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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 ed on precast beams. Pre cast columns and shear walls are used for ing the floors. The building is ed 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 . 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
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 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
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: 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 for gravity loads. Moment resisting frames provide resistance to lateral load primarily by flexural action of . 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 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 under thermal gradient) after the design of the structure we have checked the whole structure under the thermal effect. All of the have 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 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.18%
Area of steel required
= 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.43%
Area of steel required
= 4752mm2
Area of steel provided in the section
=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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.18%
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 :-
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
Modulus of Elasticity(Ec)
28 kN/mm2
Modulus of Elasticity(Es)
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
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
⎛ 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
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
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 ed 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.67%
Area of steel required
= 1152mm2
Area of steel provided in the section
=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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.3%
Area of steel required
= 576mm2
Area of steel provided in the section
=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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.17%
Area of steel required (minimum)
= 325mm2
Area of steel provided in the section
=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
Area of steel provided in the section
=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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.5%
Area of steel required
= 860mm2 28
Area of steel provided in the section
=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
Area of steel provided in the section
=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
= 0.13%
Area of steel required
= 1248mm2
Area of steel provided in the section
=8040mm2(10T32)
Hence Safe
31
Design for Shear:-
Shear force at face of
= 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 = 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:-
We have x 1 M= Asfs(d- )+ bhfct(h-x) 3 3 7340x106 = 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= 660.2mm fc
= 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)
We have x 1 M= Asfs(d- )+ bhfct(h-x) 3 3 7270x106 = 24120 x fs x (1480-
x 1 ) + x 600x1600xfctx (1600-x) ----------------------- (1) 3 3
Maximum tensile stress allowable in concrete (fct)
=
(h − x) × 0.55 (d − x)
=
(1600 − x) × 0.55 ------ (2) (1480 − x)
We have From the strain distribution fc
=
E eff x × × 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= 924.4mm fc
= 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.7%
Area of steel required
= 1484mm2/m
Area of steel provided in the section
= 2010mm2/m
Hence Safe.
Moment
= 20kNm
Mu b d2
= 0.44
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.25%
Area of steel required
= 530mm2/m
Area of steel provided in the section
= 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.77%
Area of steel required
= 1509mm2/m 37
Area of steel provided in the section
= 2010mm2/m
Hence Safe.
Moment
= 18kNm
Mu b d2
= 0.47
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.25%
Area of steel required
= 530mm2/m
Area of steel provided in the section
= 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
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 = 1.2 Internal Pressure Coefficient = ±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 ed 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
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 ed on precast beams. Pre cast columns and shear walls are used for ing the floors. The building is ed 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 . 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
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 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
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: 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 for gravity loads. Moment resisting frames provide resistance to lateral load primarily by flexural action of . 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 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 under thermal gradient) after the design of the structure we have checked the whole structure under the thermal effect. All of the have 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 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.18%
Area of steel required
= 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.43%
Area of steel required
= 4752mm2
Area of steel provided in the section
=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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.18%
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 :-
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
Modulus of Elasticity(Ec)
28 kN/mm2
Modulus of Elasticity(Es)
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
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
⎛ 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
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
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 ed 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.67%
Area of steel required
= 1152mm2
Area of steel provided in the section
=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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.3%
Area of steel required
= 576mm2
Area of steel provided in the section
=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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.17%
Area of steel required (minimum)
= 325mm2
Area of steel provided in the section
=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
Area of steel provided in the section
=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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.5%
Area of steel required
= 860mm2 28
Area of steel provided in the section
=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
Area of steel provided in the section
=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
= 0.13%
Area of steel required
= 1248mm2
Area of steel provided in the section
=8040mm2(10T32)
Hence Safe
31
Design for Shear:-
Shear force at face of
= 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 = 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:-
We have x 1 M= Asfs(d- )+ bhfct(h-x) 3 3 7340x106 = 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= 660.2mm fc
= 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)
We have x 1 M= Asfs(d- )+ bhfct(h-x) 3 3 7270x106 = 24120 x fs x (1480-
x 1 ) + x 600x1600xfctx (1600-x) ----------------------- (1) 3 3
Maximum tensile stress allowable in concrete (fct)
=
(h − x) × 0.55 (d − x)
=
(1600 − x) × 0.55 ------ (2) (1480 − x)
We have From the strain distribution fc
=
E eff x × × 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= 924.4mm fc
= 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.7%
Area of steel required
= 1484mm2/m
Area of steel provided in the section
= 2010mm2/m
Hence Safe.
Moment
= 20kNm
Mu b d2
= 0.44
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.25%
Area of steel required
= 530mm2/m
Area of steel provided in the section
= 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
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.77%
Area of steel required
= 1509mm2/m 37
Area of steel provided in the section
= 2010mm2/m
Hence Safe.
Moment
= 18kNm
Mu b d2
= 0.47
From Chart No.2 BS 8110 Part 3 Percentage of steel required
= 0.25%
Area of steel required
= 530mm2/m
Area of steel provided in the section
= 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
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 = 1.2 Internal Pressure Coefficient = ±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 ed 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
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