Lecture 4 Slabs(Important)

8/12/2019 Lecture 4 Slabs(Important)

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Practical Design to Eurocode 2

Slabs

Outline Week 4

We will look at the following topics:

Designing for shear including punching shear

Serviceability cracking and deflection

Detailing Solid slabs

Workshop serviceability cracking & deflection

Flat Slab Design includes flexure workshop

Tying systems


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2

Shear

Shear

There are three approaches to designing for shear:

When shear reinforcement is not required e.g. slabs

When shear reinforcement is required e.g. Beams

Punching shear requirements e.g. flat slabs

The maximum shear strength in the UK should not exceed thatof class C50/60 concrete


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Shear resistance withoutshear reinforcement

where:

k = 1 + (200/d) 2.0

l =Asl/bwd

Asl = area of the tensile reinforcement,

bw = smallest width of the cross-section in the tensile area [mm]

cp = NEd/Ac < 0.2fcd [MPa]

NEd = axial force in the cross-section due to loading or pre-stressing [in N]

Ac = area of concrete cross section [mm2]

VRd,c = [0.12k(100lfck)1/3 + 0.15cp] bwd (6.2.a)

with a minimum of

VRd,c = (0.035k1.5fck

0.5 + 0.15 cp) bwd (6.2.b)

Without Shear ReinforcementCl. 6.2.2 7.2


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ShearTable 7.1

vRd,c resistance of members without shear reinforcement, MPa

%

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 0.36

0.50 0.59 0.57 0.56 0.55 0.54 0.52 0.51 0.49 0.48 0.47 0.45

0.75 0.68 0.66 0.64 0.63 0.62 0.59 0.58 0.56 0.55 0.53 0.51

1.00 0.75 0.72 0.71 0.69 0.68 0.65 0.64 0.62 0.61 0.59 0.57

1.25 0.80 0.78 0.76 0.74 0.73 0.71 0.69 0.67 0.66 0.63 0.61

1.50 0.85 0.83 0.81 0.79 0.78 0.75 0.73 0.71 0.70 0.67 0.65

1.75 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 2.00 1.94 1.89 1.85 1.82 1.76 1.71 1.67 1.63 1.58 1.52

Table derived from: vRd,c = 0.12 k (100Ifck)(1/3) 0.035 k1.5fck

0.5 where k = 1 + (200/d) 2 and I =As/(bd) 0.02

Note: This table has been prepared forfck = 30. Where I exceeds 0.4% the following factors may be used:

fck 25 28 32 35 40 45 50

factor 0.94 0.98 1.02 1.05 1.10 1.14 1.19

Shear in Slabs

Most slabs do not require shear

reinforcement

Check VEd < VRd,c

Where VRd,c is shear resistance of

members without reinforcement

vRd,c = 0.12 k (100 Ifck)1/3

0.035 k1.5fck0.5

Where VEd > VRd,c,

shear reinforcement is required

and the strut inclination method

should be used

(How-to Compendium p21)


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Punching shear

Punching shear does not use the Variable Strut inclination method

and is similar to BS 8110 methods

The basic control perimeter is set at 2dfrom the loaded area

The shape of control perimeters have rounded corners

bz

by

2d 2d 2d

2du1

u1 u1

Punching ShearCl. 6.4 Figure 8.3

Where shear reinforcement is required the shear resistance is the

sum of the concrete and shear reinforcement resistances.


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For structures where:

lateral stability does not

depend on frame action

adjacent spans do not differ

by more than 25%

the approximate values for

shown may be used:

The applied shear stress should be taken as:

vEd = VEd/ui d where=1 + k(MEd/VEd)u1/W1

Punching Shear (2)

For a rectangular internal column

with biaxial bending the following

simplification may be used:

= 1 + 1.8{(ey/bz)2

+ (ez/by)2

}0.5

where by and bz are the dimensions

of the control perimeter

For other situations there is plenty of guidance on determining given in Cl 6.4.3 of the Code.

Where the simplified arrangement is not applicable then can becalculated:

c1

c2

2d

2d

y

z

Punching Shear (3)


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kd

Outer controlperimeter

Outer perimeter of shear

reinforcement

1.5d (2d if > 2d from

column)

0.75d

0.5dA A

Section A - A

0.75d

0.5d

Outer control

perimeter

kd

The outer control perimeter at

which shear reinforcement is not

required, should be calculated

from:

uout,ef = VEd / (vRd,c d)

The outermost perimeter of

shear reinforcement should be

placed at a distance not

greater than kd( k = 1.5)

within the outer controlperimeter.

Punching Shear Reinforcement (1)Cl. 6.4.5 Figures 12.5 & 12.6

1,5d

2d

d

d

> 2d

1,5d

uout

uout,ef

Where proprietary systems are used the control perimeter at which

shear reinforcement is not required, uout or uout,ef (see Figure) should be

calculated from the following expression:

uout,ef = VEd / (vRd,c d)

Punching Shear Reinforcement (2)Cl. 6.4.5 Figure 8.10


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Where shear reinforcement is required it should be calculated in

accordance with the following expression:

vRd,cs = 0.75 vRd,c + 1.5 (d/sr)Aswfywd,ef(1/(u1d)) sin

Asw = area of shear reinforcement in each perimeter around the col.

sr = radial spacing of shear reinforcement = angle between the shear reinforcement and the plane of slab

fywd,ef = effective design strength of the punching shear reinforcement,

= 250 + 0.25 d fywd (MPa.)

d = mean effective depth of the slabs (mm)

Vv v

u dEd

Ed Rd,max

0

= = 0.5 fcdMax. shear stress at column face,

Punching Shear Reinforcement (3)Cl. 6.4.5 8.5

Check vEd 2 vRdc at basic control perimeter

Note: UK NA says first control perimeter, but the paper* on which

this guidance is based says basic control perimeter

The minimum area of a link leg (or equivalent),Asw,min, is given by the

following expression:Asw,min (1.5 sin+ cos)/(sr st) (0,08 (fck))/fyk equ 9.11

*FRASER, AS & JONES, AEK. Effectiveness of punching shear reinforcement to EN1992-1-1:2004. The Structural Engineer ,19 May 2009.

Punching Shear Reinforcement (4)


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Punching shearWorked example

Punching shear

At C2 the ultimate column

reaction is 1204.8 kN


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Punching shear

uoutu1

u0

2d

Solution

1. Check shear at the perimeter of the column

vEd = VEd/(u0d) < vRd,max

= 1.15

u0 = 4 x 400 = 1600 mm

d = (260 + 240)/2 = 250 mm

vEd = 1.15 x 1204.8 x 1000/(1600 x 250)

= 3.46 MPa

vRd,max = 0.5 fcd

= 0.5 x 0.6(1-fck/250) x ccfck/m

= 0.5 x 0.6(1-30/250) x 1.0 x 30 /1.5 = 5.28 MPa

vEd < vRd,max ...OK

= 1,4

= 1,5

= 1,15

C

B A


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Solution2. Check shear at the basic control perimeter

vEd = VEd/(u1d) < vRd,c

u1 = 2(cx + cy) + 2x 2d

= 2(400 + 400) + 2x 2 x 250 = 4742 mm

vEd = 1.15 x 1204.8 x 1000/(4742 x 250) =

= 1.17 MPa

vRd,c = 0.12 k(100lfck)1/3

k = 1 + (200/d)1/2

= 1 + (200/250)1/2 = 1.89

l = (lylx)1/2 = (0.0085 x 0.0048) 1/2 = 0.0064

vRd,c = 0.12 x 1.89(100 x 0.0064 x 30)1/3 = 0.61 MPa

vEd > vRd,c ...Punching shear reinforcement required

Solution

3. Perimeter at which punching shear no longer required

uout = VEd/(dvRd,c)

= 1.15 x 1204.8 x 1000/(250 x 0.61)

= 9085 mm

Rearrange: uout = 2(cx + cy) + 2rout

rout = (uout - 2(cx + cy))/(2)

= (9085 1600)/(2) = 1191 mm

Position of outer perimeter of reinforcement from column face:

r = 1191 1.5 x 250 = 816 mm

Maximum radial spacing of reinforcement:

sr,max = 0.75 x 250 = 187 mm, say 175 mm


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Solution4. Area of reinforcement

Asw (vEd 0.75vRd,c)sru1/(1.5fywd,ef)

fywd,ef = (250 + 0.25d) = 312 MPa

Asw (1.17 0.75 x 0.61) x 175 x 4741/(1.5 x 312)

1263 mm2/perim.

Solution


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Serviceability

What does Eurocode 2 Cover?Cl. 7.2 10.1

Stress limitation (7.2)

Stress checks in reinforced concrete members have not been

required in the UK for the past 50 years or so and there has been

no known adverse effect. Provided that the design has been carried

out properly for ultimate limit state there will be no significant

effect at serviceability in respect of longitudinal cracking PD6687 Cl. 2.20

Control of cracking (7.3)

Control of deflections (7.4)


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Crack control

Control of Cracking

In Eurocode 2 cracking is controlled in the following ways:

Minimum areas of reinforcement cl 7.3.2 & Equ 7.1

As,mins = kckfct,effAct this is the same as

Crack width limits (Cl. 7.3.1 and National Annex). These

limits can be met by either:

direct calculation (Cl. 7.3.4) crack width is wk

deemed to satisfy rules (Cl. 7.3.3)

Note: slabs 200mm depth are OK if As,min is provided.


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Minimum Reinforcement AreaThe minimum area of reinforcement for slabs (and beams) is given by:

dbf

dbfA

t

yk

tctm

min,s013.0

26.0

Crack Width Limits - use Table NA.4

Recommended values of wmax

Exposure class RC or unbonded PSC

members

Prestressed members

with bonded tendons

Quasi-permanent load Frequent load

X0,XC1 0.3* 0.2

XC2,XC3,XC4 0.3

XD1,XD2,XS1,

XS2,XS3

Decompression

* Does not affect durability, may be relaxed where appearance

is not critical (eg use 0.4 mm)


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Crack Control Without Direct

CalculationCrack control may be achieved in two ways:

limiting the maximum bar diameter using Table 7.2

limiting the maximum bar spacing using Table 7.3

Maximum Bar DiametersCl. 7.3.3 Table 10.1

0

10

20

30

40

50

100 150 200 250 300 350 400 450 500

Reinforcement stress, s (N/mm2)

maximumb

ardia

meter(mm)

wk=0.3 mm

wk=0.2 mm

wk= 0.4

(Stress due to quasi-permanent actions)

Crack Control


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Maximum Bar SpacingsCl. 7.3.3 Table 10.2

0

50

100

150

200

250

300

150 200 250 300 350 400

stress in reinforcement (MPa)

Max

imumb

arspacing(mm) wk= 0.4

wk = 0.3

wk= 0.2

(Stress due to quasi-permanent actions)

Crack Control

Deflection control


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Deflection LimitsThe deflection limits are:

Span/250 under quasi-permanent loads to avoid impairment of

appearance and general utility

Span/500 after construction under the quasi-permanent loads to

avoid damage to adjacent parts of the structure.

Deflection requirements can be satisfied by the following methods:

Direct calculation (Eurocode 2 methods considered to be an

improvement on BS 8110) See How toDeflection calculations

Limiting span-to-effective-depth ratios

Deflection calculations


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Basic span/effective Depth Ratios

Table 7.4(N) use Table NA.5 Table 10.3

Structural system K =

1.5%

=

0.5%

S.S. beam or slab 1.0 14 20

End span 1.3 18 26

Interior span 1.5 20 30

Flat slab 1.2 17 24

Cantilever 0.4 6 8

Limiting Span-effective-depth ratios

Cl 7.4.2 & Exp (7.16a & b) 10.5.2

K factor taking account of the different structural systems

0 reference reinforcement ratio = fck 10-3

required tension reinforcement ratio at mid-span to resist the momentdue to the design loads (at support for cantilevers)

required compression reinforcement ratio at mid-span to resist themoment due to design loads (at support for cantilevers)

++=

23

0ck

0ck 12,35,111

ffK

d

l if0

+

+=

0

ck0

ck

'

12

1

'

5,111

ffK

d

lif>0

There are adjustments to these expressions in cl 7.4.2 (2) for the

steel stress, flanged sections and long spans with brittle finishes.


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Graph of Exp. (7.16)- Figure 15.2

EC2 Span/Effective Depth Ratios

18.5

Percentage of tension reinforcement (As,reqd/bd)

Spantodepthratio(l/d)

This graph has beenproduced for K = 1.0

StructuralSystem

K

Simplysupported

1.0

End span 1.3

Interior Span 1.5

Flat Slab 1.2

How to guide Figure


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Factors to be appliedEC2: cl 7.4.2 & NA Concise 10.5.2

F1 - Flanged sections

where the ratio of the flange breadth to the rib breadth exceeds3, the values of l/d given by Expression (7.16) should bemultiplied by 0.8.

F2 Brittle partitions

For slabs (other than flat slabs), with spans exceeding 7.0 m,which support partitions liable to be damaged by excessivedeflections, the values of l/d given by Expression (7.16) should bemultiplied by 7.0/ leff(leffin metres, see 5.3.2.2 (1)).

For flat slabs, with spans exceeding 8.5 m, which support

partitions liable to be damaged by excessive deflections, thevalues of l/d given by Expression (7.16) should be multiplied by8.5 / leff(leffin metres, see 5.3.2.2 (1)).

F3 s Steel stress under service loadMay be adjusted by 310/ s 1.5 or As,prov/As,req 1.5where s calculated using characteristic loads.

Flow Chart

Is basic l/dx F1 x F2 x F3 >Actual l/d?

Yes

No

Factor F3 accounts for stress in the reinforcementF3 = As,prov /As,reqd 1.5 or 310/s 1.5 (UK NA)

Check complete

Determine basic l/d

Factor F2 for spans supporting brittle partitions > 7m

F2 = 7/leff

Factor F1 for ribbed and waffle slabs onlyF1 = 1 0.1 ((bf/bw) 1) 0.8

IncreaseAs,provor fck

No


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Detailing - Solid slabs

Where partial fixity exists, not taken into account in design:

Internal supports:As,top 0,25As for Mmax in adjacent spanEnd supports: As,top 0,15As for Mmax in adjacent span

This top reinforcement should extend 0,2 adjacent span

2h

h

Reinforcement at free edges should include u bars and longitudinal

bars

Detailing Solid slabs

Rules for one-way and two-way solid slabs


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Workshop - serviceability

Introduction to workshopproblem

This is example 3.4

of Worked examples

to Eurocode 2:Volume 1.


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Week 4 - Workshop problems

Design information

Design strip along grid line C

Assume strip is 6 m wide

As,req = 1324 mm2

Check deflection and

cracking slab along

grid line C.


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Deflection

Check: basic l/d x F1 x F2 x F3 actual l/d

1. Determine basic l/d

The reinforcement ratio,=As,req/bd = 1324 x 100/(1000 x 260)= 0.51%

Basic Span-to-Depth Ratios(for simply supported condition)

20.5


Spantodepthratio(l/d)

This graph has beenproduced for K = 1.0

StructuralSystem

K

Simply supported 1.0

End span 1.3

Interior Span 1.5

Flat Slab 1.2


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Deflection



The reinforcement ratio,=As,req/bd = 1324 x 100/(1000 x 260)

= 0.51%

From graph basic l/d = 20.5 x 1.2 = 24.6 (K= 1.2 for flat slab)

2. Determine Factor F1

F1 = 1.0

3. Determine Factor F2(Assume no brittlepartitions)

F2 = 1.0

For flanged sections where the ratio of the flangebreadth to the rib breadth exceeds 3, the values of l/dgiven by Expression (7.16) should be multiplied by 0.8.

For flat slabs, with spans exceeding 8.5 m, whichsupport partitions liable to be damaged by excessivedeflections, the values of l/d given by Expression (7.16)should be multiplied by 8.5 / leff (leff in metres, see5.3.2.2 (1)).

Deflection


As,req = 1324 mm2 (ULS)

Assume we require H16 @ 100 c/c (2010 mm2) to control deflection

F3 =As,prov /As,req = 2010 / 1324 = 1.52 1.5

24.6 x 1.0 x 1.0 x 1.5 9500 / 260

36.9 36.5


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Cracking

Action 0 1 2Imposed loads in buildings,

Category A : domestic, residentialCategory B : office areasCategory C : congregation areas

Category D : shopping areas

Category E : storage areas

0.70.7

0.7

0.7

1.0

0.50.5

0.7

0.7

0.9

0.30.30.6

0.6

0.8

Category F : traffic area, 30 kNCategory G : traffic area, 30160 kN

Category H : roofs

0.7

0.7

0.7

0.5

0

0.6

0.3

0Snow load: H 1000 m a.s.l. 0.5 0,2 0Wind loads on buildings 0.5 0,2 0

0.7

Determination of Steel Stress

252

Ratio Gk/Qk

Unmodifiedsteelst

ress,su

Ratio Gk/Qk = 8.5/4.0 = 2.13


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Crack Widths

From graph su = 252 MPa

s = (suAs,req) / (As,prov)

s = (252 x 1324)/(1.0 x 2010)

= 166 MPa

For H16 @ 100 c/c

Design meets both criteria

Maximum bar size or spacing to limitcrack width

Steelstress(s) MPa

wmax = 0.3 mm

Maximumbar size(mm)

OR

Maximumbar spacing(mm)

160 32 300

200 25 250

240 16 200

280 12 150320 10 100

360 8 50

For loading

or restraintFor loading

only

Workshop problem


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Workshop problem

For the edge stripindicated checkdeflection iswithin designlimits and ensurethe crack widthsare also limited.

As,req = 959 mm2/m

d= 240 mm

Design for this span

Deflection



The reinforcement ratio,=As,req/bd = 959 x 100/(1000 x 240)

= 0.40%


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Basic Span-to-Depth Ratios(for simply supported condition)

26.2


Span

todepthratio(l/d)

This graph has beenproduced for K= 1.0

StructuralSystem

K

Simply supported 1.0

End span 1.3

Interior Span 1.5

Flat Slab 1.2

Deflection7.4.2 EN 1992-1-1



The reinforcement ratio,=As,req/bd = 959 x 100/(1000 x 240) =0.40%

From graph basic l/d = 26.2 x 1.2 = 31.4 (K= 1.2 for flat slab)


F1 = 1.0


F2 = 1.0

For flanged sections where the ratio of the flangebreadth to the rib breadth exceeds 3, the values of l/dgiven by Expression (7.16) should be multiplied by 0.8.

For flat slabs, with spans exceeding 8.5 m, whichsupport partitions liable to be damaged by excessivedeflections, the values of l/d given by Expression (7.16)should be multiplied by 8.5 / leff (leff in metres, see5.3.2.2 (1)).


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Deflection


As,req = 959 mm2 (ULS)

Assume we require H16 @ 200 c/c (1005 mm2) to control deflection

F3 =As,prov /As,req = 1005 / 959 = 1.05 1.5

31.4 x 1.0 x 1.0 x 1.05 5900 / 240

33.0 24.5

Cracking

Action 0 1 2Imposed loads in buildings,

Category A : domestic, residentialCategory B : office areasCategory C : congregation areas

Category D : shopping areasCategory E : storage areas

0.70.7

0.7

0.71.0

0.50.5

0.7

0.70.9

0.30.30.6

0.60.8

Category F : traffic area, 30 kNCategory G : traffic area, 30160 kNCategory H : roofs

0.7

0.7

0.7

0.50

0.6

0.30

Snow load: H 1000 m a.s.l. 0.5 0,2 0Wind loads on buildings 0.5 0,2 0

0.7


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Determination of Steel Stress

252

Ratio Gk/Qk

Unmod

ifiedsteelstress,su

Ratio Gk/Qk = 8.5/4.0 = 2.13

Crack Widths

From graph su = 252 MPa

s = (suAs,req) / (As,prov)

s = (252 x 959) /(1.0 x 1005)

= 240 MPa

For H16 @ 200 c/c

Design meets both criteria

Maximum bar size or spacing to limitcrack width

Steelstress(s) MPa

wmax = 0.3 mm

Maximumbar size(mm)

OR

Maximumbar spacing(mm)

160 32 300

200 25 250

240 16 200

280 12 150

320 10 100

360 8 50

For loading

or restraintFor loading

only


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Flat Slab Design

Flat Slab Design

EC2 particular rules for flat slabs

Initial sizing

Analysis methods - BMs and Shear Force

Design constraints Punching shear Deflection Moment transfer from slab to column


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Particular rules for flat slabs

EC2 sections relevant to Flat Slabs:

Section 6 Ultimate Limit States cl 6.4 Punching (shear) & PD 6687 cl 2.16, 2.17 & 2.1.8

Section 9 Detailing of members and particular rules Cl 9.4 Flat slabs

9.4.1 Slab at internal columns

9.4.2 Slab at edge and corner columns

9.4.3 Punching shear reinforcement

Annex I (Informative) Analysis of flat slabs and shear walls

I.1 Flat Slabs

I.1.1 GeneralI.1.2 Equivalent frame analysisI.1.3 Irregular column layout

The Concrete Society, Technical Report 64 - Guide to the Design and

Construction of Reinforced Concrete Flat Slabs

Distribution of moments

EC2: Figure I.1Concise Figure 5.11



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Distribution of moments

EC2: Table I.1Concise: Table 5.2


Arrangement of reinforcement should reflect behaviourunder working conditions.

At internal columns 0.5At should be placed in a width =0.25 panel width.

At least two bottom bars should pass through internal

columns in each orthogonal directions.


EC2: Cl. 9.4 Concise: 12.4.1


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Design reinforcement at edge and corner reinforcement

should be placed within the be

cz

cy

y

be = cz + y

A

cz

cyy

A

be = z + y/2

z

A

Particular rules for flat slabsEC2: Figure 9.9 Concise Figure 5.12

The maximum moment that can be transferred from theslab to the column should be limited to 0.17bed

2fck

Elastic Plane Frame Equivalent Frame Method, Annex I

Apply in both directions X and Y

Method of Analysis for Bending Moments & SFs

Equivalent Frame - the Beams are the Slab width

Kslab = use full panel width for vertical loads.

Kslab = use 40% panel width for horizontal loads. AnnexI.1.2.(1)

Analysis Methods


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Analysis Methods

Load cases

NA can use single load case provided:

Variable load 1.25 x Permanent load

Variable load 5.0 kN/m2

Condition of using single load case is that Support BMs should be

reduced by 20% except at cantilever supports

TR 64 Figure 14

Reduction in maximum hogging moment

at columns

Analysis Methods


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Analysis Methods Equi Frame

Distribution of Design Bending Moments, Annex I

Table I.1 Column Strip Middle Strip

Negative 60 - 80% 40 - 20%

Positive 50 - 70% 50 - 30%

At = Reinforcement area to resist full negative moment. Cl 9.4.1

Analysis Methods Equi Frame

400 mm2/m

100 mm2/m

100 mm2/m

200 mm2/m

200 mm2/m

Distribution of Design Bending Moments - Example

Table I.1 Column Strip Middle Strip

Negative 75% 25%

At = Reinforcement area to resist full negative moment. Cl 9.4.1

= 1600 mm2

Column strip = 1200 mm2 Middle strip = 400 mm2


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Introduction to workshop

problem

This is example 3.4

of Worked examples

to Eurocode 2:

Volume 1.

Determine Cover


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Design strip

Assume strip is 6 m wide.

Slab is 300 mm deep


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From analysis

From analysis

(Using Concise table 15.5)

z = d[ 1 + (1 - 3.529K)0.5]/2 = 260[1 + (1 3.529 x 0.069)0.5]/2 = 243 mm


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Workshop problemNow determine the reinforcement for the hogging

moments

Hint: You will need to work out reinforcement for both

column and middle strips and then work out how it is

distributed.

Solution

(Using Concise table 15.5)

z = d[ 1 + (1 - 3.529K)0.5]/2 = 260[1 + (1 3.529 x 0.109)0.5]/2 = 232 mm


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Solution

(Using Concise table 15.5 )

z = d[ 1 + (1 - 3.529K)0.5]/2 = 260[1 + (1 3.529 x 0.069)0.5]/2= 243 mm < 0.95d< 247 mm

0.047

Reinforcement distribution

Total area of reinforcement:

As,tot = 2213 x 3 + 887 x 3 = 9300 mm2

50% As,tot = 9300/2 = 4650 mm2

This is spread over a width of 1.5m

As,req

= 4650/1.5 = 3100 mm2/m

Use H20 @ 100 ctrs (3140 mm2/m)

Remaining column strip:

As,req = (2213 x 3 4650)/1.5 = 1326 mm2/m

Use H16 @ 100 ctrs (1540 mm2/m)

Middle strip: As,req = 887 mm2/m

Use H12 @ 100 ctrs (1130mm2/m)


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Tying systems

Peripheral ties (9.10.2.2) & NA:Ftie,per = (20 + 4n0)li 60kNwhere n0 is the number of storeys and li is the length of the end span.

Internal ties (including transverse ties) (9.10.2.3) & NA :F

tie,int= ((g

k+ q

k)/7.5)(l

r/5)F

t F

tkN/m

Where (gk + qk) is the sum of the average permanent and variable floor loads (kN/m2), lr is the

greater of the distances (m) between the centres of the columns, frames or walls supporting any two

adjacent floor spans in the direction of the tie under consideration and Ft = (20 + 4n0) 60kN.

Maximum spacing of internal ties = 1.5 lr

Horizontal ties to columns or walls (9.10.2.4) & NA :Ftie,fac = Ftie,col (2 Ft (ls/2.5)Ft) and 3% of NEdNEd = the total design ultimate vertical load carried by the column or wall at that level. Tying of external

walls is only required if the peripheral tie is not located within the wall. Ftie,fac in kN per metre run of wall,

Ftie,col in kN per column and ls is the floor to ceiling height in m.

All Concrete Structures


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Tying systems (3)

Vertical ties (9.10.2.5):In panel buildings of 5 storeys or more, ties should be provided incolumns and/or walls to limit damage of collapse of a floor.

Normally continuous vertical ties should be provided from the lowestto the highest level.

Where a column or wall is supported at the bottom by a beam or slabaccidental loss of this element should be considered.

Continuity and anchorage ties (9.10.3):Ties in two horizontal directions shall be effectively continuous and

anchored at the perimeter of the structure.

Ties may be provided wholly in the insitu concrete topping or atconnections of precast members.

Lecture 4 Slabs(Important)

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Transcript of Lecture 4 Slabs(Important)