1Shear and Bond

4Shear capacity of section

aggregate interlock

dowel action

Vs

Vcz

Vd

Va

shear resistance in uncracked concrete

four sources of shear resistance

Take a look at this region when shear failure occurs

Principal tensile stress causes an

inclined crack

2Shear and Bond

4Vcz = Shear resistance from the uncracked concrete in the compression zone

Va = vertical component of aggregate interlock action at the crackVd = shear resistance from dowel bar action of the longitudinal

reinforcement (This means ONLY the bottom tensile reinforcement will contribute towards this resistance. Compression reinforcement, if any, will not count as the dowel bar action will increase the danger of local buckling of the compression steel bar. Also any tensile reinforcement at the top of the section, i.e. tensile steel at the interior support of a continuous beam, will have no contribution towards this resistance.)

Vc = shear capacity of concrete

Vc = Vcz + Vd + Va

20~40% 15~25% 35~50% contribution

Will not contribute

aggregate interlock

dowel action

Vs

Vcz

Vd

Va

shear resistance in uncracked concrete

3Shear and Bond

4Shear resistance in beam comes from the shear resistance in concrete (from web ONLY) and resistance from steel reinforcement (shear links and inclined bars) is

V = (Vcz + Vd + Va) + Vs

= Vc + Vs

V = design ultimate shear resistance

If expressed as average stress over the shear area, we have

Vc = vc bd

where vc is designated as the nominal shear stress from Table 6.3of the design code.

We called this theory the Average Shear Theory

Shear resistance of section

from concrete

from link

4Shear and Bond

4

Consider the general case with an inclined web reinforcement at angle to the horizontal. The compression members of the truss are at an angle to the horizontal.

Truss Analogy to explain the behaviourForce

This is an overlapping double truss

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A Single Truss with angle = 90 showing thediagonal compression concrete member

Diagonal compression in concrete members

Compression components

6Shear and Bond

4Observation from laboratory test on 10/2/09

A very clear truss action overlapping with the beam action in resisting the applied load is observed.

front view

back view

No crack observed close to support due to Enhanced Shear effect.Enhanced Shear Theory

7Shear and Bond

4Suspected case of Shear Crack in Real Structure

The portal frame supporting the Ngau Tau Kwok MTR station.

8Shear and Bond

4For the cut section A-A in the truss, the number of web reinforcement intercepted by the section A-A is

Shear resistance provided by the web reinforcement

where Asv is the total area of a web reinforcement.

fyv is the yield strength of shear steel.=(250 N/mm2 for mild steel and 500 N/mm2 for H.T. steel)

vsddddn cot)'(cot)'(

( ) sin

( ' ) cot ( ' )cot sin

'cos sin cot

s yv

sv yvv

sv yvv

V Area of steel f

d d d dA fs

d dA fs

9Shear and Bond

4Vertical linksFor = 90

Angle can be taken as 45 for most of the shear cracks, therefore

Since Vs = V Vc and V = vbd in term of average stressVc = vcbd

cotcot'v

yvsvv

yvsvs sdfA

sddfAV

vyvsvs s

dfAV

v

yvsvc

vyvsvc

sfAbvv

sdfAdbvv

)(

)(

10

Shear and Bond

4Including the m=1.15 for the shear steel, (Table 2.3, Concrete 2013)

Concrete2013 limits the ultimate shear stress (v) to 0.8 fcu or 7 N/mm2

and the need to provide minimum links to give an equivalent minimum steel resisting stress of (v vc) = 0.4 N/mm2 of concrete.

2( 0.4 / ) 0.87v yv svv

dN mm b d f As

0.40.87

v vsv

yv

b sAf

( )0.87

c v vsv

yv

v v b sA

f

for fcu 40 MPa or 0.4(fcu/40)2/3 for 40 < fcu < 80 MPa)

Minimum shear link

area

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Shear and Bond

4Design Procedure

Find v c from Table 6.3

Design shear stressv = V / (bv d)

Provide links

Revise Section

NO

Yes

v

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13

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For d > 400, use values in column for 400 mm

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Shear and Bond

4Enhanced Shear close to support - (Enhanced Shear Theory) Section close to support has an enhanced shear resistance

owing to the induced compressive stress from the reaction and the steeper angle of failure plane (usually 30 to the horizontal).

Within a distance of 2d from support or a concentrated load, the design concrete shear stress vc may be increased to (2d/av)vcwhere av is measured from support or load to the section being designed.

This behaviour can be observed in the laboratory test. This enhancement is useful when designing flexural members

with concentrated loads near a support.

This method will not be taught as the shear link arrangement will be different. However, please refer to Chapter 5 of the textbookfor self-learning.

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Shear and Bond

4Bent-up bars

For =45

For =45 bent-up bars

usually sb = (d - d)

Similarly you can also get the formula on the capacity of a bent-up bar system with =60

b

yvsbb sddfAV 'cotsincos)87.0(

b

yvsbb sddfAV 'sincos)87.0(

byvsbb s

ddfAV ')414.1()87.0(

Similar to the shear capacity of an inclined steel link, the shear capacity of a bent-up bar is

compression strut

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Shear and Bond

4Maximum spacing of shear links and bent-up barsVertical shear links

Bent-up barsMaximum spacing sb 1.5dResistance provided by stirrups > resistance provided by

bent-up bars (Table 6.2)

Bent-up bars

Svor Sb

span

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Shear and Bond

4Arrangement of vertical shear links

h

usually used when the fabrication of closed links is difficult. In most case it would be found in slabs.

Open link

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single link

double links

Asv for one set of links

Closed link

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Shear and Bond

4Example on shear resistance at a section - Ex. 5.1

fyv=250 N/mm2 for stirrups

fyv=500 N/mm2 for bent-up bars

fcu=40 N/mm2

The bent-up of one bar at a time is wrong!

Should bend up two bars in a system instead of a single bar

As=1964mm2

(==45)

4T25

1T252T25

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Shear and Bond

4At the face of support, it has been checked that v=V/bwd< 7 N/mm2 or 0.8fcu.

Since d=650, vc=0.50 N/mm2 by interpolation. and As=113 mm2 for a 12mm diameter bar or =226 mm2 for a shear link.

Take spacing sv=100 mm,

Shear resistance of cross-section

100 100 982 0.43350 650

sAbd

3

0.87 2.26 0.87 250 650 350 0.50 650

(319.5 114) 10 433.5

ss yv v c

v

AV f d b v ds

N kN

From concrete

From shear links

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Shear and Bond

4For the single bent-up bar system (with two sets of bent up bars and

sb = (d d) ), the shear resistance is

Total shear resistance of the concrete, stirrups and bent-up bars is

Finally, the resistance provided by the stirrups is checked to be larger than that from the bent-up bars. Satisfactory (Table 6.2)

1.414 0.87

1.23 500 491301.97

b yv sbV f A

kN

(433.5 301.97)735.44

s bV V V kNkN

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Shear and Bond

4Shear Steel arrangement in a continuous beamdesign shear force

Figure 7.18: Typical arrangement of shear reinforcement

S.F Envelope

H8@250 H10@150 H8@250H10@ 200

H8@ 200

H10@ 150

Zone with shear resistance from conc

and min shear link

Zone with shear resistance from conc

and min shear link

shear resistance from concsection and minimum shear

link = (vc + 0.4)bvdStep 1:-

Step 3:-

Step 2:-

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Shear and Bond

4Effect of Steel curtailment on vc from concrete

We need As to calculate vc from Table 6.3.

This As is the area of steel bar at the bottom of section. 0% should be used for the section at the root of cantilever beam.

As a conclusion, the real % of reinforcement as shown in the figures on the left should be used in Table 6.3.

Cl. 9.2.1 of Concrete2013 is referred.

0.08L 0.08LL

0.1L 0.15LL

c=0.15Lc=0.25L

c

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Shear and Bond

4

25

Shear and Bond

4Anchorage of ReinforcementAll reinforcing bars shall be anchored to have safe

transfer of forces to concrete avoiding longitudinal cracking or spalling.

The forces in each bar should be developed by an appropriate embedment length or other end anchorage, and local bond stress may be ignored if properly done.

Force

Anchorage Bond stress fb

embedment length or anchorage length

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Shear and Bond

4Anchorage bond stress fb, is assumed to be constant over the effective anchorage length.

The design anchorage bond stress fb is also assumed to be constant over the anchorage length and is given by

fb = Fs/( lb)where Fs is the force in bar with a maximum of 0.87fy; is the effective bar size;

lb is the anchorage bond length;

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Shear and Bond

4Design ultimate anchorage bond stressThe design ultimate anchorage bond stress fbu (a property of steel bar surface) may be obtained from

fbu = fcuwhere is the coefficient dependent on the bar type and given in Table 8.3.

Cl 8.4.4 is referred.

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Shear and Bond

4How to ensure sufficient anchorage of reinforcement ?Proper anchorage of reinforcement is ensured by

providing a minimum ultimate anchorage bond lengths or anchorage lengths as

Values of these minimum bond lengths for fs=0.87fy(at steel yield strength) are given in Table 8.4 as multiples of bar diameter

4s

bbu

flf

This is obtained from

fbu( lb) = Fslb fbu = fs2/4

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Shear and Bond

4Hooks and bends

Bend or hook do not contribute to compression anchorages

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Shear and Bond

4Laps in reinforcement

Force Force

lap length

Examples of failure due to poor lapping

The force in steel bar can be transferred to another one throughlapping of the bars in concrete over a lap length

Laps should preferably be staggered and be away from sections with high stresses

Minimum lap length >15 or 300mm for bars (not structural) Tension lap length design ultimate anchorage length in Table 8.4 Compression lap length 1.25 of the compression anchorage length

or compression embedment length in Table 8.4

Basic lap lengths are shown in Table 8.5

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Shear and Bond

4Example on anchorage bond lengthA simply supported reinforced concrete beam is reinforced with 2T25 bars.

The steel is of deformed Type 2 bars, fcu=35 N/mm2; fy=500 N/mm2. Calculate the ultimate anchorage bond length required at midspan.

Maximum force in each bar = 0.87fy area of barUltimate anchorage bond force = L fbu

fbu = fcu = 0.525 = 2.5 N/mm2

Equating forces, L fbu = 0.87fy area of barThe required ultimate anchorage bond length is

L = 0.87 500 25/(42.5) = 40 25 = 1087.5 mm

Table 8.3 gives =0.5

~~~ End ~~~

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Shear and Bond

4End bearing action of compression bar

Force

Bond stress fb

embedment length

bearing force at end of bar

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35

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(cl. 8.7.3.2)

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