Formulas for Solid Round Bar

Check on your instructions. It should be D² (DxD) and not Dx2.

Your formula also requires a length.

What units do they want you to use?

Hello Hendrik

Agree about the D².

A couple of points - he doesn't say what material or give a density, but if it's steel at 7850kg/m³ his first formula (with D²) is about right if d is in mm. Second formula is rubbish. Assuming he means .....3.1428/(4x.00785) -- 0.785 = pi/4 so pi cancels and you get 100*D², which is clearly wrong.

But don't agree with comment Your formula also requires a length. No - he's trying to work out mass per meter.

Cheers.....Codey

Correction to earlier post, just spotted the second formula isn't rubbish, my mistake!

Reading it as .....(3.1428/4)x.00785, and noting that 0.00785 is the density of steel (with a few powers of 10 to get the decimal point in the answer in the right place), it agrees with the first formula. Coincidence that pi/4 = 0.785.

Apologies......Codey

You are right about the length =- I saw it and I missed it.

I also had the other one wrong at first.

I also divided by 4 x .00785.

Don't worry too much about the formulae. For any material, just calculate the volume per whatever length you want and multiply by the density (grams per cubic centimeter, etc.) of the material. Just remember that all dimensions must be in the same units.

For example, assume a 2.54 centimeter diameter solid round carbon steel bar. The weight will be 3.983 kilograms per meter. Calculated as follows:

A = ∏/4 D² = 5.067 cm² x 100 cm x 7.861 grams per cm³ (density of carbon steel) = 3.983 kg per meter.

If you replace the 3.1428 in the second formula with 3.14159, then the two formulas produce the same answer.

I agree with others that the diameter would have to be squared rather than multiplied by 2. (A 50 mm bar will weight 4 times as much, per meter, as a 25mm bar)

3.14159 = pi (approximately)

D² x pi /4 = pi r² which is the common formula for the area of a circle. In the second formula, all that remains is the .00785 factor, which accounts for the density of the material, and the units used, which I would assume to be mm for diameter. So a 50mm bar of this material would weigh 15.4 kg, which sounds about right for steel. For the formula to be used with a different material, the .00785 factor would need to be different.

Measure the diameter, and convert to centimeters. If you are measuring in millimeters, divide your measurement by 10. I you are measuring in inches, multiply your measurement by 2.54.

Divide the diameter by 2, then multiply this number by itself. For instance, if you measure the diameter as 0.5 inches, multiply this by 2.54 to equal 1.27 centimeters. Divide this by 2 and you now have a radius of 0.635 centimeters. Now, multiply this number by itself. 0.635 times 0.635 equals 0.403. Multiply this by the value of Pi. Generally you can use just 3.14, but for more accuracy, you can use 3.141592654. The area of a cross-section of your solid rod therefore is 1.267 square centimeters.

Since you are working to learn the weight of a meter of this material, multiply this by 100 (since there are 100 centimeters in a meter). Now, in this example, you know there are 126.7 cubic centimeters in a meter of half-inch stock.

You then need to know the specific gravity of different materials, and this is available in many references. Generally, you can use the following as a guideline:

Aluminum = 2.7

Brass = 8.2 to 8.7

Bronze = 8.8

Copper = 8.9

Steel = 7.6 to 7.8

Zinc = 7.1

(these are not exact, and not specific for different alloys, but they can get you close)

So, if your solid round bar is made of steel, you can multiply your cubic centimeter answer by 7.7. 126.7 times 7.7 equals 975 grams or 0.98 kilos.

As long as you pay attention to the UNITS OF MEASURE, the formula is:

Diameter divided by 2, times itself, times PI, times the length, times the specific gravity of the material.

=((D/2)^2)*3.14*L*Specific_Gravity

If I over simplified, please don't take offense. I just wanted to break it down to the the simplest form. I hope this helps.

The equation to calculate the total weight W (in kg) of a bar with diameter D (in m), length L (in m) and density Ρ (kg/m³) is:

W = Volume of bar x Density = (Π/4) D² x L x Ρ

The weight of bar per unit length (kg/m) = (Π/4) D² x 1 x Ρ (L is considered as unity, i.e. 1 m)

Note. Density of steel, Ρ = 7850 kg/m³ & Π = 3.14 or (22/7).

Dear Guest,

The formula you specified is 1).Diameter X 2 X 0.0061654 , 2) D X 2 X 3.1428/4 X 0.00785.

Both the formulas is wrong, because Diameter X 2 is suposed to be DX D

If you considered D x2 to D X D

).0061654 = (Pie/4) xD X D = area

(Pie) = 22/7 = 3.1428

Pie/4 = 3.1428 /4 = 0.7857

Your formula = D X D X (3.1428/4 X 0.00785) = D X D X 0.006167.

Both the formulas are same provided you consider the density of steel is same in both cases.

I will explain you the thumb rule of calculating the weight of 1 meter length of any metal/non metal/ tube or any section.

Assume the dia of the M.S Rod = 80 mm. you should find the weight in Kgs per one meter length.

Mass = Volume X Density.

= Volume = Area X Length

= Area = Pie/ 4 X D x D

Pie = 22/7

Pie/4 = 22/28

= Area = (22/28) X D X D

= (0.7857) X (80/1000) m x (80/1000) m =

= 0.7857 X 0.080 m X 0.080 m = 0.005028 m2

= Volume = area X Length

= 0.005028 X 1 m = 0.005028 m3

mass = Volume X density

density = 7.857 MT/m3

mass = 0.005028 m3 X 7.857 MT/m3

= 0.03950 m3 X MT/ m3

= 0.03950 MT

convert into kilograms

= 0.03950/ 1000 = 39.50 kilogram( Kg)

SIMILARLY YOU CAN CALCULATE ANY SHAPE OF METAL OR NON METAL ONLY YOU SHOULD KOW THE DENSITY OF THOSE METALS.

with best regards,

Srinivas kommineni

What I was implying in my earlier response (when I said not to worry too much about formulae) was never just accept a formula for anything at face value. Formulae are important but it is far more important to understand the formula and how it was derived. Too often folks simply memorize a formula to pass a test, handle a job assignment, etc. without really understanding the formula. However, if one understands the formula, how it is derived and what it really means, one can always reconstruct it (even if you forget the formula).