Lecture_6

# Lecture_6 - 2.3 Singly Reinforced Rectangular Section in...

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Unformatted text preview: 2.3 Singly Reinforced Rectangular Section in Bending 2.3.1 Design formulae The design formulae are derived using the BS 8110 simplified stress block. z = la d Fst εs Fig. 2.3-1 Singly reinforced section with rectangular stress block (1) Design formulae For equilibrium, the ultimate design moment M = Fcc ⋅ z = Fst ⋅ z (2.15) where z is the lever arm (the distance between Fcc and Fst) o Concrete compressive force Fcc = 0.45fcu b(0.9x) (2.16a) Fst = 0.95fy As (2.16b) z = d – s/2 = d − 0.45x (2.16c) o Steel tensile force o Lever arm 43 Substituting Eqs (2.16a) and (2.16c) into Eq. (2.15) gives M = 0.9 f cu b(d − z ) z (2.17) Let K= M bd 2 f cu (2.18) Eq. (2.17) becomes ( z / d ) 2 − ( z / d ) + K / 0 .9 = 0 (2.19) Solving this quadratic equation gives z = d [0.5 + 0.25 − K / 0.9 ] (2.20) which is the equation presented in BS 8110 for determining the lever arm of a singly reinforced section. o By substituting Eq. (2.16b) into Eq. (2.15), the tension steel As = M 0.95 f y ⋅ z (2.21) o Both Eqs (2.20) and (2.21) are used to design the area of tension steel to resist an ultimate moment, M. In BS 8110: Part 1, o Upper limit of the lever arm (Clause 3.4.4.1): z max = 0.95d ( xmin = 0.11d ) for maintaining a minimum area of concrete in compression o Lower limit of the lever arm (Clause 3.4.4.4): z min = 0.775d ( xmax = 0.5d ) 44 i.e. 0.11d ≤ x ≤ 0.5d Fig. 2.3-2 Lever-arm example o With x = xmax = 0.5d (z = 0.775d), Eq. (2.17) becomes M = 0.9 f cu b(d − 0.775d ) × 0.775d (2.22) M = 0.156 bd 2 f cu (2.23) or This is the maximum moment of resistance for a singly reinforced section. As compared to Eq. (2.18), K = 0.156. o When M = K > 0.156 bd 2 f cu compression reinforcement is required to supplement the moment of resistance of concrete. 45 (2) Design procedure (i) Calculate K value (Eq. 2.18) K = M /(bd 2 f cu ) (ii) Determine lever arm (Eq. 2.20) z = d [0.5 + 0.25 − K / 0.9 ] (iii) Determine area of steel reinforcement (Eq. 2.21) As = M 0.95 f y ⋅ z (iv) Select suitable bar sizes Table 2.3-1 Areas of groups of bars Number of bars in groups Diameter (mm) 1 3 4 5 6 7 8 28 50 78 113 201 314 490 804 6 8 10 12 16 20 25 32 2 56 100 157 226 402 628 981 1608 84 150 235 339 603 942 1472 2412 113 201 314 452 804 1256 1963 3216 141 251 392 565 1005 1570 2454 4021 169 301 471 678 1206 1884 2945 4825 197 351 549 791 1407 2199 3436 5629 226 402 628 904 1608 2513 3927 6433 (v) Check reinforcement ratio ρ = i. ρ max = 4% As bd (In BS 8110: Part 1, Clause 3.12.6.1: As /Ac or As′ / Ac ≤ 4% ). ii. ρ min = ⎧0.13% (high-yield steel) ⎨ ⎩0.24% (mild steel) 2 In Clause 3.12.5.3: As /Ac ≥ 0.13% (for f y = 460 N/mm ) , 2 and As /Ac ≥ 0.24% (for f y = 250 N/mm ) . 46 Example 2.3-1 Design of a singly reinforced rectangular section The ultimate design moment to be resisted by the section in the following figure is 185 kNm. Determine the area of tension reinforcement (As) required, given that characteristic material strengths are fy = 460 N/mm2 and fcu = 30 N/mm2. K= M bd 2 f cu 185 × 10 6 = 0.122 260 × 440 2 × 30 < 0.156 = Therefore compression steel is not required. Design example – singly reinforced section Lever arm: ⎡ K⎤ z = d ⎢0.5 + 0.25 − ⎥ 0.9 ⎦ ⎣ ⎡ 0.122 ⎤ = 440⎢0.5 + 0.25 − ⎥ 0.9 ⎦ ⎣ = 369 mm < 0.95d = 418 mm. O.K. (Or alternatively, the value of z = la d may be obtained form the lever-arm diagram, Fig. 2.3-2.) The area of tension reinforcement: As = M 0.95 f y z 185 ×106 0.95 × 460 × 369 = 1147 mm 2 = Provide 4T20, As = 1256 mm2 (from Table 2.3-1) 47 Example 2.3-2 A simply supported rectangular beam of 8 m span carries a uniformly distributed dead load including an allowance for self-weight of 7kN/m and an imposed load of 5 kN/m. The beam dimensions are breadth b = 250 mm and effective depth d = 400 mm. Use grade 30 concrete and high-yield steel reinforcement. i) Design load = (1.4×7) + (1.6×5) = 17.8 kN/m Design moment = 17.8 × 82/8 = 142.4 kN-m ii) K = 142.4 x 10 6 M = = 0.119 bd 2 f cu 250 x 400 2 x 30 iii) Lever arm z = 400 [0.5 + (0.25 - 0.119/0.9)1/2 ] = 337 mm < 0.95d = 380 mm. O.K. iv) Steel area As = M 0.95 f y z 142.4 × 10 6 0.95 × 460 × 337 = 967 mm 2 = Provide 2T25, As = 981 mm2 v) Reinforcement ratio ρ= = As bd 981 = 0.98% 250 × 400 ρmin = 0.13% < 0.98% < ρmax = 4%. O.K. 48 (3) Design charts ο Design charts for singly reinforced beams in BS 8110: Part 3. ο Design procedure: First, calculate M/(bd2); then directly determine 100As /(bd) from the charts with the given value of fcu . 49 ο In Example 2.3-1, 185 ×10 6 M = 3.67 = bd 2 260 × 440 2 Using the Chart No. 2, 100 As = 1.07 bd ∴ As = 1.07 / 100 × 260 × 440 = 1220 mm 2 (compare to the result As = 1147 mm2) ο In Example 2.3-2, M 142.4 × 10 6 = 3.56 = 250 × 40 2 bd 2 Using the Chart No. 2, 100 As = 1.05 bd ∴ As = 1.05 / 100 × 250 × 400 = 1 050 mm 2 (compare to the result As = 967 mm2) 50 2.3.2 Checking (Section analysis) • Aim at determining the moment of resistance of a given section with a known area of steel reinforcement. • Moment of resistance The tension reinforcement is assumed to have yielded; then z = la d εs Q Fcc = Fst or Fs t 0.45fcu bs = 0.95fy As ∴ The depth of stress block s= and 0.95 f y As (2.24) 0.45 f cu b x = s / 0 .9 (2.25) ∴ The moment of resistance of the section is M R = Fst ⋅ z = 0.95 f y As (d − s / 2) or M R = 0.95 f y As (d − 0.95 f y As 0.9 f cu b ) (2.26) Question: If the tension steel has not yielded, how to determine the moment of resistance of the section? 51 Example 2.3-3 Analysis of singly reinforced rectangular section in bending Determine the ultimate moment of resistance of the cross-section shown in the following figure, given that the characteristic strengths are fy = 460 N/mm2 for the reinforcement and fcu = 30 N/mm2 for the concrete. Analysis example – singly reinforced section Fst For equilibrium of the compressive and tensile forces on the section, Fcc = Fst therefore 0.45fcubs = 0.95fyAs i.e. 0.45×30×300×s = 0.95×460×1470 Then s = 159 mm and x = s / 0.9 = 159/0.9 = 177 mm Moment of resistance of the given cross-section is M = Fst × z = 0.95 f y As (d − s / 2) = 0.95 × 460 × 1470(520 − 145 / 2) × 10 −6 = 287 kNm 52 ...
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