4. Force on +θ and −θ planes

We will again use the approximate method. The coordinates of the center of the +θ and −θ planes are (r, θ + ∆θ, z) and (r, θ - ∆θ, z) respectively. The areas of the +θ and −θ planes are the same but we will still get extra terms here because of difference in basis vectors on the two planes: the two planes have different θ coordinates.

In this chunk, we are looking at how forces act on the +θ and -θ planes of the cylindrical element. The point where we measure these forces is slightly shifted in the angular direction (θ) by a small increment (∆θ). This increment means that while the areas of the two planes are the same, the way we calculate the forces acting on them will be influenced by the differences in the orientation of their basis vectors.

Think of this like holding two identical flat plates at a slight angle. Even though the plates are the same size, the direction in which you push them will affect how they respond, due to their angle. This is similar to how the basis vectors change based on the angle θ.

From Figure 2, we can note the areas of the θ faces (Aθ) is: Aθ = ∆r∆z.

Here we derive the area of the +θ and -θ planes. The area is calculated by multiplying the radial increment (∆r) by the height (∆z) of the cylindrical element. This equation gives us a clear picture of how much surface area is available for forces to act upon at those θ planes.

Imagine slicing a cylindrical cake into two even slices and measuring the flat surface area of each slice. If the height of the cake is constant and the thickness of each slice varies, the area available for icing (or in our case, force) will change accordingly based on how thick you cut each slice.

The total force on +θ and −θ planes will therefore be ... (the equation is not provided), and the traction on the +θ plane will be ... (the equation is not provided).

This chunk introduces the calculations for determining the overall force acting on the +θ and -θ planes. We need to account for the traction intensity at these specific planes. The specific traction values at these planes contribute to the total force calculated, highlighting that this calculation takes into consideration the unique positioning of the planes in relation to the basis vectors.

Consider holding a balloon with a string; the tension on the string can be thought of as traction. The upward force on the balloon relates to the amount of pull you exert on the string and varies depending on your angle. Just like the force on the planes, differing angles will produce different effects.

Now, e and e also have to be expanded using Taylor’s series. For the Taylor’s expansion, we will consider σe and τe as the functions to be expanded.

In this chunk, we discuss how to expand the components of the traction along the θ direction using Taylor's series. This expansion will help us account for changes in traction due to the angle θ. By expanding these functions, we can express forces in a way that considers slight variations in angle, thereby making our model more accurate.

Think about a car's speedometer. If you're traveling at a steady speed, but you hit a slight incline, the speed apparatus will react (or expand its reading) depending on the angle of the road – similar to how we adjust our force calculations with angle expansions.

Upon adding the two contributions, we finally obtain ... (the equation is not provided). We can see from this equation that the product rule will give us extra terms in the form of the derivatives of basis vectors with respect to θ.

This final portion shows the result of adding together the contributions from both the +θ and -θ planes. The product rule indicates that as we derive these functions concerning θ, we're introduced to additional terms that represent changes in the basis vectors along with changes in σ and τ. This further emphasizes the complexity of forces acting in different coordinate systems.

Imagine two friends pushing a seesaw from different angles. As they push harder (like applying more force), the reaction of the seesaw isn’t just affected by their strength but also by their angles and the way the seesaw's structure can shift under those forces, illustrating how changing angles affect overall force distributions.

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