Deformation of axially loaded bar

[vanquish]

Homework Statement

Determine the extension, due to its own weight, of the conical bar show in Fig 5.13. The bar is made of aluminum alloy [E=10,600ksi] and [tex]\gamma[/tex]=.100 lb/in³
The bar has a 2in radius at its upper end and a length of L=20ft assume the taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.

So its basically an cone hanging off the ceiling and I need to the extension of it.

Homework Equations

[tex]\delta[/tex]=[tex]\int[/tex] [tex]\frac{Fdy}{AE}[/tex]

The Attempt at a Solution

Since the only force acting on the cone is its own weight:
F=weight=[tex]\gamma[/tex]*A
because the density times the area

Since the it's a cone, the radius changes and therefor so does the area
Area=[tex]\pi[/tex]*r²
To find r, I used the similar triangles which gave me:
2/10=r/y => r=y/10
so: Area=[tex]\pi[/tex]*(y/10)²

So when I plug it all in I get:
[tex]\delta[/tex]=[tex]\int[/tex] [tex]\frac{ \gamma Aydy}{AE}[/tex]

And the area's cancel out. I continued the problem even though I'm quite sure that the area's should not cancel out. I ended up with:

[tex]\delta[/tex]=[tex]\frac{ \gamma y^2}{2E}[/tex]

Upon plugging in the numbers i got an answer of: 2.7*10^-4inches
When the correct answer (according to the book) is: 9.06*10^-5

EDIT: I figured out what I did wrong, silly mistake

 

[LightHero]

on my part. The Area's do not cancel out. I should have kept the A in the denominator and gotten: \delta=\frac{ \gamma y^2}{2EA}Which gave me the correct answer.

 

[Mene]

. I used the wrong density, I used the density of water instead of the given density of the aluminum alloy. The correct answer using the given density would be 9.06*10-5 inches.