764
Drilling and Well Completions
where c
=
maximum permissible dog-leg severity in "/lo0 ft
ob
=
maximum permissible bending strength in psi
G~
=
tensile stress due to the weight of the drill string suspended below a
E
=
Young's modulus,
E
=
30
x
lo6
psi
D,,
=
outside diameter of drill pipe in in.
L
=
half the distance between tool joints, L
=
180 in. for Range 2 drill
pipe. Equation 4-75 does not hold true for Range
3
drill pipe.
T
=
weight of drill pipe suspended below the dog-leg in lb
A
=
cross-sectional area of drill pipe in in.*
dog-leg in psi
I
=
drill pipe moment of inertia with respect to its diameter in in.4
By
intelligent application of these formulas, several practical questions can be
answered, both at the borehole design state and while drilling.
Example
Calculate the maximum permissible hole curvature for data as below:
New EU 444.. Range 2 drill pipe, nominal weight 16.6 lb/ft, steel grade
S-135, with NC50
(IF)
tool joint
Drill collars,
7
x
2$ in., unit weight 117 lb/ft
Length of drill collars, 550 ft
Drilling fluid density, 12 lb/gal
Anticipated length of the hole below the dog-leg, 8,000 ft
Assume the hole is vertical below the dog-leg
Solution
From Table 4-79:
Ddp
=
4.5 in., ddp
-
3.826 in.,
A
=
4.4074 in.; and from Table
Weight of drill collar string is
4-100: unit weight of drill pipe adjusted for tool joint, WdP
=
18.8 lb/ft.
(550)( 117) 1
-
-
=
52,543 lb
(
6i24)
Weight of drill pipe is
(8,000-550)(18.8) 1--
=
114,3611b
(
6i24)
Weight suspended below the dog-leg, T
=
166,904 lb,
166,904
4.4074
Tensile stress
tst
=
-
=
37,869 psi
Maximum permissible bending stress,
(2,
=
20,000 1
-
-
=
14,7761b
(
1345
,EO)