Question 9.4: During the drilling of an oil well, the section of the drill......
During the drilling of an oil well, the section of the drill pipe at A (above ground level) is under combined loading due to a tensile force P = 70 kips and a torque T = 6 kip · ft, as illustrated in Fig. 1. The drill pipe has an outside diameter of 4.0 in. and an inside diameter of 3.640 in.
Determine the maximum shear stress at point A on the outer surface of the drill pipe. The radial stress at this point is zero. The yield strength in tension of this drill pipe is 95 ksi.
Plan the Solution We can use Mohr’s circle to combine the normal stress σ due to force P and the shear stress τ due to T, but we may also need to consider the three-dimensional aspect of the stress at A to determine the absolute maximum shear stress.
Stress Resultants: The stress resultants are given in the problem statement:
F = P = 70 kips, T = 6 kip · ft (1)
Individual Stresses: From Eq. 2.4, we get the normal stress
σ(x) = \frac{F(x)}{A(x)} Axial Stress Equation (2.4)
σ = \frac{F}{A}= \frac{70 kips}{\pi [(2 in.)^2 – (1.820 in.)^2 ]} = 32.41 ksi (2)
From Eq. 4.13, we get the torsional shear stress
τ_{max} = \frac{T_{max}r}{I_p} Maximum Shear-Stress Formula (4.13)
τ = \frac{Tr_o}{I_p}= \frac{(6 kip · ft)(12 in./ft)(2 in.)}{\frac{\pi}{2}[ (2 in.)^4 – (1.820 in.)^4]} = 18.23 ksi (3)
Figure 2 summarizes the “in-plane” stresses on the surface of the drill pipe at point A. The radial stress, normal to the surface, is zero.
Superposition of Stresses: Mohr’s circle may be used to combine the in-plane stresses in Fig. 2. From the Mohr’s circle in Fig. 3,
R = \sqrt{(16.20 ksi)^2 + (18.23 ksi)^2} = 24.39 ksi (4)
Then,
σ_1 = σ_{avg} + R = 16.20 + 24.39 = 40.6 ksi
σ_3 = σ_{avg} – R = 16.20 – 24.39 = -8.2 ksi (5)
The in-plane principal stresses are labeled σ_1 and σ_3, since the out-of-plane principal stress, σ_r = 0, is the intermediate principal stress. Then, from Eq. 8.32,
τ_{\begin{matrix} abs \\max \end{matrix} } = \frac{σ_{max} – σ_{min}}{2} (8.32)
τ_{\begin{matrix} abs \\max \end{matrix} } = \frac{σ_{max} – σ_{min}}{2}= \frac{40.6 ksi – (-8.2 ksi)}{2} (6a)
or
τ_{\begin{matrix} abs \\max \end{matrix} } = 24.4 ksi (6b)
Review the Solution The calculations in Eqs. (2) and (3) should be rechecked. Points X and Y are plotted correctly, so σ_1 and σ_3 appear to be correct. Finally, since the working stresses in this example should not produce yielding of the drill pipe, the absolute maximum shear stress should be much less than half the tensile yield strength. Therefore, the answer in Eq. (6b) seems reasonable.
