Solid Slabs - Dr. ALaa Bashandy.pdf

j;12

1. ميحرلا نمحرلا ﷲ مسب
ALB
د.م . يدنشب ىلع ءلاع
2014
Dr. Eng. ALaa Ali Bashandy
Assoc. Prof. at Civil Dep., F. of Engineering, Menoufia Unv.
ةيناسرخلا تاطلابلا ميمصت
Design of R. C. Solid Slabs
ﺔﻴﻧﺎﺳﺮﺧ تﺂﺸﻨﻣ ﻢﻴﻤﺼﺗ 1

2. ALB
أ.م.د . ىدنشب ىلع ءلاع Assoc. Prof. ALaa A. Bashandy
معلا اذھ للاخ نم هيلا فدھأ ام حيضوت ىلع ىنتدعاس ةمولعم وأ ةروص هنم تخسن وأ ترعتسا نم لكل ركشلا ليزج ل

3. ةرضاحملا تايوتحم Lecture Contents ةرضاحملا تايوتحم Lecture Contents
ةياھنب هذھ ةرضاحملا
بجي نا نوكت ًاـمِھفَتُم ًاــمِلـُمو ِطاقنلاب ةيلاتلا :
عاونأ تاطلابلا ةيناسرخلا . تاطلابلا ةيناسرخلا ةتمصملا )تاذ هاجتلاا دحاولا - تاذ نيھاجتلاا (. ةيفيك باسح لامحأ تاطلابلا ةيناسرخلا . ميمصت تاطلابلا ةيناسرخلا ةتمصملا :
 موھفم ميمصتلا .
 ةيفيك ضرف داعبلاا ةيئدبملا باسحو لامحلأا .
 باسح داعبا تاعاطقلا .
 باسح حيلستلا .
 مسر ليصافت حيلستلا .
ALB
Dr. Eng. ALaa Ali Bashandy

4. ALB
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
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

5. فقسلأا تاطلاب نم امادختسإ رثكلاا عاونلاا
ALB
باصعلاا تاذ ةتمصملا تاطلابلا
Ribbed Slabs
غرفملا بوطلا تاطلاب
Slabs Hollow Block
ةتمصملا تاطلابلا
Solid Slabs
ةيرمك لالا تاطلابلا
Flat Slabs
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

6. ةتمصملا تاطلابلا Solid Slabs
ALB
ىف ابلاغ عونلا اذھ مدختسي :
ةينكسلا ةيداعلا ىنابملا – بتاكملا – ساردملا – تايفشتسملا .....
خلا
جاتحي اذھ عونلا ىلا تارمك ةيلخاد ةيجراخو زاكترلال اھيلع لقنل لمح طلابلا تا
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

7. Solid Slabs
One-way S. Slab
Two-way S. Slab
ALB
د. ىدنشب ءلاع
L
long
/ L
short
≥2
L
long
/ L
short
≤2
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

8. Design
s
t Depth
ALB
د. ىدنشب ءلاع
ـل ىندلاا دحلا ددحي
s
t

ثيحب :
.1 ميخرتلا دودح نع لقت لا deflection
.2 دوكلل اعبت لامحلاا ةمواقمل ىندلاا دحلاب ىفت :
ةيكيتاتسا لامحا ≤8
مس
- ةيكيمانيد لامحأ ≤12
مس
s
A Rfmt Reinforcement
حيلستلا ةميق باسح متي
s
A

ثيحب :
.1 نع لقي لا 0.25 % نم
concrete
A
ةبولطملا .
.2 نع لقي لا 0.15 % نم
concrete
A
اھرايتخا مت ىتلا .
.3 دشلا قطانم عيمج ىطغي .
.4 دتمي
3
/
1
ةزيكرلل ةزيكرلا نم حيلستلا .
.5 ىوناثلا ديدحلا ) ليوطلا هاجتلاا ىف ( نع لقيلا 20 % ىسيئرلا ديدحلا نم ) ريصقلا هاجتلاا (.6 خايسلاا نيب ةفاسم ىصقأ 20 وأ مس
s
t2
.7 نع لقيلا ىلوطلا رتملا ىف خايسلاا ددع 5 نع ديزيلاو 10 خايسا /م
/
.8 حيلستلا خايسلا رطق لقا 6
مم و ةميقتسملا خايسلال 8
مم
ةحسكملا خايسلال .
.9 ةيولع ةكبش عضوت كمس تاذ تاطلابلل <16مس نع لقتلا 20 % نم
ىندا دحبو هاجتا لك ىف ىسيئرلا حيلستلا 5Ф8 مم /م
/
نم حيسكتلا ءدب متي ةحسكم خايسا مادختسا ةلاح ىف
5
/
1
خيسلا دتميو رحبلا سمخ
نع لقت لا ةفاسمب ةرواجملا ةطلابلا ىلا ¼ نيترواجتملا نيتطلابلل ربكلاا رحبلا
Drawing Details
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

9. فقسلا حيلست صر
ALB
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

10. ALB
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

11. 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
it is required to check deflection if the span > 10 m
Simple Slab
Continuous Slab
from tow side
Cantilever Slab
Continuous Slab
from one side
For Mild Steel
For deflection requirements
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

12. Generally t
s
for one-way solid slabs
10 /
c
L 40 /
s
L 35 /
s
L 30 /
s
= L
min s
t
it is required to check deflection if the span > 10 m
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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

13. W
u
S
(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 wei
g
ht -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
W
u
S
(t/m
’)
= 1.4 D.L
(for slab)
+ 1.6 L.L.
or
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

14. Live Load Values
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

15. 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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
One direction
2-sides
One direction
2-sides
One direction
1-side

16. S
L
\
t/m
u
W
T h is imag e c an n o t c u r r en tly b e d isp lay ed .
/24 WL
2
/24 WL
2
/24 WL
2
/24 WL
2
/24 WL
2
/8 WL
2
/8 WL
2
/10 WL
2
/10 WL
2
/10 WL
2
/10 WL
2
/12 WL
2
/12 WL
2
=
ve+
M
min
8
L x
u
W
2
Moment Values
Simply supportedcontinuous two spans
continuous more than two spans
Empirical values for B.M
(Max difference in load& span≤20% and D.L >L.L )
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

17. Moment Values
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
1
L
2
L
1
L
2
L
1
M
2
M
1
M
2
M
1
M
2
M

18. In case of heavy L.L.
D.L → g
L.L → P
in Egyptian Code of practice,
IF P > 2 g → (M
min- ve
) in the middle of the span must be taken in
to consideration as;
24
L
2
p
- g
M
2
min
ve-







ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

19. y
, F
cu
cm , F 100 , b =
s
, t
u
M Given :
s
A Req. :

d = t
s
- c
(cover) c = 15 - 25 mm
A
S
min = 0.15 % A
c
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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

20. & J
1
C
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

21. 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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

22. 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,
حيلستلا ليصافت ............................. Reinforcing Details
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

23. sec.
As
sec.
As
m / 0.5A
S

m / 0.5A
S

m / 0.5A
S

m / 0.5A
S

-Ve
Rfmt
is extended to
0.25 L
larger
each side
Note,
for slabs of t
s
≥ 16 cm,
upper steel mesh must be
added with
A
s
≥ 20 % of main steel
min 5Ø8/m’
Using straight bars
In case of two or more spans,
حيلستلا ليصافت ............................. Reinforcing Details
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

24. m /8 5
As 0.25 As
main sec.  


m / 0.5A
S

m / 0.5A
S

sec.
As
clear
L 0.1
L
S
L
When we use bent bars,
In case of simply supported span,
حيلستلا ليصافت ............................. Reinforcing Details
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

25. 1
L
2
L
m / 0.5A
S
m / 0.5A
S

m / 0.5A
S

m / 0.5A
S

sec.
As
sec.
As
Using bent bars
In case of two or more spans,
حيلستلا ليصافت ............................. Reinforcing Details
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

26. =
s
t
8 cm 300/35 = 8.57 cm
)1 EXAMPLE (
F
cu
= 25 N/mm
2
Steel 240/350 and360/520
2
kg/m 300 =
2
N/mm 30 L.L =
F.C = 15 N/mm
2
= 150 kg/m
2
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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
01.1
3 5.2
3 x 1.17 m 2.5 x 0.813
L L
Lx M Lx M
M
2 1
2 2 1 1
support








27. 100 x 250
10x01.1
C1 5.8
5

C1 = 4.22 & J= 0.81
/m 10 Ф6=
2
cm 4.44 =
4.449 x 0.25 =
s
% A 25=
\
s
A
= 0.96 cm
2
= 5Ф8 /m .. Secondary Dir.
5.8 3600 81.0
10 01.1
A
5
S  


As = 3.133 cm
2
= 5Ф10/m
A
s
\
= 25% A
s
= 0.25x3.133
= 0.96 cm 2
= 5Ф8 /m .. Secondary Dir.
cm 100 b
cm 8.5 1.5 – t d
:(1) SEC
s

 
Rfmt.
Detailin
g
m 6.0
m 3.0
m 3.0
m/10 5 
m/10 5 
m/10 5 
m/85 
m/85 
m/85 m/85 
10
10
cm 100 b
cm 8.5 1.5 – t d
:(2) SEC
s

 
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. En
g
. ALaa Ali Bashand
y
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

28. cm 100 b
cm 8.5 1.5 – t d
:(2) SEC
s

 
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
100 x 250
10 .936x 0
C1 5.8
5

C1 = 4.39 & J= 0.815
\
/m 10 Ф5=
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 .650
C1 5.8
5

C1 = 5.27 & J= 0.826
\
/m8 Ф8=
2
cm 3.86 =
Not recommended
5.8 x 0042 x 826.0
10 x 936.0
A mm 8 φ use
5
S

\
/m 10 Ф5→
\
/m 10 Ф4=
2
cm 2.57 =
s
% A 25=
\
s
A
= 5Ф8 /m .. Secondary Dir.
5.8 x 0063 x 826.0
10 x 936.0
A mm 10 φ use
5
S


29. ALB
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

30. 4
S
1
S
5
S
6
S
9
S
8
S
7
S
2
S
3
S
ةتمصملا تاطلابلا لامحا عيزوت
ALB
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

31. Slab thickness t
s
for two-way solid slabs
t
smin
= L
s
/ 35 L
s
/ 40 L
s
/ 45
it is required to check deflection if the span > 10 m
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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. En
g
. ALaa Ali Bashand
y

32. Effective Span
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

33. The load is distributed in two
direction by the values (α,β)
α→short direction
→W
α
=αxW
u
s
β→long direct
→W
β
=βxW
u
s
Two Way Solid Slab
2<
s
r = L / L
The values of (α) and (β) are Calculated by
3 methods
βα
direction long in w
direction short in w
β
α
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

34. 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/m
2
•
2 way H.B with L.L > 5 KN/m
2
•
Paneled beam slab & Ribbed Slab
Calculation of α& β

ة
حلسملا
ة
يناسرخلا
ة
تمصملا تاطلابلا ميمص
ت

Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

35. 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/m
2
3- Code of Practice:
•
Solid slab with L.L ≤ 5 KN/m
2
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.

36. 1.0
L m
Lm
r
s




= 0.87
= 0.76
Calculation of r
1.0 =
\
Where m & m
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

37. Reinforcement Details for TWO-WAY Solid Slab
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

38. حيلستلا خايسا صر ةيفيك
Design of R.C. Beams د. ىدنشب ءلاع

39. ةتمصملا تاطلابلا ناكرا حيلست ليصافت
For spans ≥ 5m
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

40. ALB
د. ىدنشب ءلاع
s
A Rfmt Reinforcement
حيلستلل ةبسنلاب ىتلاا ىعاري :
.1 نع لقي لا 0.25 % نم
concrete
A

ةبولطملا .
.2 نع لقي لا 0.15 % نم
concrete
A

اھرايتخا مت ىتلا .
.3 دشلا قطانم عيمج ىطغي .
.4 دتمي
3
/
1
ةزيكرلل ةزيكرلا نم حيلستلا .
.5 ىوناثلا ديدحلا ) ليوطلا هاجتلاا ىف ( نع لقيلا 20 % ىسيئرلا ديدحلا نم ) ريصقلا هاجتلاا (.6 خايسلاا نيب ةفاسم ىصقأ 20 وأ مس
s
t2
.7 نع لقيلا ىلوطلا رتملا ىف خايسلاا ددع 5 نع ديزيلاو 10 خايسا /م
/
.8 حيلستلا خايسلا رطق لقا 6
مم و ةميقتسملا خايسلال 8
مم
ةحسكملا خايسلال .
.9 ةيولع ةكبش عضوت كمس تاذ تاطلابلل <16مس نع لقتلا 20 % نم
ىندا دحبو هاجتا لك ىف ىسيئرلا حيلستلا 5Ф8 مم /م
/
نم حيسكتلا ءدب متي ةحسكم خايسا مادختسا ةلاح ىف
5
/
1
خيسلا دتميو رحبلا سمخ
نع لقت لا ةفاسمب ةرواجملا ةطلابلا ىلا ¼ نيترواجتملا نيتطلابلل ربكلاا رحبلا
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

41. ) : 2 Example (
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

42. 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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
By Code Method:
using code Method is sufficient safe

43. 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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
By Grashoff Method:
using Grashoff Method is more more safe

44. 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 or360/520
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 =
ةحلسملا ةيناسرخلا تارمكلا ميمصت
Design of R. C. Beams
Dr. Eng. ALaa Ali Bashandy
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

45. 0.156 844.0 526.1
0.3x 76.0
0.4x 87.0
   
 
r
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
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
0.4x 87.0
0.6x 87.0
   
 
r
m)6 x 4 (
2
Slab S
0.219 781.0 374.1
0.5x 76.0
0.6x 87.0
   
 
r
m)6 x 5 (
3
Slab S
0.114 886.0 67.1
0.3x 76.0
0.5x 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

46. 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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

47. 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 Wβ
M
12
L Wα
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
 

 

 
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

48. 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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

49. 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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

50. 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
t
s 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 .41 Wu
2
2 3
S
 
 
 
   

III
II
I
50.1
50.4
00.4
2.00 6.00
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
According to code

51. 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.
826.0J 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
821.0J 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 5.21 25.1
50.1
50.4
m t/ 1.11
m t/ 577.0 11.1 x 52.0 
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

52. 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
826.0J
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
 
   
 

  

 


ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
m t/ 11.1
00.2
KN.m 5.55 m.t 0.555

53. 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 
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
W
2
L
2
2
/8
W
1
L
1
2
/8
or
W
1
L
1
2
/10
W
2
L
2
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
Mor M of larger the e tak
:3 Method
right left


54. 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
826.0J 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
  
    
    

  

 


ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

55. 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
826.0J 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
  
  
    
    

  

  
 


ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

56. 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
826.0J 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
  
    
    

  

 


ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

57. m / 8 5 cm 2.133
2400 x 10.5 x 826.0
10 * 0.444
A
M.S. 240/350 mm 8 use
826.0J 88.7 C1

250x100
10 * 0.444
C1 10.5
cm 10.5 2.5 - t d
3 SEC.
2
5
s
5
s
   
  

 


ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

58. 6.00 m 2.00 m
12 cm
/
/ m 8 Ф2.5
/
/ m 8 Ф5
/
/ m 8 Ф5
4.50 m 4.50 m 1.50 m
/
/ m 8 Ф5
/
/ m 10 Ф3
/
/ m 10 Ф3
/
/ m 8 Ф5
/
/ m 10 Ф2.5
/
/ m 10 Ф3
/
/ m 8 Ф3
/
/ m 8 Ф5
/
/ m 10 Ф2.5
/
/ m 8 Ф5
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy
12
12

59. 1-one way
1
t
2
t
1
s
1
s
tc
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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

60. 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)
eB
2- Two way
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

61. ..... .....
) length inclined the is L : (where
* m
L* m

L
S
 




 
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.
Mcosθ

J & get
b.
Mcosθ
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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

62. 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 50.100.5
00.5
00.2
WαWβ
II
I
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

63. 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
 
 
    
   
   

 

ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

64. 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
00.550.1 50.1
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

65. 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
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

66. PLAN OF
RFT. DETAILING
12
5.00
5.00
1.50
1.50
58/m`
58/m`
58/m`
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

67. B.M.D
Typical Rft. Detailing for Some Slabs
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

68. B.M.D
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

69. B.M.D
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

70. B.M.D
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
Dr. Eng. ALaa Ali Bashandy

71. ALB
Dr. Eng. ALaa Ali Bashandy
ةحلسملا ةيناسرخلا ةتمصملا تاطلابلا ميمصت
Design of Reinforced Concrete Solid Slabs
عجارملا :
.1 دوكلا ىرصملا ةناسرخلل ةحلسملا ECP 203-2007 - دھعم ثوحب ءانبلا – ةرازو ناكسلاا – ج.م.ع.
.2 قحلملا لولاا دوكلل ىرصملا ةناسرخلل ” تادعاسم ميمصتلا “ - دوك 203 ةنسل 2007 .
.3 قحلملا ىناثلا دوكلل ىرصملا ةناسرخلل ” ليلد ليصافتلا ةيئاشنلاا “ - دوك 203 ةنسل 2007 .
.4 Design of R. C. Structures – -Vol. 1, 2, 3 أ.د
. روھشم مينغ – أ.د. دومحم ىمليھملا - ةيلك ةسدنھلا –
ةعماج ةرھاقلا .
.5 Design of Halls – د. دمحم للاھ .