BAR BENDING SCHEDULE
BAR BENDING SCHEDULE
BAR BENDING SCHEDULE OF BEAM (5.1 x 0.3)
Main steel
bar 16 mm ^ Main bent up bar 16 mm ^ Anchor bar 12 mm ^ Stirrups 8 mm ^
1. Length of main
steel bar 16mm ^ No. of bar = 2
Length of bar , L = Clear span + bearing
= 4.8 + 0.3 = 5.1 m
Total length of bar
= L + 2 hooks (2x9 D) = 5.1 + (18 x 0.016) = 5.088 m
2. Main
bent up bar 16 mm ^
No. of bar = 1
Length
of bar = L = Clear span + bearing = 5.1 m
Total length of bar = L + 2 hooks + 0.84 for
two- bent up = 5.1 + (18x 0.016) + (0.84x 0.10) = 5.472 m
3. Anchor
bar 12 mm ^
No. of bar = 2 Length of
bar = L = 5.1m Total length of bar = L + 2 hooks
= 5.1+ (2(9) x 0.012) =
5.316 m
4.
Stirrups 8 mm ^ @ 180 c/c
L = 5.1
m
No. of stirrups at 5 cm
c/c at end = 2 No. No. of stirrups at 18 cm c/c in b/w = 510-10/18 = 28
Total No. of stirrups = 2 + 28= 30
Length of one stirrups = 2(44+ 22) + 30 = 162 cm
= 1.62 m
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Schedule of Bars (R.C.C. Beam).
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Main steel bar 12 mm ^
Anchor bar 12 mm ^ Stirrups 8 mm ^
1.
Length of main steel bar 12mm ^
No. of
bar = 2
Length
of bar , L = Clear span + bearing = 2.8 + 0.3 = 3.1 m Total length of bar = L +
2 hooks (2x9 D) = 3.1 + (18 x 0.012) = 3.32 m
2.
Anchor bar 12 mm ^
No. of bar = 2 Length of
bar = L = 3.1m Total length of bar = L + 2 hooks
= 3.1+ (2(9) x 0.012) = 3.32 m
3. Stirrups 8 mm ^
@ 180 c/c L = 3.1 m
No. of stirrups at 5 cm
c/c at end = 2 No. No. of stirrups at 18 cm c/c in b/w = 300-10/18 = 16 Total
No. of stirrups = 2 + 16 = 18 Length of stirrups = 2 (27 cm + 25 cm) + 30 extra
= 134
cm = 1.34 m So total length of 16 nos. stirrups each.
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Sr. no.
|
Description
|
Shape
|
Length
|
No.
|
Total
length (m)
|
Wt./m
length
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Total
wt. (kg)
|
|
1.
|
Main steel bar12 mm
l
|
|
3.32
|
2
|
6.64
|
0.89
|
5.91
|
|
2.
|
Anchor bar 12mm |
|
|
3.32
|
2
|
6.64
|
0.89
|
5.91
|
|
3.
|
Stirrups 8 mm|
|
*
|
1.34
|
16
|
21.44
|
0.4
|
8.6
|
|
Total
20.4 kg
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1.
14 mm | bar in base footing
2.
18 mm | bar vertical
3.
6 mm | ties @ 290 mm c/c
1. 11 bar 14 mm |
C/c spacing = 1900/11 =
172 mm
Total No. of bar = 11+ 11
= 22 no. Total length of bar = 1.6 + 2 hooks
= ( 1.6 + ( 18 x 0.014) =
1.85 m
Total bar length = 22x 1.85 = 40.7 m
2. 16 mm 18 ^ vertical bar No. of bar = 4 no.
Total length of bar = 3+ 0.4 + 0.20 + 12d +0.20 = 4 m
3. 6 mm ^ tie bar @ 290
C/c
No. of tie = (300+ 40)/2+ 1 = 12.7 4(250) + 24
x 0.08 = = say 13 no.
Length = 4 C + 24
D = 1001.9 mm = 1.01
m
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Schedule of Bars (R.C.C. square column)
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1. Main bar of 10 mm ^ @ 190 mm c/c L = Length - 2 cover = 12.6 -
2x 0.04 = 12.52 m Total length of main bar = 12.52 + 2 hooks = 12.52 + ( 18x
0.010) = 12.7 m
No. of
bar = (51.1/.38) + 1 = (51100/ 380) + 1 = 135.4 no. say 136 no.
2. Bent -up bar of 10 mm ^ @ 190 mm c/c
= 12.6 + bearing - 2 cover
+ 2 bent-up + 2 hooks
= 12.6 + 0.20 - (2x 0.04)
+ ( 2 x.0.42 x 0.06) + (2 x 0.010) = 12.8 m
No. of bent-up bar = 51.1/ .38 = 134.4 say 135 no.
3.
Distribution bar 8 mm ^ @ 270 mm c/c
Length of
straight bar = 51.1 - 2 end cover + 2 hooks = 51.1 - 0.08 + ( 18 x 0.008) =
51.2 m
No.
of bars = 12.6/ .27 = 1260 / 270
= 46.6 say 47 no.
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Sr.
no.
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Description
|
Shape
|
Length
|
No.
|
Total length (m)
|
Wt./m length
|
Total
wt. (kg)
|
|
1.
|
Main bar 10 mm 4> @ 190
mm c/c
|
|
12.7
|
1 3 6
|
1727.2
|
0.62
|
1107.1
|
|
2.
|
Main bent- up bar10 mm
4> @ 190 c/c
|
|
12.8
|
1 3 5
|
1728
|
0.62
|
1072.1
|
|
3.
|
Distribution
bars 8 mm 4> @ 270
mm c/c
|
|
51.2
|
47
|
358.4
|
0.39
|
139.8
|
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Total
2318 kg = 23.18 quintal
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BAR BENDING SCHEDULE FOR LINTEL
Main straight bar 8 mm |
Anchor bar 8 mm | Stirrups 6 mm | @ 150 mm c/c
1. Main
straight bar of 8 mm |
No. of bars = 2
Length of bar = L + bearing - end cover
= 1.2 + 0.20 - 0.08 = 0.92 m
= 92 cm
Total length of bar = L +
2 hooks
= 0.92 + (18 x 0.082) = 1.06 m = 106 cm So
total length of main bar 8 mm | = 1.14 m each
2.
Anchor bar 8 mm |
No. of bars = 2
Length of bar L= 0.92 m
Total length of bar = 106 cm
3.
Stirrups 6 mm | @ 150 c/c
No.
of stirrups at 5 cm c/c at the end = 2 no.
No.
of stirrups at 8 cm in b/w = 92 - 10 ^ 8 = 8.25 say 9
Total
no. of stirrups = 2+9 = 11 no.
Length
of one stirrups = 2(22+90) + 30 extra
= 72 cm = 0.72 m Total length of 11 no.
stirrups 6 mm ^ = 0.72 each
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Sr.
no.
|
Description
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Shape
|
Length
|
No.
|
Total length (m)
|
Wt./m length
|
Total wt. (kg)
|
|
1.
|
Main st. bar 8 mm ^
|
|
1.06
|
2
|
2.12
|
0.4
|
0.85
|
|
2.
|
Anchor
bar 8 mm ^
|
|
1.06
|
2
|
2.12
|
0.4
|
0.85
|
|
3.
|
Stirrups
6 mm ^
|
|
0.72
|
11
|
7.92
|
0.4
|
3.2
|
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Total 4.9 kg
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About Author:
I am Thomas Britto here to share my experiences in the civil engineering field to all my readers.Today many students are struggling to buy books at high prices. So I decided to start a blog and share my experience and knowledge with all my readers.
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