Question 1 |
A perfectly flexible and inextensible cable is shown in the figure (not to scale). The external loads at F and G are acting vertically.
The magnitude of tension in the cable segment FG (in kN, round off to two decimal places) is _______
The magnitude of tension in the cable segment FG (in kN, round off to two decimal places) is _______
4.68 | |
3.78 | |
5.64 | |
8.25 |
Question 1 Explanation:
\begin{aligned} \Sigma M_F=0\\ \left ( 10.667+\frac{H}{6} \right )2-H \times 3=0\\ H=8\; kN\\ V_E=10.667+\frac{8}{6}\\ V_E=12\; kN\\ V_H=10\; kN \end{aligned}
Consider left side of section (1)-(1)
\begin{aligned} \Sigma H=0\\ T \cos \theta =H\\ T \cos \theta =8\\ T \sin \theta =2\\ \therefore \;\; T^2 \cos ^2\theta + T^2 \sin ^2 \theta =8^2+2^2\\ T=8.246\; kN \end{aligned}
Tension in segment GF is 8.246 kN.
Question 2 |
A prismatic fixed-fixed beam, modelled with a total lumped-mass of 10 kg as a single degree of freedom (SDOF) system is shown in the figure.
If the flexural stiffness of the beam is 4 \pi^{2} \mathrm{kN} / \mathrm{m}, its natural frequency of vibration (in Hz, in integer) in the flexural mode will be _______
If the flexural stiffness of the beam is 4 \pi^{2} \mathrm{kN} / \mathrm{m}, its natural frequency of vibration (in Hz, in integer) in the flexural mode will be _______
8 | |
16 | |
10 | |
18 |
Question 2 Explanation:
Given Data:
\begin{aligned} m&=10\; kg\\ K&=4 \pi^2 \times 10^3\\ &\text{Natural frequency,}\\ f_n&=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\\ &=\frac{1}{2 \pi} \sqrt{\frac{4 \pi ^2 \times 10^3}{10}}\\ &=\frac{2 \pi \times 10}{2 \pi}=10 \end{aligned}
\begin{aligned} m&=10\; kg\\ K&=4 \pi^2 \times 10^3\\ &\text{Natural frequency,}\\ f_n&=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\\ &=\frac{1}{2 \pi} \sqrt{\frac{4 \pi ^2 \times 10^3}{10}}\\ &=\frac{2 \pi \times 10}{2 \pi}=10 \end{aligned}
Question 3 |
A frame EFG is shown in the figure. All members are prismatic and have equal flexural rigidity. The member FG carries a uniformly distributed load w per unit length. Axial deformation of any member is neglected.
Considering the joint F being rigid, the support reaction at G is
Considering the joint F being rigid, the support reaction at G is
0.375 wL | |
0.453 wL | |
0.482 wL | |
0.500 wL |
Question 3 Explanation:
Compatibility condition
\begin{array}{l} \qquad \qquad \qquad \qquad \qquad \frac{\partial U}{\partial R}=0 \\ \Rightarrow \qquad \qquad \frac{\partial}{\partial R}\left[\int \frac{M^{2}}{2 E \mid} d x\right]=0 \\ \Rightarrow \frac{\partial}{\partial R}\left[\int_{0}^{L} \frac{\left(R x-\frac{w x^{2}}{2}\right)^{2}}{2 E I} d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)^{2}}{2 E I} d x\right]=0 \\ \Rightarrow \int_{0}^{L} \frac{2\left(R x-\frac{w x^{2}}{2}\right)}{2 E I}(x) d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)}{E I}(L) d x=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{W}{2} \times \frac{L^{4}}{4}+R L^{2}(2 L)-\frac{W^{3}}{2}(2 L)=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{w^{4}}{8}+2 R L^{3}-w L^{4}=0 \\ \Rightarrow 2 R L^{3}+\frac{R L^{3}}{3}=\frac{w^{4}}{8}+w^{4}=\frac{9}{8} w^{4} \\ \Rightarrow \frac{7 R L^{3}}{3}=\frac{9}{8} w L^{4} \\ \Rightarrow R=\frac{27}{56} w L=0.482 w L \end{array}
Alternative
\begin{aligned} \frac{w L^{4}}{8 E I}+\frac{w L^{2}(2 L)}{2 E I} \times L &=\frac{R L^{3}}{3 E I}+\frac{R L \times 2 L \times L}{E I} \\ \frac{w^{4}}{8 E I}+\frac{w^{4}}{E I} &=\frac{R L^{3}}{3 E I}+\frac{2 R L^{3}}{E I} \\ \frac{9 W^{4}}{8 E I} &=\frac{7 R L^{3}}{3 E I} \\ R &=\frac{27 w L}{56}=0.482 \mathrm{wL} \end{aligned}
\begin{array}{l} \qquad \qquad \qquad \qquad \qquad \frac{\partial U}{\partial R}=0 \\ \Rightarrow \qquad \qquad \frac{\partial}{\partial R}\left[\int \frac{M^{2}}{2 E \mid} d x\right]=0 \\ \Rightarrow \frac{\partial}{\partial R}\left[\int_{0}^{L} \frac{\left(R x-\frac{w x^{2}}{2}\right)^{2}}{2 E I} d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)^{2}}{2 E I} d x\right]=0 \\ \Rightarrow \int_{0}^{L} \frac{2\left(R x-\frac{w x^{2}}{2}\right)}{2 E I}(x) d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)}{E I}(L) d x=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{W}{2} \times \frac{L^{4}}{4}+R L^{2}(2 L)-\frac{W^{3}}{2}(2 L)=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{w^{4}}{8}+2 R L^{3}-w L^{4}=0 \\ \Rightarrow 2 R L^{3}+\frac{R L^{3}}{3}=\frac{w^{4}}{8}+w^{4}=\frac{9}{8} w^{4} \\ \Rightarrow \frac{7 R L^{3}}{3}=\frac{9}{8} w L^{4} \\ \Rightarrow R=\frac{27}{56} w L=0.482 w L \end{array}
Alternative
\begin{aligned} \frac{w L^{4}}{8 E I}+\frac{w L^{2}(2 L)}{2 E I} \times L &=\frac{R L^{3}}{3 E I}+\frac{R L \times 2 L \times L}{E I} \\ \frac{w^{4}}{8 E I}+\frac{w^{4}}{E I} &=\frac{R L^{3}}{3 E I}+\frac{2 R L^{3}}{E I} \\ \frac{9 W^{4}}{8 E I} &=\frac{7 R L^{3}}{3 E I} \\ R &=\frac{27 w L}{56}=0.482 \mathrm{wL} \end{aligned}
Question 4 |
A propped cantilever beam XY, with an internal hinge at the middle, is carrying a uniformly distributed load of 10 kN/m, as shown in the figure.
The vertical reaction at support X ( in kN, in integer) is _____
The vertical reaction at support X ( in kN, in integer) is _____
25 | |
30 | |
35 | |
40 |
Question 4 Explanation:
\begin{aligned} B M &=0 \text { at hinge } \\ R_{Y} \times 2-10 \times 2 \times 1 &=0 \\ R_{Y} &=10 \mathrm{kN} \\ R_{X}+R_{Y} &=10 \times 4 \\ R_{X}+10 &=40 \\ R_{X} &=30 \mathrm{kN} \end{aligned}
Question 5 |
Employ stiffness matrix approach for the simply supported beam as shown in the figure to calculate unknown displacements/rotations. Take length, L=8 m; modulus of elasticity, E=3 \times 10^{4} \mathrm{~N} / \mathrm{mm}^{2}; moment of inertia, I=225 \times 10^{6} \mathrm{mm}^{4}.
The mid-span deflection of the beam (in mm,round off to integer) under P=100 kN in downward direction will be ___________
The mid-span deflection of the beam (in mm,round off to integer) under P=100 kN in downward direction will be ___________
186 | |
91 | |
119 | |
192 |
Question 5 Explanation:
By stiffness matrix method
Step-1 : Generation of stiffness matrix
Column-1
\begin{aligned} \mathrm{k}_{11}&=\frac{3 E(2 I)}{(L / 2)^{3}}+\frac{3 E(I)}{(L / 2)^{3}}=\frac{72 \mathrm{E} I}{L^{3}}\\ k_{21}&=-\frac{3 E(2 I)}{(L / 2)^{2}}+\frac{3 E(I)}{(L / 2)^{2}}=-\frac{12 E I}{L^{2}} \end{aligned}
Column-2
\begin{aligned} k_{22}&=\frac{3 E(2 I)}{(L / 2)}+\frac{3 E(I)}{(L / 2)}=18 \frac{E I}{L} \\ \text { Stiffness matrix }[k]&=\left[\begin{array}{cc} 72 \frac{E I}{L^{3}} & -\frac{12 E I}{L^{2}} \\ -\frac{12 E I}{L^{2}} & \frac{E I}{L} \end{array}\right] \end{aligned}
Step-2: Calculation of unknown Nodal displacements \left(\Delta_{\mathrm{B}}, \theta_{\mathrm{B}}\right)
\begin{aligned} \text{Using} \qquad \qquad \qquad \qquad [P]&=[k][\Delta]\\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \quad & \quad\left[\begin{array}{l} P \\ O \end{array}\right]=\left[\begin{array}{cc} \frac{72 E I}{L^{2}} & -\frac{12 E I}{L^{2}} \\ -\frac{12 E I}{L^{2}} & \frac{8 E I}{L} \end{array}\right]\left[\begin{array}{l} \Delta_{B} \\ \theta_{B} \end{array}\right]\\ \text{On solving} \qquad \qquad \qquad \quad\\\Delta_{\mathrm{B}}&=\frac{P L^{3}}{64 E I}(\downarrow) \\ \therefore \qquad \qquad \qquad \qquad \qquad \theta_{\mathrm{B}}&=\frac{P L^{2}}{96 E I}(\mathrm{CW}) \\ \Delta_{\mathrm{B}}&=\frac{\left(100 \times 10^{3}\right) \times(8000)^{3}}{64 \times 3 \times 10^{4} \times 225 \times 10^{6}}\\ &=118.519 \mathrm{~mm} \simeq 119 \mathrm{~mm} \end{aligned}
Question 6 |
Refer the truss as shown in the figure (not to scale).
If load, F=10 \sqrt{3} \mathrm{kN}, moment of inertia, I=8.33 \times 10^{6} \mathrm{~mm}^{4}, area of cross-section, A=10^{4} \mathrm{~mm}^{2}, and length, L=2 m for all the members of the truss, the compressive stress (in \mathrm{kN} / \mathrm{m}^{2},in integer) carried by the member Q-R is _________
If load, F=10 \sqrt{3} \mathrm{kN}, moment of inertia, I=8.33 \times 10^{6} \mathrm{~mm}^{4}, area of cross-section, A=10^{4} \mathrm{~mm}^{2}, and length, L=2 m for all the members of the truss, the compressive stress (in \mathrm{kN} / \mathrm{m}^{2},in integer) carried by the member Q-R is _________
370 | |
650 | |
760 | |
500 |
Question 6 Explanation:
V_{P}=V_{S}=5 \sqrt{3} \mathrm{kN}
Considering equilibrium of LHS of section (1 )-(1)
Taking moment about 'T'
\begin{aligned} \Sigma M_{T}(C W)&=0\\ (5 \sqrt{3} \times a)+F_{Q R}\left(\frac{\sqrt{3} a}{2}\right)&=0\\ \Rightarrow \qquad \qquad \qquad \qquad \quad F_{O A}&=-10 \mathrm{kN} or 10 \mathrm{kN} (C) \\ \text{ Compressive} &\text{ stress in member QR} \left(\sigma_{\mathrm{C}}\right)\\ \sigma_{\mathrm{C}}&=\frac{F_{Q A}}{2 A} \\ &=\frac{10 \mathrm{kN}}{2 \times\left(10^{4} \times 10^{-6}\right) \mathrm{m}^{2}} \\ &=500 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}
Question 7 |
A truss EFGH is shown in the figure, in which all the members have the same axial rigidity R. In the figure, P is the magnitude of external horizontal forces acting at joints F and G.
If R=500 \times 10^{3} \mathrm{kN},P=150 \mathrm{kN} and L=3 m, the magnitude of the horizontal displacement of joint G (in mm,round off to one decimal place) is ________
If R=500 \times 10^{3} \mathrm{kN},P=150 \mathrm{kN} and L=3 m, the magnitude of the horizontal displacement of joint G (in mm,round off to one decimal place) is ________
0.4 | |
0.6 | |
0.9 | |
0.2 |
Question 7 Explanation:
Note: No need to calculate 'I< force in all members because 'P force is zero fore all members except FG
By unit load method
\begin{aligned} \text { 1. } \Delta_{\mathrm{HG}}=& \sum_{i=1}^{n} \frac{P_{i} k_{i} L_{i}}{A_{i} E_{i}} \\ \Delta_{\mathrm{HG}}=& \text { Horizontal deflection at joint } \mathrm{G} \\ \therefore \qquad & \qquad \qquad\Delta_{\mathrm{HG}} =\underbrace{\frac{P \times 1 \times L}{A E}}_{\text {FG-member }} \times \underbrace{Q^{\prime}}_{\text {(For all other members) }} \\ & \qquad \qquad\Delta_{\mathrm{HG}} =\frac{P L}{A E}=\left(\frac{150 \times 3}{500 \times 10^{3}} \times 10^{3}\right) \mathrm{mm} \\ & \qquad \qquad \qquad=0.90 \mathrm{~mm} \end{aligned}
Question 8 |
A propped cantilever beam EF is subjected to a unit moving load as shown in the figure (not to scale). The sign convention for positive shear force at the left and right sides of any section is also shown.
The CORRECT qualitative nature of the influence line diagram for shear force at G is
The CORRECT qualitative nature of the influence line diagram for shear force at G is
 | |
 | |
 | |
 |
Question 8 Explanation:
As per Muller Breslau principle ILD for stress function (shear -V_{G}) will be a combination of curves (3^{\circ} curves).
Question 9 |
The plane truss has hinge supports at P and W and is subjected to the horizontal forces
as shown in the figure.
Representing the tensile force with '+' sign and the compressive force with '-' sign, the force in member XW (in kN, round off to the nearest integer), is _________.
Representing the tensile force with '+' sign and the compressive force with '-' sign, the force in member XW (in kN, round off to the nearest integer), is _________.
30 | |
-30 | |
40 | |
-40 |
Question 9 Explanation:
Force in PQ
Considering the section above (1) - (1)
Taking moment about 'R'
\begin{aligned} \Sigma M_{RL}&=0\\ (10 \times 4)+(10 \times 8)&+F_{PQ} \times 4=0\\ F_{PQ}&=-\frac{120}{4}\\ &=-30kN=30kN (Comp.) \end{aligned}
Considering the section above (1) - (1)
Taking moment about 'R'
\begin{aligned} \Sigma M_{RL}&=0\\ (10 \times 4)+(10 \times 8)&+F_{PQ} \times 4=0\\ F_{PQ}&=-\frac{120}{4}\\ &=-30kN=30kN (Comp.) \end{aligned}
Question 10 |
The planar structure RST shown in the figure is roller-supported at S and pin-supported
at R. Members RS and ST have uniform flexural rigidity (EI) and S is a rigid joint. Consider
only bending deformation and neglect effects of self-weight and axial stiffening
When the structure is subjected to a concentrated horizontal load P at the end T, the magnitude of rotation at the support R, is
When the structure is subjected to a concentrated horizontal load P at the end T, the magnitude of rotation at the support R, is
\frac{PL^3}{12EI} | |
\frac{PL^2}{12EI} | |
\frac{PL^2}{6EI} | |
\frac{PL}{6EI} |
Question 10 Explanation:
\theta _R=\frac{(PL/2)L}{6EI}
\theta _R=\frac{PL^2}{12EI}
There are 10 questions to complete.
