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

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SUMMARY

The discussion focuses on calculating the percentage saving in mass when using a hollow shaft compared to a solid shaft. The user initially attempted to use an incorrect equation but later derived a more accurate method involving the areas of both shafts. The correct approach involves calculating the volume of both shafts and using the formula: Percentage saving = (Volume of solid shaft - Volume of hollow shaft) / Volume of solid shaft * 100. The user provided specific dimensions for the shafts: solid shaft diameter (Ds) = 0.177 m, external hollow shaft diameter (D) = 0.23 m, and internal hollow shaft diameter (d) = 0.198 m.

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  • Understanding of geometric volume calculations
  • Familiarity with the concept of hollow versus solid shafts
  • Knowledge of basic algebra for manipulating equations
  • Ability to compute areas and volumes of circular shapes
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  • Study the formula for the volume of a hollow cylinder
  • Research the implications of material savings in engineering design
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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 that's 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.
 
Physics news on Phys.org
1. Compute the volume of your solid shaft.
2. Compute the volume of the hollow part.

Subtract 2 from 1. Percentage saving = 2/1. Oh yeah, times 100.
 

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