Lecture 2 Beams Jb Apr 2013 [Compatibility Mode] (1)

Practical Design to Eurocode 2 Apr 2013 1

Jenny Burridge

Head of Structural Engineering

MPA – The Concrete Centre

Practical Design to Eurocode 2

Course Outline

17th April 2013Basics

EC0, EC1, Materials, Cover

24th April 2013Beams

Bending, Shear, Detailing

1st May 2013Columns

Axial load, Column Moments, Buckling

8th May 2013Slabs

Serviceability, Punching Shear

15th May 2013Foundations

Pads, Piles, Retaining Walls


Bending

Beams

Section Design: Bending

• In principle flexural design is generally the same

as BS8110

• EC2 presents the principles only

• Design manuals will provide the standard

solutions for basic design cases.

• There are modifications for high strength

concrete

Basics


As

d

η fcd

Fs

λx

εs

x

εcu3

Fc Ac

η = 1,0 for fck ≤ 50 MPa

= 1,0 – (fck – 50)/200 for 50 < fck ≤ 90 MPa

400

)508,0 ck −

−=(f

for 50 < fck ≤ 90 MPa

λ = 0,8 for fck ≤ 50 MPa

Rectangular Concrete Stress Block (3.1.7, figure 3.5)

Remember

this from last

week?

Singly Reinforced BeamsEN 1992-1-1Cl 3.1.7

Design equations can be derived as follows:

For grades of concrete up to C50/60,

εs= 500/(1.15 200000) = 0.0022 εcu= 0.0035,

η = 1 and λ = 0.8.

fyd = fyk/1.15 fcd= 0.85fck/1.5,

Fst = 0.87As fyk Fc = (0.85 fck / 1.5) b (0.8 x) = 0.453 fck b x

M

b


Take moments about the centre of the tension force

M = 0.453 fck b x z (1)

Now z = d - 0.4 x

∴ x = 2.5(d - z)

& M = 0.453 fck b 2.5(d - z) z

= 1.1333 (fck b z d - fck b z2)

Let K = M / (fck b d 2)

(K may be considered as the normalised bending resistance)

∴ 0 = 1.1333 [(z/d)2 – (z/d)] + K

0 = (z/d)2 – (z/d) + 0.88235K

Section analysis

==

2

2

22 - 1.1333

bdf

bzf

bdf

bdzf

bdf

MK

ck

ck

ck

ck

ck

M

0 = (z/d)2 – (z/d) + 0.88235K

Solving the quadratic equation:

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

M


Take moments about the centre of the compression force

M = 0.87As fyk z

Rearranging

As = M /(0.87 fyk z)

The required area of reinforcement can now be:

• calculated using these expressions

• obtained from Tables of z/d (eg Table 5 of How to

beams or Concise Table 15.5 )

• obtained from graphs (eg from the ‘Green Book’)

Tension steel, As

Design aids for flexureConcise: Table 15.5

Besides limits on

x/d, traditionally

z/d was limited to

0.95 max to avoid

issues with the

quality of

‘covercrete’.


Beams with Compression Reinforcement, As2

According to Cl 5.5(4) the depth of the neutral axis is limited, viz:

δ ≥ k1 + k2 xu/d

where

k1 = 0.4

k2 = 0.6 + 0.0014/ εcu2 = 0.6 + 0.0014/0.0035 = 1

xu = depth to NA after redistribution

δ = Redistribution ratio

∴ xu = d (δ - 0.4)

Therefore there are limits on K.

The limiting value for K (denoted K’) can be calculated as follows:

As before M = 0.453 fck b x z (1)

leading to K = M / (fck b d 2)

Substituting xud for x in eqn (1) and rearranging:

M’ = b d2 fck (0.6 δ – 0.18 δ 2 - 0.21)

∴ K’ = M’ /(b d2 fck) = (0.6 δ – 0.18 δ 2 - 0.21)

c.f. from BS 8110 rearranged K’ = (0.55 β – 0.18 β 2 – 0.19)

Some engineers advocate taking x/d < 0.45, and ∴K’ < 0.168. It is often

considered good practice to limit the depth of the neutral axis to avoid

‘over-reinforcement’ to ensure a ductile failure. This is not an EC2

requirement and is not accepted by all engineers (but is by TCC).

K’


Asfor beams with Compression Reinforcement,

The concrete in compression is at its design

capacity and is reinforced with compression

reinforcement. So now there is an extra force:

Fsc = 0.87As2 fyk

The area of tension reinforcement can now be considered in two

parts.

The first part balances the compressive force in the concrete

(with the neutral axis at xu).

The second part is balances the force in the compression steel.

The area of reinforcement required is therefore:

As = K’ fck b d 2 /(0.87 fyk z) + As2

where z is calculated using K’ instead of K

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

tension force:

M = K’ fck b d 2 + 0.87 fyk As2 (d - d2)

Rearranging

As2 = (K - K’) fck b d 2 / (0.87 fyk (d - d2))

As2


The following flowchart outlines the design procedure for rectangular

beams with concrete classes up to C50/60 and grade 500 reinforcement

Determine K and K’ from:

Note: δδδδ =1.0 means no redistribution and δδδδ = 0.8 means 20% moment redistribution.

Compression steel needed -doubly reinforced

Is K ≤ K’ ?

No compression steelneeded – singly reinforced

Yes No

ck

2 fdb

MK ==== 21.018.06.0'& 2 −−−−−−−−==== δδδδδδδδK

Carry out analysis to determine design moments (M)

It is often recommended in the UK that K’ is limited to 0.168 to ensure ductile failure

δδδδ K’

1.00 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

Design Flowchart

Calculate lever arm z from:

* A limit of 0.95d is 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 > Øbar > 20 > Agg + 5

Check max spacing between bars

Calculate tension steel required from:zf

MA

yd

s====

Singly-reinforced Beam


Doubly-reinforced Beam

Calculate lever arm z from: [[[[ ]]]]'53.3112

Kd

z −−−−++++====

Calculate excess moment from: (((( ))))'' 2 KKfbdMck

−−−−====

Calculate compression steel required from:

(((( ))))2yd2s

'

ddf

MA

−−−−====

Calculate tension steel required from:

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

2syd

2

s'

Azf

bdfKA ck ++++====

Flexure Worked Example(Doubly reinforced)


Worked Example 1

Design the section below to resist a sagging moment of 370 kNm

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

Take fck = 30 MPa and fyk = 500 MPa.

d

Initially assume 32 mm φ for tension reinforcement with 30 mm

nominal cover to the link (allow 10 mm for link) and 20mm φ for

compression reinforcement with 25 mm nominal cover to link.

Nominal side cover is 35 mm.

d = h – cnom - Ølink - 0.5Ø

= 500 – 30 - 10 – 16

= 444 mm

d2 = cnom + Ølink + 0.5Ø

= 25 + 10 + 10

= 45 mm

444


∴ provide compression steel

[ ]

[ ]mm363

168.053.3112

444

'53.3112

=

×−+=

−+= Kd

z

'. K

fbd

MK

>=××

×=

=

2090

30444300

103702

6

ck

2

1680.' =K δδδδ K’

1.00 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

( )

kNm7.72

10)168.0209.0(30444300

''

62

2

=

×−×××=

−=−

KKfbdM ck

( )

2

6

2yd

2s

mm 419

45) – (444435

10 x 72.7

'

=

×=

−=

ddf

MA

2

6

2s

yd

s

mm2302

419363435

10)7.72370(

'

=

+×

×−=

+−

= Azf

MMA


Provide 2 H20 for compression steel = 628mm2 (419 mm2 req’d)

and 3 H32 tension steel = 2412mm2 (2302 mm2 req’d)

By inspection does not exceed maximum area or maximum spacing of

reinforcement 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

Flexure –High strength concrete


As

d

η fcd

Fs

λx

εs

x

εcu3

Fc Ac

fck ≤≤≤≤ 50 MPa 50 < fck ≤≤≤≤ 90 MPa

λλλλ 0.8 = 0.8 – (fck – 50)/400

ηηηη 1.0 = 1.0 – (fck – 50)/200

fcd = αcc fck /γc

= 0.85 fck /1.5

Higher Concrete StrengthsRectangular Concrete Stress Block (3.1.7, Figure 3.5)

fck λ η

50 0.8 1

55 0.79 0.98

60 0.78 0.95

70 0.75 0.9

80 0.73 0.85

90 0.7 0.8

Higher Concrete Strengths

fck ≤ 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 −+=


Design aids for flexureConcise: Table 15.5

Besides limits on

x/d, traditionally

z/d was limited to

0.95 max to avoid

issues with the

quality of

‘covercrete’.

Valid up to C50/60

Factors for NA depth (x) and lever arm (z) for concrete grade ≤≤≤≤ 50 MPa

0.00

0.20

0.40

0.60

0.80

1.00

1.20

M/bd 2fck

Fa

cto

r

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)


Factors for NA depth (x) and lever arm (z) for concrete grade 70 MPa

0.00

0.20

0.40

0.60

0.80

1.00

1.20

M/bd 2fck

Factor

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)

Shear

Beams


Shear in Beams

• Shear design is different from BS8110

• Shear strength should be limited to the value for C50/60

Definitions:

•VRd,c – Resistance of member without shear reinforcement

•VRd,s - Resistance of member governed by the yielding of shear

reinforcement

•VRd,max - Resistance of member governed by the crushing of

compression struts

•VEd - Applied shear force

• For predominately UD, shear may be checked at d from face of support

Strut inclination method

θθθθcotswsRd, ywdfz

s

AV ====

32

21.8°°°° < θθθθ < 45°°°°

Eurocode 2 – Beam shear


Members Requiring Shear Reinforcement (6.2.3.(1))

θ

s

d

V(cot θ - cotα )

V

N Mα ½ z

½ zVz = 0.9d

Fcd

Ftd

compression chord compression chord

tension chordshear reinforcement

α angle between shear reinforcement and the beam axis

θ 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 = 0,9d may

normally be used.

Shear Design: Links

Variable strut method allows a shallower strut angle – hence

activating more links.

Strut angle increases as ‘shear stress’ increases

Angle = 45°(=max)

V carried on 3 linksAngle = 21.8°(=min)

V carried on 6 links

d

Vhigh

z

x

d

x

Vlow

θz

s


35

Shear reinforcement

density

Asfyd/s

Shear Strength, VR

BS8110: VR = VC + VS

Test results VR

Eurocode 2:

VRmax

Minimum links

Fewer links(but more critical)

Safer

Eurocode 2 vs BS8110:

Shear

shear reinforcement control

VRd,s = Asw z fywd cot θ /s Exp (6.8)

concrete strut control

VRd,max = z bw ν1 fcd /(cotθ + tanθ) = 0.5 z bw ν1 fcd sin 2θ Exp (6.9)

where ν1 = ν = 0.6(1-fck/250) Exp (6.6N)

1 ≤ cotθ ≤ 2,5

Basic equations

d

V

z

x

d

x

V

θz

s

Shear Resistance of Sections with Shear Reinforcement


37

We can manipulate the

Expressions for concrete

struts so that when

vEd < vRd,cot θ =2.5,

then

cot θ = 2.5 (θ = 21.8°)

and

Asw/s = vEd bw/(2.5fywd)

ShearEurocode 2 – Beam shear links

fck vRd, cot θθθθ = 2.5 vRd, cot θθθθ = 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 6.02

40 4.63 6.72

45 5.08 7.38

50 5.51 8.00

EC2 – Shear Flow Chart for vertical links

Yes (cot θθθθ = 2.5)

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

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

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

Is vRdcot θθθθ = 2.5 > vEd?No

Check maximum spacing of shear reinforcement :

s,max = 0.75 d

Determine θθθθ from:θθθθ = 0.5 sin-1[(vEd/(0.20fck(1-fck/250))]

Is vRdcot θθθθ = 1.0 > vEd?

Yes (cot θθθθ > 1.0)

NoRe-size


Beam Example 1

Cover = 40mm to each face

fck = 30

Determine the flexural and shear

reinforcement required (try 10mm

links and 32mm main steel)

Gk = 75 kN/m, Qk = 50 kN/m , assume no redistribution and use

equation 6.10 to calculate ULS loads.

8 m

450

1000

Beam Example 1 – Bending

ULS load per m = (75 x 1.35 + 50 x 1.5) = 176.25

Mult = 176.25 x 82/8

= 1410 kNm

d = 1000 - 40 - 10 – 32/2

= 934

120.030934450

1014102

6

ck

2====

××××××××××××

========fbd

MK

K’ = 0.168 (c.f.0.208)

K < K’⇒ No compression reinforcement required

[[[[ ]]]] [[[[ ]]]] dKd

z 95.0822120.0x53.3112

93453.311

2≤≤≤≤====−−−−++++====−−−−++++====

2

6

yd

smm3943

822x435

10x1410============

zf

MA

Provide 5 H32 (4021) mm2)


Beam Example 1 – Shear

Shear force, VEd = 176.25 x 8/2 = 705 kN say (could take 505 kN @ d from face)

Shear stress:

vEd = VEd/(bw 0.9d) = 705 x 103/(450 x 0.9 x 934)

= 1.68 MPa

vRdcot θ = 2.5 = 3.64 MPa

vRdcot θ = 2.5 > vEd

∴ cot θ = 2.5

Asw/s = vEd bw/(fywd cot θ)

Asw/s = 1.68 x 450 /(435 x 2.5)

Asw/s = 0.70 mm

Try H8 links with 3 legs.

Asw = 151 mm2

s < 151 /0.70 = 215 mm

⇒ provide H8 links at 200 mm spacing

fckvRd, cot θθθθ =

2.5

vRd, cot θθθθ =

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 6.02

40 4.63 6.72

45 5.08 7.38

50 5.51 8.00

Beam Example 1

Provide 5 H32 (4021) mm2)

with H8 links at 200 mm spacing


Beam Example 2 – High shear

Find the minimum area of

shear reinforcement

required to resist the

design shear force using

EC2.

Assume that:

fck = 30 MPa and

fyd = 500/1.15 = 435 MPa

UDL not dominant

Find the minimum area of shear reinforcement required to resist

the design shear force using EC2.

Assume that:

fck = 30 MPa and

fyd = 500/1.15 = 435 MPa

Shear stress:

vEd = VEd/(bw 0.9d)

= 312.5 x 103/(140 x 0.9 x 500)

= 4.96 MPa



vRdcot θ = 2.5 < vEd < vRdcot θ = 1.0

∴ 2.5 > cot θ > 1.0 ⇒ Calculate θ

fckvRd, cot θθθθ =

2.5

vRd, cot θθθθ =

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 6.02

40 4.63 6.72

45 5.08 7.38

50 5.51 8.00



Calculate θ

(((( ))))°°°°====

====

−−−−====

−−−−

−−−−

0.35

250 / 30 -130x20.0

96.4sin5.0

)250/1(20.0sin5.0

1

ckck

Ed1

θθθθ

θθθθff

v

43.1cot ====∴∴∴∴ θθθθ

Asw/s = vEd bw/(fywd cot θ )

Asw/s = 4.96 x 140 /(435 x 1.43)

Asw/s = 1.12 mm

Try H10 links with 2 legs.

Asw = 157 mm2

s < 157 /1.12 = 140 mm

⇒ provide H10 links at 125 mm spacing


Where av ≤ 2d the 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)


Summary

• Flexural principles similar

• Shear approach different – should result in less shear

reinforcement

• SLS and detailing rules later

Workshop Problem


Workshop Problem

Cover = 35 mm to each face

fck = 30MPa

Design the beam in flexure and shear

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

8 m

300

450

Exp (6.10)

Remember

this from

last week?

Aide memoire

OrConcise Table 15.5


Workings:- Load, Mult, d, K, (z/d,) z, As, VEd, Asw/s

Summary

• Flexural principles similar

• Shear approach different – should result in less shear

reinforcement

• We will look at the SLS and detailing rules later


Detailing

UK CARES (Certification - Product & Companies)

1. Reinforcing bar and coil

2. Reinforcing fabric

3. Steel wire for direct use of for further

processing

4. Cut and bent reinforcement

5. Welding and prefabrication of reinforcing steel

www.ukcares.co.uk www.uk-bar.org


Identification of bars

Class A

Class B

Class C

Reinforced Concrete Detailing to Eurocode 2

Section 8 - General Rules Anchorage

Laps

Large Bars

Section 9 - Particular RulesBeams

Slabs

Columns

Walls

Foundations

Discontinuity Regions

Tying Systems

Cover – Fire

Specification and Workmanship


Spacing of bars (8.2)

• Clear horizontal and vertical distance ≥ φ, (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

• To avoid damage to bar use

Bar dia ≤ 16mm Mandrel size 4 x bar diameter

Bar dia > 16mm Mandrel size 7 x bar diameter

The bar should extend at least 5 diameters beyond a bend

Bending of reinforcement – minimum mandrel size, φφφφm (8.3)

• To avoid concrete failure: φφφφ m,min ≥ Fbt ((1/ab) +1/(2 φφφφ)) / 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 φφφφ /2

• Mandrel size need not be checked to avoid concrete failure if :

– anchorage does not require more than 5φ past end of bend

– bar is not the closest to edge face and there is a cross bar ≥φinside bend

– mandrel size is at least equal to the recommended minimum value

Section 8 - General Rules


The design value of the ultimate bond stress, fbd = 2.25 η1η2fctdwhere fctd should be limited to C60/75

η1 =1 for ‘good’ and 0.7 for ‘poor’ bond conditions

η2 = 1 for φ ≤ 32, otherwise (132- φ)/100

a) 45º ≤≤≤≤ αααα ≤≤≤≤ 90º c) h > 250 mm

h

Direction of concreting

≥ 300

h


b) h ≤≤≤≤ 250 mm d) h > 600 mm

unhatched zone – ‘good’ bond conditions

hatched zone - ‘poor’ bond conditions

α


250


Ultimate bond stressEC2: Cl. 8.4.2 Concise: 11.5

lb,rqd = (φ φ φ φ / 4) (σσσσsd / fbd)

where

σsd = the design stress of the bar at the position from

where the anchorage is measured.

Usually taken as 500/1.15 = 435 MPa

That is safe but may be too conservative!

Basic required anchorage lengthEC2: Cl. 8.4.3 Concise: 11.4.3

• For bent bars lb,rqd should be measured along the

centreline of the bar


lbd = α1 α2 α3 α4 α5 lb,rqd ≥≥≥≥ lb,min

However:

(α2 α3 α5) ≥≥≥≥ 0.7

lb,min > max(0.3lb,rqd ; 10φφφφ, 100mm)

Design Anchorage Length, lbdEC2: Cl. 8.4.4 Concise: 11.4.2

Alpha valuesEC2: Table 8.2 Concise: 11.4.2


Concrete coverEC2: Figures 8.3 and 8.4

Figure 8.4

Concise: Figure 11.3

Figure 8.3 re α1 and α2

Figure 8.3 re α3

l0 = α1 α2 α3 α5 α6 lb,rqd ≥≥≥≥ l0,min

α6 = (ρ1/25)0,5 but between 1.0 and 1.5

where ρ1 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%

α6 1 1.15 1.4 1.5

Note: Intermediate values may be determined by interpolation.

α1 α2 α3 α5 are as defined for anchorage length

l0,min ≥ max{0.3 α6 lb,rqd; 15φ; 200}

Design Lap Length, l0 (8.7.3)EC2: Cl. 8.7.3; Table 8.3 Concise: 11.6.2


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

Bond Length in bar diameters

conditions fck /fcu25/30

fck /fcu28/35

fck /fcu30/37

fck /fcu32/40

Full tension and

compression anchorage

length, lbd

‘good’ 40 37 36 34

‘poor’ 58 53 51 49

Full tension and

compression lap length, l0

‘good’ 46 43 42 39

‘poor’ 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.

Anchorage /lap lengths for slabsManual for the design of concrete structures to Eurocode 2

Alternative approach How to design concrete structures using Eurocode 2


Where possible laps staggered and not located in regions of high stress, the

arrangement of lapped bars should comply with the following:

The clear distance between lapped bars should not be greater than 4Ø or

50 mm otherwise the lap length should be increased by a length equal to

the clear space where it exceeds 4Ø or 50 mm

1. The longitudinal distance between two adjacent laps should not be

less than 0,3 times the lap length, l0;

2. In case of adjacent laps, the clear distance between adjacent bars

should not be less than 2Ø 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 Laps (8.7.3)




• Where the diameter, φφφφ, of the lapped bars ≥ 20 mm, the transverse

reinforcement should have a total area, ΣAst ≥ 1,0As of one spliced bar. It

should 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 φφφφ transverse bars should

be 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 /30ΣA /2

st

ΣA /2st

l /30

FsFs

≤150 mm

l0

Transverse Reinforcement at Laps (8.7.4) only if bar Ø > 20mm or laps > 25%

Figure 8.9


Detailing of members and particular rules

SECTION 9

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

• As,max = 0,04 Ac

• Section at supports should be designed for a

hogging moment ≥ 0,25 max. span moment

• Any design compression reinforcement (φ) should be held by transverse reinforcement with spacing ≤15 φ

Beams (9.2)


• Tension reinforcement in a flanged beam at

supports should be spread over the effective width

(see 5.3.2.1)

Beams (9.2)

(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 [Ftd may 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 θ - cot α)/2 = 0.5 z cot θ for vertical shear links

z= lever arm, θ = angle of compression strut

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

Curtailment (9.2.1.3)


21.8°°°° < θθθθ < 45°°°°

Strut Inclination Method

Horizontal component of

diagonal shear force

= (V/sinθ) . cosθ = V cotθ

Applied

shear V

Applied

moment M

M/z + V cotθ/2= (M + Vz cotθ/2)/z

∴ ∆M = Vz cotθ/2

dM/dx = V

∴ ∆M = V∆x

∴ ∆x = z cotθ/2 = al

z

V/sinθ

θ

M/z - V cotθ/2

al

Curtailment of longitudinal tension reinforcement

‘Shift’ Rule for Shear


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

a l

∆Ftd

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 it is always conservative to use

al = 1.125d

EC2: Cl. 9.1.3 Concise: 12.2.2

‘Shift’ Rule for Shear

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

Simple support (indirect) Simple support (direct)

• As bottom steel at support ≥ 0.25 As provided in the span

• Transverse pressure may only be taken into account with

a ‘direct’ support.

Shear shift rule

al

Tensile Force Envelope

Anchorage of Bottom Reinforcement at End Supports (9.2.1.4)


Simplified Rules for Beams How to….EC2

Detailing section

c) Simple support, bottom reinforcement

≤ 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

Supporting Reinforcement at ‘Indirect’ Supports (9.2.5)

Plan view


• Curtailment – as beams except 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

Solid Slabs 9.3

Detailing Comparisons

Beams EC2 BS 8110

Main Bars in Tension Clause / Values Values

As,min 9.2.1.1 (1): 0.26 fctm/fykbd ≥0.0013 bd

0.0013 bh

As,max 9.2.1.1 (3): 0.04 bd 0.04 bh

Main Bars in Compression

As,min -- 0.002 bh

As,max 9.2.1.1 (3): 0.04 bd 0.04 bh

Spacing of Main Bars

smin 8.2 (2): dg + 5 mm or φ or 20mm dg + 5 mm or φ

Smax Table 7.3N Table 3.28

Links

Asw,min 9.2.2 (5): (0.08 b s √fck)/fyk 0.4 b s/0.87 fyv

sl,max 9.2.2 (6): 0.75 d 0.75d

st,max 9.2.2 (8): 0.75 d ≤ 600 mm

9.2.1.2 (3) or 15φ from main bar

d or 150 mm from main bar



Slabs EC2 Clause / Values BS 8110 Values

Main Bars in Tension

As,min 9.2.1.1 (1):

0.26 fctm/fykbd ≥ 0.0013 bd

0.0013 bh

As,max 0.04 bd 0.04 bh

Secondary Transverse Bars

As,min 9.3.1.1 (2):

0.2As for single way slabs

0.002 bh

As,max 9.2.1.1 (3): 0.04 bd 0.04 bh

Spacing of Bars

smin 8.2 (2): dg + 5 mm or φ or 20mm

9.3.1.1 (3): main 3h ≤ 400 mm

dg + 5 mm or φ

Smax secondary: 3.5h ≤ 450 mm 3d or 750 mm

places of maximum moment:

main: 2h ≤ 250 mm

secondary: 3h ≤ 400 mm


Punching Shear EC2Clause / Values BS 8110 Values

Links

Asw,min 9.4.3 (2):Link leg = 0.053sr st √(fck)/fyk Total = 0.4ud/0.87fyv

Sr 9.4.3 (1): 0.75d 0.75d

St 9.4.3 (1):

within 1st control perim.: 1.5d

outside 1st control perim.: 2d

1.5d

Columns

Main Bars in Compression

As,min 9.5.2 (2): 0.10NEd/fyk ≤ 0.002bh 0.004 bh

As,max 9.5.2 (3): 0.04 bh 0.06 bh

Links

Min size 9.5.3 (1) 0.25φ or 6 mm 0.25φ or 6 mm

Scl,tmax 9.5.3 (3): min(12φmin; 0.6b; 240 mm) 12φ

9.5.3 (6): 150 mm from main bar 150 mm from main bar


Detailing IssuesEC2 Clause Issue Possible resolve in 2013?

8.4.4.1 Lap lengths assume

4φ centres in 2 bar

beams

α7 factor for spacing e.g. 0.63 for 6φcentres in slabs or 10φ centre in two bar beams

Table 8.3 α6 varies depending

on amount staggered

α6 should always = 1.5.

8.7.2(3)

& Fig 8.7

0.3 lo gap between

ends of lapped bars is

onerous.

For ULS, there is no advantage in staggering

bars. For SLS staggering at say 0.5 lo might

be helpful.

Table 8.2 α2 for compression

bars

Should be the same as for tension.

8.7.4.1(4)

& Fig 8.9

Requirements for

transverse bars

impractical

No longer requirement for transverse bars

to be between lapped bar and cover.

Requirement only makes 10-15% difference

in strength of lap

Fig 9.3 lbd anchorage into

support

May be OTT as compression forces increase

bond strength. Issue about anchorage

beyond CL of support

End of Beams

Practical Design to Eurocode 2


Next Week

17th April 2013Basics

EC0, EC1, Materials, Cover

24th April 2013Beams

Bending, Shear, Detailing

1st May 2013Columns

Axial load, Column Moments, Buckling

8th May 2013Slabs

Serviceability, Punching Shear

15th May 2013Foundations

Pads, Piles, Retaining Walls

Lecture 2 Beams Jb Apr 2013 [Compatibility Mode] (1)

Documents

Transcript of Lecture 2 Beams Jb Apr 2013 [Compatibility Mode] (1)