Beam Practical Design to Eurocode 2_Sep 2013

Beams

Lecture 2

19th September 2013

Practical Design to Eurocode 2

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:

ck

2fbd

M K 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

As

d

fcd

Fs

x

s

x

cu3

Fc Ac

fck 50 MPa 50 < fck 90 MPa

0.8 = 0.8 (fck 50)/400

1.0 = 1,0 (fck 50)/200

fcd = cc fck /c = 0.85 fck /1.5

Rectangular Concrete Stress Block

& at failure concrete strain, cu= 0.0035

EC2: Cl 3.1.7, Fig 3.5

For fck 50 MPa:

fck

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

ud

fyd/Es

fyk

kfyk

fyd = fyk/s

kfyk/s

Idealised

Design

uk

Reinforcement

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

In UK fyk = 500 MPa

fyd = fyk/s = 500/1.15 = 435 MPa

Es may be taken to be 200 GPa

Steel yield strain = fyd/Es (s at yield point) = 435/200000 = 0.0022

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:

For grades of concrete up to C50/60, cu= 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

M

b

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

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

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

2

2

22 - 1.1333

bdf

bzf

bdf

bdzf

bdf

MK

ck

ck

ck

ck

ck

M

Analysis of a singly reinforced beam

Determine As Direct method

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

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

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)

Tension steel, As

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

= Redistribution ratio

xu = d ( - 0.4)

Therefore there are limits on K and

this limit is denoted K

Moment Bending Elastic

Moment Bending tedRedistribu

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

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

and 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) = (0.6 0.18 2 - 0.21)

c.f. from BS 8110 rearranged K = (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).

K

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

Compression steel needed -

doubly reinforced

Is K K ?

No compression steel

needed singly reinforced

Yes No

ck

2 fdb

MK 21.018.06.0'& 2 K

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

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 > bar > 20 > Agg + 5

Check max spacing between bars

Calculate tension steel required from:

zf

MA

yd

s

Flow Chart for Singly-reinforced

Beam

Flow Chart for 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 > bar > 20 > Agg + 5

2syd

2

s

'A

zf

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. = 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 - link - 0.5

= 500 30 - 10 16

= 444 mm

d2 = cnom + link + 0.5

= 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 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 reqd)

and 3 H32 tension steel = 2412mm2 (2302 mm2 reqd)

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

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

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)

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))

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 = 0,9d may normally

be used.

cotswsRd, ywdfzs

AV

tancot

1maxRd,

cdwcw

fzbV

21.8 < < 45

Strut Inclination Method

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

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))]

fck vRd, cot =

2.5

vRd, cot = 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 6.2.3 EN 1992-1-1

Shear Design: Links

Variable strut method allows 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

38

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 /s Exp (6.8)

concrete strut control

VRd,max = z bw 1 fcd /(cot + tan) = 0,5 z bw 1fcd sin 2 Exp (6.9)

where 1 = 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

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 w,min and max spacing, sl,max = 0.75d w,min = (0.08 fck)/fyk cl 9.2.2

Shear Resistance with Shear

Reinforcement

EC2 Shear Flow Chart for vertical links

Yes (cot = 2.5)

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

Calculate area of shear

reinforcement:

Asw/s = vEd bw/(fywd cot )

Determine vEd where:

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

Is vRdcot = 2.5 > vEd? No

Check maximum spacing of shear

reinforcement :

s,max = 0.75 d

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

Is vRdcot = 1.0 > vEd?

Yes (cot > 1.0)

No Re-size

Design aids for shear Concise Fig 15.1 a)

Design aids for shear Concise Fig 15.1 b)

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:

d d

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

Beam examples

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 = 0.208

K < K No compression reinforcement required

dKdz 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 = 2.5 = 3.64 MPa

vRdcot = 2.5 > vEd

cot = 2.5


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

fck vRd, cot =

2.5

vRd, cot = 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 = 2.5 = 3.64 MPa

vRdcot = 1.0 = 5.28 MPa

vRdcot = 2.5 < vEd < vRdcot = 1.0

2.5 > cot > 1.0 Calculate

fck vRd, cot =

2.5

vRd, cot = 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

0.35

250 / 30 -130x20.0

96.4sin5.0

)250/1(20.0sin5.0

1

ckck

Ed1

ff

v

43.1cot


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



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

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

1370

30389300

101862

6

ck

2.

fbd

MK

K < K No compression reinforcement required

dKdz 950334389x8601370x533112

38953311

2.....

26

yd

s mm1280334x435

10x186

zf

MA

Provide 3 H25 (1470 mm2)

K = 0.168

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 = 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


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 Rules

Beams

Slabs

Columns

Walls

Foundations

Discontinuity Regions

Tying Systems

Cover Fire

Specification and Workmanship

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 bars

EC2: Cl. 8.2 Concise: 11.2

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

Minimum mandrel size, m

Min. Mandrel Dia. for bent bars EC2: Cl. 8.3 Concise: 11.3

Minimum mandrel size, m

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

Min. Mandrel Dia. for bent bars EC2: Cl. 8.3 Concise: 11.3

Bearing stress

inside bends

Anchorage of reinforcement

EC2: Cl. 8.4

The design value of the ultimate bond stress, fbd = 2.25 12fctd

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

a) 45 90 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


250


Ultimate bond stress EC2: Cl. 8.4.2 Concise: 11.5

300

lb,rqd = (/ 4) (sd / fbd)

where sd is the design stress of the bar at the position

from where the anchorage is measured.

Basic required anchorage length EC2: Cl. 8.4.3 Concise: 11.4.3

For bent bars lb,rqd should be measured along the centreline of the bar

EC2 Figure 8.1

Concise Fig 11.1

lbd = 1 2 3 4 5 lb,rqd lb,min

However: (2 3 5) 0.7

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

Design Anchorage Length, lbd

EC2: Cl. 8.4.4 Concise: 11.4.2

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

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

Table 8.2 - Cd & K factors

Concise: Figure 11.3 EC2: Figure 8.3

EC2: Figure 8.4

Table 8.2 - Other shapes


Alpha values EC2: Table 8.2 Concise: 11.4.2

Anchorage of links Concise: Fig 11.2 EC2: Cl. 8.5

Laps

EC2: Cl. 8.7

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

6 = (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 cross-

section area

< 25% 33% 50% >50%

6 1 1.15 1.4 1.5

Note: 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}

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

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 1 2 fctd EC2 Equ. 8.2

1 = 1.0 Good bond conditions

2 = 1.0 bar size 32

fctd = ct fctk,0,05/c EC2 cl 3.1.6(2), Equ 3.16

ct = 1.0 c = 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 = ct fctk,0,05/c = 1.795/1.5 = 1.197

fbd = 2.25 x 1.197 = 2.693 MPa

Basic anchorage length, lb,req

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

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 EC2: Table 8.2 Concise: 11.4.2

Table 8.2 - Cd & K factors


EC2: Figure 8.4

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


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

Design anchorage length, lbd

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


b) Tension anchorage Other shape bars

1 = 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

Worked example - summary

H16 Bars Concrete class C25/30 25 Nominal cover

Tension anchorage straight bar lbd = 36.97 = 592mm

Tension anchorage Other shape bars lbd = 40.36 = 646mm

lbd is measured along the centreline of the bar

Compression anchorage (1 = 2 = 3 = 4 = 5 = 1.0)

lbd = 40.36 = 646mm

Anchorage for Poor bond conditions = Good/0.7

Lap length = anchorage length x 6

Anchorage & lap lengths How to design concrete structures using 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.

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

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.

Arrangement of Laps EC2: Cl. 8.7.2 Concise: Cl 11.6

Any Transverse reinforcement provided for other reasons will be

sufficient if bar < 20mm or laps< 25%

If bar 20mm then additional transverse reinforcement may be needed. It should be positioned at the outer sections of the lap as shown

below.

l /30A /2

st

A /2st

l /30FsFs

150 mm

l0

Transverse Reinforcement at Laps

Bars in tension EC2: Cl. 8.7.4, Fig 8.9

Concise: Cl 11.6.4

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

Transverse Reinforcement at Laps

Bars in compression EC2: Cl. 8.7.4, Fig 8.9

Concise: Cl 11.6.4

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

EC2 Section 9

Cl 9.2 Beams

Detailing of members and

particular rules

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)

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 - 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 (9.2.1.3)

Shift Rule for Shear Horizontal component of diagonal shear force

= (V/sin) . cos = V cot

Applied

shear V

Applied

moment M M/z + V cot/2 = (M + Vz cot/2)/z

M = Vz cot/2

dM/dx = V

M = Vx x = z cot/2 = al

z

V/sin

M/z - V cot/2

al

Curtailment of longitudinal tension reinforcement

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

Shift Rule Curtailment of reinforcement

EC2: Cl. 9.2.1.3, Fig 9.2 Concise: 12.2.2

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

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 Detailing Rules for

Beams

How to.EC2

Detailing section

Concise: Cl 12.2.4

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

Plan view

EC2: Cl. 9.2.5

Concise: Cl 12.2.8

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

Solid slabs EC2: Cl. 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 or 20mm dg + 5 mm or

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 15 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 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


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) 0.25 or 6 mm 0.25 or 6 mm

Scl,tmax 9.5.3 (3): min(12min; 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 10centre 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

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 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

Documents

Beam Practical Design to Eurocode 2_Sep 2013