# Percentage Saving in mass/(weight) of a hollow shaft - equation?

## Homework Statement

I'm looking for an equation that i can use to work out the percentage saving in mass gained by using a hollow shaft rather than a solid shaft.

The question is simply that.

I've done research and found a dodgy equation that doesn't work for me, this is below, and I know there is area involved somehow but can't find anything legit at the moment and this is urgent, I need this sorted by lunchtime tomorrow.

## Homework Equations

Dodgy Equation use: [Ds^2-(D^2-d^2)/Ds^2]*100

## The Attempt at a Solution

Ds = Diameter of Solid Shaft = 0.177 m (This was worked out but it is correct)

D = External Diameter of Hollow Shaft = 0.23 m (This was given in the orginial question)

d = Internal Diameter of Hollow Shaft = 0.198 m (Also figured out but correct)

Using the above 'dodgy' equation that i've now come to call, I got the following answer: 56.28 %

Which is way too much I was predicting something along the lines of 10 - 15 %

Ok I just found something else and will do it all here whilst I'm looking at it:

Area of Ds^2 - Area of Dh^2 / Ds^2 = % Saving

Now the Area of a Solid Shaft from what it tells me is:

pi/4 * Ds^2

= pi/4 * (0.177)^2

= 0.0246 m^2

Area of Hollow Shaft:

pi/4 * (D^2 - d^2)

= pi/4 * (0.23^2 - 0.198^2)

= 0.0108 m^2

% = Ds^2 - Dh^2 / Ds^2

= (0.0246^2) - (0.0108^2)

= 0.8073

Well yes I didn't have high hopes.

See I have a length thats equal to both shafts which is: 10.6 m

Surely, this is used in the area somewhere?? My researching skills recently a very poor and my concentration is becoming a joke.

I don't need any workings out, just an equation to work with.

I feel like a regular to this site now, and I keep asking for help, i'm going to have to some day help someone else for the sake of karma haha.

Thanks for taking the time to read through this and for any help or advice you wish to give.