Question 1 |

A thin-walled cylinder of radius r and thickness t is open at both ends, and fits snugly
between two rigid walls under ambient conditions, as shown in the figure.

The material of the cylinder has Young's modulus E, Poisson's ratio v, and coefficient of thermal expansion \alpha. What is the minimum rise in temperature \Delta T of the cylinder (assume uniform cylinder temperature with no buckling of the cylinder) required to prevent gas leakage if the cylinder has to store the gas at an internal pressure of p above the atmosphere?

The material of the cylinder has Young's modulus E, Poisson's ratio v, and coefficient of thermal expansion \alpha. What is the minimum rise in temperature \Delta T of the cylinder (assume uniform cylinder temperature with no buckling of the cylinder) required to prevent gas leakage if the cylinder has to store the gas at an internal pressure of p above the atmosphere?

\Delta T=\frac{3vpr}{2\alpha tE} | |

\Delta T=\left ( v-\frac{1}{4} \right )\frac{pr}{\alpha tE} | |

\Delta T=\frac{vpr}{\alpha tE} | |

\Delta T=\left ( v+\frac{1}{2} \right )\frac{pr}{\alpha tE} |

Question 1 Explanation:

since cylinder is open at both end.

\begin{aligned} \therefore \qquad \sigma_{L}&=0\\ \text{For no leakage,} \quad \varepsilon_{L_{P_{r}}}&=\varepsilon_{L_{\text {temp }}} \\ \frac{v \sigma_{n}}{E} &=\alpha \Delta T \\ \frac{v P r}{t E} &=\alpha \Delta T \\ \Delta T &=\frac{v p r}{\alpha t E} \end{aligned}

\begin{aligned} \therefore \qquad \sigma_{L}&=0\\ \text{For no leakage,} \quad \varepsilon_{L_{P_{r}}}&=\varepsilon_{L_{\text {temp }}} \\ \frac{v \sigma_{n}}{E} &=\alpha \Delta T \\ \frac{v P r}{t E} &=\alpha \Delta T \\ \Delta T &=\frac{v p r}{\alpha t E} \end{aligned}

Question 2 |

A steel bar is held by two fixed supports as shown in the figure and is subjected to an increases of temperature \Delta T=100^{\circ}C.If the coefficent of thermal exapnasion and young's modules of elasticity of steel are 11\, \times \, 10^{-6} \, /^{\circ}C and 200GPa, respectively, the magnitude of thermal stress (in MPa) induced in the bar is _____

220 | |

225 | |

230 | |

235 |

Question 2 Explanation:

We know that for completely restricted expansion, thermal stress developed in bar is given by

\begin{aligned} \sigma_{t h} &=\alpha(\Delta T) E \\ &=11 \times 10^{-6} \times 100 \times 200 \times 10^{3}\\ &=220 \mathrm{MPa}\text{(comp. in native)} \end{aligned}

\begin{aligned} \sigma_{t h} &=\alpha(\Delta T) E \\ &=11 \times 10^{-6} \times 100 \times 200 \times 10^{3}\\ &=220 \mathrm{MPa}\text{(comp. in native)} \end{aligned}

Question 3 |

A cantilever beam of length L and flexural modulus EI is subjected to a point load P at the free end. The elastic strain energy stored in the beam due to bending (neglecting transverse shear) is

\frac{P^{2}L^{3}}{6EI} | |

\frac{P^{2}L^{3}}{3EI} | |

\frac{PL^{3}}{3EI} | |

\frac{PL^{3}}{6EI} |

Question 3 Explanation:

Let U is the S.E. in the beam due to B.M. (M)

U=\int_{0}^{L}\frac{(M_{x-x})^2 dx}{2(EI)_{x-x}}=\int_{0}^{L}\frac{(Px)^2 dx}{2EI_{NA}}

U=\frac{P^2}{2EI_{NA}}\int_{0}^{L}x^2dx=\frac{P^2}{2EI}\left ( \frac{L^3}{3} \right )

U=\frac{P^2L^3}{6EI_{NA}}

U=\int_{0}^{L}\frac{(M_{x-x})^2 dx}{2(EI)_{x-x}}=\int_{0}^{L}\frac{(Px)^2 dx}{2EI_{NA}}

U=\frac{P^2}{2EI_{NA}}\int_{0}^{L}x^2dx=\frac{P^2}{2EI}\left ( \frac{L^3}{3} \right )

U=\frac{P^2L^3}{6EI_{NA}}

Question 4 |

Two threaded bolts A and B of same material and length are subjected to identical tensile load. If the elastic strain energy stored in bolt A is 4 times that of bolt B and the mean diameter of bolt A is 12 mm, the mean diameter of bolt B in mm is

16 | |

24 | |

36 | |

48 |

Question 4 Explanation:

Given:

\begin{aligned} P_{1}&=P_{2}=P\\ &\text{(identical tensile load on bolt A \& B )} \\ &\qquad \text{(same length)}\\ L_{1} &=L_{2}=L \\ d_{A} &=12 \mathrm{mm} \\ U_{A} &=\text { strain energy in bolt } A \\ U_{B} &=\text { Strain energy in bolt } B \\ U_{A} &=4 U_{B}(\text { Given }) \qquad \cdots(i)\\ \therefore \quad d_{B} &=? \end{aligned}

Strain energy is given by

U_{A}=\frac{1}{2} P \times \delta=\frac{1}{2} \frac{P^{2} L}{A E}

\therefore Eq. (i)

\begin{aligned} \frac{1 P_{1}^{2} L_{1}}{2} &=4 \times \frac{1}{2} \frac{P_{2}^{2} L_{2}}{A_{2} E} \\ \frac{1}{A_{1}} &=4 \times \frac{1}{A_{2}} \\ \frac{4}{\pi(12)^{2}} &=4 \times \frac{4}{\pi\left(d_{B}\right)^{2}} \\ d_{B}^{2} &=576 \\ d_{B} &=24 \mathrm{mm} \end{aligned}

\begin{aligned} P_{1}&=P_{2}=P\\ &\text{(identical tensile load on bolt A \& B )} \\ &\qquad \text{(same length)}\\ L_{1} &=L_{2}=L \\ d_{A} &=12 \mathrm{mm} \\ U_{A} &=\text { strain energy in bolt } A \\ U_{B} &=\text { Strain energy in bolt } B \\ U_{A} &=4 U_{B}(\text { Given }) \qquad \cdots(i)\\ \therefore \quad d_{B} &=? \end{aligned}

Strain energy is given by

U_{A}=\frac{1}{2} P \times \delta=\frac{1}{2} \frac{P^{2} L}{A E}

\therefore Eq. (i)

\begin{aligned} \frac{1 P_{1}^{2} L_{1}}{2} &=4 \times \frac{1}{2} \frac{P_{2}^{2} L_{2}}{A_{2} E} \\ \frac{1}{A_{1}} &=4 \times \frac{1}{A_{2}} \\ \frac{4}{\pi(12)^{2}} &=4 \times \frac{4}{\pi\left(d_{B}\right)^{2}} \\ d_{B}^{2} &=576 \\ d_{B} &=24 \mathrm{mm} \end{aligned}

Question 5 |

A solid steel cube constrained on all six face is heated so that the temperature rises uniformly by \DeltaT. If the thermal coefficient of the material is \alpha, young's modulus is E and the Poisson's ratio is v, the thermal stress developed in the cube due to heating is

-\frac{\alpha (\Delta T)E}{(1-2v)} | |

-\frac{2\alpha (\Delta T)E}{(1-2v)} | |

-\frac{3\alpha (\Delta T)E}{(1-2v)} | |

-\frac{\alpha (\Delta T)E}{3(1-2v)} |

Question 5 Explanation:

Let the side of cube be a

Now since the cube is uniformly constrained to

expand, the stress produced in all the three

directions will be same

\therefore Strain in x direction

\begin{aligned} &=-\alpha(\Delta T)=\frac{\sigma_{x}}{E}-\frac{v \sigma_{y}}{E}-\frac{v \sigma_{x}}{E} \\ \sigma_{x}&=\sigma_{y}=\sigma_{z}=\sigma \\ \therefore \sigma&=-\frac{\alpha(\Delta T) E}{(1-2 v)} \end{aligned}

Now since the cube is uniformly constrained to

expand, the stress produced in all the three

directions will be same

\therefore Strain in x direction

\begin{aligned} &=-\alpha(\Delta T)=\frac{\sigma_{x}}{E}-\frac{v \sigma_{y}}{E}-\frac{v \sigma_{x}}{E} \\ \sigma_{x}&=\sigma_{y}=\sigma_{z}=\sigma \\ \therefore \sigma&=-\frac{\alpha(\Delta T) E}{(1-2 v)} \end{aligned}

Question 6 |

A stepped steel shaft shown below is subjected to 10 Nm torque. If the modulus of rigidity is 80 GPa, the strain energy in the shaft in N mm is

4.12 | |

3.46 | |

1.73 | |

0.86 |

Question 6 Explanation:

\begin{array}{l} \text { Strain energy }=\frac{T^{2} L}{2 G J_{1}}+\frac{T^{2} L}{2 G J_{2}} \\ \qquad \frac{T^{2} I}{2 G}\left(\frac{1}{J_{1}}+\frac{1}{J_{2}}\right)=\frac{(10 \times 1000)^{2} \times 100}{2 \times 80 \times 1000}\\ \qquad\left[\left(\frac{32}{\pi \times(50)^{4}}+\frac{32}{\pi \times(25)^{4}}\right)\right] \\ =\frac{10^{8} \times 100 \times 32}{2 \pi \times 80 \times 1000}\left[\frac{1}{(50)^{4}}+\frac{1}{(25)^{4}}\right]=1.73 \mathrm{N}-\mathrm{mm} \end{array}

Question 7 |

A steel rod of length L and diameter D, fixed at both ends, is uniformly heated to a temperature rise of \Delta T . The Young's modulus is E and the co efficient of linear expansion is \alpha. The thermal stress in the rod is

0 | |

\alpha \Delta T | |

E\alpha \Delta T | |

E\alpha \Delta T L |

Question 7 Explanation:

Thermal strain,

\begin{aligned} \varepsilon_{t} &=\frac{\text { Change in length prevented }}{\text { original length }}\\ &=\frac{L \alpha \Delta T}{L} \\ &=\alpha \Delta T\end{aligned}

Thermal stress,

\begin{aligned} \sigma_{t} &=\varepsilon_{t} \times E \\ &=\alpha \Delta T E \end{aligned}

\begin{aligned} \varepsilon_{t} &=\frac{\text { Change in length prevented }}{\text { original length }}\\ &=\frac{L \alpha \Delta T}{L} \\ &=\alpha \Delta T\end{aligned}

Thermal stress,

\begin{aligned} \sigma_{t} &=\varepsilon_{t} \times E \\ &=\alpha \Delta T E \end{aligned}

Question 8 |

A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. The rod rests on a frictionless surface. The rod is heated uniformly. If the radial and longitudinal thermal stresses are represented by \sigma _{r}
and \sigma _{z}, respectively, then

\sigma _{r}=0,\sigma _{z}=0 | |

\sigma _{r}\neq 0,\sigma _{z}=0 | |

\sigma _{r}= 0,\sigma _{z}\neq 0 | |

\sigma _{r}\neq 0,\sigma _{z}\neq 0 |

Question 8 Explanation:

If a body is allowed to expand or contract freely

with rise or fall in temperature then no stress are

induced in body. i.e.

\sigma_{r}=0 \text { and } \sigma_{z}=0

with rise or fall in temperature then no stress are

induced in body. i.e.

\sigma_{r}=0 \text { and } \sigma_{z}=0

There are 8 questions to complete.

Question 2 has a typing mistake. Coefficient of thermal expansion should be 11×10^(-6) rather than 11*10^(-16).

Thank You DURGA SINGH,

We have updated the questions.

this is very helpful for aspirants ,thanks for this, sir !!!!!!!