Calculating The Weight of Iron

Practical formulation to calculate the weight of iron
In general formulation to calculate the weight of iron is :
Vb x Bjb = … .. Kg
where: Vb = volume of iron (m3)
Bjb = density of iron = 7850 (kg/m3)
Example:
1. Iron plate with a size (1m x 1m) with 1 mm thick plate, calculate the weight?
weight of iron = (1 x 1 x 0001) m3 x 7850 kg/m3 = 7.85 kg
(Note: 1 mm = 0.001 m)
2. Base plate with size (25 cm x 30cm) with 12 mm thick plate, calculate the weight?
heavy base plate = (0.25 x 0:30 x 0012) m3 x 7850 kg/m3 = 7065 kg
Up here is quite easy to understand right? … .. Well now how its formulation to calculate the weight of steel reinforcement for concrete?.
The answer:
Same way there is no difference, the point is the volume multiplied by the weight of iron.
Example:

1. Calculate the weight of steel reinforcement diameter 16 with a length of 12 meters?
Ø16-sectional area = 1 / 4 (π) d2 = 1 / 4 (3.14) (0,016) 2 = 0.00020096 m2
Ø16 volume = trunk cross-sectional area x length = 0.00020096 m 2 x 12 m = 0.002411 m3
iron weight Ø16 x 7850 kg/m3 = Volume m3 x 0.002411 = 7850 kg/m3 = 18.93 kg
easy enough right?, from the way I have described above, there is again a faster way to calculate the weight of steel reinforcement, namely by using the formulation:
Heavy iron bars = 0.006165 x d2 x L … (Kg)
where: d = diameter of reinforcement (mm)
L = length of reinforcing rod (m)
Example:

2. Calculate the weight of iron from the sample questions # 1, with the formulation of the above?
heavy iron Ø16 = 0.006165 x 162 x 12 = 18.93 kg
same right result, .. please you count yourself by trial and error is another measure of iron reinforcement, and I make sure that the two ways above will produce the same results, see for yourself … brow, God willing, must be the same.
Well … now the question is “where the origin of the formulation above number 0.006165?”.
The following are penjabarannya:
As I’ve described above, the formula to find the weight of iron is: Vb x Bjb
where Vb = Volume of iron and iron Bjb = density = 7850 kg/m3
So the heavy iron bars (round cross-section):
= Vb x 7850 kg/m3
= (1 / 4 x π x d2 x L) x 7850 kg/m3
= 1 / 4 x 3.1415 x d2 x L x 7850 kg/m3
because d = diameter of reinforcement specified in millimeters (mm), then we had the conversion to meters (m),
d2 = (d x d) … … … … … … … …. … … mm2
converted to meters (1 mm = 0.001 m)
= (X 0.001d 0.001d)
= (1x 10-6) d2 … … … … … … … m2
Thus,
= 1 / 4 x 3.1415 x (1x 10-6) d2 x L x 7850
D2 = 0.006165 x L
So the formulation to calculate the weight of iron is d2 = 0.006165 x L