Weight - Steel Tube

[Guest]

The length of a hollow steel tube is 30 ft. The outer diameter is 18 in., inner diameter is 10 in. Determine weight of the tube (tons).
Steel weighs .283 lbs. per cu. in.

Check my calculations:

Volume of hollow cylinder: pi(R^2 - r^2)L R = outer radius r = inner radius

3.1416 x 9^2 - 5^2 x 30
3.1416 x 81 - 25 x 30
3.1416 x 56 x 30 = 5277.888 = 5278 cu. in.
5278 x .283 = 1493.674 = 1494 lbs. / 2000 = 0.747 tons.

[Guest]

Yes you have got the correct answer but you used the brackets in your calculation OK but you ignored the brackets when typing in you worked solution
When reading a solution...most read from left to right and use the BODMAS or BOMDAS ruling....putting in brackets enforces that.

3.1416 x 9^2 - 5^2 x 30 ....this reads as (3.1416 x 9^2) minus (5^2 x 30) ....wrong
3.1416 x 81 - 25 x 30 ......now you have worked it as though the brackets had been there....
3.1416 x 56 x 30 = 5277.888 = 5278 cu. in. ......and subtracted as though the brackets had been there....
5278 x .283 = 1493.674 = 1494 lbs. / 2000 = 0.747 tons. .....and got right answer because you wrote it down differently from the way you calculated it.


3.1416 x(9^2 - 5^2) x 30 ....Work what in brackets first...take the squares .......
3.1416 x (81 - 25) x 30 .....and subtract them.......
3.1416 x( 56) x 30 = 5277.888 = 5278 cu. in. .....brackets not needed here but shows where the subtraction answer is.
5278 x .283 = 1493.674 = 1494 lbs. / 2000 = 0.747 tons. .......now the right answer and all reads correctly.

[Guest]

Thanks.

Question: Can the weight be determined by a formula using the diameter instead of radius ?

[Guest]

Yes, look at another thread on this forum "Amount in feet"

We have to find the area of the crossection and multiply by length to get volume.

There are various forms for the area of an annulus or area between 2 concentric circles as in the crossection of a pipe.

Area = Pi(R^2 - r^2) this is Pi times big R squared minus little r squared.

OR in another form using the fact that the difference of 2 squares is their sum times their difference
Area = Pi[(R + r)(R - r)]

OR in another form noting that (R + r) = (D - t) .....the outside diameter "D" minus the thickness "t"......and....(R - r) is just the thickness "t".
Area = Pi[(D - t)(t)] = Pi(D - t)(t)

Also write any of the above in terms of "D"
Area = Pi((D/2)^2 - (d/2)^2) = Pi((D^2/4) -(d^2/4))
where "D" is the outside diameter and "d" is the inside diameter

etc etc.....

[Guest]

Thanks again.

You stated I have the correct answer. I looked at the problem again and noticed the length is in feet and diameters are in inches. Should I have all in the same units ?

[Guest]

Yes all dimensions should be converted to inches so volume ic calculated as cubic inches.
So should be...... 3.1416 x(9^2 - 5^2) x 30 x 12

[Guest]

" So should be...... 3.1416 x(9^2 - 5^2) x 30 x 12 "

3.1416 x ( 81 - 25 ) x 360
3.1416 x (56) x 360
3.1416 x 20160 = 63334.656 = 63335 cu .in.
63335 x .283 = 17923.805 = 17924 lbs. / 2000 = 8.962 = 9 tons.

[Guest]

Yes its big and heavy

[Guest]

Thanks.