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Transcript of Final
Chapter 1
INTRODUCTION
1.1 Background
The sky rocketing population and haphazard land use has decreased the land availability
for construction of any structure requiring large plinth area. With due consideration to
this fact, a High-Rise building seemed to be one of the best options. Taking into account
this fact, we have come up with a project work on “Computer Aided Structural Analysis
and Design of High-Rise Hospital Building” located in Kathmandu Valley. It is intended
to design the hospital so as to provide health facilities to the general public. The hospital
will surely aid in the development of the society and hence country as a whole. The
people will also be benefitted by the quality service provided by the hospital.
A designer has to deal with various structures ranging from simple to more complex ones
like a multistoried frame building. All the structural members are subjected to various
loads like concentrated loads, uniformly distributed loads, live loads, earthquake load,
wind load, etc. The structure transfers the loads acting on it to the supports and ultimately
to the ground. While transferring the loads acting on the structure, the members of the
structure are subjected to the internal forces like axial forces, shear forces, bending and
torsion moments.
Structural Analysis deals with analyzing these internal forces in the structural member
developed as a result of various loading conditions or combinations.
Structural Design deals with sizing various members of the structures to resist the
internal forces to which they are subjected in the course of their life cycle. Unless the
proper structural detailing method is adopted the structural design will be no more
effective. A Standard Code of practice (Indian Standard code in our case) should be
1
thoroughly followed and implemented for proper analysis, design and detailing with
respect to safety, economy, stability, strength.
Besides the scarcity of land, earthquake is one of the dominant constraints while
designing the multistory frame building in earthquake prone zone like Kathmandu.
According to IS1893: 2002, Kathmandu lies on V zone, the severest one, hence the effect
of earthquake is predominant to wind load. So, the building is analyzed for earthquake as
lateral load. The seismic coefficient design method as stipulated in IS 1893:2002 is
applied to analyze the building for earthquake. The 3- dimensional moment resistance
frame with shear wall is considered as the main structural system of the building.
This project work has been undertaken for the partial fulfillment of requirements for the
Bachelor’s Degree in Civil Engineering. This project work contains structural analysis,
design and detailing of a hospital building located in Kathmandu Valley. All the
theoretical knowledge on analysis and design acquired during the course works are
utilized with practical application. The main objective of the project work is to acquaint
us in the practical aspects of Civil Engineering.
1.2 Theme of Project work
This group under the project work has undertaken the Computer Aided Structural Analysis
and Design of High-Rise Hospital Building. The main aim of the project work under the
title is to acquire knowledge and skill with an emphasis on practical application. Besides
the utilization of analytical methods and design approaches, exposure and application of
various available codes of practices is another aim of the work.
1.3 Objectives
The specific objectives of the project work are:
i. Preparation of the plan of the building to meet the requirements for its intended
use.
ii. Identification of the structural arrangement of the plan.
iii. Modeling the building for structural analysis.
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iv. Analyzing the structure using structural analysis program.
v. Sectional design of the structural members.
vi. Preparation of detail structural drawing of the design.
1.4 Building Description
Building Type : High-Rise Hospital Building
Structural System : RCC Space Frame with Shear Wall
Plinth area covered : 1418 m2
Type of Foundation : Mat Foundation
No. of Storey : 16 + 1 (basement)
Floor Height : 3m
Type of Sub-Soil : Medium Soil
Seismic zone : V
Expansion Joints : Expansion joints are provided
According to IS 456-2000, Clause 27, structures in which changes in plan dimensions
take place abruptly shall be provided with expansion joints at the section where such
changes occur. Reinforcement shall not extend across an expansion joints and the break
between the sections shall be completed. Normally structure exceeding 45m in length is
designed with one or more expansion joints.
The design is intended to serve quality service to the general public round the clock.
1.5 Identification of Loads
i. Dead loads are calculated as per IS: 875( Part I) – 1987
ii. Seismic load is calculated according to IS :1893 (Part I) – 2002 considering
Kathmandu located at Zone V
iii. Imposed loads according to IS : 875(Part II) – 1987
3
1.6 Code of Practices
Following codes of practices developed by Bureau of Indian Standards were followed in
the analysis and design of building:
i. IS 456:2000 (Code of practice for plain and reinforced concrete)
ii. IS 1893 (part 1):2002 (Criteria for earthquake resistant design of structures)
iii. IS 13920: 1993 (Code of practice for ductile detailing of reinforced concrete
structures subjected to seismic forces)
iv. IS 875 (part 1):1987 (to assess dead loads)
v. IS 875 (part 2):1987 (to assess live loads)
vi. IS 875 (part 5):1987 (for load combinations)
vii. SP 16 and SP 34 (design aids and hands book)
1.7 Idealization and Assumption in Analysis and Design
Various assumptions have been made in analysis and design of the structures, for consideration of simplicity and economy, viz.:
1. Tensile strength of concrete is ignored.
2. Shrinkage and temperature strength are negligible.
3. Adhesion between concrete and steel is adequate to develop full strength.
4. Seismic and wind load do not occur simultaneously.
5. Centerlines of beams, columns and shear walls are concurrent everywhere.
1.8 Method of Analysis
The building is modeled as a space frame. SAP2000 V12 is adopted as the basic tool for
the execution of analysis. SAP2000 program is based on Finite Element Method. Due to
possible actions in the building, the stresses, displacements and fundamental time periods
4
are obtained using SAP2000 which are used for the design of the members. Mat
foundation, staircase and slabs are analyzed separately.
1.9 Design
The following materials are adopted for the design of the elements:
i. Concrete Grade :
M30 for the beams, columns, slabs and shear walls
M20 for the foundation
ii. Reinforcement Steel :
Fe500 for staircase
Fe415 for all other structural members
The design of RC elements is carried out using Limit State Method. The design is based
on various Indian Standards Code of Practice for Plain and Reinforced Concrete IS 456-
2002, Design Aids for Reinforced Concrete to IS 456:1987(SP-16), Criteria for
Earthquake Resistant Design Structures IS 1893-2002, Ductile Detailing of Reinforced
concrete Structures Subjected to Seismic Forces IS13920:1993, Handbook on Concrete
Reinforcement and Detailing SP-34. Various handbooks like Reinforced Concrete
Designer’s Handbook – Charles E. Reynolds and James C. Stedman, are extensively used
in the progress of design.
The design moments, shear forces, axial forces and torsions are taken as computed by
computer software program “SAP2000 V12” for the worst possible combination and a
number of hand calculations are done so as to verify the reliability of the design results
suggested by the software.
1.10 Detailing
The space frame is considered as a Special Moment Resisting Frame (SMRF) with a
special detailing to provide ductile behavior and comply with the requirements given in
5
IS 13920:1993. Handbook on Concrete Reinforcement and Detailing SP-34 has been
extensively used for the detailing of beams and columns.
1.11 Scope
The project work is limited to the structural analysis and design only.
Design and Detailing of following structural elements is performed:
1. Slab
2. Beam
3. Column
4. Staircase
5. Mat plus Foundation
Design and layout of the building services like pipeline, electrical appliances, sanitary
and sewage system are not covered.
The required parking facilities are assumed to be provided in a separate parking structure,
however its design is not concerned with this project.
The project is not concerned with the existing soil condition of the locality.
The bearing capacity of the soil is assumed.
The environmental, social and economical condition of the locality is not taken into
consideration.
The project work is only related with the practical application of the studied courses in
the field.
Detail cost estimate of the project is not included in this report.
6
Chapter 2
STRUCTURAL SYSTEM AND PRELIMINARY DESIGN
2.1 Structural System
Any structure is made up of structural elements (load carrying, such as beams and
columns and non-structural elements (such as partitions, false ceilings, doors). The
structural elements put together, constitute the structural systems. Its function is to resist
effectively the action of gravitational and environmental loads, and to transmit the
resulting forces to the supporting ground without significantly disturbing the geometry,
integrity and serviceability of the structure.
2.2 Structural Arrangement Plan
The planning of the building has been done by the group and is not an existing plan. The
final plan of the building was a result of review of various literatures like Timesavers
Standard, various codes of practice and some other reference books suggested by our
Supervisor. The positioning of the columns, staircases, toilets and bathrooms, lifts are
appropriately done and accordingly beam arrangement is carried out so that the whole
building will be aesthetically, functionally and economically feasible.
The aim of the design is the achievement of an acceptable probability that structures
being design will perform satisfactorily during their intended life with an appropriate
degree of safety, they should sustain all the loads and deformations of normal
construction and use and have adequate durability and resistance to the effect of misuse
and fire.
The building consists of two blocks separated by an expansion joint.
7
2.3 Need of Preliminary Design
It is necessary to know the preliminary section of the structure for the detail analysis. As
the section should be given initially while doing analysis in almost all software, the need
of preliminary design is vital. Only dead loads and live loads are considered while doing
preliminary design. Preliminary design is carried out to estimate approximate size of the
structural members before analysis of structure.
2.4 Preliminary Design
Slab
Each floor slabs are divided into different types as per end conditions defined in IS
456:2000.
Design of General use Slab type –S1
Size of slab = 6.7 X 6.1 m2
Ly/Lx = 6.7/6.1 = 1.098 <2
Therefore, the slab is two-way slab.
Calculating, depth of slab
d = span/αβγδλ
where,
α depends on support condition
β depends on span
γ depends on % tension steel
δ depends on compression steel
λ is considered for flanged beam
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or, d = 6100/(26*1*1.8*1*1)
d = 124.9 mm
So take D= 150 mm
Thus effective depth of 125 mm is adopted. So, take overall depth, D = 150 mm.
(Clear cover = 25mm)
Beam
The design is as per IS-456:2000. Same section of beam is provided throughout the
building.
Span of beam = 6.1m
Since Depth = span/(12~15)
= 6100/13
= 470 mm
Adopt 500 mm
Column
It is designed on the basis of IS 456:2000 & IS 456:1978.
Column size: grade M25
Pu = 0.4 *fck* Ac + 0.67*fy * Ast
= 0.4*25*(Ag-0.02Ag)+ 0.67*415*Ag*0.02 (Taking pt=2%)
= 15.61Ag
~ 16 Ag
9
Table 1: Calculation of preliminary loads on column
Load(KN) Size of column (mm2)
948 250*250
1887 360*360
3545 510*510
3785 510*510
7089 690*690
7075 690*690
2862 440*440
6285 650*650
2600 420*420
Thus, adopting the sizes as listed below:
450*450 mm2
500*500mm2
700*700mm2
Staircase design
15 cm X 30 cm
Width of flight=1.2m
Height of each flight =3/2=1.5m
10
No. of riser = 1.5/.15=10
Tread: 9
Space for tread: 9X30 = 270
Fig.1: Staircase Plan
Passage= 6-2.7-1.2=2.1
Room Size=6 m X 6.7m
Stair case Load
Live load = 5 KN/m2
R X T=15 cm X 30 cm
Height of floor: 3.0 m
Riser = 11
Tread = 10
10 X 30=300 cm = 3m
11
Passage = 6.7-3.2=1.7m
Effective span:
Leff = c/c distance between support
= 6.7 + .25/2 = 0.25/2 = 6.95 m
Assume slab thickness= L/26
= 6950/26*1.4
= 200 mm
Assume overall depth= 200 mm
Load on going:
√(T2 + R2) = √0.30 2 + 0.15 2
= 0.335 m
i. Self weight of slab = 25 * 0.2 * 0.335/0.3 = 5.6 KN/m2
ii. Step weight of steps = 25 * (1/2*0.15) = 1.9 KN/m2
iii. Finish (thickness = 25 mm) = 0.67 KN/m2
iv. Live load = 5 KN/m2
_______________
13.15 KN/m2
Load on landing:
i. Self weight of slab = 25 X 0.150 = 3.75 KN/m2
ii. Finish = 0.67 KN/m2
iii. Live load = 5 KN/m2
______________________
9.42 KN/m2
12
Determination of Stiffness Centre (SC) and Mass Centre (MC)
Calculation of Centre of Stiffness ( block 1)
For First floor:
From IS:456-2000,
Modulus of Elasticity of Concrete (Ec) = 5000√ f ck =5000√25= 25000 N/mm2
=2.5E10 N/m2
Moment of Inertia (I1) =0.45*0.45**3/12 =3.417E-3 m4
Moment of Inertia (I2) = 0.5*0.5**3/12 =5.208E-3 m4
Moment of Inertia (I3) = 0.7*0.7**3/12=2.0E-2 m4
Height (h) =3m
Stiffness (k1) =12EI1/h3=3.797E7=k (say)
Stiffness (k2) =5.787E7=1.524k
Stiffness (k3) =2.223E8=5.855k
_X = (k2+4k3+k1)*20+ (k1+k2+4k3)*40+ (k1+k2+5k3)*60+(k1+k2+5k3)*80
+ (k1+k2+5k3)*100+ (k1+6k2)*120
(7k1+16k2+23k3)
= 62.66f t= 19.105m
_Y= (2k2+5k3)*8+(2k2+5k3)*30+ (2k2+5k3)*40+(2k2+5k3)*62+ (4k2+3k3)*72+4k2*94
(7k1+16k2+23k3)
13
=40.963 ft
=12.49 m
Similarly, centre of stiffness of block2 is = (12.19, 8.52) m.
From excel sheet data centre of mass of block1 is = (20.38, 12.29) m.
And centre of mass of block2 is = (12.21, 9.22) m.
For block 1
Eccentricity ex= 1.275 m
ey= 0.2 m
For block 2
Eccentricity ex = 0.02 m
ey = 0.7 m
14
CHAPTER 3
LOAD ASSESSMENT
3.1 Introduction
As described earlier, the building is a RCC framed structure, located in the
Kathmandu valley. Thus wind loads, snow loads, and other special types of loads
described by IS 875 (part 5):1987 can be taken as negligible as compared to the
dead, live and seismic loads.
3.1.1 Dead Loads:
According to the IS 875:1964:
The dead load in a building shall comprise the weights of all walls,
partitions, floors and roofs and shall include the weights of all other
permanent features in the building.
3.1.2 Live Loads:
It means the load assumed or known resulting from the occupancy or use
of a building and includes the load on balustrades and loads from movable
goods, machinery and plant that are not an integral part of the building.
3.1.3 Seismic Loads:
These are the load resulting from the vibration of the ground underneath
the superstructure during the earthquake. The earthquake is an
unpredictable natural phenomenon. Nobody knows the exact timing and
magnitude of such loads. Seismic loads are to be determined essentially to
produce an earthquake resistant design.
Seismic loads on the building may be incorporated by-
15
1. Response Spectrum Method: In this method the design earthquake
forces are determined adopting IS 1893:2002. These design forces for the
buildings located along two perpendicular directions may be assumed to
act separately along each of these two horizontal directions.
2. Time History Analysis: In it the ground is subjected to a
predetermined acceleration and subsequent stress in the structural
elements are determined by appropriate methods.
3.2 Unit Weights for Dead Load Calculation
1. RCC: (IS 875 (part 1) :1987 table 1)
a) For slabs and shear walls:
γRCC = 25 KN/m3
b) For columns:
γRCC = 25 KN/m3
c) For Beams:
γRCC = 25 KN/m3
2. Plaster (20mm thickness):
γplaster = 20.40 KN/m3
3.3 Live Loads
1. On Floors: (IS875(part2):1987 table1)
2. On Partition walls: Live Load=1KN/m2
(Assuming a minimum live load as per IS 875(part2):1987, 3)
3. On roofs: Live Load=1.5 KN/m2
(Assuming access not provided except for the case of maintenance),
(IS875(part2):1987)
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3.4 Seismic Load
Seismic weight is the total dead load plus appropriate amount of specified
imposed load. While computing the seismic load weight of each floor, the weight
of columns and walls in any story shall be equally distributed to the floors above
and below the storey. The seismic weight of the whole building is the sum of the
seismic weights of all the floors. It has been calculated according to IS: 1893(Part
I) – 2002.
IS: 1893(Part I) – 2002 states that for the calculation of the design seismic forces
of the structure the imposed load on roof need not be considered
Base Shear Calculation
According to IS 1893 (Part I): 2002 Cl. No. 6.4.2 the design horizontal seismic
coefficient Ah for a structure shall be determined by the following expression:
Ah=Z I Sa
2 R g
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 high rise hospital 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 Fundamental
natural period of vibration (Ta).
According to IS 1893 (Part I): 2002 Cl. No. 7.6.1
The approximate fundamental natural period of vibration (Ta) in second of
moment resisting frame building without brick infill panel may be estimated by
empirical expression:
Ta = 0.075h0.75 for RC frame building
Where,
17
h= height of building in metre. This excludes the basement storeys, where
basement walls are connected with the ground floor deck or fitted between the
building columns. But, it includes the basement storeys, when they are not so
connected.
For h= 51 m
Ta= 1.43 sec.
And Sa/g = 0.95 (From IS Code 1893:2000) Fig 2
Now,
A h= z2
×IR
×Sag
=0.362
×15
× 0.95=.0513
According to IS 1893 (Part I) : 2002 Cl. No. 7.5.3 the total design lateral force or
design seismic base shear (VB) along any principle direction is given by
VB = Ah x W
Where, W = Seismic weight of the building
Wblock 1= 149259.12KN
Wblock 2 = 81603.644 KN
Then,(Vb)block 1=0.513*149259.12KN=7656.99KN
(Vb)block 2=0.513*81603.644KN=4186.27KN
The total base shear is firstly distributed horizontally in basement in proportion to
the stiffness. Then according to IS 1893 (part I):2002 C.L. No. 7.7.1 the design
base shear (VB) computed above shall be distributed along the height of the
building as per the following expression:
Qi=Vb∗(Wi∗h i2)
∑j=1
n
Wi∗hi2
Where, Qi=Design lateral force at floor i
18
W1
W2
W3
W4
W5
W1
W2
W3
W4
W5
Wi=Seismic weight of floor i
hi=Height of floor I measured from base
n=no. of stories in the building
Additional Shear Calculation Due to Torsion in Building
Center of Rigidity (CR) - A point through which a horizontal force is applied
resulting in translation of the floor without any rotation
Determination of Stiffness Centre (SC) and Mass Centre (MC)
Calculation of Centre of Stiffness ( block1)
For First floor:
From IS:456-2000,
Modulus of Elasticity of Concrete (Ec) = 5000√ f ck =5000√25= 25000 N/mm2
=2.5E10 N/m2
19
Fig.2 Storey Shear in X direction Fig.3 Storey Shear in Y direction
Moment of Inertia (I1) =0.45*0.45**3/12 =3.417E-3 m4
Moment of Inertia (I2) = 0.5*0.5**3/12 =5.208E-3 m4
Moment of Inertia (I3) = 0.7*0.7**3/12=2.0E-2 m4
Height (h) =3m
Stiffness (k1) =12EI1/h3=3.797E7=k (say)
Stiffness (k2) =5.787E7=1.524k
Stiffness (k3) =2.223E8=5.855k
_
X = (k2+4k3+k1)*20+ (k1+k2+4k3)*40+ (k1+k2+5k3)*60+(k1+k2+5k3)*80
+ (k1+k2+5k3)*100+ (k1+6k2)*120
(7k1+16k2+23k3)
= 62.66f t= 19.105m
_
Y=(2k2+5k3)*8+(2k2+5k3)*30+(2k2+5k3)*40+(2k2+5k3)*62+ (4k2+3k3)*72+
4k2*94
(7k1+16k2+23k3)
=40.963 ft
=12.49 m
Similarly, centre of stiffness of block2 is = (12.19, 8.52) m.
From excel sheet data centre of mass of block1 is = (20.38, 12.29) m.
20
And centre of mass of block 2 is = (12.21, 9.22) m.
For block 1
Eccentricity ex= 1.275 m
ey= 0.2 m
For block 2
Eccentricity ex = 0.02 m
ey = 0.7 m
Response spectrum Analysis:
The response history analysis provides structural response r(t) as a function of
time, but the structural design is usually based on the peak values of forces and
deformations over the duration of earthquake induced response. The peak
response can be determined directly from the response spectrum for the ground
motion in case of single degree of freedom. The peak response of multi-degree
freedom systems can be calculated from the response spectrum.
The exact peak value of Nth mode response rn(t)=-rnstAn
Where, An is the ordinate of the pseudo acceleration spectrum corresponding to
the natural period Tn and Damping ratio ε.
The peak value ro of the total response can be estimated by combining the modal
peaks rno according to one of the modal combination rules. Because the natural
frequencies of transverse vibration of a beam are well separated, the SRSS
combination is satisfactory. Thus,
Ro=¿)
21
3.5 Load Combination
In the course of analysis, different load cases and combinations are considered to
obtain the most critical stresses in the element of the structure. The load cases
considered for the structural analysis are:
i. Dead Load (DL)
ii. Live Load (LL)
iii. Earthquake load in X (EQx) static
iv. Earthquake load in Y (EQy) static
v. Eathquake load in X (Rx) response spectrum method
vi. Eathquake load in y (Ry) response spectrum method
Following load combination as per IS 1893(Part I):2002 are adopted:
i. 1.5(DL + LL)
ii. 1.2(DL + LL + EQx)
iii. 1.2(DL + LL - EQx)
iv. 1.2(DL + LL + EQy)
v. 1.2(DL + LL - EQy)
vi. 1.5(DL + EQx)
vii. 1.5(DL - EQx)
viii. 1.5(DL + EQy)
ix. 1.5(DL - EQy)
x. 0.9DL+1.5EQx
xi. 0.9DL-1.5EQx
xii. 0.9DL+1.5EQy
xiii. 0.9DL-1.5EQy
xiv. 1.5(DL + Rx)
xv. 1.5(DL - Rx)
xvi. 1.5(DL + Ry)
22
xvii. 1.5(DL - Ry)
xviii. 1.2(DL + LL + Rx)
xix. 1.2(DL + LL - Rx)
xx. 1.2(DL + LL + Ry)
xxi. 1.2(DL + LL - Ry)
After checking the results, it was found that the stress developed are most critical
for the following load combinations:
i. 1.5 (DL + LL)
ii. 1.2 (DL + LL + EQx)
iii. 1.2 (DL + LL - EQx)
iv. 1.2 (DL + LL + EQy)
v. 1.2 (DL + LL - EQy)
vi. 1.2(DL + LL + Rx)
vii. 1.2(DL + LL - Rx)
viii. 1.2(DL + LL + Ry)
ix. 1.2(DL + LL - Ry)
The characteristic loads considered in the design of foundation are:
i. Dead Load plus Live Load
ii. Dead Load plus Earthquake Load
iii. Dead Load minus Earthquake Load
To find the stress at the various points of the foundation, depth of footing and
reinforcements most critical factored loads are taken into account.
23
Chapter 4
MODELING AND STRUCTURAL ANALYSIS
4.1 Salient Feature of Sap2000
SAP2000 represents the most sophisticated and user-friendly release of SAP series of
computer programs. Creation and modification of the model, execution of the analysis,
and checking and optimization of the design are all done through this single interface.
Graphical displays of the results, including real-time display of time-history
displacements are easily produced.
The finite element library consists of different elements out of which the three
dimensional FRAME element was used in this analysis. The Frame element uses a
general, three dimensional, beam-column formulation which includes the effects of
biaxial bending, torsion, axial deformation, and biaxial shear deformations. Structures
that can be modeled with this element include:
i. Three-dimensional frames
ii. Three-dimensional trusses
iii. Planar frames
iv. Planar grillages
v. Planar trusses
A Frame element is modeled as a straight line connecting two joints. Each element has its
own local coordinate system for defining section properties and loads, and for
interpreting output.
Each Frame element may be loaded by self-weight, multiple concentrated loads, and
multiple distributed loads. End offsets are available to account for the finite size of beam
and column intersections. End releases are also available to model different fixity
conditions at the ends of the element. Element internal forces are produced at the ends of
each element and at a user specified number of equally-spaced output stations along the
length of the element.
24
Loading options allow for gravity, thermal and pre-stress conditions in addition to the
usual nodal loading with specified forces and or displacements. Dynamic loading can be
in the form of a base acceleration response spectrum, or varying loads and base
accelerations.
4.2 Structural Analysis
The structural analysis is done using the SAP 2000(V12). Since the design is aimed for
the adequate resistance to the earthquake, the analysis is done by both methods of
analysis namely Static and Dynamic analysis.
For the static analysis, the design lateral force is distributed to the various floor levels
which is then distributed to individual lateral load resisting elements depending on floor
diaphragm action. In SAP 2000, the modeling for rigid floor diaphragm is done. A floor
diaphragm is modeled as rigid horizontal plane parallel to each other in X – Y plane.
Each floor diaphragm is established by a joint in the plane of the diaphragm called
Master Joint of the diaphragm. The location of the master joint on each floor diaphragm
is arbitrary and is selected by the user. For the convenience, the master joint is located at
the center of mass of each floor. All the joints that exists on the diaphragm are connected
to the master joint by rigid links and their displacement are dependent on the
displacement of the master joint. Therefore these points are called dependent joint or
slave joint. The lateral forces calculated earlier are applied to each master joint located
on each floor of the building.
IS 1893:2002 specifies that for the zone IV & V, the dynamic analysis shall be done for
the regular buildings greater than 40 m in height and for the irregular buildings greater
than 12 m. So the dynamic analysis is also done for the building by the use of response
spectrum method in SAP 2000. The design base shear (Vb) is compared with the base
shear (Vb’) calculated using fundamental period (Ta). Where Vb is less than Vb’,all the
response quantities like member forces, store shears and base reactions shall be
multiplied by Vb’/Vb
The natural period of the building calculated is 1.43 seconds whereas the period obtained
from the analysis result of the SAP 2000 is 1.14 seconds.
25
4.2.1 Storey Drift
It is the displacement of one level relative to the other level above or below.
According to IS 1893:2002 Clause 7.11.1, the storey drift due to the minimum
specified design lateral force with partial load factor of 1.0 shall not exceed 0.004
times the storey height which comes to be 12 mm. However, the storey drift value
obtained from SAP analysis of the building is found out to be 11.3 mm which is
lower than the limited value as specified by code. Hence, our building is safe
against the storey drift.
4.2.2 Stress Envelope
Stress envelope is the diagrammatic representation of maximum and minimum
stress values in any given structural member resulting from different load
combinations assigned for the analysis. The stresses in the structural elements are
interpreted with the stress envelope which represents the maximum values of the
stresses in the member.
4.3 Inputs and Outputs
The design of earthquake resistant structure should aim at providing appropriate dynamic
and structural characteristics so that acceptable response level results under the design
earthquake. The aim of design is the achievement of an acceptable probability that
structures being designed will perform satisfactorily during their intended life. With an
appropriate degree of safety, they should sustain all the loads and deformations of normal
construction and use and have adequate durability and adequate resistance to the effects
of misuse and fire.
For the purpose of seismic analysis of our building we used the structural analysis
program SAP2000. SAP2000 has a special option for modeling horizontal rigid floor
diaphragm system. A floor diaphragm is modeled as a rigid horizontal plane parallel to
global X-Y plane, so that all points on any floor diaphragm cannot displace relative to
each other in X-Y plane. This type of modeling is very useful in the lateral dynamic
26
analysis of building. The base shear and earthquake lateral force are calculated as per
code IS 1893(part1)2002 and are applied at each master joint located on every storey of
the building.
Table 2: Sample output of SAP2000
FrameDesignTyp
e LocationFTopComb
o FTopAreaFBotComb
o FBotArea VCombo VRebarText Text m Text m2 Text m2 Text m2/m
4668 Beam 0.0000 envelope 0.0013 envelope 0.0009 envelope 0.0000
4668 Beam 0.4877 envelope 0.0008 envelope 0.00061.5DL+1.5LL 0.0003
4668 Beam 0.9754 envelope 0.0006 envelope 0.00061.5DL+1.5LL 0.0003
4668 Beam 1.4630 envelope 0.0006 envelope 0.0006 envelope 0.00004668 Beam 1.9507 envelope 0.0006 envelope 0.0006 envelope 0.00004668 Beam 2.4384 envelope 0.0007 envelope 0.0009 envelope 0.0000
4669 Beam 0.0000 envelope 0.0026 envelope 0.00141.5DL+1.5LL 0.0003
4669 Beam 0.4790 envelope 0.0019 envelope 0.00131.5DL+1.5LL 0.0003
4.4 Joint Displacement at Expansion Joint
After the analysis of structure using SAP2000 the maximum displacement of nodes at the
expansion joint was found out. It is clear from table below that the available gap for
expansion joint is much greater relative displacement of the nodes at joint. In order to
reduce the pounding effect between the two units, the adequate spacing is provided. The
separation between the adjacent units of the same buildings in between shall be separated
by a distance equal to the amount R times the sum of the calculated storey displacements
to avoid the damaging contact when the two units deflect towards each other. Since the
elevation levels of both units are same in our case the factor R is replaced by R/2. Hence
the building will not collide at the expansion joint during earthquake condition.
27
Drift Calculation:
Table 3: Drift calculation of Block 1
storey drift Drift ratiofloor U1 U2 D1 D2 DR1 DR2roof 0.1165 0.1548 15th 0.1127 0.1498 0.0038 0.005 0.001267 0.00166714th 0.108 0.1429 0.0047 0.0069 0.001567 0.002313th 0.1026 0.1355 0.0054 0.0074 0.0018 0.00246712th 0.0964 0.1271 0.0062 0.0084 0.002067 0.002811th 0.0896 0.1178 0.0068 0.0093 0.002267 0.003110th 0.082 0.1076 0.0076 0.0102 0.002533 0.00349th 0.074 0.0968 0.008 0.0108 0.002667 0.00368th 0.0655 0.0855 0.0085 0.0113 0.002833 0.0037677th 0.0576 0.0737 0.0079 0.0118 0.002633 0.0039336th 0.0478 0.0619 0.0098 0.0118 0.003267 0.0039335th 0.0388 0.05 0.009 0.0119 0.003 0.0039674th 0.0299 0.0383 0.0089 0.0117 0.002967 0.00393th 0.0214 0.0271 0.0085 0.0112 0.002833 0.0037332th 0.0134 0.0116 0.008 0.0113 0.002667 0.0051671th 0.0066 0.0077 0.0068 0.0039 0.002267 0.0013ground 0.0015 0.0012 0.0051 0.0065 0.0017 0.002167basement0 0 0 0.0015 0.0012 0.0005 0.0004
Table 4: Drift calculation of Block 2
storey drift Drift ratiofloor U1 U2 D1 D2 DR1 DR2roof -0.1201 -0.1417 15th -0.1147 -0.1404 -0.0054 -0.0013 -0.0018 -0.0004314th -0.1083 -0.1331 -0.0064 -0.0073 -0.00213 -0.0024313th -0.1021 -0.125 -0.0062 -0.0081 -0.00207 -0.002712th -0.095 -0.1163 -0.0071 -0.0087 -0.00237 -0.002911th -0.0874 -0.1069 -0.0076 -0.0094 -0.00253 -0.0031310th -0.0793 -0.0968 -0.0081 -0.0101 -0.0027 -0.003379th -0.0708 -0.0863 -0.0085 -0.0105 -0.00283 -0.00358th -0.062 -0.0754 -0.0088 -0.0109 -0.00293 -0.00363
28
7th -0.0531 -0.0643 -0.0089 -0.0111 -0.00297 -0.00376th -0.0441 -0.0532 -0.009 -0.0111 -0.003 -0.00375th -0.0353 -0.0432 -0.0088 -0.01 -0.00293 -0.003334th -0.0319 -0.0319 -0.0034 -0.0113 -0.00113 -0.003773th -0.0189 -0.022 -0.013 -0.0099 -0.00433 -0.00332th -0.0118 -0.0134 -0.0071 -0.0086 -0.00237 -0.002871th -0.0058 -0.0062 -0.006 -0.0072 -0.002 -0.0024ground -0.0015 -0.0011 -0.0043 -0.0051 -0.00143 -0.0017basement0 0 0 -0.0015 -0.0011 -0.0005 -0.00037
Minimum spacing of expansion joint = Maximum deflection of Block 1 in
positive X-direction + Maximum deflection of Block 2 in negative X- direction
= 0.0113 + 0.0113
= 0.0226 m
So we provided expansion joint of 30 cm.
29
CHAPTER 5
SECTION DESIGN
5.1 Limit State Method
` In the method if design based on limit state concept, the structure shall be designed to
withstand safely all loads liable to act on it throughout its life; it shall also satisfy the
serviceability requirements, such as limitations on deflection and cracking. The
acceptable limit for the safety and serviceability requirements before failure occurs is
called a ‘limit state’. The aim of design is to achieve acceptable probabilistic that the
structure will not become unfit for the use for which it is intended, that is, that it will not
reach a limit state.
Assumptions for flexural member:
i) Plane sections normal to the axis of the member remain plane after bending.
ii) The maximum strain in concrete at the outermost compression fiber is 0.0035.
iii) The relationship between the compressive stress distribution in concrete and the
strain in concrete may be assumed to be rectangle, trapezoidal, parabola or any
other shape which results in prediction of strength in substantial agreement with
the result of test. For design purposes, the compressive strength of concrete in the
structure shall be assumed to be 0.67 times the characteristic strength. The partial
safety factor γm = 1.5 shall be applied in addition to this.
iv) The tensile strength of concrete is ignored.
v) The design stresses in reinforcement are derived from representative stress-strain
curve for the type of steel used. For the design purposes the partial safety factor γm
=1.15 shall be applied.
vi) The maximum strain in the tension reinforcement in the section at failure shall not
be less than: fy/1.15Es + 0.002
30
Where, fy = characteristic strength of steel
Es = modulus of elasticity of steel
Limit state of collapse for compression:
Assumption:
In addition to the assumptions given above from i) to v), the following shall be assumed:
i.) The maximum compressive strain in concrete in axial compression is taken
as 0.002.
ii.) The maximum compressive strain at highly compressed extreme fiber in concrete
subjected to axial compressive and bending and when there is no tension on the
section shall be 0.0035 minus 0.75 times the strain at the least compressed
extreme fiber.
The limiting values of the depth of neutral axis for different grades of steel based
on the assumptions are as follows:
Table 5: Limiting values of depth of neutral axis
Fy Xu, max
250 0.53
415 0.48
500 0.46
Materials adopted in our design:
M20 (1:1.5:3)
M30 (proportion according to Mix Design)
Fe415
Use of SP16, IS456-2000, IS1893-2002, IS13920-1993, SP34:
After analyzing the given structure using the software SAP2000 the structural
elements are designed by Limit state Method. Account should be taken of
accepted theories, experiment, experience as well as durability.
The code we use for the design is IS456-2000; IS1893-2002, IS13920-1993 and
Design aids are SP16 and SP34. Suitable material, quality control, adequate
31
detailing and good supervision are equally important during implementation of
the project.
Use of different handbook for the design:
The structural elements (special staircases, lift wall, basement wall) which are not
described by the above mentioned codes and design aids were handled with the
help of the handbooks viz. Reinforced concrete Designer’s Handbook – Charles
E. Reynolds.
5.2 Design Of Structural Elements
The aim of design is the achievement of an acceptable probability that structures being
designed will perform satisfactorily during their intended life. With an appropriate degree
of safety, they should sustain all the loads and deformations of normal construction and
use and have adequate durability and adequate resistance to the effects of misuse and fire.
After analyzing the structure using SAP 2000, the structural elements are designed by
Limit State Method. In the limit state design concept, the structure shall be designed to
withstand safely all loads liable to act on it throughout its life; it shall also satisfy the
serviceability requirements such as limitation on deflection and cracking. The acceptable
limit for the safety and serviceability requirements before failure occurs is called a limit
state. The aim of limit state design is to achieve acceptable probabilities that the structure
will not become unfit for the use for which it is intended, i.e. it will not reach a limit
state. The design is in compliance with clearly defined standards for materials,
production, workmanship, and maintenance and use of the structure in service.
Following are the sample calculation of the design process of single element as
i. Slab
ii. Beam
iii. Column
iv. Staircase
v. Shear Wall
vi. Basement Wall
vii. Mat Foundation
32
5.2.1 Design of slab
Grade of concrete = M30 Grade of steel = Fe415
Ref Step Calculations Output
Slab ID S1
IS456:2000
Clause
23.2.1
1
Interior Panel
Thickness of slab and durability consideration
Clean spans
Lx=6.1 m
Ly=6.7 m
Depth (d)= Shorter span/(α*β*γ*δ*λ)
= 6100/(26 *1*1.6*1*1)
= 146.6 mm
Provide D= 150 mm
And clear cover = 15mm
Provide 10 mm Φ
d = 150 – 15 - 5 = 130 mm
Since,
Ly/Lx = 6700/6100
= 1.1 < 2
(Design as two way slab)
D=
150
mm
d =
130
mm
2 Design load
Self weight of slab= 0.15* 25 = 3.75 KN/m2
Plaster = 1 KN/m2
Others = 1 KN/m2
__________________ 5.75 KN/m2
Live load = 3.5 KN/m2
Design load = 1.5 *(DL + LL)
33
= 1.5 *(5.75 + 3.5)
= 13.875 KN/m2
Considering unit length of slab
w = 13. 875 KN/m
w =
13.
875
KN/m
IS 456,
Table 26
IS 456,
Annex
Clause
D.1.0
3. Moment calculation
For Ly/Lx= 1.1
-ve BM coefficient at continuous edge
αx=- 0.037
αy =- 0.032
+ve moment at mid span
αx = 0.028
αy= 0.024
For shorter span
Support Moment Ms = - αx wLx2
= -.037 * 13.875* 6.12
= -19.1 KNm
Mid span moment Mm = αx w Lx2
= 0.028 * 13.875* 6.12
= 14.456 KNm
For larger span
Support Moment Ms=- αy *w*Ly 2
= -0.032 * 13.875 * 6.72
= - 19.93 KNm
Mid span moment Mm = αy *w*Ly 2
= 0.024 * 13.875 * 6.72
= 14.948 KNm
34
IS
456 :2000
Annex
G.1.1
4 Check for depth from moment consideration
Mu= 0.36 * xu,max /d*(1 – 0.42 xu,max /d)* bd2 *fck
19.93 X 10^6 = 0.36 * 0.48 *(1-0.42 * 0.48) * 1000 * d2*
30
d = 69.4 mm < 130 mm OK
IS 456
Clause
26.5.2.1
IS
456 :2000
Annex
G.1.1
5 Calculation of Area of steel
Minimum area of steel (Ast,min)= 0.12 % of bD
= 180 mm2
For shorter span,
Area of steel at support ((Top bars)
Mu= 0.87 * fy* Ast * d(1- Ast*fy/bd*fck)
19.1 X 106 = 0.87 * 415 * Ast * 130 *(1-
Ast*415/(1000*130*30))
Ast= 426.3 mm2 > Ast
Provide 10 mm Φ bars
Ab= 78.5 mm2
Spacing Sv= 184.14
Providing 10 mm Φ @ 180 mm c/c
Then Area of steel Ast = 436 mm2
Pt= 0.29 %
Again,
Area of steel at mid span (bottom bar)
14.456X10^6 = 0.87*415*Ast * 130 *(1- Ast*415/
(1000*130*30))
Ast = 318. 8 mm2 > Ast min
10
mm Ǿ
@
180
mm
c/c
Area
of
steel
Ast =
436
mm2
Pt=
35
IS
456 :2000
Annex
G.1.1
Providing 10 mm Φ bars
Ab= 78.5 mm2
Spacing of bars Sv= 246.2 mm
Providing 10 mm Φ @ 240 c/c
Then Actual Area Ast= 327 mm2
Pt = 0.22 %
For longer span
Area of steel at support ( Top bars)
Mu= 0.87 * fy* Ast * d(1- Ast*fy/bd*fck)
19.93X 10^6 = 0.87 * 415*Ast * 120 *(1- Ast *
415/(1000*120*30))
Ast=446 mm2 >Ast min
Providing 10 mm Φ bars
Ab = 78.5 mm2
Spacing of bar, Sv= Ab/Ast * 100
= 78.5 *1000/446
= 176 mm
Providing 10 mm Φ @ 170 mm c/c
Actual Ast = Ab/ Sv * 1000
= 446 mm2
Pt= 0.3 %
Again,
Area of steel at mid span ( Bottom bars)
Mu= 0.87 * fy* Ast * d(1- Ast*fy/bd*fck)
14.948X 10^6 = 0.87* 415*Ast *120*(1- Ast *
415/(1000*120*30))
Ast=330 mm2 >Ast min
Providing 10 mm Φ bars
Ab = 78.5 mm2
0.29 %
36
Spacing of bar, = Sv = Ab/Ast * 100
= 78.5 *1000/330
=238 mm
Providing 10 mm Φ @ 230 mm c/c
Actual Ast = Ab/ Sv * 1000
= 341.3 mm2
Pt= 0.23 %
IS 456,
Table 19
Clause
40.2.1.1
6 Check for shear
For short span,
Shear force at the face of support
V = w Ly
= 13.875 * 6.7
= 46.48 KN
Shear at critical section
46.48/3350 = x/3220
X = 44.67 KN
Hence tension reinforcement of slab contribute in shear
For Pt = 0.29 %
τc = 0.39 N/mm2
And K = 1.3
K τc *bd = 1.3 * 0.39 * 1000* 130/1000
37
= 65.91 KN > Vu OK .
7 Check for deflection
Along the shorter span
Since, both ends are continuous the basic value may be
taken as 26.
fs= 0.58 fy*Ast,req / Ast,prov
= 0.58 * 415 * 426/436
= 235.2 N/mm2
Pt = 0.26 %
Modification factor (MF) = 1.85
dper= 6100/1.85*26 = 126.82 < 130 mm OK.
IS 456,
Clause
26.2.1
IS 456,
Clause
26.2.3.2
8 Check for development length
Ld= Φ σs/1.6*4 *τbd
= 10 * 0.87 * 415/1.6*4*1.5
= 350 mm
Ld< M1/V + Lo
For short span
Ld< M1/V + Lo
= (19.93/2)/44.67 + 0.15
= 373 mm > Ld OK.
IS 456,
ANNEX D,
clause
D.1.10
9 Torsion reinforcement:
Since this is interior panel, no need of torsion
reinforcement
38
Ref Ste
p
Calculations Outpu
t
Slab ID S2
IS456:200
0
Clause
23.2.1
1
Exterior Panel (three edges continuous and one long edge
discontinuous)
Thickness of slab and durability consideration
Clean spans
Lx=6.1 m
Ly=6.7 m
Depth (d)= Shorter span/(α*β*γ*δ*λ)
= 6100/(26 *1*1.6*1*1)
= 146.6 mm
Provide D= 150 mm
And clear cover = 15mm
Provide 10 mm Φ
d = 150 – 15 - 5 = 130 mm
Effective length= Ly/Lx
= 6700/6100
= 1.1 < 2
(Design as two way slab)
2 Design load
Self weight of slab= 0.15* 25 = 3.75 KN/m2
Plaster = 1 KN/m2
Others = 1 KN/m2
____________ 5.75 KN/m2
Live load = 3.5 KN/m2
Design load = 1.5 *(DL + LL)
= 1.5 *(5.75 + 3.5)
39
= 13.875 KN/m2
Considering unit length of lab
w = 13. 875 KN/m
IS 456,
Table 26
IS 456,
Annex
Clause
D.1.0
3. Moment calculation
For Ly/Lx= 1.1
-ve BM coefficient at continuous edge
αx= -0.037
αy = -0.037
+ve moment at mid span
αx = 0.028
αy= 0.028
For shorter span
Support Moment Ms = - αx wLx2
= -.037 * 13.875* 6.12
= -19.1 KNm
Mid span moment Mm = αx w Lx2
= 0.028 * 13.875* 6.12
= 14.456 KNm
For larger span
Support Moment Ms=- αy *w*Ly 2
= -0.037 * 13.875 * 6.72
= - 23.045 KNm
Mid span moment Mm = αy *w*Ly 2
= 0.028 * 13.875 * 6.72
= 17.44 KNm
4 Check for depth from moment consideration
Mu= 0.36 * xu,max /d*(1 – 0.42 xu,max /d)* bd2 *fck
40
IS
456 :2000
Annex
G.1.1
Or, 23.045 X 106 = 0.36 * 0.48 *(1-0.42 * 0.48) * 1000 *
d2* 30
Or, d = 74.62 mm < 130 mm. OK.
IS 456,
Clause
26.5.2.1
IS
456 :2000
Annex
G.1.1
5 Calculation of Area of steel
Minimum area of steel ( Ast,min) = 0.12 % of bD
= 180 mm2
For shorter span
Area of steel at support ((Top bars)
Mu= 0.87 * fy* Ast * d(1- Ast*fy/bd*fck)
19.1 X 106 = 0.87 * 415 * Ast * 130 *(1-
Ast*415/(1000*130*30))
Ast= 426.3 mm2 > Ast
Provide 10 mm Φ bars
Ab= 78.5 mm2
Spacing Sv= 184.14
Providing 10 mm Φ @ 180 mm c/c
Then Area of steel Ast = 436 mm2
Pt= 0.29 %
Again,
Area of steel at mid span (bottom bar)
14.456X106 = 0.87*415*Ast * 130 *(1- Ast*415/
(1000*130*30))
Ast = 318. 8 mm2 > Ast min
Providing 10 mm Φ bars
Ab= 78.5 mm2
Spacing of bars Sv= 246.2 mm
Providing 10 mm Φ @ 240 c/c
Then Actual Area Ast= 327 mm2
Pt = 0.22 %
41
IS
456 :2000
Annex
G.1.1
For longer span
Area of steel at support ( Top bars)
Mu= 0.87 * fy* Ast * d(1- Ast*fy/bd*fck)
Or, 23.045X 10^6 = 0.87 * 415*Ast * 120 *(1- Ast *
415/(1000*120*30))
Or, Ast=519 mm2 >Ast min
Providing 10 mm Φ bars
Ab = 78.5 mm2
Spacing of bar, = Sv = Ab/Ast * 100
= 78.5 *1000/446
= 151 mm
Providing 10 mm Φ @ 150 mm c/c
Actual Ast = Ab/ Sv * 1000
= 523.33 mm2
Pt= 0.35 %
Again,
Area of steel at mid span ( Bottom bars)
Mu= 0.87 * fy* Ast * d(1- Ast*fy/bd*fck)
Or, 17.44 X 106 = 0.87* 415*Ast *130*(1- Ast *
415/(1000*130*30))
or, Ast=387.6 mm2 >Ast min
Providing 10 mm Φ bars
Ab = 78.5 mm2
Spacing of bar, Sv= Ab/Ast * 1000
= 78.5 *1000/387.6
=202.58 mm
Providing 10 mm Φ @ 200 mm c/c
Actual Ast = Ab/ Sv * 1000
42
= 392.5 mm2
Pt= 0.26 %
IS 456,
Table 19
Clause
40.2.1.1
6 Check for shear
For short span,
Shear force at the face of support
V = w Ly
= 13.875 * 6.7/2
= 46.48 KN
Shear at critical section
46.48/3350 = x/3220
X = 44.67 KN
Hence tension reinforcement of slab contribute in shear
For Pt = 0.29 %
τc = 0.39 N/mm2
And K = 1.3
K τc *bd = 1.3 * 0.39 * 1000* 130/1000
= 65.91 KN > Vu OK.
43
7 Check for deflection
Along the shorter span
Since, both ends are continuous the basic value may be
taken as 26
fs= 0.58 fy*Ast,req / Ast,prov
= 0.58 * 415 * 519/523.33
= 238 N/mm2
Pt = 0.35 %
Modification factor (MF) = 1.85
dper= 6100/1.85*26 = 126.82 < 130 mm OK.
IS 4565,
Clause
26.2.1
IS 456,
Clause
26.2.3.2
8 Check for development length
Ld= Φ σs/1.6*4 *τbd
= 10 * 0.87 * 415/1.6*4*1.5
= 350 mm
Ld< M1/V + Lo
For short span
Ld< M1/V + Lo
= (23.045/2)/44.67 + 0.12
= 378 mm > Ld OK.
IS
456,Annex
D, D.1.8
9 Torsion reinforcement
Area of torsion reinforcement
= 0.75 /2 *area required for maximum mid span
moment
= 0.75/2* 387.6
=145.35 mm2
So, provide 10 mm Φ @ 300mm c/c at a distance 0.2lx
from the edge.
44
45
Fig.4 Design of Slab
Design of Beams
Grade of concrete = M30 Grade of steel = Fe415
46
47
Ref. Calculation
Beam (4687)
ID
IS456:2000
Annex 6.4.
IS456,
clause38.1,
2000
IS456: 2000
Annex G-1.1
At section L=0
D=600 mm
d = 560 mm
b= 300 mm
L=0
M3 V2 T P
292.115 50.013 4.9247 5.393
-416.214 -190.135 -6.448 -3.54
Mt = T(1+D/b)/1.7
= ( 4.9247(1 + 600/300))/1.7
= 8.690 KNm (+ve)
Mt = 6.448 (1 + 600/300)/1.7
= 11.38 KN-m (-ve)
Maximum moment= 292.115 + 8.690 = 300.805 KNm (+ve)
Minimum moment = -416.214 – 11.38 = -427.59 KNm (-ve)
Moment of Resistance of the beam
Mu,lim = 0.36 x u,lim /d*(1-0.42*xu,max/d)*bd2 fck
= 389.38 X 106 Nmm
For +ve moment,
Since Md< Mu, beam is singly reinforced.
For singly reinforced beam
Md= 0.87 * 415*Ast * 560 * (1- Ast * 415/300*560*30)
300.805 X 106 = 0.87* 415*Ast * 560 * (1- Ast * 415/300*560*30)
Ast = 1735.8 mm2
Assuming 12 mm Φ bar
Asb= π * 122 / 4 = 113.04 mm2
No. of bar = 1735.8 mm2 / 113.04 mm2
= 15.35 Say 16
Again, Take 25 mm Φ bar
Asb= π * 252 / 4 = 490.87 mm2
The above design also fulfill the standards of IS 13920 : 1993, clause 6.1.1, clause
6.1.2,clause 6.1.3, clause 6.1.4, clause 6.2.1, clause 6.2.2, clause 6.2.3 and detailing is
done also considering the standards of code IS 13920 : 1993, clause 6.2 and 6.3.
Design of column
Grade of concrete M30 Grade of steel Fe415
48
49
Reference Calculation Rema
rks
IS
456:2000
(cl.25.4)
Design of Longitudinal Reinforcement:
Column of size 450 X 450
Sample Calculation: For Column ID -5374
Ultimate Load, Pu = 180.739KN
Mux = 178.2433KNm
Muy = 193.2118KNm
Column Section =450mm × 450mm
Grade of concrete =M30
Grade of Steel = Fe 415
Design Steps:
Assume following data
Clear cover = 40mm
Diameter of longitudinal Reinforcement Φ = 20mm
So, effective cover =40+20/2 =50mm
Effective length =.65*3 =1.95m
Effective length/least lateral dimension =1.95*1000/450 =4.33<12(short
column)
Minimum eccentricity =L/500+Least dimension/30≥20mm
eminx =3000/500+450/30 =21mm>20mm
eminy=3000/500+450/30 =21mm>20mm
Moment due to minimum eccentricity
Mux=Pu*eminx =180.739*.021 =3.795 KNm<178.243KNm
Muy= Pu*eminy =180.739*.021 =3.795 KNm<193.212KNm
Therefore,
Mux = 178.2433 KN-m
(FRO
M
SAP)
The above design also fulfill the standards of IS 13920 : 1993, clause 7.1.1, clause
7.1.2,clause 7.1.3, and detailing is done also considering the standards of code IS 13920 :
1993, clause 7.2, clause 7.3 and clause 7.4.
Design of Shear wall
Sample detailing of 2.6 m structural wall with boundary element at lower
basement according to IS13920:1993
Materials: Grade of concrete = M30
Grade of steel = Fe415
Ref. Step Calculations Output
Shell
element :1079(
1 Direction Dimension no. Mu(kNm) Pu(kN)
y 2.6 2 89.44 8271.05
50
57) Description: yz plane.
IS:13920:1993
Cl.9.4.1
2 Check whether boundary elements are required:
Extreme fibre compressive stress in the wall should
not exceed 0.2 fck
L = 3050 mm
tw = 200 mm
Ag = 610000 mm2
Ix = 4.73*E11 mm4
(200*30503/12)
y = 1525 mm
0.2 fck = 6 N/mm2
fc = Pu /Ag + Mu.y/Iy
= 13.55mm2>0.2 fck
Hence,
boundar
y
elements
required.
3 Check whether two curtains are required:
IS:13920:1993
Cl.9.1.5
Two curtains of bars is required if thickness of wall is
more than 200mm or if the factored shear stress is
more than 0.25√ fck.
1.25 fck = 37.50 > Vu
Shear stress = Vu = 16.05kN
Providing two curtains.
Provide
two
curtains.
IS 13920:1993
Cl. 9.1.4
4 Minimum reinforcement required in longitudinal
and transverse direction:
0.25 % of gross area
5 Shear strength requirement:
51
IS 456:2000
Cl. 40.1
Tb. 19
Tb.20
IS 13920:1993
Cl.9.1.7
Vu = 16.05kN
τ v = Vu / twdw
=0.0328 N/mm2
τc = 0.37 N/mm2 for 0.25% of steel
τc,max = 4 N/mm2
τ v < τc
Hence, providing minimum reinforcement in each
direction.
Maximum spacing of reinforcement:
Description lw/5 3tw 450
max. spacing for
3.05wall 488 600 450
max. spacing for 6.1m
wall 976 600 450
max. dia. Of bar 20 provide 12mm
Gross area per meter of wall = tw * 1000 mm2 (per
meter of wall)
= 200 * 1000 mm2
= 200000 mm2
The required area of reinforcement area of
reinforcement in each direction per meter of wall is,
0.0025 * 200000 = 500 mm2 per meter.
Required spacing of 12mm dia. Bar (in two curtains,
As = 2 * 113.04 mm2) :
S (required) = (2 * 113.04) * 1000/ 500
= 452.16 mm
dw=
effective
depth of
wall
section
taken as
0.8 * lw
lw=.8L
52
Hence,
Providing spacing of 400 mm for 3.05m shear
wall.
S > max.
spacing
as
specified
by code.
IS 13920:1993
Annex A
6 Flexure strength requirement :
Mu = 89.44 kNm
Pu = 8271.059kN
d' = 26
d'/D = 0.1
λ = Pu/fck.tw.lw = 0.636375
p/ fck = 0.0000625
φ = (0.87*fy*p ) / fck
= 0.002256
xu/lw = (φ +λ)/(2 φ + 0.36)
= 1.626
xu*/lw = 0.69
Moment of resistance = Muv/fck tw lw2
β = 0.5157
α1 = 0.3548
α2 = 0.4166
α3 = -0.01739
therefore,
Mux/fck tw lw2 = 0.1116
Muv = 6036.08985 kNm
Mux >
Mu
Safe in
flexure
53
7 Checking adequacy of boundary element:
a. Calculations for the longitudinal bars:
Let us assume that the size of column is adequate for
boundary element.
Size of boundary element :
Length = 800 mm
Breathe = 800 mm
Take clear cover = 40 mm
Dia. of bar = 25 mm
Farzad Naiem
IS 13920:1993
Cl. 9.4.4
SP 16
Axial load on boundary element is calculated as,
Axial load on wall/2 + moment / width of wall
Ultimate axial load on boundary wall = 6652 kN
Percentage of vertical reinforcement in the boundary
elements shall not be less than 0.8%, nor greater than
6%. In order to avoid the congestion, practical limit
would be 4%.
Taking 2.5% of gross area as reinforcement bar.
Ultimate load capacity of boundary element:
Pu, = 0.4fck Ag + (0.67 fy -0.4fck) As
where,
Ag = 800 * 800 mm2
= 640000 mm2
As = 2.5% of Ag
= 0.0025 * 640000
= 16000 mm2
Therefore,
Pu,max = 11936.8 kN ( > 6652 kN )
Pu,capacity
> Pu
54
Hence, provided size of boundary element is adequate
for the ultimate load.
Then,
Dia. of bar = 25mm
Ast = 491.07 mm2
no. of bars = As / Ast
= 16000 / 491.07
= 32
Actual area of steel = 16198.84 mm2
Actual % of steel = 2.53%
Hence,
Provide 32 bars of 25 mm diameter distributed on the
four sides of section .
Min. spacing = dia. of bar = 25mm
= 5 mm more than the nominal
maximum size of aggregate.
Spacing in 800mm length= (800 – 2*40 – 25 )/(8– 1)
= 99 mm
Hence, provide:
Dimensio
n 800 mm 800 mm
no. of
bars 8 8
clear
cover 40 40
spacing 99 mm 99 mm
No. of
bars
provided
= 32
S j 100
mm
55
IS 13920:1993
Cl. 9.4.5
Cl.7.4.6
Cl. 7.4.8
b. Calculations for the confinement bars:
Boundary elements, where required, as per Cl.9.4.1
(already defined above), shall be provided throughout
their height, with special confining reinforcement.
Spacing of hoops shall not exceed 1/4 th of min.
member dimension but need not be less than 75mm
nor more than 100 mm
¼ of min. dimension = ¼ *800
= 200 mm
Hence, Providing 100mm spacing.
The area of cross section , Ash of the bar forming
rectangular hoop, to be used as special confining
reinforcement shall not be less than,
Ash = 0.18 S h fck / fy [ Ag / Ak – 1 ]
Where,
S = 100 mm
h = 272 mm
Providing 12mm Φ rectangular hoops
Ak = area of concrete core
= 744 * 744 mm2
=553536 mm2
Ag = 800 * 800
= 640000 mm2
Therefore,
Ash = 0.18 S h fck / fy [ Ag / Ak – 1 ]
Spacing
of
confine
ment
bars =
100 mm
Ash =
113.143
mm2
800 –
2*40
+2*12 =
56
= 55.28mm2 < 113.143mm2
Hence, providing 12mmφ confining bars at 100mm
spacing as confining reinforcement.
744 mm2
O.K
Design of Basement wall
Ref. Step Calculation Output
1 Design constants
Concrete grade M20
Steel Fe415
Clear height of the basement wall , H =3
Specific weight of soil, γsoil = 16.5 kN/m3
Angle of internal friction, φ = 30.
Surcharge load , = 19 kN/m2
Safe bearing capacity, qs = 150 kN/m2
2 Approximate design of section
57
IS
456:2000
Cl. 32.3.4
Let effective depth of wall, d = H/15 = 3000/15 =
200 mm
Take clear 25 mm, diameter of bar = 16 mm
Provide overall depth, D = 233 mm
Effective depth, d = 200 mm
Slenderness ratio = H/d = 15 < 30
D = 240 mm,
d = 207 mm
O.K
3 Moment Calculations
Coeff. of earth pressure Ka = (1 – sin φ )/(1 +sin φ
) = 0.3333
Lateral load due to soil pressure pa = 1/2kaγH2 =
24.50 kN/m
Lateral load due to surcharge load ps = kawH =
18.81 kN/m
Characteristic moment at the base of the wall, Mc
= paH/3 + psH/2 = 52.715 kNm
Design moment Md = 1.5 Mc = 79.07 kNm
pa = 24.50 kN/m
ps = 18.81 kN/m
Md = 79.07 kNm
4 Check for the depth
Considering unit length of wall,
Mu,lim = 0.138fckbd2 = 118.26 kNm > Md
Depth of wall from moment consideration, d =
(Md/0.138fckb)0.5 = 169.26 mm < d provided
O.K
O.K
58
IS
456:2000
Cl.32.5
Cl.26.5.2.2
5 Calculation of main steel reinforcement
Ast = 0.5*fck/fy*b*d*(1-(1-4.6Md/fck/b*d2)0.5) =
1147.22mm2
Min. area of tensile reinforcement = 0.12% of bd
= 248.4 mm2
Max. spacing = 3d = 621 mm or 450 mm
Max. dia of reinforcement = D/8 = 30 mm
Providing 16 mm Φ bar, Ab = 201.143 mm2
Spacing of bar Sv = 175.25
Actual Ast = 1183.19mm2
Actual pt = 0.4605%
For front face provide nominal vertical
reinforcement Φ 12mm @ 300mm c/c
Provide Φ
16mm@ 170 mm
c/c
Front face Φ
12mm @ 300mm
c/c
IS
456:2000
Cl.31.6.2.1
table 19
6 Check for shear
Shear force V = 0.5kaγH2 = 23.98 kN
Design shear force Vu = 1.5V = 35.96 kN
Nominal shear stress τv =Vu/bd = 0.174 N/mm2
Permissible shear stress τc = 0.469 N/mm2 >τv No shear
reinforcement
IS
456:2000
cl.23.2
7 Check for deflection
Leff = H+d = 3.207 m
Allowable deflection = Leff/250 = 12.83mm
Actual deflection = Psleff4/8EI+Paleff
4/30EI =
11.5mm < 12.83
O.K
IS 8 Calculation of horizontal steel bars
59
456:2000
cl 32.5
i.
ii.
Area of horizontal steel reinforcement =
0.0012DH = 864mm2
As the temperature changes occurs at the front
face of the basement wall,2/3 of the horizontal
reinforcement Is provided at the front face and
remaining at inner face.
So front face horizontal reinforcement = 2/3 of
area of ho. Reinforcement = 576mm2
Providing 8mm dia bar
Ab = 50.286 mm2
No of bars N = 576/50.286 = 12
Spacing = (H – clear cover – Φ )/(N-1) = 269
mm
Max spacing 3d = 621or 450 mm
Inner face reinforcement= 1/3*1176 = 288 mm2
Providing 8mm bars
No of bars N = 6
Spacing = 166.7 mm
Max spacing 3d = 621 or 450
Provide Φ 8mm @
260mmc/c
O.K
Provide Φ 8mm
@160c/c
O.K
9 Curtailment of Reinforcement
No bars can be curtailed in less than Ld distance
from the bottom of the stem,
Ld = 564mm
The curtailment of bars can be done in two layers
1/3 and 2/3 height of steam above the base.
Let us curtail bars at 1/3 distance i.e. 1000mm
from base,
Lateral load due to soil pressure,
60
Pa = 10.65KN/m
Lateral load due to surcharge load,
Ps = 12.65KN/m
Characteristics BM at base of the wall is,
Mc = paH/3 + psH/2 = 29.625 kNm
Design moment ,M=1.5 Mc=44.44KNm
Ast = 620mm2
No. of bars = 8
Spacing = 142.8
Provide Φ
10mm@140mm c/c
Design of Mat foundation
Required data:
Case considered = 1.5(DL + LL)
Total design load (P) = 403065.4 KN (From SAP)
From soil investigation report,
Safe soil bearing capacity = 150 KN/m2
61
Grade of concrete = M20
Grade of steel = Fe415
Calculation of centre of gravity of plan area
Taking moment of column load about outer face of the mat,
X = = 95.31 ft
Y = = 36.96 ft
Geometric centre of mat foundation,
X = = 100 ft
Y = = 37 ft
ex = X –X = 100 - 95.31 = 4.69 ft =1.43 m
ey = Y – Y = 37 – 36.96 = 0.04 ft ≈ 0 m
Calculation of MOI
Ixx = 82190.29 m4
Iyy = 505873.19 m4
Area coverage of mat = 1532.17 m2
Mxx = ∑P × ey = 403065.4 × 0 = 0
Myy = ∑P × ex = 403065.4 × 1.43 = 576383.52 KNm
= 263.07 KN/m2
Soil pressure calculation at different points:
At corner A-2
A-2 = 263.07 + × (18.79 - 30.48) = 249.18 KN/m2
At corner A-3
A-3 = 263.07 + × (37.08 - 30.48) = 270.6 KN/m2
At corner B-1
62
B-1 = 263.07 + × (0 - 30.48) = 228.33 KN/m2
At corner B-4
B-4 = 263.07 + × (66.57 - 30.48) = 304.21 KN/m2
At corner C-1
C-1 = 263.07 + × (0 - 30.48) = 228.33 KN/m2
At corner C-4
C-4 = 263.07 + × (66.57 - 30.48) = 304.21 KN/m2
In x-direction the raft is divided into three strips, that is three equivalent beam
(i) Beam A-A with 5.35m width and soil pressure of 270.6 KN/m2
(ii) Beam B-B with (5.35+19.9/2) = 15.3 m width and soil pressure of 0.5(270.6+304.21) = 287.41 KN/m2
(iii) Beam C-C with 9.95 m width and soil pressure of 0.5(304.21+304.21) = 304.21 KN/m2
The bending moment is obtained by using a coefficient of 1/10 as a centre of column,
+M = -M =wl2/2
For strip A-A,
Maximum moment = 270.6*6.12/10= 1006.25 KNm per meter
For strip B-B,
Maximum moment = 287.41*6.12/10 = 1069.45 KNm per meter
For strip C-C,
Maximum moment = 304.21*6.12/10 = 1131.96 KNm per meter
For any strip in Y-direction, taking coefficient of 1/10 with maximum centre to centre length of 6.7m
For strip 1-1,
Maximum moment = 228.33*6.72/10 = 1024.97KNm per meter
For strip 2-2,
63
Maximum moment = 0.5*(228.33+249.13)*6.72/10 = 1071.61KNm per meter
Similarly, for strip 3-3, maximum moment = 1166.53 KNm per meter
For strip 4-4, maximum moment = 1290.16 KNm per meter
(i) Calculation of depth of foundation
Depending upon moment criteria, taking maximum moment in either direction,
Maximum moment is occurring for strip
M = 1920.16 KNm per meter
Mu,lim /(fck* b*d2) = 0.138
Mu,lim = 0.138*30*1000*d2
1290.16*10^6 = 0.138*30*1000*d2
d = 558.24 m
(ii) Calculation of depth from two way shear,
The depth of the raft will be governed by two-way shear at one of the exterior column, in case location of critical shear Is not obvious it may be necessary to check all possible location. When shear reinforcement is not provided, the calculated shear stress at critical section shall not excess Ksτc
i.e, τv = Ksτc (clause 31.6.3, IS 456, 2006)
Ks = 0.5 + β3 but not greater than 1
Then, β3 = (19.90+10.75)/66.57 = 0.46
Then, Ks =0.5+0.46 = 0.96
Shear strength of concrete,
τc = 0.25 *√fck (clause 31.6.3, IS 456, 2000)
= 0.25 *√30
= 1.37 N/mm2
64
= 1369.30 KN/m2
For corner, column say A-1
Column load = 4918KN (from SAP)
Perimeter = Po = 2(d/2 + 750) = d + 1500
τv = column load/(Po*d)
1.37 = 4918*1000/(d + 1500)*d
d = 1287.71mm
For side column say A-2
Column load = 6954.92 KN (from sap)
Perimeter = 2(0.5d + 750) + (d+560)
= 2d + 2000
τv = column load/(Po*d)
Solving, d = 1191.32mm
Adopt, effective depth of 1300mm
Total depth= 1350mm
Calculation of reinforcement in foundation:-
Alone x- direction
Maximum negative moment = maximum positive moment=1131.96KNm/m
Then, area of steel, Ast = 2476.97 mm2
Minimum reinforcement = 0.12 * 1000 * 1300 = 1560 mm2
65
` ok
Provide 20mm bar, area of single bar = 314.16 mm2
Providing 20 mm Ǿ @ 140mm c/c in X-direction
Alone Y- direction
Moment = 1290.16 KNm/m
Then, area of steel, Ast =2834.21 mm2
So, providing 20 mm Ǿ @ 110mm c/c in Y-direction
Table 6: Summary of Calculation of reinforcement in foundation
SN Description Value Remarks1 Mux (moment in X-direction) 1131.96 KNm Max –ve moment=
Max +ve moment2 Muy (moment in Y-direction) 1290.96 KNm Max –ve moment=
Max +ve moment3 Area of Steel in X- direction(Astx) 20mm Φ @ 140
mm c/c Provided in per metre width
4 Area of Steel in Y-direction(Asty) 20 mm Φ @ 110 mm c/c
Provided in per metre width
5 Number of bar in X-direction 8 Provided in per metre width
6 Number of bar in Y-direction 10 Provided in per metre width
66