Development Lenth 2

7/24/2019 Development Lenth 2

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FLEXURAL DESIGN, _~ê

development, hooked bars

UNIVERSITY OF WISCONSIN STOUTCOLLEGE OF SCIENCE, TECHNOLOGY, ENGINEERING, ANDMATHEMATICS

LECTURE IV

Dr. Jason E. Charalambides

Identifying The Problem

! The forensics:

! We often see concrete reinforcement becoming exposed with concrete slicedand disengaged from the rest of the structural element. Besides the case of poorly cast concrete (e.g. honeycombing etc.), geometric formation andphysics/statics may be the reason behind the result.



How Does DevelopmentLength Work?

! But as the form of the concretelocks the bars in place, with everyapplication of forces, these formsexperience stresses that generatereactions. Thus a significant amountof surfaces of both steel andconcrete need to be in contact inorder to allow those stresses to betransferred to the volume of thebeam. The deeper the rebar, themore surface will be in contact, the

better the dissipation of the stress(generated by the tension applied onthe steel) within the concrete.

Insufficient development length at the top of column. Source:

http://www.koeri.boun.edu.tr/depremmuh/eqspecials/bingol/bingol_eq.htm Oct.2009



!

Nevertheless, the depth of the rebar is notthe only factor. Think of this issue in termsof physical scale, as well as numerical scalethat pertains to the strength of thematerials.

" What if we have very strong rebars andvery mild concrete, or vice versa, and whatabout the location of the bars within thebeam (top rebars vs bottom rebars)?

" !=location factor=1 or 1.3 if top bars! !=bar coating factor=1.0 if not coated, or

1.2 or 1.5 (accordingly) if bars are coated! !=concrete density factor=1 or 1.3 if

lightweight concrete

Default conditions

see next slide for definition)

How Does DevelopmentLength Work?

More Detail in Formula

! Favorable bar placement conditionsrequire that:

" Clear cover and clear spacing of barsis db or greater when there aretransverse ties or stirrups as ithappens in most beams, or

" Clear cover is more than db andspacing between bars is greater than2db where ties and stirrups are notpresent as in the case of slabs

! For default formula:

" c=dimension from center of bar toedge of section.

" Ktr=transverse reinforcement ratio.



Graphic Method

Graphic Method

! The “Shortest possible” graph yields its results from the

highest permissible value of [(c+Kv)/db]=2.5

! The central value referred as “Usual” is applicable where

bars are placed inside stirrups or ties, and the clear spacing

between bars as well as edge cover distance is at least as

large as a bar diameter and more than 1in. If no stirrups or

ties are present but the clear spacing between bars is at

least 2 bd, the “Usual” graph is applicable again.

! The “Worst” case gives values 50% higher than the “Usual”.



Hooked Bars

! Framing conditions may limit the space to developtensile strength along a straight bar, imposing a

condition where bars may be bent 90º or hooked.

! ldh can be much shorter than ld.

! ldh is measured from outer edge of hook.

! See following chart:

Hooked Bars Cont (Grade 50, & 4ksi conc.)



Development of Bars inContinuous Beams

! Inside bend radii must be no less than 6db for bars #3through #8, 8db for bars #9 through #11, and 10db for

larger bars.

! Compression bar development length can be much

shorter:

! It can be noted that ldc=ldh for the materialsstrengths f`c=4ksi and fy=60ksi

Development of BarsThrough Lap Splices

1. Lap splices are restricted to #11 or smaller

2. Class B Tension Splice length shall be 1.3 ld

3. Class A Tension Splice length=1.0ld may be used if:! Fewer than half of the bars are spliced at the same location, or

! Bar stress developed is less than fy/2

1. Compression Splice Length shall be 30 db but not lessthan 12 in.

2. Compression splice length for bars higher than Grade

60 shall be (0.9 fy-24)db with fy expressed in ksi.

3. Compression splices within spiral columns may be75% as long but no less than 12 in.



Where is Anchorage Required?

Bar Development Anchorage

! At least 1/3 of bottom bars required for +ve moment must extendat least 6 in into supports of simply supported beams.

! At least 1/4 of bottom bars required for +ve moment must extendat least 6 in into supports of continuous beams.

! At least 1/3 of top bars required for -ve moment must extendbeyond the point of inflection associated with –ve moment.

!

Continuing bars for flexure must extend beyond the point at whichthey are required for a distance at least 12db or the depth of themember d.



No Hassle Bar DetailingExterior Spans

No Hassle Bar DetailingExterior Spans



No Hassle Bar DetailingInterior Spans

In Class Example:

! Steel Grade is 75 andf`c=5ksi. Calculate the ldfor both top and bottombars.



Reading

! Reading:" Req: Furlong, Chapt. 5

" Recom: McCormac & Nelson, Chapter 7 for this lecture.

Documents

Development Lenth 2