E4-E5 Technical Architecture Rev date: 14-08-12 ©BSNL India For Internal Circulation Only Page: 1 Chapter-9 UNDERSTANDING STRUCTURAL DESIGN OF RCC BUILDING COMPONENTS Rajendra Mathur Dy. Dir(BS-C) 09412739 232(M) e-mail–[email protected]
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Chapter-9
UNDERSTANDING
STRUCTURAL DESIGN
OF
RCC BUILDING COMPONENTS
Rajendra Mathur Dy. Dir(BS-C) 09412739 232(M) e-mail–[email protected]
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Understanding Structural Design of
RCC Building Components
1.0 Introduction
The procedure for ana lysis and design of a given building depends on
the t ype of building, it s complexit y, the number of sto ries etc. Based
on preliminar y architec tural drawings of the building, structu ral system
is f inalized & sizes of structural members are decided and conve yed to
the concerned architect.
Befo re start ing the structural design of bu ild ing, the fo llowing
informat ion /data are required –
( i) A set of architectural drawings;
( ii) Soil Investigat ion report (SIR) or soil data in lieu thereof;
( iii) Locat ion of the p lace or cit y in order to decide on wind and
seismic loadings;
( iv) Data for lifts, water tank capac it ies on top, special roof
features or loadings, etc.
Choice of an appropriate structural system for a given bu ilding is
vital for its econom y and safet y. There are two type of building
systems:-
(a) Load Bearing Masonry Buildings.
(b) Framed Build ings.
(a) Load Bearing Masonry Buildings:-
Small buildings like houses with small spans of beams & slabs
are generally constructed as load bearing brick wall structures with
reinforced concre te slab & beam roofing. This system is su itable for
build ing up to four or less stories (as shown in fig. below). In such
build ings crushing strength of b ricks sha ll be 100 kg/cm2 minimum for
four stories. This system is adequate for vertica l loads it also serves to
resis ts horizontal loads like wind & ear thquake by box action. It is
necessary to provide RCC Bands in horizonta l & vert ical reinforcement
in brick wall as per IS: 4326-1967(Indian Standards Code of Practice
for Earthquake Resistant Construction of Buildings) to ensure the
box act ion for resistance against earthquake. In some Build ings,
115mm thick b rick walls are provided. Since these walls are incapable
of supporting vert ical loads, beams have to be provided along their
lengths to support ad jo ining s lab & the weight of 115mm thick brick
wall of upper storey. These beams are made to rest on 230 mm thick
brick walls or re info rced concrete columns if required . The design of
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Load Bear ing Masonry Bu ild ings are done as per IS:1905-1980 (Indian
Standards Code o f Practice for Structura l Safety o f Buildings:
Masonry Walls(Second Revision).
Load bearing brick wall
Structural system
(b) RCC Framed Buildings:-
In these types of buildings, re inforced concrete frames are
provided in both principal d irect ions to res ist vertica l loads and the
vertica l loads are transmitted to Foundations through vertical framing
system of columns and beams in both principal direct ions. This t ype of
system is e ffect ive in resist ing both vert ical & horizonta l loads. The
brick walls are to be regarded as non load bearing filler walls only.
This system is suitab le for mult i-storied building as it is very e ffect ive
in resist ing horizonta l loads due to earthquake. In this system the f loor
slabs, generally 100-150 mm thick with spans ranging from 3.0m to
7.0m. In certa in earthquake prone areas, even single or double storey
build ings are made framed structures fo r safet y reasons. Also the
single storey build ings of large storey heights (5.0m or more ) , l ike
electric substat ion etc. are made framed structure as brick walls of
large heights are slender and load carrying capacit y of such walls
reduces due to slenderness.
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RCC Framed
Structural system 2.0 Basic Codes for Design . The design should be carried so as to conform to the fo llowing
Indian code for reinforced concrete design, published by the Bureau of
Indian Standards, New Delhi:
Purpose o f Codes
National building codes have been formulated in d ifferent
countries to lay down gu idelines for the design and construction of
structu re. The codes have evolved from the co llect ive wisdom of expert
structu ral engineers, gained over the years. These codes are
periodically revised to bring them in line with current research, and
often, current trends.
The codes serve at least four distinct functions.
Firs t ly, the y ensure adequate structural sa fet y, b y specifying certain
essent ial minimum requirement for design .
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Secondly, the y render the task of the designer relat ive ly simple; often,
the result of sophisticate analyses is made ava ilable in the form of a
simple formula or chart.
Third ly, the codes ensure a measure of consistenc y among different
designers.
Finally, the y have some legal va lid ity in that they protect the structu ral
designer from any liab ilit y due to structural failu res that are caused by
inadequate supervis ion and/o r fau lty material and construction.
(i)IS 456 : 2000 – Plain and reinforced concrete – code o f practice
(fourth revision)
(ii) Loading Standards
These loads to be considered for structural design are specified in the
fo llowing loading standards:
IS 875 (Part 1-5) : 1987 – Code of practice for design loads
(other than earthquake) for buildings and structures ( second
revision)
Part 1 : Dead loads
Part 2 : Imposed (Live) loads
Part 3 : Wind loads
Part 4 : Snow loads
Part 5 : Special loads and load combinations
IS 1893 : 2002 – Criteria for earthquake resistant design of
structure (fourth revision) . IS 13920 : 1993 – Ductile detailing of reinforced concrete structure
subject to seismic forces.
Design Handbooks
The Bureau of Ind ian standards has a lso published the fo llowing
handbooks, which serve as useful supp lement to the 1978 vers ion of the
codes. Although the handbooks need to be updated to bring them in
line with the recent ly revised (2000 vers ion) of the Code, many o f the
provis ions cont inue to be valid (especia lly with regard to structural
design provisions).
SP 16 : 1980 – Design Aids (for Reinforced Concrete) to IS 456 :
1978
SP 24 : 1983 – Explanatory handbook on IS 456 : 1978
SP 34 : 1987 – Handbooks on Concrete Reinforced and Detailing .
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3.0 General Design Consideration of IS: 456-2000.
The general design and construction of reinforced concrete buildings is
governed by the provisions of IS 456 –2000 & some of the important provisions are
given below –
AIM OF DESIGN
The aim of design is achievement of an acceptable probability that structures being
designed shall, with an appropriate degree of safety –
� Perform satisfactorily during their intended life.
� Sustain all loads and deformations of normal construction & use
� Have adequate durability
� Have adequate resistance to the effects of misuse and fire.
METHOD OF DESIGN –
� Structure and structural elements shall normally be designed by Limit State
Method.
� Where the Limit State Method cannot be conveniently adopted, Working
Stress Method may be used
MINIMUM GRADE OF CONCRETE
The minimum grade of concrete for plain & reinforced concrete shall be as per table
below –
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26.4 Nominal Cover to Reinforcement
26.4.1 Nominal Cover
Nominal cover is the design depth of concrete cover to all steel
reinforcements, including links. It is the dimension used in design and
indicated in the drawings. It shall be not less than the diameter of the bar.
26.4.2 Nominal Covers to Meet Durability Requirement
Minimum values for the nominal cover of normal weight aggregate concrete
which should be provided to all reinforcement, including links depending on
the condition of exposure described in 8.2.3 shall be as given in Table 16.
Table 16 Nominal Cover to Meet Durability Requirements
(Clause 26.4.2)
Exposure Nominal Concrete Cover in mm not Less Than
Mild 20
Moderate 30
Severe 45
Very Severe 50
Extreme 75
NOTES
1. For main reinforcement up to 12 mm diameter bar for mild exposure
the nominal cover may be reduced by 5 mm.
2. Unless specified otherwise, actual concrete cover should not deviate
from the required nominal cover by + 10 mm
3. For exposure condition ‘severe’ and ‘very severe’, reduction of 5 mm
may be made, where concrete grade is M35 and above.
26.4.2.1 However for a longitudinal reinforcing bar in a column nominal cover shall
in any case not be less than 40 mm, or less than the diameter of such bar. In
the case of columns of minimum dimension of 200 mm or under, whose
reinforcing bars do not exceed 12 mm, a nominal cover of 25 mm may be
used.
26.4.2.2 For footing minimum cover shall be 50 mm.
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26.4.3 Nominal Cover to Meet Specified Period of Fire Resistance
Minimum values of nominal cover of normal-weight aggregate concrete to be
provided to all reinforcement including links to meet specified period of fire
resistance shall be as given in Table 16A.
21.4 Minimum Dimensions of RC members for specified Period of Fire
Resistance
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DESIGN LOAD
Design load is the load to be taken for use in appropriate method of design. It is –
� Characteristic load in case of working stress method &
� Characteristic load with appropriate partial safety factors for limit state
design.
LOAD COMBINATIONS
As per IS 1893 (Part 1): 2002 Clause no. 6.3.1.2, the following load cases have to be
considered for analysis:
� 1.5 (DL + IL)
� 1.2 (DL + IL ± EL)
� 1.5 (DL ± EL)
� 0.9 DL ± 1.5 EL
� Earthquake load must be considered for +X, -X, +Z and –Z directions.
� Moreover, accidental eccentricity during earthquake can be such that it causes
clockwise or anticlockwise moments. So both clockwise & anticlockwise
torsion is to be considered.
� Thus, ±EL above implies 8 cases, and in all, 25 cases must be considered.
It is possible to reduce the load combinations to 13 instead of 25 by not using negative
torsion considering the symmetry of the building.
STRUCTURAL FRAMES 22.4 The simplifying assumptions as given in 22.4.1 to 22.4.3 may be used in
the analysis of frames.
ARRANGEMENT OF LIVE LOAD
22.4.1 a) Consideration may be limited to combinations of:
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1) Design dead load on all spans with full design live load on two
adjacent spans; and
2) Design dead load on all spans with dull design live load on alternate
spans.
22.4.1 b) When design live load does not exceed three-fourths of the design dead
load, the load arrangement may be design dead load and design live load
on all the spans.
Note: For beams continuous over support 22.4.1 (a) may be assumed. 22.4.2 Substitute Frame: For determining the moments and shears at any floor
or roof level due to gravity loads, the beams at that level together with
columns above and below with their far ends fixed may be considered to
constitute the frame.
22.4.3 For lateral loads, simplified methods may be used to obtain the moments
and shears for structures that are symmetrical. For unsymmetrical or very
tall structures, more rigorous methods should be used.
CRITICAL SECTIONS FOR MOMENT AND SHEAR
22.6.1 For monolithic construction, the moments computed at the face of the
supports shall be used in the design of the members at those sections. For
non-monolithic construction the design of the member shall be done
keeping in view 22.2.
22.6.2 Critical Section for Shear The shears computed at the face of the Support shall be used in the design
of the member at that section except as in 22.6.2.1
22.6.2.1 When the reaction in the direction of the applied shear introduces
compression into the end region of the member, sections located at a
distance less than d from the face of the support may be designed for the
same shear as that computed at distance d.
EFFECTIVE DEPTH 23.0 Effective depth of a beam is the distance between the centroid of the area
of tension reinforcement and the maximum compression fibre, excluding
the thickness of finishing material not placed monolithically with the
member and the thickness of any concrete provided to allow for wear.
This will not apply to deep beams.
CONTROL OF DEFLECTION
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23.2 The deflection of a structure or part thereof shall not adversely affect the
appearance or efficiency of the structure or finishes or partitions. The
deflection shall generally be limited to the following:
a) The final deflection due to all loads including the effects of temperature,
creep and shrinkage and measured from the as-cast level of the supports of
floors, roofs and all other horizontal members, should not normally exceed
span/250.
b) The deflection including the effects of temperature, creep and shrinkage
occurring after erection of partitions and the application of finishes should
not normally exceed span/350 or 20mm whichever is less.
23.2.1 For beams, the vertical deflection limits may generally be assumed to be
satisfied provided that the span to depth ratio are not greater than the value
obtained as below:
a) Basic values of span to effective depth ratios for spans up to 10m:
Cantilever 7
Simply supported 20
Continuous 26
b) For spans above 10m, the values in (a) may be multiplied by 10/span in
metres, except for cantilever in which case deflection calculations should
be made.
c) Depending on the area and the type of steel for tension reinforcement, the
value in (a) or (b) shall be modified as per Fig. 4
d) Depending on the area of compression reinforcement, the value of span to
depth ratio be further modified as per Fig. 5
e) For flanged beams, the value of (a) or (b) be modified as per Fig. 6 and the
reinforcement percentage for use in fig. 4 and 5 should be based on area of
section equal to bf d.
Note: When deflections are required to be calculated, the method given
Annexure ‘C’ of IS 456-2000 may be used.
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CONTROL OF DEFLECTION – SOLID SLABS
24.1 General The provisions of 32.2 for beams apply to slabs also. NOTES
1. For slabs spanning in two directions, the shorter of the two spans should be
used for calculating the span to effective depth rations.
2. For two-way slabs of shorter spans (up to 3.5 m) with mild steel
reinforcement, the span to overall depth rations given below may generally
be assumed to satisfy vertical deflection limits for loading class up to 3
kN/m2.
Simply supported slab 35
Continuous slabs 40
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For high strength deformed bars of grade Fe 415,the values given above
should be multiplied by 0.8.
Simply supported slab 28
Continuous slabs 32
23.3 Slabs Continuous Over Supports
Slabs spanning in one direction and continuous over supports shall be
designed according to the provisions applicable to continuous beams.
23.4 Slabs Monolithic with Supports
Bending moments in slabs (except flat slabs) constructed monolithically with
the supports shall be calculated by taking such slabs either as continuous over
supports and capable of free, or as members of a continuous frame work with
the supports, taking into account the stiffness of such support. If such supports
are formed due to beams which justify fixity at the support of slabs, then the
effects on the supporting beam, such as, the bending of the web in the
transverse direction of the beam, wherever applicable, shall also be considered
in the design of the beams.
23.4.1 For the purpose of calculation of moment in slabs in a monolithic structure, it
will generally be sufficiently accurate to assume direct members connected to
the ends of such slab are fixed in position and direction at the end remote from
their connection with the slab.
26.5 REQUIREMENT OF REINFORCEMENT FOR STRUCTURAL
MEMBER
26.5.1 BEAMS
26.5.1.1 Tension reinforcement
(a) Minimum reinforcement:- The minimum area of tension reinforcement shall
not be less than that given by the following:-
As = 0.85 bd fy
where
As = minimum area of tension reinforcement.
b = breadth of beam or the breadth of the web of T-beam.
d = effective depth, and
fy = characteristic strength of reinforcement in M/mm2
(b) Maximum reinforcement:- the maximum area of tension reinforcement shall
not exceed 0.04bD.
26.5.1.2 Compression reinforcement
The maximum area of comparison reinforcement shall not exceed 0.04 bd.
Comparison reinforcement in beams shall be enclosed by stirrups for effective lateral
restraint.
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26.5.1.3 Side face reinforcement
Where the depth of the web in the beam exceeds 750mm, side face reinforcement
shall be provided along the two faces. The total area of such reinforcement shall be
not less than 0.1 % of the web area and shall be distributed on the equally on the two
face at spacing not exceeding 300mm or web thickness whichever is less.
26.5.1.4 Transverse reinforcement in beam for shear torsion
The transverse reinforcement in beam shall be taken around the outer most tension &
compression bars. In T-beams and I-beams, such reinforcement shall pass around
longitudinal bars located close to the outer face of the flange.
26.5.1.5 Maximum spacing of shear reinforcement
Maximum spacing of shear reinforcement means long by axis of the member shall not
exceed 0.75 d for vertical stirrups and d for inclined stirrups at 45” where d is the
effective depth on the section under consideration. In no case shall be spacing exceed
300mm.
26.5.1.6 Minimum shear reinforcement
Minimum shear reinforcement in the form of stirrups shall be provided such that:
Asv 0.4 bsv 0.87 fy Where Asv = total cross-sectional area of stirrups legs effective in shear.
Sv = stirrups spacing along the length of the member
B = breadth of the beam or breadth of the web of flange beam, and
fy = characteristic strength of the stirrups reinforcement in N/mm2 which shall not
taken greater than 415 N/mm2
Where the maximum shear stress calculated is less than half the permissible value in
member of minor structure importance such as lintels, this provision need not to be
complied with.
26.5.1.7 Distribution of torsion reinforcement
When a member is designed for torsion torsion reinforcement shall be provided as
below:
a) the transverse reinforcement for torsion shall be rectangular closed stirrups
placed perpendicular to the axis of the member. The spacing of the stirrups
shall not exceed the list of x1, x1+y1/4 and 300 mm, where x1, y1 are
respectively the short & long dimensions of the stirrup.
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b) Longitudinal reinforcement shall be place as closed as is practicable to the
corner of the cross section & in all cases, there shall be at least one
longitudinal bar in each corner of the ties. When the cross sectional dimension
of the member exceed 450 mm additional longitudinal bar shall be provided to
satisfy the requirement of minimum reinforcement & spacing given in
26.5.1.3.
26.3.2 Minimum Distance between Individual Bars
(a) The horizontal distance between two parallel main reinforcing bars shall
usually be not-less than the greatest of the following:
(i) Dia of larger bar and
(ii) 5 mm more than nominal maximum size of coarse aggregate.
(b) When needle vibrators are used it may be reduced to 2/3rd of nominal
maximum size of coarse aggregate,
Sufficient space must be left between bars to enable vibrator to be immersed.
(c) Where there are two or more rows of bars, bars shall be vertically in line and
the minimum vertical distance between bars shall be 15 mm, 2/3rd of nominal
maximum size of aggregate or the maximum size of bars, whichever is greater.
26.5.2 Slabs
The rule given in 26.5.2.1 and 26.5.2.2 shall apply to slabs in addition to those given
in the appropriate clause.
26.5.2.1 Minimum reinforcement
The mild steel reinforcement in either direction in slabs shall not be less than 0.15
percent of the total cross-sectional area. However, this value can be reduced to 0.12
percent when high strength deformed bars or welded wire fabric are used.
26.5.2.2 Maximum diameter
The diameter of reinforcing bars shall not exceed one eight of the total thickness of slab.
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26.3.3 Maximum distance between bars - Slabs
1) The horizontal distance between parallel main reinforcement bars shall not be
more than three times the effective depth of solid slab or 300 mm whichever is
smaller.
2) The horizontal distance between parallel reinforcement bars provided against
shrinkage and temperature shall not be more than five times the effective depth of a
solid slab or 300 mm whichever is smaller.
Torsion reinforcement - Slab
Torsion reinforcement is to be provided at any corner where the slab is simply
supported on both edges meeting at that corner. It shall consist of top and bottom
reinforcement, each with layers of bars placed parallel to the sides of the slab and
extending from the edges a minimum distance of one-fifth of the shorter span. The
area of reinforcement in each of these four layers shall be three-quarters of the area
required for the maximum mid-span moment in the slab.
D-l.9 Torsion reinforcement equal to half that described in D-l.8 shall be provided at a
corner contained by edges over only one of which the slab is continuous.
D-1.10 Torsion reinforcements need not be provided at any comer contained by edges
over both of which the slab is continuous.
26.5.3 COLUMNS
A. Longitudinal Reinforcement
a. The cross sectioned area of longitudinal reinforcement shall be not less than
0.8% nor more than 6% of the gross sectional area of the column. Although it is
recommended that the maximum area of steel should not exceed 4% to avoid
practical difficulties in placing & compacting concrete.
b. In any column that has a larger cross sectional area than that required to support
the load, the minimum percentage steel must be based on the area of concrete
resist the direct stress & not on the actual area.
c. The bar should not be less than 12 mm in diameter so that it is sufficiently rigid
to stand up straight in the column forms during fixing and concerting.
d. The minimum member of longitudinal bars provided in a column shall be four
in rectangular columns & six in circular columns.
e. A reinforced concrete column having helical reinforcement must have at least
six bars of longitudinal reinforcement with the helical reinforcement. These
bars must be in contact with the helical reinforcement & equidistance around its
inner circumference.
f. Spacing of longitudinal should not exceed 300 mm along periphery of a
column.
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g. In case of pedestals, in which the longitudinal reinforcement is not taken into
account in strength calculations, nominal reinforcement should be not be less
than 0.15% of cross sectional area.
B. Transverse Reinforcement
a. The diameter of lateral ties should not be less than ¼ of the diameter of the
largest longitudinal bar in no case should not be less than 6 mm.
b. Spacing of lateral ties should not exceed least of the following:-
• Least lateral dimension of the column.
• 16 times the smallest diameter of longitudinal bars to be tied.
• 300mm.
SHEAR
40.1 Nominal Shear Stress
The nominal shear stress in beams of uniform depth shall be obtained by the
following equation:
τv = Vu/ b.d where Vu = shear force due to design loads;
b = breadth of the member, which for flanged section shall be taken as the breadth of
the web, bw; and
d = effective depth.
40.2.3 With Shear Reinforcement
Under no circumstances, even with shear reinforcement, shall the nominal shear stress
in beams should not exceed given in Table 20.
40.2.3.1 For solid slabs, the nominal shear stress shall not exceed half the appropriate
values given in Table 20.
40.3 Minimum Shear Reinforcement When τv, is less than τc given in Table 19, minimum shear reinforcement shall be provided in accordance with 26.5.1.6.
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40.4 Design of Shear Reinforcement
When τv, is exceeds τc , given in Table 19, shear reinforcement shall be provided in
any of the following forms:
a) Vertical stirrups,
b) Bent-up bars along with stirrups, and Where bent-up bars are provided, their
contribution towards shear resistance shall not be more than half that of the total shear
reinforcement.
Shear reinforcement shall be provided to carry a shear equal to Vu – τ c b d. the
strength of shear reinforcement Vus shall be calculated as below:
a) For Vertical Stirrups:
0.87 fy Asv d
Vus = ___________ Sv
b) For inclined stirrups or a series of bars bent up at different cross –section:
0.87 fy Asv d
Vus = ___________ (Sin ά + Cos ά) Sv
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c) For single bar or single group of parallel bars, all bent up at the same cross sections:
Vus = 0.87 fy Asv Sin ά
Where Asv = total cross –sectional area of stirrups legs or
bent-up bar within a distance Sv,
Sv = spacing of the stirrups or bent-up bars along the
length of the member.
τ v = nominal shear stress, τ c = design shear strength of the concrete, b = breadth of the member which for flanged beams,
shall be taken as the breadth of the web bw. fy = characteristic strength of the stirrup or bent-up reinforcement which shall not be taken greater
than 415 N/mm2,
ά = angle between the inclined stirrup or bent up bar
and the axis of the member not less than 45o,
and
d = effective depth
DEVELOPMENT LENGTH OF BARS
26.2 Development of Stress in Reinforcement
The calculated tension or compression in any bar at any section shall be developed on
each side of the section by an appropriate development length or end anchorage or by
a combination thereof.
Development length Ld is given by
Ld = φσst /4τbd
φ = nominal diameter of bar, τbd = design bond stress
σst = stress in bar at the section considered at design load
� Design bond stress in limit state method for plain bars in tension is given in
clause 26.2.1.1
� For deformed bars conforming to IS 1786 these values are to be increased by
60 %.
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� For bars in compression, the values of bond stress for bars in tension is to be
increased by 25 percent
B. Shear reinforcement (STIRRUPS)
Development length and anchorage requirement is satisfied, in case of stirrups and
transverse ties, when Bar is bent –
• Through an angle of at least 90 degrees (round a bar of at least its own dia) &
is continued beyond for a length of at least 8 φ, or
• Through an angle of 135 degrees & is continued beyond for a length of at least
6 φ or
• Through an angle of 180 degrees and is continued beyond for a length of at
least 4 φ
DUCTILE DETAILING AS PER IS: 13920
• Provisions of IS 13920-1993 shall be adopted in all reinforced concrete
structures which are located in seismic zone III, IV or V
The provisions for reinforced concrete construction given in IS 13920-1993 shall
apply specifically to monolithic reinforced concrete construction. Precast and/or
prestressed concrete members may be used only if they can provide the same level of
ductility as that of a monolithic reinforced concrete construction during or after an
earthquake.
The definition of seismic zone and importance factor are given in IS 1893-2002.
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CODAL PROVISIONS OF IS 13920
5.2 For all buildings which are more than 3 storeys in height, the minimum grade of
concrete shall be M20 (fck = 20 MPa ).
5.3 Steel reinforcements of grade Fe 415 (see IS 1786 : 1985 ) or less only shall be
used. However, high strength deformed steel bars, produced by the thermo-
mechanical treatment process, of grades Fe 500 and Fe 550, having elongation more
than 14.5 percent and conforming to other requirements of IS 1786 : 1985 may also be
used for the reinforcement.
Flexure Members
6.1.2 The member shall preferably have a width-to-depth ratio of more than 0.3.
6.1.3 The width of the member shall not be less than 200 mm.
6.1.4 The depth D of the member shall preferably be not more than 1/4 of the clear
span.
6.2 Longitudinal Reinforcement
6.2.1 a) The top as well as bottom reinforcement shall consist of at least two bars
throughout the member length.
b) The tension steel ratio on any face, at any section, shall not be less than ρmin =
0.24(fck)1/2 /fy ; where fck and fy are in MPa.
6.2.2 The maximum steel ratio on any face at any section, shall not exceed ρmax =
0.025.
6.2.3 The positive steel at a joint face must be at least equal to half the negative steel
at that face.
6.2.4 The steel provided at each of the top and bottom face of the member at any
section along its length shall be at least equal to one-fourth of the maximum negative
moment steel provided at the face of either joint
6.2.6 The longitudinal bars shall be spliced, only if hoops are provided over the entire
splice length, at a spacing not exceeding 150 mm 6.3
Web Reinforcement
6.3.1 Web reinforcement shall consist of vertical hoops. A vertical hoop is a closed
stirrup having a 135° hook with a 10 diameter extension (but not < 75 mm) at each
end that is embedded in the confined core
6.3.2 The minimum diameter of the bar forming a hoop shall be 6 mm. However, in
beams with clear span exceeding 5 m, the minimum bar diameter shall be 8 mm.
6.3.4 The contribution of bent up bars and inclined hoops to shear resistance of the
section shall not be considered.
6.3.5 The spacing of hoops over a length of 2d at either end of a beam shall not
exceed (a) d/4, and (b) 8 times the diameter of the smallest longitudinal bar; however,
it need not be less than 100 mm. Elsewhere, the beam shall have vertical hoops at a
spacing not exceeding d/2.
Columns
7.1.2 The minimum dimension of the member shall not be less than 200 mm.
However, in
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frames which have beams with centre to centre span exceeding 5 m or columns of
unsupported length exceeding 4 m, the shortest dimension of the column shall not be
less than 300 mm.
7.1.3 The ratio of the shortest cross sectional dimension to the perpendicular
dimension shall preferably not be less than 0.4.
7.2 Longitudinal Reinforcement
7.2.1 Lap splices shall be provided only in the central half of the member length. It
should be proportioned as a tension splice. Hoops shall be provided over the entire
splice length at spacing not exceeding 150 mm centre to centre. Not more than 50
percent of the bars shall be spliced at one section.
7.3 Transverse Reinforcement
7.3.1 Transverse reinforcement for circular columns shall consist of spiral or circular
hoops. In rectangular columns, rectangular hoops may be used. A rectangular hoop is
a closed stirrup, having a 135° hook with a 10 diameter extension (but not < 75 mm)
at each end, that is embedded in the confined core.
7.3.3 The spacing of hoops shall not exceed half the least lateral dimension of the
column, except where special confining reinforcement is provided, as per 7.4.
7.4 Special Confining Reinforcement
This requirement shall be met with, unless a larger amount of transverse
reinforcement is required from shear strength considerations.
7.4.1 Special confining reinforcement shall be provided over a length lo from each
joint face, towards midspan, and on either side of any section, where flexural yielding
may occur under the effect of earthquake forces. The length ‘lo’ shall not be less than
(a) larger lateral dimension of the member at the section where yielding occurs, (b)
1/6 of clear span of the member, and (c) 450 mm.
7.4.2 When a column terminates into a footing or mat, special confining
reinforcement shall extend at least 300 mm into the footing or mat.
7.4.6 The spacing of hoops used as special confining reinforcement shall not exceed
1/4 of minimum member dimension but need not be less than 75 mm nor more than
100 mm.
8 JOINTS OF FRAMES
8.1 The special confining reinforcement as required at the end of column shall be
provided through the joint as well, unless the joint is confined as specified by 8.2.
8.2 A joint which has beams framing into all vertical faces of it and where each beam
width is at least 3/4 of the column width, may be provided with half the special
confining reinforcement required at the end of the column. The spacing of hoops shall
not exceed 150 mm.
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As per IS-875(Part-1)-1987
0.16-0.23STEEL WORK -ROOFING
0.65MANGLORE TILES
0.15GI SHEET -ROOFING
0.16AC SHEET -ROOFING
78.5STEEL
18LIME -PLASTER
21CEMENT-PLASTER
6-10TIMBER
21-27STONE MASONRY
19-20BRICK MASONRY
25REINFORCED CONCRETE
24PLAIN CONCRETE
kN/m2kN/m3
UNIT WEIGHT
MATERIAL
DEAD LOADS – UNIT WEIGHTS OF SOME MATERIALS/BUILDING COMPONENTS
4.0�STAIRS-NOT LIABLE TO OVER CROWDING
5.0- LIABLE TO OVER CROWDING
7.5–HEAVY VEHICLES
4.0�GARRAGES –LIGHT VEHICLES
10.0�STACK ROOM IN LIBRARIES ,BOOK STORES
5.0-10� FACTORIES & WAREHOUSES
5.0- WITHOUT FIXED SEATING
4.0- WITH FIXED SEATING
� SHOPS,CLASS ROOMS,WAITINGS ROOMS,
RESTAURANTS,WORK ROOMS,THEATRES ETC
4.0– WITHOUT SEPARATE STORAGE
2.5� OFFICIAL – WITH SEPARATE STORAGE
2.0� RESIDENTIAL
LIVE LOAD
(kN/m2)TYPE OF FLOOR USAGE
LIVE LOADS ON FLOORS AS PER IS-875(Part-2)-1987
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0.75� ROOF WITHOUT ACCESS
1.5� ROOF WITH ACCESS
LIVE LOADS ON ROOFS
12.0� WEATHER MAKER
10.0� MDF ROOM
6.0� OMC ROOM,DDF ROOM,POWER PLANT,
BATTERY ROOM
6.0� SWITCH ROOM(NEW TECHNOLOGY)
LIVE LOAD (kN/m2)TYPE OF FLOOR USAGE
LIVE LOADS ON FLOORS OF T.E.BLDGS
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3.0 Steps for Design of a Multi-Storeyed Building:-
Manual Method of Analysis & Design:-
Step1: Study of architectural Drawings:- Before proceeding for structural
design of any building it is ensure that approved working drawings are available
in the office. All working drawings i.e. each floor plan, elevations, sections, are
studied thoroughly & discrepancy if any brought to the notice of concern Architect
for rectification/correction. The problems coming in finalization of structural
configuration may also be intimated to concern Architect for
rectification/correction if any.
Step2: Finalization of structural Configuration. After receiving corrected
working drawing from the architectural wing, the structural system is finalized. The
structural arrangements of a building is so chosen as to make it efficient in resisting
vertical as well as horizontal loads due to earthquake. The span of slabs co chosen
that thickness of slab 100-150mm and slab panels, floor beams, and columns, are
all marked and numbered on the architectural plans. Now the building is ready for
structural design to start.
ISOMETRIC VIEW OF FRAMED STRUCTURE
Step3: Load Calculation and analysis. For each floor or roof, the loading
intensity of slab is calculated taking into account the dead load of the slab, finish
plaster, etc. including partitions and the live load expected on the floor, depending on
the usage of the floor or roof. The linear loading of beams, columns, walls, parapets,
etc. also calculated.
Step3 (a): Preliminary Sizes of structural members. Before proceeding for
load calculation preliminary sizes of slabs, beams,& columns decided. In manual load
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calculation preliminary sizes of structural members should be judicially fixed as once
load calculation & analysis is done it is not easy to revise the same. But in computer
aided analysis & design it can be revised easily.
• Slab:- The thickness of the slab decided on the basis of span/d ratio
assuming appropriate modification factor.
• Beam : The width the beam generally taken as the width of wall i.e
230 or 300 mm. The width of beam is help full in placement of
reinforcement in one layer & more width is help full in resisting shear
due to torsion. The depth of beam is generally taken as 1/12 th (for
Heavy Loads) to 1/15 th (for Lighter Loads) of span.
• Column:- Size of column depends upon the moments from the both the
direction and the axial load. Preliminary Column size may be finalized
by approximately calculation of axial load & moments.
Procedure for vertical load calculation on Columns: Step(i): First, the load from slab (including Live load & Dead Load) is transferred on
to the adjoining beams using formulas given below|:- For computation of shear force on beams & reactions on columns, an
equivalent uniformly distributed load per linear meter of beam may be taken as : Equivalent u.d.l. on short beam of slab panel = w B/4.0 Equivalent u.d.l. on long beam of slab panel = w B/4 x [2-(B/L)] Where w is the total load on the slab panel in Kn/Sqm & L & B are long span & short spans of slab panel respectively. Step(ii): Over this load, the weight of wall (if any), self weight of beam etc. are added
to get the load on beam (in running metre). Step(iii):The load (in running metre) on each beam is calculated as in Step 1 & Step 2.
Step(iv):Then the loads from the beams are transferred to the columns. Step(v):Step (i) to Step (v) is repeated for each floor. Step(vi):These loads at various floors on each column are then added to get the total
loads on each column, footing and the whole building.
Step4: HORIZONTAL (SEISMIC) LOAD CALCULTAION:
The Horizontal Load Calculation or the Load Calculations for Seismic case is carried
out as per the Indian Standard Code IS:1893-2002.
The loads calculated in Para-II above at various floor levels are modified as per the
requirement of Para 7.3.1 of IS:1893-2002.
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The Seismic Shear at various floor levels is calculated for the whole Building using
the values from IS 1893-2002.
Calculation of horizontal loads on buildings (As per is-1893-2002)
Sample example for horizontal load calcula tion
(I) BUILDING IS ON SEISMIC ZONE-IV (II) FOUNDATION TYPE ISOLATED FOOTINGS
As per clause 7.5 .3 of IS-1893-2002 Design base shear vb V
b = Ah W (F) Where A h = Design Horizontal acceleration spectrum value as per 6.4.2
of the code
= (Z/2) (I/R) (Sa /g)
Where Z = Zone factor as per table 2 of IS Code (1893-2002) = 0.24 (in this case) I = Importance factor as per table 6 of IS-1893-2002)
= 1.5 (Assuming that the bldg. is T.E. Bldg.)
R = Response reduction factor as per table 7 of IS code
= 3.0 (for ordinary R.C. Moment res isting frame (OMRF)
(Sa /g) = Average response acce lerat ion coefficient for soil t ype
& appropriate natura l periods and lamping of the
structu re.
For calculat ing of (Sa/g) va lue as above we have to calcu late value of
T i.e. Fundamental Nat ional Period (Seconds) (Clause 7.6 of IS Code)
T = 0 .075 h 0 . 75 (For RC Frame building)
= 0.0 85 h 0 . 75 (For Steel frame building)
h = Height of bu ilding in Meter
In case o f building with brick in f ills walls.
T = 0.09 h /d 1 /2
Where h = height of build ing in Meter
and d = Base dimension of the building at the plinth level in
Meter along the considered directio n of the lateral
force.
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Value of (Sa/g) is to be read from fig 2 on page 16 of IS Code
depending upon Soil cond it ion & Fundamental Natu ra l period T.
Or the value of (Sa/g) ma y be calculated on the basis of
Following.
Formulas:- ( i) For rocky, or hard soil s ites
(Sa/g) = 1+15 T if 0 .00≤T≤ 0 .10
= 2.50 if 0 .10≤T≤ 0 .40
=1.00/T if 0 .40≤T≤ 4 .00
(ii) For medium soil s ites (Sa/g) = 1+15 T if 0.00≤T≤ 0 .10
= 2.5 0 if 0 .10≤T≤0 .55
= 1.36/T if 0 .55≤T≤ 4 .00
(iii) For so ft soil s ites (Sa/g) = 1+15 T if 0.00≤T≤
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A
B
C
2 b
ays @
7.5
m C
/C
TYPICAL FLOOR PLAN
Bldg. is three storey with Each
storey of 5.0m height
4 bays @ 4.0 m C/C
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164.9
157.99
47.01
Frame with EQ Loads
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Calculating Seismic weight o f building per frame for frame (B)
Length b ldg = 16.00 M
C/C distance of frames = 7 .50 M
Densit y o f R.C.C = 25 KN/m3
Floor slab = 0 .15×16.00×7 .50×25 = 450 KN (A)
Column below slab = 0.300×(0.70-0.15)×16.00×25
= 0.30×0.55×16 .00×25=66 KN (B)
Columns = 0.30×0.60×(5.00+5.00)/2 ×25× 5 = 112.5 KN (C) Live load = 600 kg/m2 = 6.00 kn/m2
As per table 8 of code when live load is above 3.00 kn/m2
50% of live load to be considered for lamp mass calculat ion.
Lump mass at First Floor = 0.50×6.00×16×7.50 = 360 KN (D) To tal lamp mass f irst f loor & second floor (Assuming same L.L. on S.F.)
(A)+(B)+(C)+(D) = 450+66+112.50+360
= 988.50 KN
( ii) Wight lamped at terrace Floor slab :-
= 0.13×16.00×7.50×25 = 390 KN (E)
Beam below slab = 0.23×(0.60-0.15)×16 .00×25
= 0.23×0.45×16.00×25
= 41.4 .KN (F)
Columns = 0.30×0.600×5.00/2×25×5 = 56.25 KN (G) L.L. = Nil During Earthquake = 0. ( As per the clause 7.3.2 of the
code the imposed load on roof need no t to be considered )
To tal lamped mass at terrace level = (E)+(F)+(G)
= 390+41.40+56.25=487.65 KN
Total weight of bu ilding per framed per inner frame F.F = 988.50 KN S.F. = 988.50 KN Terrace = 487.65 KN 2464.65 KN
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Pu tting all values in Formulas (F) Vb = Design base shear = (z/z)(I/R(Sa/g) w Value of T = 0.09 h/Vd
H = Height o f b ldg. = 15.00 m (3x5.0=15.00m )
d = 16.00 m
T = 0.09×15.00/V 16.00 = 0.3375 = 0.34
For med ium soils For T = 0 .34 Sa/g = 2.50 Vb = 0.24/2×1.50/3 .00×2.50×2465.65 = 369.85 KN Distribution base shear is done using formula (clause 7 .7) F i = w i h i
2 / ∑ w j h j 2 x Vb
Where Fi = Design lateral force at f loor i
W i = Seismic weight of floor i
h i = he ight of f loor in m from base.
n = number of story’s in the build ing is equal to number of levels at
which masses are located.
Vb = 369.85 KN
Floor W i KN h i (m) W i h i 2 F i
F.F. S.F.
Terrace level
988.50 988.50 487.65
6.00 11.00 16.00
35586 119608.5 124800
47.01 KN 157.99 164.85
∑w i hi 2 = 279994.5 ∑ = 369.85 KN
Step5. VERTICAL LOAD ANALYSIS:
a ) GENERAL:
The skeleton frame work of a multi storied R.C.C. framed
structure is made up of a system of columns, beams and slabs. It is
p resumed that the re inforcements are always so arranged that al l
jo ints of the frame are monolithic.
In view of the uncerta in property o f material creep, shr inkage
and a number of approximate s implifying assumptions made in the
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detailed analysis of mult i sto ried framed structures (such as
cond itions o f end restra ints etc.) it is considered suffic ient to ob tain
reasonable accurac y of ana lysis for the design of structure. If the
normal moment d istr ibution is applied to all joints, the work
invo lved is enormous. However with certain assumptions, it is
possible to ana lyze the frames and get results which will be adequate
for design purposes.
To simplify ana lysis the three dimensional mult isto ried R.C.C.
framed structure are considered as combinat ions of planer framed in
two direct ions. It is assumed that each of these p laner frames act
independent ly of the frames.
Procedure for Frame analysis for calculation of moments in
Columns & beams:
Step(i) : Fir st, the load from slab (including Live load & Dead Load)
is transferred on to the ad joining beams using formulas
given below|:-
For computation of Bend ing Moments in beams , an equiva lent
uniformly distr ibuted load per linear meter of beam may be taken as
:
Equivalent u.d.l. on short beam of slab panel = w B/3.0
Equivalent u .d.l. on long beam of s lab panel = w B/6 x [ 3-
(B/L)2 ]
where w is the tota l load on the s lab panel in Kn/Sqm & L & B are
long span & short spans of slab panel respect ive ly.
Step(ii) : Over this load , the weight of wall ( if any), self weight of
beam etc. are added to get the load on beam (in running Meter).
Step(iii) :The load (in running Meter) on each beam is calcu lated as
in Step 1 & Step 2.
Step(iv):Step (i) to Step (iii) is repeated for each floor Step(v):Then these loads are used as u.d.l on a particu lar frame for
analysis by moment distr ibution method as descr ibed in the next
sect ion.
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b) METHOD OF ANALYSIS:
Analysis o f large framed structu res is too Cumbersome with
the classical method of structure analysis such a Clapeyron’s
theorem of three moments, Cast ingiliano’s therefore of least work,
Po ison’s method of vir tual work etc. Therefore, it become necessar y
to evolve s impler methods.
Some of these are :-
a.) Hardy cross method o f moment distr ibution.
b.) Kani’s method of iteration.
c) HARDY CROSS METHOD OF MOMENT DISTRIBUTION:
In this method , the ‘balancing’ and ‘carry-over’ const itute one
c ycle and it has been found the ‘carry-over’ values converge fast
enough to become qu ite insignificant after four cyc le of operation. it
is, therefo re, often adequate to stop the computation after four
c ycles.
The frame is analyzed by this method either:
i. Floor-wise assuming the co lumns to be fixed for ends.
or
i i. Taking the frame as a whole. The whole frame analysis can
be carried out for several alternat ive loading arrangements
for ob taining maximum posit ive and negative bending
moment. Generally frames are ana lyzed floor-wise for the
worst condit ions of loading.
Step6. HORIZONTAL LOAD ANALYSIS:-
Frame analysis for horizonta l loads calcu lated in step 4 is carr ied out by using :-
(a)Approximate Method:-
i) Cant ilever method.
ii) Portal method.
Approximate methods are used for preliminar y designs only. For f ina l design we may use exact method i.e ( i) Slope deflect ion or matrix methods ( ii) Factor method . We will not discuss these methods in detail as now modern
computer package as STAAD PRO is ava ilab le for analysis.
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Step7: DESIGN OF COULMN, FOUNDATIONS, BEAMS &
SLABS:
After load calculatio n & ana lysis for vertical & horizo ntal
loads, design of Columns ,Foundations, Beams, Slabs and are to be
carried out as per the various clauses of IS codes, IS 456-2000,
IS:1893-2002, IS:13920-1993 etc.
The Design of Co lumn, Foundation, Beams and Slabs are discussed
in deta ils in following sect ion.
A. DESIGN OF COLUMNS: -
With the knowledge of ( i) Vert ica l load (ii) Moments due to
horizonta l loads on either axis;( iii) Moments due to vert ical loads on
either axis, act ing o n each column, at all f loor levels of the bu ild ing,
columns are designed by charts o f SP-16(Design Aids) with a load
factor of 1.5 for vert ical load effect and with a load factor of 1.2 for
the combined effects of the vert ical and the horizontal loads. The
step confirms the s ize o f columns assumed in the architectural
d rawings. The design of each co lumn is carried out from the top of
foundation to the roof, var ying the amount o f stee l re inforcement for
suitable groups for ease in design. Further, slenderness e ffects in
each sto rey are considered for each co lumn group .
Important Considerations in design o f Columns:-
( i)Effective height o f column:- The effective height of a co lumn is
defined as the height between the points of contra f lexure of the
buckled column. For effect ive column height refer table 28
(Annexure E) of IS: 456-2000.
For framed structure effect ive height of column depends
upon relat ive st iffness of the column & various beams framing into
the co lumn at it s two ends. (Refer Annexure E o f IS: 456-2000 .)
( ii)Unsupported Length: - The unsupported length l, of a
compression member shall be taken as the c lear d istance between
end restraints excep t that :-
In beam & slab construction, it shall be the c lear distance
between the floor & under side of the shallower beam framing into
the co lumns in each direction at the next higher f loor level.
( iii) Slenderness limits for columns: - The unsupported length
between end restraints shall not exceed 60 times the least lateral
d imension of a column.
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( iv) Minimum Eccentricity: - All co lumns sha ll be designed for
minimum eccentr icit y equal to unsupported length of co lumn/500
p lus least lateral d imension/30, subject to a minimum of 20 mm.
Or emi n ≥ l/500+ D/30 ≥ 20 mm Where l= unsupported length of column in mm.
D=Lateral d imension of column in the direction under
considerat ion in mm.
(v)Design Approach: - The design of column is complex since it is
subjected to axial loads & moments which may very independent ly.
Column design requ ired:-
I. Determinat ion o f the cross sect iona l dimension.
II. The area of longitudinal steel & it s distr ibution. III. Transverse stee l.
The maximum axial load & moments act ing along the length of
the column are considered for the design of the column sect ion
either by the working stress method or limit state method .
The transverse reinfo rcement is provided to impart effect ive
lateral support against buckling to every longitud inal bar. It is e ither
in the form of circular r ings of polygonal links ( lateral ties) .
C. DESIGN OF FOUNDATIONS: -
With the knowledge of the co lumn loads and moments at base and
the soil data, foundations for columns are designed
The following is a list of different t ypes o f foundations in
o rder to p reference with a view to economy: ( i) Ind ividual footings
( ii) Combinat ion of ind ividual and combined footings (iii) Strip
foo tings with reta ining wall act ing as s tr ip beam wherever
applicab le; ( iv) Raft foundations of the t ypes (a) Slab (b) beam-slab.
The brick wall footings are a lso designed at this stage. Often,
p linth beams are provided to support brick walls and a lso to act as
earthquake t ies in each p rincipa l direct ion. Plinth beams, reta ining
wall if any, are also designed at this stage, being considered as part
o f foundations.
Important Considerations in design o f Foundations:-
a) Introduction: - Foundations are s tructural elements that transfer loads from the building or individ ual column to the earth. If these loads are to be properly transmitted , foundations must be designed to prevent excessive sett lement or rotation, to minimize
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different ia l settlement and to provide adequate safet y against slid ing and over turning.
b) Depth of foundation:- Depth of foundation below ground level may be obtained by using
Rankine's formula
2 p 1 – Sin Ø h = - - - - - - - - - - - - - - - γ 1 + Sin Ø Where h = Minimum dep th of foundation
p = Gross bearing capacit y
γ = Densit y o f soil
Ø = Angle of Repose of soil
c) Recommendations of IS 456 -2000, limit sta te design, bending,
shear, cracking & development
i) To determine the area required for proper transfer of total load on the soil,
the total load (the combination of dead, live and any other load without
multiplying it with any load factor) need be considered.
To tal Load includ ing Self Weight Plan Area of footing = ------------------------------------- Allowable bearing capacit y of soil i i) IS 1904 – 1978, Code of Practice for Structural Safety of
Buildings: shallow foundation, sha ll govern the genera l details.
i i i) Thickness o f the edge o f footing :-(Reference clause 34.1.2) The thickness at the edge shall not be less than 15 cm for footing on soils. iv) Dimension o f pedestal:-
In the case of plain Cement Concrete pedestals, the angle between the plane passing through the bottom edge of the pedestal and the corresponding junction edge of the column with pedestal and the horizontal plane shall be governed by the expression.
100 qo Tan α (should not be less than) 0.9 x ----------- + 1 F ck
Where q o = Calculated maximum bearing p ressure at the base of the
pedestal/ footing in N/mm2 fck = Characteris t ic strength of concrete at 28 days in N/mm2
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(v) Bending Moment
(Reference Clauses- 34.2.3.1 & 34 .2.3.2)
ISOLATED COLUMN FOOTING
FAC E OF PEDESTA
P LAIN CONCRETE PEDESTAL
α
Co lumn
X Y
X Y
COLUMN
PEDESTAL BASE
FACE OF COLUMN
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The bending Moment will be considered at the face of column, Pedestal or wall
and shall be determined by passing through the section a vertical place which
extends completely across the footing, and over the entire area of the footing
or, one side of the said plane.
(vi)Shear
(Reference Clause 33.2.4.1)
The shear strength of footing is governed by the following two factors:-
a) The footing acting essentially as a wide beam, with a potential diagonal
crack intending in a plane across the entire width, the critical section for
this condition shall be assumed as a vertical section located from the
face of the column, pedestal or wall at a distance equal to the effective
depth of the footing in case of footings on soils.
FOR ONE WAY BENDING ACTION
For one way shear action, the nominal shear stress is calculated as follows:-
Vu
τv = -------
b.d
Where
τv = Shear stress
Vu = Factored vertical shear force
b = Breadth of critical section
d = Effective depth
τv < τc ( τc = Design Shear Strength of Concrete Based on % of longitudinal tensile reinforcement refer Table 61 of SP-16)
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CRITICAL SECTION FOR ONE -WAY SHEAR
(FOR TWO WAY BENDING ACTION)
For two may bending action, the following should be checked in punching shear. Punching shear shall be around the perimeter 0.5 time the effective depth away from the face of column or pedestal. For two way shear action, the nominal shear stress is calculated in accordance with lause 31.6.2 of the code as follows:-
Vu
τv = ----------
b0.d
Where
τv = Shear stress
b0 = Periphery of the critical section
d = Effective depth
Vu = Factored vertical shear force
B
A
C R IT IC AL SE C T ION
d
d
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When shear reinforcement is not provided, the nominal shear stress at the critical section should not exceed [Ks. τc]
Where
Ks = 0.5 + Bc (But not greater than 1)
Short dimension of column or pedestal
Bc = ----------------------------------------------------
Long dimension of column or pedestal
τc = 0.25 fek N/mm2
Note:-It is general practice to make the base deep enough so that shear reinforcement is not required.
(vii)Development Length
(Reference Clause 34.2.4.3)
The critical section for checking the development length in a footing shall be assumed at the same planes as those described for bending moment in clause 34.2.3 of code (as discussed 4.5 of the handout) and also at all other vertical planes where abrupt changes of section occur.
(viii) Reinforcement:- The Min % of steel in footing slab should be 0.12% & max spacing should not be more than 3 times effective depth or 450 mm whichever is less. (Reference Clause 34.3) Only tensile reinforcement is normally provided. The total
reinforcement shall be laid down uniformly in case of square footings. For rectangular footings, there shall be a central band, equal to the width of the footings. The reinforcement in the central band shall be provided in accordance with the following equation.
Reinforcement in central Band width 2
-------------------------------------------------- = ------
Total reinforcement in short direction B + 1
Where Long side of footing
B = --------------------------- Short side of footing
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(ix)Transfer of Load at the Base of Column
(Reference Clause 34.4)
The compressive stress in concrete at the base of column or pedestal shall
be transferred by bearing to the top of supporting pedestal or footing.
The bearing pressure on the loaded area shall not exceed the permissible
bearing stress in direct
A1
Compression multiplied by a value equal to ------
A2
but not greater than 2
Where
A1 = Supporting area for bearing of footing, which is sloped or
stepped footing may be taken as the area of the lower base
of the largest frustum of a pyramid or cone contained wholly
with in the footing and having for its upper base, the area
actually loaded and having side slope of one vertical to two
horizontal.
A2 = Loaded area at the column base.
For limit state method of design, the permissible bearing stress shall be = 45 fek
4.91 If the permissible bearing stress is exceeded either in column concrete or in
footing concrete, reinforcement must be provided for developing the excess
force. The reinforcement may be provided either by extending the longitudinal
bars into the footing or by providing dowels in accordance with the code as
give in the following:-
1) Minimum area of extended longitudinal bars or dowels must be 0.5% of
cross sectional area of the supported column or pedestal.
2) A minimum of four bars must be provided.
3) If dowels are used their diameter should not exceed the diameter of the
column bars by more than 3 mm.
4) Enough development length should be provided to transfer the
compression or tension to the supporting member.
5) Column bars of diameter larger than 36 mm, in compression only can
be dowelled at the footing with bars of smaller diameters. Te dowel
must extend into the column a distance equal to the development length
of the column bar. At the same time, the dowel must extend vertically
into the footing a distance equal to the development length of the
dowel.
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C. DESIGN OF FLOOR SLABS:-. Design of f loor slabs and beams is taken up with the First Floor & upwards .The s labs are designed as one-way or two-way panels, taking the edge condit ions of the supporting edges in to account, with the loading already decided as per funct ional use of s lab panel. The design of f loor slab is carr ied out as per clause 24 .4 & 37.1.2 & Annexure D of IS: 456-2000. The Bending moment coefficients are to be taken from table- 26 of the code depend ing upon the support cond itio n & bending moment calcu lated & reinforcement stee l ma y be calculated from the charts of SP-16. The s lab design for particular f loor may be done in tabular form as shown below.
SLAB DESIGN
Name of project:-
Level of s lab
Slab
ID
Edge co ndit i
on
Tot al load in KN/Sq.m w
Short
span lx m
long spa
n ly m
ly/lx
1 .5 *w* lx
* lx
s lab thic
kness in
mm
Shor t span Mome nt KN-M
Steel in
shor t span
Lo ng span mome nt KN-M
Stee l in
lo ng span
αx
(+)
Αx
(-)
mux
+
mux
-
Stee
l
Αy
(+) αy(-)
muy
+
muy
- Steel
1 2 3 4 5 6 7 8 9 1 0 11=
7 x 9
12=
7
x10
13 14 15
16=
7
x14
17=
7
x15
1 8
S1
Two Adj . Edge. Discont . (Case No.4)
8 .50 3.50
5 .25
1.5
156.80
120 0 .056
0 .075
8. 78 11. 76
0.035
0 .047
5 .49 7.37
S2
Method of calculation of steel from Tables of SP-16 for slab design
Determine the main reinforcement required for a slab with the following data:
Factored moment Mu 9.60 kN.mper Metre width
Depth of slab 10 cm
Concrete mix M 20
Characteristic strength a) 415 N/mm2
METHOD OF REFERRING TO TABLES FOR SLABS Referring to table 35 (for fck=20 & fy = 415 N/mm
2), directly we get the following
reinforcement for a moment of resistance of 9.60 kN.m per Metre width:
8 mm dia at 13 cm spacing or 10 mm dia at 210 cm spacing Reinforcement given in the table is based on a cover of 15 mm or bar diameter which-
ever is greater.
Check for Deflection:-Slab is also checked for control of Deflection as per clause
23.2.1, 24.1 & Fig 4. of the IS:456-2000.
D. Design of floor Beams:-. The beams are designed as continuous beams,
monolithic with reinforced concrete columns with their far ends assumed fixed. The
variation in the live load position is taken into account by following the two-cycle
moment distribution. the moments are applied a face correction to reduce them to the
face of the members. The moments due to horizontal loads are added to the above
moments. Each section of the beam is designed for load factor of 1.5 for vertical load
effect and with a load factor of 1.2 for the combined effects of the vertical and the
horizontal loads.
The effect of the shear due to vertical and horizontal loads is also similarly
taken care of. It may be noted that the shear component due to wind or earthquake
may be significant and it may affect the size and the range of shear stirrups. Bent- up
bars are not effective for earthquake shear due to its alternating nature. The beam
design can be easily done by a computer program which will give reinforcement at
various critical sections along the length of the beam and also shear stirrups required
it saves considerable time and labour of a designer.
In manual method span of a beam is generally designed at three sections i.e
at two supports & at Mid span. The each section is designed for factored Moment,
Shear & equivalent shear for Torsion if any at a section.
Two examples of beam design are given below illustrating calculation of
steel reinforcement with help of SP-16.
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Example1.Singly Reinforced Beam
Determine the main tension reinforcement required for a
rectangular beam sect ion with the following data:
Size o f beam 30 X 60 cm
Concrete mix M 20
Character ist ic strength 415 N/mm2 o f reinfo rcement
Facto red moment 170 kN.m
Assuming 20 mm dia bars with 25 mm clear cover, Effect ive depth= 600 – 25 – 20 /2 = 565 mm From Tab le D for fy = 415 N/ mm
2 and f c k = 20 N/mm
2
Mu, l im/bd2 = 2.76 N/ mm2
= 2.76/1000 X (1000)2
= 2.76 X 103 kN/m2
Mu, l im = 2.76 X 103 bd 2
= 2.76 X 103 X 0.300 X0.565X0.565
= 264.32 kN.m
Actual moment of 170 kN.m is less than Mu, l im. The sect ion is therefo re
to be designed as a s ingly reinfo rced (under-reinforced) rectangu lar
sect ion.
Referring to table 2 of SP-16 we have to ca lculate Mu/bd2 Mu/bd2 = 170 x106 /(300x 565 x565) = 1.78 From Tab le 2. p t =0.556 A st=0.556 x 300x 565/100 =942.42mm
2=9.42 cm2
Example2 .Doubly Reinforced Beam (i)Determine the main reinforcements required for a rectangular
with the following data :
Size o f beam 30×60cm
Concrete mix M 20
Character ist ic strength of 415/Nmm2
Reinforcement
Facto red moment 320Kn.m
Assuming 20 mm dia bars with 25 mm clear cover,
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D=600-25 – 20/ 2 =565 mm From tab le D, fo r fy= 415 N/mm
2 and fc k = 20N/mm2
Mu2 lim/bd
2= 2.76 N/mm2 = 2.76 × 103 KN/m2 Mu2 lim= 2.76 × 10
3×0.300×0.565×0.565 = 264.32 KN-M Actual moment of 320 Kn.M is greater than Mu2 lim hence the sect ion
is to be designed as a doubly reinforced section .
Reinforcement from Table 50 Mu/bd2= 320 × 106 / (300×5652 ) = 3.34 N/mm2
d’/d = (25+10)/565 = 0.062
Next higher value of d1 /d = 0.1 , will be used for referr ing to Table 50 For Mu/bd2 = 3.34 and d ’ /d = 0.10, p t
= 1.152, p c = 0.207 Ast = 1.152 x 300x 565 /100 =1952.64 mm
2 =19.52 cm2 And Asc = 0.207 x 300x 565/100 =350.86 mm
2 =3.51 cm2 (ii) Determine the Shear reinforcement (vertical stirrups ) required for the same beam section if factored shear force is Vu =250 KN. Shear stress τ v = Vu/bd = 250 × 10
3 /(300× 565) =1.47 N/mm2
τv < τma x( 2.8 N /mm2) hence sect ion is adequate regarding shear
stress. From table 61 for p t=1.15 τc=0.65 N/mm
2
Shear capac it y of co ncrete sect ion = τc × b× d = 0.65 × 300 ×565/1000=110 .18 kN Shear to be carried b y st irrups Vu s=Vu - τc × b× d = 250 - 110.18 =139.82 kN Vus/d = 139.82/56.5 = 2.47 kN/cm Referr ing to tab le 62 for steel fy = 415 N/mm
2 Provide 8 mm diameter two legged vert ical st irrups at 140 mm spacing.
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TABLES
FOR
DESIGN
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DETAILING AS PER IS 13920
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HOOP SPACING
HOOP SPACING
< d/4 and 8 db
B = BREADTH OF BEAM
db = DIAMETER OF LONGITUDINAL BAR
2d
d
2d
db
50 m max 50 m max
MIN 2 BARS FOR FULL LENGTH
ALONG TOP AND BOTTOM FACE
AS > MIN. Bd
AS < MAX Bd
> d /2
BEAM REINFORCEMENT37
Questions:-
1. Which are the important BIS Codes/handou ts used for structu ral
design of RCC build ings?
2. In which seismic zones provis ions of IS 13920 is to be adopted
fo r all re inforced concrete structu res?
3. (a) What are the basic values o f span to effect ive depth rat ios for
beams as per IS 456 fo r span upto 10meter for –
( i) Cant ilever
( ii) Simply supported
(iii) Cont inuous
(b) What are the basic values of span to overall depth ratios for
two-way s labs upto 3 .5 m span & with Fe415 steel re inforcement
and loading class upto 3KN/m2?
4. What are the provisions o f IS 456 for nominal cover to meet
durabilit y requirements? As per IS 456 how much minimum
cover should be provided for –
a) Co lumn b) Footing
5. (a) What are the minimum reinforcement provis ion of IS 456 for
beams in respect of:
(i) Tension re inforcement
(ii) Shear re inforcement
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(b) What are the IS 456 provisions for maximum reinforcement
in beams fo r:-
(i) Compression reinfo rcement
(ii) Tension re inforcement
6. What is the maximum permissible spacing for shear
reinforcement in beams? Expla in IS 456 provis ions for side face
reinforcement in beams.
7. How much minimum reinforcement must be provided in s labs?
8. As per IS1893 give formulae for calculating –
a) Design Base Shear (Vb)
b) Design Hor izonta l accelerat ion (Ah)
9. Give formulae for calculat ing t ime period as per IS1893 for –
a) RCC Frame Building b) RCC Building Br ick in f ill walls
10. How vert ical loads on co lumns are ca lculated? Give names of
s impler methods of ana lysis of structures.
11. What is the minimum eccentr icit y for which all columns should
be designed? List out minimum and maximum longitudina l
reinfo rcement required to be provided in co lumns? Give in
brief provisions for maximum spacing of lateral t ies in a
column?
12. What are the cr itical sect ions in isolated footing design fo r the
following:-
a) Bending moment
b ) One way shear
c) Two way shear
13. How many minimum longitudinal reinforcement bars should be
provided in:-
a) Circular co lumn
b) Rectangular column
14. What is the minimum diameter bar that can be used in
longitudina l reinforcement in column?
--------------------
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