Formula Estimate

December 23, 2017 | Author: Ramil S. Artates | Category: N/A
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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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