Assignment-7.solution.Un-printable

Assignment-7.solution.Un-printable - CIVL 232 Design of...

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Unformatted text preview: CIVL 232 Design of Structural Concrete Spring 2006 Solutions of Assignment 7 Question 7.1 (a) The ultimate axial load N = 1.4Gk + 1.6Qk = 1.4×1000 + 1.6×1000 = 3000 kN Q N = 0.4fcubh + Asc (0.8fy – 0.4fcu) Thus 3000×103 = 0.4×40×350×350 + Asc (0.8×460 – 0.4×40) Asc = 2955 mm2 Provide 4T32, Asc = 3216 mm2. (b) Check reinforcement ratio 3216 Asc = bh 350 × 350 = 2.6% ρ= 0.4% < 2.6% < 6% OK (c) Links Minimum diameter = ¼ × 32 = 8 mm Maximum spacing = 12 × 32 = 384 mm Provide R10@300 links. 350 50 46 The concrete cover = 25 mm for mild exposure. 4T32 R10-300 R8-350 1 Question 7.2 (a) Slenderness = l 5000 = = 14.3 < 15 . h 350 Thus, the column is a short braced column due to the approximately symmetrical arrangements of beams. (b) The ultimate axial load N = 1.4Gk + 1.6Qk = 1.4×1000 + 1.6×1000 = 3000 kN As the approximately symmetrical arrangement of beams, N = 0.35fcubh + Asc (0.7fy – 0.35fcu) Thus, 3000×103 = 0.35×40×350×350 + Asc (0.7×460 – 0.35×40) Asc = 4172 mm2 Provide 4T32 + 2T25, Asc = 4200 mm2. (c) Check reinforcement ratio ρ= Asc 4200 = = 3.4% < 6% bh 350 × 350 OK (d) Provide T8@300 links 2T32 6T32 2T25 T8-300 R8-350 2T32 2 Question 7.3 (a) Assume 25 mm diameter bars for the main reinforcing bars and 8 mm diameter links. The cover on the links is 25 mm. d = 300 − 25 − 8 − 12.5 = 255 mm d / h = 255 / 300 = 0.85 Using the chart in BS 8110, where d / h = 0.85 , N 1500 × 103 = = 16.7 bh 3002 M 70 × 106 = = 2.6 bh 2 3003 From the design chart, 100 Asc / bh = 2.5 . Then Asc = 2.5 × 3002 /100 = 2250 mm 2 Provide 4T25 + 2T20 (Asc = 2588 mm2) Links 1 × 25 = 6.25 mm 4 Maximum spacing = 12 × 25 = 300 mm Minimum size = Provide R10@300 links. 2T25 6T32 2T20 R8-350 R10-300 2T25 3 (b) Mx 50 = = 0.2 h′ ( 300 − 50 ) My 35 = = 0.14 b′ ( 300 − 50 ) ∴ Mx My > h′ b′ Therefore the increased single axial design moment about x-x axis is ′ Mx = Mx + β h′ My b′ N 1500 × 103 = = 0.56 bhf cu 300 × 300 × 30 From Table 7.5-1 of the Notes, β = 0.35 . M x′ = 50 + 0.35 × 250 × 35 = 62.3 kNm 250 N 1500 × 103 = = 16.7 bh 3002 M 62.3 × 106 = = 2.3 bh 2 300 × 3002 From the design chart, 100 Asc / bh = 2.3 . Then Asc = 2.5 × 3002 /100 = 2070 mm 2 Provide 4T25 + 2T20 (Asc = 2588 mm2) Provide R10@300 links The reinforcement detailing is the same as that in Question 7.3(a). 4 ...
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