Pls. let me know that how the 160 (or) 160000 is used to find the weight of s.steel blank?.
Thx.
Thx.
I saw a methode to find weight of steel blank.
This is the formula: Dia x Dia x Thick/160
How did 160 come?
Pakirmohd, what units do you work in for length and weight?
The equation for weight of a cylinder (i.e. circular blank) is:
Weight = (3.14*D*D/4)*(thickness)*(density)
The 160 probably comes from simplification of:
a) pi (3.14)
b) 4
c) the density of steel
d) any unit conversions.
Interesting. Thx.
I do not believe that the given equation can be assumed to be a valid general equation for determining tubing weights; because the substitution of D*d to replace "(D^2 - d^2)" in a the standard equation: (D^2 - d^2) * pi/4 * t (thickness of the plate) for the volume of metal in a cylinder is not valid.
The only way I can see the above equation being valid is if it is for a specific O.D. I.D. ratio tube; where, for example, d = D/2 and the 2 is one of the factors included in the 160 divisor value.
The D*D in above equation is only the equivalent of D^2 in the standard formula: D^2 * pi/4 * t for the volume of metal in a solid round bar.
I also performed trials using the standard density for a number of metals in the equation and none resulted in the 160 divisor value.
The equation I presented is only valid for a solid steel blank, see attached picture.
Pakirmohd, what units are you using for this equation? Meters? centimeters? millimeters? Inches? And for weight... Pounds? kilograms?
Thanks,
Will
Dear Will,
It's mm.
Pakirmohd, please forgive my confusion, while I realize the correct geometrical term for a solid cylindrical shape is a "cylinder", in the ********** in which I have worked this term is principally utilized to indicate a tubular item and "round bar" is the term used to identify a solid cylinder of material.
1/160 = 0.0063 as a multiplier and I cannot find a single material units of density that will convert this to the generally acceptable density value of .283 lb/ft^3 or 489 lb/ft^3 for basic steel.
In addition to Will1234's request for the units, can you also attach a copy of the original document giving this equation?
It's great answer with perfect calculation.