Lecture 4 Slabs

7/27/2019 Lecture 4 Slabs

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Practical Design to Eurocode 2

Week 4 - Slabs


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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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Designing for Shear


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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 without shearreinforcement


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where:

k = 1 + (200/d) 2.0l = Asl/bwdAsl = area of the tensile reinforcement,

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

cp = NEd/Ac < 0.2 fcd [MPa] Compression +veNEd = 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(100 lfck)1/3 + 0.15cp] bwd (6.2.a)

with a minimum ofVRd,c = (0.035k

3/2fck1/2 + 0.15 cp) bwd (6.2.b)

Without Shear ReinforcementCl. 6.2.2 7.2


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Shear- Table 7.1

vRd,c resistance of members without shear reinforcement, MPa

As(bd) %

Effective depth, d(mm)

200 225 250 275 300 350 400 450 500 600 7500.25 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 0.68

2.00 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 k1.5fck0.5 where k = 1 + (200/d) 2 and I =As/(bd) 0.02

Note: This table has been prepared forfck = 30. Where I exceeds 0.40% 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


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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


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Shear resistance with shearreinforcement

See Week 2 Beams

Variable strut inclination method


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EC2: Cl 6.2.3


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cotswsRd, ywdfzs

AV

tancot

1maxRd,

cdwcw fzbV

21.8 < < 45

Variable Strut Inclination MethodCl. 6.2.3


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fck

vRd,cot =2.5

vRdcot =1.0

20 2.54 3.68

25 3.10 4.50

28 3.43 4.97

30 3.64 5.28

32 3.84 5.58

35 4.15 6.0240 4.63 6.72

45 5.08 7.38

50 5.51 8.00

Variable strut inclination method

Cl. 6.2.3

vRd,max


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Design Flow Chart for Shear

Yes (cot = 2.5)

Determine the concrete strut capacity vRd when cot = 2.5vRd = 0.138fck(1-fck/250)

Calculate area of shear reinforcement:Asw /s = vEd bw/(fywd cot )

Determine vEd where:

vEd = design shear stress [vEd = VEd/(bwz) = VEd/(bw 0.9d)]

Determine from: = 0.5 sin-1[(vEd/(0.20fck(1-fck/250))]Is vRd > vEd?

No

Check maximum spacing of shear reinforcement :

s,max = 0.75 dFor vertical shear reinforcement


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Punching shear


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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 notdepend 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)


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For a rectangular internal columnwith 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 be

calculated

c1

c2

2d

2d

y

z

Punching Shear (3)


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kd

Outer control

perimeter

Outer perimeter of shear

reinforcement

1.5d(2dif > 2dfrom

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 calculatedfrom:

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

The outermost perimeter ofshear reinforcement should be

placed at a distance not

greater than kd( k = 1.5)

within the outer control

perimeter.

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


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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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EC 2: Concise:

NA - Check vEd 2 vRdc at basic control perimeter

The minimum area of a link leg:

Asw,min (1.5 sin+ cos)/(sr st) (0.08 (fck))/fyk equ 9.11

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

sr = radial spacing of layers 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 dfywd (MPa.)

d = mean effective depth of the slabs (mm)

V

v vu dEd

Ed Rd,max0

= 0.5 f

cdMax. shear stress at column face,

Punching Shear Reinforcement (3)Cl. 6.4.5 8.5


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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

Asw,min (0,053 sr st (fck)) /fyk

Punching Shear Reinforcement (4)


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Punching shearWorked example

From Worked Examples to EC2: Volume 1 Example 3.4.10


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Punching shear

At C2 the ultimate column

reaction is 1204.8 kN


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Solution1. 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 MPavEd < 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(c

x+ c

y) + 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 = (lylz)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

NA check vEd 2vRd,c at basic control perimeter


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Solution3. 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.

Minimum area of a link leg:

Asw,min (0.053 sr st (fck)) /fyk = 0.053 x 175 x 350 x 30 / 500

36 mm2


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Solution


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Outline Week 4


Designing for shear

Serviceability


Workshop - serviceability

Flat Slabs

Tying systems


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Serviceability


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EC 2: Concise:

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 beenno 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 PD

6687 Cl. 2.20

Control of cracking (7.3)

Control of deflections (7.4)


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Crack control


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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.

Equ 9.1N


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

dbf

dbfA t

yk

tctmmin,s 013.0

26.0

EC2: Cl 9.2.1.1

9.1N


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Crack Width Limits - use Table NA.4

Recommended values ofwmax

Exposure class RC or unbonded PSC

members

Prestressed

members with

bonded tendons

Quasi-permanent load Frequent load

X0,XC1 0.3* 0.2XC2,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

Note: For cracking due to restraint use only max bar size


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EC 2: Concise:

Maximum Bar Diameters

Cl. 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

ar

diameter(mm

)

wk=0.3 mm

wk=0.2 mm

wk = 0.4

(Stress due to quasi-permanent actions)

Crack Control


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EC 2: Concise:

Maximum Bar Spacings

Cl. 7.3.3 Table 10.2

0

50

100

150

200

250

300

150 200 250 300 350 400

stress in reinforcement (MPa)

Maximumb

ar

spacing(mm) wk = 0.4

wk = 0.3

wk = 0.2

(Stress due to quasi-permanent actions)

Crack Control


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Deflection control


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EC 2: Concise:

Deflection Limits (7.4.1)

Cl. 7.4.1 -

Span/250 under quasi-permanent loads to avoid impairment ofappearance and general utility

Span/500 after construction under the quasi-permanent loads toavoid damage to adjacent parts of the structure.


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EC 2: Concise:

Deflection Control

Cl. 7.4 10.5

Deflection control may be achieved by the followingmethods:

Direct calculation (Eurocode 2 methods considered to

be an improvement on BS 8110) See How

toDeflection calculations

Using simplified span-to-effective depth limits from

the code (control of deflection without calculation)


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Deflection calculations


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Deflection: L/d check


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EC 2: Concise:

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 if 0

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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EC 2: Concise:

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


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EC 2: Concise:

Graph of Exp. (7.16)

- Figure 15.2


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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

Simply

supported

1.0

End span 1.3

Interior Span 1.5

Flat Slab 1.2

How to guide Figure


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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 > 7mF2 = 7/leff

Factor F1 for ribbed and waffle slabs only

F1 = 1 0.1 ((bf/bw) 1) 0.8

IncreaseAs,provor fck

No

F t t b li d


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Factors to be applied

EC2: 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 excessive

deflections, the values of l/d given by Expression (7.16) should bemultiplied by 7.0/ leff(leff in metres, see 5.3.2.2 (1)).

For flat slabs, with spans exceeding 8.5 m, which supportpartitions 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.


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Outline Week 4


Designing for shear

Serviceability



Flat Slabs

Tying systems


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Detailing - Solid slabs


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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 slabsRules for one-way and two-way solid slabsEC2 9.3


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Outline Week 4


Designing for shear

Serviceability



Flat Slab Design

Other slabs

Tying systems


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Workshop serviceability

www.eurocode2.info

Introduction to workshop


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problem

This is example 3.4

ofWorked examples

to Eurocode 2:

Volume 1.

Week 4 - Workshop problems


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Week 4 Workshop problems

Design information


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Worked example

For the slab strip along grid line C check deflection is

within design limits and ensure the crack widths inthe bottom of the slab are also limited.

Design strip along grid line C


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Assume strip is 6 m wide

As,req

= 1324 mm2 B G

= 1.25

Check deflection andcracking slab along

grid line C.

Deflection


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Deflection

Check: basic l/dx 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


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Basic Span to Depth Ratios

(for simply supported condition)

20.5

Percentage of tension reinforcement (As,reqd/bd)

Spantodept

hratio(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

Deflection


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Deflection




= 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 ofl/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


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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

Crack Control Without Direct


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CalculationEC2: Cl. 7.3.3

Example Check cracking in bottom of slab.Gk = 8.5 kN/m

2 Qk = 4.0 kN/m2

2 = 0.3 (office loading) G = 1.25

As,req = 1324 mm2/m

Try H16 @ 100 As,prov = 2010 mm2/m

Cracking


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g

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

0.8Category F : traffic area, 30 kNCategory G : traffic area, 30160 kN

Category 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 00.7

Determination of Steel Stress


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252

RatioG

k/Q

k

Un

modifiedste

elstress,su

Ratio Gk/Qk = 8.5/4.0 = 2.13

Crack Widths


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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 t o l imit

cr ack wid t h

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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www.eurocode2.info

Workshop problem

Deflection & Cracking

Workshop problem


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For the edge stripindicated checkdeflection is within

design limits andensure the crackwidths in the bottomare also limited.

As,req = 959 mm2/m B

d= 240 mm

G = 1.25

Design for this span

Deflection


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= 0.40%

Basic Span-to-Depth Ratios


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(for simply supported condition)

26.2

Percentage of tension reinforcement (As,reqd

/bd)

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

Deflection


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7.4.2 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

3. Determine Factor F2F2 = 1.0

For flanged sections where the ratio of the flangebreadth to the rib breadth exceeds 3, the values ofl/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


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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


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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

0.8Category F : traffic area, 30 kNCategory G : traffic area, 30160 kN

Category H : roofs

0.7

0.70.7

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

Determination of Steel Stress


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252

Ratio Gk

/Qk

Unmodifiedste

elstress,su

Ratio Gk/Qk = 8.5/4.0 = 2.13

Crack Widths


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From graph su = 252 MPa

s

= (su

As,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 t o l imit

cr ack wid t h

Steel

stress(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

Outline Week 4


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Designing for shear

Serviceability

Detailing Solid slabs Workshop - serviceability

Flat Slab Design

Tying systems


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Flat Slab Design

Paul Gregory

Flat Slab Design


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g

Flat slabs - Introduction

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

Flat Slabs - Introduction


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What are flat slabs?

Solid concrete floors of constant thickness

They have flat soffits



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Column Head

Waffle Slab

Drop Panel



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1. HANSON COBIAX2. BUBBLEDECK

VOIDED SLABS



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Hybrid (PC & In-situ) flat slabs



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Hybrid (PC & In-situ) flat slabs

Particular rules for flat slabs


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EC2 sections relevant to Flat Slabs:


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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 General

I.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



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Distribution of momentsEC2: Figure I.1 Concise Figure 5.11

Di ib i f



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Distribution of moments

EC2: Table I.1 Concise: Table 5.2


EC2 Cl 9 4 C i 12 4 1


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Arrangement of reinforcement should reflect behaviour

under 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 internalcolumns in each orthogonal directions.

EC2: Cl. 9.4 Concise: 12.4.1


EC2 Fig 9 9 C i Fig 5 12


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Design reinforcement at edge and corner reinforcement

should be placed within the becz

cy

y

be = cz + y

A

cz

cyy

A

be = z + y/2

z

A

EC2: Figure 9.9 Concise Figure 5.12

The maximum moment that can be transferred from the

slab to the column should be limited to 0.17bed2fck

Initial sizing


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3 methods:

1. Simple span to depth table

2. Use Economic Concrete Frame Elements

Imposed Load, Qk (kN/m2) 2.5 5 7.5 10

Multiple Span 28 26 25 23

3 methods:

Initial sizing


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3 methods:

1. Simple span to depth table

2. Use Economic Concrete Frame Elements

3. Use Concept.xls

Initial sizing


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Initial sizing


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8mEquivalent frame method

Analysis Methods


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Elastic Plane Frame Equivalent Frame Method, Annex I

Tabular Method - Equivalent Frame Method, Annex I

Grillage

Yield Line Plastic method of design

Finite Element Analysis

Elastic method

Elastic Plane Frame Equivalent Frame Method Annex I

Analysis Methods


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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. Annex

I.1.2.(1)

Analysis Methods

L d


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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

Analysis Methods


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TR 64 Figure 14

Reduction in maximum hogging moment

at columns

Analysis Methods Equi Frame

Distribution of Design Bending Moments, Annex I


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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

Distribution of Design Bending Moments - Example


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400 mm2

/m

100 mm2/m

100 mm2/m

200 mm2/m

200 mm2/m

Distribution of Design Bending Moments ExampleTable 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

Equivalent frame method


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Equivalent frame method


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Equivalent frame method - Elastic Plane Frame

Analysis Methods


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q

Computer software normally used to assess bending moments and

shear forces

Design forfull load in both directions RC spreadsheet TCC33.xls will carry out the analysis and design

Tabular MethodSimplest method

Analysis Methods


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Simplest method

Use coefficients from Concise Tables 15.2 and 15.5 to determine

bending moments and shear forces

Design forfull load in both directions

Frame lateral stability must not be dependent on slab-col connections

There must be at least three approx equal spans.

Method uses single load case. Note: No column BM given in table.

Yield Line Method

Analysis Methods


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Equilibrium and work methods.

work method

External energy expended by

the displacement of loads

Internal energy dissipated by

the yield lines rotating

=

Analysis Methods


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Yield Line Method

Suitable for:

irregular layouts

Slabs supported on 2 or 3

edges only

Detailed guidance and numerous

worked examples contained in:

Practical Yield Line Design

Deflection design to simplified rules

Finite Element Method

Analysis Methods


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Suitable for:

irregular layouts

slabs with service openings

post tensioned design

(specialist software)

Common pitfalls:

Use long term E-values (typically 1/3 to 1/2 short term value)

Use cracked section properties (typically 1/2 gross

properties) by adjusting E-value to suit

Therefore appropriate E-values are usually 4 to 8 kN/mm2

Finite Element Method

Analysis Methods


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Design moment is the integral of the section through the contour

plot or the bay width

Peak

moment

Integral for

bay width

Finite Element - Design moments


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Distribution of top reinforcementAssume a total area ofsteel, At = 8000 mm

2

Distribution 75% Column strip & 25% Middle strip


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Distance (m)

600

100

200

300

400

500

0

Bendingm

oment(kNm/m)

0 1 2 3 4 5 6 7 8

Centre column strip:4000 mm2 or 2000 mm2/m

Outer column strip:0.75 x 8000 4000= 2000 mm2 or 1000 mm2/m

Middle strip:0.25 x 8000 = 2000 mm2

or 500 mm2/m

p p6000 mm2 2000 mm2

Punching Shear - EC2: cl 6.4 and cl 9.4.3

Design Constraints


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g

Traditional links

Shear Rails

Deflection:

Design Constraints


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Wherever possible use the span/effective depth ratios, cl 7.4.2 (2)

Span is based on the longer span and the K factor is 1.2

Reduction factor for brittle finishes for spans greater than 8.5m

Moment Transfer from slab to column:

Design Constraints


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Edge and corner columns have limited capacity to transfer moments

from slab redistribution may be necessary (Annex I.1.2 (5), EC2 cl

9.4.2 & TR 64)

Mt max = 0.17 be d2 fck

Effectivewidth, be.


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Flat slab Workshop

Cover and Flexure


problem


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This is example 3.4ofWorked examples

to Eurocode 2:

Volume 1.


Determine the cover andreinforcement slab

along grid line C


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Assume strip is 6 m wide

along grid line C.


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Flat slab

Workshop cover

Worked example

Determine Cover


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Flat slab

Workshop Flexure

Sagging reinforcement Worked example

Hogging reinforcement Workshop problem

Design strip


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From analysis


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From analysis


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(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

Alt

ernativ

e

Workshop problem

Now determine the reinforcement for the hogging

moments


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moments

Hint: You will need to work out reinforcement for both

column and middle strips and then work out how it is

distributed.

Solution


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(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

Alte

rnativ

e

Solution


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(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

Alte

rnativ

e

0.047

Reinforcement distribution

Total area of reinforcement:

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


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As,tot 2213 x 3 887 x 3 9300 mm

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 mm

2/m

Use H20 @ 200 CTRS (1570 mm2/m)

Middle strip: As,req = 887 mm2/m

Use H16 @ 200 CTRS (1010mm2/m)

Outline Week 4



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Designing for shear

Serviceability



Flat Slab Design

Tying systems


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Tying systems

Tying systems (1) ALL concrete structures

Peripheral ties (9.10.2.2) & NA:

Ftie per = (20 + 4n0)l i 60kN where n0 is the number of storeys


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Ftie,per (20 4n0)l i 60kN where n0 is the number of storeys

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

Ftie,int = ((gk + qk)/7.5)(lr/5)Ft Ft kN/mWhere (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.

Internal Ties: EC2 specifies a

Tying systems (2)

ALL concrete structures


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20kN/m requirement which is

significantly less than BS8110.

UK NA requirements similar to BS 8110

Tying systems (3)

Vertical ties (9.10.2.5):In panel buildings of 5 storeys or more, ties should be provided in

columns and/or walls to limit damage of collapse of a floor


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columns and/or walls to limit damage of collapse of a floor.

Normally continuous vertical ties should be provided from the lowest

to the highest level.

Where a column or wall is supported at the bottom by a beam or slab

accidental 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 at

connections of precast members.

Week 4


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END

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