Lecture Notes in Civil Engineering
STRUCTURAL
DESIGN
PART I – REINFORCED CONCRETE STRUCTURES
Chapters
1.
Basic Properties of Materials
2.
Limit State Method
3.
Design for Flexure
4.
Design for Shear
5.
Design for Bond
6.
Design of Slabs & Staircases
KIRAN S. R.
Lecturer
Department of Civil Engineering
Central Polytechnic College Trivandrum
CHAPTER 1
BASIC PROPERTIES OF MATERIALS
Concrete consists of the following ingredients:
1.
2.
3.
4.
5.
Cement – Binding material; Micro-void filler
Fine Aggregate – Void filler (use Sand)
Coarse Aggregate – Strength provider (use Hard blasted granite chips)
Water – Hydrates cement; Controls workability and strength
Admixture – modifies the property of concrete in fresh and hardened state
1. Cement [Cl. 5 of IS456]
Manufactured by mixing Calcareous (limestone) & Argillaceous (clay) materials in definite
proportions. It contains the following oxides:
Constituent
Approximate
Remarks
%
composition
by volume
CaO
62
Gives strength & soundness to cement. If
excess, cause unsoundness.
SiO2
22
Gives strength. If excess, delays setting
Al2O3
5
Causes
setting
&
lowers
clinkering
temperature
Fe2O3
3
Gives color & hardness to cement
MgO
2
Gives color & hardness to cement. If excess,
cause unsoundness.
SO3
2
If excess, cause unsoundness
Alkalies (K2O,
1
If excess, cause alkali-aggregate reaction,
Na2O)
efflorescence & staining
It is fed into a rotary kiln and blasted at 1500oC. Clinker (of nodular shape) is obtained, which is
cooled and ground into fine powder. To this 3-5% Gypsum is added to prevent quick setting of
cement.
The cement, thus obtained, contains four major compounds known as Bogue’s Compounds:




tricalcium silicate (C3S)
dicalcium silicate (C2S)
tricalcium aluminate (C3A)
tetracalcium aluminoferrite (C4AF)
~ 50% by volume of cement
~ 25%
~ 15%
~ 10%
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 Hydration of cement: When Cement mixed with water, the Bogue’s compounds react
with water and heat is liberated (Exothermic Reaction). This heat is called Heat of
Hydration.
Order of hydration is as follows:
Bogue’s
Compounds
Heat of Hydration
Remarks
C4AF
C3A
C3S
C2S
100 cal/gm
200 cal/gm
120 cal/gm
60 cal/gm
Responsible for initial strength development of
cement (<3 days)
Early strength development of cement (3-7 days)
Late strength development of cement (7-28 days)
 Performance Characteristics of Ordinary Portland Cement:
Property
Fineness
Definition
Tests
Measure of size Sieve test
of
cement
particles
Air
Permeability
Test
Consistency determination of Vicat
the quantity of apparatus
water required to (mould
of
prepare cement 80mm dia &
paste
of 40mm depth
standard
+ plunger of
consistency (P)
10mm dia &
50mm long)
Setting time
Vicat
(a)
Initial the time when apparatus
Setting
the
paste (mould
+
Time
becomes
1mm
dia
unworkable .
needle)
(b)
Final
Setting
the time to reach Vicat mould
Time
a
state
of + needle with
complete
circular
solidification .
arrangement)
Soundness Whether
the Le Chatelier
cement does not test
undergo
appreciable
change
in Autoclave
volume (causing test
concrete
to
crack)
Specification
Only <10% by mass of cement shall be
retained on 90μ.
Specific surface
>2250cm2/gm.
of
cement
should
be
The cement paste is consistent, if the plunger
penetrates through 33-35mm from the top of
the mould.
Generally, consistency is 26-33%.
To prepare paste, water used= 0.85 P
Should be greater than 30minutes.
Should be less than 10 hours.
To prepare paste, water used= 0.78 P
The distance between indicators should be
<10mm.
% increase in length of specimen <0.8%
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Property
Strength
Definition
Average 28-day
compressive
strength of 3
cement-mortar
cubes
Tests
Specification
Use cement- Should be >53MPa, for 53grade cement.
mortar
1:3 Should be >43MPa, for 43grade cement.
using
Should be >33MPa, for 33grade cement.
(P/4+3)%
water
to
make cubes
of 7cm size.
Types of cement – refer Cl. 5.1 of IS 456
2. Aggregates [Cl. 5.3 of IS456]
 Fine Aggregates
 particle size between 0.075mm and 4.75mm
 generally used – river sand, manufactured sand





Coarse Aggregates
particle size > 4.75mm
generally used – Hard blasted granite chips
The nominal maximum size of aggregate should be < (1/4)th the minimum thickness of
the member
In case of heavily reinforced sections, nominal maximum size of aggregate should be
 Clear distance between bars minus 5mm
whichever is smaller
 Minimum cover to reinforcements minus 5mm
 Grading of Aggregates
 ‘Grading’ is the particle size distribution of aggregate; it is measured by sieve analysis,
and is generally described by means of a grading curve, which depicts the ‘cumulative
percentage passing’ against the standard IS sieve sizes.
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




In figure, standard grading curves 1, 2, 3 and 4 are shown (plotted per Indian Standards
IS383). The particle size distribution of the given sample of aggregates shall conform to
any of the zones A, B or C. Curve 1 represents the coarsest grading, while curve 4
represents the finest grading.
The grading (as well as the type and size) of aggregate is a major factor which influences
the workability of fresh concrete, and its consequent degree of compaction.
This is of extreme importance with regard to the quality of hardened concrete, because
incomplete compaction results in voids, thereby lowering the density of the concrete and
preventing it from attaining its full compressive strength capability; furthermore, the
impermeability and durability characteristics get adversely affected.
It is seen from the figure given below that with 95% density (i.e., with 5 percent of voids),
there is only 68% strength (i.e., 32% strength lost). This shows that the presence of just
5% voids can lower the strength by 32%.
Presence of more “fines” (i.e., cement & sand) in a concrete mix would improve both
workability and resists segregation (segregation means separation of grout from
aggregates in a concrete mix due to addition of excess water to concrete)
3. Water [Cl. 5.4 of IS456]



For mixing of fresh concrete
For curing of concrete, while hardening
Water used for concrete should be potable
Parameters
pH
Organic
Inorganic
Sulphate
Chlorides
Suspended solids
Permissible limit
≥6
< 200 mg/L
< 3000 mg/L
< 400 mg/L
< 500 mg/L for RCC
< 2000 mg/L for PCC
< 2000 mg/L
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



Excess water in concrete tends to rise up to the surface of the mix, as the solid
constituents settle downwards. This is called Bleeding.
Use of seawater for mixing or curing of concrete is not recommended due to the
presence of harmful salts
Generally, water content used in concrete is 180-200 L / m3 of concrete.
Abrams’ Law: Compressive strength of hardened concrete is inversely proportional to
the water-cement ratio (see figure below). A reduction in water-cement ratio generally
results in an increased quality of concrete, in terms of density, strength, impermeability,
reduced shrinkage and creep etc.
4. Admixtures [Cl. 5.5 of IS456]



Added to concrete to modify its properties in fresh & hardened state
Broadly, two types
o Chemical Admixtures (liquid in form)
o Mineral admixtures (fine granular in form)
Types of Chemical Admixtures
o Accelerators – accelerates the hardening process (early strength development)
o Retarders – delays the setting of concrete, to reduce the heat generation
o Superplasticizers – high range water reducers; increase flowability without
increasing water content; produce high strength concrete
o Air-entraining Agents – introduce microscopic air bubble cavities in concrete;
minimize damage due to alternate freezing & thawing
o Bonding admixtures – used to improve adherence of fresh concrete to old
concrete
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
Types of Mineral Admixtures (Puzzolonas) [Cl. 5.2 of IS456] – Mineral admixtures are
generally used as partial replacement of cement in concrete. They react with Calcium
hydroxide in the presence of water to form cementitious compounds.
o Fly Ash
o Ground Granulated BlastFurnace Slag (GGBS)
o Silica Fume
o Rice Husk Ash
o Metakaoline
6. PROPERTIES OF FRESH CONCRETE
 Workability
 Workability is the ease with which the concrete can be mixed, placed,
consolidated and finished.
 Workable concrete is the one which exhibits very little internal friction
between particle and particle or which overcomes the frictional resistance
offered by the formwork surface or reinforcement contained in the concrete.
 The factors affecting workability are given below:

Water Content

Mix Proportions

Size of Aggregates

Shape of Aggregates

Surface Texture of Aggregate

Grading of Aggregate

Use of Admixtures
 The following tests are commonly employed to measure workability.

Slump Test

Compacting Factor Test

Flow Test

Kelly Ball Test

Vee Bee Consistometer Test
 Segregation
 Segregation can be defined as the separation of the constituent materials of
concrete.
 A good concrete has all its constituents properly distributed to form a
homogenous mixture. To ensure this, optimum grading, size, shape and
surface texture of aggregates with optimum quantity of cement & water
makes a mix cohesive. Such a concrete does not exhibit the tendency for
segregation.
 Prime cause of segregation is the difference in specific gravity of
constituents of concrete.
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

Segregation may be one of the following types:
o Coarse aggregate separating out of the rest
o Cement paste or cement-fine aggregate matrix separating out from
coarse aggregate
o Water separating out of the rest
The conditions that favour segregation are:
o Bad mix proportion
o Inadequate mixing
o Excessive compaction by vibration of wet mix
o Large height of dropping of concrete for placement
o Long distance conveyance of mix
 Bleeding
 Here, water from the concrete comes out to the top surface of the concrete
after casting.


The conditions that favour bleeding are:
o Highly wet mix
o Bad mix proportion
o Inadequate mixing
Sometimes, the bleeding water is accompanied to the surface by certain
quantity of cement, which forms a cement paste (known as Laitance) at the
surface.
7. PROPERTIES OF HARDENED CONCRETE
 Grade of concrete
 Designated in terms of letter ‘M’ followed by a number. ‘M’ refers to mix; the
number represents the 28-day characteristic compressive strength of concrete
cubes (150mm) expressed in MPa.
Eg: M20 denotes the concrete mix with 28-day characteristic compressive strength
of 20MPa.

Minimum grade of concrete used is dictated by durability (the environment to
which the structure is exposed to, expressed in terms of exposure conditions)
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Exposure
Condition
Mild
Moderate
Severe
Very
severe
Extreme

Minimum
grade
of
concrete
for
RCC works
M20
M25
M30
M35
M40
Classification
o Ordinary concrete – M10 to M20
o Standard concrete – M25 to M55
o High Strength concrete – M60 and above
 Stress-strain curve of concrete
 Stress-Strain curves of concrete for various grades obtained from uniaxial
compression tests are shown in above figure
 Maximum stress is attained by concrete at an approximate strain of 0.002
 The strain at failure is in the range 0.003 to 0.005
 The curves are linear within the initial portion of the curve. This is approximately
true upto one-third of the maximum stress level, beyond which the non-linearity
continues
 For higher grades of concrete, the initial portion of the stress-strain curve is
steeper, but the failure strain is low. For low strength concrete, the initial slope of
curve is gentle but has high failure strain. (observe the above figure)
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

Poisson Effect: Failure of concrete subject to uniaxial compression is primarily
initiated by longitudinal cracks (cracks developed parallel to direction of loading)
formed due to lateral expansion (because lateral fibres experience tensile stress)
and finally lateral strain exceeds limiting tensile strain of concrete of 0.0001 to
0.0002. These longitudinal cracks generally occur at coarse aggregate-mortar
interface.
The descending part of Stress-Strain curve is attributed to the extensive
microcracking in mortar. This is called Strain-Softening of Concrete.
 Modulus of Elasticity of concrete
 Young’s Modulus of Elasticity (equal to ratio of stress to strain, when the material
is loaded within the linearly elastic limit) for concrete subjected to uniaxial
compression, has validity only within the initial portion of the Stress-Strain curve.
For concrete, there are 3 ways to determine the Modulus of Elasticity. This is
shown in figure.
o Initial Tangent Modulus – Slope of tangent at origin of curve; measure of
Dynamic Modulus of Elasticity of concrete
o Tangent Modulus – Slope of tangent at any point on the curve
o Secant Modulus – slope of line joining origin & one-third of maximum stress
level; measure of Static Modulus of Elasticity of concrete
 Static Modulus of Elasticity of concrete – is applicable to static system of loads on
structures
 Dynamic Modulus of Elasticity of concrete – is applicable when structure is subject
to dynamic loads (wind & earthquake loads)
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum


Secant Modulus at one-third of maximum stress level represents the “Short-term
Static Modulus of Elasticity of Concrete (Ec)”. “Short-term” means the long term
effects of creep & shrinkage are not considered.
According to IS456,
Ec = 5000 f
where fck is the 28-day characteristic compressive strength of 150mm concrete
cubes. Thus, it should be noted that Modulus of Elasticity of concrete is a function
of its strength.
8. PROPERTIES OF STEEL

Stress-strain curve of reinforcing steel
 Reinforcing steel may be categorized broadly into:
o
Plain Mild steel bars

has well-defined yield point

Eg: Fe250 – Yield strength= 250MPa; Ultimate strength= 412MPa;
Min % elongation= 22%
o
High Yield Strength Deformed bars

doesnot have well-defined yield point

these are cold-worked bars (involves stretching and twisting of mild
steel bars)

Eg: Fe415 – Yield strength= 415MPa; Ultimate strength= 485MPa;
Min % elongation= 14.5%

Eg: Fe500 – Yield strength= 500MPa; Ultimate strength= 545MPa;
Min % elongation= 12%
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

Characteristic strength of reinforcing steel =
o yield strength of steel– for those with well-defined yield point (Fe250)
o 0.20% Proof Stress – for those without well-defined yield point (Fe415&Fe500)
0.2% Proof Stress is measured as shown below
9. PERMISSIBLE STRESSES IN CONCRETE AND STEEL [Refer Table21 & 22 in
Annex B of IS456]
 Permissible Stresses in Concrete
Grade of
Direct
concrete Tension
(N/mm2)


Bending
compression
(N/mm2)
Direct
compression
(N/mm2)
Average bond
for plain bars in
tension
(N/mm2)
5
0.8
6
0.9
8
1
9
1.1
10
1.2
in compression, increase the values of stress for
M20
2.8
7
M25
3.2
8.5
M30
3.6
10
M35
4
11.5
M40
4.4
13
For bond stress for plainbars
bars in tension by 25%
For bond stress for HYSD bars, increase the values of stress for bars in tension by
60%
 Permissible Stresses in Steel Reinforcement
Type of stress in
steel bars
Tension
(a) ≤ 20mm dia bar
(b) > 20mm dia bar
Compression
in
column bars
Fe250
(N/mm2)
140
130
130
Fe415
Fe500
(N/mm2) (N/mm2)
230
230
190
275
275
190
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum

NOTE: When effects of temperature, shrinkage and wind (or earthquake) are
considered, in addition to effects of dead, live & impact loads, then the above
values of permissible stresses in concrete & steel may be exceeded upto a limit of
33.33%
10. DESIGN CODES, HANDBOOKS & OTHER REFERENCES
Basic codes for Design
:
1) IS 456 : 2000 — Plain and reinforced concrete – Code of practice
Loading Standards :
1) IS 875 (Parts 1-5) : 1987 — Code of practice for design loads (other than earthquake) for
buildings and structures (second revision)
a. Part 1 : Dead loads
b. Part 2 : Imposed (live) loads
c. Part 3 : Wind loads
d. Part 4 : Snow loads
e. Part 5 : Special loads and load combinations
2) IS 1893 : 2002 — Criteria for earthquake resistant design of structures (fourth revision).
Design Handbooks :
1)
2)
3)
4)
SP 16 : 1980 — Design Aids (for Reinforced Concrete) to IS 456-1978
SP 24 : 1983 — Explanatory Handbook on IS 456 : 1978
SP 34 : 1987 — Handbook on Concrete Reinforcement and Detailing
SP 23 : 1982 — Design of Concrete Mixes
Other Relevant Codes
:
1) IS 13920 : 1993 — Ductile detailing of reinforced concrete structures subjected to
seismic forces.
2) IS 10262 : 2009 — Guidelines for Concrete Mix Proportioning
Other Reference Books
1) Unnikrishna Pillai & Devdas Menon, “Reinforced Concrete Design”, 3rd Edition, Tata
McGraw Hill Publishers, 2009
2) Ashok K. Jain, “Reinforced Concrete – Limit State Design”, 6th Edition, Nem Chand &
Bros Publishers, 2002
.
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
CHAPTER 2
LIMIT STATE METHOD
[Refer Section 5 – Page 67 of IS456]


The acceptable limit for safety and serviceability requirements before failure occurs is
known as Limit State.
LSM involves underestimation of the material strength and overestimation of
external loads. For this, the method uses partial safety factor format.
The design of any structure should satisfy the following 2 conditions:
SAFETY
SERVICEABILITY
 With due consideration to strength,
 Satisfactory performance of structure
stability & structural integrity.
under service loads. Ensures no
discomfort to the user
 If this condition is satisfied, the
likelihood for “collapse” is acceptably
 If this condition is satisfied, the
low under service loads (usual or
likelihood for “user discomfort” is
expected loads) as well as probable
acceptably low under service loads.
overloads
(extreme
winds,
 User discomfort may occur due to:
earthquake etc.)
o Deflection
 Collapse may occur due to:
o Cracking
o Exceeding of strength of
o Vibrations
material or load bearing
o Durability
capacity of material.
o Impermeability
o Sliding
o Thermal Insulation (or Fire
o Overturning
resistance)
o Buckling
 Limit states involved in user comfort
o Fatigue
are
called
“Limit
state
of
o Fracture
serviceability”, which are defined for,
o Deflection
 Limit states involved in collapse are
called “Limit State of Collapse” or
o Cracking
“Ultimate Limit State”, which are
o Durability
defined for the following,
o Fire Resistance
o Flexure
o Compression
o Shear
o Torsion
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IMPORTANT TERMINOLOGIES
1. Characteristic Strength
 It is the strength of the material below which not more than 5% of the test results
are expected to fall.
 Characteristic strength of concrete = 28-day characteristic compressive strength of
150mm concrete cubes (fck). For the design of structures, only 67% of fck is
considered.
 Characteristic strength of reinforcing steel = yield strength of steel (fy) or 0.20%
Proof Stress.
2. Characteristic Loads
 Loads which have 95% probability of not being exceeded during the life of the
structure.
 Magnitudes of these loads are enlisted in:
o IS 875 (Part 1) – Dead Loads
o IS 875 (Part 2) – Live loads
o IS 875 (Part 3) - Windloads
o IS 875 (Part 4) - Snowloads
o IS 875 (Part 5) – Load Combinations
o IS 1893 – Seismic loads
3. Design strength

Design strength =


By this, the strength of material is underestimated
Characteristic strength of material
= 0.67 fck
(for concrete)
= fy
(for reinforcing steel)
Partial safety factor for material = 1.5
(for concrete)
= 1.15
(for reinforcing steel)
.
Therefore, Design strength of concrete =
= 0.447 fck ≈ 0.45 fck
.


and, Design strength of steel
=
.
= 0.87 fy
4. Design Loads
 Design Load = Characteristic Load x Partial safety factor for corresponding load
 By this, the loads on structure are overestimated
 Partial safety factor for different loading conditions is given in Table 18 of IS456
(Page 68)
 Eg: For Limit State for Collapse
 For structures subjected to Dead loads & Live loads
Design load on the structure = 1.5 DL + 1.5 LL
 For structures subjected to Dead loads & Wind loads
Design load on the structure = 1.5 DL + 1.5 WL
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
 For structures subjected to Dead loads, Live loads & Windloads
Design load on the structure = 1.2 DL + 1.2 LL + 1.2 WL
DESIGN STRESS-STRAIN CURVE FOR MATERIALS
a) For concrete in flexural compression
b) For Reinforcing steel in tension or compression
o In case of Fe 250 or Mild Steel:
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
o In case of Cold worked bars (Fe415 and Fe500):
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
CHAPTER 3
DESIGN FOR FLEXURE
FLEXURAL BEHAVIOUR OF REINFORCED CONCRETE MEMBERS





Flexure or Bending occurs in structural elements such as beams/slabs, subjected to
loads acting transverse (i.e., perpendicular) to the axis/plane of the element.
Structural design involves two kinds of problemso Analysis:
 Material properties and complete cross-sectional dimensions, including the
details of steel reinforcement, are known.
 It is required to compute the Loads/Moments that can be resisted by the
member. Further, the stresses in the material, deflections, crackwidth etc
may be computed.
o Design:
 Loads/Moments to be resisted by the member is known.
 It is required to select appropriate grades of materials and compute the
section dimensions, including details of steel reinforcement.
Concrete is very weak in tension, but strong in compression. Its tensile strength is only 715% of compressive strength. Hence in a structural member subjected to flexure (known
as beam), steel bars are embedded in the tension zone of concrete due to the incapacity
of concrete to resist tensile stress.
Modulus of Rupture (fcr) is the theoretical maximum flexural tensile stress reached in
the extreme tension fibre of a beam. When the stress in the extreme tension fibre of a
RCC beam reaches fcr, the concrete cracks for the first time. The corresponding value of
bending moment at that section is known as Moment at first crack or Cracking
moment (Mcr).
If a reinforced concrete section is subjected to gradually increasing load, it shall go
through three distinct stages before complete failure.
o Uncracked phase:
 Applied moment at any section is < Mcr
 Maximum tensile stress developed in concrete is < fcr
 Here, the entire section is effective in resisting the moment
o Linearly elastic Cracked phase:
 Applied moment at any section is > Mcr
 Maximum tensile stress developed in concrete is > fcr , but stress in
concrete is within the linearly elastic region.
 Hence, cracks are initiated at the extreme tension fibres of concrete. With
increased loading, they get widened and propagates towards neutral axis.
The cracked portion of concrete thus becomes completely ineffective in
resisting tensile stress. Hence this tensile stress is completely borne by
steel reinforcement on the tension side. This is why, the concrete on the
tension side of section is completely ignored.
 Analysis of this phase is performed in Working Stress Method (WSM) and
in Limit State of Serviceability.
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o Non-linear Cracked phase or Ultimate Strength phase:
 Applied moment at any section is >> Mcr
 Maximum tensile stress developed in concrete is >> fcr , and stress in
concrete enters non-linear range.
 Analysis of this phase is performed in Limit State of Collapse or Ultimate
Limit State.
 Behaviour of beam in this phase depends on the amount of reinforcing steel
provided. Based on this, the steel reinforcement may or maynot yield before
the final crushing of concrete. Such a section is termed as under-reinforced
section and over-reinforced section respectively (This is elaborated in the
next section).
TYPES OF R.C.C. SECTIONS
Any Reinforced Cement Concrete section may be classified as one of the following types:
1) Balanced section
o Both concrete & steel in the given section reaches ultimate stress simultaneously.
o Yielding of steel and crushing of concrete occurs simultaneously.
o Therefore, strain in
o Concrete (at the extreme compression fibre) = εcu = 0.0035
o Steel (on tension side) = εst = εy =
.
+ 0.002
o Depth of Neutral axis (where the strain is zero) for a balanced section is
designated as Xu,max.
2) Under-reinforced section
o Steel reaches ultimate stress before concrete.
o Here, Yielding of steel occurs first. This is accompanied by wider tensile cracks
and increased curvatures and deflection of the beam. Thus it gives clear warning
to the user about failure. Finally, failure occurs by crushing of concrete.
o In other words, the steel would already have yielded by the time the concrete
crushes.
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
o Therefore, strain in
o Concrete (at the extreme compression fibre) = εcu = 0.0035
o Steel (on tension side)
εst >
.
+ 0.002
o Depth of Neutral axis (Xu) for an under-reinforced section is less than Xu,max
o This type of failure is called “Ductile Failure”.
3) Over-reinforced section
o Concrete reaches ultimate stress first.
o Here, crushing of concrete occurs first. It gives no warning to the user about
failure. Hence, it is a catastrophic or disastrous failure.
o In other words, when the concrete at extreme compression fibre crushes, the steel
would not have reached the yield point.
o Therefore, strain in
o Concrete (at the extreme compression fibre) = εcu = 0.0035
o Steel (on tension side)
εst <
.
+ 0.002
o Depth of Neutral axis (Xu) for an over-reinforced section is greater than Xu,max
o This type of failure is called “Brittle Failure”.
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In short, the above three types of sections can be summarized from the strain distribution at
ultimate limit state (i.e. at failure) as shown below.
TYPES OF RCC BEAMS
Generally three types of beams are studied here
1) Singly Reinforced Beam
o A rectangular beam with steel bars placed on the tension side only.
2) Doubly Reinforced Beam
 A rectangular beam with steel bars placed on both tension & compression sides.
3) Flanged Beam
o A rectangular beam which is cast integrally with slab (monolithic construction).
o Includes T-beams & L-beams.
STUDY OF RELEVANT PROVISIONS OF CODE IS456-2000
1) Exposure conditions
[Refer Cl. 8.2.2; Table 3; 6.1.2; Table 5]
 The general environment to which the concrete will be exposed during its working life is
classified into five levels of severity, as given below. Also given alongside is the
requirements for RCC work with aggregate of 20mm nominal size.
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2) Dimensions of beam
Following are the important parameters of a beam
cross-section
b = width of beam
D = overall depth of beam
d’ = effective cover to the tension steel
d = effective depth of beam (= D – d’)
xu = depth of neutral axis (measured from the
extreme compression fibre)
Ast = Area of steel on the tension side of crosssection
3) Effective span
[Refer Cl. 22.2]
 = clear span + d
= span of beam measured between centers of supports
 = clear span + (d/2)
(For Cantilever Beams)
whichever is minimum
(For simply supported
beam)
4) Concrete cover
[Refer Cl. 26.4; Table 16]
 Clear cover or Nominal cover is the distance measured from the exposed concrete
surface (without plaster and other finishes) to the nearest surface of the reinforcing
bar, including links.
 This cover is required
o to protect the reinforcing bars from corrosion and fire
o to give the reinforcing bars sufficient embedment for sufficient bond with concrete
(prevent slippage between concrete & steel).
 There is a minimum clear cover value specified for each exposure condition (see
previous table), as this satisfies the durability requirement of RCC structure. The
maximum deviation permitted from these values are “+10 mm and –0mm”.
 Minimum clear cover requirements
o for columns = 40 mm
o for footings = 50 mm
5) Spacing of bars
[Refer Cl. 26.3]
 Both minimum & maximum limits of spacing between steel bars are specified in code:
o Minimum Limits
=> ensure that concrete can be placed easily between
bars
o Maximum Limits
=> ensure that concrete crack-widths get controlled
and bond between concrete & steel is improved.
 Codal provisions are summarized in the following figure:
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(a) Singly Reinforced Beams
(b) One way Slabs
6) Area of steel bars
[Refer Cl. 26.5]
 Both minimum & maximum limits of area of steel bars are specified in code:
o Minimum Limits => helps control cracking in concrete due to shrinkage &
temperature change
o Maximum Limits => helps avoid congestion so that concrete can be placed easily
between bars
 Minimum tension reinforcement in beams (Ast,min)
o Ast,min =

.
Minimum compression reinforcement in beams (Asc,min)
o Asc,min = 0
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


Minimum flexural reinforcement in slabs (Ast,min)
o Ast,min = 0.15 % of Ag (for Fe250)
o Ast,min = 0.12 % of Ag (for Fe415)
Where Ag is the gross area of section = b x D
 Note: Even the distribution bars of one-way slab to be provided this
minimum steel.
Maximum tension & compression reinforcement in beams
o Ast,max = 0.04 bD
o Asc,max = 0.04 bD
Side Face Reinforcement
o To be provided if the depth of the beam (or depth of web) exceeds 750mm
o Helps control cracking in concrete due to shrinkage & temperature change
o Helps improve resistance against lateral buckling of web
o Provide a minimum of 0.1% of Ag, where Ag is the gross area of section (or
web area), to be distributed equally on two faces
o Spacing between bars should be ≤ 300mm or beam width (or web
thickness), whichever is minimum.
7) Deflection control by Limiting Span-to-Depth Ratios
[Refer Cl. 23.2]
 Deflection control is a serviceability criterion.
 Deflection in structures is limited as follows:
o Final deflection due to all loads should be ≤
(Including effects of creep, shrinkage & temperature)
𝐬𝐩𝐚𝐧
𝟐𝟓𝟎
o But, Deflection due to all loads should be restricted to
(Including effects of creep, shrinkage & temperature)
≤
𝐬𝐩𝐚𝐧
𝟑𝟓𝟎
or 20mm, whichever is less
(inorder to prevent damages to partitions & finishes)



Since vertical deflection of structures is a function of the ratio
()
( )
,
deflection control is ensured by limiting the (l/d) ratio.
Deflection criteria is satisfied if the (l/d) ratio is
o ≤7
(for Cantilever)
upto 10m spans
o ≤ 20 (for Simply supported)
(in beams & one-way slabs)
o ≤ 26 (for Continuous)
Above values of (l/d) ratio shall be modified in accordance with the code, depending
on the following factors:
[Refer Cl. 23.2.1 (b) to (e)]
o If span > 10m
o Area & stress in tension reinforcement
o Area of compression reinforcement
o For flanged beams
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LIMIT STATE OF COLLAPSE – FLEXURE
[Refer Cl.38 – Page 69 of IS456]
 Assumptions
o Plane sections normal to the beam axis remain plane even after bending, i.e., in an
initially straight beam, normal strain varies linearly over the depth of the section.
o
o
o
The maximum compressive strain in concrete (at the outermost fibre) εcu = 0.0035.
This is so, because regardless of whether the beam is under−reinforced or overreinforced, collapse invariably occurs by the crushing of concrete.
The tensile strength of the concrete is ignored.
For the section to be balanced or under-reinforced, the strain εst in the tension
reinforcement shall not be less than εy:
=>
εst ≥
.
+ 0.002
 Analysis of Singly reinforced Beams
The distribution of stress and strain across the section is shown above.
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From this, the following may be deduced.

The general expression for Depth of Neutral axis (Xu) is obtained from strain diagram,
considering similar triangles.

The limiting value of depth of Neutral axis (Xu,max), which corresponds to a balanced
section, is obtained by substituting
εst
=
.
+ 0.002, in the above equation.
Therefore, we get,

Substituting suitable values of yield strength of steel, the limiting depth of Neutral axis
(Xu,max) may be obtained as follows:

The Concrete Stress block (Compressive stress distribution in concrete at ultimate limit
state) is analysed as follows:
 Compressive force C = 0.36 fck b xu
 Tensile force T = 0.87 fy Ast
 Lever arm (i.e., the perpendicular distance between line of action of compressive
force and tensile force) z = d – 0.42 xu
 For any given section, since the forces are in equilibrium, C = T
Therefore, depth of neutral axis is obtained as:
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
Note: If Xu > Xu,max, then the section is over-reinforced. In such a case, since steel may
not have yielded, the stress in steel shall have not reached 0.87fy.
o Assume a suitable initial (trial) value of Xu
o Determine εst by considering strain compatibility
o Determine the design stress fst corresponding to εst using the design stress-strain
curve
o Derive the value of corresponding to fst by considering T = fst Ast and applying the
force equilibrium condition C = T, whereby
o Compare this value of Xu with the initial value assumed in the first step. If the
difference between the two values is acceptably small, accept this value of Xu.
Otherwise, repeat previous steps with an improved (say, average) value of Xu,
until convergence.

Ultimate moment of resistance of the beam section (Mu) is given by:
Mu
=C.z
Mu
=T.z
OR
= 0.36 fck b xu (d – 0.42 xu)
= fst Ast (d – 0.42 xu)
where fst = 0.87fy if Xu ≤ Xu,max

Limiting moment of resistance of the beam section (Mu,lim) corresponds to the condition
Xu = Xu,max is given by:
Mu,lim = 0.36 fck b xu,max (d – 0.42 xu,max)
OR
Mu,lim = 0.87fy Ast (d – 0.42 xu,max)
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 Analysis of Doubly reinforced Beams
Doubly reinforced beams are resorted to under following circumstances:



Where cross sectional dimensions of the beam are restricted, due to architectural or
other considerations
Where singly reinforced beam is inadequate to resist moment
Where reversal of moments is likely to occur
Presence of compression reinforcement helps reduce long-term deflection & cracking due to
shrinkage & creep.
The distribution of stress and strain across the section is shown above. From this, the following
may be deduced.
Let fsc = stress in compression steel
Asc = area of compression steel
d’ = distance between centroid of compression steel to the extreme compression fibre




Compressive force in Concrete Cc = 0.36 fck b xu
Compressive force in steel Cs = fsc Asc - 0.45 fck Asc
Tensile force in steel T = 0.87 fy Ast
Depth of neutral axis Xu is obtained from Cc + Cs = T, as:
Xu =
.
.
.

In the above equation, fsc is taken from Stress-Strain curve of steel for the corresponding
value of strain εsc computed from:

In general practice, the value of ratio (d’/d) ranges between 0.05 to 0.20.
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

Value of fsc is given by:
o = 0.87 fy = 217.5 N/mm2, for Fe250 grade steel
o If Fe415 or Fe500 is used, the compression steel would not have yielded as
Fe250. Hence, fsc is obtained from εsc.
Value of fsc for a balanced section is given below (in N/mm2)

Ultimate moment of resistance of this section is:
Mu
= Cc (d - 0.42 Xu)
+
Cs (d – d’)
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DESIGN EXAMPLES
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 Analysis of Singly reinforced Flanged Beams


Involves T-sections & L-sections in Beam-supported slab floor system
Flanged beams have 2 parts
o Web: The fully rectangular portion of the beam other than the over hanging parts
of slab
o Flange: The overhanging parts of slab

Flexural compressive stress distribution developed in flange is not uniform along its width
(see the figure below). To ease the analysis, this non-uniform stress distribution is
replaced by an equivalent uniform distribution (of magnitude = the peak stress in actual
distribution) over a reduced width known as “equivalent flange width” or “effective
flange width”, represented as bf.

As per IS 456 (Cl. 23.1.2),
where lo
bf
=
+ bw + 6Df
(for T-beams)
=
+ bw + 3Df
(for L-beams)
= effective distance between points of zero moments in beam
= 0.7 x effective span (for continuous beams)
bw
= breadth of web
Df
= overall depth of flange
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
The value of bf should be restricted as shown:

Analysis:
 Case 1: Neutral axis lies within flange, i.e., Xu ≤ Df




Since all concrete on the tension side is considered ineffective, the T-beam
may be analysed as a rectangular beam of width bf and effective depth d.
Compressive force in Concrete C = 0.36 fck bf xu
Tensile force in steel T = 0.87 fy Ast
Depth of neutral axis Xu is obtained from C = T, as:


Xu =
.
.
Ultimate moment of resistance of the beam section (Mu) is given by:
Mu
Mu
=C.z
OR
=T.z
= 0.36 fck bf xu (d – 0.42 xu)
= fst Ast (d – 0.42 xu)
where fst = 0.87fy if Xu ≤ Xu,max
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 Case 2: Neutral axis lies within web, i.e., Xu > Df

𝟑
Subcase 2(a): If Xu ≥ Df
𝟕
o
o
o
o
o
Compressive force in flange Cf = 0.45 fck (bf – bw) Df
Compressive force in web Cw = 0.36 fck bw xu
Tensile force in steel T = 0.87 fy Ast
Depth of neutral axis Xu is obtained from Cf + Cw = T
Ultimate moment of resistance of this section is:
Mu

= Cf (d –
𝐃𝐟
𝟐
)
+
Cw (d – 0.42 Xu)
𝟑
Subcase 2(b): If Xu < Df
𝟕
o
o
o
o
Here in this case, an equivalent rectangular stress block with uniform
stress 0.45 fck and reduced depth yf is assumed for convenience, as
shown in figure.
Compressive force in web Cw = 0.36 fck b xu
Compressive force in flange Cf = 0.45 fck (bf – bw) yf
where yf = (0.15 xu + 0.65 Df)
≤ Df
Tensile force in steel T = 0.87 fy Ast
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o
o
Depth of neutral axis Xu is obtained from Cw + Cf = T, as:
Ultimate moment of resistance of this section is:
Mu
𝐲
= Cf (d – 𝐟 )
𝟐
+
Cw (d – 0.42 Xu)
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CHAPTER 4
DESIGN FOR SHEAR
STRESSES IN HOMOGENOUS RECTANGULAR BEAMS
From basic mechanics of materials, it is known that the flexural (bending) stress fx and the shear
stress Ƭ at any point in the section, located at a distance y from the neutral axis, are given by:
where I is the second moment of area of the section about the neutral axis, Q the first moment
of area about the Neutral Axis of the portion of the section above the layer at distance y from the
NA, and b is the width of the beam at the layer at which Ƭ is calculated.
Consider an element at a distance y from the Neutral Axis (NA).
The combined flexural and shear stresses on that element can be resolved into equivalent
principal stresses f1 and f2 acting on orthogonal planes.
As a result, the stress on the beam is depicted in terms of the principal stress trajectories as
shown.
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In a material like concrete which is weak in tension, tensile cracks would develop in a direction
that is perpendicular to that of the principal tensile stress. Thus the compressive stress
trajectories in the above figure indicate potential crack patterns, as shown below.
Potential crack patterns
MODES OF CRACKING
1) Flexural cracks
 Occurs in reinforced concrete beams of usual proportions, subjected to relatively high
flexural stresses fx and low shear stresses Ƭ.
 Maximum principal tensile stress occurs in the outer fibre at the bottom face of the
concrete beam at the peak moment locations. As a result, cracks are formed, which are
termed as flexural cracks.
o
 These are formed at 90 from the extreme tension fibre towards neutral axis.
 These are controlled by the tension bars.
2) Diagonal Tension Cracks / Web shear cracks
 Occurs in beams which are subjected to high shear stresses Ƭ (due to heavy
concentrated loads) and relatively low flexural stresses fx (such as, short−span beams
which are relatively deep and have thin webs).
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


It is likely that the maximum principal tensile stress is located at the neutral axis level at
an inclination α= 45o (to the longitudinal axis of the beam)
Cracks occur near the supports (where shear force is generally maximum) near neutral
axis and inclined at 45o to the longitudinal axis of the beam. These are termed as web
shear cracks or diagonal tension cracks.
These can be resisted by providing shear reinforcements or stirrups.
3) Flexure-Shear cracks
 When a ‘flexural crack’ occurs in combination with a ‘diagonal tension crack’, the crack is
termed as a flexure-shear crack.
 Occurs in beam subjected to both flexure and shear.
 Note: The presence of shear stress reduces the strength of concrete in compression as
well as tension. Accordingly, the tensile strength of the concrete in a reinforced concrete
beam subjected to both flexure & shear will be less than that subjected to flexure only.
 Here, flexural crack usually forms first, and extends into a diagonal tension crack.
4) Secondary cracks / Splitting cracks
 The Flexure-Shear cracks may sometimes propagate along the tension reinforcement
towards the support. These are referred to as secondary cracks or splitting cracks.
 These are attributed to
o the wedging action of the tension bar deformations. As the tension reinforcement
o the tension bars serve as dowels across the Flexure-Shear cracks. As the beam
segments on either sides of the crack displaces, secondary cracks may propagate
along tension bars. This is known as dowel action.
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SHEAR PARAMETERS FOR DESIGN
1) Nominal Shear Stress
 For prismatic members of rectangular (or flanged) sections, the Code (Cl. 40.1) uses
the term nominal shear stress Ƭv, defined at the ultimate limit state, as
where Vu is the factored shear force at the section under consideration, b is the width of
the beam (taken as the web width bw in flanged beams), and d the effective depth of the
section.

In the case of members with varying depth, the nominal shear stress, defined above,
needs to be modified, to account for the contribution of the vertical component of the
flexural tensile force Tu which is inclined at an angle β to the longitudinal direction.
Accordingly, the nominal shear stress (Cl. 40.1.1 of the Code), is obtained as
where Vu and Mu are the applied factored shear force and bending moment at the section
under consideration. The negative sign applies where Mu increases in the same direction
as the depth increases and the positive sign applies where Mu decreases in this direction,
as shown below.
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2) Critical Sections for shear
Critical sections are those sections at which shear force is maximum.
Location of critical sections for different cases are shown below. [Refer Cl. 22.6.2]
3) Design Shear Strength of Concrete in Beams (Ƭc)
 It is the average shear strength of concrete in reinforced concrete beams without shear
reinforcement. It is the stress that corresponds to the load at which the first inclined crack
develops.
 If the shear stress in beam Ƭv is less than Ƭc, shear reinforcements are not to be
designed (only minimum shear reinforcements shall be provided). But if Ƭv > Ƭc, shear
reinforcements shall be designed.
 Therefore, Ƭc is the safe limiting value below which the beam is safe even without shear
reinforcement.
 Ƭc depends on grade of concrete (fck) and the percentage tension steel pt = 100Ast/(bd).
The values of Ƭc are given in the Code (Refer Table 19).
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
Shear strength of slabs is higher than that of beams, owing to small thickness. The
thinner the slab, the greater is the increase in shear strength. The Code (Cl. 40.2.1.1)
suggests an increased shear strength for slabs, equal to k Ƭc, where the multiplication
factor k ranges between 1 and 1.3. In general, slabs subjected to normal distributed
loads satisfy the requirement Ƭv < Ƭc, and hence do not need shear reinforcement.
4) Types of Shear Reinforcement

Shear reinforcement, also known as web reinforcement may consist of any one of the
following systems (Cl. 40.4 of the Code)
a) stirrups perpendicular to the beam axis;
b) stirrups inclined (at 45° or more) to the beam axis; and
c) longitudinal bars bent-up (usually, not more than two at a time) at 45° to 60° to the
beam axis, combined with stirrups.


By far, the most common type of shear reinforcement is the two-legged stirrup, comprising
a closed or open loop, with its ends anchored properly around longitudinal bars/stirrup
holders (to develop the yield strength in tension). It is placed perpendicular to the member
axis (‘vertical stirrup’), and may or may not be combined with bent-up bars.
Where bent-up bars are provided, their contribution towards shear resistance shall not be
more than half that of the total shear reinforcement.
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5) Limiting Ultimate Shear Strength of beam (Ƭc,max)
 The nominal shear stress (Ƭv) on the beam should not exceed the limiting total shear
strength of beam including shear reinforcement (Ƭc,max).
 Such a limit is set to the shear stress in beam Ƭv because : if the shear reinforcement
provided in the section is excessive, failure may occur by crushing of concrete (known
as shear-compression failure which occurs due to crushing of the reduced concrete section
after formation of flexure-shear crack), even before yielding of shear reinforcements. Since



this is a brittle fracture, such a failure is undesirable.
Thus by limiting the shear stress in beam Ƭv to less than Ƭc,max, shear-compression
failures can be prevented.
Values of Ƭc,max is given in Table 20 of IS456. It may also be obtained from the
following approximate relation.
Ƭc,max ≈ 0.62 𝑓
In the case of solid slabs, the Code (Cl. 40.2.3.1) specifies that Ƭv should not exceed
0.5 Ƭc,max .
6) Design of shear reinforcement
 If Ƭv > Ƭc
 Design as per Cl. 40.4 of IS456
 Provide shear reinforcements in any of the following forms – vertical
stirrups, Inclined Stirrups and Bent-up bars with stirrups
 Shear force to be resisted by stirrups Vus = Vu - Ƭc b d
 If vertical stirrups are used, center-to-center spacing of the stirrups along
the length of the member, Sv is determined from:
Where Asv is the cross-sectional area of stirrup legs or bent-up bars.
For n-legged stirrups of diameter φ (where n = 2, 4, 6),
Asv =n ( φ2)
From above equation, the required spacing sv of ‘vertical stirrups’ for a
selected diameter φ is given by:
 For vertical stirrups, the maximum spacing between stirrups is limited as
follows:
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 If Ƭv < Ƭc
 If Ƭv < 0.5 Ƭc
No shear reinforcement is required.
 If Ƭv > 0.5 Ƭc
The Code (Cl. 26.5.1.6) specifies a minimum shear reinforcement to be
provided in the form of stirrups.
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
CHAPTER 5
DESIGN FOR BOND
CURTAILMENT OF FLEXURAL TENSION REINFORCEMENT
 A beam is designed for the section of maximum bending moment (Mu,max). Although the
bending moment progressively decreases away from these critical sections, the same
reinforcement (Ast) is provided throughout the beam for convenience. But for economical
consideration, the tension reinforcement may be cutoff (“curtailment”) progressively with
decrease in bending moment.
 Theoretical Bar Cut-off points
The ‘theoretical cut-off point’ for a bar in a flexural member is that point beyond which it is
(theoretically) no longer needed to resist the design moment.
o Since it is found that Ast
o
o
o
o
α Mu,max,
it can be seen that at any section where the moment is, say, 60 percent of Mu,max, the
reinforcement area required is only 60 percent of the designed area Ast, and the
remaining 40 percent may be ‘cut off’ — as far as the flexural requirement is concerned
Bars to be cut off are selected in terms of numbers rather than percentage of areas.
If there are n bars provided at the critical section, and if the bars are all of the same
diameter, then the strength per bar = Mu,max /n .
Theoretical cut-off point for the 1st bar occurs at a section where Mu = (n-1) x Mu,max /n
Theoretical cut-off point for the 2nd bar occurs at a section where Mu = (n-2) x Mu,max /n
Theoretical cut-off point for the 3rd bar occurs at a section where Mu = (n-3) x Mu,max /n
…………………………………………etc etc
Theoretical bar cut-off points may be modified because of the following:
 Development Length Requirements
 Stress developed at the end of a steel bar is zero.
 Stress develops in a steel bar only through bond with surrounding concrete.
 Hence, inorder to develop full design stress in a bar (=0.87fy), sufficient
length of bar must be embedded in concrete on either side of the theoretical
cut-off point. This is called Development Length (Ld).
 Therefore, tension bars should be extended beyond its theoretical
cutoff point by a distance = Development Length (Ld).
 Formation of premature diagonal tension cracks

Cutting off bars in the tension zone lowers substantially the shear strength
of the beams.

The discontinuity at the cut end of the bar introduces stress concentration
which can cause premature flexural cracks that may further develop into
diagonal tension cracks — particularly if the shear stress at this section is
relatively high.
Page 66 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum

Thus, tension steel required at any section must be extended beyond
the theoretical cutoff point by a distance = effective depth (d).
 Codal provisions:
 Tension steel required at any section must be extended beyond the
theoretical cutoff point by a distance = 12 x Bar Diameter.
DESIGN FOR BOND

‘Bond’ in reinforced concrete refers to the adhesion between the reinforcing steel and
the surrounding concrete. It is this bond which is responsible for the transfer of axial
force from a reinforcing bar to the surrounding concrete (therefore, ensures strain
compatibility).

Bond stress : Development of tangential (shear) stress components along the
interface (contact surface) between the reinforcing bar and the surrounding concrete.
The stress so developed at the interface is called bond stress, and is expressed in
terms of the tangential force per unit nominal surface area of the reinforcing bar.

Bond resistance in reinforced concrete is achieved through the following mechanisms:
 Chemical adhesion — due to a gum-like property in the products of hydration
(formed during the making of concrete).
 Frictional resistance — due to the surface roughness of the reinforcement and the
grip exerted by the concrete shrinkage.
 Mechanical interlock — due to the surface protrusions or ‘ribs’ (oriented
transversely to the bar axis) provided in deformed bars.

In general, bond strength is enhanced when the following measures are adopted:
o
o
o
o
o
o
o
deformed (ribbed) bars are used instead of plain bars;
smaller bar diameters are used;
higher grade of concrete (improved tensile strength) is used;
increased cover is provided around each bar;
increased length of embedment, bends and /or hooks are provided;
mechanical anchorages are employed;
stirrups with increased area, reduced spacing and/or higher grade of steel are
used;
Page 67 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
o termination of longitudinal reinforcement in tension zones is avoided;
o any measure that will increase the confinement of the concrete around the bar is
employed.

There are two types of loading situations which induce bond stresses, and accordingly
‘bond’ is characterised as:
 Flexural bond stress / Local Bond stress is that which arises in flexural
members on account of shear or a variation in bending moment. Evidently,
flexural bond is critical at points where the shear (V= dM/dx) is significant.
o Flexural (local) bond stress uf , uniformly distributed over the steel-concrete
interface, is given by:
where Σo is the total perimeter of the bars at the beam section under
consideration, and z is lever arm.
o Actual flexural bond stress is affected by Flexural cracking, local slipping and
splitting, which are not taken care of by the above equation.
Page 68 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
 Anchorage bond stress / Development bond stress is stress developed in
steel reinforcement bars at the extreme ends or at cutoff points. See the following
example, where steel bar is provided on the tension side of a cantilever beam.
o The steel bar in the beam is designed based on the maximum bending moment in
the beam that occurs at section D. Since the bar is under tensile stress fs, it should
not be terminated at D, but extended further to section C, such that the bond
stress between concrete and the additional length of bar L resists the tensile
stress fs in the bar. Therefore, average anchorage bond stress ua
where φ is the diameter of the bar. L is known as Embedment Length.
o Development length / Anchorage Length (Ld) is that a certain minimum length
of the bar (under fully stressed condition, i.e., fs = 0.87fy) additionally required on
either sides of the point of curtailment, to prevent the bar from pulling out under
tension (Refer Cl. 26.2.1).
where τbd is the ‘design bond stress’, which is the permissible value of the average
anchorage bond stress ua, and fs = 0.87fy



τbd values for plain bars in tension are given in Cl. 26.2.1.1.
τbd values are increased by 60%, for deformed bars.
τbd values are increased by 25%, for bars in compression.
Page 69 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
o However, when the required bar embedment cannot be conveniently provided due
to practical difficulties, bends, hooks and mechanical anchorages can be used.
[Refer Cl. 26.2.2]
 Bends:
 The anchorage value (or the equivalent anchorage length) of a
bend shall be taken as 4 times the diameter of the bar for each
45o bend.
 Maximum anchorage value of a bend = 16φ
 A ‘standard 90o bend’ has anchorage value = 8φ (including a
minimum extension of 4φ.)

Hooks:



When the bend is turned around 180o, it has anchorage value =
16φ (including a minimum extension of 4φ), it is called a standard
U-type hook.
The minimum internal turning radius (r) specified for a hook =
o 2φ for plain mild steel bars
o 4φ for cold-worked deformed bars
Hooks are generally considered mandatory for plain bars in
tension
Page 70 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Note: Bends and hooks introduce bearing stress in the concrete that they bear against. To
ensure that these bearing stresses are not excessive, the turning radius r should be sufficiently
large. The Code (Cl. 26.2.2.5) recommends a check on the bearing stress fb inside any bend or
hook, calculated as follows:
where Fbt is the design tensile force in the bar, r is the internal radius of the bend, and φ is the
bar diameter. Bearing stress fb should be less than the limiting value stipulated in the code.

Mechanical anchorages:
o Mechanical anchorages in the form of welded plates, nuts and
bolts, etc. can be used, provided they are capable of developing
the strength of the bar without damage to concrete (to be
ascertained through tests).
Page 71 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
SPLICING OF REINFORCEMENTS




Splices are required when
o bars required are longer than available lengths, and hence needs to be extended.
o the bar diameter has to be changed along the length.
The purpose of ‘splicing’ is to transfer effectively the axial force from the terminating bar
to the connecting (continuing) bar. This invariably introduces stress concentrations in the
surrounding concrete.
As per IS456,
o splices in flexural members should not be at sections where the bending moment
is more than 50 percent of the moment of resistance
o not more than half the bars shall be spliced at a section”
There are 3 ways of splicing of bars:
o Lap Splices [Refer Cl. 26.2.5.1]
 Only for bars of dia φ ≤ 36 mm

Lap length L =
 ≥ Ld or 24 φ, for bars in compression
 ≥ 2Ld or 30 φ, for bars in tension
o Welded splices [Refer Cl. 26.2.5.2]
 For bars of dia φ > 36 mm
 Butt welding is generally adopted.
Page 72 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
CHAPTER 6
DESIGN OF SLABS & STAIRCASES
Slabs are generally classified into two:
1) One-way slabs
These have either one of the following definitions:
 Slabs which are supported only on two opposite sides

Slabs which are supported on all four sides and its length is atleast greater than
twice the width, i.e.,
>2
2) Two-way slabs

Slabs which are supported on all four sides and its length is comparable to the
width, i.e.,
<2
Page 73 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
DESIGN OF ONE-WAY SLABS


Slabs under flexure behave in the same way as beams.
A slab of uniform thickness subject to a bending moment uniformly distributed over its
width may be treated as a wide shallow beam for the purpose of analysis and design.

Main reinforcing bars are uniformly spaced over the width of the slab. For convenience,
computations are generally based on a typical one-metre wide strip of the slab
considered as a beam, i.e., Take b = 1000 mm.

If s is the centre-to-centre spacing of bars in mm, then the number of bars in the 1m wide
strip is given by 1000/s. Accordingly, denoting Ab as the cross-sectional area of one bar
(equal to πφ2/4) and the required area of tensile steel in 1m wide strip (Ast), expressed in
units of mm2/m, it may concluded that
=
=> Spacing s = 1000

Reinforced concrete slabs are generally under-reinforced and singly reinforced.
Page 74 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum

Apart from main bars, one-way slabs are generally provided with reinforcement in the
transverse direction also. These are called Secondary Reinforcements or Distribution
Bars or Distributors or Temperature Reinforcements. These are provided due to the
following reasons:
o The portion of the section above the neutral axis is under compression and hence
subjected to a lateral expansion due to the Poisson effect. Similarly, the part below
the NA is subjected to a lateral contraction.
This effect is resisted by the remainder of the slab. These give rise to secondary
moments in the transverse direction
o Also, secondary moments are also generated locally in slabs due to concentrated
loads.
o Further, shrinkage and temperature stresses developed in slabs shall also be
resisted by secondary reinforcements.
Q1. Design a simply supported RCC slab for an office floor having clear dimensions 4m x 10m
with 230mm thick masonry walls all around. The slab carries 4kN/m2 liveload & 1kN/m2 floor
finish. Use M25 concrete & Fe415 steel.
Ans:
Given, L = 10m; B = 4m
Determining type of slab:
=
= 2.5 >2
=> One-way slab
Determining depth of slab:
Effective span l = c/c distance b/w supports of shorter span = 4 + 0.23 = 4.23m
Take Modification factor = 1.25 for one-way slabs (generally for tension steel 0.4-0.5%)
Page 75 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Hence,
≤ 20 x 1.25 = 25
=>
≤ 25
=> d ≥ 169.2 mm
Therefore, Take d = 175mm
and
D = 175+25 = 200mm
Finding Effective Span:
Effective span
= smaller of (1) Clear span + d = 4.00 + 0.175
= 4.175m
(2) c/c distance between supports = 4 + 0.230 = 4.230m
= 4.175m
Because slab is considered as
a one-metre wide strip beam
Determination of Moments:
Loads:
Dead load
= 25kN/m
Live load
= 4kN/m
Finish load
= 1kN/m
3
x 1m x D = 25 x 1 x 0.2
= 5 kN/m
2
x 1m
= 4 kN/m
2
x 1m
= 1 kN/m
Total load w = 10 kN/m
Factored load wu = w x 1.5 = 10 x 1.5 = 15kN/m2
Factored Moment Mu =
=
.
.
= 32.68 kNm
Check with Mulim:
Mulim
= 0.36 fck b xumax (d – 0.42xumax)
= 0.36 x 25 x 1000 x (0.48 x 175) (175 – 0.42 x 0.48 x 175)
6
= 105.63 x 10 Nmm
= 105.63 kNm
Since Mu < Mulim, the section is safe in flexure
Check for depth of Neutral axis:
Mu = 0.36 fck b xu (d – 0.42 xu)
6
=> 32.68 x 10 = 0.36 x 25 x 1000 xu (175 – 0.42 xu)
2
6
=> 3780 xu – 1575000 xu + 32.68 x 10 = 0
Page 76 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Solving this quadratic equation, taking smaller of the roots, we get xu = 21.90 mm
Also, xumax = 0.48 x 175 = 84mm
Since xu < xumax => Underreinforced section => Safe
Determining area of main steel:
Since xu =
Ast =
.
.
.
.
= 546mm2
.
Assuming 12mm dia bars,
spacing s = 1000
= 1000 x
.
/
= 207mm
=> Provide 12mm dia main bars @ 200 mm c/c
Check for spacing of main steel:
Max spacing in main bars = minimum of
Cl. 26.3.3 (b) (1)
(1) 3d
= 3 x 175
= 525mm
(2) 300mm
= 300mm
Since s < 300mm, therefore Safe
Determining area of distribution steel:
Ast (distr)
= Ast,min
Cl. 26.5.2.1
= 0.12% of gross area
=
.
x 1000 x 200 = 240mm2
Assuming 8mm dia bars,
spacing s = 1000
= 1000 x
.
/
= 209mm
=> Provide 8mm dia distribution bars @ 200 mm c/c
Page 77 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Check for spacing of distribution steel:
Max spacing in distributors = minimum of
Cl. 26.3.3 (b) (2)
(1) 5d
= 3 x 175
= 875mm
(2) 450mm
= 450mm
Since s < 450mm, therefore Safe
____________________________________________________________________________
Q2. Design a cantilever RCC slab……
Ans: Same as above design. Only differences are :

Take ,

Mu =

Provide reinforcement on the top
≤ 7 x Modification factor
____________________________________________________________________________
Q3. Design a RCC sunshade slab……
Ans: Same as above design. Only differences are:

Take ,

Mu =

Provide reinforcement on the top
≤ 7 x Modification factor
Page 78 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
DESIGN OF STAIRCASE











Staircase is the means of access between the various floors in the building.
It consists of a flight of steps, usually with one or more intermediate landings (horizontal
slab platforms) provided between the floor levels.
The horizontal top portion of a step (where the foot rests) is termed tread and the vertical
projection of the step (i.e., the vertical distance between two neighbouring steps) is called
riser.
Generally, values of tread & rise are adopted in the range of 300 mm and 150 mm
respectively — particularly in public buildings.
Also, Generally, number of risers in a flight should not exceed about 12 in number.
Number of Treads = Number of Risers - 1
Width of the stair is generally around 1.1 – 1.6m, and in any case, should normally not be
less than 0.85.
The horizontal projection (plan) of an inclined flight of steps, between the first and last
risers, is termed going.
The steps in the flight can be designed in a number of ways (see figure on next page):
o with waist slab
o with tread-riser arrangement (without waist slab)
o with isolated tread slabs
Types of Stair case based on Geometric Configuration (see figure on next page):
o Straight flight stairs
o Quarter-turn stairs
o Dog-legged stairs
o Open-well stairs
o Spiral stairs
o Helicoidal stairs
See Cl.33 of IS456 for design of stairs
Page 79 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
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Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Relevant provisions for staircase design:


Effective span of staircase = centre-to-centre distance between the beam/wall supports
For staircase slab (or waist slab) spanning longitudinally, the staircase is designed similar
to a simply supported one-way slab of one metre width, as shown below:


Here loads are calculated separately for Landings and Going.
Waist slab thickness is determined from limiting values of (l/d) ratios, as in case of oneway slabs. It may be taken equal on both Going & Landing.
Self-weight of waist slab on Going (in kN/m)

= Unit weight of concrete x Width of slab x Depth of slab x

Self-weight of waist slab on Landing (in kN/m)
= Unit weight of concrete x Width of slab x Depth of slab


Maximum Bending moment occurs at midspan of this simply supported beam. This is
computed.
Remaining steps of design is similar to that of one-way slab
Q4. Design a simply supported dog-legged staircase for an office building, given the following
data. Use M20 concrete & Fe415 steel.




height between floors = 3.20m
rise = 160mm; tread = 270mm
width of flight = width of landing = 1.25m
Liveload = 5kN/m2; Finish load = 0.60kN/m2
Page 82 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Ans:
Proportioning the staircase dimensions:
Given, height of room = 3.20m
Rise = 160mm = 0.16m
=> No. of risers =
.
.
= 20nos
Therefore, no of risers in each flight = 20/2 = 10nos.
and, no of treads in each flight = no of risers – 1 = 10 – 1 = 9nos.
Length of landing = 1.25m
Length of going = 9 x 0.27 = 2.43m
Assume 230mm thick support walls on both sides.
Page 83 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Determining depth of waist slab:
Effective span l = c/c distance b/w supports = 2.43 + 1.25 + 1.25 + 0.23 = 5.16m
Take Modification factor = 1.25 for one-way slabs
Hence,
≤ 20 x 1.25 = 25
=>
≤ 25
=> d ≥ 206 mm
Therefore, Take d = 235mm
and
D = 235+25 = 260mm
Finding Effective Span:
Effective span
= smaller of (1) Clear span + d = (2.43 + 1.25 + 1.25) + 0.235= 5.165m
(2) c/c distance between supports
= 5.160m
= 5.160m
Determination of Moments:
Loads on going:
Because slab is
considered as a onemetre wide strip
Self weight of waist slab = 25kN/m
3
x 1m x 0.26 x
Self weight of steps
= 25kN/m
3
x 1m x Avg thickness = 25 x 1 x
Live load
= 5kN/m
Finish load
= 0.6kN/m x 1m
2
=7.56 kN/m
.
x 1m
= 2 kN/m
= 5 kN/m
2
= 0.6 kN/m
Total load w = 15.16 kN/m
Factored load on Going wu,going = w x 1.5 = 15.16 x 1.5 = 22.74kN/m
Page 84 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Loads on landing:
3
Self weight of waist slab
= 25kN/m
Live load
= 5kN/m
Finish load
= 0.6kN/m
2
2
x 1m x 0.26
= 6.5 kN/m
x 1m
= 5 kN/m
x 1m
= 0.6 kN/m
Total load w = 12.1 kN/m
Factored load on each landing wu,landing = w x 1.5 = 12.10 x 1.5 = 18.15kN/m
Maximum Bending Moment occurs at midspan and it is found to be :
2
Mu
= (52.4 x 2.58) – (18.15 x 1.365 x (2.58 – 1.365/2)) – (22.74 x (2.58 – 1.365) /2)
= 71.40 kNm
Check with Mulim:
Mulim
= 0.36 fck b xumax (d – 0.42xumax)
= 0.36 x 20 x 1000 x (0.48 x 235) (235 – 0.42 x 0.48 x 235)
6
= 152.40 x 10 Nmm
= 152.40 kNm
Since Mu < Mulim, the section is safe in flexure
Check for depth of Neutral axis:
Mu = 0.36 fck b xu (d – 0.42 xu)
6
=> 71.40 x 10 = 0.36 x 20 x 1000 xu (235 – 0.42 xu)
2
6
=> 3024 xu – 1692000 xu + 71.40 x 10 = 0
Solving this quadratic equation, taking smaller of the roots, we get xu = 46.00 mm
Also, xumax = 0.48 x 235 = 112.80mm
Since xu < xumax => Underreinforced section => Safe
Page 85 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Determining area of main steel:
Since xu =
Ast =
.
.
.
= 917mm2
.
Assuming 12mm dia bars,
spacing s = 1000
= 1000 x
.
/
= 123mm
=> Provide 12mm dia main bars @ 120 mm c/c
Check for spacing of main steel:
Max spacing in main bars = minimum of
(1) 3d
= 3 x 235
= 705mm
(2) 300mm
= 300mm
Since s < 300mm, therefore Safe
Determining area of distribution steel:
Ast (distr)
= Ast,min
= 0.12% of gross area
=
.
x 1000 x 260 = 312mm2
Assuming 8mm dia bars,
spacing s = 1000
= 1000 x
.
/
= 161mm
=> Provide 8mm dia distribution bars @ 150 mm c/c
Check for spacing of distribution steel:
Max spacing in distributors = minimum of
(1) 5d
= 3 x 235
= 705mm
(2) 450mm
= 450mm
Since s < 450mm, therefore Safe
Page 86 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
DESIGN OF TWO-WAY SLABS
[Refer Annex D of IS456 – Page 90]



Slabs that deform with significant curvatures in two perpendicular directions must be
designed as two-way slabs
Effective span = minimum of
(1) clear span + d
(2) centre-to-centre distance between supports
Effective span is represented as:
o lx
:
shorter effective span
o
ly
:
longer effective span

The corners of these slabs have a tendency to lift up, unless otherwise it is restrained
(held down by force). Such slabs, the corners of which are free to lift up, are called
Torsionally unrestrained two-way slabs [Refer Annex D-2 – Page 90]

If corners are prevented from lifting up, then such slabs are called Torsionally restrained
two-way slabs. This is because they develop Twisting Moments, which are significant
along diagonals. Hence the corners have to be suitably reinforced at the top & bottom, in
order to resist the potential cracks that may develop at corners of the slabs due to
twisting moments. Such reinforcements are called Torsion Reinforcements. [Refer Annex
D-1], shown below.
Page 87 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Q5. Design a simply supported slab for a room of internal dimensions 4m x 5m. The slab is
subjected to 3kN/m2 liveload & 1kN/m2 Finishload. Use M20 concrete & Fe415 steel. Assume
the slab corners are free to lift up.
Ans:
Determining type of slab:
=
= 1.25 < 2
=> Two-way slab
Determining depth of slab:
Assume 230mm thick masonry walls surrounding the room
Shorter Effective span lx = c/c distance b/w supports of shorter span = 4 + 0.23 = 4.23m
Take Modification factor = 1. 5 for two-way slabs
Hence,
≤ 20 x 1.5 = 30
=>
≤ 30
=> d ≥ 141 mm
Therefore, Take d = 145mm
and
D = 145+25 = 170mm
Page 88 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Finding Effective Span:
Shorter Effective span lx = smaller of (1) Clear span + d = 4.00 + 0.145
= 4.145m
(2) c/c distance between supports = 4 + 0.230 = 4.230m
= 4.145m
Longer Effective span ly = smaller of (1) Clear span + d = 5.00 + 0.145
= 5.145m
(2) c/c distance between supports = 5 + 0.230 = 5.230m
= 5.145m
Because slab is considered as
a one-metre wide strip beam
Determination of Loads:
Loads:
Dead load
= 25kN/m
Live load
= 3kN/m
Finish load
= 1kN/m
3
x 1m x D = 25 x 1 x 0.170 = 4.25 kN/m
2
x 1m
= 3 kN/m
2
x 1m
= 1 kN/m
Total load w = 8.25 kN/m
Factored load wu = w x 1.5 = 8.25 x 1.5 = 12.375kN/m2
Determination of Moments: Since slab corners free to lift up, see Cl.D-2 of IS456 – Page 90
Design moments per unit width:
Mux = αx wu lx 2
Muy = αy wu lx 2
and
From Table 27 (page 91),
Since
=
.
.
= 1.24,
Moment coefficients
αx = 0.087
αy = 0.057
Therefore,
2
= 18.50 kNm
2
= 12.12 kNm
Mux
=
0.087 x 12.375 x 4.145
Muy
=
0.057 x 12.375 x 4.145
Page 89 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Check for depth of Neutral axis:
Shorter span:
Mux = 0.36 fck b xu (d – 0.42 xu)
6
=> 18.50 x 10 = 0.36 x 20 x 1000 xu (145 – 0.42 xu)
2
6
=> 3024 xu – 1044000 xu + 18.50 x 10 = 0
Solving this quadratic equation, taking smaller of the roots, we get xu = 18.73 mm
Also, xumax = 0.48 x 145 = 69.60mm
Since xu < xumax => Underreinforced section => Safe
Longer span:
Muy = 0.36 fck b xu (d – 0.42 xu)
6
=> 12.12 x 10 = 0.36 x 20 x 1000 xu (145 – 0.42 xu)
2
6
=> 3024 xu – 1044000 xu + 12.12 x 10 = 0
Solving this quadratic equation, taking smaller of the roots, we get xu = 12.03 mm
Since xu < xumax => Underreinforced section => Safe
Determining area of main steel along shorter span:
Since xu =
Ast =
.
.
.
.
.
= 374mm2
Assuming 12mm dia bars,
spacing s = 1000
= 1000 x
.
/
= 302mm
=> Provide 12mm dia main bars @ 300 mm c/c
Page 90 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Check for spacing of main steel:
Max spacing in main bars = minimum of
Cl. 26.3.3 (b) (1)
(1) 3d
= 3 x 145
= 435mm
(2) 300mm
= 300mm
Since s ≤ 300mm, therefore Safe
Determining area of main steel along longer span:
Since xu =
.
.
.
Ast =
.
.
= 240mm2
Assuming 12mm dia bars,
spacing s = 1000
= 1000 x
.
/
= 470mm
But s ≤ 300mm should be satisfied
=> Provide 12mm dia main bars @ 300 mm c/c
____________________________________________________________________________
Q6. Design a simply supported slab for a room of internal dimensions 4m x 5m. The slab is
subjected to 3kN/m2 liveload & 1kN/m2 Finishload. Use M20 concrete & Fe415 steel. Assume
the slab corners are held down (or prevented from lifting up).
Ans:
Determining type of slab:
=
= 1.25 < 2
=> Two-way slab
Determining depth of slab:
Assume 230mm thick masonry walls surrounding the room
Shorter Effective span lx = c/c distance b/w supports of shorter span = 4 + 0.23 = 4.23m
Take Modification factor = 1. 5 for two-way slabs
Hence,
=>
≤ 20 x 1.5 = 30
≤ 30
=> d ≥ 141 mm
Page 91 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Therefore, Take d = 145mm
and
D = 145+25 = 170mm
Finding Effective Span:
Shorter Effective span lx = smaller of (1) Clear span + d = 4.00 + 0.145
= 4.145m
(2) c/c distance between supports = 4 + 0.230 = 4.230m
= 4.145m
Longer Effective span ly = smaller of (1) Clear span + d = 5.00 + 0.145
= 5.145m
(2) c/c distance between supports = 5 + 0.230 = 5.230m
= 5.145m
Because slab is considered as
a one-metre wide strip beam
Determination of Loads:
Loads:
Dead load
= 25kN/m
Live load
= 3kN/m
Finish load
= 1kN/m
3
x 1m x D = 25 x 1 x 0.170 = 4.25 kN/m
2
x 1m
= 3 kN/m
2
x 1m
= 1 kN/m
Total load w = 8.25 kN/m
Factored load wu = w x 1.5 = 8.25 x 1.5 = 12.375kN/m2
Determination of Moments: Since slab corners held down, see Cl.D-1 of IS456 – Page 90
Design moments per unit width:
Mux = αx wu lx 2
Muy = αy wu lx 2
and
From Table 26 (page 91) – all four edges discontinuous case,
Since
=
.
.
= 1.24,
Moment coefficients
αx = 0.075
αy = 0.056
Therefore,
2
= 15.95 kNm
2
= 11.91 kNm
Mux
=
0.075 x 12.375 x 4.145
Muy
=
0.056 x 12.375 x 4.145
Page 92 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
Check for depth of Neutral axis:
Shorter span:
Mux = 0.36 fck b xu (d – 0.42 xu)
6
=> 15.95 x 10 = 0.36 x 20 x 1000 xu (145 – 0.42 xu)
2
6
=> 3024 xu – 1044000 xu + 15.95 x 10 = 0
Solving this quadratic equation, taking smaller of the roots, we get xu = 16 mm
Also, xumax = 0.48 x 145 = 69.60mm
Since xu < xumax => Underreinforced section => Safe
Longer span:
Muy = 0.36 fck b xu (d – 0.42 xu)
6
=> 11.91 x 10 = 0.36 x 20 x 1000 xu (145 – 0.42 xu)
2
6
=> 3024 xu – 1044000 xu + 11.91 x 10 = 0
Solving this quadratic equation, taking smaller of the roots, we get xu = 11.80 mm
Since xu < xumax => Underreinforced section => Safe
Determining area of main steel along shorter span:
Since xu =
Ast =
.
.
.
.
= 319mm2
Assuming 12mm dia bars,
spacing s = 1000
= 1000 x
.
/
= 355mm
Check for spacing of main steel:
Max spacing in main bars = minimum of
Cl. 26.3.3 (b) (1)
(1) 3d
= 3 x 145
= 435mm
(2) 300mm
= 300mm
Page 93 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum
But, since s ≤ 300mm condition should be satisfied
=> Provide 12mm dia main bars @ 300 mm c/c
Determining area of main steel along longer span:
Since xu =
Ast =
.
.
.
.
.
= 235mm2
Assuming 12mm dia bars,
spacing s = 1000
= 1000 x
.
/
= 481mm
But s ≤ 300mm should be satisfied
=> Provide 12mm dia main bars @ 300 mm c/c
Provision of Torsion Reinforcement:
Since slab corners are held down & edges discontinuous, Torsion reinforcement must be
provided.
According to Cl D-1.8 of IS456 (Page 90),
Torsion reinforcement shall be provided in Four Layers
Width of each layer =
x lx = =
x 4.145 = 0.83m
Area of Torsion reinforcement in each of the layers
=
x Ast for shorter span
=
x 319 = 240mm2
Assuming 8mm dia bars,
spacing s = width of each layer x
= 830 x
.
/
= 173mm
=> Provide 8mm dia Torsion reinforcements @ 170 mm c/c in
4 Layers at each slab corners
Page 94 of 95
Prepared by Kiran S. R., Lecturer, Department of Civil Engineering, Central Polytechnic College Trivandrum