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Lecture_10 - 2.4.2 Condition for yielding of compression...

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Unformatted text preview: 2.4.2 Condition for yielding of compression steel • In the analysis, it has been assumed that the compression steel has yielded, so that the steel stress f s′ = 0.95 f y . From the proportions of the strain diagram shown below, ε ’s εs ε s′ x − d′ = 0.0035 x x − d′ ε s′ = 0.0035 x (2.2) d′ ε s′ =1− x 0.0035 or (2.42) From Eq. (2.2), At yielding with fy = 460 N/mm2, the strain of compression steel ε's = εy = 0.00219. From Eq. (2.42), ε s′ d′ 0.00219 = 1− = 1− = 0.37 x 0.0035 0.0035 ∴ Condition for yielding of compression steel (high-yield steel) is d′ ≤ 0.37 x (2.43) 77 • When x = d/2 (i.e., fully utilise the concrete compression), d′ ≤ 0.185 d (2.44) [It is noted that the ratio of d’/d for the yielding of other grades of steel reinforcement and for different values of x (considering moment redistribution) can be determined using Eq. 2.42]. • When d′ > 0.37 , the compression steel has not yielded; thus x f s′ = Esε s′ = 200,000ε s′ (2.45) • The design formulae (2.37) and (2.38) should then be modified as ( K − K ′) f cu bd 2 f sc ( d − d ′) (2.46) K ′f cu bd 2 f s′ + As′ 0.95 f y z 0.95 f y (2.47) ′ As = As = where f s′ ≤ 0.95 f y , and z = 0.775d. Example 2.4-2 Design a section with moment redistribution applied (βb = 0.8) The section shown in the following figure is subject to an ultimate design moment of 228 kNm. fy = 460 N/mm and fcu = 30 N/mm. Determine the reinforcement required. Figure of example 2.4-2 with moment redistribution, βb = 0.8 78 I. Solution from first principles i. Depth of neutral axis x = (βb – 0.4)d = (0.8 – 0.4)d = 0.4d Depth of stress block s = 0.9x = 0.36d Lever arm z = d – s/2 = 0.82d Moment of resistance of the concrete Mcc = Fcc·z = 0.45fcubs·z = 0.45 × 30 × 260 × 0.36 × 0.82 × 4402 × 10-6 = 201 kNm < 228 kNm (the applied moment) therefore compression steel is required. d’/x = 50/176 = 0.28 < 0.37 The compression steel has yielded. ii. Compression steel: A s′ = = M − M cc 0 .95 f y ( d − d ′) (228 − 201) x10 6 = 158 mm 2 0.95x460(4 40 − 50) Provide 2T12 (A’s = 226 mm2) iii. Tension steel: As = M cc ′ + As 0.95 f y z 201 x 10 6 + 158 0.95 x 460 x 0.82 x 440 = 1275 + 158 = 1433 mm 2 = Provide 3T25 (As = 1472 mm2) 79 II. Alternative solution using BS 8110 design equations i. From Eq. (2.41), K’ = 0.405(βb – 0.4) – 0.18(βb – 0.4)2 = 0.405(0.8 – 0.4) – 0.18(0.8 – 0.4)2 = 0.132 K = M/(bd2fcu) 228 x 10 6 260 x 440 2 x 30 = = 0.151 > K’ = 0.132 therefore compression steel is required. ii. Compression steel (Eq. 2.37): As′ = ( K − K ′) f cu bd 2 0.95 f y (d − d ′) = (0.151 − 0.132)30 × 260 × 440 2 0.95 × 460(440 − 50) = 168 mm 2 Tension steel: ( = d (0.5 + z = d 0.5 + 0.25 − K ′ / 0.9 ) 0.25 − 0.132 / 0.9 ) = 0.82d K ′f cu bd 2 ′ + As As = 0.95 f y z = 0.132 × 30 × 260 × 440 2 + 168 0.95 × 460 × 0.82 × 440 = 1265 + 168 = 1433 mm2 80 2.4.3 Design charts – (typical design chart for doubly reinforced beams) Procedure: Calculate M/(bd2) and choose a value of x/d (according to the moment redistribution) from the chart determine A’s and As . Example: M/(bd2) = 7, x/d = 0.5, d’/d = 0.1 From the design chart: A’s / bd = 0.7%, As / bd = 2.25% 81 2.4.4 Checking (Analysis of given sections) • Example 2.4-3: Analysis of a doubly reinforced rectangular section Determine the ultimate moment of resistance of the cross-section shown in the figure given that fy = 460 N/mm2 and fcu = 30 N/mm2. Figure of Example 2.4-3 analysis of doubly reinforced section For equilibrium: Fst = Fcc + Fsc Assuming that f s and f s′ are the design yield values, then ′ 0.95 f y As = 0.45 f cu bs + 0.95 f y As Therefore, ′ 0.95 f y ( As − As ) s= 0.45 f cu b = 0.95 × 460(2410 − 628) = 206 mm 0.45 × 30 × 280 x = s / 0 .9 = 229 mm Q x / d = 229 / 510 = 0.45 < 0.615 ∴ The tension steel will have yielded. Also d ′ / x = 50 / 229 = 0.22 < 0.37 ∴ The compression steel will also have yielded, as assumed. Taking moments about the tension steel, M = Fcc (d − s / 2) + Fsc (d − d ′) ′ = 0.45 f cu bs(d − s / 2) + 0.95 f y As (d − d ′) = 0.45 × 30 × 280 × 206(510 − 206 / 2) + 0.95 × 460 × 620(510 − 50) = 441× 106 Nmm = 441 kNm 82 • Discussion on the case of When d′ > 0.37 x d′ > 0.37 , the compressive steel will not have yielded; then, x ′ Fsc ≠ 0.95 f y As . Equilibrium condition: Fcc + Fsc = Fst , 0.45 f cu b(0.9 x) + f s′As′ = 0.95 f y As or (2.48) where f s′ = Esε s′ . εcu ε's d' ε s′ = ε cu x − d′ x εs Eq. (2.48) can then be written as 0.405 f cu bx + E s ε cu x − d′ ′ As = 0.95 f y As x or ′ ′ 0.405 f cubx2 + ( Esε cu As − 0.95 f y As ) x − Esε cu d ′As = 0 (2.49) The value of x can be determined by solving Eq. (2.49). By taking moments about the tension steel, the moment of resistance of the cross-section is given by M = Fcc ( d − 0.9 x / 2) + Fsc (d − d ′) (2.49) where Fcc = 0.45fcub(0.9x) and Fsc = f s′ As′ = ( Esε s′ ) As′ = Esε cu x − d′ As′ x (2.50) 83 ...
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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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