Different BWG Sheet Unit Weight


 

 

SOME IMPORTANT INFORMATION ABOUT ESTIMATING.

1.Ms Flat bar weight determine equation=0.00785A
2.Brick require for 1 square meter one brick flat soling=32nos.
3.Brick require for 1 squre meter one layer herring bone bond =52nos.
4.Cement require for 9.3 squre meter neat cement finishing=1/2bag or 25kg
5.A head mason can work 25-30 cum brick work in a day.

 

HOW MANY BRICK , CEMENT,SAND REQUIRE FOR 150 CUM CC WORK.

How many brick , cement,sand require for 150 cum cc work? Ratio 1:3:6.
Answer:
1+3+6=10,
Dry volume=150x1.54=231, 1.Brick needed=(231x6x300)/10 = 41580 nos, 2.Sand needed=(231x3)/10 =69.30 cum,
3.Cement needed=(231x1)/(10x0.0347) = 665.71bags

 

HOW MUCH CONSTRUCTION MATERIAL REQUIRE FOR 10 CUM BRICK WORK.

How much construction material require for 10 cum brick work?Ratio 1:6

Answer:

Dry morter=10x35%=3.5cum, 1+6=7, Brick needed=10/(0.254x0.127x0.076)=4079 nos, Cement needed=3.5(1/7)=0.50cum=0.50(1440/50)=14.4bags. Sand needed=(3.5x6)/7=3.00cum

 

HOW MUCH CONSTRUCTION MATERIALS REQUIRE FOR 12 MM THICK OF 100 SQM CEMENT PLASTERING WORK.

How much construction materials require for 12 mm thick of 100 sqm cement plastering work?Ratio 1:6

Answer:

Dry volume={(100x12x1.5)/1000}=1.8,
Cement needed={(1.8x1x30)/7}=7.71,Nos
Sand needed={(1.8×6)/7}=1.54 cum

 

A SOAK WELL 25 CM THICK WHOSE INNER DIA IS 150CM AND DEPTH IS 600 CM.HOW MUCH CUM BRICK WORK REQUIRE FOR THIS SOAK WELL CONSTRUCTION.

A soak well 25 cm thick whose inner dia is 150cm and depth is 600 cm.How much cum brick work require for this soak well construction?

Answer:
inner length=[{3.14(inner dia+outer dia)}/2] =[{3.14x(150+200)}/2]=549.5cm=5.495m
brick work needed=5.495x0.25×6=8.24cum.

 

25CM THICK 1.5M INNER DIA AND 6M DEPTH OF A SOAK WELL.HOW MUCH CONSTRUCTION MATERIALS REQUIRE.

25cm thick 1.5m inner dia and 6m depth of a soak well.How much construction materials require?Ratio 1:6

Answer:
Brick work needed=1[3.14x{(1.5+2)/2}]x0.25x6=8.24 cum,
1+6=7,
Dry morter=8.24x35%=2.884cum, Brick needed=8.24/(0.254x0.127x0.076)=3361 nos, sand needed=2.884(6/7)=2.47cum,
Cement needed=(2.8841)/(7x0.0347)=11.87bags.

 

HOW TO DETERMINE HOW MANY ROD REQUIRES FOR A 10’6″ X 12’6″ FOOTING.

How to determine how many Rod requires for a 10’6″ x 12’6″ footing?Rod spacing 3inch in both side.

Answer:
Rod required=10x12/spacing=120/3=40 nos, Rod required=12x12/spacing=144/3=48 nos.
(3″ covering deduction from both side)

 

HOW MANY ROD REQUIRE FOR 1000 SFT AND 5″ THICK SLAB CASTING.


How many Rod require for 1000 sft and 5″ thick Slab Casting?

Answer:
Total Concrete’s 1% or 1.5% or 2% of Rod should consider in slab casting.
We may consider here 1.5%
Formula is
{(Area x Thickness) } X {(1.5%) X Unit weight of Rod}
So,
[{(1000) X (5/12)}] X {(1.5/100) X 222}
[ note:Unit weight of Rod is 222kg/cft]
=(416.66) X (3.33)
=1387.48kg

 

CI SHEET ESTIMATE

 

ABOUT BEAM.







 

HOW TO DETERMINE CRANK LENGTH?



 

DEFINITION OF A CANTILEVER BEAM.


It is fixed in a wall or column at one end and the other end is free, it is called cantilever beam. It has tension zone in the top side and compression zone in the bottom side.Its maximum bending moment occurs at the support and zero moment occurs at the free end.No Rod is require at bottom.At cantilever beam rod requires only at top.


 

FM OF SAND.

Local sand FM limit should be 0.5 to 0.8

Toak sand FM limit should be 1.2 to 1.8

Coarse sand or shylhet sand FM should be 2.5

At improved sub grade 0.5 to 0.8 FM sand should be used.

 

FORMULA FOR LAPPING.



  1. Formula for lapping in reinforcement at tensile zone is 48d+600 mm (where d as dia).
  2. Formula for lapping in reinforcement at compression zone is 40d.
    Where d is the diameter of Rod.
    If we consider 12mm dia rod at column section for lapping,
    Then the lapping length at column section should be = 40X12 = 480mm = 0.48m = 19 inch.

 

FORMULA FOR MEASURE EXTRA TOP ROD’S LENGTH FROM SUPPORT.



 

HANGER ROD OF A BEAM.



 

HOW MANY BAG OF CEMENT SHOULD STORE VERTICALLY?

Answer:
Not More than 15 bags.

 

HOW TO DETERMINE SLAB THICKNESS?

Answer:
The Maximum Span length in inch /35
Example:
The maximum span length is 14.5′ then
The slab thickness is = (14.5×12)/35=174/35=4.97
So, you can assume the slab thickness is 5″

 

ANGLE’S UNIT WEIGHT AS PER METER.

(20mmx20mmx3mm=0.9kg) (20mmx20mmx4mm=1.1kg)
(25mmx25mmx3mm=1.1kg) (25mmx25mmx4mm=1.4kg)
(25mmx25mmx5mm=1.8kg) (30mmx30mmx3mm=1.4kg)
(30mmx30mmx4mm=1.8kg) (30mmx30mmx5mm=2.2kg)
(35mmx35mmx3mm=1.6kg) (35mmx35mmx4mm=2.1kg)
(35mmx35mmx5mm=2.6kg) (35mmx35mmx6mm=3kg)
(40mmx40mmx3mm=1.8kg) (40mmx40mmx4mm=2.4kg)
(40mmx40mmx5mm=3kg) (40mmx40mmx6mm=3.5kg)
(45mmx45mmx3mm=2.1kg) (45mmx45mmx4mm=2.7kg)
(45mmx45mmx5mm=3.4kg) (45mmx45mmx6mm=4kg)
(50mmx50mmx3mm=2.3kg) (50mmx50mmx4mm=3kg)
(50mmx50mmx5mm=3.4kg) (50mmx50mmx6mm=4.5kg)
(55mmx55mmx5mm=4.1kg) (55mmx55mmx6mm=4.9kg)
(55mmx55mmx8mm=6.4kg) (55mmx55mmx10mm=7.9kg)
(60mmx60mmx5mm=4.5kg) (60mmx60mmx6mm=5.4kg)
(60mmx60mmx8mm=7.0kg) (60mmx60mmx10mm=8.6kg)
(65mmx65mmx5mm=4.9kg) (65mmx65mmx6mm=5.8kg)
(65mmx65mmx8mm=7.7kg) (65mmx65mmx10mm=9.4kg)
(70mmx70mmx5mm=5.3kg) (70mmx70mmx6mm=6.3kg)
(70mmx70mmx8mm=8.3kg) (70mmx70mmx10mm=10.2kg)
(75mmx75mmx5mm=5.7kg) (75mmx75mmx6mm=6.8kg)
(75mmx75mmx8mm=8.9kg) (75mmx75mmx10mm=11kg).
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