Example: Cylinder. Find the polar moment of inertia and radius of gyration of the thick-walled cylinder with the cross-section shown below.
The Correct Answer and Explanation is:
Here is the solution for the problem.
Correct Answer:
The polar moment of inertia (J) and the radius of gyration (k) for the thick-walled cylinder’s cross-section are:
- Polar Moment of Inertia (J):
J = (π/2) * (r₂⁴ – r₁⁴) - Radius of Gyration (k):
k = √[ (r₁² + r₂²) / 2 ]
Explanation
This problem requires finding two important geometric properties for the cross-section of a thick-walled or hollow cylinder: the polar moment of inertia and the radius of gyration. The cross-section is an annulus, which is a ring-shaped area defined by an inner radius (r₁) and an outer radius (r₂).
The polar moment of inertia, denoted as J, is a measure of a cross-section’s ability to resist twisting or torsional deformation. To calculate J for the hollow circular cross-section, we can use the principle of superposition. This involves imagining the shape as a large, solid circle of radius r₂ and then subtracting the smaller, hollow circle of radius r₁. The formula for the polar moment of inertia of a solid circle about its center is (π/2)r⁴. Applying this principle, we subtract the moment of inertia of the inner circle from that of the outer circle. This results in the equation J = (π/2)r₂⁴ – (π/2)r₁⁴, which simplifies to J = (π/2)(r₂⁴ – r₁⁴).
The radius of gyration, denoted as k, represents the radial distance from the axis of rotation at which the entire cross-sectional area could be concentrated to produce the same polar moment of inertia. The relationship between these three quantities is given by the formula J = A * k², where A is the cross-sectional area. To find k, we first calculate the area of the annulus, which is the area of the outer circle minus the area of the inner circle: A = πr₂² – πr₁² = π(r₂² – r₁²). By rearranging the formula to solve for k, we get k = √(J/A). Substituting the expressions for J and A and simplifying the resulting fraction leads to the final expression for the radius of gyration: k = √[(r₁² + r₂²)/2].
