Design Calculations

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DESIGN CALCULATION FOR

COMMERCIAL CENTRE, CORPORATE OFFICE.

Client: JUMEIRA GOLF ESTATES

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CONTENTS

1) INTRODUCTION 03

2) GEOMETRY 03

3) GEOTECHNICAL DATA 03

4) MATERIAL 03

5) CODES AND DESIGN STANDARDS 03

6) LOADINGS 03

7) MODELLING 10

8) DESIGN OF RAFT FOUNDATION 10

9) DESIGN OF RETAINING WALL 15

10) DESIGN OF SHEAR WALL 18

11) DESIGN OF RAMP 27

12) DESIGN OF INSITU BEAMS 30

13) DESIGN OF INSITU SLABS 38

14) DESIGN OF STEEL ROOF 39

REFERENCES.

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1. INTRODUCTION M/s. Jumeirah Golf Estates is setting up a commercial centre in Jumeirah. This report

deals with the analysis and design of raft, shear wall and some in-situ beams and slabs

2. GEOMETRY The building utilizes a reinforced concrete structure. All floors are made by precast

hollow core slab supported on precast beams. Pre cast columns and shear walls are used

for supporting the floors. The building is supported by raft foundation. The length of the

building is 110m and breadth is 82.35 m. This has 1 basement floor and 3 floors above

that.

3. GEOTECHNICAL DATA As per the soil investigation report, the average bearing pressure of the soil is taken as

150 kN/m2 and modulus of subgrade reaction as 7500kN/m3.

4. MATERIAL

M40 grade concrete and Fy 460 steel (conforming to BS: 4449-1997) with moderate

exposure condition as per BS: 8110-1- 1997 is assumed.

5. CODES AND DESIGN STANDARDS

• BS 8110 Part 1:1997 Code of Practice for design and construction

• BS 6399 Part 1: 1996 Code of practice for dead and imposed loads

• BS 6399 Part 2: 1997 Code of practice for wind loads

• BS 6399 Part 3: 1988 Code of practice for imposed loads

• UBC 1997 Uniform Building Code.

6. LOADINGS

6 .1 Dead Load & Live Load (BS 6399 Part I, Part II)

In addition to the self-weight of the structure the following dead & Live loads are taken

into account.

a) Pitched Roof

Concrete Roof Tile = 0.51 kN/m2

200 mm Thick Slab = 5 kN/m2

Ceiling and Services = 0.80 kN/m2

Live Load =1.5 kN/m2

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b) Flat Roof

Hollow Core = 3.75 kN/m2

Fill = 0.19 kN/m2

Screed = 0.7 kN/m2

Ceiling and Services =1.8 kN/m2

Live Load =1.5 kN/m2

c) 2nd Floor Hollow Core = 3.75 kN/m2 Fill = 0.19kN/m2 Ceiling and Services = 0.8 kN/m2

Raised Floor = 0.7 kN/m2

Partition Wall = 3 kN/m2 Marble = 0.7 kN/m2 Screed = 0.7 kN/m2 Live Load = 3 kN/m2 d) 1st Floor Hollow Core =3.75 kN/m2 Fill = 0.19 kN/m2 Ceiling and Services = 0.8 kN/m2

Raised Floor =0.7 kN/m2

Partition Wall = 3 kN/m2

Marble = 0.7 kN/m2

Screed = 0 .7 kN/m2 Live Load = 3 kN/m2 e) Ground Floor Hollow Core = 3.75 kN/m2 Fill = 0.19kN/m2 Ceiling and Services = 0.8 kN/m2

Raised Floor = 0.7 kN/m2

Partition Wall = 3 kN/m2

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Marble = 0.7 kN/m2 Screed = 0.7 kN/m2 Live Load = 3 kN/m2 f) Ground Floor (Grid A TO B1) Slab (350mm) = 8.3 kN/m2 Fill (0.75m) =15kN/m2 Ceiling and Services = 0.8 kN/m2

Live Load = 3 kN/m2

g) Ramp

Slab (250mm) = 6.25 kN/m2

Finishing = 2 kN/m2

Live Load = 5 kN/m2

h) Stair Case

Slab (250mm) = 6.25 kN/m2

Steps = 2.05kN/m2

Finishing = 2 kN/m2

Live Load = 5 kN/m2

6.2 Wind Load Wind load corresponding to basic wind speed of 25 m/s is considered as per BS: 6399-

Part II

Data available

Height of building = 20m

Location = Dubai

Basic wind speed = 25 m/s

Longest side = 110.3m

Shortest side = 39.15m

Site Altitude = 0m

The dynamic pressure is given by

qs = 0.613Ve²

Ve = Effective wind speed (Clause 2.2.3, BS: 6399- Part II)

Ve =Vs×Sb

Vs = Site speed from (Clause 2.2.2, BS: 6399- Part II)

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Sb = Terrain and building factor (Clause 2.2.3.3, BS: 6399- Part II)

Vs= Vb×Sa×Sd×Ss×Sp

Where

Vb=Basic wind speed = 25m/s (Clause 2.2.1, BS: 6399- Part II)

Sa=Altitude factor = 1+0.001∆s (Clause 2.2.2.2, BS: 6399- Part II)

Sa =1 Sd=Directional factor =1

Ss=Seasonal factor =1(Clause 2.2.2.4, BS: 6399- Part II)

Sp=Probability factor =1(Clause 2.2.2.5, BS: 6399- Part II)

Then

Vs= Vb×Sa×Sd×Ss×Sp

= 25×1×1×1×1

= 25m/s

Ve = Vs × Sb

Where Sb =1.77(Table 4 BS: 6399- Part II) with respect to He = 20m

Ve = 25×1.77

= 44.25 m/s

Therefore qs = 0.613× Ve²

=0.613×44.25²

= 1.2 KN/m²

6.3 Earthquake load The earthquake forces are considered as per UBC 1997. The loads are applied in two

horizontal directions.

CRITERIA FOR SELECTION:

1) 1629.2 Occupancy Criteria:

The structure shall be placed in one of the standard occupancy category and an

importance factor of 1.0 shall be assigned I=1.0

2) 1629.4 Site Seismic Hazard Characteristics

Seismic hazards characteristics for the site shall be established based on the seismic zone

and proximity of the site to active seismic source site soil profile characteristics and the

structure is importance factor. The site shall be assigned a seismic zone and each

structure shall be assigned a seismic zone of factor Z

Z=2A

3) 1629.5 Configuration Requirement

The structure has no significant physical discontinuities in plan or vertical configuration

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or in their lateral force resisting system. Therefore the structure has regular and simple

with clear and direct path for transmission of seismic forces.

4) 1629.6 Moment Resisting Frame System:

Structural system with an essential complete space frame providing support for gravity

loads. Moment resisting frames provide resistance to lateral load primarily by flexural

action of members.

5) 1629.7 Height Limits:

The structure is in seismic zone 2A, there is no limit.

6 ) 1629.8 Calculation Lateral Force :

The static lateral force procedure shall be used in accordance with section 1630

7) 1630.1 Earthquake Loads:

The structure shall be designed for ground motion producing structural response

and seismic forces in any horizontal direction. Seismic design shall be carried out in

accordance with Uniform Building Code 1997, volume 2, Chapter 16 division IV

Building Criteria:

As per table 16 –k, UBC 1997 we have chosen standard occupancy for the building.

Seismic Importance Factor I =1.0

Wind Importance Factor Iw = 1.0

Seismic Importance Factor (for panel connections) Ip =1.0

Soil Profile Type = SC

Dubai is situated in a low seismic zone region. However seismic zone 2A is taken for

design.

TABLE 16-I, Seismic zone Factor = 0.15

TABLE 16-Q, Seismic Factor Ca = 0.18

TABLE 16-R, Seismic Factor Cv = 0.25

Structural Configuration:

The structure has no significant physical discontinuities in plan or vertical configuration

or in their lateral force resisting system. Therefore the structure is regular and simple

with clear and direct paths for transmission of seismic forces.

R (numerical coefficient representative of the inherent over strength and global ductility

capacity of lateral force resisting systems as per 16-N or 16-P) =5.5

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Lateral force procedure

Simplified static approach is applicable

Structural period

T = Ct (hn) 3/4 hn=20m

Ct = 0.0731(in SI units)

T =0.0731(20)3/4 = 0.69 Seconds

b) Calculation of Base Shear

The total design base shear in a given duration

V = WTRICv ×⎟⎠⎞

⎜⎝⎛

×× < W

RICa ×⎟⎠⎞

⎜⎝⎛ ×5.2

V = WTRICv ×⎟⎠⎞

⎜⎝⎛

×× > 0.11Ca×I×W

Where

W =Total load of structure =229570kN

Total Design Base shear = WTRICv ×⎟⎠⎞

⎜⎝⎛

×× = 229570

69.05.5125.0

×⎟⎠⎞

⎜⎝⎛

××

= 15123kN

The distribution of base shear

along vertical direction ( )∑ =

×

××−= n

i ii

xxt

hwhwFV

1

Where Ft = 0 since T<0.7secs

Table-1: Base shear distribution at different storey levels.

Storey Label Height(hx)

In metre

Seismic

weight(Wx)

In kN

Base shear(Fx)

( )∑ =

×

××−= n

i ii

xxt

hw

hwFV

1

Roof 4.16 20998 1034kN

2nd floor 4.55 40647 2190.4kN

First floor 4.55 44564 2401kN

Ground floor 6.5 123361 9496kN

6.4 Temperature load (As per UBC 1997)

With reference to the size of the building it is necessary to consider the thermal

effect of the environment on the whole structure. In order to avoid additional self-

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straining (creep and shrinkage and additional curvature in the members under thermal

gradient) after the design of the structure we have checked the whole structure under the

thermal effect. All of the members have pass safely the additional stress due to new load

combinations employed the thermal effect as a new load case except some perimeter

columns and beams which needed to be modified in terms of No. of reinforcements.

6.5 Load Combinations The following load combinations are considered for the analysis and the critical

load combination is taken for the design of the structure.

1. 1.4Dead load + 1.6Live load

2. 1.4Dead load ± 1.4Wind load(X)

3. 1.4Dead load ± 1.4Wind load(Y)

4. 1.2Dead load + 1.2Live load± 1.2Wind load(X)

5. 1.2Dead load + 1.2Live load± 1.2Wind load(Y)

6. 1.32Dead load + 0.55Live load± 1.11EQ(X)

7. 1.32Dead load + 0.55Live load± 1.11EQ(Y)

8. Dead load ± 1/4EQ(X)

9. Dead load ± 1/4EQ(Y)

10. 1.4Dead load + 1.6Live load ± 1.2Temperature Load

11. 1.4Dead load ± 1.4Wind load(X) ± 1.2Temperature Load

12. 1.4Dead load ± 1.4Wind load(Y) ± 1.2Temperature Load

13. 1.2Dead load + 1.2Live load± 1.2Wind load(X) ± 1.2Temperature Load

14. 1.2Dead load + 1.2Live load± 1.2Wind load(Y) ± 1.2Temperature Load

15. 1.32Dead load + 0.55Live load± 1.11EQ(X) ± 1.2Temperature Load

16. 1.32Dead load + 0.55Live load± 1.11EQ(Y) ± 1.2Temperature Load

17. Dead load ± 1/4EQ(X) ± 1.2Temperature Load

18. Dead load ± 1/4EQ(Y) ± 1.2Temperature Load

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7. MODELLING

The proposed building is modeled as a three dimensional structure using a

standard finite element software “Etabs” as shown in the Fig.1. The beams and columns

are modeled as frame elements and the slabs & walls were modeled as shell elements. At

the bottom of the columns raft foundation were modeled and soil spring value was given

as per the soil investigation report. Now the appropriate loadings were given and a static

earth quake analysis was carried out to obtain the design forces.

Fig.1:-Finite Element model of Building

8.0 DESIGN OF RAFT FOUNDATION

The rafts were modeled throughout the area of the building. The soil parameters

used in the model were as per the soil investigation report. The Safe Bearing Capacity of

the soil assumed was 150kN/m2. The soil springs were modeled below the raft

considering the spring value of 7500kN/m3. The Fig2 and Fig3 shows the bending

moment diagrams along X and Y direction respectively from the SAFE analysis. The

sample calculation for the design of raft is given below.

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Negative Positive

Fig.2:-Bending Moment X-Direction (M11)

Fig.3:-Bending Moment Y-Direction (M22)

Positive Negative X

Y

X

Y

Maximum Sagging Moment

Maximum Hogging Moment

Maximum Sagging Moment

Maximum Hogging Moment

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Sample Calculation:-

Assume M40 grade concrete and Fy-460 steel.

From Fig. 2,

The maximum sagging bending moment in X direction in slab

M11 = 1700 kNm/m

Maximum hogging bending moment in X direction in slab

M11 = 690kNm/m

From Fig. 3,

The maximum sagging bending moment in Y direction in slab

M22 = 1960kNm/m

Maximum hogging bending moment in Y direction in slab

M22 = 700 kNm/m

Design of Bottom Reinforcement in X Direction:-

Depth of slab provided = 1200mm

Clear cover assumed = 75mm

Effective depth(d) = 1112.5mm

Moment = 1700kNm

u2

Mb d

= 1.37

From Chart No.2 BS 8110 Part 3

Percentage of steel required = 0.38%

Area of steel required = 4247mm2

Area of steel provided in the section =5359mm2>4247

Hence Safe

Design of Top Reinforcement in X Direction:-



Effective depth (d) = 1112.5mm

Moment = 690kNm

u2

Mb d

= 0.56



Area of steel required = 2011.5mm2

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Area of steel provided in the section =3266mm2>2011.5

Hence Safe

Design of Bottom Reinforcement in Y Direction:-



Effective depth (d) = 1100mm

Moment = 1960kNm

u2

Mb d

= 1.7




Area of steel provided in the section =5359mm2>4752

Hence Safe

Design of Top Reinforcement in X Direction:-



Effective depth (d) = 1100mm

Moment = 700kNm

u2

Mb d

= 0.58



Area of steel required = 2011.5mm2

Area of steel provided in the section =3266mm2 >2011.5

Hence Safe

Design for shear:-

Check for punching shear:-

a) At the face of support:-

The maximum axial load from analysis = 7665kN

Breadth of column = 600mm

Depth of column = 800mm

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Perimeter = 2 x 600 + 2 x 800

= 2800mm

Shear stress (ν) = 5.11122800

107665 3

××

= 2.46N/mm2<0.8 fck =5N/mm2

Hence safe

a) The critical section for shear is 1.5 x effective depth = 1.5 x 1112.5 =1668.75mm

from the column face, thus the length of the perimeter

= 2(600+1668.75 x2)+2(800 +1668.75 x 2) = 16150mm

Shear stress (ν) = 5.111216150

107665 3

××

= 0.42N/mm2

dAs100

= 0.48%

From Table3.8 BS8110-1:1997

Allowable shear stress νc = 0.49N/mm2>0.42N/mm2

Hence Safe

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9.0 DESIGN OF RETAINING WALL

9(a). Height=5m

The retaining wall is analysed as fixed at bottom and free at top with a surcharge

load of 5kN/m2 and soil pressure of height 5m as shown in Fig-4

Unit weight of soil (γ) =18kN/m3

Angle of repose = 330

Height of soil fill (h) = 5m

Surcharge Load = 5kN/m2

Equivalent height of soil = 5/γ =0.278m

Soil pressure due to surcharge = 5/γ x γ x (1-sinφ)/ (1+sinφ)

= 1.476kN/m2

Soil Pressure (at bottom of retaining wall) due to 5m height of soil

= (1-sinφ)/ (1+sinφ) x γ x h

= 26.55kN/m2

Fig-4

Strength of Concrete(fcu) 40 N/mm2

Strength of Steel(fy) 460 N/mm2

Modulus of Elasticity(Ec) 28 kN/mm2

Modulus of Elasticity(Es) 200 kN/mm2

B 1000 mm

Over all Depth 300 mm

Cover(Cmin) 50 mm

1.476kN/m2 26.55kN/m2

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d 240 mm

Moment 116 kNm

dia of bar 20 mm

Spacing 150 mm

area 2093.333 mm2

neutral axis depth(Xu) 94 mm

Stress in steel(Fs) 265 N/mm2

Strain in steel 0.001327 mm

Srain in Concrete at Y1((d+x/2) from

top) 6.63E-04 mm

Srain in Concrete at Y2(bottom face) 1.87E-03 mm

Em at Y1((d+x/2) from top) 6.E-04 mm

Em at Y2(bottom face) 1.64E-03 mm

acr for Y1((d+x/2) from top) 85 mm

acr for Y2(bottom face) 75 mm

Crack width at Y1((d+x/2) from top) 0.10 mm

Crack width at Y2(bottom face) 0.29 mm

Since the crack width is less than 0.3mm, the provided reinforcement (T20-150) is safe.

9(b). Height=3m

The retaining wall is analysed as fixed at bottom and free at top with a surcharge

load of 5kN/m2 and soil pressure of height 3m as shown in Fig-5

Unit weight of soil (γ) =18kN/m3

Angle of repose = 330

Height of soil fill(h) = 3m

Surcharge Load = 5kN/m2

Equivalent height of soil = 5/γ =0.278m

Soil pressure due to surcharge = 5/γ x γ x (1-sinφ)/ (1+sinφ)

= 1.476kN/m2

Soil Pressure (at bottom of retaining wall) due to 5m height of soil

= (1-sinφ)/ (1+sinφ) x γ x h

= 16kN/m2

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Fig-5

Strength of Concrete(fcu) 40 N/mm2

Strength of Steel(fy) 460 N/mm2

Modulus of Elasticity(Ec) 28 kN/mm2

Modulus of Elasticity(Es) 200 kN/mm2

B 1000 mm

Over all Depth 300 mm

Cover(Cmin) 50 mm

d 242 mm

Moment 32 kNm

dia of bar 16 mm

Spacing 150 mm

area 1339.733 mm2

neutral axis depth(Xu) 79 mm

Stress in steel(Fs) 111 N/mm2

Strain in steel 0.000554 mm

Srain in Concrete at Y1((d+x/2) from

top) 2.77E-04 mm

Srain in Concrete at Y2(bottom face) 7.51E-04 mm

Em at Y1((d+x/2) from top) 1.E-04 mm

Em at Y2(bottom face) 3.78E-04 mm

acr for Y1((d+x/2) from top) 92 mm

acr for Y2(bottom face) 74 mm

Crack width at Y1((d+x/2) from top) 0.03 mm

Crack width at Y2(bottom face) 0.07 mm

Since the crack width is less than 0.3mm, the provided reinforcement (T16-150) is safe.

1.476kN/m2 16kN/m2

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10.0 DESIGN OF SHEAR WALL

The shear wall is modeled as pier element (See Etabs Model) and was labeled as shown

in fig. Each area object that makes up a part of a wall is assigned as one pier label. The

walls are designed as compression elements under the combined action of in-plane

bending and axial forces. The design of the shear wall was done based on BS 8110-1997.

One sample design calculation for the shear wall (Pier P2) is given below.

P15 P16

P1

P2

P3

P4 P5

P6

P7

P8

P9

P13

P10

P12

P11 P14

Fig-6:- Labeling of shear wall

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Sample Calculation (Shear wall-Pier P2)

Datas

Strength of Concrete(fcu) = 40N/mm2

Strength of Steel(fy) = 460N/mm2

Modulus of Elasticity(Ec) = 28N/mm2

Modulus of Elasticity(Es) = 200N/mm2

Modular Ratio(m) = 7.14

Length(L) = 5000mm

Thickness(t) =200mm

From Etab Analysis,

Maximum Axial Load(Ultimate)- Nu1 =3400kN

Minimum Axial Load(Ultimate)-Nu2 =575kN

Maximum Moment (Ultimate )-Mu1 =3827kNm

Maximum Axial Load(Service)-N1 =2430kN

Minimum Axial Load(Service)-N2 =480kN

Maximum Moment (Service)-M1 = 2734kNm

Reinforcement Ratio Provided(r) = 0.0136

Check for Ultimate Strengths

a) Ultimate Compressive Strength

Nu=(0.4fcu + 0.72fy × r) × t × L = 20504.32kN > 3400kN

Hence Safe

b)Ultimate moment

For maximum Compression

= ⎟⎟⎠

⎞⎜⎜⎝

⎛

u

u

NN 1 = ⎟

⎠⎞

⎜⎝⎛

32.205043400

From Chart-1

Then

Mumax = 0.145×5×20504.32 =

Hence Safe

⎟⎟⎠

⎞⎜⎜⎝

⎛

uNN = 0.17

14865.63kNm >3827kNm

⎟⎟⎠

⎞⎜⎜⎝

⎛× LN

M

u

u max = 0.145

20

For minimum Compression

⎟⎟⎠

⎞⎜⎜⎝

⎛

u

u

NN

= ⎟⎟⎠

⎞⎜⎜⎝

⎛

u

u

NN 1 = 03.0

32.20504575

=⎟⎠⎞

⎜⎝⎛

From Chart-1

Mumax = 0.11×5×20504.32

Hence Safe

Check for stress limits:-

As per BS 8110-1:1997

Max:Permissible Stress in Concrete = cuf4.0 = 16 N/mm2

Max:Permissible Stress in Steel = 0.87fy = 400.2 N/mm2

For Max: Compression

e = ⎟⎟⎠

⎞⎜⎜⎝

⎛

1

1

NM =1125mm

Le =

50001125 =0.23

For Min: Compression

e = ⎟⎟⎠

⎞⎜⎜⎝

⎛

2

1

NM =5695mm

Le =

50005695 =1.14

For e/L = 0.23

From Chart-2 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛LN

NN

N/max

0

max 2.45

Then Nmax = 2.45×L

Nu1

= 2.45×5

2430

⎟⎟⎠

⎞⎜⎜⎝

⎛× LN

M

u

u max = 0.11

= 11277.38kNm>3837kNm

= 1190.7kN/m

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Compressive Stress in Concrete =

22maxc /16/43.5

0136.014.71(2007.1190

)1( mmNmmN

mrtN

<=⎟⎟⎠

⎞⎜⎜⎝

⎛×+×

=⎟⎟⎠

⎞⎜⎜⎝

⎛+×

=σ

Hence Safe

For e/L = 1.14

we have

Solve for (x/L) a = 1

b = 1.92

a(x/L)3-b(x/L)2-c(x/L)+d = 0 c = 0.66

d = -0.38

From Trial and Error Method X/L=0.2915

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛LN

NN

N/max

0

max

= 4.79

Then Nmax = 4.79×L

Nu1

= 4.79×5

480

Maximum Stress in Concrete(σc)

22max /16/1.20136.014.71(200

460)1(

mmNmmNmrt

N<=⎟⎟

⎠

⎞⎜⎜⎝

⎛×+×

=⎟⎟⎠

⎞⎜⎜⎝

⎛+×

=

Maximum Stress in steel =

mxL

c )1( −×σ 22 /2.400/43.3614.7)143.3(1.2 mmNmmN <=×−×=

Hence Safe

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

+)/5.01(/5.0

)1(LxmrLx

mr

⎟⎟⎠

⎞⎜⎜⎝

⎛×−××+×

×+=

)2915.05.01(0136.014.72915.05.0)0136.014.71(

= 460kN/m

22

Chart-1(N/N0--Mu/NuL)

Chart-2(Nmax/N0--e)

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Check For Shear

Shear Force(From Analysis) = 1700kN

Shear stress(τ) = =×××

50002008.0101700 3

2.13N/mm2

From Table-3.8BS 8110-1:1997

Shear stress of concrete(τc) = 0.82N/mm2

Area of steel required = 449mm2/m

Minimum area of steel required = 500mm2/m

Area of steel provided = 2103mm2/m

Hence Safe

The design result from Etabs is shown in Table-2

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Table-2 – Shear wall Design output- Etabs

Story Pier

Label Location Edge Bar

End Bar

End Spacing

Required Ratio of

Reinforcement

Provided Ratio of

Reinforcement Shear Reinforcement

ROOF P1 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0025 0.0143 500

SF P1 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0026 0.0143 500

FF P1 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0048 0.0143 500

GF P1 Top 16d 16d 150 0.0037 0.0143 500 Bottom 16d 16d 150 0.0025 0.0143 500



FF P2 Top 16d 16d 150 0.0025 0.0137 630.4 Bottom 16d 16d 150 0.0025 0.0137 624








FF P4 Top 16d 16d 150 0.0025 0.0136 824 Bottom 16d 16d 150 0.0067 0.0136 844.9







25


















FF P10 Top 16d 16d 150 0.0025 0.0135 897.4 Bottom 16d 16d 150 0.0045 0.0135 894






26







FF P13 Top 16d 16d 150 0.0025 0.0136 671.2 Bottom 16d 16d 150 0.0061 0.0136 663.1








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11.0 DESIGN OF RAMP

Fig-7: Finite Element Model of Ramp

The ramp is modeled as shown in Fig.7. The ramp is assumed to be supported on wall on

the two sides.

Design of Ramp slab

From the analysis,

The Maximum Sagging Moment in shorter direction= 70kNm

The Maximum Hogging Moment in shorter direction= 40kNm

The Maximum Sagging Moment in longer direction = 16kNm

The Maximum Hogging Moment in shorter direction= 0kNm

Design of Bottom Reinforcement in Shorter Direction:-



Effective depth = 172mm

Moment = 70kNm

u2

Mb d

= 2.37




Area of steel provided in the section =1340mm2

Design of Top Reinforcement in Shorter Direction:-




Moment = 40kNm

u2

Mb d

= 1.08

X

Z

Y

28





Design of Bottom Reinforcement in Longer Direction:-




Moment = 16kNm

u2

Mb d

= 0.54



Area of steel required (minimum) = 325mm2


|Design of Top Reinforcement in Longer Direction:-




Moment = 0kNm


Area of steel required (minimum) = 325mm2


Design of Ramp wall:-

Design of Vertical Reinforcement:-

Thickness of wall provided = 250mm



Moment (from analysis) = 30kNm

u2

Mb d

= 1




29


Design of Horizontal Reinforcement:-

Thickness of wall provided = 250mm



Since there is no horizontal moment

We have to provide minimum area of reinforcement



30

12. DESIGN OF INSITU BEAMS (Span-17.4m)

The analysis of the beam was done by modeling it as a frame as shown in Fig-8. The

moment at the ends of beam is released. The Bending Moment and shear Force

Diagrams are shown in Fig-8 (a), Fig-8 (b) respectively.

17400

6500

1750kN

120kN/m

11732kNm

1150kNm 1150kNm

2200kN

2200kN

250k

N

250k

N

890kN

(Ultimate)

(Ultimate)

Hinge Hinge

Fig-8: 2D Frame

Fig-8(a): Bending Moment Diagram

Fig-8(b): Shear Force Diagram

31

Grade of concrete = 60 N/mm2

Grade of steel = 460 N/mm2

Clear cover to reinforcement = 30mm

Width of the beam = 600 mm

Depth of the beam = 1600 mm

Design for mid-span moment:-

Diameter of bar = 32 mm


Moment (from analysis) = 11732kNm

Check for Compression Reinforcement:-

u2

Mb d

= 8.92< 0.156×60=9.36

Therefore we don’t require compression reinforcement

We have lever arm (z) = d x ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛ −+

9.025.05.0 k

= 1480 x ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛ −+

9.0148.025.05.0

= 1172.9mm

Area Tension steel required = zf

M

y95.0

= 9.117246095.0

1011732 6

×××

= 22889mm2

Area of steel provided in the section =24120mm2(30T32)

Hence Safe

Minimum Percentage of steel required at support = 0.13%


Area of steel provided in the section =8040mm2(10T32)

Hence Safe

32

Design for Shear:-

Shear force at face of support = 2200kN

Shear stress (v) = 1480600

102200 3

××

=2.47N/mm2<0.8 cuf =6.19N/mm2

Hence safe

Shear force at a distance‘d’ from the face of support = 1950kN

Shear stress = 1480600101950 3

××

= 2.19N/mm2

bdAs100

= 0.9%

From Table3.8 BS8110-1:1997

νc = 0.72N/mm2

We have v

vs

SA

= y

c

fvvb

95.0)( −

= 46095.0

)72.054.2(600×−

= 2.5

Spacing of 6 legged T12 stirrup required = 5.2

1136× = 270mm

Spacing of 6 legged T12 stirrup provided = 200mm

Hence safe

Check for Deflection (serviceability) (As per BS8110-2:1985):-

We have

As =24120mm2

h = 1600mm

b = 600mm

d = 1480mm

Total load (including total live load)

Concentrated load = 1190kN

Uniformly distributed load = 80kN/m

Permanent (1ncluding 25% live load only) = 966kN, 60kN/m

Concentrated load = 966kN

Uniformly distributed load = 60kN/m

33

Moment due to Total load = 8250kNm

Moment due to permanent load = 7340kNm

Short term deflection due to total load:-

We have

M= Asfs(d-3x )+

31 bhfct(h-x)

8250x106 = 24120 x fs x (1480- 3x ) +

31 x 600x1600xfctx (1600-x) ----------------------- (1)

Maximum tensile stress allowable in concrete (fct) = 1)()(×

−−

xdxh

= 1)1480()1600(×

−−

xx ------ (2)

We have

From the strain distribution

fc = ss

c fEE

xdx

××− )(

---------------(3)

and by equating tension and compression

21

cbxf = Asfs+ 21 bfct(h-x)-----------------(4)

By solving the above 4 equations using trial and error method

We have x= 658.7mm

fc = 33.37N/mm2

Short term curvature br1 =

c

c

Exf×

= 310327.65837.33××

= 1.58x10-6/mm

Short term deflection due to creep = 21 Lkrb

××

From Table3.1, BS8110-2:1985, k= 0.083

Deflection = 1.58x10-6 x 0.083 x 174002

= 39.7mm

34

Short term deflection due to Permanent load:-

We have

M= Asfs(d-3x )+

31 bhfct(h-x)

7340x106 = 24120 x fs x (1480- 3x ) +

31 x 600x1600xfctx (1600-x) ----------------------- (1)

Maximum tensile stress allowable in concrete (fct) = 1)()(×

−−

xdxh

= 1)1480()1600(×

−−

xx ------ (2)

We have


fc = ss

c fEE

xdx

××− )(

---------------(3)


21

cbxf = Asfs+ 21 bfct(h-x)-----------------(4)


We have x= 660.2mm

fc = 29.66N/mm2

Short term curvature br1 =

c

c

Exf×

= 310322.66066.29××

= 1.4x10-6/mm

Short term deflection due to creep = 21 Lkrb

××

From Table3.1, BS8110-2:1985, k= 0.083

Deflection = 1.4x10-6 x 0.083 x 174002

= 35.6mm

Long term deflection due to permanent loads:-

Reduced modulus of elasticity Eeff = )1( φ+

cE

Effective section thickness = perimetre

areasctheTwice /

35

= )6001600(2

16006002+×××

= 436.36mm

The value of creep coefficient (Ф) From Fig-7.1, BS8110-2:1985 for loading at 28 days

with indoor exposure condition is approximately 2

Eeff = )21(

32+

=10.67N/mm2

We have

M= Asfs(d-3x )+

31 bhfct(h-x)

7270x106 = 24120 x fs x (1480- 3x ) +

31 x 600x1600xfctx (1600-x) ----------------------- (1)

Maximum tensile stress allowable in concrete (fct) = 55.0)()(×

−−

xdxh

= 55.0)1480()1600(×

−−

xx ------ (2)

We have


fc = ss

eff fEE

xdx

××− )(

---------------(3)


21

cbxf = Asfs+ 21 bfct(h-x)-----------------(4)


We have x= 924.4mm

fc = 22.63N/mm2

Long term curvature br1 =

eff

c

Exf×

= 31067.104.92463.22

××= 2.29x10-6/mm

Long term deflection due to creep = 21 Lkrb

××

= 2.29x10-6 x 0.083 x 174002

= 57.54mm

36

Deflection due to shrinkage:-

csr1 =

ISsccs ××αε

cα = ffe

s

EE

=67.10

200 = 18.74

I = ( )223

)2

(12

xdAxbxbxsc −++ α

= 2.978x 1011mm4

Ss = ( )xdAs −

= 24120 x (1480-924.4)

= 13.41 x 106mm3

From Fig-7.2, BS8110-2:1985

csε = 327 x 10-6

Thus

csr1 = 11

66

1097.21041.1374.1810327

××××× −

=2.76 x 10-7/mm

Long term deflection due to shrinkage = 21 Lkrb

××

= 2.76x10-7 x 0.083 x 174002

= 6.93mm

Short term deflection due to total loads = 39.7mm

Short term deflection due to permanent loads = 35.6mm

Short term deflection due to non permanent loads = 39.7-35.6 =4.1mm

Long term deflection (permanent loads) = 57.54mm

Long term deflection (shrinkage) = 6.93mm

Total long term deflection = 68.57mm

Permissible deflection = l/250 =69.6mm

Hence safe.

37

13. DESIGN OF IN-SITU GROUND FLOOR SLAB (b/w grids 7, 8, E&F)

Material constants:-

Concrete fck = 40 N/mm2

Steel fy = 460 N/mm2

Shorter Span = 8100mm

Longer Span = 8700mm

Shorter span:-

Thickness of slab = 250mm

Diameter of bar = 16mm


Mid-span Moment = 62kNm

u2

Mb d

= 1.37




Area of steel provided in the section = 2010mm2/m

Hence Safe.

Support Moment = 20kNm

u2

Mb d

= 0.44





Hence Safe.

Longer span:-

Thickness of slab = 250mm

Diameter of bar = 16mm


Mid-span Moment = 60kNm

u2

Mb d

= 1.56




38


Hence Safe.

Support Moment = 18kNm

u2

Mb d

= 0.47





Hence Safe.

Check for Deflection:-

The total short term deflection from analysis = 8.86mm

The long term deflection from analysis = 19mm

The Total Deflection = 27.86mm

Permissible deflection = span/250 = 32.4mm

Hence safe

39

14. Design of Roof:-

The analysis of the roof was done by using modeling it as a frame as shown in Fig-9.

The frame is spaced at 4.05m apart.

Fig-9- Staad Model

Loadings:

Dead load

Concrete Roof Tile = 0.5 kN/m2

Ceiling and Services = 0.80 kN/m2

Live Load =1 kN/m2

Wind load:-

Wind pressure = 1.2kN/m2 (See Section 6.2)

From BS 6399 Part 2

External Pressure Coefficient Cp = 1.2

Internal Pressure Coefficient Cp = ±0.2

Max Wind Pressure = 1.2 x (1.2+0.2)

= 1.68kN/m2

Design of Purlins:

Provide purlins at 1.5m c/c

The purlins are designed as a simply supported beam with the following loads.

Total Dead Load coming in the Purlin = (0.5+0.8) x 1.5 = 1.95kN/m

Live Load = 1 x 1.5 = 1.5kN/m

Wind Load = 1.68 x 1.5 =2.52kN/m

From Staad analysis the section required is UB 125×65× 15mm

(As per BS 5950-2000)

40

Design of Main Beam:

Total Dead Load coming in the frame = (0.5+0.8) x 4.05 = 5.26kN/m

Live Load = 1 x 4.05 = 4.05kN/m

Wind Load = 1.68 x 4.05 =6.8kN/m

From Staad analysis the section required is UB 533×201× 102mm

(As per BS 5950-2000)

41

15. REFERENCES:

1. British standards, “Structural Use of Concrete”- Code of Practice for design and

construction (BS 8110-1:1997)

2. British standards, “Structural Use of Concrete”- Code of Practice for special

circumstances” (BS 8110-2:1997)

3. British standards, “Structural Use of Concrete”- Design Charts” (BS 8110-3:1997)

4. British standards, “Loading for Buildings”- (BS 6399-1,2,3:1996)

5. British standards, “Structural Use of Steel work”- (BS 5950:2000)

6. Universal Building Code -1997

7. A.W. Irvin ”Design of Shear wall Buildings,CIRIA(Construction Industry Research

and Information Association) Report.

8. L.J Morris and D.R Plum ”Structural Steel work Design to BS 5950”

9. Devdas Menon & S Unnikrisna Pillai, Reinforced Concrete Design , Tata Mc Graw-

hill publishing company Ltd. , Delhi

10. Prof S.Ramamruthum”Design of Reinforced Concrete Structures ,Dhanpat Rai

Publishing Company(P) Ltd,New Delhi

11. Joseph E Bowles, Foundation analysis and Design , Tata Mc-Graw Hill,

International edition, New Delhi 1988

12. Geotechnical report

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Design Calculations