7_Slabs

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CE 432 1st Semester 14/15

Dr. Nadim Shbeeb

Slabs

One Way Slabs


Dr. Nadim Shbeeb

Introduction• Reinforced concrete building construction

commonly has floor slabs, beams, girders and

columns continuously placed to form a

monolithic system.

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Introduction• B1,B2, and B3 are supported on the Girders

• B4,B5, and B6 are supported on the Columns

• G1,G2, and G3 are supported on the Columns


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Types of Slabs• One-Way Slabs

– Solid Slabs

– Ribbed Slabs (Ribbed Joist floor)

2SpanShort

Span Long>

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Types of Slabs

• Two-Way Slabs

– Solid Slabs

– Ribbed Slabs (Ribbed Joist floor)


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Types of Slabs

• Flat Slabs

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Example S1

• A one-way single span reinforced concrete

slab has a simple span of 5.0 m. The service

live load is 15 kN/m2, the dead load is 10

kN/m2 (including self weight) and 28 MPa

concrete is specified for use with steel with a

yield stress equal to 420 MPa. Design the slab,

following the provisions of the ACI code.


Dr. Nadim Shbeeb

Example S1• First determine the thickness of the slab:

– From Table 9.5a of the ACI code

– Assume a 1.0 m thickness strip (b=1.0 m)

m 250.020

5

20===

lh

mm 217.5d and mm 250h use thus250 mm 88.163

28

420*0123857.0588.011000*420*0123857.0*9.010*112

0123857.06.0

m-kN 5.1128

kN/m 3615*6.110*2.1

26

max

2

==<=⇒

−=

=

==

=+=

d

d

lwM

w

uu

u

ρ

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Dr. Nadim Shbeeb

Example S12mm 89.26935.217*1000*0123857.0 ==sA

Use 5φ25

m-kN 70.181

0113.01000*186

51.3694

90.00980.096.50

96.505.217003.0

mm 50.96cmm 31.431000*28*85.0

420*37.2454a

mm 5.2175.1220250

min

=

>==

=⇒=

−=

=⇒==

=−−=

u

t

M

d

ρρ

φε

Using ρ formula As=1451.88 mm2 3 φ25

Mu=113.84 kN-m


Dr. Nadim Shbeeb

Example S2

• A reinforced concrete slab is built integrally

with its supports and consists of two equal

spans, each with clear span of 4.5 m. The

service live load is 5 kN/m2, and 28 MPa

concrete (Density =23.5 kN/m3) is specified for

use with steel with a yield stress equal to 420

MPa. Design the slab, following the provisions

of the ACI code.

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Dr. Nadim Shbeeb

Example S2

• First determine the thickness of the slab:

– From Table 9.5a of the ACI code

– Use h=165 mm, thus

• Calculate the moments using the ACI coefficients listed in section 8.3.3 of the ACI code

m 161.028

5.4

28===

lh

2

2

kN/m 656.125*6.188.3*2.1

kN/m 88.3165.*5.23

=+=

==

uw

DL


Dr. Nadim Shbeeb

Example S28.3.3 — As an alternate to frame analysis, the following approximate

moments and shears shall be permitted for design of continuous

beams and one-way slabs (slabs reinforced to resist flexural stresses

in only one direction), provided (a) through (e) are satisfied:

(a) There are two or more spans;

(b) Spans are approximately equal, with the larger of two

adjacent spans not greater than the shorter by more than 20

percent;

(c) Loads are uniformly distributed;

(d) Un-factored live load, L, does not exceed three times un-

factored dead load, D; and

(e) Members are prismatic.

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Example S2

kN.m 679.105.4*656.12*24

1

24 :supportexterior At

kN.m 306.185.4*656.12*14

1

14 :midspanAt

kN.m 476.285.4*656.12*9

1

9 :supportinterior At

22

22

22

===−

===+

===−

n

n

n

wlM

wlM

wlM

24

1−

14

1+

9

1−


Dr. Nadim Shbeeb

Example S2

mm 115d thusmm 165h Use

mm 40.70

)59.01(

0181.0

'

_

0.005

==

=

−

=

=

c

y

y

Interioru

f

fbf

Md

ρφρ

ρ

mEach 145 Use

mm 733.693

)36.2(85.0

85.0

2

'2''

φ

ρ

φ

φφρ

==⇒

−−=

bdA

df

Mfbd

b

f

f

f

s

y

ucc

y

c

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Dr. Nadim Shbeeb

Example S2

m-kN 44.29

0071.01000*108

69.769

90.0173.098.15

98.15108003.0

mm 98.151000*28*85.0

420*69.769

85.0

1c

mm 108750165

_

min

=

>==

=⇒=

−=

==

=−−=

Interioru

t

M

d

ρρ

φε

Similarly

m)per 143 (Use mm 360

m)per 143 (Use mm 437

2_

2_

φ

φ

=

=

Exteroirs

midspans

A

A


Dr. Nadim Shbeeb

Example S2

Shrinkage bars=0.0018bh=297 mm2 use 1φ20

c

unu

u

V

dwlw

V

φ2

1kN 38.31108.*656.12

2

5.4*656.12*15.1

215.1

<=−=

−=

Shear Consideration

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Reinforcement Details

L2L1

L1/5 L1/3 L2/3 L2/3

L2L1

L1/5 L1/3 L2/3 L2/3

L1/7 L1/5 L2/4 L2/4

Straight bars

Bent bars


Dr. Nadim Shbeeb

Ribbed Slab

40 cm

36 cm

2 cm

14, 18 ,24 , 32 cm

7 cm

9 cm

3.5 cm

2.5 cm

12-15 cm20 cm 20 cm

Main

Reinforcement

Shrinkage and

Temperature

ب�ط

مونه

حصمه

قصاره

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Ribbed Slab

Design as T-beam

52 cm

12 cm

7 cm

14,18, 24,32 cm

Review ACI provisions 8.13


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3.0 m

6.5 m

0.5 m

0.5 m

0.5 m

0.5 m

Design the Ribbed slab below according

to the ACI code provisions fy=420 MPa and fc’=21 MPa

B1

B3

B2

B4

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Dr. Nadim Shbeeb

Ribbed Slab

15 cm20 cm 20 cm

2 cm

18cm

7 cm

9 cm

3.5 cm

2.5 cmب�ط

مونه

حصمه

قصاره

cm 75.1816

3

16===

lhTo estimate bw assume:

Cover =25 mm,

Stirrups φ8,

Main reinforcement 2 φ20,

distance between reinforcement=30mm mmmm

bw

150 Use136

30)20825(*2

=

+++=


Dr. Nadim Shbeeb

Ribbed Slab

15 cm20 cm 20 cm

2 cm

18cm

7 cm

9 cm

3.5 cm

2.5 cmب�ط

مونه

حصمه

قصاره

2kN/m 34.10.55

0.73kN/m 73.024*18.0*)19.015.0(

2

1===+

To estimate DL:

Rib :

ب�ط : 2kN/m 625.025*025.0 =

:مونه 2kN/m 77.022*035.0 =

حصمه :2kN/m 8.120*09.0 =

قصاره : 2kN/m 44.022*020.0 =

Block:2kN/m 36115

20

1

550

1.*.

.*

.=

hf:2kN/m 68124070 .*. =

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Dr. Nadim Shbeeb

Ribbed Slab

2kN/m 64.150.2*6.137.10*2.1 =+=⇒ uw

DLSlab=1.8+.625+.77+.44+1.36+1.68=8.02 kN/m2

Assume Partitions DL=2.35 kN/m2

DLTotal=10.37 kN/m2

LL=2.0 kN/m2

Thus load per rib is

kN/m 60864.15550 .*.wu ==⇒


Dr. Nadim Shbeeb

Ribbed Slab

kN.m 17.138

5.3*81.8

8

22

===lw

M uu

Design as T-beam

26

mm 563.193

)2

70215(*420*9.0

10*17.13=

−

=sA

Assume stress block depth equal to flange thickness

fha <=××

×=

>==

mm 28.85502185.0

420563.193

006.0215*150

563.193minρρ

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Dr. Nadim Shbeeb

Ribbed SlabUse 2φ14

u

t

1

M kN.m 26.24

)2

17.13215(42088.3079.0

00955.0215*150

88.307

005.00386.049.15

49.15215003.0

mm 49.1585.0

17.13ac

mm 17.13550*21*85.0

420*88.307

>=

−×××=

==

>=−

=

===

==

u

w

M

a

ρ

ε

β


Dr. Nadim Shbeeb

Ribbed Slab

2 cm

18 cm

7 cm

9 cm

3.5 cm

2.5 cm

15 cm20 cm 20 cm

2φ14 1φ8

ب�ط

مونه

حصمه

قصاره

Design for Shear as before

Design B3 & B4 using wu=15.63*1.5=23.4 kN/m

Design Columns with axial Compression load of

23.4*7/2=81.9 kN/m

7_Slabs

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