1. COMPANY PROFILEUNICO Associates was established in 2012 as structural engineering consultancy, with the purpose of working with imaginative architect and developers to make outstanding structures. Company’s portfolio includes a wide range of structure including residential, commercial and projects varied from small renovation to school building to design and construction of residential building.

Managing director Manjunath has worked with some of the innovative and emerging architects and structural engineer in Bangalore. Based on this diverse portfolio and experience company can guarantee an excellent outcome of the project. Company enjoy the technical , logical, engineering principles that reinforce engineering profession; applying them creatively to help architects and clients bridge the gap between concept and reality, to meet our clients individual needs.

Most recently company is engaged in the design of various multi-storey residential building. Structural engineer in the consultancy is heavily involved constantly evolving technology around new design methodologies and strategies.

Company experience with providing feasibility assessment and in working with multi disciplinary teams enables them to make a constructive contribution at an early stage in the design of the project. When engineering input can often help to achieve a sophisticated solution that is also financially beneficial. Company are always striving to find better ways to tackle the challenges of each project.

Company Provides structural design and construction services for a variety of agencies and private sector clients.

3. TASK PERFORMEDAny project can be designed only after a combined effort of various professionals.

First of all an architect prepares a conceptual plans. He then circulates the conceptual plans to various other related professionals like structural engineer, sanitary engineer, electrical engineer, interior designers and site engineer. After getting feedback from those professionals he finalizes the plans and elevations and starts making working drawings.

During consultation with structural engineer, the structural engineer works out the structural schemes and also gives the sizes of the structural members like beam, column, slab, etc. the architect. After working drawings have been prepared, detailed structural designs are carried out. Safety is the prime concern of the structural design. Serviceability and economy are other basic requirements that need to be considered during structural designing. A building needs to be designed for all loads acting on it. On keeping that things in mind I have started making the detail analysis of the building given to me.

The building should be designed not only for vertical loads but also for horizontal loads such as wind and earthquake loads. The structural engineer is to take care of the safety of the building against the loads mentioned above. He is concerned in designing both superstructure and the substructure of the building. A building must be strong enough to transfer all the loads acting on it safely to the ground. It should be able to withstand all loads acting on it.

Designing building against earthquake forces does not mean that we are making the building proof against it. Although, we can design a building earthquake proof, it would be too expensive to build. It would be far cheaper to rebuild the building again instead of making it proof against earthquake. Hence, the buildings are made seismic resistant rather than seismic proof. According to this philosophy, no matter how much severe earthquake occurs, the building won’t collapse although it may be irreparably damaged. The building will be able to resist minor to moderate earthquakes without any damage.

The contractor who is selected by the client does construction of the building. The role of the contractor is of paramount importance as he is the one who will execute the construction work at the site. He is required to execute the work according to the drawing supplied by the consultant to him. Design and construction are inter-related jobs. If the construction is not done as per design requirement the objective of design will not be fulfilled. A large percentage of failure of the building is attributed to poor quality of construction. Past experiences from damages have shown that quality of material and workmanship plays an important role in good seismic behavior of the buildings. Hence, quality assurance in both design and construction is foremost important.

Followings are the task performed in design of building

Studying the given project thoroughly.

Understanding the soil and site condition by geotechnical report if given.

Understanding the architectural plan, levels.

Defining the structural difficulties in architectural plan and discussing

the same with guide.

Calculation of dead load, live load and earthquake or wind load

based on codal specifications.

First column positions are done for any given building plan. Once the

column positions are finalised then the centreline of all floors are done

as well beam layout. Import the centre line in the design software

(ETABS) or directly the plan is modelled in ETABS.

Modelling of whole structure, assigning respective supports,

sectional properties, assigning loads, load combinations.

Defining specifications such as releases, truss, compression, tension, cable, etc.(if any)

Design of concrete or steel sections.

Performing the analysis and studying the output file.

Using column reactions and moments the column and footings are

designed using design sheets and appropreiate sizes are fixed.

Then plinth and floor beams are been designed.

Finally slab are designed.

Detailing of sections in AUTOCAD based on IS code specifications.

ABOUT THE PROJECT

It is a Seven-storied building. The typical floor-to-floor height is 3.5052 m and the floor height for basement floor is 2.4384 m. The site is located at Mysore. The building has been designed as Ductile Moment Resisting RCC framed structure. The footings consist of Isolated Footings. Beam Size of 228.6 mm X 228.6 mm for support of stair landing and 228.6 mm X304.8 mm for floor support has been adopted. Column size of 228.6 mm by 457.2 mm and 228.6mm by 381mm has been adopted. The slab thickness has been calculated as 127 mm thick. The bearing capacity of soil is taken as 250 kN/m².

Fig: Floor plan

Fig: Elevation

Fig:Section

MATERIALS

Reinforced Concrete of grade M25

Reinforcement bars of grade TORSTEEL (FE 415).

Unit weight = 25 kN/m3

Unit mass = 2.55 tonnes/m3

Young’s Modulus of Elasticity = 5000 √fck = 25 x 106 kN/m2

Poisson’s Ratio = 0.15

MODELLING

Since this is normal moment resisting frame structure, main components to be modeled are: Beams and Columns, Slabs, Foundation and Staircases, Walls, plinth beams. No parapet wall is modeled.

For the purpose of analysis, following material properties are assumed for concrete:

Grade of Concrete = M25

Grade of Steel = Fe415

Unit weight = 25 kN/m

Unit mass = 2.55 tonnes/m

Young’s Modulus of Elasticity = 5000 √fck = 25 x 106 kN/m2

Poisson’s Ratio = 0.15

LOADS

Live Load : 2 kN/m² at typical floor

: 1.5 kN/m² on terrace

Floor Finish : 1.5 kN/m² on terrace

: 1 KN/m² on typical floor

Wind Load : As per IS:875- Not designed for wind load, since earthquake load exceed wind load

Earthquake load

As per IS-1893(part-I)-2002

Zone Factor Z 0.1 Mandya

Importance Factor I 1.50

Response reduction factor R 5.00

LOAD CASE

Following loads have been considered in the analysis of the building as per IS

456-2000 and IS: 1893:2002.

1. Dead Load (DL)

2. Live load (LL)

3. Earthquake load in +ve X-direction (EQX)

4. Earthquake load in +ve Y-direction (EQY)

LOAD COMBINATION

Following load combinations have been adopted as per IS1893:2003

1.5(DL+LL)

1.2(DL+LL ± EL)

1.5(DL ± EL)

0.9 DL ± 1.5 EL

SEISMIC LOAD CALCULATION

Earthquake analysis has been done using Seismic Coefficient method. According to this method base shear is calculated using the formula given in IS1893:2002, which is as, follows:

V = Ah* W

C is given by:

Ah = Z I Sa/ 2Rg

This calculated base shear is then distributed at each floor level according to the formula given below:

F i= VWihi²/Wihi²

After distributing at each floor level the floor forces are then distributed to individual frames according to their stiffness.

ANALYSIS AND DESIGN

ANALYSIS

A two dimensional linear analysis has been carried out using the standard software ETABS. The Structure is assumed to be fixed at the foundation level. The brick wall is considered as the filler wall only. The beams are modelled as rectangular beams. The flange effect of the beams has been neglected. Centre to centre dimension of the structure has been considered. The rigid end effect has also been considered in the analysis.

Fig: Floor plan

Fig: Isometric view of whole structure

Fig: Load assign

Fig:Bending moment distribution

Fig: Shear force diagram

Fig: Deformed view

DESIGN

The design of the members has been done as per philosophy of limit state method. For the design of the members IS 456:2000 and design aid SP 16 has been used extensively. Footings have been designed for vertical loads and moments developed at the base due to dead load and live load only. Square footings have been adopted from seismic point of view that reversal stress may occur. And footing beams are provided for column at foundation for more rigidity of building and also need for the column located at boundary. Longitudinal reinforcement in columns has been calculated based on critical load combination among the five load combinations. A symmetric arrangement of the reinforcement has been adopted from seismic point of view that reversal stress may occur. Longitudinal reinforcement in beams is also based on critical load combination. It is calculated from the envelope of bending moment diagram. Spacing of the shear reinforcement has been calculated as per the ductility principle. IS 13920 -1993 have been used for this purpose.

Calculation of the reinforcement for the typical members has been included in the Sample Calculation Section of this report.

SAMPLE CALCULATION

Load Calculation

Member self weightColumn(350*350) 3.0625 kN/mcolumn rectangular(230*600) 3.45 kN/m

Main Beam(230*550) 3.1625 kN/mSecondary beam(230*300) 1.725 kN/mSlab(150mm thick) 3.75 kN/m²Brick wall external(230mm thick) 4.85 kN/m²Floor wall(height 3m) 14.55 kN/mBrick wall internal(100mm thick) 7.14 kN/mGround Floor wall(height 0.533m) 2.58505 kN/mTerrace Parapet(height 1m) 5.4805 kN/m

Slab load calculation

On RoofSelf 3.75 kN/m²Floor Finish 1.5 kN/m²Live load 1.5 kN/m²Factored Load 10.125 kN/m²

On FloorSelf 3.75 kN/m²Floor Finish 1 kN/m²Live load 2 kN/m²Factored Load 10.125 kN/m²

Design of Beam

Categorization of Beam:

Category-I: Both end continuous

Beam number

Category II: One end simply supported and other continuous

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,18,19,20,21,22

25,26,27,28,29,31,32,33,34,35,36

17,24,31,23,30,37

Location from

Left support

Mu

(kNm)

B

(mm)

d

mm

Mu/bd²

N/mm²

type Pt Pc Ast

mm²

Asc

mm²

0.2 -33.83 230 515 0.5545 Singly 0.34 - 402.12 -

2.0356 21 230 515 0.344 Singly 0.34 - 402.12 -

4.33 -27.51 230 515 0.45 Singly 0.34 - 402.12 -

Summary of longitudinal reinforcement provided in beam

At Supports Top bars

Bottom bars

2-16mmφ,Ast=402.12mm², with 160mm internal radius at bend

2-16mmφ

At centre Top bars

Bottom bars

2-16mmφ

2-16mmφ

Check for reinforcement

(IS13920; clause 6.2.1)

1.10.3.2 (a) Minimum two bars should be continuous at top and bottom.

Here 2-12mmφ bars continuous throughout at top and 3-12mmφ are continuous at bottom. Hence Ok.

(b) Pt,min=0.24√fck/fy =0.24*√25/415

=0.00289 i.e 0.289%

Ast,min= (0.289*230*515)/100 =365.585 mm²

Provided reinforcement is more. Hence ok.

(IS 13920; clause 6.2.2)

Maximum steel ratio on any face at any section should not exceed 2.5, i.e,

Ast,max= (2.5*230*515)/100 =2961.25mm².

Provided reinforcement is less. Hence ok.

(IS 13920; clause 6.2.3)

The positive steel at a joint face must be at least equal to half the negative steel at that face

Joint A

Half the negative steel =402.12/2 =201.06mm²

Positive Steel = 402.12>201.06 mm²

Hence Ok.

(IS 13920; clause 6.2.4)

Along the length of the beam,

Ast at top or bottom>0.25 Ast at top at joint support

Ast at top or bottom >0.25*402.12

>100.53 mm²

Hence ok.

(IS 13920; clause 6.2.5)

At the external joint, anchorage at top and bottom bars=Ld in tension + 10db

Ld for Fe415 M25 concrete is 40.3db

Here minimum anchorage= 40.3 db+10db= 50.3db

(i.e 50.3*16 = 804.8 mm ) into the column.

1.10.4 Web reinforcement

Vertical hoops shall be used as shear reinforcement (IS 13920; clause 6.3.1 and 3.4)

Hoop dia≥6mm

≥ 8mm for span more than 5m

Clear span= 4445-200-115 =4130mm

Use 6 mm dia 2 legged hoops

The moment capacity of the beam at the support are 74.77kNm

1.2(DL+LL) for UDL load on the beam 1.2(11.3145+1.666)=15.5766KN/m

Va(DL+LL)=Vb(DL+LL)=0.5*15.5766*4.445= 34.61899 KN

Beam B17

Sway to right

Vu,a=VaD+L-1.4[Mur,maxAS+Mur,maxBH]/Lab

=34.618-1.4[74.77+74.77]/4.445

=-12.48KN

Vu,b=81.717 KN

Sway to left

Vu,a=VaD+L+1.4[Mur,maxAH+Mur,maxBS]/Lab

=81.717 KN

Vu,b=-12.48KN

Maximum design shear at A= 81.717KN

Maximum design shear at B=81.717 KN

B17 A B

Distance(mm) 0 2.035 4445

36.1088 12.31 -34.38

Shear from analysis(KN)

Shear due to yielding(KN) 81.717 81.717 -81.717

Design shears(KN) 81.717 81.717 -81.717

The shear force to be resisted by hoops shall be greater of:

-calculated shear force as per analysis

-Shear force due to formation of plastic hinges at the both ends of the beam plus the factored gravity load in the span.

As per clause 6.3.5 of IS 13920:1993, the first stirrups shall be within 50mm from the joint face.

Spacing s, of hoops within 2d(2*515=1030mm) from the support shall not exceed:

1. d/4 =128.75mm2. 8 times diameter of the smallest longitudinal bar= 8*16 =128mm.

Hence spacing of 128 mm c/c governs.

Elsewhere in the span, spacing

s≤d/2 =257.5 mm

Maximum nominal shear stress in the beam

Tc= 81.717*1000/(230*515) =0.689 N/mm²< 3.1 N/mm² (Tc,max for M25 mix)

DESIGN OF COLUMN

Since all the column are laterally supported by the wall, they can be assumed to be braced and column end not free to sway. For the top column Leff may be taken equal to L and for intermediate storeysLeff may be taken approx. as 0.8L. Thus, depending on the ratio of Leff/b for column in different stroreys having width 400mm or 230 mm the column type is decided.Details of column type for slenderness for columns in different storeys

Top storey Middle storey bottom storey

Column length L mm 3000 3000 3533

Beam depth D mm 550 550 550

Unsupported legth L mm 2450 2450 2983

Leff 2450 1960 2386.4

Width of column b mm

Leff/b

230/400

10.65/6.125

230/400

8.52/4.9

230/400

10.375/5.966

Column type Short/short Short/short Short/short

Categorization of Column

Category 1: Axially loaded column: All interior column

Category 2: Column subjected to axial compression and uniaxial bending: the entire column on the side between corners.

Category 3: Column under axial compression and biaxial bending: corner column.

Design of Column C2

This column is subjected to biaxial bending.

Column C2Size of column (400*400)Design loads Pu=71.0996 kNm

Mux=5.9894 kNmMuy=17.853 kNm

Clear cover 40 mmDiameter of bar to be used 16 mmd' 48 mm

Check for short ColumnUnsupported length Lx = 3 m

Ly = 3 m

Effective length Lex= 2.4m Ley= 2.4 mLex/D 6<12 ShortLey/B 6<12 ShortLet us assume Pt= 1%Puz 2779.2 kNMoment due eccentricityex ey Mex Mey

0.02 0.02 1.42 1.42Moment to be considered in the designPu/fck b D 0.018P/fck 0.04As per chart 45 of SP16 For d'/D = 0.15Mux1/fck b D² 0.088

As per chart 45 of SP16 For d'/D = 0.15Muy1/fck b D² 0.088Mux1 140.8 kNmMuy1 140.8 kNmPu/Puz 0.031

n 1

(Mux/Mux1)n + (Muy/Muy1)n < 1 IS 456-2000 Cl 39.6

0.17

Result < 1 okProvide

Reinf. details at support

Nos. dia Asc p prov.

mm mm2 %4 16

4 16 1608.50 1.01

Design of transverse reinforcement

As per IS 13920:1993

A.Design of general hoops

Rectangular hoop may be used in rectangular column. Here rectangular hoop of 8mm dia are used.

Here h=400-40-8(using 8mm dia ties) =352mm>300mm (clause 7.3.1 IS13920:1993)

Spacing of bars =352/4 =88mm which is more than 75 mm .Thus cross ties on all bar

are required.(Clause IS456:26.5.3.2(b))

Provide 3 open cross tie in X and Y direction.

B. Spacing of hoops

As per IS 456 clause 26.5.3.2(c) , the pitch of ties shall not exceed

a) 400mmb) 16*16= 256mm…………………1c) 300 mm

The spacing of Hoops can also be checked from clause 26.5.1.5

b*d= 400*352 Using 8mm dia hoops.

Asv=4*50 =200mm²

The spacing should not exceed

0.87fy*Asv/(0.4b) = 564.14 mm

0.75d= 0.75*352 =264mm

256 mm

As per IS 13920:1993

Spacing of hoops ≤b/2 of column = 400/2 =200mm

Maximum spacing of stirrups is 200mm.

Design Shear

As per clause 7.3.4 IS13920:1993, the design shear force for column shall be maximum of:

a. Calculated factored shear force as per analysisb. Factored shear force given by:

Vu=1.4[(Mur,max,bL+Mur,max,bR)/Hst]

From analysis shear force is calculated as 25.7955 KN

Mur,max,bL= 81.717 KN Mur,max,bR= 81.717 KN

Hst =3m

Vu=1.4[(81.717+81.717)/3] =76.2692 KN

So design shear force is 76.26 KN.Consider column as doubly reinforced beam with

b=400 mm and d=352mm

As=0.5Asc = 804.25mm²

For the member subjected to axial compression Pu,

Δ= 1 + (3*Pu)/(Ag*fck) = 1.05

≤ 1.5 (As per IS456 Clause 40.2.2)

Take δ=1.05

100As/bd = 1.142

tc = 0.68 N/mm² and δtc= 0.714N/mm²

Since tv<tc minimum shear reinforcement will be provided. Spacing of hoops is 200 mm.

Design of special confining hoops

As per clause 7.4.1 of IS13920:1993 special confining reinforcement shall be provided over the

length L0,where flexural yielding may occur. Lo shall not be less than

i)D of member i.e 500 mm

ii)Lc/6 = (3000-550)/6 =408.33 mm

iii)450mm

Provide the confining reinforcement from top of column upto 460 mm.

As per clause 7.4.2 IS13920:1993 special confining reinforcement shall extend minimum

of 300mm into the footing.

AS per clause 7.4.8 IS13920:1993 area of special confining reinforcement is given is given as:

Ash= 0.18s h fck/fy [(Ag/Ak)-1]

Here h= longer dimension of rectangular confining hoop measured to its outer face but not

grater than 300mm.

S=62.57mm≤100mm

Provide 4 legged 8mm dia confining hoops in both direction @100mm c/c.

Design of Slab

A reinforced concrete slab covers a relatively large area compared to a beam or column. Therefore,

volume of concrete and hence, dead load is large in case of slab. A small reduction in the depth of

slab therefore, leads to considerable economy. But, care has to be taken to see that its performance

(serviceability) is not affected due to excessive deflection and cracking.

In this Project, As per IS 456-2000 slabs are considered as one-way slab if ‘ly/lx’ ratio is

greater than two, and two way slabs if ‘ly/lx’ is less than or equal two. And have designated

as the slabs as:

One way slabs are S14,S15,S16,S17,S18,S19,S20,S21,S22,S23.

Two way slabs are S1,S2,S3,S4,S5,S6,S7,S8,S9,S10,S11,S12,S13.

DESIGN OF ONE-WAY SLAB(S2)Effective SpansLx = 3.652 m.

Assmed overall thickness of slab = 150 mm.Effective depth of the slab = 125 mm.DL = 3.75 kN/m^2Floor Finishes = 1.5 kN/m^2

Live Load = 1.5 kN/m^26.75 kN/m^2

Load 6.75 kN/m^2

Factored load = 10.125 kN/m^2

Bending moment wl^2/8 11.25 kNm/m

Mu/b*d^2 0.72 N/mm^2Pt

1 N/mm^2 0.3271.15 N/mm^2 0.3430.72 N/mm^2 0.297

Reinforcement required: 371.25 mm^2.Dia = 10 mm 79 mm^2.Spacing = 213 mm.Reinforcement provided

10 Tor @ 150 mm C/C.Area Provided = 527 mm^2.% provided = 0.422

Disribution steel

For Fe-415, pt.min = 0.12%

Ast = 180 mm2

Dia of bars = 8 mm

Spacing, srqrd = 279.11 mmReinforcement provided

8 Tor @ 200 mm C/C.

Check for deflection Fs = 170From fig. 4 of code, modification factor = 1.86Minimum effective depth required = 76

OK

DESIGN OF TWO-WAY SLAB (S1)Ly = 5.182 m.Lx = 4.445 m.Ly/Lx = 1.17

Edge conditions = twoadjacent edges discontinuous

Assmed overall thickness of slab = 150 mm.Effective depth of the slab = 126 mm.DL = 3.75Floor Finishes = 1.5Live Load = 1.5

6.75 kN/m^2Load 6.75 kN/m^2Factored w*(Lx^2) = 200.05

Short span moment coefficients:(Table 26 of IS 456 - 2000)Ly/Lx= 1.1 0.053Ly/Lx= 1.2 0.06Ly/Lx= 1.17 0.058

(-)Mx = 11.58 kNm/mMu/b*d^2 0.73 N/mm^2

1.05 N/mm^2 0.3111.1 N/mm^2 0.327

Pt 0.73 N/mm^2 0.209Reinforcement required: 263.34 mm^2.Dia = 8 mm 50 mm^2.Spacing = 190 mm.

Tor @ 170 mm C/C.Area Provided = 294 mm^2.For Positive moment at mid-span:

Ly/Lx= 1.1 0.04Ly/Lx= 1.2 0.045Ly/Lx= 1.17 0.044

Positive moment at mid-span (+)Mx = 8.7 kNm/mMu/b*d^2 0.55 N/mm^2

0.8 N/mm^2 0.2330.85 N/mm^2 0.248

Pt 0.55 N/mm^2 0.158Reinforcement required: 199.08 mm^2.Spacing = 251 mm.

Tor @ 240 mm C/C.Area Provided = 208 mm^2.% provided 0.17Long span moment coefficients

y 0.037 (-)My = 7.4 kNm/mMu/b*d^2 0.47 N/mm^2

0.7 N/mm^2 0.2030.75 N/mm^2 0.218

Pt 0.47 N/mm^2 0.134Reinforcement required: 168.84 mm^2.Dia = 8 mm 50 mm^2.Spacing = 296 mm.

Tor @ 290 mm C/C.Area Provided = 172 mm^2.

y 0.028 (+)My = 5.6 kNm/mMu/b*d^2 0.35 N/mm^2

0.55 N/mm^2 0.1580.6 N/mm^2 0.172

Pt 0.35 N/mm^2 0.102Reinforcement required: 128.52 mm^2.Spacing = 389 mm.

Tor @ 120 mm C/C.Area Provided = 417 mm^2.Check for deflection Fs = 230From fig. 4 of code, modification factor = 2.10Minimum effective depth required = 81

OK