Question 1 |

A rigid beam AD of length 3a = 6 m is hinged at
frictionless pin joint A and supported by two strings
as shown in the figure. String BC passes over two
small frictionless pulleys of negligible radius. All
the strings are made of the same material and have
equal cross-sectional area. A force F = 9 kN is
applied at C and the resulting stresses in the strings
are within linear elastic limit. The self-weight of the
beam is negligible with respect to the applied load.
Assuming small deflections, the tension developed
in the string at C is _________ kN (round off to 2
decimal places).

2.57 | |

3.65 | |

1.45 | |

4.65 |

Question 1 Explanation:

F.B.D of beam

\begin{aligned} \delta l_1:\delta l_2:\delta l_3&=T_1:T_2:T_3\\ 1:2:3&=T_1:T_2:T_3\\ \text{Let }T_1&=T\\ \therefore \; T_2&=2T,T_3=3T\\ \Sigma M_A&=0\\ \therefore \; -T \times a-2T &\times 2a-3T \times 3a+F \times 2a=0\\ \therefore \;\; F \times 2a &=14 Ta\\ 2F&=14T\\ T&=\frac{F}{7}=\frac{9}{7}kN\\ T_2&=2T=\frac{18}{7}=2.57kN \end{aligned}

\begin{aligned} \delta l_1:\delta l_2:\delta l_3&=T_1:T_2:T_3\\ 1:2:3&=T_1:T_2:T_3\\ \text{Let }T_1&=T\\ \therefore \; T_2&=2T,T_3=3T\\ \Sigma M_A&=0\\ \therefore \; -T \times a-2T &\times 2a-3T \times 3a+F \times 2a=0\\ \therefore \;\; F \times 2a &=14 Ta\\ 2F&=14T\\ T&=\frac{F}{7}=\frac{9}{7}kN\\ T_2&=2T=\frac{18}{7}=2.57kN \end{aligned}

Question 2 |

A linear elastic structure under plane stress condition
is subjected to two sets of loading, I and II. The
resulting states of stress at a point corresponding
to these two loadings are as shown in the figure
below. If these two sets of loading are applied
simultaneously, then the net normal component of
stress \sigma _{xx} is ________.

\frac{3\sigma }{2} | |

\sigma \left (1+\frac{1}{\sqrt{2}} \right ) | |

\frac{\sigma }{2} | |

\sigma \left (1-\frac{1}{\sqrt{2}} \right ) |

Question 2 Explanation:

\sigma _x=0,\sigma _y=0,\tau _{xy}=0,\theta =-45^{\circ}

\begin{aligned} \sigma _{xx}&=\sigma +\sigma _\theta \\ \sigma _\theta&=\left [ \frac{\sigma _x +\sigma _y}{2} \right ] +\left [ \frac{\sigma _x -\sigma _y}{2} \right ] \cos 2\theta +\tau _{xy} \sin 2\theta \\ \sigma _\theta &= \frac{0+\sigma }{2} +\left [\frac{0-\sigma }{2} \right ] \cos 2(-45)+0\\ \sigma _\theta &= \frac{\sigma }{2}\\ \sigma _{xx} &=\sigma +\sigma _\theta =\sigma +\frac{\sigma }{2}=\frac{3\sigma }{2} \end{aligned}

Question 3 |

Which one of the following is the definition of
ultimate tensile strength (UTS) obtained from a
stress-strain test on a metal specimen?

Stress value where the stress-strain curve transitions from elastic to plastic behavior | |

The maximum load attained divided by the original cross-sectional area | |

The maximum load attained divided by the corresponding instantaneous crosssectional area | |

Stress where the specimen fractures |

Question 3 Explanation:

Tensile Strength: The tensile strength, or ultimate
tensile strength (UTS), is the maximum load
obtained in a tensile test, divided by the original
cross-sectional area of the specimen.

\sigma _u=\frac{P_{max}}{A_o}

where, \sigma _u= Ultimate tensile strength, kg/mm^2

P_{max}= Maximum load obtained in a tensile test, kg

A_o= Original cross-sectional area of gauge length of the test piece, mm^2

The tensile strength is a very familiar property and widely used for identification of a material. It is very easy to determine and is a quite reproducible property. It is used for the purposes of specifications and for quality control of a product. Tensile strength can be empirically correlated to other properties such as hardness and fatigue strength. For brittle materials, the tensile strength is a valid criterion for design.

\sigma _u=\frac{P_{max}}{A_o}

where, \sigma _u= Ultimate tensile strength, kg/mm^2

P_{max}= Maximum load obtained in a tensile test, kg

A_o= Original cross-sectional area of gauge length of the test piece, mm^2

The tensile strength is a very familiar property and widely used for identification of a material. It is very easy to determine and is a quite reproducible property. It is used for the purposes of specifications and for quality control of a product. Tensile strength can be empirically correlated to other properties such as hardness and fatigue strength. For brittle materials, the tensile strength is a valid criterion for design.

Question 4 |

A thin-walled cylindrical pressure vessel has mean
wall thickness of t and nominal radius of r. The
Poisson's ratio of the wall material is 1/3. When it
was subjected to some internal pressure, its nominal
perimeter in the cylindrical portion increased by
0.1% and the corresponding wall thickness became
\bar{t}. The corresponding change in the wall thickness
of the cylindrical portion, i.e. 100\times (\bar{t}-t)/t, is
________%(round off to 3 decimal places).

0.06 | |

-0.06 | |

0.12 | |

-0.12 |

Question 4 Explanation:

\pi(d+\delta d)=1.001 \pi d

\varepsilon _1=\frac{\delta d}{d}=0.001=\frac{Pd}{4tE}(2-v)

\frac{Pd}{4tE}=\frac{0.001}{(2-v)}=6 \times 10^{-4}

Radial strain

\varepsilon _2=\frac{\delta t}{t}=\frac{1}{E}\left [ \sigma _r-v(\sigma _h-\sigma _L) \right ]

\frac{\delta t}{t}=\frac{1}{E}\left [ -v\frac{Pd}{4t}(2+1) \right ]

\frac{\delta t}{t}=-\frac{Pd}{4tE}v(3)=-6 \times 10^{-4} \times \frac{1}{3}(3)=-6 \times 10^{-4}

Therefore, The corresponding change in the wall thickness of the cylindrical portion

=100 \times \left [ \frac{\bar{t}-t}{t} \right ]=100 \times \frac{\delta t}{t}=100 \times (-6) \times 10^{-4}=-0.06 \%

Question 5 |

An L-shaped elastic member ABC with slender
arms AB and BC of uniform cross-section is
clamped at end A and connected to a pin at end C.
The pin remains in continuous contact with and is
constrained to move in a smooth horizontal slot.
The section modulus of the member is same in both
the arms. The end C is subjected to a horizontal
force P and all the deflections are in the plane of the
figure. Given the length AB is 4a and length BC is
a, the magnitude and direction of the normal force
on the pin from the slot, respectively, are

3P/8, and downwards | |

5P/8, and upwards | |

P/4, and downwards | |

3P/4, and upwards |

Question 5 Explanation:

No vertical deflection allowed

\delta _{VC}=\delta _{VB}=0 \text{ (Vertical deflection) }

\begin{aligned} \delta _B&=\frac{ML^2}{2EI}-\frac{NL^2}{3EI}=0\\ \frac{M}{2}&\frac{NL}{3}\\ N&=\frac{3M}{2L}\;\;(\because L=4a)\\ N&=\frac{3Pa}{2 \times 4 a}\\ N&=\frac{3}{8}P \text{ (downward)} \end{aligned}

Question 6 |

Assuming the material considered in each statement
is homogeneous, isotropic, linear elastic, and the
deformations are in the elastic range, which one or
more of the following statement(s) is/are TRUE?

**MSQ**A body subjected to hydrostatic pressure has
no shear stress | |

If a long solid steel rod is subjected to tensile
load, then its volume increases. | |

Maximum shear stress theory is suitable for
failure analysis of brittle materials. | |

If a portion of a beam has zero shear force,
then the corresponding portion of the elastic
curve of the beam is always straight. |

Question 6 Explanation:

(c) Wrong : Maximum shear stress theory good for
ductile material.

(d) If shear force = 0, M = C, But elastic curve is always non-linear.

(d) If shear force = 0, M = C, But elastic curve is always non-linear.

Question 7 |

A uniform light slender beam AB of section modulus
EI is pinned by a frictionless joint A to the ground
and supported by a light inextensible cable CB to
hang a weight W as shown. If the maximum value
of W to avoid buckling of the beam AB is obtained
as \beta \pi ^2 EI, where \pi is the ratio of circumference to
diameter of a circle, then the value of \beta is

0.0924\; m^{-2} | |

0.0713\; m^{-2} | |

0.1261\; m^{-2} | |

0.1417\; m^{-2} |

Question 7 Explanation:

Draw FBD of AB

\Sigma M_A=0

W \times 2.5=T \sin 30^{\circ} \times 2.5

T=2W

Compressive load acting on AB =T \cos 30^{\circ}=2W \times \frac{\sqrt{3}}{2}=\sqrt{3}W

Buckling happens when \sqrt{3}W=P_{cr}=\frac{\pi ^2 EI}{L_e^2}

\sqrt{3}W=\frac{\pi ^2 EI}{L^2} \;\;(\because L_e=L \text{as both ends hinged})

W=\frac{1 \times \pi ^2 EI}{\sqrt{3} \times (2.5)^2}=0.0924 \pi^2 EI

W0.0924 \pi^2 EI=\beta \pi ^2 EI

\beta =0.0924 m^{-2}

\Sigma M_A=0

W \times 2.5=T \sin 30^{\circ} \times 2.5

T=2W

Compressive load acting on AB =T \cos 30^{\circ}=2W \times \frac{\sqrt{3}}{2}=\sqrt{3}W

Buckling happens when \sqrt{3}W=P_{cr}=\frac{\pi ^2 EI}{L_e^2}

\sqrt{3}W=\frac{\pi ^2 EI}{L^2} \;\;(\because L_e=L \text{as both ends hinged})

W=\frac{1 \times \pi ^2 EI}{\sqrt{3} \times (2.5)^2}=0.0924 \pi^2 EI

W0.0924 \pi^2 EI=\beta \pi ^2 EI

\beta =0.0924 m^{-2}

Question 8 |

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 8 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 9 |

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 9 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 10 |

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 10 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.

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.