6. Solution for u and u

As u is only a function of z and axial strain u' is a constant, integrating axial strain leads to (37) Here we assumed that u vanishes when z=0 because the axial displacement of the cylindrical mid-section (z=0) is zero by symmetry.

In this chunk, we discuss how the displacement 'u' depends only on the axial position 'z'. The axial strain, represented as 'u'', is constant across the length of the cylinder. When we integrate this strain with respect to 'z', we derive a relationship for 'u'. Additionally, we make an assumption that at the midpoint of the cylinder (where z = 0), there is no displacement due to symmetry. This means that any movement is balanced out at the center of the cylinder.

Imagine a perfectly balanced seesaw where the center does not move while the ends go up and down. The midpoint's position remains unchanged even if the ends are lifted. Similarly, in our cylinder, the midpoint has no axial displacement even as other points may experience strain.

A typical cross-section of the hollow cylinder is shown in Figure 3. The cross-section normal points in the z direction. So, σ, τ and τ act on it. We had found τ to be zero. Thus, we need to analyze σ and τ only. If the traction on any point on the cross-section is represented by t, then the moment due to this traction about the cross-section center 'O' will be given by the integration of r × t for each small area element in the cross-section (r represents the position vector of the area element from the center). If Ω denotes the area of the cross-section, the moment M will be given by (39) The torque T is simply the component of moment along the axis, i.e., (using (8)) (40) The term is the polar moment of area (a geometrical quantity) and is denoted by J.

Here, we are discussing how the torque (T) in the cylinder is related to the applied traction at any point in its cross-section. The moment (M) is derived from the traction times the distance from the center. Importantly, since we found that shear stress (τ) isn't present, our analysis focuses mainly on the normal stress (σ) and the resulting torque. The polar moment (J) is a measure of the cylinder's geometric resistance to twisting, which plays a critical role in determining how much torque is required to achieve a desired end-to-end rotation (Ω).

Think of twisting a soda can. When you twist it, the forces on different points of its circular cross-section create moments trying to cause rotation about the center. The further out you apply force (imagine pressing downwards on the edge), the more effective your twist is, much like how torque functions in the cylinder.

To find ϵ, let us obtain the axial force in the cross-section through the integration of σ, i.e., (42) Here, A denotes the cross-sectional area. As we know the value of C from equation (29), we are finally able to relate axial force F with axial strain ϵ. In the special case when P=0, the expression of C becomes simpler which yields (43) where E is the Young’s modulus of elasticity.

In this chunk, we focus on deriving the relationship between axial force (F) and axial strain (ϵ) using the integration of normal stress (σ) over the cylinder's cross-section area (A). By applying our earlier findings, especially related to the constant 'C', we establish how the force is linked to strain. In situations where there is no internal pressure (P=0), we simplify our earlier equations to arrive at a clearer expression that involves material properties represented by Young's modulus (E), a measure of stiffness.

Consider stretching a rubber band. The more force you apply (the axial force), the more it elongates (the strain). The relationship between the amount you can stretch it and the force you apply is characterized by the rubber band’s stiffness, similar to how we relate axial force and strain using Young’s modulus in our equations.

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