Beam Practical Design to Eurocode 2_Sep 2013

Practical Design to Eurocode 2 Beams Lecture 2

19th September 2013

Contents - Beams • Bending/ Flexure – – – –

Section analysis, singly and doubly reinforced Tension reinforcement, As neutral axis depth limit & K’ Compression reinforcement, As2

• Shear in beams – variable strut method

• Detailing – Anchorage & Laps – Members & particular rules Shift rule for curtailment

Bending/ Flexure

Section Design: Bending • In principal flexural design is generally the same as BS8110 • High strength concrete ( fck > 50 MPa ) can be designed. • EC2 presents the principles only • Design manuals will provide the standard solutions for basic design cases.

Note: TCC How to guide equations and equations used on this course are based on a concrete fck ≤ 50 MPa

Section Analysis to determine Tension & Compression Reinforcement EC2 contains information on: • Concrete stress blocks • Reinforcement stress/strain curves • The maximum depth of the neutral axis, x. This depends on the moment redistribution ratio used. • The design stress for concrete, fcd and reinforcement, fyd

In EC2 there are no equations to determine As and As2 for a given ultimate moment, M, on a section. Equations, similar to those in BS 8110, are derived in the following slides. As in BS8110 the terms K and K’ are used:

K 

M bd 2 f ck

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

Rectangular Concrete Stress Block cu3

EC2: Cl 3.1.7, Fig 3.5 Ac

fcd Fc

x

x d

As

Fs

s fck  50 MPa

50 < fck  90 MPa



0.8

= 0.8 – (fck – 50)/400



1.0

= 1,0 – (fck – 50)/200

For fck  50 MPa:

fcd = cc fck /c = 0.85 fck /1.5

fck 50 55 60 70 80 90

λ 0.8 0.79 0.78 0.75 0.73 0.7

η 1 0.98 0.95 0.9 0.85 0.8

& at failure concrete strain, εcu= 0.0035

Reinforcement Design Stress/Strain Curve EC2: Cl 3.2.7, Fig 3.8



Idealised

kfyk

In UK fyk = 500 MPa

kfyk/s

fyk fyd = fyk/s

fyd = fyk/γs = 500/1.15 = 435 MPa

Design

Es may be taken to be 200 GPa Steel yield strain (s at yield point)

= fyd/Es = 435/200000 = 0.0022

fyd/Es

 ud uk

At failure concrete strain is 0.0035 for fck  50 MPa. If x/d is 0.6 steel strain is 0.0023 and this is past the yield point. Design steel stress is 435 MPa if neutral axis, x, is less than 0.6d.



Analysis of a singly reinforced beam Cl 3.1.7 EN 1992-1-1 Design equations can be derived as follows: b

M

For grades of concrete up to C50/60, εcu= 0.0035,  = 1 and  = 0.8. fcd = 0.85fck/1.5, fyd = fyk/1.15 = 0.87 fyk Fc = (0.85 fck / 1.5) b (0.8 x) = 0.453 fck b x Fst = 0.87As fyk Methods to find As: • Iterative, trial and error method – simple but not practical • Direct method of calculating z, the lever arm, and then As

Analysis of a singly reinforced beam Determine As – Iterative method b

M

For horizontal equilibrium Fc= Fst 0.453 fck b x = 0.87As fyk

Guess As

Solve for x

z = d - 0.4 x

M = Fc z

Analysis of a singly reinforced beam Determine As – Direct method Take moments about the centre of the tension force M

= 0.453 fck b x z

Now z

= d - 0.4 x



x

= 2.5(d - z)

&

M

= 0.453 fck b 2.5(d - z) z

(1)

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

K = M / (fck b d 2)

 fckbdz M fckbz 2   K  1.1333  2 2 2  fckbd fckbd   fckbd (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

M

0

= (z/d)2 – (z/d) + 0.88235K

Solving the quadratic equation: z/d z

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

Higher Concrete Strengths fck ≤ 50MPa

z  d[1 (1 3,529K )]/2

fck = 60MPa

z  d[1 (1 3,715K )]/2

fck = 70MPa

z  d[1 (1 3,922K )]/2

fck = 80MPa

z  d[1 (1 4,152K )]/2

fck = 90MPa

z  d[1 (1 4,412K )]/2

Tension steel, As 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’ or Fig B.3 in Concrete Buildings Scheme Design Manual)

Design aids for flexure Concise: 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’.

Design aids for flexure TCC Concrete Buildings Scheme Design Manual, Fig B.3

Design chart for singly reinforced beam

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



 k1 + k2 xu/d where k1 = 0.4 k2 = 0.6 + 0.0014/ cu2 = 0.6 + 0.0014/0.0035 = 1 xu = depth to NA after redistribution  

Redistributed Bending Moment Elastic Bending Moment

 xu = d ( - 0.4) Therefore there are limits on K and this limit is denoted K’

= Redistribution ratio

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

M = 0.453 fck b x z

(1)

K = M / (fck b d 2)

Substituting xu for x in eqn (1) and rearranging: M’ = b d2 fck (0.6  – 0.18  2 - 0.21) 

K’ = M’ /(b d2 fck)

c.f. from BS 8110 rearranged K’

= (0.6  – 0.18  2 - 0.21) = (0.55  – 0.18  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).

As

for 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 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

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))

Design Flowchart The following flowchart outlines the design procedure for rectangular beams with concrete classes up to C50/60 and grade 500 reinforcement Carry out analysis to determine design moments (M) Determine K and K’ from: M K & K '  0.6  0.18 2  0.21 2 b d fck Note:  =1.0 means no redistribution and  = 0.8 means 20% moment redistribution.

Yes

Is K ≤ K’ ?

No compression steel needed – singly reinforced

No



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

Compression steel needed doubly reinforced

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

Flow Chart for Singly-reinforced Beam 



Calculate lever arm z from: z  d 1  1  3.53K  0.95d *

2

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

Calculate tension steel required from:

As 

M fyd z

Check minimum reinforcement requirements:

As,min

0.26 fctm bt d   0.0013 bt d fyk

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

Flow Chart for DoublyReinforced Beam 

d z  1  1  3.53K ' 2

Calculate lever arm z from: Calculate excess moment from:



M '  bd 2fck K  K '

Calculate compression steel required from: M' As2  fydd  d 2  Calculate tension steel required from:

K ' fck bd 2 As   As2 fyd z

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

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.  = 0.85). Take fck = 30 MPa and fyk = 500 MPa.

d

Initially assume 32 mm  for tension reinforcement with 30 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 - Ølink - 0.5Ø = 500 – 30 - 10 – 16 = 444 mm d2 = cnom + Ølink + 0.5Ø = 25 + 10 + 10 = 45 mm

444

K '  0.168 K

M bd 2 f ck

370  10 300  444 2  30  0.209  K ' 6



 provide compression steel





d 1  1  3.53K ' 2 444  1  1  3.53  0.168 2  363 mm

z







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

M '  b d2fck K  K '  300  4442  30  (0.209  0.168)  106  72.7 kNm As 2 

M' fyd d  d 2 

72.7 x 106  435  (444 – 45)  419 mm2

As 

M M'  As 2 fyd z

(370  72.7)  106   419 435 363  2302 mm2

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

Simplified Factors for Flexure (1) Factors for NA depth (x) and lever arm (z) for concrete grade  50 MPa 1.20

1.00

lever arm

Factor

0.80

0.60

0.40

NA depth 0.20

0.00

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

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

M/bd 2fck

Simplified Factors for Flexure (2) Factors for NA depth (x) and lever arm (z) for concrete grade 70 MPa 1.20

lever arm

1.00

Factor

0.80

0.60

0.40

NA depth 0.20

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.10

0.11

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

M/bd 2fck

0.12

0.13

0.14

0.15

0.16

0.17

Shear in Beams

Shear There are three approaches to designing for shear: • When shear reinforcement is not required e.g. slabs Shear check uses VRd,c

• When shear reinforcement is required e.g. Beams Variable strut method is used to check shear in beams Strut strength check using VRd,max Links strength using VRd,s • Punching shear requirements e.g. flat slabs The maximum shear strength in the UK should not exceed that of class C50/60 concrete

Shear in Beams Shear design is different from BS8110. EC2 uses the variable strut method to check a member with shear reinforcement.

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 UDL, shear may be checked at d from face of support

Members Requiring Shear Reinforcement (6.2.3.(1)) compression chord

compression chord

V(cot  - cot

Fcd d



½z



z = 0.9d V

s

shear reinforcement

½z

N V

Ftd tension chord



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 = 0,9d may normally be used.

M

Strut Inclination Method VRd,max

 b z  1 fcd  cw w cot   tan 

VRd,s

Asw  z f ywd cot  s 21.8 <  < 45

Shear 6.2.3 EN 1992-1-1 We can use the following expressions from the code to calculate shear reinforcement for a beam (Assumes shear reinforcement is always provided in a beam) VRd,s = Asw z fywd cot  /s

…1

fck

VRd,max = 0.5 z bw fcd sin 2

…2

where 0.6 (1- fck/250) When cot = 2.5 (= 21.8°)

VRd = 0.138 bw z fck (1 - fck/250) Or in terms of stress: vRd = 0.138 fck (1 - fck/250) Rearranging equation 2 in terms of stress: 

 = 0.5 sin-1[vRd /(0.20 fck(1 - fck/250))]

vRd, cot 

=

vRd, cot 

2.5

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

=

Shear Design: Links Variable strut method allows a shallower strut angle – hence activating more links.

As strut angle reduces concrete stress increases

V z

s

V

d

x Angle = 45° V carried on 3 links



z

d

x Angle = 21.8° V carried on 6 links

Eurocode 2 vs BS8110: Shear Safer Shear reinforcement density

Asfyd/s

Fewer links

Eurocode 2: BS8110: VR = VC + VS

VRmax

Test results VR

(but more critical)

Minimum links

Shear Strength, VR 38

Shear Resistance of Sections with Shear Reinforcement V

z x

s

d

V

 1  cot   2,5

Basic equations shear reinforcement control VRd,s = Asw z fywd cot /s

z

x

Exp (6.8)

concrete strut control VRd,max = z bw 1 fcd /(cotθ + tanθ) = 0,5 z bw 1fcd sin 2       Exp (6.9) where 1 = 0,6(1-fck/250)

Exp (6.6N)

d

Shear Resistance with Shear Reinforcement Procedure for design with variable strut 1. Determine maximum applied shear force at support, VEd 2. Determine VRd,max with cot = 2.5 3. If VRd,max > VEd

cot = 2.5, go to step 6 and calculate required shear reinforcement

4. If VRd,max < VEd

calculate required strut angle:  = 0.5 sin-1[(vEd/(0.20fck(1-fck/250))]

5. If cot is less than 1 re-size element, otherwise

6. Calculate amount of shear reinforcement required Asw/s = vEd bw/(fywd cot ) = VEd /(0.78 d fyk cot ) 7. Check min shear reinforcement, Asw/s ≥ bw ρw,min and max spacing, sl,max = 0.75d ρw,min = (0.08 √fck)/fyk cl 9.2.2

EC2 – Shear Flow Chart for vertical links Determine vEd where: vEd = design shear stress [vEd = VEd/(bwz) = VEd/(bw 0.9d)] Determine the concrete strut capacity vRd when cot  = 2.5 vRdcot  = 2.5 = 0.138fck(1-fck/250)

Is vRdcot  = 2.5 > vEd? Yes

No

(cot  = 2.5)

Calculate area of shear reinforcement: Asw/s = vEd bw/(fywd cot ) Check maximum spacing of shear reinforcement : s,max = 0.75 d

Is vRdcot  = 1.0 > vEd? Yes

No

Re-size

(cot  > 1.0)

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

Design aids for shear Concise Fig 15.1 a)

Design aids for shear Concise Fig 15.1 b)

Short Shear Spans with Direct Strut Action (6.2.3)

d av

d av

• 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: − The longitudinal reinforcement is fully anchored at the support. − Only that shear reinforcement provided within the central 0.75av is included in the resistance. Note: see PD6687-1:2010

Cl 2.14 for more information

Beam examples

Beam Example 1 Gk = 75 kN/m, Qk = 50 kN/m , assume no redistribution and use equation 6.10 to calculate ULS loads.

8m Cover = 40mm to each face

1000

fck = 30 Determine the flexural and shear reinforcement required

450

(try 10mm links and 32mm main steel)

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

M 1410  106 K   0.120 2 2 bd fck 450  934  30 K’ = 0.208 K

< K’  No compression reinforcement required









d 934 z  1  1  3.53K  1  1  3.53 x 0.120  822  0.95d 2 2 M 1410 x 106 As    3943 mm2 fyd z 435 x 822

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 vRd, cot  = vRd, cot  fck vRdcot  = 2.5 = 3.64 MPa 2.5 1.0 vRdcot  = 2.5 > vEd 20 2.54 3.68  cot  = 2.5 25 3.10 4.50 Asw/s = vEd bw/(fywd cot ) 28 3.43 4.97 Asw/s = 1.68 x 450 /(435 x 2.5) 30 3.64 5.28 Asw/s = 0.70 mm 32 3.84 5.58 Try H8 links with 3 legs. 35 4.15 6.02 Asw = 151 mm2 40 4.63 6.72 s < 151 /0.70 = 215 mm 45 5.08 7.38 50 5.51 8.00  provide H8 links at 200 mm spacing

=

Beam Example 1

Provide 5 H32 (4021) mm2) with H8 links at 200 mm spacing

Beam Example 2 – High shear

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

Beam Example 2 – High shear Find the minimum area of shear reinforcement required to resist the design shear force using EC2. Assume that: vRd, cot  = vRd, cot  fck = 30 MPa and fck 2.5 1.0 fyd = 500/1.15 = 435 MPa 20 2.54 3.68 Shear stress: 25 3.10 4.50 vEd = VEd/(bw 0.9d) 28 3.43 4.97 = 312.5 x 103/(140 x 0.9 x 500) 30 3.64 5.28 = 4.96 MPa 32 3.84 5.58 vRdcot  = 2.5 = 3.64 MPa 35 4.15 6.02 vRdcot  = 1.0 = 5.28 MPa 40 4.63 6.72 vRdcot  = 2.5 < vEd < vRdcot  = 1.0 45 5.08 7.38  2.5 > cot  > 1.0  Calculate 

50

5.51

8.00

=

Beam Example 2 – High shear Calculate 



 v Ed   0.20 fck (1  fck / 250 ) 

  0.5 sin 1  

 4.96   0.20 x 30 1 - 30 / 250 

  0.5 sin 1   35.0  cot   1.43

Asw/s = vEd bw/(fywd cot  ) 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

Workshop Problem

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

8m Cover = 35 mm to each face

450

fck = 30MPa Design the beam in flexure and shear

300

Aide memoire Exp (6.10) Remember this from last week?

Or Concise Table 15.5

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

Solution - Flexure

ULS load per m

= (10 x 1.35 + 6.5 x 1.5) = 23.25 kN/m

Mult = 23.25 x 82/8

= 186 kNm

d

= 450 - 35 - 10 – 32/2 = 389 mm

M 186  10 6 K 2   0.137 2 bd f ck 300  389  30

K’ = 0.168

K

< K’  No compression reinforcement required

z

d 389 1 1 3.53 K  1 1 3.53 x 0.137  0.86 x 389  334  0.95d 2 2







M 186 x 10 6 As    1280 mm 2 f yd z 435 x 334



Provide 3 H25 (1470 mm2)

Solution - Shear Shear 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  = 2.5

Asw/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

Detailing

www.ukcares.co.uk www.uk-bar.org UK CARES (Certification - Product & Companies)

1. 2. 3. 4. 5.

Reinforcing bar and coil Reinforcing fabric Steel wire for direct use of for further processing Cut and bent reinforcement Welding and prefabrication of reinforcing

steel

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 Rules Beams Slabs Columns Walls Foundations Discontinuity Regions Tying Systems Cover – Fire Specification and Workmanship

Section 8 - General Rules Spacing of bars EC2: Cl. 8.2

Concise: 11.2

• 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.

Min. Mandrel Dia. for bent bars EC2: Cl. 8.3

Concise: 11.3

Minimum mandrel size, m • To avoid damage to bar is 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

Min. Mandrel Dia. for bent bars EC2: Cl. 8.3

Concise: 11.3

Minimum mandrel size, m Bearing stress inside bends • To avoid failure of the concrete inside the bend of the bar:  m,min  Fbt ((1/ab) +1/(2 )) / fcd Fbt 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  /2 Mandrel 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

Anchorage of reinforcement EC2: Cl. 8.4

Ultimate bond stress EC2: Cl. 8.4.2

Concise: 11.5

The design value of the ultimate bond stress, fbd = 2.25 12fctd where fctd should be limited to C60/75 1 =1 for ‘good’ and 0.7 for ‘poor’ bond conditions 2 = 1 for   32, otherwise (132- )/100 Direction of concreting



a) 45º    90º Direction of concreting

Direction of concreting

250

c) h > 250 mm Direction of concreting  300 300 h

h

b) h  250 mm

d) h > 600 mm

unhatched zone – ‘good’ bond conditions hatched zone - ‘poor’ bond conditions

Basic required anchorage length EC2: Cl. 8.4.3

Concise: 11.4.3

lb,rqd = (/ 4) (sd / fbd)

where sd is the design stress of the bar at the position from where the anchorage is measured. • For bent bars lb,rqd should be measured along the centreline of the bar

EC2 Figure 8.1 Concise Fig 11.1

Design Anchorage Length, lbd EC2: Cl. 8.4.4

Concise: 11.4.2

lbd = α1 α2 α3 α4 α5 lb,rqd  lb,min However: (α2 α3 α5)  0.7 lb,min > max(0.3lb,rqd ; 10, 100mm)

Alpha values EC2: 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

λ

= (∑Ast – ∑ Ast,min)/ As Where Ast is area of transverse reinf.

Table 8.2 - Cd & K factors EC2: Figure 8.3

EC2: Figure 8.4

Concise: Figure 11.3

Table 8.2 - Other shapes EC2: Figure 8.1

Concise: Figure 11.1

Alpha values EC2: Table 8.2

Concise: 11.4.2

Anchorage of links EC2: Cl. 8.5

Concise: Fig 11.2

Laps EC2: Cl. 8.7

Design Lap Length, l0 (8.7.3) EC2: Cl. 8.7.3

Concise: 11.6.2

l0 = α1 α2 α3 α5 α6 lb,rqd  l0,min

α1 α2 α3 α5 are as defined for anchorage length α6 = (r1/25)0,5 but between 1.0 and 1.5 where r1 is the % of reinforcement lapped within 0.65l0 from the centre of the lap

Percentage of lapped bars relative to the total crosssection area

α6

< 25%

33%

50%

>50%

1

1.15

1.4

1.5

Note: Intermediate values may be determined by interpolation. l0,min  max{0.3 α6 lb,rqd; 15; 200}

Arrangement of Laps EC2: Cl. 8.7.3, Fig 8.8

Worked example Anchorage and lap lengths

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

Bond stress, fbd fbd = 2.25 η1 η2 fctd

EC2 Equ. 8.2

η1 = 1.0 ‘Good’ bond conditions η2 = 1.0 bar size ≤ 32

fctd = αct fctk,0,05/γc αct = 1.0

EC2 cl 3.1.6(2), Equ 3.16 γc = 1.5

fctk,0,05 = 0.7 x 0.3 fck2/3

EC2 Table 3.1

= 0.21 x 252/3 = 1.795 MPa fctd = αct fctk,0,05/γc = 1.795/1.5 = 1.197 fbd = 2.25 x 1.197 = 2.693 MPa

Basic anchorage length, lb,req lb.req

= (Ø/4) ( σsd/fbd)

EC2 Equ 8.3

Max stress in the bar, σsd = fyk/γs = 500/1.15

= 435MPa. lb.req

= (Ø/4) ( 435/2.693) = 40.36 Ø For concrete class C25/30

Design anchorage length, lbd lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min lbd = α1 α2 α3 α4 α5 (40.36Ø)

For concrete class C25/30

Alpha values EC2: Table 8.2

Concise: 11.4.2

Table 8.2 - Cd & K factors EC2: Figure 8.3

EC2: Figure 8.4

Concise: Figure 11.3

Design anchorage length, lbd lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min

lbd = α1 α2 α3 α4 α5 (40.36Ø)

For concrete class C25/30

a) Tension anchorage – straight bar

α1 = 1.0 α3 = 1.0

conservative value with K= 0

α4 = 1.0

N/A

α5 = 1.0

conservative value

α2 = 1.0 – 0.15 (Cd – Ø)/Ø α2 = 1.0 – 0.15 (25 – 16)/16 = 0.916 lbd = 0.916 x 40.36Ø = 36.97Ø = 592mm

Design anchorage length, lbd lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min

lbd = α1 α2 α3 α4 α5 (40.36Ø)

For concrete class C25/30

b) Tension anchorage – Other shape bars α1 = 1.0

Cd = 25 is ≤ 3 Ø = 3 x 16 = 48

α3 = 1.0

conservative value with K= 0

α4 = 1.0

N/A

α5 = 1.0

conservative value

α2 = 1.0 – 0.15 (Cd – 3Ø)/Ø ≤ 1.0

α2 = 1.0 – 0.15 (25 – 48)/16 = 1.25 ≤ 1.0 lbd = 1.0 x 40.36Ø = 40.36Ø = 646mm

Worked example - summary H16 Bars – Concrete class C25/30 – 25 Nominal cover Tension anchorage – straight bar

lbd = 36.97Ø = 592mm

Tension anchorage – Other shape bars

lbd = 40.36Ø = 646mm

lbd is measured along the centreline of the bar Compression anchorage (α1 = α2 = α3 = α4 = α5 = 1.0) lbd = 40.36Ø = 646mm

Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7 Lap length = anchorage length x α6

Anchorage & lap lengths How to design concrete structures using Eurocode 2

Anchorage /lap lengths for slabs Manual for the design of concrete structures to Eurocode 2 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 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.

Arrangement of Laps EC2: Cl. 8.7.2

Concise: Cl 11.6

Laps between bars should normally be staggered and not located in regions of high stress. Arrangement of laps should comply with Figure 8.7:

All bars in compression and secondary (distribution) reinforcement may be lapped in one section.

Transverse Reinforcement at Laps Concise: Cl 11.6.4 Bars in tension EC2: Cl. 8.7.4, Fig 8.9 •

Transverse reinforcement is required in the lap zone to resist transverse tension forces.

•

Any Transverse reinforcement provided for other reasons will be sufficient if

•

bar Ø < 20mm or laps< 25%

If bar Ø ≥ 20mm

then additional transverse reinforcement may be needed. It should be positioned at the outer sections of the lap as shown below.

Ast /2

Ast /2

l 0 /3

l 0 /3

150 mm F s

Fs

l0

Transverse Reinforcement at Laps Bars in compression Concise: Cl 11.6.4 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

Detailing of members and particular rules EC2 Section 9

Cl 9.2 Beams

Beams (9.2) • 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 () should be held by transverse reinforcement with spacing 15 

Beams (9.2) • 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) (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, ΔFtd, should be calculated according to 6.2.3 (7). For members without shear reinforcement ΔFtd 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 of longitudinal tension reinforcement ‘Shift’ Rule for Shear Horizontal component of diagonal shear force

M/z - V cot/2

= (V/sin) . cos = V cot

V/sin 

z

M/z + V cot/2

al

Applied

Applied

shear V

moment M

= (M + Vz cot/2)/z

 M = Vz cot/2 dM/dx = V

 M = Vx  x = z cot/2 = al

“Shift Rule” Curtailment of reinforcement

Concise: 12.2.2

EC2: Cl. 9.2.1.3, Fig 9.2 Envelope of (MEd /z +NEd)

lbd lbd

Acting tensile force Resisting tensile force

lbd

al

lbd

Ftd

al Ftd lbd

lbd lbd

lbd

• For members without shear reinforcement this is satisfied with al = d • For members with shear reinforcement: al = 0.5 z Cot  But it is always conservative to use al = 1.125d

Anchorage of Bottom Reinforcement at End Supports (9.2.1.4) Tensile Force Envelope al

Shear shift rule

S i m p l e s u p p o r t ( i n d i r e c t )

S i m p l e s u p p o r t ( d i r e c t )

• As bottom steel at support  0.25 As provided in the span • lbd is required from the line of contact of the support. • Transverse pressure may only be taken into account with a ‘direct’ support.

Simplified Detailing Rules for Beams Concise: Cl 12.2.4

How to….EC2 Detailing section

Supporting Reinforcement at ‘Indirect’ Supports Concise: Cl 12.2.8 EC2: Cl. 9.2.5 A

supporting beam with height h1

B

supported beam with height h2 (h1  h2)

B  h 2 /3

 h 2 /2

Plan view • The supporting reinforcement is in addition to that required for other reasons

 h 1 /3

A

 h 1 /2

• The supporting links may be placed in a zone beyond the intersection of beams

Solid slabs EC2: Cl. 9.3

• 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 Beams

EC2

BS 8110

Main Bars in Tension

Clause / Values

Values

As,min

9.2.1.1 (1): 0.0013 bd

0.26 fctm/fykbd 

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 dg + 5 mm or  or 20mm

smin

8.2 (2):

Smax

Table 7.3N

dg + 5 mm or  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

d or 150 mm from main bar

9.2.1.2 (3) or 15 from main bar

Detailing Comparisons 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 bh

0.04 bd

Spacing of Bars

smin

8.2 (2): dg + 5 mm or  or 20mm 9.3.1.1 (3): main 3h  400 mm

dg + 5 mm or 

Smax

secondary: 3.5h  450 mm

3d or 750 mm

places of maximum moment: main: 2h  250 mm secondary: 3h  400 mm

Detailing Comparisons 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

St

9.4.3 (1): within 1st control perim.: 1.5d outside 1st control perim.: 2d

0.75d

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):

0.06 bh

0.04 bh

Links Min size

9.5.3 (1) 0.25 or 6 mm

0.25 or 6 mm

Scl,tmax

9.5.3 (3): min(12min; 0.6b; 240 mm)

12

9.5.3 (6): 150 mm from main bar

150 mm from main bar

Detailing Issues EC2 Clause Issue

Possible resolve in 2013?

8.4.4.1

Lap lengths assume 4 centres in 2 bar beams

7 factor for spacing e.g. 0.63 for 6 centres in slabs or 10centre in two bar beams

Table 8.3

6 varies depending on amount staggered

6 should always = 1.5.

8.7.2(3) & Fig 8.7

0.3 lo gap between For ULS, there is no advantage in staggering ends of lapped bars is bars. For SLS staggering at say 0.5 lo might onerous. be helpful.

Table 8.2

2 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