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DESIGN OF A MODERN HIGH-RISE BUILDING IN ABU-DHABI United Arab Emirates University Faculty of Engineering Department of Civil and Environmental Engineering Graduation Project II Fall 2010 Thursday 13 January 2010 Student Name ID Number Abdulrahman Abdulla Alili 200409918 Mohammed Amer Al-Ameri 200416269 Wadah Abdulla Ahmed 200540613 Amr Ezzat Abdel-Havez 200540677 Advisor : Dr. Aman Mwafy Examination Committee: Dr. Hany Maximos (Faculty) Dr. Amr Sweedan (Dept.) Dr. Bilal El-Ariss (Dept.)

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Page 1: DESIGN OF A MODERN HIGH-RISE BUILDING IN ABU …libvolume3.xyz/civil/btech/semester6/designdrawingofrcstructures/... · DESIGN OF A MODERN HIGH-RISE BUILDING IN ABU-DHABI United Arab

DESIGN OF A MODERN HIGH-RISE BUILDING

IN ABU-DHABI

United Arab Emirates University

Faculty of Engineering

Department of Civil and Environmental Engineering

Graduation Project II

Fall 2010

Thursday 13 January 2010

Student Name ID Number

Abdulrahman Abdulla Alili 200409918

Mohammed Amer Al-Ameri 200416269

Wadah Abdulla Ahmed 200540613

Amr Ezzat Abdel-Havez

200540677

Advisor : Dr. Aman Mwafy

Examination Committee:

Dr. Hany Maximos (Faculty)

Dr. Amr Sweedan (Dept.)

Dr. Bilal El-Ariss (Dept.)

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Objectives

� The Graduation Project is divided into two main phases, namely

GPI and GPII. The three-dimensional (3D) analytical modeling of

the 60-story building, load calculations and verifications of the

analytical model were performed in GPI. Tasks and results of this

phase are briefly presented in this report.

� The second phase of the project focus on designing different

structural members of the high-rise building such as floor slabs,

beams, columns, shear walls and foundations using latest analysis

software and modern design codes.

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Original Building

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Building After Modifications

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Design process

Requirements Specifications Conseptual Design

Embodiment Design

(Simulation Model)

Detailed Design

Specification:

Building should be design according to:

–ACI 318-05 Code.

–IBC 2009/ASCE 7-05 for calculating the wind loads.

Structural Analysis programs:

–CSI ETABS

–CSI SAFE

–PROKON

Requirement:

–Plan area: 1750.2 m2

–Total number of stories: 60

– Typical Story height: 3.5 m

–Use of the building: residential building

–Used materials: normal and high strength

material and reinforcing steel

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Summary of GPI

� Geometric and Load Modeling

� Structural elements modeling (beams, slabs etc..)

� Load modeling

� Hand Calculations

� Actions and deformations

� Preliminary Cost Estimate

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Summary of GPI

0

5000

10000

15000

20000

25000

30000

EarthquakeLoadWind Load E-W

Hand Calculation ETABS Results

0

200000

400000

600000

800000

1000000

1200000

Dead Load

Live Load

Hand ETABS Results

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GPII Tasks

� Verified three dimensional analytical model of a typical 60-story building,

representing the modern tall buildings in the UAE.

� Design different structural elements, including the complete design of suitable

floor slab systems, such as solid slabs and flat slabs, at different story levels of

the high-rise building using the SAFE and PROKON programs.

� Optimized design of shear walls using the latest version of the ETABS program

with different load combinations and different cross section sizes and

reinforcement ratios to arrive at the most cost-effective design.

� Design of columns and different types of beams.

� Design of stairs and the piles foundation system.

� Final Cost Estimates.

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Design of slab systems

Flat Slab

Hollow Block Slab Solid Slab

Diaphragms and slabs can be defined as a structural system that resist, collect and

distribute the lateral forces, either earthquake or wind, in the horizontal planes of a structure

then transmit them to the vertical bearing elements (shear walls, frames) then to the

foundation and the ground.

In GPII we have designed three types of slab systems

1- Flat Slab

2- Hollow Block Slab

3- Solid Slab

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Design of Hollow Block slabs

Design as a T-Section

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Hollow Block slabs � Different alternatives of the slab dimensions are considered.

� We started with a depth equals to 250 mm, which led to a deflection that

exceeds the

maximum allowed value.

� To overcome this issue, we increased the depth of the slab to 350 mm by having

two

layers of blocks each has 150 mm height, which produced safe deflection

Increasing the width of ribs

Increasing the depth & number of blocks

Increasing of the reinforcement

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Solid Slabs

Wu=1.2WD+ 1.6WL == 16.98 kN/m2

Mu = 13.262 kN.m/m

Rn=[Mu/ ( = 2.62 � ρ = 0.0065

As= ρbd=487.5 mm2 � use 7 bars #10

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Flat slabs Exporting three slabs:

� The following three slabs are exported from ETABS to SAFE:

� Ground story slab.

� 17th story slab.

� 37th story slab.

� Two critical load combinations are selected to extract results from SAFE,

1.4 SDL+1.4 O.W+ L.L+ 1.4 EQX) a

1.4 SDL+1.4 O.W+ L.L+ 1.4 EQY).

�The design process of flat slabs is started by

determining the most optimum thickness of the slab.

�The optimum thickness can be determined by

selecting the slab thickness and verifying the

deflection

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Slab Optimum thickness �∆L < L/360

�∆L + ∆add < L/360

�∆add = λ∆ * ∆D

�λ∆ = ξ / 1+ 50 ρ

�ρ` = As`/ bd

Ground story slab thickness equals to 200 mm:

L/360 = 22.2 mm

L/240 = 33.33 mm

d = 200 – 25 = 175 mm

ρ ̀= As /̀ bd = 0.003686

λ∆ = ξ / 1+ 50 ρ ̀= 1.6887

where � ξ = 2

∆L < L/360 � 5.22 < 22.22 � O.K

∆D = 13.5 + 13 = 26.5

∆add = λ∆ * ∆D = 44.71

∆L+ ∆add < L/360 � 49.93 > 33.33 �

unsafe, we have to increase the thickness.

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Design of flat slabs

Strip # Direction Strip width (m) Strip type

S1 x X direction 1 Column strip

S2 y Y direction 4 Middle strip

S3 y Y direction 4 Middle strip

�Three strips are defined for each slab, two middle strips and one column

strip, as shown below.

�The defined strips are used to calculate the maximum bending moment, as

shown in Figures

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Design of flat slabs � Example :Ground story

� After extracting the maximum positive and negative moment, each is divided by the strip width to get the bending moment per unit length and calculate the corresponding reinforcement

Strip No. +M -M +M/width -M/width

S1x 123.3 240.9 123.3 240.9

S2y 229.13 600.3 57.28 150.1

S3y 298.78 196.9 74.695 49.2

Strip No. +M/width ρ As = ρbd Mesh top & bot.

S1x 123.3 0.00457 1188.2 6Ф16

Strip No. -M/width As = ρbd Amesh As - Amesh Additional Reinf.

S1x 240.9 2410 1188.2 1221.8 6Ф16

S2y 150.1 1457.64 1188.2 269.44 4Ф12

S3y 49.2 463.46 1188.2 ---- ----

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Strip No. -M/width As = ρbd Amesh

As -

Amesh Additional Reinf.

S1x 240.9 2410 1188.2 1221.8 6Ф16

S2y 150.1 1457.64 1188.2 269.44 4Ф12

S3y 49.2 463.46 1188.2 ---- ----

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Check of Punching Shear

12

')2(

fc

b

d

o

s

+α bod

3

'fc bod

6

')

21(

fc

cβ+ bod Vc1 =

Vc2 =

Vc3 =

Vu = Ф Vc + Ф Vs

Element Vc (KN) f Vc (KN) Vu (KN) fVu (KN) Decision

Col-1 595.9 446 161.88 242.82 Safe

Col-2 1115.94 836.96 275.08 357.6 Safe

Col-3 2147.66 1610.745 151.8 197.34 Safe

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Best Alternative

Flat Slabs

Hollow Block Slabs

Decision Based

on

Efficiency/

Construction

Cost

0.00

10.00

20.00

30.00

40.00

50.00

60.00

Floor slabs cost Hollow block slab

Cost in Millions AED

- The choice of the slab system for the 60-

stor building is based on the cost and

performance

- The hollow block slab system is more

expensive than flat slabs. Moreover, flat

slabs are much easier in construction, and

therefore the flat slabs are selected.

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Stairs

Wu = 16.08kN/m2

Wu X 0.3 load of each stair = 16.08 X 0.3 = 4.824 kN/m

(M max)= (Wu x L2)/2= (4.824 x 1.452) /2 = 5.07 kN .m

.

STEP 1: Determine steel ratio (ρ)

Rn = [Mu/ (= (5.07*106)/ (0.9*300*(195)2) = 0.49

ρ = (1- ) = 0.00117

ρmin = 0.003521

STEP 2: Determine As

As= ρ×b×d = 0.003521*300*195= 206 mm2

[ use 3 bars#10]

Stairs

�Design of stairs using hand calculations

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Stairs

Reinforcement details of stairs

3#10 /step

3#10 /step

�Reinforcement distribution

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Beams �Design of hidden beam

�Design of edge beam

�Design of connecting beam B1

Connecting beams

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Beams

(M max) = (Wu x L2)/8 =168.44 (KN.m)

STEP 1: Determine steel ratio (ρ)

Rn= = = 0.975

ρ = (1- ) = 0.002361

ρmin = 0.003521 from Eqn. 13

ρmax= 0.0216 from

Then ρmin ≤ ρ ≤ ρmax � use ρmin

STEP 2: Determine As

d= Beam height – cover = 450 -50= 400 mm

As = ρ × b × d = 0.003521 × 1200 × 400= 1690.1 mm2

From Ref [11] , Table B.4 � use 9 bars # 16

From Ref [11], Table B.5 � One layer

Beam cross section

�Design of hidden beam using hand calculations

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Beams

Bending moment diagram of the hidden beam

Beam deflection

�Design of hidden beam using Prokon

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Beams

Slab Load:

Dead load from slab = 48.03 kN/m

Live load from slab = 8.5 kN/m

Beam own weight

hb= 620 mm

O.W Beam= bw× hb× γ c = ((0.25×0.24) + (0.81 × 0.38)) × (25) = 9.195 kN/m)

Wall own weight

O.W wall = bw× hw× γ = 0.25 × 2.88 × 10 = 7.2 (KN/m)

Effective Length

bw + 6t= 0.25+ (6×0.38)=2.53

Effective width (bE)= Smaller of bw + L / 12=0.25 +(6.7/12)=0.81m

bw + b0= 0.25+ 4= 4.25 m

Then, Effective length (bE)= 0.81 m = 810 mm

Edge beam cross section

�Design of edge beam using hand calculations

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Beams

Long-term deflection

Moment x-x

�Design of edge beam using Prokon

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Reinforcement distribution

Beams �Reinforcement distribution

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Beams

� Check if it’s coupling or conventional beams:

�Design of connecting beams

Length divided by

height

Greater than 2

From ACI code

Conventional beams

� [3/1]= 3 greater than 2 �.ok

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Beams

Input data in prokon

�Design of connecting beams using ETAB’s & Prokon

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Beams

Level Beams No. Flange Width Bending Moment Shear Force

ground

story levels

B1_1 750mm M=1270.42 KN.m V=865.2 (KN)

Reinforcement 8 T 25 4 T 8@ 120 mm

B1_2 625mm M=1660.9 KN.m V= 1047.6 (KN)

Reinforcement 10 T 25 4 T 10 @ 150 mm

B1_3 625mm M=1537.34 KN.m V= 1072.71 (KN)

Reinforcement 9 T 25 4 T 10@ 120 mm

17th story

levels

B2_1 750mm M=763.82 KN.m V= 620.4 (KN)

Reinforcement 5 T 25 4 T 8@ 150 mm

B2_2 625mm M=895.1 KN.m V= 716.71 (KN)

Reinforcement 6 T 25 4 T 8@ 120 mm

B2_3 625mm M=767.92 KN.m V= 619.9 (KN)

Reinforcement 5 T 25 4 T 8@ 150 mm

36th story

levels

B3_1 750mm M=760.1 KN.m V= 630.37 (KN)

Reinforcement 5 T 25 4 T 8@ 150 mm

B3_2 625mm M=820.9 KN.m V= 637.1 (KN)

Reinforcement 5 T 25 4 T 8@ 150 mm

B3_3 625mm M=679.5 KN.m V= 564.4 (KN)

Reinforcement 4 T 25 4 T 8@ 200 mm

Data outcome form Prokon & ETAB’S

�Design of connecting beams using ETAB’s & Prokon

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Shear Walls

�Design shear walls using ETAB’s

Selected shear walls

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Shear Walls

o Define pier section.

o Pier section data.

o Section designer.

o Assign pier section.

o Assign general Reinforcing pier section.

o Start design of section.

�Design process

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Shear Walls

� The D/C ratio indicates the demand over capacity:

◦ D/C Ratio less than 1 � section is safe in flexure

◦ D/C Ratio greater than 1� Section is unsafe

�General reinforcing Pier Section

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Shear Walls

� As(min) = (0.25/100) × Ag

�Optimization

Level Wall Reinforcement

Layout (1,2 and3)

P3S mm167@12#6

P2SS mm167@12#6

P1SS mm167@12#6

P4S mm167@12#6

P5 mm167@12#6

P6 mm200@12#5

P7 mm167@16#6

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Columns

Columns name Col(1-20) Col(2-20) Col(2-40)

No. of columns 8 4 4

Columns dimension (1000x300) (1500x400) (1500x300)

�Design columns using ETAB’s

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Columns

�Columns reinforcement detail

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Columns

� If columns un-safe:

�Check safety of columns

Increase columns dimension

Increase concrete strength

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Design of Foundation

among the raft area.

�Using Safe program 709 piles have been distributed

among the raft area.

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Foundation Results

�Point loads representation due to applied loads.

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Foundation Results

�Deformed shape.

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Foundation Results

Strip # 1 in X-direction Strip # 2 in Y-direction Strip # 3 in Y-direction

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Foundation Reinforcement

�Reinforcement detailing for the raft foundation

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�Beams Beams (cost) = width × length × depth × number of stories × cost of one m3

�Columns Columns (cost) = width × length × height × number of stories × cost of one m3

Cost Estimate

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Cost Estimate

�Shear Walls Shear walls (cost) = thickness × length × height × number of stories × cost of one m3

�Floor slab Flat slab (cost) =thickness × net area × cost of one m3× number of stories

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�Stairs

Stairs (cost) = Area Stairs × thickness × Factor (1.2)× number of stories × cost of one

m3

= [(3 m ×7 m)×0.23 m×1.2× 60 stories ×2500 (Dhs/ m3)]×2= 1,738,800 AED

Cost Estimate

�Excavation Excavations (cost) = Depth × Area Raft × cost of one m3

= 11 m × 1750.2 m2 × 50 (Dhs/m3) = 962,610 AED

�Plain Concrete Concrete Plan(cost)= Area Raft × Thickness × Cost of one m3

= (1750.2) m2 × 0.4 m × 700 (Dhs/ m3)= 490,056 AED

�Raft Foundation Raft foundation(cost)= Area Raft × Thickness × Cost of one m3

= (1750.2) m2 × 3.2 m× 2200 (Dhs/m3) = 12,321,408AED

�Pile Foundation Piles foundation (cost)= Number of piles × length of pile × Cost of one meter

= 709 piles × 25 m × 2800 (Dhs/m) = 49,630,000 AED

Foundations

170,969,819

AED

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Cost Estimate

0

10

20

30

40

50

60

70

Cost

in M

illi

on (

AE

D)

�Final cost estimate of structural system

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Project Management

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Outcomes and Deliverable

�Verified three dimensional analytical model for a typical

60-story building, representing the modern tall buildings in

Dubai and Abu Dhabi.

�Design different structural elements and establish full

design of suitable floor slab systems, such as solid slabs and

flat slabs, at different story levels of the high-rise building

using SAFE and PROKON programs.

�Design of columns and different types of beams such as

conventional.

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Conclusion

�The analytical model and modern design provisions

have been employed in the second phase of the project

(GPII) to fully design different structural members of the

60-story high-rise building.

�Work in a group and write technical reports.