lecture 2 beams - oct 12 - end

8/12/2019 Lecture 2 Beams - Oct 12 - End

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Practical Design to Eurocode 2

Beams

Bending / Flexure Shear

Detailing

Anchorage & Laps

Members & particular rules


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Beams

Flexure


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Section Design: Bending

Flexural design is generally the same as BS8110 in

principle

Modified for high strength concrete

EC2 presents the principles only

Design manuals will provide the standard solutions for

basic design cases.


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As

d

fcd

Fs

x

s

x

cu3

FcAc

fck50 MPa 50 < fck90 MPa 0.8 = 0.8 (f ck 50)/400 1.0 = 1,0 (fck 50)/200fcd = ccfck /c

= 0.85fck /1.5

Rectangular Concrete StressBlock (3.1.7, Figure 3.5)


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Maximum neutral axis depth

M

b

MomentBendingElastic

MomentBendingtedRedistribu du21 xkk EC2 Equ 5.10a

xu = Neutral axis depth after redistribution

EC2 NA gives k1 = 0.4 and k2 = 1.0

dux0.4

0.4-d

x u

ck

2fbdMK

Value of K for maximum value of M

with no compression steel and

when x is at its maximum value.If K > K Compression steel required


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Analysis of a Singly ReinforcedBeam3.1.7 EN 1992-1-1

Design equations can be derived as follows:

For grades of concrete up to C50/60, cu= 0.0035, = 1 and = 0.8.fcd = 0.85fck/1.5,

fyd =fyk/1.15

Fc = (0.85 fck/ 1.5) b (0.8x) = 0.453 fckb x

Fst = 0.87As fyk

Take moments about the centre of the tension force

M = 0.453 fckb x z 1

M

b


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0 = (z/ d)2 (z/ d) + 0.88235K

Solvi ng t he quadrat ic equat ion:

z/ d= [1 + (1 - 3.529K)0.5]/2

z = d[ 1 + (1 - 3.529K)0.5]/2

Rearranging

z = d[ 0.5 + (0.25 K / 1.134)0.5]

This compares to BS 8110

z = d[ 0.5 + (0.25 K / 0.9)0.5]

The lever arm for an applied moment is now known


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Higher Concrete Strengthsfck 50MPa )]/23,529K(1d[1z

)]/23,715K(1d[1z fck = 60MPa

fck = 70MPa

fck = 80MPa

fck = 90MPa

)]/23,922K(1d[1z

)]/24,152K(1d[1z

)]/24,412K(1d[1z


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Take moments about the centre of the compression force

M = 0.87Asfyk z

Rearranging

As = M /(0.87fyk z)

The required area of reinforcement can now be:

calculated using these expressions

obtained from Tables (eg Table 5 of How t o beams and ConciseTable 15.5 )

obtained from graphs (eg from the Green Book )

However, it is often considered good practice to limit the depth of the

neutral axis to avoid over-reinforcement (ie to ensure a ductile failure).

This is not an EC2 requirement and is not accepted by all engineers. A

limiting value for Kcan be calculated (denoted K) as follows.


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Design aids for flexureConcise: Table 15.5


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k1 + k2xu/d

where

k1 = 0.4

k2 = 0.6 + 0.0014/ cu2 = 0.6 + 0.0014/0.0035 = 1

xu = depth to NA after redistribution

= Redistribution ratio xu = d (- 0.4)

Substituting forx in eqn 1 above and rearranging:

M = b d2fck (0.6 0.18 2 - 0.21)

K = M /(b d2fck) = (0.6 0.18 2 - 0.21)

From BS 8110 K = (0.55 0.182 0.19) rearranged

Some engineers advocate takingx/d< 0.45, and K < 0.168


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Beams with Compression

Reinforcement

There is now an extra force Fsc

= 0.87As2

fyk

The area of tension reinforcement can now be

considered in two parts, the first part to balance the

compressive force in the concrete, the second part is to

balance the force in the compression steel. The area of

reinforcement required is therefore:

As

= Kfck

b d 2 /(0.87fyk

z) +As2

where z is calculated using K instead of K


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As2 can be calculated by taking moments about the centre of the tensionforce:

M = Kfck b d2 + 0.87fykAs2 (d - d2)

Rearranging

As2 = (K- K)fck b d2 / (0.87fyk (d - d2))


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Calculate lever arm zfrom:

* A limit of 0.95dis considered good practice, it is not a requirement of Eurocode 2.

*95.053.3112

dKd

z

Check minimum reinforcement requirements:

dbf

dbfA t

yk

tctmmin,s 0013.0

26.0

Check max reinforcement provided As,max 0.04Ac (Cl. 9.2.1.1)Check min spacing between bars > bar > 20 > Agg + 5

Check max spacing between bars

Calculate tension steel required from: zf

MA

yd

s

Flow Chart for Singly-reinforced

Beam


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Flow Chart for Doubly-

Reinforced Beam

Calculate lever arm zfrom: '53.3112

Kd

z Calculate excess moment from: '' 2 KKfbdM

ck

Calculate compression steel required from:

2yd2s'

ddfA

Calculate tension steel required from:

Check max reinforcement provided As,max 0.04Ac (Cl. 9.2.1.1)Check min spacing between bars >

bar> 20 > A

gg+ 5

2syd

2

s' A

zfbdfKA ck


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Flexure Worked Example


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Worked Example 1Design the section below to resist a sagging moment of 370 kNm assuming

15% moment redistribution (i.e. = 0.85).

Takefck

= 30 MPa andfyk

= 500 MPa.

d


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Initially assume 32 mm for tension reinforcement with 25 mm nominal cover

to the link (allow 10 mm for link) and 20mm for compression reinforcement

with 25 mm nominal cover to link.Nominal side cover is 35 mm.

d = h cnom -link - 0.5

= 500 25 - 10 16= 449 mm

d2 = cnom +link + 0.5

= 25 + 10 + 10

= 45 mm

449


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provide compression steel

mm368

168.053.3112

449

'53.3112

Kd

z

'204.0

30449300

103702

6

ck2

K

fbd

MK

168.0'K K1.00 0.208

0.95 0.195

0.90 0.182

0.85 0.168

0.80 0.153

0.75 0.137

0.70 0.120


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kNm

KKfbdMck

3.65

10)168.0204.0(30449300

''

62

2

2

6

2yd

2s

mm372

45)(449435

10x65.3

'

ddf

MA

2

6

2s

yd

s

mm2293

372364435

10)3.65370(

'

Azf

MMA

368


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Provide 2 H20 for compression steel = 628mm2 (372 mm2 reqd)

and 3 H32 tension steel = 2412mm2 (2296 mm2 reqd)

By inspection does not exceed maximum area or maximum spacing ofreinforcement rules

Check minimum spacing, assuming H10 links

Space between bars = (300 35 x 2 - 10 x 2 - 32 x 3)/2

= 57 mm > 32 mm OK


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Factors for NA depth (=nd) and lever arm (=zd) for concrete grade 50 MPa

0.00

0.20

0.40

0.60

0.80

1.00

1.20

M/bd2fck

Factor

n 0.02 0.04 0.07 0.09 0.12 0.14 0.17 0.19 0.22 0.24 0.27 0.30 0.33 0.36 0.39 0.43 0.46

z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.84 0.83 0.82

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17

lever arm

NA depth

Simplified Factors for Flexure

(1)


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Factors for NA depth (=nd) and lever arm (=zd) for concrete grade 70 MPa

0.00

0.20

0.40

0.60

0.80

1.00

1.20

M/bd2fck

Facto

r

n 0.03 0.05 0.08 0.11 0.14 0.17 0.20 0.23 0.26 0.29 0.33

z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.91 0.90 0.89 0.88

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17

lever arm

NA depth

Simplified Factors for Flexure

(2)


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Shear in Beams

Shear design is different from BS8110

Shear strength should be limited to the value for

C50/60

The shear effects in links and longitudinal steel have to

be considered explicitly


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Definitions Resistance of member without shear reinforcement

VRd,c

Resistance of member governed by the yielding ofshear reinforcement - VRd,s

Resistance of member governed by the crushing ofcompression struts - VRd,max

Applied shear force - VEd


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Members Requiring Shear

Reinforcement (6.2.3.(1))

s

d

V(cot - cot

V

N M z

zVz = 0.9d

Fcd

Ftd

compression chord compression chord

tension chordshear reinforcement

angle between shear reinforcement and the beam axis

angle between the concrete compression strut and the beam axis

z inner lever arm. In the shear analysis of reinforced concrete

without axial force, the approximate value z= 0,9dmay normallybe used.


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cotswsRd, ywdfzs

AV

tancot

1maxRd,

cdwcw fzbV

21.8 < < 45

Strut Inclination Method


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Shear Design: LinksVariable strut methodallows a shallower strut angle

hence activating more links.

As strut angle reduces concrete stress increases

Angle = 45 V carried on 3 links Angle = 21.8 V carried on 6 links

d

V

z

x

d

x

V

z

s


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shear reinforcement control

VRd,s =Asw z fywd cot/s Exp (6.8)

concrete strut controlVRd,max = z bw1fcd /(cot + tan) = 0,5zbw fcd sin2

Exp (6.9)

where = 0,6(1-fck/250) Exp (6.6N)

1 cot 2,5

Basic equations

d

V

z

x

d

x

V

z

s

Shear Resistance of Sections with

Shear Reinforcement


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fck vRd, cot = 2.5 vRd, cot = 1.0

20 2.54 3.6825 3.10 4.5

28 3.43 4.97

30 3.64 5.28

32 3.84 5.58

35 4.15 6.02

40 4.63 6.72

45 5.08 7.38

50 5.51 8.00


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Shear - Variable strut method Concise Fig 15.1 a)


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Shear - Variable strut method Concise Fig 15.1 b)


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Where av 2dthe applied shear force, VEd, for a point load(eg, corbel, pile cap etc) may be reduced by a factor av/2d

where 0.5 av 2d provided:

dd

av av

The longitudinal reinforcement is fully anchored at the support. Only that shear reinforcement provided within the central 0.75av is

included in the resistance.

Short Shear Spans with Direct

Strut Action (6.2.3)

Note: see PD6687-1:2010 Cl 2.14 for more information


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Shift Rule For members without shear reinforcement this is satisfied with al = d

a lFtd

a l

Envelope of (MEd /z+NEd)

Acting tensile force

Resisting tensile force

lbd

lbd

lbd

lbd

lbd lbd

lbd

lbdFtd

For members with shear reinforcement: Conservatively al = 1.125d


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Shear Example

Design of shear reinforcement using

Eurocode 2


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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 = vEdbw/(fywd cot )

DeterminevEd 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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Calculate

0.35

250/30-130x20.0

96.4sin5.0

)250/1(20.0sin5.0

1

ckck

Ed1

ff

v

43.1cot Asw/ s = vEdbw/ (fywdcot )Asw/ s = 4.96 x 140 / (435 x 1.43)

Asw/ s = 1.12 mm

Tr y H10 l i nks wi t h 2 legs.

Asw = 157 mm2

s < 157 / 1. 12 = 140 mm

pr ovid e H10 l i nks at 125 mm spacing


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Beam Example 1

Cover = 40mm to each face

fck = 30

Determine the flexural and shearreinforcement required

(try 10mm links and 32mm main steel)

Gk = 75 kN/m, Qk = 50 kN/m , assume no redistribution and useequation 6.10 to calculate ULS loads.

8 m

450

1000


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Beam Example 1 BendingULS load per m = (75 x 1.35 + 50 x 1.5) = 176.25

Mult = 176.25 x 82/8

= 1410 kNmd = 1000 - 40 - 10 32/2

= 934

120.030934450

1014102

6

ck

2 fbdMKK = 0.208

K < KNo compression reinforcement required dKdz 95.0822120.0x53.311

293453.311

2

2

6

yd

smm3943

822x435

10x1410 zf

MA

Provide 5 H32 (4021 mm2)


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Workshop Problem


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Workshop Problem

Cover = 35 mm to each face

fck = 30MPa

Check the beam in flexure and shear

Gk = 10 kN/m, Qk = 6.5 kN/m (Use eq. 6.10)

8 m

300

450


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Solution - FlexureULS load per m = (10 x 1.35 + 6.5 x 1.5) = 23.25Mult = 23.25 x 8

2/8

= 186 kNm

d = 450 - 35 - 10 32/2

= 389

137.030389300

101862

6

ck

2 fbdMK

K = 0.208

K < KNo compression reinforcement required dKdz 95.0334137.0x53.311

2

38953.311

2

2

6

yds mm1280

334x435

10x186 zf

MA

Provide 3 H25 (1470 mm2)


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Solution - ShearShear force, VEd = 23.25 x 8 /2 = 93 kN

Shear stress:

vEd

= VEd

/(bw

0.9d) = 93 x 103/(300 x 0.9 x 389)

= 0.89 MPa

vRd = 3.64 MPa

vRd > vEd

cot = 2.5Asw/s = vEd bw/(fywd cot )

Asw/s = 0.89 x 300 /(435 x 2.5)

Asw/s = 0.24 mm

Try H8 links with 2 legs.

Asw = 101 mm2

s < 101 /0.24 = 420 mm

Maximum spacing = 0.75 d= 0.75 x 389 = 292 mm

provide H8 links at 275 mm spacing

fck

vRd (whencot = 2.5)

20 2.54

25 3.1028 3.43

30 3.64

32 3.84

35 4.15

40 4.63

45 5.08

50 5.51


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Detailing

www ukcares co uk


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UK CARES (Certification - Product & Companies)

1. Reinforcing bar and coil2. Reinforcing fabric

3. Steel wire for direct use of for further

processing

4. Cut and bent reinforcement

5. Welding and prefabrication of reinforcingsteel

www.ukcares.co.uk

www.uk-bar.org


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Reinforced Concrete Detailing


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Reinforced Concrete Detailing

to Eurocode 2

Section 8 - General RulesAnchorage

LapsLarge Bars

Section 9 - Particular RulesBeams

SlabsColumns

Walls

Foundations

Discontinuity Regions

Tying Systems

Cover Fire

Specification and Workmanship

Section 8 - General Rules


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Clear horizontal and vertical distance , (dg +5mm) or 20mm

For separate horizontal layers the bars in each layer should be

located vertically above each other. There should be room to allow

access for vibrators and good compaction of concrete.

Section 8 General Rules

Spacing of barsEC2: Cl. 8.2 Concise: 11.2

Min Mandrel Dia for bent bars


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To avoid damage to bar is

Bar dia 16mm Mandrel size 4 x bar diameterBar dia > 16mm Mandrel size 7 x bar diameterThe bar should extend at least 5 diameters beyond a bend

Minimum mandrel size, mMin. Mandrel Dia. for bent bars

EC2: Cl. 8.3 Concise: 11.3

Min. Mandrel Dia. for bent bars


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Minimum mandrel size, m

To avoid failure of the concrete inside the bend of the bar: m,min Fbt ((1/ab) +1/(2 )) /fcdFbt ultimate force in a bar at the start of a bend

ab for a given bar is half the centre-to-centre distance between bars.

For a bar adjacent to the face of the member, ab should be taken as

the cover plus /2Mandrel size need not be checked to avoid concrete failure if :

anchorage does not require more than 5 past end of bend bar is not the closest to edge face and there is a cross bar inside bend mandrel size is at least equal to the recommended minimum value

Min. Mandrel Dia. for bent bars

EC2: Cl. 8.3 Concise: 11.3

Bearing stressinside bends


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Anchorage of reinforcementEC2: Cl. 8.4

Ultimate bond stress


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The design value of the ultimate bond stress,fbd = 2.25 12fctdwherefctd should be limited to C60/75

1 =1 for good and 0.7 for poor bond conditions

2 = 1 for 32, otherwise (132-)/100

a) 45 90 c) h> 250 mm

h

Direction of concreting

300

h


b) h 250 mm d) h > 600 mmunhatched zone good bond conditions

hatched zone - poor bond conditions


250


EC2: Cl. 8.4.2 Concise: 11.5

300


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Design Anchorage Length, lbd


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lbd = 1 2 3 4 5 lb,rqdlb,minHowever: (2 3 5) 0.7lb,min > max(0.3lb,rqd ; 10, 100mm)

g g g , bd

EC2: Cl. 8.4.4 Concise: 11.4.2

Alpha valuesEC2: Table 8.2

Table requires values for:

Cd Value depends on cover and bar spacing, see Figure 8.3

K Factor depends on position of confinement reinforcement,

see Figure 8.4

= (Ast

Ast,min

)/ As

Where Ast

is area of transverse reinf.

Table 8.2 - Cd & K factors


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d

Concise: Figure 11.3EC2: Figure 8.3

EC2: Figure 8.4


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Alpha values


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EC2: Table 8.2 Concise: 11.4.2

Anchorage of links


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g

Concise: Fig 11.2EC2: Cl. 8.5


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LapsEC2: Cl. 8.7

Design Lap Length, l0 (8.7.3)


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l0 = 1 2 3 5 6 lb,rqdl0,min

6 = (1/25)0,5 but between 1.0 and 1.5

where1 is the % of reinforcement lapped within 0.65l0 from the

centre of the lap

Percentage of lapped bars

relative to the total cross-

section area

< 25% 33% 50% >50%

6 1 1.15 1.4 1.5Note: Intermediate values may be determined by interpolation.

1 2 3 5 are as defined for anchorage length

l0,min max{0.3 6 lb,rqd; 15; 200}

EC2: Cl. 8.7.3 Concise: 11.6.2

Arrangement of Laps


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EC2: Cl. 8.7.3, Fig 8.8


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Worked example

Anchorage and lap lengths

Anchorage Worked Example


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Anchorage Worked Example

Calculate the tension anchorage for an H16 bar in the

bottom of a slab:

a) Straight bars

b) Other shape bars (Fig 8.1 b, c and d)

Concrete strength class is C25/30

Nominal cover is 25mm

Assume maximum design stress in the bar


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Basic anchorage length, lb,req


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lb.req = (/4) ( sd/fbd) EC2 Equ 8.3

Max stress in the bar, sd = fyk/s = 500/1.15

= 435MPa.

lb.req = (/4) ( 435/2.693)

= 40.36

For concrete class C25/30

Design anchorage length, lbd


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g g g , bd

lbd = 1 2 3 4 5 lb.req lb,min

lbd = 1 2 3 4 5 (40.36) For concrete class C25/30

Alpha values


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EC2: Table 8.2 Concise: 11.4.2

Table 8.2 - Cd & K factors


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Concise: Figure 11.3EC2: Figure 8.3

EC2: Figure 8.4



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a) Tension anchorage straight bar

1 = 1.0

3 = 1.0 conservative value with K= 0

4 = 1.0 N/A

5 = 1.0 conservative value

2 = 1.0 0.15 (Cd )/

2 = 1.0 0.15 (25 16)/16 = 0.916

lbd = 0.916 x 40.36 = 36.97 = 592mm



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b) Tension anchorage Other shape bars1 = 1.0 Cd = 25 is 3 = 3 x 16 = 48

3 = 1.0 conservative value with K= 0

4 = 1.0 N/A

5 = 1.0 conservative value

2

= 1.0 0.15 (Cd

3)/ 1.0

2 = 1.0 0.15 (25 48)/16 = 1.25 1.0

lbd = 1.0 x 40.36 = 40.36 = 646mm


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Anchorage /lap lengths for slabs


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Table 5.25: Typical values of anchorage and lap lengths for slabs

Bond Length in bar diameters

conditions fck /fcu

25/30

fck /fcu

28/35

fck /fcu

30/37

fck /fcu

32/40

Full tension and

compression anchorage

length, lbd

good 40 37 36 34

poor 58 53 51 49

Full tension andcompression lap length, l0

good 46 43 42 39poor 66 61 59 56

Note: The following is assumed:

- bar size is not greater than 32mm. If >32 then the anchorage and lap lengths should be

increased by a factor (132 - bar size)/100

- normal cover exists- no confinement by transverse pressure

- no confinement by transverse reinforcement

- not more than 33% of the bars are lapped at one place

Lap lengths provided (for nominal bars, etc.) should not be less than 15 times the bar size

or 200mm, whichever is greater.

Manual for the design of concrete structures to Eurocode 2

Arrangement of LapsEC2: Cl. 8.7.2 Concise: Cl 11.6


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Laps between bars should normally be staggered and not located in regions

of high stress, the arrangement of lapped bars should comply with the

following (see Figure 8.7 on next slide):

The clear distance between lapped bars should not be greater than4 or 50 mm otherwise the lap length should be increased by a length

equal to the clear space where it exceeds 4 or 50 mm

The longitudinal distance between two adjacent (staggered?) laps

should not be less than 0,3 times the lap length, l0; In case of adjacent laps, the clear distance between adjacent bars

should not be less than 2 or 20 mm.

When the provisions comply with the above, the permissible percentage of

lapped bars in tension may be 100% where the bars are all in one layer.Where the bars are in several layers the percentage should be reduced to

50%.

All bars in compression and secondary (distribution) reinforcement may be

lapped in one section.

Arrangement of LapsEC2: Cl. 8.7.2, Fig 8.7 Concise: Cl 11.6.2


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g

Transverse Reinforcement at LapsBars in tension Concise: Cl 11.6.4


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Where the diameter, , of the lapped bars 20 mm, the transversereinforcement should have a total area, Ast 1,0As of one spliced bar. Itshould be placed perpendicular to the direction of the lapped

reinforcement and between that and the surface of the concrete.

If more than 50% of the reinforcement is lapped at one point and the

distance between adjacent laps at a section is 10 transverse bars shouldbe formed by links or U bars anchored into the body of the section.

The transverse reinforcement provided as above should be positioned at

the outer sections of the lap as shown below.

l /30A /2

st

A /2st

l /30FsFs

150 mm

l0

EC2: Cl. 8.7.4, Fig 8.9 Rules apply if bar diameter 20mm

Figure 8.9 bars in tension

Transverse Reinforcement at LapsBars in compression

EC2 Cl 8 7 4 Fi 8 9

Concise: Cl 11.6.4


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EC2: Cl. 8.7.4, Fig 8.9

In addition to the rules for bars in tension one bar of the transverse

reinforcement should be placed outside each end of the lap length.

Figure 8.9 bars in compression


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SECTION 9

Detailing of members and particular rules

Beams (9.2)


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As,min = 0,26 (fctm/fyk)btd but 0,0013btd

As,max = 0,04Ac

Section at supports should be designed for a

hogging moment 0,25 max. span moment

Any design compression reinforcement () should be

held by transverse reinforcement with spacing 15

Beams (9.2)


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Tension reinforcement in a flanged beam at

supports should be spread over the effective width

(see 5.3.2.1)

Curtailment (9.2.1.3)


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(1) Sufficient reinforcement should be provided at all sections to resist the

envelope of the acting tensile force, including the effect of inclined cracks

in webs and flanges.

(2) For members with shear reinforcement the additional tensile force, Ftd,

should be calculated according to 6.2.3 (7). For members without shear

reinforcement Ftdmay be estimated by shifting the moment curve a

distance al = d according to 6.2.2 (5). This "shift rule may also be used

as an alternative for members with shear reinforcement, where:

al = z (cot - cot )/2 = 0.5 z cot for vertical shear links

z= lever arm, = angle of compression strut

al = 1.125 d when cot = 2.5 and 0.45 d when cot = 1

Curtailment of reinforcementEC2: Cl. 9.2.1.3, Fig 9.2 Concise: 12.2.2


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For members without shear reinforcement this is satisfied with al = d

a lFtd

a l

Envelope of (MEd /z+NEd)

Acting tensile force

Resisting tensile force

lbd

lbd

lbd

lbd

lbd lbd

lbd

lbd Ftd

Shift rule

For members with shear reinforcement: al = 0.5 zCot But it is always conservative to use al = 1.125d

Anchorage of BottomReinforcement at End Supports

(9 2 1 4)


93/99

lbd is required from the line of contact of the support.

Simple support (indirect) Simple support (direct)

As bottom steel at support 0.25As provided in the span

Transverse pressure may only be taken into account with

a direct support.

Shear shift rule

al

Tensile Force Envelope

(9.2.1.4)

Simplified Detailing Rules for

BeamsConcise: Cl 12 2 4


94/99

Beams

How to.EC2

Detailing section

Concise: Cl 12.2.4

Supporting Reinforcement atIndirect Supports

EC2: Cl 9 2 5

Concise: Cl 12.2.8


95/99

h /31

h /21

B

A

h /32 h /22

supporting beam with height h1

supported beam with height h2 (h1 h2)

The supporting reinforcement is in

addition to that required for other

reasons

A

B

The supporting links may be placed in a zone beyond

the intersection of beams

Plan view

EC2: Cl. 9.2.5

C t il t b t f th Shift l d

Solid slabsEC2: Cl. 9.3


96/99

Curtailment as beams except for the Shift rule al = d

may be used

Flexural Reinforcement min and max areas as beam

Secondary transverse steel not less than 20% main

reinforcement

Reinforcement at Free Edges

Detailing Comparisons


97/99

d or 150 mm from main bar9.2.2 (8): 0.75 d 600 mm

9.2.1.2 (3) or 15 from main bar

st,max

0.75d9.2.2 (6): 0.75 dsl,max

0.4 b s/0.87 fyv9.2.2 (5): (0.08 b s fck)/fykAsw,min

Links

Table 3.28Table 7.3NSmax

dg + 5 mm or 8.2 (2): dg + 5 mm or or 20mmsmin

Spacing of Main Bar s

0.04 bh9.2.1.1 (3): 0.04 bdAs,max

0.002 bh--As,min

Main Bar s i n Compr essi on

0.04 bh9.2.1.1 (3): 0.04 bdAs,max

0.0013 bh9.2.1.1 (1): 0.26fctm/fykbd 0.0013 bd

As,min

ValuesClause / Val uesMain Bar s i n Tensi on

BS 8110EC2Beams


98/99

Detailing Comparisons

BS 8110EC2P hi Sh


99/99

Columns

150 mm from main bar9.5.3 (6): 150 mm from main bar

129.5.3 (3): min (12min; 0.6 b;240 mm)Scl,tmax

0.25 or 6 mm9.5.3 (1) 0.25 or 6 mmMin size

Links

0.06 bh9.5.2 (3): 0.04 bhAs,max

0.004 bh9.5.2 (2): 0.10NEd/fyk 0.002bhAs,min

Main Bar s i n Comp r ession

1.5d9.4.3 (1):

within 1st control perim.: 1.5d

outside 1st control perim.: 2d

St

0.75d9.4.3 (1): 0.75dSr

Spacing of Li nks

Total = 0.4ud/0.87fyv9.4.3 (2): Link leg = 0.053 sr st(fck)/fyk

Asw,min

ValuesClause / Val uesLinks

BS 8110EC2Punching Shear

lecture 2 beams - oct 12 - end

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