Unformatted text preview: CIVL 232 Design of Structural Concrete
Spring 2006
Solutions of Assignment 6
Question 6.1
(a) Design for the secondary beam S1S2S3S4S5S6
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 S1S2S3S4S5S6 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 [email protected] (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 [email protected] (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 [email protected] (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 (Hangup bars: 2T16)
Fig. 1 Longitudinal reinforcement
ii) Shear reinforcement [email protected] C
L (800) [email protected] [email protected] [email protected] Fig. 2 Transverse reinforcement 4 (b) Design for slabs
i) Design loads and bending moments in slabs
Consider a 1m 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 simplysupported. 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 2.41, 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 [email protected] (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 [email protected] (As = 201 mm2)
iii) Distribution steel
As,min = 0.13 × 1000 × 120 / 100 = 156 mm2 Q Spacing ≤ 3d = 270 mm
Provide [email protected] (As = 186 mm2 > As,min = 156 mm2)
iv) Outer support
As = 0.5 × 331 = 166 mm2
Provide [email protected] (As = 186 mm2) T8270 T8270 T8150 T8150 T8250 T8250 T8250 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 [email protected] (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 [email protected] (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 [email protected] (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 [email protected] (201 mm2/m)
iii) Edge strip reinforcement ρ min = 0.13%
As ,min = 0.13 × 1000 × 120 100 = 156 mm 2 / m
Provide [email protected] (168 mm 2 m )
iv) Negative reinforcement at discontinuous edges • Support 5 (in short span) As = 0.5 × 312 = 156 mm 2 / m
Provide [email protected] (168 mm 2 / m ) • Support 4 (in long span) As = 0.5 × 204 < As ,min = 156 mm 2 / m
Provide [email protected] (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 [email protected] (252 mm 2 / m ) Corner Y: As ,Y = 234 × 0.5 = 117 mm 2 / m < As ,min = 156 mm 2 / m
Provide [email protected] (168 mm 2 / m ) 8 2×3T8300 vi) Sketch of reinforcement Bottom 2×
2× 43T8120 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 [email protected] (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 [email protected] (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 [email protected] (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 [email protected] (168 mm 2 m ) iii) Edge strip reinforcement ρ min = 0.13%
As ,min = 156 mm 2 m
Provide [email protected] (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 midspan 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 [email protected] (167 mm 2 m )
(c) Reinforcement in slab
9T8180 15T8300
for the landing and stair slabs
9T8180
9T8180 14 ...
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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.
 Spring '06
 JSKUANG

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