# Deformation of axially loaded bar

1. Sep 22, 2009

### vanquish

1. The problem statement, all variables and given/known data
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 $$\gamma$$=.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.

2. Relevant equations
$$\delta$$=$$\int$$ $$\frac{Fdy}{AE}$$
3. The attempt at a solution
Since the only force acting on the cone is its own weight:
F=weight=$$\gamma$$*A
because the density times the area

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

So when I plug it all in I get:
$$\delta$$=$$\int$$ $$\frac{ \gamma Aydy}{AE}$$

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:

$$\delta$$=$$\frac{ \gamma y^2}{2E}$$

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

Last edited: Sep 23, 2009
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