STANDARD HANDBOOK OF PETROLEUM & NATURAL GAS ...

744 Drilling and Well Completions
(text continued from page 739)
Consider a case in which the drill pipe is exposed to an axial load (P) and a
torque (T). The axial stress (0.) and the shear stress (2) are given by the following
formulas:
P
OZ = -
A
(4-55)
T
2=-
Z
(4-56)
where P = axial load in lb
A = cross-sectional area of drill pipe in.2
T = torque in in-lb
Z = polar section modulus of drill pipe, in.3
Z = 2J/Dd
J = (~/327(D$~ - d:,) polar moment of inertia, in.4
D = outside diameter of drill pipe in in.
dP
ddp = inside diameter of drill. pipe in in.
Substituting Equation 4-55 and Equation 4-56 into Equation 4-54 and putting
oe = Y, 6, = 0 (tangential stress equals zero in this case), the following formulas
are obtained:
(4-57)
(4-58)
where P, = Y,A = tensile load capacity of drill pipe in uniaxial tensile stress in lb
Equation 4-58 permits calculation of the tensile load capacity when the pipe
is subjected to rotary torque (T).
Example
Determine the tensile load capacity of a 4 +in., 16.6-lb/ft, steel grade X-95 new
drill pipe subjected to a rotary torque of 12,000 ft-lb if the required safety factor
is 2.0.
Solution
From Table 4-71, cross-sectional area body of pipe A = 4.4074 in.2 and Polar
section modulus Z = 8.542 in.3
From Table 4-80, tensile load capacity of drill pipe at the minimum yield
strength P, = 418,700 lb (P, = 4.4074 x 95,000 = 418,703 lb).
Using Equation 4-58,