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
P_cr Solid =117.132 KN
P_cr Hol = 561.59KN
Thin walled tube design is better
To determine the Euler buckling load for each type of tube, we need to use the formula:
Pcr = π² * E * I / (L / n)²
where:
- Pcr is the Euler buckling load
- E is the elastic modulus of the material (in this case, aluminum with E = 69 GPa)
- I is the moment of inertia of the cross-sectional area of the tube
- L is the length of the tube
- n is a constant related to the boundary conditions (in this case, n = 1 for pin-pin ends)
Now let's calculate the Euler buckling load for each tube.
For the solid tube:
- The mass of the solid tube is given as 1 kg, and the density of aluminum is 2700 kg/m³. Therefore, the volume of the solid tube is 1 kg divided by 2700 kg/m³, which equals approximately 0.00037037 m³.
- The moment of inertia of a solid cylindrical tube is given by the formula I = π * r^4 / 4, where r is the radius of the tube.
- The radius of the tube can be calculated using the mass and density of aluminum, since the volume of a cylinder is given by V = π * r² * h, where V is the volume, r is the radius, and h is the height or length of the cylinder.
- Rearranging the formula for the volume of the solid tube, we get r = sqrt(mass / (density * π * h)).
- Plugging in the values, we have r = sqrt(1 kg / (2700 kg/m³ * π * 0.5 m)).
- Evaluating this expression, we find r ≈ 0.09315 m.
Substituting these values into the Euler buckling load formula for the solid tube, we have:
Pcr,sol = π² * 69 GPa * (π * 0.09315 m^4 / 4) / (0.5 m / 1)²
Calculating this expression, Pcr,sol ≈ 165.15 kN.
For the thin-walled tube:
- The formula for the moment of inertia of a thin-walled cylindrical tube is given by the formula I = 2 * π * r³ * t, where r is the outer radius and t is the wall thickness.
- The outer radius can be obtained by adding the wall thickness to the inner radius of the solid tube, so r = 0.09315 m + 0.005 m (0.5 cm converted to meters).
- Plugging in the values and substituting into the Euler buckling load formula for the thin-walled tube, we have:
Pcr,thin-walled = π² * 69 GPa * (2 * π * (0.09315 m + 0.005 m)³ * 0.005 m) / (0.5 m / 1)²
Calculating this expression, Pcr,thin-walled ≈ 255.33 kN.
Comparing the results, we can see that the Euler buckling load for the thin-walled tube (Pcr,thin-walled ≈ 255.33 kN) is higher than that of the solid tube (Pcr,sol ≈ 165.15 kN). Therefore, the thin-walled tube design is better in terms of resisting Euler buckling.