Calculating the weight of prefabricated items based on volume is an often used skill. Weighing items on a scale is plain easier but before you lift that load onto the scale, just how many people do you need for a safe lift? If slings are to be used, how strong? Which one? Determining prefabricating sizes to stay within load capacity of transport vehicles is also where these math skills are applied.
This weight calculation combines the volume of cylinder with the density of steel. Of course, other combinations are also possible such as:
 Weight of steel plate calculation uses the volume of a rectangular prism and the density of steel
 Weight of a pipe calculation: volume of a hollow cylinder and the density of PVC
 Weight of a scaffold frame: volume of all hollow cylinders and the density of aluminum
 Weight of a precast manhole cover: volume of a cylinder and density of cast iron or concrete
 Weight of a footing calculation: volume of forms and density of concrete
 Weight of a roof rafter: volume of an oblique rectangular prism and the density of wood
 Weight of a bronze spacer: volume of a cone frustum and the density of bronze
The shapes and materials in this list are found everywhere in construction and manufacturing.
Get a $20orso dualdisplay scientific calculator, even if you can`t take it with you into the exam hall. A calculator is also a learning tool, Technology Use is one of the 9 Essential Skills you can`t do without.
Check out the ADVANCED tabs. The exercises will further your understanding and help with building a bigger mental picture. It builds links between numbers and algebra. It is important to sketch the shapes, label the shapes with the given numbers, tracking changes to units of measure, visualizing to crosslink the concepts of volume, weight and density. We use the numbers from this rebar example.
A Short Review
To calculate weight, the volume of an object and the density of its material needs to be multiplied together.
W = V · ρ
The symbol for Density is the Greek letter “rho”, it looks like this: ρ It is not a lowercase “p”.
Density is often measured in kg/m^{3} or lb/in^{3}.
Density of materials needs to be found from tables, published by science communities. For example, the density of gold is 19.32 g/cm^{3}, concrete is 150 lb/ft^{3}, air is 1.29 kg/m^{3}.
Temperature further affects these density figures.
Volume can be calculated if the object is a geometric solid, but complex shapes are often submersed in a tank of water or other noncorrosive liquid to determine their volume.
Worksheet: Level 1
Worksheet: Level 1 Sample Questions
Example:
Density is a property of matter that describes how compacted matter is of a given size.
From the table below find the densities of:
 Naval Brass (lb/ft³): _____
Further examples:
What is the density of steel? Silver? Crushed rock? Find metric and Imperial figures.
Worksheet: Level 1 Answer Key
Answer:
 Naval Brass = 546.25 lb/ft³
Download Worksheet: Level 1
Worksheet: Level 2
Worksheet: Level 2 Sample Questions
Example:
Calculate the weight of the following objects. Answer in decimals.
 Calculate a steel cylinder‘s weight if its volume = 10 cm³, steel density is 7.85 g/cm³
Further example:
What is the mass of a steel exit staircase, V = 0.25 m³, density = 7850 kg/m³?
Worksheet: Level 2 Sample Answer Key
Answer:
Calculate the weight of the following objects. Answer in decimals.
 Calculate a steel cylinder‘s weight if its volume = 10 cm³, steel density is 7.85 g/cm³
V × ? = m
volume x density = mass/weight
10 × 7.85 = 78.5 g
Download Worksheet: Level 2
Worksheet: Level 3
Worksheet: Level 3 Sample Questions
Example:
 Calculate a steel cylinder‘s weight if its radius is 4 in, height is 5 in, steel density is 490 lb/ft³.
Further example:
Calculate the mass of a granite cone, d = 5 cm, h = 8 cm, ρ = 2.7 g/cm³.
Worksheet: Level 3 Sample Answer Key
Answer:
r²? ⋅ h = V
radius² × ? × height = volume of cylinder
4² × ? × 5 = 251.327 in³ ÷ 1728 = 0.14544 ft^{3}
volume x density = mass (weight)
0.14544 × 490 = 71.26 lb
Note: Mass and weight are different in physics, but not for doing applied math with them in trades. A pound is both a unit of force and unit of mass in nonmetric calculations. In the Metric System Kg is a unit of mass, N is a unit of force.
Download Worksheet: Level 3
Infosheet: Advanced
Additional Information: Another Way to Get the Answer
Another Way to Get the Answer
Task:
Calculate the weight of a steel reinforcing bar (long cylinder) if length is 50 feet, the crosssectional area is 0.785 in^{2 }(Bar diameter is 1 inch.) The density of mild steel is either 0.2836 lb/in^{3 }(= 490 lb/ft^{3}).
One way to get the answer, using the formulas V = A · L and Weight = density x V
Layout your calculations neatly, so you can review, track changes, correct or learn from them. One way of layout can look like this:
Calculate the weight of a steel reinforcing bar (long cylinder) if length is 50 feet, the crosssectional area is 0.785 in^{2 }(Bar diameter is 1 inch.) The density of mild steel is either 0.2836 lb/in^{3 }(= 490 lb/ft^{3}). Another way to get the answer, using the formulas V = A · L and Weight = density x V
1. Identify that the weight of a steel cylinder is to be calculated.
2. Make a sketch of the cylinder:
3. Label the sides of the cylinder:
Use either full words or can abbreviate. The idea is to visualize the problem before calculation.
For now, don`t sketch or label the 1 inch diameter dimension mentioned in the brackets, it`s just there to help visualize the steel bar. The density amount is impossible to sketch, just leave it for now.
Look at the 2 formulas, there is one to calculate volume and one to calculate weight. The first formula is for volume. We need an answer for volume to be filled into the second formula. Volume of the cylinder must be calculated first:
4. Write the formula: A · L = V
5. Rewrite the formula: change all of the letters into words, according to their meaning: Area of cross section x length = volume of cylinder
6. Look at the units of measure for area and length. Length is feet, and area is based on inches.
Recognize that all measurements have to be in the same unit. At this point, the area of 0.785 in^{2 }will be converted to ft^{2}.
7. Recognize that the conversion factor to convert from in^{2 }to ft^{2 }is 144. It comes from 12 inches times 12 inches making 1 square foot, which 12 x 12 equals 144.
8. Recognize that when converting from smaller unit to bigger unit, that is from in^{2 }to ft^{2}, the conversion factor will be a divisor because many of the smaller units (in^{2}) are to be shared equally by groups of the bigger unit (ft^{2}).
9. Set up the problem:
in^{2 }÷ conversion factor = ft^{2}
10. Calculate: 0.785 ÷ 144 = 0.005451388
11. Take a pencil and write down “A = 0.00545” or “rebar crosssection = 0.00545”
12. Write ft^{2 }after this number.
13. Decide that the converted cross sectional area of a single rebar is 0.00545 ft^{2}
Now the volume can be calculated:
14. Rewrite the formula: change all of the words for actual given numbers: Leave out units of measure words. (eg. feet, cm)
0.00545 x 50 = volume of cylinder
15. Calculate. Use mental math, paper & pencil, or electronic calculator: 0.00545 x 50 = 0.2725
16. Take a pencil and write down “V = 0.2725” or “rebar volume = 0.2725”
17. Look at the units of measure. ft^{2 }was multiplied with ft. The unit for volume is ft^{3}. Take a pencil and write ft^{3 }after the 0.2725
18. Decide that the correct way to write the answer is 0.2725 ft^{3}
Now the weight can be calculated:
19. Write the formula: V · density = Weight
20. Rewrite the formula: change all of the letters into words, according to their meaning:
volume x density of steel = weight of steel bar
21. Rewrite the formula: change all of the words for actual given numbers:
Leave out units of measure words. (eg. feet, cm):
490 x 0. 2725 = weight of steel bar
22. Calculate: 490 x 0.2725 = 133.525
23. Take a pencil and write down 133.5
24. Write lb after the number 133.5 Now you`re done with the math.
25. The last step is to check your work: make sure everything was copied and written correctly then determine that the correct way to write the answer is 133.5 lb
We don`t think Vaughn can lift that. In addition, long and thin steel bars bend a lot, so even if he lifts in the middle, the ends will still be on the ground, making walking impossible.
Additional Information: Another Layout
Layout your calculations neatly, so you can review, track changes, correct or learn from them. One way a layout can look like is this:
Additional Information: Advanced Part 1
ADVANCED 1
This calculation is somewhat lengthy. Here is a way how you can make your very own formula for future reference to speed up calculations:
In the animated calculation, the sequence of calculations were:
We started with 50 This amount was:
 multiplied by 12,
 then the result was multiplied by 0.785,
 then the result was multiplied by 0. 2836 and we got 133.6 lb
It is possible to enter everything in the calculator in one line (on a calculator with a multiline display):
50 x 12 x 0.785 x 0.2836 = 133.5756 lb
Now, we can make a formula by replacing the numbers we can with words, according to their meaning:
length (feet) x 12 x area (in^{2}) x density (lb/in^{3}) = weight (lbs)
Now we need to replace the words with letter symbols:
L x 12 x A x D = weight
Multiplications are interchangeable. We can rearrange the letters a bit to make it easier to memorize:
12 · L · A · D = weight
Enter the amounts in your calculator in the order shown in this formula. Do you still get 133.5756?
Meticulously compare this homemade formula with the next calculation`s formula. Note similarities and differences. Both formulas determine the weight of cylindrical steel objects in pounds. Why are they different?
ANSWER:
Similarities:
 Results are the same. That`s reassuring.
 Both calculations use length, area, density to end up with weight.
 Both calculations take the same number of steps.
Differences:
 One calculation is based on in and has an in^{2} to ft^{2} conversion in it.
 The calculations use 2 different figures for density. This is why the formulas are different.
 One has a division, the other is all multiplications.
Additional Information: Advanced Part 2
ADVANCED 2 (based on the second version)
This calculation is somewhat lengthy. Here is a way how you can make your very own formula for future reference to speed up calculations:
In the animated calculation, the sequence of calculations were: We started with 0.785 This amount was:
 divided by 144,
 then the result was multiplied by 50,
 then the result was multiplied by 490
and we got 133.5 lb
It is possible to enter everything in the calculator in one line (on a calculator with a multiline display):
0.785 ÷ 144 x 50 x 490 = 133.5590278
Now, we can make a formula by replacing the numbers we can with words, according to their meaning:
area (in^{2}) ÷ 144 x length (feet) x density (lb/ft^{3}) = weight (lbs)
Now we need to replace the words with letter symbols:
A ÷ 144 x L x D = weight
Multiplications are interchangeable. We can rearrange the letters a bit to make it easier to memorize:
L · A · D ÷ 144 = weight
Enter the amounts in your calculator in the order shown in this formula. Do you still get 133.5590278?
Meticulously compare this homemade formula with the previous calculation`s formula. Both formulas determine the weight of cylindrical steel objects in pounds. Note similarities and differences. Why are they different?
ANSWERS:
Similarities:
 Results are the same. That`s reassuring.
 Both calculations use length, area, density to end up with weight.
 Both calculations take the same number of steps.
Differences:
 One calculation is based on in and has an in^{2} to ft^{2} conversion in it.
The other is based on feet and has a ft to in conversion in it.  The calculations use 2 different figures for density. This is why the formulas are different.
 One has a division, the other is all multiplications.
Additional Information: Advanced Part 3
ADVANCED 3
Print out and compare the volume of cylinder calculations in episodes 27 & 29.
Volume of a cylinder was calculated by “V = A · h” in the animated version of Episode 27 and “V = A · L” in both versions of Episode 29. One includes “height” the other does not. One includes “length” the other does not. The formulas look different. But they are really not.
1. Why? Is there still a way that explains how these 2 different formulas are the same, consistent and valid?
Look for clues in episode 09 & 26 ADVANCED.
In Episode 27b the volume of the cylinder was calculated by the formula V = r²? · h, which looks totally different from either “V = A · h” or “V = A · L” Hmmm. Looks messed up. What`s going on?
The volume of cylinders can be reliably and correctly calculated either way.
2. How is the longer cylinder formula “V = r²? · h” the same as “V = A · h” ?
3. Based on clues from Episode 20, area of circles, come up with another formula for calculating the
volume of cylinders that`s based on diameter.
4. What if your cylinder is oblique, leaning at a 45° angle to the left? (As with a solid bar, end cut on an angle.) How would you calculate its volume? Write a formula to do that.
ANSWERS:
 Length and height are synonyms, interchangeable. Laid horizontally, the height of a cylinder becomes the length of the steel bar.
 “r²?” is replaced by the letter A
 2 examples are here:
\[\text{V = }\frac{d^2 \pi}{4}\cdot{h}\]or\[\text{V = }d^2\frac{\pi}{4}\cdot{h}\]…, but any of the 6 circle area formulas in Episode 20 ADVANCED can be either before or after the multiplication by “ h ”. So that`s 12 possible volume of cylinder formulas.
 Same formulas as above work. It doesn’t matter how much the cylinder leans. Height is a perpendicular measurement between the 2 parallel circle bases.
Download Infosheet: Advanced
Additional Information
ADVANCED PLUS
Look at the final answers using different calculations: 133.525 133.5756 133.5590278
There are some slight differences in the decimal digits. This is due to rounding from step to step. This rounding introduces a level of uncertainty into the calculations, undermining the degree of accuracy of the final answer.
There are 3 ways to eliminate some of the rounded numbers. The decimals for area (0.785) and density (0.2836) are both already rounded amounts.
 We can eliminate inaccuracy by using either whole numbers,
 or unrounded & noninfinite decimals.
 We can also enter more terms per calculation into the calculator without pressing the = button.
The rebar with the A = 0.785 in^{2} actually is a 1 inch rebar. Its diameter, 1 inch, is a nice whole number.
 Given the 1 inch diameter rebar size, we can calculate a more accurate cross sectional area:
Diameter = 1 inch, so radius = ½ inch
A = r^{2 }?, so 0.5^{2} x ? = 0.785398163 in^{2} This new amount is more accurate than the 0.785 in^{2}
The 0.2836 figure for density has also been rounded. We can calculate more accurate density:
 Mild steel density is 490 lb/ft^{3}, this is a whole number, the weight of mild steel for a cubic foot. Since there are 1728 in^{3} in a ft^{3}, mild steel`s density is 490 ÷ 1728 = 0.283564814 lb/in^{3}The figure 0.283564814 lb/in^{3} is more accurate than 0.2836 lb/in^{3}.
For the most accurate result:
 replace the 0.758 by entering 0.5^{2} x ? for area in the formula.
also replace 0.2836 by entering 490 ÷ 1728 for density in the formula. length (feet) x 12 x area (in^{2}) x density (lb/in^{3}) = weight (lbs)
Now enter in your calculator: 50 x 12 x 0.5^{2} x ? x 490 ÷ 1728 = 133.6267709 lb
This level of accuracy is important if you have hundreds or thousands of bars to calculate with.
This is still a mathematical model of a rebar based on a smooth cylinder. Real rebars have ridges, which may add more weight to the calculated figure. Rebar can also be rusty or dirty, adding even more weight.
There is a solution to all this puzzle. Manufacturers actually specify and publish the real weight of manufactured materials per foot length, eg:
Imperial Bar Size

Nominal Diameter (in)

"Soft" Metric Size

Weight per unit length (lb/ft)

Mass per unit length (kg/m)

Nominal Diameter (mm)

Nominal Area(in^{2})

Nominal Area (mm^{2})

#8

1.000

#25

2.67

3.982

25.4

0.79

509

http://www.harrissupplysolutions.com/8rebar.html
The manufacturer`s numbers are the best choice to work with. Weight = 50 feet x 2.67 lb/ft = 133.5 lb
Note: "Nominal" means "in name only". Actual measurements vary slightly during shifts, due to environmental or minute physical differences in identical models of machines.