Question 1 |

Consider two concentric circular cylinders of different materials M and N in contact with each other at r=b, as shown below. The interface at r=b is frictionless. The composite cylinder system is subjected to internal pressure P. Let (u_r^M,u_\theta ^M)
and (\sigma_{rr}^M, \sigma_{\theta \theta}^M) denote the radial and tangential displacement and stress components, respectively, in material M. Similarly, (u_r^N,u_\theta ^N)
and (\sigma_{rr}^N,\sigma_{\theta \theta}^N) denote the radial and tangential displacement and stress components, respectively, in material N. The boundary conditions that need to be satisfied at the frictionless interface between the two cylinders are:

u_r^M=u_r^N and \sigma _{rr }^M=\sigma _{rr}^N | |

u_r^M=u_r^N and \sigma _{rr }^M=\sigma _{rr}^Nand u_{\theta }^M=u_{\theta }^N and \sigma _{\theta \theta }^M=\sigma _{\theta \theta }^N | |

u_{\theta }^M=u_{\theta }^N and \sigma _{\theta \theta }^M=\sigma _{\theta \theta }^N | |

\sigma _{rr }^M=\sigma _{rr}^N and \sigma _{\theta \theta }^M=\sigma _{\theta \theta }^N |

Question 1 Explanation:

As the contact is frictionless, one cylinder can rotate freely with respect other. The displacement in
tangential directions need not be same at a point contact for two cylinders (i.e., u_{\theta}^{M} \neq u_{\theta}^{N} ). Similarly

the Hoop stress at point of contact need not be same (i.e., \sigma_{\theta \theta}^{M} \neq \sigma_{\theta \theta}^{N} ). As the interface will be always in contact the displacement in radial direction and stress in radial directions must be same for two cylinders. i.e., u_{r}^{M}=u_{r}^{N} and \sigma_{r}^{M}=\sigma_{r r}^{N}

the Hoop stress at point of contact need not be same (i.e., \sigma_{\theta \theta}^{M} \neq \sigma_{\theta \theta}^{N} ). As the interface will be always in contact the displacement in radial direction and stress in radial directions must be same for two cylinders. i.e., u_{r}^{M}=u_{r}^{N} and \sigma_{r}^{M}=\sigma_{r r}^{N}

Question 2 |

A thin cylindrical pressure vessel with closed-ends is subjected to internal pressure. The ratio of circumferential (hoop) stress to the longitudinal stress is

0.25 | |

0.5 | |

1 | |

2 |

Question 2 Explanation:

For thin cylinder

Circumferential stress

\sigma_{h}=\frac{p d}{2 t}

Longitudinal stress,

\begin{aligned} \sigma_{L} &=\frac{p d}{4 t} \\ \therefore \quad \frac{\sigma_{h} }{\sigma_{L}} &=2 \end{aligned}

Circumferential stress

\sigma_{h}=\frac{p d}{2 t}

Longitudinal stress,

\begin{aligned} \sigma_{L} &=\frac{p d}{4 t} \\ \therefore \quad \frac{\sigma_{h} }{\sigma_{L}} &=2 \end{aligned}

Question 3 |

A cylindrical tank with closed ends is filled with compressed air at a pressure of 500 kPa. The inner radius of the tank is 2 m, and it has wall thickness of 10 mm. The magnitude of maximum in-plane shear stress (in MPa) is ________

25MPa | |

34MPa | |

65MPa | |

45MPa |

Question 3 Explanation:

Maximum in plane shear stress (in MPa)

\begin{aligned} \tau &=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{\frac{p d}{2 t}-\frac{p d}{4 t}}{2} \\ &=\frac{500 \times 10^{3} \times 4000}{8 \times 10}=25 \times 10^{6} \mathrm{Pa}=25 \mathrm{MPa} \end{aligned}

\begin{aligned} \tau &=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{\frac{p d}{2 t}-\frac{p d}{4 t}}{2} \\ &=\frac{500 \times 10^{3} \times 4000}{8 \times 10}=25 \times 10^{6} \mathrm{Pa}=25 \mathrm{MPa} \end{aligned}

Question 4 |

A gas is stored in a cylindrical tank of inner radius 7 m and wall thickness 50 mm. The gage pressure of the gas is 2 MPa. The maximum shear stress (in MPa) in the wall is

35 | |

70 | |

140 | |

280 |

Question 4 Explanation:

Maximum shear stress in the wall

=\frac{\sigma_{1}}{2}=\frac{p d}{4 t}=\frac{2 \times 14 \times 1000}{4 \times 50}=140 \mathrm{MPa}

=\frac{\sigma_{1}}{2}=\frac{p d}{4 t}=\frac{2 \times 14 \times 1000}{4 \times 50}=140 \mathrm{MPa}

Question 5 |

A thin gas cylinder with an internal radius of 100 mm is subject to an internal pressure of 10 MPa. The maximum permissible working stress is restricted to 100 MPa. The minimum cylinder wall thickness (in mm) for safe design must be _______

10mm | |

20mm | |

30mm | |

40mm |

Question 5 Explanation:

\begin{aligned} \sigma_{\text {wothing }} &=\frac{p d}{2 t} \quad(d=2 r=200 \mathrm{mm}) \\ t &=\frac{p d}{2 \sigma_{w}}=\frac{10 \times 200}{2 \times 100}=10 \mathrm{mm} \end{aligned}

Question 6 |

A long thin walled cylindrical shell, closed at both the ends, is subjected to an internal pressure. The ratio of the hoop stress (circumferential stress) to longitudinal stress developed in the shell is

0.5 | |

1 | |

2 | |

4 |

Question 6 Explanation:

Hoop stress: \sigma_{h}=\frac{p d}{2 t}

Longitudinal stress: \sigma_{L}=\frac{p d}{4 t}

\therefore \frac{\sigma_{h}}{\sigma_{L}}=2

Longitudinal stress: \sigma_{L}=\frac{p d}{4 t}

\therefore \frac{\sigma_{h}}{\sigma_{L}}=2

Question 7 |

A thin walled spherical shell is subjected to an internal pressure. If the radius of the shell is increased by 1% and the thickness is reduced by 1%, with the internal pressure remaining the same, the percentage change in the circumferential (hoop) stress is

0 | |

1 | |

1.08 | |

2.02 |

Question 7 Explanation:

Hoop stress =\frac{p d}{4 t}

p= internal pressure

d= diameter of shell

t= thickness of shell

\begin{array}{l} \text { Now } \sigma=\frac{p d^{\prime}}{4 t^{\prime}} \\ \begin{aligned} d^{\prime}=1.01 d &\; \text { and } t^{\prime}=0.99 t \\ \therefore \quad \sigma^{\prime} &=\frac{p \times 1.01 d}{4 \times 0.99 t}=1.0202 \sigma \\ \text { \% change } &=\frac{\sigma^{\prime}-\sigma}{\sigma} \times 100 \\ &=\frac{(1.0202-1) \sigma}{\sigma} \times 100=2.02 \% \end{aligned} \end{array}

p= internal pressure

d= diameter of shell

t= thickness of shell

\begin{array}{l} \text { Now } \sigma=\frac{p d^{\prime}}{4 t^{\prime}} \\ \begin{aligned} d^{\prime}=1.01 d &\; \text { and } t^{\prime}=0.99 t \\ \therefore \quad \sigma^{\prime} &=\frac{p \times 1.01 d}{4 \times 0.99 t}=1.0202 \sigma \\ \text { \% change } &=\frac{\sigma^{\prime}-\sigma}{\sigma} \times 100 \\ &=\frac{(1.0202-1) \sigma}{\sigma} \times 100=2.02 \% \end{aligned} \end{array}

Question 8 |

A thin cylinder of inner radius 500mm and thickness 10mm is subjected to an internal pressure of 5MPa. The average circumferential (hoop) stress in MPa is

100 | |

250 | |

500 | |

1000 |

Question 8 Explanation:

Circumferential or hoop stress.

\begin{aligned} \sigma_{h}&=\frac{p d}{2 t}\\ \text{where }p&=5 \mathrm{MPa} \\ d &=1000 \mathrm{mm} \\ t &=10 \mathrm{mm} \\ \therefore \quad \sigma_{H} &=\frac{5 \times 1000}{2 \times 10}=250 \mathrm{MPa} \end{aligned}

\begin{aligned} \sigma_{h}&=\frac{p d}{2 t}\\ \text{where }p&=5 \mathrm{MPa} \\ d &=1000 \mathrm{mm} \\ t &=10 \mathrm{mm} \\ \therefore \quad \sigma_{H} &=\frac{5 \times 1000}{2 \times 10}=250 \mathrm{MPa} \end{aligned}

There are 8 questions to complete.