# Strength of Materials

 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 $\pi= 3.14$.
 A 238.64 B 254.35 C 632.25 D 456.35
GATE ME 2023      Torsion of Shafts
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 $g=9.8 m/s^2$

 A 2.08 B 3.21 C 5.36 D 8.25
GATE ME 2023      Strain Energy and Thermal Stresses
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}

 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)
 A 145.6 B 325.6 C 206.6 D 125.2
GATE ME 2023      Stress and Strain
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 $\sigma_{\text {ultimate }}=\frac{\sigma_{T}}{1+\varepsilon}+\frac{278.74}{1+0.35}=206.55 \mathrm{MPa}$
 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?

 A The interface between P and Q moves to the left after heating B The interface between P and Q moves to the right after heating C $\sigma _1 \lt \sigma _2$ D $\sigma _1 = \sigma _2$
GATE ME 2023      Strain Energy and Thermal Stresses
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 $\mathrm{Q}=2 \alpha \ell \Delta \mathrm{T}-\frac{\mathrm{P} \ell}{2 \mathrm{AE}}=\Delta_{\mathrm{Q}}$
$\Delta_{\mathrm{Q}} \gt \Delta_{\mathrm{P}}$
 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 $y$?

 A A B B C C D D
GATE ME 2023      Thin Cylinder
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.

### 9 thoughts on “Strength of Materials”

1. Diagram of q12 is wrong.

• Dear Saket Wahane
Thank you for your suggestions. We have updated the correction suggested by You.

2. Q32, both the criterion statements are same?

• Dear Saqib
Thank you for your suggestions. We have updated the correction suggested by You.

3. Q67, values given in options do not match the answer.