10. Reinforced Concrete Flat Slabs • Introduction • Reinforced concrete flat slabs • simplified method, in seven steps
• Idealized frame method • comments only
• Design implications and practices © Universit University y of Western Australia Australia School of Civil and Resource Engineering Engineering 2004
INTRODUCTION
What is a ‘flat slab’? Compare and contrast:
Two-way slab system, with edge-supported panels - NOT a flat slab Two-way slab system, with no edge supports to interior panels - IS a flat slab
So definition of a flat slab? . . .
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. AS3600 cl. 7.1.1 : (c) RC two-way multiple spans (flat slabs).
Simplified method
(d) RC and PC frames with two-way slab systems.
Idealized frame method cl. 7.5
Suitable generally, but particularly for un-braced frames, in which sway action imposes moment actions in slab to column connections.
cl. 7.4
This lecture Comments only
We’ll focus our attention on this method.
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? . . .
Example: 3 spans each way, shown without drop panels, i.e flat plate B
A
A
B N O I T C E S
PLAN
B
SECTION A
. . . and with drop panels . . .
Example: 3 spans each way, with drop panels B
A
A
B N O I T C E S
PLAN
B
SECTION A
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:
- ive yield line
- ive yield line + ive yield line
-ive yield lines
+ 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 Localised circumferential and radial topside cracking at column heads
All this assumes that punching shear fracture has not occurred.
What about deflections at working load? . . . .
Deflected shape at working load of internal panel of a flat slab :
Severe curvature at columns (umbrella-like), and moderate deflection at mid-span
SECTION AT COLUMNS
SECTION MID-WAY BETWEEN COLUMNS
Much smaller curvature, but greater deflection at mid-span
Punching Shear? . . .
Punching Shear Problem (shown for square column):
d om
Flat Plate : no capital or drop panel. c fracture surface critical perimeter u = 4 (c + dom) Vuo is function of u AND d om Also V u involves M*v
Now Vu must exceed V* where V* is load from floor slab If not, what can we do? 1. Increase column dimensions, or 2. Provide a capital to the column, or 3. Provide a drop panel, etc.
What are these? . . .
Flat Slab with Capital. d om
fracture surface
Flat Slab with Drop Panel. d om
capital
critical perimeter, much larger than for flat plate of same slab thickness and column.
Increased strength comes from increased critical perimeter.
2 possible fracture surfaces
drop panel
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 . . .
Step 1. Slab Thickness and Minimum Steel.
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 2. Calculate Fd and Lo each way. 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 L o is estimated thus:
as is the length of span support this depends on the geometry of the support.
Step 3. . .
Step 3. Assess Need for Column Head. 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. 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.
V* = Fd x (contributory area of slab)
Vuo = u dom f cv 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 4. Allocate design strips, and calculate Mo A design strip is a uniform width of slab, extending from one extreme to the other, which carries the loading virtually as a oneway slab, except for some transfer of bending moment to the supporting columns.
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 E-W Interior Design Strip
Typical N-S 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 effective span length of subject span m. width of design strip m. ultimate load on slab kPa
. . . from which the positive and negative moments are estimated . . .
Step 5. . .
Step 5. Apportion positive and negative moments to each span.
0.65 Mo
0.65 Mo
Mo 0.35 Mo
Interior Span Table 7.4.3(B)
MNE Exterior Moment Exterior Span
MNI Interior Moment
Mo MM Lo
Moments depend on restraint offered by support, eg simple support, integral column, column and edge beam, or ful restraint. Table 7.4.3(A)
These moments are not evenly distributed across the strip. Step 6 . . .
Step 6. Apportion moments to column strips and middle strips, and design rebar for each. Width of design strip Edge of column
Bending moment is not distributed uniformly across a design strip : Negative moment is highly concentrated within the critical perimeter at column.
Negative moment at column Half Middle Strip
Edge of critical perimeter Half Column Strip Middle Strip
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.
Moment per unit width
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 . . .
top rebar
bottom rebar
Column and Middle Strips and the same in E-W direction.
Let’s review what we have just done . . .
Review of Steps 4, 5 and 6: EXAMPLE: 3-SPAN FLAT SLAB, WITH DROP PANELS
Difference in moments assigned to the column
BENDING MMTS IN DESIGN STRIP
Design rebar for maximum moment at the column
Bending mmts in Column Strip Bending mmts in Half-Middle Strip
Finally, punching shear
Step 7. . .
Step 7. Check Punching Shear Vuo = u dom f cv
critical surface
f cv = 0.17(1+2/ h) (f’c)0.5 <= 0.34 (f’c)0.5
dom = mean effective depth
h=
Y/X: c1 X
u = 2 (c1 + c2) + 2 dom
u=
(c + dom)
Y
c2
c2 c1 RECTANGULAR COLUMN
c CIRCULAR COLUMN
BUT there is always some moment to be transferred to column, so V u < V uo ! . . .
Dealing with the out-of-balance moment:
M’
x - direction M
M M’
Out of balance moment Mv = M - M’
BMD x - direction NOTE: There is an absolute minimum value of Mv prescribed in AS3600:
Mv.min = 0.06 [(1.25g + 0.75q) L t Lo2 - 1.25g Lt L’o2] where L’o is minimum value of L o for adjacent spans.
R V2 V1
T1 dom /2
c2 c1
V = 2 V1 + 2 V2
M’2
M2 V2
T1
The vertical force V is transferred by shear along the surfaces of the critical perimeter, so that
V1
R+V
Critical shear stresses occur here, where stresses due to vertical shear and torsional shear are additive.
The out-of-balance moment Mv is transferred partly by moments M2 and M’2, and partly by torsional moments T1 Vu is diminished to Vu =
Vuo M*v u 1+ 8 V* dom a
IDEALIZED FRAME METHOD Applicable to an unbraced frame, eg a 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.
