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, P_cr,sol (in kN), for the solid tube?
P_cr,sol [kN] = ?
What is the Euler buckling load, P_cr,sol (in kN),for the thin-walled tube?
P_cr,sol [kN] = ?
To find the Euler buckling load for both the solid tube and the thin-walled tube, we can use the following formula:
P_cr = ((n^2 * π^2 * E * I) / L^2)
where:
P_cr is the Euler buckling load
n is the number of half-waves in the mode shape (given as 1)
π is a mathematical constant approximately equal to 3.14159
E is the Young's modulus of aluminum (given as 69 GPa, which can be converted to N/m^2 by multiplying by 10^9)
I is the second moment of area for the cross-sectional shape of the tube
L is the length of the tube
First, let's calculate the Euler buckling load for the solid tube:
To find the second moment of area (I) for the solid tube, we can use the formula:
I = (π * r^4) / 4
where:
r is the radius of the tube
Since the tube is cylindrical, the radius is half of the diameter, which is 0.5m.
So, the radius (r) of the solid tube is 0.5m / 2 = 0.25m.
Now, let's calculate I for the solid tube:
I = (π * (0.25m)^4) / 4
≈ 0.02454 m^4
Next, let's substitute the given values into the Euler buckling load formula for the solid tube:
P_cr,sol = ((1^2 * π^2 * (69 * 10^9 N/m^2) * 0.02454 m^4) / (0.5m)^2)
P_cr,sol = ((π^2 * 69 * 10^9 * 0.02454) / 0.25)
P_cr,sol ≈ 21.56 kN
Therefore, the Euler buckling load for the solid tube is approximately 21.56 kN.
Now, let's calculate the Euler buckling load for the thin-walled tube:
To find the second moment of area (I) for the thin-walled tube, we can use the formula:
I = (π * (R^4 - r^4)) / 4
where:
R is the outer radius of the tube
r is the inner radius of the tube
Since the wall thickness (t) of the thin-walled tube is given as 0.5 cm, the inner radius (r) is equal to the radius of the solid tube, which is 0.25 m.
To find the outer radius (R) of the thin-walled tube, we can use the formula:
R = r + t
R = 0.25 m + (0.005 m) (converting 0.5 cm to meters)
R ≈ 0.255 m
Now, let's calculate I for the thin-walled tube:
I = (π * ((0.255 m)^4 - (0.25 m)^4)) / 4
≈ 0.03493 m^4
Next, let's substitute the given values into the Euler buckling load formula for the thin-walled tube:
P_cr,thin = ((1^2 * π^2 * (69 * 10^9 N/m^2) * 0.03493 m^4) / (0.5m)^2)
P_cr,thin = ((π^2 * 69 * 10^9 * 0.03493) / 0.25)
P_cr,thin ≈ 33.12 kN
Therefore, the Euler buckling load for the thin-walled tube is approximately 33.12 kN.
To calculate the Euler buckling load for both the solid tube and the thin-walled tube, we need to use the formula:
P_cr = (π^2 * E * I) / (L^2)
where:
P_cr is the critical buckling load
E is the modulus of elasticity of aluminum (69 GPa)
I is the area moment of inertia of the tube
L is the length of the tube
First, let's calculate the area moment of inertia (I) for both the solid tube and the thin-walled tube.
For the solid tube:
The area moment of inertia for a solid cylinder is given by the formula:
I = (π/4) * (d^4)
where d is the diameter of the solid cylinder.
Given that the mass of the solid tube is 1 kg and the density of aluminum is 2700 kg/m^3, we can calculate the diameter of the solid cylinder using the formula:
m = ρ * V
1 kg = 2700 kg/m^3 * (π/4) * (d^2) * L
d^2 = (4 * 1 kg) / (2700 kg/m^3 * (π/4) * 0.5 m)
d^2 ≈ 0.2963
d ≈ √0.2963
d ≈ 0.544 m
Now, we can substitute the values into the Euler buckling load formula:
P_cr,sol = (π^2 * E * (π/4) * (d^4)) / (L^2)
P_cr,sol = (π^3 * E * d^4) / (16 * L^2)
P_cr,sol = (π^3 * 69 GPa * (0.544 m)^4) / (16 * (0.5 m)^2)
P_cr,sol ≈ 10.06 kN (rounded to two decimal places)
Therefore, the Euler buckling load for the solid tube is approximately 10.06 kN.
For the thin-walled tube:
The area moment of inertia for a thin-walled tube is given by the formula:
I = 2 * t * (d/2)^3
where t is the wall thickness and d is the outer diameter of the thin-walled tube.
Given that the wall thickness (t) is 0.5 cm and the outer diameter (D) is equal to the diameter of the solid tube, we can calculate the area moment of inertia (I) for the thin-walled tube as follows:
I = 2 * (0.005 m) * ((0.544 m)/2)^3
I ≈ 2.55316 × 10^-5 m^4
Substituting the values into the Euler buckling load formula:
P_cr,thin = (π^2 * E * I) / (L^2)
P_cr,thin = (π^2 * 69 GPa * 2.55316 × 10^-5 m^4) / (0.5 m)^2
P_cr,thin ≈ 5.04 kN (rounded to two decimal places)
Therefore, the Euler buckling load for the thin-walled tube is approximately 5.04 kN.