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Formula Estimate
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RamilArtates
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DETAILED OF THE CONSTRUCTION ESTIMATE CONCRETE MATERIALS :
FOOTING 1
CLASS "A" MIXTURE FOOTING DEMINSION LENGTH:
1.2 m
WIDTH:
1m
THICK:
0.35 m
VOL. OF FOOTING
0.42
cu.m.
TOTAL VOLUME OF THE VOLUME= CEMENT:
12
FOOTING
5.04 cu.m.
5.04
x
9
=
45.36 bags
SAND:
5.04
x
0.5
=
2.52 cu.m.
GRAVEL:
5.04
x
1
=
5.04 cu.m.
REINFORCE BARS MATERIALS :
FOOTING 1
NET LENGTH : 1.2
-
0.075 +
0.075
=
1.05 m
TOTAL NO. OF CUT BARS IN ONE FOOTING BY DIRECT COUNTING 6x
2
=
12
pcs
TOTAL NO. OF BARS FOR THE 12 12
x
12
=
144
FOOTING
pcs
@
1.05
m
SELECT THE STEEL BARS WHOSE LENGTH IS ECONOMICALLY CUT INTO 6
/
1.05
=
5.71429 pcs
Divide the result: 144
=
5
29
pcs of 16mm diameter bars
TIE WIRE NUMBER OF THE BAR INTERSECTION IN ONE FOOTING 6
x
TOTAL TIES FOR
12 12
Using
6
x
0.25 m
=
36
ties
432
ties
FOOTINGS 36
=
length per tie =
432
One kilo of no. 16 G.I. wire is approximately 108
/
30 =
3.6
30 m kilos
x
0.250
=
108
DETAILED OF THE CONSTRUCTION ESTIMATE
1.05
M LONG
meters
CONCRETE MATERIALS :
FOOTING 2
CLASS "A" MIXTURE FOOTING DEMINSION LENGTH:
1m
WIDTH:
1m
THICK:
0.35 m
VOL. OF FOOTING
0.35 cu.m.
TOTAL VOLUME OF THE
4
VOLUME= CEMENT:
FOOTING
1.4 cu.m.
1.4
x
9
=
12.6 bags
SAND:
1.4
x
0.5
=
0.7 cu.m.
GRAVEL:
1.4
x
1
=
1.4 cu.m.
REINFORCE BARS MATERIALS :
FOOTING 2
NET LENGTH : 1
-
0.075 +
0.075
=
0.85 m
TOTAL NO. OF CUT BARS IN ONE FOOTING BY DIRECT COUNTING 5x
2
=
10
pcs
TOTAL NO. OF BARS FOR THE 10
x
4
=
40
4
FOOTING
pcs
@
0.85
m
SELECT THE STEEL BARS WHOSE LENGTH IS ECONOMICALLY CUT INTO 6
/
0.85
=
0.85
7.05882 pcs
Divide the result: 40
=
7
6
pcs of 16mm diameter bars
TIE WIRE NUMBER OF THE BAR INTERSECTION IN ONE FOOTING 5
x
TOTAL TIES FOR
4
4 Using
5
x
0.25 m
=
25
ties
100
ties
FOOTINGS 25
=
length per tie =
100
One kilo of no. 16 G.I. wire is approximately 25
/
30 =
0.8
30 m kilos
x
0.250
=
25
meters
M
meters
LONG
COLUMN 1 TIES :
HEIGHT FROM GROUND LINE TO BEAM HT. =
6.6
-
Mtrs.
@
0.2 m. spacing
Divide: 6.6 0.2
=
44 pcs
- Total ties of the columns=
12
- Multiply: 12 x 44
=
528
- Length of one lateral ties 1.45 m. long cut from a 6.00 m. steel bars 6 = 1.45
w= 0.275
5
pcs
- Result d= 0.45
528 5
10 pcs tiesl bars
= 104 pcs- 10 mm ties bars
- Tie wire no. of ties in a column 6 x 44 =
264
total wire for the 264 x 12 =
12 3168
multiply: 3168 x
One kilo of no. 16 G.I. wire is approximately 792
/
35 =
22.6
35 m kilos
0.25 =
792
EAM
m. spacing
pcs
6.00 m. steel bars
10 mm ties bars
columns pcs
m
w= 0.275
12
A= d=
0.45
7.5
6 bars
B= TOTAL =
7.6
0.200 mtrs. Length
Result: no. of vertical bars Total of vertical bars =
8
pcs of one column 99.94 pcs
16mm vertical bars
column post
rtical bars
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