Question 1 |
A cylindrical transmission shaft of length 1.5 m and diameter 100 mm is made
of a linear elastic material with a shear modulus of 80 GPa. While operating at
500 rpm, the angle of twist across its length is found to be 0.5 degrees.
The power transmitted by the shaft at this speed is _______kW. (Rounded off to two decimal places)
Take \pi= 3.14.
The power transmitted by the shaft at this speed is _______kW. (Rounded off to two decimal places)
Take \pi= 3.14.
238.64 | |
254.35 | |
632.25 | |
456.35 |
Question 1 Explanation:
Given: Length, L=1.5 \mathrm{~m}, Dia, d=100 \mathrm{~mm}=0.1 \mathrm{~mm}
Shear modulus, \mathrm{G}=80 \mathrm{GPa}
N=500 \mathrm{rpm} \text {, }
angle of twist,
\theta=0.5^{\circ}
\theta=0.5 \times \frac{\pi}{180^{\circ}} \mathrm{rad}
\because As, We know,
\frac{\mathrm{T}}{\mathrm{J}}=\frac{\mathrm{G} \theta}{\mathrm{L}}
where \mathrm{J} for cylinder solid shaft =\frac{\pi \mathrm{d}^{4}}{32}
Torque,
T=\frac{G \theta \cdot J}{L}=\frac{\left(80 \times 10^{9}\right) \times\left(0.5 \times \frac{\pi}{180}\right)}{1.5} \times \frac{\pi}{32} \times(0.1)^{4} =4.56 \times 10^{3} \mathrm{~N}-\mathrm{m}
Now, power,
\quad P=T \times \omega=T \times \frac{2 \pi N}{60} =4.56 \times 10^{3} \times \frac{2 \pi \times 500}{60}
P =238.64 \mathrm{~kW}
Question 2 |
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 2 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 3 |
Ignoring the small elastic region, the true stress (\sigma) - true strain ( \varepsilon) variation of a
material beyond yielding follows the equation \sigma =400 \varepsilon ^{0.3} MPa. The engineering
ultimate tensile strength value of this material is ________ MPa.
(Rounded off to one decimal place)
145.6 | |
325.6 | |
206.6 | |
125.2 |
Question 3 Explanation:
Given
\begin{aligned} \sigma_{\mathrm{T}} & =400 \varepsilon^{0.3}=400 \varepsilon_{\mathrm{T}}^{0.3} \;\;\;...(i)\\ \varepsilon & =\text { true strain }=\varepsilon_{\mathrm{T}} \\ \mathrm{n} & =0.3=\text { strain hardening exponent } \end{aligned}
The true strain \left(\varepsilon_{T}\right) at the onset of necking equal to the strain hardening exponent i.e. at ultimate tensile point.
So, \quad \varepsilon_{\mathrm{T}}=\mathrm{n}=0.3 at ultimate point
So, from equation (i)
\sigma_{\mathrm{T}} =400 \times(0.3)^{0.3} =278.74 \mathrm{MPa} \text { at ultimate point }
As \varepsilon_{\mathrm{T}}=\ln (1+\varepsilon)
So, \varepsilon=\mathrm{e}^{\varepsilon} \mathrm{T}-1=\mathrm{e}^{0.3}-1=0.35
and \sigma_{\mathrm{T}}=\sigma(1+\varepsilon)
So \sigma_{\text {ultimate }}=\frac{\sigma_{T}}{1+\varepsilon}+\frac{278.74}{1+0.35}=206.55 \mathrm{MPa}
\begin{aligned} \sigma_{\mathrm{T}} & =400 \varepsilon^{0.3}=400 \varepsilon_{\mathrm{T}}^{0.3} \;\;\;...(i)\\ \varepsilon & =\text { true strain }=\varepsilon_{\mathrm{T}} \\ \mathrm{n} & =0.3=\text { strain hardening exponent } \end{aligned}
The true strain \left(\varepsilon_{T}\right) at the onset of necking equal to the strain hardening exponent i.e. at ultimate tensile point.
So, \quad \varepsilon_{\mathrm{T}}=\mathrm{n}=0.3 at ultimate point
So, from equation (i)
\sigma_{\mathrm{T}} =400 \times(0.3)^{0.3} =278.74 \mathrm{MPa} \text { at ultimate point }
As \varepsilon_{\mathrm{T}}=\ln (1+\varepsilon)
So, \varepsilon=\mathrm{e}^{\varepsilon} \mathrm{T}-1=\mathrm{e}^{0.3}-1=0.35
and \sigma_{\mathrm{T}}=\sigma(1+\varepsilon)
So \sigma_{\text {ultimate }}=\frac{\sigma_{T}}{1+\varepsilon}+\frac{278.74}{1+0.35}=206.55 \mathrm{MPa}
Question 4 |
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 4 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 5 |
The figure shows a thin-walled open-top cylindrical vessel of radius r and wall
thickness t. The vessel is held along the brim and contains a constant-density
liquid to height h from the base. Neglect atmospheric pressure, the weight of the
vessel and bending stresses in the vessel walls.
Which one of the plots depicts qualitatively CORRECT dependence of the magnitudes of axial wall stress (\sigma _1) and circumferential wall stress (\sigma _2) on y?
Which one of the plots depicts qualitatively CORRECT dependence of the magnitudes of axial wall stress (\sigma _1) and circumferential wall stress (\sigma _2) on y?
A | |
B | |
C | |
D |
Question 5 Explanation:
Unit weight of liquid =\gamma
\text { Weight of liquid }=\gamma \times \text { volume }
W=\gamma \times\left(\pi r^{2}\right)(h)
\sigma_{1}=\frac{W_{T}}{(2 \pi r \times t)}=\frac{\gamma \times \pi r^{2}(h)}{2 \pi r t}=\frac{\gamma r h}{2 t}
\text { Pressure at } y^{\prime}=\gamma \times y
\sigma_{2}(\text { circumterential stress })=\frac{P d}{2 t}=\frac{\gamma \times y \times 2 r}{2 t}=\frac{\gamma r y}{t}
There are 5 questions to complete.
Diagram of q12 is wrong.
Dear Saket Wahane
Thank you for your suggestions. We have updated the correction suggested by You.
Q32, both the criterion statements are same?
Dear Saqib
Thank you for your suggestions. We have updated the correction suggested by You.
Q67, values given in options do not match the answer.
Dear Saqib
Thank you for your suggestions. We have updated the correction suggested by You.
Q170) correct answer is given as wrong… and explaination also wrong
que 19 is wrong printed
Question no 8 is actually wrong.. Let me explain.. In the explanation of this question as shown in this page, we get c=2L assuming Mz=0 at x=L… Now, if the question would be correct, then we would also get c=2L assuming Mz=M at x=0 since the expression is given to be valid at 0≤x≤L . But we actually get different value of ‘c’ in this assumption. Hence, it is proved that the question is wrong.