Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Your Ultimate AI Essay Writer & Assistant.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Your Ultimate AI Essay Writer & Assistant.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 11

Q. 11.10

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.

Step-by-Step

Verified Solution

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.

11.30