Chapter 7 بسم الله الرحمن الرحيم Design of Concrete Structure I University of Palestine Instructor: Eng. Mazen Alshorafa

Chapter 7

الرحمن الله بسمالرحيم

Design of Concrete Structure I

University of Palestine

Instructor:

Eng. Mazen Alshorafa

Page 1



Bond, Development Lengths, and Splices

Instructor:


Concept of Bond Stress

Concrete

Reinforcing bar

C

T

T=Asfy

Bond stress

T2=fs2Ab

dbµ=Bond stress

T1=fs1Ab

l

M

fs2=fs1+∆fs

Bond stresses are existent whenever the stress or force in a reinforcing bar changes from point to point along the length of the bar in order to maintain equilibrium.

Page 2




Instructor:


Concept of Bond Stress

C

T

T

Bond stress

M

L

forceBondTF 0.0

ldffd

bavgssb )(

4 12

2

avg

bss dffl

412

)(,' bdfkfck

T2=fs2Ab

dbµ=Bond stress

T1=fs1Ab

l

fs2=fs1+∆fs

Equilibrium Condition for Rebar

µavg= bond stress (coefficient of friction)

Page 3




Instructor:


Mechanism of Bond Transfer

(a )Forces on bar

(b )Forces on concrete

Radial longitudinal

(c )Components of forces on concrete

(d )Splitting stresses

A smooth bar embedded in concrete develops bond by Adhesion

between concrete & reinforcement, and a small amount Friction.

Note: These bonds are quickly lost when the bar is loaded in tension

On the other hand, a deformed bar generates bond by friction and by

bearing on the deformations of the bar against the concrete

Page 4




Instructor:



Splitting cracks result in loss of bond transfer.

Reinforcement can be used to restrain these cracks.

The load at which splitting failure develops is a function of :

1. The minimum distance from the bar to the surface of the concrete

or to the next bar. The smaller the distance, the smaller is the

splitting load.

2. The tensile strength of the concrete.

3. The average bond stress. The higher the average bond stress, the

higher is the splitting resistance.

Page 5




Instructor:



If the concrete cover and bar spacing are large compared to the bar

diameter, a pullout failure can occur.

Page 6




Instructor:


Development Length

The development length ld is shortest length of bar in which the bar

stress can increase from zero to the yield strength, fy

The development Lengths are different in tension and compression,

because a bar loaded in tension is subject to in-and-out bond stresses

and hence requires a considerably longer development length

failurebondatvaluetheiswheredf

l avguavguavg

byd

,,

,4

Page 7




Instructor:


Development Length of Deformed Bars in Tension

5.2d

kcin which 30cm,d

d

kc

λγβα

fc'

f

7

2l

b

trb

b

tr

yd

'fc

According to ACI Code, the development length for deformed bars in tension is given by

to safeguard against pullout type failure

where,

ld = development length, cm

db = nominal diameter of bar, cm

fy = specified yield strength of reinforcement, kg/cm2

= square root of specified compressive strength of concrete, kg/cm2

C = spacing or cover dimension, cm

Ktr = transverse reinforcement index

Page 8




Instructor:


Development Length of Deformed Bars in Tension [contd.]

C is the smaller of

(a) the smallest distance measured from the center of the bar to the nearest concrete surface

(b) one-half the center-to-center spacing of bars being developed.

α is a bar location factor (a) Horizontal reinforcement so placed that more than 30 cm of fresh concrete is cast in the member below the development length or

splice……………………………………………………………………. 1.3

(b) Other

reinforcement………………………………………………………….. 1.0

β is a coating factor that reflects the adverse effects of epoxy coating

(a) Epoxy-coated bars or wires with cover less than 3db

or clear spacing less than 6db …………………………………………..

1.5 (b) All other epoxy-coated bars or wires………………………………...

1.2 (c) Uncoated

reinforcement……………………………………………………… 1.0

Page 9




Instructor:



However, the product αβ is not to be greater than 1.7

γ is a reinforcement size factor that reflects better performance of the smaller diameter reinforcement (a) Φ20mm and smaller bars.……………………………………………….. 0.8 (b) Φ22mm and larger bars.…………..…….………………………………. 1.0

λ is a lightweight aggregate concrete factor that reflects lower tensile

strength of lightweight concrete, & resulting reduction in splitting

resistance. (a) When lightweight aggregate concrete is used…….……..…….

0.8 (b) When normal weight concrete is used…………………..………….

1.0

Page 10




Instructor:



ns

fAk yttr

tr 100

Atr

Potential plane of splitting

Ktr is a transverse reinforcement index that represents the contribution of confining reinforcement

WhereAtr = total cross sectional area of all transverse reinforcement within

the spacing s, which crosses the potential plane of splitting along the reinforcement being developed with in the development length cm2

fyt = specified yield strength of transverse reinforcement, kg/cm2

s = maximum center-to-center spacing of transverse reinforcement within development length ld , cm.

n = number of bars being developed along the plane of splitting.

Note: It is permitted to use Ktr= 0.0 as

design simplification even if transverse reinforcement is present.

Page 11




Instructor:



Excessive ReinforcementAccording to ACI Code, reduction in development length is allowed

Where As provided > As required. the reduction is given by

-Except as required for seismic design

-Good practice to ignore this factor, since use of structure may

change over time.

providedA

requiredA factorReduction

s

s

Page 12




Instructor:


Example # 1

Determine the development length required for the uncoated bottom bars as shown in figure.Use fc’ = 250 kg/cm2 normal weight concrete and

fy = 420 kg/cm2

Solution:

α=1.0 for bottom bars, β=1.0 for uncoated bars

α β =1.0 <1.7 OK

γ=0.8 for Φ20mm,

λ=1.0 for normal weight concrete

C the smallest of 4.0+1.0+1.0=6 cm

[40-2(4.0)-2(1.0)-2.0]/(3)(2)=4.67 cm

i.e., C is taken as 4.67 cm

4Φ20

40 cm

60

cm

Φ10@20

Page 13




Instructor:


Example # 1 [contd.]

cml

OKd

KC

cml

d

KCuseei

d

KC

cmns

fAK

d

b

tr

d

b

tr

b

tr

yttrtr

1.520.233.2

)0.1)(8.0)(0.1)(0.1(

250

4200

7

2

5.233.220

07.46

0.0Kassuming

6.480.25.2

)0.1)(8.0)(0.1)(0.1(

250

4200

7

2

5.2.,.

5.275.20.2

83.067.4

83.0)4)(20(100

4200)79.0(2

100

tr

4Φ20

40 cm

60

cm

Φ10@20

Page 14




Instructor:


Development Length of Deformed Bars in Compression

Shorter development lengths are required for compression than for tension since flexural tension cracks are not present for bars in compression.

According to ACI Code, the development length ld , for deformed bars

in compression is computed as the product of the basic development

length ldc and applicable modification factors, but ld is not to be less

than 20 cm.

ld = ldc x applicable modification factors ≥ 20 cm.

The basic development length ldb for deformed bars in compression is

given as

by

c

bydc df

f

dfl 044.0

'

073.0

Page 15




Instructor:


Development Length of Deformed Bars in Compression [contd.]Applicable Modification Factors

1. Excessive reinforcement factor =As required / As provided

2. Spirals or Ties: the modification factor for reinforcement, enclosed

with spiral reinforcement ≥ 6mm in diameter and ≤ 10 cm pitch or

within Φ12mm ties spaced at ≤ 10 cm on center is given as 0.75

Development Lengths for Bundled Bars

Based on ACI Code, development length of individual bars within a

bundle, in tension or compression, is taken as that for individual bar,

increased 20% for three-bar bundle, and 33% for four-bar bundle.

For determining the appropriate modification factors, a unit of bundled bars is treated as a single bar of a diameter derived from the

equivalent total area of bars.

Page 16




Instructor:


Development of Standard Hooks in Tension

b

c

yhb d

f

fl

'

073.0

Hooks are used to provide additional anchorage

when there is insufficient length available

to develop a bar.

According to ACI Code development length ldh , for deformed bars in

tension terminating in a standard hook is computed as the product of

the basic development length lhb and applicable modification factors,

but ldh is not to be less than 8db, nor less than 15 cm.

ldh = lhb x applicable modification factors ≥ 15 cm or 8db.

The basic development length lhb for hooked bars is given as

Page 17




Instructor:


Development of Standard Hooks in Tension [contd.]

Applicable Modification Factors

1. Concrete cover: for db ≤ Φ36mm, side cover (normal to plane of

hook) ≥ 6.35 cm, and for 90 degree hook, cover on bar extension

beyond hook ≥ 5.0 cm, the modification factor is taken as 0.7.

2. Excessive reinforcement factor =As required / As provided

3. Spirals or Ties: for db ≤ Φ36mm, hooks enclosed vertically or

horizontally within ties or stirrups spaced along the full development

length ldh not greater than 3 db , where db is diameter of hooked bar, is

taken as 0.8.

4. Lightweight aggregate concrete: the modification factor is 1.3.

5. Epoxy-coated reinforcement: the modification factor for hooked

bars with epoxy coating is taken as 1.2.

Page 18




Instructor:


Development of Standard Hooks in Tension [contd.]

Part (a)

Part (b)

Φ10 through Φ25

Φ28 through Φ36

Φ44 through Φ56

4db or 64mm

ldh

ldh

Development length ldh is measured

from the critical section of the bar

to the out-side end or edge of the

hooks. Either a 90 or a 180-degree

hook, shown in Figure, may be used

Development of reinforcement- General* ACI code notes that hooks are not considered effective in compression and may not be used as anchorage.* The values of used in this Lecture Shall not exceed 26.5 kg/cm2.

'fc

Page 19




Instructor:


Splices of Reinforcement

Forces on bar at splice

Splicing of reinforcement bars is necessary, either because the available bars are not long enough, or to ease construction, in order to guarantee continuity of the reinforcement according to design requirements.

Types of Splices

(a) Welding (b) Mechanical connectors

(c) Lap splices (simplest and most economical method)

In a lapped splice, the force in one bar is transferred to the concrete,

which transfers it to the adjacent bar.

Splice length is the distance the two bars are overlapped.

Page 20




Instructor:


Splices of Reinforcement

Important note: Lap splices have a number of disadvantages, including congestion

of reinforcement at the lap splice and development of transverse cracks due to stress concentrations. It is recommended to locate splices at sections where stresses are low.

Types of lap Splices

1. Direct Contact Splice as figure a

2. Non-Contact Splice (spaced) the distance between two bars cannot be greater than 1/5 of the splice length nor 15 cm

Bars are spaced

ls

ls

s

Direct contact

T

T

T

T

Page 21




Instructor:


Splices of Deformed Bars in Tension

ACI code divides tension lap splices into two classes, A and B. the class of splice used is dependent on the level of stress in the reinforcing and on the percentage of steel that is spliced at particular location.

The splice lengths for each class of splice are as follow

Class A splice: 1.0 ld

Class B splice: 1.3 ld

Tension Lap Splices

Maximum Percent of As

spliced within required lap length

As providedAs required

50 100

Equal to or greater than 2

Less than 2

Class A Class B

Class B Class B

Page 22




Instructor:


Splices of Deformed Bars in Compression

Bond behavior of compression bars is not complicated by the problem of transverse tension cracking and thus compression splices do not require provisions as strict as those specified for tension

Based on the ACI code the Compression lap splice length shall be

0.007 fy db ≥ 30.0 cm for fy ≤ 4200 kg/cm2

(0.013 fy-24) db ≥ 30.0 cm for fy > 4200 kg/cm2

The computed splice length should be increase by 33% if fc’<210kg/cm2

According to ACI code when bars of different size are lap spliced in

compression, splice length shall be the larger of either development

length of larger bar, or splice length of smaller bar.

Page 23




Instructor:


Example # 2

Determine the development or embedment length required for the epoxy-coated top bars of the beam as shown in figure.(a)If the bars are straight(b)If a 180 hook is used(c)If a 90 hook is used

Use fc’ = 280 kg/cm2 and fy = 4200 kg/cm2

Solution: (a) Straight Bars

α=1.3 for top bars, β=1.5 for coated bars

α β =1.3x1.5 = 1.95 > 1.7 use 1.7

γ=1.0 for Φ32mm, λ=1.0 for normal weight concrete

C the smallest of 4.0+1.2+1.6=6.8 cm

[40-2(4.0)-2(1.2)-3.2]/(3)(2)=4.4 cmi.e., C is taken as 4.4 cm

4Φ32

40 cm

50

cm

Φ12@15

4Φ32

Page 24




Instructor:



(b) Using 180 hook

ldh = lhb x applicable modification factors ≥ 150 mm or 8db.

applicable modification factors =1.2 for epoxy-coated hooks

cml

OKd

KC

cmns

fAK

d

b

tr

yttrtr

1672.387.1

)0.1)(8.0)(7.1(

280

4200

7

2

5.287.12.3

58.14.4

58.1)4)(15(100

4200)13.1(2

100

cmdf

fl b

c

yhb 3.582.3

280

4200073.0

'

073.0

20815702.13.58 bdh dandcml

4Φ32

40 cm

50

cm

Φ12@15

Page 25




Instructor:



(c) Using 90 hook

ldh=70 cm

5db =16 4db =12.8

Critical section

180o hook

Ldh=70 cm

Critical section

12

db=

38

.4

90o hook

Page 26




Instructor:


Example # 3

To facilitate construction of a cantilever retaining wall, the vertical reinforcement shown in Figure, is to be spliced to dowels extending from the foundation. Determine the required splice length when all reinforcement bars are spliced at the same location.


Solution: Class B splice is required where ls = 1.3 ldα=1.0, β=1.0 → α β =1.0 < 1.7 OK

γ=0.8, λ=1.0 for normal weight concrete

C the smallest of 7.5+0.8=8.3 cm

25/2=12.5 cm

i.e., C is taken as 8.3 cm

Ktr =0.0, since no stirrups are used

ls

Φ16 @ 250

Φ16 @ 250

Page 27




Instructor:



OKcm

cml

d

KCei

d

KC

d

b

tr

b

tr

3001.46)3.1(5.35llengthspliceRequired

5.356.15.2

)0.1)(8.0)(0.1(

300

4200

7

2

5.2.,.5.219.56.1

03.8

s

Ls=50 cm

Φ16 @ 25

Φ16 @ 25

Page 28




Instructor:


Example # 4

Design a compression lap splice for a tied column whose cross section is shown in Figure when:

(a) Φ16 mm bars are used on both sides of the splice.(b) Φ 16 mm bars are lap spliced with φ 18 mm bars.


Solution: (a) For bars of similar diameter

splice length in compression and for fy =4200 kg/cm2

is equal to 0.07 fy db

= 0.07 (4200)(1.6) = 47 cm >30 cm

taken as 47 cm

Page 29




Instructor:



(b) For bars of different diameters

splice length in compression shall be the larger of either development

length of larger bar, or splice length of smaller bar.

The development length of larger bar

ld = ldb x applicable modification factors

applicable modification factors =1.0

Splice length of smaller diameter bar is evaluated in part (a) as 470mm. Thus, the splice length is taken as 470 mm.

cm33.31.0ll

cm33.30)1.80.0044(420fd0.0044thanlessnotbut

cm31.8300

1.842000.073

'f

d0.073fl

dcd

yb

c

bydc

Documents

Chapter 7 بسم الله الرحمن الرحيم Design of Concrete Structure I University of Palestine Instructor: Eng. Mazen Alshorafa