APPENDIX G – Pumps and Turbines
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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-m 2 as well as N-m 2 . Compute radius of gyration of this plate with respect to both x and z axes.
Area moment of inertia I x = (π/4) r 4
I x = (πr 2 ) (1/4) r 2 = Area * (1/4) r 2
Mass moment of inertia I xm = Mass m * (1/4) r 2
Mass = m = Area * density ρ * plate thickness t = (πr 2 ρ t)
Mass moment of inertia about principal x-axis I xm = (1/4) m r 2 = (1/4) (πr 2 ρ t) r 2 = (1/4) ρt πr 4
I xm = 0.0155 kg-m 2 = 0.1521 N-m 2
Radius of gyration R xm = (I xm /Mass) 0.5
Mass = 2.7567 kg, R xm = 0.075 m
Polar mass moment of inertia I zm = 2 I x = (1/2) m r 2
I zm = (1/2) (πr 2 ρ t) r 2 = (1/2) ρt πr 4
I zm = 0.0310 kg-m 2 = 0.3042 N-m 2
Radius of gyration R zm = (I zm /Mass) 0.5 = 0.1061m
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