Download presentation

1
**Lecture 15- Bar Development**

July 11, 2003 CVEN 444

2
**Lecture Goals Bar Cut-off Points Splice Tension Splice**

Compression Splice

3
**Determining Locations of Flexural Cutoffs**

Given a simply supported beam with a distributed load.

4
**Determining Locations of Flexural Cutoffs**

Note: Total bar length = Fully effective length + Development length

5
**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..)

6
**Determining Locations of Flexural Cutoffs**

Development of flexural reinforcement in a typical continuous beam. ACI 318R for flexural reinforcement

7
**Bar Cutoffs - General Procedure**

1. 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 )

8
**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 )

9
**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 )

10
**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 ).

11
**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 ).

12
**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 )

13
**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).

14
**Bar Cutoffs - General Rules**

Positive Moment Bars Rule 4.

15
**Bar Cutoffs - General Rules**

Negative Moment Bars Rule 5. Negative moment reinforcement must be anchored into or through supporting columns or members (ACI Sec ).

16
**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 ).

17
**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 ).

18
**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.

19
**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.

20
**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.

21
**Moment Resistance Diagrams**

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

22
**Moment Resistance Diagrams**

The moment diagram is

23
**Moment Resistance Diagrams**

The moment resistance of one bar is

24
**Moment Resistance Diagrams**

The moment diagram and crossings

25
**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

26
**Moment Resistance Diagrams**

Compute the bar development length is

27
**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.

28
**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.

29
**Moment Resistance Diagrams**

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

30
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.

31
Example – Cutoff Determine the moment capacity of the bars.

32
Example – Cutoff Determine the location of the bar intersections of moments.

33
Example – Cutoff Determine the location of the bar intersections of moments.

34
Example – Cutoff Determine the location of the bar intersections of moments.

35
Example – Cutoff The minimum distance is

36
Example – Cutoff The minimum amount of bars are As/3 or two bars

37
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”

38
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”

39
Example – Cutoff The moment diagram is the blue line and the red line is the envelope which encloses the moment diagram.

40
**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

41
**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.

42
**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.

43
**Types of Splices Class B Splice (ACI 12.15.2)**

All tension lay splices not meeting requirements of Class A Splices

44
**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.

45
**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 )

46
**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 )

47
**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.

48
**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

49
**Example – Splice Tension**

For #8 bars, db =1.0 in and a = b = g = l =1.0

50
**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.

51
**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

52
**Example – Splice Compression**

For #10 bars, db =1.27 in. Check ls > db fy = 38.1 in. So ls = 39 in.

53
**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.

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google