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EC2 Webinar – Autumn 2016 Lecture 3/1
Bending and Shear in Beams
Lecture 3
5th October 2016
Contents – Lecture 3
• Bending/ Flexure
– Section analysis, singly and doubly reinforced
– Tension reinforcement, As
– neutral axis depth limit & K’
– Compression reinforcement, As2
• Flexure Worked Example – Doubly reinforced
• Shear in Beams - Variable strut method
• Beam Examples – Bending, Shear & High shear
• Exercise - Design a beam for flexure and shear
EC2 Webinar – Autumn 2016 Lecture 3/2
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.
• There are modifications for high strength concrete
( fck > 50 MPa )
Note: TCC How to guide equations and equations used on
this course are based on a concrete fck ≤ 50 MPa
EC2 Webinar – Autumn 2016 Lecture 3/3
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, tension steel, and As2,
compression steel, 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
Concise:
Fig 6.1
Remember this
from last week?
EC2 Webinar – Autumn 2016 Lecture 3/4
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
For no compression
reinforcement Fsc = 0
EC2 Webinar – Autumn 2016 Lecture 3/5
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, Fst:
M = Fc z = 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
EC2 Webinar – Autumn 2016 Lecture 3/6
0 = (z/d)2 – (z/d) + 0.88235K
Solving the quadratic equation:
z/d = [1 + (1 - 3.53K)0.5]/2
z = d [ 1 + (1 - 3.53K)0.5]/2
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
EC2 Webinar – Autumn 2016 Lecture 3/7
Take moments about the centre of the compression force, Fc:
M = Fst z = 0.87As fyk z
Rearranging
As = M /(0.87 fyk z)
The required area of reinforcement can now be found using three
methods:
a) calculated using these expressions
b) obtained from Tables of z/d (eg Table 5 of How to beams or
Concise Table 15.5, see next slide)
c) obtained from graphs (eg from the ‘Green Book’ or Fig B.3 in
Concrete Buildings Scheme Design Manual, next slide but one)
Tension steel, AsConcise: 6.2.1
Design aids for flexure - method (b)Concise: Table 15.5K = M / (fck b d 2)
‘Normal’ tables and
charts are only valid
up to C50/60
.Traditionally z/d was
limited to 0.95 max to
avoid issues with the
quality of ‘covercrete’.
EC2 Webinar – Autumn 2016 Lecture 3/8
Design aids for flexure- method (c)TCC Concrete Buildings Scheme Design Manual, Fig B.3
Design chart for singly reinforced beam
K=
M / (
f ck
b d
2)
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’
For K > K’ Compression steel needed
Moment Bending Elastic
Moment Bending tedRedistribu =δ
Concise:
Table 6.1
EC2: Cl 5.5 Linear elastic analysis with limited redistribution
EC2 Webinar – Autumn 2016 Lecture 3/9
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 & xu = d (δ – 0.4)
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 δδδδ – 0.18 δδδδ 2 - 0.21)
Min δ = 0.7 (30% redistribution). Steel to be either Class B or C for 20% to
30% redistribution.
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.
Note: For plastic analysis xu/d must be ≤ 0.25 for normal strength concrete,
EC2 cl 5.6.2 (2).
Concise: 6.2.1
K’ and Beams with Compression Reinforcement, As2
For K > K’ compression reinforcement As2 is required.
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))
Compression steel, As2Concise: 6.2.1EC2: Fig 3.5
EC2 Webinar – Autumn 2016 Lecture 3/10
Tension steel, 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
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
EC2 Webinar – Autumn 2016 Lecture 3/11
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/Slab K ≤ K’
(Cl.9.2.1.1)Exp. (9.1N)
Minimum Reinforcement Area
The minimum area of reinforcement for beams and slabs is given by:
EC2: Cl. 9.2.1.1, Exp. 9.1N
dbf
dbfA t
yk
tctmmin,s 0013.0
26.0≥≥
EC2 Webinar – Autumn 2016 Lecture 3/12
Flow Chart for Doubly-Reinforced Beam K > K’
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 ++++====
Flexure Worked Example(Doubly reinforced)
EC2 Webinar – Autumn 2016 Lecture 3/13
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
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
EC2 Webinar – Autumn 2016 Lecture 3/14
∴ provide compression steel
[ ]
[ ]mm363
168.053.3112
444
'53.3112
=
×−+=
−+= Kd
z
'. K
fbd
MK
>=
××
×=
=
2090
30444300
103702
6
ck
2
1680.' =Kδδδδ 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
( )
kNm7.72
10)168.0209.0(30444300
''
62
2
=
×−×××=
−=
−
KKfbdM ck
( )
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
EC2 Webinar – Autumn 2016 Lecture 3/15
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 6 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
For H type bar reinforcement what is fyd?
Poll Q1:Design reinforcement strength, fyd
a. 435 MPa
b. 460 MPa
c. 476 MPa
d. 500 MPa
EC2 Webinar – Autumn 2016 Lecture 3/16
A beam section has an effective depth of 500mm and the
ultimate elastic bending moment has been reduced by 30%.
What is the maximum depth of the neutral axis, xu?
Poll Q2:Neutral axis depth, x
a. 150 mm
b. 250 mm
c. 300 mm
d. 450 mm
Shear in Beams
Variable strut method
EC2 Webinar – Autumn 2016 Lecture 3/17
Shear
There are three approaches to designing for shear:
• When shear reinforcement is not required e.g. slabs, week 5Shear 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, week 5
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
Shear in Beams
Shear design is different from BS8110. EC2 uses the variable strut
method to check a member with shear reinforcement.
Definitions:
VEd - Applied shear force. For predominately UDL, shear may be checked
at d from face of support
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
EC2: Cl 6.2.3 Concise: 7.3
Whilst Eurocode 2 deals in Resistances (capacities), VRd,c ,VRd,s ,VRd,max and Effect of actions, VEd in kN, in practice, it is often easier to consider shear strengths vRd, vRd,max and shear stresses, vEd, in MPa.
EC2 Webinar – Autumn 2016 Lecture 3/18
Members Requiring Shear Reinforcement
θ
s
d
V(cot θ - cotα )
V
N Mα ½ z
½ zVz = 0.9d
Fcd
Ftd
compression chord compression chord
tension chordshear reinforcement
α angle between shear reinforcement and the beam axis
Normally links are vertical. α = 90o and cot α is zero
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
compression strut
Members Requiring Shear Reinforcement
EC2: 6.2.3(1)Concise: Fig 7.3
bw is the minimum width
Asw Area of the shear reinforcement
fywd design yield strength = fyk/1.15
fcd design compressive strength = αccfck/1.5
= fck/1.5 (αcc = 1.0 for shear)
αcw = 1.0 Coefficient for stress in compression chord
ν1 strength reduction factor concrete cracked in shear
ν1 = ν = 0.6(1-fck/250) Exp (6.6N)
EC2 Webinar – Autumn 2016 Lecture 3/19
θθθθcotswsRd, ywdfz
s
AV ====
θθθθθθθθ
νννναααα
tancot
1maxRd,
++++==== cdwcw fzb
V21.8°°°° < 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
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
EC2 Webinar – Autumn 2016 Lecture 3/20
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
shear reinforcement control – Exp (6.8)
VRd,s = Asw z fywd cot θ /s
concrete strut control – Exp (6.9)
VRd,max = αcwz bw ν1 fcd /(cotθ + tanθ) = 0.5 z bw ν1 fcd sin 2θ
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
where:
fywd = fyk/1.15
ν1 = ν = 0.6(1-fck/250)
αcw = 1.0
EC2 Webinar – Autumn 2016 Lecture 3/21
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
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 minimum area, cl 9.2.2:Asw/s ≥ bw ρw,min
ρw,min = (0.08 √fck)/fyk ≈ 0.001
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
Check maximum spacing of shear reinforcement, cl 9.2.2:
sl,max = st,max = 0.75 d
EC2 Webinar – Autumn 2016 Lecture 3/22
Design aids for shearConcise Fig 15.1 a)
• 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
EC2 Webinar – Autumn 2016 Lecture 3/23
Beam examples
Bending, Shear and High Shear
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 EC0 equation 6.10 to calculate ULS loads.
8 m
450
1000
EC2 Webinar – Autumn 2016 Lecture 3/24
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 (We 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 H10 links with 4 legs.
Asw = 314 mm2
s < 314 /0.70 = 448 mm
⇒ provide H10 links at 450 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
EC2 Webinar – Autumn 2016 Lecture 3/25
Beam Example 1
Provide 5 H32 (4021) mm2)
with
H10 links at 450 mm spacing
Max spacing is 0.75d = 934 x 0.75
= 700 mm
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
EC2 Webinar – Autumn 2016 Lecture 3/26
Find the minimum area of shear reinforcement required to resist
the design shear force using EC2.
fck = 30 MPa and
fyd = 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
EC2 Webinar – Autumn 2016 Lecture 3/27
Exercise
Lecture 3
Design a beam for flexure and shear
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 EC0 eq. 6.10)
8 m
300
450
Beam Exercise – Flexure & Shear
EC2 Webinar – Autumn 2016 Lecture 3/28
Exp (6.10)
Remember
this from the
first week?
Aide memoire
OrConcise Table 15.5
Workings:- Load, Mult, d, K, 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
EC2 Webinar – Autumn 2016 Lecture 3/29
Working space
Working space
EC2 Webinar – Autumn 2016 Lecture 3/30
Working space
End of Lecture 3
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