ANALYSIS AND DESIGN OF COMMERCIAL BUILDING WITH FLAT SLAB
BATCHLOR OF TECHNOLOGY IN CIVIL ENGINEERING By RAHIL ABRAR (12D95A0117 ) SAYED ZAHED (11D91A01A5 ) MOHD FARHAN (11D91A0164 )
DEPARTMENT OF CIVIL ENGINEERING AURORA’S SCIENTIFIC, TECHOLOGICAL & RESEARCH ACADEMY Bandlaguda, near Chandrayanagutta, Hyderabad. APRIL-2015
ANALYSIS AND DESIGN OF COMMERCIAL BUILDING WITH FLAT SLAB A MAIN PROJECT REPORT IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE AWARD OF DEGREE FOR BACHLOR OF TECHNOLOGY IN CIVIL ENGINEERING By RAHIL ABRAR (12D95A0117 )
DEPARTMENT OF CIVIL ENGINEERING AURORA’S SCIENTIFIC, TECHOLOGICAL & RESEARCH ACADEMY Bandlaguda, near Chandrayanagutta, Hyderabad. APRIL-2015
AURORA’S SCIENTIFIC, TECHOLOGICAL & RESEARCH ACADEMY Bandlaguda, near Chandrayanagutta, Hyderabad. Department of Civil Engineering
CERTIFICATE
Certified that the project work entitled “ANALYSIS AND DESIGN OF COMMERCIAL BUILDING WITH FLAT SLAB” carried out by Mr. RAHIL ABRAR, HT. No 12D95A0117, a bonafide student of IV year in partial fulfillment
for
the
award
of Bachelor
of
Technology in Civil
Engineering of the AURORA’S SCIENTIFIC, TECHOLOGICAL & RESEARCH ACADEMY during the year Academic year 2014-15. The project report has been approved as it satisfies the academic requirements in respect of Project work prescribed for the said Degree.
Name & Signature of the internal Guide Signature of the HOD G.VenkataRatnam (M.E)
External Viva: Name and Signature of the examiners with date
Name &
AURORA’S SCIENTIFIC, TECHOLOGICAL & RESEARCH ACADEMY Bandlaguda, near Chandrayanagutta, Hyderabad. Department of Civil Engineering
DECLARATION BY THE STUDENT
I, RAHIL ABRAR bearing Hall Ticket No.12D95A0117, hereby declare that the project report entitled “ANALYSIS AND DESIGN OF COMMERCIAL BUILDING WITH FLAT SLAB” under the guidance of Mr. G.VenkataRatnam, Department of Civil Engineering, Aurora’s Scientific, Technological and Research Academy, Bandlaguda is submitted in partial fulfillment of the requirements for the award of the degree of B.Tech in Civil Engineering. This is a record of bonafide work carried out by me and the results embodied in this project have not been reproduced or copied from any source. The results embodied in this project report have not been submitted to any other university or institute for the award of any other degree or diploma.
Name: RAHIL ABRAR, H.T.No.: 12D95A0117.
INDEX
ACKNOWLEDGMENT...............................................................i ABSTRACT............................................................................ii NOTATIONS...........................................................................v LIST OF FIGURES.................................................................vii LIST OF TABLES....................................................................ix CHAPTER – 1.........................................................................1 INTRODUCTION.....................................................................2 1.1 General..........................................................................................2 1.2 Flat Slab.........................................................................................2 1.2.1 Basic Definition of Flat Slab.....................................................3 1.3 Components of Flat Slabs..............................................................3 1.4 Advantages & Disadvantages........................................................5 1.4.1 Advantages..............................................................................5 1.4.2 Disadvantages:........................................................................6 1.5 Key Messages:...............................................................................7 1.6 Proprietary Punching Shear Reinforcement Systems:....................7 1.7 Objective.......................................................................................8 1.8 Scope of Work................................................................................8 CHAPTER – 2.........................................................................9 MODELLING AND ANALYSIS..................................................10 2.1 Materials and Properties:.............................................................10
2.1.1 Building Materials:.................................................................10 2.2 Loads:..........................................................................................10 2.2.1 Dead loads & Live loads:.......................................................10 2.3 Load Combination:.......................................................................11 CHAPTER-3.........................................................................16 DESIGN PHILOSOPHIES........................................................17 3.1 Methods of Design:......................................................................17 3.1.1 Working Stress Method:.........................................................17 3.1.2 Ultimate Load Method:...........................................................18 3.1.3 Limit State Design:................................................................18 CHAPTER - 4........................................................................ 20 STRUCTURAL PLANNING.......................................................21 4.1 Structural Planning Of Reinforced Concrete Framed Building:.....21 4.1.1 Column Positions...................................................................21 4.1.2 Orientation of columns:.........................................................21 4.1.3 Beam Locations.....................................................................23 4.2 My Project Plans..........................................................................24 CHAPTER - 5........................................................................ 34 LOADINGS...........................................................................35 5.1 Load Conditions and Structural System Response:.....................35 5.1.1 Building Loads Categorized by Orientation:...........................35 5.1.2 Horizontal (Lateral) Loads:.....................................................35 5.2 LOAD CALCULATIONS................................................................37 5.2.1 INTRODUCTION......................................................................37 5.3 Assumptions................................................................................42
Table 1 : Density of Materials Used...................................................43 Table 2 : Design Constants................................................................43 5.4 Assumptions Regarding Design:..................................................43 5.4.1 Assumptions on design:.........................................................44 5.5 Data Collection............................................................................44 CHAPTER-6.........................................................................45 DESIGNING.......................................................................... 46 6.1 Flat Slab.......................................................................................46 6.2 Design of Beams..........................................................................53 6.2.1 Members - B31, B32 & B39 (from ETABS)..............................57 6.3 Column........................................................................................66 6.5 Footing.........................................................................................74 6.6 Stair Case....................................................................................77 CHAPTER-7.........................................................................84 CONCLUSIONS.....................................................................85 REFERENCES.......................................................................86
ACKNOWLEDGMENT I am very pleased to present this main project work. This period of my student life has been truly rewarding a number of people were of immense help to me during the course of my project work and preparation of this book. First, I wish to thank God Almighty who created heavens and earth, who helped me in completing this project. I thank my parents, encouraged me all along to complete this big task and to my friends who has given good support along the way. I would like to thank all my department’s lecturers, Aurora’s Scientific, Technological and Research Academy, Bandlaguda, my project internal guide, for his guidance and help. His insight during the course of my research and regular guidance were invaluable to me. And also I thank Sri G.Venkata Ratnam, Head of the Department, Civil Engineering, Aurora’s Scientific, Technological and Research Academy,
Bandlaguda,
for
his
encouragement
and
cooperation
throughout the project.
I would also thank Smt. Ch.Srilatha, Principal of our college, for extending his help.
Rahil Abrar
i
ABSTRACT With the increase in population and development of civilization, the demand for HOUSING is increasing at a peak rate. Especially in towns due to rapid industrialization, the demand is very high. Adapting the construction of Multi-storied Building not only matches with demand but also decreases the price of the single house. Hence an Engineer to be knowledgeable about the planning and designing of such Multi-storied Buildings. Advancements of computer packages have given many tools to the designer towards achieving the best and accuracy in their work. The aim of our project is to design a G+13 building with flat slabs instead of conventional slab. It is designed by using M25 grade concrete and Fe415 steel. The dead load, live load and seismic load are applied and the design for beams, columns, footing is obtained. Analysis & Design of the building with flat slab is done by using ETABS software.
ii
PLANNING Planning is the first step of project management philosophy of planning, organizing and controlling the execution of the projects. Project planning and project scheduling is two separate and distinct function of the project management. At its inception a building normally begins as an inspiration or an idea in someone’s mind. Once the person or the client has a clear concept of what he/she wants, in order for that thought to be a reality, the idea must then be converted into a Construction Project. All Construction Projects have 4 major phases: Initiation, Planning & Design, Implementation and Completion. TYPES: There are several types of project planning. The three major types of construction project planning are: 1. Strategic planning: This involves the high-level selection of the project objectives and it is done by the owner’s corporate planners, 2. Operational planning: This involves the detailed planning required to meet the strategic objectives and it is done by construction teams. They ask certain questions before making operational plan for the project, 3. Scheduling: This puts the detailed operational plan on a time scale set by the strategic objectives.
iii
DESIGN Design is the creation of a plan or convention for the construction of an object or a system so designing can be done manually or by software such as Etabs and Staadpro. The components to design are foundation, plinth beam, columns, beams, slab and staircases. COMMERCIAL BUILDING A commercial
building is
a building that
is
used
for
commercial use. Such as office buildings, banks, hotels, super market, shopping mall, restaurant, warehouses, or retail In our project, a G+13 Structure with flat slab is analyzed and designed individually for Gravity loads Lateral loads. The complete process of Modeling, Analysis of whole structure is carried by using ETABS Packages and the designs of typical structural elements (beam, column, and slab) are done by manually. Punching shear reinforcement is an efficient method to increase not only the strength but also the deformation capacity of flat slabs supported by columns. Especially, the increase in deformation capacity is desired so that the load can be distributed to other supports preventing a total collapse of the structure in the case of the occurrence of a local failure
iv
NOTATIONS A
area
b
breadth of beam ,shorter dimension of a rectangular column
bef
effective width of slab
br
effective width of flange
bw
breadth of web or rib
D
overall depth of beam or slab or diameter of column, dimension of rectangular column in the direction under consideration
Df
thickness of flange/slab in flanged beam
d
effective depth of beam or slab
d1
depth of compression reinforcement from the highly
compressed face DL
dead load
e
eccentricity
fck
characteristic compressive strength of concrete
fy
characteristic strength of steel
ld
development length
LL L
live load or imposed load length of a column or beam between lateral restraints or the
unsupported length of a
column
lef
effective span of beam or slab or effective length of column
lx
breadth or shorter side of slab
ly
length or longer side of slab
lo
distance between points of zero moments in a beam
M
bending moment
Mu
factored bending moment
P
axial load on a compressive member or a pull
Pu
factored load
s
spacing of stirrups, standard deviation
T
tensional moment
V
shear force v
W
total load
w
distributed load unit area
wd
distributed dead load per unit area
x
depth of neutral axis
Z
modulus section
z
lever arm
σsc
permissible stress in steel in compression
σst
permissible stress in steel in tension permissible
τ
tensile stress in shear reinforcement shear stress
τbd
design bond stress
τc
shear stress in concrete
τv
nominal shear stress
Ac
area of concrete
As
area of minimum reinforcement or longitudinal tension
reinforcement for columns Asc
area of compression reinforcement or area of longitudinal
reinforcement for columns Ast
area of tension reinforcement
Asv
total cross sectional area of stirrup legs or bent- up bars within a
distance equal to Spacing of stirrup or bent- up bar C1
constant
G
going of stairs
hi
height measured from base of building to any floor1
I
moment of inertia
IS
Indian standards
KNm Kilo Newton meter L
span of beam or slab
w/c
water- cement ratio
αx,αy bending moment coefficient for two –way slabs α
tensile stress
τve
equivalent nominal shear stress vi
LIST OF FIGURES Figure 1 : Cellar Plan.............................................................................24 Figure 2 : Ground Floor Plan..................................................................25 Figure 3 : Mazzanine Floor Plan............................................................26 Figure 4 : Typical Floor Plan..................................................................27 Figure 5 : Terrace Plan..........................................................................28 Figure 6 : Section at X-X........................................................................29 Figure 7 : Section at Y-Y.........................................................................30 Figure 8 : Elevations.............................................................................31 Figure 9: Slabs with columns..................................................................3 Figure 10 : Column Positions.................................................................22 Figure 11 : Beams Location...................................................................23 Figure 12 : flat slab representation.......................................................46 Figure 13 : flat slab sizes......................................................................46 Figure 14 : Beams representation.........................................................53 Figure 15 : Beams sizes........................................................................54 Figure 16 : Beams Area.........................................................................55 Figure 17 : B.M Diagram.......................................................................56 Figure 18 : Area B39.............................................................................57
vii
Figure 19 : Steel Detail - B39................................................................59 Figure 20 : Area B31.............................................................................60 Figure 21 : Steel Detail - B31................................................................62 Figure 22 : Area B32.............................................................................63 Figure 23 : Steel detail - B32................................................................65 Figure 24 : Center line Diagram............................................................66 Figure 25 : Columns representation......................................................67 Figure 26 : Columns B.M.......................................................................68 Figure 27 : Area C25.............................................................................69 Figure 28 : Steel Detail - C5, C12 & C25...............................................73
viii
LIST OF TABLES Table 1 : Density of Materials Used......................................................33 Table 2 : Design Constants...................................................................33
ix
CHAPTER – 1
1
INTRODUCTION 1.1 General Now days, there is an increase in housing requirement with increased population and urbanization. Therefore, building sector has gained increasing prominence. However, the fact that the suitable lands for building/construction- especially in the areas in which people live intensively- are limited and expensive shows that there is a necessity
for
optimal
evaluation
of
these
lands.
Additionally,
continuously increasing prices leads to increase in building costs; so, both dimensional and cost optimization becomes necessary and even indispensable. When a building is projected, geometrical dimensions of elements belonging to carrier system of the structure are usually determined by using engineering capability and experiences gained over time. In sizing, the tensile forces to which the material to be subjected to should comply with the specifications. In the building design, the pre-sizing details provided
are generally not changed
much; sizes obtained in second or – at most third solution are taken as carrier system sizes. In fact, carrier system can be sized in infinite possibilities in a manner to ensure all the necessary conditions; and the cost of each carrier system alternative can be different from each other. The basic aim in the engineering is to find a design having lowest cost, and ensuring predicted limitations. 1.2 Flat Slab Flat slabs system of construction is one in which the beams used in the conventional methods of constructions are done away with. The slab directly rests on the column and load from the slab is directly transferred to the columns and then to the foundation. To support heavy loads the thickness of slab near the support with the column is increased and these are called drops, or columns are generally provided with enlarged heads called column heads or capitals. 2
Absence of beam gives a plain ceiling, thus giving better architectural appearance and also less vulnerability in case of fire than in usual cases where beams are used. 1.2.1 Basic Definition of Flat Slab In general normal frame construction utilizes columns, slabs & Beams. However it may be possible to undertake construction without providing beams, in such a case the frame system would consist of slab and column without beams. These types of Slabs are called flat slab, since their behavior resembles the bending of flat plates. A reinforced concrete slab supported directly by concrete columns without the use of beams
Figure 1: Slabs with columns 1.3 Components of Flat Slabs a. Drops: To resist the punching shear which is predominant at the contact of slab and column Support, the drop dimension should not be less than one -third of panel length in t hat direction. b. Column
Heads:
Certain
amount
of
negative
moment
is
transferred from the slab to the column at the support. To resist this negative moment the area at the support needs to be increased .this is facilitated by providing column capital/heads. 3
Flat slabs are appropriate for most floor situations and also for irregular column layouts, curved floor shapes, ramps etc. The benefits of choosing flat slabs include a minimum depth solution, speed of construction, flexibility in the plan layout (both in terms of the shape and column layout), a flat soffit (clean finishes and freedom of layout of services) and scope and space for the use of flying forms. The flexibility of flat slab construction can lead to high economy and yet allow the architect great freedom of form. Examples are; solid flat slab, solid flat slab with drop panel, solid flat slab with column head, coffered flat slab, coffered flat slab with solid panels, banded coffered flat slab.
Fig: 1.2 Solid Flat Slab
Fig: 1.3
Coffered Flat Slab
Fig: 1.4 Solid Flat slab with Drop Panels A flat slab is a flat section of concrete. These slabs are classically used in foundations, although they can also be used in the construction of roadways, paths, and other structures. Depending on the size and 4
complexity of a flat slab, it may need to be designed by an engineer who is familiar with the limitations and needs of slabs, or it may be possible for a handy do it yourself to make one in an afternoon for a simple project. Typically, a flat slab is made with reinforced concrete, in which rebar is criss-crossed in the forms to provide support and reinforcement once the concrete is poured and hardened. The slab design is designed to be reinforced in several directions so that it can withstand stresses such as shifting ground, earthquakes, frost, and so forth. Failure to fully reinforce a flat slab can cause it to crack or give along weak lines in the concrete, which will in turn cause instability. For some sites, a flat slab is poured in situ. In this case, the site is prepared, forms for the concrete are set up, and the reinforcing rebar or other materials are laid down. Then, the concrete is mixed, poured, and allowed to cure before moving on to the next stage of construction. The time required can vary considerably, with size being a major factor; the bigger the slab, the more complex reinforcement needs can get, which in turn adds to the amount of time required for set up. Once poured, the slab also has to be examined and tested to confirm that the pour was good, without air pockets or other problems which could contribute to a decline in quality. In other cases, a flat slab may be prefabricated off site and transported to a site when it is needed. This may be done when conditions at the site do not facilitate an easy pour, or when the conditions for the slab's construction need to be carefully controlled. Transportation of the slab can be a challenge if it is especially large. Barges, cranes, and flatbed trucks may be required to successfully move it from the fabrication site to the site of the installation. The flat slab foundation is not without problems. It can settle on uneven ground, allowing the structure to settle as well, for example, and during seismic activity, a slab foundation cannot hold up if the soils 5
are subject to liquefaction. A flat slab can also become a major source of energy inefficiency, as structures tend to lose heat through the concrete. 1.4 Advantages & Disadvantages 1.4.1 Advantages Advantages of flat-slab reinforced concrete structures are widely known but there are also known the disadvantages concerning their earthquake resistance. It is remarkable that both Greek codes, Reinforced Concrete Code and Seismic Code do not forbid the use of such
structural
systems
however
both
Codes
provide
specific
compliance criteria in order such structures to be acceptable. The advantages of these systems are:
The ease of the construction of formwork.
The ease of placement of flexural reinforcement.
The ease of casting concrete
The free space for water, air pipes, etc between slab and a possible furred ceiling.
The free placing of walls in ground plan.
The use of cost effective pressurizing methods for long spans in order to reduce slab thickness and deflections as also the time needed to remove the formwork.
The reduction of building height in multi-storey structures by saving one storey height in every six storey’s thanks to the elimination of the beam height.
These structural systems seem to attract global interest due to their advantages mainly in countries in which the seismicity is low. The 6
application of flat-slab structures is restrained due to the belief that such structures are susceptible to seismic actions. Moreover, it is known that in Central America, at the beginning of 1960’s, flat-slab structures displayed serious problems during earthquake actions. 1.4.2 Disadvantages: There are two main failure modes of flat slabs: a. Flexural Failure b. Punching Shear Failure Slabs are designed to fail by flexural failure, the failure mode is ductile therefore giving relatively large deflections under excessive loading and also cracks will appear on the bottom surface before failure occurs. These signs allow the problem to be addressed before failure occurs. Punching shear failure by comparison is a brittle failure mode when shear reinforcement is not added, meaning failure will occur before significant deflections take place, in addition to this any cracks that will develop before failure will propagate from the top surface. Since this surface is typically covered, it is unlikely that there will be sufficient warning
available
before
failure
occurs.
However, Thornsteinsson noted that it can be difficult to classify a failure mode to be an ideal representation of either flexural or punching shear failure and instead these modes often interact.
7
Punching shear failure in flat slabs
1.5 Key Messages: 1. For spans from 5 to 9 m, thin flat slabs are the preferred solution for the construction of in-situ concrete frame buildings where a square or near-square grid is used. For spans over 9 m post-tensioning should be considered. 2. Eliminating drops results in simpler false work and formwork arrangements, enabling rapid floor construction and giving maximum flexibility to the occupier. 3. The benefits associated with flat slab construction may well outweigh those of other structural solutions, which could be more complicated, time-consuming and ultimately more costly. 1.6 Proprietary Punching Shear Reinforcement Systems: Thin flat slab construction will almost certainly require punching shear reinforcement at columns. This has traditionally taken the form of a large number of individual shear links arranged on a series of 8
perimeters from the edge of the column. However, proprietary shear reinforcement systems are now available, which can greatly speed up the fixing process. These are described in a companion Best Practice Guide: Prefabricated punching shear reinforcement or reinforced concrete flat slabs. The savings in labour and time make these systems almost always worthwhile.
1.7 Objective The main objective of this study is to identify various parameters that affected the ANALYSIS AND DESIGN OF MULTI-STORY BUILDING FOR FLAT FLOOR SYSTEM USING ETABS. The ETABS stands for extended 3D (Three-Dimensional) Analysis of Building Systems. This is based on the stiffness matrix and finite element based software. The analysis and design is done to satisfy all the checks as per Indian standards.
1.8 Scope of Work ANALYSIS AND DESIGN OF MULTI-STORY BUILDING FOR FLAT FLOOR SYSTEM USING ETABS.The structure is analyzed for both gravity and lateral loads (seismic and wind load). The individual structural elements are designed for worst load combinations.
9
CHAPTER – 2
10
MODELLING AND ANALYSIS The analysis and design of RCC building was carried out using the software ETABS. It is the most popular structural engineering software product for 3D model generation, analysis and multi-material design. It has an intuitive, user-friendly GUI, visualization tools, powerful analysis and design facilities and seamless integration to several other modeling and design software products. For static or dynamic analysis of bridges, containment structures, embedded structures (tunnels and culverts), pipe racks, steel, concrete, aluminum or timber buildings, transmission towers, stadiums or any other simple or complex structure, has been the choice of design professionals around the world for their specific analysis needs. 2.1 Materials and Properties: 2.1.1 Building Materials: The required material properties like mass, weight density, modulus of elasticity, shear modulus and design values of the material used can be modified as per requirements or default values can be accepted. Beams and column members have been defined as ‘frame elements’ with the appropriate dimensions and reinforcement. Soil structure interaction has not been considered and the columns have been restrained at the base. The height of all the stories is 3m. The modulus of elasticity and shear modulus of concrete has been taken as E = 2.55 ×107 KN/m³ and G = 1.06 ×107 kN/m². 2.2 Loads: 2.2.1 Dead loads & Live loads: After having modeled the structural components, all possible load cases are assigned. In this study we are primarily concerned with observing the deformations, forces and moments induced in the structure due to dead, live loads and earthquake loads. The load case 11
‘Dead Load (DL) takes care of the self-weight of the frame members and the area sections. The wall loads have been defined under the case ‘Live load (LL)’. 1. Floor finish is assigned as 1 kN/m². 2. Live load is assigned as 2 kN/m² As per Table 8, Percentage of Imposed load to be considered in the Seismic weight calculation, IS 1893 (Part 1): 2002, since live load class is up to 5kN/m², 0.5% of imposing load has been considered.
2.3 Load Combination: The
structure
has
been
analyzed
for
load
combinations
considering all the previous loads in proper ratio. In the first case a combination of self-weight, dead load, live load and wind load was taken in to consideration. In the second combination case instead of wind load seismic load was taken into consideration. All the load cases are tested by taking load factors and analyzing the building in different load combination as per IS456 and analyzed the building for all the load combinations and results are taken and maximum load combination is selected for the design load factors as per IS456-2000. Select Define from menu bar, select load combinations. Then specify the following load combinations:
1.5(DL+LL)
1.5(DL+LL+WX)
1.5(DL+LL-WX)
1.5(DL+LL+EQX)
1.5(DL+LL-EQX) 12
1.2(DL+LL+WX)
1.2(DL+LL-WX)
1.2(DL+LL+EQX)
1.2(DL+LL-EQX)
DL+0.25LL
0.9DL+1.5LL
Method of analysis of statistically indeterminate portal frames: A. Method of flexibility coefficients. B. Slope displacements methods(iterative methods) C. Moment distribution method D. Kane’s method E. cantilever method F. Portal method G. Matrix method H. STAAD Pro A. Method of Flexibility Coefficients: The method of analysis is comprises reducing the hyper static structure to a determinate structure form by: Removing the redundant support (or) introducing adequate cuts (or) hinges. Limitations: 13
It is not applicable for degree of redundancy > 3 B. Slope displacement equations: It
is
advantageous
when
kinematics
indeterminacy
indeterminacy. This procedure was first formulated by axle bender in 1914 based on the applications of compatibility and equilibrium conditions. The method derives its name from the fact that support slopes and displacements
are
explicitly
comported.
Set
up
simultaneous
equations is formed the solution of these parameters and the joint moment in each element or computed from these value. Limitations: A solution of simultaneous equations makes methods tedious for manual computations. This method is not recommended for frames larger than too bays and two storeys. C. Iterative methods: These methods involve distributing the known fixed and moments of the structural member to adjacent members at the joints in order satisfy the conditions of compatibility. Limitations It presents some difficulties when applied to rigid frame especially when the frame is susceptible to side sway. The method cannot be applied to structures with intermediate hinges D. Kani’s method: This method over comes some of the disadvantages of hardy cross method. Kani’s approach is similar to H.C.M to that extent it also involves repeated distribution of moments at successive joints in 14
frames and continues beams. However there is a major difference in distribution process of two methods. H.C.M distributes only the total joint moment at any stage of iteration. The most significant feature of Kani’s method is that process of iteration is self corrective. Any error at any stage of iterations corrected in subsequent steps consequently skipping a few steps error at any stage of iteration is corrected in subsequent consequently skipping a few steps of iterations either by over sight of by intention does not lead to error in final end moments. Advantage It is used for side way of frames. Limitations: The rotational of columns of any storey should be function a single rotation value of same storey. The beams of storey should not undergo rotation when the column undergoes translation. That is the column should be parallel. Frames with intermediate hinges cannot be analysis.
APPLICABLE
15
NON-APPLICABLE
E. Approximate Method: Approximate analysis of hyper static structure provides a simple means of obtaining a quick solution for preliminary design. It makes some simplifying assumptions regarding structural behavior so to obtain a rapid solution to complex structures. The usual process comprises reducing the given indeterminate configuration to determine structural system by introducing adequate no of hinges. It is possible to sketch the deflected profile of the structure for the given loading and hence by locate the point inflection since each point of inflection corresponds to the location of zero moment In the structure. The inflection points can be visualized as hinges for the purpose of analysis. The solution of structure is sundered simple once the inflection points are located. The loading cases are arising in multistoried frames namely horizontal and vertical loading. The analysis carried out separately for those two cases. Horizontal cases:
16
The behavior of a structure subjected to horizontal forces depends upon its heights to width ratio among their factor. It is necessary to differentiate between low rise and high rise frames in this case. Low rise structures: Height < width It is characterized predominately by shear deformation. High rise buildings Height > width It is dominated by bending action F. Matrix analysis of frames: The individual elements of frames are oriented in different directions unlike those of continues beams so their analysis is more complex .never the less the rudimentary flexibility and stiffness methods are applied to frames stiffness method is more useful because its adaptability to computer programming stiffness method is used when degree of redundancy is greater than degree of freedom. However stiffness method is used degree of freedom is greater than degree of redundancy especially for computers.
17
CHAPTER-3
DESIGN PHILOSOPHIES 3.1 Methods of Design: Some of the popular design methods are: 3.1.1 Working Stress Method. 3.1.2 Ultimate Load Method. 3.1.3 Limit State Method. 3.1.1 Working Stress Method: This is also known as MODULAR RATIO METHOD, F.O.S. METHOD and ELASTIC METHOD.
18
In this method, analysis is based on the elastic theory assuming that both materials obey Hook’s Law. It is a traditional method which is used for the design of reinforced concrete design where it is assumed that concrete and steel act together and are perfectly elastic at all stages and relationship between the loads and stresses is linear upto the collapse of the structure. It is based on the criteria that the actual stresses developed in the material under the action of the working loads is limited to a set of allowable values. Thus, the sections are designed in such a way that the stresses are within the permissible limits. This leads to un-economical sections, as the method doesn’t utilize the full strength of the material resulting in heavier sections. Design Loads
= working /service loads.
Design Stresses
= characteristic values /F.O.S
F.O.S For concrete
= 3 ---- for bending 4 ---- for shear / compression
F.O.S for Steel
= 1.78 ---- for bending, shear & compression.
DEFECTS: It neither shows its real strength nor gives true factor of safety of structure against failure.
It results in larger % of compressive steels then limit state design. 3.1.2 Ultimate Load Method: This is also known as LOAD FACTOR METHOD. In this method, inelastic behavior of concrete is taken into account. At the failure the material tends to behave elastically, the strain increases many times beyond those in the elastic theory and 19
stress distribution adjusts itself to enable member to develop maximum capacity. In this method, service loads are proportioning the section to carry up to the ultimate strength of the material. Design Load
= Working load * load factor
Design Stress
= characteristic value / Load factor.
Load Factor = 1.5 ----- Concrete = 1.15 ----- Steel. DEFECTS: This method gives slender sections but larger deflections and larger cracks. Thus, in this method serviceability is not taken care off. 3.1.3 Limit State Design: It is an ideal method of design which takes into consideration not only
ultimate
strength
but
also
serviceability
and
durability
requirement. It includes merits of both elastic and ultimate theories. When a structure or apart of the structure becomes unfit for It is said to have reached its “LIMIT STATE”. This method is to provide an acceptable probability that the structure will not reach any limit state during its services life time. Design loads = working loads * P.S.F. Design stress = characteristic values/ P.S.F. P.S.F. depends on the load combinations as per cl. 36.4.1, IS-456-2000. It consists of: Limit state of collapse. Limit state of serviceability. 20
Limit state of collapse: It is the limit state on attainment of which the structure is likely to collapse. It relates to strength and stability of the structures. Design to this limit ensures safety of structure against collapse Limit state of collapse includes:
Bending
Shear
Compression
Torsion
Limit state of serviceability: It relates to performance and behavior of structure at working loads and is based on causes affecting serviceability of the structure. It concerns with cracking and deflection of the structure. Limit state of serviceability includes:
Deflections
Vibrations
Cracking
Durability
21
CHAPTER - 4
STRUCTURAL PLANNING Structural planning is first stage in any structural design. It involves the determination of appropriate form of structure, material to be used, the structural system, the layout of its components and the method of analysis. As the success of any engineering project measured in terms of safety and economy, the emphasis today is being more on economy.
22
Structural planning is the first step towards successful structural design. 4.1 Structural Planning Of Reinforced Concrete Framed Building: Structural planning of R.C framed building involves determination of 4.1.1 Column Positions Following are some of the guidelines principles for positioning of columns. Column should be preferably located at or near the corner of the building and at intersection of the walls, because the function of the column is to support beams which are normally placed under walls to support them. The columns, which are near to property line, can be exception from above consideration as the difficulties are encountered in providing footing for such columns. When center to center distance between the intersection of the walls is large or where there are no cross walls, the spacing between two columns is governed by limitations on spans of supported beams because spacing of column beside the span of the beams. As the span of the beam increase as the required depth increase and hence its self weight. On the other hand increase in total load is negligible in case of column due to increase in length. Therefore, columns are generally cheaper compared to beams on basis of unit cost. Therefore, large spans of beam should be avoided for economy reasons. 4.1.2 Orientation of columns: Column normally provided in the building are rectangular width of the column not less than the width of support for effective load transfer. As far as possible, the width of the column shall not exceed 23
the thickness of the walls to avoid the offsets. Restrictions on the width of the column necessitate the other side (the depth) of the column to be larger the desired load carrying capacity. This leads to the problems of orientation of columns.
Figure 2 : Column Positions 24
4.1.3 Beam Locati0ons Following are some of the guiding principles for the positioning of beams: a.
Beams shall, normally be provided under the walls and below a every concentrated load to avoid these loads directly coming on slabs. Basic principle in deciding the layout of a component member is that heavy loads should be transferred to the foundation along the shortest path.
b.
Since beams are primarily provided to support slabs, its spacing shall be decided by the maximum spans of slabs which decide the spacing of beams is governed by loading and limiting thickness.
The
maximum
practical
thickness
for
Residential/Office/Public building is 200mm, while minimum is 100mm.
25
Figure 3 : Beams Location
26
4.2 My Project Plans
Figure 4 : Cellar Plan 27
Figure 5 : Ground Floor Plan 28
Figure 6 : Mazzanine Floor Plan 29
Figure 7 : Typical Floor Plan 30
Figure 8 : Terrace Plan 31
Figure 9 : Section at X-X 32
Figure 10 : Section at Y-Y
33
Figure 11 : Elevations 34
35
36
CHAPTER - 5
37
LOADINGS 5.1 Load Conditions and Structural System Response: The concepts presented in this section provide an overview of building loads and their effect on the structural response of typical wood-framed homes. As shown in Table, building loads can be divided into types based on the orientation of the structural action or forces that
they
induce:
vertical
and
horizontal
(i.e.,
lateral)
Classifications of loads are described in the following sections: 5.1.1 Building Loads Categorized by Orientation: Types of loads on a hypothetical building are as follows. Vertical Loads Dead (gravity) Live (gravity) Snow(gravity) Wind(uplift on roof) Seismic and wind (overturning) Seismic( vertical ground motion) 5.1.2 Horizontal (Lateral) Loads: Direction of loads is horizontal w.r.t to the building. Wind Seismic(horizontal ground motion) Flood(static and dynamic hydraulic forces
38
loads.
Soil(active lateral pressure) Vertical Loads: Gravity loads act in the same direction as gravity (i.e., downward or vertically) and include dead, live, and snow loads. They are generally static in nature and usually considered a uniformly distributed or concentrated load. Thus, determining a gravity load on a beam or column is a relatively simple exercise that uses the concept of tributary areas to assign loads to structural elements, including the dead load (i.e., weight of the construction) and any applied loads (i.e., live load). For example, the tributary gravity load on a floor joist would include the uniform floor load (dead and live) applied to the area of floor supported by the individual joist. The structural designer then selects a standard beam or column model to analyze bearing connection forces (i.e., reactions) internal stresses (i.e., bending stresses, shear stresses, and axial stresses) and stability of the structural member or system a for beam equations. The selection of an appropriate analytic model is, however no trivial matter, especially if the structural system departs significantly from traditional engineering assumptions are particularly relevant to the structural systems that comprise many parts of a house, but to varying degrees. Wind uplift forces are generated by negative (suction) pressures acting in an outward direction from the surface of the roof in response to the aerodynamics of wind flowing over and around the building. As with gravity loads, the influence of wind up lift pressures on a structure or assembly (i.e., roof) are analyzed by using the concept of tributary areas and uniformly distributed loads. The major difference is that wind pressures act perpendicular to the building surface (not in the direction of gravity) and that pressures vary according to the size of the tributary area and its location on the building, particularly proximity to changes in geometry (e.g., eaves, corners, and ridges).Even though the wind loads are dynamic and highly variable, the design approach is based on a maximum static 39
load (i.e., pressure) equivalent. Vertical forces are also created by overturning reactions due to wind and seismic lateral loads acting on the overall building and its lateral force resisting systems, Earthquakes also produce vertical ground motions or accelerations which increase the effect of gravity loads. However, Vertical earthquake loads are usually considered to be implicitly addressed in the gravity load analysis of a light-frame building. Lateral Loads: The primary loads that produce lateral forces on buildings are attributable to forces associated with wind, seismic ground motion, floods, and soil. Wind and seismic lateral loads apply to the entire building. Lateral forces from wind are generated by positive wind pressures on the windward face of the building and by negative pressures on the leeward face of the building, creating a combined push and-pull effect. Seismic lateral forces are generated by a structure’s dynamic inertial response to cyclic ground movement. The magnitude of the seismic shear (i.e., lateral)load depends on the magnitude of the ground motion, the buildings mass, and the dynamic structural response characteristics (i.e. dampening, ductility, natural period of vibration ,etc) for houses and other similar low rise structures, a simplified seismic load analysis employs equivalent static forces based on fundamental Newtonian mechanics (F=ma) with somewhat subjective (i.e., experience-based) adjustments to account for inelastic, ductile response characteristics of various building systems. Flood loads are generally minimized by elevating the structure on a properly designed foundation or avoided by not building in a flood plain. Lateral loads from moving flood waters and static hydraulic pressure are substantial. Soil lateral loads apply specifically to foundation wall design, mainly as an “out-of-plane” bending load on the wall. Lateral loads also produce an overturning moment that must be offset by the dead load and connections of the building.
40
Therefore, overturning forces on connections designed to restrain components from rotating or the building from overturning must be considered. Since wind is capable of the generating simultaneous roof uplift and lateral loads, the uplift component of the wind load exacerbates
the overturning
tension
forces
due to
the lateral
component of the wind load. Conversely the dead load may be sufficient to offset the overturning and uplift forces as is the case in lower design wind conditions and in many seismic design conditions. 5.2 LOAD
CALCULATIONS
5.2.1 INTRODUCTION Loads and properties of materials constitute the basic parameter of a R.C structures. Both of them are basically of a varying nature .For such a quality of varying nature, it is necessary to arrive of a single representative value. Such value is known as characteristic value. The value to be taken in design which provides appropriate or designed margin of safety is known as design values. The loads are taken as per IS-875 and the material properties like characteristic value are taken from IS-456. Types of Loads The various types of loads acting on the structure which needs consideration in building design as follows:
Dead loads
Live loads
Wind loads
Seismic loads
Dead Load (As per IS 875 part 1)
41
This load is due to its own self weight or other structural element present on it. This includes a) self weight
b) weight of finishes c)
weight of partition, walls etc Dead loads shall be calculated on the basis of unit weights of materials. Calculation of DL on beams Self weight of beams
= 0.23 x 0.3 x 25 = 1.725 kN/m
Weights due to walls on beam
= (3 x 0.23 x 20) = 13.8 kN/m
Total
= 15.525 kN/m
Amount of distributed load coming from slab either in the form of triangular load or Trapezoidal load = {w Lx ( 3 – (Lx / Ly)
2
}/6
or { w Lx / 3 }
And loads from cantilever slabs ie = w Lx Here w = self wt of slab, Lx = shorter dimension, Ly= longer dimension of slab panel Calculation of DL on slab Self weight of the slab = 0.15 x 25
= 3.75 kN/m²
Floor finish on the slab
= 1.5 kN/m²
Water proofing load
= 2.0 kN/m²
Total
=
Live Load (As per IS 875 part 2) Live Load on beams:-
42
7.25 kN/m²
This is the live loads of slab which comes on beams in form of triangular or trapezoidal variation. Live load on slab:This are assumed to be On floors for residential buildings
= 3.0 kN/sq.m
On terrace
=1.5 kN/sq.m
On terrace
=3.0 kN/sq.m
Wind Load (As per IS 875 part 3) These are the lateral loads which are caused by the wind normal to the structure. Calculation Design wind speed = Vb x k1 x k2 x k3
(as per clause
5.3) Here, Vb = basic wind speed m/sec k1 = probability risk factor (from table 1) k2 = depends terrain height and structure height factor (from table2) k3 = depends on topography (clause 5.3.3.1) Basic wind speed for Hyderabad region = 44 m/sec (clause 5.2) For all general buildings k1 = 1 As per clause 5.3.2.1 our structure fall under terrain category 3 and class B
43
So,
k2 = for 0 m = 0 (GL) For 9 m = 0.98 For 18 m = 1.038 For 24 m = 1.07 For 30 m = 1.1 For horizontal topography
k3 = 1
After calculation we get values of Vz at different heights Vz = 43.12 m/sec at 9 m 45.67 m/sec at 18m 47.08 m/sec at 24m 48.40 m/sec at 30m Pressure intensity = 0.6 Vz ²
(as per clause 5.4)
So we calculated the pressure intensity at different heights of our structure from the above relation …. Pressure = 1.11 kN / m²
at 9 m
1.25 kN / m²
at 18m
1.32 kN / m²
at 24m
1.40 kN / m²
at 30m
We calculated the c/c of the column in both the direction Pressure intensity x c/c distance = load per unit length acting on different heights 44
This udl is applied on the structure normal to the columns with different magnitudes obtained at different heights ….. Earthquake Loads (As per IS 1893-2002) This force is defined as product of mass and accelerations. During earthquake, the mass is imparted by the building whereas the acceleration is imparted by the ground motion. Base shear = ah* w ah = (Z/2) * (I/R) * (Sg/g) Where
ah = Design horizontal acceleration spectrum value Z = Zone factor for MCE conditions R = Response reduction factor Sg = Spectrum acceleration depending upon period of
vibration and damping g = Acceleration due to gravity. Z= zone factor for Hyderabad
zone 2 = 0.1
Importance factor I = 1 for our structure `
Response reduction factor R = 3 Tx= (0.09 x h/ (d^0.5))
Where,
Tz = (0.09 x h / (d^0.5))
Tx and Tz are time periods (As per IS 1893 clause 7.6.2) h is height of structure d is dimension of structure in that direction
After calculation we get
Tx= 0.472 sec 45
Ty= 0.744 sec
We get the value of Sa/g from graph based on time periods (fig2 pg 16) We get value as 2.5 Finally ah value can be calculated ah = 0.0417 W= seismic weight The seismic weight includes the dead weight of the building and reduced live load on the building For calculating the design seismic forces of the structure, the imposed loads on roof need
not be considered (As per clause 7.3.2 pg 17)
Damping ratio was taken as 5% 5.3 Assumptions 1. Using partial safety factor for loads in accordance with clause 36.4 of IS-456 2000 as ϒt=1.5 2. Partial safety factor for material in accordance with clause 36.4.2 is IS-456-2000 is taken as 1.5 for concrete and 1.15 for steel. 3. Using partial safety factors in accordance with clause 36.4 of IS-4562000 combination of load. D.L+L.L. D.L+L.L+W.L Table 1 : Density of Materials Used MATERIAL
DENSITY 24.0KN/m3
i) Plain concrete
25.0KN/m3
ii) Reinforced 46
iii) Flooring material(c.m)
1.0KN/m2
iv) Brick masonry
19.0KN/m3
LIVE LOADS: (In accordance with IS.875) i) Live load on slabs
2.0KN/m2
ii) Live load on passage
4.0KN/m2
iii)Live load on stairs
4.0KN/m2
Using M25 and Fe 415 grade of concrete and steel for beams, slabs, footings, columns. Table 2 : Design Constants Therefore: fck
Characteristic strength for M25-25N/mm2
fy
Characteristic strength of steel-415N/mm2
5.4 Assumptions Regarding Design: 1
Slab is assumed to be continuous over interior support and partially fixed on edges, due to monolithic construction and due to construction of walls over it.
2
Beams are assumed to be continuous over interior support and they frame in to the column at ends.
5.4.1 Assumptions on design: 1.
M25 grade is used in designing unless specified.
2.
Tor steel Fe 415 is used for the main reinforcement.
3.
Tor steel Fe 415 and steel is used for the distribution reinforcement.
4.
Mild steel Fe 250 is used for shear reinforcement. 47
5.5 Data Collection The building models are 15 storey’s located in zone II. Tables 4.0 and Table 4.2present a summary of the building parameters. Table 3: General data collection and condition assessment of building
Sl.No .
Description
Information
Remarks
Building height 1
a) 11-storey
Including the
33 m
foundation level
2
Number
of
basements
below ground
----
0
3
Open ground storey
Yes
----
4
Special hazards
None
----
Regular/Irregul 5
Type of building
ar
Space
frames 6
Horizontal floor system
7
Software used
Beams slabs Etabs2013
48
and
IS 1893:2002 Clause 7.1
-------
CHAPTER-6
49
DESIGNING 6.1 Flat Slab
Figure 12 : flat slab representation
50
Figure 13 : flat slab sizes Given data: Interior panel = 12.11 x 10.21 m Live load
= 4 KN/ m2
Floor finished load = 1 KN/ mm2 fck = 25 N/mm2 fy = 415 N/mm2 Column size = 750 x 750 mm Thickness of slab: Thickness of slab = 40, if mild steel = 32, if fy415 or fy500 Thickness of slab, d = span/32 51
= 12116/32 d = 378.62 mm ~ 380 mm Take cover as 40mm D = 380 + 40 = 420 mm D = d + 40 D= 380 + 40= 420 mm Drop = 1/3 x span = 1/3 x 12.116 =4m Provide drop of 4 m x 4 m Provide a drop of 150 mm thick Total thickness = slab + drop D = 420 + 150 = 570 mm Self weight of slab = 0.57 x 25 = 14.25 KN/m2 Floor finished load = 1 KN/m2 Live load = 4 KN/m2 Total load = 14.25 + 1 + 4 = 19.25 KN/m2 Design factored load, Wu = 1.5 x 19.25 = 28.87 KN/m2 52
Clear span = 12.116 - 0.75 = 11.366 m Design load, Wo = Wu x le x lx = 28.87 x 12.11 x 11.36 Wo = 3975.7 KN Design total moment, Mo = Wlx / 8 = (3975.7 x 11.36) / 8 = 5648.47 KNm Negative design moment = 0.65 x Mo Positive design moment = 0.35 x Mo Total negative moment = 0.65 x 5648.47 = 3671.51 KNm Total positive moment = 0.35 x 5648.47 = 1976.96 KNm Width of column strip = width of middle strip = 4000 mm = 4 m
Column strip Negative moment
0.75 x 3671.51 = 2753 0.25 KNm
Positive moment
Middle strip
0.60
x
3671.51
=
917.87 KNm x
1976.96
1186.18 KNm
= 0.4x1976.96 = 790.784 KNm
53
Mulimit = 0.138fckbd2 = 0.138 x 25 x 10211 x 5302 = 9895.5 x 106 Nmm = 9895.5 KNm Mulimit = 9895.5 KNm Mu = 5648.47 KNm Mulimit > Mu Hence thickness is safe and sufficient. Check for shear: Critical section is at distance d/2 = 530/2 = 265 mm It is square in size = column size + 265 + 265 = 750 + 265 + 265 = 1280 mm V = Total load – Wo x 0.810 x 0.810 = (28.87x12.11x10.21) – (28.87 x 1.28 x 1.28) = 3524.4 KN Nominal shear, v = 3524.4 x (1000/1280 x 4 x 530) = 1.298 N/mm2 Shear strength = Ksc 54
Ks = 0.5 + c c = L1/L2 = 12.11/10.21 = 1.19 Ks = 0.5 + 1.19 =1.7 c = 0.25fck = 0.2525 =1.25 N/mm2 v > c Hence slab is not safe in shear. Reinforcement: For negative moment in column strip: Mu=2753.6KNm d = 530mm Mu = 0.87fyAstd [1-Astfy/bdfck] 2753.6 x 106 = 0.87 x 415 x Ast x 530 [1 – (Ast x 415) / (10211 x 530 x 25)] = 191356.5 Ast [1 - Ast / 135.3x106] Ast = 14389.9 mm2 Width = 10.211 mm Using 16mm bar spacing requirement is 55
S = (/4 x 162 / 14390) x 10211 = 142.7 mm Provide 16mm bars at 200 mm C/C. For positive moment column strip: Mu=1186.18KNmm d = 380 mm Mu = 0.87fyAstd [1-Astfy/bdfck] 1186.18x106 = 0.87 x 415 x Ast x 380 [1-Astx415 / 10211x380x25] = 137199 Ast [1 – 415Ast/97x106] Ast = 8991.6 mm2 Using 16mm bar spacing requirement is S = (/4x162 / 8991.6) x 10211 =288.3 mm Provide 16mm bars at 250 mm C/C. For negative moment in middle strip: Mu=917.87KNm d = 380 mm Mu = 0.87fyAstd [1-Astfy/bdfck] 917.87x106 = 0.87 x 415 x Ast x 380 [1 - Astx415 / 10211x380x25] = 137199 Ast [1 - 415Ast / 97x106] Ast = 6893.37 mm2 56
Width = 10211 mm Using 16mm bar spacing requirement is S = (/4 x 162 / 6893.37) x 10211 = 297.8 mm Provide 16mm bars at 300 mm C/C. For positive moment middle strip: Mu=790.784x106KNmm d = 380 mm Mu = 0.87fyAstd [1-Astfy/bdfck] 790.78x106 = 0.87 x 415 x Ast x 380 [1 - Astx415 / 10211x380x25] = 137199Ast [1 – 20.75Ast / 97x106] Ast = 5913.4 mm2 Using 16mm bar spacing requirement is S = (/4 x 162 / 5913.4) x 10211 = 347.2 mm Provide 16mm bars at 400 mm C/C.
57
6.2 Design of Beams
Figure 14 : Beams representation
58
Figure 15 : Beams sizes
59
Figure 16 : Beams Area 60
Figure 17 : B.M Diagram 61
6.2.1 Members - B31, B32 & B39 (from ETABS) B39:
Figure 18 : Area B39 Size of beam Effective cover
= 230 x 650 mm. = 25 mm.
Effective depth (d) = 650-25 = 625 mm. Using M25 grade concrete and Fe 415 grade steel. Required area of beam is shown in fig. below 3252 3487
62
╠ ╣ 2468 Required area of steel (top left of beam); Ast = 3252 mm2
(from Etabs) Provided area of steel; ast
Provide 25 mm dia. Bar;
ast =
П x 252 / 4
= 491 mm2 Number of bars in a beam
=
Ast / a st
230 = 3252 / 491 =6.6≈7No.s 650 Hence provide 3 no.s of main bars and 4 no.s of extra bars at sides.
Required area of steel (whole bottom of beam); Ast = 2468 mm2 (from Etabs) Provided area of steel; ast
63
Provide
25 mm dia. Bar;
П x 25 2
a st =
/
4
230 = 491 mm2 Number of bars in a beam
=
Ast / a st
650 = 2468 / 491 = 5 No.s Hence provide 3 no.s of main bars and 2 no.s of extra bars at sides. Required area of steel (top right of beam); Ast = 3487 mm2
(from Etabs) Provided area of steel; ast Provide
25 mm dia. Bar;
a st = П x 252
/
4
230 = 491 mm2 Number of bars in a beam
=
Ast / a st
650 = 3487 / 491 = 7 No.s Hence provide 3 no.s of main bars and 4 no.s of extra bars at sides.
64
Figure 19 : Steel Detail - B39 B31:
65
Figure 20 : Area B31 Size of beam Effective cover
= 230 x 450 mm. = 25 mm.
Effective depth (d) = 450-25 = 425 mm. Using M25 grade concrete and Fe 415 grade steel. Required area of beam is shown in fig. below 248 397 ╠ ╣ 66
261
Required area of steel (top left of beam); Ast = 248 mm2
(from
Etabs) Provided area of steel; ast ast = П x 122 / 4
Provide 12 mm dia. Bar;
= 113.1 mm2 Number of bars in a beam
=
Ast / a st
230 = 248 / 113.1 =2.2≈
3
No.s
450 Hence provide 2 no.s of main bars and 1 no.s of extra bars.
Required area of steel (whole bottom of beam); Ast = 261 mm2 (from Etabs) Provided area of steel; ast Provide
12 mm dia. Bar;
a st =
230 = 113.1 mm2
450 67
П x 122
/
4
Number of bars in a beam
=
Ast / ast
= 261 / 113.1 = 2.3 ≈ 3 No’s Hence provide 2 no.s of main bars and 1 no.s of extra bars.
Required area of steel (top right of beam); Ast = 397 mm2
(from Etabs) Provided area of steel; ast Provide 12 mm dia. Bar;
ast = П x 122 / 4 = 113.1 mm2
Number
of
bars
in
a
beam
=
Ast
/
a st
230 = 397 / 113.1 =3.5 ≈ 4 No.s 450 Hence provide 2 no.s of main bars and 2 no.s of extra bars.
68
Figure 21 : Steel Detail - B31 B32:
69
Figure 22 : Area B32 Size of beam Effective cover
= 230 x 450 mm. = 25 mm.
Effective depth (d) = 450-25 = 425 mm. Using M25 grade concrete and Fe 415 grade steel. Required area of beam is shown in fig. below 377 248 ╠ ╣
70
248 Required area of steel (top left of beam); Ast = 377 mm2
(from
Etabs) Provided area of steel; ast Provide 12 mm dia. Bar;
ast = П x 122 / 4 = 113.1 mm2
Number
of
bars
in
a
beam
=
Ast
/
a st
230 = 377 / 113.1 =3.33≈4No.s 450 Hence provide 2 no.s of main bars and 2 no.s of extra bars.
Required area of steel (whole bottom of beam); Ast = 248 mm2 (from Etabs) Provided area of steel; ast Provide
12 mm dia. Bar;
ast =
230 = 113.1 mm2
450
71
П x 122 /
4
Number of bars in a beam
=
Ast / ast
= 248 / 113.1 = 2.2 ≈ 3 No.s Hence provide 2 no.s of main bars and 1 no.s of extra bars. Required area of steel (top right of beam); Ast = 248 mm2
(from Etabs) Provided area of steel; ast Provide
12 mm dia. Bar;
П x 122
a st =
/
4
230 = 113.1 mm2 Number of bars in a beam
=
450 = 248 / 113.1 = 3.2 ≈ 3 No.s Hence provide 2 no.s of main bars and 1 no.s of extra bars.
72
Ast / a st
Figure 23 : Steel detail - B32
73
6.3 Column
Figure 24 : Center line Diagram
74
Figure 25 : Columns representation
75
Figure 26 : Columns B.M Column name: C25 76
Figure 27 : Area C25 Given data: Column size
= 381 x 229 mm
M25 grade of concrete Fe 415 grade of steel Reinforcement is distributed equally on all four sides Factored axial load, Pu
= 17.97 kN. (From Etabs)
Factored moment about X-axis, Mux = 10.69 kN-m. (From ETABS) Factored moment about Y-axis, Muy = ETABS) 77
4.99
kN-m.
(from
; Ast = 697 mm2
Area of steel Effective length:
Assuming effectively held in position and restrained against rotation at one end , and at the other restrained against rotation but not held in position. Unsupported length of column (L)
= 3000 mm
D
= 229mm
3000 / 229 = 13 Slenderness ratio, Le/b = >12 Therefore, it is designed as long column Provide 12 mm dia. Bar;
ast = П x 122 / 4 = 113.1 mm2
Number
of
bars
in
a
column
=
Ast
/
a st
229 = 697 / 113.1 =6.16 ≈ 6 No.s 381 Hence provide 3 no.s of bars at one side and 3 no.s of bars at other side. Column name: C12 Colum
= 762 x 762 mm 78
M25 grade of concrete Fe 415 grade of steel Reinforcement is distributed equally on all four sides Factored axial load, Pu
= 2863.98 kN. (from Etabs)
Factored moment about X-axis, Mux = 80.66 kN-m. (from ETABS) Factored moment about Y-axis, Muy = 14.02 kN-m. (from ETABS) ; Ast = 5161 mm2
Area of steel Effective length:
Assuming effectively held in position and restrained against rotation at one end , and at the other restrained against rotation but not held in position. Unsupported length of column (L)
= 3000 mm
D
= 762 mm
3000 / 762 = 3.9 Slenderness ratio, Le/b = <12 Therefore, it is designed as short column Provide 25 mm dia. Bar;
ast = П x 252 / 4 = 490.8 mm2
Number
of
bars
in
a
column
762 79
=
Ast
/
a st
= 5161 / 490.8 =10.5≈12 No.s 762 Hence provide 6 no.s of bars at one side and 6 no.s of bars at other side. Column name : C5 Colum
= 762 x 762 mm
M25 grade of concrete Fe 415 grade of steel Reinforcement is distributed equally on all four sides Factored axial load, Pu
= 7700.45 kN. (From Etabs)
Factored moment about X-axis, Mux =
75.6
kN-m.
(From
ETABS) Factored moment about Y-axis, Muy = 55.47 kN-m. (From ETABS) ; Ast = 5161 mm2
Area of steel Effective length:
Assuming effectively held in position and restrained against rotation at one end , and at the other restrained against rotation but not held in position. Unsupported length of column (L)
= 3000 mm
D
= 762 mm 80
3000 / 762 = 3.9 Slenderness ratio, Le/b = < 12 Therefore, it is designed as short column Provide 25 mm dia. Bar;
ast = П x 252 / 4 = 490.8 mm2
Number
of
bars
in
a
column
=
Ast
/
a st
762 = 5161 / 490.8 =10.5≈12No.s 762 Hence provide 6 no.s of bars at one side and 6 no.s of bars at other side.
81
Figure 28 : Steel Detail - C5, C12 & C25
82
6.5 Footing Footing no-3 1) Type of footing
= Square footing
Size of the column
= 750 x 750 mm
2) Load on footings: Load on footing (from ETABS.) (P) = 2863.98 KN Moment arrived from ETABS analysis Self-weight of footing
= 1.588 kN-m
= 10% of axial load. = 286.4 KN
Total load transmitted to the soil
= 3150.38 kN = 300kN/m2
S.B.C of soil 3) Size of footing: Area of footing (A)
= Total load/SBC of soil = 3150.38 / 300 = 10.5 m2
Design of rectangular footing B / L = b / a =750 / 750 = 1 A=LxB 10.5 = L x L L2 = 10.5 L = 3.24 83
B = 3.24 x 1 = 3.24 m A = 3.5 x 3.5 = 12.25 mm2 Net upward soil pressure Foundation
= 1.5 x 2863.98 / (3.5 x 3.5) = 350.7 KN/m2
Therefore footing design for a maximum pressure of 350kN/m 2 4) Determination of minimum depth required from B.M: Mx-x
=
Qu(L-a)2/8L
= My-y
=
[350x(1.37)2/2] 3.5
=
1035 kN-m 1035 kN-m
Depth of footing: Mu 1035 x 106 d
0.138 fckBd2
= = =
0.138 x 25 x 3500 x d2 293 mm
Therefor provide an effective depth of footing shear consideration up to two times as 500 mm Assume effective cover
= 50 mm
Overall depth
= 500 + 50 = 550 mm
5) Check for one way shear: Critical section for one way shear is‘d’ from face of the column. 84
Shear Force Vu = Qu (l - d) x B Shear Force Vu = 350 (1.37 - .5) x 3.5 Shear Force Vu = 1065 kN. Nominal Shear stress
v = 1065/ (3.5x.5) = 0.6 N/mm2
From IS: 456 Table 19, Design shear strength for pt(= 1%) is 0.64N/mm2
v< c
Hence safe.
6) Check for two-way shear: Shear Force Vu = Pu [B2-(b+d) 2] = 350 [(3.5)2 - (.75+0.5)] Shear Force Vu = 3850 kN.
v v
= v
Vu 4[(b d )]d
= (3850 x 103 ) / 4(750+500)x500
v = 1.54N / mm2 c = 0.25
fck
x Ks
c = 0.25√25 x 1.5 v< c
= 1.8
Hence safe.
7) Area Of Reinforcement Along X and Y Direction: 85
c
=
d = 550 -16 / 2 - 50 = 492mm Mu = .87fy Ast [d- (fy Ast / bfck )] 1035 x 103 = 0.87 x 415 x Ast [500 – (415 Ast / 3500 x 25)] Ast = 5453.23 mm2 Provide 16mm ϕ bars, area of single bar = 201.06 mm2 Spacing of bars = (201.6 / 5453.23) x 3500= 129.4 mm Provide 16mm ϕ bars at a spacing of 130 mm c/c.
6.6 Stair Case Design of Flight Slab: No of flights for each floor = 2 Height of the floor = 3.05 m Height of each flight = 3.05 / 2 = 1.53m Assume Raise as 150mm and Tread as 300mm No of Raisers = 1.53 / 0.15 = 10.2 Say 10 Say Raise = 155mm No of Treads = 10-1 = 9 Treads Width of Stair = 1.68m 86
1.68m 1.68m
2.64m
2.81m
Effective Span = 5.22m Span / Overall depth = 20 for deflection criteria Let modification factor = 5220 / (25) = 208mm Say 200mm Effective depth = 200-25-5 = 170mm 1.
Loads:
Per meter width of stair case Dead load of slab (On slope) = 0.2 x 1.00 x 25 = 5 KN/m Finished load (On slope)
= 1.50 KN/m Total load =
6.5 KN/m
Dead load of slab / horizontal meter run 0.302 + 0.152 = 6.5
0.3
=
KN /m Dead load of one step
= ½ x 0.15 x 0.3 x 25 = 0.562KN
Dead load of steps / horizontal meter run = 0.56/0.3 = 1.87 KN/m Live load
3 KN / m2
= 3.00 KN/m 87
3.9
Total working load Factored Load
= 8.85 KN/m = 1.50 x 8.85
= 13.27 KN/m
2. Bending Moment: Maxi. BM = WL2 / 8 = (13.27 x 3.12) / 8 = 15.9 KN-m 3. Effective depth of slab (For max.BM): Mu lim = 0.138fckbd2 d =
( Mu lim ) / ( 0.138fckb )
15.9 x 106 d =
= 67.9 mm <
available depth 200mm 0.138 x 25 x 1000 Hence OK 4.Calculation of Area of steel: Main steel: MD = 15 KN-m
0.50 fCK Ast
=
4.6 x Mu 1-
1-
bd fCK bd2
fy 88
4.6 x 15.9 x 106
0.50 x 25 Ast
=
1-
1-
x 1000 x 170 25 x 1000 x 1702
415
= 266.09 mm2 > 204 mm2 (Min.Ast 0.12% of bD as per Clause 26.5.2.1
of IS 456-2000)
Provide 12 mm dia. Tor steel
ast = П x 122 / 4 = 113 mm2
ast Spacing of Steel =
X 1000 Ast
113 Spacing of Steel =
x 1000 = 424.6 mm 266.09
89
But maximum spacing of Tension Reinforcement as per Clause 26.3.3 b 1 from IS 456 – 2000 is 3d or 300 mm whichever is less, 1) 3d = 3 x 170 = 510 mm
or
2) 300 mm
Hence provide 12 mm dia. @ 300 mm C/C Revised Astxm = (1000 x 170) / 300 = 566.6 mm2
Distribution Steel: Area of steel = 0.12% of gross sectional area = (0.12 x 1000 x 170) / 100 = 204 mm Provide 8 mm dia. Tor steel
ast = П x 82 / 4 = 50.26 mm2
ast Spacing of Steel =
X 1000 Ast
50.26 Spacing of Steel =
x 1000 = 246.3 mm 204
90
But maximum spacing of Tension Reinforcement as per Clause 26.3.3 b 2 from IS 456 – 2000 is 5d or 450 mm whichever is less, 1) 5d= 5 x 170 = 850 mm
or
2) 450 mm
Hence provide 8 mm dia. @ 250 mm C/C
5. Check for Shear: Shear Force (Vu) = WL/2 = (13.27 x 3.1) / 2 = 20.56 KN Nominal shear stress ( V) = Vu / bd = (20.56 x 106) / (1000 x 170) = 0.12 N/mm2 Area of steel available Ast = 560mm2 Pt% = (100 x 560 / 1000 x 170) = 0.32% From Table 19 of IS 456 – 2000
c = 0.464 N/ mm2
c > V Hence shear stress is within limits 7) Check for Bond: Development length Ld for 12 mm ǿ = (ǿ x 0.87fy) / 4bd = (12 x 0.87 x 415) / 4 x 1.40 = 773.67mm From 26.2.1.1 of IS456 – 2000 bd = 1.40 for M25 grade 91
Tension bars crossing at bends should be extended by 773.67mm beyond crossing point. 8) Check for Serviceability: Basic l/d Ratio = 20 Pt = 0.32%
Service stress = (0.58 x 415 x 270)/ 560 =
116N/mm2 Modification factor = 1.20 Modified value of l/d ratio = 20 x 1.20 = 24 Actual l/d ratio = 3100/170 = 18 < 24 Actual l/d ratio is < modified value of l/d ratio Hence the thickness of the slab is safe.
92
CHAPTER-7
CONCLUSIONS Flat-slab building structures possesses major advantages over traditional slab-beam-column structures because of the free design of space,
shorter
construction
time,
architectural
–functional
and
economical aspects. Because of the absence of deep beams and shear walls, flat-slab structural system is significantly more flexible for lateral loads then traditional RC frame system and that make the system more vulnerable under seismic events. The purely flat-slab RC structural system is considerably more flexible for horizontal loads than the traditional RC frame structures which contributes to the increase of its vulnerability to seismic effects. The critical moment in design of these systems is the slab-column 93
connection, i.e., the penetration force in the slab at the connection, which
should
retain
its
bearing
capacity
even
at
maximal
displacements. The ductility of these structural systems is generally limited by the deformability capacity of the column-slab connection. To increase the bearing capacity of the flat-slab structure under horizontal loads, particularly when speaking about seismically prone areas and limitation of deformations, modifications of the system by adding structural elements are necessary.
REFERENCES Design of Reinforced Concrete Structures by A. K. Jain, Surya Prakash S. Krishna Murthy,
Design of R.C.C structural elements by S.S. Bhavikatti,
Design of R.C.C slabs by K.C.Jain,
Design of R.C.C structures by prof.N.Krishna Raju,
Design of R.C.C structures by prof.S.Ramamrutham,
94
The code books referred for this project are: 1. 2. SP 16 (design aids for IS 456), 3. IS 1893 (part-1) 2002 criteria for earthquake resistant design of structures Part-1 general provision & buildings, 4. IS 456:2000 for RCC design, 5. IS 875 – part I for weight and density of materials ( RCC = 25 kN/m3, PCC = 24 kN/m3, Brick = 18-20 kN/m3 ect.), 6. IS 875 – part II for live load, 7. IS 875 – part III for wind load and 8. IS 875 – part V for load combination.
95
