Solid Slabs - Dr. ALaa Bashandy.pdf

‫الرحيم‬ ‫الرحمن‬ ‫ﷲ‬ ‫بسم‬
ALB
‫د‬
.
‫م‬
.
‫بشندي‬ ‫على‬ ‫عالء‬
2014
Dr. Eng. ALaa Ali Bashandy
Assoc. Prof. at Civil Dep., F. of Engineering, Menoufia Unv.
‫الخرسانية‬ ‫البالطات‬ ‫تصميم‬
Design of R. C. Solid Slabs
‫ﺧﺮﺳﺎﻧﻴﺔ‬‫ﻣﻨﺸﺂت‬ ‫ﺗﺼﻤﻴﻢ‬
1

ALB
‫أ‬
.
‫م‬
.
‫د‬
.
‫بشندى‬ ‫على‬ ‫عالء‬
Assoc. Prof. ALaa A. Bashandy
‫العم‬ ‫ھذا‬ ‫خالل‬ ‫من‬ ‫اليه‬ ‫أھدف‬ ‫ما‬ ‫توضيح‬ ‫على‬ ‫ساعدتنى‬ ‫معلومة‬ ‫أو‬ ‫صورة‬ ‫منه‬ ‫نسخت‬ ‫أو‬ ‫استعرت‬ ‫من‬ ‫لكل‬ ‫الشكر‬ ‫جزيل‬
‫ل‬

‫المحاضرة‬ ‫محتويات‬
Lecture Contents ‫المحاضرة‬ ‫محتويات‬
Lecture Contents
‫بنھاية‬
‫ھذه‬
‫المحاضرة‬
‫يجب‬
‫ان‬
‫تكون‬
ً‫ا‬‫مـ‬ِ‫ھ‬‫ف‬َ‫ت‬ُ‫م‬
ً‫ا‬‫ِمــ‬‫ل‬‫ـ‬ُ‫م‬‫و‬
ِ‫بالنقاط‬
‫التالية‬
:
‫أنواع‬
‫البالطات‬
‫الخرسانية‬
.
‫المصمتة‬
)
‫ذات‬
‫االتجاه‬
‫الواحد‬
-
‫ذات‬
‫االتجاھين‬
(
.
‫كيفية‬
‫حساب‬
‫أحمال‬
.
‫تصميم‬
‫المصمتة‬
:

‫مفھوم‬
‫التصميم‬
.

‫كيفية‬
‫فرض‬
‫االبعاد‬
‫المبدئية‬
‫وحساب‬
‫األحمال‬
.

‫حساب‬
‫ابعاد‬
‫القطاعات‬
.

‫حساب‬
‫التسليح‬
.

‫رسم‬
‫تفاصيل‬
.
ALB Dr. Eng. ALaa Ali Bashandy

ALB
‫المسلحة‬ ‫الخرسانية‬ ‫المصمتة‬ ‫البالطات‬ ‫تصميم‬
Design of Reinforced Concrete Solid Slabs
Concrete
Members
Slabs
Beams
Column
Footing
Shallow found.
Deep found.
Solid Slabs
Hollow Block Slabs
Flat Slabs
Paneled Beam system
Simple Beam Continuous Be. Cantilever Beam
Dropped Beam Inverted Beam Hidden Beam
Axially Loaded Col. Eccentric Loaded Col.
Long Column Short Column
Rectangular Column Circular Column
Isolated Footing
Combined Footing
Strip Footing
Raft/ Mat found.
Friction Pile
Bearing Pile

‫األسقف‬ ‫بالطات‬ ‫من‬ ‫إستخداما‬ ‫االكثر‬ ‫االنواع‬
ALB
‫االعصاب‬ ‫ذات‬ ‫المصمتة‬ ‫البالطات‬
Ribbed Slabs
‫المفرغ‬ ‫الطوب‬ ‫بالطات‬
Slabs
Hollow Block
‫المصمتة‬ ‫البالطات‬
Solid Slabs
‫كمرية‬ ‫الال‬ ‫البالطات‬
Flat Slabs

‫المصمتة‬ ‫البالطات‬
Solid Slabs
ALB
‫فى‬ ‫غالبا‬ ‫النوع‬ ‫ھذا‬ ‫يستخدم‬
:
‫السكنية‬ ‫العادية‬ ‫المبانى‬
–
‫المكاتب‬
–
‫المدراس‬
–
‫المستشفيات‬
.....
‫الخ‬
‫يحتاج‬
‫ھذا‬
‫النوع‬
‫الى‬
‫كمرات‬
‫داخلية‬
‫وخارجية‬
‫لالرتكاز‬
‫عليھا‬
‫لنقل‬
‫حمل‬
‫البالط‬
‫ات‬

Solid Slabs
One-way S. Slab
Two-way S. Slab
ALB
‫د‬
.
‫بشندى‬ ‫عالء‬
Llong / Lshort ≥ 2
Llong / Lshort ≤ 2

Design
s
t
Depth
ALB
‫د‬
.
‫لـ‬ ‫االدنى‬ ‫الحد‬ ‫يحدد‬
s
t
‫بحيث‬
:
.1
‫الترخيم‬ ‫حدود‬ ‫عن‬ ‫تقل‬ ‫ال‬
deflection
.2
‫للكود‬ ‫تبعا‬ ‫االحمال‬ ‫لمقاومة‬ ‫االدنى‬ ‫بالحد‬ ‫تفى‬
:
‫استاتيكية‬ ‫احمال‬
≤
8
‫سم‬
-
‫ديناميكية‬ ‫أحمال‬
≤
12
‫سم‬
s
A
Rfmt
Reinforcement
‫التسليح‬ ‫قيمة‬ ‫حساب‬ ‫يتم‬
s
A
‫بحيث‬
:
.1
‫عن‬ ‫يقل‬ ‫ال‬
0.25
%
‫من‬
concrete
A
‫المطلوبة‬
.
.2
0.15
%
‫من‬
concrete
A
‫اختيارھا‬ ‫تم‬ ‫التى‬
.
.3
‫الشد‬ ‫مناطق‬ ‫جميع‬ ‫يغطى‬
.
.4
‫يمتد‬
3
/
1
‫للركيزة‬ ‫الركيزة‬ ‫من‬ ‫التسليح‬
.
.5
‫الثانوى‬ ‫الحديد‬
)
‫الطويل‬ ‫االتجاه‬ ‫فى‬
(
‫عن‬ ‫اليقل‬
20
%
‫الرئيسى‬ ‫الحديد‬ ‫من‬
)
‫القصير‬ ‫االتجاه‬
(
.6
‫االسياخ‬ ‫بين‬ ‫مسافة‬ ‫أقصى‬
20
‫أو‬ ‫سم‬
s
t
2
.7
‫عن‬ ‫اليقل‬ ‫الطولى‬ ‫المتر‬ ‫فى‬ ‫االسياخ‬ ‫عدد‬
5
‫عن‬ ‫واليزيد‬
10
‫اسياخ‬
/
‫م‬
/
.8
‫التسليح‬ ‫السياخ‬ ‫قطر‬ ‫اقل‬
6
‫مم‬
‫و‬ ‫المستقيمة‬ ‫لالسياخ‬
8
‫مم‬
‫المكسحة‬ ‫لالسياخ‬
.
.9
‫علوية‬ ‫شبكة‬ ‫توضع‬
‫سمك‬ ‫ذات‬ ‫للبالطات‬
<
16
‫سم‬
‫عن‬ ‫التقل‬
20
%
‫من‬
‫ادنى‬ ‫وبحد‬ ‫اتجاه‬ ‫كل‬ ‫فى‬ ‫الرئيسى‬ ‫التسليح‬
5
Ф
8
‫مم‬
/
‫م‬
/
‫من‬ ‫التكسيح‬ ‫بدء‬ ‫يتم‬ ‫مكسحة‬ ‫اسياخ‬ ‫استخدام‬ ‫حالة‬ ‫فى‬
5
/
1
‫السيخ‬ ‫ويمتد‬ ‫البحر‬ ‫خمس‬
‫عن‬ ‫تقل‬ ‫ال‬ ‫بمسافة‬ ‫المجاورة‬ ‫البالطة‬ ‫الى‬
¼
‫المتجاورتين‬ ‫للبالطتين‬ ‫االكبر‬ ‫البحر‬
Drawing Details

‫السقف‬ ‫تسليح‬ ‫رص‬
ALB

ALB

For High Tensile Steel (H.T.S)
s
t
Estimation of
Simple Slab
Continuous Slab
from tow side
Cantilever Slab
Continuous Slab
from one side
10
/
c
L
28
/
s
L
24
/
s
L
20
/
s
= L
min
s
t
it is required to check deflection if the span > 10 m
12
/
c
L
35
/
s
L
30
/
s
L
25
/
s
= L
min
s
t
Simple Slab
Continuous Slab
from tow side Cantilever Slab
Continuous Slab
from one side
For Mild Steel
For deflection requirements

Generally ts for one-way solid slabs
10
/
c
L
40
/
s
L
35
/
s
L
30
/
s
= L
min
s
t
S
L
1.00m
L
strip
Simple Slab
Continuous Slab
From tow side Cantilever Slab
Continuous Slab
From one side
Static load
→
cm
8
=
min
s
t
= 12 cm → Dynamic load
but, not less than

WuS (t/m’) = 1.5 (D.L (for slab) + L.L.) L.L/D.L < 75%
1. Dead Load (D.L)
2
2
3
P.C
3
R.C
R.C
Concrete
R.
Concrete
t/m
0.15
kg/m
150
C.
FL.
t/m
2.2
γ
&
t/m
2.5
γ
γ
*
1.0
x
1.0
x
t
x γ
V
W
)
F.C
(
cover
floor
-
2
(O.Wt)
slab
of
Own weight
-
1
s






2. Live Load ( L .L )
)
S
u
Total Load (W
.
3
Load Values
1.00 m
s
t
1.00 m
. + FL.C.
O.Wt
D.L =
According to the CODE for loads
WuS (t/m’) = 1.4 D.L (for slab) + 1.6 L.L.
or

Live Load Values

Different Cases for One-way Slab
Cantilever one-way slab
2.0
L
L
S

S
L
L
S
L C
L
Load Distribution
One-way slab
One direction
2-sides
One direction
2-sides
One direction
1-side

S
L

t/m
u
W
This image cannot currently be displayed.
/24
WL2
/24
WL2
/24
WL2
/24
WL2
/24
WL2
/8
WL2
/8
WL2
/10
WL2
/10
WL2
/10
WL2
/10
WL2
/12
WL2
/12
WL2
=
ve
+
M
min
8
L
x
u
W 2
Moment Values
Simply supported continuous two spans
continuous more than two spans
Empirical values for B.M (Max difference in load & span ≤ 20% and D.L >L.L )

Moment Values
1
L 2
L
1
L 2
L
1
M 2
M 1
M 2
M 1
M 2
M

In case of heavy L.L.
D.L → g
L.L → P
in Egyptian Code of practice,
IF P > 2 g → (Mmin- ve ) in the middle of the span must be taken in
to consideration as;
24
L
2
p
-
g
M
2
min ve
-








y
, F
cu
cm , F
100
, b =
s
, t
u
M
Given :
s
A
Req. :
d = ts - c (cover) c = 15 - 25 mm
AS min = 0.15 % Ac H.T.S
for Mild steel
c
% A
0.25
=
1.0m
d
s
t
.....
J
.....
C
b
.
F
Mu
C
d 1
cu
1 


m

 /
cm
........
f
.
d
.
J
Mu
As 2
y
 
 
H.T.S
for
A
%
0.15
but
Ac
Φ
f
φ
f
0.25% c
y
y



* d
100
=
c
A
Design of Section

& J
1
C

s
max (in the design) comparing to t
Ф
8 mm → 8 cm
10 mm → 10 cm
14 mm → 12 cm
16 mm → 14 cm
Max. spacing between bars = 20 cm
Min. spacing between bars = 10 cm
10
=

Max .number of bars / m
5
=

Min .number of bars / m

A)
-
(A
SEC.
B)
-
(B
SEC.
B
L
S
L
B
A
A
Details of Reinforcement R.F.T
When we use straight bars,
In case of simply supported span,
‫تفاصيل‬
.............................
Reinforcing
Details

sec.
As
sec.
As
m
/
0.5AS

m
/
0.5AS

m
/
0.5AS

m
/
0.5AS

- Ve Rfmt is extended to
0.25 L larger each side
Note,
for slabs of ts ≥ 16 cm,
upper steel mesh must be
added with
As ≥ 20 % of main steel
min 5Ø8/m’
Using straight bars In case of two or more spans,
‫تفاصيل‬
.............................
Reinforcing
Details

m
/
8
5
As
0.25
As main
sec.





m
/
0.5AS

m
/
0.5AS

sec.
As
clear
L
0.1
L
S
L
When we use bent bars, In case of simply supported span,
‫تفاصيل‬
.............................
Reinforcing
Details

1
L
2
L
m
/
0.5AS
 m
/
0.5AS

m
/
0.5AS

m
/
0.5AS

sec.
As
sec.
As
Using bent bars In case of two or more spans,
‫تفاصيل‬
.............................
Reinforcing
Details

=
s
t
8 cm
300/35 = 8.57 cm
)
1
EXAMPLE (
Fcu = 25 N/mm2 Steel 240/350 and 360/520
2
kg/m
300
=
2
N/mm
30
L.L =
F.C = 15 N/mm2 = 150 kg/m2
cm
10
=
s
t
deflection must be checked
2
s t/m
04
.
1
0.30
*
1.6
0.15)
2.50
*
0.1
(
1.4
Wu 



m
1.0
m
6.0
m
2.5 m
3.0
t/m'
1.04
m
3.0
m
2.5
1
2
3
KN.m
2.7
t.m
27
.
0
KN.m
3.9
t.m
39
.
0
KN.m
6.5
t.m
65
.
0
KN.m
9.36
t.m
936
.
0
KN.m
10.1
t.m
.01
1
01
.
1
3
5
.
2
3
x
1.17
m
2.5
x
0.813
L
L
L
x
M
L
x
M
M
2
1
2
2
1
1
support 







100
x
250
10
x
01
.
1
C1
5
.
8
5

C1 = 4.22 & J= 0.81
/m
10
Ф
6
=
2
cm
4.44
=
4.449
x
0.25
=
s
% A
25
=

s
A
= 0.96 cm2 = 5Ф8 /m .. Secondary Dir.
5
.
8
3600
81
.
0
10
01
.
1
A
5
S




As = 3.133 cm2 = 5Ф10/m
As
= 25% As = 0.25x3.133
= 0.96 cm2 = 5Ф8 /m .. Secondary Dir.
cm
100
b
cm
8.5
1.5
–
t
d
:
(1)
SEC
s



Rfmt. Detailing
m
6.0
m
3.0
m
3.0
m
/
10
5 

m
/
10
5 

m
/
10
5 

m
/
8
5 

m
/
8
5 

m
/
8
5 
 m
/
8
5 

10 10
cm
100
b
cm
8.5
1.5
–
t
d
:
(2)
SEC
s



KN.m
2.7
t.m
27
.
0
KN.m
3.9
t.m
39
.
0
KN.m
6.5
t.m
65
.
0
KN.m
9.36
t.m
936
.
0
KN.m
10.1
t.m
.01
1

cm
100
b
cm
8.5
1.5
–
t
d
:
(2)
SEC
s



100
x
250
10
.936x
0
C1
5
.
8
5

C1 = 4.39 & J= 0.815

/m
10
Ф
5
=
2
cm
3.753
=
3.753
x
0.25
=
s
% A
25
=

s
A
= 5Ф8 /m .. Secondary Dir.
5
.
8
x
3600
x
815
.
0
10
x
936
.
0
A
5
S 
cm
100
b
cm
8.5
1.5
–
t
d
m.t
0.65
M
:
span
mid
left
At
s
u




100
x
250
10
x
.65
0
C1
5
.
8
5

C1 = 5.27 & J= 0.826

/m
8
Ф
8
=
2
cm
3.86
=
Not recommended
5
.
8
x
00
4
2
x
826
.
0
10
x
936
.
0
A
mm
8
φ
use
5
S 

/m
10
Ф
5
→

/m
10
Ф
4
=
2
cm
2.57
=
s
% A
25
=

s
A
= 5Ф8 /m .. Secondary Dir.
5
.
8
x
00
6
3
x
826
.
0
10
x
936
.
0
A
mm
10
φ
use
5
S 

4
S
1
S
5
S
6
S
9
S
8
S
7
S
2
S
3
S
‫المصمتة‬ ‫البالطات‬ ‫احمال‬ ‫توزيع‬
ALB

Slab thickness ts for two-way solid slabs
t s min = Ls / 35 Ls / 40 Ls / 45
Simple Slab
Continuous Slab
From tow side
Continuous Slab
From one side
Static load
→
cm
8
=
min
s
t
= 12 cm → Dynamic load
but, not less than

Effective Span

The load is distributed in two
direction by the values (α , β)
α → short direction
→ Wα = α x Wus
β → long direct
→ Wβ = β x Wus
Two Way Solid Slab
2
<
s
r = L / L
The values of (α) and (β) are Calculated by 3 methods
β
α 
direction
long
in
w
direction
short
in
w
β
α

1- Grashoff Method:
0
.
1
2
1







Grashoff
1
2


 





 1
1
1
&
1 4
4
4
r
r
r
17
-
6
See code
Assumption of Grashoff Method :
1. Neglect effect of plate action of slab.
2. Neglect corner effect.
3. Neglect torsion rigidity.
(2)
STRIP
(1)
STRIP
1.00m
1.00m
1

L
Ls
Wα
Wβ
• 2 way S.S with L.L > 5 KN/m2
• 2 way H.B with L.L > 5 KN/m2
• Paneled beam slab & Ribbed Slab
Calculation of α & β

2- Marcus Method:
10
–
6
see code
8
.
0

 

• 2 way S.S. resting on masonry walls
• 2 way H.B. with L.L. ≤ 5 KN/m2
3- Code of Practice:
• Solid slab with L.L ≤ 5 KN/m2
2
r
0.35
β
0.15
0.5r
α



In design of solid slab we use the distribution of code of practice
Assumption of marcus Method :
1. Neglect effect of plate action of slab.
2. Neglect corner effect.

1.0
L
m
L
m
r
s





= 0.87
= 0.76
Calculation of r
1.0
=

Where m & m

Reinforcement Details for TWO-WAY Solid Slab

‫التسليح‬ ‫اسياخ‬ ‫رص‬ ‫كيفية‬
Design of R.C. Beams
‫د‬
.

‫المصمتة‬ ‫البالطات‬ ‫اركان‬ ‫تسليح‬ ‫تفاصيل‬
For spans ≥ 5m

ALB
‫د‬
.
s
A
Rfmt
Reinforcement
‫للتسليح‬ ‫بالنسبة‬ ‫االتى‬ ‫يراعى‬
:
.1
0.25
%
‫من‬
concrete
A
‫المطلوبة‬
.
.2
0.15
%
‫من‬
concrete
A
‫اختيارھا‬ ‫تم‬ ‫التى‬
.
.3
‫الشد‬ ‫مناطق‬ ‫جميع‬ ‫يغطى‬
.
.4
‫يمتد‬
3
/
1
‫للركيزة‬ ‫الركيزة‬ ‫من‬ ‫التسليح‬
.
.5
‫الثانوى‬ ‫الحديد‬
)
‫الطويل‬ ‫االتجاه‬ ‫فى‬
(
‫عن‬ ‫اليقل‬
20
%
‫الرئيسى‬ ‫الحديد‬ ‫من‬
)
‫القصير‬ ‫االتجاه‬
(
.6
‫االسياخ‬ ‫بين‬ ‫مسافة‬ ‫أقصى‬
20
‫أو‬ ‫سم‬
s
t
2
.7
‫عن‬ ‫اليقل‬ ‫الطولى‬ ‫المتر‬ ‫فى‬ ‫االسياخ‬ ‫عدد‬
5
‫عن‬ ‫واليزيد‬
10
‫اسياخ‬
/
‫م‬
/
.8
‫التسليح‬ ‫السياخ‬ ‫قطر‬ ‫اقل‬
6
‫مم‬
‫و‬ ‫المستقيمة‬ ‫لالسياخ‬
8
‫مم‬
‫المكسحة‬ ‫لالسياخ‬
.
.9
‫علوية‬ ‫شبكة‬ ‫توضع‬
‫سمك‬ ‫ذات‬ ‫للبالطات‬
<
16
‫سم‬
‫عن‬ ‫التقل‬
20
%
‫من‬
‫ادنى‬ ‫وبحد‬ ‫اتجاه‬ ‫كل‬ ‫فى‬ ‫الرئيسى‬ ‫التسليح‬
5
Ф
8
‫مم‬
/
‫م‬
/
‫من‬ ‫التكسيح‬ ‫بدء‬ ‫يتم‬ ‫مكسحة‬ ‫اسياخ‬ ‫استخدام‬ ‫حالة‬ ‫فى‬
5
/
1
‫السيخ‬ ‫ويمتد‬ ‫البحر‬ ‫خمس‬
‫عن‬ ‫تقل‬ ‫ال‬ ‫بمسافة‬ ‫المجاورة‬ ‫البالطة‬ ‫الى‬
¼
‫المتجاورتين‬ ‫للبالطتين‬ ‫االكبر‬ ‫البحر‬

) :
2
Example (

0.198
515
.
0
33
.
1
0
.
3
87
.
0
0
.
4
87
.
0
S1







 

r
0.164
578
.
0
46
.
1
0
.
3
87
.
0
0
.
5
76
.
0
S2







 

r
0.26
433
.
0
16
.
1
0
.
3
87
.
0
5
.
3
87
.
0
S3







 

r
1.0)
(r
97
.
0
5
.
4
87
.
0
0
.
5
76
.
0
S4







 r
0.35
35
.
0
03
.
1
0
.
5
76
.
0
5
.
4
87
.
0






 

r
By Code Method:
using code Method is sufficient safe

0.242
758
.
0
33
.
1
0
.
3
87
.
0
0
.
4
87
.
0
S1







 

r
0.18
82
.
0
46
.
1
0
.
3
87
.
0
0
.
5
76
.
0
S2







 

r
0.356
644
.
0
16
.
1
0
.
3
87
.
0
5
.
3
87
.
0
S3







 

r
1.0)
(r
97
.
0
5
.
4
87
.
0
0
.
5
76
.
0
S4







 r
0.47
53
.
0
03
.
1
0
.
5
76
.
0
5
.
4
87
.
0






 

r
By Grashoff Method:
using Grashoff Method is more more safe

Example:
For the given plan it is required to:
Calculate loads for slabs & Beams
Data:
2
kg/m
150
FL.C =
2
kg/m
300
L.L =
Steel Grade 240/350 or 360/520
Solution:
Slabs
2
t/m
0.45
=
2
t/m
0.15
)+
3
t/m
2.5
m *
0.12
D.L = (
2
t/m
0.3
L.L =
‫المسلحة‬ ‫الخرسانية‬ ‫الكمرات‬ ‫تصميم‬
Design of R. C. Beams
2
t/m
1.11
=
2
t/m
0.3
x
1.6
+
2
t/m
0.45
x
1.4
=
s
u
W
s
To have slab thickness t
cm
10
=
40
/
400
=
s
t
→
1
S
cm
6.67
=
45
/
300
=
s
t
→
2
S
cm
11.1
=
45
/
500
=
s
t
→
3
S
cm
12
=
s
take t

0.156
844
.
0
526
.
1
0
.
3
x
76
.
0
0
.
4
x
87
.
0



 

r
m)
4
x
3
(
1
Slab S
t/m’
0.173
=
β
W
t/m’
0.937
=
α
W
0.165
835
.
0
.
0
5
.
1
0
.
4
x
87
.
0
0
.
6
x
87
.
0



 

r
m)
6
x
4
(
2
Slab S
0.219
781
.
0
374
.
1
0
.
5
x
76
.
0
0
.
6
x
87
.
0



 

r
m)
6
x
5
(
3
Slab S
0.114
886
.
0
67
.
1
0
.
3
x
76
.
0
0
.
5
x
76
.
0



 

r
m)
5
x
3
(
4
Slab S
m)
9
x
1
(
5
Slab S
t/m’
0.183
=
β
W
t/m’
0.927
=
α
W
t/m’
0.243
=
β
W
t/m’
0.867
=
α
W
t/m’
0.127
=
β
W
t/m’
0.983
=
α
W
t/m’
1.11
=
S
Wu
=
5
S
W
m)
6
x
1.5
(
5
Slab S t/m’
1.11
=
S
Wu
=
6
S
W
way slab
-
One
way slab
-
One

Loads:
D.L
-
1
Own weight ( O.W )
-
-
Estimation of thickness:
mm static load
80
min =
s
t
= 120 mm dynamic load
min =
s
T
35
Ls
min =
s
t
min =
s
T
40
Ls
45
Ls
Slab simply supported
Continuous from one end
Continuous from two end
Deflection must be checked
deflection
check
t
don'
we
if



9B
36
))
1500
Fy
(
(0.8
Ln
t




= clear span
n
L
s
B = L / L
2
in N/mm
y
F

3
R.C.
R.C.
KN/m
25
when
1.0
1.0
t
o.w./m
γ
γ





2
KN/m
1.5
F.C. assumed
10
L
W
β
M
12
L
W
α
M
end
one
from
continous
is
slab
If
-
:
follows
as
taken
be
may
moment
bending
of
values
the
simplicity
For
-
slab
way
one
as
values
emperical
same
with the
method
strip
using
calculated
are
forces
Internal
-
:
forces
Internal
L.L.
1.6
D.L.
1.4
W
type
building
to
according
assumed
L.L.
-
2
2
L
2
S
L
U
S











8
L
W
8
W
supported
simply
is
slab
If
12
L
W
12
W
ends
two
from
continous
is
slab
If
2
L
2
LS
2
L
2
LS
M
M
M
M




















S
S
L
L
-
:
sections
of
Design
1.00m
Wα
Wβ
1.00m
L
Ls
Detail(A)
(A)
Detail
(long)
As
(short)
As

As
m
/
cm
F
.
d
.
J
Mu
As
....
J
....
C1
b
F
Mu
C1
d
1.5
-
t
d
8
)
(L
W
M
min
2
y
short
cu
s
Short











 
short
2
y
long
cu
s
2
long
β
As
0.25
m
/
cm
F
.
d
.
J
Mu
As
....
J
....
C1
b
F
Mu
C1
d
2.5
-
t
d
8
)
(L
*
W
M











-
:
direction
Short
-
:
direction
Long

2
2
2
2
m
t /
0.30
L.L
m
t /
0.15
F.C.
520
/
360
-
240/350
steel
kg/cm
300
N/mm
30
Fcu
-
:
)
3
(
Example




cm
8
25
/
200
)
deflection
check
(then
cm
12
cm
11.25
40
/
450
t
cm
12
2
/15
150
ts
s








 
0.20
(1.33)
0.35
β
0.52
0.15
-
0.50
*
1.33
1.33
4.5
*
0.87
6.0
*
0.87
r
t/m
1.11
0.3
x
1.6
0.15
)
t/m
2.5
x
(0.12
x
.4
1
Wu
2
2
3
S











III
II
I
50
.
1
50
.
4
00
.
4
2.00
6.00
According to code

m
/
10
5
use
m
/
8
8
cm
4.02
24600
x
10.5
x
826
.
0
10
x
0.836
A
mm
8
M.S.
rec.
not
m
/
10
4
cm
2.678
3600
x
10.5
x
826
.
0
10
x
0.836
A
mm
10
H.T.S.
826
.
0
J
74
.
5
C1
100
x
250
10
x
0.836
C1
10.5
-
:
)
2
(
SEC.
m
/
10
6
cm
4.2
3600
x
10.5
x
821
.
0
10
x
1.25
As
821
.
0
J
69
.
4
C1
100
x
250
10
x
1.25
C1
10.5
cm
10.5
2.5
-
d
-
:
)
1
(
SEC.
-
:
)
I
(
STRIP
2
5
s
2
5
s
5
2
5
5
s
t






























1
2
KN.m
8.36
0.836 
KN.m
5
.
2
1
25
.
1 
50
.
1
50
.
4
m
t/
1.11 
m
t/
577
.
0
11
.
1
x
52
.
0 


safe
ok
cm
2.04
100
x
12
x
100
0.17
As
m
/
8
6
cm
2.67
2400
x
10.5
x
826
.
0
10
*
0.58
A
M.S.
240/350
mm
8
use
cm
1.77
3600
x
10.5
x
826
.
0
10
*
0.58
A
H.T.S.
360/520
mm
10
use
826
.
0
J
047
.
7
C1
x100
250
10
*
0.58
C1
10.5
cm
10.5
2.5
-
d
-
:
)
II
(
STRIP
2
min
2
5
s
2
5
s
5
s
t

















m
t/
11
.
1 
00
.
2
KN.m
5.55
m.t
0.555 

t
.
m
0.888
M
)
0.37
1.998
(
6
-
0
)
2
6
(
M
2
0
Equ.
M
-
3
:
1
Method
M
support
at
moment
have
To
-
:
)
III
(
STRIP
supp.






m
t/
13
.
1 
m
t/
22
.
0 
00
.
2
00
.
6
m
t/
11
.
1 
m
t/
22
.
0
11
.
1
x
20
.
0 

W2L2
2/8
W1L1
2/8
or
W1L1
2/10 W2L2
2/10
M supp.= 0.888 m.t
0.444 m.t
0.799 m.t
B. M. D.
= 0.999 m.t
= 0.555 m.t
m.t
888
.
0
2
6
2
0.555
6
x
0.999
L
L
L
M
L
M
M
:
2
Method
2
1
2
2
1
1
supp.









m.t
0.999
m.t
0.555
or
m.t
99
.
0
M
or
M
of
larger
the
e
tak
:
3
Method
right
left


m
/
10
5
use
much
so
m
/
8
9
cm
2.67
2400
x
10.5
x
826
.
0
10
*
0.888
A
M.S.
240/350
mm
8
use
m
/
10
4
cm
2.91
3600
x
10.5
x
826
.
0
10
*
0.888
A
H.T.S.
360/520
mm
10
use
826
.
0
J
57
.
5
C1
250x100
10
*
0.888
C1
10.5
cm
10.5
2.5
-
t
d
1
SEC.
2
5
s
2
5
s
5
s























m
/
8
2.5
10
2.5
mm
90.5
362
x
25%
4
As
As
much
so
m
/
8
9
cm
4.24
2400
x
9.5
x
826
.
0
10
*
0.799
A
M.S.
240/350
mm
8
use
m
/
10
4
cm
2.0
3600
x
9.5
x
826
.
0
10
*
0.799
A
H.T.S.
360/520
mm
10
use
826
.
0
J
57
.
5
C1
250x100
10
*
0.799
C1
9.5
cm
9.5
cm
1.0
-
2.5
-
12
-
c
-
t
d
S
slab
of
direction
Long
2
SEC.
2
short
min
2
5
s
2
5
s
5
s
3






























m
/
8
5
cm
2.133
2400
x
10.5
x
826
.
0
10
*
0.444
A
M.S.
240/350
mm
8
use
826
.
0
J
88
.
7
C1
250x100
10
*
0.444
C1
10.5
cm
10.5
2.5
-
t
d
3
SEC.
2
5
s
5
s













6.00
m
2.00
m
12 cm
/
/
m
8
Ф
2.5
/
/ m
8
Ф
5 /
/ m
8
Ф
5
4.50 m 4.50 m 1.50 m
/
/ m
8
Ф
5
/
/ m
10
Ф
3
/
/ m
10
Ф
3
/
/
m
8
Ф
5
/
/
m
10
Ф
2.5
/
/ m
10
Ф
3
/
/ m
8
Ф
3
/
/
m
8
Ф
5
/
/
m
10
Ф
2.5
/
/
m
8
Ф
5
12 12

1- one way
1
t
2
t
1
s
1
s
t
c
α
(Be)
width
Effective
t
C
2
t
S 1
1 


t
C
2
t
S 2
1 


 
 
m
2.0
S1
B
(2/3)
As
/
As
of
ratio
max
L
As
As
S1
B
e
(main)
(sec.)
main
sec.
e






Concentrated line load on solid slab (wall)
2
or S
1
Loads is taken as distributed on a length S

L
L
-
2
L
0.4
S
Be
L
0.4
S
Be
L
L
L
P
P
L
L
L
P
P
way
one
1.5
L
L
way
two
1.5
L
L
if
S
S
1
(L)
S
2
(L)
S
L
S
S
LS
S
S
















only
loads
ed
concentrat
for
L
s
L
2
S
1
S
)
(L
B s
e
(L)
e
B
2- Two way

.....
.....
)
length
inclined
the
is
L
:
(where
*
m
L
*
m
LS








r
f
w
y
s
s
J.d.
M
As
J
&
C1
get
2.5
t
d
8
L
.
L
.
β.
B.M.
)
1
(
strip





I
II
L
LS
 
f
A
c
f
c
t
L
w
y
S
1
cu
1
s
2
S
s
J.d.
Mcos θ
J
&
get
b.
Mcos θ
d
2.5
d
design
in
Mcos
Take
8
.
β.
B.M.(M)
)
II
(
strip






100
ts
L
S
L
θ
M
cosθ
M
sinθ
M
S
W
*
α
S
W
*
α
Inclined Slabs

2
2
2
2
2
)
2
(
*
(5)
L
projection
horizontal
on
t/m
0.30
L.L.
t/m
0.15
F.C.
520
/
360
STELL
/mm
N
25
Fcu
:
(6)
EXAMPLE





50
.
1
50
.
1 00
.
5
00
.
5
00
.
2
Wα
Wβ
II
I

s =
t
150 / 15 + 2 = 12 cm
500 / 35 = 14.3 cm
cm
14
t s 

0.24
1.22
0.35
β
0.46
0.15
-
0.50
*
1.22
α
1.22
r
1
1.0
0.82
5.00
*
1.00
5.39
*
0.76
r
m
t/
1.17
)
0.93
*
0.3
0.15
2.50
*
0.14
(
1.5
u
)
(w
m
t/
1.2
)
0.3
0.15
2.50
*
0.14
(
1.5
u
)
(w
0.93
θ
cos
21.8
)
2/5
(
tan
θ
2
2
(inclined)
s
2
slab)
(cont.
s
-1



















 

m
/
8
5
mm
233.8
A
take
mm
233.8
1000
*
140
*
100
0.167
As
cm
1.26
mm
126
As
-
:
)
2
(
SEC.
m
/
10
5
cm
3.63
mm
363.2
3600
.
125
.
0.826
10
*
13.5
As
0.826
J
5.38
25
1000
10
*
13.5
105
mm
125
25
-
d
-
:
)
1
(
SEC.
-
:
)
I
(
STRIP
2
S
2
min
2
2
2
2
6
1
6
1
s
c
c
t





















3
2
1
B.M.D
KN.m
13.5
KN.m
13.5
t/m`
1.2
t/m`
1.2
00
.
5 50
.
1
50
.
1

m
/
8
φ
5
2.38cm
mm
238
3600
.
115
.
0.826
10
*
8.14
As
0.826
J
6.37
c
25
1000
10
*
8.14
c
115
t
.
m
8.14
0.93
*
8.75
θ
cos
M
mm
115
25
-
d
-
:
)
II
(
STRIP
2
2
6
1
6
1
s
t














t/m`
28
.
0
17
.
1
*
24
.
0 
00
.
5
B.M.D
KN.m
75
.
8
875
.
0 

PLAN OF
RFT. DETAILING
12
5.00
5.00 1.50
1.50
58/m
`
58/m`
58/m`

B.M.D
Typical Rft. Detailing for Some Slabs

B.M.D

ALB Dr. Eng. ALaa Ali Bashandy
‫المراجع‬
:
.1
‫الكود‬
‫المصرى‬
‫للخرسانة‬
‫المسلحة‬
ECP 203-2007
-
‫معھد‬
‫بحوث‬
‫البناء‬
–
‫وزارة‬
‫االسكان‬
–
‫ج‬
.
‫م‬
.
‫ع‬
.
.2
‫الملحق‬
‫االول‬
‫للكود‬
‫المصرى‬
”
‫مساعدات‬
‫التصميم‬
“
-
‫كود‬
203
‫لسنة‬
2007
.
.3
‫الملحق‬
‫الثانى‬
‫للكود‬
‫المصرى‬
”
‫دليل‬
‫التفاصيل‬
‫االنشائية‬
“
-
‫كود‬
203
‫لسنة‬
2007
.
.4
Design of R. C. Structures
–
- Vol. 1, 2, 3
‫أ‬
.
‫د‬
.
‫مشھور‬
‫غنيم‬
–
‫أ‬
.
‫د‬
.
‫محمود‬
‫المھيلمى‬
-
‫كلية‬
‫الھندسة‬
–
‫جامعة‬
‫القاھرة‬
.
.5
Design of Halls
–
‫د‬
.
‫محمد‬
‫ھالل‬
.

Solid Slabs - Dr. ALaa Bashandy.pdf

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