Beam Practical Design to Eurocode 2_Sep 2013

Beams

Lecture 2

19

th

_{September 2013}

Contents - Beams

•

Bending/ Flexure

–

Section analysis, singly and doubly reinforced

–

Tension reinforcement, A

_s

–

neutral axis depth limit & K’

–

Compression reinforcement, A

_s2

•

Shear in beams

–

variable strut method

•

Detailing

–

Anchorage & Laps

–

Members & particular rules

Section Design: Bending

•

In principal flexural design is generally the same as

BS8110

•

High strength concrete ( f

_ck

> 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 f

_ck

≤ 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, f_cd and reinforcement, f_yd

In EC2 there are no equations to determine A_s and A_s2 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 2

f

bd

M

K



Value of K for maximum value of M

with no compression steel and when x is at its maximum value.

As d fcd Fs x s x cu3 Fc Ac f_ck  50 MPa 50 < f_ck  90 MPa  0.8 = 0.8 – (f_ck – 50)/400  1.0 = 1,0 – (f_ck – 50)/200 f_cd = _cc f_ck /_c = 0.85 f_ck /1.5

Rectangular Concrete Stress Block

& at failure concrete strain, ε_cu= 0.0035

EC2: Cl 3.1.7, Fig 3.5 For f_ck  50 MPa: f_ck 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

_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 f_yk = 500 MPa

f_yd = f_yk/γ_s = 500/1.15 = 435 MPa

E_s may be taken to be 200 GPa

Steel yield strain = f_yd/E_s

(



_s at yield point) = 435/200000 = 0.0022

At failure concrete strain is 0.0035 for f_ck  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, ε_cu= 0.0035,  = 1 and  = 0.8.

fcd = 0.85fck/1.5, fyd = fyk/1.15 = 0.87 fyk F_c = (0.85 f_ck / 1.5) b (0.8 x) = 0.453 f_ck b x F_st = 0.87A_s f_yk M b

Methods to find As:

• Iterative, trial and error method – simple but not practical

Analysis of a singly reinforced beam

Determine A

_s

– Iterative method

For horizontal equilibrium F_c= F_st

0.453 f_ck b x = 0.87A_s f_yk

Guess A_s Solve for x z = d - 0.4 x M = F_c z

M b

Take moments about the centre of the tension force M = 0.453 f_ck b x z (1) Now z = d - 0.4 x  x = 2.5(d - z) & M = 0.453 f_ck b 2.5(d - z) z = 1.1333 (f_ck b z d - f_ck b z2₎ Let K = M / (f_ck 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

1.1333

₂

-

2₂

bd

f

bz

f

bd

f

bdz

f

bd

f

M

K

ck ck ck ck ck M

Analysis of a singly reinforced beam

Determine A

_s

– 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

Higher Concrete Strengths

f_ck ≤ 50MPa

_z



_d[1



₍₁



_3,529K

_)]/2

)]/2

3,715K

(1

d[1

z







f_ck = 60MPa f_ck = 70MPa f_ck = 80MPa f_ck = 90MPa

)]/2

3,922K

(1

d[1

z







)]/2

4,152K

(1

d[1

z







)]/2

4,412K

(1

d[1

z







Take moments about the centre of the compression force M = 0.87A_s f_yk z

Rearranging

A_s = M /(0.87 f_yk 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

Maximum neutral axis depth

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

  k₁ + k₂ x_u/d where

k₁ = 0.4

k₂ = 0.6 + 0.0014/ _cu2 = 0.6 + 0.0014/0.0035 = 1 x_u = depth to NA after redistribution

= Redistribution ratio

 x_u = d ( - 0.4)

Therefore there are limits on K and this limit is denoted K’

Moment Bending Elastic Moment Bending ted Redistribu  

The limiting value for K (denoted K’) can be calculated as follows: As before M = 0.453 f_ck b x z (1)

and K = M / (f_ck b d 2₎

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

M’ = b d2 _f ck (0.6  – 0.18  2 - 0.21)  K’ = M’ /(b d2 _f ck) = (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).

A

_s

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: F_sc = 0.87A_s2 f_yk

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 x_u).

The second part balances the force in the compression steel. The area of reinforcement required is therefore:

A_s = K’ f_ck b d 2 _{/(0.87 f}

yk z) + As2

A_s2 can be calculated by taking moments about the centre of the tension force: M = K’ f_ck b d 2 _{+ 0.87 f} yk As2 (d - d2) Rearranging A_s2 = (K - K’) f_ck b d 2 _{/ (0.87 f} yk (d - d2))

A

_s2

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 f d b M K  & ' 0.6 0.182 0.21 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.



1 1 3.53



0.95 *

2 K d

d

z    

Check minimum reinforcement requirements:

d

b

f

d

b

f

A

_t yk t ctm min , s

0

.

0013

26

.

0

_



Check max reinforcement provided A_s,max  0.04A_c (Cl. 9.2.1.1) Check min spacing between bars > Ø_bar > 20 > A_gg + 5

Check max spacing between bars

Calculate tension steel required from:

z f M A yd s 

Flow Chart for Singly-reinforced

Beam

Flow Chart for Doubly-

Reinforced Beam

Calculate lever arm z from:



_{1 1 3.53 '}



2 K

d

z   

Calculate excess moment from: _' 2

_

_'

_

K K

f bd

M  _ck 

Calculate compression steel required from:



₂



yd 2 s ' d d f M A  

Calculate tension steel required from:

Check max reinforcement provided A_s,max  0.04A_c (Cl. 9.2.1.1) Check min spacing between bars > Ø_bar > 20 > A_gg + 5

2 s yd 2 s ' A z f bd f K A  ck 

Flexure Worked Example

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 f_yk = 500 MPa.

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 – c_nom - Ø_link - 0.5Ø = 500 – 30 - 10 – 16 = 444 mm d₂ = c_nom + Ø_link + 0.5Ø = 25 + 10 + 10 = 45 mm

444

 provide compression steel









mm 363 168 . 0 53 . 3 1 1 2 444 ' 53 . 3 1 1 2         d K z ' . K f bd M K        209 0 30 444 300 10 370 2 6 ck 2

168

0.

'



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





kNm 7 . 72 10 ) 168 . 0 209 . 0 ( 30 444 300 ' ' 6 2 2           K K f b d M _ck





2 6 2 yd 2 s mm 419 45) – (444 435 10 x 72.7 '      d d f M A 2 6 2 s yd s mm 2302 419 363 435 10 ) 7 . 72 370 ( '          A z f M M A

Provide 2 H20 for compression steel = 628mm2 _{(419 mm}2 _req’d)

and 3 H32 tension steel = 2412mm2 _{(2302 mm}2 _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

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 Fac tor 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

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 2_f ck Fac to 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

Shear

There are three approaches to designing for shear:

• When shear reinforcement is not required e.g. slabs

Shear check uses V_Rd,c

• When shear reinforcement is required e.g. Beams

Variable strut method is used to check shear in beams

Strut strength check using V_Rd,max Links strength using V_Rd,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:

V_Rd,c – Resistance of member without shear reinforcement

V_Rd,s - Resistance of member governed by the yielding of shear reinforcement

V_Rd,max - Resistance of member governed by the crushing of compression struts

V_Ed - 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  ½ z ½ z V z = 0.9d F_cd F_td

compression chord compression chord

tension chord shear 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.



cot sw s Rd, z fywd s A V 









tan cot 1 max Rd,  cw w_ cd f z b V 21.8 <  < 45

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) V_Rd,s = A_sw z f_ywd cot  /s …1 V_Rd,max = 0.5 z b_w f_cd sin 2 …2 where 0.6 (1- f_ck/250) When cot = 2.5 (= 21.8°) V_Rd = 0.138 b_w z f_ck (1 - f_ck/250) Or in terms of stress: v_Rd = 0.138 f_ck (1 - f_ck/250)

Rearranging equation 2 in terms of stress:

  = 0.5 sin-1[v_Rd /(0.20 f_ck(1 - f_ck/250))] f_ck 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 A_sf_yd/s Shear Strength, V_R BS8110: V_R = V_C + V_S Test results V_R Eurocode 2: V_Rmax Minimum links Fewer links

(but more critical)

Safer

Eurocode 2 vs BS8110:

Shear

shear reinforcement control

V

_Rd,s

= A

_sw

z f

_ywd

cot





/s

Exp (6.8)

concrete strut control

V

_Rd,max

= z b

_w



₁

f

_cd

/(cotθ + tanθ) = 0,5

z

b

_w



₁

f

_cd

sin

2















Exp (6.9)



where



₁



=

0,6(1-f

_ck

/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, V_Ed 2. Determine V_Rd,max with cot = 2.5

3. If V_Rd,max > V_Ed cot = 2.5, go to step 6 and calculate required shear reinforcement

4. If V_Rd,max < V_Ed calculate required strut angle:

 = 0.5 sin-1_[(v

Ed/(0.20fck(1-fck/250))]

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

6. Calculate amount of shear reinforcement required

A_sw/s = v_Ed b_w/(f_ywd cot ) = V_Ed /(0.78 d f_yk cot )

7. Check min shear reinforcement, A_sw/s ≥ b_w ρ_w,min and max spacing,

s_l,max = 0.75d ρ_w,min = (0.08 √f_ck)/f_yk 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 v_Rd when cot  = 2.5

v_{Rdcot  = 2.5} = 0.138f_ck(1-f_ck/250)

Calculate area of shear reinforcement:

A_sw/s = v_Ed b_w/(f_ywd cot ) Determine v_Ed where:

v_Ed = design shear stress [v_Ed = V_Ed/(b_wz) = V_Ed/(b_w 0.9d)]

Is v_{Rdcot  = 2.5} > v_Ed? No

Check maximum spacing of shear reinforcement : s_,max = 0.75 d Determine  from:  = 0.5 sin-1_[(v Ed/(0.20fck(1-fck/250))] Is v_{Rdcot  = 1.0} > v_Ed? Yes (cot  > 1.0) No Re-size

Design aids for shear

Design aids for shear

•

Where a

_v

 2d the applied shear force, V

_Ed

, for a point load

(eg, corbel, pile cap etc) may be reduced by a factor a

_v

/2d

where 0.5  a

_v



2d provided:

d d

a_v a_v

− The longitudinal reinforcement is fully anchored at the support.

− Only that shear reinforcement provided within the central 0.75a_v is included in the resistance.

Short Shear Spans with Direct

Strut Action (6.2.3)

Beam Example 1

Cover = 40mm to each face f_ck = 30

Determine the flexural and shear reinforcement required

(try 10mm links and 32mm main steel) G_k = 75 kN/m, Q_k = 50 kN/m , assume no redistribution and use

equation 6.10 to calculate ULS loads.

8 m

450

Beam Example 1 – Bending

ULS load per m = (75 x 1.35 + 50 x 1.5) = 176.25 M_ult = 176.25 x 82_/8 = 1410 kNm d = 1000 - 40 - 10 – 32/2 = 934 120 . 0 30 934 450 10 1410 2 6 ck 2       f bd M K K’ = 0.208

K < K’  No compression reinforcement required



K







d d z 1 1 3.53 x 0.120 822 0.95 2 934 53 . 3 1 1 2         2 6 yd s 3943 mm 822 x 435 10 x 1410 _   z f M A Provide 5 H32 (4021 mm2₎

Beam Example 1 – Shear

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

Shear stress: v_Ed = V_Ed/(b_w 0.9d) = 705 x 103_{/(450 x 0.9 x 934)} = 1.68 MPa v_{Rdcot  = 2.5} = 3.64 MPa v_{Rdcot  = 2.5} > v_Ed  cot  = 2.5 A_sw/s = v_Ed b_w/(f_ywd cot ) A_sw/s = 1.68 x 450 /(435 x 2.5) A_sw/s = 0.70 mm Try H8 links with 3 legs.

A_sw = 151 mm2

s < 151 /0.70 = 215 mm

 provide H8 links at 200 mm spacing

f_ck 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) mm

2

₎

with

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

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: v_Ed = V_Ed/(b_w 0.9d) = 312.5 x 103_{/(140 x 0.9 x 500)} = 4.96 MPa v_{Rdcot  = 2.5} = 3.64 MPa v_{Rdcot  = 1.0} = 5.28 MPa v_{Rdcot  = 2.5} < v_Ed < v_{Rdcot  = 1.0}  2.5 > cot  > 1.0  Calculate  f_ck 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 -1 30 x 20 . 0 96 . 4 sin 5 . 0 ) 250 / 1 ( 20 . 0 sin 5 . 0 1 ck ck Ed 1   f f v 43 . 1 cot    A_sw/s = v_Ed b_w/(f_ywd cot  ) A_sw/s = 4.96 x 140 /(435 x 1.43) A_sw/s = 1.12 mm

Try H10 links with 2 legs.

A_sw = 157 mm2

s < 157 /1.12 = 140 mm

 provide H10 links at 125 mm spacing

Workshop Problem

Cover = 35 mm to each face f_ck = 30MPa

Design the beam in flexure and shear G_k = 10 kN/m, Q_k = 6.5 kN/m (Use eq. 6.10)

8 m

300

Exp (6.10) Remember this from last week?

Aide memoire

Or Concise Table 15.5

Solution - Flexure

ULS load per m = (10 x 1.35 + 6.5 x 1.5) = 23.25 kN/m M_ult = 23.25 x 82_{/8 = 186 kNm} d = 450 - 35 - 10 – 32/2 = 389 mm _{0 137} 30 389 300 10 186 2 6 ck 2 _{ }  .    f bd M K

K < K’  No compression reinforcement required



K







d d z 1 1 3 53 x 0 137 0 86x389 334 0 95 2 389 53 3 1 1 2   .    . .  .   .  2 6 yd s

₄₃₅

_x

₃₃₄

1280

mm

10

x

186







z

f

M

A

_{Provide 3 H25 (1470 mm}2₎ K’ = 0.168

Solution - Shear

Shear force, V_Ed = 23.25 x 8 /2 = 93 kN Shear stress: v_Ed = V_Ed/(b_w 0.9d) = 93 x 103_{/(300 x 0.9 x 389)} = 0.89 MPa v_Rd = 3.64 MPa v_Rd > v_Ed  cot  = 2.5 A_sw/s = v_Ed b_w/(f_ywd cot ) A_sw/s = 0.89 x 300 /(435 x 2.5) A_sw/s = 0.24 mm

Try H8 links with 2 legs, A_sw = 101 mm2

s < 101 /0.24 = 420 mm

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

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Identification of bars

Class A

Class B

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

• Clear horizontal and vertical distance  , (d_g +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

• 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

Minimum mandrel size,



_m

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

 _m,min  F_bt ((1/a_b)+1/(2 )) / f_cd

F_bt ultimate force in a bar at the start of a bend

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

For a bar adjacent to the face of the member, a_b 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

The design value of the ultimate bond stress, f_bd = 2.25 ₁₂f_ctd where f_ctd 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) 45º    90º c) h > 250 mm h Direction of concreting 300 h Direction of concreting b) h  250 mm d) h > 600 mm

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



Direction of concreting

250

Direction of concreting

Ultimate bond stress

EC2: Cl. 8.4.2 _{Concise: 11.5}

l

_b,rqd

= (



/ 4) (



_sd

/ f

_bd

)

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 l

_b,rqd

should be measured along the

centreline of the bar

EC2 Figure 8.1

l_bd = α₁ α₂ α₃ α₄ α₅ l_b,rqd  l_b,min However: (α₂ α₃ α₅)  0.7

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

Design Anchorage Length, l

_bd

EC2: Cl. 8.4.4 Concise: 11.4.2

Alpha values

EC2: Table 8.2

Table requires values for:

C_d Value depends on cover and bar spacing, see Figure 8.3 K Factor depends on position of confinement reinforcement,

see Figure 8.4

Table 8.2 - C

_d

& K factors

Concise: Figure 11.3

EC2: Figure 8.3

Table 8.2 - Other shapes

Concise: Figure 11.1

Alpha values

Anchorage of links

Concise: Fig 11.2

Laps

l₀ = α₁ α₂ α₃ α₅ α₆ l_b,rqd  l_0,min

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 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

l_0,min  max{0.3 α₆ l_b,rqd; 15; 200}

Design Lap Length, l

₀

(8.7.3)

Arrangement of Laps

EC2: Cl. 8.7.3, Fig 8.8

Worked example

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

Bond stress, f

_bd

f_bd = 2.25 η₁ η₂ f_ctd EC2 Equ. 8.2

η₁ = 1.0 ‘Good’ bond conditions

η₂ = 1.0 bar size ≤ 32 f_ctd = α_ct f_ctk,0,05/γ_c EC2 cl 3.1.6(2), Equ 3.16 α_ct = 1.0 γ_c = 1.5 f_ctk,0,05 = 0.7 x 0.3 f_ck2/3 _{EC2 Table 3.1} = 0.21 x 252/3 = 1.795 MPa f_ctd = α_ct f_ctk,0,05/γ_c = 1.795/1.5 = 1.197 f_bd = 2.25 x 1.197 = 2.693 MPa

Basic anchorage length, l

_b,req

l

_b.req

= (Ø/4) ( σ

_sd

/f

_bd

)

EC2 Equ 8.3

Max stress in the bar, σ

_sd

= f

_yk

/γ

_s

= 500/1.15

= 435MPa.

l

_b.req

= (Ø/4) ( 435/2.693)

= 40.36 Ø

Design anchorage length, l

_bd

l

_bd

= α

₁

α

₂

α

₃

α

₄

α

₅

l

_b.req

≥ l

_b,min

Alpha values

Table 8.2 - C

_d

& K factors

Concise: Figure 11.3

EC2: Figure 8.3

Design anchorage length, l

_bd

l

_bd

= α

₁

α

₂

α

₃

α

₄

α

₅

l

_b.req

≥ l

_b,min

l

_bd

= α

₁

α

₂

α

₃

α

₄

α

₅

(40.36Ø) For concrete class C25/30

a) Tension anchorage – straight bar

α

₁

= 1.0

α

₃

= 1.0

conservative value with K= 0

α

₄

= 1.0

N/A

α

₅

= 1.0

conservative value

α

₂

= 1.0 – 0.15 (C

_d

– Ø)/Ø

α

₂

= 1.0 – 0.15 (25 – 16)/16 = 0.916

l

_bd

= 0.916 x 40.36Ø = 36.97Ø = 592mm

Design anchorage length, l

_bd

l

_bd

= α

₁

α

₂

α

₃

α

₄

α

₅

l

_b.req

≥ l

_b,min

l

_bd

= α

₁

α

₂

α

₃

α

₄

α

₅

(40.36Ø) For concrete class C25/30

b) Tension anchorage – Other shape bars

α

₁

= 1.0

C

_d

= 25 is ≤ 3 Ø = 3 x 16 = 48

α

₃

= 1.0

conservative value with K= 0

α

₄

= 1.0

N/A

α

₅

= 1.0

conservative value

α

₂

= 1.0 – 0.15 (C

_d

– 3Ø)/Ø ≤ 1.0

α

₂

= 1.0 – 0.15 (25 – 48)/16 = 1.25 ≤ 1.0

l

_bd

= 1.0 x 40.36Ø = 40.36Ø = 646mm

Worked example - summary

H16 Bars – Concrete class C25/30 – 25 Nominal cover

Tension anchorage – straight bar

l

_bd

= 36.97Ø = 592mm

Tension anchorage – Other shape bars l

_bd

= 40.36Ø = 646mm

l

_bd

is measured along the centreline of the bar

Compression anchorage (α

₁

= α

₂

= α

₃

= α

₄

= α

₅

= 1.0)

l

_bd

= 40.36Ø = 646mm

Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7

Lap length = anchorage length x α

₆

Anchorage & lap lengths

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

Bond Length in bar diameters

conditions f_ck /f_cu 25/30 f_ck /f_cu 28/35 f_ck /f_cu 30/37 f_ck /f_cu 32/40 Full tension and

compression anchorage length, l_bd

‘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

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

• 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. l /30 A /2_st A /2st l /30 F_s F_s 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.

EC2 Section 9

Cl 9.2 Beams

Detailing of members and

particular rules

• A

_s,min

=

0,26

(f

_ctm

/f

_yk

)b

_t

d but



0,0013b

_t

d

• A

_s,max

=

0,04

A

_c

• 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



• Tension reinforcement in a flanged beam at

supports should be spread over the effective width

(see 5.3.2.1)

(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, ΔF_td,

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

reinforcement ΔF_td may be estimated by shifting the moment curve a

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

as an alternative for members with shear reinforcement, where: a_l = z (cot θ - cot α)/2 = 0.5 z cot θ for vertical shear links

z= lever arm, θ = angle of compression strut

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

‘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 = Vx



x = z cot/2 = a_l z V/sin  M/z - V cot/2 a_l

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

al

Ftd

al

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: a_l = 0.5 z Cot But it is always conservative to use a_l = 1.125d

• l

_bd

is required from the line of contact of the support.

Simplesupport(indirect) _{Simplesupport(direct)}

• A

_s

bottom steel at support



0.25

A

_s

provided in the span

• Transverse pressure may only be taken into account with

a ‘direct’ support.

Shear shift rule

a_l

Tensile Force Envelope

Anchorage of Bottom

Simplified Detailing Rules for

Beams

How to….EC2 Detailing section

 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

• Curtailment – as beams except for the “Shift” rule a

_l

= 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 A_s,min 9.2.1.1 (1): 0.26 f_ctm/f_ykbd 

0.0013 bd

0.0013 bh

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

Main Bars in Compression

A_s,min -- 0.002 bh

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

Spacing of Main Bars

s_min 8.2 (2): d_g + 5 mm or  or 20mm d_g + 5 mm or 

S_max Table 7.3N Table 3.28

Links

A_sw,min 9.2.2 (5): (0.08 b s f_ck)/f_yk 0.4 b s/0.87 f_yv

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

s_t,max 9.2.2 (8): 0.75 d  600 mm

9.2.1.2 (3) or 15 from main bar

Detailing Comparisons

Slabs EC2 Clause / Values BS 8110 Values

Main Bars in Tension

A_s,min 9.2.1.1 (1):

0.26 f_ctm/f_ykbd  0.0013 bd

0.0013 bh

A_s,max 0.04 bd 0.04 bh

Secondary Transverse Bars

A_s,min 9.3.1.1 (2):

0.2A_s for single way slabs

0.002 bh A_s,max 9.2.1.1 (3): 0.04 bd 0.04 bh Spacing of Bars s_min 8.2 (2): d_g + 5 mm or  or 20mm 9.3.1.1 (3): main 3h  400 mm d_g + 5 mm or  S_max secondary: 3.5h  450 mm 3d or 750 mm

places of maximum moment:

main: 2h  250 mm

Detailing Comparisons

Punching Shear EC2Clause / Values BS 8110 Values

Links

A_sw,min 9.4.3 (2):Link leg = 0.053s_r s_t (f_ck)/f_yk Total = 0.4ud/0.87fyv

S_r 9.4.3 (1): 0.75d 0.75d

S_t 9.4.3 (1):

within 1st control perim.: 1.5d outside 1st control perim.: 2d

1.5d

Columns

Main Bars in Compression

A_s,min 9.5.2 (2): 0.10N_Ed/f_yk  0.002bh 0.004 bh

A_s,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

S_cl,tmax 9.5.3 (3): min(12min; 0.6b; 240 mm) 12

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 l_o gap between

ends of lapped bars is onerous.

For ULS, there is no advantage in staggering

bars. For SLS staggering at say 0.5 l_o 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 l_bd anchorage into

support

May be OTT as compression forces increase bond strength. Issue about anchorage