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Unformatted text preview: 2.4 Doubly Reinforced Rectangular Section in Bending 2.4.1 Design formulae
• Derivation of design formulae From the section dealing with the analysis of a singly reinforced
section, when the design ultimate moment exceeds the ultimate
moment of resistance,
M > 0.156 f cu bd 2 compression reinforcement is required. ε′s εs Fst Fig. 2.41 Section with compression reinforcement
o Equilibrium condition: Fst = Fcc + Fsc ′
With steel reinforcement (both As and As ) at yielding ,
′
0.95 f y As = 0.45 f cu bs + 0.95 f y As Q s = 0.9 x = 0.9 ⋅ d / 2 = 0.45d ∴ ′
0.95 f y As = 0.202 f cu bd + 0.95 f y As (2.33)
72 By taking moments about the centroid of As ,
M = Fcc z + Fsc (d − d ′)
′
= 0.202 f cu bd ⋅ 0.775d + 0.95 f y As (d − d ′) or
′
M = 0.156 f cu bd 2 + 0.95 f y As (d − d ′) (2.34) From Eq. (2.34)
M − 0.156 f cu bd 2
0.95 f y (d − d ′) (2.35) 0.156 f cu bd 2
′
As =
+ As
0.95 f y z (2.36) ′
As = From Eq. (2.33) where z = 0.775d.
o In BS 8110: Part 1, the design formulae are given as (Clause 3.4.4.4)
′
As = (2.37) As = where ( K − K ′) f cu bd 2
0.95 f y (d − d ′)
K ′f cu bd 2
′
+ As
0.95 f y z (2.38) K ′ = 0.156 K= M
bd 2 f cu It is noted that the formulae are derived for the case where the
reduction in moment due to moment redistribution is not greater
than 10%, i.e. β b ≥ 0.9 .
73 • The section with moment redistribution applied
The formation of plastic hinges requires relatively large rotations
with yielding of the tension steel.
o To ensure large strains in the tension reinforcement, BS 8110 restricts the neutral axis depth x according to the reduction of
the elastic moment:
x = ( β b − 0.4) d Note that βb = 0.9 for 0 − 10% moment redistribution.
The lever arm is given by
εcu
z = d − s / 2 = d − 0 .9 x / 2
= d − 0.45 x or
z = d − 0.45( β b − 0.4)d (2.39) εs
The moment of resistance of the concrete in compression is
M cc = Fcc z = 0.45 f cu b(0.9 x) ⋅ z
= 0.405bd 2 f cu ( β b − 0.4)[1 − 0.45( β b − 0.4)] ∴ M cc
= 0.405( β b − 0.4) − 0.18( β b − 0.4) 2
2
bd f cu The BS 8110 equation: M cc = K ′bd 2 f cu (2.40) K ′ = 0.405( β b − 0.4) − 0.18( β b − 0.4) 2 (2.41) where 74 o In Eq. (2.41), (a) when moment redistribution is equal to or smaller than 10%,
i.e. β b ≥ 0.9 (x = 0.5d), K ′ = 0.156
(b) when moment redistribution is equal to or larger than 10%,
i.e. β b ≤ 0.9 (x < 0.5d),
K ′ = 0.405( β b − 0.4) − 0.18( β b − 0.4) 2
Table 2.41 Moment redistribution design factors o Therefore, when the design ultimate moment is such that M > K ′bd 2 f cu
′
The design formulae for As
and As are identical in form with Eqs (2.37) and (2.38).
compression steel is required. • Example 2.41
Design of a rectangular section with compression reinforcement,
moment redistribution factor β b ≥ 0.9 . Figure of design example 2.41
with compression reinforcement,
β b ≥ 0.9 75 The section shown in the figure is to resist an ultimate design
moment of 285 kNm. The characteristic material strengths are fy
= 460 N/mm2 and fcu = 30 N/mm2. Determine the areas of reinforcement required.
K= M
bd 2 f cu 285 × 10 6
= 0.189
260 × 440 2 × 30
> 0.156
= Therefore compression steel is required.
Compression steel:
( K − K ′) f cu bd 2
′
As =
0.95 f y (d − d ′)
(0.189 − 0.156)30 × 260 × 440 2
=
0.95 × 460(440 − 50)
= 292 mm 2 Tension steel:
As = K ′f cu bd 2
′
+ As
0.95 f y z 0.156 × 30 × 260 × 440 2
+ 292
0.95 × 460(0.775 × 440)
= 1581 + 292
= = 1873 mm 2 Question:
If ε′s < εy (i.e., compression steel has not yielded at the ultimate limit
state), are the design formulas (2.37) and (2.38) still valid to use for
determining As′ and As?
76 ...
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This note was uploaded on 12/24/2011 for the course CIVL 232 taught by Professor Jskuang during the Spring '06 term at HKUST.
 Spring '06
 JSKUANG

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