Earthquake resistant analysis and design of multistoried building

Seismic Analysis and Design of
Multistoried RCC Building
Prepared by :
Amrit Baral (072/BCE/009)
Amrit Subedi (O72/BCE/010)
Anup Adhikari (072/BCE/016)
Bed p. Neupane (072/BCE/022)

Objectives
• The Objectives of the Project are:-
Identification of structural arrangement of plan
Modelling of building for structural analysis
Detailed structural analysis using ETABS
Design of structural components
To get real life experience with engineering practices

Background
• Civilization
• Masonry and RCC building
• Earthquake resistant building

Literature Review
• Nepal National Building Code(NBC 000-1994)
• Indian Standard (IS) codes
• Bureau of Indian Standards Special Publication (SP)
• Resources from National Information Center of Earthquake
Engineering(NICEE),Kanpur ,India
• Textbooks on RCC Design ,Earthquake Engineering and other books
• Old Reports on the same subject

Methodology
• Planning Phase
Functional Planning
Structural Planning
• Load Assessment
Gravity Load
Lateral Load

Methodology Contd…
• Preliminary Design
Slab
Beam
Column
• Idealization of structure
Idealization of support
Idealization of load
Idealization of structural system

Methodology Contd…
• Design and Detailing
Design Philosophy
• Limit State Method of Design for RCC structures
Detailing Principle for RCC structures
• Codal References
Drawings

Earthquake Resistant Building
• Earthquake-resistant building designed to prevent total collapse,
preserve life, and minimize damage in case of an earthquake or tremor.
• Earthquakes exert lateral as well as vertical forces, and a structure’s
response to their random, often sudden motions is a complex task that
is just beginning to be understood.
• Earthquake-resistant structures absorb and dissipate seismically
induced motion through a combination of means: damping decreases
the amplitude of oscillations of a vibrating structure, while ductile
materials (e.g., steel) can withstand considerable inelastic deformation.

Earthquake Resistant Building
• The objective of earthquake resistant building is to protect the
structure during vibration from collapsing and the major feature to
ensure in such a construction is the ductility.
• In the case of designing the Earthquake Resistant Building we are here
using IS 1893:2002.

Design philosophy of earthquake
resistant designs
• Under minor but frequent shaking, the main members of the buildings
that carry vertical and horizontal forces should not be damaged;
however buildings parts that do not carry load may sustain repairable
damage.
• Under moderate but occasional shaking, the main members may
sustain repairable damage, while the other parts that do not carry load
may sustain repairable damage.
• Under strong but rare shaking, the main members may sustain severe
damage, but the building should not collapse.

Concepts of reducing the effect of
earthquake
1)Ductility
• Ductility is a kind of deformation.
• Ductility can be defined as the “ability of material to undergo large
deformations without rupture before failure”.
• Ductility in concrete is defined by the percentage of steel
reinforcement with in it. Mild steel is an example of a ductile material
that can be bent and twisted without rupture.

earthquake Contd…
• Each design code recognizes the importance of ductility in design
because if a structure is ductile it ability to absorb energy without
critical failure increases.
• Ductility behavior allows a structure to undergo large plastic
deformations with little decrease in strength.

earthquake Contd…
SIGNIFICANCE OF DUCTILITY
 If ductile members are used to form a structure, the structure can
undergo large deformations before failure.
 This is beneficial to the users of the structures, as in case of
overloading, if the structure is to collapse, it will undergo large
deformations before failure and thus provides warning to the
occupants.
 This gives a notice to the occupants and provides sufficient time for
taking preventive measures. This will reduce loss of life.

earthquake Contd…
2. Strong Column-Weak Beam
• In multistory reinforced concrete buildings it is desirable to dissipate
earthquake induced energy by yielding of the beams rather than the
columns.
• The columns are responsible from overall strength and stability of the
structure, with severe consequences of failure.

earthquake Contd…
• Furthermore, columns are compression members and axial
compression reduces the ductility of reinforced concrete columns, thus
necessitating more stringent confinement reinforcement.
• Therefore, it is preferable to control inelasticity in columns, to the
extent possible, while dissipating most of the energy through yielding
of the beams. This is known as the “strong-column weak-beam
concept.

Earthquake resistant analysis and design of multistoried building

earthquake Contd…
3. Building configuration
• The behavior of building during earthquakes depends critically on its
overall shape, size and geometry.
• Hence, at planning stage itself, architects and structural engineers must
work together to ensure that the unfavorable features are avoided and a
good building configuration is chosen.
• Both shape and structural system work together to make the structure a
marvel.

Size of building
In tall buildings with large weight-
to-base size ratio the horizontal
movement of the floors during
ground shaking is large. In short but
very long buildings, the damaging
effects during earthquake shaking are
many.
And, in buildings with large plan
area, the horizontal seismic forces
can be excessive to be carried by
columns and walls.

Horizontal Layout
of Buildings
• Buildings with simple geometry
in plan perform well during
strong earthquakes. Buildings
with re-entrant corners, like U, V,
H and + shaped in plan sustain
significant damage.
• The bad effects of these interior
corners in the plan of buildings
are avoided by making the
buildings in two parts by using a
separation joint at the junction.

Vertical Layout of
Buildings
• Earthquake forces developed at
different floor levels in a building
need to be brought down along the
height to the ground by the shortest
path, any deviation or discontinuity
in this load transfer path results in
poor performance of building.
• Buildings with vertical setbacks
cause a sudden jump in earthquake
forces at the level of discontinuity.

Adjacency of Buildings
• When two buildings are close to
each other, they may pound on each
other during strong shaking.
• When building heights do not
match the roof of the shorter
building may pound at the mid-
height of the column of the taller
one; this can be very dangerous.

Functional And Structural Planning
• Functional Planning
 It is a prerequisite of any type of building
Functional planning of the building is governed by the client
requirement, site conditions, provincial by-laws,

Contd…
1. Planning of Space and Facilities
• The layout of the building plan was prepared and finalized as
per client requirements
• For vertical mobility, dog legged staircase are provided.
• Prefabricated Capsule type elevators were provided in the left
part of the building
• Washroom for ladies and gents are provided centrally in each
floor at four different location of the building

Contd…
2.Architectural planning of 3D framework of Building
• For reinforced concrete frames, a grid layout of beams is made
considering the above functional variables.
• In most of grid intersection points, columns are placed.
• This framework for each floor is then utilized with positioning
of masonry wall between the columns
• Arrangement of beams is done along the grid interconnecting
the columns at grid intersections.
• With this framework of beam and column having RCC slab in
the floor and roof, architectural planning of the building is
complete and 3D framework is thus complete

Contd…

Contd…
• Structural Planning
Structural System
• structural orientation of the building in horizontal and vertical
plane avoiding irregularities mentioned in IS 1893 (part
1):2002.
• The following types of irregularities mentioned in Table 4 & 5
of IS 1893 (part 1):2002 should be avoided

Contd…
Plan Irregularities
• Torsion Irregularity
• Re-entrant corners
• Diaphragm Discontinuity
• Out of plane Offsets
• Non-parallel Systems
Vertical Irregularities
• Stiffness Irregularity –Soft
Storey
• Mass irregularity
• Vertical Geometric Irregularity
• In-plane discontinuity in vertical
elements resisting lateral force

Torsion Irregularity
• To be considered when floor diaphragms
are rigid in their own plan in relation to
the vertical structural elements that resist
the lateral forces.
• Torsional irregularity to be considered to
exist when the maximum storey drift,
computed with design eccentricity, at one
end of the structures transverse to an axis
is more than 1.2 times the average of the
storey drifts at the two ends of the
structure

Re-entrant corners
• Plan configurations of a structure
and its lateral force resisting
system contain re-entrant corners,
where both projections of the
structure beyond the re-entrant
corner are greater than 15 percent
of its plan dimension in the given
direction

Diaphragm
Discontinuity
• Diaphragms with abrupt
discontinuities or variations in
stiffness, including those having cut-
out or open areas greater than 50
percent of the gross enclosed
diaphragm area, or changes in
effective diaphragm stiffness of
more than 50 percent from one
storey to the next

Out of plane Offsets
• Discontinuities in a lateral
force resistance path, such
as out-of-plane offsets of
vertical elements.

Non-parallel Systems
• The vertical elements
resisting the lateral force
are not parallel to or
symmetric about the major
orthogonal axes or the
lateral force resisting
elements.

Stiffness Irregularity
(soft storey )
• A soft storey is one in which
the lateral stiffness is less than
70% of that in the storey
above or less than 80% of the
average of the stiffness of the
three storeys above.

Mass irregularity
• Mass irregularity shall be
considered to exist where the
seismic weight of any storey
is more than 200 percent of
that of its adjacent storeys.
The irregularity need not be
considered in case of roofs.

Vertical Geometric
Irregularity
• Vertical geometric irregularity
shall be considered to exist
where the horizontal
dimension of the lateral force
resisting system in any storey
is more than 150 percent of
that in its adjacent storey

In-plane discontinuity
in vertical elements
resisting lateral force
• A in-plane offset of the lateral
force resisting elements
greater than the length of
those elements

Load Assessment And Preliminary
Design
• Load Assessment
• Assessment of loads on the structural system thus planned is based on
IS 875(part I-V):1987
• Gravity Load Assessment
• The gravity loads on the building are derived from IS 875 (part I) dead loads and IS 875
(part II) imposed loads.
• Total seismic weight of building is found out

Load Assessment
And Preliminary
Design
• Load Assessment
 Gravity load assessment I
 Lateral Load Assessment S
875(part II) for imposed loads

Design Contd…
• Lateral Load Assessment
Earthquake Load on Super Structures
According to IS 1893(part-I) :2002
Theory of Base Shear Calculation

Theory of Base Shear Calculation
• Theory of Base Shear Calculation
• According to IS 1893 (Part I): 2002 Cl. No. 6.4.2 the design horizontal
seismic coefficient Ah for a structure is,
• Ah =
ZISa
2Rg
Where,
Z = Zone factor given by IS 1893 (Part I): 2002 Table 2, Here for Zone V
Z=0.36
I = Importance Factor, I = 1.5 for assembly building
R = Response reduction factor given by IS 1893 (Part I): 2002 Table 7, R =
5.0
Sa
g
= Average response acceleration coefficient which depends on approximate
fundamental natural period of vibration (Ta).

Contd…
•
Sa
g
Vs Time period

Contd…
• Base Shear, VB = Ah ∗ W
Where ,W= Seismic Weight of building
• According to IS 1893 (Part I): 2002 Cl. No. 7.7.1 the design base shear (VB)
computed above shall be distributed along the height of the building as
• Qi = VB
𝑊 𝑖ℎ 𝑖
2
𝑖=1
𝑛
𝑊 𝑖ℎ 𝑖
2
Where,
Qi = Design lateral force at floor i
Wi= Seismic Weight of floor i
hi = Height of floor i measured from base
n=No. of stories in the building

Contd…

Design Contd…
• Preliminary design of RCC elements
• SLAB 127 mm thick spanning in two directions.
• BEAM = 254mm x 355mm – Primary beam
220mm x 290mm-Secondary beam
• COLUMN 650mm square sections.

Drift Calculation
• Lateral (story) drift is the
amount of sidesway between
two adjacent stories of a
building caused by lateral
(wind and seismic) loads.
• Should not exceed 0.004 times
the floor height.

Load path from the structure
slab to the ground

Design and detailing
• Design of slab
• Design of beam
• Design of column
• Design of staircase
• Design of footing
• Design of basement wall

Design of Slab
• Types of Slab
• One way
• Two way
If Ly/Lx < 2, two way slab
Ly/Lx < 2, One way slab
Where Ly & Lx are the larger and smaller dimension of slab

Design of Slab Contd…
One way Slab Two way slab

Two way slab
• Slabs are considered as
divided in each direction into
middle strips and edge
• Moments for each strips are
calculated by using this
expression
• Mx = αxwlx
2
• My = αywlx
2

• Design Steps
Calculation of thickness
Calculated from deflection control criteria

𝑠ℎ𝑜𝑟𝑡𝑒𝑟 𝑠𝑝𝑎𝑛
𝑑
=αβγδλ
Acc. to IS 456: 2000 Cl. 23.2
Thickness=127mm
Calculation of effective calculation
 Effective span of the slab is taken as minimum of C/C distance of the slab or clear span +
effective depth of the slab for x- and y- direction separately

Calculation of design Moment
Maximum Negative Moment (Mx
–) = αx
-×wf×lx
2
Maximum Positive Moment (Mx
+) = αx
+×wf×lx
2
Design of Negative and Positive Reinforcement
𝑀 𝑢,𝑙𝑖𝑚 = 0.36 ×
𝑥 𝑢,𝑚𝑎𝑥
𝑑
× 1 − 0.42 ×
𝑥 𝑢,𝑚𝑎𝑥
𝑑
𝑏𝑑2
× 𝑓𝑐𝑘
𝑀 𝑢 = 0.87 × 𝑓𝑦 × 𝐴 𝑠𝑡 × 𝑑(1 −
𝐴 𝑠𝑡
𝑏𝑑
×
𝑓𝑦
𝑓 𝑐𝑘
)
Both direction 8 mm Ø bars c/c 250mm

• Check for Shear
Maximum shear force intensity in either direction can be taken as
𝑤×𝑙 𝑥
2
where
lx is clear short span.
Nominal shear stress (τv) =
𝑉
𝑏×𝑑
Percentage of steel Pt = (Ast, provided/bD) ×100
From Table 19 IS 456:2000
Permissible shear stress (τc
’) = k× τc
from Clause 40.2.1.1 of IS 456 k I s founded for slab thickness
Table 20 of IS 456; τc, max = 3.1 N/mm2 (for M25)

• Check for Development Length
The development length (Ld) is given by (IS 456: 2000, Cl. 26.2);
Ld=
0.87×𝑓 𝑦×Ø
4×𝜏 𝑏𝑑
Also from IS 456:2000 Cl. 26.2.3.3
Ld≤
1.3×𝑀 𝑙
𝑉𝑢
+ l0
Here Ml = moment of resistant of the section assuming all the reinforcement at the section
to be stressed to fd.
𝑀𝑙 = 0.87 × 𝑓𝑦 × 𝐴 𝑠𝑡
𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
× 𝑑(1 −
𝐴 𝑠𝑡
𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
𝑏𝑑
×
𝑓𝑦
𝑓 𝑐𝑘
)
• Corner Reinforcement:
Area of each layer of reinforcement = 75% of area of the total reinforcement provided.

Design of Beam
• Beams are a structural members assigned to transmit the loads from
slab to the column through it.
• There are three types of reinforced concrete beams:
Singly reinforced beams
Doubly reinforced beams
Singly or doubly reinforced flanged beams

Design of Beam Contd…
• In singly reinforced simply supported beams, reinforcements are
placed at the bottom of the beam
• A doubly reinforced concrete beams are reinforced in both
compression and tension regions.
• A complete design of beam involves consideration of
safety under ultimate limit state in flexure,
shear, torsion
serviceability limit states of deflection,
crack width, durability.

• Design Steps
Check for member size
According to, IS13920:1993 cl.6.1.3 and cl.6.1.2
Breadth/Depth>0.3
Clear span/Depth>4
Check for limiting longitudinal Reinforcement
Acc. To IS13920:1993, cl.6.2.1b
Ast,min=
√𝑓 𝑐𝑘
𝑓𝑦
× 𝐵 × 𝐷 × 0.24
Acc. To IS13920:1993, cl.6.2.2
Max. Reinforcement, Ast,max= 0.025BD

• Design for flexure
According to IS 456:2000 Cl. 41.4.2.
Mulim =0.1336fckbd2 for Fe500.
Calculation of Positive and Negative Reinforcement
By calculating by using
𝑀 𝑢
𝑏𝑑2 and
𝑑′
𝑑
Using these values in SP 16 table 55 for Fe500 and M25
Using these values in SP 16 table 16 for Fe500 and M20
Calculation of tensile and compression reinforcement

• Check for shear
Maximum shear force intensity in either direction are taken from
ETABS Analysis.
Nominal shear stress (τv) =
𝑉
𝑏×𝑑
Percentage of steel Pt = (Ast, provided/bD) ×100
From Table 19 IS 456:2000
Permissible shear stress (τc
’) = k× τc
from Clause 40.2.1.1 of IS 456 k is founded for slab thickness
Table 20 of IS 456; τc, max = 3.1 N/mm2 (for M25)

Design shear strength of concrete= τcbd
For each end Equivalent shear force is calculated
Using 4-legged 8mm dia stirrups.
Calculation of spacing of stirrups is done acc to IS 456:2000 Cl.
40.4
by using this expression
Sv=
0.87𝑓𝑦 𝐴 𝑠𝑣 𝑑
𝑉𝑢
Providing two 2L- 8dia stirrups @100c/c
Providing two 2L- 8dia stirrups @150c/c

Design of Column
• Design Steps
1. Calculation of the Influence Area of the Columns
• The Influence Area of a column is the area of which load is being
transferred to the column to be designed for.
2. Calculation of the Loads Coming on Columns from the Influence Area
• The Loads acting are broadly classified as Dead Load (DL) and Live
Load (LL). Dead Loads are the load of objects which cannot be moved
from on place to another like the loads of Brick Work, Beams, Slabs
etc. and the Live Loads
• The Factor of Safety for Dead Load + Live Load Combination is 1.5

Design of Column Contd…
• According to I.S.: 456-2000,
The Ultimate Load of biaxially loadedColumn is given by,
Pu = 0.4.fck.Ac + 0.67.fy.Asc
Where, Pu = Ultimate Load of the Column in N/mm2
fck= Yield Strength of Concrete in N/mm2
Ac = Area of Concrete (Cross-Sectional Area) of Column in mm2
fy = Yield Strength Of Steel in N/mm2
Asc = Area of Steel (Cross-Sectional Area) in Column in mm2

3. Check For Long/Short Columns
• Effective Length /Least Lateral Dimension>12,Long Column
• Effective Length /Least Lateral Dimension<12 Short Column
• A short column mainly fails by direct compression and has a lesser chance of
failure by buckling
• All are short column

• Check For Eccentricity
Check For Eccentricity
The eccentric load cause the column to bend towards the eccentricity of the
loading and hence generates a bending moment in the column.

• IS 456-2000 says, the eccentricity which we have to consider for design
must be taken as the greater of the followings :-
i) 20mm.
ii) (lef/500) +b/30)
Where,
 (lef = Effective Length of the Column
b = Lateral Dimension of the Column
• Permissible Eccentricity:- 0.05b where b is the dimension of a side of a
column

• Calculating The Area Of Steel Required
Area of Steel Required Asc is to be calculated from the Ag as the predetermined
percentage of Ag.
• Determining The Diameter and Spacing Of The Lateral Ties
Determine the Diameter and the Spacing of the Lateral Ties
The Diameter of the Ties shall not be lesser than the Greatest of the following
two values
• 5mm
• 1/4th of the Diameter of the Largest Diameter Bar

• The Spacing of Ties shall not exceed the least of the followings three
values
 Least Lateral Dimension
16 Times of the Diameter of the Smallest Diameter Longitudinal Bar
48 Times of the Diameter of Ties
• Check

𝑀 𝑢𝑥
𝑀 𝑢𝑥,𝑙
∝ 𝑛
+
𝑀 𝑢𝑦
𝑀 𝑢𝑦,𝑙
∝ 𝑛
<1

DESIGN OF STAIRCASE
• Means of access between the various floors
• staircases may be classified largely into two categories, depending on
the predominant direction in which the slab component of the stair
undergoes flexure:
1. Stair slab spanning transversely (stair widthwise)
2. Stair slab spanning longitudinally (along the incline)

Design of Staircase Contd…
• General consideration
Between consecutive floors there should be an equal rise and going
for every parallel step.
Should have at least 2m headroom.
The sum of going of single step plus twice the rise should be
between 550-700mm.
The rise of step should not be about more than 200mm and going
not less than 240 mm.
The pitch should not be more than about 380.

• Known data
Floor height= 4m
No of flight=2
No of riser in each flight = 13
No of tread in each flight = 12
Length of riser =150mm
Length of Tread = 300mm
Nosing = 20mm
Going =280mm

• Design Steps
THICKNESS OF WAIST SLAB,W

𝑆𝑝𝑎𝑛
𝐸𝑓𝑓.𝑑𝑒𝑝𝑡ℎ
=30
Load calculation
Dead Load =
1
𝐺
(WB+
𝑅𝑇
2
) ×25 + Floor Finish
• Design Load
• 1.5(DL+LL)
• Calculation of bending moment
• Mu =
𝑤𝑙2
10

• Checking depth
• d=
𝑀 𝑢
3.34×𝑏
1
2
• Area of steel
•
𝑥
𝑑
= 1.2 − 1.22 −
6.68𝑀 𝑢
𝑓 𝑐𝑘 𝑏𝑑2
1
2
• 𝑍 = 𝑑 1 − 0.42
𝑥
𝑑
• Ast=
𝑀 𝑢
0.87×𝑓 𝑦×𝑍

• Distribution steel
• 12% of gross area
• Check for Shear
• V=
𝑤𝑙
2
• 𝜏 𝑣=V/bd
• Check for deflection
•
𝑙
𝑑
< 𝛼𝛽𝛾𝛿

Design of Foundation
• Introduction
• Substructure which interfaces the superstructure and the supporting ground.
• Transfers loads from the superstructure to the soil safely.
• Foundation must not settle sufficiently.

Design of Footing Contd…
• Selection of appropriate footing is governed by following major
factors:
Soil strata
Bearing capacity of soil
Type of structure
Type of load
Permissible differential settlement
Economy

• Raft foundation
Raft foundation is done due to
• load transmitted by the columns in a structure are so heavy
• allowable soil pressure so small that individuals footing would
either overlap
• or cover more than about one half of the area

• Design Steps
• Known data:
Size of column = 650 mm X 650 mm
Strength of concrete = M25
Grade of steel = Fe500
Allowable Bearing capacity of soil = 120 KN/m2
Vertical load, P = 41660.5 KN

• Foundation Type Selection
Area occupied by isolated footing is greater than 50% of plan
area.
• Determination of C.G. of column load on footing:
For Load Combination1.5 (DL+LL),
Data taken from EATBS

Determination of C.G. of column load on footing:
For Load Combination1.5 (DL+LL),
Data taken from EATBS

• Eccentricity Calculation
Difference between geometric C.G. of Mat and C.G. of Column
load
• ex=0.3169m
• ey=-0.1226m
• Load to Area ratio check
P/A<SBC

• Calculation of soil pressure
Soil pressure at different points is given by
σs = (P/A)±(My/Iy)×X ±(Mx /Ix) ×Y
• Calculation of Bending Moment
Moment = wl2/10
Taken maximum moment from both direction

• Calculation of depth of foundation:
Calculation of depth at the position of all column
Taken maximum depth
Two way shear consideration
• Reinforcement is given by
As per IS 456 : 2000 ANNEX G-1.1 b
Moment=0.87 × 𝑓𝑦 × 𝐴 𝑠𝑡(𝑑 −
𝐴 𝑠𝑡×𝑓𝑦
𝑓 𝑐𝑘×𝑏
)
• Minimum reinforcement in slab =0.12% of total cross-section as per
IS 456 : 2000 CL 26.5.2.1

• Calculation of spacing of bars
• Calculation of Development length
As per (IS456:2000 Cl. 26.2.1)
Ld = (0.87×fy×φ)/ (4×Ԏbd)
From IS456:2000Cl. 26.2.3.3.c
Ld≤1.3M/V+Lo
Lo=12Ф or d whichever is greater

• Summary of Design of the Mat Foundation
Strips Bottom Reinforcement Top Reinforcement
Diameter Spacing Diameter Spacing
X- direction 16 mm 100 mm 16 mm 100 mm
Y- direction 16 mm 120 mm 16 mm 120 mm

Design of Basement Wall
• Basement wall is constructed to retain the earth and to prevent
moisture from seeping into the building.
• Since the basement wall is supported by the mat foundation, the
stability is ensured
• Design of the basement wall is limited to the safe design of vertical
stem.

Design of Basement Wall Contd…
• Design Constants
• Earth height(h) =8 m
• unit weight of soil, γ = 17 KN/m3
• Angle of internal friction of the soil, ө = 30֯
• Safe bearing capacity of soil , qs= 120 KN/m2
• Moment calculation
• Lateral load due to soil pressure, Pa = Ka × 3 × γsat × h
• Characteristic Bending moment at the base of wall can be calculated as
• Mx=Pa × ℎ
3
×
2𝑎+𝑏
𝑎+𝑏

• Design of section
• Bending moment, M=0.133fckbd2
• Design of main reinforcement
• 𝑀 𝑢 = 0.87 × 𝑓𝑦 × 𝐴 𝑠𝑡 × 𝑑(1 −
𝐴 𝑠𝑡
𝑏𝑑
×
𝑓𝑦
𝑓 𝑐𝑘
)
• Check for minimum reinforcement and max. diameter
• Ast,min=0.0015*b*D
• Maximum diameter of bar =D/8

• Calculation of horizontal reinforcement
• Minimum reinforcement required=0.0025*D*h
• Check for shear
• V=WL/3
• Nominal shear stress=V/bd
• Permissible shear stress from IS 456:2000,Table-19
• Check.

Earthquake resistant analysis and design of multistoried building

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