Supplementary Material, Appendix G – G.7

APPENDIX G – Pumps and Turbines

Main – Previous – Next

Surge Analysis and the Wave Plan Method

Supplementary Material: Example Problems and Solutions

Appendix G – Problem 7

G.7 Compute mass moment of inertia of a circular steel ring of 300 mm outer diameter, 250 mm inner diameter, and  5 mm thickness with respect to its principal x-axis as well as centerline z-axis. Compute radius of gyration of this plate with respect to both x and z axes. 

Solution:

Area moment of inertia I_x = (π/4) (r₂⁴ – r₁⁴)  = (π/4) (r₂² – r₁²) (r₂² + r₁²)  = π(r₂² – r₁²) (1/4) (r₂² + r₁²) 

I_x = Area * (1/4) (r₂² + r₁²) 

Mass moment of inertia I_xm = Mass m * (1/4) (r₂² + r₁²) 

m = π(r₂² – r₁²)  ρt

Mass moment of inertia I_xm = (1/4) (π(r₂² – r₁²) ρt) (r₂² + r₁²)  = (1/4) ρt π(r₂⁴ – r₁⁴) 

I_xm= 0.00803 kg-m² = 0.0787 N-m²

Radius of gyration R_xm = (I_x/Mass)^0.5

Mass = 0.8423 kg, R_xm = 0.0976 m

Polar mass moment of inertia I_zm = 2 I_xm 

I_zm = 0.01605 kg-m² = 0.1575 N-m²

Radius of gyration R_zm = (I_zm/Mass)^0.5 = 0.138 m

---

Summary of revisions to this page:

Date/Revision