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).


12 | |
16 | |
18 | |
24 |
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}
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) ?
30 | |
40 | |
45 | |
55 |
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) ? ____


4 | |
6 | |
8 | |
10 |
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).


197 | |
125.2 | |
362.1 | |
148.2 |
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)


\sigma_{x}=26.7 \mathrm{MPa}, \alpha_{y}=172.5 \mathrm{MPa} | |
\sigma_{x}=54.0 \mathrm{MPa}, \sigma_{y}=128.5 \mathrm{MPa} | |
\sigma_{x}=67.5 \mathrm{MPa}, \sigma_{y}=213.3 \mathrm{MPa} | |
\sigma_{x}=16.0 \mathrm{MPa}, \sigma_{y}=138.5 \mathrm{MPa} |
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.