Four identical pin-pin ended solid concrete cylindrical columns that are L=10m tall need to be able to withstand 6.8*10^8 N of force before Euler buckling. What should be the diameter, d, in meters, of each column if the force is uniformly distributed? The Young’s modulus, E, for concrete is 17GPa, and 𝐼=(πr^4)/4 . Assume n=1.

I used: P_critical = (n^2*π^2*E*I)/(L^2*(πr^2)) to solve for r then multiplied by two for diameter but this isn't correct. Any advice for correcting my formula? Thank you.

To solve this problem, you can use the Euler buckling formula for columns under compression. However, it seems like there is a mistake in the formula you mentioned.

The correct formula for the critical buckling force, P_critical, is given by:

P_critical = (n^2 * π^2 * E * I) / L^2

where:
- P_critical is the critical buckling force
- n is the effective length factor (since you have pinned-pinned ends, n=1)
- π is pi (approximately 3.14159)
- E is the Young's modulus of concrete (17 GPa = 17 * 10^9 Pa)
- I is the moment of inertia of the column cross-section
- L is the height of the column (10m in this case)

Now, let's calculate the moment of inertia, I, for a solid cylindrical column with a diameter d.

The formula for the moment of inertia of a solid cylindrical column is:

I = (π * d^4) / 64

where:
- I is the moment of inertia
- π is pi (approximately 3.14159)
- d is the diameter of the column

Substituting the moment of inertia formula into the critical buckling force formula, we get:

P_critical = (n^2 * π^2 * E * (π * d^4 / 64)) / L^2

Given that P_critical = 6.8 * 10^8 N and L = 10m, we can rearrange the formula to solve for d:

d^4 = (P_critical * L^2 * 64) / (n^2 * π^2 * E * π)

d = [ (P_critical * L^2 * 64) / (n^2 * π^2 * E * π)]^(1/4)

Let's plug in the values:

d = [ (6.8 * 10^8 * (10)^2 * 64) / (1^2 * π^2 * 17 * 10^9 * π) ]^(1/4)

Calculating this using a calculator, we get:

d ≈ 0.0963 meters (or approximately 96.3 mm)

Therefore, each column should have a diameter of approximately 0.0963 meters (or 96.3 mm) to withstand the given force before Euler buckling.