lecture 2 beams - mar 14 - k - rev 8h

1

Beams

Lecture 2

18th March 2014

Practical Design to Eurocode 2

Contents – Lecture 2 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 rulesShift rule for curtailment

Note: SLS check for deflection and cracking is done in week 4 - slabs

2

Bending/ Flexure

Section Design: Bending

• In principal flexural design is generally the same as

BS8110

• EC2 presents the principles only

• Design manuals will provide the standard solutions for

basic design cases.

• High strength concrete ( fck > 50 MPa ) can be designed.

Note: TCC How to guide equations and equations used on

this course are based on a concrete fck ≤ 50 MPa

3

Section Analysis to determineTension & 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, δ.• 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

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

Rectangular Concrete Stress Block

For fck ≤ 50 MPa failure concrete strain, εcu, = 0.0035

EC2: Cl 3.1.7, Fig 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

4

εud

ε

σ

fyd/ Es

fyk

kfyk

fyd = fyk/γs

kfyk/γs

Idealised

Design

εuk

ReinforcementDesign Stress/Strain Curve

EC2: Cl 3.2.7, Fig 3.8

In UK fyk = 500 MPa

fyd = fyk/γs = 500/1.15 = 435 MPa

Es may be taken to be 200 GPa

Steel yield strain = fyd/Es

(εεεεs 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 beamEC2: Cl 3.1.7

Design equations can be derived as follows:

For grades of concrete up to C50/60, εcu= 0.0035, ηηηη = 1 and λλλλ = 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

5

Analysis of a singly reinforced beamDetermine As – Iterative method

For horizontal equilibrium Fc= Fst0.453 fck b x = 0.87As fyk

Guess As Solve for x z = d - 0.4 x M = Fc z

M

b

Stop when design applied BM, MEd, ≃ M

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 beamDetermine As – Direct method

6

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

Quadratic formula

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 −+=

Normal strength

7

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, AsConcise: 6.2.1

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’.

8

Design aids for flexureTCC 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:

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

= Redistribution ratio

∴ xu = d (δ - 0.4)

Therefore there are limits on K and

this limit is denoted K’

Moment Bending Elastic

Moment Bending tedRedistribu =δ

9

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) & z = d – 0.4 x

Substituting xu 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.

Concise: 6.2.1

K’ and Beams with Compression Reinforcement, As2

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 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

10

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))

As2Concise: 6.2.1EC2: Fig 3.5

The following flowchart outlines the design procedure for rectangularbeams 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

11

Calculate lever arm z from:

(Or look up z/d from table or from chart.)

* 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====

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 > Øbar > 20 > Agg + 5

2syd

2

s'

Azf

bdfKA ck ++++====

12

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

13

Initially assume 32 mm φ for tension reinforcement with 30 mm

nominal cover to the link all round (allow 10 mm for link) and assume

20mm φ for compression reinforcement.

d = h – cnom - Ølink - 0.5Ø

= 500 – 30 - 10 – 16

= 444 mm

d2 = cnom + Ølink + 0.5Ø

= 30 + 10 + 10

= 50 mm

= 50

= 444

Worked Example 1

∴ 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

Worked Example 1

14

( )

kNm7.72

10)168.0209.0(30444300

''

62

2

=

×−×××=

−=−

KKfbdMck

( )

2

6

2yd

2s

mm 424

50) – (444435

10 x 72.7

=

×=

−=

ddf

MA

'

2

6

2s

yd

s

mm2307

424363435

10772370

=

+×

×−=

+−

=

).(

'A

zf

MMA

Worked Example 1

Provide 2 H20 for compression steel = 628mm2 (424 mm2 req’d)

and 3 H32 tension steel = 2412mm2 (2307 mm2 req’d)

By inspection does not exceed maximum area (0.04 Ac) or maximum

spacing of reinforcement rules (cracking see week 4 notes)

Check minimum spacing, assuming H10 links

Space between bars = (300 – 30 x 2 - 10 x 2 - 32 x 3)/2

= 62 mm > 32 mm* …OK

* EC2 Cl 8.2 (2) Spacing of bars for bond:

Clear distance between bars > Ф bar > 20 mm > Agg + 5 mm

Worked Example 1

15

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)

Normal strength

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)

16

Shear in Beams

Shear

There are three approaches to designing for shear:

• When shear reinforcement is not required e.g. slabsShear check uses VRd,c

• When shear reinforcement is required e.g. Beams

Variable strut method is used to check shear in beamsStrut 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. Cl 3.1.2 (2) P and NA.

EC2: Cl 6.2.2, 6.2.3, 6.4 Concise: 7.2, 7.3, 8.0

17

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

EC2: Cl 6.2.3 Concise: 7.3

Members Requiring Shear Reinforcement

θ

s

d

V(cot θ - cotα )

V

N Mα ½ z

½ zV

z = 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.

EC2: 6.2.3(1)Concise: Fig 7.3

18

θθθθcotswsRd, ywdfz

s

AV ====

θθθθθθθθννννααααtancot

1maxRd, ++++

==== cdwcw fzbV

21.8°°°° < θθθθ < 45°°°°

Strut Inclination MethodEC2: Equ. 6.8 & 6.9 for Vertical links

Equ 6.9

Equ 6.8

VEd

Strut angle limits

Cot θ = 2.5 Cot θ = 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

Max angle - max shear resistanceAngle = 21.8° V carried on 6 links

Min strut angle - Minimum links

dz

x

d

x

θz

sVhigh Vlow

Min curtailment

for 45o strut

19

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 Vertical Shear Reinforcement Concise: 7.3.3

20

Equation 6.9 is first used to determine the strut

angle θ and then equation 6.8 is used to find the

shear link area, Asw, and spacing s.

Equation 6.9 gives VRd,max values for a given strut

angle θ

e.g. when cot θ θ θ θ = 2.5 (θ θ θ θ = 21.8°) Equ 6.9 becomesVRd,max = 0.138 bw z fck (1 - fck/250)

or in terms of stress:

vRd ,max= 0.138 fck (1 - fck/250)Values are in the middle column of the table.

Re-arranging equation 6.9 to find θ:

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

Suitable shear links are found from equation 6.8:

Asw /s = vEdbw/( fywd cot 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

Shear linksEC2: Cl 6.2.3

vRd ,max values, MPa, for

cot θ = 1.0 and 2.5

Procedure for design with variable strut

1. Determine maximum applied shear force at support, VEd

2. Determine VRd,max with cotθ = 2.5

3. If VRd,max > VEd cotθ = 2.5, go to step 6 and calculate required shear

reinforcement

4. If VRd,max < VEd calculate required strut angle:

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

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

6. Calculate amount of shear reinforcement required

Asw/s = vEd bw/(fywd cot θ) = VEd /(0.78 d fyk cot θ)

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

Shear Resistance with Shear Reinforcement

21

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) (or look up from table)

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 vRd,cot 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 vRd,cot θθθθ = 1.0 > vEd?

Yes (cot θθθθ > 1.0)

NoRe-size

Design aids for shearConcise Fig 15.1 a)

22

Design aids for shearConcise 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:

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

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

EC2: 6.2.3

23

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

24

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

[[[[ ]]]] [[[[ ]]]] 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

25

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

26

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



27

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

28

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

ΦΦΦΦmm

Area, mm2

8 50

10 78.5

12 113

16 201

20 314

25 491

32 804

29

Working space

Working space

30

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

137030389300

101862

6

ck

2.=

×××

==fbd

MK

K < K’⇒ No compression reinforcement required

[ ] [ ] dKd

z 950334389x8601370x533112

38953311

2..... ≤==−+=−+=

26

yd

s mm1280334x435

10x186===

zf

MA Provide 3 H25 (1470 mm2)

K’ = 0.208

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 θ = 2.5

Asw/s = vEd bw/(fywd cot θ)

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


31

Detailing

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

32

• 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 RulesSpacing 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, φφφφm

Min. Mandrel Dia. for bent barsEC2: Cl. 8.3 Concise: 11.3

φφφφm

33

Minimum mandrel size, φφφφm

• To avoid failure of the concrete inside the bend of the bar:

φφφφ m,min ≥ Fbt ((1/ab) +1/(2 φφφφ)) / 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 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 ≥φ inside bend

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

Min. Mandrel Dia. for bent barsEC2: Cl. 8.3 Concise: 11.3

Bearing stress

inside bends

Anchorage of reinforcement

EC2: Cl. 8.4

34

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

300

lb,rqd = (φ φ φ φ / 4) (σσσσsd / fbd)

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

from where the anchorage is measured.

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

EC2 Figure 8.1

Concise Fig 11.1

35

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

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 factorsConcise: Figure 11.3EC2: Figure 8.3

EC2: Figure 8.4

36

Table 8.2 - Other shapesConcise: Figure 11.1EC2: Figure 8.1

Alpha valuesEC2: Table 8.2 Concise: 11.4.2

37

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

Laps

EC2: Cl. 8.7

38

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 Concise: 11.6.2

Arrangement of Laps

EC2: Cl. 8.7.3, Fig 8.8

39

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

40

Bond stress, fbdfbd = 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 EC2 cl 3.1.6(2), Equ 3.16

αct = 1.0 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

41

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 valuesEC2: Table 8.2 Concise: 11.4.2

42

Table 8.2 - Cd & K factorsConcise: Figure 11.3EC2: Figure 8.3

EC2: Figure 8.4

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


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

43

Design anchorage length, lbd

lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min


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

44

Anchorage & lap lengthsHow 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 /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

45

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 LapsEC2: Cl. 8.7.2 Concise: Cl 11.6

• Any Transverse reinforcement provided for other reasons will be

sufficient if the lapped bar Ø < 20mm or laps< 25%

• If the lapped bar Ø ≥ 20mm 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. Also it should be positioned at the outer sections

of the lap as shown:

Transverse Reinforcement at LapsBars 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.

l /30

ΣA /2st

ΣA /2st

l /30

FsFs

≤150 mm

l0Figure 8.9 (a) -

bars in tension

46

Transverse Reinforcement at LapsBars in tension

EC2: Cl. 8.7.4, Fig 8.9

Concise: Cl 11.6.4

• Also, if the lapped bar Ø ≥ 20mm and more than 50% of the

reinforcement is lapped at one point and the distance between adjacent

laps at a section, a, ≤ 10φφφφ , then transverse bars should be formed by links or

U bars anchored into the body of the section.

Transverse Reinforcement at LapsBars in compressionEC2: 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 (b) – bars in compression

47

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 (φ) should be held by transverse reinforcement with spacing ≤15 φ

Beams (9.2)

48

• 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)

49

Horizontal component of diagonal shear force

= (V/sinθ) . cosθ = V cotθ

Applied

shear V

Applied

moment MM/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’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 θBut it is always conservative to use al = 1.125d

50

• 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.

Anchorage of Bottom Reinforcement at End Supports

EC2: Cl. 9.2.1.4

Simplified Detailing Rules for Beams

How to….EC2

Detailing section

Concise: Cl 12.2.4

51

≤ 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 slabsEC2: Cl. 9.3

52

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

53


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

Class A

Class B

Class C

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

Identification of bars on siteCurrent BS 4449

54

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

Identification on siteCurrent BS 4449

Detailing IssuesEC2 Clause Issue Possible resolve in 2014?

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

lecture 2 beams - mar 14 - k - rev 8h

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