Overview

Test Case

Determine the first two natural frequencies f₁ and f₂ of an oil-well drill pipe of length and polar moment at inertia l_p fixed at the upper end and terminating at the lower end to a drill collar with torsional mass inertia J_o. The drill collar length is small compared to the pipe length.

Figure 81: Drill Pipe Problem Sketch

Material Properties	Geometric Properties
G = 12 x 106 psi υ = 0.3 γ = 490 lb/ft3 ρ =  γ / g = 15.2174 lb-sec2/ft4	= 5000 ft OD = 4.5 in = (4.5/12) ft ID = 3.83 in = (3.83/12) ft Jo = 29.3 lb-ft-sec2 Ip = 0.0009226 ft4

Analysis Assumptions and Modeling Notes

The drill pipe is modeled using pipe and beam elements. Modal analysis is performed with Block Lanczos solver for (PIPE16), (PIPE288), (PIPE289), (BEAM188), and (BEAM189) elements. Young's modulus (E) is calculated as E = 2G (1 + υ)*144 = 4.4928 x 109lb/ft² and pipe thickness is calculated as (OD - ID)/2.

Results Comparison