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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 = vEd bw/(fywd cot )
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
Beam Example 2 High shear
Calculate
0.35
250 / 30 -130x20.0
96.4sin5.0
)250/1(20.0sin5.0
1
ckck
Ed1
ff
v
43.1cot
Asw/s = vEd bw/(fywd cot )
Asw/s = 4.96 x 140 /(435 x 1.43)
Asw/s = 1.12 mm
Try H10 links with 2 legs.
Asw = 157 mm2
s < 157 /1.12 = 140 mm
provide H10 links at 125 mm spacing
Beam Example 2 High shear
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 = vEd bw/(fywd cot )
Asw/s = 0.89 x 300 /(435 x 2.5)
Asw/s = 0.24 mm
Try H8 links with 2 legs, Asw = 101 mm2
s < 101 /0.24 = 420 mm
Maximum spacing = 0.75 d = 0.75 x 389 = 292 mm
provide H8 links at 275 mm spacing
Detailing
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
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
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
Concise: Figure 11.1 EC2: Figure 8.1
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
Concise: Figure 11.3 EC2: Figure 8.3
EC2: Figure 8.4
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
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
lbd = 1 2 3 4 5 (40.36) For concrete class C25/30
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
Detailing Comparisons
Slabs EC2 Clause / Values BS 8110 Values
Main Bars in Tension
As,min 9.2.1.1 (1):
0.26 fctm/fykbd 0.0013 bd
0.0013 bh
As,max 0.04 bd 0.04 bh
Secondary Transverse Bars
As,min 9.3.1.1 (2):
0.2As for single way slabs
0.002 bh
As,max 9.2.1.1 (3): 0.04 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
Detailing Comparisons
Punching Shear EC2Clause / Values BS 8110 Values
Links
Asw,min 9.4.3 (2):Link leg = 0.053sr st (fck)/fyk Total = 0.4ud/0.87fyv
Sr 9.4.3 (1): 0.75d 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