Question 11.10: A closed thin-walled tube has a mean radius of 40.0 mm and a...
A closed thin-walled tube has a mean radius of 40.0 mm and a wall thickness of 4.0 mm. It is subjected to an internal pressure of 11.0 MPa. In addition to the internal pressure, the tube is subjected to an axial load P_z = 80.0 kN, bending moments M_x = 600 Nm and M_y = 480 Nm and torque T = 3.60 kN m. If yielding is impending in the tube, calculate the yield point stress \left(\sigma_{y p}\right) of the material based on the
(a) maximum shear stress criterion of failure by yielding and
(b) maximum octahedral shear stress criterion of failure by yielding.
Let us consider Figure 11.30.
For the thin-walled tube with mean radius r and wall thickness t, we get its area A as
A=\pi\left\lgroup r+\frac{t}{2} \right\rgroup^2-\pi\left\lgroup r-\frac{t}{2} \right\rgroup^2=4 \pi r \cdot \frac{t}{2}=2 \pi r t
Area moment of inertia is
I=\frac{\pi}{4}\left\lgroup r+\frac{t}{2} \right\rgroup^4-\frac{\pi}{4}\left\lgroup r-\frac{t}{2} \right\rgroup^4
=\frac{\pi}{4}\left\{\left[\left\lgroup r+\frac{t}{2} \right\rgroup^2+\left\lgroup r-\frac{t}{2} \right\rgroup^2\right]\left[\left\lgroup r+\frac{t}{2} \right\rgroup^2-\left\lgroup r-\frac{t}{2} \right\rgroup^2\right]\right\}
=\frac{\pi}{4}\left[2\left\lgroup r^2+\frac{t^2}{4} \right\rgroup 4 r\left\lgroup \frac{t}{2} \right\rgroup\right]=\pi\left[r\left\lgroup r^2+\frac{t^2}{4} \right\rgroup t\right]
\cong \pi r^3 t
and polar moment of inertia = J = 2I = 2πr³t.
Thus, for the thin-walled tube, we obtain
A=2 \pi r t=2 \pi(40)(4) mm ^2=1005.31 mm ^2
\bar{I}=\pi r^3 t=\pi(40)^3(4) mm ^4=804.25 \times 10^3 mm ^4
and J=2 \pi r^3 t=1.6085 \times 10^6 mm ^4
Now
1. Due to internal pressure p = 11.0 MPa, circumferential stress is
\sigma_1=\frac{p r}{t}=(11) \frac{40}{4}=110 MPa (1)
Axial stress is
\sigma_2=\frac{p r}{2 t}=55 MPa (2)
2. Due to axial force P_z , we get axial stress as
\sigma^{\prime}=\frac{P_z}{A}=\frac{80 \times 10^3}{1005.31}=79.58 MPa (3)
3. Due to M_{x}, we get axial stress as
\sigma^{\prime \prime}=\frac{660\left(10^3\right)(40)}{804.25\left(10^3\right)} MPa =32.83 MPa (4)
4. Due to M_{y} , we get axial stress as
\sigma^{\prime \prime \prime}=\frac{480\left(10^3\right)(40)}{804.25\left(10^3\right)} MPa =23.87 MPa (5)
5. Due to torsional moment, shear stress is
\tau=\frac{\operatorname{Tr}}{J}=\frac{3.6\left(10^6\right)(40)}{1.6085\left(10^6\right)}=89.52 MPa (6)
Now, normal stress on any element of the pressure vessel is
\sigma_{z z}=\left(\sigma^{\prime}+\sigma_2\right)+\sqrt{\sigma^{\prime \prime 2}+\sigma^{\prime \prime \prime 2}}
=(79.58+55)+\sqrt{32.83^2+23.87^2} MPa
= 175.17 MPa
Also,
\sigma_{y y}=\sigma_1=110 MPa \text { and } \tau_{y z}=89.52 MPa
Using maximum octahedral shear stress theory [refer to Eq. (11.44)], we get
\sqrt{I_1^2-3 I_2}=\sigma_{ yp } (11.44)
\sigma_{ yp }=\sqrt{\sigma_{y y}^2+\sigma_{z z}^2-\sigma_{y y} \sigma_{z z}+3 \tau_{y z}^2}
=\sqrt{110^2+175.17^2-(110)(175.17)+3(89.52)^2} MPa
= 218.1 MPa
Again for the above stress state, we get
\sigma_{\max }=\frac{1}{2}\left(\sigma_{y y}+\sigma_{z z}\right)+\sqrt{\left\lgroup \frac{\sigma_{y y}-\sigma_{z z}}{2} \right\rgroup^2+\tau_{y z}^2}
=\frac{1}{2}(110+175.17)+\sqrt{\left\lgroup \frac{175.17-110}{2} \right\rgroup^2+89.52^2}
= 237.85 MPa
and \sigma_{\min }=\frac{1}{2}(110+175.17)-\sqrt{\left\lgroup \frac{175.17-110}{2} \right\rgroup^2+(89.52)^2}
= 47.32 MPa
According to the maximum shear stress theory
\frac{1}{2} \sigma_{ yP }=\operatorname{Max}\left\lgroup \frac{\left|\sigma_1-\sigma_2\right|}{2}, \frac{\left|\sigma_2-\sigma_3\right|}{2}, \frac{\left|\sigma_3-\sigma_1\right|}{2} \right\rgroup
or \sigma_{ yp }=\operatorname{Max}(237.85,47.32,190.53) MPa
\sigma_{ yp }=237.85 MPa
Therefore, the required yield point stresses for the material are: 237.85 MPa by maximum shear stress theory and 218.1 MPa by maximum octahedral shear stress theory.
