Structural Design of Reinforced Concrete Structures
Structural Design of Reinforced Concrete Structures
Structural Design of Reinforced Concrete Structures
STRUCTURAL
DESIGN
PART I – REINFORCED CONCRETE STRUCTURES
Chapters
KIRAN S. R.
Lecturer
CHAPTER 1
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:
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.
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Order of hydration is as follows:
Bogue’s Heat of Hydration Remarks
Compounds
C4AF 100 cal/gm Responsible for initial strength development of
C3A 200 cal/gm cement (<3 days)
C3S 120 cal/gm Early strength development of cement (3-7 days)
C2S 60 cal/gm Late strength development of cement (7-28 days)
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Strength Average 28-day Use cement- Should be >53MPa, for 53grade cement.
compressive mortar 1:3 Should be >43MPa, for 43grade cement.
strength of 3 using Should be >33MPa, for 33grade cement.
cement-mortar (P/4+3)%
cubes water to
make cubes
of 7cm size.
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)
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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
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.
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Exposure Minimum
Condition grade of
concrete for
RCC works
Mild M20
Moderate M25
Severe M30
Very M35
severe
Extreme M40
• Classification
o Ordinary concrete – M10 to M20
o Standard concrete – M25 to M55
o High Strength concrete – M60 and above
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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.
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• 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 ඥfୡ୩
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
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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
.
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CHAPTER 2
• 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.
SAFETY SERVICEABILITY
• With due consideration to strength, • Satisfactory performance of structure
stability & structural integrity. under service loads. Ensures no
• If this condition is satisfied, the discomfort to the user
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,
• Limit states involved in collapse are o Deflection
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
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.
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• 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
େ୦ୟ୰ୟୡ୲ୣ୰୧ୱ୲୧ୡ ୱ୲୰ୣ୬୲୦ ୭ ୫ୟ୲ୣ୰୧ୟ୪
• Design strength =
ୟ୰୲୧ୟ୪ ୱୟୣ୲୷ ୟୡ୲୭୰ ୭୰ ୫ୟ୲ୣ୰୧ୟ୪
• 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)
. ୡ୩
• Therefore, Design strength of concrete = = 0.447 fck ≈ 0.45 fck
ଵ.ହ
୷
and, Design strength of steel = = 0.87 fy
ଵ.ଵହ
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
• For structures subjected to Dead loads, Live loads & Windloads
Design load on the structure = 1.2 DL + 1.2 LL + 1.2 WL
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DESIGN STRESS-STRAIN CURVE FOR MATERIALS
a) For concrete in flexural compression
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• In case of Cold worked bars (Fe415 and Fe500):
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CHAPTER 3
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Analysis of this phase is performed in Working Stress Method (WSM) and
in Limit State of Serviceability.
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).
2) Under-reinforced section
• Steel reaches ultimate stress before concrete.
• 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.
• In other words, the steel would already have yielded by the time the concrete
crushes.
• Therefore, strain in
o Concrete (at the extreme compression fibre) = εcu = 0.0035
.଼౯
o Steel (on tension side) εst > + 0.002
౩
• Depth of Neutral axis (Xu) for an under-reinforced section is less than Xu,max
• This type of failure is called “Ductile Failure”.
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3) Over-reinforced section
• Concrete reaches ultimate stress first.
• Here, crushing of concrete occurs first. It gives no warning to the user about
failure. Hence, it is a catastrophic or disastrous failure.
• In other words, when the concrete at extreme compression fibre crushes, the steel
would not have reached the yield point.
• Therefore, strain in
o Concrete (at the extreme compression fibre) = εcu = 0.0035
.଼౯
o Steel (on tension side) εst < + 0.002
౩
• Depth of Neutral axis (Xu) for an over-reinforced section is greater than Xu,max
• This type of failure is called “Brittle Failure”.
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In short, the three types of sections can be summarized from the strain distribution at ultimate
limit state (i.e. at failure) as shown below.
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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
3) Effective span
[Refer Cl. 22.2] whichever is minimum
• = clear span + d (For simply supported
= span of beam measured between centers of supports beam)
• = clear span + (d/2) (For Cantilever Beams)
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
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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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• Minimum tension reinforcement in beams (Ast,min)
.଼ହ ୠୢ
o Ast,min =
౯
• Minimum compression reinforcement in beams (Asc,min)
o Asc,min = 0
• 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.
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o If span > 10m
o Area & stress in tension reinforcement
o Area of compression reinforcement
o For flanged beams
• 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 over-
reinforced, 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:
.଼౯
=> εst ≥ + 0.002
౩
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The distribution of stress and strain across the section is shown above. 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
.଼౯
section, is obtained by substituting εst = + 0.002, in the above equation.
౩
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:
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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.
• 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.
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• In general practice, the value of ratio (d’/d) ranges between 0.05 to 0.20.
• 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)
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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), bf = + bw + 6Df (for T-beams)
= + bw + 3Df (for L-beams)
ଵଶ
where lo = effective distance between points of zero moments in beam
bw = breadth of web
• 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:
.଼ ୷ ୱ୲
Xu =
.ଷ ୡ୩ ୠ
Ultimate moment of resistance of the beam section (Mu) is given by:
Mu =C.z = 0.36 fck bf xu (d – 0.42 xu)
OR
Mu =T.z = 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
Subcase 2(a): If Xu ≥ Df
ૠ
Subcase 2(b): If Xu < Df
ૠ
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CHAPTER 4
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.
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.
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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.
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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]
• 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 ඥ݂
• In the case of solid slabs, the Code (Cl. 40.2.3.1) specifies that Ƭv should not exceed
0.5 Ƭc,max .
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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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CHAPTER 5
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diagonal tension cracks — particularly if the shear stress at this section is
relatively high.
• 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.
• In general, bond strength is enhanced when the following measures are adopted:
o deformed (ribbed) bars are used instead of plain bars;
o smaller bar diameters are used;
Page 43 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
o higher grade of concrete (improved tensile strength) is used;
o increased cover is provided around each bar;
o increased length of embedment, bends and /or hooks are provided;
o mechanical anchorages are employed;
o stirrups with increased area, reduced spacing and/or higher grade of steel are
used;
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.
where Σo is the total perimeter of the bars at the beam section under
consideration, and z is lever arm.
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Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
o Actual flexural bond stress is affected by Flexural cracking, local slipping and
splitting, which are not taken care of by the above equation.
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 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 45 of 72
Prepared by Kiran S. R., Lecturer in 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 46 of 72
Prepared by Kiran S. R., Lecturer in 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 47 of 72
Prepared by Kiran S. R., Lecturer in 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”
Lap length L =
• ≥ Ld or 24 φ, for bars in compression
• ≥ 2Ld or 30 φ, for bars in tension
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Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
CHAPTER 6
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
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Prepared by Kiran S. R., Lecturer in 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
ଵ ୱ୲
=
ୗ ౘ
ౘ
=> Spacing s = 1000
ୱ୲
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Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
• Reinforced concrete slabs are generally under-reinforced and singly reinforced.
• 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:
ଵ
= = 2.5 >2 => One-way slab
ସ
Page 51 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
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%)
୪
Hence, ≤ 20 x 1.25 = 25
ୢ
ସଶଷ
=> ≤ 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
Determination of Moments: a one-metre wide strip beam
3
Loads: Dead load = 25kN/m x 1m x D = 25 x 1 x 0.2 = 5 kN/m
2
Live load = 4kN/m x 1m = 4 kN/m
2
Finish load = 1kN/m x 1m = 1 kN/m
௪ೠ మ ଵହ ୶ ସ.ଵହ ୶ ସ.ଵହ
Factored Moment Mu = = = 32.68 kNm
଼ ଼
Check with Mulim:
Page 52 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Check for depth of Neutral axis:
Solving this quadratic equation, taking smaller of the roots, we get xu = 21.90 mm
.଼ ୷ ୱ୲
Since xu =
.ଷ ୡ୩ ୠ
.ଷ ୶ ଶହ ୶ ଵ ୶ ଶଵ.ଽ
Ast = = 546mm2
.଼ ୶ ସଵହ
Assuming 12mm dia bars,
ౘ ଷ.ଵସ ୶ ଵଶ ୶ ଵଶ /ସ
spacing s = 1000 = 1000 x = 207mm
ୱ୲ ହସ
Page 53 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Assuming 8mm dia bars,
ౘ ଷ.ଵସ ୶ ଼ ୶ ଼ /ସ
spacing s = 1000 = 1000 x = 209mm
ୱ୲ ଶସ
Page 54 of 72
Prepared by Kiran S. R., Lecturer in 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 55 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Page 56 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Page 57 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Relevant provisions for staircase design:
• 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.
Page 58 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Ans:
Proportioning the staircase dimensions:
Page 59 of 72
Prepared by Kiran S. R., Lecturer in 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
୪
Hence, ≤ 20 x 1.25 = 25
ୢ
ହଵ
=> ≤ 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
Because slab is
Determination of Moments: considered as a one-
metre wide strip
Loads on going:
3 ଷଵସ
Self weight of waist slab = 25kN/m x 1m x 0.26 x =7.56 kN/m
ଶ
3 ା.ଵ
Self weight of steps = 25kN/m x 1m x Avg thickness = 25 x 1 x = 2 kN/m
ଶ
2
Live load = 5kN/m x 1m = 5 kN/m
2
Finish load = 0.6kN/m x 1m = 0.6 kN/m
Page 60 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Loads on landing:
3
Self weight of waist slab = 25kN/m x 1m x 0.26 = 6.5 kN/m
2
Live load = 5kN/m x 1m = 5 kN/m
2
Finish load = 0.6kN/m x 1m = 0.6 kN/m
= 71.40 kNm
Check with Mulim:
Solving this quadratic equation, taking smaller of the roots, we get xu = 46.00 mm
Page 61 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Also, xumax = 0.48 x 235 = 112.80mm
.଼ ୷ ୱ୲
Since xu =
.ଷ ୡ୩ ୠ
.ଷ ୶ ଶ ୶ ଵ ୶ ସ
Ast = = 917mm2
.଼ ୶ ସଵହ
Assuming 12mm dia bars,
ౘ ଷ.ଵସ ୶ ଵଶ ୶ ଵଶ /ସ
spacing s = 1000 = 1000 x = 123mm
ୱ୲ ଽଵ
Page 62 of 72
Prepared by Kiran S. R., Lecturer in 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 63 of 72
Prepared by Kiran S. R., Lecturer in 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:
ହ
= = 1.25 < 2 => Two-way slab
ସ
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
Page 64 of 72
Prepared by Kiran S. R., Lecturer in 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
Longer Effective span ly = smaller of (1) Clear span + d = 5.00 + 0.145 = 5.145m
Determination of Moments: Since slab corners free to lift up, see Cl.D-2 of IS456 – Page 90
Design moments per unit width:
Page 65 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Check for depth of Neutral axis:
Shorter span:
Solving this quadratic equation, taking smaller of the roots, we get xu = 18.73 mm
Longer span:
Solving this quadratic equation, taking smaller of the roots, we get xu = 12.03 mm
.଼ ୷ ୱ୲
Since xu =
.ଷ ୡ୩ ୠ
.ଷ ୶ ଶ ୶ ଵ ୶ ଵ଼.ଷ
Ast = = 374mm2
.଼ ୶ ସଵହ
Assuming 12mm dia bars,
ౘ ଷ.ଵସ ୶ ଵଶ ୶ ଵଶ /ସ
spacing s = 1000 = 1000 x = 302mm
ୱ୲ ଷସ
Page 66 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Check for spacing of main steel:
Max spacing in main bars = minimum of (1) 3d = 3 x 145 = 435mm
Cl. 26.3.3 (b) (1) (2) 300mm
= 300mm
Since s ≤ 300mm, therefore Safe
.଼ ୷ ୱ୲
Since xu =
.ଷ ୡ୩ ୠ
.ଷ ୶ ଶ ୶ ଵ ୶ ଵଶ.ଷ
Ast = = 240mm2
.଼ ୶ ସଵହ
Assuming 12mm dia bars,
ౘ ଷ.ଵସ ୶ ଵଶ ୶ ଵଶ /ସ
spacing s = 1000 = 1000 x = 470mm
ୱ୲ ଶସ
But s ≤ 300mm should be satisfied
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:
ହ
= = 1.25 < 2 => Two-way slab
ସ
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
Page 67 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
ସଶଷ
=> ≤ 30
ୢ
=> d ≥ 141 mm
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
Longer Effective span ly = smaller of (1) Clear span + d = 5.00 + 0.145 = 5.145m
Determination of Moments: Since slab corners held down, see Cl.D-1 of IS456 – Page 90
Design moments per unit width:
Page 68 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Therefore,
2
Mux = 0.075 x 12.375 x 4.145 = 15.95 kNm
2
Muy = 0.056 x 12.375 x 4.145 = 11.91 kNm
Longer span:
Solving this quadratic equation, taking smaller of the roots, we get xu = 11.80 mm
.଼ ୷ ୱ୲
Since xu =
.ଷ ୡ୩ ୠ
.ଷ ୶ ଶ ୶ ଵ ୶ ଵ
Ast = = 319mm2
.଼ ୶ ସଵହ
Page 69 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
Assuming 12mm dia bars,
ౘ ଷ.ଵସ ୶ ଵଶ ୶ ଵଶ /ସ
spacing s = 1000 = 1000 x = 355mm
ୱ୲ ଷଵଽ
.଼ ୷ ୱ୲
Since xu =
.ଷ ୡ୩ ୠ
.ଷ ୶ ଶ ୶ ଵ ୶ ଵଵ.଼
Ast = = 235mm2
.଼ ୶ ସଵହ
Assuming 12mm dia bars,
ౘ ଷ.ଵସ ୶ ଵଶ ୶ ଵଶ /ସ
spacing s = 1000 = 1000 x = 481mm
ୱ୲ ଶଷହ
But s ≤ 300mm should be satisfied
Page 70 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum
ଷ
Area of Torsion reinforcement in each of the layers = x Ast for shorter span
ସ
ଷ
= x 319 = 240mm2
ସ
Assuming 8mm dia bars,
ౘ ଷ.ଵସ ୶ ଼ ୶ ଼ /ସ
spacing s = width of each layer x = 830 x = 173mm
ୱ୲ ଶସ
Page 71 of 72
Prepared by Kiran S. R., Lecturer in Civil Engineering, Central Polytechnic College Trivandrum