Highrise Building design report

PG/DIP Structural Engineering Page-1

Contents

1. Introduction

2. Preliminary selection of structural elements

3. Wind Analysis

4. Earthquake Analysis

5. Element Design

6. Reference


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


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


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


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)


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


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


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


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


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


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


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



Table 2 Shear wall Thickness

Fig 4

Shear wall

arrangement


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


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


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


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


Ac = 548,372 mm2

Hence, minimum column dimension required is 740 mm x 740 mm.

Therefore 750mm x 750mm columns are used


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



N = 0.4Ac x 40 + 0.8 x 0.04 x Ac x 460

22477 x 1000= 30.72Ac

Ac = 731673 mm2



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



N = 0.4Ac x 50 + 0.8 x 0.05x Ac x 460

32055 x 1000= 38.4Ac

Ac = 834,765 mm2




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



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,


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


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


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,


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


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


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


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


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


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


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


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.


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.


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


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


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


Nos = 14

Spacing 131 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


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


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


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)


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


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


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


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.


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


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


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


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)


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,

Documents

Highrise Building design report