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Hollowblock slabs guide...

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RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

Hollow Block Slabs

Hollow Block Slabs (definitions) 

Hollow blocks are used to fill portions of the slab thickness; this results in deeper arm for the reinforcement while saving the amount of concrete and hence the own weight of the slab.

 The

2

reinforcement is located between the blocks inside the ribs.  Blocks may be concrete blocks or styro-foam

1

Monday, October 15, 2012 RC Design 2 - Dr. Hany Nessim

Hollow Block Slabs (definitions) 

When the ribs are in one direction then it is a one-way hollow block slab, regardless of the rectangularity; r.



When ribs are in both directions then it is a two-way hollow block slab.

Advantages & Disadvantages 

Advantages 

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

3

  



Reducing slab weight by reducing amount of concrete below neutral axis. Ease of construction, especially when all beams are hidden beams. Economic for spans > 5m with moderate live load: hospitals, office and residential buildings. Improved insulation for sound and heat.

Disadvantages   

Not economic for small spans Not suitable for heavy loads or dynamic loads. Difficult to repair or strengthen

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2

Monday, October 15, 2012

One-Way Hollow Block Slabs ts

RC Design 2 - Dr. Hany Nessim

t

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

5

b

e

Support condition

Simply supported

Continuous one end

Continuous both ends

cantilever

Clear span (Ln/t)

20

25

28

8



Ribs are in one direction only



For spans less than 10m or cantilevers less than 2m long.



St. 400/600; for other grades divide values by  0 . 4 +



f



650

y

  

Cross ribs 

Cross ribs are used when the live load ≥ 3kN/m2 or Span > 5m.

Span

≤ 5m

>5m

4m – 7m

Live Load

≤ 3kN/m2 ≤ 3kN/m2 > 3kN/m2

>3kN/m2

Cross Sibs

No

Three cross ribs

One cross One cross rib rib

>7m

6

3

Monday, October 15, 2012

Minimum Dimensions

RC Design 2 - Dr. Hany Nessim

ts t

b

e

e ≤ 700mm  b = t/3 ≥ 100mm  ts = e/10 ≥ 50mm 

Monday, October 15, 2012

7

Loading

RC Design 2 - Dr. Hany Nessim

ts t

b

e

Own weight including blocks’ weight.  Flooring  Live Load 

8

4



Own Weight including blocks’ weight (kN/m2) ts

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

Loading

t

b

e

Concrete Blocks Concrete Blocks Concrete Blocks 400x200x150 400x200x200 400x200x250

Foam Blocks 500x400x200

One way

Two way

One way

Two way

One way

Two way

One way

Two way

3.03

3.36

3.30

3.80

4.10

4.78

0.70

1.20

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

9

Load Distribution Factors for Two-way 

When LL ≤ 5 kN/m2 or when the compression flange is incomplete.

r

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

α

0.396

0.473

0.543

0.606

0.660

0.706

0.746

0.778

0.806

0.830

0.849

β

0.396

0.333

0.262

0.212

0.172

0.140

0.113

0.093

0.077

0.063

0.053

10

5

Monday, October 15, 2012 RC Design 2 - Dr. Hany Nessim

Load Distribution Factors for Two-way 

When LL > 5 kN/m2

r

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

α

0.500

0.595

0.672

0.742

0.797

0.834

0.867

0.893

0.914

0.928

0.941

β

0.500

0.405

0.328

0.258

0.203

0.166

0.133

0.107

0.086

0.072

0.059

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

11

Calculate Bending Moments 

Two spans When LL ≤ DL, and the difference between spans is not more than 20%; then the bending moment could be calculated as following.

w

W

= wu for limit state design method. & w = w for allowable stresses design method.

12

6

W

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

Calculate Bending Moments



More than two spans  

Spans differences are not more than 20% LL ≤ DL

Using Design Aids Charts (flexure)

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

13

14

7

Shearing stresses should be less than the concrete shear capacity and, hence, minimum shear reinforcement could be used.

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

Design for Shear

Shearing stress; q u =

Qu b.t

Concrete shear capacity; q cu = 0.16

f cu

γc

Minimum shear reinforcement; A s, min =

≥ q u ( N / mm 2 )

0.4  q u b.s fy  q cu

  ( mm 2 ) 

RC Design 2 - Dr. Hany Nessim

Monday, October 15, 2012

15

Minimum Reinforcement

Slab reinf.

Stirrups Rib reinf.

5φ6mm/m’ perpendicular to ribs.  4φ6mm/m’ parallel to ribs.

In top slab

Bottom reinforcement = rib main reinforcement.  Top reinforcement = ½ bottom reinforcement

In Cross rib





16

8

Monday, October 15, 2012 RC Design 2 - Dr. Hany Nessim

Minimum Reinforcement

Slab reinf.

Stirrups Rib reinf.

A s, min

= min.

A s, min

  ≥   

  0.225    1.3 A s

0.25 100 0.15 100

f cu bd fy



bd;

for

mild

bd;

for

deformed

1.1   fy   

steel steel

In Main rib     

(as beams)

17

Example

9

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