© University of Western AustraliaSchool of Civil and Resource Engineering 2004

10. Reinforced ConcreteFlat Slabs

• Introduction

• Reinforced concrete flat slabs • simplified method, in seven steps

• Idealized frame method• comments only

• Design implications and practices

What is a ‘flat slab’? Compare and contrast:

INTRODUCTION

Two-way slab system, with edge-supported panels - NOT a flat slabTwo-way slab system, with no edge supports to interior panels - IS a flat slab

So definition of a flat slab? . . .

(c) RC two-way multiple spans(flat slabs).

Simplified method cl. 7.4 This lecture

(d) RC and PC frames withtwo-way slab systems.

Idealized frame methodcl. 7.5

Comments only

Flat slab = a continuous two-way solid or ribbed slab, with or without drop-panels, having at least two spans in each direction, supported internally by columns without beams and supported externally by walls, or columns with or without spandrel beams, or both. AS3600 - 2001 cl. 1.6.3.21.

Flat Plate = particular type of flat slab, without either drop panels or capitals.

Suitable generally, but particularly for un-braced frames, in which sway action imposes moment actions in slab to column connections.

We’ll focus our attention on this method.

AS3600 cl. 7.1.1 :

Why use flat slabs? . . .

If supporting walls and/or supporting beams are not used, the slab must both carry its load, and transfer the loading actions to the supporting columns. There are advantages in this system, but also some concerns which must be addressed:

Advantages :1. Downward beam protrusion is eliminated, reducing ceiling

congestion, and probably reducing floor-to-floor heights.

2. Simplified formwork and construction generally.

Concerns :

1. Thicker slab is needed, hence . .

2. Heavier overall structure.

3. Serious attention required to deflection control.

4. Very serious attention required to punching shear problem at slab to column connections.

What does it look like in Plan and Section? . . .

SEC

TIO

N B

PLAN

SECTION A

A A

B

B

Example: 3 spans each way, shown without drop panels, i.e flat plate

. . . and with drop panels . . .

SEC

TIO

N B

PLAN

SECTION A

A A

B

B

Example: 3 spans each way, with drop panels

Firstly, consider bending failure . . .

Failure could be in either an E-W span or a N-S span. Potential bending failure, in an E-W span:

- iveyield line

- iveyield line

-ive yield lines

+ iveyield line

+ ive yield line

. . . or in a N-S span :

-ive yield line

-ive yield line

+ive yield line

So putting these together . . .

So failure can occur either E-W or N-S, with yield lines thus:

Negative (topside crack) yield lines -on column centre lines

Positive (underside crack) yield lines -at mid-spans

Localisedcircumferential and radial topside cracking at column heads

All this assumes that punching shear fracture has not occurred.

What about deflections at working load? . . . .

SECTION AT COLUMNS

Severe curvature at columns (umbrella-like), and moderate deflection at mid-span

SECTION MID-WAY BETWEEN COLUMNS

Deflected shape at working load of internal panel of a flat slab :

Much smaller curvature, but greater deflection at mid-span

Punching Shear? . . .

Punching Shear Problem (shown for square column):

Flat Plate : no capital or drop panel.

fracture surface

d om

critical perimeter u= 4 (c + dom)

Vuo is function of u AND dom

Also Vu involves M*v

Now Vu must exceed V* where V* is load from floor slabIf not, what can we do?1. Increase column dimensions, or2. Provide a capital to the column, or3. Provide a drop panel, etc.

What are these? . . .

c

Flat Slab with Capital.

fracture surface

d om

critical perimeter, much larger than for flat plate of same slab thickness and column.

Increased strength comes from increased critical perimeter.

Flat Slab with Drop Panel.

d om

capital drop panel

2 possible fracture surfaces

Increased strength comes from increased dom and critical perimeter for inner, and increased critical perimeter for outer.

So now for the design of a reinforced concrete flat slab . . .

REINFORCED FLAT SLAB BY SIMPLIFIED METHOD

Seven easy (?) steps:

1. Slab Thickness and Minimum Steel.2. Calculate Fd and Lo each way.3. Assess need for column head.4. Allocate design strips and calculate Mo.5. Apportion negative and positive moments to each span.6. Apportion moments to column strips and middle strips,

and design rebar for each.7. Check punching shear.

We’ll proceed step by step . . .

Ds : Recommend Prof. Rangan’s method of approximating slab thickness, but suggest use w = g + q in the formulas.

Minimum steel Ast.min :

Lower ductility 0.0025 b d

Crack control minor, moderate or major, per Code

Max. bar spacing {2.0 Ds; 300 mm}min

Thickness must be later checked for deflection (see later lecture) and for punching shear strength (Steps 3 and 7).

Step 2. . . .

Step 1. Slab Thickness and Minimum Steel.

Fd means w*, i.e. 1.2 g + 1.5 q, in kPa.

Lo is effective span, and this depends on the stiffness of the column/slab joint. So Lo is assessed by estimating the ‘span support’ as of each end of each span of the slab, from which Lo is estimated thus:

as is the length of span support -this depends on the geometry of the support.

Step 3. . .

Step 2. Calculate Fd and Lo each way.

This requires calculating the total vertical load to be transferred from the slab to the column at this floor level, and checking whether the column head zone must be strengthened by drop panel, or capital, etc.

V* = Fd x (contributory area of slab)

Approximate methods must be used for estimating the bending to be transferred to the column - this moment diminishes the punching shear capacity, as discussed later under Step 7.

Vuo = u dom fcv

Suggest:

Vu = 0.85 Vuo

Remember that this decision remains tentative until checked at Step 7, at which time some adjustment may be required.

Step 4. . .

Step 3. Assess Need for Column Head.

A design strip is a uniform width of slab, extending from one extreme to the other, which carries the loading virtually as a one-way slab, except for some transfer of bending moment to the supporting columns.

Step 4. Allocate design strips, and calculate Mo

Generally, the width of a design strip comprises the sum of one-half of the column spacing either side of the subject column.

Design strips may be

• Edge design strips, or

• Interior design strips.

Each design strip is designed to resist the positive and negative bending moments much like a one-way slab.

Design strips look like this . . .

Typical N-S Interior Design Strip

Typical E-W Interior Design Strip

Moments in design strips? . . .

The bending moments in a design strip, both positive and negative, are calculated by reference to a ‘total static moment’ Mo :

Mo = Fd Lt Lo2 / 8

ultimate load on slab kPa

width of design strip m.

effective span length of subject span m.

. . . from which the positive and negative moments are estimated . . .

Step 5. . .

Interior SpanTable 7.4.3(B)

Exterior Span Moments depend on restraint offered by support, eg simple support, integral column, column and edge beam, or ful restraint. Table 7.4.3(A)

ExteriorMoment

Interior Moment

These moments are not evenly distributed across the strip.

Step 5. Apportion positive and negative moments to each span.

Step 6 . . .

0.65 Mo Mo0.65 Mo

0.35 Mo

MNE

MNI

MMLo

Mo

Step 6. Apportion moments to column strips and middle strips, and design rebar for each.

Bending moment is not distributed uniformly across a design strip :

Negative moment at column

Width of design stripEdge of column

Edge of critical perimeter

Moment per unit widthNegative moment

is highly concentrated within the critical perimeter at column.

Positive moment is much more uniform, with max. at column centreline.

So the design strip moments are apportioned between Column Strips, and Half-middle Strips.

Half Middle Strip

Half Middle Strip

Column Strip

The actual moments sustained depends somewhat on the areas of steel allocated. So some flexibility is available in these distributions, as shown by Table 7.5.5:

DISTRIBUTION OF BENDING MOMENTS TO THE COLUMN STRIP

Bending moment under consideration Column strip moment factor

Negative moment at an interior support 0.60 to 1.00

Negative moment at an exterior support 0.75 to 1.00

Positive moment in all spans 0.50 to 0.70

The residual moments are then assigned to the half-middle strips which comprise the rest of the design strip width.

Regardless of the above, ensure that 25% of the negative steel in cloumn and edge strips is concentrated in the column head region.

. . which all leads to . . .

Column and Middle Strips and the same in E-W direction.

top rebar

bottomrebar

Let’s review what we have just done . . .

EXAMPLE: 3-SPAN FLAT SLAB, WITH DROP PANELS

BENDING MMTS IN DESIGN STRIP

Bending mmts in Column Strip

Bending mmts in Half-Middle Strip

Difference in moments assigned to the column

Design rebar for maximum moment at the column

Review of Steps 4, 5 and 6:

Finally, punching shear Step 7. . .

dom = mean effective depth

critical surfaceVuo = u dom fcv

RECTANGULAR COLUMN

c1

c2

u = 2 (c1 + c2) + 2 dom

CIRCULAR COLUMN

c

u = (c + dom)

fcv = 0.17(1+2/h) (f’c)0.5

<= 0.34 (f’c)0.5

h = Y / X :

c1

c2Y

X

BUT there is always some moment to be transferred to column, so Vu < Vuo ! . . .

Step 7. Check Punching Shear

BMD x - direction

MM’

Out of balance moment

Mv = M - M’

Dealing with the out-of-balance moment:

x - direction

M

M’

NOTE: There is an absolute minimum value of Mv prescribed in AS3600:

Mv.min = 0.06 [(1.25g + 0.75q) Lt Lo2 - 1.25g Lt L’o

2]

where L’o is minimum value of Lo for adjacent spans.

dom/2

c1

c2

R

R + V

The vertical force V is transferred by shear along the surfaces of the critical perimeter, so that

V = 2 V1 + 2 V2

V1

V2

V1

V2

The out-of-balance moment Mv is transferred partly by moments M2 and M’2, and partly by torsionalmoments T1

M2M’2

T1

T1

Critical shear stresses occur here, where stresses due to vertical shear and torsionalshear are additive.

Vu is diminished to

Vu = Vuo

1 + M*v u

8 V* dom a

IDEALIZED FRAME METHOD

Applicable to an unbraced frame, ega low-rise building where bracing is not available.

Transverse loading eg wind, earthquake, creates bending and shear actions in frame members -these actions are additive to those from vertical floor loadings.

The question is: how do the ‘beams’sustain these actions?

If the ‘beams’ are part of a flat slab, then the design strip acts as the beam, and is designed accordingly.

Otherwise, T-beam or L-beam action occurs.

DESIGN IMPLICATIONS AND PRACTICES

• Care with deflections - thoroughly investigate the basis of an ‘allowable’ deflection before commitment.

• Locate openings remote from columns, since these areas are very highly stressed in bending and shear.

• Care with crack control, especially if slab is exposed to weather.

• Care with vibrations - flat slabs may not be suitable for sensitive laboratories, or factory floors.

• Span/depth ratios are likely to be in range 25 - 32.

• For commercial work, maximum spans are about 7 - 8 metres.

• For longer spans, consider prestressed flat slabs - subject of next lecture, when we’ll summarise both reinforced and prestressed flat slabs.