Exercise F-1.1.2

1. What is the formula for calculating the normal force \(N\) at any location x in a section perpendicular to the axis of the cone?

$$ N(x) = \dfrac{\varrho \cdot r_0^2 \cdot \pi \cdot g}{3 \cdot l^2} \cdot x^3 $$

If desired, one can further simplify \(r_0^2 \cdot \pi=A_0\) and obtain the final result as follows:

$$ N(x) = \dfrac{\varrho \cdot A_0 \cdot g}{3 \cdot l^2} \cdot x^3 $$

1. What is the formula for calculating the normal stress \(\sigma\) at any location x in a section perpendicular to the axis of the cone?

$$ \sigma(x) = \dfrac{\varrho \cdot g}{3} \cdot x $$

1. What is the magnitude of the normal stress \(\sigma\) at the location \(x=70~\mathrm{mm}\) in a section perpendicular to the axis of the cone?

$$ \sigma(x=70~\mathrm{mm}) = 2,083 \cdot 10^{-4}~\frac{\mathrm{N}}{\mathrm{mm^2}} $$

Preliminary Considerations

So, you want to crack the stress state in the cone? Cool stuff! But don't worry, it's not as tough as it seems.

First things first: Chill out! The cone is sturdily hanging in there; it won't just tumble down. While gravity is pulling downward, the support reaction, that is, the gutter-icicle connection, holds its ground. This creates a uniaxial stress state – all chill, no stress.

Now, you're looking for the normal stress in a section perpendicular to the axis. For that, you need the formula:

Okay, okay, looks complicated, but it's not a big deal.

\(N_x\) is the normal force acting in the section. It depends on the weight force of the cut cone. And that, my friend, depends on the position x in the cone: The higher you cut, the more cone mass pulls downward, the greater \(N_x\).

\(A\) is the cross-sectional area of the cone at point \(x\). The higher you cut, the bigger the circle, the larger \(A\).

See, both \(N_x\) and \(A\) change with \(x\). That's why you need to use formula 1.3b to calculate stress at every point \(x\).

But no need to panic! With a bit of math and patience, you got this.

Tip: Imagine slicing the cone into lots of thin discs. Calculate stress for each disc, and you'll get a sense of how stress is distributed in the cone.

Good luck! And if you ever feel stuck, just peek at the complete solution below.

P.S.: Don't forget that the cone is made of a certain material. The material influences stress, so consider the material properties.

P.P.S.: Oh, and units! Be careful with the right units; otherwise, the result won't add up in the end.

Step 1: Determine the normal force \(N(x)\) based on the equilibrium conditions

To calculate the normal force, we cut the rod at any arbitrary point \(x\) and represent the external force \(F_G\) and the internal normal force \(N\).

As a reminder, at the positive left cross-section, we represent \(N\) in the positive direction (towards the positive x-axis), while at the negative right cross-section, we represent it in the negative direction (opposite to the positive x-axis).

In this task, we need to consider that the weight of the "cut-off" cone-shaped object depends on \(x\): The larger the cut-off cone-shaped object, the greater its weight (here: \(F_{G1})\).

To solve this task, it is actually not necessary to cut out the remaining frustum of the cone from the section. But for the sake of completeness: At the right cross-section of the frustum, there are two forces acting: the reaction normal force of the section (also denoted as \(N(x)\)) and the weight of the remaining frustum (here: \(F_{G2})\).

In contrast, there is the reaction force of the gutter, which must be equal to the weight of the entire cone-shaped object before the section. This is because only this force acts as an external load, and the icicle must be in a state of rest.

---

Continue with TechMechAcademy+

Everything. Always. Everywhere.

With TechMechAcademy+ full access to all content.

---

Are you already a TechMechAcademy+ premium member? Then please log in here to enjoy full access to all content.