EC2 Webinar – Autumn 2016 Lecture 3/1 Bending and Shear in Beams Lecture 3 5 th October 2016 Contents – Lecture 3 • Bending/ Flexure – Section analysis, singly and doubly reinforced – Tension reinforcement, A s – neutral axis depth limit & K’ – Compression reinforcement, A s2 • 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
Bending and Shear in Beams
Apr 05, 2023
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Microsoft PowerPoint - Lecture 3 Bending and Shear in Beams - PHG - A8 - 2Oct16 - Print versionBending and Shear in Beams
Lecture 3
– Tension reinforcement, As
– Compression reinforcement, As2
• 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
BS8110
• Design manuals will provide the standard solutions
for basic design cases.
( 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 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, 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
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
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
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
ε ud
Es may be taken to be 200 GPa
Steel yield strain = fyd/Es
= 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 EC2: 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
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, Fst:
M = Fc z = 0.453 fck b x z (1)
Now z = d - 0.4 x
∴ x = 2.5(d - z)
= 1.1333 (fck b z d - fck b z 2)
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
Analysis of a singly reinforced beam Determine 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
The lever arm for an applied moment is now known
M
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, As Concise: 6.2.1
Design aids for flexure - method (b) Concise: Table 15.5K = M / (fck b d 2)
‘Normal’ tables and
avoid issues with the
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 / (
Maximum neutral axis depth
According to Cl 5.5(4) the depth of the neutral axis is limited, viz:
δ ≥ k1 + k2 xu/d
xu = depth to NA after redistribution
= Redistribution ratio
this limit is denoted K’
For K > K’ Compression steel needed
Moment Bending Elastic
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).
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, As2 Concise: 6.2.1EC2: Fig 3.5
EC2 Webinar – Autumn 2016 Lecture 3/10
Tension steel, 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
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: δδδδ =1.0 means no redistribution and δδδδ = 0.8 means 20% moment redistribution.
Compression steel needed - doubly reinforced
Is K ≤ K’ ?
Yes No
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
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.
[[[[ ]]]] *95.053.311 2
dK d
Check min spacing between bars > Øbar > 20 > Agg + 5
Check max spacing between bars
Calculate tension steel required from: zf
M A
(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
db f
Flow Chart for Doubly- Reinforced Beam K > K’
Calculate lever arm z from: [[[[ ]]]]'53.311 2
K d
−−−−====
(((( ))))2yd 2s
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
2s yd
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
∴ provide compression steel
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 > bar > 20 mm > Agg + 5 mm
Worked Example 1
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
Shear
There are three approaches to designing for shear:
• When shear reinforcement is not required e.g. slabs, week 5 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, 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
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.
compression strut
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
θθθθcotsw sRd, ywdfz
Strut Inclination Method EC2: Equ. 6.8 & 6.9 for Vertical links
Equ 6.9
Equ 6.8
Shear
Safer
Shear Design: Links
hence activating more links.
As strut angle reduces concrete stress increases
Angle = 45°°°° V carried on 3 links
Max angle - max shear resistance Angle = 21.8°°°° V carried on 6 links
Min strut angle - Minimum links
dz
x
d
x
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
Shear Resistance of Sections with Vertical Shear Reinforcement Concise: 7.3.3
where:
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 becomes VRd,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)
fck vRd, cot θθθθ
EC2 – Shear Flow Chart for vertical links
Yes (cot θθθθ = 2.5)
Determine the concrete strut capacity vRd when cot θθθθ = 2.5 vRdcot 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
Is vRd,cot θθθθ = 1.0 > vEd?
Yes (cot θθθθ > 1.0)
sl,max = st,max = 0.75 d
Design aids for shear Concise 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
− Only that shear reinforcement provided within the central 0.75av is
included in the resistance.
Note: see PD6687-1:2010 Cl 2.14 for more information
EC2: 6.2.3
Beam examples
Beam Example 1
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
Beam Example 1 – Bending
ULS load per m = (75 x 1.35 + 50 x 1.5) = 176.25
Mult = 176.25 x 82/8
= 934
[[[[ ]]]] [[[[ ]]]] dK d
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 > vEd
Asw/s = 1.68 x 450 /(435 x 2.5)
Asw/s = 0.70 mm
Asw = 314 mm2
⇒ provide H10 links at 450 mm spacing
fck vRd, cot θθθθ =
Beam Example 1
with
Max spacing is 0.75d = 934 x 0.75
= 700 mm
shear reinforcement
Find the minimum area of shear reinforcement required to resist
the design shear force using EC2.
fck = 30 MPa and
= 4.96 MPa
vRdcot θ = 2.5 < vEd < vRdcot θ = 1.0
∴ 2.5 > cot θ > 1.0 ⇒ Calculate θ
fck vRd, cot θθθθ =
Calculate θ
Asw/s = 4.96 x 140 /(435 x 1.43)
Asw/s = 1.12 mm
Asw = 157 mm2
⇒ provide H10 links at 125 mm spacing
Beam Example 2 – High shear
EC2 Webinar – Autumn 2016 Lecture 3/27
Exercise
Cover = 35 mm to each face
fck = 30MPa
Gk = 10 kN/m, Qk = 6.5 kN/m (Use EC0 eq. 6.10)
8 m
Exp (6.10)
Or Concise Table 15.5
Workings:- Load, Mult, d, K, K’, (z/d,) z, As, VEd, Asw/s
ΦΦΦΦ mm
Area, mm2
8 50
10 78.5
12 113
16 201
20 314
25 491
32 804
Working space
Working space
Working space
Lecture 3
– Tension reinforcement, As
– Compression reinforcement, As2
• 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
BS8110
• Design manuals will provide the standard solutions
for basic design cases.
( 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 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, 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
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
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
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
ε ud
Es may be taken to be 200 GPa
Steel yield strain = fyd/Es
= 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 EC2: 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
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, Fst:
M = Fc z = 0.453 fck b x z (1)
Now z = d - 0.4 x
∴ x = 2.5(d - z)
= 1.1333 (fck b z d - fck b z 2)
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
Analysis of a singly reinforced beam Determine 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
The lever arm for an applied moment is now known
M
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, As Concise: 6.2.1
Design aids for flexure - method (b) Concise: Table 15.5K = M / (fck b d 2)
‘Normal’ tables and
avoid issues with the
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 / (
Maximum neutral axis depth
According to Cl 5.5(4) the depth of the neutral axis is limited, viz:
δ ≥ k1 + k2 xu/d
xu = depth to NA after redistribution
= Redistribution ratio
this limit is denoted K’
For K > K’ Compression steel needed
Moment Bending Elastic
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).
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, As2 Concise: 6.2.1EC2: Fig 3.5
EC2 Webinar – Autumn 2016 Lecture 3/10
Tension steel, 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
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: δδδδ =1.0 means no redistribution and δδδδ = 0.8 means 20% moment redistribution.
Compression steel needed - doubly reinforced
Is K ≤ K’ ?
Yes No
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
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.
[[[[ ]]]] *95.053.311 2
dK d
Check min spacing between bars > Øbar > 20 > Agg + 5
Check max spacing between bars
Calculate tension steel required from: zf
M A
(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
db f
Flow Chart for Doubly- Reinforced Beam K > K’
Calculate lever arm z from: [[[[ ]]]]'53.311 2
K d
−−−−====
(((( ))))2yd 2s
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
2s yd
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
∴ provide compression steel
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 > bar > 20 mm > Agg + 5 mm
Worked Example 1
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
Shear
There are three approaches to designing for shear:
• When shear reinforcement is not required e.g. slabs, week 5 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, 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
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.
compression strut
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
θθθθcotsw sRd, ywdfz
Strut Inclination Method EC2: Equ. 6.8 & 6.9 for Vertical links
Equ 6.9
Equ 6.8
Shear
Safer
Shear Design: Links
hence activating more links.
As strut angle reduces concrete stress increases
Angle = 45°°°° V carried on 3 links
Max angle - max shear resistance Angle = 21.8°°°° V carried on 6 links
Min strut angle - Minimum links
dz
x
d
x
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
Shear Resistance of Sections with Vertical Shear Reinforcement Concise: 7.3.3
where:
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 becomes VRd,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)
fck vRd, cot θθθθ
EC2 – Shear Flow Chart for vertical links
Yes (cot θθθθ = 2.5)
Determine the concrete strut capacity vRd when cot θθθθ = 2.5 vRdcot 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
Is vRd,cot θθθθ = 1.0 > vEd?
Yes (cot θθθθ > 1.0)
sl,max = st,max = 0.75 d
Design aids for shear Concise 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
− Only that shear reinforcement provided within the central 0.75av is
included in the resistance.
Note: see PD6687-1:2010 Cl 2.14 for more information
EC2: 6.2.3
Beam examples
Beam Example 1
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
Beam Example 1 – Bending
ULS load per m = (75 x 1.35 + 50 x 1.5) = 176.25
Mult = 176.25 x 82/8
= 934
[[[[ ]]]] [[[[ ]]]] dK d
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 > vEd
Asw/s = 1.68 x 450 /(435 x 2.5)
Asw/s = 0.70 mm
Asw = 314 mm2
⇒ provide H10 links at 450 mm spacing
fck vRd, cot θθθθ =
Beam Example 1
with
Max spacing is 0.75d = 934 x 0.75
= 700 mm
shear reinforcement
Find the minimum area of shear reinforcement required to resist
the design shear force using EC2.
fck = 30 MPa and
= 4.96 MPa
vRdcot θ = 2.5 < vEd < vRdcot θ = 1.0
∴ 2.5 > cot θ > 1.0 ⇒ Calculate θ
fck vRd, cot θθθθ =
Calculate θ
Asw/s = 4.96 x 140 /(435 x 1.43)
Asw/s = 1.12 mm
Asw = 157 mm2
⇒ provide H10 links at 125 mm spacing
Beam Example 2 – High shear
EC2 Webinar – Autumn 2016 Lecture 3/27
Exercise
Cover = 35 mm to each face
fck = 30MPa
Gk = 10 kN/m, Qk = 6.5 kN/m (Use EC0 eq. 6.10)
8 m
Exp (6.10)
Or Concise Table 15.5
Workings:- Load, Mult, d, K, K’, (z/d,) z, As, VEd, Asw/s
ΦΦΦΦ mm
Area, mm2
8 50
10 78.5
12 113
16 201
20 314
25 491
32 804
Working space
Working space
Working space
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