3.4.4 Show that the dependence on tube geometry may be expressed in terms of \(I\), the second moment of area of the tube section about the tube axis, so that (3.4.5) may be written $$ \frac{M}{I}=\mu \frac{\phi}{l} $$

We are given the equation $$ \frac{M}{I}=\mu \frac{\phi}{l} $$ Our objective is to show that this equation expresses the dependence on tube geometry in terms of the second moment of area (I) of the tube section about the tube axis.

Rearrange the equation to make the second moment of area (I) the subject of the equation. $$ I = \frac{Ml}{\mu \phi} $$ Now, the equation shows the second moment of area (I) in terms of bending moment (M), length of the tube (l), the coefficient of rigidity (μ), and the angle of twist (ϕ).

In this step, we will explain how the second moment of area (I) depends on the tube's geometry. The second moment of area (I) is a geometrical property of the cross-sectional shape of a beam, which determines the stiffness of the beam in bending. In our derived equation: $$ I = \frac{Ml}{\mu \phi} $$ The second moment of area (I) is directly proportional to the bending moment (M) and the length of the tube (l) and inversely proportional to the coefficient of rigidity (μ) and the angle of twist (ϕ). As the bending moment (M) and the length of the tube (l) increase, so does the second moment of area (I), indicating a stiffer beam. Conversely, as the coefficient of rigidity (μ) and the angle of twist (ϕ) increase, the second moment of area (I) decreases, indicating a less stiff beam. Therefore, the equation shows the dependence on tube geometry in terms of the second moment of area (I) of the tube section about the tube axis.