TALL
BUILDING
DRAFT DESIGN
Joan Tresserras Pujol-Xicoy
1. Section dimensions
 Columns:
•
1 – 5 floors:
𝐺1 = 1.2𝐷 + 1.4𝐿 = 1.2 ∗ 5.5 + 1.4 ∗ 3.5 = 11.5 𝑘𝑁/𝑚2
𝐺1−4 = 11.5 ∗ 4 ∗ 8,4 ∗ 8,4 = 3245.8𝑘𝑁
𝐺1−35 = 11.5 ∗ 35 ∗ 8,4 ∗ 8,4 = 28400.4 𝑘𝑁
𝐺1−4 ∗ 1.3 3245.8 ∗ 103 ∗ 1.3
𝑏𝑐 ℎ𝑐 =
=
= 152992.7 𝑚𝑚2
𝑓𝑐
27.58
𝐺1−35 ∗ 1.3 28400.4 ∗ 103 ∗ 1.3
𝑏𝑐 ℎ𝑐 =
=
= 1338670.1 𝑚𝑚2
𝑓𝑐
27.58
𝑏
𝑐
1−4
=ℎ
= 391.14 𝑚𝑚 = 0.4𝑚
𝑐
1−35
=ℎ
= 1157.01𝑚𝑚 = 1.2𝑚
𝑏
𝑐
1−4
𝑐
1−35
1. Section dimensions
•
5 – 10 floors:
𝑏
•
𝑐
5−35
11 – 15 floors:
𝑏
𝑐
11−35
•
16 – 20 floors:
𝑏
𝑐
16−35
•
•
= 884.07𝑚𝑚 = 0.9𝑚
=ℎ
𝑐
16−35
= 790.74𝑚𝑚 = 0.8𝑚
=ℎ
26 – 30 floors:
𝑐
26−35
•
31 – 35 floors:
𝑏
𝑐
31−35
= 968.45𝑚𝑚 = 1𝑚
𝑐
11−35
𝑐
21−35
𝑏
𝑐
5−35
=ℎ
21 – 25 floors:
𝑏
=ℎ
𝑐
21−35
=ℎ
=ℎ
𝑐
26−35
𝑐
31−35
= 684.8𝑚𝑚 = 0.7𝑚
= 559.13𝑚𝑚 = 0.6𝑚
= 395.37𝑚𝑚 = 0.4𝑚
1. Section dimensions
 Beams:
1
8.4
ℎ=
𝐿=
= 0.56 𝑚 ≅ 0.6 𝑚
15
15
1
0.6
𝑏=
ℎ=
= 0.24𝑚 ≈ 0.3 𝑚
2.5
2.5
 Slabs:
𝐿𝑠 8.4
𝑡=
=
= 0.1867 𝑚 ≅ 0.2 𝑚
45 45
2. Design requirements
2. Design requirements
3. Loading
 Wind
•
loading:
According to the chinese code:
•
Return period of 50 years  Wp = 0,55KN/m2.
•
Return period of 100 years  Wp = 0,60KN/m2
•
Building situated in Tianjin, China  Roughness type C
•
When using the program we choose the Chinese 2002 code
3. Loading
 Earthquake
loading:
•
Design with a basic fortification intensity of 7.
•
In consequence a ground acceleration of 0,15g is used.
•
Site classification is a site-class IV.
•
When using the program we choose the Chinese 2002 code.
3. Loading
 Dead
•
load:
Automatically calculated the program (Self Weight
Multiplier = 1).
•
It is necessary to add dead load to the slabs inside the
shear wall areas, due to their very small thickness.
These slabs help to transfer lateral stresses.
•
Add a dead load = 7,5kN/ 𝑚2
3. Loading
 Live
•
load:
Following the chinese code:
4. Materials
•
Basically, the material used will be concrete. Default
concrete given by ETABS is a C40 concrete. For
structural reasons a new concrete C60 will be created.
•
Steel only used in the reinforcement bars.
5. Model elaboration
•
A 3-D grid with all the elevation and plan dimensions
inputed has to be created.
5. Model elaboration
•
It is convenient to create a new
set of columns every 5 stories
to
make
it
easy
when
dimensioning them.
•
In the first 4 floors the columns
that don’t have floors on top
will also be different than the
ones with 35 floors on top
5. Model elaboration
•
When drawing shear walls it’s
possible to distinguish into
two kinds of shear walls:
•
The
ones
surrounding
the
elevators and stairs.
•
In the residential floors, the
ones
dividing the different
apartments.
5. Model elaboration
•
There are two kinds of beams:
•
Primary beams that go from column to column. The moment
release for this kind of beams will be continuous.
•
Secundary beams that go from beam to beam. The chosen
distribution for this type of beams is a cross distribution. The
moment release for this kind of beams will be pinned.
•
No beams should be drawn inside the shear walls.
5. Model elaboration
•
Two kinds of slabs: 1.Normal slabs ; 2.Inside shear walls slabs
6. Model analysis
•
•
Three requirements should be met by the model:
•
First rotation period / x or y period < 0,9
•
Story drift < 1/800
•
Maximum story drift / average story drift < 1,4
To meet this three requirements a process of analysis
and later dimensioning of the structural members will
take place.
•
It will be necessary to look at the different modes of
structure to determine which one is the first to have
rotation.
6. Model analysis
1st mode
2nd mode
3rd mode
6. Model analysis
Deformed shape
3-D View Longitudinal Reinforcing
6. Model analysis
7. Requirements
 First
rotation period / x or y period < 0,9
The first rotation period is the one of the third mode = 1,85398
The first period with x or y movement is the first mode = 2,50081
1,85398 / 2,50081 = 0,741352 < 0,9
 Story
drift < 1/800
The maximum story drift is for the 35th floor:
1/811 < 1/800
 Maximum story
drift / average story drift < 1,4
The maximum story drift is for the 1st floor= 1/811
Average story drift = 9,5992*10^(-4)
(1/811) / ( 9,5992*10^(-4)) = 1,285 < 1,4
THANK YOU!
谢谢!