You have 1kg of aluminum (E=69GPa) to make a cylindrical tube with a high Euler buckling load. The tube must have a length L of 0.5m. You can either make a solid tube, or a thin walled tube with a wall thickness, t=0.5cm. The density of aluminum is 2700kg/m3. The tubes are pin-pin ended; assume n=1.
What is the Euler buckling load, Pcr,sol (in kN), for the solid tube?
Pcr,sol (in kN):
What is the Euler buckling load, Pcr,sol (in kN),for the thin-walled tube?
Which design is better?
The solid tube
The thin-walled tube
To find the Euler buckling load for the solid tube, we can use the formula:
Pcr,sol = (π² * E * A) / (L / r)²
where:
- Pcr,sol is the Euler buckling load for the solid tube,
- E is the modulus of elasticity for aluminum (69 GPa = 69 * 10⁹ Pa),
- A is the cross-sectional area of the tube, and
- r is the radius of the tube.
The cross-sectional area of a solid cylindrical tube is given by:
A = π * r²
Given that the tube has a mass of 1 kg and the density of aluminum is 2700 kg/m³, we can find the volume of the tube using:
Volume = mass / density = 1 kg / 2700 kg/m³
Since the tube is solid, the volume can also be expressed as:
Volume = π * r² * L
Using these equations, we can solve for the radius:
1 / (2700 kg/m³) = π * r² * 0.5 m
r² = (1 / (π * 2700 kg/m³ * 0.5 m))
r = √(1 / (π * 2700 kg/m³ * 0.5 m))
Once we have the radius, we can calculate the cross-sectional area:
A = π * r²
Finally, we substitute the values of E, A, L, and r into the formula for Pcr,sol to find the Euler buckling load for the solid tube.