HowTo Design Eurocode 2

8/16/2019 HowTo Design Eurocode 2

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A J Bond

O Brooker

A S Fraser

A J Harris

T Harrison

A E K Jones

R M Moss

R S Narayanan

R Webster

cement concrete



Foreword

Acknowledgements

Published by The Concrete Centre Tel: Fax: www.concretecentre.com

Tel: email:

www.concrete bookshop.com



1. Introduction to Eurocodes

2. Getting started

3. Slabs

4. Beams

5. Columns

6. Foundations

7. Flat slabs

8. Deflection calculations

9. Retaining walls

10. Detailing

11. BS 8500 for building structures

12. Structural fire design







2

2



3

3



4

4



5

5



6

6



7

7

g g



8

8

1

2

3

4

5 6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25



The design process

1

2

3

Design life

4

Actions on structures 5

6

3

O Brooker



10

Load arrangements

Load set 1. Alternate or adjacent spans loaded

g

g

Load set 2. All or alternate spans loaded

g

Load set 3. Simplified arrangements for slabs

2

2

Figure 2Adjacent spans loaded

Figure 3All spans loaded

Figure 1Alternate spans loaded

Table 1Indicative design working life (from UK National Annex to Eurocode)

Design life (years) Examples

Table 2Selected bulk density of materials (from Eurocode 1, Part 1–1)

Material Bulk density (kN/m3)

10



11

Combination of actions

Material propertiesConcrete

7

Reinforcing steel

8

9

Table 4Selected concrete properties based on Table 3.1 of Eurocode 2, Part 1–1

Table 3Selected imposed loads for buildings (from draft UK National Annex to Eurocode 1, Part 1–1)

For members supporting one variable action the ULS combination

1.25 Gk + 1.5 Qk (derived from Exp. (6.10b), Eurocode)

can be used provided the permanent actions are not greater than

4.5 times the variable actions (except for storage loads).

Symbol Description Properties

Key

a

b

Category Example use qk (kN/m2) Qk (kN)

Reinforcing steel

8

9

11



12

Structural analysis

Minimum concrete cover

D

D

Table 6Bending moment and shear co-efficients for beams

Table 7Exposure classes

Class Description

No risk of corrosion or attack

Corrosion induced by carbonation

Corrosion induced by chlorides other than from seawater

Corrosion induced by chlorides from seawater

Freeze/thaw with or without de-icing agents

Chemical attack (ACEC classes)

Table 5Characteristic tensile properties of reinforcement

Moment Shear

Keya

Notes1 2 3 4

D

Class (BS 4449) and designation (BS 8666) A B C

e

Notes1 2

3

12



13

Minimum cover for bond

Minimum cover for durability

12

13

Design for fire resistance 14

≥ f f

R

E I

Table 9Minimum column dimensions and axis distances for columns withrectangular or circular section – method A

Standard fire Minimum dimensions (mm)resistance Column width ( bmin)/axis distance (a) of the main bars

Column exposed on more Exposed on one sidethan one side (m f i = 0.7) (m f i = 0.7)

Notes

1

2 m m

*

† m

Standard Minimum dimensions (mm)fire One-way Two-way spanning slab Flat slab Ribs in a two-way spanning ribbed slabresistance spanning slab l y/ lx ≤ 1.5 1.5 < l y/lx ≤2 (bmin is the width of the rib)

Notes

1 2 3

Figure 4Sections through structural members, showing nominal axis distance, a

Table 10Minimum dimensions and axis distances for reinforced concrete slabs

13



14

Table 8Selected recommendations for normal-weight reinforced concrete quality for combined exposure classes and cover to reinforcement for at least a 50- year intended working life and 20 mm maximum aggregate size

Exposure conditions Cement/combinationdesignationsb

Nominal cover to reinforcementdTypical example Primary Secondary

15 + D cdev 20 +D cdev 25 +D cdev 30 +D cdev 35 +D cdev 40 +D cdev 45 +D cdev 50 +D cdev

Strength classc, maximum w/c ratio, minimum cement or combinationcontent (kg/m3), and equivalent designated concrete (where applicable)

X0

XC1

XC2

XC3XC4

XD1

XD3

XS1

AC-1

XF1

XF3

XF2

XF4

XF4

XF3 or XF4

XF1

XF3

Key

a

b

c

d D

e

f

g

h

j

-

14

Cement/combinationdesignationsb



15

a) Bracing system b) Floor diaphragm c) Roof diaphragm

Figure 5Examples of the effect of geometric imperfections

Stability and imperfections

y

y a a

a R

a 0.5

y

y

y

15

Steelstress(s s)MPa

OR OR

Table 11Maximum bar size or spacing to limit crack width

w max = 0.4 mm w max = 0.3 mm

Maximum Maximum Maximum Maximumbar bar bar bar size (mm) spacing (mm) size (mm) spacing (mm)

Note

s

g d

g

d

Crack control

Figure 6Determination of steel stress for crack width control

To determine stress in the reinforcement (s s), calculate the ratio Gk /Qk ,read up the graph to the appropriate curve and read across to determine s su.

s s can be calculated from the expression: s s = s su As,req

As,prov

1d( () )

15

A p p r o x i m a t e s t e e l s t r e s s a t S L S f o r A

s , r e q ,

s s u





1

2

3

4

5

6



18

18



19

f f f f

K

d K

K z/d

K z/d

d d

d

f f

z/d

19



20

7

s s

s s s

d

s

s

s

20

, s s u

A

A p p r o x i m a t e s t e e l s t r e s s a t S L S f o r A

s , r e q ,

s s u



21

8

9

b 2

b 2

b b

3

v

r d

A bd

k

r I

R r I

r I

f

r

r r

r r

r

r

r

r

r r

r

r

r

21



22

4

l l

l l

d d d

22



23

5

l

a g

d d

a

d

g g

g

r R

r

r

23





• • • • • • • 1

2

3

•••

Fire resistance

R Moss O Brooker



26

ecnadiugr ehtr uFksaTpetS

dr adnatSChapter in this publication

Note

Table 1Beam design procedure

ε

ε

ε

a a

a b a b

••••••••••••••••••••• ••••••••••••••••••• •••••••••••••••••••••• • • ••••••••••••••••••••••

a

26



27

Outside scope of thispublication

d d

d

f f

Table 4Values for K ’

% redistribution d (redistribution ratio) K ’

Key

a

K z/d

K z/d

Key

a

for singly reinforced rectangular sections

Table 6Minimum percentage of required reinforcement

f ck f ctm Minimum percentage (0.26 f ctm / f yka)

Key

a

Table 5 z/d

27



28

Determine v Ed wherev Ed = design shear stress v Ed = V Ed/(bw z ) = V Ed/(0 .9 bwd )][

Yes (cot y = 2.5)

No No

START

Determine the concrete strut capacity v Rd,max cot y = 2.5from Table 7

Redesign

section

Determine y from:

Calculate area of shear reinforcement:

Check maximum spacing for vertical shear reinforcement: s l,max = 0.75 d

= s

Asw

y = 0.5 sin-10.20 f ck (1 – f ck /250)

v Ed

f ywd cot y

v Ed bw

Isv Ed < v Rd,max cot

y = 2.5?

Isv Ed < v Rd,max coty = 1.0?

(see Table 7)

Yes

f v y v y

y

y y

2

y y

Deflection

•

Flanged beams

28



29

12

14

16

18

20

22

24

26

28

30

32

34

36

S p a n t o d e p t h r a t i o (

Percentage of tension reinforcement ( As,req’d/bd )

l / d )

0.40% 0.60% 0.80% 1.00% 1.20% 1.40% 1.60% 1.80% 2.00%

f ck

ck

ck

ck

ck

ck

ck

ck

ck

= 50

f = 45

f = 40

f = 35

f = 32

f = 30

f = 28 f = 25

f = 20

f ck

ck

ck

ck

ck

ck

ck

ck

ck

= 50

f = 45

f = 40

f = 35

f = 32

f = 30

f = 28 f = 25

f = 20

Figure 7Basic span-to-effective-depth ratios

Notes

1

2 r

3

r r

r r

r

r

r

r

r r

r

r

r

s

s s

s s

s s s

d

s

c g

c g

c g

c g

c g

c g

c g

29

For flanged sections

Where the beam span exceeds 7 m and it supports

To determine stress in the provided reinforcement (s s), calculate the ratio

Percentage of tension reinforcement (As,req /bd )

A p p r o x i m a t e s t e e l s t r e s s a t S L S f o r A s

, r e q , s s u



30

l

d d d

D

D

D

y

30

f yd (d – 0.5 hf ) f ydz z



31

Figure 14Procedure for determining longitudinal shear capacity of flanged beams

Yes

No No

No

Calculate the longitudinal shear stressfrom: v Ed = D F d/( hf D x )

(see Figure 13)

Determine the concrete strut capacityfromTable 8 or from:

v Rd = 0.160 f ck (1 – f ck /250)

Calculate area of transverse reinforcement from:

Yes (cot y f = 2.0) Yes (cot y f = 1.25)

Is v RD > v Ed? Is v RD > v Ed?

Is length of

flange under considerationin tension?

Determine y f from:

Determine the concrete

strut capacity from Table 8or from:v Rd = 0.195 f ck (1 – f ck /250)

= s

Asf

y f =0.5sin-10.2 f ck (1 – f ck /250)

v Ed

f yd cot y f

v Ed hf

Table 8Concrete strut capacity for longitudinal shear in flanged beams

f ck v Rd,max

Flange in compression Flange in tension

Minimum area of shear reinforcement

r

4 r

Longitudinal shear

Rules for spacing andquantity of reinforcementMinimum area of longitudinal reinforcement

Maximum area of longitudinal reinforcement

Minimum spacing of reinforcement

Table 9Values for r w,min

f ck 20 25 28 30 32 35 40 45 50

r

31



32

References1

2

3

4

5

7

6

r

d d

a

d

g g g

r R

r

a g

S eulaVnoitinif eDlobm y Sy eulaVnoitinif eDlobm

Selected symbols

32

6



Designing to Eurocode 2

Design procedure

Fire resistance

R Moss O Brooker



34

Table 1Column design procedure

Step Task Further guidance

Chapter in the publication Standard

Note

Table 2 Minimum column dimensions and axis distances for fire resistance

Standard fireresistance

Minimum dimensions (mm)

Column width bmin/axis distance, a, of the mainbars

Column exposed on more thanone side

Columnexposed on

one side( µfi = 0.7) µfi = 0.5 µfi = 0.7

Note

1

2

3

4

Key

a

b

Table 3 Minimum reinforced concrete wall dimensions and axis distances forload-bearing for fire resistance

Standardfireresistance


Wall thickness/axis distance, a, of the main bars

Wall exposed on one side( µfi = 0.7)

Wall exposed on twosides ( µfi = 0.7)

Notes

1

2

Key

a

Figure 1 Section through structural member, showing nominal axis distance a

h ≥ b

a

b







37

Table 4Effective length factor, F, for braced columns

k 2 k 1

0.10 0.20 0.30 0.40 0.50 0.70 1.00 2.00 5.00 9.00 Pinned

0.10 0.59 0.62 0.64 0.66 0.67 0.69 0.71 0.73 0.75 0.76 0.77

0.20 0.62 0.65 0.68 0.69 0.71 0.73 0.74 0.77 0.79 0.80 0.81

0.30 0.64 0.68 0.70 0.72 0.73 0.75 0.77 0.80 0.82 0.83 0.84

0.40 0.66 0.69 0.72 0.74 0.75 0.77 0.79 0.82 0.84 0.85 0.86

0.50 0.67 0.71 0.73 0.75 0.76 0.78 0.80 0.83 0.86 0.86 0.87

0.70 0.69 0.73 0.75 0.77 0.78 0.80 0.82 0.85 0.88 0.89 0.90

1.00 0.71 0.74 0.77 0.79 0.80 0.82 0.84 0.88 0.90 0.91 0.92

2.00 0.73 0.77 0.80 0.82 0.83 0.85 0.88 0.91 0.93 0.94 0.95

5.00 0.75 0.79 0.82 0.84 0.86 0.88 0.90 0.93 0.96 0.97 0.98

9.00 0.76 0.80 0.83 0.85 0.86 0.89 0.91 0.94 0.97 0.98 0.99

Pinned 0.77 0.81 0.84 0.86 0.87 0.90 0.92 0.95 0.98 0.99 1.00

105 kNm

105 kNm 105 kNm

105 kNm 105 kNm

r m = 1.0 r m = -1.0r m = 0

a) C = 1.7 - 1 = 0.7 b) C = 1.7 - 0 = 1.7 c) C = 1.7 + 1.0 = 2.7

0

Figure 6Calculating factor C

Slenderness

l l

l

l lim =

20 ABC ≤

15.4C

n n

h h

ω

l l

Column design resistance

outside

within

0.00175

h x

x x

hingepoint

hingepoint

0.00175 / ( /2) x x – h

h d

h/ 2

0.0035 max 0.0035 max

0.00175 min

d) General relationshipc) When x < hb) When x > ha) Pure compression

0.00175

e x e x

Figure 8Strain diagrams for columns

a) Strain diagram b) Stress diagram

h

d c

s sce cu2

e sc

e yd2

s st

n. axis

d 2

x

f cd

As2

As

Figure 7Stress block diagram for columns



38

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.3

1.2

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45

0

0.1

0.2

0.30.4

0.5

0.6

0.7

0.8

0.9

1.0

A f bhf s yk ck/

M/bh f 2

ck

N

b h f

/

c

k

K r

= 0.2

d h2/ = 0.05

Figure 9a

Column design chart for rectangular columns d 2 / h = 0.05

σ σ

σ σ

l

l

σ σ

Creep

h

Biaxial bending

MEdz

a+ MEdy

a≤ 1.0 MRdz MRdy



39

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0

0.1

0.2

0.30.4

0.5

0.7

0.9

1.0

d h2/ = 0.15

1.3

1.2

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

A f bhf s yk ck/

N

b h f

/

c k

M bh f / 2

ck

K r =1

0.6

0.8

Figure 9c


0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.3

1.2

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45

M bh f / 2

ck

d h2/ = 0.10

K r =1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

N

b h f

/

c k

A f bhf s yk/ ck

Figure 9bColumn design chart for rectangular columns d 2 / h = 0.10



40

A f bhf s yk ck/

d h2/ = 0.25

M/bh f 2

ck

N

b h f

/

c k

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.3

1.2

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0 0.05 0.10 0.15 0.20 0.25 0.30

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

K r = 1

Figure 9e


0 0.05 0.10 0.15 0.20 0.25 0.30 0.350

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.1

1.30.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

A f bhf s yk ck/

d h2/ = 0.20

M/bh f 2

ck

N

b h f

/

c k

K r = 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Figure 9dColumn design chart for rectangular columns d 2 / h = 0.20



41

Table 5 Value of a for rectangular sections

NEd /NRd 0.1 0.7 1.0

Note

Unbraced columns

Walls

Rules for spacing andquantity of reinforcementMaximum areas of reinforcement

Minimum reinforcement requirements

Symbol Definition Value

ε

h

h

α γ

h

ω

α

β λ

ε

γ γ γ

l

l

h h ∞

h

∞

ω

Selected symbols



42

References 1

2 3

4

5

6

7

8

9

10

11

Spacing requirements for columns

Particular requirements for walls



Eurocode 7: Geotechnical designScope

Limit states

EQU

STR

GEO UPL

HYD

Geotechnical Categories

R Webster O Brooker





45

Geotechnical design report

Spread foundations

Table 4Design values of actions derived for UK design, EQU ultimate limitstate – persistent and transient design situations

CombinationExpressionreference

Permanent actions Leadingvariableaction

Accompanying variableactions

Unfavourable Favourable Main(if any)

Others

c

Key

a

b g

c c

Table 5Presumed allowable bearing values under static loading (from BS 8004)

Category Type of soil Presumed allowable bearing value (kN/m2) Remarks

Note



46

Reinforced concrete pads

Figure 1Procedures for depth of spread foundations

Design foundation (structural design) using the worst ofCombinations 1 and 2 (ULS) for actions and geotechnical

material properties.

START

Design usingdirect method?

Obtain soil parameters from Ground Investigation report

Size foundation(geotechnical design) usingthe worst of Combinations

1 or 2 (ULS) for actionsand geotechnical materialproperties. Combination 2

will usually govern.

Use prescriptive method.Size foundation

(geotechnical design)

using SLS for actionsand presumed

bearing resistance

Is there anoverturning moment?

Check overturning using EQUlimit state for actions and

GEO Combination 2for material properties.

Yes No

Yes

No

MM M

P

PPe

eee = /M P

P

P

L

L

L

L1 +

e6

6e

1

2P

1.5 3L e

L = width of base

SLS pressure distributions ULS pressure distribution

or

P

2L e

P P P

Figure 2Pressure distributions for pad foundations



47

START

Determine value of factor β (β =1.0 when applied moment is zero; refer to Expressions

(6.38) to (6.42) from BS EN 1992–1–1 for other cases)

Determine value of v Ed,max (design shear stress at face of column) from:

v Ed,max = β (V Ed – DVEd) (from Exp. (6.38))

(u0d eff )

where u0 is perimeter of column(see Clause 6.4.5 for columns at base edges)

d eff = (d y + d z)/2 where d y and d z are the effective depths in orthogonal directions

Determine value of v Rd,max (refer to Table 7)

Determine concrete punching shear capacity v Rd (withoutshear reinforcement) from 2dv Rd,c/a (Refer to Table 6 for v Rd,c)

Yes

Either increase mainsteel, or providepunching shear

reinforcement required.(Not recommended

for foundations.)

No

No shear reinforcement required. Check complete.

Figure 4Procedure for determining punching shear capacity for pad foundations

Yes

Redesign foundationIs v Ed,max < v Rd,max?No

Determine value of v Ed, (design shear stress) from:v Ed = (V Ed – DV Ed)

(u1d eff )where u1 is length of control perimeter (refer to Figure 5). For

eccentrically loaded bases, refer to Exp. (6.51).

The control perimeter will have to be found through iteration;it will usually be between d and 2d

Is v Ed < v Rd atcritical perimeter?

Design for punching shear

Table 6v Rd,c resistance of members without shear reinforcement, MPa

r l Effective depth, d (mm)

300 400 500 600 700 800 900 1000a

0.25% 0.47 0.43 0.40 0.38 0.36 0.35 0.35 0.34

0.50% 0.54 0.51 0.48 0.47 0.45 0.44 0.44 0.43

0.75% 0.62 0.58 0.55 0.53 0.52 0.51 0.50 0.49

1.00% 0.68 0.64 0.61 0.59 0.57 0.56 0.55 0.54

1.25% 0.73 0.69 0.66 0.63 0.62 0.60 0.59 0.58

1.50% 0.78 0.73 0.70 0.67 0.65 0.64 0.63 0.62

1.75% 0.82 0.77 0.73 0.71 0.69 0.67 0.66 0.65

≥2.00% 0.85 0.80 0.77 0.74 0.72 0.70 0.69 0.68

k 1.816 1.707 1.632 1.577 1.535 1.500 1.471 1.447

Key

a

Notes1 r

√ r √ r r r r

2 r

f 25 28 32 35 40 45 50

Factor

Beam shear faces

2d

d

d

h

Bends mayberequired

Punching shear perimeters,(load within deducted from V )Ed

Figure 3Shear checks for pad foundations

b z

2d 2d

b y

u1

u1

Figure 5Typical basic control perimeters around loaded areas



48

y

y

Raft foundations

Piled foundations

Table 7Values for v Rd, max

f ck v Rd,max

a a

bF

hF

Figure 8Dimensions for plain foundations

Stress zone

45o

As contributing to shear capacity

Figure 7Shear reinforcement for pilecaps

Punching shear 5 2d from column face

f /5

f /5

f

Beam shear 5 d from column face

Figure 6Critical shear perimeters for piles



49

Table 8Minimum percentage of reinforcement required

f ck f ctm Minimum % (0.26 f ctm / f yka )

Key

a

Plain concrete foundations

a

s

Rules for spacing and

quantity of reinforcementCrack control

Minimum area of principal reinforcement

Maximum area of reinforcement


Deep elements


a g

d d

a

b

d

g

r

r

c c

c

Selected symbols



50

References 1

2

3

4

5

6

7

8

9



Designing to Eurocode 2

Analysis

Design procedure

R Moss O Brooker



52

Table 1Flat slab design procedure

Step Task Further guidance

Chapter in this publication Standard

Note

Table 2Minimum dimensions and axis distances for reinforced concrete slabs

Standard fire resistance Minimum dimensions (mm)

Slab thickness, hs Axis distance, a

Notes

1

2

3

4

5

Key

a

Fire resistance

f f

Flexure



53

Table 4Values for K ’

% redistribution d (redistribution ratio) K ’

Keya

Table 5 z/d for singly reinforced rectangular sections

K z/d K z/d

Keya

Table 6Minimum percentage of reinforcement required

f ck f ctm Minimum % (0.26 f ctm / f yka )

Keya

Figure 1Procedure for determining flexural reinforcement

Table 3Bending moment coefficients for flat slabs

End support/slab connection Firstinteriorsupport

Interiorspans

Interiorsupports

Pinned Continuous

Endsupport

Endspan

Endsupport

Endspan

Notes

1

2 3

4

Carry out analysis of slab to determine design moments(M) (Where appropriate use coefficients from Table 3).

No compression reinforcement required

Check maximum reinforcement requirements. As,max = 0.04 Ac for tension or compression

reinforcement outside lap locations

Determine K ’ from Table 4 orK ’ = 0.60d – 0.18d2 – 0.21 where d ≤ 1.0

START

Yes

Yes

Outside scope ofthis publication

Concrete class≤C50/60?

No

Compressionreinforcementrequired – not

recommended fortypical slabs

Is K ≤ K ’ ?No

Determine K from: K =Determine K from: K =bd 2 f ck

M

Obtain lever arm z from Table 5 or

1 + ≤ 0.95d z =2d

1 – 3.53 K[ ]

Calculate tension reinforcement required from

As = f yd z

M

Check minimum reinforcement requirements (see Table 6)

As,min = where f yk ≥ 25 f yk

0.26 f ctm bt d



54

Deflection

Punching shear

b b

Figure 2Simplified rectangular stress block for concrete up to class C50/60from Eurocode 2

Figure 4Basic span-to-effective-depth ratios for flat slabs

Figure 3Procedure for assessing deflection

Is basic l/d x F1 x F2 x F3 ≥ Actual l/d ?

Increase As,prov

Determine Factor 3 (F3)F3 = 310/s s

Where s s = Stress in reinforcement at serviceabilitylimit state (see Figure 5)

s s may be assumed to be 310 MPa (i.e. F3 = 1.0)

Note: As,prov ≤ 1.5 As,req’d (UK National Annex)

Determine basic l/d from Figure 4

Check complete

Determine Factor 1 (F1)For ribbed or waffle slabs

F1 = 1 – 0.1 ((bf /bw) – 1) ≥ 0.8† (bf is flange breadth and bw is rib breadth)

Otherwise F1 = 1.0

Determine Factor 2 (F2)Where the slab span exceeds 8.5 m and it supports

brittle partitions, F2 = 8.5/leff

Otherwise F2 = 1.0

No

Yes

START

† The Eurocode is ambiguous regarding linear interpolation. It is understood that thiswas the intention of the drafting committee and is in line with current UK practice.

Notes

1

2 r

3

r r

r r

r r r

r

r

[ ]

r

r r

r [ ( ) ]

f ck = 50

f ck = 45

f ck = 40

f ck

= 35

f ck = 32

f ck = 30

f ck = 28

f ck = 25

f ck = 20

Percentage of tension reinforcement ( As,req/bd )

39

37

35

33

31

29

27

25

23

21

19

17

150.40% 0.60% 0.80% 1.20% 1.80%

Span-to-effective-depth

ratio

( l /

d )

1.00% 1.40% 1.60% 2.00%



55

Figure 5Determination of steel stress

Figure 6Procedure for determining punching shear capacity

START

Determine value of factor β (refer to Figure 7 or Expressions (6.38) to (6.46)

of the Eurocode)

Determine value of v Ed (design shear stress at face of column) from:

v Ed = β V Ed /(ui d eff )where ui is perimeter of column

d eff = (d y + d z)/2 (d y and d z are the effective depths in orthogonal directions)

Determine value of v Rd,max from Table 7

Determine value of v Ed, (design shear stress) from:

v Ed = β V Ed /(ui d eff )where u1 is length of control perimeter (see Figure 8)

Yes

Redesign slabIs v Ed, v Rd?No

Determine concrete punching shear capacity(without shear reinforcement),v RD,c from Table 8

where r l = (r ly r lz)0.5

(r ly, r lz are the reinforcement ratios in two orthogonaldirections for fully bonded tension steel, taken over a

width equal to column width plus 3d each side.)

Is v Ed > v Rd,c?

Yes

Punching shearreinforcement not

required

No

Determine area of punching shear reinforcement per perimeterfrom:

Asw = (v Ed – 0.75v Rd,c) sr u1/(1.5 f ywd,ef )where

sr is the radial spacing of shear reinforcement (see Figure 9) f ywd,ef = 250 + 0.25 d eff ≤ f ywd (see Table 9)

Determine the length of the outer perimeter where shearreinforcement not required from:

uout,ef = b V Ed/(v Rd,c d )

Determine layout of punching shear reinforcement(see ‘Spacing of punching shear reinforcement’

Section and Figure 9).

To determine stress in the provided reinforcement (s s), calculate the ratioGk /Qk , read up the graph to the appropriate curve and read across todetermine s su.

s s can be calculated from the expression: s s = s su As,req

As,prov

1

d( ) )

Ratio Gk /Qk

A p p r o x i m a t e s t e e l s t r e s s a t

S L S f o r A

s , r e q

180

200

220

240

260

280

300

320

1.0 2.0 3.0 4.0

c 2

= 0.8, g G

= 1.35

c 2 = 0.6, g G = 1.35

c 2 = 0.3, g G = 1.35

c 2 = 0.2, g G = 1.35

c 2 = 0.6, g

G = 1.25

c 2 = 0.3, g G = 1.25

c 2 = 0.2, g G = 1.25

(c

2 is the factor for quasi-permanent value of a variable action. For furtherexplanation refer to How to design concrete structures using Eurocode 2:Introduction to Eurocodes3.

A p p r o m x i m a t e s t e e l s t r e s s a t S L S f o r A

s , r e q , s s u



56

Rules for spacing andquantity of reinforcement

Minimum area of reinforcement

Maximum area of reinforcement


Maximum spacing of main reinforcement

b = 1.5

b = 1.4 b = 1.15

Internal column

Corner column

Edge column

Figure 7Recommended standard values for b

Table 8v Rd,c resistance of members without shear reinforcement, MPa

r I Effective depth, d (mm)

≤200 225 250 275 300 350 400 450 500 600 750

0.25%

0.50%

0.75%

1.00%

1.25%

1.50%

1.75%

≥ 2.00%

k

Notes

1 r I

R r I Rr r r r

2 r

I

f 25 28 32 35 40 45 50

Factor

Figure 8Typical basic control perimeters around loaded areas

b z

2d 2d

b y

u1

u1

Table 7Values for v Rd,max

Table 9Values for f ywd,ef

f ck v Rd, max d eff f ywd,ef



57

Table 10Factor, F , for determining Asw, min

f ck Factor, F

Note

Spacing of punching shear reinforcement


a g

d d

a

d

g g g

r R

r

r

Selected symbols

Figure 9Punching shear layout

sr

Outer controlperimeter

s 0.75r ≤ d

≤ 0.5 d

kd

Section A - A

st

Outer perimeter of shear reinforcement

Outer controlperimeter uout

0.5 d

≤ 0.75d

≤1.5

(2 if > 2from column)

d

d d

A A≤1.5d



58

References 1

2

3 4

5

6

7

8

9



Methods for checking deflection

Overview

What affects deflection?There are numerous factors

that affect deflection. These

factors are also often time-

related and interdependent,

which makes the prediction

of deflection difficult.

The main factors are:

Concrete tensile strength

Creep

Elastic modulus

Other factors include:

Degree of restraint

Magnitude of loading

Time of loading

Duration of loading

Cracking of the concrete

Shrinkage

Ambient conditions

Secondary load-paths

Stiffening by other elements

R Webster O Brooker



60

Factors affecting deflection

Tensile strength

Creep

h

Elastic modulus

Figure 1Typical floor layouts

a) Favourable layout of restraining walls (low restraint)

b) Unfavourable layout of restraining walls (high restraint)



61

Cracking

K f W ctm= 0.5^ h

z

h

h h

Loading sequence

Figure 2Loading history for a slab – an example

0

2

4

6

8

10

12

14

0 50 100 150 200 250 300

Duration (days)

L o a d ( k N / m ) a

b

c

de

f

gh

Loading sequence

a Slab struck

b 1st slab above cast

c 2nd slab above cast

d 3rd slab above cast

e Floor finishes applied

f

g Quasi-permanent variable actions

h Frequent variable actions

Partitions erected



62

z

Shrinkage curvature

Methods forcalculating deflections

Rigorous method

Panel 1Determining long term elastic modulus of elasticity

E E

W

E

W

E

W

E

W

E

W

eff,1

1

eff, 2

2

eff, 3

3

eff, 4

4

eff, 5

5LT = + + + +R W c m

h

h

Figure 3Outline of rigorous method for calculating deflection

Collate input data

Determine the curvature of the slab

Assess whether the element has flexural cracking

h a a

z z z

z

] g11r E I

M

fl e c E Ie u

QPMQP–= +g g

R e p e a t

a t 1 / 2 0 p o i n t s f o r a l l t h r e e l o a d i n g s t a

g e s



63

Table 1Concrete properties

f ck MPa 320 325 328 330 332 335 340 350

e

e

e

e

e

e

e ∞

Key

a

Panel 2Useful Expressions for a rectangular section

2 1 x

bh A d d Ase s2 2

2

u =

+ - +a] ]g1bh A Ae s s2- +a] ]g g

g+

bh+ + + x - 1I 12bh

2h

A d x x d u

3

u u

2

s Ae 2u s2= - - -aa ] ] ]k g g 22 g6 @

1 2 A b A d d bs 2s2 x A s2 As2 Ac e

2

s e

0.5= + +- +a As eaea a] 1-ea]^ ^ ^g 1- - +ea]h g h g h7 A# -

g g g3 1Ibx

A d x cc3

e s As2 d 22 2

c x c= + - - -+a ea^ ^ ^1r I

S

cscs e

u

u= + -1g g f a cs ef a^ h

I

S

c

c

a

100 300 500 700 900 1100 1300

ho (mm)h ?( , )t 0

C20/25C25/30

C30/37C35/45

C40/50C45/55

C50/60

100 500 900 1300h0 (mm)h

100 100

N NR RS S

t 0 t 0

1 1

2 2

3 35 5

10 10

20 20

50 50

30 30

7.0 7.06.0 6.05.0 5.04.0 4.03.0 3.02.0 2.01.0 1.00 0( ,? t )0

A

E B

D

C

a) Inside conditions - RH = 50% b) Outside conditions - RH = 80%

1100700300

Notes1

2

3 Intersection point between lines D & E can also be above point A4 For t 0 > 100 it is sufficiently accurate to assume t = 100

t 0 = age of concrete at time of loading

h0 = 2 Ac/u

KeyHow to use Nonogram

Figure 4Method for determining creep coefficient h (∞,t0)



64

8

Loading Bending moment diagram

M M

M

M

al W

l

0.125

M Wa a l= (1- )

0.0625

al alW /2 W /2

0.104

0.102

M =Wal

2

q

ql

q

q

q

0.125

3 4a48 (1-a)

If a = , K =

MA

MA

MC

MC

MC MC

MB

MB

WalW al

al

al al

a a(4 )

if = = 0.25a l , K

K = 0.083 (1 )

b =

1

80Wl 2

24 (3 4 )a

2

qa l2 2

2

K = 0. 104 ( 1 )

b =M MA + B

2

15.6

ql 2

K

a6

112

12

2

2

End deflection(3 )a a6

load at end = 0.333K

12

b

4

M + MA B

(5 4 )a

3 4a

2 2

=

b

10

Simplified method

Figure 5Simplified method for calculating deflection

Figure 6Values for K for various bending moment diagrams

Calculate the moment, MQP, due to quasi-permanent actions atthe critical section (i.e. mid-span or at support for cantilever)

Calculate creep coefficient, h (∞,t 0), using either Figure 4or Annex B (in which case look-up f cm in Table 1)

START

Calculate flexural curvature 1r E I

M1

n ef f c E Ieff u

QP MQP= + –g g ^ h

Obtain concrete properties, f ctm, and E c28 from Table 1

1 Calculate long term elastic modulus, E eff from: E eff = E c28/[1+h (∞,t 0)]2 Calculate effective modulus ratio, ae from ae = E s/E eff , where E s is

elastic modulus for reinforcement (200 GPa)3 Calculate depth to neutral axis for uncracked condition, x u4 Calculate second moment of area for uncracked condition, Iu

Calculate depth to neutral axis for crackedcondition, x c and calculate second moment of area

for cracked condition, Ic

Yes

FinishNo

Calculate the deflection that will occur at the time of application ofthe load due to partitions and/or cladding.

1 Calculate the creep coefficient h (t,t 0), where t is the age whenpartition/cladding loads are applied and t 0 is the age of striking.h (t,t 0) ≈ h (∞,t 0) b c(t,t 0). For b c(t,t 0) refer to Figure 7, alternativelyrefer to Annex B of Eurocode 2.

2 Calculate the moment due to self-weight, partitions/cladding and anyother loads which have been applied prior to the installation of thecladding/partition, Mpar and use in place of MQP

3 Recalculate the section properties, curvature and hence deflection,dpar , using h (t,t 0) or equivalent instead of h (∞,t 0)

4 The approximate deflection affecting cladding and partitions isd = dQP – dpar

Calculate cracking moment, Mcr from:0.9

M h

f Icr

x u

ctm u=

–(Note the factor 0.9 has been introduced into this method

because the loading sequence is not considered)

Yes No

Section is uncrackedz = 0

Is Mcr > MQP?

Section is crackedz = 1 – 0.5(Mcr /MQP)2

Calculate total shrinkage strain ecs from ecs = ecd + eca where:

ecd = k h ecd,0 = Drying shrinkage straink h = Coefficient based on notional size, see Table 2ecd,0 = Nominal unrestrained drying shrinkage, see Table 1eca = b as(t ) e ca(∞) = e ca(∞) for long-term deflection, see Table 1

Calculate curvature due to shrinkage strain 1/r cs (see Panel 2)

Calculate total curvature = +1

r t, csnQP

1 1r r

Calculate quasi-permanent deflection from 1

KLQP2=d r t,QP

where K can be obtained from Figure 6 and L is the span.

Do you needto calculate deflectiondue to cladding and

partitions?



65

100 300 500 700 9000.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

Coefficient,

b c(

t ,

t 0)

h0 (mm)

t = 90, t 0 = 7

t = 60, t 0 = 7

t = 28, t 0 = 7

t = 90, t 0 = 3

t = 60, t 0 = 3

t = 28, t 0 = 3

Precamber

Flat slabs

Accuracy

Figure 7Coefficient for development of creep with time after loading

Table 2Values for K h

h0 k h

Notes

Notes

Figure 8Precambering of slabs

Deflection affecting partitions

Just before installationof partitionsPrecamber

Deflection due to

frequent combinationDeflection due toquasi-permanentcombination

Figure 9Recommended acceptance criteria for flat slabs

a

X

Notes

d d



66

References 1

2

3

4

5

6

7

8

9

10

Cladding tolerances



Introduction

Geotechnical Categories

Limit states

A J Bond O Brooker A J Harris



682

g c

g

c c c c

g

g

h h h h g h

Calculation models forstrength limit states

favourable

unfavourable

■

■

■

■

■

Figure 1Ultimate limit states for reinforced concrete retaining walls

a) Overall stability

b) Sliding c) Toppling

d) Bearing e) Structural failure

Table 1Partial factors to be used for retaining wall design according todesign approach 1 (UK National Annex)

Combin-ation

Partial factors onactions

Partial factors on materialproperties of soil

g Ga g G,fav g Q g ϕ

b g c g cu g γ

Keya g

b g h h

68



693

Calculation model A

c = b c h h

Figure 2Calculation model A

h

a

Figure 3Calculation model B

h

c c h h

Calculation model B

69



704

inclined

y d

y

d

h

Figure 4Overall design procedure

Design procedure

both combinations.

Overall stability of the site

Initial sizing

8

Figure 5Symbols for initial sizing

70



715

Material properties

g

h

Figure 6Procedure for determining material properties, geometry and actions

From Figure 4

Determine characteristic material properties g h

g

h

g

Determine initial geometry and actions (see Panel 1)1

(y 2 3

Σ

Σ

Carry out separately for both Combinations 1 and 2

Determine design material properties and earthpressures (see Panel 2)1

g g g g g g γ h h g φ

g

2

h b

To Figure 4

Panel 1General expressions for geometry and actions

g

g

For calculation model A:

b

g

&

For calculation model B:

& g

y

b

W b

W y

Panel 2General expressions for material properties and earth pressures

For calculation model A:

b R h

b

b R h

b b

g

g g

For calculation model B:

h

b

b

h

h

h

h

b h y h

b

g

b b y

g

g g

g

71





73

Figure 8Effects of shear key

0

Panel 4Expressions for bearing resistance

g / W W

W

Undrained bearing resistance:

p

R

W

Drained bearing resistance:

g

g

g g

W

g /

W

g

g

Figure 11Bearing capacity factors, N, from ground properties

Angle

gc

,

B e a r i n g

c a p a c i t y

f a c t o r ,

N

significant

Design against bearing failure

Figure 10Procedure for design against bearing failure

From Figure 4

Ω

To Figure 4

73



748

Structural design

10

Compaction earth pressures

s

Figure 12Effective base width, B’

At-rest earth pressures

h R b

b

Figure 13Design procedure for structural design

From Figure 4

11

To Figure 4

Figure 14Compaction earth pressures for structural design of cantileverretaining walls

z J

z K

z D

74



759

Figure 15Pressure diagram for design of reinforced concrete base

Panel 5General expressions for designing walls for compaction pressures

Forces for stem design (see Figure 14)

Forces for heel design (see Figure 15b)

&

&

Forces for toe design (see Figure 15a)

g

R

R

s g

s

R

s

g

pg

pg

g

p

b R h

b

b R

h

b b

b R h

b

b R h

b b

Table 2Typical compaction equipment12

Description of compactionequipment

Mass (kg) Centrifugalvibratorforce (kN)

Designforce (kN)

Panel 6Expressions for the structural design of basement walls for ‘at-rest’pressures

h

R b

s

g

s

s

s

g

s

s

s

s

Forces for stem design

75

Soil pressure abovetoe ignored



7610

s

s

Detailing

Control of cracking

13

14

3

restraint

Figure 16Earth and pore water pressures for structural design of retaining

walls subject to 'at-rest' conditions

z

W

z D

W

■

■

■

■

loading

s

1

Table 3Maximum bar size or spacing to lmit crack width (mm)

Steelstress(s s)MPa

Wmax = 0.3 Wmax = 0.2

Maximumbar size(mm)

Maximumbar spacing(mm)

Maximumbar size(mm)


Note

s

g d

g

d

■

76



7711

s

Large radius bends

f

f

f

f

f

15

Rules for spacing and quantity ofreinforcement

Vertical reinforcement

Figure 17Typical drainage layout for a retaining wall

300 mm widegranular backfill

Weep hole

Drainage pipe

Large radius bendif required

f

f

Horizontal reinforcement

Practical issuesDesign for movement

16

2

3

1

Drainage

77



7878

Construction

References

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16



Introduction

Type and grade of reinforcement

O Brooker



80

Table 1

Notation for steel reinforcement

Type of steel reinforcement Notation

Note

Figure 1Description of bond conditions

a) 45º < a < 90º

Key

b) h < 250 mm d) h > 600 mm

c) h > 250 mm

a

Cover

D

D

D

Anchorage and lap lengths a a

Anchorage of bars and links

Table 2Anchorage and lap lengths for concrete class C25/30 (mm)

Bondcondition,(see Figure 1)

Reinforcement in tension, bar diameter, f (mm) Reinforcementincompression8 10 12 16 20 25 32 40

f

f

f

f

a

f

f

a

f

f

Notes

1 a

2 a a a a

3 s s

4

5

6

Concrete class C20/25 C28/35 C30/37 C32/40 C35/45 C40/50 C45/55 C50/60



81

Bars in compression

f

Figure 2Design anchorage length lbd, for any shape measured along thecentreline

f lbd

Figure 3Anchorage of links

a) Bend angle > 150º b) Bend angle 150º

f f

f

f

Figure 4Percentage of lapped bars in one lapped section

a

Figure 5Arranging adjacent lapping bars

f

f

f

Arrangement of laps

1.

f

2.

3.

f

Transverse reinforcementBars in tension

f

f

Σ

f



82

BeamsCurtailment

y

y

Figure 6Transverse reinforcement for lapped splices

a) Bars in tension

b) Bars in compression

f f

Table 3Bar sizes for transverse reinforcement

Lap length(mm), fortransversebars at 150mm centresa

Numberof barsat eachlap

Bar size (mm)

20 25 32 40

As = 314 As = 491 As = 804 As = 1260

Key

a

Figure 7Illustration of curtailment of longitudinal reinforcement

D

D

Figure 8Simplified detailing rules for beams

a) Continuous member, top reinforcement

l a l

l a

b) Continuous member, bottom reinforcement

l a

l

c) Simple support, bottom reinforcement

l

Notes

1

2

3

4

5 6

7

8



83

Reinforcement in end supports

s

Flanged beams

Minimum area of longitudinal reinforcement


Figure 9

Placing of tension reinforcement in flanged cross section

beff

beff1 beff2

As

bw

hf


Shear reinforcement

r r

SlabsCurtailment

Reinforcement in end supports

Table 4

Minimum percentage of reinforcement required

f ck f ctm Minimim percentage(0.26 f ctm / f yka)

r w, min

x 10-3

Keya



84

Minimum spacing requirements

Maximum spacing of reinforcement

s s

s

Figure 10Simplified detailing rules for slabs

a) Continuous member, top reinforcement

l l

l

b) Continuous member, bottom reinforcement

l

c) Simple support, bottom reinforcement

l

Notes

1

2

3

4

5 6

7

Minimum areas of reinforcement


Edge reinforcement

Flat slabs

Figure 11Edge reinforcement for slab

≥ 2h

h

Table 5Values for r w,min

Steelstress

(e s) MPa

w max = 0.4 mm w max = 0.3 mm

Maximumbar size(mm)

OR


Maximumbar size(mm)

OR


Table 6Apportionment of bending moments in flat slabs – equivalent frame method

Location Negative moments Positive moments

Notes



85

Figure 12Division of panels in flat slabs

lx > l y

l y/4 l y/4

l y

l y/4

l y/4

Middle strip = lx – l y /2

Middle strip = l y /2

Column strip = l y/2

Punching shear reinforcement

√

Figure 13

Effective width, be of a flat slab

a) Edge column b) Corner columnNote

Figure 14Punching shear layout

s d

d

kd

u

d

d

d

d

d

d



86

Columns and wallsMaximum areas of reinforcement

Minimum reinforcement requirements

Particular requirements for walls

Table 7

Factor, F , for determining Asw, min

f ck

Factor, F

Note

Table 8Requirements for column reinforcement

Bar dia. (mm) 12 16 20 25 32 40

Keya

b

c

Lapping fabric

f

f

f

f

Tolerances

Table 9Minimum area of vertical reinforcement in walls (half in each face)

Wall thickness (mm) As,min/m length of wall (mm2)

Table 10Tolerance

Cutting and bending processes Tolerance (mm)

Bending:

Table 11Deductions to bar dimensions to allow for deviations between twoconcrete faces

Distance betweenconcrete faces (mm)

Type of bar Total deduction(mm)



87

Figure 15Lapping of welded fabric

a) Intermeshed fabric (longitudinal section)

b) Layered fabric (longitudinal section)

Tying requirements

Minimum radii and endprojections

Table 12Minimum scheduling radii and bend allowances

r P≥ 5d

Nominalsize of bar,d (mm)

Minimumradius forscheduling, r (mm)

Mimimum end projection, P

General (min 5d straight), includinglinks where bend >150° (mm)

Links where bend< 150° (min 10d straight) (mm)

Key

a

References 1

2

3

4

5

6

7

8

9

10

11

12

13





89

Bond condition(see Figure 1)

Reinforcement in tension, bar diameter, f (mm) Reinforcementin compression

8 10 12 16 20 25 32 40

Concrete class C35/45

l

f

f

f

f

l

a

f

f

a

f

f


l

f

f

f

f

l

a

f

f

a

f

f


l

f

f

f

f

l

a

f

f

a

f

f


l

f

f

f

f

l

a

f

f

a

f

f

Notes

1 a

2 a a a a

3 s s

4

5



9090

Table 14Sectional areas of groups of bars (mm2)

Bar size(mm)

Number of bars

1 2 3 4 5 6 7 8 9 10

Table 15Sectional areas per metre width for various spacings of bars (mm2)

Bar size(mm) Spacing of bars (mm)75 100 125 150 175 200 225 250 275 300

Table 16Mass of groups of bars (kg per metre run)

Bar size(mm)

Number of bars

1 2 3 4 5 6 7 8 9 10

Table 17Mass in kg per square metre for various spacings of bars (kg per m 2)

Bar size(mm)

Spacing of bars (mm)

75 100 125 150 175 200 225 250 275 300



91



92

92





94

94



95

95



96

96





98

98



99

Introduction

A S Fraser A E K Jones

Start

Finish

Can thetabular method conditions

be met?

Is the element abraced column?

Is the element aslab or beam?

Is there anacceptable solution?

Is there anacceptable solution?

Use simplified methods

Use tabular method

Use 500oc isotherm methodor zone method

Use Annex C of Part 1–2:Buckling of columns under fire

Use Annex E of Part 1–2:Simplified calculation method

for beams and slabs

NoNo

No

No

Yes

Yes

Yes

No

Yes

Yes



100

Basic concepts

Fire types

Level of protection

R

E

I

Material factors

g

y

y y

Combinations of actions

n n

n c

Coefficient k c(y ( f ck) of concrete

1.0

0.8

0.6

0.4

0.2

00 200 400 600 800 1000 1200

Temperature, ( C)o

Coefficient,

(

)

kc

y

Calcareousaggregates

Siliceousaggregates

y

Coefficient k s(y f ck)of tension and compression reinforcement (class N)

1. 0

0.8

0.6

0.4

0.2

00 200 400 600 800 1000 1200

Hot-rolled tensionreinforcement, 2%s,fi

Cold-worked tensionreinforcement,

Compression

reinforcement andtension reinforcement,where < 2%

Temperature, ( C)o

y

Coefficient,

(

)

k s

y

e

2%s,fie

s,fie

Coefficient k p(y (b f pk

1.0

0.8

0.6

0.4

0.2

00 200 400 600 800 1000 1200

Quenched and temperedprestressing steel (bars)

Cold-worked prestressingsteel (wires and strands)Class A

Cold-workedprestressingsteel (wires andstrands) Class B

Temperature, ( C)o

y

Coefficient,

(

)

k p

y



101

c c

n

n

n c c

n c

c

c

c

Explosive spalling

Concrete falling off the section

Tabular method

n

m n m

n fi

0.90

0.80

0.70

0.60

0.50

0.40

0.30

0.200 0.5 1.0 1.5 2.0 2.5 3.0

Reductionfactor,

f i

1, 1 = 0.2

1, 1 = 0.9

1, 1 = 0.5

1, 1 = 0.7

Ratio,

c n

c

c

c

n fi

0.90

0.80

0.70

0.60

0.50

0.40

0.30

0.20

0 0.5 1.0 1.5 2.0 2.5 3.0

0, 1 = 0. 5 ; 1, 1 = 0. 2

0, 1 = 0.7 ; 1, 1 = 0.5

0, 1 = 0.7 ; 1, 1 = 0.7

0, 1 = 1.0 ; 1,1 = 0.9c c

c c

c c

c c

Ratio,

Reductionfactor,

f i

n

a

a a

b

b

asd

h b





103

Method B

l

w

w

Walls

Table 2

Standard fire resistance Mechanicalreinforcement ratio, w

Minimum dimensions (mm). Column width bmin a

n n n n

a a a a

a a a a

a a a a

a a a a

a a a a

a a a

a a a a

a a a

a a a

a a a

a a a

a a

a a b

a a

a a b

a a b

a a b

a b

Key

a

b



104

Minimum wall thicknesses for walls

Standardfireresistance

Non-wallthickness(mm)

Minimum dimensions (mm) a

mfi mfi

Number of sides of

Number of sides of

One Two One Two

b b b b b

b b b b b

b b b

b

b

b

Key

a

b



Simply supported beams Continuous beams

Possible combinations of a and bmin where a is the bmin is the width of beam

Webthickness bw

Possible combinations of a and bmin where a is the bmin is the width of beam

Webthickness bw

2 6 9

a

a

a

a

a

110

130

150

170

Key

a

Notes

2 a a a



105

Tensile members

Slabs

Simply supported slabs

Continuous slabs

0.3 leff 0.3 leff 0.4 leff

BM from Exp. (5.11)

BM in fire location

Design BMaccording toBS EN 1992–1–1

BM when

= 0t

Two-way slabs

a

prestressed solid slabs



One-wayslab

a Flat slab

l y/l ≤ l y/l ≤ 2 d ≤ c d c

b b b b b

b b b

b

Key

a

b

c d

Notes

2

b b b

bw

(a) Constant width (b) Variable width (c) -SectionI



106

Table 6

supported ribbed slabs in reinforced or prestressed concrete



Possible combinations of width ofribs bmin a

Slab thickness hs distance a inSimply supported

restrained

a a a

a a a a

a a

Key

a

Notes

2

Simplified calculation method for beamsand slabs

n

n

Simply supported members

g g y

Flat slabs

Ribbed slabs

Simplified calculationmethods



107

g

g

y

y

y

Continuous members

g g

Minimum width of cross-section as function of fire resistance

Fire resistance

Minimum width ofcross-section (mm)

Flow chart for simplified calculation method for beams and slabs

Finish

Is the element asimply supported?

Is MEd, fi. ≤ MRd, fi?

Are the support

moments exceeded?

Calculate the support designmoment of resistance,

MRd, fi, support

‘Fit’ the ‘free’ bending

moment so that MEd,fi = MRd,fi

Redesign section or usealternative methods

Start

Calculate MRd, fi

Calculate MEd, fi.

Determine k s(y ) from Figure 12

Determiney

, using temperature profiles in Annex A of Part 1-2.

Yes

No No

Yes

Yes

No

1.0

0.8

0.6

0.4

0.2

00 200 400 600 800 1000 12

Coefficient,

k s

( y c

r)or

k p

( y c

r)

Reinforcing steel

Prestressing steel (bars)

Prestressing steel(wires and strands)

Temperature, ( C)o

y



108

Reduced cross-section of reinforced concrete beam and column

c) Fire exposure on four sides (beam or column)

500 Co

500 Co500 Co

h hfi

bfi

bfifi

b

bb

b

a) Fire exposure on three sideswith tension zone exposed

d d fi =d fi

b) Fire exposure on three sides withthe compression zone exposed

Compression Tension

Tension Compression

900

100

200

300

400

500

600

700

800

240

220

200

180

160

140

120

100

80

60

40

20

00 20 40 60 80 100 120 140

Distance from bottom left corner of element (mm)

Distancefrom

bottom

leftcornerofelement(mm)

cross-section with compression reinforcement.

As1

As

z ’ d 1

b1

x

Mu1z z’

F A f s )s1 scd,fi m

Mu2

A f s1 sd,fi( )m

f cd, 1(20)

xb f n cd, 1(20) x l

l

n

y

y = (

g

y

y

y

y

Note: y y

l n

y

y



109

Finish

Is MEd,fi ≤ MRd,fi?Redesign section or use

alternative methods

Start

Calculate MEd, fi (see simplified calculation method for beams and slabs)

Check the minimum dimensions exceed the values in Table 7

Determine reduced section size (bfi d fi) using Figure 13 andtemperature profiles in Annex A of Part 1–2

No

Yes

Determine y , using temperature profiles in Annex A of Part 1–2

Determine k s (y ), from Figure 3 or Figure 4

Calculate Mu, using stress distribution shown in Figure 15. Mu = Mu1 + Mu2

Finish

Does the sectionalso resist torsion?

Calculate the referencetemperature at points P alongthe line A–A − see Figure 20

Calculate the torsion resistanceand interaction with shear usingsection 6.3 of Eurocode 2, Part 1–1

Start

Determine the reduced cross-section using either 500°Cisotherm or zone methods

Calculate the compressive and tensile concrete strengths:

For isotherm method, f cd,fi = f cd,fi(20) = f ck and f ctd,fi = f ctd,fi(20) = f ctk For zone method, f cd,fi = k c(y m) f cd,fi(20) and f ctd,fi = k ct(y m) f ctd,fi(20),

where k c(y m) and k ct(y m) may be taken as k c(y ) and can be determined

from Figure 2

Determine position P, the point at which the reference temperature,

y p is calculated. P is located along section A–A, which isdetermined from hc,ef (see Figure 19)

Yes

No

Determine y p using temperature profiles in Annex A of Part 1–2

Calculate the reduced design strength of the shear reinforcement, f sd,fi,from: f sd,fi = k s(y ) f sd(20) = k s(y ) f ywd

where k s(y ) can be determined from Figure 3 or Figure 4

Calculate the shear resistance using the methods given for ambienttemperature design, see Chapter 4 Beams11

Coefficient k c,t(y f ck,t) of concrete at elevated temperatures

1.0

0.8

0.6

0.4

0.2

00 100 200 300 400 500 600

Temperature, ( C)o

C o e

f f i c i e

n t

,

(

)

k c

, t

y

y

Calculation methods for shear and torsion

y



110

Unbraced structures

References

2

6

9

temperature y p at point P

A

AA

c,eff

c,ef = MIN {2.5 ( ); (

hd

x

hc,ef

2 = 0

1

e

e

The reference temperature y p the calculation of torsion resistance

A

AA

A

A

AA

A

p in linksy



Members of the Steering Group

Members of the Concrete Industry Eurocode 2 Group (CIEG)



How to Design Concrete Structures using Eurocode 2

Dr Andrew Bond

Owen Brooker

Dr Andrew Fraser

Andrew Harris

This publication brings together in one place “How to...”

guidance for the design, specification and detailing of a broad

range of concrete elements.

Date post:	05-Jul-2018
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HowTo Design Eurocode 2

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