Lecture 20 Bar Development - PowerPoint PPT Presentation

1 / 54
About This Presentation
Title:

Lecture 20 Bar Development

Description:

Determine the development length of the bars if the beam is made of normal ... To develop full tensile force, the bars should extend a distance x2 before they ... – PowerPoint PPT presentation

Number of Views:283
Avg rating:3.0/5.0
Slides: 55
Provided by: ericsa3
Category:

less

Transcript and Presenter's Notes

Title: Lecture 20 Bar Development


1
Lecture 20 - Bar Development
  • October 16, 2002
  • CVEN 444

2
Lecture Goals
  • Bar Development
  • Hook development

3
Example - Development
For the cross section of a simply supported beam
reinforced with 4 8 bars that are confined with
3 stirrup spaced at 6 in. Determine the
development length of the bars if the beam is
made of normal weight concrete fc 3 ksi and fy
60 ksi
4
Example - Development
Check if conditions for spacing and concrete
cover are met
For 8 bars, db 1.0 in. Clear cover 2.5 in -
0.5 in. 2.0 in. db
Clear spacing between bars
5
Example - Development
Bars are confined with 3 stirrups. The
conditions are met. Determine the factorsa
1.0 (bottom bars), b 1.0 (no coating) and l
1.0 (normal weight concrete) and 54.8
psi 6
Example - Development
So ld 54.8(1.0 in.) 54.8 in. 55 in.
Using the more general formula Ktr 0.0
7
Example - Development
c 1.17 in. controls
8
Example - Development
So ld 55 in.
9
Example - Development
If the same beam is made of light weight
aggregate concrete and the bars are epoxy coated
and As required for analysis is 2.79 in2
10
Example - Development
11
Example - Development
12
Critical Sections in Flexural Members
13
Critical Sections in Negative Moment Reinforcement
Three sections are critical for the negative
moment reinforcement Section 1 is at the face of
the support, where the negative moment as well as
stress are at maximum value. Two development
lengths, x1 and x2 must be checked.
14
Critical Sections in Negative Moment Reinforcement
Section 2 is the section where part of the
negative reinforcement bar can be terminated. To
develop full tensile force, the bars should
extend a distance x2 before they can be
terminated Once part of the bars are terminated
the remaining bars develop maximum stress.
15
Critical Sections in Negative Moment Reinforcement
Section 3 is a point of inflection The bars shall
extend a distance x3 beyond section 3 x3 must
be equal to or greater than the effective depth
d, 12db or 1/16 the span, which ever is greater.
At least 1/3 of the total reinforcement provided
for negative moment at support shall extend a
distance x3 beyond the point of inflection.
16
Critical Sections in Positive Moment Reinforcement
Section 4 is that of maximum positive moment and
maximum stresses. Two development lengths x1 and
x2 have to be checked. The length x1 is the
development length ld specified by the ACI Code
Section 12.11. The length x2 is equal to or
greater than the effective depth d, 12db .
17
Critical Sections in Positive Moment Reinforcement
18
Critical Sections in Positive Moment Reinforcement
Section 6 is at the points of inflection limits
are according to section 12.11.3 of the ACI Code.
19
Example Problem
A continuous beam has the bar details shown. The
bending moments for maximum positive and negative
moments are given. We must check the development
lengths at all critical
sections. fc 3ksi fy 60 ksi , b 12 in. d
18 in. and the span L 24 ft.
20
Example Problem
The critical sections are at the face of the
support for tension and compression
reinforcement, at points where tension bars are
terminated within the span and at point of
inflection and at midspan.
Development length for negative moment x1 4.5
ft from face of support where as 3 bars extend to
a distance of 6 ft.
21
Example Problem
The development length is (d 1.128 in.)
(if
spacing and cover
conditions are met)
22
Example Problem
The bars are at the top so a 1.3 and the
development length is
So ld 54 in. 4.5 ft 12 in. (minimum)
23
Example Problem
24
Example Problem
25
Example Problem
26
Example Problem
It can not be less than 8 in. The length 15 in.
controls. For 8 bars db 1 in. ld provided is
15 in. which is greater than that required.
27
Example Problem
The development length for positive moment
reinforcement 3 8 bars extend 6 ft beyond the
centerline, and the other bars extend to the
support. The development length x6 from center
line is
28
Example Problem
The ld is 37 in. But it can not be less than 12
in.. So x6 provided is 6 ft 72 in. 37 in. The
length x7 is equal to d or 12 db, and 18 in. is
provided.
29
Standard Hooks
A hook is used at the end of a bar when its
straight embedment length is less than the
necessary length, ld. The minimum diameter of
bend, measured on the inside of the main bar of a
standard hook D is
30
Standard Hooks
The stress distribution for a 90o hook under a
force P is shown.
31
Standard Hooks
The basic development length lhb must be
multiplied one of the following factors.
32
Standard Hooks
The basic development length lhb must be
multiplied one of the following factors.
33
Standard Hooks
The basic development length lhb must be
multiplied one of the following factors.
34
Standard Hooks
The basic development length lhb must be
multiplied one of the following factors.
35
Example- Hook
Compute the development length required for the
top 8 bar of the cantilever beam that extend
into the column support if the bars are
  • Straight.
  • Have a 90o hook
  • Have a 180o hook

The bars are confined by 3 stirrups spaced at 6
in., and clear cover 1.5 in., and clear
spacing 2.0 in. and fc 4 ksi and fy 60 ksi
36
Example- Hook
Straight bars For 8 bar db 1.0 in., because
the clear spacing 2db and clear cover is greater
than db with the bars confined condition (a) and
(b) are met. For top bars a 1.3
37
Example- Hook
Bar with 90o hook For 8 bar db 1.0 in., the
basic equation is No modification apply, than
ldh 19 in. 8db 8 in. or 6 in. The factor a
1.3 does not apply to hooks.
38
Example- Hook
Bar with 180o hook For 8 bar db 1.0 in., the
basic equation is No modification apply, than
ldh 19 in. 8db 8 in. or 6 in. The factor a
1.3 does not apply to hooks.
39
Example- Hook
The basic summary of the results for the two
hooks
40
Example Development Length
Design anchorage of 4 8 top bars in column. 4
11 bars in the transverse direction.
41
Example Development Length
Try a straight bar for the development Use
12.2.2 Spacing between bars 48
42
Example Development Length
The spacing 2.75 in. db 1.0 in. So use
43
Example Development Length
The coefficients need to be determined for the
reinforcement coefficient, a. There is more than
12 in. of concrete below the bar.
Uncoated bar so the coating factor is
44
Example Development Length
The coefficient for aggregate is
The development length of the bar
45
Example Development Length
The development of the bar going out into the
beam. Check the development length into the
column.
46
Example Development Length
The clear space between the bars are
Edge
Center to center
47
Example Development Length
The transverse steel area of the column is
comprised of 411 bars (As 1.56 in2 db1.41
in.) and the coefficient Ktr is defined as
There are 4 transverse bars
48
Example Development Length
The transverse coefficient is
There are 4 transverse bars
49
Example Development Length
The transverse coefficient is
The coefficient
50
Example Development Length
The development length is
51
Example Development Length
Use a hook anchorage
Compute the coefficients for the column. Side
coverage
52
Example Development Length
The side coverage for a 90o hook.
Top coverage is
The multiplier is 1.0 ACI 12.5
53
Example Development Length
Additional ties per ACI 12.5.4 not require only
min ties spacing 3 db. Multiplier is 1.0
The available length
The hook extension
54
Example Development Length
The side coverage for a 180o hook.
The multiplier is 0.7
Write a Comment
User Comments (0)
About PowerShow.com