Design for RC Flat Slabs

Ir Prof Dr J Chiang

Outline

• Definitions• Punching shear problem• Code provisions: BS8110 & EC 2• Analysis procedures• Design preliminaries• Design procedures• Yield line method

Definitions

• In BS8110 cl 1.3.2.1: a slab with or without drops, supported generally without beams and columns with or without column heads.

• Next slide shows various types of RC flat slabs commonly designed and built.

Diagrams of types of flat slabs

Flat slab with drop panel and capital Flat plate (with no drops, etc)

c.f. conventional beam-slab design

Other possible flat slab arrangements

• Waffle slab• Band beam slab

Flat slab plan and elevation

• Typical layout and sectional view

As built RC flat plate

• Refer to photograph– Note absence of beams

What is punching shear failure?

• Punching shear is a type of two-way shear, often found in flat slabs or pad footings.

• Members which fail due to punching are often examined at regions near to the columns, with a pyramid-like diagonal crack occurring at the distance surrounding the column.

• This type of failure is catastrophic because it gives no warning or visible signs prior to occurrence of the failure.

Punching shear effect

• Refer to diagram below

Incidents of punching shear failure• A catastrophic structural failure happened in June 1995, where a five

storey Sampoong department store in South Korea collapsed due to punching shear failure. The disaster has killed 500 people, leaving almost 1000 others injured.

Code provisions for RC flat slabs - 1

• BS8110 cl 3.7.1 refers:• (1) Ratio of longer to shorter span < 2• (2) Design moments by equiv frame, simplified

or finite element methods• (3) Effective dimension of column head is

specified, where column head is shown in next few slides ahead…

• (4) Effective diameter of column is specified, again discussed in next few slides…

Code provisions for RC flat slabs - 2

• (5) Use of drop panels – influence moment distribution, i.e. smaller dimension of drop ≥ ⅓ smaller panel dimension

• (6) Panel thickness is generally controlled by deflection, where thickness > 125 mm

Arrangement for column head

• Where necessary a thickening (in width and depth) around the column support is allowed for to mitigate punching shear effect.

Column head dimensions• lh = effective dimension of column head

• lc = column dimension (along direction of lh )

• lho = actual dimension of column head

• lh = lesser of lc or lh max = lc + 2(dh – 40)

Considerations for punching shear locations in a flat slab

Analysis of loads on flat slab

• Sufficient to use ULS loads, 1.4G + 1.6Q on all spans.

• Two methods are specified in BS8110:– (1) Frame analysis method– (2) Simplified method

• Both methods require the division of strips on the panel along column and middle strips.

Column and middle strips for analysis – 1

• The floor plan is divided into column strips and middle strips

Column and middle strips for analysis – 2

• Red outline = column strips

• Blue outline = middle strips

Column and middle strips for analysis – 3

• Blue outline = middle strips

Frame analysis method – flat slab

• Table 3.12 (BS 8110) refers internal panels:

• Note: F = total design load on slab strip, between column supports l = full panel length in span direction

At first interior support Middle of interior span Interior supports

Moment -0.086 Fl +0.063 Fl -0.063 Fl

Shear 0.6F 0.5F

Simplified method

• Table 3.18 (BS8119) refers moment distribution:

• Some conditions apply:– Only one single load case, i.e. 1.4G + 1.6Q– At least three rows of panels approx of equal span– Moments prescribed in Table 3.12 be reduced by 0.15Fhc, where hc = effective

diameter of column head

Distribution between column and middle strip as percentage of total negative or positive moments

Column strip Middle strip

Negative 75% 25%

Positive 55% 45%

Punching shear perimeter

• As defined in BS8110

Perimeter – another interpretation

Shear forces & resistance – 1

• Design effective shear force, at punching shear perimeter around column:– where Vt = design shear transferred to column

• Checks for shear resistance (Table 5.1 BS8110):– At face of column:

– At perimeter 1.5d from column face:

teff VV 15.1

2max /58.0 mmNorf

du

Vcu

o

eff

ms

ceff

dbd

A

ud

V /

40010079.0

41

31

Shear forces & resistance – 2• Conditions applied and terms used:

– c = design concrete shear stress (i.e shear capacity)

– As = area of bottom tension steel reinforcement along mid-span of column strip

– m = partial safety factor for material = 1.25

– fcu = concrete grade for compressive strength (N/mm2)

• If is ok, then no shear reinforcement is required. m

cusc

eff f

dbd

A

ud

V /

25

40010079.0

31

41

31

1400

;3100

/40,/2525

223

1

dbd

Aand

mmNfbutmmNfiff

s

cucucu

c

Case for shear reinforcement• When then shear

reinforcement is required to resist punching shear.• The shear reinforcement located within the checked shear

perimeter is given by

• where,– = angle between shear reinforcement and plane of slab– u = length of outer perimeter of the zone– Asv = total area of shear reinforcement (in mm2)

mcus

ceff f

dbd

A

ud

V /

25

40010079.0

31

41

31

2/460,95.0

4.0sin

26.195.0

7.05sin

6.195.0

sin

mmNfandf

udA

caseeitherin

wheref

udA

orwheref

udA

yvyv

sv

cyv

csv

cyv

csv

Provision of shear reinforcement

Successive perimeters for punching shear check

• Plan of slab around a concentrated load showing successive perimeters for punching shear check, e.g.

dbaU

dbaU

dbaU

122

92

62

3

2

1

Typical arrangement of shear reinforcement

Distribution of shear reinforcement

• In three successive perimeters around loaded area (or column support)

Worked example 1

Worked example 1 – solution p1

Worked example 1 – solution p2

• Capacity of slab to take punching shear is ok

Worked example 2

Worked example 2 – solution p1

Worked example 2 – solution p2

Worked example 2 – solution p3

Worked example 2 – solution p4

Worked example 2 – solution p5

Worked example 2 – solution p6

• Plan view of punching shear steel reinforcement layout

Worked example 3 – 1

Worked example 3 – 2

Worked example 3 – 3

Worked example 3 – 4

Worked example 3 – Solution p1

Worked example 3 – Solution p2

Worked example 3 – Solution p3

Worked example 3 – Solution p4

Worked example 3 – Solution p5

Worked example 3 – Solution p6

Worked example 3 – Solution p7

Worked example 3 – Solution p8

Worked example 3 – Solution p9

Worked example 3 – Solution p10

Worked example 3 – Solution p11

Worked example 3 – Solution p12

Worked example 3 – Solution p13

Design flat slab to Eurocode 2

• Using Eurocode 2 for the analysis of flat slabs is similar to using BS 8110. The following methods may be used:– Equivalent frame method– Finite element analysis– Yield line analysis– Grillage analogy

• The Eurocode gives further advice on the equivalent frame method in Annex I and designers used to BS 8110 will find this very familiar.

Difference in EC 2 concrete grade

• The Eurocode gives recommendations for the design of concrete up to class C90/105.

• However, for concrete strength greater than class C50/60, the stress block is modified.

• It is important to note that concrete strength is based on the cylinder strength and not the cube strength (i.e. for class C28/35 the cylinder strength is 28 MPa, whereas the cube strength is 35 MPa).

BM coeffs for flat slabs in EC 2

End support/slab connection First interior support

Interior spans

Interior supportsPinned Continuous

End support

End span

End support

End span

Moment 0 0.086Fl -0.04Fl 0.075Fl -0.086Fl 0.063Fl -0.063Fl

Notes1 Applicable to slabs where the area of each bay exceeds 30 m2, Qk, ≤ 1.25 Gk and qk ≤ 5 kN/m2

2 F is the total design ultimate load, l is the effective span3 Minimum span > 0.85 longest span, minimum 3 spans4Based on 20% redistribution at supports and no decrease in span moments

Punching shear resistance (w/o shear reinforcement)

Table 8vRd,c resistance of members without shear reinforcement, MPa

•Table derived from: where

•For fck = 30 N/mm2l Effective depth, d (mm)200 225 250 275 300 350 400 450 500 600 750

0.25% 0.54 0.52 0.50 0.48 0.47 0.45 0.43 0.41 0.40 0.38 0.360.50% 0.59 0.57 0.56 0.55 0.54 0.52 0.51 0.49 0.48 0.47 0.450.75% 0.68 0.66 0.64 0.63 0.62 0.59 0.58 0.56 0.55 0.53 0.511.00% 0.75 0.72 0.71 0.69 0.68 0.65 0.64 0.62 0.61 0.59 0.571.25% 0.80 0.78 0.76 0.74 0.73 0.71 0.69 0.67 0.66 0.63 0.611.50% 0.85 0.83 0.81 0.79 0.78 0.75 0.73 0.71 0.70 0.67 0.651.75% 0.90 0.87 0.85 0.83 0.82 0.79 0.77 0.75 0.73 0.71 0.682.00%

0.94 0.91 0.89 0.87 0.85 0.82 0.80 0.78 0.77 0.74 0.71

k 2.000 1.943 1.894 1.853 1.816 1.756 1.707 1.667 1.632 1.577 1.516 ckckcRd fkfk 5.131

1, 035.010012.0

02.0;2/2001 1 lzlydk bdAbdA szlzsyly /,/

Basic control perimeter for punching shear in EC 2

• u1 = basic control perimeter

• Control perimeters at a distance less than 2d should be considered where the concentrated force is opposed by a high distributed pressure (e.g. soil pressure on a base), or by the effects of a load or reaction within a distance 2d of the periphery of area of application of the force.

2d

u1

u12d

bz

by 2d

2d

u1

Control perimeter near an opening• For loaded areas situated near openings, if the shortest distance between the

perimeter of the loaded area and the edge of the opening does not exceed 6d, that part of the control perimeter contained between two tangents drawn to the outline of the opening from the centre of the loaded area is considered to be ineffective

l2

2d

- openingA

6dl1 l2 6d

(l2 · l1)

l1 > l2

A A

B B

- ineffective control perimeterB

Control perimeters for loaded areas close to or at edge or corner• For loaded areas situated near or on an edge

or corner, i.e. at a distance smaller than d, special edge reinforcement should always be provided.

u1

2d

2d

u1

2d

2d

u1

2d

2d

Slab with enlarged column head

• where IH < 2.0 hH

A

= arctan (½)

= 26.6

cIH < 2.0 hH

d

rcont

hH

rcont

IH < 2.0 hH

hH

B

B - loaded area AloadA - basic control section

Slab with enlarged column head• where IH > 2(d + hH)

A = arctan (½)

= 26.6

cIH > 2(d + hH)

d

rcont,int

hH

rcont,int

IH > 2(d + hH)

hH

B

B - loaded area AloadA - basic control sections for circular columns

dHdH

rcont,extrcont,ext

d

Shear distribution

• due to an unbalanced moment at a slab-internal column connection

c1

c2

2d

2d

0.5 1.0 2.0 3.0

k 0.45 0.60 0.70 0.80

2

1

c

c

2

1

c

c

Punching shear formula - 1

• The design punching shear stress is given by:

cpmincp3

1

cklcRd,cRd, 100100100 ..fkC

fck is the concrete characteristic compressive strength in MPa.

ka value given by:

l is the steel ratio for longitudinal reinforcement given by:ly, lz relate to the bonded tension steel in y- and z-directions respectively. The values ly and lz should be

calculated as mean values taking into account a slab width equal to the column width plus 3d each side.

cp compressive stress in the concrete from axial load or prestressing, and it is given as:cp = ½(cy + cz)where

cy, cz are the normal concrete stresses in the critical section in y- and z-directions (in MPa), positive if compression:

NEdy, NEdz are the longitudinal forces across the full bay for internal columns and the longitudinal force across the control section for edge columns. The force may be from a load or prestressing action.

Ac is the area of concrete according to the definition of NEd

Punching shear formula - 2

• Where shear reinforcement is required:

sin1

517501

efywd,swr

cRd,csRd,

dufA

s

d..

Asw is the area of one perimeter of shear reinforcement around the column.

sr is the radial spacing of perimeters of shear reinforcement.fywd,ef is the effective design strength of the punching shear

reinforcement according to:fywd,ef = 250 + 0.25d fywd (MPa)

d is the mean of the effective depths in the orthogonal directions (mm).

the angle between the shear reinforcement and the plane of the slab.