Question 1 |

A cylindrical bar has a length L=5m and cross section area S=10m^2 . The bar
is made of a linear elastic material with a density \rho =2700kg/m^3
and Young's
modulus E=70GPa . The bar is suspended as shown in the figure and is in a state
of uniaxial tension due to its self-weight.

The elastic strain energy stored in the bar equals ____ J. (Rounded off to two decimal places)

Take the acceleration due to gravity as g=9.8 m/s^2

The elastic strain energy stored in the bar equals ____ J. (Rounded off to two decimal places)

Take the acceleration due to gravity as g=9.8 m/s^2

2.08 | |

3.21 | |

5.36 | |

8.25 |

Question 1 Explanation:

Given data length L=5 \mathrm{~m}

cross-sectional area A=10 \mathrm{~m}^{2}

Density of bar, \rho=2700 \mathrm{~kg} / \mathrm{m}^{3}

Young's modulus \mathrm{E}=70 \mathrm{GPa}

Bar is in a state of uniaxial tension due to its self weight so the elastic energy stored in the bar.

\begin{aligned} U & =\int_{0}^{L} \frac{P_{e}^{2} d x}{2 A E}=\int_{0}^{L} \frac{\left(\frac{w}{L} x\right)^{2} d x}{2 A E} \\ & =\frac{w^{2} L^{3}}{L^{2} 6 A E}=\frac{\gamma^{2} L^{3} A}{6 E} \end{aligned}

where, \gamma= weight/volume =\frac{\mathrm{W}}{\mathrm{AL}}

weight of the bar W=\rho g V

So,

\begin{aligned} \gamma & =\frac{\rho g V}{A L} \quad \because V=A \cdot L \\ \gamma & =\rho g \\ V & =\rho g \frac{(\rho g)^{2} L^{3} \mathrm{~A}}{6 \mathrm{E}} \\ & =\frac{(2700 \times 9.8)^{2} \times(5)^{3} \times 10}{6 \times 70 \times 10^{9}} \\ & V=2.08 \end{aligned}

cross-sectional area A=10 \mathrm{~m}^{2}

Density of bar, \rho=2700 \mathrm{~kg} / \mathrm{m}^{3}

Young's modulus \mathrm{E}=70 \mathrm{GPa}

Bar is in a state of uniaxial tension due to its self weight so the elastic energy stored in the bar.

\begin{aligned} U & =\int_{0}^{L} \frac{P_{e}^{2} d x}{2 A E}=\int_{0}^{L} \frac{\left(\frac{w}{L} x\right)^{2} d x}{2 A E} \\ & =\frac{w^{2} L^{3}}{L^{2} 6 A E}=\frac{\gamma^{2} L^{3} A}{6 E} \end{aligned}

where, \gamma= weight/volume =\frac{\mathrm{W}}{\mathrm{AL}}

weight of the bar W=\rho g V

So,

\begin{aligned} \gamma & =\frac{\rho g V}{A L} \quad \because V=A \cdot L \\ \gamma & =\rho g \\ V & =\rho g \frac{(\rho g)^{2} L^{3} \mathrm{~A}}{6 \mathrm{E}} \\ & =\frac{(2700 \times 9.8)^{2} \times(5)^{3} \times 10}{6 \times 70 \times 10^{9}} \\ & V=2.08 \end{aligned}

Question 2 |

Cylindrical bars P and Q have identical lengths and radii, but are composed of
different linear elastic materials. The Young's modulus and coefficient of thermal
expansion of Q are twice the corresponding values of P. Assume the bars to be
perfectly bonded at the interface, and their weights to be negligible.

The bars are held between rigid supports as shown in the figure and the temperature is raised by \Delta T. Assume that the stress in each bar is homogeneous and uniaxial. Denote the magnitudes of stress in P and Q by \sigma _1 and \sigma _2, respectively.

Which of the statement(s) given is/are CORRECT?

The bars are held between rigid supports as shown in the figure and the temperature is raised by \Delta T. Assume that the stress in each bar is homogeneous and uniaxial. Denote the magnitudes of stress in P and Q by \sigma _1 and \sigma _2, respectively.

Which of the statement(s) given is/are CORRECT?

The interface between P and Q moves to the left after heating | |

The interface between P and Q moves to the right after heating | |

\sigma _1 \lt \sigma _2 | |

\sigma _1 = \sigma _2 |

Question 2 Explanation:

Given:
\alpha_{Q}=2 \alpha_{P}

Given \ell_{1}=\ell_{2}=\beta \ell

E= young's modulus

Change in temperature =\Delta \mathrm{T}

where A= cross sectional area of bar and is same for both due to same radii.

So, \sigma_{1}=\sigma_{2}

Deformation in \mathrm{P}=\ell \alpha \Delta \mathrm{T}-\frac{\mathrm{P} \ell}{\mathrm{AE}}=\Delta_{\mathrm{P}}

Deformation in \mathrm{Q}=2 \alpha \ell \Delta \mathrm{T}-\frac{\mathrm{P} \ell}{2 \mathrm{AE}}=\Delta_{\mathrm{Q}}

\Delta_{\mathrm{Q}} \gt \Delta_{\mathrm{P}}

Given \ell_{1}=\ell_{2}=\beta \ell

E= young's modulus

Change in temperature =\Delta \mathrm{T}

where A= cross sectional area of bar and is same for both due to same radii.

So, \sigma_{1}=\sigma_{2}

Deformation in \mathrm{P}=\ell \alpha \Delta \mathrm{T}-\frac{\mathrm{P} \ell}{\mathrm{AE}}=\Delta_{\mathrm{P}}

Deformation in \mathrm{Q}=2 \alpha \ell \Delta \mathrm{T}-\frac{\mathrm{P} \ell}{2 \mathrm{AE}}=\Delta_{\mathrm{Q}}

\Delta_{\mathrm{Q}} \gt \Delta_{\mathrm{P}}

Question 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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}}

There are 5 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 !!!!!!!