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
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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.