APPENDIX G – Pumps and Turbines
Surge Analysis and the Wave Plan Method
Supplementary Material: Example Problems and Solutions
Appendix G – Problem 6
G.6 Compute mass moment of inertia of a circular steel plate of 300 mm diameter and 5 mm thickness with respect to its principal x-axis as well as centerline z-axis. Express the results in both kg-m2 as well as N-m2. Compute radius of gyration of this plate with respect to both x and z axes.
Solution:
Area moment of inertia Ix = (π/4) r4
Ix = (πr2) (1/4) r2 = Area * (1/4) r2
Mass moment of inertia Ixm = Mass m * (1/4) r2
Mass = m = Area * density ρ * plate thickness t = (πr2 ρ t)
Mass moment of inertia about principal x-axis Ixm = (1/4) m r2 = (1/4) (πr2 ρ t) r2= (1/4) ρt πr4
Ixm = 0.0155 kg-m2 = 0.1521 N-m2
Radius of gyration Rxm = (Ixm/Mass)0.5
Mass = 2.7567 kg, Rxm = 0.075 m
Polar mass moment of inertia Izm = 2 Ix = (1/2) m r2
Izm = (1/2) (πr2 ρ t) r2= (1/2) ρt πr4
Izm = 0.0310 kg-m2 = 0.3042 N-m2
Radius of gyration Rzm = (Izm/Mass)0.5 = 0.1061m
Summary of revisions to this page:
Date/Revision
