Euler Buckling Load of Spaghetti: Length & Diameter Impact

[pisiks]

Homework Statement

A piece of spaghetti 5 cm in length and 2mm in diameter has a weight of 0.2 gms. It will break under tension at a load of 5 kg (force). And its Euler buckling load is 500 gms (force). What length of the spaghetti could be suspended vertically from one end before it broke under its own weight?

What if the diameter were 4mm?

Homework Equations

EI(d2y / dx2) + Py = 0.

 

[PhanthomJay]

Have you made an attempt at a solution? Note that buckling criteria applies to members subject to compressive stresses. Is the hanging spaghetti subject to tension forces and/or compression forces?

 

[Mene]

(Euler's buckling equation)

The Euler buckling load of a material is determined by its elastic modulus (E), moment of inertia (I), and length (L). The diameter (d) is also a factor, as it affects the moment of inertia (I). As the diameter increases, the moment of inertia increases, resulting in a higher Euler buckling load. Therefore, if the diameter of the spaghetti is increased to 4mm, the Euler buckling load will also increase. This means that a longer length of spaghetti could be suspended vertically before it breaks under its own weight.

Using the equation, we can calculate the new Euler buckling load for a 5 cm length and 4mm diameter spaghetti:

EI(d2y / dx2) + Py = 0

Where:
E = elastic modulus of spaghetti
I = moment of inertia of spaghetti
d = diameter of spaghetti
y = deflection
x = length of spaghetti
P = load on the spaghetti

Since the length and weight of the spaghetti remain the same, we can focus on the change in diameter. The moment of inertia of the spaghetti is directly proportional to the diameter, so we can use the following equation to calculate the new moment of inertia (I'):

I' = (pi/64) * d^4

Where:
I' = new moment of inertia
d = new diameter

Substituting the values, we get:

I' = (pi/64) * (0.004m)^4 = 1.6 * 10^-10 m^4

Now, we can use the original Euler buckling equation and solve for the new load (P'):

EI'(d2y/dx2) + P'y = 0

500 g = E * 1.6 * 10^-10 m^4 * (d2y/dx2) + 0.2 g * 9.8 m/s^2 (since 500 g = 0.5 kg and 0.2 g = 0.0002 kg)

Solving for (d2y/dx2), we get:

(d2y/dx2) = -0.6 * 10^11 m^-2

Now, we can use this value to calculate the new length of spaghetti that can be suspended before it breaks under its own weight:

(d2y/dx2) = -0.6 *