reodata 4.0 - OneSteel Reinforcing

stress Development & lap splicing of
straight Deformed Bars in tension
In AS 3600 : 2009, Clause 13.1.2.2 requires that the basic development
length (Lsy.tb) to develop the yield stress (fsy) of a straight deformed bar in
tension shall be calculated as follows:
Lsy.tb = 0.5k1k3fsydb k2 √f’ c
k1 = 1.3 for horizontal bars with
more than 300 mm of
concrete cast below the bars; or
= 1.0 for all other bars
k2 = (132 - db) /100
k3 = [0.7 ≤ {1.0 - 0.15(cd-db)/db}≤ 1.0]
fsy = characteristic yield stress of the
reinforcing bars (500 MPa)
db = nominal bar diameter (mm)
≥ 29k 1d b where
cd = the least clear concrete cover
(mm) to the bars (c), or half the
clear distance between adjacent
parallel bars developing stress (a),
whichever is the lesser (mm), noting
that the upper and lower bounds
on k3 mean that db ≤ cd ≤ 3db when
substituted into the formula for k3
f’c = the concrete compressive strength
grade, but not to exceed 65 MPa
when substituted into above formula
The value of Lsy.tb so calculated shall be multiplied by 1.3 if lightweight
concrete (as defined in AS 3600) is used and/or by 1.3 for structural
elements built with slip forms.
In accordance with Clause 13.1.2.3 of AS 3600 : 2009, the refined
development length (Lsy.t) shall be determined as follows:
Lsy.t = k4k5Lsy.tb with 0.7≤ k3k4k5≤1.0 where
k4 = [0.7≤ {1.0 - Kλ} ≤ 1.0]
k5 = [0.7≤ {1.0 - 0.04ρ p} ≤ 1.0]
K & λ account for transverse reinforcement -
see fig. 13.1.2.3(B) of AS 3600 : 2009
ρ p = transverse compressive pressure at ultimate load (MPa)
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