Lecture_9

# Lecture_9 - 2.4 Doubly Reinforced Rectangular Section in Bending 2.4.1 Design formulae • Derivation of design formulae From the section dealing

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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. 2.4-1 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 2.4-1 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 2.4-1 Design of a rectangular section with compression reinforcement, moment redistribution factor β b ≥ 0.9 . Figure of design example 2.4-1 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.

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