What Determines the Maximum Height of a Self-Supporting Column?

[Jacob87411]

There is a maximum height of a uniform vertical column made of any material that can support itself without buckling, and it is independent of the cross sectional area (why?). Calculate this height for a) steel (density 7.8x10^3) b) Granite (density 2.7x10^3)

I'm just really confused on how to even start this, what equation to use. I am guessing it has something to do with compressive strength but not sure at all

 

[Andrew Mason]

There is a maximum height of a uniform vertical column made of any material that can support itself without buckling, and it is independent of the cross sectional area (why?).

A material has a maximum compressive strength expressed as a force /unit area or pressure. What is the the force/unit area at the base of a column of material of uniform cross-section with density $\rho$? Is it related to anything other than the height?

Calculate this height for a) steel (density 7.8x10^3) b) Granite (density 2.7x10^3)[/QUOTE] Find the pressure at the base of a column of steel of height h and set the pressure to the yield pressure for steel (compressive strength). What is h? Do the same for granite.

AM

 

[wais]

The maximum height of a uniform vertical column is determined by its ability to resist buckling, which is a form of instability where the column fails under compressive forces. This maximum height is known as the critical buckling height and it is independent of the cross-sectional area because it is primarily influenced by the material properties and the column's slenderness ratio.

The slenderness ratio is the ratio of a column's length to its cross-sectional dimension and it plays a crucial role in determining the critical buckling height. As the slenderness ratio increases, the column becomes more susceptible to buckling.

To calculate the critical buckling height for a given material, we can use the Euler's buckling formula:

H = (π²EI)/(KL)²

Where:
H = critical buckling height
E = Young's modulus of the material
I = moment of inertia of the cross section
K = effective length factor
L = length of the column

For a) steel:
Given density = 7.8x10^3
Young's modulus of steel is approximately 200 GPa (2x10^11 Pa)
Assuming a square cross section with side length of 1 m, the moment of inertia (I) = (1/12)(1m)^4 = 1/12 m^4
The effective length factor (K) for a fixed-fixed column is 0.5 (assuming both ends are fixed)
Substituting these values into the formula, we get:
H = (π² x 2x10^11 x 1/12 m^4)/(0.5 x 1m)² = 157.9 m

For b) granite:
Given density = 2.7x10^3
Young's modulus of granite is approximately 50 GPa (5x10^10 Pa)
Assuming a square cross section with side length of 1 m, the moment of inertia (I) = (1/12)(1m)^4 = 1/12 m^4
The effective length factor (K) for a fixed-fixed column is 0.5 (assuming both ends are fixed)
Substituting these values into the formula, we get:
H = (π² x 5x10^10 x 1/12 m^4)/(0.5 x 1m)² = 39.5 m

In summary, the maximum height of