BBS FOR TWO WAY SLAB
Figure shows the two way slab
Required
data:
Dia
of all bars = 12mm @ 10mm c-c spacing
Length
of longer span (lx) = 6m
Length
of shorter span (ly) = 4m
1
< 1.5
Hence
it is two way slab
Take
section A:
Cutting
length of distribution bars:
Heare
two types bars there one is cranked bar and another one is straight bar
Cutting
length of cranked bar = span length + developing length + cranked bar – bends
Span
length = 4000m
developing length = 50×d
=
2×( 50d ) for two sides
Crank
length =0.42 × H
=
(0.42H) × 2 for two crank
Bend
= 2d for 90˚
Bend
= 1d for 45˚
Cutting length = 4000 + (50d)×2 + (0.42H) × 2 – (2d)
× 2 – (1d) × 4
Were H = 200 -2× clear cover – dia of bar
=
200 - 2×30 – 12
=
200 – 60 – 12
=128mm
Cutting length = 4000 + (50×12)×2 + (0.42×128) × 2 –
(2×12) × 2 – (1×12) × 4
=
4000 + 1200 + 107.52 – 48 -48
=5211.52
mm
=5.211m
Cutting
length of straight main bar:
Cutting
length = span length + developing length
=
4000 + (50d) × 2
=
4000 + (50×12) × 2
=
4000 + 1200
=
5200 mm
=
5.2 m
No
of bars =
1
=
+
1
=
60 + 1
=
61 nos
Take
31 nos as cranked bars and 30 bars as straight bar
Top or extra reinforcement:
Cutting
length = L – 2(
)+
(2×100)
Were
100 is lapping distance
Cutting
length = 4000 -2(
)+
(2×100)
=
4000 –2( 800) + 200
=
4000 – 1600 + 200
=
2600 mm
=
2.600m
No
of bars =
1
Length
=
=
=
1200
No
of bars =
=
12
= 2×12
= 24 nos
Take
section B:
Cutting
length of distribution bars:
Hear two types bars
there one is cranked bar and another one is straight bar
Cutting
length of cranked bar = span length + developing length + cranked bar – bends
Span
length = 6000m
Developing
length = 50×d
=
2×( 50d ) for two sides
Crank
length =0.42 × H
=
(0.42H) × 2 for two crank
Bend
= 2d for 90˚
Bend = 1d for 45˚
Cutting
length = 6000 + (50d)×2 + (0.42H) × 2 – (2d) × 2 – (1d) × 4
Were H = 200 -2× clear cover – dia of bar
=
200 - 2×30 – 12
=
200 – 60 – 12
=128mm
Cutting length = 6000 + (50×12)×2 + (0.42×128) × 2 –
(2×12) × 2 – (1×12) × 4
= 6000 + 1200 + 107.52 – 48 -48
=7211.52 mm
=7.21m
Cutting
length of straight main bar:
Cutting
length = span length + developing length
=
6000 + (50d) × 2
=
6000 + (50×12) × 2
=
6000 + 1200
=
7200 mm
=
7.2 m
No
of bars =
1
=
+
1
=
40 + 1
=
41 nos
Take
21 nos as cranked bars and 20 bars as straight bar
Top or extra reinforcement:
Cutting
length = L – 2(
)+
(2×100)
Were
100 is lapping distance
Cutting
length = 6000 -2(
)+
(2×100)
=
6000 –2( 1200) + 200
=
6000 – 22400 + 200
=
3800 mm
=
3.800m
No
of bars =
1
Length
=
=
=
800
No
of bars =
=8+1
= 9
=2×9
= 18 nos
SI.NO
|
DISCRIPTION
|
DIA OF BAR
|
CUTTING LENGTH
|
NO.OF BARS
|
TOTAL LENGTH
|
WEIGHT
|
A
|
MAIN BAR
|
|||||
1
|
CRSNKED BAR
|
12MM
|
5.211M
|
31 NOS
|
161.541M
|
142.156KG
|
2
|
STRAIGHT BAR
|
12MM
|
5.2M
|
30 NOS
|
156M
|
137.28KG
|
3
|
TOP OR EXREA BAR
|
12MM
|
2.6M
|
24NOS
|
62M
|
54.912KG
|
B
|
DISRTIBUTION BAR
|
|||||
1
|
CRSNKED BAR
|
12MM
|
7.211M
|
21 NOS
|
157.431M
|
133.259KG
|
2
|
STRAIGHT
BAR
|
12MM
|
7.2M
|
20 NOS
|
144M
|
26.72KG
|
3
|
TOP OR EXREA BAR
|
12MM
|
3.8M
|
18 NOS
|
68.4M
|
60.192KG
|
TOTAL
|
654.519KG
|
|||||
ADDING OF 5% WASTAGE
= 0.05×654.519
= 32.725 KG
TOTAL WEIGHT OF STEEL = 654.519 + 32.72
= 687.544 KG
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