bond, anchor and shear

7/28/2019 bond, anchor and shear

1/15

3. Bond , Anchorage and Shear

This chapter will discuss the following topics:

Outline the theory of calculating the anchorage bond

length.,

length and compression lap length by using Table 8.4

and Table 8.5 of the code.

Outline the design formulae for the design of shear

reinforcement and the use of Table 6.3 of the Code.

RC Design and Construction HKC 2004 (2ndEdition)3-1

3. Bond, Anchorage and Shear

Minimum Distance Between Bars

The min. distance between bars are given in terms of

agg

.

( hagg + 5mm) or

.



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3.1 Minimum and Maximum Percentages of

Reinforcement in Beams, Slabs and Columns

Minimum % : Table 9.1 of the code

Maximum %

(a) Beam (9.2.1.3)Neither the area of tension reinforcement nor the area of com ression

reinforcement should exceed 4% of the gross cross-sectional area of concrete.

(b) Column (9.5.1)

The longitudinal reinforcement should not exceed the followingamounts, calculated as percentages of the gross cross-sectional area of theconcrete :

~ vertically cast columns : 6%

~ horizontally cast columns : 8%

~ laps in the vertically or horizontally cast column : 10%

(c) Wall (9.6.2)

The area of vertical reinforcement should not exceed 4% of the concrete


cross-sectional area of the wall.

3.2 Anchorage Bond

The steel bar subjected to tension shown in Fig.

anchorage length (embedded length) depends on

the bond between the bar and the concrete and the

contact area.

L

F

Fi . 3.1 Anchora e Bond



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3.2 Anchorage Bond

L = Min. anchorage length to prevent pull out.

= .

fbu= Ultimate anchorage bond stress.fs = Direct tensile or compressive stress in the bar.

Consider the forces on the bar,

Tensile pull-out force = Area of bar * Stress of bar

=

4

fsRC Design and Construction HKC 2004 (2ndEdition)

3-5

3.2 Anchorage Bond

Anchorage force = Contact Area * Bond Stress

bu

= (L)fbu 2

4

0.87f=

bubu ffL

4=

4

The ultimate anchorage bond stress, .

Values of are iven in table 8.3.

f fbu cu

anchorage length L can be written as


= A see a e . o e o e


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3.2 Anchorage Bond

Table 8.3 - Values of the bond coefficient ar ype

Bars in tension Bars in compression

Plain bars 0.28 0.35

Type 2 : deformed bars 0.50 0.63

Fabric 0.65 0.81

Normally deformed type 2 bars are used in

construction works.

Furthermore anchorage may be provided by hooks

. . . .


Fig. 8.2 Anchorage of Links

Reproduce from HK Code



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3.3 Lapping of Reinforcement

Lapping of reinf. may be required if the reinf. is of

insufficient len th or there is curtailment cuttin

off of reinf. when it is no longer required in order

to save mone of reinf.

u es or app ng are:-

The laps should be staggered and be away from sections

. . . .

with large bending moment as it induces large tensile stress

in reinforcement) Min. lap length should be not less than the greater of 15

or 300mm.



The tension lap length should be at least equal to the design tension

anchorage length. Lap lengths for unequal size bars may be based

upon the diameter of the smaller bar. The following provisions also

apply:

minimum cover is less than twice the diameter of the lapped

reinforcement, the lap length should be increased by a factor

of 1.4

where a lap occurs at a corner of a section and the minimum

lapped reinforcement, the lap length should be increased by a

factor of 1.4

where both conditions above apply, the lap length should beincreased by a factor of 2.0


Figure 8.5 and table 8.5 are referred.


6/15


Rules for lapping are:-

Com ression la s should be at least 25% reater than the

compression anchorage length.

Lap lengths for unequal size bars may be based on thesmaller bar.

x 2.0x 1.4

If Cover < 2, multiple lap length by

1.4 or2.0

x .x 1.0

(Lap length = anchorage length x 1.0)

Fi . 3.3 La in of tension reinforcement


3.4 Shear

Shear reinforcement can be in the form of shear

-

inclined bars are less frequently used in

construction toda .

ear n s rrups

Links can be broadly classified into shear link and

nominal link (minimum link) and the design

formulae for these types of links are as follows:



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Design Formulae for Shear Reinforcementor cu a-

( )csv vvbA yvv fs 87.0

Minimum (nominal) Links:-

yvv

sv

f

b

s

A

87.0

4.0


Design Formulae for Shear Reinforcement

Where:-

sv -

link.

= rea o eamfyv = Characteristic strength of the shear reinf. = fy

sv = ong u na spac ng o s ear n s

v = Design shear stress

= es gn s ear orce =

vc = The ultimate shear stress that can be resisted by the

concrete.



8/15

Design Formulae for Shear Reinforcement

The values of vc is given in table 6.3 of the Code.

It shall be noted that the values given in table 6.3 is forgrade 25 conc. For other grades of conc., adjustment shall

be made by multiplying fcu

1

3 The values of v increases for shallow members and those

25

with larger percentages of tensile reinforcement.


3.5 Enhanced Shear Resistance

Within a distance of 2d from a support or a

vc may be increased to

c v

The distance av is measured from the support or

concentrated load to the section being designed.

Avera e shear stress should never exceed the

lesser of or 7 N/mm2.0 8. fcu



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Figure 6.3 Shear failure near supports


epro uce rom o e

Table 6.3 - Values of vc Design Concrete Shear Stress


epro uce rom o e


10/15

Table 9.1 - Minimum Percentage of Reinforcement


epro uce rom o e

Table 8.4 Ultimate Anchorage Bond Lengths (lb)




11/15

Table 8.5 Ultimate Lap Lengths



Figure 8.5 - Factors for lapping bars




12/15

Self-Assessment Questions

Q1. Determine the anchorage bond length from the first

principle for a T32 bar in tension (deformed type 2)

and fcu = 35 N/mm2. the bars as listed below:

Choices:

(a) 1000 mm

(b) 1082 mm

(c) 1360 mm



Q2. Determine the anchorage bond length by using table 8.4

of HKC 2004 for a R16 bar in compression and fcu = 30

N/mm2 .

Choices:(a) 464 mm

(b) 528 mm

(c) 540 mm



13/15


Q3. Determine the ultimate shear resistance of concrete vcfor the singly reinforced concrete section shown in Fig.

Q3, without shear enhancement. Given that fcu = 35

N/mm2 and fy = 460 N/mm2.

050

403

2T32

Choices:

(a) 0.70 N/mm2

Fig. Q3

(b) 0.90 N/mm2

(c) 1.10 N/mm2


Assignment No. 3

AQ1 Fig. AQ1 shows the lapping of steel reinforcement of a beam.

Determine the lap length for the bar groups 1, 2, 3 and 4. Given

and fcu = 35 N/mm2. The spacing between group 1 and 2 bars is

greater than s and the spacing between group 3 & 4 bars is less

t ans.

(a) The lapping of steel reinforcement is at tension zone.

(b) The lapping of steel reinforcement is at compression zone

3T16

(1) (2)

3T323T32

(3) (4)


Fi . A 1


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Assignment No. 3

AQ2 Determine the nominal cover and breadth of the web for the

simply supported beams whose sections are shown in Fig.

AQ2a, AQ2b & AQ2c. The maximum size of the aggregate

in each case is 20 mm, f cu = 40 N/mm2

and the link size is3. -

cement ratio = 0.55).

a xposure con on: severe

Fire resistance: 2 hours

(b) Exposure condition: moderate

Fire resistance: 1 hours

(c) Exposure condition: severe


re res s ance: ours

Assignment No. 3

Fi . A 1a

3T40

Fi . A 1b

4T20

. .

2T32


Fig. AQ1cFig. AQ2c


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Assignment No. 3

AQ3. Determine the average shear stress for the section

shown in Fig. AQ3a, AQ3b & AQ3c, and state whether

shear reinforcement is required. Given that fcu = 30

N/mm2 and fy = 460 N/mm2.

225 300 200

350

300

2T20450

400

260

300

2T16

Ultimate Shear = 150 kN

Fi . A 2a

3T25 Ultimate Shear = 250 kN

Fig. AQ2c

Fig. AQ3c Ultimate Shear = 60 kNFig. A 2b

.

Fig. AQ3b


Assignment No. 3

AQ4 The sections in Fig. AQ4a & AQ4b are subject to shear

forces at the ultimate limit state as shown. Determine

the spacing of R10 links. Given that fcu = 35 N/mm2, fy

= 460 N/mm2 and fyv = 250 N/mm2.

225 300

350

310

2T204

50

400


Fi . A 3a

3T25


Fi . A 3b

g. a

Fig. AQ4b


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bond, anchor and shear