RESOURCE SCHEDULING OF CONSTRUCTION PROJECT USING GENETIC ALGORITHM

ANALYSIS AND DESIGN OF IRREGULAR SHAPED BUILDINGPROJECT MEMBERS BHARATH KUMAR.P GUHANATHAN.K RAM PRASAD.P LOGESH.RGUIDED BYK.SATHYA PRABHAASSISTANT PROFESSORDEPT. OF CIVIL ENGG.CARE SCHOOL OF ENGG.TRICHY

Contents

IntroductionObjectiveMethodologyLiterature reviewImportance of the studyAnalysis and ResultsDesign of shear wallFuture scope of the projectConclusion

Objective The main objective of this project is to analyse and design the plan irregular multi-storey building using STAAD.Pro software The above software can also be used for modelling of asymmetrical building.Design of shear wall as per IS (1893- Part 1:2002)NecessityThese building structures may possess irregular distributions in their mass, stiffness and strength along the height of building.When such buildings are located in a high seismic zone, then it is very essential to design structures resistant to these buildings

MethodologyThe work is carried out using STADD. Pro software. The methodology is mentioned belowLiterature collectionLoad CalculationModeling of the structureAnalysis and DesignConclusion

Literature Review

A.M. Chandler a, P.A. Mendis (1998) studied about Performance of reinforced concrete frames using force and displacement based seismic assessment methods.C.J. Athanassiadou (2007) studied about Seismic performance of R/C plane frames irregular in elevation.

Importance of the study

Improper load distribution leads to failure of irregular buildings.Load calculationBRICK WORK LOAD Unit weight of brick = 19 KN/m3 (IS 875 part I) Height of the wall = 3.1 m (10) Thickness of the wall = 0.23 m Brick work load = 193.10.23 = 13.55 KN/mReduction in brick work load: Brick work Under beam 0.61 m death = (3.1-0.61)0.2319 = 10.88 KN/m

Brick work load on floor Slab thickness = 0.150 m Slab self weight = 10.1525 =3.75 KN/m2Floor finish and ceiling self weight = 0.5+0.5 (IS 875 Part I - Table 2) = 1 KN/m2 Therefore total dead load in floor = 4.75 KN/m2

Live load

Sl.noIn buildingLive load unit1Floors4KN/m22Roof2KN/m2Analysis and Results

GROUPING OF BEAMS AND COLUMNS

Under ground floorGround floor

First floor

Second floor

INTERMEDIATE BEAMS

3m span4m span

6m span5m span7m span

COLUMN GROUPING

Under ground floor

Ground & first floor

Design

Manual check for Beam No: 8 (Ground Floor Beam)

Span = 6mfck = 25N/mm2fy = 415 N/mm2d = 600mmb = 300mm

Mu = 305.898 kNm (From Staad.)Mu = 0.87 fy Astd (1-(fyAst / bdfck)305.898106 = 0.87415Ast600 (1-(415Ast/60030025))Ast = 5965mm2Therefore Ast required = 5065mm2Ast Provided = 6606mm2Hence Safe.

COLUMN NO 573

Manual Check for Column no: 573 (Ground Floor)

b = 600mmd = 600mmfck = 25N/mm2fy = 415N/mm2Pu = 91.48kNMux = 146.346kNm, Muy =7.524kNmAssume d = 50mm(d/D) = 0.10

Mu = 1.15 =1.15 = 168.52 KN.m(Pu / fck bd) = (91.48103 / 50600600) = 0.01(Mu / fck bD2) = (168.52106 / 506006002) = 0.03As per SP-16 chart -44 (P/fck) = 0.02P = 500.02 = 1

As per SP-16 chart -44 (P/fck) = 0.02P = 500.02 = 1Asc = (Pbd/100) = (1600600/100) = 3600mm2Area required = 2973mm2Area provided = 3600mm2Hence safe.

DESIGN OF SHEAR WALL

Load calculationWeight of slab = 36260.225 = 4860 KNWeight of beam = 0.6(277+366)25 = 1822.5 KNWeight of column = 0.60.6 3.12536 = 1004.4 KNWeight of shear wall = 6.73.10.25252 = 259.6 KNLive load on floors = 36271.5(25/100) = 364.5 KN

Seismic weight calculation of the buildingFloorSlab(KN)Beam (KN)Column(KN)Shear wall(KN)Live load(KN)Total load(KN)448601822.5864259-7805.5348601822.510042593648309.5248601822.510042593648309.5148601822.510042593648309.5 32734 KNCalculation of seismic base shear (VB) VB = W Ah Ah = (Z/2) (I/R) (Sa/g) = (0.1/2) (1.5/5) (2.77) = 0.04155 = 0.0415532734 VB = 1360.09 KN

Shear strength requirements as per IS 456:2000

The nominal shear stress, v shall be calculated asv = Vu/twlwWhere,Vu = factored shear forcetw = thickness of walldw = effective depth of wall

Vu = 1.51360.09/2 = 1020.07 KNdw = 0.86700 = 5360 mmpermissible shear stress = 0.25 = 0.25 = 0.36 N/mm2v = 1020.07/(2505360)v = 0.76 N/mm2c,max = 2.8 N/mm2v < c,max = 2.8 N/mm2when v > c (0.36 N/mm2)

Vus = ( v - c)twdw = (0.76-0.36)2505360 = 536 KN Use 8mm dia bars, Vus = 0.87fyAhdw/Sv Sv = (0.874151605360)/(5360103) Sv = 280 mm

Provide 8mm dia bar at 280mm c/c in 2 curtains as horizontal reinforcementHence provide 8mm dia bar ok at 280mm c/c in 2 curtains as vertical reinforcement.

FUTURE SCOPE OF THE PROJECTThe project can be helpful in designing structures that are capable of resisting damages and failure during earthquakes.

In future ductile detailing and seismic joints can be done to these structures with reference to IS13920 -1993 and IS 1893-2002 and hence create structures resistant to earthquakesSUMMARY AND CONCLUSION

The analysis and design is completed, the study is carried out on force, shear, bending, torsion and displacement.

The Design of Column and Beam for Plan Irregular Structure is carried out as per IS 456 2000The design of shear wall is carried out as per IS (1893- Part 1:2002)

THANK YOU