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 = 3.64 MPa
vRdcot θ = 1.0 = 5.28 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
Beam Example 2 – High shear
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
⇒ provide H10 links at 125 mm spacing
Beam Example 2 – High shear
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
⇒ provide H8 links at 275 mm spacing
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
Direction of concreting
b) h ≤≤≤≤ 250 mm d) h > 600 mm
unhatched zone – ‘good’ bond conditions
hatched zone - ‘poor’ bond conditions
α
Direction of concreting
250
Direction of concreting
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
lbd = α1 α2 α3 α4 α5 (40.36Ø) For concrete class C25/30
a) Tension anchorage – straight bar
α1 = 1.0
α3 = 1.0 conservative value with K= 0
α4 = 1.0 N/A
α5 = 1.0 conservative value
α2 = 1.0 – 0.15 (Cd – Ø)/Ø
α2 = 1.0 – 0.15 (25 – 16)/16 = 0.916
lbd = 0.916 x 40.36Ø = 36.97Ø = 592mm
43
Design anchorage length, lbd
lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min
lbd = α1 α2 α3 α4 α5 (40.36Ø) For concrete class C25/30
b) Tension anchorage – Other shape bars
α1 = 1.0 Cd = 25 is ≤ 3 Ø = 3 x 16 = 48
α3 = 1.0 conservative value with K= 0
α4 = 1.0 N/A
α5 = 1.0 conservative value
α2 = 1.0 – 0.15 (Cd – 3Ø)/Ø ≤ 1.0
α2 = 1.0 – 0.15 (25 – 48)/16 = 1.25 ≤ 1.0
lbd = 1.0 x 40.36Ø = 40.36Ø = 646mm
Worked example - summary
H16 Bars – Concrete class C25/30 – 25 Nominal cover
Tension anchorage – straight bar lbd = 36.97Ø = 592mm
Tension anchorage – Other shape bars lbd = 40.36Ø = 646mm
lbd is measured along the centreline of the bar
Compression anchorage (α1 = α2 = α3 = α4 = α5 = 1.0)
lbd = 40.36Ø = 646mm
Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7
Lap length = anchorage length x α6
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
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 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
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) 0.25φ 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
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