4.3. Final Solution

Upon plugging the boundary condition (26) in equation (20), we get the following set of equations: (27) solving which we get . (28)

In this chunk, we start with the application of the boundary conditions derived earlier to the equation from the previous sections. We substitute boundary conditions into our mathematical model to derive specific equations. The new set of equations we get will help in analyzing the behavior of the hollow cylinder under stress. These equations will eventually aid in calculating the stresses at various sections of the cylinder.

Imagine trying to fit a lid onto a jar. You know that the lid should fit snugly at the top (boundary condition). When you apply that condition and adjust, you determine how tightly it should screw on. Similarly, applying our boundary conditions lets us find how stresses distribute within the cylinder.

For a positive internal pressure P, we always have A > 0 and B < 0. Using (20), we can now plot the variation in both σ and σ through the tube’s thickness as shown in Figure 2.

This chunk discusses the implications of the results we derived. When we apply a positive internal pressure, the constants A and B we found give us insight into how stress behaves within the material. A positive A indicates tensile stress at certain points, while a negative B suggests compressive stress at others. Plotting these stresses helps visualize how they change from the inner to the outer surface of the cylinder.

Think of a balloon. When you blow air into it (akin to applying pressure), the inside stretches while the outside pushes against the air inside. Depending on how much air you put, the stresses within the balloon change: it might bulge more at one point than another. That’s similar to how our cylindrical model predicts stress distribution.

As r → ∞, both σ and σ approach which is a positive number for positive pressure P although r is feasible only between r1 and r2. From the boundary condition, we also know directly from boundary condition that σ is −P at r = r1 and 0 at r = r2.

This part elaborates on the behavior of stresses as one moves radially outward in the cylinder. As we approach infinity radially, the stresses stabilize at a certain level, indicating the material’s response to internal pressure. The boundary conditions give us fixed points: the stress at the inner surface is equal to the negative pressure, and at the outer surface, the stress is zero. This establishes how the cylinder resists internal pressure by showing that internal pressure leads to negative stress.

Imagine a tightly packed soda can. When you pressure the can from the inside (the soda), at one point, called the bottom (r1), it feels a lot of pressure pushing out (−P). As you move upwards (out to r2, the top), the can feels less pressure until at the very top (r2), it feels none, indicating the can can bear pressures differently based on its shape and material.

Thus, we infer that extension and torsion cannot generate σ and σ. This happens because the cross section is free to relax during extension and torsion of a circular cylinder.

In this chunk, the discussion revolves around the effects of extension and torsion on stress. It highlights a critical observation: simply pulling (extension) or twisting (torsion) a circular cylinder does not inherently create internal stresses σrr and σθθ. The circular shape allows it to accommodate these actions without generating additional stresses, indicating the special role of form in stress behavior.

Think about stretching a rubber band. When you pull it, it can stretch easily without tearing. Now, if you twist a bottle cap, the cap is designed to let you twist without breaking—its shape allows it to handle the torque. This is similar to our cylindrical model; its shape helps it to withstand certain forces without internal stress.

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