STANDARD HANDBOOK OF PETROLEUM & NATURAL GAS ...

734 Drilling and Well Completions
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in vertical holes has been studied by A. Lubinski [171] and the weight on the
bit that results in first and second order buckling can be calculated as
wcr, = 1.94(~1p~)‘~ (4-51)
Wcrll = 3.75( EIp2)‘/’ (4-52)
where E = module of elasticity for drill collars in Ib/ft2 (for steel, E = 4320 x
lo6 lb/ft2)
p = unit weight of drill collar in drilling fluid in lb/ft
I = moment of inertia of the drill collar cross-section with respect to its
diameter, in ft
I = (~/64)(D:~- d4,J
Ddc = outside diameter of drill collars in ft
ddc = inside diameter of drill collars in ft
Example
Find the magnitude of the weight on bit and corresponding length of drill
collars that result in second order of buckling. Drill collars: 6Q in. x 2+ ft, mud
density = 12 lb/gal.
Solution
Moment of inertia:
Unit weight of drill collar in drilling fluid:
p = 108 1 -- = 88.181b/ft
( 6i24)
For weight on the bit that results in the second order of buckling, use
Equation 4-61:
Wcr,l, = 3.75(4320 X lo6 x 4.853 x
x 88.182)’/3 = 20,468 lb
Corresponding length of drill collars:
L, =-- 20’ 468 - 232ft
88.18
If the total length of drill collar string would be, for example, 330 ft, then
the number “232 ft” would indicate the distance from the bit to the neutral point.