Lecture_15

Lecture_15 - 3.2.4 Procedures for design of shear...

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Unformatted text preview: 3.2.4 Procedures for design of shear reinforcement (1) Calculate the design shear stress at critical sections: v = V/(bvd). (2) Check v < 0.8 f cu or 5 N/mm2, whichever is the lesser. (3) Determine the design concrete shear stress vc using Eq. (3.2) or directly from Table 3.8 of BS 8110: Part 1 (Notes p.114). (4) Determine Asv using Eq. (3.11). sv (5) Calculate the spacing of minimum links using Eq. (3.13). (6) Determine the shear resistance of concrete plus minimum links from Eq. (3.9); then plot the envelope of this shear resistance. (7) Space the web reinforcement (i.e. stirrups arrangement). Asv for varying stirrup diameter and spacing sv 123 Example 3.2-1: Design of shear reinforcement Design load = 75.2 kN/m d = 550 b = 300 2R10 2T25 300 fcu = 25 N/mm2 6000 214.3 kN 214.3 kN Vcritical-section shear force diagram (1) Maximum shear stress Shear force at the face of support 6 0.3 V = 75.2 × − 75.2 × = 214.3 kN 2 2 The maximum shear stress v max V 214.3 × 10 3 = 1.30 N mm 2 < 0.8 f cu = 4 N mm 2 = = bv d 300 × 550 OK (2) Design concrete shear stress vc 2T25: As = 982 mm 2 100 As 100 × 982 = = 0.59 bv d 300 × 550 From BS8110: Part 1, Table 3.8 (Notes p. 114) vc = 0.52 N mm 2 124 (3) Links (stirrups) The shear force of the section at a distance d from the support face Vcritical-section = 214.3 – 75.2d = 214.3 – 75.2 × 0.55 = 172.9 kN Shear stress v= Vcritical-section 172.9 × 103 = = 1.05 N mm 2 300 × 550 bv d Shear links (stirrups) required Asv bv (v − vc ) 300 × (1.05 − 0.52) = = = 0.67 0.95 f yv 0.95 × 250 sv For R10 links ( Asv = 157 mm2 ) sv = 157 = 234 mm < 0.75d = 413 mm 0.67 Provide R10 links at 230 mm centres (R10@230) (4) Minimum links ⎛ Asv ⎞ 0.4bv 0.4 × 300 = = 0.51 ⎜ ⎟= ⎝ sv ⎠ min 0.95 f yv 0.95 × 250 For R10 links sv = 157 = 308 mm < 0.75d = 413 mm 0.51 Provide R10@300 ( Asv = 157/300 = 0.52) sv (5) Shear resistance of concrete plus minimum links Vn = bv vc d + Asv 0.95 f yv d sv = 300 × 0.52 × 550 + 0.52 × 0.95 × 250 × 550 = 153.7 kN 125 (6) Envelope of the shear resistance of concrete plus minimum links 214.3 kN Required links Minimum links 153.7 kN 153.7 kN s = 806 mm The distance s = 214.3 − 153.7 = 806 mm 75.2 214.3 kN (7) Stirrup arrangement 5R@230 12R@300 5R@230 2T12 hanger 2T25 50 Note: In design practice, the enhanced shear strength of sections near supports may not be considered. The shear force at the centre line of support is often taken as the design shear force. In this example, the design shear force will be V = 75.2 × 6/2 = 225.6 kN The values of V = 214.3 kN and V = Vcritical-section = 172.9 kN are not used. 126 3.2.5 Shear resistance of bent-up bars • To resist the shearing force, bars may be bent up near the supports as shown in Fig. 3.2-3. The bent-up bars and the concrete in compression are considered to act as an analogous lattice girder. Fig. 3.2-3 Bent-bar systems • Single system − The diagonal crack only intercepts one bent-up bar. From Fig. 3.2-3(a), the spacing of the bent-up bars is sb = (d − d ′)(cot α + cot β ) (3.16) where α = angle between bent-up bar and axis of the beam β = angle between compression strut and axis of the beam The shear resistance of the single bar is Vb = 0.95 f yv Asb sin α (3.17) where Asb is the cross-section area of the bent-up bar. 127 • Multiple system − The diagonal crack intercepts more than one bent-up bar. o From Fig. 3.2-3(b), the shear resistance is given by Vb = 0.95 f yv Asb sin α where (d − d ′)(cot α + cot β ) sb (3.18) (d − d ′)(cot α + cot β ) can be considered as the number sb of bent-up bars intercepting the diagonal crack. o The truss should be arranged as follows: Both α and β are greater than or equal to 45°; and Spacing sb has a maximum value of 1.5d [referring to Eq. (3.16): When α = β = 45°, sb,max = 2(d − d'). Take sb,max = 1.5d] o With α = β = 45° (this is a common practice) and sb = d – d’, Eq. (3.18) becomes V = 1.34 f yv Asb (3.19) This arrangement is commonly referred to as a double system. This is because the number of bent-up bars crossing the diagonal crack is (d − d ′)(cot α + cot β ) (d − d ′)(cot 45° + cot 45°) = =2 sb d − d′ • BS 8110: Part1, Clause 3.4.5.6 states that at least 50% of shear resistance provided by the steel should be in the form of links. 128 Example 3.2-2 Shear resistance of a beam section with both stirrups and bent-up bars 100 As 100 x 982 = = 0.43 bd 350 x 650 From BS 8110 Table 3.8 (Notes p.114), vc = 0.5 N/mm2 by interpolation. Crosssectional area of a size 12 bar = 113 mm2. Vs = Asv /sv·0.95fyvd + bvcd = 349 + 114 = 463 kN 303 kN V = Vs + Vb = 463 + 303 = 766 kN 129 ...
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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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