Question 8.4: The hollow pipe casing for a production oil well (see Fig. 8...
The hollow pipe casing for a production oil well (see Fig. 8-24) is 200 mm in outer diameter and 18 mm in thickness. The internal pressure due to gas and oil is 15 MPa. At some point above the blowout preventer, the compressive force in the pipe (due to the weight of the pipe) is 175 kN, and the torque is 14 kN.m . Determine the maximum tensile, compressive, and shear stresses in the pipe casing.
The stresses in the well casing are produced by the combined action of the axial force P, the torque T, and internal pressure p (Fig. 8-24b). Therefore, the stresses at any point on the surface of the shaft at some depth consist of circumferential stress \sigma_{x}, longitudinal stress \sigma_{y}, and shear stresses \tau_{xy}, as shown on the stress element on the surface of the casing in Fig. 8-24b. Note that the y-axis is parallel to the longitudinal axis of the casing.
The circumferential stress \sigma_{x} is due to the internal pressure of oil and gas and is computed using Eq. (8-11) \sigma_{1}=\frac{p r}{t} as
\sigma_{x}=\frac{p r}{t}=\frac{[15 \mathrm{~MPa} \times(100 \mathrm{~mm})]}{18 \mathrm{~mm}}=83.3 \mathrm{~MPa}The longitudinal stress \sigma_{y} is caused by the axial compressive force P (due to self-weight) and is divided by casing cross sectional area A. The longitudinal tensile stress \sigma_{L} is due to internal pressure [see Eq. (8-12) \sigma_{2}=\frac{p r}{2 t} to find \sigma_{L}, which is non-zero when the well is capped and not operational]. Here we assume that oil and gas are flowing, so \sigma_{L} is zero and \sigma_{y} is computed as
\sigma_{y}=\frac{-P}{A}=\frac{-(175 \mathrm{~kN})}{\pi\left[r^{2}-(r-t)^{2}\right]}=-17 \mathrm{~MPa}The shear stress \tau_{xy} is obtained from the torsion formula [see Eq. (3-13) of Section 3.3]:
\tau_{x y}=\frac{T r}{l_{p}}=\frac{(14 \mathrm{~kN} \cdot \mathrm{~m}) \times(100 \mathrm{~mm})}{8.606\left(10^{-5}\right) \mathrm{~m}^{4}}=16.3 \mathrm{~MPa}The shear stress is positive in accordance with the sign convention established in Section 1.7.
Knowing the stresses \sigma_{x}, \sigma_{y}, and \tau_{xy}, we now can obtain the principal stresses and maximum shear stresses by the methods described in Section 7.3. The principal stresses are obtained from Eq. (7-17):
Substituting \sigma_{x}=83.3 \mathrm{~MPa}, \sigma_{y}=-17 \mathrm{~MPa} \text {, and } \tau_{x y}=16.3 \mathrm{~MPa} , we get
These are the maximum tensile and compressive stresses in the drill casing. The maximum in-plane shear stresses from Eq. (7-25) are
\tau_{\max }=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}}=52.7 \mathrm{~MPa}Because the principal stresses \sigma_{1} and \sigma_{2} have opposite signs, the maximum in-plane shear stresses are larger than the maximum out-of-plane shear stresses [see Eqs. (7-28a, b, and c) and the accompanying discussion]. Therefore, the maximum shear stress in the drill casing is 52.7 MPa.
\begin{aligned}&\left(\tau_{\max }\right)_{x_{1}}=\pm \frac{\sigma_{2}}{2} \quad\left(\tau_{\max }\right)_{y_{1}}=\pm \frac{\sigma_{1}}{2} \\&\left(\tau_{\max }\right)_{z_{1}}=\pm \frac{\sigma_{1}-\sigma_{2}}{2}\end{aligned} (7-28 a,b,c)
