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Transcript of Design Calculations
1
DESIGN CALCULATION FOR
COMMERCIAL CENTRE, CORPORATE OFFICE.
Client: JUMEIRA GOLF ESTATES
2
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.
3
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
4
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
5
Marble = 0.7 kN/m2 Screed = 0.7 kN/m2 Live Load = 3 kN/m2 f) Ground Floor (Grid A TO B1) Slab (350mm) = 8.3 kN/m2 Fill (0.75m) =15kN/m2 Ceiling and Services = 0.8 kN/m2
Live Load = 3 kN/m2
g) Ramp
Slab (250mm) = 6.25 kN/m2
Finishing = 2 kN/m2
Live Load = 5 kN/m2
h) Stair Case
Slab (250mm) = 6.25 kN/m2
Steps = 2.05kN/m2
Finishing = 2 kN/m2
Live Load = 5 kN/m2
6.2 Wind Load Wind load corresponding to basic wind speed of 25 m/s is considered as per BS: 6399-
Part II
Data available
Height of building = 20m
Location = Dubai
Basic wind speed = 25 m/s
Longest side = 110.3m
Shortest side = 39.15m
Site Altitude = 0m
The dynamic pressure is given by
qs = 0.613Ve²
Ve = Effective wind speed (Clause 2.2.3, BS: 6399- Part II)
Ve =Vs×Sb
Vs = Site speed from (Clause 2.2.2, BS: 6399- Part II)
6
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
7
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
8
Lateral force procedure
Simplified static approach is applicable
Structural period
T = Ct (hn) 3/4 hn=20m
Ct = 0.0731(in SI units)
T =0.0731(20)3/4 = 0.69 Seconds
b) Calculation of Base Shear
The total design base shear in a given duration
V = WTRICv ×⎟⎠⎞
⎜⎝⎛
×× < W
RICa ×⎟⎠⎞
⎜⎝⎛ ×5.2
V = WTRICv ×⎟⎠⎞
⎜⎝⎛
×× > 0.11Ca×I×W
Where
W =Total load of structure =229570kN
Total Design Base shear = WTRICv ×⎟⎠⎞
⎜⎝⎛
×× = 229570
69.05.5125.0
×⎟⎠⎞
⎜⎝⎛
××
= 15123kN
The distribution of base shear
along vertical direction ( )∑ =
×
××−= n
i ii
xxt
hwhwFV
1
Where Ft = 0 since T<0.7secs
Table-1: Base shear distribution at different storey levels.
Storey Label Height(hx)
In metre
Seismic
weight(Wx)
In kN
Base shear(Fx)
( )∑ =
×
××−= n
i ii
xxt
hw
hwFV
1
Roof 4.16 20998 1034kN
2nd floor 4.55 40647 2190.4kN
First floor 4.55 44564 2401kN
Ground floor 6.5 123361 9496kN
6.4 Temperature load (As per UBC 1997)
With reference to the size of the building it is necessary to consider the thermal
effect of the environment on the whole structure. In order to avoid additional self-
9
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
10
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.
11
Negative Positive
Fig.2:-Bending Moment X-Direction (M11)
Fig.3:-Bending Moment Y-Direction (M22)
Positive Negative X
Y
X
Y
Maximum Sagging Moment
Maximum Hogging Moment
Maximum Sagging Moment
Maximum Hogging Moment
12
Sample Calculation:-
Assume M40 grade concrete and Fy-460 steel.
From Fig. 2,
The maximum sagging bending moment in X direction in slab
M11 = 1700 kNm/m
Maximum hogging bending moment in X direction in slab
M11 = 690kNm/m
From Fig. 3,
The maximum sagging bending moment in Y direction in slab
M22 = 1960kNm/m
Maximum hogging bending moment in Y direction in slab
M22 = 700 kNm/m
Design of Bottom Reinforcement in X Direction:-
Depth of slab provided = 1200mm
Clear cover assumed = 75mm
Effective depth(d) = 1112.5mm
Moment = 1700kNm
u2
Mb d
= 1.37
From Chart No.2 BS 8110 Part 3
Percentage of steel required = 0.38%
Area of steel required = 4247mm2
Area of steel provided in the section =5359mm2>4247
Hence Safe
Design of Top Reinforcement in X Direction:-
Depth of slab provided = 1200mm
Clear cover assumed = 75mm
Effective depth (d) = 1112.5mm
Moment = 690kNm
u2
Mb d
= 0.56
From Chart No.2 BS 8110 Part 3
Percentage of steel required = 0.18%
Area of steel required = 2011.5mm2
13
Area of steel provided in the section =3266mm2>2011.5
Hence Safe
Design of Bottom Reinforcement in Y Direction:-
Depth of slab provided = 1200mm
Clear cover assumed = 75mm
Effective depth (d) = 1100mm
Moment = 1960kNm
u2
Mb d
= 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
u2
Mb d
= 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 support:-
The maximum axial load from analysis = 7665kN
Breadth of column = 600mm
Depth of column = 800mm
14
Perimeter = 2 x 600 + 2 x 800
= 2800mm
Shear stress (ν) = 5.11122800
107665 3
××
= 2.46N/mm2<0.8 fck =5N/mm2
Hence safe
a) The critical section for shear is 1.5 x effective depth = 1.5 x 1112.5 =1668.75mm
from the column face, thus the length of the perimeter
= 2(600+1668.75 x2)+2(800 +1668.75 x 2) = 16150mm
Shear stress (ν) = 5.111216150
107665 3
××
= 0.42N/mm2
dAs100
= 0.48%
From Table3.8 BS8110-1:1997
Allowable shear stress νc = 0.49N/mm2>0.42N/mm2
Hence Safe
15
9.0 DESIGN OF RETAINING WALL
9(a). Height=5m
The retaining wall is analysed as fixed at bottom and free at top with a surcharge
load of 5kN/m2 and soil pressure of height 5m as shown in Fig-4
Unit weight of soil (γ) =18kN/m3
Angle of repose = 330
Height of soil fill (h) = 5m
Surcharge Load = 5kN/m2
Equivalent height of soil = 5/γ =0.278m
Soil pressure due to surcharge = 5/γ x γ x (1-sinφ)/ (1+sinφ)
= 1.476kN/m2
Soil Pressure (at bottom of retaining wall) due to 5m height of soil
= (1-sinφ)/ (1+sinφ) x γ x h
= 26.55kN/m2
Fig-4
Strength of Concrete(fcu) 40 N/mm2
Strength of Steel(fy) 460 N/mm2
Modulus of Elasticity(Ec) 28 kN/mm2
Modulus of Elasticity(Es) 200 kN/mm2
B 1000 mm
Over all Depth 300 mm
Cover(Cmin) 50 mm
1.476kN/m2 26.55kN/m2
16
d 240 mm
Moment 116 kNm
dia of bar 20 mm
Spacing 150 mm
area 2093.333 mm2
neutral axis depth(Xu) 94 mm
Stress in steel(Fs) 265 N/mm2
Strain in steel 0.001327 mm
Srain in Concrete at Y1((d+x/2) from
top) 6.63E-04 mm
Srain in Concrete at Y2(bottom face) 1.87E-03 mm
Em at Y1((d+x/2) from top) 6.E-04 mm
Em at Y2(bottom face) 1.64E-03 mm
acr for Y1((d+x/2) from top) 85 mm
acr for Y2(bottom face) 75 mm
Crack width at Y1((d+x/2) from top) 0.10 mm
Crack width at Y2(bottom face) 0.29 mm
Since the crack width is less than 0.3mm, the provided reinforcement (T20-150) is safe.
9(b). Height=3m
The retaining wall is analysed as fixed at bottom and free at top with a surcharge
load of 5kN/m2 and soil pressure of height 3m as shown in Fig-5
Unit weight of soil (γ) =18kN/m3
Angle of repose = 330
Height of soil fill(h) = 3m
Surcharge Load = 5kN/m2
Equivalent height of soil = 5/γ =0.278m
Soil pressure due to surcharge = 5/γ x γ x (1-sinφ)/ (1+sinφ)
= 1.476kN/m2
Soil Pressure (at bottom of retaining wall) due to 5m height of soil
= (1-sinφ)/ (1+sinφ) x γ x h
= 16kN/m2
17
Fig-5
Strength of Concrete(fcu) 40 N/mm2
Strength of Steel(fy) 460 N/mm2
Modulus of Elasticity(Ec) 28 kN/mm2
Modulus of Elasticity(Es) 200 kN/mm2
B 1000 mm
Over all Depth 300 mm
Cover(Cmin) 50 mm
d 242 mm
Moment 32 kNm
dia of bar 16 mm
Spacing 150 mm
area 1339.733 mm2
neutral axis depth(Xu) 79 mm
Stress in steel(Fs) 111 N/mm2
Strain in steel 0.000554 mm
Srain in Concrete at Y1((d+x/2) from
top) 2.77E-04 mm
Srain in Concrete at Y2(bottom face) 7.51E-04 mm
Em at Y1((d+x/2) from top) 1.E-04 mm
Em at Y2(bottom face) 3.78E-04 mm
acr for Y1((d+x/2) from top) 92 mm
acr for Y2(bottom face) 74 mm
Crack width at Y1((d+x/2) from top) 0.03 mm
Crack width at Y2(bottom face) 0.07 mm
Since the crack width is less than 0.3mm, the provided reinforcement (T16-150) is safe.
1.476kN/m2 16kN/m2
18
10.0 DESIGN OF SHEAR WALL
The shear wall is modeled as pier element (See Etabs Model) and was labeled as shown
in fig. Each area object that makes up a part of a wall is assigned as one pier label. The
walls are designed as compression elements under the combined action of in-plane
bending and axial forces. The design of the shear wall was done based on BS 8110-1997.
One sample design calculation for the shear wall (Pier P2) is given below.
P15 P16
P1
P2
P3
P4 P5
P6
P7
P8
P9
P13
P10
P12
P11 P14
Fig-6:- Labeling of shear wall
19
Sample Calculation (Shear wall-Pier P2)
Datas
Strength of Concrete(fcu) = 40N/mm2
Strength of Steel(fy) = 460N/mm2
Modulus of Elasticity(Ec) = 28N/mm2
Modulus of Elasticity(Es) = 200N/mm2
Modular Ratio(m) = 7.14
Length(L) = 5000mm
Thickness(t) =200mm
From Etab Analysis,
Maximum Axial Load(Ultimate)- Nu1 =3400kN
Minimum Axial Load(Ultimate)-Nu2 =575kN
Maximum Moment (Ultimate )-Mu1 =3827kNm
Maximum Axial Load(Service)-N1 =2430kN
Minimum Axial Load(Service)-N2 =480kN
Maximum Moment (Service)-M1 = 2734kNm
Reinforcement Ratio Provided(r) = 0.0136
Check for Ultimate Strengths
a) Ultimate Compressive Strength
Nu=(0.4fcu + 0.72fy × r) × t × L = 20504.32kN > 3400kN
Hence Safe
b)Ultimate moment
For maximum Compression
= ⎟⎟⎠
⎞⎜⎜⎝
⎛
u
u
NN 1 = ⎟
⎠⎞
⎜⎝⎛
32.205043400
From Chart-1
Then
Mumax = 0.145×5×20504.32 =
Hence Safe
⎟⎟⎠
⎞⎜⎜⎝
⎛
uNN = 0.17
14865.63kNm >3827kNm
⎟⎟⎠
⎞⎜⎜⎝
⎛× LN
M
u
u max = 0.145
20
For minimum Compression
⎟⎟⎠
⎞⎜⎜⎝
⎛
u
u
NN
= ⎟⎟⎠
⎞⎜⎜⎝
⎛
u
u
NN 1 = 03.0
32.20504575
=⎟⎠⎞
⎜⎝⎛
From Chart-1
Mumax = 0.11×5×20504.32
Hence Safe
Check for stress limits:-
As per BS 8110-1:1997
Max:Permissible Stress in Concrete = cuf4.0 = 16 N/mm2
Max:Permissible Stress in Steel = 0.87fy = 400.2 N/mm2
For Max: Compression
e = ⎟⎟⎠
⎞⎜⎜⎝
⎛
1
1
NM =1125mm
Le =
50001125 =0.23
For Min: Compression
e = ⎟⎟⎠
⎞⎜⎜⎝
⎛
2
1
NM =5695mm
Le =
50005695 =1.14
For e/L = 0.23
From Chart-2 =⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛LN
NN
N/max
0
max 2.45
Then Nmax = 2.45×L
Nu1
= 2.45×5
2430
⎟⎟⎠
⎞⎜⎜⎝
⎛× LN
M
u
u max = 0.11
= 11277.38kNm>3837kNm
= 1190.7kN/m
21
Compressive Stress in Concrete =
22maxc /16/43.5
0136.014.71(2007.1190
)1( mmNmmN
mrtN
<=⎟⎟⎠
⎞⎜⎜⎝
⎛×+×
=⎟⎟⎠
⎞⎜⎜⎝
⎛+×
=σ
Hence Safe
For e/L = 1.14
we have
Solve for (x/L) a = 1
b = 1.92
a(x/L)3-b(x/L)2-c(x/L)+d = 0 c = 0.66
d = -0.38
From Trial and Error Method X/L=0.2915
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛LN
NN
N/max
0
max
= 4.79
Then Nmax = 4.79×L
Nu1
= 4.79×5
480
Maximum Stress in Concrete(σc)
22max /16/1.20136.014.71(200
460)1(
mmNmmNmrt
N<=⎟⎟
⎠
⎞⎜⎜⎝
⎛×+×
=⎟⎟⎠
⎞⎜⎜⎝
⎛+×
=
Maximum Stress in steel =
mxL
c )1( −×σ 22 /2.400/43.3614.7)143.3(1.2 mmNmmN <=×−×=
Hence Safe
⎟⎟⎠
⎞⎜⎜⎝
⎛−+
+)/5.01(/5.0
)1(LxmrLx
mr
⎟⎟⎠
⎞⎜⎜⎝
⎛×−××+×
×+=
)2915.05.01(0136.014.72915.05.0)0136.014.71(
= 460kN/m
22
Chart-1(N/N0--Mu/NuL)
Chart-2(Nmax/N0--e)
23
Check For Shear
Shear Force(From Analysis) = 1700kN
Shear stress(τ) = =×××
50002008.0101700 3
2.13N/mm2
From Table-3.8BS 8110-1:1997
Shear stress of concrete(τc) = 0.82N/mm2
Area of steel required = 449mm2/m
Minimum area of steel required = 500mm2/m
Area of steel provided = 2103mm2/m
Hence Safe
The design result from Etabs is shown in Table-2
24
Table-2 – Shear wall Design output- Etabs
Story Pier
Label Location Edge Bar
End Bar
End Spacing
Required Ratio of
Reinforcement
Provided Ratio of
Reinforcement Shear Reinforcement
ROOF P1 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0025 0.0143 500
SF P1 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0026 0.0143 500
FF P1 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0048 0.0143 500
GF P1 Top 16d 16d 150 0.0037 0.0143 500 Bottom 16d 16d 150 0.0025 0.0143 500
ROOF P2 Top 16d 16d 150 0.0025 0.0137 500 Bottom 16d 16d 150 0.0025 0.0137 500
SF P2 Top 16d 16d 150 0.0025 0.0137 500 Bottom 16d 16d 150 0.0025 0.0137 500
FF P2 Top 16d 16d 150 0.0025 0.0137 630.4 Bottom 16d 16d 150 0.0025 0.0137 624
GF P2 Top 16d 16d 150 0.0025 0.0137 500 Bottom 16d 16d 150 0.0025 0.0137 500
ROOF P3 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0025 0.0143 500
SF P3 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0025 0.0143 500
FF P3 Top 16d 16d 150 0.0025 0.0143 500 Bottom 16d 16d 150 0.0036 0.0143 500
GF P3 Top 16d 16d 150 0.0045 0.0143 500 Bottom 16d 16d 150 0.0025 0.0143 500
ROOF P4 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
SF P4 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
FF P4 Top 16d 16d 150 0.0025 0.0136 824 Bottom 16d 16d 150 0.0067 0.0136 844.9
GF P4 Top 16d 16d 150 0.0038 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
ROOF P5 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
SF P5 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
FF P5 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.005 0.0136 500
GF P5 Top 16d 16d 150 0.0037 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
ROOF P6 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
25
SF P6 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
FF P6 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 507
GF P6 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
ROOF P7 Top 16d 16d 150 0.0025 0.0145 500 Bottom 16d 16d 150 0.0025 0.0145 500
SF P7 Top 16d 16d 150 0.0025 0.0145 500 Bottom 16d 16d 150 0.0025 0.0145 500
FF P7 Top 16d 16d 150 0.0025 0.0145 500 Bottom 16d 16d 150 0.0029 0.0145 500
GF P7 Top 16d 16d 150 0.0025 0.0145 500 Bottom 16d 16d 150 0.0025 0.0145 500
ROOF P8 Top 16d 16d 150 0.0025 0.014 500 Bottom 16d 16d 150 0.0025 0.014 500
SF P8 Top 16d 16d 150 0.0025 0.014 500 Bottom 16d 16d 150 0.0025 0.014 500
FF P8 Top 16d 16d 150 0.0025 0.014 500 Bottom 16d 16d 150 0.0025 0.014 500
GF P8 Top 16d 16d 150 0.0025 0.014 500 Bottom 16d 16d 150 0.0025 0.014 500
ROOF P9 Top 16d 16d 150 0.0025 0.0145 500 Bottom 16d 16d 150 0.0025 0.0145 500
SF P9 Top 16d 16d 150 0.0025 0.0145 500 Bottom 16d 16d 150 0.0025 0.0145 500
FF P9 Top 16d 16d 150 0.0025 0.0145 500 Bottom 16d 16d 150 0.0025 0.0145 500
GF P9 Top 16d 16d 150 0.0025 0.0145 500 Bottom 16d 16d 150 0.0025 0.0145 500
ROOF P10 Top 16d 16d 150 0.0025 0.0135 500 Bottom 16d 16d 150 0.0025 0.0135 500
SF P10 Top 16d 16d 150 0.0025 0.0135 500 Bottom 16d 16d 150 0.0025 0.0135 500
FF P10 Top 16d 16d 150 0.0025 0.0135 897.4 Bottom 16d 16d 150 0.0045 0.0135 894
GF P10 Top 16d 16d 150 0.0042 0.0135 500 Bottom 16d 16d 150 0.0025 0.0135 500
ROOF P11 Top 16d 16d 150 0.0025 0.0141 500 Bottom 16d 16d 150 0.0025 0.0141 500
SF P11 Top 16d 16d 150 0.0025 0.0141 500 Bottom 16d 16d 150 0.0025 0.0141 500
FF P11 Top 16d 16d 150 0.0025 0.0141 500 Bottom 16d 16d 150 0.0025 0.0141 500
GF P11 Top 16d 16d 150 0.0025 0.0141 500 Bottom 16d 16d 150 0.0025 0.0141 500
26
ROOF P12 Top 16d 16d 150 0.0025 0.0141 500 Bottom 16d 16d 150 0.0025 0.0141 500
SF P12 Top 16d 16d 150 0.0025 0.0141 500 Bottom 16d 16d 150 0.0025 0.0141 500
FF P12 Top 16d 16d 150 0.0029 0.0141 500 Bottom 16d 16d 150 0.0063 0.0141 500
GF P12 Top 16d 16d 150 0.0052 0.0141 500 Bottom 16d 16d 150 0.0036 0.0141 500
ROOF P13 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
SF P13 Top 16d 16d 150 0.0025 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
FF P13 Top 16d 16d 150 0.0025 0.0136 671.2 Bottom 16d 16d 150 0.0061 0.0136 663.1
GF P13 Top 16d 16d 150 0.0036 0.0136 500 Bottom 16d 16d 150 0.0025 0.0136 500
ROOF P14 Top 16d 16d 150 0.0025 0.0146 500 Bottom 16d 16d 150 0.0025 0.0146 500
SF P14 Top 16d 16d 150 0.0025 0.0146 500 Bottom 16d 16d 150 0.0025 0.0146 500
FF P14 Top 16d 16d 150 0.0025 0.0146 500 Bottom 16d 16d 150 0.0025 0.0146 500
GF P14 Top 16d 16d 150 0.0025 0.0146 500 Bottom 16d 16d 150 0.0025 0.0146 500
GF P15 Top 16d 16d 150 0.0025 0.0138 500 Bottom 16d 16d 150 0.0025 0.0138 500
GF P16 Top 16d 16d 150 0.0025 0.0137 500 Bottom 16d 16d 150 0.0025 0.0137 500
27
11.0 DESIGN OF RAMP
Fig-7: Finite Element Model of Ramp
The ramp is modeled as shown in Fig.7. The ramp is assumed to be supported on wall on
the two sides.
Design of Ramp slab
From the analysis,
The Maximum Sagging Moment in shorter direction= 70kNm
The Maximum Hogging Moment in shorter direction= 40kNm
The Maximum Sagging Moment in longer direction = 16kNm
The Maximum Hogging Moment in shorter direction= 0kNm
Design of Bottom Reinforcement in Shorter Direction:-
Depth of slab provided = 250mm
Clear cover assumed = 70mm
Effective depth = 172mm
Moment = 70kNm
u2
Mb d
= 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
u2
Mb d
= 1.08
X
Z
Y
28
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
u2
Mb d
= 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
u2
Mb d
= 1
From Chart No.2 BS 8110 Part 3
Percentage of steel required = 0.5%
Area of steel required = 860mm2
29
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
30
12. DESIGN OF INSITU BEAMS (Span-17.4m)
The analysis of the beam was done by modeling it as a frame as shown in Fig-8. The
moment at the ends of beam is released. The Bending Moment and shear Force
Diagrams are shown in Fig-8 (a), Fig-8 (b) respectively.
17400
6500
1750kN
120kN/m
11732kNm
1150kNm 1150kNm
2200kN
2200kN
250k
N
250k
N
890kN
(Ultimate)
(Ultimate)
Hinge Hinge
Fig-8: 2D Frame
Fig-8(a): Bending Moment Diagram
Fig-8(b): Shear Force Diagram
31
Grade of concrete = 60 N/mm2
Grade of steel = 460 N/mm2
Clear cover to reinforcement = 30mm
Width of the beam = 600 mm
Depth of the beam = 1600 mm
Design for mid-span moment:-
Diameter of bar = 32 mm
Effective depth = 1480mm
Moment (from analysis) = 11732kNm
Check for Compression Reinforcement:-
u2
Mb d
= 8.92< 0.156×60=9.36
Therefore we don’t require compression reinforcement
We have lever arm (z) = d x ⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎟⎠⎞
⎜⎝⎛ −+
9.025.05.0 k
= 1480 x ⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎟⎠⎞
⎜⎝⎛ −+
9.0148.025.05.0
= 1172.9mm
Area Tension steel required = zf
M
y95.0
= 9.117246095.0
1011732 6
×××
= 22889mm2
Area of steel provided in the section =24120mm2(30T32)
Hence Safe
Minimum Percentage of steel required at support = 0.13%
Area of steel required = 1248mm2
Area of steel provided in the section =8040mm2(10T32)
Hence Safe
32
Design for Shear:-
Shear force at face of support = 2200kN
Shear stress (v) = 1480600
102200 3
××
=2.47N/mm2<0.8 cuf =6.19N/mm2
Hence safe
Shear force at a distance‘d’ from the face of support = 1950kN
Shear stress = 1480600101950 3
××
= 2.19N/mm2
bdAs100
= 0.9%
From Table3.8 BS8110-1:1997
νc = 0.72N/mm2
We have v
vs
SA
= y
c
fvvb
95.0)( −
= 46095.0
)72.054.2(600×−
= 2.5
Spacing of 6 legged T12 stirrup required = 5.2
1136× = 270mm
Spacing of 6 legged T12 stirrup provided = 200mm
Hence safe
Check for Deflection (serviceability) (As per BS8110-2:1985):-
We have
As =24120mm2
h = 1600mm
b = 600mm
d = 1480mm
Total load (including total live load)
Concentrated load = 1190kN
Uniformly distributed load = 80kN/m
Permanent (1ncluding 25% live load only) = 966kN, 60kN/m
Concentrated load = 966kN
Uniformly distributed load = 60kN/m
33
Moment due to Total load = 8250kNm
Moment due to permanent load = 7340kNm
Short term deflection due to total load:-
We have
M= Asfs(d-3x )+
31 bhfct(h-x)
8250x106 = 24120 x fs x (1480- 3x ) +
31 x 600x1600xfctx (1600-x) ----------------------- (1)
Maximum tensile stress allowable in concrete (fct) = 1)()(×
−−
xdxh
= 1)1480()1600(×
−−
xx ------ (2)
We have
From the strain distribution
fc = ss
c fEE
xdx
××− )(
---------------(3)
and by equating tension and compression
21
cbxf = Asfs+ 21 bfct(h-x)-----------------(4)
By solving the above 4 equations using trial and error method
We have x= 658.7mm
fc = 33.37N/mm2
Short term curvature br1 =
c
c
Exf×
= 310327.65837.33××
= 1.58x10-6/mm
Short term deflection due to creep = 21 Lkrb
××
From Table3.1, BS8110-2:1985, k= 0.083
Deflection = 1.58x10-6 x 0.083 x 174002
= 39.7mm
34
Short term deflection due to Permanent load:-
We have
M= Asfs(d-3x )+
31 bhfct(h-x)
7340x106 = 24120 x fs x (1480- 3x ) +
31 x 600x1600xfctx (1600-x) ----------------------- (1)
Maximum tensile stress allowable in concrete (fct) = 1)()(×
−−
xdxh
= 1)1480()1600(×
−−
xx ------ (2)
We have
From the strain distribution
fc = ss
c fEE
xdx
××− )(
---------------(3)
and by equating tension and compression
21
cbxf = Asfs+ 21 bfct(h-x)-----------------(4)
By solving the above 4 equations using trial and error method
We have x= 660.2mm
fc = 29.66N/mm2
Short term curvature br1 =
c
c
Exf×
= 310322.66066.29××
= 1.4x10-6/mm
Short term deflection due to creep = 21 Lkrb
××
From Table3.1, BS8110-2:1985, k= 0.083
Deflection = 1.4x10-6 x 0.083 x 174002
= 35.6mm
Long term deflection due to permanent loads:-
Reduced modulus of elasticity Eeff = )1( φ+
cE
Effective section thickness = perimetre
areasctheTwice /
35
= )6001600(2
16006002+×××
= 436.36mm
The value of creep coefficient (Ф) From Fig-7.1, BS8110-2:1985 for loading at 28 days
with indoor exposure condition is approximately 2
Eeff = )21(
32+
=10.67N/mm2
We have
M= Asfs(d-3x )+
31 bhfct(h-x)
7270x106 = 24120 x fs x (1480- 3x ) +
31 x 600x1600xfctx (1600-x) ----------------------- (1)
Maximum tensile stress allowable in concrete (fct) = 55.0)()(×
−−
xdxh
= 55.0)1480()1600(×
−−
xx ------ (2)
We have
From the strain distribution
fc = ss
eff fEE
xdx
××− )(
---------------(3)
and by equating tension and compression
21
cbxf = Asfs+ 21 bfct(h-x)-----------------(4)
By solving the above 4 equations using trial and error method
We have x= 924.4mm
fc = 22.63N/mm2
Long term curvature br1 =
eff
c
Exf×
= 31067.104.92463.22
××= 2.29x10-6/mm
Long term deflection due to creep = 21 Lkrb
××
= 2.29x10-6 x 0.083 x 174002
= 57.54mm
36
Deflection due to shrinkage:-
csr1 =
ISsccs ××αε
cα = ffe
s
EE
=67.10
200 = 18.74
I = ( )223
)2
(12
xdAxbxbxsc −++ α
= 2.978x 1011mm4
Ss = ( )xdAs −
= 24120 x (1480-924.4)
= 13.41 x 106mm3
From Fig-7.2, BS8110-2:1985
csε = 327 x 10-6
Thus
csr1 = 11
66
1097.21041.1374.1810327
××××× −
=2.76 x 10-7/mm
Long term deflection due to shrinkage = 21 Lkrb
××
= 2.76x10-7 x 0.083 x 174002
= 6.93mm
Short term deflection due to total loads = 39.7mm
Short term deflection due to permanent loads = 35.6mm
Short term deflection due to non permanent loads = 39.7-35.6 =4.1mm
Long term deflection (permanent loads) = 57.54mm
Long term deflection (shrinkage) = 6.93mm
Total long term deflection = 68.57mm
Permissible deflection = l/250 =69.6mm
Hence safe.
37
13. DESIGN OF IN-SITU GROUND FLOOR SLAB (b/w grids 7, 8, E&F)
Material constants:-
Concrete fck = 40 N/mm2
Steel fy = 460 N/mm2
Shorter Span = 8100mm
Longer Span = 8700mm
Shorter span:-
Thickness of slab = 250mm
Diameter of bar = 16mm
Effective depth = 212mm
Mid-span Moment = 62kNm
u2
Mb d
= 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.
Support Moment = 20kNm
u2
Mb d
= 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
u2
Mb d
= 1.56
From Chart No.2 BS 8110 Part 3
Percentage of steel required = 0.77%
Area of steel required = 1509mm2/m
38
Area of steel provided in the section = 2010mm2/m
Hence Safe.
Support Moment = 18kNm
u2
Mb d
= 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
39
14. Design of Roof:-
The analysis of the roof was done by using modeling it as a frame as shown in Fig-9.
The frame is spaced at 4.05m apart.
Fig-9- Staad Model
Loadings:
Dead load
Concrete Roof Tile = 0.5 kN/m2
Ceiling and Services = 0.80 kN/m2
Live Load =1 kN/m2
Wind load:-
Wind pressure = 1.2kN/m2 (See Section 6.2)
From BS 6399 Part 2
External Pressure Coefficient Cp = 1.2
Internal Pressure Coefficient Cp = ±0.2
Max Wind Pressure = 1.2 x (1.2+0.2)
= 1.68kN/m2
Design of Purlins:
Provide purlins at 1.5m c/c
The purlins are designed as a simply supported beam with the following loads.
Total Dead Load coming in the Purlin = (0.5+0.8) x 1.5 = 1.95kN/m
Live Load = 1 x 1.5 = 1.5kN/m
Wind Load = 1.68 x 1.5 =2.52kN/m
From Staad analysis the section required is UB 125×65× 15mm
(As per BS 5950-2000)
40
Design of Main Beam:
Total Dead Load coming in the frame = (0.5+0.8) x 4.05 = 5.26kN/m
Live Load = 1 x 4.05 = 4.05kN/m
Wind Load = 1.68 x 4.05 =6.8kN/m
From Staad analysis the section required is UB 533×201× 102mm
(As per BS 5950-2000)
41
15. REFERENCES:
1. British standards, “Structural Use of Concrete”- Code of Practice for design and
construction (BS 8110-1:1997)
2. British standards, “Structural Use of Concrete”- Code of Practice for special
circumstances” (BS 8110-2:1997)
3. British standards, “Structural Use of Concrete”- Design Charts” (BS 8110-3:1997)
4. British standards, “Loading for Buildings”- (BS 6399-1,2,3:1996)
5. British standards, “Structural Use of Steel work”- (BS 5950:2000)
6. Universal Building Code -1997
7. A.W. Irvin ”Design of Shear wall Buildings,CIRIA(Construction Industry Research
and Information Association) Report.
8. L.J Morris and D.R Plum ”Structural Steel work Design to BS 5950”
9. Devdas Menon & S Unnikrisna Pillai, Reinforced Concrete Design , Tata Mc Graw-
hill publishing company Ltd. , Delhi
10. Prof S.Ramamruthum”Design of Reinforced Concrete Structures ,Dhanpat Rai
Publishing Company(P) Ltd,New Delhi
11. Joseph E Bowles, Foundation analysis and Design , Tata Mc-Graw Hill,
International edition, New Delhi 1988
12. Geotechnical report