STANDARD HANDBOOK OF PETROLEUM & NATURAL GAS ...

746 Drilling and Well Completions
Substituting Equations 459, 4-60, 456, and 4-55 into Equation 4-54 and solving
for the tensile load capacity of drill pipe yields
(4-61)
Example
Find the tensile load capacity of 5-in., nominal weight 19.5-lb/ft, steel grade
E, premium class drill pipe exposed to internal drill pipe pressure P,, = 3,000 psi
and rotary torque T = 15,000 ft-lb.
Solution
From Table 4-79 Nominal D, = 5 in., nominal ddp = 4.276 in., nominal wall
thickness t = 0.362. Reduced wall thickness for premium class drill pipe =
(0.8)(0.362) = 0.2896 in. Reduced D, for premium class = 4.276 + (2)(0.2896) =
4.8552 in. Cross-sectional area for premium class = Area based on reduced
Ddp - Area based on nominal ddp:
.It
d,, = 2(4.8552)2--(4.276)2 = 4.1538in.*
4 4
Section modulus for premium class:
Dip - df
’=:( D, )=E( 4.8552
From Table 4-81, PI = 311,400 lb (using Equation 4-61),
= 260,500 lb
(3,000)( 4.8552)( 4.1538) ( 4.1538)( 180,000)
[ 0.2896 8.9526
The reduction in the tensile load capacity of the drill pipe is 311,400 -
260,500 = 50,900 lb. That is about 17% of the tensile drill pipe resistance
calculated at the minimum yield strength in uniaxial state of stress. For practical
purposes, depending upon drilling conditions, a reasonable value of safety factor
should be applied.
During DST operations, the drill pipe may be affected by a combined effect
of collapse pressure and tensile load. For such a case,
or
(4-62)