Question 1 |

The linearly elastic planar structure shown in the figure is acted upon by two
vertical concentrated forces. The horizontal beams UV and WX are connected
with the help of the vertical linear spring with spring constant k = 20 kN/m. The
fixed supports are provided at U and X. It is given that flexural rigidity
EI = 10^5 kN-m^2, P = 100 kN, and a = 5 m. Force Q is applied at the center of beam WX such that the force in the spring VW becomes zero.

The magnitude of force Q (in kN) is ________. (round off to the nearest integer)

The magnitude of force Q (in kN) is ________. (round off to the nearest integer)

640 | |

458 | |

251 | |

325 |

Question 1 Explanation:

If force in spring is zero, there will be no
deformation in spring i.e., deflection of point V
will be equal to deflection of point W

\begin{aligned} \delta _v&=\delta _w\\ \frac{P(2a)^3}{3EI}&=\frac{Q(a)^3}{3(2EI)}+\frac{Q(a)^2}{2(2EI)} +a\\ \frac{8}{3}P&=\left ( \frac{1}{6}+\frac{1}{4} \right )Q\\ Q&=\frac{32}{5}P=640kN \end{aligned}

\begin{aligned} \delta _v&=\delta _w\\ \frac{P(2a)^3}{3EI}&=\frac{Q(a)^3}{3(2EI)}+\frac{Q(a)^2}{2(2EI)} +a\\ \frac{8}{3}P&=\left ( \frac{1}{6}+\frac{1}{4} \right )Q\\ Q&=\frac{32}{5}P=640kN \end{aligned}

Question 2 |

Which of the following statement(s) is/are correct?

If a linearly elastic structure is subjected to a set of loads, the partial derivative
of the total strain energy with respect to the deflection at any point is equal to
the load applied at that point. | |

If a linearly elastic structure is subjected to a set of loads, the partial derivative
of the total strain energy with respect to the load at any point is equal to the
deflection at that point. | |

If a structure is acted upon by two force system Pa and Pb , in equilibrium
separately, the external virtual work done by a system of forces Pb during the
deformations caused by another system of forces Pa is equal to the external
virtual work done by the Pa system during the deformation caused by the Pb
system. | |

The shear force in a conjugate beam loaded by the M/EI diagram of the real
beam is equal to the corresponding deflection of the real beam. |

Question 3 |

Consider a beam PQ fixed at P, hinged at Q, and subjected to a load F as shown in figure (not drawn to scale). The static and kinematic degrees of
indeterminacy, respectively, are

2 and 1 | |

2 and 0 | |

1 and 2 | |

2 and 2 |

Question 3 Explanation:

Static indeterminacy, SI=r-3=(3+2)-3=2

Kinematic indeterminacy=0+1=1

Question 4 |

Consider a simply supported beam PQ as shown in the figure. A truck having
100 kN on the front axle and 200 kN on the rear axle, moves from left to right.
The spacing between the axles is 3 m. The maximum bending moment at point
R is ______ kNm. (in integer)

124 | |

180 | |

147 | |

582 |

Question 4 Explanation:

\frac{ab}{L}=\frac{1 \times 4}{5}=0.8m

To get maximum BN at R

\begin{aligned} BM_{max}&=200 \times \frac{ab}{L}+100 \times y\\ \frac{ab/l}{b}&=\frac{y}{4-3}\Rightarrow \frac{0.8}{4}=\frac{y}{1}=0.2m\\ BM_{max}&=200 \times 0.8+100 \times 0.2=180kNm \end{aligned}

Question 5 |

Consider the linearly elastic plane frame shown in the figure. Members HF, FK
and FG are welded together at joint F. Joints K, G and H are fixed supports. A
counter-clockwise moment M is applied at joint F. Consider flexural rigidity
EI=10^5kN-m^2
for each member and neglect axial deformations.

If the magnitude (absolute value) of the support moment at H is 10 kN-m, the magnitude (absolute value) of the applied moment M (in kN-m) to maintain static equilibrium is ___________. (round off to the nearest integer)

If the magnitude (absolute value) of the support moment at H is 10 kN-m, the magnitude (absolute value) of the applied moment M (in kN-m) to maintain static equilibrium is ___________. (round off to the nearest integer)

60 | |

25 | |

85 | |

45 |

Question 5 Explanation:

\frac{M}{6}=10kNm\Rightarrow M=60kNm

Question 6 |

The plane truss shown in the figure is subjected to an external force P. It is given that P = 70 kN, a = 2 m, and b = 3 m.

The magnitude (absolute value) of force (in kN) in member EF is _______. (round off to the nearest integer)

The magnitude (absolute value) of force (in kN) in member EF is _______. (round off to the nearest integer)

12 | |

30 | |

87 | |

92 |

Question 6 Explanation:

\begin{aligned} \Sigma M_J&=0\\ V_A \times 8-H_A \times 1-70 \times 4&=0\\ 8V_A-H_A&=280 \;\;\;(i) \end{aligned}

To find H_A (Cut the truss by 1-1)

Consider left hand side

\begin{aligned} \Sigma M_E&=0\\ V_A \times 4-H_A \times 4&=0\\ V_A&=H_A \;\;\;(ii)\\ \text{using (i) and (ii)}&\\ 8V_A-V_A&=280\\ 7V_A&=280\\ V_A&=40kN\\ H_A&=40kN\\ H_J&=40kN\\ V_J&=70-40=30kN\\ \end{aligned}

To find force in member EF (Cit the truss by 2-2)

Consider right hand side

Force in member EF

F_{EF}=V_J=30kN

Question 7 |

A semi-circular bar of radius R m, in a vertical plane, is fixed at the end G, as
shown in the figure. A horizontal load of magnitude P kN is applied at the end
H. The magnitude of the axial force, shear force, and bending moment at point
Q for \theta =45 ^{\circ}, respectively, are

\frac{P}{\sqrt{2}}kN,\frac{P}{\sqrt{2}}kN \text{ and } \frac{PR}{\sqrt{2}}kNm | |

\frac{P}{\sqrt{2}}kN,\frac{P}{\sqrt{2}}kN \text{ and } 0 \; kNm | |

0 \; kN,\frac{P}{\sqrt{2}}kN \text{ and } \frac{PR}{\sqrt{2}}kNm | |

\frac{P}{\sqrt{2}}kN,0 \; kN \text{ and } \frac{PR}{\sqrt{2}}kNm |

Question 7 Explanation:

\begin{aligned} (F_Q)&=P\sin \theta =\frac{P}{\sqrt{2}}&(\; at\; \theta =45^{\circ})\\ (S_Q)&=P\cos \theta =\frac{P}{\sqrt{2}}&(\; at\; \theta =45^{\circ})\\ M_Q&=PR \sin \theta =\frac{PR}{\sqrt{2}}&( \; at\; \theta =45^{\circ}) \end{aligned}

Question 8 |

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 8 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 9 |

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 9 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 10 |

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 10 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}

There are 10 questions to complete.

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