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Contents
1. Introduction
2. Preliminary selection of structural elements
3. Wind Analysis
4. Earthquake Analysis
5. Element Design
6. Reference
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Design of a Forty Story Building for ABC bank head quarts
1.0 Introduction
It is decided to design a forty storied building for the use of ABC bank at Colombo 7. Whole
building is to be used as Bank head quarts and rent some space to be used as office.
Designed will be done as a part of the course work of Pg Dip /MEng (Structural Design )
program.
Key features of the building
1- 40 storied building
2- 8 floors to be used as car park
3- Lift system is designed as “Hard Zoning”
4- Building length 48m
5- Building width 36m
6- Super structure will be reinforced concrete and sub structure will be pile foundation.
Following designs will be considered in this Report.
Decide grid arrangement and building services
Selecting the arrangement of strong elements
Selection of suitable loads
Selection of dynamic parameters
Determination of wind loads and induced acceleration
Determination of earth quake loads
Develop computer model and determination of member forces
Design of slab beam wall column and pile foundation
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1.1 Calculation of basic parameters of the building
Let Column spacing in longest direction = 8m
Therefore no of columns 48/8+1 = 7
Let column spacing in shortest direction = 6m
Therefore no of columns 36/6+1 = 7
Let beam height (hb) 6000/14+25 = 450mm
Slab Thickness (Ts) 6000/44+25 = 160 mm
Let space for services c = 300 mm
Let head clearance h = 2.5 m
Total Floor height h+c+Ts+hb = 2.5+.3+.45
H = 3.25m
Height of the Building 3.25x40 = 130m
Height/Width Ratio 130/36 = 3.6
One Floor Area 36x48 = 1728 m2
Lift Core Area 24x16 = 384 m2
Usable Area = 1344 m2
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7
6
5
4
3
2
16.00 m
6.00 m
6.00 m
6.00 m
6.00 m
6.00 m
36.00 m
48.00 m
8.00 m
A B C D E F G8.00 m 8.00 m 8.00 m 8.00 m 8.00 m
Fig 1 Building Layout
1.2 Calculation of Number of Lifts
Let population density = 10 m2
Number of people per floor 1344/10 = 134.4 per floor
Let = 135 persons per floor
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8th Floor (Car park)
10th Floor
20th Floor
30th Floor
40th Floor
Exp
ress R
un
Exp
ress R
un
Hard zoning arrangement will be used as follows.
Fig 2 Lift Arrangement
1.2.1 Ground floor level to 19th floor level
Population calculation Ground floor to 19th floor = 135 x 12
= 1620 persons
Let consider serving for 5 minute peak capacity.
Peak capacity is assumed as 14% of the total population .
Hence, population handled in 5 minute period = 1620x14/100
= 226.8persons
Let 227 persons
Let select 24 capacity lift, Average interval = 28 sec (Table 25 7.1)
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Let select 4m/s speed lift system, hence round trip time = 165 s
Hence, number of lifts required Round Trip Time
Average Interval
= 165 / 28
= 5.89
Hence, 6 lift required between ground floor level and 19th floor level.
1.2.2 20th floor level to 29th floor level
Population calculation 20th floor to 29th floor = 135 x 10
= 1350 persons
Let consider serving for 5 minute peak capacity.
Peak capacity is assumed as 14% of the total population
Hence, population handled in 5 minute period = 1350 x 14/100
= 189 persons
Let select 24 capacity lift ,Average interval = 32 sec (Table 25 7.1)
Let select 4m/s speed lift system, for 18 floors hence round trip time = 160 s
Lift will travel express 6 ms-1 from 8th floor level to 19th floor level and serve 20th floor to 29th
floor.
Hence, total distance traveled = 12 x 3.25 x 2
= 78 m
Hence, time taken = 78 / 6
= 13 s
Total round trip time = 160 + 13 s
= 173 s
Hence, number of lifts required = Round Trip Time
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Average Interval
= 173 / 32
= 5.4
Say, 5 lifts
Hence, 5 lift cars to serve between 20th floor level and 29th floor level.
1.2.3 30th floor level to 39th floor level
Population calculation 30th floor to 39th floor = 135 x 10
= 1350 persons
Let consider serving for 5 minute peak capacity.
Peak capacity is assumed as 14% of the total population
Hence, population handled in 5 minute period = 1350 x 14/100
= 189 persons
Let select 24 capacity lift ,Average interval = 32 sec (Table 25 7.1)
Let select 4m/s speed lift system, hence round trip time = 160 s
Lift will travel express 6 ms-1 from 8th floor level to 29th floor level and serve 30th floor to 39th
floor.
Hence, total distance traveled = 22 x 3.25 x 2
= 143 m
Hence, time taken = 143 / 6
= 23.8 s
Total round trip time = 160 + 24 s
= 184 s
Hence, number of lifts required = Round Trip Time
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Average Interval
= 184 / 32
= 5.7
Say 6 lifts
Hence, 6 lift cars can be used to serve between 30th floor level and 39th floor level.
Summary of the lift arrangement as follows.
Location No of lifts
Ground floor level to 19th floor level 6
20th floor level to 29th floor level 5
30th floor level to 39th floor level 6
Total 17
Table 1 Required lift at each zone
1.2.4 Lift cabin arrangement
From Table 25.3
Cabin size of the 24 passenger lift = 2100 x 1650
Shaft size = 2550 x 2400
1.2.5 Stair case arrangement
Width = 2m
Length = 3.35m
Landing length = 4m
Landing width = 1.5m
Required opening area = (3.35+1.5)x 4 m2
= 19.4m2
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1.2.6 Wash rooms arrangement
There will be two washrooms for men and women, 7.5m x7.5m each
1.2.7 Service Core Arrangement
Length = 24m
Width = 16m
Fig 3 Service Core Arrangements
men
wash room
ladies
wash room
7
6
5
4
3
2
16.00 m
6.00 m
6.00 m
6.00 m
6.00 m
6.00 m
36.00 m
48.00 m
8.00 m
A B C D E F G8.00 m 8.00 m 8.00 m 8.00 m 8.00 m
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men
wash room
ladies
wash room
7
6
5
4
3
2
16.00 m
6.00 m
6.00 m
6.00 m
6.00 m
6.00 m
36.00 m
48.00 m
8.00 m
A B C D E F G8.00 m 8.00 m 8.00 m 8.00 m 8.00 m
1.2.8 Shear wall arrangement
Lateral loads are to be resisted by a shear wall arrangement.
Assumed Shear wall thicknesses are as follows.
Zone Selected Shear wall
thickness / (mm)
Ground to 9th floor level 300
10th floor to 19th floor level 300
20th floor to 29th floor level 300
30th floor to 39th floor level 300
Table 2 Shear wall Thickness
Fig 4
Shear wall
arrangement
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2.0 Preliminary selection of structural elements.
Slab Thickness
Selected slab thickness = 175 mm
Cover = 25 mm
Diameter of the reinforcement = 10 mm
Hence, effective depth = 175 - 25 – 10/2
= 145 mm
Span / Effective depth = 6000 / 145
= 41.4
Basic span / Effective depth = 26 ( T 3.9 BS8110)
Let Assume a modification factor of 1.6
Modified span / Effective depth = 26 x 1.6
= 41.6
Hence, 175 mm slab thickness is capable of meeting the deflection criteria satisfactorily.
2.1 Selection of Beam Dimensions
Selected beam depth = 600 mm
Hence, effective depth = 550 mm
Span / Effective depth = 8000 / 550
= 14.55<26
Hence, depth is a reasonable value.
Selected width of the beam = 400 mm
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2.2 Determination of Column Dimensions
Initial column size can be selected with trial approximate loads
Assume column carries load from 8 x 6 panel area of the building. Trial size of column will be
selected according to axial load capacity
2.2.1 Dead loads calculation
Slab self-weight = 8 x 6 x 0.175 x 24
= 201.6 kN
Assuming Finishes of 0.75 kN / m2 = 0.75 x 8 x 6
= 36kN
Assuming partition of 1.0 kN / m2 = 1.0 x 8 x 6 (BS 6399 5.1.4)
= 48 kN
Assuming services of 1.0 kN / m2 = 1.0 x 8 x 6
= 48 kN
Beams Weight = (8 + 6) x 0.4 x 0.6 x 24
= 80.64 kN
Live load on slab of2.5 kN / m2 = 2.5 x 8 x 6 (BS 6399)
= 120 kN
Calculation of design load
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Consider 50% maximum reduction in impose load (BS 6399)
Assume 40% reduction in impose load in our design
1.4 x (201.6 + 36 + 48 + 48 +80.64) + 1.6 x 120 x 0.60 = 695.14 kN
2.3 Selection of Column Sizes
Select Trial column size from 30th floor level to 39th floor= 700mm x 700mm
Axial load at 30th floor level [695 x 10 + (0.7 x 0.7 x 3.25 x 24) x 10 x 1.4]
= 7485.1kN
Axial load of Short columns resisting moment and axial forces given by,
N = 0.4fcuAc + 0.8Ascfy (BS 8110 -1997 equation 38) --------------------- > Eq(A)
Let assume fcu= 40 N/mm2 and Asc =3%Ac
N = 0.4Ac x 40 + 0.8 x 0.03 x Ac x 460
N = 27.04 Ac
N= 7485 at 30th floor
7485x103 = 27.04 Ac
Ac = 276,812mm2
Therefore required column size = 526 x 526
Hence, selected size is too large and 600 x 600 will be selected
Select Trial column size from 20th floor level to 29th floor= 700mm x 700mm
Axial load at 20th floor level [695 x 10 + (0.7 x 0.7 x 3.25 x 24) x 10 x 1.4]+7343
=14828
From Eq(A) , Let assume fcu= 40 N/mm2 and Asc =3%Ac
Same as previous calculation
14828x103=27.04Ac
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Ac = 548,372 mm2
Hence, minimum column dimension required is 740 mm x 740 mm.
Therefore 750mm x 750mm columns are used
Select Trial column size from 10th floor level to 19th floor= 800mm x 800mm
Axial load at 10th floor level [695 x 10 + (0.8 x 0.8 x 3.25 x 24) x 10 x 1.4]+ 14828
=22477
From Eq(A) , Let assume fcu= 40 N/mm2 and Asc =4%Ac
Same as previous calculation
N = 0.4Ac x 40 + 0.8 x 0.04 x Ac x 460
22477 x 1000= 30.72Ac
Ac = 731673 mm2
Hence, minimum column dimension required is 855 mm x 855 mm.
Therefore 900mm x 900mm columns are used
Select Trial column size from ground floor level to 9th floor= 1000mm x 1000mm
Additional impose load of 2.5kN/m2 will be considered up to 8th floor to equate car park
load
Axial load at Ground floor level [695 x 10 + (1 x 1 x 3.25 x 24) x 10 x 1.4]+ 22477+2.5x8x6x8
=32055
From Eq(A) , Let assume fcu= 50 N/mm2 and Asc =5%Ac
Same as previous calculation
N = 0.4Ac x 50 + 0.8 x 0.05x Ac x 460
32055 x 1000= 38.4Ac
Ac = 834,765 mm2
Hence, minimum column dimension required is 914 mm x 914 mm.
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Therefore 1000mm x 1000mm columns are used
Selected element sizes are given in following table.
Element
Zone Shear Wall Column beam (6m) Beam (8m) Slab
Ground floor to 9th floor 0.3 1.0x1.0 0.45 0.6 0.175
10th floor to 19th floor 0.3 0.9x0.9 0.45 0.6 0.175
20th floor to 29th floor 0.3 0.7x0.7 0.45 0.6 0.175
30th floor to 39th floor 0.3 0.6x0.6 0.45 0.6 0.175
Table 3 Summary of element sizes.
3.0 Wind Analysis
3.1 Determination of Wind Induced Forces of the Structure
Hourly mean wind speed is calculated using following equation.
Vz = V. M (z.cat). Ms. Mt. Mi
Vz = Design hourly mean wind speed at height z in ms-1
V= Basic wind speed
Basic wind is taken as 38 ms-1 . (Zone-2)
Ms= 1.0 (It is assumed the building spacing > 12-Table-4.2.7)
Mt = 1.0
Ms= 1.0 (It is assumed the building as normal structure-Table-4.2.9)
Hence, Vz = 38M (z.cat)
Fz= ∑Cp,e.qz.Az
Hence dynamic wind pressure
qz= 0.6vz2 x 10-3
Let Cp,e = Pressure coefficients for both windward and leeward surfaces of rectangular
buildings.
Az= Area of a structure at height z
X- Direction,
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d/b = 48 / 36
= 1.33
Y- Direction,
d/b = 36 / 48
= 0.75
X- Direction,
(Cp,e)x = 0.8 + 0.3 = 1.1
Y- Direction,
(Cp,e) y= 0.8 + 0.5 = 1.30
Fx = (Cp,e)x.qz. (Az)x
Hence, Fx = 1.1qz
Fy = (Cp,e)y qz. (Az)y
Hence, Fy = 1.30 qz
Total wind force acting on each level is assumed to act on two frames which are connected to shear walls . Nodel forces are calculated based on total affecting force divided by 2.
Level
Floor-Floor
height (m)
Z (m) M(z,cat) Vz
(ms-
1)
qz
(kPa)
Fx
(kNm-
1)
Fy
(kNm-
1)
X nodal force
Y Nodal Force
G 3.25 0 - - - -
1 3.25 3.25 0.35 13.3 0.106 0.12 0.14 2.34 3.64
2 3.25 6.5 0.35 13.3 0.106 0.12 0.14 2.34 3.64
3 3.25 9.75 0.35 13.3 0.106 0.12 0.14 2.34 3.64
4 3.25 13 0.35 13.3 0.106 0.12 0.14 2.34 3.64
5 3.25 16.25 0.35 13.3 0.106 0.12 0.14 2.34 3.64
6 3.25 19.5 0.35 13.3 0.106 0.12 0.14 2.34 3.64
7 3.25 22.75 0.358 13.6 0.111 0.13 0.15 2.535 3.9
8 3.25 26 0.368 13.98 0.117 0.13 0.16 2.535 4.16
9 3.25 29.25 0.378 14.36 0.124 0.14 0.17 2.73 4.42
10 3.25 32.5 0.385 14.63 0.128 0.15 0.17 2.925 4.42
11 3.25 35.75 0.39 14.82 0.132 0.15 0.18 2.925 4.68
12 3.25 39 0.398 15.12 0.137 0.16 0.18 3.12 4.68
13 3.25 42.25 0.4 15.2 0.139 0.16 0.19 3.12 4.94
14 3.25 45.5 0.41 15.58 0.146 0.17 0.19 3.315 4.94
15 3.25 48.75 0.417 15.85 0.151 0.17 0.2 3.315 5.2
16 3.25 52 0.42 15.96 0.153 0.17 0.2 3.315 5.2
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17 3.25 55.25 0.438 16.64 0.166 0.19 0.22 3.705 5.72
18 3.25 58.5 0.45 17.1 0.175 0.2 0.23 3.9 5.98
19 3.25 61.75 0.462 17.56 0.185 0.21 0.25 4.095 6.5
20 3.25 65 0.474 18.01 0.195 0.22 0.26 4.29 6.76
21 3.25 68.25 0.485 18.43 0.204 0.23 0.27 4.485 7.02
22 3.25 71.5 0.497 18.89 0.214 0.24 0.28 4.68 7.28
23 3.25 74.75 0.508 19.3 0.224 0.25 0.3 4.875 7.8
24 3.25 78 0.514 19.53 0.229 0.26 0.3 5.07 7.8
25 3.25 81.25 0.52 19.76 0.234 0.26 0.31 5.07 8.06
26 3.25 84.5 0.525 19.95 0.239 0.27 0.32 5.265 8.32
27 3.25 87.75 0.53 20.14 0.243 0.27 0.32 5.265 8.32
28 3.25 91 0.535 20.33 0.248 0.28 0.33 5.46 8.58
29 3.25 94.25 0.54 20.52 0.253 0.28 0.33 5.46 8.58
30 3.25 97.5 0.546 20.75 0.258 0.29 0.34 5.655 8.84
31 3.25 100.75 0.551 20.94 0.263 0.29 0.35 5.655 9.1
32 3.25 104 0.555 21.09 0.267 0.3 0.35 5.85 9.1
33 3.25 107.25 0.56 21.28 0.272 0.3 0.36 5.85 9.36
34 3.25 110.5 0.564 21.43 0.276 0.31 0.36 6.045 9.36
35 3.25 113.75 0.569 21.62 0.281 0.31 0.37 6.045 9.62
36 3.25 117 0.574 21.81 0.285 0.32 0.38 6.24 9.88
37 3.25 120.25 0.578 21.96 0.289 0.32 0.38 6.24 9.88
38 3.25 123.5 0.583 22.15 0.294 0.33 0.39 6.435 10.14
39 3.25 126.75 0.587 22.31 0.299 0.33 0.39 6.435 10.14
40 3.25 130 0.592 22.5 0.327 0.36 0.43 3.51 5.59
Table 4 Wind load on each floor
3.2 Assignment of Wind Force to the Modal
Wind forces calculated above were assigned to two frames of the structure to ensure
symmetrical distribution of wind forces. Grid 2,6 and C,E were selected for applying wind
force
3.3 Lateral Deflection of the Structure due to Wind (Along X direction)
Maximum drift index for normal building 1/500 .However tall buildings the drift index value
is limited to 1/1000.
Height of the building = 130 m
The maximum top deflection obtained from SAP 2000 under load combination-2 is 59 mm.
Drift index = 59 / (130 x 1000)
= 1 / 2203 < 1 / 1000
Hence structure satisfy Drift index
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3.4 Calculation of Gust factor
The gust factor G can be calculated using the equation,
G = 1 + r [ gv2.B. (1 + w)2 + (gf
2.S.E / ε) ]
Let assume Damping ratio ε = 0.01
gv = 3.7
Turbulence length
Lh = 1000 x (h / 10)0.25
= 1000 x (130/ 10)0.25
= 1898 m
Background factor
B = 1/ {1 + [ (36h2 + 64b2) 1/2/ Lh ] }
= 1/ {1 + [ (36x1302 + 64x482) 1/2/ 1898] }
= 0.685
r- Roughness factor
r = 2(σv/ v) / Mt
σv/ v =0.219 (From Table 4.2.5.3)
r = 2 x 0.219 / 1.0
= 0.438
w-Factor to account for the second order effects of turbulence intensity
w = gv.r.(B)1/2 / 4
= 3.7 x 0.438 x (0.685)1/2 / 4
= 0.335
gf- Peak factor
gf = (2 log(3600na))1/2
Natural period of vibration along Y direction obtained after analyzing the modal in SAP 2000
and it is assumed T = 4 sec
Hence, natural frequency of the first mode vibration is,
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na = 1 / 4
= 0.25
gf = (2 log(3600 x 0.25))1/2
= 2.43
S-Size factor
S = 1/ {[1 + (3.5 nah /Vz)] x [1 + (4 nab /Vz)]}
Vz = V.x Mzcat
= 38 x 0.652
= 24.78 m/s
S = 1/ {[1 + (3.5 x 0.25 x 130/ 24.78)] x [1 + (4 x 0.25 x 48 / 24.78)]}
= 0.061
N-Effective reduced frequency
N = na x Lh / Vz
= 0.25 x 1898 / 24.78
= 19.14
E- Spectrum of turbulence in the approaching wind stream
E = 0.47N / (2+ N2)5/6
= 0.47 x 16.64 / (2+ 16.64x16.64)5/6
= 0.065
G = 1 + r [ gv2.B. (1 + w)2 + (gf
2.S.E / ε) ]1/2
= 1 + 0.438 [ 3.72 x 0.685 x (1 + 0.335)2 + (2.432 x 0.061 x 0.065 / 0.01) ]
G = 2.91
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3.5 Calculation of Wind Induced Acceleration (Along X direction)
Generally accelerations above 0.05ms−2 are perceived by the people. A peak acceleration of
0.15ms −2 to 0.20ms −2 is often considered as the limiting value at the suitable design wind
speeds.
Along wind acceleration,
αa = ( 2πna)2gf.r(SE / ε)1/2 (Δ/G)
=(2 x 22/7 x 0.25)2 x 2.91 x 0.438 (0.061 x 0.065/0.01)1/2 (Δ/2.91)
= 0.681 Δ
=0.681x59/1000
=0.04 ms−2 Marginaly Satisfied
Across wind acceleration,
αc = [1.5 gf . qh .b(0.76 + 0.24k)(πCfs/ ε)1/2] / m0
k = 1.0 (Building with a central core and moment resisting facade)
qh = 0.6 Vz2
gf = (2 log(3600nc))1/2
Natural period of vibration along Y direction obtained after analyzing the modal in SAP 2000
and it is assumed T = 4 sec
Hence, natural frequency of the first mode vibration is,
nc = 1 / 4
= 0.25
gf = (2 log(3600 x 0.25))1/2
= 2.43
h:b:d =130:48:36
=3.6 : 1.33 :1
Vh/ncb =24.78/(0.25 x48)
= 2.065
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Cfs = 0.0022 (Cross wind force spectrum coefficient)
m0 = 262 kg / m
αc = [1.5 gf . qh .b(0.76 + 0.24k)(πCfs/ ε)1/2] / m0
= [1.5 x 2.43 x 0.303 x 48 (0.76 + 0.24 x 1.0 )(22/7 x 0.0022/ 0.01)1/2] /262
= 0.168 ms-2
4.0 Earthquake Analysis
4.1 Determination of the Weight of the Stricture
Number of Columns =34 Nos.
Shear Wall Length per floor =136 m
Beam Length per floor 600mm =336m
Beam Length per floor 450mm =336m
Building weight Zone
Amount Level 0-9 (kN) Amount
Level 10-19 (kN) Amount
Level 20-29 (kN) Amount
Level 30-40 (kN)
Columns 1x1 2652 0.9x0.9 21481 0.7x0.7 12994 0.6x0.6 9547
Beams 600mm 336m 10282 336m 10282 336m
10281.6 336m 10281.6
Beams 450mm 252m 4989.6 252m 4989.6 252m 4989.6 252m 4989.6
Salb 1599m2 67158 1599m2 67158 1599m2 67158 1599m2 67158
Shear walls 136m 1326 136m 1326 136m 1326 136m 1326
Impose load 4.5kNm
2 71955
2.5kNm2
(60%) 23985 2.5kNm
2
(60%) 23985 2.5kNm
2
(60%) 23985
Super impose Dead load 2.75kNm
2 43973 2.75kNm
2 43973 2.75kNm
2 43972.5 2.75kNm
2 43972.5
Total Load 202335 173194 164707 161260
Table 5 Load Summary
Total building load 701495 kN
Building volume 224640 m3
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Density of the building = 3.12 KN/ m3
4.2 Determination of the Base Shear
V = ZIKCSW
V-Base shear
Z-Seismic probability zone factor
Z = 3 / 16 (Minor earthquake)
I-Occupancy importance factor
I = 1.00 (Office building)
K-Building type factor
K = 1.33
T- Fundamental natural period of vibration
T =4 sec
C = 1/15(T)1/2
= 1/15(T)1/2
= 0.13 < 0.2
S-Soil interaction factor
S = 1.2 (It is assumed cohesionless or stiff clay soil conditions overlying rock at a depth
greater than 200ft in Colombo 03)
W – Total dead load and appropriate portions of the live load
W = 701,495 kN
V = ZIKCSW
V = (3 / 16) x 1.00 x 1.33 x 0.03 x 1.2 x 701,495
= 6,297 kN
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4.3 Determination of Earthquake Forces
X Direction:
The base shear is to act with a minimum eccentricity of 5% of the maximum building
dimension.
Eccentricity, e = 0.05Dmax
= 0.05 x 48
= 2.4m
Base shear will be assigned to frames 2 & 6 as follows.
V2 = 6297 x (12-2.4)/24
V2 = 2518 KN
V6 = 6297-2518 KN
= 3779 KN
Ft = 0.07 TV
=0.07 x 4 V
=0.28V >0.25V then Ft=0.25V is selected
Ft = 1574 kN
Ft2 =629 kN
Ft6 =944 kN
Let assume weight at the floor levels is constant.
FX = (V- Ft) hx / ∑ hx
(V- Ft) =6297-1574
=4723 kN
∑ hx =3.25(40/2x(1+40)m
=2665m
Fx(2) = 0.75x2518/2665 hx
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Fx(2) =0.709 hx
Fx(6) = 0.75x3779/2665 hx
Fx(6) = 1.06 hx
Y Direction:
The base shear is to act with a minimum eccentricity of 5% of the maximum building
dimension.
Eccentricity, e = 0.05Dmax
= 0.05 x 48
= 2.4 m
Base shear can be assigned to frames along gridlines C & F as follows.
VC = 6297 x (8+2.4)/16
VC = 4093 KN
VE = 6297-4093 KN
= 2204 KN
Ft = 0.07 TV
=0.07 x 4 V
=0.28V >0.25V then Ft=0.25V is selected
Ft =1574
Ftc =1023
FtE =550
FX = (V- Ft)hx / ∑ hx
FC = (0.75 x 4093/2665) hx
=1.15 hx
FE = (0.75 x 2204/2665) hx
=0.62 hx
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Calculation of Earthquake Load at each level
Level hx (m)
Earthquake Load
X Direction (kN) Y Direction (kN)
F2 F6 FC FE
G 0 0 0 0 0
1 3.25 2.30425 3.445 3.7375 2.015
2 6.5 4.6085 6.89 7.475 4.03
3 9.75 6.91275 10.335 11.2125 6.045
4 13 9.217 13.78 14.95 8.06
5 16.25 11.52125 17.225 18.6875 10.075
6 19.5 13.8255 20.67 22.425 12.09
7 22.75 16.12975 24.115 26.1625 14.105
8 26 18.434 27.56 29.9 16.12
9 29.25 20.73825 31.005 33.6375 18.135
10 32.5 23.0425 34.45 37.375 20.15
11 35.75 25.34675 37.895 41.1125 22.165
12 39 27.651 41.34 44.85 24.18
13 42.25 29.95525 44.785 48.5875 26.195
14 45.5 32.2595 48.23 52.325 28.21
15 48.75 34.56375 51.675 56.0625 30.225
16 52 36.868 55.12 59.8 32.24
17 55.25 39.17225 58.565 63.5375 34.255
18 58.5 41.4765 62.01 67.275 36.27
19 61.75 43.78075 65.455 71.0125 38.285
20 65 46.085 68.9 74.75 40.3
21 68.25 48.38925 72.345 78.4875 42.315
22 71.5 50.6935 75.79 82.225 44.33
23 74.75 52.99775 79.235 85.9625 46.345
24 78 55.302 82.68 89.7 48.36
25 81.25 57.60625 86.125 93.4375 50.375
26 84.5 59.9105 89.57 97.175 52.39
27 87.75 62.21475 93.015 100.9125 54.405
28 91 64.519 96.46 104.65 56.42
29 94.25 66.82325 99.905 108.3875 58.435
30 97.5 69.1275 103.35 112.125 60.45
31 100.75 71.43175 106.795 115.8625 62.465
32 104 73.736 110.24 119.6 64.48
33 107.25 76.04025 113.685 123.3375 66.495
34 110.5 78.3445 117.13 127.075 68.51
35 113.75 80.64875 120.575 130.8125 70.525
36 117 82.953 124.02 134.55 72.54
37 120.25 85.25725 127.465 138.2875 74.555
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38 123.5 87.5615 130.91 142.025 76.57
39 126.75 89.86575 134.355 145.7625 78.585
40 130 92.17 137.8 149.5 80.6
Table 6 Earth quake loads
4.4 Application of the Earthquake Load to the SAP 2000 Modal
A new load case is defined as Earthquake.
A new load combination is defined.
Then the earthquake forces calculated above are assigned to the frames along grid 2 & 6 on
x direction and along C & F on y direction.
5.0 Element Design
Following elements are designed in this model analysis.
1- Pile & Pile Cap
2- Beam
3- Column
4- Slab
5.1Design of Pile
5.1.1 Estimation of Carrying Capacity of a Single Pile
Consider foundation at F-2 column location
Design load on foundation = 21, 550 kN
Assuming bed rock level is 20 m below existing ground level and skin friction values are
assumed as follows. Assumed allowable bearing capacity of rock is 4 N/mm2.
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Pu – Column Load W –Self Weignt
P si -Skin Friction
Ppu -End Bearing
Pu = Ppu + ∑ Psi – W
Tu = Psi + W
Where,
Pu = Ultimate compressive load on pile
Tu = Ultimate tensile load on pile
∑ Psi = Skin Friction Resistance
W =Weight of pile
Ppu = End bearing resistance
Assuming 1500mm diameter bored cast in-situ R/C pile.
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Average depth of a pile is 20 m.
Hence, End bearing resistance = π x 1.52 x 4000/4
Ppu = 7068kN
Assuming ultimate skin friction values as follows
Depth Aerage
SPT value
Skin Friction (kN/m2)
(Ultimate) From To
0,00 10 20 20
10 20 25 25
20 25 30 30
25 30 >50 100
Table 6:1 Skin friction at each depth
∑ Psi = π x 1.5x [20x10+25x10+30x5 +1.5x150]
= 3887kN
Factor of safety 3
Therefore total allowable skin friction = 3887kN
Total Pile Resistance = (7068+ 3887)
= 10955 kN
Ultimate compressive load on pile,
Pu = 10955 – π x 1.5 2 x 24 x 25/4
= 9895kN
Hence, allowable load on the pile 9895 kN
Design load of the internal column from SAP analysis = 18805 kN
Therefore number of piles required for an internal column location
= 18805/9895 = 1.9
Therefore allow two piles for one internal column location.
Therefore load acting on one pile = 18805/2
= 9402 kN
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Design load on pile < Allowable load on pile
Hence pile carrying capacity for axial load is satisfied.
Let consider Grade 30 concrete
Working stress of the concrete at pile toe = (9402 + 1050) / (π x 1.5 2 /4)
= 5.9 N/mm2
< 0.25fcu = 7.5 N/mm2
Required area of reinforcement 1% of Concreter Area = 17671mm2
Required number of 25mm bars = 17671/490
Number of bars = 36 Nos
Reinforcement provided = 36Y25 +Y12 @200
Piles should be socketed 1500mm in to parent rock and capacity should be verified by a full
scale load test.
Piled raft can be used for connection of piles to the column. Individual pile caps are not
suitable for this building because the pile caps will be arranged in very congested pattern.
5.2 Determination of Pile Cap Dimensions
Centre to centre distance of the pile (2.5 ϕ ) = 2.5 x 1500
= 3750 mm
Distance from the edge of the pile cap = 400 mm
length of the pile cap = 3750 + 1500 + 400 + 400
= 6050 mm
Effective depth of the pile cap (2.5 ϕ /2) = 1875 mm
Hence, pile cap dimension is 6050 mm x 2300 mm x 1875 mm
5.2.2 Design of a Pile Cap
pile spacing = 3750 mm
Edge distance = 400 mm
Length of pile cap = 6050 mm
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Width of pile cap = 2100 mm
Pile cap main reinforcement calculation
Ultimate load on pile cap (N) = 30500 kN
Assume depth of pile cap = 1800 mm
cover = 75 mm
Effective depth (d) = 1665 mm
Total tensile force in each direction without considering column size
= N l / 2d
= 20608 kN
As required = T /0.87 fy
= 51495 mm2
Layer 1 Diameter = 40 mm
Nos = 14
Spacing 131 mm
Layer 2 Diameter = 40 mm
Nos = 14
Spacing 131 mm
Layer 3 Diameter = 40 mm
Nos = 14
Spacing 131 mm
As Provided = 52769 mm2
Check for punching shear
Punching shear around column perimeter = 3.82 N/mm2
0.8√fcu = 4.38 N/mm2
Check Satisfied
Since spacing of piles = 3φ no further check is required
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Check for line shear
Column side dimension = 1000 mm
av = 1300 mm
V at critical section = 15250 kN
v = 5.09 N/mm2
2d/av = 2.56
100As/bd = 1.76
vc = 2.08 N/mm2
Shear r/f need not to be provided
Determination of Distribution steel
Min Steel 100 As/ Ac = 0.13
As = 4212 mm2
Diameter = 32 mm
Spacing = 175
As = 4593 mm2
Determination of Horizontal binders
Use 25% of main steel
As = 13192 mm2
Diameter = 32 mm
Nos = 15
As Provided = 12061 mm2
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5.3 Design of a Beam Fig 7
Design
moments of the beam at first floor level
Max Support moment = 278 kNm
Max Bending moment at span = 189 kNm
5.3.1 Design of Beams
Depth of the Beam = 600 mm
Width of the Beam = 300 mm
Concrete Strength = 30 N/mm2
Assume moderate exposure conditions, for outdoor exposure. Assume a fire resistance of 2 hours.
Cover of the Beam = 25 mm
Assume a link diameter of 10 mm and reinforcement size of 25mm
Effective Depth of the Beam = 600-25-10-12.5
= 553mm
Design for Flexure
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Sample calculation is done for hoggin moment at the exterior column
Max. hogging Bending Moment = 278 kNm
moment, exterior support moment will be re-distributed by 10%
Then design support moment
=278x0.9
=250.2 kNm
Reduction in moment =27.8 kNm
At the exterior support ,
K = M/bd2fcu
= 250.2 x 1000 x 1000/300x5532x30
= 0.091 < 0.156
Section is singly reinforced
Z = d 0.5 + 0.25 - K
0.9
= d 0.5 + 0.25 - 0.091
0.9
= d x 0.886
= 0.89 d < 0.90d
Z = 0.89 d
= 489 mm
As = M/(0.87fyZ)
= 250 x 1000 x 1000
0.87x460x489
= 1278 mm2
Minimum reinforcement = 0.13/100bh
= 0.13/(300x600x100)
= 234 mm2
Therefore As = 1278 mm2
Provide 3T25 ( 1470 mm2)
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Design for Shear
From Sap2000 Results
Vmax = 268kN
Vmax = V/ bd
vmax = 268 x 1000 /300 x 553
= 1.62 N/mm2
100As/bd = 0.887
Use R8 @ 100mm spacing
5.4 Design of a Column
Ultimate
Axial
Load
= 26785 kN
Majour B/M = 64 kNm
Miner B/M = 18 kNm
Column Axial Force Column Major Moment Column Miner moment
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Column Size 1000mm x 1000mm
Concrete Grade 50 N/mm2
Top Condition
= Condition 2
Bottom Condition = Condition 1
5.4.1 Effective Height Calculation
For X-X / Y-Y Direction
q
L=3.25 m
p
So ,
βpq
= 1.3
ℓo x
= 3.25 - 0.6
= 2.65 m
ℓex
= β x ℓo
= 1.3 x 2.65
ℓex , ℓey = 3.45 m
5.4.2 Check for slenderness
ℓex /h = ℓey/h
= 3.45 / 1
= 3.4
< 10
The column is a short column.
60 b
= 60 x 1200 / 1000
= 72 m
> ℓo
Slenderness limit is not exceeded.
Considering 40% live load reduction ,Critical load combination 1.4 Gk + 1.6 (0.6) Qk
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5.4.3 Design of Longitudinal R/F
N
= 26785 kN
Critical section for the design lies at the top.
Mx
= 68 kNm
My
= 18 kNm
Minimum moment (MMin) = N emin
= N x (0.05 h or 20mm or lesser)
= 26785 x (0.05 x 1000 or 20)
= 536 kNm
Mxx and Myy
< MMin
Uniaxial bending about Minor Axis
Design bending moment about y-y,
My = 536 kNm
h
= b
= 1000 mm
d
= 1000 - 45 - 10 - (32/2)
=
929
mm
M
= 536 x 10
6
b h2
1000 x 1000 x 1000
= 0.54
N
= 26785 x 10
3
bh
1000 x 1000
= 26.79
d/h
= 929 / 1000
= 0.94
fcu
= 50 N/mm
2
Chart No 39
100 Asc / bh
= 1.3
Asc
= 1.3 x 1000 x
1000
100
= 13000 mm
2
Steel Provided 24 No. Y 32 bars.
Asc provided = 24 x π x (32/2)
2
= 19301 mm
2 > Required Asc
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5.4.4 Design for seismic effects
Asc provided
= 24 x π x (32/2)
2 x 100
1000 x 1000
= 1.93%
1% < Asc provided < 4%
Longitudinal R/F O.K.
Asc at lap
= 48 x π x (32/2)
2 x 100
1000 x 1000
=
3.86
%
< 6%
Lap R/F is O.K.
Check for shear:
M/N
= 35 / 21765 = 0.002 m
0.75h
= 0.75 m >M/N
From Sap2000
V
= 15 kN
v
= 15 x 1000
1000 x 929
= 0.02 N/mm
2
0.8 = 5.0 N/mm
2 > v
Hence shear r/f is not required.
5.4.5 Check for Deflection:
For unbraced columns:
le / h = 5.58 / 1
= 5.58 < 30
Hence no check is required for deflection.
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5.5 Design of a Slab
Assume Following parameters
Slab thickness = 175 mm
Slab self-weight = 4.20 kN/m2
Finishes = 1.00 kN/m2
Partitions = 1.00 kN/m2
Services = 1.00 kN/m2
Total DL = 7.20 kN/m2
Slab Live load = 2.50 kN/m2
Design Load N =1.4 Gk +1.6 Qk =14.08 kN/m2
Lx = 6m
6
Ly = 8m
Ly / Lx = 1.3 < 2
Two way spanning slab
Select Corner panel Two adjacent edges discontinues
Moment Coefficients.
Short span, continues edge
0.069
Short span, at mid-span
0.051
Long span ,at continues edge
0.045
Long span, at mid-span
0.034
Maximum design moment per unit width
Msx = B sx n lx
2
Msy = Bsy n lx
2
Short span, continues edge = 34.97 kNm/m
Short span, at mid-span = 25.85 kNm/m
Long span ,at continues edge = 22.81 kNm/m
Long span, at mid-span = 17.23 kNm/m
5.5.1 Design of reinforcement
Let use 10 mm RF with 25 mm cover , fcu =30 N/mm2, d=145mm
Short way, mid span RF
K= M/bd
2 fcu = 0.041
Z = d {0.5 + SQRT (0.25- K/0.9)} = 0.96 d > 0.95d
Z = 137.75 mm
As(req) = M /0.87 fy Z = 469 mm
2
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Use T 10 @ 150
As provided = 520> As Req
Maximum spacing between bars < 3d
3d = 435 > 150 mm
Minimum R/F
100 As/ Ac = 0.13
As min=227 < As Provided
5.5.2 Deflection check
Maximum moment at mid span
Mmax = 25.9 kNm/m
M/bd2 = 1.23
Service stress at steel Fs = 343 N/mm2
Modification factor for tension reinforcement
T 3.11
Ft = 1.81
Basic (span / d ) = 26
Allowable (span / d) = 47
Actual (span / d) = 41 < 47
5.5.3 Short way, support (top R/F)
K= M/bd2 fcu = 0.055
Z = d {0.5 + SQRT (0.25- K/0.9)} =0.953 d > 0.95d
Z = 138 mm
As = M /0.87 fy Z = 634 mm
2
Use T 12 @ 175
(As) provided =645 > 634 ,(As, required)
5.5.54 Long way, mid span (bottom R/F)
Effective depth = 135 mm
K= M/bd
2 fcu = 0.027
Z = d {0.5 + SQRT (0.25- K/0.9)} = 0.966 d > 0.95d
Z = 128 mm
As = M /0.87 fy Z = 336 mm
2
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Use T 10 @ 200
(As) provided = 393 > 336, As, req
5.5.6 Long way, support (top R/F)
K= M/bd
2 fcu = 0.042
Z = d {0.5 + SQRT (0.25- K/0.9)} = 0.94 d
Z = 127 mm
As = M /0.87 fy Z = 448 mm
2
Use T 10 @ 150
(As) provided =523 > 442, As, req
100 As /Ac = 0.13
Minimum As = 228 mm2/m
For T 10 @ 250
(As) provided = 314 mm2/m
5.5.7 Check for shear
Shear force induced at support V = Bv n lx
Short way support V =
0.4 x 13.03
x 6 = 31.27 kN/m
v = V/bd = 0.216
100 As/bd =0.271
Vc =0.52 N/mm
2 ( T 3.9 )
Vc >V T 3.17
No shear r/ f required
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5.5 Design Shear wall
Wall is classified as braced column
Top condition
= 2
Bottom condition
= 1
Value of β for braced wall = 0.75
ℓo
= 3.25
ℓe
= 0.95 x 3.25
= 3.1 m
Wall thickness at graound floor is 300mm.
ℓe/h
= 3.1 / 0.3
= 10.3 < 15
Therefore the wall is short.
For all load combinations wall is subjected to compression only.
Calculation of Compression Capacity
Maximum axial load on wall = 3384kN/m x 16
= 54144 kN
fcu
= 40
fy
= 460 N/mm2
Assume 1% of reinforcement in the wall.
Total design axial load on wall,
nw
0.35 fcu Ac + 0.67 Asc fy
(0.35x40x300x16000) +
(0.67x0.01x300x16000x460)
81993kN
Hence compression capacity of wall with 1% of reinforcement is
satisfactory.
Design of vertical reinforcement
Required reinforement
= 1% of Ac
= 1x300x16000 / 100
= 48000mm
2
= 2000 mm
2/m/per layer
Provide T16 bars at 100 mm spacing. (Aspro = 2010 mm
2/m)
Design of horizontal reinforcement
Required reinforcement
= 0.25% of Ac
= 0.25x300x1000 / 100
= 750 mm
2/m
= 375 mm
2/m/per layer
Provide T12 bars at 200 mm spacing. (Aspro = 786 mm
2/m)
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6.0. References
1. B.S.8110 Part1:1985, The Structural Use of Concrete (Code of Practice for Design and Construction)
2. B.S.6399 Part1:1996, Loadings on buildings( Code of Practice for dead and Imposed
loads)
3. AS1170.2-1989, “Minimum Design Loads on Structures – part 2: Wind Loads”, Standards Australia, New South Wales.
4. AS 1170.4-1993,” Structural design actions Part 4: Earthquake actions “, Standards
Australia, New South Wales.
5. Jayasinghe, M.T.R, “Wind loads for tall buildings in Sri lanka”, Seminar on structural design for wind loading, Society of Structural Engineers, Sri lanka,2008.
6. Wijeratne, M.D., Jayasinghe, M.T.R., “Wind loads for high-rise buildings constructed in Sri Lanka”, Transactions Part 2- Institution of Engineers, Sri Lanka, 1998,