STANDARD HANDBOOK OF PETROLEUM & NATURAL GAS ...

Drill String: Composition and Design 745
P = (418700)' - 3
[ ( 8.542
Due to the safety factor of 2.0 the tensile load capacity of the drill pipe is
398432/2 = 199,216 lb.
Example
Calculate the maximum value of a rotary torque that may be applied to the drill
pipe as specified in Example 5 if the actual working tension load P = 300,000 lb.
(For instance, pulling and trying to rotate a differentially stuck drill string.)
Solution
From Equation 4-58, the magnitude of rotary torque is
so
(418,700' - 300,000)2
4.4074 3
= 353,571 in-lb or 29,464 ft-lb
Caution: No safety factor is included in this example calculation. Additional
checkup must be done if the obtained value of the torque is not greater than
the recommended makeup torque for tool joints.
During normal rotary drilling processes, due to frictional pressure losses, the
pressure inside the drill string is greater than that of the outside drill string.
The greatest difference between these pressures is at the surface.
If the drill string is thought to be a thin wall cylinder with closed ends, then
the drill pipe pressure produces the axial stress and tangential stress given by
the following formulas:
(For stress calculations, the pressure loss in the annulus may be ignored.)
0, =
'dpDdp
4t
(4-59)
2t
(4-60)
where oa = axial stress in psi
ot = tangential stress in psi
PdP = internal drill pipe pressure in psi
t = wall thickness of drill pipe in psi
Ddp = outside diameter of drill pipe in in.