ĐỒ ÁN TỐT NGHIỆP NGÀNH XÂY DỰNG ( TIẾNG ANH ) part 2

GRADUATION THESIS

PAGE 1

CHAPTER 1

INSTRUCTOR

DESIGN STAIRS FOR FLOORS
2 - 15
1.1. PLAN VIEW AND
SECTION VIEW OF
STAIR

1.1.1.1. PLAN VIEW AND SECTION VIEW
We have plan view and section view of stair floor 2-3 are shown in Table 3.1

Table 3.1 Plan view, section view of stair floor 2 - 3

1.1.2. Preliminary size
The stairs type from floor 3 to floor 15 of this building is 3-step stair case form:
Stair case 1 has 7 steps

- Stair case 3 has 8 steps

- Stair case 2 has 7 steps


GRADUATION THESIS

PAGE 2

INSTRUCTOR

Totally we have 22 steps
Size of steps:

h

3400
 150 mm
22
; l = 300 mm

Preliminary steps thickness by this fomulation:
Lo
4700
hb 

 (156 �120)  mm  .
30 �40 30 �40
With L is the design bar in Table 3.1: L o  1300 �2  2100  4700 (mm)
o

So we choose

h b  130  mm 

HeigLL of beam:

hd 

Lo
4700

 470 �365 mm
10 �13 10 �13

Choose hd = 400 mm

hd
400

 200 �100 mm
2 �4 2 �4

Choose bd = 200 mm

Width of beam:
bd 

1.2. TYPES OF
LOADS :
1.2.1. Load by layers of staircase break:
We have load by layers of staircase break is shown in Table 3.2
Granite, thickness-20mm

Đáhoa cương, dà
y 20 mm
Lớ
p vữ

a ló
t, dà
y 20mm
Reinforcement concrete, thickness-130mm
Bả
n sà
n bêtô
ng, dà
y 130 mm
Plaster, -thickness-30mm
Lớ
p vữ
a xi mă
ng, dà
y 15mm

Mortar lining , thickness-30mm

Table 3.2 Staircase break’s layers


GRADUATION THESIS

PAGE 3

INSTRUCTOR

1.2.1.1. Dead load by layers of staircase break
n

g s1  � i h i n i

We definite dead load by this fomulation :
With :

1

 i : is the weigLL of i layer ; n i : is the load factor i layer;
th
th
h i : is the thickness of i layer
th
We have dead load by layers is shown in Table 3.4:
Thicknes



s

(kN/m3

(m)

)

Granite

0.02

24

1.2

0.58

Mortar lining

0.03

18

1.3

0.70

Reinforcement

0.13

25

1.1

3.58

0.025

18

1.3

0.59

Load type

Layers

Dead load

Design
n

load
(kN/m2)

concrete
Plaster
Total

5.72

Table 3.4. Dead load by layers of staircase break
1.2.1.2. Live load
As “TCVN 2737-2012”: ptc = 3kPa, n = 1.2
Live load:

ps  n �p tc  1.2 �3  3.6  kN / m 2 

1.2.1.3. Load on the slant slab
Dead load is calculate by this famulation:

n

g  � i h tdi n i
'
s2

1

With :

 kN / m 
2

 i : is the weigLL of i layer ; n i : is the load factor i layer ;
th
th
h tdi : is the thickness of i layer as slant side
th


GRADUATION THESIS

PAGE 4

INSTRUCTOR

Calculation the equivalent thickness of layers:

cos α 
We have :

lb



l2b  h 2b

0.3
0.32  0.152

 0.894

Granite and Mortar lining ( use the same method for Granite )
 l  h b  i cos   (300  150) �25 �0.894  33.52(mm).
h td  b
lb
300
Calculation the equivalent thickness of steps:
h td 

h b cos  150 �0.894

 67.05 (mm).
2
2

We have dead load by layers on slant slab is shown in Table 3.5:

Load
Type

Layers

Thicknes

Thickness

s

equivalent

(m)

(mm)

Design


(kN/m3)

n

load
(kN/m2)

Dead

Granite

0.02

26.82

24

1.2

0.77

load

Mortar lining

0.025

33.52

18

1.3

0.96

Steps

0.15

67.05

18

1.2

1.45

Reinforcement

0.13

0.13

25

1.1

3.58

0.025

0.025

18

1.3

0.58

concre
Plaster

Total
Table 3.5 Dead load on slant slab

Vertical live load on slab:

g s2 

'
g s2
7.61

 8.51  kN / m 2 
cos α 0.894

Note: With weigLL of handrails is 0.30 kN/m.
1.2.1.4. Total load
Staircase break:

7.61


GRADUATION THESIS

PAGE 5

INSTRUCTOR

q s1  gs1  ps  5.72  3.6  9.32 (kN/m2)
Stair case 1, 2, 3:
q s2  g s2  ps 

Stair case 3 :

g lc
0.3
 8.51  3.6 
 12.34
A1
1.3
(kN/m2)

q s2 cos   12.34 �0.894  11.03

 kN / m 
2

1.3. USING “SAP2000”
FOR CALCULATION
STAIR CASE 1 , 2
We cut a strip wide 1m , then calculate for the stairs which is shown in Table
3.6, Table 3.7 and Table 3.8:

Table 3.6. Distributed load on stairs

Table 3.7. Moment chart for stairs


GRADUATION THESIS

PAGE 6

INSTRUCTOR

Table 3.8. Joint reaction forces chart for stairs

 kN.m / m 
M sup port  0.4M max  0.4 �6.93  2.78  kN.m / m 
M span  0.7M max  0.7 �6.63  4.64

Correction load :

1.4. REINFORCEMEN
T CALCULATION
FOR STAIRS
Assume that the distance from the reinforced concrete edge to the center of
tensile reinforcement group is: a  25 mm
Steel ratio :  min  0.05%
 max 

R  b R b 0.604 �0.9 �14.5

�100  2.16%
Rs
365
for reinforcement group AIII

 max 

R  b R b 0.651 �0.9 �14.5


�100  3.78%
Rs
225
for reinforcement group AI

1.4.1. Reinforcement calculation for supports in slant slab (Stair case 1, 2)
Working heigLL of section : h o  h  a  130  25  105 (mm)
Definite coefficient  m :

m 

M
2.78

 0.02
2
 b R b bh o 0.9 �14.5 �103 ��
1 0.1052

�   1  1  2 m  1  1  2 �0.02  0.02


GRADUATION THESIS

PAGE 7

INSTRUCTOR

Reinforcement :
 b R b bh o 0.02 �0.9 �14.5 �1000 �105


 122 (mm 2 / m)
Rs
225

Astt 

c
2
Choose: 8a200 , A s  250 (mm / m).

Checking steel ratio :

 min

Asch
250


�100  0.22%
bh o 1000 �115
 0.05%    0.22%   max  3.78%
Eligible.

1.4.2. Reinforcement calculation for span in slant slab (Stair case 1, 2)
Working heigLL of section : h o  h  a  130  25  105 (mm)
M
4.64
m 


 0.027
2
3
2


R
bh
0.9
�
14.5
�
10
��
1
0.105
m
b
b
o
Definite coefficient
:
�   1  1  2 m  1  1  2 �0.027  0.027

Reinforcement :
Astt 

 b R b bh o 0.027 �0.9 �14.5 �1000 �115

 180(mm 2 / m)

Rs
225

c
2
Choose 8a200 , A s  250 (mm / m).

Checking ratio :

Asch
250

�100  0.22%
bh o 1000 �115
 0.05%    0.22%   max  3.78% Eligible.


 min

Horizontal reinforcement we choose 6a250 for supports, 6a200 for span.
1.4.3. Calculation reinforcement for stair case 3:
�d
� � 2.1
�
;A 2 � �
 2.35m ; 1.3m �
�
�= �0.894
�
We have stair case 3 is a slab with size : �cos 

GRADUATION THESIS

PAGE 8

INSTRUCTOR

d
�
 2A 2  2.6 m �
cos 
�
��
h d 400

 3.1  3 �
h s 130
�
�

We have:
Two-way slab; link between segment and
staircase break’s beam is fixed supports, 2 edges which ink with staircase break is
q cos   12.34 �0.894  11.03  kN / m 2 
joint , others is free joint , have load s2
Follow “sơ đồ 2 – Phụ lục 13 ( Sách BTCT tập 3 – Võ Bá Tầm)”
d
L2 
 2.35  m 

cos

We have
; L = A = 1.3 (m)
1

2

x
�
�
y
L1 1.30
�

 0.553 � �
L 2 2.35
x
�
�
 ry
�

 0.0021
 0.0149
 0.2785
 0.0366

And :
d

A 2  q s2dA 2  12.34 �2.1 �1.3  33.68  kN.m / m 
cos 
�
M x  q x  33.68 �0.0021  0.07  kN.m / m 
�
M y  q y  33.68 �0.0149  0.501  kN.m / m 
�
��
M x  q  x  33.68 �0.2785  9.379  kN.m / m 
�
� r
M x  q r x  33.68 �0.0366  1.232  kN.m / m 
�
Reinforcement calculation:
b �h s 
We calculate reinforcement as bending component has section 
with b = 1m,
q  q s2 cos 

hs = 0.13 (m).
Choose a = 25 (mm) � h 0  0.13  0.025  0.105(m)
Definite :

m 
Checking:

M
 b R b bh 02

GRADUATION THESIS

PAGE 9

INSTRUCTOR

 m � R �   1  1  2 m
Diện tích cốt thép yêu cầu trong phạm vi bề rộng b = 1m:

 b R b bh 0
Rs
We have reinforement calculation is shown in Table 3.9:
As
b
Positio
M
h0


(m
(mm2/m
m
n
(kN.m/m)
(m)
)
)
Span L1
0.07
1

0.10 0.0005 0.0005
3
5
Span L2
0.501
1
0.10
0.003
0.003
18
5
Fixed
9.379
1
0.10
0.07
0.073
444
support
5
Consol
1.232
1
0.10
0.009
0.009
55
e
5
Table 3.9: Reinforcement calculation for staircase 3

As 

Achs
6a200
6a200
10a170
6a200

1.5. REINFORCEMEN
T CALCULATION
FOR STAIRCASE
BREAK
1.5.1. Load on beam of staircase break ( using “sap2000”)
 WeigLL of beam it self:
Horizontal segment :
g d1  bd (h d  h s )n btct  0.2 �(0.4  0.13) �1.1 �25  1.485 (kN / m)
Slant segment :
g d2 

bd (h d  h s )n  btct 1.485

 1.66 (kN / m)
cos 
0.894

 WeigLL of wall :
Horizontal segment ( staircase break 1):
g t1  b t h t1n  t  0.2 �(3.4  7 �0.15) �1.1 �16.5  8.53(kN / m)
Horizontal segment ( staircase break 2)
g t3  b t h t2 n  t  0.2 �(3.4  (7 �0.15  8 �0.15)  0.4) �1.1 �16.5  2.72 (kN / m)

Slant segment :


GRADUATION THESIS

PAGE 10

INSTRUCTOR

�h  h t2 �
�2.35  0.75 �
g t2  nb t � t1
 t  1.1�0.2 ��
�16.5  5.62 (kN / m)
�
�
2
�
�
� 2
�
With : LL1, LL2 – wall’s heigLLs of staircase breaks 1,2
 Stairs transmiDEADe to:
Staircase break 1, 2 has RC = RD = 12.30 (kN/m)
Load of slant slab by staircase 3 : RG = qs2A2 = 12.34 �1.3 = 16.042 (kN/m)
 Total load on the beam is:
Staircase break 1:

g d1  gd1  g t1  R C  1.485  8.53  12.30  22.32  kN / m 

Staircase break 2:

g d3  g d1  g t3  R D  1.485  2.72  12.30  16.505  kN / m 

Slant slab :

g d2  g d2  g t 2  R G  1.66  5.62  16.042  23.32  kN / m 

We have calculation chart of load on staircase breaks are shown in Table 3.10,
Table 3.11 and Table 3.12:

Table 3.10 - Load chart on staircase break


GRADUATION THESIS

PAGE 11

INSTRUCTOR

Table 3.11 – Moment chart (kN.m/m)

Table 3.12 – Axis force chart (kN)

1.5.2. Reinforcement calculation for staircase break
Vertical reinforcement calculation:
Choose a = 50 (mm) � h 0  h d  a  0.4  0.05  0.35 (m)
We have :

m 

M
 b R b bh 02



51.36
0.9 �14.5 �103 �0.2 �0.352

 0.171

And:
  1  1  2 m  1  1  2 �0.171  0.188

Diện tích cốt dọc:
As 

� Choose



 b R b bh 0 0.19 �0.9 �14.5 �200 �350

 620 (mm 2 )
Rs
280

222 Asch  628 mm 2

.

GRADUATION THESIS

PAGE 12

INSTRUCTOR

Checking reinforcement ratio :
 min  0.05% � 

As
628

 0.0089% � max  3.27%
bh 0 200 �350

Stirrup calculation:
By Table 3.12 we have axis force Qmax = 18.65 kN
Q b   b3 (1   n )R bt bh 0
 0.6 �1.05 �200 �350 �10 3  44.1(kN)  Q max  18.65( kN)
� The concrete not effect by axis force, so we only need arrange constructive steel
bars.
We choose 6a200 .


GRADUATION THESIS

PAGE 13

INSTRUCTOR

CHAPTER 4 DESIGN ROOF WATER TANK
2.1. WATER USE
REQUIREMENTS
2.1.1. Determination of using water
We preliminarily caculate the water demand: the building has 15 floors with
most of rooms is office .The working area for office is 6m 2/ person (most of Việt
Nam’s enterprise use), plan floor area is

S  30.5 �22.3  680.15m 2 with

calculation of people :
N

15 �680.15
 1700
6
people

Standard water use average according to “TCVN 4513-1988”:
qtb =10 m3/person/day.nigLL
Harmonize coefficient according to “TCVN 33:1985” : kday = (1.35 �1.5) , choose
kday = 1.5
We calculate the using water of the building :
q �N �k d 10 �1700 �1.5
Qsh  tb

 25.5
1000

1000
(m3/day.nigLL)
2.1.2. Determination for size of water roof tank
We assume the water roof tank be used and pumped in one day. So the water
volume for one day is Wd.n = 25.5 m3
Harmonize volume :
Wdh  30%Wd.n  0.3 �25.5  7.7  m3 
Fire resistant volume:

Wcc  0.6 �q cc �n cc  0.6 �10 �1  18  m 3 

qcc = 15 (l/s) – according to “ Table 13. TCVN 2622-1995”
ncc = 2 Number of hydrant using in time
Calculation of water tank capacity :
Wb  k  Wdh  Wcc   1.2 � 7.7  18   30.84 �31 m 3 
From the required water volume for the tank, we design the size of the tank (7.5 x 7
x 1)m; Using reinforcement concrete . Tank cap visit places in the corner has size
(600x600) mm.


GRADUATION THESIS

PAGE 14

INSTRUCTOR

We have ratio :
a 7.5
h
1


 1.07  3 ; 
 0.13  2 �
b
7
a 7.5
Low water tank
2.2. DESIGN SIZE
2.2.1. DESIGN STRUCTURE
We have size of the water roof tank is shown in Fig 4.1:

Fig 4.3 Size of water roof tank

2.2.2. Determinate the size:
We have the premilinary of sizes are shown in Table 4.1:
Slab’s thickness

Beam sections

Column

(mm)

(mm)

(mm)

Top slab

100

Main beam:

Bottom slab

130

+ Top beam

200 x 450

Side slab

100

+ Bottom beam

200 x 450
400 x 400

Secondary
beam:

200 x 400

+ Top beam

200 x 400

+ Bottom beam

GRADUATION THESIS

PAGE 15

INSTRUCTOR

Table 4.1 Roof water tank sizes

2.3. TOP SLAB
We have top slab plan is shown in Fig 4.2:

Fig 4.2 Beams arrangement plan of top slab

2.3.1. Dead load
We have dead load on top slab is shown in Table 4.2:
Thicknes
Load

Layers

s
(m)

γ
(kN/m3)

Standard
load

n

`(kN/m2)

Design
load
(kN/m2)

Cemaric

0.01

20

0.20

1.1

0.22

Plaster

0.03

18

0.54

1.3

0.70

0.10

25

2.5

1.1

2.75

Water prooffing

-

-

0.02

1.3

0.026

Mortar lining

0.02

18

0.36

1.3

0.468

Dead

Reinforcement

load

concrete


GRADUATION THESIS

PAGE 16

Total load

INSTRUCTOR

3.62

4.16

Table 4.2 Dead load on top slab

2.3.2. Live load
tc
2
According to “TCVN 2737-1995” , we have live load: p  0.75 (kN / m )
tt
tc
2
Design live load : p  n �p  1.3 �0.75  0.98 (k N / m )

 Total load:

q  g tt  p tt  4.16  0.98  5.14 (k N / m 2 )
2.3.3. Reinforcement calculation for top slab
2.3.3.1. Calculation of slabs ( 3.75m x 3.5 m)
L 2 3.75

   2 �
L
3.5
Ratio 1
Two-way slab
We have: h d / h s  450 / 100  4.5 �3 � Fixed supports ( slab number 9 )
Moment in mid-span:
Short demension :
M1  m 9n qL1L 2  0.0190 �5.14 �3.5 �3.75  1.29 (kNm/m)
Long demension :
M 2  m9d qL1L 2  0.0166 �5.14 �3.5 �3.75  1.12 (kNm/m)
Moment in support:
Short demension :
M I  K 9n q L1 L 2  0.0441 �5.14 �3.5 �3.75  2.98 (kNm/m)

Long demension :
M II  K 9d qL1L 2  0.0385 �5.14 �3.5 �3.75  2.6 (kNm/m)
 Reinforcement calculation for short demension
In supports
Working heig

LL : h o  h s  a  100  25  75 (mm)

m 

MI
2.98

 0.041   R  0.439
2
 b R b bh o 0.9 �14.5 �103 ��
1 0.0752


GRADUATION THESIS

PAGE 17

INSTRUCTOR

We have
�   1  1  2 m  1  1  2 �0.041  0.042

Reinforcement:

Astt 

 b R b bh 0 0.042 �0.9 �14.5 �1000 �75

 181  mm 2 / m 
Rs
225

Choose 8a200 , A s  250 (mm / m)
Checking steel ratio:
A sc
250


�100  0.33 %
bh o 1000 �75
 min  0.1%    0.33%   max  3.78%
=> Eligible.
c

2

Use the same method for others position we have reinforcement design is shown
in Table 4.3:

Sla
b

Reinforcement
calculation

Moment
αm

ζ

0.01
8
0.01
9
0.04
1
0.03
5

0.01
8
0.01
9
0.04
2
0.03
6

(kN.m/m)
M1
All

M2
MI

MII

1.29
1.12
2.98
2.60

Design reinforcement

Asds

Ø

a

(mm2/m)

(mm)

(mm)

77

6

75

Asc
(mm2/m
)

Ratio
 bt (%)

200

141

0.19

6

200

141

0.21

181

8

200

250

0.33

157

6

180

157

0.21

Table 4.3 Reinforcement calculation for top slab
2.3.3.2. Reinforcement for tank cap visit (600x600):
Reinforcement lost for cuDEADing the hole ( assume that short dimension as long
dimension) :
250 �600
Ascat 
 150 mm 2
1000


GRADUATION THESIS

PAGE 18

INSTRUCTOR

Choose 38 As = 151 mm2 and reinforcement anchor is �30  240 mm from the
edge of tank cap visit; arrange for 4 edges.
2.4. SIDE SLAB
2.4.1. Loads on side slab
2.4.1.1. Dead load:
We have dead load on side slab is shown in Table 4.4

Thicknes
Load

Layers

s
(m)

Dead
load

γ
(kN/m3)

Standard
load
`(kN/m2)

Design
n

load
(kN/m2)

Cemaric

0.01

20

0.2

1.1

0.22

Plaster

0.03

18

0.54

1.3

0.70

Water prooffing

-

-

0.02

1.3

0.026

Reinforcement

0.10

25

2.5

1.1

2.75

Mortar lining

0.02

18

0.36

1.3

0.468

Total load

3.62

4.16

Table 4.4 Dead load on side slab
For easy at calculation, we skip the self load of side slab.
2.4.1.2. Live load:
Water pressure : is liner type, effect the most in foot of side slab
p n   n �h �n  10 ��
1 1.1  11  kN / m 2 
:  n - specific weigLL of water,  n =10kN/m3
h – Side slab heigLL
n – over load coefficent , n = 1.1
Wind load :
With the heigLL is high from 58.5 m to 60.5m , the building is located in Hồ
Chí Minh so we have it is in II.A region, togographic type C.
With


GRADUATION THESIS

PAGE 19

INSTRUCTOR

W  n �W0 �k �c  kN / m 2 
With :
n = 1.2 : coefficent of wind load W0 : Standard wind force
k: coefficent about the changing of wind pressure depend on heigLL
and togographic according to “Table 5 TCVN 2737 – 1995”
c: aerodynamic coefficient . IIA region, togographic type C =>
W0 = 83 (kN/m2)
c = +0.8 for pushed wind and c = -0.6 for absorb wind

h = 58.5 m , togographic type C => k = 1.073
+ absorb wind:
Wh  n �W0 �k �c  1.2 �0.83 �1.073 �0.6  0.64  kN / m 2 
+ pushed wind:
Wd  n �W0 �k �c  1.2 �0.83 �1.073 �0.8  0.85  kN / m 2 
2.4.2. Reinforcement of side slab
2.4.2.1. Calculation by using “SAP 2000”:
a 7
  7  2 
We have : h 1
one-way slab
Skip the weigLL of slab is self , we assume the slab as a bending component is
effected by horizontal load inclue wind and water pressure at the same time.
We have the load on side slab by using “SAP 2000”is shown in Fig 4.3 and Fig 4.4:

Fig 4.3. Load charts by water pressure and absorb wind


GRADUATION THESIS

PAGE 20

INSTRUCTOR

Fig 4.4. Moment and force

We have : Msupport = 0.81 kN.m; Mspan = 0.36 kN.m
Choose a = 25 mm => h0 = h – a = 100 – 25 = 75 mm
We using the same method in “4.3.3 Reinforcement calculation for top slab”

m 

M
 b R b bh 02

Check ratio:
 m � R �   1  1  2 m

Reinforcement in the 1m width:
 R bh
As  b b 0
Rs
We have the reinforcement calculation is shown in Table 4.4:
M
(kN.m/m)
MSP

0.81

MSPAN 0.36

m



0.011

0.011

0.005

0.005

A stt

Ach
s

(mm2/m)
48

(mm2/m)

22

141
141

Choose
steel
6a200
6a200

Ratio
0.188
0.188

Table 4.4 Reinforcement for side slab
2.5. BODEADOM
SLAB

We have beams arrangement plan of boDEADom slab is shown in Fig 4.5:


GRADUATION THESIS

PAGE 21

INSTRUCTOR

Fig 4.5 – Beams arrangement plan of boDEADom slab

2.5.1. Load on boDEADom slab:
We have the dead load on boDEADom slab is shown in Table 4.5:

Load

Dead
load

Thicknes



Standard

s

(kN/m3

load

(m)

)

`(kN/m2)

Cemaric

0.01

20

0.2

1.1

0.22

Plaster

0.03

18

0.54

1.3

0.70

Water prooffing

-

-

0.02

1.3

0.026

Reinforcement

0.13

25

2.

1.1

5.50

Mortar linging

0.02

18

0.36

1.3

0.468

Layers

Total load

6.12

Design
n

load
(kN/m2)

6.91


GRADUATION THESIS

PAGE 22

INSTRUCTOR

Table 4.5 Dead load on boDEADom slab
Water load : pn = 11 (kN/m2)

Slabs S3 and S4 have L2/L1 = 3750/3500 = 1.071 < 2 => Two-way slab
h mb 600

3
hs
150
=> the link between slab and beam is fixed supporter

h sb 500

3
h s 150
=> the link between slab and beam is fixed supporter
We use calculation chart number

9 to definite coefficent  91;  92 ; 91 ; 92

as we calculated in “4.3.3.1 calculation of slab”
Choose a = 20 mm => h0 = h – a = 150 – 20= 130 mm:
We have:

m 

M
 b R b bh 02

We have:
 m � R �   1  1  2 m

Reinforcement calculation in 1m width :

 R bh
As  b b 0
Rs
We have reinforcement calculation of boDEADom slab is shown in Table 4.6:
M
(kN.m/m)
M1

4.48

M2

3.90

MI

10.40

MII

9.06

m



0.021

0.021

0.020

0.020

0.047

0.048

A stt

Choose

A cs

(mm2/m)

(mm2/m)

155

steel
6a180

144

6a180

157

364

10a200

157

392

10a200
316
0.041 0.042
392
Table 4.6. Reinforcement calculation of boDEADom slab

Ratio
0.12
0.13
0.30
0.30


GRADUATION THESIS

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INSTRUCTOR

2.6. TOP SLAB BEAM
We have transmission diagram of top slab beam is shown in Fig 4.6:

Fig 4.6 Transmission diagram of top slab beams

BEAM TB1:

+ The slab dead load transmis to beam is triangle shape with value:
L
3.5
q1  g tt 1  4.16 �  7.28  kN / m 
2
2
+ Live load transmis to beam is triangle shape with value :
L
3.5
q 2  p tt 1  0.98 �  1.715  kN / m 
2
2
+ The water pressure load on side slab transmis to beam is :

q 3  1.35  kN / m 

+ Wind load :
h
1
 0.64 �  0.32  kN / m 
2
2
h
1
q 5  Wd  0.85 �  0.43  kN / m 
2
2
q 4  Wh

Use the same method for all beams we have maximum load on all beams is shown
in Table 4.5:
( Note : wind load and load by pressure of water is the same at all beams)


GRADUATION THESIS

q1

PAGE 24

TB1 + TB3
q2
q3
q4

7.28 1.715 1.35

0.32

INSTRUCTOR

TB2
q5
0.43

q1
14.5

q2

TB4 + TB6
q1
q2

3.43
7.28
6
Table 4.5 Load on top slab beams

1.715

TB5
q1
14.5
6

2.7. BODEADOM
SLAB BEAMS
We have transmission diagram of top slab beam is shown in Fig 4.7:

Fig 4.7 Transmission diagram of boDEADom slab beams
BEAM BB1:

+ The slab load transmis to beam is triangle shape with value:
L
3.50
q 6  g tt 1  6.91 �
 12.1  kN / m 

2
2
+ Pressure load of water:
q7  pn

L1
3.5
 11�  19.25  kN / m 
2
2

+ Dead load by side slab:
q8  g tt h b  4.16 �1  4.16  kN / m 
+ Wind load :

q2
3.43


GRADUATION THESIS

PAGE 25

INSTRUCTOR

h
1
 0.64 �  0.32  kN / m 
2
2

h
1
q10  Wd  0.85 �  0.43  kN / m 
2
2
q 9  Wh

+Pressure load of water on side slab transmis to beam:

q11  4.79  kN / m 

Use the same method for all beams we have maximum load on all beams is shown
in Table 4.6:
( Note : wind load, load by side slab and load by pressure of water is the same at all
beams)
q6
12.

q7
19.2

1

5

BB1 + BB3
q8
q9

BB2

BB4 + BB6
q10
q11
q6
q7
q6
q7
4.7
19.2
4.16 0.32 0.43
24.2 38.5 12.1
9
5
Table 4.6 Load on boDEADom slab beams

BB5
q6
q7
24. 38.
2

5

2.8. “USING SAP 2000”
FOR CALCULATION
2.8.1. Load charts :
We have all the load charts on beams is shown from Fig 4.8 to Fig 4.21:

Fig 4.8. Dead load on main beam axis B ,C