Design Calculations

DESIGN CALCULATION FOR COMMERCIAL CENTRE, CORPORATE OFFICE. Client: JUMEIRA GOLF ESTATES

1

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.

2

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

3

b) Flat Roof Hollow Core

= 3.75 kN/m2

Fill

= 0.19 kN/m2

Screed

= 0.7 kN/m2


=1.8 kN/m2

Live Load

=1.5 kN/m2

c) 2nd Floor Hollow Core

= 3.75 kN/m2

Fill

= 0.19kN/m2


= 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


= 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


= 0.8 kN/m2

Raised Floor

= 0.7 kN/m2

Partition Wall

= 3 kN/m2 4

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


= 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: 6399Part 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) 5

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 6

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

7

Lateral force procedure Simplified static approach is applicable Structural period T

= Ct (hn) 3/4

Ct

= 0.0731(in SI units)

T =0.0731(20)3/4

= 0.69 Seconds

hn=20m

b) Calculation of Base Shear The total design base shear in a given duration V

⎛C ×I ⎞ ⎛ 2.5Ca × I ⎞ =⎜ v ⎟ ×W < ⎜ ⎟ ×W R ⎝ R ×T ⎠ ⎝ ⎠

V

⎛C ×I ⎞ =⎜ v ⎟ × W > 0.11Ca×I×W ⎝ R ×T ⎠

Where W

=Total load of structure =229570kN

⎛C ×I ⎞ Total Design Base shear = ⎜ v ⎟ ×W ⎝ R ×T ⎠

⎛ 0.25 × 1 ⎞ =⎜ ⎟ × 229570 ⎝ 5.5 × 0.69 ⎠ = 15123kN

The distribution of base shear along vertical direction =

(V − Ft )× wx × hx

∑

n

i =1

wi × hi

Where Ft = 0 since T<0.7secs Table-1: Base shear distribution at different storey levels.

Storey Label

Height(hx) In metre

Base shear(Fx)

Seismic weight(Wx) In kN

=

(V − Ft ) × wx × hx

∑

n

i =1

wi × hi

Roof

4.16

20998

1034kN

nd

2 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 self8

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 9

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.

10

Maximum Sagging Moment

Y

Maximum Hogging Moment

X Negative

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

Maximum Hogging Moment Maximum Sagging Moment

Y

X Negative

Positive

Fig.3:-Bending Moment Y-Direction (M22) 11

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

Mu b d2

= 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:Depth of slab provided

= 1200mm

Clear cover assumed

= 75mm

Effective depth (d)

= 1112.5mm

Moment

= 690kNm

Mu b d2

= 0.56


= 0.18%


= 2011.5mm2

12

=3266mm2>2011.5

Area of steel provided in the section Hence Safe

Design of Bottom Reinforcement in Y Direction:Depth of slab provided

= 1200mm

Clear cover assumed

= 75mm

Effective depth (d)

= 1100mm

Moment

= 1960kNm

Mu b d2

= 1.7


= 0.43%


= 4752mm2


=5359mm2>4752

Hence Safe

Design of Top Reinforcement in X Direction:Depth of slab provided

= 1200mm

Clear cover assumed

= 75mm

Effective depth (d)

= 1100mm

Moment

= 700kNm

Mu b d2

= 0.58


= 0.18%


= 2011.5mm2


=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 13

Perimeter

= 2 x 600 + 2 x 800 = 2800mm

Shear stress (ν)

=

7665 × 10 3 2800 × 1112.5

= 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)

7665 × 10 3 = 16150 × 1112.5 = 0.42N/mm2

100 As d

= 0.48%

From Table3.8 BS8110-1:1997 Allowable shear stress

νc = 0.49N/mm2>0.42N/mm2

Hence Safe

14

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

1.476kN/m2

26.55kN/m2

Fig-4

Strength of Concrete(fcu) Strength of Steel(fy)

40 N/mm2 460 N/mm2

Modulus of Elasticity(Ec)

28 kN/mm2

Modulus of Elasticity(Es)

200 kN/mm2

B Over all Depth Cover(Cmin)

1000 mm 300 mm 50 mm 15

d

240 mm

Moment

116 kNm

dia of bar

20 mm

Spacing

150 mm 2093.333 mm2

area neutral axis depth(Xu)

94 mm 265 N/mm2

Stress in steel(Fs) 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

16

16kN/m2

1.476kN/m2 Fig-5

Strength of Concrete(fcu) Strength of Steel(fy)

40 N/mm2 460 N/mm2


28 kN/mm2


200 kN/mm2

B Over all Depth Cover(Cmin) d

1000 mm 300 mm 50 mm 242 mm

Moment

32 kNm

dia of bar

16 mm

Spacing area neutral axis depth(Xu) Stress in steel(Fs) Strain in steel

150 mm 1339.733 mm2 79 mm 111 N/mm2 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) Em at Y2(bottom face)

1.E-04 mm 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. 17

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.

P3

P2

P1 P5 P4

P12 P9 P14

P6

P11

P8

P10

P13

P7

P16 P15

Fig-6:- Labeling of shear wall

18

Sample Calculation (Shear wall-Pier P2) Datas

Strength of Concrete(fcu)

= 40N/mm2

Strength of Steel(fy)

= 460N/mm2


= 28N/mm2


= 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

⎛ N ⎜⎜ ⎝ Nu

⎛ N ⎞ ⎛ 3400 ⎞ ⎞ ⎟⎟ = ⎜⎜ u1 ⎟⎟ = ⎜ ⎟ ⎝ N u ⎠ ⎝ 20504.32 ⎠ ⎠

= 0.17

From Chart-1

⎛ M u max ⎞ ⎟⎟ ⎜⎜ ⎝ Nu × L ⎠

= 0.145

Then Mumax = 0.145×5×20504.32 = 14865.63kNm >3827kNm Hence Safe

19

For minimum Compression ⎛ Nu ⎜⎜ ⎝ Nu

⎞ ⎛ N u1 ⎞ ⎛ 575 ⎞ ⎟⎟ = ⎜⎜ ⎟⎟ = ⎜ ⎟ = 0.03 N ⎠ ⎝ u ⎠ ⎝ 20504.32 ⎠

From Chart-1

⎛ M u max ⎞ ⎟⎟ ⎜⎜ N × L ⎠ ⎝ u

= 0.11

Mumax = 0.11×5×20504.32 = 11277.38kNm>3837kNm Hence Safe Check for stress limits:-

As per BS 8110-1:1997 Max:Permissible Stress in Concrete

= 0.4 f cu = 16 N/mm2

Max:Permissible Stress in Steel

= 0.87fy = 400.2 N/mm2

For Max: Compression ⎛M ⎞ e = ⎜⎜ 1 ⎟⎟ =1125mm ⎝ N1 ⎠

e = L

1125 =0.23 5000

For Min: Compression ⎛M ⎞ e = ⎜⎜ 1 ⎟⎟ =5695mm ⎝ N2 ⎠

e = L

5695 =1.14 5000

For e/L = 0.23 ⎛N From Chart-2 ⎜⎜ max ⎝ N0

Then Nmax = 2.45×

⎞ ⎛ N max ⎞ ⎟⎟ = ⎜ ⎟ = 2.45 ⎠ ⎝ N /L⎠

N u1 L

= 2.45×

2430 = 1190.7kN/m 5

20

Compressive Stress in Concrete = ⎛

N

⎞

⎛

⎞

1190.7

max ⎟⎟ = ⎜⎜ ⎟⎟ = 5.43N / mm 2 < 16 N / mm 2 σ c = ⎜⎜ ⎝ t × (1 + mr ) ⎠ ⎝ 200 × (1 + 7.14 × 0.0136 ⎠

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

⎛ N max ⎜⎜ ⎝ N0

⎞ ⎛ N max ⎞ ⎟⎟ = ⎜ ⎟= ⎠ ⎝ N /L⎠

X/L=0.2915

⎞ ⎛ (1 + mr ) ⎟⎟ ⎜⎜ ⎝ 0.5 x / L + mr (1 − 0.5 x / L) ⎠ ⎛ ⎞ (1 + 7.14 × 0.0136) ⎟⎟ = ⎜⎜ ⎝ 0.5 × 0.2915 + 7.14 × 0.0136 × (1 − 0.5 × 0.2915) ⎠

= 4.79 Then Nmax = 4.79×

N u1 L

= 4.79×

480 = 460kN/m 5

Maximum Stress in Concrete(σc) ⎛ N max ⎞ ⎛ ⎞ 460 ⎟⎟ = ⎜⎜ ⎟⎟ = 2.1N / mm 2 < 16 N / mm 2 = ⎜⎜ × + × + × t ( 1 mr ) 200 ( 1 7 . 14 0 . 0136 ⎝ ⎠ ⎝ ⎠

Maximum Stress in steel =

L x

σ c × ( − 1)m = 2.1× (3.43 − 1) × 7.14 = 36.43N / mm 2 < 400.2 N / mm 2 Hence Safe

21

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

Chart-2(Nmax/N0--e)

22

Check For Shear

Shear Force(From Analysis)

= 1700kN

Shear stress(τ)

=

1700 × 103 = 2.13N/mm2 0.8 × 200 × 5000

From Table-3.8BS 8110-1:1997 Shear stress of concrete(τc)

= 0.82N/mm2


= 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

23

Table-2 – Shear wall Design output- Etabs

Story

Pier Label

ROOF

P1

SF

P1

FF

P1

Location

Edge Bar

End Bar

End Spacing

Required Ratio of Reinforcement

Provided Ratio of Reinforcement

Shear Reinforcement

Top

16d

16d

150

0.0025

0.0143

500

Bottom

16d

16d

150

0.0025

0.0143

500

Top

16d

16d

150

0.0025

0.0143

500

Bottom

16d

16d

150

0.0026

0.0143

500

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

ROOF

P2

Bottom Top

16d 16d

16d 16d

150 150

0.0025 0.0025

0.0143 0.0137

500 500

Bottom

16d

16d

150

0.0025

0.0137

500

Top

16d

16d

150

0.0025

0.0137

500

Bottom

16d

16d

150

0.0025

0.0137

500

Top

16d

16d

150

0.0025

0.0137

630.4

Bottom

16d

16d

150

0.0025

0.0137

624

SF FF

P2 P2

GF

P2

Top

16d

16d

150

0.0025

0.0137

500

ROOF

P3

Bottom Top

16d 16d

16d 16d

150 150

0.0025 0.0025

0.0137 0.0143

500 500

Bottom

16d

16d

150

0.0025

0.0143

500

Top

16d

16d

150

0.0025

0.0143

500

Bottom

16d

16d

150

0.0025

0.0143

500

Top

16d

16d

150

0.0025

0.0143

500

SF

P3

FF

P3

GF

P3

ROOF SF

P4 P4

FF

P4

GF

P4

ROOF SF

P5 P5

FF

P5

GF

P5

ROOF

P6

Bottom

16d

16d

150

0.0036

0.0143

500

Top

16d

16d

150

0.0045

0.0143

500

Bottom

16d

16d

150

0.0025

0.0143

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

824

Bottom

16d

16d

150

0.0067

0.0136

844.9

Top

16d

16d

150

0.0038

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.005

0.0136

500

Top

16d

16d

150

0.0037

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

24

SF

P6

FF

P6

GF ROOF

P6 P7

SF

P7

FF

P7

GF ROOF

P7 P8

SF

P8

FF

P8

GF ROOF

P8 P9

SF

P9

FF

P9

GF ROOF

P9 P10

SF

P10

FF

P10

GF ROOF

P10 P11

SF

P11

FF

P11

GF

P11

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

507

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0145

500

Bottom

16d

16d

150

0.0025

0.0145

500

Top

16d

16d

150

0.0025

0.0145

500

Bottom

16d

16d

150

0.0025

0.0145

500

Top

16d

16d

150

0.0025

0.0145

500

Bottom

16d

16d

150

0.0029

0.0145

500

Top

16d

16d

150

0.0025

0.0145

500

Bottom

16d

16d

150

0.0025

0.0145

500

Top

16d

16d

150

0.0025

0.014

500

Bottom

16d

16d

150

0.0025

0.014

500

Top

16d

16d

150

0.0025

0.014

500

Bottom

16d

16d

150

0.0025

0.014

500

Top

16d

16d

150

0.0025

0.014

500

Bottom

16d

16d

150

0.0025

0.014

500

Top

16d

16d

150

0.0025

0.014

500

Bottom

16d

16d

150

0.0025

0.014

500

Top

16d

16d

150

0.0025

0.0145

500

Bottom

16d

16d

150

0.0025

0.0145

500

Top

16d

16d

150

0.0025

0.0145

500

Bottom

16d

16d

150

0.0025

0.0145

500

Top

16d

16d

150

0.0025

0.0145

500

Bottom

16d

16d

150

0.0025

0.0145

500

Top

16d

16d

150

0.0025

0.0145

500

Bottom

16d

16d

150

0.0025

0.0145

500

Top

16d

16d

150

0.0025

0.0135

500

Bottom

16d

16d

150

0.0025

0.0135

500

Top

16d

16d

150

0.0025

0.0135

500

Bottom

16d

16d

150

0.0025

0.0135

500

Top

16d

16d

150

0.0025

0.0135

897.4

Bottom

16d

16d

150

0.0045

0.0135

894

Top

16d

16d

150

0.0042

0.0135

500

Bottom

16d

16d

150

0.0025

0.0135

500

Top

16d

16d

150

0.0025

0.0141

500

Bottom

16d

16d

150

0.0025

0.0141

500

Top

16d

16d

150

0.0025

0.0141

500

Bottom

16d

16d

150

0.0025

0.0141

500

Top

16d

16d

150

0.0025

0.0141

500

Bottom

16d

16d

150

0.0025

0.0141

500

Top

16d

16d

150

0.0025

0.0141

500

Bottom

16d

16d

150

0.0025

0.0141

500

25

ROOF

P12

SF

P12

FF GF

P12 P12

ROOF

P13

SF

P13

FF GF

P13 P13

ROOF

P14

SF

P14

FF GF

P14 P14

GF

P15

GF

P16

Top

16d

16d

150

0.0025

0.0141

500

Bottom

16d

16d

150

0.0025

0.0141

500

Top

16d

16d

150

0.0025

0.0141

500

Bottom

16d

16d

150

0.0025

0.0141

500

Top

16d

16d

150

0.0029

0.0141

500

Bottom

16d

16d

150

0.0063

0.0141

500

Top

16d

16d

150

0.0052

0.0141

500

Bottom

16d

16d

150

0.0036

0.0141

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0136

671.2

Bottom

16d

16d

150

0.0061

0.0136

663.1

Top

16d

16d

150

0.0036

0.0136

500

Bottom

16d

16d

150

0.0025

0.0136

500

Top

16d

16d

150

0.0025

0.0146

500

Bottom

16d

16d

150

0.0025

0.0146

500

Top

16d

16d

150

0.0025

0.0146

500

Bottom

16d

16d

150

0.0025

0.0146

500

Top

16d

16d

150

0.0025

0.0146

500

Bottom

16d

16d

150

0.0025

0.0146

500

Top

16d

16d

150

0.0025

0.0146

500

Bottom

16d

16d

150

0.0025

0.0146

500

Top

16d

16d

150

0.0025

0.0138

500

Bottom

16d

16d

150

0.0025

0.0138

500

Top

16d

16d

150

0.0025

0.0137

500

Bottom

16d

16d

150

0.0025

0.0137

500

26

11.0 DESIGN OF RAMP

Z Y X 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:Depth of slab provided

= 250mm

Clear cover assumed

= 70mm

Effective depth

= 172mm

Moment

= 70kNm

Mu b d2

= 2.37


= 0.67%


= 1152mm2


=1340mm2

Design of Top Reinforcement in Shorter Direction:Depth of slab provided

= 250mm

Clear cover assumed

= 50mm

Effective depth

= 192mm

Moment

= 40kNm

Mu b d2

= 1.08 27


= 0.3%


= 576mm2


=1340mm2

Design of Bottom Reinforcement in Longer Direction:Depth of slab provided

= 250mm

Clear cover assumed

= 70mm

Effective depth

= 172mm

Moment

= 16kNm

Mu b d2

= 0.54


= 0.17%

Area of steel required (minimum)

= 325mm2


=753mm2

|Design of Top Reinforcement in Longer Direction:Depth of slab provided

= 250mm

Clear cover assumed

= 70mm

Effective depth

= 172mm

Moment

= 0kNm

Percentage of steel required

= 0.13%

Area of steel required (minimum)

= 325mm2


=753mm2

Design of Ramp wall:-

Design of Vertical Reinforcement:Thickness of wall provided

= 250mm

Clear cover assumed

= 70mm

Effective depth

= 172mm

Moment (from analysis)

= 30kNm

Mu b d2

=1


= 0.5%


= 860mm2 28


=1340mm2

Design of Horizontal Reinforcement:Thickness of wall provided

= 250mm

Clear cover assumed

= 70mm

Effective depth

= 164mm

Since there is no horizontal moment We have to provide minimum area of reinforcement Area of steel required

= 635mm2


=753mm2

29

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. 1750kN(Ultimate) 120kN/m (Ultimate)

Hinge

6500

Hinge

17400

Fig-8: 2D Frame

1150kNm

1150kNm

11732kNm

Fig-8(a): Bending Moment Diagram 2200kN

250kN

890kN

250kN

2200kN

Fig-8(b): Shear Force Diagram 30

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 Effective depth = 1480mm Moment (from analysis)

= 11732kNm

Check for Compression Reinforcement:Mu b d2

= 8.92< 0.156×60=9.36

Therefore we don’t require compression reinforcement We have

⎧⎪ k ⎞ ⎫⎪ ⎛ lever arm (z) = d x ⎨0.5 + ⎜ 0.25 − ⎟⎬ 0.9 ⎠ ⎪⎭ ⎪⎩ ⎝ ⎧⎪ 0.148 ⎞ ⎫⎪ ⎛ = 1480 x ⎨0.5 + ⎜ 0.25 − ⎟⎬ 0.9 ⎠ ⎪⎭ ⎪⎩ ⎝ = 1172.9mm

Area Tension steel required

=

M 0.95 f y z

11732 × 10 6 = 0.95 × 460 × 1172.9 = 22889mm2 Area of steel provided in the section

=24120mm2(30T32)

Hence Safe Minimum Percentage of steel required at support

= 0.13%


= 1248mm2


=8040mm2(10T32)

Hence Safe

31

Design for Shear:-

Shear force at face of support

= 2200kN

Shear stress (v)

=

2200 × 10 3 600 × 1480

=2.47N/mm2<0.8

f cu =6.19N/mm2

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

=

1950 × 10 3 600 × 1480

= 2.19N/mm2 100 As bd

= 0.9%

From Table3.8 BS8110-1:1997 νc We have

= 0.72N/mm2

As v b (v − v c ) = Sv 0.95 f y

=

600(2.54 − 0.72) 0.95 × 460

= 2.5

6 × 113 = 270mm 2.5

Spacing of 6 legged T12 stirrup required

=

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 32

Moment due to Total load

= 8250kNm

Moment due to permanent load

= 7340kNm

Short term deflection due to total load:-

We have x 1 M= Asfs(d- )+ bhfct(h-x) 3 3 8250x106 = 24120 x fs x (1480-

x 1 ) + x 600x1600xfctx (1600-x) ----------------------- (1) 3 3

Maximum tensile stress allowable in concrete (fct)

=

( h − x) ×1 (d − x)

=

(1600 − x) × 1 ------ (2) (1480 − x)

We have From the strain distribution fc

=

E x × c × f s ---------------(3) ( d − x) E s

and by equating tension and compression 1 1 bxf c = Asfs+ bfct(h-x)-----------------(4) 2 2 By solving the above 4 equations using trial and error method We have x= 658.7mm fc

= 33.37N/mm2 fc 1 = rb x × E c

Short term curvature

Short term deflection due to creep

=

33.37 = 1.58x10-6/mm 3 658.7 × 32 × 10

=

1 × k × L2 rb

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

= 1.58x10-6 x 0.083 x 174002 = 39.7mm

33

Short term deflection due to Permanent load:-


x 1 ) + x 600x1600xfctx (1600-x) ----------------------- (1) 3 3


=

( h − x) ×1 (d − x)

=

(1600 − x) × 1 ------ (2) (1480 − x)


=

E x × c × f s ---------------(3) ( d − x) E s


= 29.66N/mm2 fc 1 = rb x × E c

Short term curvature

Short term deflection due to creep

=

29.66 = 1.4x10-6/mm 660.2 × 32 ×10 3

=

1 × k × L2 rb

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 Effective section thickness

Eeff

=

Ec (1 + φ )

=

Twice the c / s area perimetre 34

=

2 × 600 × 1600 2 × (1600 + 600)

= 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

=

32 =10.67N/mm2 (1 + 2)


x 1 ) + x 600x1600xfctx (1600-x) ----------------------- (1) 3 3


=

(h − x) × 0.55 (d − x)

=

(1600 − x) × 0.55 ------ (2) (1480 − x)


=

E eff x × × f s ---------------(3) (d − x) E s


= 22.63N/mm2

Long term curvature

Long term deflection due to creep

fc 1 = rb x × E eff =

22.63 = 2.29x10-6/mm 3 924.4 × 10.67 × 10

=

1 × k × L2 rb

= 2.29x10-6 x 0.083 x 174002 = 57.54mm 35

Deflection due to shrinkage:1 ε cs × α c × S s = I rcs

αc =

Es 200 = 18.74 = E e ff 10.67

bx 3 x 2 + bx( ) 2 + α c As (d − x ) I= 12 2

= 2.978x 1011mm4 Ss

= As (d − x ) = 24120 x (1480-924.4) = 13.41 x 106mm3

From Fig-7.2, BS8110-2:1985

ε cs

= 327 x 10-6

Thus 1 327 × 10 −6 × 18.74 × 13.41 × 10 6 =2.76 x 10-7/mm = 11 rcs 2.97 × 10 Long term deflection due to shrinkage

=

1 × k × L2 rb

= 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.

36

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 Effective depth = 212mm Mid-span Moment

= 62kNm

Mu b d2

= 1.37


= 0.7%


= 1484mm2/m


= 2010mm2/m

Hence Safe.

Support Moment

= 20kNm

Mu b d2

= 0.44


= 0.25%


= 530mm2/m


= 753mm2/m

Hence Safe.

Longer span:Thickness of slab

= 250mm

Diameter of bar = 16mm Effective depth = 196mm Mid-span Moment

= 60kNm

Mu b d2

= 1.56


= 0.77%


= 1509mm2/m 37


= 2010mm2/m

Hence Safe.

Support Moment

= 18kNm

Mu b d2

= 0.47


= 0.25%


= 530mm2/m


= 753mm2/m

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

38

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


= 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)

39

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)

40

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 Grawhill 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

41

Design Calculations

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