Question 1 |
A cantilever beam with a uniform flexural rigidity (EI = 200 \times 10^6 N.m^2) is loaded with a concentrated force at its free end. The area of the bending moment diagram corresponding to the full length of the beam is 10000 \;N.m^2. The magnitude of the slope of the beam at its free end is ________micro radian (round off to the nearest integer).
42 | |
50 | |
65 | |
84 |
Question 1 Explanation:
Assume:
\begin{aligned} &A=10000 \mathrm{~N}-\mathrm{m}^{2}\\ &\mathrm{El}=200 \times 10^{6} \mathrm{~N}-\mathrm{m}^{2} \end{aligned}
As per moment area first theorem.
\begin{aligned} \theta_{\mathrm{B}}-\theta_{A} &=\left(\frac{A}{E I}\right) A B \\ \theta_{\mathrm{B}}-0 &=\frac{10000}{200 \times 10^{6}}=0.5 \times 10^{-4} \mathrm{radian} \\ \theta_{\mathrm{B}} &=50 \mu \text { radians } \end{aligned}
Question 2 |
A plane frame PQR (fixed at P and free at R) is shown in the figure. Both members (PQ and QR) have length, L, and flexural rigidity, EI. Neglecting the effect of axial stress and transverse shear, the horizontal deflection at free end, R, is
\frac{5FL^3}{3EI} | |
\frac{4FL^3}{3EI} | |
\frac{2FL^3}{3EI} | |
\frac{FL^3}{3EI} |
Question 2 Explanation:
\begin{aligned} U&=U_{P Q}+U_{Q P} \\ U&=\frac{M^{2} L}{2 E I}+\int_{0}^{L} \frac{\left(M_{x-x}\right)^{2}(d x)}{2 E I} \\ U&=\frac{(F L)^{2} L}{2 E I}+\int_{0}^{L}\left(\frac{(F x)^{2}(d x)}{2 E I}\right) \\ U&=\frac{F^{2} L^{3}}{2 E I}+\frac{F^{2} L^{3}}{6 E I}=\frac{2 F^{2} L^{3}}{3 E I} \end{aligned}
By Castigliano's theorem:
\left(\delta_{H}\right)_{R}=\frac{\partial U}{\partial F}=\frac{4 F L^{3}}{3 E I}
Question 3 |
A steel cubic block of side 200 mm is subjected to hydrostatic pressure of 250 N/mm^2. The elastic modulus is 2 \times 10^5 \; N/mm^2 and Poisson ratio is 0.3 for steel. The side of the block is reduced by _____mm (round off to two decimal places).
0.18 | |
0.22 | |
0.1 | |
0.05 |
Question 3 Explanation:
\begin{aligned} E &=200 \mathrm{GPa} \\ \sigma &=250 \mathrm{MPa} \\ \mu &=0.3 \\ \epsilon_{x} &=\epsilon_{y}=\epsilon_{z}=\frac{\delta a}{a} \\ \frac{1}{E}\left[\sigma_{x}-\mu\left(\sigma_{y}+\sigma_{z}\right)\right] &=\frac{(\delta a)}{a} \\ \left(\frac{-\sigma}{E}\right)(1-2 \mu)&=\frac{\delta a}{a}\\ \delta a &=-\frac{(250)(1-0.6)(200)}{200 \times 10^{3}} \\ \delta a &=(-) 0.10 \mathrm{~mm} \end{aligned}
Reduction in side of cube is 0.10mm.
Question 4 |
A column with one end fixed and one end free has a critical buckling load of 100 N. For the same column, if the free end is replaced with a pinned end then the critical buckling load will be ________N (round off to the nearest integer).
800 | |
450 | |
1020 | |
680 |
Question 4 Explanation:
For a given material in 0.1 and length,
\mathrm{Pe} \propto \frac{1}{\alpha^{2}}
where, \quad \alpha= length fixity coefficient
Pe= Buckling or critical load.
\begin{aligned} \frac{(P e)_{I I}}{(P e)_{I}} &=\left(\frac{\alpha_{I}}{\alpha_{I I}}\right)^{2}=\left(\frac{2 L}{1 / \sqrt{2}}\right)^{2}=8 \\ \left(P_{e}\right)_{I I} &=8\left(P_{e}\right)_{I}=800 \mathrm{~N} \end{aligned}
Question 5 |
An overhanging beam PQR is subjected to uniformly distributed load 20 kN/m as shown in the figure.
The maximum bending stress developed in the beam is ________MPa (round off to one decimal place).
The maximum bending stress developed in the beam is ________MPa (round off to one decimal place).
125 | |
250 | |
325 | |
450 |
Question 5 Explanation:
\begin{aligned} M_{D}-M_{A} &=\frac{1}{2} \times 15 \times 0.75 \\ M_{D} &=5.625 \mathrm{kN}-\mathrm{m}(\mathrm{s}) \\ \Sigma M_{A}&=60 \times 1.5-R_{B} \times 2=0 \\ R_{B} &=45 \mathrm{kN}(\uparrow) ; \quad R_{A}=15 \mathrm{kN}(\uparrow) \end{aligned}
Location of D:
\begin{aligned} \frac{15}{x} &=\frac{25}{2-x} \\ 6-3 x &=5 x \\ &=\frac{3}{4} m \end{aligned}
\begin{aligned} M_C-M_B &=\frac{1}{2} \times 20 \times 1 \\ M_B&=-10 kN- m \\ M_B &= 10kN - m \;(+1)\\ \text{Max B.M.}&=\text{Larger of }(M_B \text{ and } M_D) \\ M_B&=10kN- m\; (+1) \\ (\sigma _b)_{max} &=\frac{M_{max}}{Z_{N.A.}}\\&=\frac{6 \times 10 \times 10^6}{224 \times 100^2}=250MPa \end{aligned}
Question 6 |
A right solid circular cone standing on its base on a horizontal surface is of height H and base radius R. The cone is made of a material with specific weight w and elastic modulus E. The vertical deflection at the mid-height of the cone due to self-weight is given by
\frac{wH^2}{8E} | |
\frac{wH^2}{6E} | |
\frac{wRH}{8E} | |
\frac{wRH}{6E} |
Question 6 Explanation:
\begin{aligned} P_{\mathrm{x}-\mathrm{x}}&=\frac{(-) \mathrm{W} A_{x-x}(x)}{3}\\ \text{Contraction of small strip }&=(\delta l)_{\text {strip }}\\ &=\frac{P_{x-x} d x}{(A E)_{x-x}} \\ &=\frac{w\left(A_{x-x}\right)(x) d x}{3\left(A_{x-x}\right) E}=\frac{w x}{3 E} d x \end{aligned}
Contraction of conical bar at mid height ( i.e. x=\frac{H}{2} )
\begin{aligned} &=\int_{H / 2}^{H}(\delta l)_{\operatorname{stn} p} \\ &=\frac{W}{3 E} \int_{H / 2}^{H} x d x \\ &=\frac{w}{6 E}\left[H^{2}-\frac{H^{2}}{4}\right]=\frac{w H^{2}}{8 E} \end{aligned}
Question 7 |
A prismatic bar PQRST is subjected to axial loads as shown in the figure. The segments having maximum and minimum axial stresses, respectively, are
QR and PQ | |
ST and PQ | |
QR and RS | |
ST and RS |
Question 7 Explanation:
\begin{aligned} P_{\max }&=P_{S T}=25 \mathrm{kN} \\ P_{\min }&=P_{R S}=5 \mathrm{kN} \end{aligned}
Hence, maximum and minimum axial stresses are in ST and RS portions because of prismatic bar.
Question 8 |
A cantilever beam of length, L, and flexural rigidity, EI, is subjected to an end moment, M, as shown in the figure. The deflection of the beam at x=\frac{L}{2} is
\frac{ML^2}{2EI} | |
\frac{ML^2}{4EI} | |
\frac{ML^2}{8EI} | |
\frac{ML^2}{16EI} |
Question 8 Explanation:
\begin{aligned} Y_{C}-Y_{A}&=\left(\frac{A \bar{X}}{E I}\right)_{A C} \\ Y_{C}-0&=\frac{1}{E I}\left[\frac{-ML}{2} \times \frac{L}{4}\right]\\ Y_{C}&=\frac{ML^{2}}{8 E I} \text { (downward) } \end{aligned}
Question 9 |
The loading and unloading response of a metal is shown in the figure. The elastic and plastic strains corresponding to 200 MPa stress, respectively, are
0.01 and 0.01 | |
0.02 and 0.01 | |
0.01 and 0.02 | |
0.02 and 0.02 |
Question 9 Explanation:
Elastic strain : Which can be recovered = 0.03 - 0.01 = 0.02
Plastic strain : Permanent strain = 0.01
Plastic strain : Permanent strain = 0.01
Question 10 |
Uniaxial compression test data for a solid metal bar of length 1 m is shown in the figure.
The bar material has a linear elastic response from O to P followed by a non-linear response. The point P represents the yield point of the material. The rod is pinned at both the ends. The minimum diameter of the bar so that it does not buckle under axial loading before reaching the yield point is _______ mm (round off to one decimal place).
The bar material has a linear elastic response from O to P followed by a non-linear response. The point P represents the yield point of the material. The rod is pinned at both the ends. The minimum diameter of the bar so that it does not buckle under axial loading before reaching the yield point is _______ mm (round off to one decimal place).
16.56 | |
56.94 | |
78.25 | |
68.42 |
Question 10 Explanation:
For both end pin,
\begin{aligned} P &=\frac{\pi^{2} E I}{L^{2}}=\frac{\pi^{2} E A \cdot \frac{d^{2}}{16}}{L^{2}} \\ \frac{P}{A E} &=\frac{\pi^{2} d^{2}}{16 L^{2}} \\ \varepsilon_{y} &=\frac{\pi^{2} d^{2}}{16 L^{2}} \\ d &=\sqrt{\frac{16 L^{2} \varepsilon_{y}}{\pi^{2}}} \\ &=\sqrt{\frac{16 \times 1000^{2} \times 0.002}{\pi^{2}}} \\ &=56.94 \mathrm{mm} \end{aligned}
\begin{aligned} P &=\frac{\pi^{2} E I}{L^{2}}=\frac{\pi^{2} E A \cdot \frac{d^{2}}{16}}{L^{2}} \\ \frac{P}{A E} &=\frac{\pi^{2} d^{2}}{16 L^{2}} \\ \varepsilon_{y} &=\frac{\pi^{2} d^{2}}{16 L^{2}} \\ d &=\sqrt{\frac{16 L^{2} \varepsilon_{y}}{\pi^{2}}} \\ &=\sqrt{\frac{16 \times 1000^{2} \times 0.002}{\pi^{2}}} \\ &=56.94 \mathrm{mm} \end{aligned}
There are 10 questions to complete.
Diagram of q12 is wrong.
Dear Saket Wahane
Thank you for your suggestions. We have updated the correction suggested by You.
Q32, both the criterion statements are same?
Dear Saqib
Thank you for your suggestions. We have updated the correction suggested by You.
Q67, values given in options do not match the answer.
Dear Saqib
Thank you for your suggestions. We have updated the correction suggested by You.
Q170) correct answer is given as wrong… and explaination also wrong
que 19 is wrong printed
Question no 8 is actually wrong.. Let me explain.. In the explanation of this question as shown in this page, we get c=2L assuming Mz=0 at x=L… Now, if the question would be correct, then we would also get c=2L assuming Mz=M at x=0 since the expression is given to be valid at 0≤x≤L . But we actually get different value of ‘c’ in this assumption. Hence, it is proved that the question is wrong.