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Chapter - 2

STRUCTURAL DESIGN

OF

RCC BUILDING

COMPONENTS

Rajendra Mathur Dy. Dir(BS-C) 09412739 232(M)

e-mail – [email protected]


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 sys tem

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 ava ilable 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;(i i)

Soil Investigation report (SIR) of soil data in l ieu thereof; (iii)

Location of the place or city in order to decide on wind and seismic

loadings;(iv) Data for l ifts, water tank capacit ies on top, special roof

features or loadings, etc.

Choice of an appropriate structural system 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

strength of bricks shall be 100 kg/cm2

minimum for four stories. This

system is adequate for vertical lo ads 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 & vertical reinforcement in brick wall as per IS: 4326-

1967( Indian Standards Code of Practice for Earthquake Resistant

Construction 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 along 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 of Load Bearing


Masonry Buildings are done as per IS:1905-1980 (Indian Standards

Code of Practice for Structural Safety of Buildings: Masonry

Walls(Second Revision).

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 10 0-150 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 of large

storey heights (5.0m or more ) ,l ike electric substation etc. are made


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 collective wisdom of expert

structural engineers, gained over the years. These codes are

periodically revised to bring them in line with current rese arch, and

often, current trends.

The codes serve at least four distinct functions .


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 analyses is made available in the form of a

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 l iabili ty due to structural failures that are caused by

inadequate supervision and/or faulty material and construction.

(i)IS 456 : 2000 – Plain and reinforced concrete – code of practice

(fourth revision)

(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 – Handbooks on Concrete Reinforced and Detailing.

General Design Consideration of IS: 456-2000.


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 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.

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

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.

STIFFNESS


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) Gross

Section

The cross-section of the member ignoring

reinforcement

b) Transformed

Section

The concrete cross-section plus the area of

reinforcement transformed on the basis of modular

ratio

c) Cracked

Section

The area of concrete in compression plus the area of

reinforcement transformed on the basis of modular

ratio

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

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.

MOMENT AND SHEAR COFFICIENTS FOR CONTINUOUS BEAMS


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 1 1 1 1

(not fixed) +-- +-- (- )-- (- )--

10 12 9 9

Note: For obtaining the bending moment, the coefficient shall be multiplied by

the total design load and effective span.

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

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

SHEAR FORCE COFFICIENTS

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


Type of Load At End At Support Next At All Other

Support 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 0.45 0.60 0.60 0.60

(not fixed)

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

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.

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


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.


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

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 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


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 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.

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 2/3rd

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.

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.

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.

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

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 = 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


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 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

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.


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


4.0STAIRS-NOT LIABLE TO OVER CROWDING

5.0- LIABLE TO OVER CROWDING

7.5–HEAVY VEHICLES

4.0GARRAGES –LIGHT VEHICLES

10.0STACK 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

0.75 ROOF WITHOUT ACCESS

1.5 ROOF WITH ACCESS

LIVE LOADS ON ROOFS

12.0WEATHER MAKER

10.0MDF 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


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


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.


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 calculation

(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 v b

V

b = Ah W (F)

Where Ah = 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 resist ing frame (OMRF)

(Sa/g) = Average response acceleration coefficient for soil type

& appropriate natural periods and lamping of the

structure.

For calculating of (Sa/g) value as above we have to calculate value of

T i.e. Fundamental National Period (Seconds) ( Clause 7.6 of IS Code)

T = 0.075 h 0 .7 5

(For RC Frame building)

= 0.0 85 h 0 .7 5

(For Steel frame building)

h = Height of building in Meter

In case of building with brick in fills walls.

T = 0.09 h /d 1 / 2

Where h = height of building in Meter

and d = Base dimension of the building at the plinth level in

Meter along the considered direction of the lateral

force.


Value of (Sa/g) is to be read from fig 2 on page 16 of IS Code

depending upon Soil condition & Fundamental Natural period T.

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 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 soft soil si tes

(Sa/g) = 1+15 T if 0.00≤T≤<0.10

= 2.50 if 0.10≤T≤0.67

=1.67/T i f 0.67≤T≤ 4.00

W= Seismic weight of the building as per clause 7.4.2 of the code .


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


164.9

157.99

47.01

Frame with EQ Loads


Calculating Seismic weight of building per frame for frame (B)

Length bldg = 16.00 M

C/C distance of frames = 7.50 M

Density of 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/m

2

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 calculation.

Lump mass at First Floor = 0.50×6.00×16×7.50 = 360 KN (D)

Total lamp mass first floor & 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 not to be considered )

Total lamped mass at terrace level = (E)+(F)+(G)

= 390+41.40+56.25=487.65 KN

Total weight of building per framed per inner frame

F.F = 988.50 KN

S.F. = 988.50 KN

Terrace = 487.65 KN

2464.65 KN

LL-- FFRRAAMMEE


Putting 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 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

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 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.

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 ih i 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

presumed that the reinforcements are always so arranged that al l

joints of the frame are monolithic.

In view of the uncertain property of material creep, shrinkage

and a number of approximate simplifying assumptions made in the


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 assumptio ns, 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 for calculation of moments in

Columns & beams:

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 u.d.l. on short beam of slab panel = w B/3.0

Equivalent u.d.l. on long beam of slab panel = w B/6 x [ 3 -

(B/L)2

]

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

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 (ii i) is repeated for each floor

Step(v):Then these loads are used as u.d.l on a particular frame for

analysis by moment distribution method as described in the next

section.


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 ry

to evolve simpler methods.

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

i i . 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 of all members. Enter them in the

calculation scheme.

Step2: Calculated the distribution factor at all joints from the

stiffness. Enter them in the calculation scheme.

Step3: Look the joints and calculate the fixed -end moments.

Enter them in the calculation scheme.


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

Meeting at the joint in proportion to their relative

stiffness and enter these value in the scheme. This

operation is called balancing.

Step5: Enter the carry-over moments at the far in the scheme.

Step6: Repeat steps 4 & 5, ti ll the carry-over moments

become insignificant .

Step7: Balance the unbalanced moment obtained from the last

carry-over operation.

Step8: Add the init ial fixed-end moments, balancing moments

and carry-over moments to get the final end moments in

beam & columns.

A sample of moment distribution method is shown on next two

pages.


CALCULATION OF DISTRIBUTION FACTOR FOR FRAME ANALYSIS

S.NO. JOINT MEMBE

R Size in Cm

Mo ment o f

Iner t i a

Cm4 ( I )

Leng

th of

mem

ber

Cm

K=I /L Su m K D.F .

1

B D

A-I I I

Right

beam 30 45 227812.50 600 379 .69

1030 .58

0 .37

Lo wer

Col . 30 45 227812.50 350 650 .89 0 .63

2

B-II I

Left

beam 30 45 227812.50 600 379 .69

1802 .01

0 .21

Right

beam 30 60 540000.00 700 771 .43 0 .43

Lo wer

Col . 30 45 227812.50 350 650 .89 0 .36

3 C-II I

Left

beam 30 60 540000.00 700 771 .43

1422 .32

0 .54

Lo wer

Col . 30 45 227812.50 350 650 .89 0 .46

4

A-II

Upper

co l . 30 45 227812.50 350 650 .89

1681 .47

0 .39

Right

beam 30 45 227812.50 600 379 .69 0 .23

Lo wer

Col . 30 45 227812.50 350 650 .89 0 .39

5

B-I I

Left

beam 30 60 540000.00 600 900 .00

2973 .21

0 .30

Upper

co l . 30 45 227812.50 350 650 .89 0 .22

Right

beam 30 60 540000.00 700 771 .43 0 .26

Lo wer

Col . 30 45 227812.50 350 650 .89 0 .22

6

C-I I

Left

beam 30 60 540000.00 700 771 .43

2073 .21

0 .37

Upper

co l . 30 45 227812.50 350 650 .89 0 .31

Lo wer

Col . 30 45 227812.50 350 650 .89 0 .31

7

A-I

Upper

co l . 30 45 227812.50 350 650 .89

1572 .99

0 .41

Right

beam 30 45 227812.50 600 379 .69 0 .24

Lo wer

Col . 30 45 227812.50 420 542 .41 0 .34

8

B-I

Left

beam 30 45 227812.50 600 379 .69

2344 .42

0 .16

Upper

co l . 30 45 227812.50 350 650 .89 0 .28

Right

beam 30 60 540000.00 700 771 .43 0 .33

Lo wer

Col . 30 45 227812.50 420 542 .41 0 .23

9

C-I

Left

beam 30 60 540000.00 700 771 .43

1964 .73

0 .39

Upper

co l . 30 45 227812.50 350 650 .89 0 .33

Lo wer

Col . 30 45 227812.50 420 542 .41 0 .28


FRME ANALYSIS BY MOMENT DISTRIBUTION METHOD

I I I

U.d. l .

28 6 .00

U.D.L

24 7.00

0 .63 0.37 30X45 0.21 0.43 0.36 30X60 0 .54 0.46

0 -84.00 Int .FEM 84.00 0 -98.00 Int .FEM 98.00 0

bal . 52.92 31.08 2 .94 6.02 5.04 -52.92 -45.08

C.O. 14.63 1.47 15.54 0.74 -26.46 2 .52 -12.66

BAL -10.14 -5 .96 2 .14 11.06 3.66 5 .48 4.66

Total 57.41 -57.41 104.62 17.82

-

115.76 53.08 -53.08

Total 38.28 4 .06 -34.87

BAL -17.43 BAL -0.42 BAL 12.99

C.O. 26.46 C.O. 3 .01 C.O. -22.54

29.25 1 .47 -25.32

0 0 0

0 .39 0 .22 0 .31

II 25 6 .00 20 7 .00

0 .39 0.23 0 .3 0 .22 0.26 0 .37 0.31

0 -75.00 Int .FEM 75.00 0 -81.67 Int .FEM 81.67 0

29.25 17.25 2 1 .47 1.73 -30.22 -25.32

C.O. 17.22 1 8 .63 5.39 -15.11 0 .87 -20.22

BAL -17.43 -10.28 -0 .58 -0 .42 -0 .5 15.5 12.99

29.04 -67.03 85.05 6.44 -95.55 67.82 -32.55

Total 41.81 15.18 -51.01

BAL -7.26 3 .66 2 .08

C.O. 14.63 0 .74 -12.66

34.44 10.78 -40.43

0 0 0

0 .41 0 .28 0 .33

I 28 6 .00 30 7 .00

0 .34 0.24 0 .16 0.23 0.33 0 .39 0.28

0 -84.00 Int .FEM 84.00 0

-

122.50 Int .FEM 122.50 0

28.56 20.16 6 .16 8.86 12.71 -47.78 -34.3

C.O. 0 3 .08 10.08 0 -23.89 6 .36 0

BAL -6.02 -4 .25 2 .09 3.01 4.31 2 .46 1.76

22.54 -65.01 102.33 11.87

-

129.37 83.54 -32.54

A B C


Step6. HORIZONTAL LOAD ANALYSIS:-

Frame analysis for horizontal loads calculated in step 4 is carried

out by using : -

(a)Approximate Method: -

i) Canti lever 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 (ii) Factor method.

We will not discuss these methods in detail as now modern

computer package as STAAD PRO is available for analysis.

Step7: DESIGN OF COULMN,FOUNDATIONS, BEAMS &

SLABS:

After load calculation & anal ysis for vertical & horizontal

loads, design of Columns ,Foundations, Beams, Slabs and are to be

carried out as per the various cl auses 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 vert ical 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 t he

amount of steel reinforcement for suitable groups for ease in design.

Further, slenderness effects in each storey are considered for each

column group.

Important Considerations in design of Columns: -

(i)Effective height of column :- 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.


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.)

(ii)Unsupported Length : - The unsupported length l , of a

compression member shall be taken as the clear distance between

end restraints except that: -

In beam & slab construction, i t 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.

(iii) Slenderness l imits for columns : - The unsupported length

between end restraints shall not exceed 60 times the least lateral

dimension of a column.

(iv) Minimum Eccentricity : - 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

Where l= unsupported length of column in mm.

D=Lateral dimension of column in the direction under

consideration in mm.

(v)Design Approach : - The design of column is complex since it is

subjected to axial loads & moments which may very independently.

Column design required:-

I. Determination of the cross sectional dimension.

II. The area of longitudinal steel & its distribution.

III. 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 (late ral ties).


B. Design of foundations: - 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 ec onomy: (i) Individual footings

(ii) Combination of individual and combined footings (i ii) 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.

Important Considerations in design of Foundations: -

a) Introduction: - Foundations are structural elements that

transfer loads from the building or individual 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

p 1 – Sin Ø

h = - - - - - - - - - - - - - - -

γ 1 + Sin Ø

Where

h = Minimum depth of foundation

p = Gross bearing capacity

γ = Density of soil

Ø = Angle of Repose of soil


c) Recommendations of IS 456 -2000, limit state 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.

Total Load including Self Weight

Plan Area of footing = -------------------------------------

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.

100 qo

Tan α (should not be less than) 0.9 x ----------- + 1

Fck

Where

qo = Calculated maximum bearing pressure at the base

of the

pedestal/footing in N/mm2

fck = Characteristic strength of concrete at 28 days in

N/mm2


(v) Bending Moment

(Reference Clauses- 34.2.3.1 & 34.2.3.2)

ISOLATED COLUMN FOOTING

FACE OF

PEDESTA

L

PLAIN

CONCRETE

PEDESTAL

α

Column

X Y

X Y

COLUMN

PEDESTAL BASE

FACE OF

COLUMN


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)


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

CR IT IC A

L SECT IO

N

d

d


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


(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.


C. Design of Floor slabs:-. Design of floor slabs and beams 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 decided 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.


SLAB DESIGN

Name of project: -

Level of s lab

Sla

b

ID

Edge

condi t i

on

Total

load in

KN/Sq

.m w

Shor

t

span

lx

m

long

spa

n

l y

m

ly/

lx

1 .5

*w* lx

*lx

s lab

th ic

knes

s in

mm

Shor t span Mo ment KN -M

Stee l

in

shor t

span

Lon g sp an mo ment KN-M

Stee l

in

long

span

αx

(+)

Αx

( - )

mux

+

mux

-

Stee

l

Αy

(+) αy( - )

muy

+

muy

- Stee l

1 2 3 4 5 6 7 8 9 1 0 11=

7 x 9

12=

7

x10

13 1 4 15

16=

7

x14

17=

7

x15

1 8

S1

Two

Adj .

Edge.

Discon

t.

(Case

No.4)

8 .50 3 .5

0

5.2

5

1 .

5

156.8

0 120

0.05

6

0.07

5 8 .78

11.7

6

0 .03

5

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/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 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.


Example1.Singly 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/mm2 of reinforcement

Factored moment 170 kN.m

Assuming 20 mm dia bars with 25 mm clear cover,

Effective depth= 600 – 25 – 20/2 = 565 mm

From Table D for fy = 415 N/ mm2

and fck = 20 N/mm2

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 M u, l im. 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/bd2

Mu/bd2

= 170 x106/(300x 565 x565) = 1.78

From Table 2. p t =0.556

A s t=0.556 x 300x 565/100 =942.42mm2=9.42 cm

2

Example2.Doubly Reinforced Beam (i)Determine the main reinforcements required for a rectangular

with the following data :

Size of beam 30×60cm

Concrete mix M 20

Characteristic strength of 415/Nmm2

Reinforcement

Factored moment 320Kn.m

Assuming 20 mm dia bars with 25 mm clear cover,


D=600-25 – 20/ 2 =565 mm

From table D, for f y= 415 N/mm2 and fck = 20N/mm

2

Mu2 lim/bd2= 2.76 N/mm

2 = 2.76 × 10

3 KN/m

2

Mu2 lim= 2.76 × 103×0.300×0.565×0.565

= 264.32 KN-M

Actual moment of 320 Kn.M is greater than Mu2 lim hence the section

is to be designed as a doubly reinforced section .

Reinforcement from Table 50

Mu/bd2= 320 × 10

6/ (300×565

2 ) = 3.34 N/mm

2

Next higher value of d1/d = 0.1, will be used for referring to

Table 50

For Mu/bd2 = 3.34 and d

’/d = 0.10,

p t = 1.152, pc = 0.207

As t = 1.152 x 300x 565/100 =1952.64 mm2 =19.52 cm

2

And

Asc = 0.207 x 300x 565/100 =350.86 mm2 =3.51 cm

2

(ii) Determine the Shear reinforcement (vertical stirrups ) required

for the same beam section if factored shear force is V u =250 KN.

Shear stress τ v = Vu /bd = 250 × 103

/(300× 565) =1.47

N/mm2

τv < τmax( 2.8 N /mm2) hence section is adequate regarding shear

stress.

From table 61 for p t=1.15 τc=0.65 N/mm2

Shear capacity of concrete section = τc × b× d

= 0.65 × 300

×565/1000=110.18 kN

Shear to be carried by stirrups V u s=Vu - τc × b× d = 250 - 110.18

=139.82 kN

Vu s/d = 139.82/56.5 = 2.47 kN/cm

Referring to table 62 for steel f y = 415 N/mm2 Provide 8 mm

diameter two legged vertical stirrups at 140 mm spacing.


TABLES

FOR

DESIGN


DETAILING AS PER IS 13920


HOOP SPACING

HOOP SPACING< d/4 and 8 db

B = BREADTH OF BEAMdb = DIAMETER OF LONGITUDINAL BAR

2d

d

2d

db

50 m max 50 m max

MIN 2 BARS FOR FULL LENGTHALONG TOP AND BOTTOM FACEAS > MIN. BdAS < MAX Bd

> d /2

BEAM REINFORCEMENT37

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 for –

(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

and loading class upto 3KN/m2?

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


(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 slabs?

8. As per IS1893 give formulae for calculating –

a) Design Base Shear (Vb)

b) Design Horizontal acceleration (Ah)

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?

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

chapter - 2 structural design of rcc building components · 2014-05-26 · ©bsnl india for...

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