Assignment-6.CIVL232.Spring_2006.solution

Assignment-6.CIVL232.Spring_2006.solution - CIVL 232 Design...

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Unformatted text preview: CIVL 232 Design of Structural Concrete Spring 2006 Solutions of Assignment 6 Question 6.1 (a) Design for the secondary beam S1-S2-S3-S4-S5-S6 The floor plan is shown as follows: Loadi ng t r ansf er t o secondar y beam S1- S2- S3- S4- S5- S6 Loads transferred to the secondary beam S1-S2-S3-S4-S5-S6 S1 S2 S3 S4 S5 S6 For each span, the ultimate load = (1.4 × 8 + 1.6 × 5 ) × 2.5 = 48 kN/m Total ultimate load on a span F = 48 × 6 = 288 kN As the loading is uniformly distributed, Qk < Gk and the spans are all equal, the moment and shear can be estimated by Bending moments i) 1st and 5th spans M = 0.09 FL = 0.09 × 288 × 6 = 155.5 kNm Assume the beam section bh = 300×450 mm. Then the effective depth d = 450 – 50 = 400 mm (bd = 300×400 mm) K= 155.5 × 106 M = = 0.108 bd 2 f cu 300 × 4002 × 30 z = d [0.5 + 0.25 − K / 0.9 ] ( ) = 400 × 0.5 + 0.25 − 0.108 / 0.9 = 344 mm As = 155.5 × 106 M = 1034 mm2 = 0.95 f y z 0.95 × 460 × 344 Provide 4T20 (As = 1256 mm2) ii) 2nd, 3rd and 4th spans M = 0.07 FL = 0.07 × 288 × 6 = 121 kNm K= 121 × 106 M = 0.084 = bd 2 f cu 300 × 4002 × 30 ( ) z = 400 × 0.5 + 0.25 − 0.084 / 0.9 = 358 mm As = M 121 × 106 = = 773 mm2 0.95 f y z 0.95 × 460 × 358 Provide 3T20 (As = 942mm2) iii) 1st and 4th interior supports M = 0.11FL = 0.11× 288 × 6 = 190.1 kNm K= M 190.1 × 106 = = 0.132 bd 2 f cu 300 × 4002 × 30 ( ) z = 400 × 0.5 + 0.25 − 0.132 / 0.9 = 329 mm As = M 190.1×106 = = 1322 mm 2 0.95 f y z 0.95 × 460 × 329 Provide 3T25 (As = 1472mm2) iv) 2nd and 3rd interior supports M = 0.08 FL = 0.08 × 288 × 6 = 138.3 kNm K= 138.3 × 106 M = = 0.096 bd 2 f cu 300 × 400 2 × 30 ( ) z = 400 × 0.5 + 0.25 − 0.096 / 0.9 = 351 mm As = M 138.3 ×106 = = 902 mm 2 0.95 f y z 0.95 × 460 × 351 Provide 3T20 (As = 942mm2) 2 Shear i) Check the maximum shear stress Vmax = 0.6 F = 0.6 × 288 = 172.8 kN vmax = Vmax 172.8 × 103 = = 1.44 N/mm 2 < 0.8 f cu = 4.38 N/mm 2 bd 300 × 400 OK ii) End supports V = 0.45 F = 0.45 × 288 = 129.6 kN v= V 129.6 ×103 = = 1.08 N/mm 2 bd 300 × 400 0.79 × 3 vc = 0.79 × 3 = 100 As 4 400 × bd d × 3 30 1.25 25 100 × 1256 4 400 × 300 × 400 400 × 3 30 = 0.68 N/mm 2 1.25 25 v − vc = 1.08 − 0.68 = 0.4 N/mm 2 Nominal link is needed, ⎛ Asv ⎞ 0.4b 0.4 × 300 ⎜ ⎜ s ⎟ = 0.95 f = 0.95 × 250 = 0.51 ⎟ yv ⎝ v ⎠ min Provide R10@300 (Asv / sv = 0.52) iii) Interior supports of 1st and 5th spans V = 0.6 × 288 = 172.8 kN v= V 172.8 ×103 = = 1.44 N/mm 2 bd 300 × 400 0.79 × 3 vc = 100 × 1472 4 400 × 300 × 400 400 × 3 30 = 0.72 N/mm 2 1.25 25 Asv 300 × (1.44 − 0.72) = = 0.91 sv 0.95 × 250 Provide R10@150 (Asv / sv = 1.047) iv) Other interior supports of 2nd, 3rd and 4th spans V = 0.55 × 288 = 158.4 kN 3 v= V 158.4 ×103 = = 1.32 N/mm 2 bd 300 × 400 0.79 × 3 vc = 100 × 943 4 400 × 300 × 400 400 × 3 30 = 0.62 N/mm 2 1.25 25 Asv 300 × (1.32 − 0.62) = = 0.88 sv 0.95 × 250 Provide R10@180 (Asv / sv = 0.924) v) Shear resistance of nominal links plus concrete ⎛A ⎞ Vn = ⎜ sv 0.95 f yv + bvc ⎟d ⎜s ⎟ ⎝v ⎠ = (0.51× 0.95 × 250 + 300 × 0.68) × 400 = 131.1 kN In the 1st span, the distance from the first support where shear reinforcement required other than the nominal links is s= Vmax − Vn ⎛ 172.8 − 131.1 ⎞ 3 =⎜ ⎟ × 10 = 869 mm wu 48 ⎝ ⎠ Sketch of reinforcement i) Flexural reinforcement 3T25 4T20 3T20 3T20 C L 3T20 (Hang-up bars: 2T16) Fig. 1 Longitudinal reinforcement ii) Shear reinforcement R10@300 C L (800) R10@180 R10@180 R10@150 Fig. 2 Transverse reinforcement 4 (b) Design for slabs i) Design loads and bending moments in slabs Consider a 1-m wide strip. Design load = 1.4Gk + 1.6Qk = 1.4 × 8 + 1.6 × 5 = 19.2 kN/m All the support moments calculated below will have included an allowance for 20% reduction due to the effects of moment redistribution, i.e. βb = 0.8. • End span and first interior support Total design load of the span = 19.2 × 2.725 = 52.3 kN Consider end supports are simply-supported. Bending moments near middle of the span and at first interior support = 0.086 × 52.3 × 2.725 = 12.3 kNm • Interior span and interior support Total design load = 19.2 × 2.5 = 48 kN Bending moments at middle of the span and at support = 0.063 × 48 × 2.5 = 7.56 kNm ii) Design of moment steel d = 120 − 30 = 90 mm • End span and 1st interior support K' = 0.132 (see Table 2.4-1, page 75 of the Notes) K= 12.3 × 106 M = = 0.051 bd 2 f cu 1000 × 90 2 × 30 ( ) z = 90 × 0.5 + 0.25 − 0.051 / 0.9 = 85 mm ≈ 0.95d = 86 mm As = OK M 12.3 × 106 = = 331 mm2 0.95 f y z 0.95 × 460 × 85 Provide T8@150 (As = 335 mm2) 5 • Interior span and interior support z = 0.95d = 0.95 × 90 = 86 mm As = M 7.56 × 106 = = 201 mm 2 0.95 f y z 0.95 × 460 × 86 Provide T8@250 (As = 201 mm2) iii) Distribution steel As,min = 0.13 × 1000 × 120 / 100 = 156 mm2 Q Spacing ≤ 3d = 270 mm Provide T8@270 (As = 186 mm2 > As,min = 156 mm2) iv) Outer support As = 0.5 × 331 = 166 mm2 Provide T8@270 (As = 186 mm2) T8-270 T8-270 T8-150 T8-150 T8-250 T8-250 T8-250 6 Question 6.2 Total design load = 1.4 × 6.0 + 1.6 × 2.5 = 12.4 kN/m2 = 12.4kN/m 2 Assumed effective depth d = 150 – 30 120 mm (a) Slab No. 1 (two adjacent edges discontinuous) i) Bending moment coefficients l y / l x = 6.0 / 4.5 = 1.33 X Short span 1 2 βsx___ 0.071 0.052 Long span 3 2 βsy___ 0.045 0.034 Y Y ii) Main moment reinforcement Short span • Support 1 msx = 0.071×12.4 × 4.52 = 17.83 kNm/m m sx 17.83 × 10 6 = = 1.24 bd 2 1000 × 120 2 100 As bd = 0.35 As = 0.35 × 1000 × 120 100 = 420 mm 2 / m Provide T8@120 (419 mm2/m) • Span 2 msx = 0.052 × 12.4 × 4.52 = 13.06 kNm/m m sx 13.06 × 10 6 = = 0.91 bd 2 1000 × 120 2 100 As / bd = 0.26 As = 0.26 ×1000 ×120 /100 = 312 mm 2 / m Provide T8@150 (335 mm2/m) 7 Long span • Support 3 msy = 0.045 × 12.4 × 4.52 = 11.30 kNm/m m sy bd 2 = 11.30 × 10 6 = 0.79 1000 × 120 2 100 As bd = 0.23 As = 0.23 ×1000 × 120 100 = 276 mm 2 / m Provide T8@150 (335 mm2/m) • Span 2 msy = 0.034 × 12.4 × 4.52 = 8.54 kNm/m 8.54 × 10 6 = 0.59 bd 2 1000 × 120 2 100 As / bd = 0.17 m sy = As = 0.17 ×1000 ×120 /100 = 204 mm 2 /m Provide T8@250 (201 mm2/m) iii) Edge strip reinforcement ρ min = 0.13% As ,min = 0.13 × 1000 × 120 100 = 156 mm 2 / m Provide T8@300 (168 mm 2 m ) iv) Negative reinforcement at discontinuous edges • Support 5 (in short span) As = 0.5 × 312 = 156 mm 2 / m Provide T8@300 (168 mm 2 / m ) • Support 4 (in long span) As = 0.5 × 204 < As ,min = 156 mm 2 / m Provide T8@300 (168 mm 2 / m ) v) Torsion steel Length of the bars lx /5 = 4.5/5 = 0.9 m Corner X: As , X = 312 × 3 / 4 = 234 mm 2 / m Provide T8@200 (252 mm 2 / m ) Corner Y: As ,Y = 234 × 0.5 = 117 mm 2 / m < As ,min = 156 mm 2 / m Provide T8@300 (168 mm 2 / m ) 8 2×3T8-300 vi) Sketch of reinforcement Bottom 2× 2× 43T8-120 Top 2× Top 9 (b) Slab No. 2 (one short edge discontinuous) i) The students may follow the same calculation procedure as that presented in (a) to conduct the design for this slab. ii) The amount of torsion reinforcement at the discontinuous corners = 3 Asx. 8 (c) Slab No. 3 (interior panel) i) Bending moment coefficients l y / l x = 6.0 / 4.5 = 1.33 Short span βsx___ 1,3 0.047 2 0.036 Long span βsy___ 4, 5 0.032 2 0.024 ii) Main moment reinforcement Short span • Supports 1 and 3 msx = 0.047 × 12.4 × 4.52 = 11.80 kNm/m m sx 11.80 × 10 6 = = 0.82 bd 2 1000 × 120 2 100 As bd = 0.23 As = 0.23 × 1000 × 120 100 = 276 mm 2 m Provide T8@150 (335 mm 2 m ) • Span 2 msx = 0.036 × 12.4 × 4.52 = 9.04 kNm/m m sx 9.04 × 10 6 = = 0.63 bd 2 1000 × 120 2 100 As bd = 0.18 As = 0.18 × 1000 × 120 100 = 216 mm 2 m Provide T8@200 (252 mm 2 m ) 10 Long span • Supports 4 and 5 msx = 0.032 ×12.4 × 4.52 = 8.04 kNm/m m sx 8.04 × 10 6 = 0.56 = bd 2 1000 × 120 2 100 As bd = 0.17 As = 0.17 × 1000 × 120 100 = 204 mm 2 m Provide T8@250 (201 mm 2 m ) • Span 2 msx = 0.024 ×12.4 × 4.52 = 6.03 kNm/m m sx 6.03 × 10 6 = = 0.42 bd 2 1000 × 120 2 100 As bd = 0.13 As = 0.13 ×1000 × 120 100 = 156 mm 2 m Provide T8@300 (168 mm 2 m ) iii) Edge strip reinforcement ρ min = 0.13% As ,min = 156 mm 2 m Provide T8@300 (168 mm 2 m ) 11 iv) Sketch of reinforcement Bottom Bottom Top T op 12 Question 6.3 (a) Loading and design moments Assume: waist thickness of concrete = 120 mm Top + Underside finishes = 30 mm Thus, overall thickness = 120 + 30 150 mm Effective span = 750 + 1540 + 235 = 2525 mm Slope length = i) (1540 + 235) 2 + 14402 = 2286 mm Landing slab DL = 1.4 × 0.15 × 24 = 5.04 kN m 2 LL = 1.6 × 5 = 8 kN m 2 Total load on landing slab = ( 5.04 + 8 ) × 0.75 ×1.35 = 13.02 kN ii) Stair slab Average thickness = 150 + 152 2 = 226 mm DL = 1.4 × 0.226 × 2.286 × 1.35 × 24 = 23.43 kN LL = 1.6 × 1.775 × 1.35 × 5 = 19.17 kN Total load on stair slab = 23.43 + 19.17 = 42.6 kN iii) Total load on span = 13.02 + 42.6 = 55.62 kN iv) The design moment for sagging moment of mid-span and hogging moment over the span = 55.62 × 2.525 10 = 14.04 kNm (b) Moment reinforcement d = h − 30 = 120 − 30 = 90 mm b = 1500 mm M 14.04 × 106 = = 1.16 bd 2 1500 × 902 100 As bd = 0.3 As = 0.3 ×1500 × 90 100 = 405 mm 2 Provide 9T8 (As = 452 mm 2 ) Distribution steel: 0.13 ×1000 × 90 100 = 117 mm 2 13 Provide T8@300 (167 mm 2 m ) (c) Reinforcement in slab 9T8-180 15T8-300 for the landing and stair slabs 9T8-180 9T8-180 14 ...
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