in this chapter, following a brief review of the general

behavior of reinforced concrete beams under torsion.

Torsion may be induced in a reinforced concrete member in various ways during the process of load transfer in

a structural system. In reinforced concrete design, the terms „equilibrium torsion‟ and „compatibility torsion‟ are

commonly used to refer to two different torsion – inducing situations.

In „equilibrium torsion‟, the torsion is induced by an eccentric loading, and equilibrium conditions alone suffice

in determining the twisting moments. In „compatibility torsion‟, the torsion is induced by the application of an

angle of twist and the resulting twisting moment depends on the torsional stiffness of the member.

In some (relatively rare) situations, axial force (tension or compression) may also be involved.

It must be clearly understood that this is merely a matter of terminology, and that it does not imply for instance,

equilibrium conditions need not be satisfied in cases of „ compatibility torsion‟.

There are some situations (such as circular beams supported on multiple columns) where both equilibrium

torsion and compatibility torsion coexist.

EQUILIBRIUM TORSION

This is associated with twisting moments that are developed in a structural member is maintain static

equilibrium with the external loads, and are independent of the torsional stiffness of the member. Such torsion

must be necessarily considered design. The magnitude of the twisting moment does not depend on the torsional

stiffness of the member, and is entirely determinable from statics alone. The member has to be designed for the

full torsion, which is transmitted by the member to the supports. More ever, the end(s) of the member should be

3.1DESIGN FOR TORSION

3.2 EQUILIBRIUM TORSION AND COMPATIBILITY TORSION



suitably restrained to enable the member to resist effectively the torsion induced. Typically, equilibrium torsion

is induced in beams supporting lateral over hanging projections, and is caused by the eccentricity in the loading

(Figure). Such torsion is also induced in beams curved plan and subjected to gravity loads, and in beams where

the transverse loads are eccentric with respect to the shear centre of the cross – section.

(a) Beam supporting a lateral

overhanging

(c) twisting moment

Diagram

(b)free body of beam

Compatibility Torsion

This is the name given to the type of torsion induced in a structural member rotations (twists) applied at one

or more points along the length of the member. It twisting moments induced are directly dependent on the

torsional stiffness of the member. These moments are generally statically in determine and their analysis

necessarily involves (rotational) compatibility conditions; hence the name „compatibility torsion‟. For example,

in the floor beam system has shown in figure, the flexure of the secondary beam BD results in a rotation θB at

the end B. As the primary (Spandrel) beam ABC is monolithically connected with the secondary beam BD at

the joint B., compatibility at B implies an angle of twist, equal to θB in the spandrel beam ABC, and a bending

moment will develop at the end b of beam BD. The bending moment will be equal to, and will act in a direction

Column

Cantilevered Shell Roof

Beam subjected to equilibrium torsion

Total torque = T

T / 2

T / 2

T / 2

T / 2



opposite to the twisting moment, in order to satisfy static equilibrium. The magnitude of θB and the twisting /

bending moment at b depends on the torsional stiffness of beam ABC and the flexural stiffness of beam BD.

The torsional stiffness of a reinforced concrete member is drastically reduced by torsional cracking. This results

in a very large increase in the angle of twist, and, in the case of „compatibility torsion‟, a major reduction in the

induced twisting moment. For this reasons, the code (CL.40.1) permits the designer to neglect the torsional

stiffness of reinforced concrete members at the structural analysis stage itself, so that the need for detailed

design for torsion in such cases does not arise at the design stage. With reference to figure, this implies

assuming a fictitious hinge (i.e., no rotational restraint) at the end B of the beam BD, and assuming a continuous

support (spring, support, actually)at the joint D. Incidentally, this assumption helps in reducing the degree of

static indeterminacy of the structure (typically, a grid floor), thereby simplifying the problem of structural

analysis. Thus, the code states:

In general, where the torsional resistance or stiffness of members has not been taken into account in the analysis

of a structure no specific calculations for torsion will be necessary [CL40.1 of the code].

Of course, this simplification implies the acceptance of cracking and increased deformations in the torsional

member. It also means that during the first time loading, a twisting moment up to the cracking torque of the



plain concrete section develops in the member, prior to torsional cracking. In order to control the subsequent

cracking and to impart ductility to the member, it is desirable to provide a minimum torsional reinforcement,

equal to that required to resist the „cracking torque‟. In fact one of the intentions of the minimum stirrup

reinforcement specified by the code (CL. 25.5.1.6) is to ensure some degree of control of torsional cracking of

beams due to compatibility torsion.

If, however, the designer chooses to consider „ compatibility torsion‟ in analysis and design, then it is important

that a realistic estimate of torsional stiffness is made for the purpose of structural analysis, and the required

torsional reinforcement should be provided for the calculated twisting moment.

Observed behavior of reinforced concrete members under torsion (see also section 7.3) shows that the

torsional stiffness is little influenced by the amount of torsional reinforcement in the linear elastic phase, and

may be taken as that of the plain concrete section. However, once torsional cracking occurs, there is a drastic

reduction in the torsional stiffness. The post – cracking torsional stiffness is only a small fraction (less than 10

percent) of the pre – cracking stiffness, and depends on the amount of torsional reinforcement, provided in the

form of closed stirrups and longitudinal bars. Heavy torsional reinforcement can, doubt, increase the torsional

resistance (strength) to a large extent, but this can be realized only at very large angles of twist (accompanied by

very large cracks).

Hence, even with torsional reinforcement provided, in most practical situations, the maximum twisting moment

in a reinforced concrete member under compatibility torsion is the value corresponding to the torsional cracking

of the member. The „cracking torque‟ is very nearly the same as the failure strength obtained for an identical

plain concrete section.

In the usual linear elastic analysis of framed structures, the torsional stiffness kt (torque per unit twist T/θ ) of a

beam of length l is expressed as

KT = GC / l

Where GC is the torsional rigidity, obtained as a product of the shear modulus G and the geometrical parameter

C of the section (Ref. 7.1). It is recommended in the Explanatory Handbook to the code (Ref.7.2) that G may be

taken as 0.4 times the c is a property of the section having the same relationship to the torsional stiffness of a

rectangular section as the polar moment of inertia has for a circular section

3.3 Estimation of Torsional stiffness



s d

Problem 1:

Determine the anchorage length of 4-20T reinforcing bars going into the supportof the simply supported beam shown in Fig. 6.15.5. The factored shear force Vu = 280kN, width of the column support = 300 mm. Use M 20 concrete and Fe 415 steel.

Solution 1:

τbd for M 20 and Fe 415 (with 60% increased) = 1.6(1.2) = 1.92 N/mm2

Ld =φσs =4τ bd

0.87(415)φ4(1.92)

(when σ s = 0.87 f y ) = 47.01φ ...... (1)

(Ld )whenσ = f ≤M1

V+ Lo

Here, to find M1, we need xu

3.4 Design Problems



)

xu =0.87 f y Ast =0.36 f ck b

0.87(415) (1256)

0.36(20) (300)= 209.94 mm

xu,max = 0.48(500) = 240 mm

Since xu < xu,max ; M1 = 0.87 fy Ast (d – 0.42 xu)

or M1 = 0.87(415) (1256) {500 – 0.42(209.94)} = 187.754 kNm

and V = 280 kN

We have from above, with the stipulation of 30 per cent increase assuming thatthe reinforcing bars are confined by a compressive reaction:

L ≤ 1.3 (M1 ) + Ld V o ...... (2)

From Eqs.(1) and (2), we have

47.01φ ≤ 1.3 (M1 ) + LV o

or 47.01φ ≤ 1.3187.754(106 )

{ }; if L is assumed as zero.

280(103 ) o

or φ ≤ 18.54 mm

Therefore, 20 mm diameter bar does not allow Lo = 0.

Determination of Lo:

1.3 (M1 ) + LV o

≥ 47.01φ

Minimum Lo = 47.01φ - 1.3 (M 1 )V

= 47.01(20) -187754

1.3(280

= 68.485 mm

So, the bars are extended by 100 mm to satisfy the requirement as shown inFig.



A column may be classified based on different criteria such as:

1. Based on shape

• Rectangle

• Square

• Circular

• Polygon

• L type

• T type

• + type

2. Based on slenderness ratio or height

Short column and Long column or Short and Slender Compression Members

A compression member may be considered as short when both the slenderness ratios namely

lex/D and ley/b are less than 12: Where

lex= effective length in respect of the major axis, D= depth in respect of the major axis,

ley= effective length in respect of the minor axis, and b = width of the member.

It shall otherwise be considered as a slender or long compression member.

UNIT-4 Design of Columns

4.1 Pre-requisite Discussion:A column is defined as a compression member, the effective length of which exceeds three

times the least lateral dimension. Compression members, whose lengths do not exceed three

times the least lateral dimension, may be made of plain concrete. A column forms a very

important component of a structure. Columns support beams which in turn support walls

and slabs. It should be realized that the failure of a column results in the collapse of

the structure. The design of a column should therefore receive importance.

4.2 Introduction:A column is a vertical structural member supporting axial compressive loads, with or without

moments. The cross-sectional dimensions of a column are generally considerably less than its

height. Columns support vertical loads from the floors and roof and transmit these loads to

the foundations.

The more general terms compression members and members subjected to combined axial

load and bending are sometimes used to refer to columns, walls, and members in concrete

trusses or frames. These may be vertical, inclined, or horizontal. A column is a special case of

a compression member that is vertical. Stability effects must be considered in the design of

compression members.

4.3 Classification of columns



The great majority of concrete columns are sufficiently stocky (short) that slenderness can be

ignored. Such columns are referred to as short columns. Short column generally fails by

crushing of concrete due to axial force. If the moments induced by slenderness effects

weaken a column appreciably, it is referred to as a slender column or a long column. Long

columns generally fail by bending effect than due to axial effect. Long column carry less load

compared to long column.

3. Based on pattern of lateral reinforcement

• Tied columns with ties as laterals

• columns with Spiral steel as laterals or spiral columns

Majority of columns in any buildings are tied columns. In a tied column the longitudinal bars

are tied together with smaller bars at intervals up the column. Tied columns may be square,

rectangular, L-shaped, circular, or any other required shape. Occasionally, when high strength

and/or high ductility are required, the bars are placed in a circle, and the ties are replaced by a

bar bent into a helix or spiral. Such a column, called a spiral column. Spiral columns are

generally circular, although square or polygonal shapes are sometimes used. The spiral acts to

restrain the lateral expansion of the column core under high axial loads and, in doing so,

delays the failure of the core, making the column more ductile. Spiral columns are used more

extensively in seismic regions. If properly designed, spiral column carry 5% extra load at

failure compared to similar tied column.

4. Based on type of loading

• Axially loaded column or centrally or concentrically loaded column (Pu)

• A column subjected to axial load and unaxial bending (Pu + Mux) or (P + Muy)

• A column subjected to axial load and biaxial bending (Pu + Mux + Muy)

5. Based on materials



Timber, stone, masonry, RCC, PSC, Steel, aluminium , composite column

RCC-Tied RCC spiral Composite columns

4.4 Behavior of Tied and Spiral Columns

Figure shows a portion of the core of a spiral column. Under a compressive load, the concrete

in this column shortens longitudinally under the stress and so, to satisfy Poisson’s ratio, it

expands laterally. In a spiral column, the lateral expansion of the concrete inside the spiral

(referred to as the core) is restrained by the spiral. This stresses the spiral in tension. For

equilibrium, the concrete is subjected to lateral compressive stresses. In a tied column in a

non seismic region, the ties are spaced roughly the width of the column apart and, as a result,

provide relatively little lateral restraint to the core. Outward pressure on the sides of the ties

due to lateral expansion of the core merely bends them outward, developing an insignificant

hoop-stress effect. Hence, normal ties have little effect on the strength of the core in a tied

column. They do, however, act to reduce the unsupported length of the longitudinal bars, thus

reducing the danger of buckling of those bars as the bar stress approaches yield. load-

deflection diagrams for a tied column and a spiral column subjected to axial loads is shown in

figure. The initial parts of these diagrams are similar. As the maximum load is reached,

vertical cracks and crushing develop in the concrete shell outside the ties or spiral, and this

concrete spalls off. When this occurs in a tied column, the capacity of the core that remains is

less than the load on the column. The concrete core is crushed, and the reinforcement buckles

outward between ties. This occurs suddenly, without warning, in a brittle manner. When the

shell spalls off a spiral column, the column does not fail immediately because the strength of

the core has been enhanced by the triaxial stresses resulting from the effect of the spiral

reinforcement. As a result, the column can undergo large deformations, eventually reaching a

second maximum load, when the spirals yield and the column finally collapses. Such a failure

is much more ductile than that of a tied column and gives warning of the impending failure,

along with possible load redistribution to other members. Due to this, spiral column carry

little more load than the tied column to an extent of about 5%. Spiral columns are used when

ductility is important or where high loads make it economical to utilize the extra strength.

Both columns are in the same building and have undergone the same deformations. The tied

column has failed completely, while the spiral column, although badly damaged, is still

supporting a load. The very minimal ties were inadequate to confine the core concrete. Had

the column ties been detailed according to ACI Code, the column will perform better as

shown.

4.5 Specifications for covers and reinforcement in column

For a longitudinal reinforcing bar in a column nominal cover shall in any case not be less

than 40 mm, or less than the diameter of such bar. In the case of columns of minimum

dimension of 200 mm or under, whose reinforcing bars do not exceed 12 mm, a nominal

cover of 25 mm may be used. For footings minimum cover shall be 50 mm.



• Effectively held in position and restrained against rotation in both ends

• Effectively held in position at both ends, restrained against rotation at one end

• Effectively held in position at both ends, but not restrained against rotation

• Effectively held in position and restrained against rotation at one end, and at the other

restrained against rotation but not held in position

• Effectively held in position and restrained against rotation in one end, and at the other

partially restrained against rotation but not held in position

• Effectively held in position at one end but not restrained against rotation, and at the

other end restrained against rotation but not held in position

• Effectively held in position and restrained against rotation at one end but not held in

position nor restrained against rotation at the other end

Table.Effective length of compression member

Sl.

No.

Degree of End Restraint of Compression Members Figure

Theo.

Value of

Effective

Length

Reco.

Value of

Effective

Length

1 Effectively held in position and restrained against

rotation in both ends

0.50 l 0.65l

2 Effectively held in position at both ends, restrained

against rotation at one end

0.70 l 0.80l

Nominal Cover in mm to meet durability requirements based on exposure

Mild 20, Moderate 30, Severe 45, Very severe 50, Extreme 75

Nominal cover to meet specified period of fire resistance for all fire rating 0.5 to 4 hours is 40

mm for columns only

4.6 Effective length of compression member

Column or strut is a compression member, the effective length of which exceeds three times

the least lateral dimension. For normal usage assuming idealized conditions, the effective

length of in a given plane may be assessed on the basis of Table 28 of IS: 456-2000.

Following terms are required.

Following are the end restraints:



3 Effectively held in position at both ends, but not

restrained against rotation

1.0 l 1.0l

4

Effectively held in position and restrained against

rotation at one end, and at the other restrained

against rotation but not held in position

1.0 l 1.20l

5


rotation in one end, and at the other partially


- 1.5l

6

Effectively held in position at one end but not

restrained against rotation, and at the other end


2.0 l 2.0l

7


rotation at one end but not held in position nor

restrained against rotation at the other end

2.0 l 2.0l

Unsupported Length

The unsupported length, l, of a compression member shall be taken as the clear distance

between end restraints (visible height of column). Exception to this is for flat slab

construction, beam and slab construction, and columns restrained laterally by struts (Ref.

IS:456-2000),

Slenderness Limits for Columns

The unsupported length between end restraints shall not exceed 60 times the least lateral

dimension of a column.

If in any given plane, one end of a column is unrestrained, its unsupported length, l, shall not

exceed 100b2

1. The cross-sectional area of longitudinal reinforcement, shall be not less than 0.8

percent nor more than 6 percent of the gross cross sectional area of the column.

2. NOTE - The use of 6 percent reinforcement may involve practical difficulties in

placing and compacting of concrete; hence lower percentage is recommended. Where

/D, where b = width of that cross-section, and D= depth of the cross-section

measured in the plane under consideration.

4.7 Specifications as per IS: 456-2000

Longitudinal reinforcement



bars from the columns below have to be lapped with those in the column under

consideration, the percentage of steel shall usually not exceed 4 percent.

3. In any column that has a larger cross-sectional area than that required to support the

load, the minimum percentage of steel shall be based upon the area of concrete

required to resist the direct stress and not upon the actual area.

4. The minimum number of longitudinal bars provided in a column shall be four in

rectangular columns and six in circular columns.

5. The bars shall not be less than 12 mm in diameter

6. A reinforced concrete column having helical reinforcement shall have at least six bars

of longitudinal reinforcement within the helical reinforcement.

7. In a helically reinforced column, the longitudinal bars shall be in contact with the

helical reinforcement and equidistant around its inner circumference.

8. Spacing of longitudinal bars measured along the periphery of the column shall not

exceed 300 mm.

9. In case of pedestals in which the longitudinal reinforcement is not taken in account in

strength calculations, nominal longitudinal reinforcement not less than 0.15 percent of

the cross-sectional area shall be provided.

Longitudinal Bar

Φ1 ≥ 12 mm

Spacing or pitch of

lateral ties

Lateral ties

Φ2 ≥ ¼ Φ1

≥ 5mm

Cover to Lateral

ties as per IS: 456-

2000

4.8 Transverse reinforcement

A reinforced concrete compression member shall have transverse or helical reinforcement so

disposed that every longitudinal bar nearest to the compression face has effective lateral

support against buckling.



The effective lateral support is given by transverse reinforcement either in the form of

circular rings capable of taking up circumferential tension or by polygonal links (lateral ties)

with internal angles not exceeding 135°. The ends of the transverse reinforcement shall be

properly anchored.

Arrangement of transverse reinforcement

If the longitudinal bars are not spaced more than 75 mm on either side, transverse

reinforcement need only to go round corner and alternate bars for the purpose of providing

effective lateral supports (Ref. IS:456).

If the longitudinal bars spaced at a distance of not exceeding 48 times the diameter of the tie

are effectively tied in two directions, additional longitudinal bars in between these bars need

to be tied in one direction by open ties (Ref. IS:456).

Pitch and diameter of lateral ties

1) Pitch-The pitch of transverse reinforcement shall be not more than the least of the

following distances:

i) The least lateral dimension of the compression members;

ii) Sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied; and

iii) 300 mm.

2) Diameter-The diameter of the polygonal links or lateral ties shall be not less than

onefourth of the diameter of the largest longitudinal bar, and in no case less than 6 mm.

Helical reinforcement

1) Pitch-Helical reinforcement shall be of regular formation with the turns of the helix spaced

evenly and its ends shall be anchored properly by providing one and a half extra turns of the



1. The maximum compressive strain in concrete in axial compression is taken as 0.002.

2. The maximum compressive strain at the highly compressed extreme fibre in concrete

subjected to axial compression and bending and when there is no tension on the

section shall be 0.0035 minus 0.75 times the strain at the least compressed extreme

fibre.

In addition the following assumptions of flexure are also required

3. Plane sections normal to the axis remain plane after bending.

4. The maximum strain in concrete at the outermost compression fibre is taken as 0.0035

in bending.

5. The relationship between the compressive stress distribution in concrete and the strain

in concrete may be assumed to be rectangle, trapezoid, parabola or any other shape

which results in prediction of strength in substantial agreement with the results of test.

6. An acceptable stress strain curve is given in IS:456-200. For design purposes, the

compressive strength of concrete in the structure shall be assumed to be 0.67 times the

characteristic strength. The partial safety factor y of 1.5 shall be applied in addition to

this.

7. The tensile strength of the concrete is ignored.

8. The stresses in the reinforcement are derived from representative stress-strain curve

for the type of steel used. Typical curves are given in IS:456-2000. For design

purposes the partial safety factor equal to 1.15 shall be applied.

Minimum eccentricity

As per IS:456-2000, all columns shall be designed for minimum eccentricity, equal to the

unsupported length of column/ 500 plus lateral dimensions/30, subject to a minimum of 20

mm. Where bi-axial bending is considered, it is sufficient to ensure that eccentricity exceeds

the minimum about one axis at a time.

Short Axially Loaded Members in Compression

The member shall be designed by considering the assumptions given in 39.1 and the

minimum eccentricity. When the minimum eccentricity as per 25.4 does not exceed 0.05

times the lateral dimension, the members may be designed by the following equation:

spiral bar. Where an increased load on the column on the strength of the helical reinforcement

is allowed for, the pitch of helical turns shall be not more than 7.5 mm, nor more than one-

sixth of the core diameter of the column, nor less than 25 mm, nor less than three times the

diameter of the steel bar forming the helix.

4.9 LIMIT STATE OF COLLAPSE: COMPRESSION

Assumptions



Pu = 0.4 fck Ac + 0.67 fy Asc

Pu = axial load on the member,

fck = characteristic compressive strength of the concrete,

Ac = area of concrete,

fy = characteristic strength of the compression reinforcement, and

As = area of longitudinal reinforcement for columns.

Compression Members with Helical Reinforcement

The strength of compression members with helical reinforcement satisfying the requirement

of IS: 456 shall be taken as 1.05 times the strength of similar member with lateral ties.

The ratio of the volume of helical reinforcement to the volume of the core shall not be less

than

Vhs / Vc > 0.36 (Ag/Ac – 1) fck/fy

Ag = gross area of the section,

Ac = area of the core of the helically reinforced column measured to the outside diameter of

the helix,

fck = characteristic compressive strength of the concrete, and

fy = characteristic strength of the helical reinforcement but not exceeding 415 N/mm.

Members Subjected to Combined Axial Load and Uni-axial Bending

Use of Non-dimensional Interaction Diagrams as Design Aids

Design Charts (for Uniaxial Eccentric Compression) in SP-16

The design Charts (non-dimensional interaction curves) given in the Design Handbook, SP :

16 cover the following three cases of symmetrically arranged reinforcement :

(a) Rectangular sections with reinforcement distributed equally on two sides (Charts 27 – 38):

the ‘two sides’ refer to the sides parallel to the axis of bending; there are no inner rows of

bars, and each outer row has an area of 0.5As this includes the simple 4–bar configuration.

(b) Rectangular sections with reinforcement distributed equally on four sides (Charts 39 –

50): two outer rows (with area 0.3As

each) and four inner rows (with area 0.1As

each)

have been considered in the calculations ; however, the use of these Charts can be

extended, without significant error, to cases of not less than two inner rows (with a

minimum area 0.3As in each outer row).

(c) Circular column sections (Charts 51 – 62): the Charts are applicable for circular sections

with at least six bars (of equal diameter) uniformly spaced circumferentially.

�

Corresponding to each of the above three cases, there are as many as 12 Charts available

covering the 3 grades of steel (Fe 250, Fe 415, Fe 500), with 4 values of d1/ D ratio for each

grade (namely 0.05, .0.10, 0.15, 0.20). For intermediate values of d1/ D, linear interpolation

may be done. Each of the 12 Charts of SP-16 covers a family of non-dimensional design

interaction curves with p/fck

values ranging from 0.0 to 0.26.



From this, percentage of steel (p) can be found. Find the area of steel and provide the

required number of bars with proper arrangement of steel as shown in the chart.

Typical interaction curve

Salient Points on the Interaction Curve

The salient points, marked 1 to 5 on the interaction curve correspond to the failure strain

profiles, marked 1 to 5 in the above figure.

• The point 1 in figure corresponds to the condition of axial loading with e = 0. For this

case of ‘pure’ axial compression.

• The point 11 in figure corresponds to the condition of axial loading with the

mandatory minimum eccentricity emin

prescribed by the Code.

• The point 3 in figure corresponds to the condition xu

= D, i.e., e = eD. For e < e

D, the

entire section is under compression and the neutral axis is located outside the section

(xu

> D), with 0.002 < εcu

< 0.0035. For e > eD, the NA is located within the section

(xu

< D) and εcu

= 0.0035 at the ‘highly compressed edge’.

• The point 4 in figure corresponds to the balanced failure condition, with e = eb

and xu

= xu, b

. The design strength values for this ‘balanced failure’ condition are denoted as

Pub

and Mub

.

• The point 5 in figure corresponds to a ‘pure’ bending condition (e = ∞, PuR

= 0); the

resulting ultimate moment of resistance is denoted Muo

and the corresponding NA

depth takes on a minimum value xu, min

.



Factored load and Factored moment

Assume arrangement of reinforcement: On two sides or on four sides

Assume moment due to minimum eccentricity to be less than the actual moment

Assume suitable axis of bending based on the given moment (xx or yy)

Assuming suitable diameter of longitudinal bars and suitable nominal cover

1. Find d1/D from effective cover d

1

2. Find non dimensional parameters Pu/fckbD and Mu/fckbD2

3. Referring to appropriate chart from S-16, find p/fck and hence the percentage of

reinforcement, p

4. Find steel from, As = p bD/100

5. Provide proper number and arrangement for steel

6. Design suitable transverse steel

7. Provide neat sketch

Members Subjected to Combined Axial Load and Biaxial Bending

The resistance of a member subjected to axial force and biaxial bending shall be obtained on

the basis of assumptions given in IS:456 with neutral axis so chosen as to satisfy the

equilibrium of load and moments about two axes. Alternatively such members may be

designed by the following equation:

[Mux/Mux1]αn

+ [Muy/Muy1]αn ≤ 1, where

Mux and My = moments about x and y axes due to design loads,

Mux1 and My1 = maximum uni-axial moment capacity for an axial load of Pu bending about

x and y axes respectively, and αn is related to Pu /Puz, where Puz = 0.45 fck .Ac + 0.75 fy Asc

For values of Pu /Puz = 0.2 to 0.8, the values of αn vary linearly from 1 .0 to 2.0. For values

less than 0.2 and greater than 0.8, it is taken as 1 and 2 respectively

NOTE -The design of member subject to combined axial load and uniaxial bending will

involve lengthy calculation by trial and error. In order to overcome these difficulties

interaction diagrams may be used. These have been prepared and published by BIS in SP:16

titled Design aids for reinforced concrete to IS 456-2000.

IS:456-2000 Code Procedure

1. Given Pu, M

ux, M

uy, grade of concrete and steel

2. Verify that the eccentricities ex

= Mux

/Pu

and ey

= Muy

/Pu

are not less than the

corresponding minimum eccentricities as per IS:456-2000

3. Assume a trial section for the column (square, rectangle or circular).

4.10 Procedure for using of Non-dimensional Interaction Diagrams as Design Aids to

find steel

Given:

Size of column, Grade of concrete, Grade of steel (otherwise assume suitably)



4. Determine Mux1

and Muy1

, corresponding to the given Pu

(using appropriate curve from

SP-16 design aids)

5. Ensure that Mux1

and Muy1

are significantly greater than Mux

and Muy

respectively;

otherwise, suitably redesign the section.

6. Determine Puz

and hence αn

7. Check the adequacy of the section using interaction equation. If necessary, redesign

the section and check again.

Slender Compression Members: The design of slender compression members shall be

based on the forces and the moments determined from an analysis of the structure, including

the effect of deflections on moments and forces. When the effects of deflections are not taken

into account in the analysis, additional moment given in 39.7.1 shall be taken into account in

the appropriate direction.

1. Determine the load carrying capacity of a column of size 300 x 400 mm reinforced

with six rods of 20 mm diameter i.e, 6-#20. The grade of concrete and steel are M20 and Fe 415 respectively. Assume that the column is short.

fck = 20 MPa, fy= 415 MPa

Area of steel ASC = 6 x π x 202/4 = 6 x 314 = 1884 mm

2

Percentage of steel = 100Asc/bD = 100x1884/300x400 = 1.57 %

Area of concrete Ac = Ag – Asc = 300 x 400 – 1884 = 118116 mm2

Ultimate load carried by the column


0.4x20x118116 + 0.67x415x1884

944928 + 523846 = 1468774 N = 1468. 8 kN

Therefore the safe load on the column = 1468.8 /1.5 = 979.2 kN

2. Determine the steel required to carry a load of 980kN on a rectangular column of

size 300 x 400 mm. The grade of concrete and steel are M20 and Fe 415 respectively.

Assume that the column is short.

fck = 20 MPa, fy= 415 MPa, P = 980 kN

Area of steel ASC = ?

Area of concrete Ac = Ag – Asc = (300 x 400 – ASC)



980 x 1.5 x 1000 = 0.4x20x (300 x 400 – ASC) + 0.67x415 ASC

= 960000 - 8 ASC + 278.06 ASC

ASC =1888.5 mm2,

Percentage of steel = 100Asc/bD = 100x1888.5 /300x400 = 1.57 % which is more than 0.8%

and less than 6% and therefore ok.

Use 20 mm dia. bas, No. of bars = 1888.5/314 = 6.01 say 6

4.11 Design Problems



3. Design a square or circular column to carry a working load of 980kN. The grade of

concrete and steel are M20 and Fe 415 respectively. Assume that the column is short.

Let us assume 1.0% steel (1 to 2%)

Say ASC = 1.0% Ag =1/100 Ag = 0.01Ag

fck = 20 MPa, fy= 415 MPa, P = 980 kN

Area of concrete Ac = Ag – Asc = Ag -0.01Ag = 0.99 Ag



980 x 1.5 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag

= 7.92 Ag + 2.78 Ag =10.7Ag

Ag = 137383 mm2

Let us design a square column:

B = D = √ Ag =370.6 mm say 375 x 375 mm

This is ok. However this size cannot take the minimum eccentricity of 20 mm as emin/D =

20/375 =0.053 > 0.05. To restrict the eccentricity to 20 mm, the required size is 400x 400

mm.

Area of steel required is Ag = 1373.8 mm2. Provide 4 bar of 22 mm diameter. Steel provided

is 380 x 4 = 1520 mm2

Actual percentage of steel = 100Asc/bD = 100x1520 /400x400 = 0.95 % which is more than

0.8% and less than 6% and therefore ok.

Design of Transverse steel:

Diameter of tie = ¼ diameter of main steel = 22/4 =5.5mm or 6 mm, whichever is greater.

Provide 6 mm.

Spacing: < 300 mm, < 16 x22 = 352mm, < LLD = 400mm. Say 300mm c/c

Design of circular column:

Here Ag = 137383 mm2

π x D2/4 = Ag, D= 418.2 mm say 420 mm. This satisfy the minimum eccentricity of 20m

Also provide 7 bars of 16 mm, 7 x 201 = 1407 mm2


Dia of tie = ¼ dia of main steel = 16/4 = 4 mm or 6 mm, whichever is greater. Provide 6 mm.

Spacing: < 300 mm, < 16 x16 = 256 mm, < LLD = 420mm. Say 250 mm c/c

4. Design a rectangular column to carry an ultimate load of 2500kN. The unsupported

length of the column is 3m. The ends of the column are effectively held in position



and also restrained against rotation. The grade of concrete and steel are M20 and Fe 415 respectively.

Given:

fck = 20 MPa, fy= 415 MPa, Pu = 2500kN


Say ASC = 1.0% Ag =1/100 Ag = 0.01Ag




2500 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag

= 7.92 Ag + 2.78 Ag =10.7Ag

Ag = 233645 mm2

If it is a square column:

B = D = √ Ag =483 mm. However provide rectangular column of size 425 x 550mm. The

area provided=333750 mm2

Area of steel = 2336 mm2, Also provide 8 bars of 20 mm, 6 x 314 = 2512 mm

2

Check for shortness: Ends are fixed. lex = ley = 0.65 l = 0.65 x 3000 = 1950 mm

lex /D= 1950/550 < 12, and ley /b = 1950/425 < 12, Column is short

Check for minimum eccentricity:

In the direction of longer direction

emin, x = lux/500 + D/30 = 3000/500 + 550/30 = 24.22mm or 20mm whichever is greater.

emin, x = 24.22 mm < 0.05D = 0.05 x 550 =27.5 mm. O.K

In the direction of shorter direction

emin, y= luy/500 + b/30 = 3000/500 + 425/30 = 20.17 mm or 20mm whichever is greater.

emin, x = 20.17 mm < 0.05b = 0.05 x 425 =21.25 mm. O.K


Dia of tie = ¼ dia of main steel = 20/4 = 5 mm or 6 mm, whichever is greater. Provide 6 mm

or 8 mm.




5. Design a circular column with ties to carry an ultimate load of 2500kN. The

unsupported length of the column is 3m. The ends of the column are effectively held

in position but not against rotation. The grade of concrete and steel are M20 and Fe

415 respectively.

Given:



Say ASC = 1.0% Ag =1/100 Ag = 0.01Ag




2500 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag

= 7.92 Ag + 2.78 Ag =10.7Ag

Ag = 233645 mm2

π x D2/4 = Ag, D = 545.4 mm say 550 mm.


2

Check for shortness: Ends are hinged lex = ley = l = 3000 mm



Here, emin, x = emin, y = lux/500 + D/30 = 3000/500 + 550/30 = 24.22mm or 20mm whichever is

greater.

emin = 24.22 mm < 0.05D = 0.05 x 550 =27.5 mm. O.K


Diameter of tie = ¼ dia of main steel = 20/4 = 5 mm or 6 mm, whichever is greater. Provide 6

mm or 8 mm.


Similarly square column can be designed.

If the size of the column provided is less than that provided above, then the minimum

eccentricity criteria are not satisfied. Then emin is more and the column is to be designed as



uni axial bending case or bi axial bending case as the case may be. This situation arises when

more steel is provided ( say 2% in this case).

Try to solve these problems by using SP 16 charts, though not mentioned in the syllabus.

6. Design the reinforcement in a column of size 450 mm × 600 mm, subject to an axial

load of 2000 kN under service dead and live loads. The column has an unsupported

length of 3.0m and its ends are held in position but not in direction. Use M 20

concrete and Fe 415 steel.

Solution:

Given: lu= 3000 mm, b = 450 mm, D

= 600 mm, P =2000kN, M20, Fe415

Check for shortness: Ends are fixed. lex = ley = l = 3000 mm

lex /D= 3000/600 < 12, and ley /b = 3000/450< 12, Column is short



emin, x = lux/500 + D/30 = 3000/500 + 600/30 = 26 mm or 20mm whichever is greater.

emin, x = 26 mm < 0.05D = 0.05 x 600 =30 mm. O.K


emin, y= luy/500 + b/30 = 3000/500 + 450/30 = 21 mm or 20mm whichever is greater.

emin, x = 21 mm < 0.05b = 0.05 x 450 =22.5 mm. O.K

Minimum eccentricities are within the limits and hence code formula for axially loaded short

columns can be used.

Factored Load

Pu

= service load × partial load factor

= 2000 × 1.5 = 3000 kN

Design of Longitudinal Reinforcement

Pu = 0.4 fck Ac + 0.67 fy Asc or

Pu = 0.4 fck Ac + (0.67 fy - 0.4fck) Asc

3000 × 103

= 0.4 × 20 × (450 × 600) + (0.67 × 415–0.4 × 20)Asc

= 2160×103

+ 270.05Asc



⇒ Asc

= (3000–2160) × 103

/270.05 = 3111 mm2

In view of the column dimensions (450 mm, 600 mm), it is necessary to place intermediate

bars, in addition to the 4 corner bars:

Provide 4–25φ at corners ie, 4 × 491 = 1964 mm2

and 4–20φ additional ie, 4 × 314 = 1256 mm2

⇒ Asc

= 3220 mm2

> 3111 mm2

⇒ p = (100×3220) / (450×600) = 1.192 > 0.8 (minimum steel), OK.

Design of transverse steel

Diameter of tie = ¼ diameter of main steel = 25/4 =6.25 mm or 6 mm, whichever is greater.

Provide 6 mm.

Spacing: < 300 mm, < 16 x 20 = 320 mm, < LLD = 450mm. Say 300 mm c/c

Thus provide ties 8mm @ 300 mm c/c

Sketch:

Example: Square Column with Uniaxial Bending

7. Determine the reinforcement to be provided in a square column subjected to

uniaxial bending with the following data:

Size of column 450 x 450 mm

Concrete mix M 25

Characteristic strength of steel 415 N/mm2

Factored load 2500 kN

Factored moment 200 kN.m

Arrangement of reinforcement:

(a) On two sides

(b) On four sides


Assuming 25 mm bars with 40 mm cover,



d = 40 + 12.5 = 52.5 mm

d1/D = 52.5/450- 0.12

Charts for d1/D = 0.15 will be used

Pu/fckbD = (2500 x 1000)/ (25 x 450 x 450) = 0.494

Mu/fckbD2 =200 x 10

6 /(25 x 450 x 450

2) = 0.088

a) Reinforcement on two sides,

Referring to Chart 33,

p/fck = 0.09

Percentage of reinforcement,

p = 0.09 x 25 = 2.25 %

As = p bD/100 = 2.25 x 450 x 450/100

= 4556 mm2

b) Reinforcement on four sides

from Chart 45,

p/fck = 0.10

p = 0.10 x 25 = 2.5 %

As = 2.5 x 450 x 450/100 = 5063 mm2

8. Example: Circular Column with Uniaxial Bending

Determine the reinforcement to be provided in a circular column with the following

data:

Diameter of column 500 mm

Grade of concrete M20

Characteristic strength 250 N/mm2



Lateral reinforcement :

(a) Hoop reinforcement

(b) Helical reinforcement

(Assume moment due to minimum eccentricity to be less than the actual moment).


d1 = 40 + 12.5 = 52.5 mm

d1/D – 52.5/50 = 0.105

Charts for d’/D = 0.10 will be used.

(a) Column with hoop reinforcement

Pu/fck D D = (1600 x 1000)/ (20 x 500 x 500) = 0.32

Mu/fck D x D2 =125 x 10

6 /(20 x 500 x 500

2) = 0.05



Referring to Chart 52, for fy = 250 N/mm2

p/fck = 0.87


p = 0.87 x 20 = 1.74 %

As = 1.74 x (π x 5002/4)/100 = 3416 mm

2

(b) Column with Helical Reinforcement

According to 38.4 of the Code, the strength of a compression member with helical

reinforcement is 1.05 times the strength of a similar member with lateral ties. Therefore, the,

given load and moment should be divided by 1.05 before referring to the chart.

Pu/fck D D = (1600/1.05 x 1000)/ (20 x 500 x 500) = 0.31

Mu/fck D x D2 =125/1.05 x 10

6 /(20 x 500 x 500

2) = 0.048

Hence, From Chart 52, for fy = 250 N/mm2,

p/fck = 0.078

p = 0.078 x 20 = 1.56 %

As = 1.56 x( π x 500 x 500/4 )/100 = 3063 cm2

According to 38.4.1 of the Code the ratio of the volume of helical reinforcement to the

volume of the core shall not be less than

0.36 (Ag/Ac - 1) x fck /fy

where Ag is the gross area of the section and Ac is the area of the core measured to the outside

diameter of the helix. Assuming 8 mm dia bars for the helix,

Core diameter = 500 - 2 (40 - 8) = 436 mm

Ag/AC = 500/436 = 1.315

0.36 (Ag/Ac - 1) x fck /fy = 0.36(0.315) 20/250 =0.0091

Volume of helical reinforcement / Volume of core

= Ash π x 428 /( π/4 x 4362) sh

0.09 Ash / sh

where, Ash is the area of the bar forming the helix and sh is the pitch of the helix.

In order to satisfy the coda1 requirement,

0.09 Ash / sh ≥ 0.0091



For 8 mm dia bar,

sh ≤ 0.09 x 50 / 0.0091 = 49.7 mm. Thus provide 48 mm pitch

Example: Rectangular column with Biaxial Bending

9. Determine the reinforcement to be provided in a short column subjected to biaxial

bending, with the following data:

size of column = 400 x 600 mm

Concrete mix = M15

Characteristic strength of reinforcement = 415 N/mm2

Factored load, Pu = 1600 kN

Factored moment acting parallel to the larger dimension, Mux =120 kNm

Factored moment acting parallel to the shorter dimension, Muy = 90 kNm

Moments due to minimum eccentricity are less than the values given above.

Reinforcement is distributed equally on four sides.

As a first trial assume the reinforcement percentage, p = 1.2%

p/fck = 1.2/15 = 0.08

Uniaxial moment capacity of the section about xx-axis :

d1/D = 52.5 /600 = 0.088

Chart for d’/D = 0.1 will be used.

Pu/fck b D = (1600 x 1000)/ (15 x 400 x 600) = 0.444

Referring to chart 44

Mu/fck b x D2 = 0.09

Mux1 = 0.09 x 15 x 400 x 6002) = 194.4 kN.m

Uni-axial moment capacity of the section about yy axis :

d1/D = 52.5 /400 = 0.131

Chart for d1/D =0.15 will be used.


Mu/fck b x D2 = 0.083

Mux1 = 0.083 x 15 x 600 x 4002) = 119.52 kN.m

Calculation of Puz :

Referring to Chart 63 corresponding to

p = 1.2, fy = 415 and fck = 15,

Puz/Ag = 10.3

Puz = 10.3 x 400 x 600 = 2472 kN

Mux/Mux1 = 120/194.4 =0.62

Muy/Muy1=90/119.52 = 0.75

Pu /Puz =1600/2472 = 0.65



Referring to Churn 64, the permissible value of Mux/Mux1 corresponding to Muy/Muy1 and Pu

/Puz is equal to 0.58

The actual value of 0.62 is only slightly higher than the value read from the Chart.

This can be made up by slight increase in reinforcement.

Using Boris load contour equation as per IS:456-2000

Pu /Puz = 0.65 thus, αn = 1 + [(2-1) / (0.8 - 0.2)] (0.65-0.2) = 1.75

[0.62 ]1.75

+ [0.75]1.75

= 1.04 slightly greater than 1 and slightly unsafe. This can be made up

by slight increase in reinforcement say 1.3%

Thus provide As = 1.3x400x600/100 = 3120 mm2

Provide 1.3 % of steel

p/fck = 1.3/15 = 0.086

d1/D = 52.5 /600 = 0.088 = 0.1

From chart 44

Mu/fck b x D2 = 0.095

Mux1 = 0.095 x 15 x 400 x 6002) = 205.2 kN.m


Mu/fck b x D2 = 0.085

Mux1 = 0.085 x 15 x 600 x 4002) = 122.4 kN.m

Chart 63 : Puz/Ag = 10.4

Puz = 10.4 x 400 x 600 = 2496 kN

Mux/Mux1 = 120/205.2 =0.585

Muy/Muy1=90/122.4 = 0.735

Pu /Puz =1600/2496 = 0.641

Referring to Chart 64, the permissible value of Mux/Mux1 corresponding to Muy/Muy1 and Pu


Hence the section is O.K.


Pu /Puz = 0.641 thus, αn = 1 + [(2-1) / (0.8 - 0.2)] (0.641-0.2) = 1.735

[120/205.2]1.735

+ [90/122.4]1.735

= 0.981 ≤ 1 Thus OK

As = 3120 mm2. Provide 10 bars of 20 mm dia. Steel provided is 314 x 10 = 3140 mm

2



Design of transverse steel: Provide 8 mm dia stirrups at 300 mm c/c as shown satisfying the

requirements of IS: 456-2000

10. Verify the adequacy of the short column section 500 mm x 300 mm under the

following load conditions:

Pu

= 1400 kN, Mux

= 125 kNm, Muy

= 75 kNm. The design interaction curves of SP 16

should be used. Assume that the column is a ‘short column’ and the eccentricity due to

moments is greater than the minimum eccentricity.

Solution:

Given: Dx

= 500 mm, b = 300 mm, A

s = 2946 mm

2

Mux

= 125 kNm, Muy

= 75 kNm, fck

= 25

MPa, fy = 415 MPa

Applied eccentricities

ex

= Mux

/Pu

= 125 × 103

/1400 = 89.3 mm ⇒ ex/D

x = 0.179

ey = M

uy/P

u = 75 × 10

3

/1400 = 53.6 mm ⇒ ey/D

y = 0.179

These eccentricities for the short column are clearly not less than the minimum eccentricities

specified by the Code.

Uniaxial moment capacities: Mux1

, Muy1

As determined in the earlier example, corresponding to Pu

= 1400 kN,

Mux1

= 187 kNm

Muy1

= 110 kNm

Values of Puz

and αn

Puz

= 0.45fck

Ag

+ (0.75fy – 0.45f

ck)A

sc

= (0.45 × 25 × 300 × 500) + (0.75 × 415 – 0.45 × 25)×2946

= (1687500 + 883800)N = 2571 kN

⇒ Pu/P

uz = 1400/2571 = 0.545 (which lies between 0.2 and 0.8)

⇒ αn

= 1.575

Check safety under biaxial bending

[125/187]1.575

+ [75/110]1

= 0.530 + 0.547

= 1.077 > 1.0

Hence, almost ok.



Most of the structures built by us are made of reinforced concrete. Here, the part of the structure

above ground level is called as the superstructure, where the part of the structure below the ground

level is called as the substructure. Footings are located below the ground level and are also referred

as foundations. Foundation is that part of the structure which is in direct contact with soil. The R.C.

structures consist of various structural components which act together to resist the applied loads

and transfer them safely to soil. In general the loads applied on slabs in buildings are transferred to

soil through beams, columns and footings. Footings are that part of the structure which are

generally located below ground Level. They are also referred as foundations. Footings transfer the

vertical loads, Horizontal loads, Moments, and other forces to the soil.

The important purpose of foundation are as follows;

1. To transfer forces from superstructure to firm soil below.

2. To distribute stresses evenly on foundation soil such that foundation soil neither fails nor

experiences excessive settlement.

3. To develop an anchor for stability against overturning.

4. To provide an even surface for smooth construction of superstructure.

Due to the loads and soil pressure, footings develop Bending moments and Shear forces.

Calculations are made as per the guidelines suggested in IS 456 2000 to resist the internal forces.

1. Shallow Foundations

2. Deep Foundations

Shallow Foundations are provided when adequate SBC is available at relatively short depth below

ground level. Here, the ratio of Df / B < 1, where Df

is the depth of footing and B is the width of

footing. Deep Foundations are provided when adequate SBC is available at large depth below ground

level. Here the ratio of Df / B >= 1.

• Isolated Footing

• Combined footing

• Strap Footing

• Strip Footing

• Mat/Raft Foundation

• Wall footing

5.2. Types of Foundations

Based on the position with respect to ground level, Footings are classified into two types;

5.2.1 Types of Shallow Foundations

The different types of shallow foundations are as follows:

5.1 . Pre-requisites



Jyothi

Typewritten text

UNIT VLIMIT STATE DESIGN OF FOOTING

Some of the popular types of shallow foundations

a) Isolated Column Footing

These are independent footings which are provided

when

• SBC is generally high

• Columns are far apart

• Loads on footings are less

The isolated footings can have different s

cross section Some of the popular shapes of footings are;

• Square

• Rectangular

• Circular

The isolated footings essentially consists of

stepped or sloping in nature. The bottom of the slab is reinforced with steel mesh to resist the two

internal forces namely bending moment and shear force.

The sketch of a typical isolated footing is shown

Fig. 1 Plan and section of typical isolated footing

b) Combined Column Footing

These are common footings which support the loads from

are provided when

• SBC is generally less

• Columns are closely spaced

• Footings are heavily loaded

In the above situations, the area required to provide isolated footings

overlap. Hence, it is advantageous to provide single combined footing

are located on or close to property line. In such cases footings cannot be extended on one side.

Here, the footings of exterior and interior columns are connected by the combined footing.

types of shallow foundations are briefly discussed below.

which are provided for each column. This type of footing is chosen

on footings are less

The isolated footings can have different shapes in plan. Generally it depends on the shape of column

Some of the popular shapes of footings are;

footings essentially consists of bottom slab. These bottom Slabs can be either

The bottom of the slab is reinforced with steel mesh to resist the two

internal forces namely bending moment and shear force.

The sketch of a typical isolated footing is shown in Fig. 1.

Fig. 1 Plan and section of typical isolated footing

which support the loads from 2 or more columns. Combined footings

, the area required to provide isolated footings for the columns

Hence, it is advantageous to provide single combined footing. In some cases the columns



This type of footing is chosen

depends on the shape of column

ottom Slabs can be either flat,

The bottom of the slab is reinforced with steel mesh to resist the two

. Combined footings

for the columns generally

. In some cases the columns





Fig. 2 Plan and section of typical co

Combined footings essentially consist

are generally rectangular in plan.

Combined footings can also have a connect

inverted T – beam slab.

c) Strap Footing

An alternate way of providing combined footing located close to property line is the strap footing

strap footing, independent slabs below columns

beam. The strap beam does not remain in contact with the soil

the soil. Generally it is used to combine the footing of the outer column to the adjacent one so that

the footing does not extend in the adjoining property

Fig. 3 Plan and section of typical strap footing

Fig. 2 Plan and section of typical combined footing

consist of a common slab for the columns it is supporting

. Sometimes they can also be trapezoidal in plan

have a connecting beam and a slab arrangement, which is

An alternate way of providing combined footing located close to property line is the strap footing

ndependent slabs below columns are provided which are then connected by

oes not remain in contact with the soil and does not transfer any pressure to

used to combine the footing of the outer column to the adjacent one so that

footing does not extend in the adjoining property. A typical strap footing is shown in Fig. 3.

Fig. 3 Plan and section of typical strap footing

it is supporting. These slabs

(refer Fig. 2).

is similar to an

An alternate way of providing combined footing located close to property line is the strap footing. In

nected by a strap

oes not transfer any pressure to

used to combine the footing of the outer column to the adjacent one so that

A typical strap footing is shown in Fig. 3.



d) Strip Footing

Strip footing is a continuous footing

columns is shown in Fig. 4.

Fig. 4 Plan and section of typical strip footing

e) Mat Foundation

Mat foundation covers the whole plan area of structure.

reinforced solid floor slabs or flat slabs.

a structure and supports all the walls and columns.

• Soil pressure is low

• Loads are very heavy

• Spread footings cover > 50% area

A typical mat foundation is shown in Fig.


Deep foundations are provided when a

different types of deep foundations. Some of the common types of deep foundations are

below.

• Pile Foundation

• Pier Foundation

• Well Foundation

ontinuous footing provided under columns or walls. A typical str


overs the whole plan area of structure. The detailing is similar to two way

reinforced solid floor slabs or flat slabs. It is a combined footing that covers the entire area beneath

a structure and supports all the walls and columns. It is normally provided when

Spread footings cover > 50% area

is shown in Fig. 5.

Plan and section of typical strip footing

Deep foundations are provided when adequate SBC is available at large depth below GL

different types of deep foundations. Some of the common types of deep foundations are

A typical strip footing for

etailing is similar to two way

footing that covers the entire area beneath

dequate SBC is available at large depth below GL. There are

different types of deep foundations. Some of the common types of deep foundations are listed

5.2.2 Types of Deep Foundations



The safe bearing capacity of soil is the safe extra load soil can withstand without experiencing shear

failure. The Safe Bearing Capacity (SBC) is considered unique at a particular site. But it also depends

on the following factors:

• Size of footing

• Shape of footing

• Inclination of footing

• Inclination of ground

• Type of load

• Depth of footing etc.

SBC alone is not sufficient for design. The allowable bearing capacity is taken as the smaller of the

following two criteria

• Limit states of shear failure criteria (SBC)

• Limit states of settlement criteria

Based on ultimate capacity, i.e., shear failure criteria, the SBC is calculated as

SBC = Total load / Area of footing

Usually the Allowable Bearing Pressure (ABP) varies in the range of 100 kN/m2

to 400 kN/m2. The

area of the footing should be so arrived that the pressure distribution below the footing should be

less than the allowable bearing pressure of the soil. Even for symmetrical Loading, the pressure

distribution below the footing may not be uniform. It depends on the Rigidity of footing, Soil type

and Conditions of soil. In case of Cohesive Soil and Cohesion less Soil the pressure distribution varies

in a nonlinear way. However, while designing the footings a linear variation of pressure distribution

from one edge of the footing to the other edge is assumed. Once the pressure distribution is known,

the bending moment and shear force can be determined and the footing can be designed to safely

resist these forces.

• Area of footing

• Thickness of footing

• Reinforcement details of footing (satisfying moment and shear considerations)

• Check for bearing stresses and development length

This is carried out considering the loads of footing, SBC of soil, Grade of concrete and Grade of steel.

The method of design is similar to the design of beams and slabs. Since footings are buried,

deflection control is not important. However, crack widths should be less than 0.3 mm.

The steps followed in the design of footings are generally iterative. The important steps in the design

of footings are;

• Find the area of footing (due to service loads)

• Assume a suitable thickness of footing

5.3. Bearing Capacity of Soil

5.4. Design of Isolated Column Footing

The objective of design is to determine



• Identify critical sections for flexure and shear

• Find the bending moment and shear forces at these critical sections (due to factored loads)

• Check the adequacy of the assumed thickness

• Find the reinforcement details

• Check for development length

• Check for bearing stresses

Limit state of collapse is adopted in the design pf isolated column footings. The various design steps

considered are;

• Design for flexure

• Design for shear (one way shear and two way shear)

• Design for bearing

• Design for development length

The materials used in RC footings are concrete and steel. The minimum grade of concrete to be used

for footings is M20, which can be increased when the footings are placed in aggressive environment,

or to resist higher stresses.

Cover:Cover:Cover:Cover: The minimum thickness of cover to main reinforcement shall not be less than 50 mm for

surfaces in contact with earth face and not less than 40 mm for external exposed face. However,

where the concrete is in direct contact with the soil the cover should be 75 mm. In case of raft

foundation the cover for reinforcement shall not be less than 75 mm.

Minimum reinforcement and bar diameter: The minimum reinforcement according to slab and

beam elements as appropriate should be followed, unless otherwise specified. The diameter of main

reinforcing bars shall not be less 10 mm. The grade of steel used is either Fe 415 or Fe 500.

The important guidelines given in IS 456 : 2000 for the design of isolated footings are as follows:

34.1 General

Footings shall be designed to sustain the applied loads, moments and forces and the induced

reactions and to ensure that any settlement which may occur shall be as nearly uniform as possible,

and the safe bearing capacity of the soil is not exceeded (see IS 1904).

34.1.1�In sloped or stepped footings the effective cross-section in compression shall be limited by

the area above the neutral plane, and the angle of slope or depth and location of steps shall be such

that the design requirements are satisfied at every section. Sloped and stepped footings that are

designed as a unit shall be constructed to assure action as a unit.

34.1.2 Thickness at the Edge of Footing

In reinforced and plain concrete footings, the thickness at the edge shall be not less than 150 mm for

footings on soils, nor less than 300 mm above the tops of piles for footings on piles.

5.5. Specifications for Design of footings as per IS 456 : 2000



34.1.3 In the case of plain concrete pedestals, the angle between the plane passing through the

bottom edge of the pedestal and the corresponding junction edge of the column with pedestal and

the horizontal plane (see Fig. 20) shall be governed by the expression:

tan� <≠ 0.9 ∗ � 100�� ⁄ � + 1

where

�� = calculated maximum bearing pressure at the base of the pedestal in N/mm2

�� = characteristic strength of concrete at 28 days in N/mm2.

34.2 Moments and Forces

34.2.1 In the case of footings on piles, computation for moments and shears may be based on the

assumption that the reaction from any pile is concentrated at the centre of the pile.

34.2.2 For the purpose of computing stresses in footings which support a round or octagonal

concrete column or pedestal, the face of the column or pedestal shall be taken as the side of a

square inscribed within the perimeter of the round or octagonal column or pedestal.

34.2.3 Bending Moment

34.2.3.1 The bending moment at any section shall be determined by passing through the section a

vertical plane which extends completely across the footing, and computing the moment of the

forces acting over the entire area of the footing on one side of the said plane.

34.2.3.2 The greatest bending moment to be used in the design of an isolated concrete footing

which supports a column, pedestal or wall, shall be the moment computed in the manner prescribed

in 34.2.3.1 at sections located as follows:

a) At the face of the column, pedestal or wall, for footings supporting a concrete column, pedestal or

wall;

b) Halfway between the centre-line and the edge of the wall, for footings under masonry walls; and

c) Halfway between the face of the column or pedestal and the edge of the gussetted base, for

footings under gussetted bases.

34.2.4 Shear and Bond

34.2.4.1 The shear strength of footings is governed by the more severe of the following two

conditions:

a) The footing acting essentially as a wide beam, with a potential diagonal crack extending in a plane

across the entire width; the critical section for this condition shall be assumed as a vertical

section located from the face of the column, pedestal or wall at a distance equal to the effective

depth of footing for footings on piles.

b) Two-way action of the footing, with potential diagonal cracking along the surface of truncated

cone or pyramid around the concentrated load; in this case, the footing shall be designed for

shear in accordance with appropriate provisions specified in 31.6.



34.2.4.2 In computing the external shear or any section through a footing supported on piles, the

entire reaction from any pile of diameter Dp whose centre is located DP/2 or more outside the

section shall be assumed as producing shear on the section; the reaction from any pile whose centre

is located DP/2 or more inside the section shall be assumed as producing no shear on the section, For

intermediate positions of the pile centre, the portion of the pile reaction to be assumed as producing

shear on the section shall be based on straight line interpolation between full value at DP/2 outside

the section and zero value at DP/2 inside the section.

34.2.4.3 The critical section for checking the development length in a footing shall be assumed at the

same planes as those described for bending moment in 34.2.3 and also at all other vertical planes

where abrupt changes of section occur. If reinforcement is curtailed, the anchorage requirements

shall be checked in accordance with 26.2.3.

34.3 Tensile Reinforcement

The total tensile reinforcement at any section shall provide a moment of resistance at least equal to

the bending moment on the section calculated in accordance with 34.2.3.

34.3.1 Total tensile reinforcement shall be distributed across the corresponding resisting section as

given below:

a) In one-way reinforced footing, the-reinforcement extending in each direction shall be distributed

uniformly across the full width of the footing;

b) In two-way reinforced square footing, the reinforcement extending in each direction shall be

distributed uniformly across the full width of the footing; and

c) In two-way reinforced rectangular footing, the reinforcement in the long direction shall be

distributed uniformly across the full width of the footing. For reinforcement in the short

direction, a central band equal to the width of the footing shall be marked along the length of the

footing and portion of the reinforcement determined in accordance with the equation given

below shall be uniformly distributed across the central band:

ReinforcementincentralbandwidthTotalreinforcementinshortdirection = �

( + 1

where β is the ratio of the long side to the short side of the footing. The remainder of the

reinforcement shall be uniformly distributed in the outer portions of the footing.

34.4 Transfer of Load at the Base of Column

The compressive stress in concrete at the base of a column or pedestal shdl be considered as being

transferred by bearing to the top of the supporting Redestal or footing. The bearing pressure on the

loaded area shall not exceed the permissible bearing stress in direct compression multiplied by a

value equal to

√*+√*,

but not greater than 2, where A1 = supporting area for bearing of footing, which in sloped or stepped

footing may be taken as the area of the lower base of the largest frustum of a pyramid or cone

contained wholly within the footing and having for its upper base, the area actually loaded and

having side slope of one vertical to two horizontal; and A2 = loaded area at the column base.



34.4.1 Where the permissible bearing stress on the concrete in the supporting or supported member

would be exceeded, reinforcement shall be provided for developing the excess force, either by

extending the longitudinal bars into the supporting member, or by dowels (see 34.4.3).

34.4.2 Where transfer of force is accomplished by, reinforcement, the development length of the

reinforcement shall be sufficient to transfer the compression or tension to the supporting member

in accordance with 26.2.

34.4.3 Extended longitudinal reinforcement or dowels of at least 0.5 percent of the cross-sectional

area of the supported column or pedestal and a minimum of four bars shall be provided. Where

dowels are used, their diameter shall no exceed the diameter of the column bars by more than 3

mm.

34.4.4 Column bars of diameters larger than 36 mm, in compression only can be dowelled at the

footings with bars of smaller size of the necessary area. The dowel shall extend into the column, a

distance equal to the development length of the column bar and into the footing, a distance equal to

the development length of the dowel.

34.5 Nominal Reinforcement

34.5.1 Minimum reinforcement and spacing shall be as per the requirements of solid slab.

34.5.2 The nominal reinforcement for concrete sections of thickness greater than 1 m shall be 360

mm� per metre length in each direction on each face. This provision does not supersede the

requirement of minimum tensile reinforcement based on the depth of the section.

Design an isolated footing for an R.C. column of size 230 mm x 230 mm which carries a vertical load

of 500 kN. The safe bearing capacity of soil is 200 kN/m2. Use M20 concrete and Fe 415 steel.

Solution

Step 1: Size of footing

Load on column = 600 kN

Extra load at 10% of load due to self weight of soil = 60 kN

Hence, total load, P = 660 kN

Required area of footing, - = ./01 = 223433 = 5. 564

Assuming a square footing, the side of footing is 7 = 0 =√5. 5 = 8. 946

Hence, provide a footing of size 1.85 m x 1.85 m

Net upward pressure in soil, : = ;��+.<=>+.<= = 175.3BC/E, < 200BC/E, Hence O.K.

Hence, factored upward pressure of soil, pu = 263 kN/m2 and, factored load, Pu = 900 kN.

5.6. Numerical Problems Example 1



Step 2: Two way shear

Assume an uniform overall thickness of footing, D = 450 mm.

Assuming 12 mm diameter bars for main steel, effective thickness of footing ‘d’ is

d = 450 – 50 – 12 – 6 = 382 mm

The critical section for the two way shear or punching shear occurs at a distance of d/2 from the face

of the column (See Fig. 6), where a and b are the sides of the column.

Fig. 6 Critical section in two way shear

Hence, punching area of footing = (a + d)2

= (0.23 + 0.382)2 = 0.375 m

2

here a = b =side of column

Punching shear force = Factored load – (Factored upward pressure x punching area of footing)

= 900 – (263 x 0.375)

= 801.38 kN

Perimeter of the critical section = 4 (a+d) = 4 (230+ 382)

= 2448 mm

Therefore, nominal shear stress in punching or punching shear stress ζV is computed as

FG = .HIJKLIMNKOPQRSQJOTOQL6OUOQVORROJULWOUKLJXIONN

=938. 59 × 83334ZZ9V594 = 3. 92[/664

Allowable shear stress = kS . ζC

where F1 = 3. 4\�RJX = 3. 4\√43 ≈ 8. 84[/664

and, XN = 3. \ +^1� = _3. \ +3.453.45` = 8.3 ; Hence, adopt ks=1

Thus, Allowable shear stress = kS .ζC = 1 x 1.12 = 1.12 N/mm2

Since the punching shear stress (0.86 N/mm2) is less than the allowable shear stress (1.12 N/mm

2),

the assumed thickness is sufficient to resist the punching shear force.

Hence, the assumed thickness of footing D = 450 mm is sufficient.

The effective depth for the lower layer of reinforcement, d = 450 – 50 – 6 = 396 mm, and

the effective depth for the upper layer of reinforcement, d = 450 – 50 – 12 – 6 = 382 mm.



Step 3: Design for flexure

The critical section for flexure occurs at the face of the column (Fig. 7).

Fig. 7 Critical section for flexure

The projection of footing beyond the column face is treated as a cantilever slab subjected to

factored upward pressure of soil.

Factored upward pressure of soil, pu = 263 kN/m2

Projection of footing beyond the column face, l = (1850 – 230)/2 = 810 mm

Hence, bending moment at the critical section in the footing is

aH = THb44 = 425V3.9844 = 92. 49X[ −6/m width of footing

The area of steel Ast can be determined using the following moment of resistance relation for under

reinforced condition given in Annex G – 1.1 b of IS 456 :2000.

ad = 3. 9eRf-NUg h8 − Rf-NUigRJXj Considering 1m width of footing,

92. 49k832 = 3. 9ekZ8\klmnk594 o8 − Z8\klmn8333k594k43p

Solving the above quadratic relation, we get

Ast = 648.42 mm2 and 17,761.01 mm

2

Selecting the least and feasible value for Ast, we have

Ast = 648.42 mm2

The corresponding value of pt = 0.17 %

Hence from flexure criterion, pt = 0.17 %

Step 4: One way shear

The critical section for one way shear occurs at a distance ‘d’ from the face of the column (Fig. 8).



Fig. 8 Critical section for one way shear

For the cantilever slab, total Shear Force along critical section considering the entire width B is

Vu = pu B (l – d) = 263 x 1.85 x (0.81 – 0.382)

= 208.24 kN

The nominal shear stress is given by

FG = GH0g = 439. 4ZV833389\3V594

& 3. 53q/664

From Table 61 of SP 16, find the pt required to have a minimum design shear strength ζC = ζV = 0.30

N/mm2 with fck = 20 N/mm

2.

For pt = 0.175 % the design shear strength ζC is 0.30 N/mm2 = ζV = 0.30 N/mm

2.

Hence from one way shear criterion, pt = 0.175 %

Comparing pt from flexure and one way shear criterion, provide pt = 0.175 % (larger of the two

values)

Hence, -NU = rn833 st = 3.8e\

8338333k594 & 22uvv4

Provide φ 12 mm dia bars at 140 mm c/c.

Therefore, Ast provided = 808 mm2 > Ast required (609 mm

2). Hence O.K.

Step 5: Check for development length

Sufficient development length should be available for the reinforcement from the critical section.

Here, the critical section considered for Ld is that of flexure.

The development length for 12 mm dia bars is given by

Ld = 47 ф = 47 x 12 = 564 mm.

Providing 60 mm side cover, the total length available from the critical section is

84 w − x� − 23 = 84 89\3 − 453� − 23 = e\3vv y wt. Hence O.K.

Step 6: Check for bearing stress

The load is assumed to disperse from the base of column to the base of footing at rate of 2H : 1V.

Hence, the side of the area of dispersion at the bottom of footing = 230 + 2 (2 x 450) = 2030 mm.

Since this is lesser than the side of the footing (i.e., 1850 mm)

A1 = 1.85 x 1.85 = 3.4225 m2



The dimension of the column is 230 mm x 230 mm. Hence, A2= 0.230 x 0.230 = 0.0529 m2

zl8l4 = z5. Z44\3. 3\4u = 9. 3Z > 2

Hence, Limit the value of {l8l4 = 4

∴ Permissible bearing stress = 3. Z\}~�{l8l4

= 0.45 x 20 x 2 = 18 N/mm2

l~n�x�s�x��mn��mm = �x~n��t��xtl��xxn~��v�sxm� = u33k8333453k453 = 8e. 38 q

vv4

Since the Actual bearing stress (17.01 N/mm2) is less than the Permissible bearing stress (18 N/mm

2),

the design for bearing stress is satisfactory.

Appropriate detailing should be shown both in plan and elevation for the footing as per the

recommendations given in SP 34.

Example 2


of 800 kN together with an uniaxial moment of 40 kN-m. The safe bearing capacity of soil is 250

kN/m2. Use M25 concrete and Fe 415 steel.

Solution





Let us provide a square isolated footing, where L=B

Equating the maximum pressure of the footing to SBC of soil,

�l + �� = ��

i.e., 99304 +Z3V205 = 4\3

On solving the above equation, and taking the least and feasible value, B = 2 m

Hence, provide a square footing of size 2 m x 2 m

The maximum and minimum soil pressures are given by

T6PV = 93344 +Z3V245 = 453X[64 < 250 X[64 �.�.

T6LI = 93344 −Z3V245 = 8e3X[64 y ��. �. Hence, factored upward pressures of soil are,

pu,max = 345 kN/m2

and pu,min = 255 kN/m2



Further, average pressure at the center of the footing is given by

pu,avg = 300 kN/m2

and, factored load, Pu = 900 kN, factored uniaxial moment, Mu = 60 kN-m


Assume an uniform overall thickness of footing, D = 450 mm


d = 450 – 50 – 16 – 8 = 376 mm


of the column (Fig. 9), where a and b are the dimensions of the column.



= (0.30 + 0.376)2 = 0.457 m

2

where a = b = side of column

Punching shear force = Factored load – (Factored average pressure x punching area of footing)

= 1200 – (300 x 0.457)

= 1062.9 kN

Perimeter along the critical section = 4 (a+d) = 4 (300+ 376)

= 2704 mm



= 8324. u × 83334e3ZV5e2 = 8. 3\[/664


where F1 = 3. 4\�RJX = 3. 4\√4\ = 8. 4\[/664

and, XN = 3. \ +^1� = _3. \ + 3.533.53` = 8. 3 ; Hence, adopt ks=1



2),





The effective depth for the lower layer of reinforcement, , d = 450 – 50 – 8 = 392 mm, and

the effective depth for the upper layer of reinforcement, d = d = 450 – 50 – 16 – 8 = 376 mm.






Factored maximum upward pressure of soil, pu,max = 345 kN/m2

Factored upward pressure of soil at critical section, pu = 306.75 kN/m2


Bending moment at the critical section in the footing is

�� = ��nx�}��~��k��mnx�~��}��}��v~��n�~x�m�~n�� = ��5Z\ + 532. e\

4 � 3. 9\� k ��4k5Z\ + 532. e\5Z\ + 532. e\ � k 3. 9\5 �

Mu = 119.11 kN-m/ m width of footing



ad = 3. 9eRf-NUg h8 − Rf-NUigRJXj

Considering 1m width of footing,

88u. 88k832 = 3. 9ekZ8\klmnk5e2 o8 − Z8\klmn8333k5e2k4\p

Solving the quadratic equation,

Ast = 914.30 mm2 and 21,735.76 mm

2

Selecting the least and feasible value, Ast = 914.30 mm2






The critical section for one way shear occurs at a distance of ‘d’ from the face of the column (Fig. 11).




For the cantilever slab, total Shear Force along critical sectionconsidering the entire width B is � = ��nx� }��~� � k � � − t� k �� = �5Z\ + 54e. 84 � k � 3. 9\ − 3. 5e2� k 4� Vu = 318.58 kN


¡ = ��t = 589. \9k83334333k5e2 = 3. Z4q/vv4

From 19 of IS 456 :2000, find the pt required to have a minimum design shear strength ζC = ζV = 0.42


2.


2.



values)

Hence, lmn = rn833 st = 3.52\

833 8333k5e2 = 85e4. Zvv4


Therefore, Ast provided = 1436 mm2 > Ast required (1372.4 mm

2). Hence O.K.







Ld = 47 ф = 47 x 16 = 752 mm.


84 w − x� − 23 = 84 4333 − 533� − 23 = eu3vv y wt Hence O.K.



Hence, the side of the area of dispersion at the bottom of footing = 300 + 2 2 x 450� = 2100 mm. Since this is lesser than the side of the footing i.e., 2000 mm�, A1 = 2 x 2 = 4 m2 The dimension of the column is 300 mm x 300 mm.

Hence, A2= 0.30 x 0.30 = 0.09 m2

zl8l4 = z Z3. 3u = 2. 2e > 2



= 0.45 x 25 x 2 = 22.5 N/mm2

Actual bearing stress = �x~n��t��xt

l��xxn~��v�sxm� = 8433k8333533k533 = 85. 55q/vv4

Since the Actual bearing stress (13.33 N/mm2) is less than the Permissible bearing stress (22.5

N/mm2), the design for bearing stress is satisfactory.





UNIT – 1

Part A

1. What are the assumptions made in the working stress method? (NOV-DEC 2012)

a) At any cross-section, plane sections before bending remain plain after bending.

b) All tensile stresses are taken up by reinforcement and none by concrete, except

as otherwise specifically permitted.

c) The stress-strain relationship of steel and concrete, under working loads, is a

straight line.

d) The modular ratio m has the value -280/3σbc.

2. Difference between Elastic method and limit state method. (NOV-DEC 2010)

Advantages of limit state method over the other methods

a. In the limit state method of analysis, the principles of both elastic as well as

plastic theories used and hence suitable for concrete structures.

b. The structure designed by limit state method is safe and serviceable under

design loads and at the same time it is ensured that the structure does not collapse

even under the worst possible loading conditions.

c. The process of stress redistribution, moment redistribution etc., are considered

in the analysis and more realistic factor of safety values are used in the design.

Hence the design by limit state method is found to be more economical.

d. The overall sizes of flexural members (depth requirements) arrived by limit

state method are less and hence they provide better appearance to the structure

e. Because of the modified assumptions regarding the maximum compressive strains in

concrete and steel, the design of compressive reinforcement for double reinforced

beams and eccentrically loaded columns by limit state method gives realistic

valued which is not so in other methods.

3. Draw stress-strain curve for concrete in working stress design and mention the

salient points. (NOV-DEC 2010)



4. Define characteristic strength in limit state method. (NOV-DEC 2009) (APRIL

MAY 2012)

The term ‘characteristic strength’ means that value of the strength of the

material below which not more than 5 percent of the test results are expected to

fall.

5. What is meant by balanced section? (NOV-DEC 2009) (NOV-DEC 2012)

When the maximum stress in steel and concrete simultaneously reach their

allowable values, the section is said to be balanced section.in this section the

actual neutral axis depth is equal to the critical neutral axis.

6. Define : Limit state". (APRIL MAY 2012)

The acceptable limit for the safety and serviceability requirements before

failure occurs is called a ‘limit state’. The aim of design is td achieve acceptable

probabilities that the structure will not become unfit for the use for which it is

intended, that is, that it will not reach a limit state.

7. What are the expressions recommended by the IS 456-2000 for Modulus of

Elasticity and Flexural Strength? (MAY JUNE 2009)

lexural strength f 0.7.fck /mm2

Where ‘fck’ is the characteristic cube compressive strength of concrete in N/mm2.

Where, E, is the short term static modulus of elasticity in N/mm2



8. Write the formula for the neutral axis depth factor 'K in working stress design.

(MAY JUNE 2009)

Neutral axis depth factor ‘K’ σbc.m/(σbc.m + σst)

Where σbc permissible stress in concrete. σbc permissible stress in steel.

M = modular ratio.

Part B

1. Explain the limit state philosophy as detailed in the current IS code. (NOV-DEC

2012)

The Answer is in Page No.67 of IS 456:2000.

In the method of design based on limit state concept, the structure shall be

designed to withstand safely all loads liable to act on it throughout its life; it shall

also satisfy the serviceability requirements, such as limitations on deflection and

cracking. The acceptable limit for the safety and serviceability requirements

before failure occurs is called a ‘limit state’. The aim of design is td achieve

acceptable probabilities that the structure will not become unfit for the use for

which it is intended, that is, that it will not reach a limit state. 351.1 All relevant

limit states shall be considered in design to ensure an adequate degree of safety

and serviceability. In general, the structure shall be designed on the basis of the

most critical limit state and shall be checked for other limit states.



35.1.2 For ensuring the above objective, the design should be based on

characteristic values for material strengths and applied loads, which take into

account the variations in the material strengths and in the loads to be supported.

The characteristic values should be based on statistical data if available; where

such data are not available they should be based on experience. The ‘design

values’ are derived from the characteristic values through the use of partial safety

factors, one for material strengths and the other for loads. In the absence of special

considerations these factors should have the values given in 36 according to the

material, the type of loading and the limit state being considered.

2. Design a R.C beam to carry a load of 6 kN/m inclusive of its own weight on an

effect span of 6m keep the breath to be 2/3 rd of the effective depth .the permissible

stressed in the concrete and steel are not to exceed 5N/mm2 and 140 N/mm2.take

m=18. (NOV-DEC 2012)







M = (w.l2)/8 = (6x62)/8 = 27kNm

M = Qbd2

d2 = M/Qb = (27x106)/ (0.84x2/3xd)

d = 245mm.




= Qbd2 = 0.84x245x3652 = 27.41kNm. > M. it can be designed as singlyMbal

reinforced section.


Ast = Mbal / (σst.j.d) 616.72mm2




3. Design a doubly reinforced beam of section 240X500mm to carry a bending

moment of 80kNm.Assume clear cover at top a bottom as 30mm and take

m=18.adopt working stress method. (NOV-DEC 2010)


5N/mm2 and 140 N/mm2.







M = 80kNm

M = Qbd2

D = 500mm, b = 240mm

d = 500-30mm = 470mm

.Step 3: Balanced Moment.



= Qbd2 = 0.84x240x4702 = 44.53kNm. < M. it can be designed as doublyMbal

reinforced section.

Step 4: Area of Tension steel.

Ast = Ast1 + Ast2

Ast1 = Mbal / (σst.j.d) (44.53x106)/(140x0.87x470) = 777.87mm2



Ast2 = (M-Mbal) / (σst.(d-d1)) = (80x106-44.53x106)/(140x(470-30)) = 575.8mm2




Asc = (M-Mbal) / (σsc.(d-d1)) = (80x106-44.53x106)/(51.8x(470-30))=1580.65 mm2




4. Design a beam subjected to a bending moment of 40kNm by working stress

design. Adopt width of beam equal to half the effective depth. (NOV-DEC 2010)


5N/mm2 and 140 N/mm2.take m=18.









M = 40kNm

M = Qbd2

d2 = M/Qb = (40x106)/ (0.84x1/2xd)

d = 456.2 say 460 mm.

b = ½ d = 0.5x460 = 230mm



reinforced section.


Ast = Mbal / (σst.j.d) (40.88x106)/(140x0.87x460) = 729.64mm2




5. Determine the moment of resistance of a singly reinforced beam 160X300mm

effective section, if the stress in steel and concrete are not to exceed 140N/mm2

and 5N/mm2.effectve span of the beam is 5m and the beam carries 4 nos of 16mm

dia bars. Take m=18.find also the minimum load the bam can carry. Use WSD

method. (NOV-DEC 2009)



Step 1: Actual NA.


160. xa2/2 = 18 X 804.24(300 –xa)

Xa = 159.42mm


xc σbc.d/(σst/.m + σcbc) 117.39mm < Xa 159.42mm

it is Over reinforced Section.



Step 4: Safe load.

M = (w.l2)/8

W = (8 x 15.74)/52 = 5.03 kN/m

6. Design an interior panel of RC slab 3mX6m size, supported by wall of 300mm

thick. Live load on the slab is 2.5kN/m2.the slab carries 100mm thick lime

concrete (density 19kN/m3).Use M15 concrete and Fe 415 steel. (NOV-DEC

2009)


ly/lx = 6/3 = 2 = 2.it has to be designed as two way slab.




D = 270 + 20 + 10/2 = 295mm


For shorter span:


Le =c/c distance b/w supports = 3000 + 2(230/2) =3.23m




For longer span:





Live load = 2.5kN/m2







Mx αx . w . lx 0.103x18.225x3.23 = 9.49kNm

My αy . w . lx 0.048 x18.225x3.23 = 4.425kNm


M = Qbd2

d2 = M/Qb = 9.49/2.76x1 = 149.39mm say 150mm.



For longer span:


4.425x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150))

Ast = 180mm2

Use 10mm dia bars



For shorter span:




9.49x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150))

Ast = 200mm2

Use 10mm dia bars



7. Differentiate between working stress method and limit state method. (APRIL

MAY 2012)

In the limit state method of analysis, the principles of both elastic as well as

plastic theories used and hence suitable for concrete structures.

The structure designed by limit state method is safe and serviceable under

design loads and at the same time it is ensured that the structure does not

collapse even under the worst possible loading conditions.

The process of stress redistribution, moment redistribution etc., are considered

in the analysis and more realistic factor of safety values are used in the design.

Hence the design by limit state method is found to be more economical.

The overall sizes of flexural members (depth requirements) arrived by limit

state method are less and hence they provide better appearance to thestructure

Because of the modified assumptions regarding the maximum compressive

strains in concrete and steel, the design of compressive reinforcement for double

reinforced beams and eccentrically loaded columns by limit state method gives

realistic valued which is not so in other methods.

8. Explain the following terms :

a. characteristic strength and characteristic loads.


b. partial safety factors.


c. Balanced section and under reinforced section. (APRIL MAY 2012)



When the maximum stress in steel and concrete simultaneously reach their

allowable values, the section is said to be balanced section.in this section the

actual neutral axis depth is equal to the critical neutral axis.

When the percentage of steel in the section is less than that required for a

balanced section. In this section the actual neutral axis depth is equal to the critical

neutral axis.

9. Derive the expressions for the depth of Neutral axis and Moment of resistance of a

Rectangular Singly reinforced balanced beam section under flexure and obtain the

design constants K, j and Q for M 20 grade concrete and Fe 415 grade steel. Use

working stress method. (MAY JUNE 2009)

10. A reinforced concrete rectangular section 300 mm wide and 600 mm overall depth

is reinforced with 4 bars of 25 mm diameter at an effective cover of 50 mm on the

tension side. The beam is designed with M 20 grade concrete and Fe 415 grade

steel. Determine the allowable bending moment and the stresses developed in steel

and concrete under this moment. Use working stress method. (MAY JUNE 2009)

Step 1: Actual NA.


300. xa2/2 = 18 X 1963.50(550 –xa)

Xa = 117.81mm


xc σbc.d/(σst/.m + σcbc) = 194.66mm > Xa = 117.81mm

it is Under reinforced Section.


For steel:

M = (Ast.σst )(d- xa/3) = (1963.5x230)(550-117.81/3) = 230.64kNm

For concrete:




UNIT –2

Part A

1. Explain the check for deflection control in the design of slabs? (NOV-DEC 2012)

The deflection of a structure or part thereof shall not adversely affect the

appearance or efficiency of the structure or finishes or partitions. The

deflection shall generally be limited to the following:

a) The final deflection due to all loads including the effects of temperature,

creep and shrinkage and measured from the as-cast level of the , supports of

floors, roofs and all other horizontal members, should not normally exceed

span/250.

b) The deflection including the effects of temperature, creep and shrinkage

occurring after erection of partitions and the application of finishes should

not normally exceed span/350 or 20 mm whichever is less.

2. When do you do for doubly reinforced beams? (NOV-DEC 2012) (NOV-DEC2010) (APRIL MAY 2012)

The section reinforced in both tension and compression zone is known as

doubly reinforced section. The doubly reinforced beams are adopt when

the balanced moment is smaller than the Actual moment.

3. What type of slabs are usually used in practice, under reinforced or over

reinforced? (NOV-DEC 2009)

The depth of slab chosen from deflection requirements will be usually

greater than the depth required for balanced design. Hence the area of steel

required will be less than the balanced amount. So, the slab is designed as under

reinforced section.

4. Why is necessary to provide transverse reinforcement in one way slab? (APRIL

MAY 2012)

Since the one way slab bends in one direction and also in shorter direction,



so it is necessary to provide transvers reinforcement in one way slabs. These slabs

adopted when availability of two supports in one direction.

5. Distinguish between under reinforced and over reinforced sections. (MAY JUNE

2009)

A beam reaches its permissible stress in steel under the working moment

before concrete reaches its stress is called as Under reinforced section.

A beam reaches its permissible stress in concrete under the working

moment before steel reaches its stress is called as Over reinforced section.

6. Sketch the edge and middle strips of a two way slab. (MAY JUNE 2009)

Part B

1. Design a one way slab with a clear span of 5m, simply supported on 230mm

thick masonry walls and subjected to a live load of 4kN/m2 and a surface

finish of 1kN/mm2.Assume Fe 415 steel. Assume that the slab is subjected

to moderate exposure conditions. (NOV-DEC 2012)


ly/lx = 5/1 = 5>2.it has to be designed as one way

slab.





D = 270 + 20 + 10/2 = 295mm


Le = clear span + effective depth = 5000 + 270 = 5.27m

(or) Le =c/c distance b/w supports = 5000 + 2(230/2) =

5.23m Adopt effective span = 5.23m least value.


Live load = 4kN/m2






M = wl2/8 = (17.625x5.232)/8 = 60.26kNm


M = Qbd2

d2 = M/Qb = 60.26/2.76x1 = 149.39mm say 150mm.

For design consideration adopt d =

150mm.



60.26x106 = 087x415xAstx150(1-(415

Ast)/(20x1000x150)) Ast = 300mm2

Use 10mm dia bars



2. Design a simply supported RC beam having an effective span of 5m.the beam has



to carry a load of 25 kN/m. sketch the reinforcement details. (NOV-DEC 2010)

(NOV-DEC 2012)

Step 1: Effective length.

Effective span,le = 5m

Step 2: Size of the beam.

Effective depth = le/10 = 5000/10 = 500mm

Assume, b = 2/3d = 2/3x500 = 333.2mm say 340mm

Step 3: Load Calculation

Live load = 25kN/m

Dead load = 1x.340x.500x25 = 4.25kN/m

Total load = 29.25kN/m

Factored load = 29.25x1.5 = 43.85kN/m

Step 4: Moment Calculation.

M = wl2/8 = (43.85x52)/8 = 137.08kNm


M = Qbd2

d2 = M/Qb = 137.08/2.76x.340 = 382.2mm say 380mm.

d = 380mm > 500mm

Hence it is safe.


Mbal = Qbd2 = 2.97x340x5002 = 252.06kNm > M

Hence it can be designed as singly reinforced beam section.

Step 6: Area of Steel


137.08x106 = 087x415xAstx500(1-(415 Ast)/(20x340x500))

Ast = 846.15mm2

Use 20mm dia bars

No of bars = Ast/ast = 846.15/314.15 = 2.45 say 3nos

Provide 3#20mm dia as tension reinforcement.



3. Design a RC beam 350X700mm effective section, subjected to a bending moment

of 300kNm.Adopt M20concrete and Fe415 steel. (NOV-DEC 2009)


b = 350mm & D = 700mm

d = 700-25-20/2 =665mm


M = 300kNm


Mbal = Qbd2 = 2.97x350x6652 = 459kNm > M

Hence it can be designed as singly reinforced beam section.



459x106 = 087x415xAstx665(1-(415 Ast)/(20x350x665))

Ast = 369.38 mm2

Use 20mm dia bars

No of bars = Ast/ast = 369.39/314.15 = 1.45 say 2nos

Provide 2#20mm dia as tension reinforcement.

4. Design a one way slab for a clear span 4m simply supported on 230mm thick wall.

Subjected to a live load of 4kN/m2 and floor finish of 1kN/m2.use M20 concrete

and F415 steel. (NOV-DEC 2009)


ly/lx = 4/1 = 4>2.it has to be designed as one way slab.



D = 270 + 20 + 10/2 = 295mm





Le =c/c distance b/w supports = 4000 + 2(230/2) = 4.23m Adopt

effective span = 4.23m least value.


Live load = 4kN/m2






M = wl2/8 = (17.625x4.232)/8 = 60.26kNm


M = Qbd2

d2 = M/Qb = 60.26/2.76x1 = 149.39mm say 150mm.




60.26x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast

= 300mm2

Use 10mm dia bars



5. Deign a rectangular beam of cross section 230 x 600 mm and of effective span

6m.imposed load on the beam is 40 kN/m. Use M20 concrete and Fe415 steel.

(APRIL MAY 2012)


b = 230mm & D = 600mm d

= 600-25-20/2 =565mm


Live load = 40kN/m2



Dead load = 1x.23x.565x25 = 3.245kN/m2




M = wl2/8 = (64.86x62)/8 = 291.9kNm


Mbal = Qbd2 = 2.97x230x5652 = 218kNm < M

Hence it can be designed as Doubly reinforced beam section.


Ast = Ast1 + Ast2


218x106 = 087x415xAstx565 (1-(415 Ast)/ (20x230x565))

Ast = 1365mm2

Use 20mm dia bars, ast = π/4 (202) = 314.15mm2

No. of bars = Ast/ast = 1365/314.15 = 4.47 say 5nos.

Ast2 = (M-Mbal)/(0.87fy(d-d1)) = (291x106-218x106)/(361x(565-20)) =371.03mm2




Asc = (M-Mbal) / (fsc.(d-d1)) = (291x106-218x106) / (351.8x(470-30))= 1580.65

mm2






6. A hall has clear dimensions 3 m x 9m with wall thickness 230 mm the live load on

the slab is 3kN/m2 and a finishing load of 1kN/m2 may be assumed. Using M20

concrete and Fe415 steel, design the slab. (APRIL MAY 2012)


ly/lx = 9/3 = 3>2.it has to be designed as one way slab.



D = 270 + 20 + 10/2 = 295mm






Live load = 4kN/m2






M = wl2/8 = (17.625x3.232)/8 = 60.26kNm


M = Qbd2

d2 = M/Qb = 60.26/2.76x1 = 149.39mm say 150mm. For

design consideration adopt d = 150mm.




= 300mm2

Use 10mm dia bars





7. Design a two way slab panel for the following data.

Size = 7mx5m

Width of Supports = 230 mm

Edge condition = interior

Live load = 4kN/m2

Floor finish = 1kN/m2

Consider M 20 grade concrete and Fe 415 grade steel. (MAY JUNE 2009)


ly/lx = 7/5 = 1.4>2.it has to be designed as two way slab.




D = 270 + 20 + 10/2 = 295mm


For shorter span:

Le = clear span + effective depth = 5000 + 270 = 5.27m (or) Le

=c/c distance b/w supports = 5000 + 2(230/2) = 5.23m

Adopt effective span = 5.23m least value. For

longer span:

Le = clear span + effective depth = 7000 + 270 = 7.27m (or) Le

=c/c distance b/w supports = 7000 + 2(230/2) = 7.23m Adopt

effective span = 7.23m least value.


Live load = 4kN/m2








Mx = αx . w . lx = 0.103x17.625x5.23 = 9.49kNm My =

αy . w . lx = 0.048 x17.625x5.23 = 4.425kNm


M = Qbd2

d2 = M/Qb = 9.49/2.76x1 = 149.39mm say 150mm. For

design consideration adopt d = 150mm.


For longer span:



= 180mm2

Use 10mm dia bars



For shorter span:


9.49x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150))

Ast = 200mm2

Use 10mm dia bars





s d

16 MARKS

Problem 1:

Determine the anchorage length of 4-20T reinforcing bars going into the supportof the simply supported beam shown in Fig. 6.15.5. The factored shear force Vu = 280kN, width of the column support = 300 mm. Use M 20 concrete and Fe 415 steel.

Solution 1:

τbd for M 20 and Fe 415 (with 60% increased) = 1.6(1.2) = 1.92 N/mm2

Ld =φσs =4τ bd

0.87(415)φ4(1.92)

(when σ s = 0.87 f y ) = 47.01φ ...... (1)

(Ld )whenσ = f ≤M1

V+ Lo

Here, to find M1, we need xu



)

xu =0.87 f y Ast =0.36 f ck b

0.87(415) (1256)

0.36(20) (300)= 209.94 mm

xu,max = 0.48(500) = 240 mm

Since xu < xu,max ; M1 = 0.87 fy Ast (d – 0.42 xu)

or M1 = 0.87(415) (1256) {500 – 0.42(209.94)} = 187.754 kNm

and V = 280 kN

We have from above, with the stipulation of 30 per cent increase assuming thatthe reinforcing bars are confined by a compressive reaction:

L ≤ 1.3 (M1 ) + Ld V o ...... (2)

From Eqs.(1) and (2), we have

47.01φ ≤ 1.3 (M1 ) + LV o

or 47.01φ ≤ 1.3187.754(106 )

{ }; if L is assumed as zero.

280(103 ) o

or φ ≤ 18.54 mm

Therefore, 20 mm diameter bar does not allow Lo = 0.

Determination of Lo:

1.3 (M1 ) + LV o

≥ 47.01φ

Minimum Lo = 47.01φ - 1.3 (M 1 )V

= 47.01(20) -187754

1.3(280

= 68.485 mm

So, the bars are extended by 100 mm to satisfy the requirement as shown inFig.



Problems

1. Determine the load carrying capacity of a column of size 300 x 400 mm reinforced

with six rods of 20 mm diameter i.e, 6-#20. The grade of concrete and steel are M20 and Fe 415 respectively. Assume that the column is short.

fck = 20 MPa, fy= 415 MPa

Area of steel ASC = 6 x π x 202/4 = 6 x 314 = 1884 mm

2

Percentage of steel = 100Asc/bD = 100x1884/300x400 = 1.57 %

Area of concrete Ac = Ag – Asc = 300 x 400 – 1884 = 118116 mm2



0.4x20x118116 + 0.67x415x1884

944928 + 523846 = 1468774 N = 1468. 8 kN

Therefore the safe load on the column = 1468.8 /1.5 = 979.2 kN

2. Determine the steel required to carry a load of 980kN on a rectangular column of

size 300 x 400 mm. The grade of concrete and steel are M20 and Fe 415 respectively.

Assume that the column is short.

fck = 20 MPa, fy= 415 MPa, P = 980 kN

Area of steel ASC = ?

Area of concrete Ac = Ag – Asc = (300 x 400 – ASC)



980 x 1.5 x 1000 = 0.4x20x (300 x 400 – ASC) + 0.67x415 ASC

= 960000 - 8 ASC + 278.06 ASC

ASC =1888.5 mm2,

Percentage of steel = 100Asc/bD = 100x1888.5 /300x400 = 1.57 % which is more than 0.8%

and less than 6% and therefore ok.

Use 20 mm dia. bas, No. of bars = 1888.5/314 = 6.01 say 6



3. Design a square or circular column to carry a working load of 980kN. The grade of

concrete and steel are M20 and Fe 415 respectively. Assume that the column is short.


Say ASC = 1.0% Ag =1/100 Ag = 0.01Ag

fck = 20 MPa, fy= 415 MPa, P = 980 kN




980 x 1.5 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag

= 7.92 Ag + 2.78 Ag =10.7Ag

Ag = 137383 mm2

Let us design a square column:

B = D = √ Ag =370.6 mm say 375 x 375 mm

This is ok. However this size cannot take the minimum eccentricity of 20 mm as emin/D =

20/375 =0.053 > 0.05. To restrict the eccentricity to 20 mm, the required size is 400x 400

mm.

Area of steel required is Ag = 1373.8 mm2. Provide 4 bar of 22 mm diameter. Steel provided

is 380 x 4 = 1520 mm2

Actual percentage of steel = 100Asc/bD = 100x1520 /400x400 = 0.95 % which is more than

0.8% and less than 6% and therefore ok.


Diameter of tie = ¼ diameter of main steel = 22/4 =5.5mm or 6 mm, whichever is greater.

Provide 6 mm.

Spacing: < 300 mm, < 16 x22 = 352mm, < LLD = 400mm. Say 300mm c/c

Design of circular column:

Here Ag = 137383 mm2

π x D2/4 = Ag, D= 418.2 mm say 420 mm. This satisfy the minimum eccentricity of 20m

Also provide 7 bars of 16 mm, 7 x 201 = 1407 mm2


Dia of tie = ¼ dia of main steel = 16/4 = 4 mm or 6 mm, whichever is greater. Provide 6 mm.


4. Design a rectangular column to carry an ultimate load of 2500kN. The unsupported

length of the column is 3m. The ends of the column are effectively held in position



and also restrained against rotation. The grade of concrete and steel are M20 and Fe 415 respectively.

Given:



Say ASC = 1.0% Ag =1/100 Ag = 0.01Ag




2500 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag

= 7.92 Ag + 2.78 Ag =10.7Ag

Ag = 233645 mm2

If it is a square column:

B = D = √ Ag =483 mm. However provide rectangular column of size 425 x 550mm. The

area provided=333750 mm2


2

Check for shortness: Ends are fixed. lex = ley = 0.65 l = 0.65 x 3000 = 1950 mm




emin, x = lux/500 + D/30 = 3000/500 + 550/30 = 24.22mm or 20mm whichever is greater.

emin, x = 24.22 mm < 0.05D = 0.05 x 550 =27.5 mm. O.K


emin, y= luy/500 + b/30 = 3000/500 + 425/30 = 20.17 mm or 20mm whichever is greater.

emin, x = 20.17 mm < 0.05b = 0.05 x 425 =21.25 mm. O.K


Dia of tie = ¼ dia of main steel = 20/4 = 5 mm or 6 mm, whichever is greater. Provide 6 mm

or 8 mm.




5. Design a circular column with ties to carry an ultimate load of 2500kN. The

unsupported length of the column is 3m. The ends of the column are effectively held

in position but not against rotation. The grade of concrete and steel are M20 and Fe

415 respectively.

Given:



Say ASC = 1.0% Ag =1/100 Ag = 0.01Ag




2500 x 1000 = 0.4x20x 0.99 Ag + 0.67x415 x 0.01Ag

= 7.92 Ag + 2.78 Ag =10.7Ag

Ag = 233645 mm2

π x D2/4 = Ag, D = 545.4 mm say 550 mm.


2

Check for shortness: Ends are hinged lex = ley = l = 3000 mm



Here, emin, x = emin, y = lux/500 + D/30 = 3000/500 + 550/30 = 24.22mm or 20mm whichever is

greater.

emin = 24.22 mm < 0.05D = 0.05 x 550 =27.5 mm. O.K


Diameter of tie = ¼ dia of main steel = 20/4 = 5 mm or 6 mm, whichever is greater. Provide 6

mm or 8 mm.


Similarly square column can be designed.

If the size of the column provided is less than that provided above, then the minimum

eccentricity criteria are not satisfied. Then emin is more and the column is to be designed as



uni axial bending case or bi axial bending case as the case may be. This situation arises when

more steel is provided ( say 2% in this case).

Try to solve these problems by using SP 16 charts, though not mentioned in the syllabus.

6. Design the reinforcement in a column of size 450 mm × 600 mm, subject to an axial

load of 2000 kN under service dead and live loads. The column has an unsupported

length of 3.0m and its ends are held in position but not in direction. Use M 20

concrete and Fe 415 steel.

Solution:

Given: lu= 3000 mm, b = 450 mm, D

= 600 mm, P =2000kN, M20, Fe415

Check for shortness: Ends are fixed. lex = ley = l = 3000 mm

lex /D= 3000/600 < 12, and ley /b = 3000/450< 12, Column is short



emin, x = lux/500 + D/30 = 3000/500 + 600/30 = 26 mm or 20mm whichever is greater.

emin, x = 26 mm < 0.05D = 0.05 x 600 =30 mm. O.K


emin, y= luy/500 + b/30 = 3000/500 + 450/30 = 21 mm or 20mm whichever is greater.

emin, x = 21 mm < 0.05b = 0.05 x 450 =22.5 mm. O.K

Minimum eccentricities are within the limits and hence code formula for axially loaded short

columns can be used.

Factored Load

Pu

= service load × partial load factor

= 2000 × 1.5 = 3000 kN

Design of Longitudinal Reinforcement

Pu = 0.4 fck Ac + 0.67 fy Asc or

Pu = 0.4 fck Ac + (0.67 fy - 0.4fck) Asc

3000 × 103

= 0.4 × 20 × (450 × 600) + (0.67 × 415–0.4 × 20)Asc

= 2160×103

+ 270.05Asc



⇒ Asc

= (3000–2160) × 103

/270.05 = 3111 mm2

In view of the column dimensions (450 mm, 600 mm), it is necessary to place intermediate

bars, in addition to the 4 corner bars:

Provide 4–25φ at corners ie, 4 × 491 = 1964 mm2

and 4–20φ additional ie, 4 × 314 = 1256 mm2

⇒ Asc

= 3220 mm2

> 3111 mm2

⇒ p = (100×3220) / (450×600) = 1.192 > 0.8 (minimum steel), OK.

Design of transverse steel

Diameter of tie = ¼ diameter of main steel = 25/4 =6.25 mm or 6 mm, whichever is greater.

Provide 6 mm.

Spacing: < 300 mm, < 16 x 20 = 320 mm, < LLD = 450mm. Say 300 mm c/c

Thus provide ties 8mm @ 300 mm c/c

Sketch:

Example: Square Column with Uniaxial Bending

7. Determine the reinforcement to be provided in a square column subjected to

uniaxial bending with the following data:

Size of column 450 x 450 mm

Concrete mix M 25

Characteristic strength of steel 415 N/mm2



Arrangement of reinforcement:

(a) On two sides

(b) On four sides





d = 40 + 12.5 = 52.5 mm

d1/D = 52.5/450- 0.12

Charts for d1/D = 0.15 will be used

Pu/fckbD = (2500 x 1000)/ (25 x 450 x 450) = 0.494

Mu/fckbD2 =200 x 10

6 /(25 x 450 x 450

2) = 0.088

a) Reinforcement on two sides,


p/fck = 0.09


p = 0.09 x 25 = 2.25 %

As = p bD/100 = 2.25 x 450 x 450/100

= 4556 mm2

b) Reinforcement on four sides

from Chart 45,

p/fck = 0.10

p = 0.10 x 25 = 2.5 %

As = 2.5 x 450 x 450/100 = 5063 mm2

8. Example: Circular Column with Uniaxial Bending

Determine the reinforcement to be provided in a circular column with the following

data:

Diameter of column 500 mm

Grade of concrete M20

Characteristic strength 250 N/mm2



Lateral reinforcement :

(a) Hoop reinforcement

(b) Helical reinforcement

(Assume moment due to minimum eccentricity to be less than the actual moment).


d1 = 40 + 12.5 = 52.5 mm

d1/D – 52.5/50 = 0.105

Charts for d’/D = 0.10 will be used.

(a) Column with hoop reinforcement

Pu/fck D D = (1600 x 1000)/ (20 x 500 x 500) = 0.32

Mu/fck D x D2 =125 x 10

6 /(20 x 500 x 500

2) = 0.05



Referring to Chart 52, for fy = 250 N/mm2

p/fck = 0.87


p = 0.87 x 20 = 1.74 %

As = 1.74 x (π x 5002/4)/100 = 3416 mm

2

(b) Column with Helical Reinforcement

According to 38.4 of the Code, the strength of a compression member with helical

reinforcement is 1.05 times the strength of a similar member with lateral ties. Therefore, the,

given load and moment should be divided by 1.05 before referring to the chart.

Pu/fck D D = (1600/1.05 x 1000)/ (20 x 500 x 500) = 0.31

Mu/fck D x D2 =125/1.05 x 10

6 /(20 x 500 x 500

2) = 0.048

Hence, From Chart 52, for fy = 250 N/mm2,

p/fck = 0.078

p = 0.078 x 20 = 1.56 %

As = 1.56 x( π x 500 x 500/4 )/100 = 3063 cm2

According to 38.4.1 of the Code the ratio of the volume of helical reinforcement to the

volume of the core shall not be less than

0.36 (Ag/Ac - 1) x fck /fy

where Ag is the gross area of the section and Ac is the area of the core measured to the outside

diameter of the helix. Assuming 8 mm dia bars for the helix,

Core diameter = 500 - 2 (40 - 8) = 436 mm

Ag/AC = 500/436 = 1.315

0.36 (Ag/Ac - 1) x fck /fy = 0.36(0.315) 20/250 =0.0091

Volume of helical reinforcement / Volume of core

= Ash π x 428 /( π/4 x 4362) sh

0.09 Ash / sh

where, Ash is the area of the bar forming the helix and sh is the pitch of the helix.

In order to satisfy the coda1 requirement,

0.09 Ash / sh ≥ 0.0091



For 8 mm dia bar,

sh ≤ 0.09 x 50 / 0.0091 = 49.7 mm. Thus provide 48 mm pitch

Example: Rectangular column with Biaxial Bending

9. Determine the reinforcement to be provided in a short column subjected to biaxial

bending, with the following data:

size of column = 400 x 600 mm

Concrete mix = M15

Characteristic strength of reinforcement = 415 N/mm2

Factored load, Pu = 1600 kN

Factored moment acting parallel to the larger dimension, Mux =120 kNm

Factored moment acting parallel to the shorter dimension, Muy = 90 kNm

Moments due to minimum eccentricity are less than the values given above.

Reinforcement is distributed equally on four sides.

As a first trial assume the reinforcement percentage, p = 1.2%

p/fck = 1.2/15 = 0.08

Uniaxial moment capacity of the section about xx-axis :

d1/D = 52.5 /600 = 0.088

Chart for d’/D = 0.1 will be used.

Pu/fck b D = (1600 x 1000)/ (15 x 400 x 600) = 0.444

Referring to chart 44

Mu/fck b x D2 = 0.09

Mux1 = 0.09 x 15 x 400 x 6002) = 194.4 kN.m

Uni-axial moment capacity of the section about yy axis :

d1/D = 52.5 /400 = 0.131

Chart for d1/D =0.15 will be used.


Mu/fck b x D2 = 0.083

Mux1 = 0.083 x 15 x 600 x 4002) = 119.52 kN.m

Calculation of Puz :

Referring to Chart 63 corresponding to

p = 1.2, fy = 415 and fck = 15,

Puz/Ag = 10.3

Puz = 10.3 x 400 x 600 = 2472 kN

Mux/Mux1 = 120/194.4 =0.62

Muy/Muy1=90/119.52 = 0.75

Pu /Puz =1600/2472 = 0.65



Referring to Churn 64, the permissible value of Mux/Mux1 corresponding to Muy/Muy1 and Pu


The actual value of 0.62 is only slightly higher than the value read from the Chart.

This can be made up by slight increase in reinforcement.


Pu /Puz = 0.65 thus, αn = 1 + [(2-1) / (0.8 - 0.2)] (0.65-0.2) = 1.75

[0.62 ]1.75

+ [0.75]1.75

= 1.04 slightly greater than 1 and slightly unsafe. This can be made up

by slight increase in reinforcement say 1.3%

Thus provide As = 1.3x400x600/100 = 3120 mm2

Provide 1.3 % of steel

p/fck = 1.3/15 = 0.086

d1/D = 52.5 /600 = 0.088 = 0.1

From chart 44

Mu/fck b x D2 = 0.095

Mux1 = 0.095 x 15 x 400 x 6002) = 205.2 kN.m


Mu/fck b x D2 = 0.085

Mux1 = 0.085 x 15 x 600 x 4002) = 122.4 kN.m

Chart 63 : Puz/Ag = 10.4

Puz = 10.4 x 400 x 600 = 2496 kN

Mux/Mux1 = 120/205.2 =0.585

Muy/Muy1=90/122.4 = 0.735

Pu /Puz =1600/2496 = 0.641

Referring to Chart 64, the permissible value of Mux/Mux1 corresponding to Muy/Muy1 and Pu


Hence the section is O.K.


Pu /Puz = 0.641 thus, αn = 1 + [(2-1) / (0.8 - 0.2)] (0.641-0.2) = 1.735

[120/205.2]1.735

+ [90/122.4]1.735

= 0.981 ≤ 1 Thus OK

As = 3120 mm2. Provide 10 bars of 20 mm dia. Steel provided is 314 x 10 = 3140 mm

2



Design of transverse steel: Provide 8 mm dia stirrups at 300 mm c/c as shown satisfying the

requirements of IS: 456-2000

10. Verify the adequacy of the short column section 500 mm x 300 mm under the

following load conditions:

Pu

= 1400 kN, Mux

= 125 kNm, Muy

= 75 kNm. The design interaction curves of SP 16

should be used. Assume that the column is a ‘short column’ and the eccentricity due to

moments is greater than the minimum eccentricity.

Solution:

Given: Dx

= 500 mm, b = 300 mm, A

s = 2946 mm

2

Mux

= 125 kNm, Muy

= 75 kNm, fck

= 25

MPa, fy = 415 MPa

Applied eccentricities

ex

= Mux

/Pu

= 125 × 103

/1400 = 89.3 mm ⇒ ex/D

x = 0.179

ey = M

uy/P

u = 75 × 10

3

/1400 = 53.6 mm ⇒ ey/D

y = 0.179

These eccentricities for the short column are clearly not less than the minimum eccentricities

specified by the Code.

Uniaxial moment capacities: Mux1

, Muy1

As determined in the earlier example, corresponding to Pu

= 1400 kN,

Mux1

= 187 kNm

Muy1

= 110 kNm

Values of Puz

and αn

Puz

= 0.45fck

Ag

+ (0.75fy – 0.45f

ck)A

sc

= (0.45 × 25 × 300 × 500) + (0.75 × 415 – 0.45 × 25)×2946

= (1687500 + 883800)N = 2571 kN

⇒ Pu/P

uz = 1400/2571 = 0.545 (which lies between 0.2 and 0.8)

⇒ αn

= 1.575

Check safety under biaxial bending

[125/187]1.575

+ [75/110]1

= 0.530 + 0.547

= 1.077 > 1.0

Hence, almost ok.



16 Mark Question and Answers




of 500 kN. The safe bearing capacity of soil is 200 kN/m2. Use M20 concrete and Fe 415 steel.

Solution





Required area of footing, - = ./01 = 223433 = 5. 564

Assuming a square footing, the side of footing is 7 = 0 =√5. 5 = 8. 946

Hence, provide a footing of size 1.85 m x 1.85 m

Net upward pressure in soil, : = ;��+.<=>+.<= = 175.3BC/E, < 200BC/E, Hence O.K.

Hence, factored upward pressure of soil, pu = 263 kN/m2 and, factored load, Pu = 900 kN.

2.




Assume an uniform overall thickness of footing, D = 450 mm.


d = 450 – 50 – 12 – 6 = 382 mm


of the column (See Fig. 6), where a and b are the sides of the column.



= (0.23 + 0.382)2 = 0.375 m

2

here a = b =side of column

Punching shear force = Factored load – (Factored upward pressure x punching area of footing)

= 900 – (263 x 0.375)

= 801.38 kN

Perimeter of the critical section = 4 (a+d) = 4 (230+ 382)

= 2448 mm



=938. 59 × 83334ZZ9V594 = 3. 92[/664


where F1 = 3. 4\�RJX = 3. 4\√43 ≈ 8. 84[/664

and, XN = 3. \ +^1� = _3. \ +3.453.45` = 8.3 ; Hence, adopt ks=1



2),



The effective depth for the lower layer of reinforcement, d = 450 – 50 – 6 = 396 mm, and

the effective depth for the upper layer of reinforcement, d = 450 – 50 – 12 – 6 = 382 mm.








Factored upward pressure of soil, pu = 263 kN/m2


Hence, bending moment at the critical section in the footing is

aH = THb44 = 425V3.9844 = 92. 49X[ −6/m width of footing



ad = 3. 9eRf-NUg h8 − Rf-NUigRJXj Considering 1m width of footing,

92. 49k832 = 3. 9ekZ8\klmnk594 o8 − Z8\klmn8333k594k43p

Solving the above quadratic relation, we get

Ast = 648.42 mm2 and 17,761.01 mm

2

Selecting the least and feasible value for Ast, we have

Ast = 648.42 mm2




The critical section for one way shear occurs at a distance ‘d’ from the face of the column (Fig. 8).




For the cantilever slab, total Shear Force along critical section considering the entire width B is

Vu = pu B (l – d) = 263 x 1.85 x (0.81 – 0.382)

= 208.24 kN


FG = GH0g = 439. 4ZV833389\3V594

& 3. 53q/664

From Table 61 of SP 16, find the pt required to have a minimum design shear strength ζC = ζV = 0.30


2.


2.



values)

Hence, -NU = rn833 st = 3.8e\

8338333k594 & 22uvv4


Therefore, Ast provided = 808 mm2 > Ast required (609 mm

2). Hence O.K.





Ld = 47 ф = 47 x 12 = 564 mm.


84 w − x� − 23 = 84 89\3 − 453� − 23 = e\3vv y wt. Hence O.K.



Hence, the side of the area of dispersion at the bottom of footing = 230 + 2 (2 x 450) = 2030 mm.

Since this is lesser than the side of the footing (i.e., 1850 mm)

A1 = 1.85 x 1.85 = 3.4225 m2



The dimension of the column is 230 mm x 230 mm. Hence, A2= 0.230 x 0.230 = 0.0529 m2

zl8l4 = z5. Z44\3. 3\4u = 9. 3Z > 2



= 0.45 x 20 x 2 = 18 N/mm2

l~n�x�s�x��mn��mm = �x~n��t��xtl��xxn~��v�sxm� = u33k8333453k453 = 8e. 38 q

vv4

Since the Actual bearing stress (17.01 N/mm2) is less than the Permissible bearing stress (18 N/mm

2),

the design for bearing stress is satisfactory.




of 800 kN together with an uniaxial moment of 40 kN-m. The safe bearing capacity of soil is 250

kN/m2. Use M25 concrete and Fe 415 steel.

Solution





Let us provide a square isolated footing, where L=B

Equating the maximum pressure of the footing to SBC of soil,

�l + �� = ��

i.e., 99304 +Z3V205 = 4\3

On solving the above equation, and taking the least and feasible value, B = 2 m

Hence, provide a square footing of size 2 m x 2 m

The maximum and minimum soil pressures are given by

T6PV = 93344 +Z3V245 = 453X[64 < 250 X[64 �.�.

T6LI = 93344 −Z3V245 = 8e3X[64 y ��. �. Hence, factored upward pressures of soil are,

pu,max = 345 kN/m2

and pu,min = 255 kN/m2

``

3.



Further, average pressure at the center of the footing is given by

pu,avg = 300 kN/m2

and, factored load, Pu = 900 kN, factored uniaxial moment, Mu = 60 kN-m


Assume an uniform overall thickness of footing, D = 450 mm


d = 450 – 50 – 16 – 8 = 376 mm


of the column (Fig. 9), where a and b are the dimensions of the column.



= (0.30 + 0.376)2 = 0.457 m

2

where a = b = side of column

Punching shear force = Factored load – (Factored average pressure x punching area of footing)

= 1200 – (300 x 0.457)

= 1062.9 kN

Perimeter along the critical section = 4 (a+d) = 4 (300+ 376)

= 2704 mm



= 8324. u × 83334e3ZV5e2 = 8. 3\[/664


where F1 = 3. 4\�RJX = 3. 4\√4\ = 8. 4\[/664

and, XN = 3. \ +^1� = _3. \ + 3.533.53` = 8. 3 ; Hence, adopt ks=1



2),





The effective depth for the lower layer of reinforcement, , d = 450 – 50 – 8 = 392 mm, and

the effective depth for the upper layer of reinforcement, d = d = 450 – 50 – 16 – 8 = 376 mm.









Bending moment at the critical section in the footing is

�� = ��nx�}��~��k��mnx�~��}��}��v~��n�~x�m�~n�� = ��5Z\ + 532. e\

4 � 3. 9\� k ��4k5Z\ + 532. e\5Z\ + 532. e\ � k 3. 9\5 �

Mu = 119.11 kN-m/ m width of footing



ad = 3. 9eRf-NUg h8 − Rf-NUigRJXj

Considering 1m width of footing,

88u. 88k832 = 3. 9ekZ8\klmnk5e2 o8 − Z8\klmn8333k5e2k4\p

Solving the quadratic equation,

Ast = 914.30 mm2 and 21,735.76 mm

2

Selecting the least and feasible value, Ast = 914.30 mm2






The critical section for one way shear occurs at a distance of ‘d’ from the face of the column (Fig. 11).




For the cantilever slab, total Shear Force along critical sectionconsidering the entire width B is � = ��nx� }��~� � k � � − t� k �� = �5Z\ + 54e. 84 � k � 3. 9\ − 3. 5e2� k 4� Vu = 318.58 kN


¡ = ��t = 589. \9k83334333k5e2 = 3. Z4q/vv4

From 19 of IS 456 :2000, find the pt required to have a minimum design shear strength ζC = ζV = 0.42


2.


2.



values)

Hence, lmn = rn833 st = 3.52\

833 8333k5e2 = 85e4. Zvv4


Therefore, Ast provided = 1436 mm2 > Ast required (1372.4 mm

2). Hence O.K.







Ld = 47 ф = 47 x 16 = 752 mm.


84 w − x� − 23 = 84 4333 − 533� − 23 = eu3vv y wt Hence O.K.



Hence, the side of the area of dispersion at the bottom of footing = 300 + 2 2 x 450� = 2100 mm. Since this is lesser than the side of the footing i.e., 2000 mm�, A1 = 2 x 2 = 4 m2 The dimension of the column is 300 mm x 300 mm.

Hence, A2= 0.30 x 0.30 = 0.09 m2

zl8l4 = z Z3. 3u = 2. 2e > 2



= 0.45 x 25 x 2 = 22.5 N/mm2

Actual bearing stress = �x~n��t��xt

l��xxn~��v�sxm� = 8433k8333533k533 = 85. 55q/vv4

Since the Actual bearing stress (13.33 N/mm2) is less than the Permissible bearing stress (22.5

N/mm2), the design for bearing stress is satisfactory.





REFERENCES

BOOKS:

KRISHNA RAJU, N., “DESIGN OF RC STRUCTURES”, CBS PUBLISHERS ANDDISTRIBUTORS, DELHI, 2006

VARGHESE, P.C., “LIMIT STATE DESIGN OF REINFORCED CONCRETE STRUCTURES”PRENTICE HALL OFINDIA PVT LTD NEW DELHI, 2007.

WEB:

CE IIT, KHARAGPUR

VTU EDUSAT PROGRAMMES

IS CODES:

IS456:2000, CODE OF PRACTICE FOR PLAIN AND REINFORCED CONCRETE, BUREAUOF INDIANSTANDARDS, NEW DELHI, 2000

SP16, IS456:1978 “DESIGN AIDS FOR REINFORCED CONCRETE TO BUREAU OFINDIAN STANDARDS, NEW DELHI, 1999

ce6505 design of reinforced concrete elements l t p c …

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