Reinforced Concrete Analysis and Flexural Design – Factors for … · –λ=concrete density factor=1 and varies to 0.85 for sand-lightweight and to 0.75 for all-lightweight concrete

Reinforced Concrete Analysis and Flexural Design – Factors for

Safety and Strain Comfort

MORGAN STATE UNIVERSITY

SCHOOL OF ARCHITECTURE AND PLANNING

LECTURE IV

Dr. Jason E. Charalambides

Copyright J. CharalambidesCopyright J. Charalambides

Identifying The Problem

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


But as the form of the concrete locks the bars in place, with every application of forces, these forms experience stresses that generate reactions. Thus a significant amount of surfaces of both steel and concrete need to be in contact in order to allow those stresses to be transferred to the volume of the beam. The deeper the rebar, the more surface will be in contact, the better the dissipation of the stress (generated by the tension applied on the 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

How Bar Development Works


How Does Development Length Work?

Nevertheless, the depth of the rebar is not the only factor. Consider this issue in terms of physical scale, as well as numerical scale that pertains to the strength of the materials.

– What if we have very strong rebars and very mild concrete, or vice versa.

– What about the location of the bars within the beam (top rebars vs bottom rebars)?

– ψt=location factor=1 or 1.3 if top bars.– ψe=bar coating factor=1.0 if not coated, or

1.2 or 1.5 (accordingly) if bars are coated.– λ=concrete density factor=1 and varies to

0.85 for sand-lightweight and to 0.75 for all-lightweight concrete.

Default conditions(see next slide for definition)

ld6=db⋅( f y⋅ψ t⋅ψ e)25⋅λ⋅√ f ' c

Note: The terms ld6 and ld7 are used specifically by the instructor in order to distinguish between the formulae tor rebars smaller or equal to #6 and larger or equal to #7.

ld6=db⋅( f y⋅ψ t⋅ψ e)20⋅λ⋅√ f ' c

Table 25.4.2.2


Favorable bar placement conditions require that:

– Clear cover and clear spacing of bars is db or greater when there are transverse ties or stirrups as it happens in most beams, or

– Clear cover is more than db and spacing between bars is greater than 2db where ties and stirrups are not present as in the case of slabs.

For default formula:– c=dimension from center of bar to

edge of section.– Ktr=factor of contribution of confining

reinforcement across potential splitting planes.

Special conditions formulae apply when default conditions on left are not satisfied

More Detail on Formula

ld=d b⋅(3f y⋅ψ t⋅ψ e⋅ψ s)

40⋅λ√ f ' c[ c+K tr

d b ]

ld6=db⋅(3f y⋅ψ t⋅ψe)50⋅λ⋅√ f ' c

ld7=db⋅(3f y⋅ψ t⋅ψe)40⋅λ⋅√ f ' c

Table 25.4.2.2

25.4.2.3a


Graphic Method


The “Shortest possible” graph yields its results from the highest permissible value of [(c+Kv)/db]=2.5The 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”.

Graphic Method


Hooked Bars

- Framing conditions may limit the space to develop tensile 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 in Continuous Beams

Inside bend radii must be no less than 6db for bars #3 through #8, 8db for bars #9 through #11, and 10db for larger bars.Compression bar development length can be much shorter:

Where ψr is a factor that relates to closely placed ties near the bend portion of a hooked bar.

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

ldc=db⋅ψr⋅( f y)50√ f ' c

≥0.0003db⋅f y⋅ψ r>8inch

25.4.9.2.


Development of Bars Through Lap Splices

1. Lap splices are restricted to #11 or smaller2. Class B Tension Splice length shall be 1.3 ld3. Class A Tension Splice length=1.0 ld may be used if:

! Fewer than half of the bars are spliced at the same location, or

! Bar stress developed is less than fy/21. Compression Splice Length shall be 30 db but not

less than 12”.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 be

75% as long but no less than 12”.


Where Is Anchorage Required?


Bar Development Anchorage

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

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

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

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


No Hassle Bar DetailingExterior Spans


No Hassle Bar DetailingExterior Spans


In class example:

Steel Grade is 75 and f`c=5ksi. Calculate the ld for both top and bottom bars.

Documents

Reinforced Concrete Analysis and Flexural Design – Factors for … · –λ=concrete density factor=1 and varies to 0.85 for sand-lightweight and to 0.75 for all-lightweight concrete