Solid Mechanics

 Question 1
An idealized frame supports a load as shown in the figure. The horizontal component of the force transferred from the horizontal member PQ to the vertical member $\mathrm{RS}$ at $\mathrm{P}$ is $\mathrm{N}$ (round off to one decimal place).

 A 12 B 16 C 18 D 24
GATE CE 2023 SET-2      Shear Force and Bending Moment Shear Force and Bending Moment
Question 1 Explanation:
As member UT is a link member, it will carry axial load only (assuming as R).
FBD of $\mathrm{PQ}^{\prime}$

$\theta=\tan ^{-1} \left(\frac{1}{2}\right)$
$\theta=39.805 ^{\circ}$
$\sum M_{P}=0$
$\Rightarrow-\mathrm{R} \sin \theta * 1.2+10 * 1.8=0$
$\Rightarrow R \sin \left(39.805^{\circ}\right)=\frac{10 * 1.8}{1.2}$
$\Rightarrow \quad \mathrm{R}=23.43 \mathrm{~N}$

Now,
\begin{aligned} & \sum F_{H}=0 \\ & \Rightarrow \mathrm{H}_{\mathrm{P}}-\mathrm{R} \cos \left(39.805^{\circ}\right)=0 \\ & \Rightarrow \quad \mathrm{H}_{\mathrm{P}}=23.43 * \cos \left(39.805^{\circ}\right) \\ & =17.988 \mathrm{~N} \end{aligned}
 Question 2
A $\quad 2 \mathrm{D}$ thin plate with modulus of elasticity, $E=1.0$ $\mathrm{N} / \mathrm{m}^{2}$, and Poisson's ratio, $\mu=0.5$, is in plane stress condition. The displacement field in the plate is given by $\mu=C^{2} y$ and $v=0$, where us and $v$ are displacements (in $m$ ) along the $X$ and $Y$ directions, respectively, and $C$ is constant (in $\mathrm{m}^{-2}$ ). The distance $x$ and $y$ along $X$ an $Y$, respectively, are in $\mathrm{m}$. The stress in the $X$ direction is $\sigma_{x x}=40 x y \mathrm{~N} / \mathrm{m}^{2}$, and the shear stress is $\tau_{x y}=a x^{2} \mathrm{~N} / \mathrm{m}^{2}$. What is the value of $\alpha$ (in $\mathrm{N} / \mathrm{m}^{4}$, in integer) ?
 A 30 B 40 C 45 D 55
GATE CE 2023 SET-2      Bending and Shear Stresses
Question 2 Explanation:
\begin{aligned} & \epsilon_{x}=\frac{\sigma_{x}}{E}-\frac{\mu \sigma_{y}}{E} \\ & \epsilon_{y}=\frac{\sigma_{y}}{E}-\frac{\mu \sigma_{x}}{E} \\ & \tau_{x y}=\frac{\gamma_{x y}}{G_{1}} \\ & G=\frac{E}{2(1+\mu)} \\ & \epsilon_{x}+\mu \epsilon_{y}=\frac{\sigma_{x}}{E}-\frac{\mu \sigma_{y}}{E}+\frac{\mu \sigma_{y}}{E}-\frac{\mu^{2} \sigma_{x}}{E} \\ & \epsilon_{x}+\mu \epsilon_{y}=\frac{\sigma_{x}}{E}\left(1-\mu^{2}\right) \\ & \sigma_{x}=\frac{E}{1-\mu^{2}}\left[\epsilon_{x}+\mu \epsilon_{y}\right] \\ & \sigma_{x}=\frac{E}{1-\mu^{2}}\left[\frac{\partial u}{\partial x}+\mu \frac{\partial v}{\partial y}\right] \\ & \sigma_{x}=\frac{E}{1-\mu^{2}}[2(x y+0)] \\ & \sigma_{x}=\frac{2 \mathrm{ECxy}}{1-\mu^{2}} \\ & \sigma_{x}=40 x y \text { (Given) } \\ & \Rightarrow \frac{2 \mathrm{ECxy}}{1-\mu^{2}}=40 x y \\ & \Rightarrow \quad \frac{2 \mathrm{EC}}{1-\mu^{2}}=40 \\ & Y_{x y}=\frac{\tau_{x}}{G} \\ & G \gamma_{x y}=\tau_{x y} \\ & \frac{E}{2(1+\mu)}\left[\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right]=\tau_{x y} \\ & \frac{E}{2(1+\mu)}\left[C^{2}+0\right]=\alpha x^{2} \\ & \alpha=\frac{C E}{2(1+\mu)} \\ & \alpha=\frac{40\left(1-\mu^{2}\right)}{2 E} \frac{E}{2(1+\mu)} \\ & \alpha=10(1-\mu)=10(1-0.5)=5 \end{aligned}

 Question 3
For the frame shown in the figure (not to scale), all members $(A B, B C, C D, G B$, and $C H)$ have the same length, $L$ and flexural rigidity, El. The joints at $B$ and $C$ are rigid joints, and the supports $A$ and $D$ are fixed supports. Beams GB and $\mathrm{CH}$ carry uniformly distributed loads of $w$ per unit length. The magnitude of the moment reaction at $A$ is $\mathrm{wL}^{2} / \mathrm{k}$. What is the value of $\mathrm{k}$ (in integer) ? ____

 A 4 B 6 C 8 D 10
GATE CE 2023 SET-2      Deflection of Beams
Question 3 Explanation:

Due to symmetry, slope at mid of $B C$ is zero and can be taken as slider at point $O$

Distribution factors:
\begin{aligned} \mathrm{K}_{\mathrm{BA}} & =\frac{4 \mathrm{EI} / \mathrm{L}}{4 \mathrm{EI} / \mathrm{L}+\mathrm{EI} /(\mathrm{L} / 2)}=\frac{2}{3} \\ \mathrm{K}_{\mathrm{BO}}&= 1-\mathrm{K}_{\mathrm{BA}} \\ & =1-2 / 3=1 / 3 \\ \therefore \quad \mathrm{M}_{\mathrm{BA}} & =\frac{\mathrm{WL}^{2}}{2} \times \frac{2}{3}=\frac{\mathrm{WL}^{2}}{3} \\ \mathrm{M}_{\mathrm{AB}} & =\frac{\mathrm{M}_{\mathrm{BA}}}{2}=\frac{\mathrm{WL}^{2} / 3}{2}=\frac{\mathrm{WL}^{2}}{6} \\ & \therefore \mathrm{K}=6 \end{aligned}
 Question 4
A beam is subjected to a system of coplanar forces as shown in the figure. The magnitude of vertical reaction at Support $\mathrm{P}$ is $\mathrm{N}$ (round off to one decimal place).

 A 197 B 125.2 C 362.1 D 148.2
GATE CE 2023 SET-2      Shear Force and Bending Moment
Question 4 Explanation:

Taking components of inclined $500 \mathrm{~N}$ load along horizontal and vertical direction.

Now, using equilibrium equations.
$\Rightarrow \quad \mathrm{H}_{\mathrm{P}}=350 \mathrm{~N}$
$\left.\sum \mathrm{M}_{\mathrm{Q}}\right)=0$ $\Rightarrow R_{P} * 6-\left(500 \sin 60^{\circ}\right) * 4+200 * 2.5+100 * 0.5=0$
$\Rightarrow \quad \mathrm{R}_{\mathrm{P}}=197 \mathrm{~N}$
 Question 5
In a two-dimensional stress analysis, the state of stress at a point is shown in the figure. The values of length of PQ, QR, and RP are 4, 3, and 5 units, respectively. The principal stresses are (round off to one decimal place)

 A $\sigma_{x}=26.7 \mathrm{MPa}, \alpha_{y}=172.5 \mathrm{MPa}$ B $\sigma_{x}=54.0 \mathrm{MPa}, \sigma_{y}=128.5 \mathrm{MPa}$ C $\sigma_{x}=67.5 \mathrm{MPa}, \sigma_{y}=213.3 \mathrm{MPa}$ D $\sigma_{x}=16.0 \mathrm{MPa}, \sigma_{y}=138.5 \mathrm{MPa}$
GATE CE 2023 SET-2      Principal Stress and Principal Strain
Question 5 Explanation:

$\Rightarrow \quad \tan \theta=\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{3}{4}$
$\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{3}{4}\right)=36.87^{\circ}$
Using transformation equations:
$\sigma_{x^{\prime}}=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta$
$120=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos \left(2 * 36.87^{\circ}\right)$
$\tau_{x^{\prime} y^{\prime}}=-\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \sin \left(2 * 36.87^{\circ}\right)$
From above equations, we get
\begin{aligned} & \sigma_{x}=67.5 \mathrm{MPa} \\ & \sigma_{y}=213.3 \mathrm{MPa} \end{aligned}

There are 5 questions to complete.