# Engineering Mechanics

 Question 1
The cross-section of a girder is shown in the figure (not to scale). The section is symmetric about a vertical axis $(Y-Y)$. The moment of inertia of the section about the horizontal axis $(X-X)$ passing through the centroid is _____ $\mathrm{cm}^{4}$ (round off to nearest integer). A 468810 B 124587 C 325469 D 458765
GATE CE 2023 SET-1
Question 1 Explanation: As section is symmetric about given $y-y$axis, hence centroid will lie on this axis.
To calculate location of C.G. from reference axis (bottom horizontal line); $\bar{y}$
$\bar{y}=\frac{A_{1} \bar{y}_{1}+A_{2} \bar{y}_{2}}{A_{1}+A_{2}}$
$\bar{y}=\frac{(40 \times 10) \times 55+(20 \times 50) \times(25)}{40 \times 10+20 \times 50}$
$\bar{y}=33.5714 \mathrm{~cm}$ (from bottom)
MOI of section about $x x$ (centroidal horizontal axis)
$I=I_{I, x x}+I_{I I, x x}$
$I_{1, x x}=\frac{b_{1} d_{1}^{3}}{12}+\left(b_{1} d_{1}\right) \times d^{2}$ \begin{aligned} & \mathrm{I}_{1, x x}=\frac{40 \times 10^{3}}{12}+(40 \times 10)(55-33.57)^{2} \\ & \mathrm{I}_{1, x x}=187031.29 \mathrm{~cm}^{4} \end{aligned} $\mathrm{d}^{\prime}=\overline{\mathrm{y}}-25$
$\mathrm{d}^{\prime}=8.5714 \mathrm{~cm}$
$\mathrm{I}_{\mathrm{II}, x x}=\frac{\mathrm{b}_{2} \mathrm{~d}_{2}^{3}}{12}+\left(\mathrm{b}_{2} \times \mathrm{d}_{2}\right) \times \mathrm{d}^{\prime 2}$
$I_{I I, x x}=\frac{20 \times 50^{3}}{12}+(20 \times 50) \times(8.57)^{2}$
$\mathrm{I}_{\mathrm{II}, \mathrm{xx}}=281778.23 \mathrm{~cm}^{4}$

Hence, MOI of whole section
$\mathrm{I}_{x \mathrm{x}}=\mathrm{I}_{\mathrm{I}, x \mathrm{x}}+\mathrm{I}_{\mathrm{Il}, \mathrm{xx}}$
$\mathrm{I}_{x x}=468809.52 \mathrm{~cm}^{4}$
 Question 2
An undamped spring-mass system with mass $m$ and spring stiffness $k$ is shown in the figure. The natural frequency and natural period of this system are $\omega$ rad/s and $T \;s$, respectively. If the stiffness of the spring is doubled and the mass is halved, then the natural frequency and the natural period of the modified system, respectively, are A $2\omega$ rad/s and T/2 s B $\omega /2$ rad/s and 2T s C $4\omega$ rad/s and T/4 s D $\omega$ rad/s and T s
GATE CE 2022 SET-2
Question 2 Explanation:
This is question of undamped free vibration of sinle degree of freedom system. For the above figure it is given that natural frequency and natural period are w rad/s and T s, respectively. The equation for soof system is
$\omega =\sqrt{\frac{k}{m}}, T=\frac{2\pi}{\omega }$
For the modified system, stiffness of the spring is doubled and the mass is halved.
Let us assume k'=2k and m'=m/2, here k' and m' represent the stiffness and mass of the modified system respectively.
$\omega '=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{\left ( \frac{m}{2} \right )}}=\sqrt{\frac{4k}{m}}=2\sqrt{\frac{k}{m}}=2\omega \;\;\; \left ( \because \omega = \sqrt{\frac{k}{m}}\right )$
$T'=\frac{2\pi}{\omega '}=\frac{2 \pi}{2\omega }=\frac{1}{2}\left ( \frac{2 \pi}{\omega } \right )=\frac{T}{2} \;\;\left ( T=\frac{2 \pi}{\omega } \right )$
$\omega '$ and $T'$ are natural frequency and natural period of the modified system.
hence, the natural frequency of the modified system is doubled and natural period is halved compared to the original given system. Hence option (A) is correct.

 Question 3
A wedge M and a block N are subjected to forces P and Q as shown in the figure. If force P is sufficiently large, then the block N can be raised. The weights of the wedge and the block are negligible compared to the forces P and Q. The coefficient of friction $(\mu)$ along the inclined surface between the wedge and the block is 0.2. All other surfaces are frictionless. The wedge angle is $30^{\circ}$. The limiting force P, in terms of Q, required for impending motion of block N to just move it in the upward direction is given as $P=\alpha Q$. The value of the coefficient '?' (rounded off to one decimal place) is
 A 0.6 B 0.5 C $2.0$ D 0.9
GATE CE 2021 SET-1
Question 3 Explanation: \begin{aligned} \Sigma Y&=0 \Rightarrow \\ N_{2} \sin 60^{\circ}-0.2 N_{2} \sin &30^{\circ}-Q=0 \\ Q&=0.766 \mathrm{~N}_{2} \end{aligned} \begin{aligned} \Sigma X &=0 \\ \Rightarrow \qquad 0.2 N_{2} \cos 30^{\circ}+N_{2} \cos 60^{\circ}-P &=0\\ P &=0.67 N_{2} \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad P &=0.67 \times \frac{Q}{0.766}\\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad P&=0.875 \mathrm{Q} \simeq 0.9 Q \\ P&=\alpha Q \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \alpha&=0.9 \end{aligned}
 Question 4
A simple mass-spring oscillatory system consists of a mass m, suspended from a spring of stiffness k. Considering z as the displacementof the system at any time t, the equation of motion for the free vibration of the system is $m\ddot{z}+kz=0$. The natural frequency of the system is
 A $\frac{k}{m}$ B $\sqrt{\frac{k}{m}}$ C $\frac{m}{k}$ D $\sqrt{\frac{m}{k}}$
GATE CE 2019 SET-1
Question 4 Explanation:
\begin{aligned} m\ddot{z}+kz&=0 \\ \ddot{z}+\frac{k}{m}z&=0 \\ \text{Comparing with} \\ \ddot{z}+\omega _n^2 z&=0 \\ \text{We get}\;\; \omega _n&=\sqrt{\frac{k}{m}} \end{aligned}
 Question 5
A cable PQ of length 25 m is supported at two ends at the same level as shown in the figure. The horizontal distance between the supports is 20 m. A point load of 150 kN is applied at point R which divides it into two equal parts. Neglecting the self-weight of the cable, the tension (in kN, in integer value) in the cable due to the applied load will be _____
 A 100 B 125 C 200 D 250
GATE CE 2018 SET-2
Question 5 Explanation: \begin{aligned} 12.5^{2} &=x^{2}+10^{2} \\ \Rightarrow \; \; x &=7.5\; \text{m} \end{aligned}
Bending moment at S=0 {consider the left part}
\begin{aligned} H_{P}\times 7.5 &=75\times 10 \\ H_{P} &=100\; \text{kN} \end{aligned}
Tension in cable
\begin{aligned} &=\sqrt{H_{P}^{2}+V_{P}^{2}}=\sqrt{75^{2}+100^{2}} \\ &=125\; \text{kN} \end{aligned}

There are 5 questions to complete.