Civil Technical

Chapter - 2

STRUCTURAL DESIGN OF RCC BUILDING COMPONENTS

Rajendra Mathur Dy. Dir(BS-C) 09412739 232(M) e-mail – [email protected]

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Structural Design of RCC

Building Components

1.0 Introduction

The procedure for analysis and design of a given building will depend on the type of building, its complexity, the number of stories etc. First the architectural drawings of the building are studied, structural system is finalized sizes of structural members are decided and brought to the knowledge of the concerned architect. The procedure for structural design will involve some steps which will depend on the type of building and also its complexity and the time available for structural design. Often, the work is required to start soon, so the steps in design are to be arranged in such a way the foundation drawings can be taken up in hand within a reasonable period of t ime. Further, before starting the structural design, the following information of data are required: (i) A set of architectural drawings;(ii) Soil Investigation report (SIR) of soil data in lieu thereof; (iii) Location of the place or city in order to decide on wind and seismic loadings;(iv) Data for lifts, water tank capacities on top, special roof features or loadings, etc. Choice of an appropriate structural s ystem for a given building is vital for its economy and safety. There are two type of building systems:(a) Load Bearing Masonry Buildings. (b) Framed Buildings. ( a ) Load Bearing Masonry Buildings:-

Small buildings like houses with small spans of beams, slabs generally constructed as load bearing brick walls with reinforced concrete slab beams. This system is suitable for building up to four or less stories.(as shown in fig. below). In such buildings crushing 2 strength of bricks shall be 100 kg/cm minimum for four stories. This system is adequate for vertical loads it also serves to resists horizontal loads like wind & earthquake by box action . Further, to ensure its action against earthquake , it is necessary to provide RCC Bands in horizontal &

1967( Indian Construction

IS: 4326vertical re inforcement in brick wall as per Standards Code of Practice for Earthquake Resistant of Buildings.) . In some Buildings, 115mm thick brick

walls are provided since these walls are incapable of supporting vertical loads, beams have to be provide a long their lengths to support adjoining slab & the weight of 115mm thick brick wall of upper storey. These beams are to rest on 230 mm thick brick walls or reinforced concrete columns if required. The design o f Load Bearing ©BSNL India


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Masonry Buildings are done as per Code of Practice for Structural Walls(Second Revision).

IS:1905-1980 (Indian Standards Safety of Buildings: Masonry

Load bearing brick wall Structural system

(b) Framed Buildings:In these types of buildings reinforced concrete frames are provided in both principal directions to resist vertical loads and the vertical loads are transmitted to vertical framing system i.e columns and Foundations. This type of system is effective in resisting both vertical & horizontal loads. The brick walls are to be regarded as non load bearing filler walls only. This system is suitable for multi-storied building which is also effective in resisting horizontal loads due to earthquake. In this system the floor slabs, generally 100-150 mm thick with spans ranging from 3. 0 m to 7.0 m. In certain earthquake prone areas, even single or double storey buildings are made framed structures for safety reasons. Also the single storey buildings o f large storey heights (5.0m or more ) ,like electric substation etc. are made

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framed structure as brick walls of large heights are slender and load carrying capacity of such walls reduces due to slenderness.

Framed Structural system

2 . 0 Basic Codes for Design

.

The design should be carried so as to conform to the following Indian code for reinforced concrete design, published by the Bureau of Indian Standards, New Delhi: Purpose of Codes

National building codes have been formulated in different countries to lay down guidelines for the design and construction of structure. The codes have evolved from the collecti ve wisdom of expert structural 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

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Firstly, they ensure adequate structural safety, by specifying certain essential minimum requirement for design. Secondly, they render the task of the designer relatively simple; often, the result of sophisticate ana lyses is made available in the form of simple formula or chart. Thirdly, the codes ensure a measure of consistency among different designers. Finally, they have some legal validity in that they protect the structural designer from any liability due to structural failures that are caused by inadequate supervision and/or faulty material and construction. (i)IS 456 : 2000 (fourth revision)

– Plain and reinforced concrete

a

– code of practice

(ii) Loading Standards

These loads to be considered for structural design are specified in the following 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 Indian standards has also published the following handbooks, which serve as useful supplement to the 1978 version of the codes. Although the handbooks need to be updated to bring them in line with the recently revised (2000 version) of the Code, many of the provisions continue to be valid (especially 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 Detailing. – Handbooks on Concrete Reinforced and

General Design Consideration of IS: 456-2000.

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The general design and construction of reinforced concrete buildings shall be governed by the provisions of IS 456 –2000 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 –

26.4 ©BSNL India

Nominal Cover to Reinforcement


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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. 2. 3.

For main reinforcement up to 12 mm diameter bar for mild exposure the nominal cover may be reduced by 5 mm. Unless specified otherwise, actual concrete cover should not deviate from the required nominal cover by + 10 mm 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.

26.4.3 Nominal Cover to Meet Specified Period of Fire Resistance

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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

DESIGN LOAD

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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.

STIFFNESS

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22.3.1

Relative Stiffness: The relative stiffness of the members may be based on

the moment of inertia of the section determined on the basis of any one of the following definitions: a) b)

c)

Cracked

The cross-section of the member ignoring reinforcement The concrete cross-section plus the area of reinforcement transformed on the basis of modular ratio The area of concrete in compression plus the area of

Section

reinforcement transformed on the basis of modular ratio

Gross Section Transformed Section

The assumptions made shall be consistent for all the numbers of the structure throughout any analysis. 22.3.2

For deflection calculations, appropriate values of moment of inertia as specified in Annexure of IS 456-2000 should be used.

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: 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 22.4.3

constitute the frame. 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.

MOMENT AND SHEAR COFFICIENTS FOR CONTINUOUS BEAM S

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22.5.1

Unless more exact estimates are made, for beams of uniform cross-section which support substantially uniformly distributed load over three or more spans which do not differ by more than 15 percent of the longest, the bending moments and shear forces used in design may be obtained using the coefficients given in Tables below. For moments at supports where two unequal spans meet or in case where the spans are not equally loaded, the average of the two values for the negative moment at the support may be taken for design. Where coefficients given in Table below are used for calculation of bending moments, redistribution referred to in 22.7 shall not be permitted.

22.5.2

Beams Over Free End Supports

Where a member is built into a masonry wall which develops only partial restraint, the member shall be designed to resist a negative moment at the face of the support of W1/24 where W is the total design load and 1 is the effective span, or such other restraining moment as may be shown to be applicable. For such a condition shear coefficient given in Table below at the end support may be increased by 0.05. ------------------------------------------------------------------------------------------------------BENDING MOMENT COFFICIENTS -------------------------------------------------------------------------------------------------------

Span Moments Support Moments ------------------------------------------------------------------------------------Types of Load Near Middle At Middle At Support At Other of End Span of interior next to the Interior span end support Supports ------------------------------------------------------------------------------------------------------Dead load and 1 1 1 1 imposed load +-+-(- )-(- )-(fixed) 12 16 10 12 Imposed load (not fixed)

1 +-10

1 +-12

1 (- )-9

1 (- )-9

For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span. ------------------------------------------------------------------------------------------------------Note:

------------------------------------------------------------------------------------------------------SHEAR FORCE COFFICIENTS -------------------------------------------------------------------------------------------------------

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Type of Load

At End Support

At Support Next At All Other to the end Support Interior Support Outer side Inner Side ------------------------------------------------------------------------------------------------------Dead load and imposed load 0.40 0.60 0.55 0.50 (fixed) Imposed load (not fixed)

0.45

0.60

0.60

0.60

Note:

For obtaining the shear force, the coefficient shall be multiplied by the total design load. ------------------------------------------------------------------------------------------------------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

22.6.2.1

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 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.

REDISTRIBUTION OF MOMENTS

22.7

Redistribution of moments may be done in accordance with 37.1.1 for limit state method and in accordance with B-1.2 for working stress method. However, where simplified analysis using coefficients is adopted, redistribution of moments shall not be done.

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 Simply supported Continuous

7 20 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 required calculated, the method given Annexure ‘C’ of IS are 456 -2000 maytobebeused.

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CONTROL OF DEFLECTION – SOLID SLABS 24.1 General

The provisions of 32.2 for beams apply to slabs also. NOTES

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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 For high strength deformed bars of grade Fe 415,the values given above should be multiplied by 0.8.

Simply supported slab Continuous slabs 23.3

28 32

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 assumed 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

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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. 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 bsv

0.4 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/mm 2 which shall not 2

taken greater than 415 N/mm 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: ©BSNL India


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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.

b)

Longitudinal reinforcement shall be place as closed as is practicable to the corner of the cross section & in all cases, there shall be atleast 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 satisfied 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 2/3 rd 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

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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. 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.

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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. 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.

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40.3 Minimum Shear Reinforcement When v, is less than τc given in Table 19, minimum shear reinforcement shall be

provided in accordance with 26.5.1.6.

40.4 Design of Shear Reinforcement When v, is exceeds c , 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 V u – τ strength of shear reinforcement Vus shall be calculated as below:

c

b d. the

a) For Vertical Stirrups:

Vus ©BSNL India

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Sv b) For inclined stirrups or a series of bars bent up at different cross – section: 0.87 fy Asv d ___________ (Sin ά + Cos ά) Sv c) For single bar or single group of parallel bars, all bent up at the same cross sections: Vus

=

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.

τv

=

nominal shear stress,

τc

=

design shear strength of the concrete,

b

= breadth of the member which for flanged beams, shall be taken as the breadth of the web bw.

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 45 o,

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

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Ld = υσst /4τbd υ = 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 %. 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

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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. 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 thermomechanical 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 min = 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 max = 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,

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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 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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DEAD LOADS – UNIT WEIGHTS OF SOME MATERIALS/BUILDING COMPONENTS As per IS-875(Part-1)-1987

UNIT WEIGHT MATERIAL

kN/m3

PLAIN CONCRETE

24

REINFORCED CONCRETE

25

BRICK MASONRY STONE MASONRY TIMBER CEMENT-PLASTER LIME -PLASTER STEEL

19-20 21-27 6-10 21 18 78.5

AC SHEET -ROOFING

0.16

GI SHEET -ROOFING

0.15

MANGLORE TILES

0.65

STEEL WORK -ROOFING

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kN/m2


0.16-0.23

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LIVE LOADS ON FLOORS AS PER IS-875(Part-2)-1987

LIVE LOAD (kN/m2)

TYPE OF FLOOR USAGE 

RESIDENTIAL

2.0



OFFICIAL – WITH SEPARATE STORAGE

2.5

– WITHOUT SEPARATE STORAGE 

RESTAURANTS,WORK ROOMS,THEATRESETC - WITH FIXED SEATING



4.0

SHOPS,CLASS ROOMS,WAITINGS ROOMS,

4.0

- WITHOUT FIXED SEATING

5.0

FACTORIES & WAREHOUSES

5.0-10

STACK ROOM IN LIBRARIES ,BOOK STORES

10.0

GARRAGES –LIGHT VEHICLES

4.0





–HEAVY VEHICLES

7.5

STAIRS-NOT LIABLE TO OVER CROWDING



- LIABLE TO OVER CROWDING

4.0 5.0

LIVE LOADS ON FLOORS OF T.E.BLDGS

TYPE OF FLOOR USAGE

LIVE LOAD (kN/m2)



SWITCH ROOM(NEW TECHNOLOGY)

6.0



OMC ROOM,DDF ROOM,POWER PLANT,

6.0

BATTERY ROOM 

MDF ROOM

10.0



WEATHER MAKER

12.0

LIVE LOADS ON ROOFS 



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ROOF WITH ACCESS

1.5

ROOF WITHOUT ACCESS

0.75


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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 1/15 th Heavy Loads) Lighter Loads) of span. from the both the Column:Size to of column(for depends upon the moments 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 (As per is-1893-2002)

buildings

Sample example for horizontal load calculation (I) (II)

BUILDING IS ON SEISMIC ZONE-IV

FOUNDATION TYPE ISOLATED FOOTINGS As per clause 7.5.3 of IS-1893-2002 Design base shear v V

b

= Ah W

b

(F)

Where A h = Design Horizontal acceleration spectrum value as per of the code

6.4.2

= (Z/2) (I/R) (Sa/g)

Where

Z

I R (S a / g)

= Zone factor as per = 0.24 (in this case)

table 2

of IS Code (1893-2002)

= Importance factor as per table 6 of IS-1893-2002) = 1.5 (Assuming that the bldg. is T.E. Bldg.) = Response reduction factor as per table 7 of IS code = 3.0 (for ordinary R.C. Moment resisting frame (OMRF) = Av erage respo ns e accel eration coeffi cie nt for s oil t yp e & appropriate natural periods and lamping of the structure.

For calculating of (Sa/g) value as above we have to calculate value of Clause 7.6 of IS Code T i.e. Fundamental National Period (Seconds) ( T h

)

= 0.075 h 0.75 (For RC Frame building) = 0.0 85 h 0.75 (For Steel frame building) = Height of building in Meter

In case of building with brick in fills walls. T = 0.09 h /d

Where h and d

©BSNL India

1/2

= height of building in Meter = Base dimension of the building at the plinth level in Meter along the considered direction of the lateral force.


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Value of (Sa/g) is to be r ead from fig 2 on page 16 depending upon Soil condition & Fundamental Natural period T.

of IS Code

Or the value of (Sa/g) may be calculated on the basis of Following. Formulas:(i) For rocky, or hard soil sites (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 sites (Sa/g)

= 1+15 T if = 2.5 0 if = 1.36/T if

0.00≤T≤ 0.10 0.10≤T≤0.55 0.55≤T≤ 4.00

(iii) For soft soil sites (Sa/g)

= 1+15 T if 0.00≤T≤ <0.10 = 2.50 if 0.10≤T≤0.67 =1.67/T if 0.67≤T≤ 4.00

W= Seismic weight of the building as per clause

©BSNL India


7.4.2 of the code

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.

Page: 31

A

C / C m .5 7 @ s y a b 2

B

C

4 bays @ 4.0 m C/C TYPICAL FLOOR PLAN

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Bldg. is three storey with Each storey of 5.0m height

164.9

157.99

47.01

Frame with EQ Loads

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Calculating Seism

Floor slab

ic weight of building per frame for frame

(B)

Length bldg = 16.00 M C/C distance of frames = 7.50 M 3 Density of R.C.C = 25 KN/m = 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 (C)

= 0.30×0.60×(5.00+5.00)/

Live load = 600 kg/m

2

= 6.00 kn/m

2 ×25× 5 = 112.5 KN

2

As per table 8 of code when live load is above 3.00 kn/m 50% of live load to be considered fo r lamp mass calculation.

2

Lump mass at First Floor = 0.50×6.00×16×7.50 = 360 KN (D) Total lamp mass S.F.)

first floor & second

floor

(Assuming same L.L.

on

(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 code the imposed load on roof need not to be considered ) Total lamped mass at terrace level = (E)+(F)+(G) = 390+41.40+56.25=487.65 KN

of the

Total weight of building per framed per inner frame F.F S.F. Terrace

= = =

988.50 KN 988.50 KN 487.65 KN 2464.65 KN

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Putting all values in Formulas (F) V b = Design base shear = (z/z)(I/R(Sa/g) w Value of T = 0.09 h/Vd H = Height of bldg. = 15.00 m (3x5.0=15.00m ) d = 16.00 m T = 0.09 × 15.00/V 16.00 = 0.3375 = 0.34 For medium soils For T = 0.34 Sa/g = 2.50 V b = 0.24/2 × 1.50/3.00 × 2.50 × 2465.65 = 369.85 KN Distribution base shear is done using f

Fi = w

i

h

i

2

/ ∑ w

j

h

j

2

ormula (clause 7.7)

x Vb

Where Fi = Design lateral force at floor i W i = Seismic weight of floor i h i = height of floor in m from base. n = number of story’s in the building is equal to number of levels at which masses are located. V b = 369.85 KN Floor

W

F.F. S.F. Terrace level

i

KN

988.50 988.50 487.65

h i (m)

W

i

h

i

6.00 35586 11.00 119608.5 16.00 124800 ∑ w i h i = 279994.5

F i

47.01 KN 157.99 164.85 ∑ = 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 presumed that the reinforcements are always so arranged that all joints o f the frame a re monoli thi c. In view of the uncertain property of material creep, shrinkage and a number of approximate simplifying assumptions made in the

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detailed analysis of multi storied framed structures (such as conditions of end restraints etc.) it is considered sufficient to obtain reasonable accuracy of analysis for the design of structure. If the normal moment distribution is applied to all joints, the work involved is enormous. However with certain assumptions, it is possible to analyze the frames and get results which will be adequate for design purposes. To simplify analysis the three dimensional multistoried R.C.C. framed structure are considered as combinations of planer framed in two directions. It is assumed that each of these planer frames act independently of the frames. Procedure for Frame analysis Columns & beams:

for calculation

of moments in

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 Bending Moments in beams , an equivalent uniformly distributed load per linear meter of beam may be taken as : Equivalent Equivalent

u.d.l. on short beam of slab panel = w B/3.0 u.d.l. on long beam of slab panel = w B/6 (B/L) 2 ]

x

[ 3-

where w is the total load on the slab panel in Kn/Sqm & L & B are long span & short spans of slab panel respectivel y. 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 calculated 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 o n a particular frame for analysis by moment distribution method a s described in the next section.

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b) METHOD OF ANALYSIS:

Analysis of large framed structures beams too Cumbersome with the classical method of structure analysis such a Clapeyron’s theorem of three moments, Castingiliano’s therefore of least work, Poison’s method of virtual work etc. Therefore, it become necessa to evolve simpler methods.

ry

Some of these are:a.) Hardy cross method of moment distribution. b.) Kani’s method of iteration. c) HARDY CROSS METHOD OF MOMENT DISTRIBUTION:

In this method, the ‘ balancing’ and ‘carry - over’ constitute one cycle and it has been found the ‘carry - over’ values converge fast enough to become quite insignificant after four cycle of operation. it is, therefore, often adequate to stop the computation after four cycles. The frame is analyzed by this method either:

i.

Floor-wise assuming the columns to be fixed for ends. or

ii.

Taking the frame as a whole. The whole frame analysis can be carried out for several alternative loading arrangements for obtaining maximum positive and negative bending moment. Generally frames are analyzed floor-wise for the worst conditions of loading.

The method is described in the following steps

Step1:

Calculate the stiffness calculation scheme.

Step2:

Calculated the distribution factor at all joints from the stiffness. Enter them in the calculation scheme. Look the joints and calculate the fixed- end moments. Enter them in the calculation scheme.

Step3:

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.


of all members. Enter them in the

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Step4: Unlock the joint one by one by applying imaginary external moments at each joint which nullifier the unbalanced moment at the joint. Distribute the imaginary external moment among all members

Step5:

Meeting at the joint in proportion to their relative stiffness and enter these value in the scheme. This operation is called balancing. Enter the carry-over moments at the far in t he scheme.

Step6:

Repeat steps 4 & 5, till the carry-over moments become insignificant .

Step7:

Balance the unbalanced carry-over operation.

Step8:

Add the initial fixed-end moments, balancing moments and carry-over moments to get the final end moments in beam & columns.

moment obtained

from the last

A sample of moment distribution method is shown on next two pages.

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CALCULATION OF DISTRIBUTION FACTOR FOR FRAME ANALYSIS

S.NO.

JOINT

MEMBE R

Size in Cm

B A-III 1

B-III 2

3

C-III

A-II 4

B-II

5

C-II 6

A-I 7

B-I

8

C-I 9

©BSNL India

Right beam Lower Col. Left beam Right beam Lower Col. Left beam Lower Col. Upper col. Right beam Lower Col. Left beam Upper col. Right beam Lower Col. Left beam Upper col. Lower Col. Upper col. Right beam Lower Col. Left beam Upper col. Right beam Lower Col. Left beam Upper col. Lower Col.

Moment of Inertia Cm4 (I)

Leng th of mem ber Cm

K=I/L

Sum K

D.F.

D

30

45

227812.50

600

379.69

30

45

227812.50

350

650.89

30

45

227812.50

600

379.69

30

60

540000.00

700

771.43

30

45

227812.50

350

650.89

30

60

540000.00

700

771.43

30

45

227812.50

350

650.89

30

45

227812.50

350

650.89

30

45

227812.50

600

379.69

30

45

227812.50

350

650.89

30

60

540000.00

600

900.00

30

45

227812.50

350

650.89

0.22

30

60

540000.00

700

771.43

0.26

30

45

227812.50

350

650.89

30

60

540000.00

700

771.43

30

45

227812.50

350

650.89

30

45

227812.50

350

650.89

30

45

227812.50

350

650.89

30

45

227812.50

600

379.69

30

45

227812.50

420

542.41

30

45

227812.50

600

379.69

30

45

227812.50

350

650.89

30

60

540000.00

700

771.43

30

45

227812.50

420

542.41

30

60

540000.00

700

771.43

30

45

227812.50

350

650.89

30

45

227812.50

420

542.41


Only

0.37 1030.58

0.63 0.21 0.43

1802.01

0.36 0.54

1422.32

0.46 0.39 0.23

1681.47

0.39 0.30

2973.21

0.22 0.37 0.31

2073.21

0.31 0.41 0.24

1572.99

0.34 0.16 0.28 0.33

2344.42

0.23 0.39 0.33

1964.73

Page: 39

0.28

FRME ANALYSIS BY MOMENT DISTRIBUTION METHOD III

bal. C.O. BAL Total

Total BAL C.O.

0.63 0 52.92 14.63 -10.14

0.37 -84.00 31.08 1.47 -5.96

57.41

-57.41

Total BAL C.O.

6.00 0.21 84.00 2.94 15.54 2.14 104.62

38.28 -17.43 26.46 29.25 0 0.39 25 0.39 0 29.25 17.22 -17.43

0.23 -75.00 17.25 1 -10.28

29.04

-67.03

17.82

Int.FEM

U.D.L 24 0.36 30X60 -98.00 Int.FEM 5.04 -26.46 3.66 115.76

7.00

4.06 -0.42 3.01 1.47 0 0.22

0.54 98.00 -52.92 2.52 5.48

0.46 0 -45.08 -12.66 4.66

53.08

-53.08

B AL C.O.

6.00

20

0.3 75.00 2 8.63 -0.58

0.22 0 1.47 5.39 -0.42

0.26 -81.67 1.73 -15.11 -0.5

85.05

6.44

-95.55

41.81 -7.26 14.63 34.44 0 0.41 28 0.24

Int.FEM

0.37 81.67 -30.22 0.87 15.5

0.31 0 -25.32 -20.22 12.99

67.82

-32.55

-51.01 2.08 -12.66 -40.43 0 0.33

6.00

30

0.16

0.23

-34.87 12.99 -22.54 -25.32 0 0.31

7.00

15.18 3.66 0.74 10.78 0 0.28

I 0.34

0.43 0 6.02 0.74 11.06

B AL C.O.

II

C.O. BAL

U.d.l. 28 30X45 Int.FEM

0.33

7.00 0.39

0.28

-

C.O. BAL

0 28.56 0 -6.02

-84.00 20.16 3.08 -4.25

22.54

-65.01

Int.FEM

84.00 6.16 10.08 2.09

0 8.86 0 3.01

122.50 12.71 -23.89 4.31

102.33

11.87

129.37

Int.FEM

122.50 -47.78 6.36 2.46

0 -34.3 0 1.76

-

A

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83.54

B


-32.54

C

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Step6. HORIZONTAL

LOAD ANALYSIS: -

Frame analysis for horizontal loads calculated in step 4 is carried out by using :(a)Approximate Method:i) Cantilever method. ii) Portal method. Approximate methods are used for preliminary designs only. For final design we may use exact method i.e (i) Slope deflection or matrix methods method. We will(ii) notFactor discuss these methods in de tail as now modern computer package as STAAD PRO is available for a nalysis.

Step7: DESIGN SLABS:

OF COULMN,FOUNDATIONS, BEAMS

&

After load calculation & analysis for vertical & horizontal loads, design of Columns ,Foundatio ns, Beams, Slabs and are to be carried out a s per the various clauses of IS codes, IS 456-2000, IS:1893-2002, IS:13920-1993 etc. The Design of Column, Foundation, Beams and Slabs are discussed in details in following section.

A. Design of columns

:

- With the knowledge of (i) Vertical

load (ii) Moments due to horizontal loads on either axis;(iii) Moments due to vertical loads on either axis, acting on each column, at all floor levels of the building, columns are designed by charts of SP-16(Design Aids) with a 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 step confirms the size of columns assumed in the architectural drawings. The design of each column is carried out from the top of foundation to the roof, varying the amount of steel reinforcement for suitable groups for ease in design. Further, slenderness effects in each storey are considered for each column group. Important Considerat

ions in design of

Columns:-

:- The effective height of a column is defined as the height between the points of contra flexure of the buckled column. For effective column height refer table 28 (Annexure E) of IS: 456-2000. (i)Effective height of column

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For framed structure effective height of column depends upon relative stiffness of the column & various beams framing into the column at its two ends. (Refer Annexure E of IS: 456-2000.) : - The unsupported length l, of a compression member shall be taken as the clear distance between end restraints except that:In beam & slab construction, it shall be the clear distance between the floor & under side of the shallower beam framing into the columns in each direction at the next higher floor level . (ii)Unsupported Length

: - The unsupported length between end restraints shall not exceed 60 times the least lateral dimension of a column. (iii) Slenderness limits for columns

: - All columns shall be designed for minimum eccentricity equal to unsupported length of column/500 plus least lateral dimension/30, subject to a minimum of 20 mm. Or emin ≥ l/500+ D/30 ≥ 20 mm (iv) Minimum Eccentricity

Where

l= unsupported length of column in mm. D=Lateral dimension of column in the direction under consideration in mm. : - The design of column is complex since it is subjected to axial loads & moments which may very independently. Column design required:(v)Design Approach

I. II. III.

Determination of the cross sectional dimension. The area of longitudinal steel & its distribution. Transverse steel.

The maximum axial load & moments acting along the length of the column are considered for the design of the column section either by the working stress method or limit state method. The transverse reinforcement is provided to impart effective lateral support against buckling to every longitudinal bar. It is either in the form of circular rings of polygonal links (lateral ties).

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foundation s: - With the knowledge of the column loads and moments at base and the soil data, foundations for columns are designed The following is a list of different types of foundations in order to preference with a view to economy: (i) Individual footings (ii) Combination of individual and combined footings (iii) Strip footings with retaining wall acting as strip beam wherever applicable; (iv) Raft foundations of the types (a) Slab (b) beam-slab. The brick wall footings are also designed at this stage. Often, plinth beams are provided to support brick walls and also to act as earthquake ties in each principal direction. Plinth beams, retaining wall if any, are also designed at this stage, being considered as part of foundations. B . Design of

Important Considerat

ions in design of

Foundatio ns:-

a ) Introduction: - Foundations are structural elements that transfer loads from the building or indiv idual column to the earth. If these loads are to be properly transmitted, foundations must be designed to prevent excessive settlement or rotation, to minimize differential settlement and to provide adequate safety against sliding and over turning. b) Depth of foundation:-

Depth of foundation below ground level may be obtained by using Rankine's formula 2 h

Where h = p = γ = Ø =

©BSNL India

=

p --γ

1 – Sin Ø --- ----- --- 1 + Sin Ø

Minimum depth of foundation Gross bearing capacity Density of soil Angle of Repose of soil


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c) Recommen dations of IS 456 -2000, limit shear, cracking & development

state design,

bending,

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. Plan Area of footing

Total Load including Self Weight -----------------------------------Allowable bearing capacity of soil

=

i i ) IS 1904 – 1978, Code of Practice for Structural Safety of Buildings : shallow foundation, shall govern the general

details. iii) Thickness of the edge of footing:-( Reference clause 34.1.2) The thickness at the edge shall not be less than 15 cm for footing on soils. iv) Dimension of 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.

Tan

100 q o ----------- + 1 Fck

(should not be less than) 0.9 x

Where qo = of the

Calculated

maximum bearing pressure at the base

pedestal/footing in N/mm fck

N/mm

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2

=

Characteristic

2

strength of concrete at 28 days in


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Column

PLAIN CONCRETE PEDESTAL

α

(v)

Bending Moment

(Reference Clauses- 34.2.3.1 & 34.2.3.2) COLUMN BASE

PEDESTAL

X

Y FACE OF PEDESTA

FACE OF COLUMN

X

Y

ISOLATED COLUMN FOOTING

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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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CRITICA L SECTIO

B

A

d d

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

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τv

=

Shear stress

b0

=

Periphery of the critical section

d

=

Effective depth

Vu

=

Factored vertical shear force


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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 =

τc

---------------------------------------------------Long dimension of column or pedestal =

0.25

N/mm2

fek

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 B

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Long side of footing --------------------------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

A2

=

=

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. Loaded area at the column base.

For limit state method of design, the permissible bearing stress shall be = 45 f ek 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 floor slabs and beam s is taken up with the First Floor & upwards .The slabs are designed as one-way or two-way panels, taking the edge conditions of the supporting edges in to account, with the loading already de cided as per functional use of slab panel.

The design of floor slab is carried 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 depending upon the support condition & bending moment calculated & reinforcement steel may be calculated from the charts of SP-16. The slab design for particular floor may be done in tabular form as shown below.

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SLAB DESIGN Name of project:Level of slab Sl a b ID

Edge conditi on

1

2

Total load in KN/Sq .m w

3

T wo Adj. Edge. S1

Discon t. (Case No.4)

8.50

Shor t span lx m

4

3. 5 0

long sp a n ly m

5

5. 2 5

l y/ lx

6

1. 5

1. 5 *w* lx *l x

7

156.8 0

slab thic knes s in mm

8

1 20

Short span Moment

αx (+ )

Αx (- )

mu x +

9

10

11 = 7 x 9

0.05 6

0.07 5

8.78

KN-M

mu x 12 = 7 x1 0

Steel in short span

Stee l

Αy (+ )

αy ( -)

13

14

15

11.7 6

S2

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Steel in long span

Long span moment KN-M

0.03 5

mu y + 16 = 7 x1 4

mu y 17 = 7 x1 5

0.047 5.49

7.37

Steel 18

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/mm2), 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 whichever 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.S ingly Reinforced

Beam

Determine the main tension reinforcement required for a rectangular beam section with the following data: Size of beam 30 X 60 cm Concrete mix M 20 Characteristic strength 415 N/mm 2 of reinforcement Factored moment 170 kN.m Assuming 20 mm dia bars with 25 mm clear c Effective depth= 600 From Table M u,lim/bd

2

– 2 5 – 20/2 = 565 mm

D for f

y

2

= 415 N/ mm

3

kN/m

= 2.76 X 10 = 2.76 X 10

ck

= 20 N/mm

2

2

= 2.76/1000 X (1000) = 2.76 X 10

and f

2

= 2.76 N/ mm

M u,lim

over,

3

2 3

bd 2

X 0.300 X0.565X0.565

= 264.32 kN.m Actual moment of 170 kN.m is l ess than M u,lim. The section is therefore to be designed as a singly reinforced (under-reinforced) rectangular section. Referring to table

2 of SP-16 we have

to calculate Mu/bd

Mu/bd 2 = 170 x10 6 /(300x 56 5 x 565 ) = 1.78 From Table 2. p t = 0.556 2 A s t =0.556 x 300x 565/100 =942.42mm =9.42 cm Example2.Dou bly Reinforced Beam (i) Determine the main reinforcements required for a with the following data :

Size of beam

2

rectangular

30×60cm

Concrete mix M 20 Characteristic strength of 415/Nmm Reinforcement Factored moment 320Kn.m Assuming 20 mm dia bars with 25 mm clear c

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2


2

over,

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D=600-25

– 20/ 2 =565 mm

From table D, for f

y=

415 N/mm

Mu 2 lim/bd 2 = 2.76 N/mm

2

2

and f

= 2.76 × 10

3

2

= 20N/mm

ck

KN/m

2

Mu 2 lim= 2.76 × 10 3 ×0.300×0.565×0.565 = 264.32 KN-M is greater than Actual moment of 320 Kn.M Mu 2 lim hence the section is to be designed as a doubly reinforced section .

Reinforcement

from Table 50

Mu/bd 2 = 320 × 10

6

/ (300×565 2 ) = 3.34 N/mm

Next higher value of d Table 50

1

2

/d = 0.1 , will b e used for refer rin g to

For Mu/bd 2 = 3.34 and d ’ /d = 0.10, p t = 1.152, p c = 0.207 A s t = 1.152 x 300x 565/100 =1952.64 mm And A s c = 0.207 x 300x 565/100 =350.86 mm

2

2

2

=19.52 cm

=3.51 cm

2

(ii) Determine the Shear reinforcement (vertical stirrups ) required for the same beam section if

2

Shear stress

N/mm τ v < τ m a x ( 2.8 stress.

factored shear v

force is

V

u

=250 KN.

= V u /bd = 250 × 10 3 /(30 0× 565 ) =1.47

N /mm 2 ) hence section is adequate

regarding shear

2 From table 61 for p τ c =0.65 N/mm t =1.15 τc × b× d Shear capacity of concrete section = = 0.65 × 300 ×565/1000=110.18 kN Shear to be carried by stirrups V u s =V u - τ c × b× d = 250 - 110.18 =139.82 kN V u s /d = 139.82/56.5 = 2.47 kN/ cm 2 Referring to table 62 for steel f Provide 8 mm y = 415 N/mm diameter two legged vertical stirrups at 140 mm spacing.

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TABLES

FOR

DESIGN

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DETAILING AS PER IS 13920

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MIN 2 BARS FOR F ULL LENGTH ALONG TOP AND BOTTOM FACE AS > MIN. Bd AS < MAX Bd 50 m max

50 m max

d

db 2d

2d

HOOP SPACING > d /2 HOOP SPACING < d/4 and 8 db B = BREADTH OF BEAM db = DIAMETER OF LONGITUDINAL BAR

BEAM REINFORCEMENT

37

Questions:1. Which are the important BIS Codes/handouts used for structural design of RCC buildings? 2. In which seismic zones provisions of IS 13920 is to be adopted for all reinforced concrete structures? 3. (a) What are the basic values of span to effective depth ratios for beams as per IS 456 for span upto 10meter fo r – (i) Cantilever (ii) Simply supported (iii) Continuous (b) What are the basic values of span to overall depth ratios for two-way slabs upto 3.5 m span & with Fe415 steel reinforcement 2 and loading class upto 3KN/m ? 4. What are the provisions of IS 456 for nominal cover to meet durability requirements? As per IS 456 how much minimum – cover should be provided for a) Column b) Footing 5. (a) What are the minimum reinforcement provision of IS 456 for beams in respect of: (i) Tension reinforcement (ii) Shear reinforcement

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(b) What are the IS 456 provisions for maximum reinforcement in beams for:(i) Compression reinforcement (ii) Tension reinforcement 6. What is the maximum permissible spacing for shear reinforcement in beams? Explain IS 456 provisions for side face reinforcement in beams. 7. How much minimum reinforcement must be provided in sl abs? – 8. As per IS1893 give formulae for calculating a) Design Base Shear (V b) b) Design Horizontal acceleration (A h) 9. Give formulae for calculating t ime period as per IS1893 for – a) RCC Frame Building b) RCC Building Brick in fill walls 10. How vertical loads on columns are calculated? Give names of simpler methods of analysis of structures. 11. What is the minimum eccentricity for which all columns should be designed? List out minimum and maximum longitudinal reinforcement required to be provided in columns? Give in brief provisions for maximum spacing of lateral ties in a column?

12. What are the critical sections in isolated footing design for 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 column b) Rectangular column 14. What is the minimum diameter bar that can be used in longitudinal reinforcement in column?

-----------------

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Civil Technical

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