This project describes the method of Analysis and Design of a Multistoried building (G+5) located at Mothinagar, Hyderabad. The scope behind presenting this project is to learn the concept of construction, and to design a safe, economical and elegant structure. CHAPTER-1 INTRODUCTION GENERAL: Engineering term applied to the profession in which a knowledge of the mathematical and natural sciences gained by the experiment and practically is applied to the effective use of materials and the forces of nature .the term engineer properly denotes a person who has received professional training in pure applied science ,but it often used to desire the operator of an engineer ,marine locative engineer or a stationary engineer .In modern terminology ,these later occupations are known as craft or trades ,civil engineering is perhaps the broadest of engineering fields, for it deals with the creation, improvement, and protection of the communal environment, providing facilities for living. Industry and transportation, include large buildings, roads, bridges, canals rail road lines, airports, water supply systems, dams, irrigation, harbors, docks, aqueducts tunnels and other engineered constructions. The Civil Engineering must have a thorough knowledge of all types of surveying of the properties and mechanics of construction materials, mechanics of structures and soils and hydraulics and fluid mechanics. Before the middle of the 18 th century, large-scale construction work usually placed in the hands of military ANALYSIS AND DESIGN OF MULTI STORIED BUILDING 1
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This project describes the method of Analysis and Design of a Multistoried building (G+5) located at Mothinagar, Hyderabad. The scope behind presenting this project is to learn the concept of construction, and to design a safe, economical and elegant structure.
CHAPTER-1
INTRODUCTION
GENERAL:
Engineering term applied to the profession in which a knowledge of the mathematical and natural sciences gained by the experiment and practically is applied to the effective use of materials and the forces of nature .the term engineer properly denotes a person who has received professional training in pure applied science ,but it often used to desire the operator of an engineer ,marine locative engineer or a stationary engineer .In modern terminology ,these later occupations are known as craft or trades ,civil engineering is perhaps the broadest of engineering fields, for it deals with the creation, improvement, and protection of the communal environment, providing facilities for living. Industry and transportation, include large buildings, roads, bridges, canals rail road lines, airports, water supply systems, dams, irrigation, harbors, docks, aqueducts tunnels and other engineered constructions. The Civil Engineering must have a thorough knowledge of all types of surveying of the properties and mechanics of construction materials, mechanics of structures and soils and hydraulics and fluid mechanics.
Before the middle of the 18th century, large-scale construction work usually placed in the hands of military engineers. Military engineering involved such work as preparation of topographically maps, their location, design and construction of road and bridges and the building of forts and docks. In 18 th century, however, the term civil Engineering came in to use to describe engineering work that was performed by civilians for non-military purpose. Concrete, artificial engineering material made from a mixture of Portland, cement, water, fine and coarse aggregates and a small amount of air. It is the most widely used construction material in the world. Concrete is the only major building material that can be delivered to the job site in a plastic state. This unique quality makes concrete desirable as a building material because it can be molded to virtually any form or a shape. concrete provide wide latitude in surface texture and colors and can be used to construct a wide variety of structures such as highways and streets ,bridges , dams ,large buildings , airport runways ,irrigation structures ,break waters ,piers and docks, sides walks, soils and farm building homes and even barges and ship. Other desired qualities of concrete as a building materials are its strength, economy and durability .depending on the mixture of the material used, concrete will support in compression, 700 kg/sq cm, (10,000 or more 1b/square cm).The tensile strength of concrete is much lower ,but by using properly designed steel reinforcing .structural member can be made that are as strong as in tension as in
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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compression .The durability of concrete is evidenced by the fact that concrete columns built by the Egyptians more than 3600 years ago are still standing .
Concrete used in more construction works reinforcement is with steel .when concrete structural member must resist extreme tensile stress, steel supplied the necessary strength .steel is embedded twisted bars or I concrete in the form of a mesh or roughened. A bond forms with both compounds. This is assembly of concrete and steel is called as REINFORCED CEMENT CONCRETE.
Reinforced concrete has a remarkable capacity to adapt to the assumptions of the designer .This has been pointed out to number of engineers .Lugi Nervy, the renowned Italians architect engineers has stated it eloquently as follows:
Mainly because of plastic flow ,a concrete structure tries with admirable ductility to adapt itself to our calculation ,which don’t always represent the most logical and spontaneous answers to the request of the forces at play and even tries to correct our deficiencies and error sections and regions to highly stressed yield and channel some of their loads to other sections or regions, which accept this additional task with commendable spirit of collaboration with the limit of their own strength .
DESIGNING
Designing of structures is an art and science of designing a safe, durable and elegant structure with economy. This is not only requires imagination but also a good knowledge of science of designing besides practical aspects, like the relevant codes and local municipal byelaw experience and judgment.
The architect whereas the requirement of safety, serviceability, durability and economy are taken care of by the structural engineer looks after the design structure of the structure and aesthetics
STAGES IN STRUCTURAL DESIGN
1. Structure planning2. Estimation of loads3. Analysis of structure4. Design of members5. Drawings of preparation of schedules
STRUCTURAL PLANNING
The planning of structure involves the determination of the structure, the materials to be used, and the structural system, the layout of its components, the method of analysis the philosophy of the structure design.
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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The Principal elements of the R.C.C building frame are:
1. Slabs to cover large areas2. Beams to support slabs and walls3. Columns to support beams4. Footing to distribute concentrated column loads over a large area of the supporting soil.
The Structural planning of the building frame involves the determination of the following:
1. Column position2. Beam location3. Slab spacing4. Planning of stairs5. Type of footing
DESIGN METHODS
Structural Design for framed R.C.C structure can be done by three methods
a. Working stress methodb. Ultimate strength methodc. Limit state method
Working Stress Method of Design
It is the earliest modified method of R.C.C structure. In this method structural element is so designed that the stress resulting from the action of service load as computed in linear elastic theory using modular ration concept does not exceed a pre-designed allowable stress which is kept as some fraction of ultimate stress, to avail a margin of safety. Since this method does not utilize full strength of the material it results in heavy section, the economy aspect cannot be fully utilized in the method.
Ultimate Strength Method of Design
This method is primarily based on strength concept in the method the structural element is proportioned to with stand the ultimate load which is obtaining by exchanging the service load of some factor referred to as load factor for giving margin of safety .Since this method is based on actual stress strain behavior of the materials, of the member as well as structures that too right up to failure, the valve calculated by this method agrees well the experiment results.
ULTIMATE STATE METHOD DESIGN:
During the past several years ,extension research works have been carried out on the different aspects of research in the actual behavior of the member and structure have
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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led to the development of design and approach of LIMIT STATE METHOD OF DESIGN .
LIMIT STATE CONCEPT: In limit state method, the working load is multiplied by the partial factor of safety. In accordance with the clause of IS-456-2000; and also the ultimate stress of material is divided by the partial safety in accordance with clause of IS-456-2000.
Partial safety factor is introduced to reduced probability of failure to about zero.
When a structure or part of structure becomes unfit for use, it is said to have reach a limit state, unfitness for use can arise in various ways and aims of limit state method of design is to provide an acceptable probability that the structure will not reach any of limit state during its service life span .
Limit state can be broadly classified in to two main categories:
1. Limit state of collapseLimit state of serviceability
LIMIT STATE OF COLLAPSE:It is the limit state on attainment of which of the structure is likely to collapse. It relates to stability and strength of the structure design to the limit ensures safety of the structure from collapse.
LIMIT STATE OF SERVICIBILITY: It relates to performance orbehavior of the structure at working loads and is based on causes effecting serviceability of the structure.
Design Principle, Assumption and Notations Assumed:
The notations adopted through the work are same as in IS 456-2000 & chart from SP16:1980.
ASSUMPTIONS IN DESIGN:
1. Using partial safety factor for loads in accordance with clause of IS 456-2000 as SF- 15
2. Partial safety factor for material is accordance with clause of IS 456-2000 is taken as 1.5 for concrete & 1.15 for steel.
3. Using partial safety factor in accordance with clause IS 456-2000 combination ofload (D.L+ L.L = 1.5).
Fck = Characteristic strength for M20 -25 N/mm2Fy = Characteristic strength for steel 415 N/mm2
NOTE: Slabs assured to be continuous over interior support & partially fixed on edges, due to monolithic construction
Design Constants:
1. Concrete density = 25KN/m32.Grade of concrete for slab = M203. Grade of concrete for columns= M204. Grade of steel =Fe415
STATEMENT OF PROJECT
Salient Features:
1) Utility of building : RESIDENTIAL BUILDING
2) No. Of Stories : G+5
3) No. of Stair cases: 5
4) Type of Construction: R.C.C structures
5) Types of walls: Brick Wall
6) Geometric Details : Floor to Floor Height : 3.0M Height of Plinth : 0.6M Depth of Foundation : 1.8 M
7) Materials:Concrete : M20Main steel : Fe415
8)Bearing Capacity of soil : 250kn/M^2Live Load where access is provided: 2KN/M^2Live Load for stair cases : 3 KN/M^2Floor Finish : 1.5 KN/m^2
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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DESIGN BASIS
The basis most important characteristic of any structural member is its actual strength ,which must be large enough to resists ,with some margin to spare , all foreseeable loads that may act on in its during the life of the structures ,without failure or distress . It is logically therefore to proportion members, i.e. to select concrete dimensions and reinforcements, so that member strengths are adequate to resists forces resulting from certain hypothetical overload stages ,significantly above loads expected actually to occur in service .This design concept is known as“ Strength Design“
For reinforced concrete structures all loads close to and as failure, one or both of materials, concrete and steel, are in variably in their non-linear in elastic range.i.e. . . . . Concrete in a structural member reaches its maximum strength and subsequently fractures at stresses and strain for beyond the initial elastic range in which stress and strains are fairly proportional .Similarly, steel close to and at failure of member is usually stressed beyond its elastic domain into and even beyond the yield region. Consequently, the nominal strength of a member must be calculated on the basis of this inelastic behavior of the materials. A member designed by strength method must also perform in a satisfactory way under normal service loading .For example, beam defections must be controlled .Serviceability limit conditions are important part of total design, although attention is focused initially on strength.
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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CHAPTER-2
LOAD CALCULATIONS
GENERAL CALCULATION:COLUMN SIZE:
COLUMN NO:9 & 12 B=375mm;D=230mm
COLUMN NO:10 & 11 B=230mm;D=230mm
BEAM SIZE:
BEAM WIDTH=230mm;D=310mm
BEAM DEPTH:
AS PER I S CODE 456-2000 = l/12 to l/15 of B
Effective depth of the slab= 120 mm Cover=20mmAssumed beam cover =40 mm
EFFECTIVE SPANS OF BEAMS:
BEAM No. 1Effective span = center to center distance between columns
= 3.47 + 0.15 + 0.1 = 3.72BEAM NO. 2Effective span = center to center distance between columns
= 1.65 + 0.1 + 0.1 = 1.85BEAM NO. 3Effective span = center to center distance between columns
= 2.21 + 0.15 + 0.1 = 2.46BEAM NO. 4Effective span = center to center distance between columns
= 3.47 + 0.15 + 0.1 = 3.72BEAM NO. 5Effective span = center to center distance between columns
= 1.65 + 0.1 + 0.1 = 1.85
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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BEAM NO. 6Effective span = center to center distance between columns
= 2.21 + 0.1 + 0.15 = 2.46BEAM NO. 7Effective span = center to center distance between columns
= 3.0 + 0.15 + 0.1 = 3.25BEAM NO. 8Effective span = center to center distance between columns
= 1.3 + 0.1 + 0.1 = 1.5
BEAM NO. 9Effective span = center to center distance between columns
= 3.0 + 0.1 + 0.15 = 3.25BEAM NO. 10Effective span = center to center distance between columns
= 3.0 + 0.15 + 0.1 = 3.25BEAM NO. 11Effective span = center to center distance between columns
= 1.3 + 0.1 + 0.1 = 1.5BEAM NO. 12Effective span = center to center distance between columns
= 3.0 + 0.1 + 0.15 = 3.25BEAM NO. 13Effective span = center to center distance between columns
= 4.27 + 0.15 + 0.1 = 4.52BEAM NO. 14Effective span = center to center distance between columns
= 4.27 + 0.15 + 0.1 = 4.52BEAM NO. 15Effective span = center to center distance between columns
= 4.27 + 0.15 + 0.1 = 4.52BEAM NO. 16Effective span = center to center distance between columns
= 3.18 + 0.1 + 0.1 = 3.38BEAM NO. 17Effective span = center to center distance between columns
= 3.26 + 0.1 + 0.1 = 3.46BEAM NO. 18Effective span = center to center distance between columns
= 3.26 + 0.1 + 0.1 = 3.46BEAM NO. 19Effective span = center to center distance between columns
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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= 4.29 + 0.1 + 0.15 = 4.54BEAM NO. 20Effective span = center to center distance between columns
= 4.29 + 0.1 + 0.15 = 4.54BEAM NO. 21Effective span = center to center distance between columns
= 4.29 + 0.1 + 0.15 = 4.54BEAM NO. 22Effective span = center to center distance between columns
= 4.29 + 0.1 + 0.15 = 4.54
LOAD ON SLAB PORTION:
Long span =
WLX6
(3−( LXLY
)2⟩
Short span =WLx /3
TOP FLOOR CALCULATION OF LOADS:
DEADLOAD: Dead load on the slab =0.12 X 1 X 25= 3 KN/m2
Floor finish ( I S -875 part – 1 ) = 1.5KN/m2
FactoredUltimate load = 4.5 X 1.5= 6.75 KN/m2
Live load = 1.5 x 1.5 = 2.25KN/m2
W = 6.75 + 2.25 = 9KN/m2
Slab load calculations:Slab portion 1(S 1):Load in Long span direction = 12.96KN/m2
Load in short span direction = 11.16KN/m2
Slab portion 2(S 2):LS = 12.96KN/m2
SS = 9.39KN/m2
Slab portion 3(S 3):LS = 9.11KN/m2
SS = 7.38KN/m2
Slab portion 4(S 4):ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
10
LS = 5.76KN/m2
SS = 4.35KN/m2
Slab portion 5(S 5):LS = 7.2KN/m2
SS = 5.25KN/m2
Slab portion 6(S 6):LS = 14.05KN/m2
SS = 10.38KN/m2
Slab portion 7(S 7):LS = 12.12KN/m2
SS = 9.75KN/m2
Slab portion 8(S 8):LS = 10.17KN/m2
SS = 9.75KN/m2
Slab portion 9(S 9):LS = 6.34KN/m2
SS = 4.5KN/m2
Slab portion 10(S 10):LS = 6.09KN/m2
SS = 4.35KN/m2
Slab portion 11(S 11):LS = 13.28KN/m2
SS = 8.19KN/m2
Total load on beams:Beam I = SS1 = 11.16KN/m2
Beam II = SS2 = 9.39KN/m2
Beam III = SS3 = 7.38KN/m2
Beam IV = SS1 + LS6 + SS5 = 30.16KN/m2
Beam V = SS2 + LS6=23.44KN/m2
Beam VI = SS3 + LS6 = 21.43KN/m2
Beam VII = SS7 + SS5 + LS6 = 29.05KN/m2
Beam VIII = SS9 + LS6 = 18.55KN/m2
Beam IX = SS8 + LS6 = 23.8KN/m2
Beam X = SS7 = 9.75KN/m2
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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Beam XI = SS9 = 4.5KN/m2
Beam XII = SS8 + LS10 = 15.08KN/m2
Beam XIII = LS1 = 12.96KN/m2
Beam XIV = LS1 + LS2 = 25.92KN/m2
Beam XV = LS2 + LS3 + SS4 = 26.42KN/m2
Beam XVI = LS3 = 9.11KN/m2
Beam XVII = SS6 + LS5 = 17.58KN/m2
Beam XVIII = SS6 + LS11 = 23.66KN/m2
Beam XIX = LS7 = 12.12KN/m2
Beam XX = LS7 + LS9 = 18.46KN/m2
Beam XXI = LS9 + LS8 = 16.51KN/m2
Beam XXII = LS8 = 10.17KN/m2
WIND LOAD:
Wind load = 3X1.5 = 4.5KN
(IS -875 part – 1)
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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CHAPTER-3
ANALYSIS & DESIGN WITH STAAD
STAAD – PRO:
STAAD-PRO is nothing but Structural Aided Analysis and Design. STAAD-
PRO is an advanced version of STAAD-III .STAAD-PRO is an analysis and design
software package for structural engineers. STAAD-PRO is simple to use and user
friendly. No prior knowledge of language of the program is necessary to get started.
There are two methods of creating the structure Data, either by using the
command file or by using graphical model generation mode or graphical user interface as
it is usually referred to (GUI) command file.
The command file is a text file which contains data for the structure being
modeled. The file consists of simple English language commands. This command file
may be created directly by the editor built into the program or for that matter, any editor
which saves data in text form such as notepad or word pad available on Microsoft
windows.
The command file also automatically created behind the screen when the
structure is generated using graphical user interface. The graphical model generation
mode and the command file are seamlessly integrated. So, at any time, you may
temporarily exit the graphical model generation mode and access command file. The text
file reflects all data entered through the graphical model generation mode. Further, if
there are any changes to command file, the GUI immediately reflects the changes made
to the structure through command file. By using both, the generation of a model can be
explained easily.
STAAD-PRO offers extensive results verification and visualization facilities. The
facilities can be accessed from post processing mode. The following are the main
features of STAAD-PRO
1) Annotation of the structure with different labels
2) Creating building model diagram.
3) Displaying the dimensions on the structure.
4) Displaying loads and member force results on the structure.
STEP-BY-STEP PROCEDURE:
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
13
Step-by-step process for creating and analyzing the space frame for seismic design is
given below.
1 Starting the program
2 Creating new structure
3 Creating joints and members
4 Specifying member properties
5 Specifying material constants
6 Specifying member offsets if necessary
7 Printing member information
8 Specifying supports
9 Specifying loads
10 Specifying the analysis type
11 Specifying post-analysis print commands
12 Specifying concrete design parameters
13 Perform analysis and design
14 Changing the mode to post processing for viewing the output graphically
15 Viewing the output file
16 Verifying results on screen, both, graphically and numerically.
17 Finish.
STAAD.PRO ANALYSIS AND DESIGN EDITOR OF MULTI STORIED BUILDING
FC 25000 MEMB 1 TO 12 15 17 19 21 TO 35 37 TO 41 44 TO 47 49 TO 53 55 TO 59 -61 TO 66 68 TO 78 80 81 85 TO 90 92 TO 106 108 TO 124 126 TO 145 147 TO 163 -165 TO 176 178 180 TO 190 193 TO 202 204 TO 208 210 TO 236 238 TO 465 467 -469 TO 679 681 683 TO 885 887 TO 895 897 TO 1107 1109 1111 TO 1308FYMAIN 415000 MEMB 1 TO 12 15 17 19 21 TO 35 37 TO 41 44 TO 47 49 TO 53 55 -56 TO 59 61 TO 66 68 TO 78 80 81 85 TO 90 92 TO 106 108 TO 124 126 TO 145 -147 TO 163 165 TO 176 178 180 TO 190 193 TO 202 204 TO 208 210 TO 236 238 -239 TO 465 467 469 TO 679 681 683 TO 885 887 TO 895 897 TO 1107 1109 1111 -1112 TO 1308DESIGN BEAM 1197DESIGN COLUMN 1 3DESIGN ELEMENT 1333 1339CONCRETE TAKEFC 25000 ALLFYMAIN 415000 ALLDESIGN BEAM 1197DESIGN COLUMN 1 3CONCRETE TAKEDESIGN ELEMENT 1330 1339FC 25000 MEMB 1 TO 12 15 17 19 21 TO 35 37 TO 41 44 TO 47 49 TO 53 55 TO 59 -61 TO 66 68 TO 78 80 81 85 TO 90 92 TO 106 108 TO 124 126 TO 145 147 TO 163 -165 TO 176 178 180 TO 190 193 TO 202 204 TO 208 210 TO 236 238 TO 465 467 -469 TO 679 681 683 TO 885 887 TO 895 897 TO 1107 1109 1111 TO 1308FYMAIN 415000 MEMB 1 TO 12 15 17 19 21 TO 35 37 TO 41 44 TO 47 49 TO 53 55 -56 TO 59 61 TO 66 68 TO 78 80 81 85 TO 90 92 TO 106 108 TO 124 126 TO 145 -147 TO 163 165 TO 176 178 180 TO 190 193 TO 202 204 TO 208 210 TO 236 238 -239 TO 465 467 469 TO 679 681 683 TO 885 887 TO 895 897 TO 1107 1109 1111 -1112 TO 1308DESIGN BEAM 1197DESIGN COLUMN 1 3DESIGN ELEMENT 1330 1339CONCRETE TAKEEND CONCRETE DESIGNFINISH
ANALYSIS RESULTS:ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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STAAD FRAME:
Load 1
XY
Z
DISPLACEMENT DIAGRAM:
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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SHEAR FORCE DIAGRAM:
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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BENDING MOMENT DIAGRAM:
DESIGN RESULTS:
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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DETAILS OF BEAMS FROM STAAD PRO
BEAM NO. 1197 DESIGN RESULTS
M25 Fe415 (Main) Fe415 (Sec.)
LENGTH: 3230.0 mm SIZE: 230.0 mm X 350.0 mm COVER: 25.0 mm
SUMMARY OF REINF. AREA (Sq.mm) ---------------------------------------------------------------------------- SECTION 0.0 mm 807.5 mm 1615.0 mm 2422.5 mm 3230.0 mm ---------------------------------------------------------------------------- TOP 150.75 150.75 150.75 150.75 150.75 REINF. (Sq. mm) (Sq. mm) (Sq. mm) (Sq. mm) (Sq. mm)
SUMMARY OF PROVIDED REINF. AREA ---------------------------------------------------------------------------- SECTION 0.0 mm 807.5 mm 1615.0 mm 2422.5 mm 3230.0 mm ---------------------------------------------------------------------------- TOP 2-10í 2-10í2-10í2-10í2-10í REINF. 1 layer(s) 1 layer(s) 1 layer(s) 1 layer(s) 1 layer(s)
SHEAR 2 legged 8í 2 legged 8í 2 legged 8í 2 legged 8í 2 legged 8í REINF.@ 110 mm c/c @ 110 mm c/c @ 110 mm c/c @ 110 mm c/c @ 110 mm c/c ---------------------------------------------------------------------------- SHEAR DESIGN RESULTS AT DISTANCE d (EFFECTIVE DEPTH) FROM
FACE OF THE SUPPORT
SHEAR DESIGN RESULTS AT 500.0 mm AWAY FROM START SUPPORT VY = 6.39 MX = -2.71 LD= 29 Provide 2 Legged 8í @ 110 mm c/c
SHEAR DESIGN RESULTS AT 430.0 mm AWAY FROM END SUPPORT VY = 6.39 MX = -2.71 LD= 29 Provide 2 Legged 8í @ 110 mm c/c
============================================================== ********************END OF BEAM DESIGN RESULTS******************* ==============================================================DETAILS OF COLUMN FROM STAAD PRO
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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COLUMN NO.1 DESIGN RESULTS
M25 Fe415 (Main) Fe415 (Sec.)
LENGTH: 3600.0 mm CROSS SECTION: 230.0 mm X 370.0 mm COVER: 40.0 mm
** GUIDING LOAD CASE: 38 END JOINT: 1TENSION COLUMN
REQD. STEEL AREA : 1090.29Sq.mm. REQD. CONCRETE AREA: 84009.72Sq.mm. MAIN REINFORCEMENT: Provide 4 - 20 dia. (1.48%, 1256.64 Sq.mm.) (Equally distributed) TIE REINFORCEMENT: Provide 8 mm dia. rectangular ties @ 230 mm c/c
SECTION CAPACITY BASED ON REINFORCEMENT REQUIRED (KNS-MET) ----------------------------------------------------------Puz: 1284.46 Muz1: 48.91 Muy1: 27.21
INTERACTION RATIO: 0.99 (as per Cl. 39.6, IS456:2000)
SECTION CAPACITY BASED ON REINFORCEMENT PROVIDED (KNS-MET) ---------------------------------------------------------- WORST LOAD CASE: 38 END JOINT: 1 Puz: 1334.37 Muz: 56.39 Muy: 30.52 IR: 0.87 ============================================================
COLUMN NO. 3 DESIGN RESULTS
M25 Fe415 (Main) Fe415 (Sec.)
LENGTH: 3600.0mm CROSS SECTION: 230.0mm X 230.0mm COVER: 40.0 mm
** GUIDING LOAD CASE: 38 END JOINT: 4 SHORT COLUMN
REQD. STEEL AREA : 423.78Sq.mm. REQD. CONCRETE AREA: 52476.22Sq.mm. MAIN REINFORCEMENT: Provide 4 - 12 dia. (0.86%, 452.39 Sq.mm.) (Equally distributed) TIE REINFORCEMENT: Provide 8 mm dia. rectangular ties @ 190 mm c/c
SECTION CAPACITY BASED ON REINFORCEMENT REQUIRED (KNS-MET) ----------------------------------------------------------Puz: 722.26 Muz: 11.96 Muy: 11.96
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
36
INTERACTION RATIO: 0.99 (as per Cl. 39.6, IS456:2000)
SECTION CAPACITY BASED ON REINFORCEMENT PROVIDED (KNS-MET) ---------------------------------------------------------- WORST LOAD CASE: 38 END JOINT: 4 Puz: 730.84 Muz: 12.47 Muy: 12.47 IR: 0.95 ============================================================ ***************END OF COLUMN DESIGN RESULTS******************* ============================================================
DESIGN ELEMENT 1330 1339
ELEMENT DESIGN SUMMARY
ELEMENT LONG. REINF MOM-X /LOAD TRANS. REINF MOM-Y /LOAD
************** END OF ELEMENT DESIGN ************* 775. CONCRETE TAKE 776. END CONCRETE DESIGN
************** CONCRETE TAKES OFF ************** (FOR BEAMS AND COLUMNS DESIGNED ABOVE)
TOTAL VOLUME OF CONCRETE = 2.27 CU.METER
BAR DIA WEIGHT (in mm) (in New) -------- -------- 8 619.16 10 234.40 12 376.14
20 1044.92------------
*** TOTAL = 2274.63
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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CHAPTER-4
DESIGN OF SLABS
TWO - WAY SLAB:
Slab Portions S1, S3, S6, S7, S8 were to be designed for TWO WAY SLAB because of the room dimensions. Design procedure is as follows for GROUPS: FLAT 1,2,3,4.
Effective span (Ly) = Clearspan + Effective depth of the slab
= 4.29- (0.20 - 0.100)+0.100
C/C distance between supports = 4.16+ 0.20 = 4.36 m
Ly = 4.26;
lylx
=4 . 262 .97
=1 . 43<2 Two-way slab
LOADS: Self-weight of the Slab = 0.12 x 1 x 25 = 3KN/m2
DESIGN MOMENTS:
Wu = 10.5 KN. Mx = axwlx2
My = αywly2
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
38
Two adjacent edges discontinuous: αx αy 1.4 1.5
-ve moment at continuousedge: 0.071 0.075
-ve moment at mid span: 0.053 0.056
-ve moment atcontinuous edge: 0.0722 0.047
-ve moment at mid span: 0.0539 0.035
Mx = αx wl x2
(-Ve) Mx= 0.0722 X 10.5 X 2.972 = 6.68 KN-m
(+Ve) Mx= 0.0539 X 10.5 X 2.972 = 4.99 K
My = αy wl x2
(-Ve) My= 0.047 X 10.5 X 2.972 = 4.35 KN-m
(+Ve) My= 0.035 X 10.5 X 2.972 = 3.24 KN-m
By considering the max.B.M .Valves
(Mx)max = 6.68 KN-m
(My)max = 4.35 KN-m
MAIN STEEL:
Mux = 0.87 fyx Ast x d (1-
Astfyb .d . fck )
6.68 x 106 = 0.87 x 415 xAstx 100 (1-
Astx 4151000 X 100 X 20 )
6.68 x 106 = 36105 x Ast(
2 X106−Astx 4152 X 106
)
1.34 x 1013= 7.22x1010Ast – Ast 214.98 x 106
14.98 x 10 6 Ast 2 – 7.22 x 1010 Ast +1.34 x1013
Ast= 193.33mm2.
SPACING:
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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As Per IS Code
(1) 3 x Effective depth of the slab. (2) 300 mm. (3) Пd 2 /4X b Ast Assuming 8mm dia bars
П(8) 2 /4X 1000 =259 mm 193.5 Lesser valve = 250 mm
LONG SPAN:
My = αy wl x2
My = 0.047 X Ast X 2.972= 4.35 KN-m(1-
Astfyb .d . fck )
My = 3.24 KN-m
Muy = 0.87 fy X Ast X d (1-
Astfyb .d . fck )
= 4.35X 106 = 0.87 X 415 X Ast X 100 (1-
Astx 4151000 X 100 X 20 )
Ast= 123.6¿ 130 mm2.
SPACING:
(1) 3X Effective depth of the slab. (2) 300 mm. (3) П(8) 2 /4X 1000 =386¿ 380 mm2.
130
C/C = 300 mm
CHECK FOR DEFLECTION:
Short Span: Pt =100 X 200 = 0.2 1000 X 100 Fs=0.58 X 415X 193.35 =232.69 200 α = 1.6
Allowable ratio = 26 X 1.6 = 41.6
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Actual ratio = Lx/d = 2970 =29.7100 29.7 < 41.6 HENCE SAFE FROM DEFLECTION
CHECK FOR SHEAR:
Nominal Shear,Tv = Vu/bd
Vu = Shear force = Wl/2 = (10.5x2.87)/2 = 30.135KN/m
For 0.2% of Steel we obtain the value of Tc
Tc’ = K x Tc
Where K is the modification factor of obtained % of steel = 1.6 from IS code provision.
Tv< Tc’
HENCE SAFE
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CHAPTER-5
THEORY OF BEAMS
Beam is an integral part of a structure which transfers the loads imposed on it, from the slab and other load carrying members acting on it to its supports, a reinforced flexural member should be able to resist tensile, compressive and shear stresses induced in it. Concrete is fairly strong in compression but very weak in tension .Plain concrete members are those limited in carrying capacity by the low tensile strength .Steel is very strong in tension. Thus the tensile weakness of concrete is overcome by the provision of reinforcing steel in the tension zone and the concrete to make a reinforced concrete member .According to yield line theory, distribution of land on slab or location of yield lines depends on Ly and Lx ratio and the support conditions.
IF Ly/Lx > 1.5 and all the supports are fixed distribution is in triangular pattern and if Ly/Lx > 1.5 and all the support are fixed then, distribution of load consists of 2 triangles and 2 trapezoids.
There are two types of reinforced concrete beams:
1. Singly reinforced beams,
2. Double reinforced beams
In single reinforced simply supported beams reinforcing beams, reinforcing beams ,reinforcing steel bars are placed near the bottom where they are most effective in resisting the tensile stress. In singly reinforced cantilever beams, reinforcing bars are placed near the top of the beam for same reason. A doubly reinforced concrete section is reinforced in both compression regions. The
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section of beam may be a rectangular , T and L section, the necessity of using steel in the compression region raised due to 2 main reasons :
When depth of the section is restricted, the strength available from a singly reinforced section is inadequate.
At a support of a continuous beam where bending moment changes sign , such a situation may also arise in the design of a beam circular in plan .
In most reinforced concrete structures, concrete slabs and beams are cast monolithic. Thus beams from part of the system together with the slab. In bending , the slab forming the top part of the beam at mid-span should be in compression for a definable width greater that the width of the rib (or a beam ), thus increasing the moment of resistance for a given rib width. At continuous support, the situation is reversed. The slab is in tension and part of the rib is in compression. Since concrete is assumed to have cracked in tension, this beam is equivalent to a rectangular section at the support.
When designing a reinforced section, the loading , span , grade of concrete and steel , and breadth of the section are usually known in advance. The section dimensions and area of reinforcing bars remain to be determined. The can be no unique section for a given set of forces. There are many possible combinations of breadth, depth and area of steel thus; Cost will govern the final design.
BENDING OF BARS:
Reinforced concrete beams are non-homogenous in that are made of two entirely different materials. The fundamental principal involved if, at any cross-section there exist internal forces which can be resolved into components normal to the section are the bending stress (tension on one side of the neutral axis and compression of the other.)their function is the resist the bending moment at the section. The tangential components are known as the shear stresses, and they resist the transverse shear forces.
REINFORCEMENT DETAILS OF BEAM:
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REINFORCEMENT DETAILS OF BEAM -2
REINFORCEMENT DIAGRAM OF BEAM-3
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CHAPTER-6
ANALYSIS AND DESIGN OF MULTI STORIED BUILDING
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THEORYOF COLUMNS:
Columns are defined as members that carry load in compression. Usually columns carry bending moments as well one or both axes of cross section and bending action may produce tensile forces over a part of cross section .Even in such cases, columns are generally referred to as compression members because the compression forces dominated their behavior. In addition to most common type of compression member, i.e. vertical elements in structures. Compression members includes arch ribs, rigid frame truss, shells or portions thereof that carry axial compression and other forms .Three types of reinforced concrete compression members are in use :
1. Members reinforced with longitudinal bars and lateral ties.2. Members reinforced with longitudinal bars and continuous ties.3. Composite compression members reinforced longitudinal with structural shapes, pipe or tubing with or without additional longitudinal bars and various types of lateral reinforcement.4. Types one or two are far the most common in use .The main reinforcement in column is longitudinal, parallel to direction of load and consist of bars arrange in squares, rectangular and circular patterns .The ratio of longitudinal steel area Ast to gross concrete cross section Ag is in the range of 0.008 to 0.06 to I.S.CODE .The lower limit is necessary bending moments not accounted for the analysis and to reduce the effect of creep and shrinkage of the concrete under sustained compression.
Ratio higher than 0.06 not only are to congestion of the reinforced particularly where the steel must be spliced .Generally the larger bars are used to reduced placement costs and to avoid unnecessary congestion .According to ISCODE minimum of 4 bars are required in square or rectangular columns and 6 in circular column.
Column may be divided in to 2 broad categories i.e., short column for which strength I governed by strength materials and the geometry of cross section and slender columns for which the strength may be significantly reduced by lateral deflections. A number of years ago an ACIASCE survey indicated that 90% of columns braced against side way and 40% of embraced columns should be designed as short columns .Shear walls , elevated and stair well shafts, diagonal bracing or a combination of this , commonly provide effective lateral bracing , which prevents relative movement of two ends of a columns. All though slender columns or most common now because of wider use of high strength materials and improved methods of dimensioning members , it is still true that most columns in ordinarily practice can be consider short columns.
LATERAL TIES AND SPIRALS
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Lateral rein forcemeat in the form of individual relatively, widely spaced ties or a continues closely spaced spiral, serves several functions. F or one such rein for cements is needed to hold to longitudinal bars in position in the form while the concrete is being placed. For this purpose, longitudinal and transverse steel or wired together to form cages, which or then moved in to the forms and properly positioned before placing the concrete. For another transverse rein for cements is needed to prevent the highly stressed, slender longitudinal bars from bucking out ward by bursting this concrete cover. Closely spaced spiral evidently serve this to functions. Ties, which can be arranged and spaced in various ways, must be so design that their requirements are to be met. This means that Spacing must be sufficiently small to prevent buckling between ties and that in any tie plane; sufficient number ties must be provider to position and hold all the bars. On the other hand, in columns, with many longitudinal bars, if the columns section is crisscrossed by too many ties. They interfere with placement of concrete in the forms. The achieve adequate tying yet hold the number of ties to a minimum, IS CODE 456 gives the rules. In order to know the placement of longitudinal bars, in general in members with large axial forces and small moments, longitudinal bars or placed more or less uniformly along the perimeter. When bending moments or large, much of longitudinal steel is concentrated at the forces of largest compression or tension i.e., at maximum distance from the axis of bending in heavily loaded columns with large steel percentages, the result of large number of bars, each of them positioned and held individually by ties, is steel congestion in the forms and differences in placing the concrete. In such cases bundled bars are frequently employed. A bundle consists of 3 or 4 bars tied in direct contact, wired or otherwise fastened together. There are usually laced in the corners. Tests have shown that adequately bundled act as 1 unit i.e., they are detailed as it a bundle constituted a single round bar of area equal to the sum of the bundled bars. The structural effect of a spiral is easily visualized by considering as a model steel drum filled with sand on the drum ,which caused hoop tension in the steel wall. The load on sand can be increased until the hoop tension becomes large enough to burst the drum. The sand pile alone.
If not confined in the drum have been able to support hardly any load.A cylindrical column, to be sure, does have a definite strength without any lateral confinement. As it is been loaded, it shortens longitudinally and expands laterally, depending on Poisson’s ratio. A closely spaced spiral confining the column counteracts the expansion, as did the steel drum in the model. This causes hoop tension in the spiral, while the carrying capacity of the confined concrete in the core is generally increased. failure only when the spiral steel yields, which greatly decreases its confining effect, or when it fractures.
REINFORCEMENT DIAGRAM OF COLUMN-1
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REIFORCEMENT DIAGRAM OF COLUMN-2
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References:
1. IS -456-2000 code of practice for plain &reinforced concrete
2. IS -875 ( PART 1,2 ) code of practice for design loads for building and structures
3. SP16:1980 for relevant charts.
4. RCC structures-B.C.Punmia
5. Limit state design of reinforced concrete by P.C.Varghese.
6. Reinforced concrete - Ashok.K.Jain
7. Limit state theory & Design of reinforced concrete by Dr. V. L.Shan & Late S. R. Karve and Shah