Question 1 |
A uniform rod KJ of weight w shown in the figure rests against a frictionless
vertical wall at the point K and a rough horizontal surface at point J. It is given
that w = 10 kN, a = 4 m and b = 3 m.
The minimum coefficient of static friction that is required at the point J to hold the rod in equilibrium is ___________. (round off to three decimal places)
The minimum coefficient of static friction that is required at the point J to hold the rod in equilibrium is ___________. (round off to three decimal places)
0.214 | |
0.258 | |
0.325 | |
0.375 |
Question 1 Explanation:
\begin{aligned} \Sigma F_v&=0\\ R_J&=0\\ \Sigma M_K&=0\\ 10 \times 1.5+\mu _sR_J \times 4-R_J \times 3&=0\\ \mu _s=\frac{10 \times 3-10 \times 1.5}{10 \times 4}&=0.375 \end{aligned}
Question 2 |
Consider two linearly elastic rods HI and IJ, each of length b, as shown in the
figure. The rods are co-linear, and confined between two fixed supports at H and
J. Both the rods are initially stress free. The coefficient of linear thermal
expansion is \alpha for both the rods. The temperature of the rod IJ is raised by \Delta T ,
whereas the temperature of rod HI remains unchanged. An external horizontal
force P is now applied at node I. It is given that \alpha ={10^{-6} } {\circ} C^{-1},\Delta T=50^{\circ}C,b = 2 m, AE = 10^6N.. The axial rigidities of the rods HI and IJ are 2AE and AE, respectively.
To make the axial force in rod HI equal to zero, the value of the external force P (in N) is _________. (round off to the nearest integer)
To make the axial force in rod HI equal to zero, the value of the external force P (in N) is _________. (round off to the nearest integer)
85 | |
23 | |
65 | |
50 |
Question 2 Explanation:
R_A+R_B=P
\because There is no axial force in rod HI \therefore R_A=0
Now check for rod IJ
\therefore R_B=P
Now as rod IJ is fixed from both end, so net deflection due to increase in temperature will be zero.
\begin{aligned} b\alpha T +\left [ -\frac{Pb}{AE} \right ]&= 0\\ b\alpha T &=\frac{Pb}{AE} \\ P &= AE\alpha T\\ P&= 10^6 \times 10^{-6} \times 50\\ P&=50N \end{aligned}
Question 3 |
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 3 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 4 |
Stresses acting on an infinitesimal soil element are shown in the figure (with \sigma _z \gt \sigma _x). The major and minor principal stresses are \sigma _1
and \sigma _3, respectively. Considering the compressive stresses as positive, which one of the
following expressions correctly represents the angle between the major
principal stress plane and the horizontal plane?
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1-\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _3-\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1+\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1+\sigma _3} \right ) |
Question 4 Explanation:
\begin{aligned} \Sigma F_x &=0 \\ \sigma _x(BC)-\tau _Z \times (AB)\sigma _1 \sin \theta &= 0\\ \sigma _x\left ( \frac{AC \sin \theta }{\cos \theta } \right )+\tau _{zx}\left ( \frac{AC \cos \alpha }{\cos \theta } \right ) &=\sigma _1 \frac{AC \sin \theta }{\cos \theta }\\ \sigma _x \tan \theta +\tau _{zx} &=\sigma _1 \tan \theta \\ \tan \theta(\sigma _1-\sigma _2) &= \tau _{zx}\\ \tan \theta &= \left ( \frac{\tau _{zx}}{\sigma _1-\sigma _x} \right ) \end{aligned}
Question 5 |
For a linear elastic and isotropic material, the correct relationship among Young's modulus of elasticity (E), Poisson's ratio (v), and shear modulus (G) is
G=\frac{E}{2(1+v)} | |
G=\frac{E}{(1+2v)} | |
E=\frac{G}{2(1+v)} | |
E=\frac{G}{(1+2v)} |
Question 5 Explanation:
E=2G(1+\mu )
G= Shear modulas
\mu =Poission's ratio
E= Young's modulus
G= Shear modulas
\mu =Poission's ratio
E= Young's modulus
Question 6 |
A horizontal force of P kN is applied to a homogeneous body of weight 25 kN,
as shown in the figure. The coefficient of friction between the body and the
floor is 0.3. Which of the following statement(s) is/are correct?
The motion of the body will occur by overturning. | |
Sliding of the body never occurs. | |
No motion occurs for P \leq 6 kN. | |
The motion of the body will occur by sliding only. |
Question 6 Explanation:
Minimum force for sliding
(P_{min})_{sliding}=(f_s)_{max} ...(i)
Applying equilibrium equation in vertical direction
Normal reaction = Weight
N=mg=25 kN ...(ii)
Using equation (i) and (ii)
(P_{min})_{sliding}=\mu N =0.3 \times 25=7.5 kN
Minimum force for overturning
At the verge of overturning
(P_{min})_{oberturning} \times 2=W \times 2
(P_{min})_{oberturning}=\frac{25 \times 0.5} {2}=6.25 kN
Here, (P_{min})_{oberturning} \lt (P_{min})_{sliding}
First overtuning will take place.
Sliding will not take place.
Question 7 |
Consider the cross-section of a beam made up of thin uniform elements having
thickness t(t \lt \lt a) shown in the figure. The (x,y) coordinates of the points along
the center-line of the cross-section are given in the figure.
The coordinates of the shear center of this cross-section are:
The coordinates of the shear center of this cross-section are:
x = 0, y = 3a | |
x = 2a, y = 2a | |
x = -a, y = 2a | |
x = -2a, y = a |
Question 7 Explanation:
Shear centre of section consisting of two intersecting narrow rectangles always lies at the intersection of centrelines of two rectangles.
Coordinate of shear centre (0, 3a).
Coordinate of shear centre (0, 3a).
Question 8 |
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 8 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 9 |
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 ________
The absolute maximum shear stress in the member (in MPa, round off to one decimal place) is ________
15.3 | |
12.6 | |
18.2 | |
11.8 |
Question 9 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 10 |
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
For the lateral deflection profile of the columns as shown in the figure, the natural frequency of the system for horizontal oscillation is
6 \sqrt{\frac{E I}{m L^{3}}} \mathrm{rad} / \mathrm{s} | |
\frac{1}{L} \sqrt{\frac{2 E I}{m}} \mathrm{rad} / \mathrm{s} | |
6 \sqrt{\frac{6 E I}{m L^{3}}} \mathrm{rad} / \mathrm{s} | |
\frac{2}{L} \sqrt{\frac{E I}{m}} \mathrm{rad} / \mathrm{s}
|
Question 10 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}
There are 10 questions to complete.