bending and shear in beams - concrete centre · pdf fileec2 webinar – autumn 2016...

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


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


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?


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


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


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


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’.


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


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


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


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≥≥


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)


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


∴ 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


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


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 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.


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)


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


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


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


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


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


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


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


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 < 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


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



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


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


Working space

Working space


Working space

End of Lecture 3

bending and shear in beams - concrete centre · pdf fileec2 webinar – autumn 2016...

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