2. Reinforced Concrete Design-I Lec-07 Bond and Development Length
By Dr. Attaullah Shah Swedish College of Engineering and Technology Wah Cantt.

3. The basic assumption of the RCC design is that the strain in concrete and reinforcing steel is the same. If the reinforcing steel slips at its ends, this is not valid. Hence it must be ensured that sufficient bond strength is developed at the interface of steel and concrete to avoid slippage of the steel.

4. Bond Strength and Development length
Two types of bond failure can be expected in reinforcing bars:
Direct pull out of the steel bars, when ample concrete confinement is provided in the form of large spacing of bars or large concrete cover
Splitting of concrete along the bar when cover confinement or bar spacing is insufficient

6. b. Actual Distribution of Flexural Bond Stress
Pure bending case
Concrete fails to resist tensile stresses only where the actual crack is located. Steel T is maximum and
T max = M / jd .
Between cracks , concrete does resist moderate amount of tension introduced by bond.
u is proportional to the rate of change of bar force, and highest where the slope of the steel force curve is greatest.
Very high local bond stress adjacent to the crack.

7. Beam under transverse loads,
According to simple crack sectional theory, T is proportional to the moment diagram and u is proportional to shear force diagram.
In actual, T is less than the simple analysis prediction everywhere except at the actual cracks.
Similarly, u is equal with simple analysis prediction only at the location where slopes of the steel force diagrams are equals .If the slope is greater than assumed, bond stress is greater; if the slope is less bond stress is less.

8. ULTIMATE BOND STRENGTH AND DEVELOPMENT LENGTH
Types of bond failure
Direct pullout of bars
(small diameter bars are used with sufficiently large concrete cover distances and bar spacing)
Splitting of the concrete along the bar (cover or bar spacing is insufficient to resist the lateral concrete tension resulting from the wedging effect of bar deformations)
Department of Civil Engineering, YTU

9. a. Ultimate Bond Strength
Direct pull out
For sufficiently confined bar, adhesive bond and friction are overcome as the tensile force on the bar is increased. Concrete eventually crushes locally ahead of the bar deformation and bar pullout results.
When pull out resistance is overcome or when splitting has spread all the way to the end of an unanchored bar, complete bond failure occurs.
Splitting
Splitting comes from wedging action when the ribs of the deformed bars bear against the concrete.
Splitting in vertical plane
Splitting in horizontal plane: frequently begins at a diagonal crack in connection with dowel action. Shear and bond failures are often interrelated.
Local bond failure
Large local variation of bond stress caused by flexural and diagonal cracks immediately adjacent to cracks leads to this failure below the failure load of the beam.
Results small slip and some widening of cracks and increase of deflections.
Harmless as long as the failure does not propagate all along the bar.
Providing end anchorage, hooks or extended length of straight bar (development length concept)

10. Consider a bar embedded in a mass of concrete
P = s * [p*db2/4]
P = t*[Lb*p*db] db Lb
t = P / [Lb*p*db] < tmax
s = P/ [p*db2/4] < smax
P < tmax * [Lb*p*db]
P < smax * [p*db2/4]
To force the bar to be the weak link: tmax * [Lb*p*db] > smax * [p*db2/4]
Lb > (smax / tmax)* [db/4]

11. Development Length Ld = development length
the shortest distance over which a bar can achieve it’s full capacity
The length that it takes a bar to develop its full contribution to the moment capacity, Mn Ld Mn Cc
Mn = (C or T)*(dist) Ts

12. Lb > (fy / tmax)* [db/4]
Steel Limit, smax
Using the bilinear assumption of ACI 318:
smax = + fy
Lb > (fy / tmax)* [db/4]
Lb > fy * db / (4*tmax)

13. Concrete Bond Limit, tmax
There are lots of things that affect tmax
The strength of the concrete, f’c
Type of concrete (normal weight or light weight)
The amount of concrete below the bar
The surface condition of the rebar
The concrete cover on the bar
The proximity of other bars transferring stress to the concrete
The presence of transverse steel

14. Concrete Strength, f’c
Bond strength, tmax, tends to increase with concrete strength.
Experiments have shown this relationship to be proportional to the square root of f’c.

15. Type of Concrete
Light weight concrete tends to have less bond strength than does normal weight concrete.
ACI introduces a lightweight concrete reduction factor, l, on sqrt(f’c) in some equations.
See ACI , for details

16. Amount of Concrete Below Bars
The code refers to “top bars” as being any bar which has 12 inches or more of fresh concrete below the bar when the member is poured.
If concrete > 12” then consolidation settlement results in lower bond strength on the bottom side of the bar
See ACI , (a)

17. Surface Condition of Rebar
All rebar must meet ASTM requirements for deformations that increase pullout strength.
Bars are often surface coated is inhibit corrosion.
Epoxy Coating  The major concern!
Galvanizing
Epoxy coating significantly reduces bond strength
See ACI , (b)

18. Proximity to Surface or Other Bars
The size of the concrete “cylinder” tributary to each bar is used to account for proximity of surfaces or other bars.

19. Presence of Transverse Steel
The bond transfer tends to cause a splitting plane
Transverse steel will increase the strength of the splitting plane.

20. b. Development Length
Development length is the length of embedment necessary to develop the full tensile strength of bar, controlled by either pullout or splitting.
In Fig., let
maximum M at a and zero at support
fs at a T = Ab fs _
Development length concept total tension force must be transferred from the bar to the concrete in the distance ‘l ‘ by bond stress on the surface.
To fully develop the strength  T = Ab fy
 ld , development length
Safety against bond failure: the length of the bar from any point of given steel stress to its nearby end must be at least equal to its development length. If the length is inadequate, special anchorage can be provided.

21. ACI CODE PROVISION FOR DEVELOPMENT OF TENSION REINFORCEMENT
Limit
(c + ktr) / db = 2.5 for pullout case
√f’c are not to be greater than 100 psi.
Department of Civil Engineering, YTU

22. For two cases of practical importance, using (c + ktr) / db = 1.5,

23. Example:

24. Continue:
Department of Civil Engineering, YTU

25. Continue:

26. ANCHORAGE OF TENSION BARS BY HOOKS
In the event that the desired tensile stress in a bar can not be developed by bond alone, it is necessary to provide special anchorage at the end of the bar.

28. b. Development Length and Modification Factors for Hooked Bars

29. Department of Civil Engineering, YTU

30. Example
Department of Civil Engineering, YTU

31. ANCHORAGE REQUIREMENTS FOR WEB REINFORCEMENT

32. DEVELOPMENT OF BARS IN COMPRESSION
Reinforcement may be required to develop its compressive strength by embedment under various circumstances.
ACI basic development length in compression
ldb = 0.02db fy/√f’c
Department of Civil Engineering, YTU

33. Determining Locations of Flexural Cutoffs
Given a simply supported beam with a distributed load.

34. Determining Locations of Flexural Cutoffs
Note:
Total bar length =
Fully effective length
+ Development length

35. Determining Locations of Flexural Cutoffs
ACI
All longitudinal tension bars must extend a min. distance = d (effective depth of the member) or 12 db (usually larger) past the theoretical cutoff for flexure (Handles uncertainties in loads, design approximations,etc..)

36. Determining Locations of Flexural Cutoffs
Development of flexural reinforcement in a typical continuous beam.
ACI 318R for flexural reinforcement

37. Bar Cutoffs - General Procedure1. 2. 3.
Determine theoretical flexural cutoff points for envelope of bending moment diagram.
Extract the bars to satisfy detailing rules (from ACI Section 7.13, 12.1, 12.10, and 12.12)
Design extra stirrups for points where bars are cutoff in zone of flexural tension (ACI )

38. Bar Cutoffs - General Rules
All Bars
Rule Rule 2.
Bars must extend the longer of d or 12db past the flexural cutoff points except at supports or the ends of cantilevers (ACI )
Bars must extend at least ld from the point of maximum bar stress or from the flexural cutoff points of adjacent bars (ACI and )

39. Bar Cutoffs - General Rules
Positive Moment Bars
Rule 3.
Structural Integrity
Simple Supports At least one-third of the positive moment reinforcement must be extend 6 in. into the supports (ACI ).
Continuous interior beams with closed stirrups. At least one-fourth of the positive moment reinforcement must extend 6 in. into the support (ACI and )

40. Bar Cutoffs - General Rules
Positive Moment Bars
Rule 3.
Structural Integrity
Continuous interior beams without closed stirrups. At least one-fourth of the positive moment reinforcement must be continuous or shall be spliced near the support with a class A tension splice and at non-continuous supports be terminated with a standard hook. (ACI ).

41. Bar Cutoffs - General Rules
Positive Moment Bars
Rule 3.
Structural Integrity
Continuous perimeter beams. At least one-fourth of the positive moment reinforcement required at midspan shall be made continuous around the perimeter of the building and must be enclosed within closed stirrups or stirrups with 135 degree hooks around top bars. The required continuity of reinforcement may be provided by splicing the bottom reinforcement at or near the support with class A tension splices (ACI ).

42. Bar Cutoffs - General Rules
Positive Moment Bars
Rule 3.
Structural Integrity
Beams forming part of a frame that is the primary lateral load resisting system for the building. This reinforcement must be anchored to develop the specified yield strength, fy, at the face of the support (ACI )

43. Bar Cutoffs - General Rules
Positive Moment Bars
Rule 4.
Stirrups
At the positive moment point of inflection and at simple supports, the positive moment reinforcement must be satisfy the following equation for ACI An increase of 30 % in value of Mn / Vu shall be permitted when the ends of reinforcement are confined by compressive reaction (generally true for simply supports).

44. Bar Cutoffs - General Rules
Positive Moment Bars
Rule 4.

45. Bar Cutoffs - General Rules
Negative Moment Bars
Rule 5.
Negative moment reinforcement must be anchored into or through supporting columns or members (ACI Sec ).

46. Bar Cutoffs - General Rules
Negative Moment Bars
Rule 6.
Structural Integrity
Interior beams. At least one-third of the negative moment reinforcement must be extended by the greatest of d, 12 db or ( ln / 16 ) past the negative moment point of inflection (ACI Sec ).

47. Bar Cutoffs - General Rules
Negative Moment Bars
Rule 6.
Structural Integrity
Perimeter beams. In addition to satisfying rule 6a, one-sixth of the negative reinforcement required at the support must be made continuous at mid-span. This can be achieved by means of a class A tension splice at mid-span (ACI ).

48. Moment Resistance Diagrams
Moment capacity of a beam is a function of its depth, d, width, b, and area of steel, As. It is common practice to cut off the steel bars where they are no longer needed to resist the flexural stresses. As in continuous beams positive moment steel bars may be bent up usually at 45o, to provide tensile reinforcement for the negative moments over the support.

49. Moment Resistance Diagrams
The nominal moment capacity of an under-reinforced concrete beam is
To determine the position of the cutoff or bent point the moment diagram due to external loading is drawn.

50. Moment Resistance Diagrams
The ultimate moment resistance of one bar, Mnb is
The intersection of the moment resistance lines with the external bending moment diagram indicates the theoretical points where each bar can be terminated.

51. Moment Resistance Diagrams
Given a beam with the 4 #8 bars and fc=3 ksi and fy=50 ksi and d = 20 in.

52. Moment Resistance Diagrams
The moment diagram is

53. Moment Resistance Diagrams
The moment resistance of one bar is

54. Moment Resistance Diagrams
The moment diagram and crossings

55. Moment Resistance Diagrams
The ultimate moment resistance is 2480 k-in. The moment diagram is drawn to scale on the basis A bar can be terminated at a, two bars at b and three bars at c. These are the theoretical termination of the bars. a b c

56. Moment Resistance Diagrams
Compute the bar development length is

57. Moment Resistance Diagrams
The ultimate moment resistance is 2480 k-in. The moment diagram is drawn to scale on the basis A bar can be terminated at a, two bars at b and three bars at c. These are the theoretical termination of the bars.

58. Moment Resistance Diagrams
It is necessary to develop part of the strength of the bar by bond. The ACI Code specifies that every bar should be continued at least a distance d, or 12db , which ever is greater, beyond the theoretical points a, b, and c. Section specify that 1/3 of positive moment reinforcement must be continuous.

59. Moment Resistance Diagrams
Two bars must extend into the support and moment resistance diagram Mub must enclose the external bending moment diagram.

60. Example – Cutoff
For the simply supported beam with b=10 in. d =17.5 in., fy=40 ksi and fc=3 ksi with 4 #8 bars. Show where the reinforcing bars can be terminated.

61. Example – Cutoff
Determine the moment capacity of the bars.

62. Example – Cutoff
Determine the location of the bar intersections of moments.

63. Example – Cutoff
Determine the location of the bar intersections of moments.

64. Example – Cutoff
Determine the location of the bar intersections of moments.

65. Example – Cutoff
The minimum distance is

66. Example – Cutoff
The minimum amount of bars are As/3 or two bars

67. Example – Cutoff
The cutoff for the first bar is 41 in. or 3 ft 5 in. and 18 in or 1 ft 6 in. total distance is 41 in.+18 in. = 59 in. or 4 ft 11 in.
Note error it is 4’-11” not 5’-11”

68. Example – Cutoff
The cutoff for the second bar is 83 in in. 101 in. or 8 ft 5 in. (37-in+5-in+18-in+41-in= 101-in.)
Note error it is 4’-11” not 5’-11”

69. Example – Cutoff
The moment diagram is the blue line and the red line is the envelope which encloses the moment diagram.

70. Bar Splices Why do we need bar splices? -- for long spans
Types of Splices
1. Butted &Welded
2. Mechanical Connectors
3. Lay Splices
Must develop 125% of yield strength ACI and ACI

71. Tension Lap Splices Why do we need bar splices? -- for long spans
Types of Splices
1. Contact Splice
2. Non-Contact Splice (distance between the bars ” and 1/5 of the splice length ACI )
Splice length (development length) is the distance the two bars are overlapped.

72. Types of Splices Class A Splice (ACI 12.15.2)
When over entire splice length.
and 1/2 or less of total reinforcement is spliced win the req’d lay length.

73. Types of Splices Class B Splice (ACI 12.15.2)
All tension lay splices not meeting requirements of Class A Splices

74. Tension Lap Splice (ACI 12.15)
where As (req’d) = determined for bending
ld = development length for bars (not allowed to use excess reinforcement modification factor)
ld must be greater than or equal to 12 in.

75. Tension Lap Splice (ACI 12.15)
Lap Splices shall not be used for bars larger than No. 11. (ACI )
Lap Splices should be placed in away from regions of high tensile stresses -locate near points of inflection (ACI )

76. Compression Lap Splice (ACI 12.16)
Lap, req’d = fy db for fy psi Lap, req’d = (0.0009fy -24) db for fy > psi Lap, req’d in
For fc psi, required lap splice shall be multiply by (4/3) (ACI )

77. Compression Lap Splice (ACI 12.17.2)
In tied column splices with effective tie area throughout splice length hs factor = 0.83
In spiral column splices, factor = 0.75
The final splice length must be in.

78. Example – Splice Tension
Calculate the lap-splice length for 6 #8 tension bottom bars in two rows with clear spacing 2.5 in. and a clear cover, 1.5 in., for the following cases a. b. c.
When 3 bars are spliced and As(provided) /As(required) >2
When 4 bars are spliced and As(provided) /As(required) < 2
When all bars are spliced at the same location fc= 5 ksi and fy = 60 ksi

79. Example – Splice Tension
For #8 bars, db =1.0 in and a = b = g = l =1.0

80. Example – Splice Tension
The As(provided) /As(required) > 2, class A splice applies; therefore lst = 1.0 ld >12 in., so lst = 43 in. > 12 in. The bars spliced are less than half the number
The As(provided) /As(required) < 2, class B splice applies; therefore lst = 1.3 ld >12 in., so lst = 1.3(42.4 in.) = 55.2 in. use 56 in. > 12 in..
Class B splice applies and lst = 56 in. > 12 in.

81. Example – Splice Compression
Calculate the lap splice length for a # 10 compression bar in tied column when fc= 5 ksi and
fy = 60 ksi
fy = 80 ksi

82. Example – Splice Compression
For #10 bars, db =1.27 in.
Check ls > db fy = 38.1 in. So ls = 39 in.

83. Example – Splice Compression
For #10 bars, db =1.27 in. The ld = 23 in.
Check ls > ( fy –24) db
=(0.0009(80000)-24)(1.27in.) = 61 in.
So use ls = 61 in.