Question 1 |
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 1 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 2 |
The components of pure shear strain in a sheared material are given in the matrix form:
\varepsilon =\begin{bmatrix} 1 &1 \\ 1& -1 \end{bmatrix}
Here, Trace(\varepsilon)=0 . Given, P=Trace(\varepsilon ^8) and Q=Trace(\varepsilon ^{11}) .
The numerical value of (P + Q) is ________. (in integer)
\varepsilon =\begin{bmatrix} 1 &1 \\ 1& -1 \end{bmatrix}
Here, Trace(\varepsilon)=0 . Given, P=Trace(\varepsilon ^8) and Q=Trace(\varepsilon ^{11}) .
The numerical value of (P + Q) is ________. (in integer)
12 | |
28 | |
32 | |
46 |
Question 2 Explanation:
\begin{aligned}
|A-\lambda I|&=0 \\
\begin{vmatrix}
1-\lambda &1 \\
1 & 1-\lambda
\end{vmatrix}&=0 \\
(1-\lambda)(-1-\lambda)-1&=0 \\
(\lambda^2-1)-1&=0 \\
\lambda&=\pm \sqrt{2}
\end{aligned}
Eigen values of \varepsilon are \sqrt{2} and -\sqrt{2}
Eigen values of \varepsilon ^8 are (\sqrt{2})^8 and ( -\sqrt{2})^8
Eigen values of \varepsilon ^{11} are (\sqrt{2})^{11} and (-\sqrt{2})^{11}
P=Trace(\varepsilon ^8) = sum of Eigen values = ( \sqrt{2})^8+( -\sqrt{2})^8=32
Q=Trace(\varepsilon ^{11}) = sum of Eigen values = ( \sqrt{2})^{11}+( -\sqrt{2})^{11}=0
P+Q=32+0=32
Eigen values of \varepsilon are \sqrt{2} and -\sqrt{2}
Eigen values of \varepsilon ^8 are (\sqrt{2})^8 and ( -\sqrt{2})^8
Eigen values of \varepsilon ^{11} are (\sqrt{2})^{11} and (-\sqrt{2})^{11}
P=Trace(\varepsilon ^8) = sum of Eigen values = ( \sqrt{2})^8+( -\sqrt{2})^8=32
Q=Trace(\varepsilon ^{11}) = sum of Eigen values = ( \sqrt{2})^{11}+( -\sqrt{2})^{11}=0
P+Q=32+0=32
Question 3 |
The hoop stress at a point on the surface of a thin cylindrical pressure vessel is
computed to be 30.0 MPa. The value of maximum shear stress at this point is
7.5 MPa | |
15.0 MPa | |
30.0 MPa | |
22.5 MPa |
Question 3 Explanation:
Given,
Hoop stress (\sigma _h)=\frac{pd}{2t}=30MPa
Maximum shear stress in plane (\tau _{max})_{\text{in plane}}=\frac{\frac{pd}{2t}-\frac{pd}{4t}}{2}=7.5MPa
Hoop stress (\sigma _h)=\frac{pd}{2t}=30MPa
Maximum shear stress in plane (\tau _{max})_{\text{in plane}}=\frac{\frac{pd}{2t}-\frac{pd}{4t}}{2}=7.5MPa
Question 4 |
The cross-section of a built-up wooden beam as shown in the figure is subjected to a vertical shear force of 8 kN. The beam is symmetrical about the neutral axis (N.A.) shown, and the moment of inertia about N.A. is 1.5 \times 10^9 mm^4. Considering that the nails at the location P are spaced longitudinally (along the length of the beam) at 60 mm, each of the nails at P will be subjected to the shear force of
60N | |
120N | |
240N | |
480N |
Question 4 Explanation:
Shear stress acting at the joint at which the nail 'P' is resisting the shear.
\begin{aligned} \tau &=\frac{FA\bar{y}}{Ib}\\ &=\frac{(8 \times 10^3)(50 \times 100)(150)}{1.5 \times 10^9 \times 50}\\ &=0.08N/mm^2 \end{aligned}
The shear force acting on nail at 'P'.
F=\tau (60\times 50)=240N
\begin{aligned} \tau &=\frac{FA\bar{y}}{Ib}\\ &=\frac{(8 \times 10^3)(50 \times 100)(150)}{1.5 \times 10^9 \times 50}\\ &=0.08N/mm^2 \end{aligned}
The shear force acting on nail at 'P'.
F=\tau (60\times 50)=240N
Question 5 |
A cantilever beam of length 2 m with a square section of side length 0.1 m is loaded vertically at the free end. The vertical displacement at the free end is 5 mm. The beam is made of steel with Young's modulus of 2.0 \times 10^{11} N/m^{2}. The maximum bending stress at the fixed end of the cantilever is
20.0 MPa | |
37.5 MPa | |
60.0 MPa | |
75.0 MPa |
Question 5 Explanation:
\begin{aligned} I&=\frac{(0.1)^4}{12} \\ \text{Deflection }\delta &=\frac{Pl^2}{3eI} \\ 5 \times 10^{-3}&=\frac{P(2)^3}{3 \times 2 \times 10^11 \times \frac{(0.1)^4}{12}} \\ P&=3125N \\ \text{Now, } M&= Pl=3125 \times 2\\ &= 6250Nm\\ \text{As, }\frac{M}{I} &=\frac{\sigma }{y}=\frac{E}{R} \\ \sigma _{max}&=\frac{M}{Z}=\frac{6250}{\frac{(0.1)^3}{6}}\\ &=37.5 \times 10^6 N/m^2=37.5MPa \end{aligned}
There are 5 questions to complete.