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Volume, Weight 2

Fri 30 Jan, 2015 09:57 pm
A hollow cylindrical iron pillar has the following dimensions:

External diameter - 6"
Internal diameter - 4"
Length - 12'

Grade of iron - 441 lbs. per cu. ft.

Determine volume and weight of pillar.

Check my calculations:

I have two formulas:

V - pi * r^2 * h

3.1416 *.4166 * .4166 * 12 = 6.54 = 7 cu. ft. ( I added the diameters)

7 * 441 = 3,087 lbs.

V - Pi (R + r) (R - r)H
R = external radius
r = internal radius
H = Height
V - Pi (3 + 2) (3 - 2) 12
V- Pi (5) (1) 12
V - Pi (5) 12
V - Pi x 60
V- 3.1416 x 60
V - 188.496 = 188 cu. ft.
W = 188 x 441 = 82,908 lbs.

Which is correct ?

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neologist

1
Sat 31 Jan, 2015 12:25 am
@Randy Dandy,
Neither
Even a solid square pillar would have no more than a 3 cubic foot volume.
.5*.5*12=3 (You could do that in your head to check the veracity of your result)
A solid round pillar:
.25*.25*(22/7)*12=2.36 cubic foot
But yours has a hollow center.
Deduct the interior volume from the exterior volume and multiply the result by 441

Explain why you added the diameters in your first example.
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raprap

1
Sat 31 Jan, 2015 08:20 am
@Randy Dandy,

What do you know?

Hint---radius is half the diameter

1) You know the outside radius (r0) and height (h) of the cylinder,

So you know the outside volume of the cylinder ( V0=pi*ro^2*h)

2) You know the inside radius (ri) and height (h) of the cylinder,

So you know the inside volume of the cylinder ( Vi=pi*ri^2*h)

3) You know the cylinder is hollow (AKA annulus) --so the volume of the hollow cylinder (Vc) is the difference between the volume of the outside less the volume of the inside.

Vc=Vo-Vi

4) Units, Units, Units---The radius and height were given in inches--so if I use those measurements the Volumes calculated in steps 1,2 & 3 are in cubic inches (in^3)--- if I want to convert these volumes to cubic feet (ft^3)--I have to convert.

As there are 1728 in^3 in 1 ft^3 (1728=12^2). To get the volumes from in^2 to ft^3 divide by 1728.

5) The density of iron (not grade) is 441 lb per ft^3

To determine the weight of the hollow cylinder lb you multiply the volume of the hollow (annulus) cylinder in ft^3 by the density in lb/ft^3.

6) Check your answer--a 6 by 12 inch inch cylinder is smaller than a breadbox--consequently even though it is iron I should be able to pick it up, so it is probably less than 100 pounds. Now this annulus is hollow, so it should be significantly less than that.

Rap
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Randy Dandy

1
Sat 31 Jan, 2015 12:25 pm
I added the diameters in the first example to get total diameters.

Recheck my calculation:

Pi * r^2 * h
3.1416 *.25 * .25 * 12 = 2.3562 cu. ft. (outside diameter - converted radius to feet)

Pi * r^2 * h
3.1416 * .166 * .166 * 12 = 1.0388 cu ft. ( inside diameter - converted radius to feet)

2.3562 - 1.0388 = 1.3274 cu. ft. ( difference)

1.3274 * 441 = 585.3834 = 585 lbs.
raprap

1
Sat 31 Jan, 2015 08:30 pm
@Randy Dandy,
pretty close--but although when you converted the outside and inside diameters from inches to feet, you forgot the height. Consequently the annulus weight you calculated, it is 12 feet tall instead of 12 inches.

Rap
Randy Dandy

1
Sat 31 Jan, 2015 08:56 pm
@raprap,
The height was 12 feet.

I have a question:

Using specific gravity to calculate weight:

2.3562 + 1.0388 = 3.395 cu. ft.

Cu. ft. of water = 62.5 lbs.

Weight of cu. ft. of iron - 62.5 x 7.8 (specific gravity of iron) = 487.5 lbs.

Weight of pillar - 487.5 x 3.395 = 1655.0625 = 1655 lbs.

If my calculation is correct, should the answer be closer to the other ?

neologist

1
Sat 31 Jan, 2015 10:11 pm
@Randy Dandy,
You were given the 441 figure as part of the problem, right?
Randy Dandy

1
Sat 31 Jan, 2015 10:38 pm
@neologist,
Right, my error.
0 Replies

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