Solid Mechanics

Question 1
A solid circular torsional member OPQ is subjected to torsional moments as shown in the figure (not to scale). The yield shear strength of the constituent material is 160 MPa.

The absolute maximum shear stress in the member (in MPa, round off to one decimal place) is ________
A
15.3
B
12.6
C
18.2
D
11.8
GATE CE 2021 SET-2      Torsion of Shafts and Pressure Vessels
Question 1 Explanation: 


\begin{aligned} \tau_{\max _{O P}}=\frac{16 T_{O P}}{\pi d_{O P}^{3}}=\frac{16 \times 3 \times 10^{3}}{\pi \times 100^{3}}=15.27 \mathrm{~N} / \mathrm{mm}^{2} \\ \tau_{\max \mathrm{PQ}}=\frac{16 T_{P Q}}{\pi d_{P Q}^{3}}=\frac{16 \times 1 \times 10^{6}}{\pi \times 80^{3}}=9.94 \mathrm{~N} / \mathrm{mm}^{2}\\ \tau_{\max }=9.94 \mathrm{~N} / \mathrm{mm}^{2} \end{aligned}
Absolute max shear stress = 15.27 \mathrm{~N} / \mathrm{mm}^{2}
Question 2
A single story building model is shown in the figure. The rigid bar of mass 'm' is supported by three massless elastic columns whose ends are fixed against rotation. For each of the columns, the applied lateral force (P) and corresponding moment (M) are also shown in the figure. The lateral deflection (\delta) of the bar is given by \delta=\frac{P L^{3}}{12 E I}, where L is the effective length of the column, E is the Young's modulus of elasticity and I is the area moment of inertia of the column cross-section with respect to its neutral axis.

For the lateral deflection profile of the columns as shown in the figure, the natural frequency of the system for horizontal oscillation is
A
6 \sqrt{\frac{E I}{m L^{3}}} \mathrm{rad} / \mathrm{s}
B
\frac{1}{L} \sqrt{\frac{2 E I}{m}} \mathrm{rad} / \mathrm{s}
C
6 \sqrt{\frac{6 E I}{m L^{3}}} \mathrm{rad} / \mathrm{s}
D
\frac{2}{L} \sqrt{\frac{E I}{m}} \mathrm{rad} / \mathrm{s}
GATE CE 2021 SET-2      Deflection of Beams
Question 2 Explanation: 


As the deflection will be same in all the 3 columns, so it represents a parallel connection.

\begin{aligned} k_{e q} &=3 k=\frac{36 E I}{L^{3}} \\ \text { Natural frequency }(\omega) &=\sqrt{\frac{k}{m}} \\ &=\sqrt{\frac{36 E I}{m L^{3}}}=6 \sqrt{\frac{E I}{m L^{3}}} \mathrm{rad} / \mathrm{s} \end{aligned}
Question 3
Strain hardening of structural steel means
A
experiencing higher stress than yield stress with increased deformation
B
strengthening steel member externally for reducing strain experienced
C
strain occurring before plastic flow of steel material
D
decrease in the stress experienced with increasing strain
GATE CE 2021 SET-2      Properties of Metals, Stress and Strain
Question 3 Explanation: 
Strain hardening is experiencing higher stress than yield stress with increased deformation
In the figure AB = Strain hardening zone
OA = Linear elastic zone
Stress corresponding to point 'A' is yield stress.

Question 4
A square plate O-P-Q-R of a linear elastic material with sides 1.0 m is loaded in a state of plane stress. Under a given stress condition, the plate deforms to a new configuration O-P'-Q'-R' as shown in the figure (not to scale). Under the given deformation, the edges of the plate remain straight.

The horizontal displacement of the point (0.5 m, 0.5 m) in the plate O-P-Q-R (in mm,round off to one decimal place) is ________
A
1.2
B
6.3
C
5.2
D
2.5
GATE CE 2021 SET-1      Properties of Metals, Stress and Strain
Question 4 Explanation: 


So horizontal displacement of the point (0.5 m, 0.5 m)
=-2.5 \mathrm{~mm}+5 \mathrm{~mm}=2.5 \mathrm{~mm}
Question 5
The equation of deformation is derived to be y=x^{2}-x L for a beam shown in the figure.

The curvature of the beam at the mid-span (in units, in integer) will be ______________
A
1
B
2
C
3
D
4
GATE CE 2021 SET-1      Deflection of Beams
Question 5 Explanation: 


\text { Given, } \qquad \qquad y=x^{2}-x l
Curvature at mid section is \begin{aligned} \frac{1}{R}&=\frac{d^{2} y}{d x^{2}}\\ \frac{d y}{d x}&=2 x-l\\ \frac{d^{2} y}{d x^{2}}&=2\\ \end{aligned}
Question 6
The state of stress in a deformable body is shown in the figure. Consider transformation of the stress from the x-y coordinate system to the X-Y coordinate system. The angle \theta, locating the X-axis, is assumed to be positive when measured from the x-axis in counter-clockwise direction.


The absolute magnitude of the shear stress component \sigma_{\mathrm{xy}} (in MPa,round off to one decimal place) in x-y coordinate system is ________________
A
96.2
B
54.6
C
48.2
D
28.7
GATE CE 2021 SET-1      Principal Stress and Principal Strain
Question 6 Explanation: 




\begin{aligned} \sigma_{x}^{\prime}&=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta+\tau_{x y} \sin 2 \theta \\ \text { Here } \theta=60^{\circ} \\ \sigma_{x}&=40 \mathrm{MPa}, \sigma_{y}=35.6, \sigma_{x}^{\prime}=120, \tau_{x^{\prime} y^{\prime}}=-50 \end{aligned}
Substituting the values in above equation, we get
\tau_{x y}=96.186 \mathrm{MPa}
Question 7
Joints I, J, K, L, Q and M of the frame shown in the figure (not drawn to the scale) are pins. Continuous members IQ and IJ are connected through a pin at N. Continuous members JM and KQ are connected through a pin at P. The frame has hinge supports at joints R and S. The loads acting at joints I, J and K are along the negative Y direction and the loads acting at joints I, M are along the positive X direction.

The magnitude of the horizontal component of reaction (in kN) at S, is
A
5
B
10
C
15
D
20
GATE CE 2020 SET-2      Shear Force and Bending Moment
Question 7 Explanation: 


Remove hinge at support S and replace it with roller support as shown in the figure.
Step One : Find coordinates of all the points where forces are acting
\begin{aligned} y_I&=\sqrt{2} \sin \theta \\ y_J&=\sqrt{2} \sin \theta \\ y_K&=\sqrt{2} \sin \theta \\ x_L&=\sqrt{2} \cos \theta \\ x_M&=5\sqrt{2} \cos \theta \\ x_S&=6\sqrt{2} \cos \theta \end{aligned}
Step Two: Find virtual displacements of all the points
\begin{aligned} \delta y_I&=\sqrt{2} \cos \theta d\theta \\ \delta y_J&=\sqrt{2} \cos \theta d\theta \\ \delta y_K&=\sqrt{2} \cos \theta d\theta \\ \delta x_L&=-\sqrt{2} \sin \theta d\theta \\ \delta x_M&=-5\sqrt{2} \sin \theta d\theta \\ \delta x_S&=-6\sqrt{2} \sin \theta d\theta \\ \end{aligned}
Step Three : Use principle of virtual work to find unknown horizontal force H_s
\Rightarrow \; \delta U=0
=[-10 \times \sqrt{2} \cos \theta d\theta ] \times 3 + [-10 \times -\sqrt{2} \cos \theta d\theta ] +[-10 \times -5\sqrt{2} \sin \theta d\theta ] - [-H_S \times -6\sqrt{2} \sin \theta d\theta ]
H_S=\frac{30\sqrt{2} \cos \theta +10\sqrt{2} \sin \theta +50\sqrt{2} \sin \theta}{6\sqrt{2} \sin \theta}
Substituting, \theta =45^{\circ},\; H_S=\frac{90}{6}=15kN
Note : Sign conventions
If a force acts along positive x or positive y-axis, take it as positive.
If a force acts along negative x or negative y-axis, take it as negative.
Question 8
A weightless cantilever beam of span L is loaded as shown in the figure. For the entire span of the beam, the material properties are identical and the cross-section is rectangular with constant width.

From the flexure-critical perspective, the most economical longitudinal profile of the beam to carry the given loads amongst the options given below, is
A
A
B
B
C
C
D
D
GATE CE 2020 SET-2      Shear Force and Bending Moment
Question 8 Explanation: 


(-PL) + (PL) + (-M_A) = 0
M_A = 0

For most economical,
Maximum cross-section is given where maximum bending moment occurs.

Question 9
The state of stress represented by Mohr's circle shown in the figure is
A
uniaxial tension
B
biaxial tension of equal magnitude
C
hydrostatic stress
D
pure shear
GATE CE 2020 SET-2      Properties of Metals, Stress and Strain
Question 9 Explanation: 
In pure shear condition, Mohr's circle has its center at origin.
Question 10
A rigid, uniform, weightless, horizontal bar is connected to three vertical members P, Q and R as shown in the figure. All three members have identical axial stiffness of 10 kN/mm. The lower ends of bars P and R rest on a rigid horizontal surface. When NO laod is applied, a gap of 2 mm exist between the lower end of the bar Q and the rigid horizontal surface. When a vertical load W is placed on the horizontal bar in the downward direction, the bar still remains horizontal and gets displayed by 5 mm in the vertically downward direction.

The magnitude of the load W (in kN, round off to the nearst integer), is ______
A
110
B
150
C
130
D
160
GATE CE 2020 SET-1      Properties of Metals, Stress and Strain
Question 10 Explanation: 


\begin{aligned} P_1 +P_1+P_2&=W \\ P_1&=P_3 \\ \frac{AE}{L}&=10kN/mm\\ \delta _1&=5mm=\frac{P_1L}{AE} \\ \delta _2&=3 mm = \frac{P_2L}{AE} \\ P_1 &=10 \times 5 =50 kN \\ P_2 &=10 \times 3 =30kN \\ W&=2(20)+30=130 kN \end{aligned}


There are 10 questions to complete.

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