Ribbed Slab Design

(1)

One-Way

Joist Floor System

l : 1 2 S l o p e , t y p e Wiclth veries 100 mm

o. rarger

. The type of slab is also called a ribbed slab.

r lt consists of a floor slab, usually 50-100 mm thick, supported by reinforced concrete ribs.

c The ribs are usually lapered and uniformly spaced at distances that do not exceed 750 mm.

J ln some fibbed slabs, the space between ribs may be filled with Dermanent fillers to orovide a horizontal slab soffit.

Top of Shb

One-Way

Joist Floor System

Advantages:

.

Longer

spans

with heavy

loads

r

Reduced

dead load due to voids

.

Electrical,

mechanical

etc. can be placed

between

voids

.

Good

vibration

resistance

(2)

. ACI Reouirements

for Joist Construction

(Sec.

8.11,

ACI 318-02)

o Slabs

and ribs

must

be cast

monolithicallv.

o Ribs must be spaced

consistently

o Ribs

may not be less

than 100 mm in width

o Depth

of ribs

may not be more

than

3.5 times

the minimum

rib width

o Clear

spacing

between

ribs

shall

notexceed

750mm.

.. Ribbed slabs not meeting these requirements

are designed

as slabs and beams.

'*

r Slab Thickness

o ( A C l

S e c . 8 . 1 1 . 6 . 1 )

t >50 mm or > (1/12

distance

between

ribs)

o Building

codes

give

minimum

fire resistance

rating:

. 1-hourfire

rating:

20mm cover,75-90

mm slab

thickness

. 2-hour

fire rating:

25mm cover,

1 15 mm slab

thickness

(3)

Pan Joist Floor Systems

o Slab

thickness

. Governed

by strength,

fire rating,

available

space

o Overall

depth and fib thickness

I Governed

by deflections

and shear

Pan Joist Floor Systems

r Lavinq

Out Pan Joist Floors

(conf.)

Typically

no stirrups

are used in joists

Reducing

Forming

Costs:

r Use constant

joist depth for entire floor

r Use same depth forjoists and beams (not always

possible)

(4)

Pan Joist Floor Systems

r Distribution

Ribs

Placed

perpendicular

to joists.

S p a n s < 6 . 0 m : N o n e

Spans

6.0 -9.0 m : Provided

a mi

o Soans

> 9.0 m : Provided

at third-ooints

o At least

one continuous

12-mm

diameter

bar is

orovided

at too and bottom

of distribution

rib.

.Note: not required by ACI Code, but typically used in

construction

. ACI provides

minimum

member

depth

and slab

thickness

requirements

that

can be used

without

a deflection

calculation

(Sec.

9.5 ACI 318)

(5)

Member

Depth

ACI 318-02:

Table

9.5a

9.5-2 - on€-w.y co.skucrlon (nonpr6l.€$.d) 9.5.2.1 - Minihlm thickless stipolar.d in Tads s 5la) shall apply lor on€ way @nslruction nol sup-poning or atlachod lo parlitims or other bnelruclion lik€ly to b€ dE@g€d by b€e d€ll€cfions, unless @m-pulalion ol deileclion ind €leB a lesse. lhidoe6s @n b€ !s6d wirhour adve60 6fi&ls.

IAAL€ 9.5(EFlllNlMUM THICKNESS OF NONPRE.

SIFESSEO BEAMS OF ONE-WAY SLAAS UNIESS

OSFLECT|oNS ARE CONPUIEO

ftr ffiruc16n r*.ry 6 b 4tuld bl ralo.

Design Steps

;t 3*

(6)

l) Flat Slab reinforcement is calculated for bending or

minimum reinforcement

for shrinkage

and

temperature.

( A C l

S e c 7 . 1 2 . 2

_{) G R 4 0 o r G R 5 0}

0 0 0 2 0 A q

GR 60 0.0018 a.

Jolst Doslgn (continued)

2) Shear Design of Joist Ribs

a) Af fowable

V"= 1.10'V"

t-r l -1 1A*1!!h ) o

b) Shear strength may be increased using shear

reinforcement

or by widening the ends of the ribs

(not typical)

_.\

. . { . . ' . -' . : , L - j ;a.ris

(7)

Jolst Deslgn

3) ACI shear and moment coefficients may bs

usod if reouirements in ACI Sec 8.3.3 are met.

4) Ribs are designed as T- Sections. Main

positive reinforcament includes at lsast 2 bars.

Example

(8)

Design the simply-supporte.t _{stab shown betow. The slab is Dart of} a floor in a typical residential building.

Use f'c = 28 MPa,fv= 420 MPa

0.20 n'

1. The effective span length = Lcr.,,+d or Lcre.!+s = 4 . 5 + 0 , 2 = 1 . 7 m 2. Live load :

From minimum design loads for structures LL=2.0 kN/m, 3. Required depth of the ribbed stab by ACt Table 9.Sa

hnh= L/16 = 4700/'16 = 294 mm

R i b b e d S l a b D e s i g n

5- Suggested Dimensions

E x a

m p l e

6. Dead Load tor r.o-m strip

Flat slab weight = 0.06'1.0t24.0= 1.44 kN/m

Ribs weight = (0.12+0.16)/2.0.24-24.0/0.52 _{= t.S5 kN/m} Own weight of blocks = 5.0-0.18/0.52 = 1.73 kN/m

(9)

R i b b e d S l a b D e s i g n E x a m p l e

6. Dead Load for 1.0-m stJip {continue}

Own weight ofthe fill = 0,1.'1.0.'12.9 = 1.29 kN/m'z Own weight of the mortar = 0-025'1.0'22.0 = 0.55 kN/m' Own weight ot the tilss = 0.025'1.0"24.0 = 0.60 kN/m'? Own weight ot plastering = 0.025'1,0*22,0 = 0.55 kN/m'? Own weight of partitions = 1.0 kN/m'z

Total Dead Load = 8.71 kN/m2

Ribbed Slab Design Example

7. Total ultimate load = 1.2DL+{.6 LL

= L2(8.7,11 + ,t .6(2.01 = | 3.65 kN/m, 8. Total ultimate load on one rib = 13.65.0.52 = 7.1 kN/m t. ,t,,a"," roagnt = w,L2l8

= 7.1'4.778 = 19.6 kN.m R M" 19 600') =0.511MPa

" dbd' 0.9(520)t270')

(10)

Ribbed Slab Design Example

A, = 0,001385 (520X270) . 19,1 nnz/ rib A.'i' = 0-co3:t(t20x270) - 107 nt|l'/ rib A,r,aaa = 2tDl2 mn I db Ch.ck N.A In tho f,ftg. .-225'420r(0.85'28'520). 7.7 nn ok 9. Shoar De3ign Ultimaierho on one db V. = t.'1'11,7124.r".27F t,1.06 kN

o r . = , . r * o . r y J 4 r t 2 0 ' 2 1 0

= 23.s7

w -+orr

Ribbed

Slab Design

Example

10. Upper

Plate

Do8ign

w, = t3.65 kN/m

ConEa tattvely, dslgn tha uppet plata ts a slmpty auppo.ted st b nu = wJ-.n = 13.65.0.4R . 0.273 kr|m

M i i 1 ! l n 6 \

R - -::+ = _{" dbd. 0.90000)(40.)}" - 1 " ' : = 0 . 1 9 . . t / P a

o _o.lsf:

( | _,f L& l=o.ooo1

4 | li 0.85r./

A. r 0.00023 _{1000X.0}= 9.2 mm, A.'r,,. o.oo18(1000x60)

- t08 mm?

(11)

Ribbed Slab Design Example

Slmpllfled Bar Cutott Polnt8

l0l rrfi.. rh.l.ro.r ot 4, rnd 4,,