# Calculate this height for aluminum column

1. Nov 11, 2004

### lmf22

There is a maximum height of a uniform vertical column made of any material that can support itself without breaking, and it is independent of the cross-sectional area.
(a) Calculate this height for aluminum (density 2.7 103 kg/m3).
(b) Calculate this height for granite (density 2.7 103 kg/m3).

How would I go about doing this?

2. Nov 11, 2004

### sal

You need the "tensile strength" of the material. Tensile strength is the maximum tension, or weight per unit area, which the material can support before it breaks. Tensile strength is an empirical property -- it's measured, not predicted, for each material.

Given the density and length of the column, you can compute the weight being supported per unit area at a particular point in the column, which is the tension on the column. Compare that to the tensile strength of the material. When they're just equal, the column is about to break.

Obviously the tension will be highest at the top of the column (assuming it's suspended from the top).

3. Nov 11, 2004

### cepheid

Staff Emeritus
I have a question...does the fact that the column is under compression (being crushed under its own weight) instead of being pulled apart mean that we need a different quantity than the tensile strength? Wasn't the tensile strength measured during a tension test of a standard sized member of that material? Would the results of a compression test be different?

4. Nov 12, 2004

### PerennialII

For most materials, especially metallic ones, their plastic behavior under tension and compression is identical (aluminum for one, porous & materials like rock have a different compression side but that probably does not factor into this problem).... as long as we're using their stress-strain response as a measure of their "strength" (if you'd be doing, say, a fracture mechanics analysis, you'd consider failure only occurring in the tension side, which sounds reasonable). In this case, once the column reaches the amount of stress in the compression side equivalent to tensile strength it will lead to structural failure, and as such the question is valid being formulated as it is.

5. Nov 12, 2004

### sal

Uh, hmm... I misread the question and just assumed the column was suspended from the top rather than sitting on the ground. In general, the "tensile strength" of building materials in compression is larger than their strength in tension, so you would need to use different numbers.

The approach to solving the problem is identical in the two cases, save that the maximum tension will be at the bottom when it's supported at the bottom, rather than when it's suspended from the top.

Note that granite and concrete are both substantially stronger in compression than in tension, and what's more, in tension stone eventually tends to crack. This is why there are no lintels on the doorways in the old Greek ruins: the Greeks used flat stones for the lintels, and over the ages, they cracked and fell. A related example is brickwork, which is strong in compression, very weak in tension, and very stiff; as a result, in an earthquake brick buildings tend to collapse much more readily than wood-frame buildings, which are both more elastic (so stresses don't go as high when they're shaken) and stronger in tension.

Last edited: Nov 12, 2004
6. Nov 12, 2004

### Gokul43201

Staff Emeritus
For most metals, strain under tension and compression are identical for small to moderate values, as PerenailII noted. For larger loads (approaching yield), the nominal compressive strength is greater than the nominal tensile strength because of an increase in the cross section area (nominal stress is defined in terms of the initial area).

For your problem, you can use the tensile strength.