CE2306
DESIGN OF RC ELEMENTS
LTPC
3104
OBJECTIVE
This course covers the different types of philosophies related to Design of Reinforced
Concrete Structures with emphasis on Limit State Method. The design of Basic elements
such as slab, beam, column and footing which form part of any structural system with
reference to Indian standard code of practice for Reinforced Concrete Structures and
Design Aids are included. At the end of course the student shall be in a position to
design the basic elements of reinforced concrete structures.
UNIT I
METHODS OF DESIGN OF CONCRETE STRUCTURES
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 flexural members and slabs by working
stress method – Principles of Design of Liquid retaining structures – Properties of uncracked section – Calculation of thickness and reinforcement for Liquid retaining
structure
12
UNIT II
LIMIT STATE DESIGN FOR FLEXURE
Analysis and design of one way and two way rectangular slab subjected to uniformly
distributed load for various boundary conditions and corner effects – Analysis and design
of singly and doubly reinforced rectangular and flanged beams
12
UNIT III
LIMIT STATE DESIGN FOR BOND, ANCHORAGE SHEAR & TORSION
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.
12
UNIT IV
LIMIT STATE DESIGN OF COLUMNS
12
Types of columns – Braced and unbraced columns – Design of short column for axial,
uniaxial and biaxial bending – Design of long columns.
UNIT V
LIMIT STATE DESIGN OF FOOTING AND DETAILING
Design of wall footing – Design of axially and eccentrically loaded rectangular footing –
Design of combined rectangular footing for two columns only – Standard method of
detailing RC beams, slabs and columns – Special requirements of detailing with
reference to erection process.
TOTAL: 60
PERIODS
TEXT BOOKS
1.
Varghese, P.C., “Limit State Design of Reinforced Concrete”, Prentice Hall of
India, Pvt. Ltd., New Delhi 2002.
2.
Krishna Raju, N., “Design of Reinforced Concrete Structures”, CBS Publishers &
Distributors, New Delhi,2003.
REFERENCES
1.
Jain, A.K., “Limit State Design of RC Structures”, Nemchand Publications,
Rourkee
2.
Sinha, S.N., “Reinforced Concrete Design”, Tata McGraw-Hill Publishing
Company Ltd., New Delhi.
3.
Unnikrishna Pillai, S., Devdas Menon, “Reinforced Concrete Design”, Tata
12
McGraw-Hill Publishing Company Ltd., New Delhi.
DESIGN OF RC ELEMENTS
UNIT – 1
METHOD OF DESIGN OF CONCRETE STRUCTURES
CONTENTS
1.1
Introduction
1.2
concept of Elastic method
1.3
concept of load factor method
1.4
Concept of limit state method
1.5
Method based on experimental approach
1.6
Advantages of LSM over other methods
1.7
Analysis of structures
1.8
Reinforced concrete slabs
1.9
Design codes and specifications
1.10 Limit state philosophy
1.11 Design of flexural members and slabs by WSM
1.11.1-One way slab design
1.11.2-Two way slab design
1.12 Design of beams
1.12.1-Singly reinforced beam
1.12.2-Doubly reinforced beam
1.12.3-Flanged beam
Technical terms
Limit state: 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 deflections and cracking. The acceptable limit of the safety and serviceability
requirements before failure occurs is called a “limit state”.
Singly reinforced section: In a R.C. section, if steel is provided to take up only tension,
the section is called singly reinforced section.
Doubly reinforced section: To take an additional B.M which is more than that it can resist as
a singly reinforced balanced section, extra reinforcement in compression and additional
reinforcement on tension side (more than required for a balanced section) are provided and such
section, which are reinforced both in tension and compression zones are called ad Doubly
Reinforced sections.
T-beam: It is economical to go for beam and slab construction, if span for slab exceeds 3
to 4 m. in this type of construction, beams are placed along shorter direction and slab is
laid continuously over the beams. Here slab and beam are cast monolithically and portion
of the slab helps the beam in taking bending moment. Here the effective portion that acts
as beam takes the shape of ‘T’ and hence it is known as ‘T’ beam or flanged beam.
Flange & rib: The slab portion of the T beam is known as flange and the rectangular
beams portion of the T beam is known as rib or web.
Dead load: dead load in a building shall include the weights of all permanent
construction in the building such as weight of walls, partitions, floors, columns, beams
and roofs etc.
Live load: live loads are all the loads placed temporarily on the floors such as movable
and immovable loads, impact and vibration on the floors.
UNIT -1
METHOD OF DESIGN OF CONCRETE STRUCTURES
1.1 INTRODUCTION OF RC MEMBER:
Reinforced concrete members are allowed to be designed according to existing
codes of practice by one of the following two methods (IS 456 (2000) clause 18)
1. The method of theoretical calculations using accepted procedures of
calculations
2. The method of experimental investigations
The theoretical methods are employed for design of commonly used structures. These
methods consist of numerical calculations based on the procedures prescribed in codes of
practices prevailing in the country. Such procedures are based on one of the following
methods of design:
1. The modular ratio or the working stress method, also known as the elastic
method
2. The load factor method or Ultimate Load Method
3. The limit state method
The experimental method are used for unusual structures and are to be carried out in a
properly equipped laboratory by (a) tests on scaled models according to model analysis
procedures, and
(b) Tests on prototype of the structure. The theoretical methods themselves are the result
of extensive laboratory tests and field investigations. Safe and universally accepted
methods of calculations based on strength of materials and applied mechanics have been
derived from these laboratory investigations and are codified into the national codes.
1.2 CONCEPT OF ELASTIC METHOD:
In this method moments and forces acting on the structure computed from the
actual load.
The section of the component member is then designed to resist these moments and
forces such that the maximum stresses developed in materials are restricted to a fraction
of their true strengths. The factors of safety used in getting maximum permissible stresses
for concrete and steel are 3 and 1.8 respectively. Where the limit state method cannot be
conveniently adopted, working stress method may be used.
1.3 CONCEPT OF LOAD FACTOR METHOD (Ultimate Load
Method):
A major defect of the modular ratio method of design is that it does not give a true
factor of the safety against failure. To overcome this, the ultimate load method of design
was introduced in R.C. design. Later this modified method is called as Load Factor
Method. In this method, the strength of the R.C. section at working load is estimated
from the ultimate strength of the section. The concept of load factor, which is defined as
the ratio of the ultimate load the section can carry to the working load it gas to carry.
Usually R.C. structures are designed for suitable separate load factors for dead loads and
for live loads with additional safety factors for strength of concrete.
1.4 CONCEPT OF LIMIT STATE METHOD:
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 failures occur is
called limit state. The aim of design is to achieve acceptable probabilities that the
structure will not become unfit for the use for which it is intended, that is, it will not
reach a limit state.
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.
1.5 METHOD BASED ON EXPERIMENTAL APPROACH:
The designer may perform experimental investigations on models or full size
structures or elements and accordingly design the structures or elements. However, the
four objectives of the structural design (sec. 1.1.2) must be satisfied when designed by
employing this approach. Moreover, the engineer in charge has to approve the
experimental details and the analysis connected therewith.
Though the choice of the method of design is still left to the designer as per cl. 18.2 of IS
456:2000, the superiority of the limit state method is evident from the emphasis given to
this method by presenting it in a full section (Section 5), while accommodating the
working stress method in Annex B of IS 456:2000, from its earlier place of section 6 in
IS 456:1978. It is expected that a gradual change over to the limit state method of design
will take place in the near future after overcoming the inconveniences of adopting this
method in some situations.
1.6 ADVANTAGES OF LIMIT STATE METHOD OVER OTHER
METHODS:
1. The working stress method is based on pure elastic theory, which is not perfectly
applicable for semi plastic materials like concrete. In the limit state method of analysis,
the principles of both elastic as well as plastic theories are used and hence suitable for
concrete structures.
2. 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.
3. 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.
4. The overall sizes of flexural members (depth requirements) arrived by limit state
method are less and hence they provide better appearance to the structures.
5. Because of the modified assumptions regarding the maximum compressive strains in
the concrete and steel, the design of compression reinforcement for doubly reinforced
beams and eccentrically loaded columns by limit state method gives realistic values
which is not so in working stress method.
1.7 ANALYSIS OF STRUCTURES
Structures when subjected to external loads (actions) have internal reactions in the
form of bending moment, shear force, axial thrust and torsion in individual members. As
a result, the structures develop internal stresses and undergo deformations. Essentially,
we analyze a structure elastically replacing each member by a line (with EI values) and
then design the section using concepts of limit state of collapse. Figure 1.1 explains the
internal and external reactions of a simply supported beam under external loads. The
external loads to be applied on the structures are the design loads and the analyses of
structures are based on linear elastic theory (vide cl. 22 of IS 456:2000).
Design Loads
The design loads are determined separately for the two methods of design as
mentioned below after determining the combination of different loads.
1. In the limit state method, the design load is the characteristic load with
appropriate partial safety factor (vide sec. 2.3.2.3 for partial safety factors).
2. In the working stress method, the design load is the characteristic load only.
What is meant by characteristic load?
Characteristic load (cl. 36.2 of IS 456:2000) is that load which has a ninety-five
per cent probability of not being exceeded during the life of the structure.
The various loads acting on structures consist of dead loads, live loads, wind or
earthquake loads etc. These are discussed in sec. 1.1.6. However, the researches made so
far fail to estimate the actual loads on the structure. Accordingly, the loads are predicted
based on statistical approach, where it is assumed that the variation of the loads acting on
structures follows the normal distribution (Fig. 1.2). Characteristic load should be more
than the average/mean load. Accordingly, Characteristic load = Average/mean load + K
(standard deviation for load)
The value of K is assumed such that the actual load does not exceed the characteristic
load during the life of the structure in 95 per cent of the cases.
Loads and Forces
The following are the different types of loads and forces acting on the structure.
As mentioned in sec. 1.1.5, their values have been assumed based on earlier data and
experiences. It is worth mentioning that their assumed values as stipulated in IS 875 have
been used successfully.
1. Dead loads
These are the self weight of the structure to be designed. Needless to mention that the
dimensions of the cross section are to be assumed initially which enable to estimate the
dead loads from the known unit weights of the materials of the structure. The accuracy of
the estimation thus depends on the assumed values of the initial dimensions of the cross
section.
The values of unit weights of the materials are specified in Part 1 of IS 875.
2. Imposed loads
They are also known as live loads and consist of all loads other than the dead loads of the
structure. The values of the imposed loads depend on the functional requirement of the
structure. Residential buildings will have comparatively lower values of the imposed
loads than those of school or office buildings. The standard values are stipulated in Part 2
of IS 875.
3 .Wind loads
These loads depend on the velocity of the wind at the location of the structure,
permeability of the structure, height of the structure etc. They may be horizontal or
inclined forces depending on the angle of inclination of the roof for pitched roof
structures. They can even be suction type of forces depending on the angle of inclination
of the roof or geometry of the buildings Wind loads are specified in Part 3 of IS 875.
4 .Snow loads
These are important loads for structures located in areas having snow fall, which gets
accumulated in different parts of the structure depending on projections, height, slope etc.
of the structure. The standard values of snow loads are specified in Part 4 of IS 875.
5 .Earthquake forces
Earthquake generates waves which move from the origin of its location (epicenter) with
velocities depending on the intensity and magnitude of the earthquake. The impact of
earthquake on structures depends on the stiffness of the structure, stiffness of the soil
media, height and location of the structure etc. Accordingly, the country has been divided
into several zones depending on the magnitude of the earthquake. The earthquake forces
are prescribed in IS 1893. Designers have adopted equivalent static load approach or
spectral method.
6 .Shrinkage, creep and temperature effects
Shrinkage, creep and temperature (high or low) may produce stresses and cause
deformations like other loads and forces. Hence, these are also considered as loads which
are time dependent. The safety and serviceability of structures are to be checked
following the stipulations of cls. 6.2.4, 5 and 6 of IS 456:2000 and Part 5 of IS 875.
7. Other forces and effects
It is difficult to prepare an exhaustive list of loads, forces and effects coming onto the
structures and affecting the safety and serviceability of them. However, IS 456:2000
stipulates the following forces and effects to be taken into account in case, they are liable
to affect materially the safety and serviceability of the structures. The relevant codes as
mentioned therein are also indicated below:
• Foundation movement (IS 1904) (Fig. 1.3)
• Elastic axial shortening
• Soil and fluid pressures (vide IS 875 - Part 5)
• Vibration
• Fatigue
• Impact (vide IS 875 - Part 5)
• Erection loads (Please refer to IS 875 - Part 2) (Fig. 1.4)
• Stress concentration effect due to point of application of load and the like.
8. Combination of loads
Design of structures would have become highly expensive in order to maintain their
serviceability and safety if all types of forces would have acted on all structures at all
times. Accordingly, the concept of characteristic loads has been accepted to ensure that in
at least 95 per cent of the cases, the characteristic loads considered will be higher than the
actual loads on the structure. However, the characteristic loads are to be calculated on the
basis of average/mean load of some logical combinations of all the loads mentioned in
sec. 1.1.6.1 to 7. These logical combinations are based on (i) the natural phenomena like
wind and earthquake do not occur simultaneously, (ii) live loads on roof should not be
present when wind loads are considered; to name a few. IS 875 Part 5 stipulates the
combination of loads to be considered in the design of structures.
1.9 DESIGN CODES AND SPECIFICATIONS:
(IS 456:2000, Plain and Reinforced Concrete - Code of Practice)
(Fourth Revision), ANNEX-B
Assumptions for Design of Members
In the methods based on elastic theory(Working Stress method), the following
assumptions shall be made:
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 -280/3𝜎cbc
Where 𝜎
cbc
permissible compressive stress due to bending in concrete in N/mm2 as
specified in Table 21 of IS 456:2000.
NOTE - The expression given for m partially takes into account long-term effects such as
creep. Therefore this m is not the same us the modular ratio derived based on the value of
Ec, given in 6.2.3.1.
PERMISSIBLE STRESSES
1. Permissible Stresses in Concrete (𝜎cbc)
Permissible stresses for the various grades of concrete shall be taken as those given in
Tables 21 and 23.
NOTE - For increase in strength with age 6.2.1 shall be applicable. The values of
permissible stress shall be obtained by Interpolation between the grades of concrete.
2. Direct Tension
For members in direct tension, when full tension is taken by the reinforcement alone, the
tensile stress shall be not greater than the values given below:
The tensile stress shall be calculated as
Ft / AC + mAst
F, =total tension on the member minus pretension in steel, if any, before concreting;
AC =cross-sectional area of concrete excluding any finishing material and reinforcing
steel
m- Modular ratio; and
Ast = cross-sectional area of reinforcing steel in tension.
3. Bond Stress for Deformed Bars
In the case of deformed bars conforming to IS 1786, the bond stresses given in
Table 21 may be increased by 60 percent.
4. Permissible Stresses in Steel Reinforcement (𝜎st)
Permissible stresses in steel reinforcement shall not exceed the values specified in
Table 22 of IS456:2000.
In flexural members the value of (T, given in Table 22 is applicable at the
centroid of the tensile reinforcement subject to the condition that when more than one
layer of tensile reinforcement is provided, the stress at the centroid of the outermost layer
shall not exceed by more than 10 percent the value given in Table 22.
5. Increase in permissible Stresses
Where stresses due to wind (or earthquake) temperature and shrinkage effects are
combined with those due to dead, live and impact load, the stresses specified in Tables
21, 22 and 23 may be exceeded up to a limit of 33 1 3 percent. Wind and seismic forces
need not be considered as acting simultaneously.
NOTES:
1 The values of permissible shear stress in concrete are given in Table 23.
2 The bond stress given in co1 4 shall be increased by 25 percent for bars in compression.
For solid slabs the permissible shear stress in concrete shall be k τC where k has the value
given below:
6. Minimum Shear Reinforcement
When τV is less than τC given in Table 23, minimum shear reinforcement shall be
provided in accordance with 26.5.1.6.
7. Design of Shear Reinforcement
When τV exceeds τC given in Table 23, shear reinforcement shall be provided in
any of the following forms:
a) Vertical stirrups,
b) Bent-up bars along with stirrups, and
c) Inclined stirrups.
Where bent-up bars are provided, their contribution towards shear resistance shall
not be more than half that of the total shear reinforcement.
Shear reinforcement shall be provided to carry a shear equal to V- τC.bd. The strength of
shear reinforcement VS, shall be calculated as-below:
1. For vertical stirrups
2. For inclined stirrups or a series of bars bent-up at different cross-sections:
3. For single bar or single group of parallel bars, all bent-up at the same crosssection
Where
ASV = total cross-sectional area of stirrup legs or bent-up bars within a distance
SV=spacing of the stirrups or bent-up bars along the length of the member
τC= design shear strength of the concrete
b= breadth of the member which for flanged beams, shall be taken as the breadth of the
web bw
σ sv =permissible tensile stress in shear reinforcement which shall not be taken greater
than 230 N/mm2
α = angle between the inclined stirrup or bent-up bar and the axis of the member, not less
than 450, and
d= effective depth.
NOTE -Where more than one type of shear reinforcement is used to reinforce the same
portion of the beam, the total shear resistance shall be computed as the sum of the
resistance for the various types separately. The area of the stirrups shall not be less than
the minimum specified in 26.5.1.6.
1.10 LIMIT STATE PHILOSOPHY
The inadequacies of the elastic and ultimate load methods of design paved the
way for the limit state method of design with a semi-probabilistic approach. Limit state
design is a method of designing structures based on a statistical concept of safety and the
associated probability of failure. Structures designed should satisfy the dual criterion of
a) Safety and
b) Serviceability
Safety may be defined as an acceptable degree of security against complete collapse
or failure, which in concrete structures can occur by various modes such as compression,
tension, flexure, shear, fatigue or their combinations.
Serviceability requirement means that the member or structure should not in its
intended lifetime deteriorate to such an extent that is fails to fulfill its function for which
it is designed. In concrete structures, this state may be reached due to excessive
deflection, cracking, vibration, corrosion of reinforcement etc.
Limit state design philosophy uses the concept of probability and is based on the
application of the method of statistics to the vibrations that occur in practice in the loads
acting on the structure and the strengths of the materials.
Assumptions in limit state method:
1. Plane section before bending will remain plane after bending.
2. Tensile strength carried by of concrete is zero.
3. Does not obey Hook`s law.
4. Ultimate strain of concrete is 0.0035
σc = 0.67 fck
Partial safety factor for concrete γ m =1.5
Hence, design strength of concrete = 0.67 fck /1.5 = 0.446 fck
5. Maximum strain in
than
0.87 f y
tensile reinforcement at
failure shall not
be less
γ m =1.15 . Thus the design strength
will be:
+ 0.002
Es
6. Partial safety factor
f y /1.15 = 0.87 f y
for steel
3
4
Balanced section:
Compression strain in concrete = 0.0035
Tensile strain in steel = yield strain
Under Reinforced Section: Compression strain in concrete < 0.0035
Tensile strain in steel = yield strain
Over Reinforced section: Tensile strain in steel < yield strain
Compression strain in concrete = 0.0035
Under Reinforced Section is Preferable
1.12 DESIGN OF FLEXURAL MEMBERS BY WORKING STRESS
METHOD:
Concrete is strong in compression and crushing strength is determined by test on standard cubes.
Large deformation is noticed at the failure load. Deformation is permanent in nature. This
suggests for the adoption of a factor of safety.
Permissible bending & direct compression are as fraction of crushing strength. For example:
M15 concrete, the value of σcbc = 5 = (15/3) and σcc = 4 = (15/3.75)
Assumptions in working stress method
1. Plane section before bending will remain plane after bending.
2. Stress-strain relations obey Hooks law (linear)
3. Tensile stresses are taken by steel only.
4. Modular Ratio m=
Es/Ec = (280/3
σ )
cbc
σcb = Compressive stress of concrete; σcbc =Permissible stress of concrete;
σst = Tensile stress in steel; σst = Permissible stress of concrete
Neutral Axis Depth Factor and Moment of Resistance of a RC Beam
Section:
The equation (6) represents the Neutral axis depth factor (k or n ) for balanced
section. So it can be written as
kbal
DESIGN OF BEAM:A Beam is any structural member which resists load mainly by bending. Therefore it is
also called flexural member. Beam may be singly reinforced or doubly reinforced. When
steel is provided only in tensile zone (i.e. below neutral axis) is called singly reinforced
beam, but when steel is provided in tension zone as well as compression zone is called
doubly reinforced beam.
The aim of design is:
 To decide the size (dimensions) of the member and the amount of reinforcement
required.
 To check whether the adopted section will perform safely and satisfactorily during
the life time of the structure.
Over all depth :The normal distance from the top edge of the beam to the bottom edge of the beam is
called over all depth. It is denoted by ‘d’.
Effective depth:The normal distance from the top edge of beam to the centre of tensile reinforcement is
called effective depth. It is denoted by ‘d’.
Clear cover:The distance between the bottom of the bars and bottom most the edge of the beam is
called clear cover.
clear cover = 25mm or dia of main bar, (which ever is greater).
Effective cover:The distance between centre of tensile reinforcement and the bottom edge of the beam is
called effective cover. Effective cover = clear cover + ½ dia of bar.
End cover:End cover = 2xdia of bar or 25mm (which ever is greater)
Neutral axis:- the layer / lamina where no stress exist is known as neutral axis. It divides
the beam section into two zones, compression zone above the neutral axis & tension zone
below the neutral axis.
Depth of netural axis:- the normal distance between the top edge of the beam & neutral
axis is called depth of neutral axis. It is denoted by ‘n’. n=kd
Lever arm:- the distance between the resultant compressive force (c) and tensile force (t)
is known as lever arm. It is denoted by ‘z’. The total compressive force (c) in concrete act
at the c.g. of compressive stress diagram i.e. n/3 from the compression edge. The total
tensile force (T) acts at c.g. of the reinforcement.
lever arm = d-n/3 = d-(kd/3)
Tensile reinforcement:The reinforcement provided tensile zone is called tensile reinforcement. It is denoted by
Ast.
compression reinforcement :The reinforcement provided compression zoneis called compression reinforcement. It is
denoted by Asc
Types of beam section:- the beam section can be of the following types:
1.balanced section
2.unbalnced section
(a) under- reinforced section
(b) over-reinforced section
1.Balanced section:- a section is known as balanced section in which the compressive
stree in concrete (in compressive zones) and tensile stress in steel will both reach the
maximum permissible values simultaneously
The neutral axis of balanced (or critical) section is known as critical neutral axis (nc). The
area of steel provided as economical area of steel. Reinforced concrete sections are
designed as balanced sections.
2. Unbalnced section:-this is a section in which the quantity of steel provided is different
from what is required for the balanced section.
Unbalanced sections may be of the following two types:
(a) under-reinforced section
(b) over-reinforced section
(a)Under-reinforced section:- if the area of steel provided is less than that required for
balanced section, it is known as under-reinforced section. Due to less reinforcement the
position of actual neutral axis (n) will shift above the critical neutral axis (nc)
I.e. n< nc. Or k<kbal
In under-reinforced section steel is fully stressed and concrete is under stressed (i.e. Some
concrete remains un-utilised). Steel being ductile, takes some time to break. This gives
sufficient warning before the final collapse of the structure. For this reason and from
economy point of view the under-reinforced sections are designed.
(b)Over-reinforced section:- if the area of steel provided is more than that required for a
balanced section, it is known as over-reinforced section. As the area of steel provided is
more, the position of NA. will shift towards steel, therefore actual axis (n) is below the
critical neutral axis (nc )
I.e. n> nc. Or k>kbal
In this section concrete is fully stressed and steel is under stressed. Under such
conditions, the beam will fail initially due to over stress in the concrete. Concrete being
brittle, this happens suddenly and explosively without any warning.
Basic rules for design of beam:1. Effective span:- In the case of simply supported beam the effective length,
l = a)Distance between the centre of support
b) Clear span + eff. Depth
eff. Span = least of i. & ii.
2. Effective depth:- The normal distance from the top edge of beam to the centre of
tensile reinforcement is called effective depth. It is denoted by ‘d’.
d= D- effect. Cover
where D= over all depth
3. Bearing :- Bearings of beams on brick walls may be taken as follow:
 Up to 3.5 m span, bearing = 200mm
 Up to 5.5 m span, bearing =300mm
 Up to 7.0 m span, bearing =400mm
4. Deflection control:- The vertical deflection limits assumed to be satisfied if (a) For
span up to 10m
Span / eff. Depth = 20
(For simply supported beam)
Span / eff. Depth = 7
(For cantilever beam)
(b) For span above 10m, the value in (a) should be multiplied by 10/span (m), except for
cantilever for which the deflection calculations should be made.
(c) Depending upon the area and type of steel the value of (a&b) modified as per
modification factor.
5. Reinforcement :(a) Minimum reinforcement:- The minimum area of tensile reinforcement shall not be
less than that given by the following:
Ast = 0.85 bd / fy
(b)Maximum reinforcement:- The maximum area of tensile reinforcement shall not be
more than 0.4bD
(c)Spacing of reinforcement bars:i. The horizontal distance between to parallel main bars shall not be less than the greatest
of the following:
 Diameter of the bar if the bars are of same diameter.
 Diameter of the larger bar if the diameter are unequal.
 5mm more than the nominal maximum size of coarse
aggregate.
ii. When the bars are in vertical lines and the minimum vertical distance between the bars
shall be greater of the following:
 15mm.
 2/3rd of nominal maximum size of aggregate.
 Maximum diameter of the bar.
6. Nominal cover to reinforcement :- The Nominal cover is provided in R.C.C. design:
 To protect the reinforcement against corrosion.
 To provide cover against fire.
 To develop the sufficient bond strength along the surface area of the steel bar.
 As per IS 456-2000, the value of nominal cover to meet durability requirements as
follow:-
Exposure conditions
Nominal cover(mm)
Not less than
Mild
20
Moderate
30
Severe
45
Very severe
50
Extreme
75
Procedure for Design of Singly Reinforced Beam by Working Stress Method
Given :
(i) Span of the beam (l)
(ii) Loads on the beam
(iii)Materials-Grade of Concrete and type of steel.
1. Calculate design constants for the given materials (k, j and R)
k = m σcbc / m σcbc + σst
where k is coefficient of depth of Neutral Axis
j = 1- k/3
where j is coefficient of lever arm.
Q= 1/2 σcbc kj
where R is the resisting moment factor.
2. Assume dimension of beam:
d = Span/10 to Span/8
Effective cover = 40mm to 50mm
b = D/2 to 2/3D
3. Calculate the effective span (l) of the beam.
4. Calculate the self weight (dead load) of the beam.
Self weight = D x b x 25000 N/m
5. Calculate the total Load & maximum bending moment for the beam.
Total load (w) = live load + dead load
Maximum bending moment, M = wl2 / 8 at the centre of beam for simply supported
beam.
M = wl2 / 2 at the support of beam for cantilever beam.
6. Find the minimum effective depth
M = Mr
= Qbd2
dreqd. = ( M / R.b )0.5
7. Compare dreqd. With assumed depth value.
(i) If it is less than the assumed d, then assumption is correct.
(ii) If dreqd. is more than assumed d, then revise the depth value and repeat steps 4, 5 & 6.
8. Calculate the area of steel required (Ast).
Ast = M / σst jd
Selecting the suitable diameter of bar calculate the number of bars required
Area of one bar = π/4 x φ2 = Aφ
No. of bars required = Ast /Aφ
9. Calculate minimum area of steel (AS) required by the relation:
AS = 0.85 bd / fy
Calculate maximum area of steel by the area relation:
Maximum area of steel = 0.04bD
Check that the actual ASt provided is more than minimum and less than maximum
requirements.
10. Check for shear and design shear reinforcement.
11. Check for development length.
12. Check for depth of beam from deflection.
13. Write summary of design and draw a neat sketch.
1.12.1 Design of singly reinforced beam
Design a rectangular reinforced concrete beam simply supported on masonry walls
300 mm thick and 6m apart (centre to centre) to support a distributed live load of
10kN/m and a dead load of 5kN/m in addition to its own weight. Assume M-20
grade of concrete and Fe-415 HYSD bars.
a)Data
Span (centre to centre) = 6m
Live load = 10kN/m
Dead load = 5kN/m
M-20 grade of concrete and Fe-415 HYSD bars
1.12.2 Design of doubly reinforced beam:
A doubly reinforced beam is to be designed having an overall cross sectional dimensions
of 250mm by 400mm with an effective span of 4m. The beam has to support an
uniformly distributed dead load of 2.5kN/m together with a live load of 20kN/m in
addition to its self weight. Adopt M-20 grade of concrete and Fe-415 HYSD bars, design
suitable reinforcements in the beam.
a)Data
Effective span = L=4m
Breadth of beam =b=250mm
Overall depth = D=400mm
Dead load =2.5kN/m
Live load = 20kN/m
M-20 grade of concrete and Fe-415 HYSD bars
1.12.3 DESIGN OF FLANGED BEAM:
Design a Tee beam for a commercial office floor to suit the following data:
a)Data
Clear span =10m
Centre to centre of wall supports = 10.5m
Spacing of tee beams = 2.5m
Live load (office floor) = 4kN/m2
Slab thickness=150mm
Materials: M-20 grade of concrete and Fe-415 HYSD bars
1.8 Reinforced Concrete Slabs
One-way Slabs
Slabs, used in floors and roofs of buildings mostly integrated with the supporting beams,
carry the distributed loads primarily by bending. The integrated slab is considered as flange of Tor L-beams because of monolithic construction. However, the remaining part of the slab needs
design considerations. These slabs are either single span or continuous having different support
conditions like fixed, hinged or free along the edges. Though normally these slabs are horizontal,
inclined slabs are also used in ramps, stair cases and inclined roofs. While square or rectangular
plan forms are normally used, triangular, circular and other plan forms are also needed for
different functional requirements. This lesson takes up horizontal and rectangular /square slabs
of buildings supported by beams in one or both directions and subjected to uniformly distributed
vertical loadings.
The other types of slabs are given below. All these slabs have additional requirements
depending on the nature and magnitude of loadings in respective cases.
(a) Horizontal or inclined bridge and fly over deck slabs carrying heavy concentrated
loads,
(b) Horizontal slabs of different plan forms like triangular, polygonal or circular,
(c) Flat slabs having no beams and supported by columns only,
(d) inverted slabs in footings with or without beams,
(e) slabs with large voids or openings,
(f) Grid floor and ribbed slabs.
One-way and Two-way Slabs
Fig 1.5 a
Fig 1.5 b
The share of loads on beams supporting solid slabs along four edges when vertical loads
are uniformly distributed, It is evident from the figures that the share of loads on beams in two
perpendicular directions depends upon the aspect ratio ly /lx of the slab, lx being the shorter
span. For large values of lY, the triangular area is much less than the trapezoidal area (Fig.1.5 a).
Hence, the share of loads on beams along shorter span will gradually reduce with increasing ratio
of ly /lx. In such cases, it may be said that the loads are primarily taken by beams along longer
span. The deflection profiles of the slab along both directions are also shown in the figure. The
deflection profile is found to be constant along the longer span except near the edges for the slab
panel of Fig.1.5 a. These slabs are designated as one-way slabs as they span in one direction
(shorter one) only for a large part of the slab when ly /lx > 2.
On the other hand, for square slabs of ly /lx = 1 and rectangular slabs of ly /lx up to 2; the deflection
profiles in the two directions are parabolic (Fig.1.5 b). Thus, they are spanning in two directions
and these slabs with ly /lx up to 2 are designated as two-way slabs, when supported on all edges.
It would be noted that an entirely one-way slab would need lack of support on short edges. Also,
even for ly /lx < 2, absence of supports in two parallel edges will render the slab one-way. In Fig.
1.5 b, the separating line at 45 degree is tentative serving purpose of design. Actually, this angle
is a function of ly /lx. .
Design of slabs:
1.11.1 One way slab (example)
Design a simply supported slab to suit the following data:
a) Data
Clear span = 3m
Slab supported on load bearing brick walls 230 mm thick.
Loading: roof load (accessible) = 1.5 kN/mm2
Materials: M-20 grade of concrete and Fe-415 HYSD bars.
`
1.11.2 Design example of two way slab
Design a two – way slab for a residential floor to suit the following data:
a)Data
Size of floor = 4m by 6m.
Edge conditions: slab simply supported on all the sides without any provision for
torsion at corners. Materials: M-20 grade concrete and Fe-415 HYSD bars.
DESIGN OF RC ELEMENTS
UNIT – 2
LIMIT STATE DESIGN FOR FLEXURE
CONTENTS
2.1 Design considerations
2.1.1 Design of one way slab
2.2 Two way slab
2.1.2 Design of two way slab
2.3 design of beams
2.3.1 Design of singly reinforced beam
2.3.2 Design of doubly reinforced beam
2.3.3 Design of flanged beam
TECHNICAL TERMS
Balanced section: If the steel provided in a section is such that the strain in concrete and the
strain in steel reaches the maximum permissible values simultaneously, then the section is
called as limiting section or balanced section.
Under reinforced section: When the percentage of steel provided is lesser than that in the
balanced condition, the neutral axis moves upward such that the force of compression is
equal to the force of tension. This process continues until the steel is stained beyond the yield
point (0.87fy/Es) + 0.002 before the concrete attains its maximum permissible value of strain
i.e., 0.0035 and the steel fails. Such a failure is known as tension failure and the section is
called as under reinforced section.
Over reinforced section: When the percentage of steel provided is more than that in the
balanced condition, the neutral axis moves upward such that the force of tension is equal to
the force of compression. This process continues until the strain in concrete reaches its
ultimate value, i.e., 0.0035 and the concrete fails. Such a failure is known as compression
failure and section is called as over reinforced section.
Flexural members: The members that will bend under load.
One way slab: The slab, which bends in one direction, is known as one way slab. One- way
slab is followed when two supports are available on opposite sides. The distance between
two opposite supports is known as span. In this case slab bends along the span. When length
to breadth ratio is more than two in case of supports available on all four sides, the slab bends
along shorter span.
Main reinforcement provided along the shorter span and distribution reinforcement
along longer direction.
Two- way slab: The slab, which bends in both the directions, is known as two way slab. Two
way slabs are followed when supports are available on all four sides and when the ratio of
larger span to shorter span does not exceed two. In two – way slabs main reinforcement are
provided in both the directions.
One-way continuous slabs: When a one – way slab is cast over more than two supports, it is
known as one way continuous slab. When span is larger than 3 to 4 m in any one direction,
we go for one-way continuous slabs over beams.
Two-way continuous slabs: When two way slabs continue in one or more directions, it is
known as two way continuous slabs. Two way continuous slabs are followed when spans is
larger in both the direction.
UNIT -2
LIMIT STATE DESIGN FOR FLEXTURE
2.1 DESIGN CONSIDERATIONS
The primary design considerations of both one and two-way slabs are strength and
deflection. The depth of the slab and areas of steel reinforcement are to be determined from
these two aspects. The detailed procedure of design of one-way slab is taken up in the next
section. However, the following aspects are to be decided first.
(a) Effective span (cl.22.2 of IS 456)
The effective span of a slab depends on the boundary condition. Table 2 gives the guidelines
stipulated in cl.22.2 of IS 456 to determine the effective span of a slab.
Table 2 Effective span of slab (cl.22.2 of IS 456)
(b) Effective span to effective depth ratio (cls.23.2.1a-e of IS 456)
The deflection of the slab can be kept under control if the ratios of effective span to effective
depth of one-way slabs are taken up from the provisions in cl.23.2.1a-e of IS 456. These
stipulations are for the beams and are also applicable for one-way slabs as they are designed
considering them as beam of unit width.
(c) Nominal cover (cl.26.4 of IS 456)
The nominal cover to be provided depends upon durability and fire resistance requirements.
Table 16 and 16A of IS 456 provide the respective values.
Appropriate value of the nominal cover is to be provided from these tables for the particular
requirement of the structure.
(d) Minimum reinforcement (cl.26.5.2.1 of IS 456)
Both for one and two-way slabs, the amount of minimum reinforcement in either direction
shall not be less than 0.15 and 0.12 per cents of the total cross-sectional area for mild steel
(Fe 250) and high strength deformed bars (Fe 415 and Fe 500)/welded wire fabric,
respectively.
(e) Maximum diameter of reinforcing bars (cl.26.5.2.2)
The maximum diameter of reinforcing bars of one and two-way slabs shall not exceed oneeighth of the total depth of the slab.
(f) Maximum distance between bars (cl.26.3.3 of IS 456)
The maximum horizontal distance between parallel main reinforcing bars shall be the lesser
of (i) three times the effective depth, or (ii) 300 mm. However, the same for
secondary/distribution bars for temperature, shrinkage etc. shall be the lesser of (i) five times
the effective depth, or (ii) 450 mm.
2.1.1 Design of One-way Slabs
The procedure of the design of one-way slab is the same as that of beams. However,
the amounts of reinforcing bars are for one meter width of the slab as to be determined from
either the governing design moments (positive or negative) or from the requirement of
minimum reinforcement. The different steps of the design are explained below.
Step 1: Selection of preliminary depth of slab
The depth of the slab shall be assumed from the span to effective depth ratios as given
in section 3.6.2.2 of Lesson 6.
Step 2: Design loads, bending moments and shear forces
The total factored (design) loads are to be determined adding the estimated dead load
of the slab, load of the floor finish, given or assumed live loads etc. after multiplying each of
them with the respective partial safety factors. Thereafter, the design positive and negative
bending moments and shear forces are to be determined using the respective coefficients
given in Tables 12 and 13 of IS 456 and explained in sec.8.18.4 earlier.
Step 3: Determination/checking of the effective and total depths of slabs
The effective depth of the slab shall be determined is given below
2
Mu, lim = R, lim bd ….
Where, the values of R, lim for three different grades of concrete and three different
grades of steel. The value of b shall be taken as one meter.
The total depth of the slab shall then be determined adding appropriate nominal cover
(Table 16 and 16A of cl.26.4 of IS 456) and half of the diameter of the larger bar if the bars
are of different sizes. Normally, the computed depth of the slab comes out to be much less
than the assumed depth in Step 1. However, final selection of the depth shall be done after
checking the depth for shear force.
Step 4: Depth of the slab for shear force
Theoretically, the depth of the slab can be checked for shear force if the design shear
strength of concrete is known. Since this depends upon the percentage of tensile
reinforcement, the design shear strength shall be assumed considering the lowest percentage
of steel. The value of cτ shall be modified after knowing the multiplying factor k from the
depth tentatively selected for the slab in Step 3. If necessary, the depth of the slab shall be
modified.
Step 5: Determination of areas of steel
Area of steel reinforcement along the direction of one-way slab should be determined
employing Eq.3.23 of sec.3.5.5 of Lesson 5 and given below as a ready reference.
Mu = 0.87 fy Ast d {1 – (Ast)(fy)/(fck)(bd)} ….
The above equation is applicable as the slab in most of the cases is under-reinforced due to
the selection of depth larger than the computed value in Step 3. The area of steel so
determined should be checked whether it is at least the minimum area of steel as mentioned
in cl.26.5.2.1 of IS 456 and explained in sec.8.18.5d.
Alternatively, tables and charts of SP-16 may be used to determine the depth of the
slab and the corresponding area of steel. Tables 5 to 44 of SP-16 covering a wide range of
grades of concrete and Chart 90 shall be used for determining the depth and reinforcement of
slabs. Tables of SP-16 take into consideration of maximum diameter of bars not exceeding
one-eighth the depth of the slab. Zeros at the top right hand corner of these tables indicate the
region where the percentage of reinforcement would exceed pt,
. Similarly, zeros at the
lim
lower left and corner indicate the region where the reinforcement is less than the minimum
stipulated in the code. Therefore, no separate checking is needed for the allowable maximum
diameter of the bars or the computed area of steel exceeding the minimum area of steel while
using tables and charts of SP-16.
The amount of steel reinforcement along the large span shall be the minimum amount of steel
as per cl.26.5.2.1 of IS 456.
Step 6: Selection of diameters and spacings of reinforcing bars
The diameter and spacing of bars are to be determined as per cls.26.5.2.2 and 26.3.3 of IS
456.
Detailing of Reinforcement
Fig 2.1
Continuous slab showing the different reinforcing bars in the discontinuous and
continuous ends (DEP and CEP, respectively) of end panel and continuous end of adjacent
panel (CAP). The end panel has three bottom bars B1, B2 and B3 and four top bars T1, T2,
T3 and T4. Only three bottom bars B4, B5 and B6 are shown in the adjacent panel. Table 8.3
presents these bars mentioning the respective zone of their placement (DEP/CEP/CAP),
direction of the bars (along x or y) and the resisting moment for which they shall be designed
or if to be provided on the basis of minimum reinforcement. These bars are explained below
for the three types of ends of the two panels.
Table Steel bars of one-way slab (Figs.2.1)
Sl.No.
Bars
Panel
Along
Resisting moment
1
B1, B2
DEP
x
+ 0.5 Mx for each,
2
B3
DEP
y
Minimum steel
3
B4, B5
CAP
x
+ 0.5 Mx for each,
4
B6
CAP
y
Minimum steel
5
T1, T2
CEP
x
- 0.5 Mx for each,
+ 0.5 M
6
T3
DEP
x
7
T4
DEP
y
x
Minimum steel
Notes: (i) DEP = Discontinuous End Panel
(ii) CEP = Continuous End Panel
(iii) CAP = Continuous Adjacent Panel
(i)
Discontinuous End Panel (DEP)
• Bottom steel bars B1 and B2 are alternately placed such that B1 bars are curtailed at a
distance of 0.25 lx1 from the adjacent support and B2 bars are started from a distance
of 0.15lx1 from the end support. Thus, both B1 and B2 bars are present in the middle
zone covering 0.6lx1, each of which is designed to resist positive moment 0.5Mx.
These bars are along the direction of x and are present from one end to the other end
of ly.
• Bottom steel bars B3 are along the direction of y and cover the entire span lx1 having the
minimum area of steel. The first bar shall be placed at a distance not exceeding s/2
from the left discontinuous support, where s is the spacing of these bars in y direction.
• Top bars T3 are along the direction of x for resisting the negative moment which is
numerically equal to fifty per cent of positive Mx. These bars are continuous up to a
distance of 0.1lx1 from the centre of support at the discontinuous end.
• Top bars T4 are along the direction of y and provided up to a distance of 0.1lx1 from the
centre of support at discontinuous end. These are to satisfy the requirement of
minimum steel.
(ii) Continuous End Panel (CEP)
• Top bars T1 and T2 are along the direction of x and cover the entire ly. They are
designed for the maximum negative moment Mx and each has a capacity of -0.5Mx.
Top bars T1 are continued up to a distance of 0.3lx1, while T2 bars are only up to a
distance of 0.15lx1.
• Top bars T4 are along y and provided up to a distance of 0.3lx1 from the support. They
are on the basis of minimum steel requirement.
(iii) Continuous Adjacent Panel (CAP)
• Bottom bars B4 and B5 are similar to B1 and B2 bars of (i) above.
• Bottom bars B6 are similar to B3 bars of (i) above Detailing is an art and hence
structural requirement can be satisfied by more than one mode of detailing each valid and
acceptable.
Numerical Problems
(a) Problem1
Design the one-way continuous slab of Fig.2.2 subjected to uniformly distributed
2
2
imposed loads of 5 kN/m using M 20 and Fe 415. The load of floor finish is 1 kN/m . The
span dimensions shown in the figure are effective spans. The width of beams at the support =
300 mm.
Fig 2.2
Solution:
Step 1: Selection of preliminary depth of slab
The basic value of span to effective depth ratio for the slab having simple support at the end
and continuous at the intermediate is (20+26)/2 = 23 (cl.23.2.1 of IS 456).
2
Modification factor with assumed p = 0.5 and fs = 240 N/mm is obtained as 1.18 from Fig.4
of IS 456.
Therefore, the minimum effective depth = 3000/23(1.18) = 110.54 mm. Let us take the
effective depth d = 115 mm and with 25 mm cover, the total depth D = 140 mm.
Step 2: Design loads, bending moment and shear force
Dead loads of slab of 1 m width = 0.14(25) = 3.5 kN/m
Dead load of floor finish =1.0 kN/m
Factored dead load = 1.5(4.5) = 6.75 kN/m
Factored live load = 1.5(5.0) = 7.50 kN/m
Total factored load = 14.25 kN/m
Maximum moments and shear are determined from the coefficients given in Tables 12 and
13 of IS 456.
Maximum positive moment = 14.25(3)(3)/12 = 10.6875 kNm/m
Maximum negative moment = 14.25(3)(3)/10 = 12.825 kNm/m
Maximum shear Vu = 14.25(3)(0.4) = 17.1 kN
Step 3: Determination of effective and total depths of slab
From Eq.3.25 of sec. 3.5.6 of Lesson 5, we have
2
2
Mu,lim = R,lim bd where R,lim is 2.76 N/mm from Table 3.3 of sec. 3.5.6 of Lesson 5. So, d
6
= {12.825(10 )/(2.76)(1000)}
0.5
= 68.17 mm
Since, the computed depth is much less than that determined in Step 1, let us keep D = 140
mm and d = 115 mm.
Step 4: Depth of slab for shear force
2
Table 19 of IS 456 gives cτ = 0.28 N/mm for the lowest percentage of steel in the slab.
Further for the total depth of 140 mm, let us use the coefficient k of cl. 40.2.1.1 of IS 456 as
2
1.3 to get c ττkc= = 1.3(0.28) = 0.364 N/mm .
2
Table 20 of IS 456 gives maxcτ = 2.8 N/mm . For this problem bdVuv/ =τ = 17.1/115 =
2
0.148 N/mm . Since, maxc cvτττ<<, the effective depth d = 115 mm is acceptable.
Step 5: Determination of areas of steel
From Eq.3.23 of sec. 3.5.5 of Lesson 5, we have
Mu = 0.87 fy Ast d {1 – (Ast)(fy)/(fck)(bd)}
(i) For the maximum negative bending moment
12825000 = 0.87(415)(Ast)(115){1 – (Ast)(415)/(1000)(115)(20)}
or - 5542.16 A2stAst + 1711871.646 = 0
Solving the quadratic equation, we have the negative Ast = 328.34 mm
2
(ii) For the maximum positive bending moment
10687500 = 0.87(415) Ast(115) {1 – (Ast)(415)/(1000)(115)(20)}
or - 5542.16 A2stAst + 1426559.705 = 0
Solving the quadratic equation, we have the positive Ast = 270.615 mm
2
Alternative approach: Use of Table 2 of SP-16
(i) For negative bending moment
2
Mu/bd = 0.9697
Table 2 of SP-16 gives: ps = 0.2859 (by linear interpolation). So, the area of negative steel =
2
0.2859(1000)(115)/100 = 328.785 mm .
(ii) For positive bending moment
2
Mu/bd = 0.8081
Table 2 of SP-16 gives: ps = 0.23543 (by linear interpolation). So, the area of positive steel =
2
0.23543(1000)(115)/100 = 270.7445 mm .
These areas of steel are comparable with those obtained by direct computation.
Distribution steel bars along longer span ly
2
Distribution steel area = Minimum steel area = 0.12(1000) (140)/100 = 168 mm . Since, both
positive and negative areas of steel are higher than the minimum area, we provide:
2
(a) For negative steel: 10 mm diameter bars @ 230 mm c/c for which A = 341 mm giving p
st
s
= 0.2965.
2
(b) For positive steel: 8 mm diameter bars @ 180 mm c/c for which Ast = 279 mm giving ps =
0.2426
(c) For distribution steel: Provide 8 mm diameter bars @ 250 mm c/c for which Ast
2
(minimum) = 201 mm .
Step 6: Selection of diameter and spacing of reinforcing bars
The diameter and spacing already selected in step 5 for main and distribution bars are
checked below:
For main bars (cl. 26.3.3.b.1 of IS 456), the maximum spacing is the lesser of 3d and 300 mm
i.e., 300 mm. For distribution bars (cl. 26.3.3.b.2 of IS 456), the maximum spacing is the
lesser of 5d or 450 mm i.e., 450 mm. provided spacing, therefore, satisfy the requirements.
Maximum diameter of the bars (cl. 26.5.2.2 of IS 456) shall not exceed 140/8 = 17 mm is
also satisfied with the bar diameters selected here.
Fig 2.3
Figure presents the detailing of the reinforcement bars. The abbreviation B1 to B3 and T1
to T4 are the bottom and top bars, respectively which are shown in Fig.2.3 for a typical oneway slab.
The above design and detailing assume absence of support along short edges. When supports
along short edges exist and there is eventual clamping top reinforcement would be necessary
at shorter supports also.
2. Design the cantilever panel of the one-way slab shown in Fig.2.4 subjected to uniformly
2
distributed imposed loads 5 kN/m using M 20 and Fe 415. The load of floor finish is 0.75
2
kN/m . The span dimensions shown in the figure are effective spans. The width of the
support is 300 mm.
Fig 2.4
Step 1: Selection of preliminary depth of slab
Basic value of span to depth ratio (cl. 23.2.1 of IS 456) = 7
Modification factor = 1.18 (see Problem 8.1)
Minimum effective depth = 1850/7(1.18) = 223.97 mm
Assume d = 225 mm and D = 250 mm.
Step 2: Design loads, bending moment and shear force
Factored dead loads = (1.5)(0.25)(25) = 9.375 kN/m
Factored load of floor finish = (1.5)(0.75) = 1.125 kN/m
Factored live loads = (1.5)(5) = 7.5 kN/m
Total factored loads = 18.0 kN/m
Maximum negative moment = 18.0(1.85)(1.85)(0.5) = 30.8025 kNm/m
Maximum shear force = 18.0(1.85) = 33.3 kN/m
Step 3: Determination of effective and total depths of slab
From Eq.3.25, Step 3 of sec. 8.18.6, we have
6
3
0.5
d = {30.8025(10 )/2.76(10 )}
2
= 105.64 mm, considering the value of R = 2.76 N/mm from
Table 3.3 of sec. 3.5.5 of Lesson 5. This depth is less than assumed depth of slab in Step 1.
Hence, assume d = 225 mm and D = 250 mm.
Step 4: Depth of slab for shear force
Using the value of k = 1.1 (cl. 40.2.1.1 of IS 456) for the slab of 250 mm depth, we have τc
2
(from Table 19 of IS 456) = 1.1(0.28) = 0.308 N/mm . Table 20 of IS 456 gives max τc = 2.8
2
N/mm . Here, the depth of the slab is safe in shear as N/mm 0.148 33.3/225 / 2=bdVuvτ.
maxccvτττ<<
Step 5: Determination of areas of steel (using table of SP-16)
Table 44 gives 10 mm diameter bars @ 200 mm c/c can resist 31.43 kNm/m >
30.8025 kNm/m. Fifty per cent of the bars should be curtailed at a distance of larger of Ld or
0.5lx. Table 65 of SP-16 gives Ld of 10 mm bars = 470 mm and 0.5lx = 0.5(1850) = 925 mm
from the face of the column. The curtailment distance from the centre line of beam = 925 +
150 = 1075, say 1100 mm.
The above, however, is not admissible as the spacing of bars after the curtailment exceeds
300 mm. So, we provide 10 mm @ 300 c/c and 8 mm @ 300 c/c. The moment of resistance
of this set is 34.3 kNm/m > 30.8025 kNm/m (see Table 44 of SP-16).
Fig 2.5
2.2 Two-way Slabs
Two-way slabs subjected mostly to uniformly distributed loads resist them primarily by
bending about both the axis. However, as in the one-way slab, the depth of the two-way slabs
should also be checked for the shear stresses to avoid any reinforcement for shear. Moreover,
these slabs should have sufficient depth for the control deflection. Thus, strength and
deflection are the requirements of design of two-way slabs.
Design Shear Strength of Concrete
Design shear strength of concrete in two-way slabs is to be determined incorporating the
multiplying factor k from Table 1
Computation of shear force
Fig 2.6
Shear forces are computed following the procedure stated below
The two-way slab of Fig. 2.6 is divided into two trapezoidal and two triangular zones by
o
drawing lines from each corner at an angle of 45 . The loads of triangular segment A will be
transferred to beam 1-2 and the same of trapezoidal segment B will be beam 2-3. The shear
forces per unit width of the strips aa and bb are highest at the ends of strips. Moreover, the
length of half the strip bb is equal to the length of the strip aa. Thus, the shear forces in both
strips are equal and we can write,
V = W (l /2)
u
x
Where W = intensity of the uniformly distributed loads.
The nominal shear stress acting on the slab is then determined from
bdVuv/ =τ
Computation of bending moments
Two-way slabs spanning in two directions at right angles and carrying uniformly distributed
loads may be analyzed using any acceptable theory. Pigeoud’s or Wester-guard’s theories are
the suggested elastic methods and Johansen’s yield line theory is the most commonly used in
the limit state of collapse method and suggested by IS 456 in the note of cl. 24.4.
Alternatively, Annex D of IS 456 can be employed to determine the bending moments in the
two directions for two types of slabs: (i) restrained slabs, and (ii) simply supported slabs. The
two methods are explained below:
1. Restrained slabs
Restrained slabs are those whose corners are prevented from lifting due to effects of torsional
moments. These torsional moments, however, are not computed as the amounts of
reinforcement are determined from the computed areas of steel due to positive bending
moments depending upon the intensity of torsional moments of different corners. This aspect
has been explained in Step 7 of sec. 8.19.6. Thus, it is essential to determine the positive and
negative bending moments in the two directions of restrained slabs depending on the various
types of panels and the aspect ratio l /l .
y x
Fig 2.7 middle and edge stirrups
Restrained slabs are considered as divided into two types of strips in each direction:
(i) one middle strip of width equal to three-quarters of the respective length of span in either
directions, and (ii) two edge strips, each of width equal to one-eighth of the respective length
of span in either direction. Figures 2.7 a and b present the two types of strips for spans lx and
ly separately.
The maximum positive and negative moments per unit width in a slab are determined from
Where α x and α
y
are coefficients given in Table 26 of IS 456, Annex D, cl. D-1.1. Total
design load per unit area is w and lengths of shorter and longer spans are represented by lx
and ly, respectively. The values of, given in Table 26 of IS 456, are for nine types of panels
having eight aspect ratios of ly/lx from one to two at an interval of 0.1. The above maximum
bending moments are applicable only to the middle strips and no redistribution shall be
made.
Tension reinforcing bars for the positive and negative maximum moments are to be
provided in the respective middle strips in each direction. Figure 8.19.2 shows the positive
and negative coefficients yxαα and The edge strips will have reinforcing bars parallel to that
edge following the minimum amount as stipulated in IS 456.
The detailing of all the reinforcing bars for the respective moments and for the minimum
amounts as well as torsional requirements are discussed in sec. 8.19.7(i).
(ii) Simply supported slabs
The maximum moments per unit width of simply supported slabs, not having adequate
provision to resist torsion at corners and to prevent the corners from lifting, are determined
from Eqs.8.3 and 8.4, where yxαα and are the respective coefficients of moments as given in
Table 27 of IS 456, cl. D-2. The notations Mx, My, w, lx and ly are the same as mentioned
below Eqs.8.3 and 8.4 in (i) above.
The detailing of reinforcing bars for the respective moments is explained below:
FIG 2.8
The coefficients α
x
and α
y
of simply supported two-way slabs are derived from the
Grashoff-Rankine formula which is based on the consideration of the same deflection at any
point P of two perpendicular interconnected strips containing the common point P of the twoway slab subjected to uniformly distributed loads.
Design Considerations
The design considerations mentioned in sec. 8.18.5 of Lesson 18 in (a), (c), (d), (e)
and (f) are applicable for the two-way slabs also. However, the effective span to effective
depth ratio is different from those of one-way slabs. Accordingly, this item for the two-way
slabs is explained below.
Effective span to effective depth ratio (cl. 24.1 of IS 456)
The following are the relevant provisions given in Notes 1 and 2 of cl. 24.1.
• The shorter of the two spans should be used to determine the span to effective depth ratio.
• For spans up to 3.5 m and with mild steel reinforcement, the span to overall depth ratios
2
satisfying the limits of vertical deflection for loads up to 3 kN/m are as follows:
Simply supported slabs -35
Continuous slabs -40
• The same ratios should be multiplied by 0.8 when high strength deformed bars (Fe 415) are
used in the slabs.
2.2.1 Design of Two-way Slabs
The procedure of the design of two-way slabs will have all the six steps mentioned in
the design of one-way slabs except that the bending moments and shear forces are
determined by different methods for the two types of slab.
While the bending moments and shear forces are computed from the coefficients
given in Tables 12 and 13 (cl. 22.5) of IS 456 for the one-way slabs, the same are obtained
from Tables 26 or 27 for the bending moment in the two types of two-way slabs and the
shear forces are computed from Eq.8.1 for the two-way slabs.
Further, the restrained two-way slabs need adequate torsional reinforcing bars at the corners
to prevent them from lifting. There are three types of corners having three different
requirements. Accordingly, the determination of torsional reinforcement is discussed in Step
7, as all the other six steps are common for the one and two-way slabs.
Step 7: Determination of torsional reinforcement
Three types of corners, C1, C2 and C3, shown in Fig.2.9, have three different
requirements of torsion steel as mentioned below.
(a) At corner C1 where the slab is discontinuous on both sides, the torsion reinforcement
shall consist of top and bottom bars each with layers of bar placed parallel to the
sides of the slab and extending a minimum distance of one-fifth of the shorter span
from the edges. The amount of reinforcement in each of the four
(b) layers shall be 75 per cent of the area required for the maximum mid-span moment in
the slab. This provision is given in cl. D-1.8 of IS 456.
Fig 2.9
(b) At corner C2 contained by edges over one of which is continuous, the torsional
reinforcement shall be half of the amount of (a) above. This provision is given in cl. D-1.9 of
IS 456.
(c) At corner C3 contained by edges over both of which the slab is continuous, torsional
reinforcing bars need not be provided, as stipulated in cl. D-1.10 of IS 456.
Detailing of Reinforcement
Areas of steel required for the maximum positive and negative moments. The two
methods are (i) employing Eq.3.23 as given in Step 5 of sec. 8.18.6 or (ii) using tables and
charts of SP-16. Thereafter, Step 7 of sec. 8.19.6 explains the method of determining the
areas steel for corners of restrained slab depending on the type of corner. The detailing of
torsional reinforcing bars is explained in Step 7 of sec. 8.19.6. In the following, the detailing
of reinforcing bars for (i) restrained slabs and (ii) simply supported slabs are discussed
separately for the bars either for the maximum positive or negative bending moments or to
satisfy the requirement of minimum amount of steel.
(i) Restrained slabs
The maximum positive and negative moments per unit width of the slab calculated by
employing Eqs.8.3 and 8.4 as explained in sec. 8.19.4.2(i) are applicable only to the
respective middle strips (Fig.8.19.2). There shall be no redistribution of these moments.
The reinforcing bars so calculated from the maximum moments are to be placed
satisfying the following stipulations of IS 456.
• Bottom tension reinforcement bars of mid-span in the middle strip shall extent in the lower
part of the slab to within 0.25l of a continuous edge, or 0.15l of a discontinuous edge (cl. D1.4 of IS 456). Bars marked as B1, B2, B5 and B6 in Figs.8.19.5 a and b are these bars.
• Top tension reinforcement bars over the continuous edges of middle strip shall extend in the
upper part of the slab for a distance of 0.15l from the support, and at least fifty per cent of
these bars shall extend a distance of 0.3l (cl. D-1.5 of IS 456). Bars marked as T2, T3, T5 and
T6 in Figs.8.19.5 a and b are these bars.
• To resist the negative moment at a discontinuous edge depending on the degree of fixity at
the edge of the slab, top tension reinforcement bars equal to fifty per cent of that provided at
mid-span shall extend 0.1l into the span (cl. D-1.6 of IS 456). Bars marked as T1 and T4 in
Figs.8.19.5 a and b are these bars.
• The edge strip of each panel shall have reinforcing bars parallel to that edge satisfying the
requirement of minimum amount as specified in sec. 8.18.15d of Lesson 18 (cl. 26.5.2.1 of IS
456) and the requirements for torsion, explained in Step 7 of sec. 8.19.6 (cls. D-1.7 to D-1.10
of IS 456). The bottom and top bars of the edge strips are explained below.
• Bottom bars B3 and B4 (Fig.8.19.5 a) are parallel to the edge along lx for the edge strip for
span ly, satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).
• Bottom bars B7 and B8 (Fig.8.19.5 b) are parallel to the edge along ly for the edge strip for
span lx, satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).
• Top bars T7 and T8 (Fig.8.19.5 a) are parallel to the edge along lx for the edge strip for span
ly, satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).
• Top bars T9 and T10 (Fig.8.19.5 b) are parallel to the edge along ly for the edge strip for
span lx, satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).
The above explanation reveals that there are eighteen bars altogether comprising eight
bottom bars (B1 to B8) and ten top bars (T1 to T10). Tables 8.4 and 8.5 present them
separately for the bottom and top bars, respectively, mentioning the respective zone of their
placement (MS/LDES/ACES/BDES to designate Middle Strip/Left Discontinuous Edge
Strip/Adjacent Continuous Edge Strip/Bottom Discontinuous Edge Strip), direction of the
bars (along x or y), the resisting moment for which they shall be determined or if to be
provided on the basis of minimum reinforcement clause number of IS 456 and Fig. No. For
easy understanding, plan views in (a) and (b) of Fig.8.19.5 show all the bars separately along
x and y directions, respectively. Two sections (1-1 and 2-2), however, present the bars shown
in the two plans. Torsional reinforcements are not included in Tables .
Table - Details of eight bottom bars
Sl.No
Bars
Into
Along
Resisting Moment
Cl.No. of IS 456
Fig.No.
.
1
B1, B2
MS
x
Max. + Mx
D-1.3,1.4
8.19.5a, c, d
2
B3
LDES
x
Min. Steel
D-1.7
8.19.5a, c
3
B4
ACES
x
Min. Steel
D-1.7
8.19.5a, c
4
B5,B6
MS
y
Max. + My
D-1.3,1.4
8.19.5b,c,d
5
B7
BDES
y
Min. Steel
D-1.7
8.19.5b,d
6
B8
ACES
y
Min. Steel
D-1.7
8.19.5b,d
Notes: (i) MS = Middle Strip
(ii) LDES = Left Discontinuous Edge Strip
(iii) ACES = Adjacent Continuous Edge Strip
(iv) BDES = Bottom Discontinuous Edge Strip
Table – Details of eight top bars
Sl.No.
Bars
Into
Along
Resisting Moment
Cl.No. of IS 456
Fig.No.
1
T1
BDES
x
+ 0.5 Mx
D-1.6
8.19.5a, d
2
T2, T3
ACES
x
- 0.5 Mx for each
D-1.5
8.19.5a, d
3
T4
LDES
y
+ 0.5 My
D-1.6
8.19.5b, c
4
T5, T6
ACES
y
-0.5 My for each
D-1.5
8.19.5b, c
5
T7
LDES
x
Min. Steel
D-1.7
8.19.5a, c
6
T8
ACES
x
Min. Steel
D-1.7
8.19.5a, c
7
T9
LDES
y
Min. Steel
D-1.7
8.19.5b, d
8
T10
ACES
y
Min. Steel
D-1.7
8.19.5b, d
Notes: (i) MS = Middle Strip
(ii) LDES = Left Discontinuous Edge Strip
(iii) ACES = Adjacent Continuous Edge Strip
(iv) BDES = Bottom Discontinuous Edge Strip
(ii) Simply supported slabs
Figures 8.19.6 a, b and c present the detailing of reinforcing bars of simply supported
slabs not having adequate provision to resist torsion at corners and to prevent corners
from lifting. Clause D-2.1 stipulates that fifty per cent of the tension reinforcement
provided at mid-span should extend to the supports. The remaining fifty per cent
should extend to within 0.1lx or 0.1ly of the support, as appropriate.
Problem 2
2
Design the slab panel 1 of above fig subjected to factored live load of 8 kN/m in addition to
2
its dead load using M 20 and Fe 415. The load of floor finish is 1 kN/m . The spans shown in
figure are effective spans. The corners of the slab are prevented from lifting.
Solution:
Step 1: Selection of preliminary depth of slab
The span to depth ratio with Fe 415 is taken from cl. 24.1, Note 2 of IS 456 as 0.8 (35 + 40) /
2 = 30. This gives the minimum effective depth d = 4000/30 = 133.33 mm, say 135 mm. The
total depth D is thus 160 mm.
Step 2: Design loads, bending moments and shear forces
Dead load of slab (1 m width) = 0.16(25) = 4.0 kN/m
Dead load of floor finish (given) = 1.0 kN/m
Factored dead load = 1.5(5) = 7.5 kN/m
Factored live load (given) = 8.0 kN/m
Total factored load = 15.5 kN/m
2
2
2
2
2
The coefficients of bending moments and the bending moments Mx and My per unit width
(positive and negative) are determined as per cl. D-1.1 and Table 26 of IS 456 for the case 4,
“Two adjacent edges discontinuous” and presented in Table 8.6. The ly / lx for this problem is
6/4 = 1.5.
Maximum bending moments
For
Short span
xα
Long span
yα
Mx
(kNm/m)
Negative
moment
at 0.075
My
(kNm/m)
18.6
0.047
11.66
13.89
0.035
8.68
continues edge
Positive moment at mid- 0.056
span
Maximum shear force in either direction is determined from Eq.8.1 (Fig.8.19.1) as
Vu = w(lx/2) = 15.5 (4/2) = 31 kN/m
Step 3: Determination/checking of the effective depth and total depth of slab
Using the higher value of the maximum bending moments in x and y directions from Table
8.6, we get from Eq.3.25 of Lesson 5 (sec. 3.5.5):
2
Mu,lim = R,lim bd
6
3
Or d = [(18.6)(10 )/{2.76(10 )}]
1/2
= 82.09 mm,
2
Where 2.76 N/mm is the value of R,lim taken from Table 3.3 of Lesson 5 (sec. 3.5.5). Since,
this effective depth is less than 135 mm assumed in Step 1, we retain d = 135 mm and D =
160 mm.
Step 4: Depth of slab for shear force
2
Table 19 of IS 456 gives the value of cτ = 0.28 N/mm when the lowest percentage of steel is
provided in the slab. However, this value needs to be modified by multiplying with k of cl.
40.2.1.1 of IS 456. The value of k for the total depth of slab as 160 mm is 1.28. So, the value
2
of cτ is 1.28(0.28) = 0.3584 N/mm .
2
Table 20 of IS 456 gives maxcτ = 2.8 N/mm . The computed shear stress vτ = Vu/bd = 31/135
2
= 0.229 N/mm .
Since, vτ < cτ < maxcτ, the effective depth of the slab as 135 mm and the total depth as 160
mm are safe.
Step 5: Determination of areas of steel
The respective areas of steel in middle and edge strips are to be determined employing
Eq.3.23 of Step 5 of sec. 8.18.6 of Lesson 18. However, in Problem 8.1 of Lesson 18, it has
been shown that the areas of steel computed from Eq.3.23 and those obtained from the tables
of SP-16 are in good agreement. Accordingly, the areas of steel for this problem are
computed from the respective Tables 40 and 41 of SP-16 and presented in Table 8.7. Table
40 of SP-16 is for the effective depth of 150 mm, while Table 41 of SP-16 is for the effective
depth of 175 mm. The following results are, therefore, interpolated values obtained from the
two tables of SP-16.
Reinforcing bars
Particulars
Short span lx
Table
Mx (kNm/m)
no.
Top steel for 40,41
Long span ly
Dia&
Table no.
My (kNm/m)
spacing
18.68>18.6
10
mm 40,41
negative
@
200
moment
mm c/c
Dia&
spacing
12.314>11.66
8 mm @
200 mm
c/c
Bottom steel 40,41
for
14.388>13.89
positive
moment
8 mm @ 40,41
9.20>8.68
8 mm @
170 mm
250 mm
c/c
c/c
The minimum steel is determined from the stipulation of cl. 26.5.2.1 of IS 456 and is
As = (0.12/100)(1000)(160) = 192 mm
2
2
and 8 mm bars @ 250 mm c/c (= 201 mm ) is acceptable. It is worth mentioning that the
areas of steel as shown in Table 8.7 are more than the minimum amount of steel.
Step 6: Selection of diameters and spacings of reinforcing bars
The advantages of using the tables of SP-16 are that the obtained values satisfy the
requirements of diameters of bars and spacings. However, they are checked as ready
reference here. Needless to mention that, this step may be omitted in such a situation.
Maximum diameter allowed, as given in cl. 26.5.2.2 of IS 456, is 160/8 = 20 mm, which is
more that the diameters used here.
The maximum spacing of main bars, as given in cl. 26.3.3(1) of IS 456, is the lesser of
3(135) and 300 mm. This is also satisfied for all the bars.
The maximum spacing of minimum steel (distribution bars) is the lesser of 5(135) and 450
mm. This is also satisfied.
Figures 8.19.8 and 9 present the detailing of reinforcing bars.
Step 7: Determination of torsional reinforcement
Torsional reinforcing bars are determined for the three different types of corners as explained
in sec. 8.19.6 (Fig.8.19.4). The length of torsional strip is 4000/5 = 800 mm and the bars are
to be provided in four layers. Each layer will have 0.75 times the steel used for the maximum
positive moment. The C1 type of corners will have the full amount of torsional steel while
C2 type of corners will have half of the amount provided in C1 type. The C3 type of corners
do not need any torsional steel. The results are presented in Table 8.8 and Figs.8.19.10 a, b
and c.
Table Torsional reinforcement bars
Dimensions along
Type
X(mm)
Y(mm)
No of bars along
Bar dia &
X(mm)
Y(mm)
spacing
C1
800
800
8 mm @ 200
Cl.no. of IS
456
5
5
D-1.8
5
8
D-1.9
8
5
D-1.9
mm c/c
C2
800
1600
8 mm @ 250
mm c/c
C2
1600
800
8 mm @ 250
mm c/c
2.3 SINGLY, DOUBLY AND FLANGED BEAMS
Types of Problems
Two types of problems are possible: (i) design type and (ii) analysis type. In the design
type of problems, the designer has to determine the dimensions b, d, D, Ast (Fig. 3.6.1) and
other detailing of reinforcement, grades of concrete and steel from the given design moment
of the beam. In the analysis type of the problems, all the above data will be known and the
designer has to find out the moment of resistance of the beam. Both the types of problems are
taken up for illustration in the following two lessons.
Design Type of Problems
The designer has to make preliminary plan lay out including location of the beam, its
span and spacing, estimate the imposed and other loads from the given functional
requirement of the structure. The dead loads of the beam are estimated assuming the
dimensions b and d initially. The bending moment, shear force and axial thrust are
determined after estimating the different loads. In this illustrative problem, let us assume that
the imposed and other loads are given. Therefore, the problem is such that the designer has to
start with some initial dimensions and subsequently revise them, if needed. The following
guidelines are helpful to assume the design parameters initially.
1 Selection of breadth of the beam b
Normally, the breadth of the beam b is governed by: (i) proper housing of reinforcing bars
and (ii) architectural considerations. It is desirable that the width of the beam should be less
than or equal to the width of its supporting structure like column width, or width of the wall
etc. Practical aspects should also be kept in mind. It has been found that most of the
requirements are satisfied with b as 150, 200, 230, 250 and 300 mm. Again, width to overall
depth ratio is normally kept between 0.5 and 0.67.
2 Selection of depths of the beam d and D
The effective depth has the major role to play in satisfying (i) the strength requirements of
bending moment and shear force, and (ii) deflection of the beam. The initial effective depth
of the beam, however, is assumed to satisfy the deflection requirement depending on the span
and type of the reinforcement. IS 456 stipulates the basic ratios of span to effective depth of
beams for span up to 10 m as (Clause 23.2.1)
1. Cantilever 7
2. Simply supported 20
3. Continuous 26
For spans above 10 m, the above values may be multiplied with 10/span in meters, except for
cantilevers where the deflection calculations should be made. Further, these ratios are to be
multiplied with the modification factor depending on reinforcement percentage and type.
Figures 4 and 5 of IS 456 give the different values of modification factors. The total depth D
can be determined by adding 40 to 80 mm to the effective depth.
3 Selection of the amount of steel reinforcement Ast
The amount of steel reinforcement should provide the required tensile force T to resist the
factored moment Mu of the beam. Further, it should satisfy the minimum and maximum
percentages of reinforcement requirements also. The minimum reinforcement As is provided
for creep, shrinkage, thermal and other environmental requirements irrespective of the
strength requirement. The minimum reinforcement As to be provided in a beam depends on
the fy of steel and it follows the relation: (cl. 26.5.1.1a of IS 456).
The maximum tension reinforcement should not exceed 0.04 bD (cl. 26.5.1.1b of IS 456),
where D is the total depth. Besides satisfying the minimum and maximum reinforcement, the
amount of reinforcement of the singly reinforced beam should normally be 75 to 80% of pt,
. This will ensure that strain in steel will be more than
lim
as the design
stress in steel will be 0.87 fy. Moreover, in many cases, the depth required for deflection
becomes more than the limiting depth required to resist Mu, lim. Thus, it is almost obligatory to
provide more depth. Providing more depth also helps in the amount of the steel which is less
than that required for Mu, lim. This helps to ensure ductile failure. Such beams are designated
as under-reinforced beams.
4 Selection of diameters of bar of tension reinforcement
Reinforcement bars are available in different diameters such as 6, 8, 10, 12, 14, 16, 18, 20,
22, 25, 28, 30, 32, 36 and 40 mm. Some of these bars are less available. The selection of the
diameter of bars depends on its availability, minimum stiffness to resist while persons walk
over them during construction, bond requirement etc. Normally, the diameters of main tensile
bars are chosen from 12, 16, 20, 22, 25 and 32 mm.
5 Selection of grade of concrete
Besides strength and deflection, durability is a major factor to decide on the grade of
concrete. Table 5 of IS 456 recommends M 20 as the minimum grade under mild
environmental exposure and other grades of concrete under different environmental
exposures also.
6 Selection of grade of steel
Normally, Fe 250, 415 and 500 are in used in reinforced concrete work. Mild steel (Fe 250)
is more ductile and is preferred for structures in earthquake zones or where there are
possibilities of vibration, impact, blast etc.
Type of Problems
It may be required to estimate the moment of resistance and hence the service load of
a beam already designed earlier with specific dimensions of b, d and D and amount of steel
reinforcement Ast. The grades of concrete and steel are also known. In such a situation, the
designer has to find out first if the beam is under-reinforced or over-reinforced. The
following are the steps to be followed for such problems.
3.7.2.1 xu, max
The maximum depth of the neutral axis xu, max is determined from Table 3.2 of Lesson 5 using
the known value of fy.
3.7.2.2 xu
The depth of the neutral axis for the particular beam is determined from Eq. 3.16 of Lesson 5
employing the known values of fy, fck, b and Ast.
3.7.2.3 Mu and service imposed loads
The moment of resistance Mu is calculated for the three different cases as follows:
(a) If xu < xu, max, the beam is under-reinforced
(b) If xu = xu, max, the Mu is obtained from Eq. 3.24
(c) If xu > xu, max, the beam is over-reinforced for which xu is taken as xu, max and then
Mu is calculated using xu = xu, max.
With the known value of Mu, which is the factored moment, the total factored load can be
obtained from the boundary condition of the beam. The total service imposed loads is then
determined dividing the total factored load by partial safety factor for loads (= 1.5). The
service imposed loads are then obtained by subtracting the dead load of the beam from the
total service loads.
Design of singly reinforced beam
1. Determine the service imposed loads of two simply supported beam of same effective
span of 8 m (Figs. 3.7.1 and 2) and same cross-sectional dimensions, but having two different
amounts of reinforcement. Both the beams are made of M 20 and Fe 415.
2
Given data: b = 300 mm, d = 550 mm, D = 600 mm, Ast = 1256 mm (4-20 T), Leff = 8 m and
boundary condition = simply supported (Fig. 3.7.1).
1 xu, max
From Table 3.2 of Lesson 5, we get x
u, max
= 0.479 d = 263.45 mm
2.3.2 Double reinforced beam
Concrete has very good compressive strength and almost negligible tensile strength.
Hence, steel reinforcement is used on the tensile side of concrete. Thus, singly reinforced
beams reinforced on the tensile face are good both in compression and tension. However,
these beams have their respective limiting moments of resistance with specified width, depth
and grades of concrete and steel. The amount of steel reinforcement needed is known as
Ast,lim. Problem will arise, therefore, if such a section is subjected to bending moment greater
than its limiting moment of resistance as a singly reinforced section.
There are two ways to solve the problem. First, we may increase the depth of the
beam, which may not be feasible in many situations. In those cases, it is possible to increase
both the compressive and tensile forces of the beam by providing steel reinforcement in
compression face and additional reinforcement in tension face of the beam without increasing
the depth (Fig. 4.8.1). The total compressive force of such beams comprises (i) force due to
concrete in compression and (ii) force due to steel in compression. The tensile force also has
two components: (i) the first provided by Ast,lim which is equal to the compressive force of
concrete in compression. The second part is due to the additional steel in tension - its force
will be equal to the compressive force of steel in compression. Such reinforced concrete
beams having steel reinforcement both on tensile and compressive faces are known as doubly
reinforced beams.
Doubly reinforced beams, therefore, have moment of resistance more than the singly
reinforced beams of the same depth for particular grades of steel and concrete. In many
practical situations, architectural or functional requirements may restrict the overall depth of
the beams. However, other than in doubly reinforced beams compression steel reinforcement
is provided when:
(i) Some sections of a continuous beam with moving loads undergo change of sign of
the bending moment which makes compression zone as tension zone or vice versa.
(ii) The ductility requirement has to be followed.
(iii) The reduction of long term deflection is needed.
It may be noted that even in so called singly reinforced beams there would be longitudinal
hanger bars in compression zone for locating and fixing stirrups.
Assumptions
(i) The assumptions of sec. 3.4.2 of Lesson 4 are also applicable here.
(ii) Provision of compression steel ensures ductile failure and hence, the limitations of
x/d ratios need not be strictly followed here.
(iii) The stress-strain relationship of steel in compression is the same as that in
tension. So, the yield stress of steel in compression is 0.87 fy.
Design type of problems
In the design type of problems, the given data are b, d, D, grades of concrete and steel. The
designer has to determine Asc and Ast of the beam from the given factored moment. These
problems can be solved by two ways: (i) use of the equations developed for the doubly
reinforced beams, named here as direct computation method, (ii) use of charts and tables of
SP-16.
(a) Direct computation method
Step 1: To determine Mu, lim and Ast, lim from Eqs. 4.2 and 4.8, respectively.
Step 2: To determine Mu2, Asc, Ast2 and Ast from Eqs. 4.1, 4.4, 4.6 and 4.7, respectively.
Step 3: To check for minimum and maximum reinforcement in compression and tension as
explained in sec. 4.8.5.
Step 4: To select the number and diameter of bars from known values of Asc and Ast.
4.8.6.2 Analysis type of problems
In the analysis type of problems, the data given are b, d, d', D, fck, fy, Asc and Ast . It is required
to determine the moment of resistance Mu of such beams. These problems can be solved: (i)
by direct computation method and (ii) by using tables of SP-16.
(a) Direct computation method
Step 1: To check if the beam is under-reinforced or over-reinforced.
The beam is under-reinforced or over-reinforced if εst is less than or more than the yield
strain.
Step 2: To determine Mu,lim from Eq. 4.2 and Ast,lim from the pt,
lim
given in Table 3.1 of
Lesson 5.
Step 3: To determine Ast2 and Asc from Eqs. 4.7 and 4.6, respectively.
Step 4: To determine Mu2 and Mu from Eqs. 4.4 and 4.1, respectively.
Problem
Design a simply supported beam of effective span 8 m subjected to imposed loads of
35 kN/m. The beam dimensions and other data are: b = 300 mm, D = 700 mm, M 20
concrete, Fe 415 steel (Fig. 4.9.1). Determine fsc from d'/d
It may be noted that Ast is provided in two layers in order to provide adequate space for
concreting around reinforcement. Also the centroid of the tensile bars is at 70 mm from
bottom.
2.3.2 Design of flanged beam:
When a reinforced concrete slab is cast monolithically with the beam as in the case of
beam supported floor slab system, the beams can considered as flanged beams with slab
acting as an effective flange on the compression side. It is important to note that continuous
T or L beams act as flanged beams only between the supports where the bending moments
are negative (sagging) and the slabs are on the compression side of the beam. In the vicinity
of the supports where the bending moments are negative (hogging), the slab is on the tension
side and hence the beam act as a rectangular beam with the tension steel located in the slab
portion of the beam. Hence at the locations of the negative moments, the beams have to be
designed as singly or doubly reinforced rectangular beams.
150mm and with the area of main reinforcement of the slab (parallel to the beam) at
its middle is 500mm2/m, the transverse reinforcement required according to the provision of
the IS:456 code will be equal to (0.6x500) = 300 mm2/m.
Design example:
A T-beam slab floor of reinforced concrete has a slab 150mm thick spanning between the Tbeams which are spaced 3m apart. The beams have a clear span of 10m and the end bearings
are 450mm thick walls. The live load on the floor is 4kN/m2. Using M-20 grade of concrete
and Fe-415 HYSD bars, design one of the intermediate T-beams.
a) Data
Clear span =10m
Bearing thickness =450mm
Live load = q= 4kN/m2
Spacing of T-beams = 3m
Fck = 20N/mm2
Fy = 415 N/mm2
Df = 150 mm
b) Cross sectional dimensions
Assuming effective depth =
Span
15
= (10x103 / 15) =666mm
Reinforcement details in T beam
DESIGN OF RC ELEMENTS
UNIT 3
LIMIT STATE DESIGN FOR SHEAR, TORSION, BOND AND
ANCHORAGE
CONTENTS
3.1 Introduction
3.2 Design of shear strength of reinforced concrete
3.3 Design of shear strength
3.4 Bond
3.4.1 Design bond stress
3.5 Development length
3.5.1 Checking of development length of bars in tension
3.6 Anchoring of reinforcing bars
3.7 Torsion
3.7.1 Shear and Torsion
TECHNICAL TERMS
Shear stress: A force that attempts to cause the internal structure of a material to slide
against itself.
Torsion: The specific type of shear stress in which one end of a part is secured while other
end is twisted.
Bond: When the concrete sets, it adheres to surface of reinforcement and tightly grips it. This
perfect adhesion between concrete and steel is known as bond.
Bond stress: the bond resists any force that rise to pull out or push the rod. The intensity of
the adhesive force is called bond stress.
Development length of bars: Whenever some reinforcing bar is to be anchored or two bars
have to be given an overlap, it is essential that they must get sufficient length of embedded
length or overlap length so that no slip takes place. The embedment or overlap is known as
development length.
UNIT -3
LIMIT STATE DESIGN FOR SHEAR, TORSION, BOND AND
ANCHORAGE
3.1 Introduction
This lesson explains the three failure modes due to shear force in beams and defines different
shear stresses needed to design the beams for shear. The critical sections for shear and the
minimum shear reinforcement to be provided in beams are mentioned as per IS 456.
Failure Modes due to Shear
Fig 3.1
Bending in reinforced concrete beams is usually accompanied by shear, the exact analysis of
which is very complex. However, experimental studies confirmed the following three
different modes of failure due to possible combinations of shear force and bending moment
at a given section (Figs 3.1a to c):
(i) Web shear (Fig. 3.1a)
(ii) Flexural tension shear (Fig 3.1b)
(iii) Flexural compression shear (Fig 3.1c)
Web shear causes cracks which progress along the dotted line shown in Fig 3.1a. Steel yields
in flexural tension shear as shown in Fig 3.1b, while concrete crushes in compression due to
flexural compression shear as shown in Fig 3.1c. An in-depth presentation of the three types
of failure modes is beyond the scope here.
Shear Stress
The distribution of shear stress in reinforced concrete rectangular, T and L-beams of
uniform and varying depths depends on the distribution of the normal stress. However, for
the sake of simplicity the nominal shear stresses τv is considered which is calculated as
follows (IS 456, cls. 40.1 and 40.1.1):
Fig 3.2
The positive sign is applicable when the bending moment Mu decreases numerically
in the same direction as the effective depth increases, and the negative sign is applicable
when the bending moment Mu increases numerically in the same direction as the effective
depth increases.
3.2 Design Shear Strength of Reinforced Concrete
Recent laboratory experiments confirmed that reinforced concrete in beams has shear
strength even without any shear reinforcement. This shear strength (τc) depends on the grade
of concrete and the percentage of tension steel in beams. On the other hand, the shear
strength of reinforced concrete with the reinforcement is restricted to some maximum value
τ
cmax
depending on the grade of concrete. These minimum and maximum shear strengths of
reinforced concrete (IS 456, cls. 40.2.1 and 40.2.3, respectively) are given below:
Design shear strength without shear reinforcement (IS 456, cl. 40.2.1)
Table 19 of IS 456 stipulates the design shear strength of concrete τc for different
grades of concrete with a wide range of percentages of positive tensile steel reinforcement. It
is worth mentioning that the reinforced concrete beams must be provided with the minimum
shear reinforcement as per cl. 40.3 even when τv is less than τc given in Table 3.1.
2
Table 3.1 Design shear strength of concrete, τ in N/mm .
c
In Table 3.1, As is the area of longitudinal tension reinforcement which continues at
least one effective depth beyond the section considered except at support where the full area
of tension reinforcement may be used provided the detailing is as per IS 456, cls. 26.2.2 and
26.2.3.
Maximum shear stress τcmax with shear reinforcement (cls. 40.2.3, 40.5.1 and 41.3.1)
Table 20 of IS 456 stipulates the maximum shear stress of reinforced concrete in
beams τcmax as given below in Table 6.2. Under no circumstances, the nominal shear stress in
beams τv shall exceed τcmax given in Table 6.2 for different grades of concrete.
2
Table 3.2 Maximum shear stress, τcmax in N/mm
Critical Section for Shear
Clauses 22.6.2 and 22.6.2.1 stipulate the critical section for shear and are as follows:
For beams generally subjected to uniformly distributed loads or where the principal
load is located further than 2d from the face of the support, where d is the effective depth of
the beam, the critical sections depend on the conditions of supports as shown in Figs 3.3 a, b
and c and are mentioned below.
(i) When the reaction in the direction of the applied shear introduces tension (Fig
3.3a) into the end region of the member, the shear force is to be computed at the face of the
support of the member at that section.
(ii) When the reaction in the direction of the applied shear introduces compression
into the end region of the member (Figs. 6.13.3b and c), the shear force computed at a
distance d from the face of the support is to be used for the design of sections located at a
distance less than d from the face of the support. The enhanced shear strength of sections
close to supports, however, may be considered as discussed in the following section.
Enhanced Shear Strength of Sections Close to Supports (cl. 40.5 of IS 456)
Figure 3.4 shows the shear failure of simply supported and cantilever beams without
o
shear reinforcement. The failure plane is normally inclined at an angle of 30 to the
horizontal. However, in some situations the angle of failure is steeper either due to the
location of the failure section closed to a support or for some other reasons. Under these
situations, the shear force required to produce failure is increased.
Such enhancement of shear strength near a support is taken into account by increasing
the design shear strength of concrete to (2dτc/av) provided that the design shear stress at the
face of the support remains less than the value of τcmax given in Table 3.2 (Table 20 of IS
456). In the above expression of the enhanced shear strength
d = effective depth of the beam,
τc = design shear strength of concrete before the enhancement as given in Table 3.1
(Table 19 of IS 456),
av = horizontal distance of the section from the face of the support (Fig 3.4).
Similar enhancement of shear strength is also to be considered for sections closed to point
loads. It is evident from the expression (2dτc /av) that when av is equal to 2d, the enhanced
shear strength does not come into picture. Further, to increase the effectivity, the tension
reinforcement is recommended to be extended on each side of the point where it is
intersected by a possible failure plane for a distance at least equal to the effective depth, or to
be provided with an equivalent anchorage.
Minimum Shear Reinforcement (cls. 40.3, 26.5.1.5 and 26.5.1.6 of IS 456)
Minimum shear reinforcement has to be provided even when τv is less than τc given in Table
3.1 as recommended in cl. 40.3 of IS 456. The amount of minimum shear reinforcement, as
given in cl. 26.5.1.6, is given below.
The minimum shear reinforcement in the form of stirrups shall be provided such that:
The above provision is not applicable for members of minor structural importance
such as lintels where the maximum shear stress calculated is less than half the permissible
value.
The minimum shear reinforcement is provided for the following:
(i) Any sudden failure of beams is prevented if concrete cover bursts and the bond to
the tension steel is lost.
(ii) Brittle shear failure is arrested which would have occurred without shear
reinforcement.
(iii) Tension failure is prevented which would have occurred due to shrinkage,
thermal stresses and internal cracking in beams.
(iv) To hold the reinforcement in place when concrete is poured.
(v) Section becomes effective with the tie effect of the compression steel.
Further, cl. 26.5.1.5 of IS 456 stipulates that the maximum spacing of shear reinforcement
measured along the axis of the member shall not be more than 0.75 d for vertical stirrups and
o
d for inclined stirrups at 45 , where d is the effective depth of the section. However, the
spacing shall not exceed 300 mm in any case.
3.3 Design of Shear Reinforcement (cl. 40.4 of IS 456)
When τv is more than τc given in Table 6.1, shear reinforcement shall be provided in any of
the three following forms:
(a) Vertical stirrups,
(b) Bent-up bars along with stirrups, and
(c) Inclined stirrups.
In the case of bent-up bars, it is to be seen that the contribution towards shear resistance of
bent-up bars should not be more than fifty per cent of that of the total shear reinforcement.
The amount of shear reinforcement to be provided is determined to carry a shear force Vus
equal to
The following two points are to be noted:
(i) The total shear resistance shall be computed as the sum of the resistance for the
various types separately where more than one type of shear reinforcement is used.
(ii) The area of stirrups shall not be less than the minimum specified in cl. 26.5.1.6.
Shear Reinforcement for Sections Close to Supports
As stipulated in cl. 40.5.2 of IS 456, the total area of the required shear reinforcement
As is obtained from:
Flanged beams
This reinforcement should be provided within the middle three quarters of av, where
av is less than d, horizontal shear reinforcement will be effective than vertical. Alternatively,
one simplified method has been recommended in cl. 40.5.3 of IS 456 and the same is given
below.
The following method is for beams carrying generally uniform load or where the
principal load is located further than 2d from the face of support. The shear stress is
calculated at a section a distance d from the face of support. The value of τc is calculated in
accordance with Table 6.1 and appropriate shear reinforcement is provided at sections closer
to the support. No further check for shear at such sections is required.
Curtailment of Tension Reinforcement in Flexural Members (cl. 26.2.3.2 of IS 456)
Curtailment of tension reinforcement is done to provide the required reduced area of steel
with the reduction of the bending moment. However, shear force increases with the reduction
of bending moment. Therefore, it is necessary to satisfy any one of following three
conditions while terminating the flexural reinforcement in tension zone:
(i) The shear stress τv at the cut-off point should not exceed two-thirds of the permitted
value which includes the shear strength of the web reinforcement. Accordingly,
(ii) For each of the terminated bars, additional stirrup area should be provided over a
distance of three-fourth of effective depth from the cut-off point. The additional stirrup area
shall not be less than 0.4 b s/fy, where b is the breadth of rectangular beams and is replaced
by bw, the breadth of the web for flanged beams, s = spacing of additional stirrups and fy is
2
the characteristic strength of stirrup reinforcement in N/mm . The value of s shall not exceed
d/(8 βb), where βb is the ratio of area of bars cut-off to the total area of bars at that section,
and d is the effective depth.
i.
For bars of diameters 36 mm and smaller, the continuing bars provide double the
area required for flexure at the cut-off point. The shear stress should not exceed
three-fourths that permitted. Accordingly,
Placement of Stirrups
The stirrups in beams shall be taken around the outer-most tension and compression
bars. In T and L-beams, the stirrups will pass around longitudinal bars located close to the
outer face of the flange. In the rectangular beams, two holder bars of diameter 10 or 12 mm
are provided if there is no particular need for compression reinforcement (Fig 3.5).
Problems
1. Determine the shear reinforcement of the simply supported beam of effective span 8
m whose cross section is shown in Fig. 3.6. Factored shear force is 250 kN. Use M 20
and Fe 415.
Fig 3.6
Solution:
3.4 Bond
The bond between steel and concrete is very important and essential so that they can
act together without any slip in a loaded structure. With the perfect bond between them, the
plane section of a beam remains plane even after bending. The length of a member required
to develop the full bond is called the anchorage length. The bond is measured by bond stress.
The local bond stress varies along a member with the variation of bending moment. The
average value throughout its anchorage length is designated as the average bond stress. In our
calculation, the average bond stress will be used.
Thus, a tensile member has to be anchored properly by providing additional length on
either side of the point of maximum tension, which is known as ‘Development length in
tension’. Similarly, for compression members also, we have ‘Development length Ld in
compression’.
It is worth mentioning that the deformed bars are known to be superior to the smooth
mild steel bars due to the presence of ribs. In such a case, it is needed to check for the
sufficient development length Ld only rather than checking both for the local bond stress and
development length as required for the smooth mild steel bars. Accordingly, IS 456, cl. 26.2
stipulates the requirements of proper anchorage of reinforcement in terms of development
length Ld only employing design bond stress τbd.
3.4.1 Design Bond Stress τbd
(a) Definition
The design bond stress τbd is defined as the shear force per unit nominal surface area of
reinforcing bar. The stress is acting on the interface between bars and surrounding concrete
and along the direction parallel to the bars.
This concept of design bond stress finally results in additional length of a bar of specified
diameter to be provided beyond a given critical section. Though, the overall bond failure may
be avoided by this provision of additional development length Ld, slippage of a bar may not
always result in overall failure of a beam. It is, thus, desirable to provide end anchorages also
to maintain the integrity of the structure and thereby, to enable it carrying the loads. Clause
26.2 of IS 456 stipulates, “The calculated tension or compression in any bar at any section
shall be developed on each side of the section by an appropriate development length or end
anchorage or by a combination thereof.”
(b) Design bond stress – values
Section 6.15.1 mentions that the local bond stress varies along the length of the
reinforcement while the average bond stress gives the average value throughout its
development length. This average bond stress is still used in the working stress method and
IS 456 has mentioned about it in cl. B-2.1.2. However, in the limit state method of design,
the average bond stress has been designated as design bond stress τbd and the values are given
in cl. 26.2.1.1. The same is given below as a ready reference.
For deformed bars conforming to IS 1786, these values shall be increased by 60 per cent. For
bars in compression, the values of bond stress in tension shall be increased by 25 per cent.
3.5 Development Length
Fig 3.7 Development length of bar
(a) A single bar
Figure 3.7 (a) shows a simply supported beam subjected to uniformly distributed
load. Because of the maximum moment, the Ast required is the maximum at x = L/2. For any
section 1-1 at a distance x < L/2, some of the tensile bars can be curtailed. Let us then assume
that section 1-1 is the theoretical cut-off point of one bar. However, it is necessary to extend
the bar for a length Ld as explained earlier. Let us derive the expression to determine Ld of
this bar.
Figure 3.7 (b) shows the free body diagram of the segment AB of the bar. At B, the tensile
2
force T trying to pull out the bar is of the value T = (π φ σs /4), where φ is the nominal
diameter of the bar and σs is the tensile stress in bar at the section considered at design loads.
It is necessary to have the resistance force to be developed by τbd for the length Ld to
overcome the tensile force. The resistance force = π φ (Ld) (τbd). Equating the two, we get
The above equation is given in cl. 26.2.1 of IS 456 to determine the development
length of bars.
The example taken above considers round bar in tension. Similarly, other sections of the bar
should have the required Ld as determined for such sections. For bars in compression, the
development length is reduced by 25 per cent as the design bond stress in compression τ is
bd
25 per cent more than that in tension (see the last lines below Table 6.4). Following the same
logic, the development length of deformed bars is reduced by 60 per cent of that needed for
the plain round bars. Tables 64 to 66 of SP-16 present the development lengths of fully
stressed plain and deformed bars (when σs = 0.87 fy) both under tension and compression. It
is to be noted that the consequence of stress concentration at the lugs of deformed bars has
not been taken into consideration.
(b) Bars bundled in contact
The respective development lengths of each of the bars for two, three or four bars in
contact are determined following the same principle. However, cl. 26.2.1.2 of IS 456
stipulates a simpler approach to determine the development length directly under such cases
and the same is given below: “The development length of each bar of bundled bars shall be
that for the individual bar, increased by 10 per cent for two bars in contact, 20 per cent for
three bars in contact and 33 per cent for four bars in contact.”
However, while using bundled bars the provision of cl. 26.1.1 of IS 456 must be satisfied.
According to this clause:
• In addition to single bar, bars may be arranged in pairs in contact or in groups of three
or four bars bundled in contact.
• Bundled bars shall be enclosed within stirrups or ties to ensure the bars remaining
together.
• Bars larger than 32 mm diameter shall not be bundled, except in columns.
Curtailment of bundled bars should be done by terminating at different points spaced
apart by not less than 40 times the bar diameter except for bundles stopping at support (cl.
26.2.3.5 of IS 456).
3.5.1 Checking of Development Lengths of Bars in Tension
The following are the stipulation of cl. 26.2.3.3 of IS 456.
(i) At least one-third of the positive moment reinforcement in simple members and
one-fourth of the positive moment reinforcement in continuous members shall be
extended along the same face of the member into the support, to a length equal to
Ld/3.
(ii) Such reinforcements of (i) above shall also be anchored to develop its design
stress in tension at the face of the support, when such member is part of the
primary lateral load resisting system.
(iii)The diameter of the positive moment reinforcement shall be limited to a diameter
such that the Ld computed for σs = fd in Eq. 6.12 does not exceed the following:
Lo = sum of the anchorage beyond the centre of the support and the equivalent
anchorage value of any hook or mechanical anchorage at simple support. At a point of
inflection, Lo is limited to the effective depth of the member or 12φ, whichever is greater, and
φ = diameter of bar.
It has been further stipulated that M1/V in the above expression may be increased by 30 per
cent when the ends of the reinforcement are confined by a compressive reaction.
Derivation of the Limiting L (Eq. 6.13)
d
Fig 3.8
At the face of simple support and at the points of inflection in continuous beams, the tensile
capacity to be developed is normally small although the rate of change of tensile stress in the
bars is high. These bond stresses are designated as flexural bond stress.
Figure 3.8 (a) shows the tensile bar AB near the support of a beam and Fig. 3.8 (b)
explains the stresses and forces in the free body diagram of the segment CD of the bar. The
tensile force at C (TC) is greater than that at D (TD). Considering z as the lever arm (distance
between the centers of gravity of tensile and compressive force), we have
3.6 Anchoring Reinforcing Bars
Section3.8 (a) mentions that the bars may be anchored in combination of providing
development length to maintain the integrity of the structure. Such anchoring is discussed
below under three sub-sections for bars in tension, compression and shear respectively, as
stipulated in cl. 26.2.2 of IS 456.
Fig 3.9
The salient points are:
• Derformed bars may not need end anchorages if the development length requirement is
satisfied.
• Hooks should normally be provided for plain bars in tension.
• Standard hooks and bends should be as per IS 2502 or as given in Table 67 of SP-16,
which are shown in Figs.3.9 a and b.
• The anchorage value of standard bend shall be considered as 4 times the diameter of the
o
bar for each 45 bend subject to a maximum value of 16 times the diameter of the bar.
• The anchorage value of standard U-type hook shall be 16 times the diameter of the bar.
(b) Bars in compression (cl. 26.2.2.2 of IS 456)
Here, the salient points are:
• The anchorage length of straight compression bars shall be equal to its development
length as mentioned in sec. 6.15.3.
• The development length shall include the projected length of hooks, bends and straight
lengths beyond bends, if provided.
(c) Bars in shear (cl. 26.2.2.4 of IS 456)
Fig 3.10 anchorage of stirrups
The salient points are:
• Inclined bars in tension zone will have the development length equal to that of bars
in tension and this length shall be measured from the end of sloping or inclined portion of the
bar.
• Inclined bars in compression zone will have the development length equal to that of
bars in tension and this length shall be measured from the mid-depth of the beam.
• For stirrups, transverse ties and other secondary reinforcement, complete
development length and anchorage are considered to be satisfied if prepared as shown in
Figs.3.10
Bearing Stresses at Bends (cl. 26.2.2.5 of IS 456)
Change in Direction of Reinforcement (cl. 26.2.2.6 of IS 456)
In some situations, the change in direction of tension or compression reinforcement
induces a resultant force. This force may have a tendency to split the concrete and, therefore,
should be taken up by additional links or stirrups. Normally, this aspect is taken care while
detailing of bars is carried out.
Reinforcement Splicing (cl. 26.2.5 of IS 456)
Reinforcement is needed to be joined to make it longer by overlapping sufficient
length or by welding to develop its full design bond stress. They should be away from the
sections of maximum stress and be staggered. IS 456 (cl. 26.2.5) recommends that splices in
flexural members should not be at sections where the bending moment is more than 50 per
cent of the moment of resistance and not more than half the bars shall be spliced at a section.
(a) Lap Splices (cl. 26.2.5.1 of IS 456)
The following are the salient points:
• They should be used for bar diameters up to 36 mm.
• They should be considered as staggered if the centre to centre distance of the splices is at
least 1.3 times the lap length calculated as mentioned below.
• The lap length including anchorage value of hooks for bars in flexural tension shall be Ld or
30φ, whichever is greater. The same for direct tension shall be 2Ld or 30φ, whichever is
greater.
• The lap length in compression shall be equal to Ld in compression but not less than 24φ.
• The lap length shall be calculated on the basis of diameter of the smaller bar when bars of
two different diameters are to be spliced.
• Lap splices of bundled bars shall be made by splicing one bar at a time and all such
individual splices within a bundle shall be staggered.
(c) Strength of Welds (cl. 26.2.5.2 of IS 456)
The strength of welded splices and mechanical connections shall be taken as 100 per cent
of the design strength of joined bars for compression splices.
For tension splices, such strength of welded bars shall be taken as 80 per cent of the design
strength of welded bars. However, it can go even up to 100 per cent if welding is strictly
supervised and if at any cross-section of the member not more than 20 per cent of the tensile
reinforcement is welded. For mechanical connection of tension splice, 100 per cent of design
strength of mechanical connection shall be taken
Problem2
Determine the anchorage length of 4-20T reinforcing bars going into the support of
the simply supported beam shown in Fig. 3.11 The factored shear force Vu = 280 kN, width
of the column support = 300 mm. Use M 20 concrete and Fe 415 steel.
Fig 3.11
3.7 Torsion
This lesson explains the presence of torsional moment along with bending moment
and shear in reinforced concrete members with specific examples. The approach of design of
such beams has been explained mentioning the critical section to be designed. Expressing the
equivalent shear and bending moment, this lesson illustrates the step by step design
procedure of beam under combined bending, shear and torsion. The requirements of IS 456
regarding the design are also explained. Numerical problems have been solved to explain the
design of beams under combined bending, shear and torsion.
Torsion in Reinforced Concrete Members
Fig 3.12
On several situations beams and slabs are subjected to torsion in addition to bending
moment and shear force. Loads acting normal to the plane of bending will cause bending
moment and shear force. However, loads away from the plane of bending will induce
torsional moment along with bending moment and shear. Space frames (Fig.3.12 a), inverted
L-beams as in supporting sunshades and canopies (Fig.3.12 b), beams curved in plan
(Fig.3.12 c), edge beams of slabs (Fig.6.16.1d) are some of the examples where torsional
moments are also present.
Skew bending theory, space-truss analogy is some of the theories developed to understand
the behavior of reinforced concrete under torsion combined with bending moment and shear.
These torsional moments are of two types:
(i) Primary or equilibrium torsion, and
(ii) Secondary or compatibility torsion.
The primary torsion is required for the basic static equilibrium of most of the
statically determinate structures. Accordingly, this torsional moment must be considered in
the design as it is a major component.
The secondary torsion is required to satisfy the compatibility condition between
members. However, statically indeterminate structures may have any of the two types of
torsions. Minor torsional effects may be ignored in statically indeterminate structures due to
the advantage of having more than one load path for the distribution of loads to maintain the
equilibrium. This may produce minor cracks without causing failure. However, torsional
moments should be taken into account in the statically indeterminate structures if they are of
equilibrium type and where the torsional stiffness of the members has been considered in the
structural analysis. It is worth mentioning that torsion must be considered in structures
subjected to unsymmetrical loadings about axes.
Clause 41 of IS 456 stipulates the above stating that, "In structures, where torsion is
required to maintain equilibrium, members shall be designed for torsion in accordance with
41.2, 41.3 and 41.4. However, for such indeterminate structures where torsion can be
eliminated by releasing redundant restraints, no specific design for torsion is necessary,
provided torsional stiffness is neglected in the calculation of internal forces. Adequate
control of any torsional cracking is provided by the shear reinforcement as per cl. 40".
Analysis for Torsional Moment in a Member
The behaviour of members under the effects of combined bending, shear and torsion
is still a subject of extensive research.
We know that the bending moments are distributed among the sharing members with the
corresponding distribution factors proportional to their bending stiffness EI/L where E is the
elastic constant, I is the moment of inertia and L is the effective span of the respective
members. In a similar manner, the torsional moments are also distributed among the sharing
members with the corresponding distribution factors proportional to their torsional stiffness
GJ/L, where G is the elastic shear modulus, J is polar moment of inertia and L is the effective
span (or length) of the respective members.
The exact analysis of reinforced concrete members subjected to torsional moments
combined with bending moments and shear forces is beyond the scope here. However, the
codal provisions of designing such members are discussed below.
Approach of Design for Combined Bending, Shear and Torsion as per IS 456.
As per the stipulations of IS 456, the longitudinal and transverse reinforcements are
determined taking into account the combined effects of bending moment, shear force and
torsional moment. Two impirical relations of equivalent shear and equivalent bending
moment are given. This fictitious shear force and bending moment, designated as equivalent
shear and equivalent bending moment, are separate functions of actual shear and torsion, and
actual bending moment and torsion, respectively. The total vertical reinforcement is designed
to resist the equivalent shear Ve and the longitudinal reinforcement is designed to resist the
equivalent bending moment Me1 and Me2, as explained in secs. 6.16.6 and 6.16.7,
respectively. These design rules are applicable to beams of solid rectangular cross-section.
However, they may be applied to flanged beams by substituting bw for b. IS 456 further
suggests to refer to specialist literature for the flanged beams as the design adopting the code
procedure is generally conservative.
Critical Section (cl. 41.2 of IS 456).
As per cl. 41.2 of IS 456, sections located less than a distance d from the face of the
support is to be designed for the same torsion as computed at a distance d, where d is the
effective depth of the beam.
3.7.1 Shear and Torsion
(a) The equivalent shear, a function of the actual shear and torsional moment is
determined from the following impirical relation:
(d) Both longitudinal and transverse reinforcement shall be provided as per cl. 41.4
and explained below in sec. 6.16.7, if veτ exceeds cτ given in Table 19 of IS 456 and Table
6.1 of Lesson 13 and is less than maxcτ, as mentioned in (b) above.
Reinforcement in Members subjected to Torsion
(a) Reinforcement for torsion shall consist of longitudinal and transverse
reinforcement as mentioned in sec. 6.16.6(d).
(b) The longitudinal flexural tension reinforcement shall be determined to resist an
equivalent bending moment Me1 as given below:
Me1 = Mu + Mt (6.24)
where Mu = bending moment at the cross-section, and
Mt = (Tu/1.7) {1 + (D/b)} (6.25)
where Tu = torsional moment,
D = overall depth of the beam, and
b = breadth of the beam.
(c) The longitudinal flexural compression reinforcement shall be provided if the numerical
value of Mt as defined above in Eq.6.25 exceeds the numerical value of Mu. Such
compression reinforcement should be able to resist an equivalent bending moment Me2 as
given below:
Me2 = Mt - Mu (6.26)
The Me2 will be considered as acting in the opposite sense to the moment Mu.
b = breadth of the member,
fy = characteristic strength of the stirrup reinforcement,
veτ= equivalent shear stress as specified in Eqs.6.22 and 6.23, and
cτ = shear strength of concrete as per Table 19 of IS 456
Requirements of Reinforcement
Beams subjected to bending moment, shear and torsional moment should satisfy the
following requirements:
(a) Tension reinforcement (cl. 26.5.1.1 of IS 456)
The minimum area of tension reinforcement should be governed by
As /(bd) = 0.85/fy (6.29)
Where As = minimum area of tension reinforcement,
b = breadth of rectangular beam or breadth of web of T-beam,
d = effective depth of beam,
2
fy = characteristic strength of reinforcement in N/mm .
The maximum area of tension reinforcement shall not exceed 0.04 bD, where D is the overall
depth of the beam.
(b) Compression reinforcement (cl. 26.5.1.2 of IS 456)
The maximum area of compression reinforcement shall not exceed 0.04 bD. They shall be
enclosed by stirrups for effective lateral restraint.
(c) Side face reinforcement (cls. 26.5.1.3 and 26.5.1.7b)
Beams exceeding the depth of 750 mm and subjected to bending moment and shear shall
have side face reinforcement. However, if the beams are having torsional moment also, the
side face reinforcement shall be provided for the overall depth exceeding 450 mm. The total
area of side face reinforcement shall be at least 0.1 per cent 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.
(d) Transverse reinforcement (cl. 26.5.1.4 of IS 456)
The transverse reinforcement shall be placed around the outer-most tension and compression
bars. They should pass around longitudinal bars located close to the outer face of the flange
in T- and I-beams.
(e) Maximum spacing of shear reinforcement (cl. 26.5.1.5 of IS 456)
The centre to centre spacing of shear reinforcement shall not be more than 0.75 d for
o
vertical stirrups and d for inclined stirrups at 45 , but not exceeding 300 mm, where d is the
effective depth of the section.
(f) Minimum shear reinforcement (cl. 26.5.1.6 of IS 456)
This has been discussed in sec. 6.13.7 of Lesson 13 and the governing equation is Eq.6.3 of
Lesson 13.
(g) Distribution of torsion reinforcement (cl. 26.5.1.7 of IS 456)
The transverse reinforcement shall consist of rectangular close stirrups placed perpendicular
to the axis of the member. The spacing of stirrups shall not be more than the least of x1, (x1 +
y1)/4 and 300 mm, where x1 and y1 are the short and long dimensions of the stirrups
(Fig.6.16.2).
Longitudinal reinforcements should be placed as close as possible to the corners of the crosssection.
(h) Reinforcement in flanges of T- and L-beams (cl. 26.5.1.8 of IS 456)
For flanges in tension, a part of the main tensile reinforcement shall be distributed over the
effective flange width or a width equal to one-tenth of the span, whichever is smaller. For
effective flange width greater than one-tenth of the span, nominal longitudinal reinforcement
shall be provided to the outer portion of the flange.
Problem 1
Determine the reinforcement required of a ring beam (Fig.6.16.3) of b = 400 mm, d =
650 mm, D = 700 mm and subjected to factored Mu = 200 kNm, factored Tu = 50 kNm and
factored Vu = 100 kN. Use M 20 and Fe 415 for the design.
DESIGN OF RC ELEMENTS
UNIT -4
LIMIT STATE DESIGN OF COLUMNS
CONTENTS
4.1 Introduction
4.2 Classification of columns based on types of reinforcement
4.3 Classification of columns based on loading
4.4 Classification of columns based on slenderness ratio
4.5 Longitudinal reinforcement
4.6 Transverse reinforcement
4.7 Pitch and diameter of lateral ties
4.8 Assumptions in the design of compression members by limit state of collapse
4.9 Design of short columns under axial load with uniaxial bending
4.10 Slender columns
4.11 Design of slender columns
TECHNICAL TERMS
Columns are the vertical members, normally subjected to compression.
Axial loaded column: when the line of action of load passes through the centre of gravity of
column, it is known as axially loaded column. In this case the column is subjected to direct
compression.
Eccentrically loaded column: when the line of action of load passes away from the centre of
gravity of column, it is known as eccentrically loaded column. In this case the column is
subjected to a bending moment in addition to direct compression.
Column bases or column footings: foundations provided for column are known as column
bases or column footings. The column bases distribute the load over a larger area so that the
intensity of pressure on soil does not exceed the safe bearing capacity of soil and settlement
of structure is kept within the permissible limit.
UNIT -4
LIMIT STATE DESIGN OF COLUMNS
4.1 Introduction
Compression members are structural elements primarily subjected to axial
compressive forces and hence, their design is guided by considerations of strength and
buckling. Figures 4.1a to c show their examples: pedestal, column, wall and strut. While
pedestal, column and wall carry the loads along its length l in vertical direction, the strut in
truss carries loads in any direction. The letters l, b and D represent the unsupported vertical
length, horizontal lest lateral dimension, width and the horizontal longer lateral dimension,
depth. These compression members may be made of bricks or reinforced concrete. Herein,
reinforced concrete compression members are only discussed.
Fig 4.1 pedestal, column and wall
This module is intended to explain the definition of some common terminologies and
to illustrate the design of compression members and other related issues. This lesson,
however, explain the definitions and classifications of columns depending on different
aspects. Further, the recommendations of IS 456 to be followed in the design are discussed
regarding the longitudinal and lateral reinforcing bars. The assumptions made in the design
of compression member by limit sate of collapse are illustrated.
Definitions
(a) Effective length: The vertical distance between the points of inflection of the
compression member in the buckled configuration in a plane is termed as effective length le
of that compression member in that plane. The effective length is different from the
unsupported length l of the member, though it depends on the unsupported length and the
type of end restraints. The relation between the effective and unsupported lengths of any
compression member is
le = k l (10.1)
Where, k is the ratio of effective to the unsupported lengths. Clause 25.2 of IS 456 stipulates
the effective lengths of compression members (vide Annex E of IS 456). This parameter is
needed in classifying and designing the compression members.
(b) Pedestal: Pedestal is a vertical compression member whose effective length le does not
exceed three times of its least horizontal dimension b (cl. 26.5.3.1h, Note). The other
horizontal dimension D shall not exceed four times of b (Fig 4.1a).
(c) Column: Column is a vertical compression member whose unsupported length l shall not
exceed sixty times of b (least lateral dimension), if restrained at the two ends. Further, its
2
unsupported length of a cantilever column shall not exceed 100b /D, where D is the larger
lateral dimension which is also restricted up to four times of b (vide cl. 25.3 of IS 456 and
Fig.4.1b).
(d) Wall: Wall is a vertical compression member whose effective height Hwe to
thickness t (least lateral dimension) shall not exceed 30 (cl. 32.2.3 of IS 456). The larger
horizontal dimension i.e., the length of the wall L is more than 4t (Fig.4.1c).
4.2 Classification of Columns Based on Types of Reinforcement
Fig 4.2
Based on the types of reinforcement, the reinforced concrete columns are classified into three
groups:
(i) Tied columns: The main longitudinal reinforcement bars are enclosed within closely
spaced lateral ties (Fig 4.2 a).
(ii) Columns with helical reinforcement: The main longitudinal reinforcement bars are
enclosed within closely spaced and continuously wound spiral reinforcement. Circular and
octagonal columns are mostly of this type (Fig 4.2 b).
(iii) Composite columns: The main longitudinal reinforcement of the composite columns
consists of structural steel sections or pipes with or without longitudinal bars (Fig.4.2c and
d).
Out of the three types of columns, the tied columns are mostly common with different
shapes of the cross-sections viz. square, rectangular, T-, L-, cross etc. Helically bound
columns are also used for circular or octagonal shapes of cross-sections. Architects prefer
circular columns in some specific situations for the functional requirement. This module,
accordingly takes up these two types (tied and helically bound) of reinforced concrete
columns.
4.3 Classification of Columns Based on Loadings
Fig 4.3
Columns are classified into the three following types based on the loadings:
(b) Columns subjected to axial loads only (concentric), as shown in Fig.4.3a.
(c) Columns subjected to combined axial load and uniaxial bending, as shown in
Fig.4.3b.
(d) Columns subjected to combined axial load and bi-axial bending, as shown in
Fig.4.3c.
Fig 4.4
Figure 4.4 shows the plan view of a reinforced concrete rigid frame having columns
and inter-connecting beams in longitudinal and transverse directions. From the knowledge of
structural analysis it is well known that the bending moments on the left and right of columns
for every longitudinal beam will be comparable as the beam is continuous. Similarly, the
bending moments at the two sides of columns for every continuous transverse beam are also
comparable (neglecting small amounts due to differences of l1, l2, l3 and b1, b2, b3, b4).
Therefore, all internal columns (C1a to C1f) will be designed for axial force only. The side
columns (C2a to C2j) will have axial forces with uniaxial bending moment, while the four
corner columns (C3a to C3d) shall have axial forces with bi-axial bending moments. Thus,
all internal columns (C1a to C1f), side columns (C2a to C2j) and corner columns (C3a to
C3d) are the columns of type (i), (ii) and (iii), respectively.
It is worth mentioning that pure axial forces in the inside columns is a rare case. Due
to rigid frame action, lateral loadings and practical aspects of construction, there will be
bending moments and horizontal shear in all the inside columns also. Similarly, side columns
and corner columns will have the column shear along with the axial force and bending
moments in one or both directions, respectively. The effects of shear are usually neglected as
the magnitude is very small. Moreover, the presence of longitudinal and transverse
reinforcement is sufficient to resist the effect of column shear of comparatively low
magnitude. The effect of some minimum bending moment, however, should be taken into
account in the design even if the column is axially loaded. Accordingly, cls. 39.2 and 25.4 of
IS 456 prescribes the minimum eccentricity for the design of all columns. In case the actual
eccentricity is more than the minimum that should be considered in the design.
4.4 Classification of Columns Based on Slenderness Ratios
Columns are classified into the following two types based on the slenderness ratios:
(i) Short columns
(ii) Slender or long columns
Fig 4.5 Modes of failure of columns
Figure 4.5 presents the three modes of failure of columns with different slenderness
ratios when loaded axially. In the mode 1, column does not undergo any lateral deformation
and collapses due to material failure. This is known as compression failure. Due to the
combined effects of axial load and moment a short column may have material failure of
mode 2. On the other hand, a slender column subjected to axial load only undergoes
deflection due to beam-column effect and may have material failure under the combined
action of direct load and bending moment. Such failure is called combined compression and
bending failure of mode 2. Mode 3 failure is by elastic instability of very long column even
under small load much before the material reaches the yield stresses. This type of failure is
known as elastic buckling.
The slenderness ratio of steel column is the ratio of its effective length l to its least
e
radius of gyration r. In case of reinforced concrete column, however, IS 456 stipulates the
slenderness ratio as the ratio of its effective length le to its least lateral dimension. As
mentioned earlier in sec. 4.2(a), the effective length le is different from the unsupported
length, the rectangular reinforced concrete column of cross-sectional dimensions b and D
shall have two effective lengths in the two directions of b and D. Accordingly, the column
may have the possibility of buckling depending on the two values of slenderness ratios as
given below:
Slenderness ratio about the major axis = lex/D
Slenderness ratio about the minor axis = ley/b
Based on the discussion above, cl. 25.1.2 of IS 456 stipulates the following:
A compression member may be considered as short when both the slenderness ratios
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 compression member.
Further, it is essential to avoid the mode 3 type of failure of columns so that all
columns should have material failure (modes 1 and 2) only. Accordingly, cl. 25.3.1 of IS 456
stipulates the maximum unsupported length between two restraints of a column to sixty times
its least lateral dimension. For cantilever columns, when one end of the column is
2
unrestrained, the unsupported length is restricted to 100b /D where b and D are as defined
earlier.
Fig 4.6 bracings of columns
It is desirable that the columns do not have to resist any horizontal loads due to wind
or earthquake. This can be achieved by bracing the columns as in the case of columns of a
water tank or tall buildings (Figs 4.6a and b). Lateral tie members for the columns of water
tank or shear walls for the columns of tall buildings resist the horizontal forces and these
columns are called braced columns. Unbraced columns are supposed to resist the horizontal
loads also. The bracings can be in one or more directions depending on the directions of the
lateral loads. It is worth mentioning that the effect of bracing has been taken into account by
the IS code in determining the effective lengths of columns (vide Annex E of IS 456).
4.5 Longitudinal Reinforcement
The longitudinal reinforcing bars carry the compressive loads along with the concrete.
Clause 26.5.3.1 stipulates the guidelines regarding the minimum and maximum amount,
number of bars, minimum diameter of bars, spacing of bars etc. The following are the salient
points:
(a) The minimum amount of steel should be at least 0.8 per cent of the gross crosssectional area of the column required if for any reason the provided area is more than the
required area.
(b) The maximum amount of steel should be 4 per cent of the gross cross-sectional
area of the column so that it does not exceed 6 per cent when bars from column below have
to be lapped with those in the column under consideration.
(c) Four and six are the minimum number of longitudinal bars in rectangular and
circular columns, respectively.
(d) The diameter of the longitudinal bars should be at least 12 mm.
(e) Columns having helical reinforcement shall have at least six longitudinal bars
within and in contact with the helical reinforcement. The bars shall be placed equidistant
around its inner circumference.
(f) The bars shall be spaced not exceeding 300 mm along the periphery of the column.
(g) The amount of reinforcement for pedestal shall be at least 0.15 per cent of the
cross-sectional area provided.
4.6 Transverse Reinforcement
Transverse reinforcing bars are provided in forms of circular rings, polygonal links
o
(lateral ties) with internal angles not exceeding 135 or helical reinforcement. The transverse
reinforcing bars are provided to ensure that every longitudinal bar nearest to the compression
face has effective lateral support against buckling. Clause 26.5.3.2 stipulates the guidelines of
the arrangement of transverse reinforcement. The salient points are:
Fig 4.7
(a) Longitudinal bars spaced at a maximum distance of 48 times the diameter of the
tie shall be tied by single tie and additional open ties for in between longitudinal bars
(Fig.4.7).
Fig 4.8
(b) For longitudinal bars placed in more than one row (Fig.4.8): (i) transverse
reinforcement is provided for the outer-most row in accordance with (a) above, and
(ii) number of bars of the inner row is closer to the nearest compression face than
three times the diameter of the largest bar in the inner row.
Fig 4.9
(c) For longitudinal bars arranged in a group such that they are not in contact and each
group is adequately tied as per (a), (b) or (c) above, as appropriate, the transverse
reinforcement for the compression member as a whole may be provided assuming
that each group is a single longitudinal bar for determining the pitch and diameter of
the transverse reinforcement as given in sec.10.21.9. The diameter of such transverse
reinforcement should not, however, exceed 20 mm (Fig.4.9).
4.7 Pitch and Diameter of Lateral Ties
(a) Pitch: The maximum pitch of transverse reinforcement shall be the least of the following:
(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.
(b) 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
(a) 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 spiral bar. The pitch of helical reinforcement shall be determined as given in
sec.10.21.9 for all cases except where an increased load on the column is allowed for on the
strength of the helical reinforcement. In such cases only, the maximum pitch shall be the
lesser of 75 mm and one-sixth of the core diameter of the column, and the minimum pitch
shall be the lesser of 25 mm and three times the diameter of the steel bar forming the helix.
(b) Diameter: The diameter of the helical reinforcement shall be as mentioned in
sec.10.21.9b.
4.8 Assumptions in the Design of Compression Members by Limit State of
Collapse
It is thus seen that reinforced concrete columns have different classifications
depending on the types of reinforcement, loadings and slenderness ratios. Detailed designs of
all the different classes are beyond the scope here. Tied and helically reinforced short and
slender columns subjected to axial loadings with or without the combined effects of uniaxial
or biaxial bending will be taken up. However, the basic assumptions of the design of any of
the columns under different classifications are the same. Furthermore, the following are the
additional assumptions for the design of compression members (cl. 39.1 of IS 456).
(i) The maximum compressive strain in concrete in axial compression is taken as 0.002.
(ii) 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.
Fig 4.10 strain profiles for different positions of neutral axis
The assumptions (i) to (v) of section 3.4.2 of Lesson 4 and (i) and (ii) mentioned
above are discussed below with reference to Fig.4.10 to c presenting the cross-section and
strain diagrams for different location of the neutral axis.
The discussion made in sec. 3.4.2 of Lesson 4 regarding the assumptions (i), (iii), (IV) and
(v) are applicable here also. Assumption (ii) of sec.3.4.2 is also applicable here when kD, the
depth of neutral axis from the highly compressed right edge is within the section i.e., k < 1.
The corresponding strain profile IN in Fig.4.10 b is for particular value of P and M such that
the maximum compressive strain is 0.0035 at the highly compressed right edge and tensile
strain develops at the opposite edge. This strain profile is very much similar to that of a beam
in flexure.
The additional assumption (i) of this section refers to column subjected axial load P only
resulting compressive strain of maximum (constant) value of 0.002 and for which the strain
profile is EF in Fig.4.10 b.The neutral axis is at infinity (outside the section).
Extending the assumption of the strain profile IN (Fig.4.10 b), we can draw another strain
profile IH (Fig.4.10 c) having maximum compressive strain of 0.0035 at the right edge and
zero strain at the left edge. This strain profile 1H along with EF are drawn in Fig.4.10 c to
intersect at V. From the two similar triangles EVI and GHI, we have
EV/GH = 0.0015/0.0035 = 3/7, which gives
EV = 3D/7 (10.2)
The point V, where the two profiles intersect is assumed to act as a fulcrum for the
strain profiles when the neutral axis lies outside the section. Another strain profile JK drawn
on this figure passing through the fulcrum V and whose neutral axis is outside the section.
The maximum compressive strain GJ of this profile is related to the minimum compressive
strain HK as explained below.
GJ = GI – IJ = GI – 0.75 HK, as we can write IJ in term of HK from two similar triangles JVI
and HVK:
IJ/HK = VE/VF = 0.75.
The value of the maximum compressive strain GJ for the profile JK is, therefore,
0.0035 minus 0.75 times the strain HK on the least compressed edge. This is the assumption
(ii) of this section (cl. 39.1b of IS 456).
Minimum Eccentricity
Section 10.21.4 illustrates that in practical construction, columns are rarely truly
concentric. Even a theoretical column loaded axially will have accidental eccentricity due to
inaccuracy in construction or variation of materials etc. Accordingly, all axially loaded
columns should be designed considering the minimum eccentricity as stipulated in cl. 25.4 of
IS 456 and given below (Fig.10.21.3c)
ex min ≥ greater of) l/500 + D/30) or 20 mm
ey min ≥ greater of) l/500 + b/30) or 20 mm
Where l, D and b are the unsupported length, larger lateral dimension and least lateral
dimension respectively.
4.9 Design of Short columns under Axial Load with Uniaxial Bending
Problem1:
Design a short spiral column subjected to Pu = 2100 kN and Mu = 187.5 kNm using M 25 and
Fe 415. The preliminary diameter of the column may be taken as 500 mm.
Solution 1:
Step 1: Selection of design chart
2
With the given fy = 415 N/mm and assuming d’/D = 0.1, the chart selected for this problem is
Chart 56.
Step 2: Determination of the percentage of longitudinal steel
2
With the given fck = 25 N/mm and assuming the given D = 500 mm, we have:
2
P /f D = 2100000/25(500)(500) = 0.336, and
u ck
3
6
Mu/fckD = 187.5(10 )/25(500)(500)(500) = 0.06
2
3
The particular point A (Fig.10.25.5) having coordinates of Pu/fckD = 0.336 and Mu/fckD =
0.06 in Chart 56 gives: p/fck = 0.08. Hence, p = 0.08(25) = 2 per cent (see Eq.10.54).
Asc = 0.02(π)(500)(500)/4 = 3928.57 mm
2
2
Provide 8-25 mm diameter bars to have Asc actually provided = 3927 mm . Marginally less
amount of steel than required will be checked considering the enhancement of strength for
spiral columns as stipulated in cl.39.4 of IS 456.
Step 3: Design of transverse reinforcement
Fig 4.11
The diameter of the helical reinforcement is taken as 8 mm (> 25 mm/4). The pitch p of the
spiral is determined from Eq.4.1 which satisfies the stipulation in cl.39.4.1 of IS 456. From
Eq.4.1, we have the pitch of the spiral p as:
2
p ≤ 11.1(Dc - spφ) asp fy/(D - f)2cDck …. (4.1)
2
2
where, Dc = 500 – 40 – 40 = 420 mm, D = 500 mm, fck = 25 N/mm , fy = 415 N/mm , spφ = 8
2
mm and asp = 50 mm .
Using the above values in Eq.4.1, we have p ≤ 25.716 mm. As per cl.26.5.3.2d1, regarding
the pitch of spiral: p >/ 420/6 (= 70 mm), p </ 25 mm and p </ 24 mm. So, pitch of the spiral
= 25 mm is o.k. Figure 4.11 presents the cross-section with reinforcing bars of the column.
Step 4: Revision of the design, if required
Providing 25 mm diameter longitudinal steel bars and 8 mm diameter spirals, we have
d’ = 40 + 8 + 12.5 = 60.5 mm. This gives d’/D = 60.5/500 = 0.121. In step 1, d’/D is assumed
as 0.1. So, the revision of the design is needed.
However, as mentioned in step 2, the area of steel required is not provided and this may be
offset considering the enhanced strength of the spiral column, as stipulated in cl.39.4 of IS
456.
We, therefore, assess the strength of the designed column, when d’/D = 0.121 and Asc = 3927
2
mm , if it can be subjected to Pu = 2100 KN and Mu = 187.5 kNm.
For the purpose of assessment, we determine the capacity Pu of the column when Mu = 187.5
kNm. Further, the revised d’/D = 0.121 needs to interpolate the values from Charts 56 (for
3
d’/D = 0.1) and 57 (for d’/D = 0.15). The value of p/fck = 0.08 and Mu/fckbD = 0.06. Table 4.1
presents the results.
2
3
Table 4.1: Value of Pu/fckbD when Mu/fckD = 0.06 and p/fck = 0.08
Sl.No.
d’/D
1
0.1
0.336 (from Chart 56)
2
0.15
0.30 (from Chart 57)
3
0.121
0.32088 (Interpolated value)
2
Pu/fckD
From Table 4.1, thus, we get,
2
P /f D = 0.32088, which gives P = (0.32088) (25)(500)(500) = 2005.5 kN.
u ck
u
Considering the enhanced strength as 1.05 times as per cl.39.4 of IS 456, the actual capacity
of this column is (1.05) (2005.5) = 2105 KN > 2100 kN.
Thus, the design is safe to carry Pu = 2100 KN and Mu = 187.5 kNm.
4.10 Slender Columns
Slender and short are the two types of columns classified on the basis of slenderness
ratios Columns having both lex/D and ley/b less than twelve are designated as short and
otherwise, they are slender, where lex and ley are the effective lengths with respect to major
and minor axes, respectively; and D and b are the depth and width of rectangular columns,
respectively. Short columns are frequently used in concrete structures; however, slender
columns are also becoming increasingly important and popular because of the following
reasons:
(i) The development of high strength materials (concrete and steel),
(ii) Improved methods of dimensioning and designing with rational and reliable
design procedures,
(iii) Innovative structural concepts – specially, the architect’s expectations for
creative structures.
Accordingly, this lesson explains first, the behavior of slender elastic columns loaded
concentrically. Thereafter, reinforced concrete slender columns loaded concentrically or
eccentrically about one or both axes are taken up. The design of slender columns has been
explained and illustrated with numerical examples for easy understanding.
4.11 Design of Slender Columns
The design of slender columns, in principle, is to be done following the same
procedure as those of short columns. However, it is essential to estimate the total moment
i.e., primary and secondary moments considering P- effects. These secondary moments and
axial forces can be determined by second-order rigorous structural analysis – particularly for
unbraced frames. Further, the problem becomes more involved and laborious as the principle
of superposition is not applicable in second-order analysis. Δ
However, cl.39.7 of IS 456 recommends an alternative simplified method of
determining additional moments to avoid the laborious and involved second-order analysis.
The basic principle of additional moment method for estimating the secondary moments is
explained in the next section.
Additional Moment Method
In this method, slender columns should be designed for biaxial eccentricities which
include secondary moments (Py of Eq.10.63 and 10.65) about major and minor axes. We first
consider braced columns which are bent symmetrically in single curvature and cause
balanced failure i.e., Pu = Pub.
(A) Braced columns bent symmetrically in single curvature and undergoing balanced
failure
For braced columns bent symmetrically in single curvature,
M = Mo + Py = Mo + P ea = Mo + Ma (4.2)
where P is the factored design load Pu, M are the total factored design moments Mux and Muy
about the major and minor axes, respectively; Mo are the primary factored moments Moux and
Mouy about the major and minor axes, respectively; Ma are the additional moments Max and
May about the major and minor axes, respectively and ea are the additional eccentricities eax
and eay along the minor and major axes, respectively. The quantities Mo and P of Eq.4.2 are
known and hence, it is required to determine the respective values of ea, the additional
eccentricities only.
Let us consider the columns of Figs.10.27.10 and 11 showing as the maximum deflection at
the mid-height section of the columns. The column of Fig.10.27.10, having a constant
primary moment MΔo, causes constant curvature φ, while the column of Fig.10.27.11, having
a linearly varying primary moment with a maximum value of Mo max at the mid-height section
of the column, has a linearly varying curvature with the maximum curvature of φ max at the
mid-height section the column. The two maximum curvatures can be expressed in terms of
their respective maximum deflection Δ as follows:
The constant curvature (Fig.10.27.10) (10.79) 2max/8 l e Δ=φ
The linearly varying curvature (Fig.10.27.11) (10.80) 2max/12 l e Δ=φ
Where, le is the respective effective lengths of the columns. We, therefore, consider the
maximum φ as the average value lying in between the two values of Eqs.10.79 and 80 as
(10.81) 2max/10 elΔ=φ
Accordingly, the maximum additional eccentricities ea, which are equal to the maximum
deflections Δ, can be written as
ea = = (10.82) Δ /10 2elφ
Fig 4.12
Assuming the column undergoes a balanced failure when Pu = Pub, the maximum
curvature at the mid-height section of the column, shown in Figs.4.12 a and b, can be
expressed as given below, assuming (i) the values of cε = 0.0035, stε = 0.002 and Dd/′ = 0.1,
and (ii) the additional moment capacities are about eighty per cent of the total moment.
φ = 80% of {(0.0035 + 0.002)/0.9D} (see Fig.4.12 c)
Or φ = 1/200D (4.3)
Substituting the value of φ in Eq.10.82,
2
ea = D(le/D) /2000 (4.4)
Therefore, the additional moment Ma can be written as,
2
Ma = Py = PΔ = Pea = (PD/2000) (le/D) (4.5)
Thus, the additional moments Max and May about the major and minor axes, respectively, are:
2
Max = (PuD/2000) (lex/D) (4.6)
2
May = (Pub/2000) (ley/b) (4.7)
Where Pu = axial load on the member,
l = effective length in respect of the major axis,
ex
ley = effective length in respect of the minor axis,
D = depth of the cross-section at right angles to the major axis, and
b = width of the member.
Clause 39.7.1 of IS 456 recommends the expressions of Eqs.4.6 and 4.7 for
estimating the additional moments Max and May for the design. These two expressions of the
additional moments are derived considering the columns to be braced and bent symmetrically
undergoing balanced failure. Therefore, proper modifications are necessary for different
situations like braced columns with unequal end moments with the same or different signs,
unbraced columns and columns causing compression failure i.e., when Pu > Pub.
(B) Braced columns subjected to unequal primary moments at the two ends
For braced columns without any transverse loads occurring in the height, the primary
maximum moment (Mo max of Eq.10.64), with which the additional moments of Eqs.10.86 and
87 are to be added, is to be taken as:
Mo max = 0.4 M1 + 0.6 M2 (4.8) and
Further Mo max ≥ 0.4 M2 (4.9)
Where M2 is the larger end moment and M1 is the smaller end moment, assumed to be
negative, if the column is bent in double curvature.
To eliminate the possibility of total moment Mu
max
becoming less than M2 for
columns bent in double curvature (see Fig.10.27.12) with M1 and M2 having opposite signs,
another condition has been imposed as
Mu max M≥2 (4.10)
The above recommendations are given in notes of cl.39.7.1 of IS 456.
(C) Unbraced columns
Unbraced frames undergo considerable deflection due to P- effect. The additional
moments determined from Eqs.10.86 and 87 are to be added with the maximum primary
moment MΔo max at the ends of the column. Accordingly, we have
Mo max = M2 + Ma (4.11)
The above recommendation is given in the notes of cl.39.7.1 of IS 456.
(D) Columns undergoing compression failure (Pu > Pub)
It has been mentioned in part A of this section that the expressions of additional
moments given by Eqs.4.6 and 4.7 are for columns undergoing balanced failure. However,
when the column causes compression failure, the e/D ratio is much less than that of balanced
failure at relatively high axial loads. The entire section may be under compression causing
much less curvatures. Accordingly, additional moments of Eqs 4.6 and 4.7 are to be modified
by multiplying with the reduction factor k as given below:
(i) For Pu > Pubx: kax = (Puz – Pu)/(Puz – Pubx) (4.12)
(ii) For Pu > Puby: kay = (Puz – Pu)/(Puz – Puby) (4.13)
With a condition that kax and kay should be ≤ 1 (4.14)
Where Pu = axial load on compression member
Puz is,
Puz = 0.45 fck Ac + 0.75 fy Ast … (4.15)
Pubx, Puby = axial loads with respect to major and minor axes, respectively,
corresponding to the condition of maximum compressive strain of 0.0035 in concrete and
tensile strain of 0.002 in outermost layer of tension steel.
It is seen from Eqs.4.12 and 4.13 that the values of k (kax and kay) vary linearly from
zero (when Pu = Puz) to one (when Pu = Pub). Since Eqs.10.92 and 10.93 are not applicable
for Pu < Pub, another condition has been imposed as given in Eq.4.14
The above recommendations are given in cl.39.7.1.1 of IS 456.
The following discussion is very important for the design of slender columns.
Additional moment method is one of the methods of designing slender columns as
discussed in A to D of this section. This method is recommended in cl.39.7 of IS 456 also.
The basic concept here is to enhance the primary moments by adding the respective
additional moments estimated in a simple way avoiding laborious and involved calculations
of second-order structural analysis. However, these primary moments under eccentric
loadings should not be less than the moments corresponding to the respective minimum
eccentricity, as stipulated in the code. Hence, the primary moments in such cases are to be
replaced by the minimum eccentricity moments. Moreover, all slender columns, including
those under axial concentric loadings, are also to be designed for biaxial bending, where the
primary moments are zero. In such cases, the total moment consisting of the additional
moment multiplied with the modification factor, if any, in each direction should be equal to
or greater than the respective moments under minimum eccentricity conditions. As
mentioned earlier, the minimum eccentricity consideration is given in cl.25.4 of IS 456.
Problem 1:
Determine the reinforcement required for a braced column against side sway with the
following data: size of the column = 350 x 450 mm (Fig.4.13); concrete and steel grades = M
30 and Fe 415, respectively; effective lengths lex and ley = 7.0 and 6.0 m, respectively;
unsupported length l = 8 m; factored load Pu = 1700 kN; factored moments in the direction of
larger dimension = 70 kNm at top and 30 kNm at bottom; factored moments in the direction
of shorter dimension = 60 kNm at top and 30 kNm at bottom. The column is bent in double
curvature. Reinforcement will be distributed equally on four sides.
Fig 4.13
Solution 1:
Step 1: Checking of slenderness ratios
lex/D = 7000/450 = 15.56 > 12,
ley/b = 6000/350 = 17.14 > 12.
Hence, the column is slender with respect to both the axes.
Step 2: Minimum eccentricities and moments due to minimum eccentricities (Eq.4.3)
ex min = l/500 + D/30 = 8000/500 + 450/30 = 31.0 > 20 mm
ey min = l/500 + b/30 = 8000/500 + 350/30 = 27.67 > 20 mm
-3
Mox (Min. ecc.) = Pu(ex min) = (1700) (31) (10 ) = 52.7 kNm
-3
Moy (Min. ecc.) = Pu(ey min) = (1700) (27.67) (10 ) = 47.04 kNm
Step 3: Additional eccentricities and additional moments
Method 1: Using Eq. 4.4
2
2
eax = D(lex/D) /2000 = (450) (7000/450) /2000 = 54.44 mm
2
2
eay = b(lex/b) /2000 = (350) (6000/350) /2000 = 51.43 mm
-3
Max = Pu(eax) = (1700) (54.44) (10 ) = 92.548 kNm
-3
May = Pu(eay) = (1700) (51.43) (10 ) = 87.43 kNm
Method 2: Table I of SP-16
For lex/D = 15.56, Table I of SP-16 gives:
eax/D = 0.1214, which gives eax = (0.1214) (450) = 54.63 mm
For l /D = 17.14, Table I of SP-16 gives:
ey
eay/b = 0.14738, which gives eay = (0.14738) (350) = 51.583 mm
It is seen that values obtained from Table I of SP-16 are comparable with those obtained by
Eq. 4.4 in Method 1.
Step 4: Primary moments and primary eccentricities (Eqs.4.8 and 4.9)
Mox = 0.6M2 – 0.4M1 = 0.6(70) – 0.4(30) = 30 kNm, which should be ≥ 0.4 M2 (= 28 kNm).
Hence, o.k.
Moy = 0.6M2 – 0.4M1 = 0.6(60) – 0.4(30) = 24 kNm, which should be ≥ 0.4 M2 (= 24
kNm). Hence, o.k.
Primary eccentricities:
3
e = M /P = (30/1700) (10 ) = 17.65 mm
x
ox
u
3
ey = Moy/Pu = (24/1700) (10 ) = 14.12 mm
Since, both primary eccentricities are less than the respective minimum eccentricities (see
Step 2), the primary moments are revised to those of Step 2. So, Mox = 52.7 kNm and Moy =
47.04 kNm.
Step 5: Modification factors
To determine the actual modification factors, the percentage of longitudinal
reinforcement should be known. So, either the percentage of longitudinal reinforcement may
be assumed or the modification factor may be assumed which should be verified
subsequently. So, we assume the modification factors of 0.55 in both directions.
Step 6: Total factored moments
Mux = Mox + (Modification factor) (Max) = 52.7 + (0.55) (92.548)
= 52.7 + 50.9 = 103.6 kNm
Muy = Moy + (Modification factor) (May) = 47.04 + (0.55) (87.43)
= 47.04 + 48.09 = 95.13 kNm
Step 7: Trial section
The trial section is determined from the design of uniaxial bending with Pu = 1700 kN
2
2 1/2
and Mu = 1.15 . So, we have M2/122) (uyuxMM+u = (1.15){(103.6) + (95.13) }
= 161.75
kNm. With these values of Pu (= 1700 kN) and Mu (= 161.75 kNm), we use chart of SP-16 for
the Dd/′ = 0.134. We assume the diameters of longitudinal bar as 25 mm, diameter of lateral
tie = 8 mm and cover = 40 mm, to get = 40 + 8 + 12.5 = 60.5 mm. Accordingly, d′Dd/′ =
60.5/450 = 0.134 and = 60.5/350 = 0.173. We have: bd/′
3
Pu/fck bD = 1700(10 )/(30)(350)(450) = 0.3598
2
6
Mu/fck bD = 161.75(10 )/(30)(350)(450)(450) = 0.076
We have to interpolate the values of p/fck for Dd/′ = 0.134 obtained from Charts 44 (for = 0.1)
and 45 (Dd/′Dd/′ = 0.15). The values of p/fck are 0.05 and 0.06 from Charts 44 and 45,
respectively. The corresponding values of p are 1.5 and 1.8 per cent, respectively. The
interpolated value of p for = 0.134 is 1.704 per cent, which gives ADd/′sc = (1.704)
2
2
(350)(450)/100 = 2683.8 mm . We use 4-25 + 4-20 (1963 + 1256 = 3219 mm ), to have p
provided = 2.044 per cent giving p/fck = 0.068.
Step 8: Calculation of balanced loads Pb
The values of P and P are determined using Table 60 of SP-16. For this purpose,
bx
by
two parameters k1 and k2 are to be determined first from the table. We have p/fck = 0.068, =
0.134 and Dd/′bd/′ = 0.173. From Table 60, k1 = 0.19952 and k2 = 0.243 (interpolated for Dd/′
= 0.134) for Pbx. So, we have: Pbx/fckbD = k1 + k2 (p/fck) = 0.19952 + 0.243(0.068) =
-3
0.216044, which gives Pbx = 0.216044(30) (350)(450)(10 ) = 1020.81 kN.
Similarly, for Pby: = 0.173, p/fbd/′ck = 0.068. From Table 60 of SP-16, k1 = 0.19048 and k2 =
0.1225 (interpolated for bd/′ = 0.173). This gives Pby/fckbD = 0.19048 + 0.1225(0.068) =
-3
0.19881, which gives Pby = (0.19881)(30)(350)(450)(10 ) = 939.38 kN.
Since, the values of Pbx and Pby are less than Pu, the modification factors are to be used.
Step 9: Determination of Puz
Method 1:
Puz = 0.45 fck Ag + (0.75 fy – 0.45 fck) Asc
= 0.45(30)(350)(450) + {0.75(415) – 0.45(30)}(3219) = 3084.71 kN
Method 2: Using Chart 63 of SP-16
2
We get Puz/Ag = 19.4 N/mm from Chart 63 of SP-16 using p = 2.044 per cent. Therefore, Puz
-3
= (19.4)(350)(450)(10 ) = 3055.5 kN, which is in good agreement with that of Method 1.
Step 10: Determination of modification factors
Method 1: From Eqs.4.12 and 4.13
kax = (Puz – Pu)/(Puz – Pubx) … (4.12)
or kax = (3084.71 – 1700)/(3084.71 – 1020.81) = 0.671 and
k = (P – P )/(P – P ) … (4.13)
ay
uz
u
uz
uby
or kay = (3084.71 – 1700)/(3084.71 – 939.39) = 0.645
The values of the two modification factors are different from the assumed value of 0.55 in
Step 5. However, the moments are changed and the section is checked for safety.
Method 2: From Chart 65 of SP-16
From Chart 65 of SP-16, for the two parameters, Pbx/Puz = 1020.81/3084.71 = 0.331 and
Pu/Puz = 1700/3084.71 = 0.551, we get kax = 0.66. Similarly, for the two parameters, Pby/Puz =
939.38/3084.71 = 0.3045 and Pu/Puz = 0.551, we have kay = 0.65. Values of kax and kay are
comparable with those of Method 1.
Step 11: Total moments incorporating modification factors
M = M (from Step 4) + (k ) M (from Step 3)
ux
ox
ax
ax
= 52.7 + 0.671(92.548) = 114.8 kNm
Muy = Moy (from Step 4) + kay (May) (from Step 3)
= 47.04 + (0.645)(87.43) = 103.43 kNm.
Step 12: Uniaxial moment capacities
The two uniaxial moment capacities Mux1 and Muy1 are determined as stated: (i) For Mux1, by
interpolating the values obtained from Charts 44 and 45, knowing the values of Pu/fckbD =
0.3598 (see Step 7), p/f = 0.068 (see Step 7), = 0.134 (see Step 7), (ii) for MDd/′ , by
ck
uy1
interpolating the values obtained from Charts 45 and 46, knowing the same values of
Pu/fckbD and p/fck as those of (i) and = 0.173 (see Step 7). The results are given below: Dd/′
2
(i) Mux1/fckbD = 0.0882 (interpolated between 0.095 and 0.085)
2
(ii) Muy1/fckbb = 0.0827 (interpolated between 0.085 and 0.08)
So, we have, Mux1 = 187.54 kNm and Muy1 = 136.76 kNm.
Step 13: Value of nα
Method 1:
We have Pu/Puz = 1700/3084.71 = 0.5511. From Eq.10.60 of Lesson 26, we have nα = 0.67 +
1.67 (Pu/Puz) = 1.59.
Method 2: Interpolating the values between (Pu/Puz) = 0.2 and 0.6
The interpolated value of nα = 1.0 + (0.5511 – 0.2)/0.6 = 1.5852. Both the values are
comparable. We use nα = 1.5852.
Step 14: Checking of column for safety
Method 1: Chart 64 of SP-16
The point having the values of (Mux/Mux1) = 114.8/187.54 = 0.612 and (Muy/Muy1) =
103.43/136.76 = 0.756 gives the value of Pu/Pz more than 0.7. The value of Pu/Puz here is
0.5511 (see Step 13). So, the section needs revision.
2
We revise from Step 7 by providing 8-25 mm diameter bars (= 3927 mm , p = 2.493 per cent
and p/fck = 0.0831) as the longitudinal reinforcement keeping the values of b and D
unchanged. The revised section is checked furnishing the repeated calculations from Step 8
onwards. The letter R is used before the number of step to indicate this step as revised one.
Step R8: Calculation of balanced loads Pb
Table 60 of SP-16 gives k1 = 0.19952, and k2 = 0.243. We have p/fck = 0.0831 now. So, Pbx =
-3
{0.19952 + (0.243)(0.0831)} (30)(350)(450)(10 ) = 1038.145 kN. Similarly, k1 = 0.19048, k2
-3
= 0.1225 and p/fck = 0.0831 give Pby = {0.19048 + (0.1225)(0.0831)} (30)(350)(450)(10 ) =
948.12 kN.
The values of Pbx and Pby are less than Pu (= 1700 kN). So, modification factors are to be
incorporated.
Step R9: Determination of Puz
P = 0.45(30)(350)(450) + {0.75(415) – 0.45(30)}(3927) = 3295.514 kN.
uz
Step R10: Determination of modification factors
kax = (3295.514 – 1700)/(3295.514 – 1038.145) = 0.707
kay = (3295.514 – 1700)/(3295.514 – 948.12) = 0.68
Step R11: Total moments incorporating modification factors
Mux = 52.70 + 0.707(92.548) = 118.13 kNm
Muy = 47.04 + 0.68(87.43) = 106.49 kNm
Step R12: Uniaxial moment capacities
Using Charts 44 and 45 for Mux1 and Charts 45 and 46 for Muy1, we get (i) the coefficient
0.1032 (interpolating 0.11 and 0.10) and (ii) the coefficient 0.0954 (interpolating 0.1 and
0.09) for Mux1 and Muy1, respectively.
-6
Mux1 = (0.1032)(30)(350)(450)(450)(10 ) = 219.429 kNm
-6
Muy1 = (0.0954)(30)(450)(350)(350)(10 ) = 157.77 kNm
Step R13: Value of nα
Pu/Puz = 1700/3295.514 = 0.5158 which gives
nα = 1 + (0.5158 – 0.2)/0.6 = 1.5263
Step R14: Checking of column for safety
1.5263
(118.13/219.424)
1.5263
+ (106.49/157.77)
= 0.3886 + 0.5488 = 0.9374 < 1.0
Hence, the revised reinforcement is safe.
DESIGN OF RC ELEMENTS
UNIT – 5
LIMIT STATE DESIGN OF FOOTING
CONTENTS
5.1 Introduction
5.2 Types of foundation structures
5.3 Safe bearing capacity of soil
5.4 Design considerations
5.5 Design problems
TECHNICAL TERMS
Column bases or column footings: foundations provided for column are known as column
bases or column footings. The column bases distribute the load over a larger area so that the
intensity of pressure on soil does not exceed the safe bearing capacity of soil and settlement
of structure is kept within the permissible limit.
Curtailment of bars: For a R.C. member under bending, steel area is obtained on the basis
of maximum moments at the critical sections. As the B.M. decreases away from critical
sections, part of bars may be cut off where they are no longer required to resist the moment.
UNIT – 5
LIMIT STATE DESIGN OF FOOTING
5.1 Introduction
The superstructure is placed on the top of the foundation structure, designated as
substructure as they are placed below the ground level. The elements of the superstructure
transfer the loads and moments to its adjacent element below it and finally all loads and
moments come to the foundation structure, which in turn, transfers them to the underlying
soil or rock. Thus, the foundation structure effectively supports the superstructure. However,
all types of soil get compressed significantly and cause the structure to settle.
Accordingly, the major requirements of the design of foundation structures are the
two as given below (see cl.34.1 of IS 456):
1. Foundation structures should be able to sustain the applied loads, moments, forces and
induced reactions without exceeding the safe bearing capacity of the soil.
2. The settlement of the structure should be as uniform as possible and it should be within the
tolerable limits. It is well known from the structural analysis that differential settlement of
supports causes additional moments in statically indeterminate structures. Therefore,
avoiding the differential settlement is considered as more important than maintaining
uniform overall settlement of the structure.
In addition to the two major requirements mentioned above, the foundation structure should
provide adequate safety for maintaining the stability of structure due to either overturning
and/or sliding (see cl.20 of IS 456). It is to be noted that this part of the structure is
constructed at the first stage before other components (columns / beams etc.) are taken up.
So, in a project, foundation design and details are completed before designs of other
components are undertaken.
However, it is worth mentioning that the design of foundation structures is somewhat
different from the design of other elements of superstructure due to the reasons given below.
Therefore, foundation structures need special attention of the designers.
1. Foundation structures undergo soil-structure interaction. Therefore, the behaviour of
foundation structures depends on the properties of structural materials and soil.
Determination of properties of soil of different types itself is a specialized topic of
geotechnical engineering. Understanding the interacting behaviour is also difficult. Hence,
the different assumptions and simplifications adopted for the design need scrutiny. In fact,
for the design of foundations of important structures and for difficult soil conditions,
geotechnical experts should be consulted for the proper soil investigation to determine the
properties of soil, strata wise and its settlement criteria.
2. Accurate estimations of all types of loads, moments and forces are needed for the present
as well as for future expansion, if applicable. It is very important as the foundation structure,
once completed, is difficult to strengthen in future.
3. Foundation structures, though remain underground involving very little architectural
aesthetics, have to be housed within the property line which may cause additional forces and
moments due to the eccentricity of foundation.
4. Foundation structures are in direct contact with the soil and may be affected due to
harmful chemicals and minerals present in the soil and fluctuations of water table when it is
very near to the foundation. Moreover, periodic inspection and maintenance are practically
impossible for the foundation structures.
5. Foundation structures, while constructing, may affect the adjoining structure forming
cracks to total collapse, particularly during the driving of piles etc.
However, wide ranges of types of foundation structures are available. It is very important to
select the appropriate type depending on the type of structure, condition of the soil at the
location of construction, other surrounding structures and several other practical aspects as
mentioned above.
5.2 Types of Foundation Structures
Foundations are mainly of two types: (i) shallow and (ii) deep foundations. The two different
types are explained below:
(A) Shallow foundations
Shallow foundations are used when the soil has sufficient strength within a short
depth below the ground level. They need sufficient plan area to transfer the heavy loads to
the base soil. These heavy loads are sustained by the reinforced concrete columns or walls
(either of bricks or reinforced concrete) of much less areas of cross-section due to high
strength of bricks or reinforced concrete when compared to that of soil. The strength of the
soil, expressed as the safe bearing capacity of the soil as discussed in sec.11.28.3, is normally
supplied by the geotechnical experts to the structural engineer. Shallow foundations are also
designated as footings. The different types of shallow foundations or footings are discussed
below.
Plain concrete pedestal footings (Fig.1) are very economical for columns of small
loads or pedestals without any longitudinal tension steel (see cls.34.1.2 and 34.1.3 of IS 456).
In Fig.1, the angle α 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 shall be
determined from Eq. 11.3.
These footings are for individual columns having the same plan forms of square,
rectangular or circular as that of the column, preferably maintaining the proportions and
symmetry so that the resultants of the applied forces and reactions coincide. These footings,
shown in Figs.2 to 4, consist of a slab of uniform thickness, stepped or sloped. Though
sloped footings are economical in respect of the material, the additional cost of formwork
does not offset the cost of the saved material. Therefore, stepped footings are more
economical than the sloped ones. The adjoining soil below footings generates upward
pressure which bends the slab due to cantilever action. Hence, adequate tensile reinforcement
should be provided at the bottom of the slab (tension face). Clause 34.1.1 of IS 456 stipulates
that the sloped or stepped footings, designed as a unit, should be constructed to ensure the
integrated action. Moreover, the effective cross-section in compression of sloped and stepped
footings shall be limited by the area above the neutral plane. Though symmetrical footings
are desirable, sometimes situation compels for unsymmetrical isolated footings (Eccentric
footings or footings with cut outs) either about one or both the axes (Figs.5 and 6).
When the spacing of the adjacent columns is so close that separate isolated footings
are not possible due to the overlapping areas of the footings or inadequate clear space
between the two areas of the footings, combined footings are the solution combining two or
more columns. Combined footing normally means a footing combining two columns. Such
footings are either rectangular or trapezoidal in plan forms with or without a beam joining the
two columns, as shown in Figs.7 and 8.
When two isolated footings are combined by a beam with a view to sharing the loads
of both the columns by the footings, the footing is known as strap footing (Fig.9). The
connecting beam is designated as strap beam. These footings are required if the loads are
heavy on columns and the areas of foundation are not overlapping with each other.
These are in long strips especially for load bearing masonry walls or reinforced
concrete walls (Fig 10). However, for load bearing masonry walls, it is common to have
stepped masonry foundations. The strip footings distribute the loads from the wall to a wider
area and usually bend in transverse direction. Accordingly, they are reinforced in the
transverse direction mainly, while nominal distribution steel is provided along the
longitudinal direction.
These are special cases of combined footing where all the columns of the building are
having a common foundation (Fig 11). Normally, for buildings with heavy loads or when the
soil condition is poor, raft foundations are very much useful to control differential settlement
and transfer the loads not exceeding the bearing capacity of the soil due to integral action of
the raft foundation. This is a threshold situation for shallow footing beyond which deep
foundations have to be adopted.
As mentioned earlier, the shallow foundations need more plan areas due to the low
strength of soil compared to that of masonry or reinforced concrete. However, shallow
foundations are selected when the soil has moderately good strength, except the raft
foundation which is good in poor condition of soil also. Raft foundations are under the
category of shallow foundation as they have comparatively shallow depth than that of deep
foundation. It is worth mentioning that the depth of raft foundation is much larger than those
of other types of shallow foundations.
However, for poor condition of soil near to the surface, the bearing capacity is very
less and foundation needed in such situation is the pile foundation (Fig 12). Piles are, in fact,
small diameter columns which are driven or cast into the ground by suitable means. Precast
piles are driven and cast-in-situ are cast. These piles support the structure by the skin friction
between the pile surface and the surrounding soil and end bearing force, if such resistance is
available to provide the bearing force. Accordingly, they are designated as frictional and end
bearing piles. They are normally provided in a group with a pile cap at the top through which
the loads of the superstructure are transferred to the piles.
Piles are very useful in marshy land where other types of foundation are impossible to
construct. The length of the pile which is driven into the ground depends on the availability
of hard soil/rock or the actual load test. Another advantage of the pile foundations is that they
can resist uplift also in the same manner as they take the compression forces just by the skin
friction in the opposite direction.
However, driving of pile is not an easy job and needs equipment and specially trained
persons or agencies. Moreover, one has to select pile foundation in such a situation where the
adjacent buildings are not likely to be damaged due to the driving of piles. The choice of
driven or bored piles, in this regard, is critical.
Exhaustive designs of all types of foundations mentioned above are beyond the scope of this
course. Accordingly, this module is restricted to the design of some of the shallow footings,
frequently used for normal low rise buildings only.
5.3 Safe Bearing Capacity of Soil
The safe bearing capacity qc of soil is the permissible soil pressure considering safety
factors in the range of 2 to 6 depending on the type of soil, approximations and assumptions
and uncertainties. This is applicable under service load condition and, therefore, the partial
safety factors fλ for different load combinations are to be taken from those under limit state
of serviceability (vide Table 18 of IS 456). Normally, the acceptable value of qc is supplied
by the geotechnical consultant to the structural engineer after proper soil investigations. The
safe bearing stress on soil is also related to corresponding permissible displacement /
settlement.
Gross and net bearing capacities are the two terms used in the design. Gross bearing
capacity is the total safe bearing pressure just below the footing due to the load of the
superstructure, self weight of the footing and the weight of earth lying over the footing. On
the other hand, net bearing capacity is the net pressure in excess of the existing overburden
pressure. Thus, we can write
Net bearing capacity = Gross bearing capacity - Pressure due to overburden soil (5.1)
While calculating the maximum soil pressure q, we should consider all the loads of
superstructure along with the weight of foundation and the weight of the backfill. During
preliminary calculations, however, the weight of the foundation and backfill may be taken as
10 to 15 per cent of the total axial load on the footing, subjected to verification afterwards.
Depth of Foundation
All types of foundation should have a minimum depth of 50 cm as per IS 1080-1962.
This minimum depth is required to ensure the availability of soil having the safe bearing
capacity assumed in the design. Moreover, the foundation should be placed well below the
level which will not be affected by seasonal change of weather to cause swelling and
shrinking of the soil. Further, frost also may endanger the foundation if placed at a very
shallow depth. Rankine formula gives a preliminary estimate of the minimum depth of
foundation and is expressed as
2
d = (qc/λ){(1 - sinφ)/(1 + sinφ)} (5.2)
Where d = minimum depth of foundation
qc = gross bearing capacity of soil
λ= density of soil
φ = angle of repose of soil
Though Rankine formula considers three major soil properties qc, λ and φ, it does not
consider the load applied to the foundation. However, this may be a guideline for an initial
estimate of the minimum depth which shall be checked subsequently for other requirements
of the design.
5.4 Design Considerations
(a) Minimum nominal cover (cl. 26.4.2.2 of IS 456)
The minimum nominal cover for the footings should be more than that of other
structural elements of the superstructure as the footings are in direct contact with the soil.
Clause 26.4.2.2 of IS 456 prescribes a minimum cover of 50 mm for footings. However, the
actual cover may be even more depending on the presence of harmful chemicals or minerals,
water table etc.
(b) Thickness at the edge of footings (cls. 34.1.2 and 34.1.3 of IS 456)
The minimum thickness at the edge of reinforced and plain concrete footings shall be
at least 150 mm for footings on soils and at least 300 mm above the top of piles for footings
on piles, as per the stipulation in cl.34.1.2 of IS 456.
For plain concrete pedestals, the angle α (see Fig.11.28.1) 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 shall be determined from the following expression
(cl.34.1.3 of IS 456)
1/2
Tanα 0.9{(100 q≤a/fck) + 1}
(5.3)
2
Where qa = calculated maximum bearing pressure at the base of pedestal in N/mm , and
2
fck = characteristic strength of concrete at 28 days in N/mm .
(c) Bending moments (cl. 34.2 of IS 456)
1. It may be necessary to compute the bending moment at several sections of the
footing depending on the type of footing, nature of loads and the distribution of pressure at
the base of the footing. However, bending moment at any section shall be determined taking
all forces acting over the entire area on one side of the section of the footing, which is
obtained by passing a vertical plane at that section extending across the footing (cl.34.2.3.1
of IS 456).
2. The critical section of maximum bending moment for the purpose of designing an
isolated concrete footing which supports a column, pedestal or wall shall be:
(i) at the face of the column, pedestal or wall for footing supporting a concrete
column, pedestal or reinforced concrete wall, (Figs. 2, 3 and 10), and
(ii) Halfway between the centre-line and the edge of the wall, for footing under
masonry wall (Fig.10). This is stipulated in cl.34.2.3.2 of IS 456.
The maximum moment at the critical section shall be determined as mentioned in 1 above.
For 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 (see cl.34.2.2 of IS 456 and Figs.13 a and b).
(d) Shear force (cl. 31.6 and 34.2.4 of IS 456)
Footing slabs shall be checked in one-way or two-way shears depending on the nature of
bending. If the slab bends primarily in one-way, the footing slab shall be checked in one-way
vertical shear. On the other hand, when the bending is primarily two-way, the footing slab
shall be checked in two-way shear or punching shear. The respective critical sections and
design shear strengths are given below:
1. One-way shear (cl. 34.2.4 of IS 456)
One-way shear has to be checked across the full width of the base slab on a vertical section
located from the face of the column, pedestal or wall at a distance equal to (Figs. 2, 3 and
10):
(i) Effective depth of the footing slab in case of footing slab on soil, and
(ii) Half the effective depth of the footing slab if the footing slab is on piles (Fig.12).
The design shear strength of concrete without shear reinforcement is given in Table 19 of
cl.40.2 of IS 456.
2. Two-way or punching shear (cls.31.6 and 34.2.4)
Two-way or punching shear shall be checked around the column on a perimeter half
the effective depth of the footing slab away from the face of the column or pedestal (Figs. 2
and 3).
The permissible shear stress, when shear reinforcement is not provided, shall not exceed kscτ,
where ks = (0.5 + cβ), but not greater than one, cβ being the ratio of short side to long side of
1/2
the column, and cτ = 0.25(fck)
in limit state method of design, as stipulated in cl.31.6.3 of
IS 456.
Normally, the thickness of the base slab is governed by shear. Hence, the necessary thickness
of the slab has to be provided to avoid shear reinforcement.
(e) Bond (cl.34.2.4.3 of IS 456)
The critical section for checking the development length in a footing slab shall be the same
planes as those of bending moments in part (c) of this section. Moreover, development length
shall be checked at all other sections where they change abruptly. The critical sections for
checking the development length are given in cl.34.2.4.3 of IS 456, which further
recommends to check the anchorage requirements if the reinforcement is curtailed, which
shall be done in accordance with cl.26.2.3 of IS 456.
(f) Tensile reinforcement (cl.34.3 of IS 456)
The distribution of the total tensile reinforcement, calculated in accordance with the moment
at critical sections, as specified in part (c) of this section, shall be done as given below for
one-way and two-way footing slabs separately.
(i) In one-way reinforced footing slabs like wall footings, the reinforcement shall be
distributed uniformly across the full width of the footing i.e., perpendicular to the direction of
wall. Nominal distribution reinforcement shall be provided as per cl. 34.5 of IS 456 along the
length of the wall to take care of the secondary moment, differential settlement, shrinkage
and temperature effects.
(ii) In two-way reinforced square footing slabs, the reinforcement extending in each
direction shall be distributed uniformly across the full width/length of the footing.
(iii) In two-way reinforced rectangular footing slabs, the reinforcement in the long
direction shall be distributed uniformly across the full width of the footing slab. In the short
direction, a central band equal to the width of the footing shall be marked along the length of
the footing, where the portion of the reinforcement shall be determined as given in the
equation below. This portion of the reinforcement shall be distributed across the central band:
Reinforcement in the central band = {2/ (β+1)} (Total reinforcement in the short
direction) (5.4)
Where, β is the ratio of longer dimension to shorter dimension of the footing slab (Fig 14).
Each of the two end bands shall be provided with half of the remaining
reinforcement, distributed uniformly across the respective end band.
Transferred to the pedestal, if any, and then from the base of the pedestal to the footing, (or
directly from the base of the column to the footing if there is no pedestal) by compression in
concrete and steel and tension in steel. Compression forces are transferred through direct
bearing while tension forces are transferred through developed reinforcement. The
permissible bearing stresses on full area of concrete shall be taken as given below from
cl.34.4 of IS 456:
brσ = 0.25fck, in working stress method, and (5.5)
brσ = 0.45fck, in limit state method (5.6)
It has been mentioned in sec. 10.26.5 of Lesson 26 that the stress of concrete is taken as
0.45fck while designing the column. Since the area of footing is much larger, this bearing
stress of concrete in column may be increased considering the dispersion of the concentrated
load of column to footing. Accordingly, the permissible bearing stress of concrete in footing
is given by (cl.34.4 of IS 456):
1/2
brσ = 0.45fck (A1/A2)
(11.7)
With a condition that
1/2
(A1/A2)
≤ 2.0 (11.8)
where A1 = maximum supporting area of footing for bearing which is geometrically similar to
and concentric with the loaded area A2, as shown in Fig.15.
A2 = loaded area at the base of the column.
The above clause further stipulates that in sloped or stepped footings, A1 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, as shown in Fig.15.
If the permissible bearing stress on concrete in column or in footing is exceeded,
reinforcement shall be provided for developing the excess force (cl.34.4.1 of IS 456), either
by extending the longitudinal bars of columns into the footing (cl.34.4.2 of IS 456) or by
providing dowels as stipulated in cl.34.4.3 of IS 456 and given below:
(i) Sufficient development length of the reinforcement shall be provided to transfer
the compression or tension to the supporting member in accordance with
Cl.26.2 of IS 456, when transfer of force is accomplished by reinforcement of column
(cl.34.4.2 of IS 456).
(ii) Minimum area of extended longitudinal bars or dowels shall be 0.5 per cent of the
cross-sectional area of the supported column or pedestal (cl.34.4.3 of IS 456).
(iii)A minimum of four bars shall be provided (cl.34.4.3 of IS 456).
(iv) The diameter of dowels shall not exceed the diameter of column bars by more than 3
mm.
(v) Column bars of diameter larger than 36 mm, in compression only can be doweled
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, as stipulated in cl.34.4.4 of IS 456 and
as shown in Fig.16.
(h) Nominal reinforcement (cl. 34.5 of IS 456)
1. Clause 34.5.1 of IS 456 stipulates the minimum reinforcement and spacing of the
bars in footing slabs as per the requirements of solid slab (cls.26.5.2.1 and 26.3.3b (2) of IS
456, respectively).
2. The nominal reinforcement for concrete sections of thickness greater than 1 m shall
2
be 360 mm per metre length in each direction on each face, as stipulated in cl.34.5.2 of IS
456. The clause further specifies that this provision does not supersede the requirement of
minimum tensile reinforcement based on the depth of section.
The foundation, assumed to act as a rigid body, is in equilibrium under the action of
applied forces and moments from the superstructure and the reactions from the stresses in the
soil. The distribution of base pressure is different for different types of soil. Typical
distributions of pressure, for actual foundations, in sandy and clayey soils are shown in Figs
17 and 18, respectively. However, for the sake of simplicity the footing is assumed to be a
perfectly rigid body, the soil is assumed to behave elastically and the distributions of stress
and stain are linear in the soil just below the base of the foundation. Accordingly, the
foundation shall be designed for the applied loads, moments and induced reactions keeping in
mind that the safe bearing capacity of the soil is within the prescribed limit. It is worth
mentioning that the soil bearing capacity is in the serviceable limit state and the foundation
structure shall be designed as per the limit state of collapse, checking for other limit states as
well to ensure an adequate degree of safety and serviceability.
In the following, the distributions of base pressure are explained for (i) concentrically loaded
footings, (ii) eccentrically loaded footings and (iii) unsymmetrical (about both the axes)
footings.
Fig 19 Isolated footing subjected to concentric loading
Figure 19 shows rectangular footing symmetrically loaded with service load P1 from
the superstructure and P2 from the backfill including the weight of the footing. The assumed
uniformly distributed soil pressure at the base of magnitude q is obtained from:
q = (P1 + P2)/A (5.9)
Where, A is the area of the base of the footing.
In the design problem, however, A is to be determined from the condition that the
actual gross intensity of soil pressure does not exceed qc, the bearing capacity of the soil, a
known given data. Thus, we can write from Eq.5.9:
A = (P1 + P2)/qc (5.10)
From the known value of A, the dimensions B and L are determined such that the maximum
bending moment in each of the two adjacent projections is equal, i.e., the ratio of the
dimensions B and L of the footing shall be in the same order of the ratio of width b and depth
D of the column.
5.5 Design problems:
Problem 1:
Design a plain concrete footing for a column of 400 mm x 400 mm carrying an axial
2
load of 400 kN under service loads. Assume safe bearing capacity of soil as 300 kN/m , at a
depth of 1 m below the ground level. Use M 20 and Fe 415 for the design.
Solution 1:
Plain concrete footing is given in secs.11.28.2(A)1 and 11.28.5(b).
Step 1: Transfer of axial force at the base of column
It is essential that the total factored loads must be transferred at the base of column without
any reinforcement. For that the bearing resistance should be greater than the total factored
load Pu.
Here, the factored load Pu = 400(1.5) = 600 kN.
Bearing stress, as per cl.34.4 of IS 456 and given in Eqs.5.7 and 5.8
1/2
brσ = 0.45 fck (A1/A2)
(5.7)
With a condition that
1/2
(A1/A2)
≤ 2.0 (5.8)
Since the bearing stress brσ at the column-footing interface will be governed by the
2
column face, we have A1 = A2 = 400(400) = 160000 mm . Using A1 = A2, in Eq.5.7, we have
-3
Pbr = Bearing force = 0.45 fck A1 = 0.45(20) (160000)(10 ) = 1440 kN > Pu (= 600 kN).
Thus, the full transfer of load Pu is possible without any reinforcement.
Step 2: Size of the footing
Let us assume the weight of footing and back fill soil as 15 per cent of Pu. Then, the base
2
2
area required = 400(1.15)/300 = 1.533 m . Provide 1250 x 1250 mm (=1.5625 m ) as shown
2
in Fig.11.29.1. The bearing pressure qa = 400(1.15)/(1.25)(1.25) = 294.4 kN/m .
Fig 20
Step 3: Thickness of footing
The thickness of the footing h is governed by below eqn
1/2
tan ≤α0.9{(100qa/fck) + 1}
…. (5.3)
1/2
≤ 0.9[{100(0.2944)/20} + 1]
≤ 1.415
We have from Fig.20a:
h = {(1250 - 400)/2}(tanα) = 601.375 mm
Provide 1250 x 1250 x 670 mm block of plain concrete.
Step 4: Minimum reinforcement
The plain concrete block 1250 x 1250 x 670 shall be provided with the minimum
reinforcement 0.12 per cent for temperature, shrinkage and tie action.
2
Minimum Ast = 0.0012(1250) (670) = 1005.0 mm .
2
Provide 9 bars of 12 mm diameter (= 1018 mm ) both ways as shown in Fig.20 b. The
spacing of bars = (1250 - 50 - 12)/8 = 148.5 mm c/c. Provide the bars @ 140 mm c/c.
Step 5: Check for the gross base pressure
3
3
Assuming unit weights of concrete and soil as 24 kN/m and 20 kN/m
Service load = 400.00 KN
Weight of footing = (0.67) (1.25) (1.25) (24) = 25.125 kN
Weight of soil = (0.33) (1.25) (1.25) (20) = 10.3125 kN
Total = 435.4375 KN
2
qa = 435.4375/(1.25)(1.25) = 278.68 kN/m < 300 kN/m
2
Hence, o.k.
2.Design one isolated footing for a column 300 mm x 450 mm, having 20 bars of 20
2
mm diameter (Ast = 4021 mm ) of Problem 1 of sec.10.25.6 of Lesson 25 carrying Pu = 1620
kN and Mu = 170 kNm using M 25 and Fe 415. Assume that the moment is reversible. The
2
safe bearing capacity of the soil is 200 kN/m at a depth of 1 meter from ground level. Use M
25 and Fe 415 for the footing.
Step 1: Size of the footing
Given Pu = 1620 kN and Mu = 170 kNm. The footing should be symmetric with
respect to the column as the moment is reversible. Assuming the weights of footing and
backfill as 15 per cent of Pu, the eccentricity of load Pu at the base is e = Mu/P (1.15) =
6
3
170(10 )/1620(1.15) (10 ) = 91.25 mm. This eccentricity may be taken as < L/6 of the
footing.
2
The factored bearing pressure is 200(1.5) = 300 kN/m . For the footing of length L
and width B, we, therefore, have:
2
Pu/BL + 6M/BL ≤ 300
2
Or, 1620(1.15)/BL + 6(170)/BL ≤ 300
2
Or, BL – 6.21L – 3.4 ≤ 0 (1)
For the economic proportion, let us keep equal projection beyond the face of the column in
the two directions. This gives
(L – 0.45)/2 = (B – 0.3)/2
Or, B = L – 0.15 (2)
Using Eq.(2) in Eq.(1), we have
2
(L – 0.15) L – 6.212 – 3.4 ≤ 0
3
2
Or L – 0.15 L – 6.21 L – 3.4 ≤ 0
We have L = 2.8 m and B = 2.65 m. Let us provide L = 2.85 m and B = 2.70 m
(Fig.11.29.4b). We get the maximum and minimum pressures as
2
1620(1.15)/(2.85)(2.70) 170(6)/(2.7)(2.85)(2.85) = 242.105 ±± 46.51 = 288.615 kN/m and
2
2
195.595 kN/m , respectively (Fig.11.29.4c). Both the values are less than 300 kN/m . Hence,
o.k.
Step 2: Thickness of footing slab based on one-way shear
The critical section (sec.11 of Figs.11.29.4a and b) is at a distance d from the face of the
column. The average soil pressure at sec.11 is {288.615 – (288.615 – 195.595)(1200 –
d)/2850} = 249.449 + 0.0326d.
The one-way shear force at sec.11 = (2.7)(1.2 – 0.001d((249.449 + 0.0326d) kN. Assuming
0.15 per cent reinforcement in the footing slab, the shear strength of M 25 concrete = 0.29
2
3
N/mm . Hence, the shear strength of the section = 2700(d)(0.29)(10 ) kN. From the condition
that shear strength has to be shear force, we have ≥
-3
2700(d)(0.29)(10 ) = (2.7)(1.2 – 0.001d)(249.449 + 0.0326d)
This gives,
2
d + 15347.51534d – 9182171.779 = 0
Solving, we get d = 576.6198. Let us assume d = 600 mm
Step 3: Checking for two-way shear
At the critical section 2222 (Figs.11.29.4a and b), the shear resistance is obtained cl.31.6.31
of IS 456, which gives cτ = (0.5 + 450/300) (0.25) (25)
1/2
450/300) >/ 1.0. So, we have cτ = 0.25(25)
1/2
but the multiplying factor (0.5 +
2
= 1.25 N/mm . Hence, the shear resistance =
(1.25) (2) {(300 + 600) + (450 + 600)}(600) = 2925 kN.
Actual shear force is determined on the basis of average soil pressure at the centre line of the
2
cross-section which is (195.595 + 288.615)/2 = 242.105 kN/m (Fig.11.29.4c). So, the actual
shear force = Vu = (242.105) {(2.7)(2.85) – (0.3 + 0.6)(0.45 + 0.6)} = 1634.209 kN < shear
resistance (= 2925 kN). Hence, the depth of the footing is governed by one-way shear. With
effective depth = 600 mm, the total depth of footing = 600 + 50 (cover) + 16 (bar dia) + 8
(half bar dia) = 674 mm.
Step 4: Gross bearing capacity
3
3
Assuming the unit weights of concrete and soil as 25 kN/m and 18 kN/m , respectively, we
have the bearing pressure for (i) Pu = 1620 KN, (ii) Mu = 170 kNm and (iii) self weight of
footing and backfill soil.
(i) Due to Pu = 1620 kN: pressure = 1620/ (2.7) (2.85) = 210.53 kN/m
2
(ii) Due to Mu = 170 kNm: pressure = ± 170(6)/ (2.7) (2.85)(2.85) = 46.51 kN/m±
2
(iii) Self weight of footing of depth 674 mm and soil of (1000 – 674) = 326 mm: pressure =
0.674(25) + 0.326(18) = 22.718 kN/m
2
Thus, the maximum and minimum pressures are = 210.53 + 22.718 46.51 = 279.758 kN/m±
2
2
2
and 186.738 kN/m < 300 kN/m . Hence, o.k.
Step 5: Bending moment
(i) In the long direction (along the length = 2850 mm)
Bending moment at the face of column (sec.33 of Figs.11.29.4a and b) is determined where
2
the soil pressure = 288.615 – (288.615 – 195.595) (1200)/2850 = 249.45 kN/m . So, the
bending moment = 249.45(2.7) (1.2) (0.6) + (288.615 – 249.45)(2.7)(1.2)(2)/(2)(3) = 527.23
kNm.
2
6
2
2
M/Bd = 527.23(10 )/(2700)(616)(616) = 0.515 N/mm < 3.45 N/mm for M 25 concrete.
Table 3 of SP-16 gives p = 0.1462 < 0.15 per cent as required for one-way shear.
2
Thus, Ast = 0.15(2700) (616)/100 = 2494.8 mm . Provide 13 bars of 16 mm diameter (area =
2
2613 mm ), spacing = (2700 – 100 – 16)/12 = 215.33 mm, say 210 mm c/c.
(ii) In the short direction (B = 2700 mm)
The average pressure on soil between the edge and centre of the footing = (288.615 +
2
242.105)/2 = 265.36 kN/m . The bending moment is determined with this pressure as an
approximation.
Bending moment = (265.36)(1.2)(0.6)(2.85) kNm = 544.519 kNm
2
6
M/Ld = 544.519(10 )/(2850)(600)(600) = 0.531
Table 3 of SP-16 gives p = 0.15068, which gives area of steel = 0.15068(2850)(600)/100 =
2
2
2576.628 mm . Provide 13 bars of 16 mm diameter (area = 2613 mm ) @ 210 mm c/c; i.e.
the same arrangement in both directions.
Step 6: Development length
Development length of 16 mm diameter bars (M 25 concrete) = 0.87(415) (16)/4(1.6)(1.4) =
644.73 mm.
Length available = 1200 – 50 – 8 = 1142 mm > 644.73 mm. Hence, o.k.
Step 7: Transfer of force at the base of the column
Since the column is having moment along with the axial force, some of the bars are in
tension. The transfer of tensile force is not possible through the column-footing interface. So,
the longitudinal bars of columns are to be extended to the footing. The required development
length of 20 mm bars = 0.87(415)/4(1.4) (1.6) = 805.92 mm. Length available = 600 mm <
o
805.92 mm. The bars shall be given 90 bend and then shall be extended by 200 mm
horizontally to give a total length of 600 + 8(20) (bend value) + 200 = 960 mm > 805.92 mm
The arrangement of reinforcement is shown in Figs
(2)
Design a combined footing for two columns C1, 400 mm x 400 mm with 8 bars of 16
mm diameter carrying a service load of 800 KN and C2, 300 mm x 500 mm with 8 bars of 20
mm diameter carrying a service load of 1200 KN (Figs.11.29.5a and b). The column C1 is
flushed with the property line. The columns are at 3.0 m c/c distance. The safe bearing
2
capacity of soil is 200 kN/m at a depth of 1.5 m below the ground level. Use M 20 and Fe
415 for columns and footing.
Solution:
Step 1: Size of the footing
Assuming the weight of combined footing and backfill as 15 per cent of the total loads of the
2
columns, we have the required base area, considering qc = 200 kN/m ,
2
Area of the base = (800 + 1200)(1.15)/200 = 11.5 m .
It is necessary that the resultant of the loads of two columns and the centroid of the footing
coincide so that a uniform distribution of soil pressure is obtained. Thus, the distance of the
centroid of the footing y from column C1 (Fig.11.29.5b) is:
y = 800(0) + 1200(3)/2000 = 1.8 m (Fig.11.29.5b). Since y is greater than half the c/c
distance of column, a rectangular footing has to be designed. Let us provide 4 m x 3 m and
the dimensions are shown in Fig.11.29.5b coinciding the centroid of the footing and the
resultant line of action of the two loads, i.e. at a distance of 2 m from the left edge.
Step 2: Thickness of footing slab based on one-way shear
Considering the footing as a wide beam of B = 3 m in the longitudinal direction, the
uniformly distributed factored load = (800 + 1200)(1.5)/4 = 750 kN/m. Figures 11.29.6a, b
and c present the column loads, soil pressure, shear force and bending moment diagrams.
The critical section of one-way shear is sec.11 (at point K) of Figs.11.29.5a and 11.29.6a, at a
distance of d + 250 mm from G (the location of column C2). The one-way shear force is
Shear force = (1600 – d – 250)1200/1600 = (1012.5 – 0.75d) KN
Assuming p = 0.15 per cent reinforcement in the footing slab, the shear strength of M 20
2
-3
concrete = 0.28 N/mm . Hence, the shear strength of section 11 = (3000) d (0.28) (10 ) kN.
-3
From the condition that shear strength shear force, we have ≥ (3000) d (0.28) (10 ) 1012.5 –
0.75d, which gives d 636.79 mm. Provide d = 650 mm and the total depth = 650 + 50 + 16 +
8 = 724 mm (assuming cover = 50 mm and the diameter of bars = 16 mm).
Step 3: Checking for two-way shear
(i) Around column C2
The effective depth along 4.0 m is 650 + 16 = 666 mm. The critical section for the two-way
shear around column C2 is at a distance of 666/2 = 333 mm from the face of the column and
marked by 2222 line in Fig.11.29.5b. The two-way punching shear force, considering the soil
2
pressure = 750/3 = 250 kN/m , is
Vu = 1800 – (1.166) (0.966) (250) = 1518.411 kN
As per cl.31.6.3.1 of IS 456, here ks = 0.5 + (500/300) but >/ 1.0; so, ks = 1.0. Therefore,
shear strength of concrete = 0.25(20)
1/2
(2) {(300 + 666) + (500 + 666)}(666) = 3174.92 kN >
1518.411 kN. Hence, o.k.
(ii) Around column C1
The effective depth of footing is 666 mm. The critical section is marked by 3333 in
Fig.11.29.5b. The two-way punching shear = 1200 – (1.066) (0.733) (250) = 1004.65 kN.
The resistance to two-way shear = 0.25(20)
1/2
(2) {(1066 + 733)} (666) = 2679.1 KN >
1004.65 kN. Hence, o.k.
Thus, the depth of the footing is governed by one-way shear.
Step 4: Gross bearing capacity
3
3
Assuming unit weights of concrete and soil as 25 kN/m and 18 kN/m , respectively, the
gross bearing capacity under service load is determined below.
(i) Due to two loads: (800 + 1200)/ (3) (4) = 166.67 kN/m
2
(ii) Due to weight of the footing: With a total depth of the footing = 724 mm, the pressure =
2
0.724(25) = 18.1 kN/m .
2
(iii) Due to backfill of 1500 – 724 = 776 mm, the pressure = 0.776(18) = 13.968 kN/m .
2
2
The total pressure = 166.67 + 18.1 + 13.968 = 198.738 kN/m < 200 kN/m . Hence, o.k.
Step 5: Bending moments (longitudinal direction)
(i) Maximum positive moment
Figure 11.29.6c shows the maximum positive bending moment = 720 kNm at a distance of
1.4 m from the column C1 (at point J). With effective depth d = 666 mm, we have
2
6
M/Bd = 720(10 )/ (3000) (666) (666) = 0.541 N/mm
2
Table 2 of SP-16 gives p = 0.1553 per cent.
Ast = 0.1553(3000) (666)/100 = 3102.894 mm
2
2
Provide 16 bars of 16 mm diameter (area = 3217 mm ), spacing = (3000 – 50 – 16)/15 =
195.6 mm c/c, say 190 mm c/c.
Development length of 16 mm bars = 47.01(16) = 752.16 mm
Length available = 1600 – 50 – 16 = 1534 mm > 752.16 mm
Hence, o.k.
(ii) Maximum negative moment
Figure 11.29.6c shows the maximum negative moment = 240 kNm at a distance of 800 mm
from the right edge. With the effective depth = 666 mm, we have
2
6
M/Bd = 240(10 )/ (3000) (666)(666) = 0.018 N/mm
2
It is very nominal. So, provide 0.15 per cent steel, which gives Ast = 0.15(3000) (666)/100 =
2
2
2997 mm . Provide 27 bars of 12 mm (area = 3053 mm ) at spacing = (3000 – 50 – 12)/26 =
113 mm c/c; say 110 mm c/c.
Development length = 47.01(12) = 564 mm
Length available = 800 – 50 – 12 = 738 mm > 564 mm
Hence, o.k.
Step 6: Design of column strip as transverse beam
Figure 11.29.7 shows the two column strips under columns C1 and C2.
(i) Transverse beam under column C1
The width of the transverse beam is 0.75d from the face of column C1. The effective depth is
666 – 6 – 8 = 652 mm, as the effective depth in the longitudinal direction = 666 mm, bottom
bar diameter in longitudinal direction = 12 mm and assuming the bar diameter in the
transverse direction as 16 mm. We have to check the depth and reinforcement in the
transverse direction considering one-way shear and bending moment.
(A) One-way shear
The factored load for this transverse strip = 1200/3 = 400 kN/m. The section of the
one-way shear in sec.44 (Fig.11.29.7) at a distance of d = 652 mm from the face of column
C1. The width of the transverse strip = 400 + 0.75(652) = 889 mm.
-3
One-way shear force in sec.44 = (1500 – 652 – 200) (400) (10 ) = 259.2 kN
3
Shear stress developed = 259.2(10 )/ (889) (652) = 0.447 N/mm
2
2
To have the shear strength of concrete 0.447 N/mm≥ , the percentage of reinforcement is
determined by linear interpolation from Table 19 of IS 456, so that the depth of footing may
remain unchanged. Table 19 of IS 456 gives p = 0.43125. Accordingly, Ast = 0.43125(889)
2
2
(652)/100 = 2499.65 mm . Provide 13 bars of 16 mm (area = 2613 mm ), spacing = 889/12 =
74.08 mm c/c, say 70 mm c/c. However, this area of steel shall be checked for bending
moment consideration also.
(B) Bending moment at the face of column C1 in the transverse strip under column
2
Bending moment = (1.3) (1.3) (400)/2 = 338 kNm. We, therefore, have M / (width of strip)d
6
= 338(10 )/(889)(652)(652) = 0.89 N/mm
2
Table 2 of SP-16 gives p = 0.2608 < 0.43125. Hence, the area of steel as determine for oneway shear consideration is to be provided. Provide 13 bars of 16 mm @ 70 mm c/c in the
column strip of width 889 mm under the column C1.
Development length of 16 mm bars = 47.01(16) = 752.16 mm
Length available = 1300 – 50 – 16 = 1234 mm > 752.16 mm
Hence, o.k.
(ii) Transverse beam under column C2
Figure 11.29.7 shows the strip of width = 500 + 0.75d + 0.75 d = 500 + 1.5(652) = 1478 mm,
considering the effective depth of footing = 652 mm.
(A) One-way shear
The factored load for this transverse strip = 1800/3 = 600 kN/m. The one-way shear section
is marked by sec.5.5 in Fig.11.29.7 at a distance of d = 652 mm from the face of the column
C2 .
-3
One-way shear in sec.55 (of width = 1478 mm) = (1500 – 652 – 150) (600) (10 ) = 418.8
KN
3
The shear stress developed = 418.8(10 )/ (1478) (652) = 0.434 N/mm
2
The corresponding percentage of area of steel, as obtained from Table 19 of IS 456 is, p =
2
0.404 per cent. Accordingly, Ast = 0.404(1478) (652)/100 = 3893.17 mm . Provide 20 bars of
2
16 mm (area = 4021 mm ), spacing = 1478/19 = 77.78 mm c/c, say 75 mm c/c. However, this
area of steel shall be checked for bending moment consideration also.
(B) Bending moment at the face of column C2 in the transverse strip under column C2
The bending moment = (1.35) (1.35) (600)/2 = 546.75 kNm. We, therefore, have
2
6
M /(width of strip)d = 546.75(10 )/(1478)(652)(652) = 0.87 N/mm
2
Table 2 of SP-16 gives: p = 0.2544 < 0.404 per cent as required for one-way shear. So,
provide 20 bars of 16 mm diameter @ 75 mm c/c in the column strip of width 1478 under
column C2.
Development length as calculated in Step (B) of column strip C1 = 752.16 mm
The length available = 1350 – 50 – 16 = 1284 mm > 752.16 mm
Hence, o.k.
Step 7: Transfer of forces at the base of the columns
(A) For column C1
The limiting bearing stress at the column face governs where the bearing stress = 0.45fck = 9
2
N/mm , since the column C1 is at the edge of the footing.
-3
The force that can be transferred = 9(400) (400) (10 ) = 1440 KN > 1200 KN, load of
column C1. Hence full transfer of force is possible without the need of any dowels. However,
four 16 mm nominal dowels are provided as shown in Figs.11.29.5a and b.
(B) For column C2
Figure 11.29.8 shows the dimensions to determine A1 and A2 for the column C2 and the
footing. Accordingly,
2
A1 = (1600) (1400) mm , and
A2 = (300) (500) mm
1/2
(A1/A2)
2
= {(16) (14) / (3) (5)}
1/2
= 3.86 but limited to 2
2
The bearing stress at the column face = 0.45fck = 9 N/mm , and the bearing stress at the
2
2
footing face = 0.45f (2) = 18 N/mm . However, the bearing stress of 9 N/mm governs.
ck
The force that can be transferred through the column C2 = 9(300) (500) = 1350 KN < 1800
kN. For the excess force (1800 – 1350) = 450 KN, dowels shall be provided. The area of
3
2
dowels = 450(10 )/0.67(415) = 1618.414 mm . The minimum area of dowels = 0.5(300)
2
2
(500)/100 = 750 mm . Provide 6 dowels of 20 mm diameter (area = 1885 mm ).
The development length required in compression
= 0.97(415) (20) /4 (1.6) (1.2) (1.25) = 752.2 mm.
The length available = 652 – 20 = 632 mm
o
Therefore, the dowels shall be given a 90 bend and shall be extended horizontally by 100
mm to have a total length of 632 + 8(20) + 100 = 892 mm > 752.2 mm (Figs.5.9a and b).
Step 8: Distribution reinforcement
Nominal distribution reinforcement shall be provided at top and bottom where the main
reinforcement bars are not provided. The amount @ 0.12 per cent comes to 0.12(1000)
2
(652)/100 = 782.4 mm /meter. Provide 12 mm diameter bars @ 140 mm c/c (area = 808
2
mm /m).
SNS COLLEGE OF ENGINEERING
COIMBATORE
DEPARTMENT OF CIVIL ENGINEERING
CE2306 – DESIGN OF RC ELEMENTS
QUESTION BANK
YEAR/SEM: III/V
PART-A
UNIT-I
5. What are the advantages of limit state method over working stress and ultimate load
methods?
6. How do you find the moment of resistance of a beam section?
7. Discuss the merits of working stress method?
8. What is modular ratio? Determine the modular ratio at M20 grade concrete.
9. What do you understand by limit state of collapse?
10. Draw stress-strain curve for various grades of steel.
11. State the assumptions made in working stress method.
12. Draw the transformed section of a singly reinforced RC beams, in the uncracked stage.
13. What are the expression recommended by the IS 456-2000 for modules of elasticity and
flexural strength?
14. Write the formula for neutral axis depth factor ’K’ in working stress design.
15. Explain the terms ‘balanced’, ‘over reinforced’ and ‘under reinforced’ sections in
bending
UNIT – II
1. Distinguish between one –way and two way slabs.
2. Discuss the different limit state to be considered in reinforced concrete design?
3. Why is it necessary to provide transverse reinforcement in a one way slab?
4. What are the three basic methods using factor of safety to achieve safe workable
structures?
•
Explain maximum depth of neutral axis?
•
Find the depth of neutral axis in terms of ‘d’ for a balanced section using Fe 415 steel, in
limits method?
•
What is the difference in the design of one way slab and two way slabs?
•
Distinguish between under reinforcement and over reinforcement section?
•
Sketch edge and middle strips of a two way slab?
•
How Limit State Method aims for a comprehensive and rational solution to the design
problem?
•
Differentiate between WSD and LSD.
UNIT-III
a) What are the types of reinforcements used to resist shear?
b) Explain the difference between primary and secondary torsion. Give two examples each?
c) Under what circumstances are doubly reinforced beams used?
d) Reinforced concrete slab are generally safe in shear and do not require shear
reinforcement? Why?
e) Mention the difference in design principles for L Beam and T Beam?
f) When shear reinforcement is necessary in a beam?
g) What is bond stress? Write an expression for it?
h) What are the types of reinforcement used to resit shear force?
i) What you understand by development length of bar?
j) Differentiate between flexural bond and development bond.
k) What will be minimum and maximum area of tension reinforcement in a beam?
l) How shear reinforcement improves the strength of beam?
UNIT-IV
1. What is the minimum and maximum percentage of steel allowed in R.C.Column? Explain
why it is necessary to specify the minimum and maximum percentage?
2. Give example of columns that are in practice subjected to uniaxial and biaxial bending.
3. Explain (a) Equilibrium torsion (b) Compatibility torsion
a) How do you classify a column as short or long?
b) Write the procedure for the design of an axially loaded short column.
c) What are braced columns?
d) State the methods recommend by IS 456 to estimate the effective length of columns.
e) Write any two function of lateral ties in a RC column.
f) Differentiate between long and short column.
g) Differentiate between uniaxial and biaxial
UNIT-V
a) Sketch the placement of steel in rectangular footing with a non-central load.
b) What are the situations in which combined footings are preferred over isolated footings?
c) Draw a neat sketch of a masonry footing.
d) What is slenderness ratio for a masonry wall? State the maximum values?
e) Compare the behavior of tied and spirally reinforced column.
f) How do you classify one-way footing and two-way footing in foundation?
g) Under what circumstances a trapezoidal footing become necessary?
h) What is a punching shear in a RCC footing?
i) Sketch the reinforcement detailing for the cantilever slab.
PART-B
UNIT-I
i) A singly reinforced concrete beam is of width 450mm and effective depth 715mm. It is
reinforced with 8Nos.20mm mild steel bars. Assuming M20 concrete, determine its
moment of resistance according to the working stress method. Determine also the stress
in steel when the beam is subjected to the above moment.
ii) Design a rectangular slab supported on its all four edges (600mm thick) over a classroom
of size 4.8m x6.2m. Two adjacent edges of the slab are discontinuous and the remaining
two edges are continuous. A finishing surface of cement concrete of 20mm.
iii) Design of roof slab for an interior panel of size 5mx6m. Live load is 5.0KN/m2. Use M30
Concrete and Fe 415 Steel.
1. A doubly reinforced concrete beam is 250mm wide and 510mm depth the center of
tensile steel reinforcement. The compression reinforcement consists of 4 Nos. of 18mm
dia bars placed at an effective cover of 40mm from the compression edge of the beam.
The tensile reinforcement consists of 4Nos. of 20mm diameter bar. If the beam section is
subjected to a BM of 85kNm, calculate the stresses in concrete and tension steel
2. Design a simply supported R.C.C.SLAB for a roof of a hall 4mx10m (inside dimensions)
with 230mm walls all around. Assume a live load of 4kN/m2 and finish 1KN/m2.Use
grade 25 concrete and Fe 415 steel.
3. A rectangular beam of width 300mm and effective depth 500mm reinforced with 4 bars
of 12mm diameter. Find the moment of resistance and stresses in the top compression
fiber of concrete and tension steel. Use concrete M20 and steel Fe415. Adopt working
stress method.
4. Derive the expression 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 M20 grade concrete and Fe 415 grenade steel. Use Working
Stress Method.
UNIT-II
1. A rectangular beam has b=200mm, d=400mm if steel used is Fe 415 and grade of
concrete is M25. Find the steel required to carry a factored moment of 12kNm.
2. A T beam continuous over several supports has to carry a factored negative support
moment of 1000kNm. Determine the area of steel at supports if bW = 400MM,
bfy =1600mm, df = 100mm, D=610mm, d’ = 60mm, fck = 30N/mm2, f = 415 N/mm2.
3. Design a smallest concrete section of a RC beam to resist an ultimate moment of 62kNm,
assuming width 230mm, concrete grade M20 and HYSD bars of grade Fe415.
4. Design the interior span of a continuous one way slab for an office floor continuous over
tee beams spaced at 3 meters. Live load = 4kN/m, Floor finish = 1kN/m2.Use concrete
M20 and steel Fe415. Adopt limit state method. Sketch the steel reinforcement.
1. Determine the reinforcement for a T beam with flange width = 1500mm, web width =
300mm, thickness of slab = 100mm, effective depth 735mm, to carry a moment of
380kNm due to characteristic loads. Use M25 concrete and Fe 415 steel..
2. Design a rectangular beam section subjected to an ultimate moment of 120kNm. Use
concrete M20 and steel Fe415. Adopt limit state method.
3. Analyse a T-Beam section of 250 mm wide of web, 1200 mm width of flanged, 100 mm
thickness flange and 450 mm effective depth to determine the ultimate Moment of
resistance of for the two cases of reinforcements
4 Nos of 20 mm diameter
4 Nos of 25 mm diameter.
Consider M 20 grade concrete and Fe415 grade steel
8. Design a two way slab panel for the following data.
Size
=
7 m x 5m
Width of supports
=
300mm
Edge condition two short edges discountiounes
Live Load
=
4kN/m2
1kN/m2
Floor Finish
=
Consider M20 grade concrete and Fe415 grade steel.
UNIT-III
1. A rectangular beam width b=350mm and d=550mm has a factored shear of 400kN at the
critical section near the support. The steel at the tension side of the section consists of
four 32mm bars which are continued to support. Assuming fck=25 and fy=415(N/mm2)
design vertical stirrups for the section.
2. A reinforced concrete rectangular beam has a breadth of 350mm and effective depth of
800mm. It has a factored shear of 105kN at section XX. Assuming that fck=25,
fy=415(N/mm2) and percentage of tensile steel at that section is 0.5percent, determine the
torsional moment the section can resist if no additional reinforcement for torsion is
provided. Workout the problem according to IS456 principles of design for torsion.
3. A simply supported beam is 5m in span and carries a characteristic load at 75kN/m. If
6Nos. of 20mm bras are continued into the supports. Check the development length at the
supports assuming grade M20 concrete and Fe415steel.
4. A rectangular RCC beam is 400×900mm in size. Assuming the use of grade M25
concrete and Fe415 steel, determine the maximum ultimate torsional moment at the
section can take it.
i.
No torsion reinforcement is provided and
ii.
Maximum torsion reinforcement is provided.
5. A rectangular beam width b = 250mm and effective depth 500mm reinforced with 4 bars
of 20mm diameter. Determine the shear reinforcement required to resist a shear force of
150kN. Use concrete M20 and steel Fe415.
6. Design a rectangular beam section of width 250mm and effective depth 500mm,
subjected to an ultimate moment of 160kNm, ultimate shear force of 30kN and ultimate
torsional moment of 10kNm. Use concrete M20 and steel 415.
7. A RC beam 300×450mm in cross section in reinforced with 3 Nos. 20mm diameter of
grade Fe250, with an effective cover of 50mm. The ultimate shear at the section of
138kn.Design the shear reinforcement (i)Using only vertical strips without bending any
bar for resisting. (ii) Bending 1 bar dia 20mm at 45 degree to resist shear at the section.
Assume concrete of grade M20.
8. A reinforced concrete beam 500mm deep and 230mm wide is reinforced with
8Nos.20mm diameter bars at mid span to carry a UDL of 22.5kn/m (inclusive of its own
weight) over simple span of 8m. Assuming concrete grade M20, steel grade Fe415, load
factor 1.5 and width of support 230mm (i) determine the minimum development length
required for 20mm diameter bar to develop full strength (ii) apply check for flexural
development length at support assuming all bar to continue at support (iii) determine the
minimum number of bars required at support for development length of flexure.
UNIT-IV
1. A rectangular column of effective height of 4m is subjected to a characteristics axial load
of 800kN and bending moment of 100kNm about the major axis of the n. Design a
suitable section for the column so that the width should not exceed 400mm. Use the
minimum percentage of longitudinal steel. Assume fy=415N/mm2 and fck=20N/mm2.
2. An R.C.Column 500×400mm is subjected to an axial ultimate load of 2500kN and bent
in single curvature about the minor axis with My(top)=90knm and My(bottom)=120knm
as ultimate moments. If L0=7.2m and Le=5.75m on both axes, calculate the design
moments for the column.
3. Design the reinforcement in a spiral column of 400mm diameter subjected to a factored
load of 1500kN.The column has an supported length of 3.4m and is braced against side
way. Use M20 concrete and Fe415 steel.
4. A column 300×400mm has an unsupported length of 3m and effective length of 3.6m.If it
is subjected to pu=1100kNm and u=230kNm about the major axis, determine the
longitudinal steel using fck=25N/mm2.
5. Calculate the ultimate strength in axial compression of column 400mm in diameter and
reinforced with 8Nos. of 20mm dia. of grade Fe250 when the column in helically
reinforced by 8mm dia at (i) 60mm pitch, (ii) 30mm pitch. Assume concrete of grade
M20. Assume clear cover equal to 40mm.
6. Design an axially loaded tied column 400mm×400mm pinned at both ends with an
unsupported length of 3m for carrying a factored load of 2300kN.Use M20 concrete and
Fe415 steel.
7. Design a circular column with helical reinforcement of 400mm diameter and 4m in
length to carry factored load of 1000kN.The column is hinged at both ends. Use concrete
M25 and steel Fe415
8. A column 300mm×400mm has an unsupported length of 4m and fixed at both ends. It is
subjected to a factored load of 1000KN and an ultimate moment of 200kNm about the
major axis. Determine the longitudinal reinforcement and lateral ties. Use concrete M25
and steel Fe415 d’=60mm.
UNIT-V
1. A rectangular column 300mm×400mm reinforced with 20mm diameter bars carries a
load of 1400kN. Design a suitable footing for the column. The safe bearing capacity of
the soil is 200kN/m2.Use concrete M20 and steel Fe415.
2. Design a combined rectangular footing for two columns spaced at 5 centers. The first
column 400mm×400mm carries a load of 1200kN and the second column 50mm×450mm
carries a load of 1800kn at service state. Weight of Soil = 20kN/m2, angle of repose=300
and safe bearing capacity of soil = 150kN/m2. Use concrete M20 and steel Fe415.
3. Design a interior wall of a single storied workshop of height 5.4m surrounding a RCC
roof. The bottom of the wall rests over a foundation block. Assume roof load equal to
45kN/m. A pier provided at a spacing of 3.6m along length of wall.
4. Design a compound wall of height 1.8m to the top of 100mm thick coping. Assume wind
pressure is equal to 1kN/m2 and is UDL. The safe bearing pressure of soil is 120kN/m2.
5. A solid footing has to transfer a dead load of 1000kn and an imposed load of 400kn from
a square column 400mm×400mm. Assuming fck=20N/mm2 and fy=415N/mm2 and safe
bearing capacity to be 200KN/m2, Design the footing.
6. Design a combined rectangular footing for two columns spaced at 500cm centers. The
first column 300mm×300mm carried load of 1000kn.and second column 300mmx300mm
carries a load of 1500kn at service state. Weight of Soil = 20kN/m2, angle of repose=300
and safe bearing capacity of soil = 150kN/m2. Use concrete M25 and steel Fe415.
7. A solid footing has to transfer a dead load of 1000kN and an imposed load of 400kN
from a square column 400×400mm (with 16mm bars). Assuming fy=415 and
fCK=20N/mm2 and safe bearing capacity to be 200kN/m2. Design the footing.
8. Design a plain concrete footing for a 450mm wall carrying 300kN per meter length.
Assume grade 20 concrete and the bearing capacity of soil to be 200kN/m2.
SNS COLLEGE OF ENGINEERING
DEPARTMENT OF CIVIL ENGINEERING
B.E DEGREE EXAMINATIONS: August- 2014
Fifth Semester-Internal Assessment Exam – I
CE2306 – Design of RC Elements
Time: 3 hours
Maximum: 100 marks
Answer ALL Question-PART A – (10 x 2 = 20 Marks)
1.
2.
3.
4.
5.
6.
7.
8.
What are the advantages of limit state method over working stress and ultimate load methods?
What is modular ratio? Determine the modular ratio at M25 grade concrete.
State the assumptions made in working stress method.
Draw stress-strain curve for steel and concrete.
Differentiate between nominal mix and design mix of concrete.
Explain the design consideration for effective span of beam.
Why is greater safety adopted for concrete than for steel in working stress method?
What are the advantages of limit state method over working stress method and ultimate load
method?
9. What are the three basic methods using factor of safety to achieve safe workable structures?
10. Weather cover is necessary for reinforcement? Explain with reasons.
PART B – (5 X 16 = 80 Marks)
1. Derive the expression for the depth of neutral axis, moment of resistance of a rectangular singly
reinforced balanced beam section under flexure and obtain the design constants k, j, Q for M20
grade of concrete and Fe 415 grade steel. Use working stress method.
(OR)
Differentiate between working stress method and Limit state method.
(8)
What are properties of cracked and uncracked sections?
(8)
2. Design a rectangular beam section subjected to an ultimate moment of 60kNm. Use concrete M20
and steel Fe415. Adopt working stress method.
(OR)
A rectangular beam of size 250mm width and 500mm effective depth. Determine the allowable
moment of resistance and stresses induced in extreme compression fiber of concrete, compression
steel and tension steel reinforced with 3-16mm diameter tension steel and 2-16mm diameter
compression steel. Consider effective cover to compression reinforcement equal to 35mm.
Consider M20 concrete and Fe 415 steel.
3. Determine the reinforcement for a T beam with flange width = 1500mm, web width = 300mm,
thickness of slab = 100mm, effective depth 500mm, to carry a moment of 250kNm due to
characteristic loads. Use M25 concrete and Fe 415 steel. Using Working Stress Design.
(OR)
Analyze a T beam with flange width = 1500mm, web width = 300mm, thickness of slab =
100mm, effective depth 450mm subjected to a moment of M=200kNm. Determine the stresses
induced in extreme compression fiber of concrete, compression steel and tension steel reinforced
with 6-25mm diameter tension steel and 4-12mm diameter compression steel. Consider M20
concrete and Fe 415 steel. Consider d’= 50mm.
4. Design a simply supported slab supported on masonry walls to the following requirements. Clear
span between supports=3m. Live load= 3kN/m2. M20 and Fe415 combinations are used.
(OR)
Design a cantilever slab supported to carry a live load of 2.5kN/m2. The projection of the slab is
1.5m. M20 and fe415 combinations are used.
5. Design a RC beam to carry a load of 6kN/m inclusive of its own weight on an effective span of
6m keep the bredth to be 2/3 of the effective depth. The permissible stresses in concrete and steel
are not to exceed 5N/mm2 and 1.4 N/mm2 . Take m=18.
(OR)
Analyze a T- beam section of 300 mm width and 1000mm width of flange and 500mm effective
depth and 120mm flange thickness. Determine the stresses induced in extreme compression fiber
of concrete and tension steel. 6-25mm diameter tension steel is provided. Consider M20 concrete
and Fe 415 steel. Consider d’= 50mm.
SNS COLLEGE OF ENGINEERING
DEPARTMENT OF CIVIL ENGINEERING
B.E DEGREE EXAMINATIONS: September- 2014
Fifth Semester-Internal Assessment Exam – II
CE2306 – Design of RC Elements
Time: 3 hours
Maximum: 100 marks
Answer ALL Question-PART A – (10 x 2 = 20 Marks)
11. What are the purposes of providing shear reinforcement in a beam?
12. Why shear reinforcements are not required for RC slab?
13. Explain the difference between primary and secondary torsion. Give two examples.
14. Under what circumstances doubly reinforced beams resorted to?
15. What are assumptions made in Limit state of flexure?
16. What is Singly and doubly reinforced beam?
17. What is meant by characteristic strength of concrete?
18. Write down the formulae for calculating effective width of flange in a flanged beam.
19. What are the loads and forces generally encountered in RCC structures?
20. What are the purposes of partial safety factor? Write the partial safety factor for steel and
concrete.
PART B – (5 X 16 = 80 Marks)
6. Design the interior span of a continuous one way slab for an office floor continuous over
tee beams spaced at 3 meters. Live load = 4kN/m, Floor finish = 1kN/m2.Use concrete
M20 and steel Fe415. Adopt limit state method. Sketch the steel reinforcement.
(OR)
Design the reinforcements required in a simply supported RC slab for a floor to carry a
Live load of 5KN/m2. The clear dimension of the room is 5m x 8m with 300 mrn walls
all around. Assume a floor finish of 1.5KN/m2. Materials used are M20 grade concrete
and Fe415 steel. Adopt Limit state method
7. A doubly reinforced beam 250 mm wide and 500 mm overall depth. It is reinforced with
2 bars of 20 mm dia compression reinforcement with the effective cover of 40mm and 4
bars of 25mm tension steel. Determine the ultimate moment of resistance of the beam.
M25 and Fe 415 grades are used.
(OR)
Design a rectangular beam section of 300mm width and 550mm overall depth which is
subjected to a ultimate moment of 300kNm. M20 and Fe 415 grades are used.
8. T-beam of 300mm width of web, 1200mm width of flange, 100mm thickness of flange
and 500 mm effective depth. Ii is reinforced with 6-25mm diameter bars. Determine the
ultimate moment of resistance of the beam. M20 and Fe 415 grades are used.
(OR)
A series of beams placed at 2.5m centres are supported on masonry walls and the
effective span of beam is 5m. The slab thickness is 100mm. and the webs below the slab
are 200mm wide and 250mm deep. If the slab and beams are so cast as to act together,
determine the reinforcement at midspan for T-beam to carry a imposed load of 5kN/m2. .
Use M20 and Fe 415 grades.
9. Determine the shear reinforcement for T-beam and slab system are made of beams
spaced at 2.4m with clear span of 7.5m between masonry walls of 300mm thick. For Tbeam Df= 10mm, bw= 300mm, D=600mm, fck= 20N/mm2, fy=415N/mm2. Live load=
8kN/m2.. Assume that 2 nos of 28mm dia bars are continued to support.
(OR)
A beam of rectangular section of 350mm and 550mm depth reinforced with 6nos of
20mm dia bars out of which 3 no of bars have been bent up at 45o . Determine he shear
resistance and additional shear reinforcement required if it s subjected to an ultimate
shear of 300kN.
10. Design a rectangular beam section of 250mm width and 500mm overall depth subjected
to ultimate values of bending moment of 40kNm, shear force of 40 kN, Torsion moment
of 30kNm. Adopt effective cover of 50mm top and bottom. Use M20 and Fe 415 grades.
(OR)
Design the transverse reinforcement using 2 legged stirrups of 10mm dia for following
data.
Size of the beam : 300mm X 600mm
Factored Torsion : 45kNm
Factored Shear : 95 kN
Tension reinforcement : 4 nos 20mm dia
Compression Reinforcement : 2 no of 12mm dia
Clear cover on all four sides( beyond the stirrups) : 15 mm
M20 and Fe415 steel
Reg.No. :
SNS COLLEGE OF ENGINEERING
DEPARTMENT OF CIVIL ENGINEERING
B.E DEGREE EXAMINATIONS: October- 2014
Fifth Semester-Internal Assessment Exam – III
CE2306 – Design of RC Elements
Time: 3 hours
Maximum: 100 marks
Answer ALL Question-PART A – (10 x 2 = 20 Marks)
21. Distinguish between barced column and unbraced column.
22. What is the importance to provide lap and anchorage length?
23. Write three assumptions made for the limit state design of columns failing in pure
compression.
24. What are the factors affecting behavior of long columns?
25. State the Rankine’s equation to determine the minimum depth of foundation.
26. Under what circumstances combined footings are provided?
27. What are assumptions made in Design of Foundation?
28. What are the different types of foundation?
29. Explain and sketch the one way shear and two way shear on footing.
30. What is punching shear in RCC footing?
PART B – (5 X 16 = 80 Marks)
11. An Overhanging beam has 5m from the support to support and 1.5m overhanging. The cross
section of the beam is 300 mm x 500 mm and the design load applied through was 40kN/m. 4
bars of 20mm diameter plain bars are provided with 50mm effective cover. What is the maximum
bond stress developed and find the anchorage length required for the over hanging portion.
(OR)
A Simply supported beam of 8m span is reinforced with 5 bars of 25mm diameter at
center of span and 3 bars are continued into supports. Check the development length at
supports. The beam supports a characteristic total load of 50kN/m.
12. Design a column of size 450mm x 300mm. Use M30 concrete and Fe415 steel. Take lex = 6m
ley = 5.5m, Pu = 1600kN, Mux = 45kNm at top and 40kNm at bottom. Muy= 40kNm at top and
25kNm at bottom. The column is bent in double curvature and assume a cover of 50mm.
(OR)
Design the longitudinal reinforcement in a short column 400mm x 600mm subjected to
an ultimate axial load of 1600kN together with ultimate moments of 120kNm and 90
kNm about the major and minor axis respectively. The reinforcements are distributed
equally on all four sides. Adopt M20 grade concrete and Fe415 steel bars.
13. A rectangular column of effective height 4m is subjected to a load of 1800kN and
bending moment of 100kNm about major axis of the column. Design a suitable section
for the column so that the width should not exceed 400 mm. Use M25 concrete and Fe
415 steel. (OR)
Design the reinforcements in a circular column of diameter 350mm with helical
reinforcement of 8mm diameter to support a factored load of 1400 kN. The column has
an unsupported length of 3.5m and is braced against sway. Adopt M20 grade concrte and
Fe 415 steel.
14. A reinforced rectangular column of size 300x 450 mm carries axial dead load of 670kN
and imposed load of 330kN. Moment acting on the column 100kNm.Safe bearing
capacity of the soil is 150kN/m2 . Adopt M20 grade concrete and Fe 415 steel. Design a
suitable foundation.
(OR)
A reinforced rectangular column of size 300x 450 mm carries axial load of 600kN with
an eccentricity of 100mm along its length. Safe bearing capacity of the soil is 150kN/m2 .
Adopt M20 grade concrete and Fe 415 steel. Design a suitable foundation to maintain
uniform base pressure.
15. Write down the design steps in detail of a rectangular combined footing.
Sketch the standard detailing of Two span one way continuous slab with curtailment
details.
(OR)
A solid footing has to transfer a dead load of 900kN and an imposed load of 600kN for a
RC column of size 300mm x 450mm. SBC of soil is 120kN/m2. Adopt M20 grade
concrete and Fe 415 steel. Design a suitable foundation.
MULTIPLE CHOICE QUESTIONS
1. In a combined footing if shear stress exceeds 5 kg/cm2, the nominal stirrups provided are:
A.6 legged
B. 8 legged
C. 10 legged
D.12 legged
E. none of these.
Answer
Answer: Option D
2. The maximum area of tension reinforcement in beams shall not exceed
A.0.15%
B. 1.5%
C. 4%
D.1%
Answer & Explanation
Answer: Option C
3. As per I.S. 456 - 1978, the pH value of water shall be
A.less than 6
B. equal to 6
C. not less than 6
D.equal to 7
Answer & Explanation
Answer: Option C
4. The minimum number of main steel bars provided in R.C.C.
A.rectangular columns is 4
B. circular columns is 6
C. octagonal columns is 8
D.all the above.
Answer & Explanation
Answer: Option D
5. Post tensioning system
A.was widely used in earlier days
B. is not economical and hence not generally used
C. is economical for large spans and is adopted now a days
D.none of these.
Answer & Explanation
Answer: Option D
6. An R.C.C. column is treated as long if its slenderness ratio is greater than
A.30
B. 35
C. 40
D.50
E. 60
Answer & Explanation
Answer: Option D
7. The width of the flange of a T-beam should be less than
A.one-third of the effective span of the T-beam
B. distance between the centres of T-beam
C. breadth of the rib plus twelve times the thickness of the slab
D.least of the above.
Answer & Explanation
Answer: Option D
8. A prestressed rectangular beam which carries two concentrated loads W at L/3 from either end,
is provided with a bent tendon with tension P such that central one-third portion of the tendon
remains parallel to the longitudinal axis, the maximum dip h is
A.
B.
C.
D.
E.
Answer & Explanation
Answer: Option C
9. Pick up the correct statement from the following:
A.A pile is a slender member which transfers the load through its lower end on a strong strata
B. A pile is a slender member which transfers its load to the surrounding soil
C. A pile is a slender member which transfers its load by friction
A pile is a cylindrical body of concrete which transfers the load at a depth greater than its
D.
width.
Answer & Explanation
Answer: Option B
10. Cantilever retaining walls can safely be used for a height not more than
A.3 m
B. 4 m
C. 5 m
D.6 m
E. 8 m
Answer & Explanation
Answer: Option D
11. If W is the load on a circular slab of radius R, the maximum circumferential moment at the
centre of the slab, is
A.
B.
C.
D.zero
E. none of these.
Answer & Explanation
Answer: Option C
12. If a bent tendon is required to balance a concentrated load W at the centre of the span L, the
central dip h must be at least
A.
B.
C.
D.
E.
Answer & Explanation
Answer: Option D
13. For M 150 mix concrete, according to I.S. specifications, local bond stress, is
A.5 kg/cm2
B. 10 kg/cm2
C. 15 kg/cm2
D.20 kg/cm2
E. 25 kg/cm2
Answer & Explanation
Answer: Option B
14. If the average bending stress is 6 kg/cm2 for M 150 grade concrete, the length of embedment
of a bar of diameter d according to I.S. 456 specifications, is
A.28 d
B. 38 d
C. 48 d
D.58 d
E. 95 d
Answer & Explanation
Answer: Option D
15. Bottom bars under the columns are extended into the interior of the footing slab to a distance
greater than
A.42 diameters from the centre of the column
B. 42 diameters from the inner edge of the column
C. 42 diameters from the outer edge of the column
D.24 diameter from the centre of the column
Answer & Explanation
Answer: Option C
16. The diameter of longitudinal bars of a column should never be less than
A.6 mm
B. 8 mm
C. 10 mm
D.12 mm
E. none of these.
Answer & Explanation
Answer: Option D
17. The design of a retaining wall assumes that the retained earth
A.is dry
B. is free from moisture
C. is not cohesives
D.consists of granular particles
E. all the above.
Answer & Explanation
Answer: Option E
18. Dimensions of a beam need be changed if the shear stress is more than
A.10 kg/cm2
B. 15 kg/cm2
C. 20 kg/cm2
D.25 kg/cm2
Answer & Explanation
Answer: Option C
19. The thickness of base slab of a retaining wall generally provided, is
A.one half of the width of the stem at the bottom
B. one-third of the width of the stem at the bottom
C. one fourth of the width of the steam at the bottom
D.width of the stem at the bottom
E. twice the width of the steam at the bottom.
Answer & Explanation
Answer: Option D
20. For a circular slab carrying a uniformly distributed load, the ratio of the maximum negative
to maximum positive radial moment, is
A.1
B. 2
C. 3
D.4
E. 5
Answer & Explanation
Answer: Option B
21. Thickened part of a flat slab over its supporting column, is technically known as
A.drop panel
B. capital
C. column head
D.none of these.
Answer & Explanation
Answer: Option A
22. An R.C.C. beam not provided with shear reinforcement may develop cracks in its bottom
inclined roughly to the horizontal at
A.25°
B. 35°
C. 45°
D.55°
E. 60°
Answer & Explanation
Answer: Option C
23. The effective span of a simply supported slab, is
A.distance between the centres of the bearings
B. clear distance between the inner faces of the walls plus twice the thickness of the wall
C. clear span plus effective depth of the slab
D.none of these.
Answer & Explanation
Answer: Option B
24. Pick up the incorrect statement from the following:
A.In the stem of a retaining wall, reinforcement is provided near the earth side
B. In the toe slab of a retaining wall, rein forcement is provided at the bottom of the slab
C. In the heel slab of a retaining wall, rein forcement is provided at the top of the slab
D.None of these.
Answer & Explanation
Answer: Option D
25. The minimum cube strength of concrete used for a prestressed member, is
A.50 kg/cm2
B. 150 kg/cm2
C. 250 kg/cm2
D.350 kg/cm2
E. 400 kg/cm2
Answer & Explanation
Answer: Option D
26. The number of treads in a flight is equal to
A.risers in the flight
B. risers plus one
C. risers minus one
D.none of these.
Answer & Explanation
Answer: Option C
27. A short column 20 cm x 20 cm in section is reinforced with 4 bars whose area of cross
section is 20 sq. cm. If permissible compressive stresses in concrete and steel are 40 kg/cm2
and 300 kg/cm2, the Safe load on the column, should not exceed
A.4120 kg
B. 41, 200 kg
C. 412, 000 kg
D.none of these.
Answer & Explanation
Answer: Option B
28. The reinforced concrete beam which has width 25 cm, lever arm 40 cm, shear force 6t/cm2,
safe shear stress 5 kg/cm2 and B.M. 24 mt,
A.is safe in shear
B. is unsafe in shear
C. is over safe in shear
D.needs redesigning.
Answer & Explanation
Answer: Option B
29. In a beam the local bond stress Sb, is equal to
A.
B.
C.
D.
E. None of these.
Answer
Answer: Option A
30. According to I.S. : 456 specifications, the safe diagonal tensile stress for M 150 grade
concrete, is
A.5 kg/cm2
B. 10 kg/cm2
C. 15 kg/cm2
D.20 kg/cm2
E. 25 kg/cm2
Answer
Answer: Option A
31. A foundation rests on
A.base of the foundation
B. subgrade
C. foundation soil
D.both (b) and (c)
Answer & Explanation
Answer: Option D
32. For initial estimate for a beam design, the width is assumed
A.1/15th of span
B. 1/20th of span
C. 1/25th of span
D.1/30th of span
E. 1/40th of span.
Answer & Explanation
Answer: Option D
33. If R and T are rise and tread of a stair spanning horizontally, the steps are supported by a wall
on one side and by a stringer beam on the other side, the steps are designed as beams of
width
A.R + T
B. T-R
C. R2 + T2
D.R-T
Answer & Explanation
Answer: Option C
34. The advantage of a concrete pile over a timber pile, is
A.no decay due to termites
B. no restriction on length
C. higher bearing capacity
D.not necessary to cut below the water mark
E. all the above.
Answer & Explanation
Answer: Option E
35. If the permissible compressive stress for a concrete in bending is C kg/m2, the modular ratio
is
A.2800/C
B. 2300/2C
C. 2800/3C
D.2800/C2
Answer: Option C
1. If permissible working stresses in steel and concrete are respectively 1400 kg/cm2 and 80
kg/cm2 and modular ratio is 18, in a beam reinforced in tension side and of width 30 cm and
having effective depth 46 cm, the lever arms of the section, is
A.37 cm
B. 38 cm
C. 39 cm
D.40 cm
Answer & Explanation
Answer: Option D
2. If the maximum bending moment of a simply supported slab is M Kg.cm, the effective depth
of the slab is (where Q is M.R. factor)
A.
B.
C.
D.
E.
Answer & Explanation
Answer: Option E
3. Though the effective depth of a T-beam is the distance between the top compression edge to
the centre of the tensile reinforcement, for heavy loads, it is taken as
A.
B.
C.
D.
E.
Answer & Explanation
Answer: Option C
4. Design of R.C.C. cantilever beams, is based on the resultant force at
A.fixed end
B. free end
C. mid span
D.mid span and fixed support.
Answer & Explanation
Answer: Option A
5. If the length of a wall on either side of a lintel opening is at least half of its effective span L,
the load W carried by the lintel is equivalent to the weight of brickwork contained in an
equilateral triangle, producing a maximum bending moment
A.
B.
C.
D.
E.
Answer & Explanation
Answer: Option C
6. Piles are usually driven by
A.diesel operated hammer
B. drop hammer
C. single acting steam hammer
D.all the above.
Answer & Explanation
Answer: Option D
7. In favourable circumstances a 15 cm concrete cube after 28 days, attains a maximum crushing
strength
A.100 kg/cm2
B. 200 kg/cm2
C. 300 kg/cm2
D.400 kg/cm2
E. 500 kg/cm2
Answer & Explanation
Answer: Option D
8. The angle of internal friction of soil mass is the angle whose
tangent is equal to the rate of the maximum resistance to sliding on any internal inclined
A.
plane to the normal pressure acting on the plane
sine is equal to the ratio of the maximum resistance to sliding on any internal inclined plane
B.
to the normal pressure acting on the plane
cosine is equal to the ratio of the maximum resistance sliding on any internal inclined plane
C.
to the normal pressure acting on the plane
D.none of these.
Answer & Explanation
Answer: Option A
9. The anchorage value of a hook is assumed sixteen times the diameter of the bar if the angle of
the bend, is
A.30°
B. 40°
C. 45°
D.60°
E. all the above.
Answer & Explanation
Answer: Option E
10. Top bars are extended to the projecting parts of the combined footing of two columns L
distance apart for a distance of
A.0.1 L from the outer edge of column
B. 0.1 L from the centre edge of column
C. half the distance of projection
D.one-fourth the distance of projection.
Answer & Explanation
Answer: Option B
11. According to the steel beam theory of doubly reinforced beams
A.tension is resisted by tension steel
B. compression is resisted by compression steel
C. stress in tension steel equals the stress in compression steel
D.no stress is developed in compression concrete as well as in tension concrete
E. all the above.
Answer & Explanation
Answer: Option E
12. The thickness of the flange of a Tee beam of a ribbed slab is assumed as
A.width of the rib
B. depth of the rib
C. thickness of the concrete topping 0d) half the thickness of the rib
D.twice the width of the rib.
Answer & Explanation
Answer: Option C
13. In a singly reinforced beam, if the permissible stress in concrete reaches earlier than that in
steel, the beam section is called
A.under-reinforced section
B. over reinforced section
C. economic section
D.critical section.
Answer & Explanation
Answer: Option B
14. The length of the lap in a compression member is kept greater than bar diameter x
(Permissible stress in bar / Five times the bond stress) or
A.12 bar diameters
B. 18 bar diameters
C. 24 bar diameters
D.30 bar diameters
E. 36 bar diameters
Answer & Explanation
Answer: Option C
15. If At is the gross area of steel in tension, d is the effective depth of the beam and y is the
depth of the centre of gravity of the resultant compression, the moment of resistance M of the
beam, is given by
A.M = At(d - y)
B. M = Atf(d - y)
C. M = Atf(d + y)
D.
M=
E. none of these.
Answer & Explanation
Answer: Option C
16. A pre-stressed concrete member
A.is made of concrete
B. is made of reinforced concrete
C. is stressed after casting
D.possesses internal stresses.
Answer & Explanation
Answer: Option D
17. A part of the slab may be considered as the flange of the T-beam if
A.flange has adequate reinforcement transverse to beam
B. it is built integrally with the beam
C. it is effectively bonded together with the beam
D.all the above.
Answer & Explanation
Answer: Option D
18. An R.C.C. roof slab is designed as a two way slab if
A.it supports live loads in both directions
B. the ratio of spans in two directions is less than 2
C. the slab is continuous over two supports
D.the slab is discontinuous at edges.
Answer & Explanation
Answer: Option B
19. If C is creep coefficient, f is original prestress in concrete, m is modular ratio, E is Young's
modulus of steel and e is shrinkage strain, the combined effect of creep and shrinkage is:
A.(1 - C)mf - eE
B. (C - 1)mf + eE
C. (C - 1)mf - eE
D.(1 - C)mf + eE
Answer & Explanation
Answer: Option B
20. In a combined footing for two columns carrying unequal loads, the maximum hogging
bending moment occurs at
A.less loaded column
B. more loaded column
C. a point equidistant from either column
D.a point of the maximum shear force
E. a point of zero shear force.
Answer & Explanation
Answer: Option E
Assignment-1
1. A reinforced concrete beam of rectangular section 300 mm wide × 650 mm
deep is reinforced with 4 bars of 25 mm diameter at an effective depth of
600 mm. Calculate the neutral axis depth and estimate the safe moment of
resistance of the section adopting M25 grade concrete and Fe415 HYSD
bars.
2. A reinforced concrete beam of rectangular section 350 mm wide × 750 mm
overall depth is reinforced with 3 bars of 20 mm diameter at an effective
depth of 700 mm. Adopting M30 grade concrete and FE500 grade steel
reinforcement, calculate the safe moment of resistance of the section. If
the beam spans over 5 m, estimate the safe permissible live load on the
beam.
3. A reinforced concrete beam of rectangular section having a width of 400
mm and overall depth 850 mm is reinforced with 4 bars of 25 mm diameter
both on the compression and tension sides at an effective cover of 50 mm.
Using M20 grade concrete and Fe415 HYSD bars, compute (a) the actual
neutral axis; (b) the critical neutral axis; and (c) the safe moment of
resistance of the section.
4. A reinforced concrete rectangular section 300 mm wide × 600 mm overall
depth is reinforced with 4 bars of 25 diameter at an effective cover of 50
mm on the tension side. Assuming M20 grade concrete and FE415 HYSD
bars, determine the allowable bending moment and the stresses in steel
and concrete corresponding to this moment.
5.
6.