Question 1 |
For the frame shown in the figure (not to scale), all members (A B, B C, C D, G B, and C H) have the same length, L and flexural rigidity, El. The joints at B and C are rigid joints, and the supports A and D are fixed supports. Beams GB and \mathrm{CH} carry uniformly distributed loads of w per unit length. The magnitude of the moment reaction at A is \mathrm{wL}^{2} / \mathrm{k}. What is the value of \mathrm{k} (in integer) ? ____


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Question 1 Explanation:

Due to symmetry, slope at mid of B C is zero and can be taken as slider at point O
Distribution factors:
\begin{aligned} \mathrm{K}_{\mathrm{BA}} & =\frac{4 \mathrm{EI} / \mathrm{L}}{4 \mathrm{EI} / \mathrm{L}+\mathrm{EI} /(\mathrm{L} / 2)}=\frac{2}{3} \\ \mathrm{K}_{\mathrm{BO}}&= 1-\mathrm{K}_{\mathrm{BA}} \\ & =1-2 / 3=1 / 3 \\ \therefore \quad \mathrm{M}_{\mathrm{BA}} & =\frac{\mathrm{WL}^{2}}{2} \times \frac{2}{3}=\frac{\mathrm{WL}^{2}}{3} \\ \mathrm{M}_{\mathrm{AB}} & =\frac{\mathrm{M}_{\mathrm{BA}}}{2}=\frac{\mathrm{WL}^{2} / 3}{2}=\frac{\mathrm{WL}^{2}}{6} \\ & \therefore \mathrm{K}=6 \end{aligned}
Question 2 |
When a simply-supported elastic beam of span \mathrm{L} and flexural rigidity EI ( E is the modulus of elasticity and \mathrm{I} is the moment of inertia of the section) is loaded with a uniformly distributed load w per unit length, the deflection at the mid-span is
\Delta_{0}=\frac{5}{384} \frac{w L^{4}}{\mathrm{El}}.
If the load on one half of the span is now removed, the mid-span deflection ____
\Delta_{0}=\frac{5}{384} \frac{w L^{4}}{\mathrm{El}}.
If the load on one half of the span is now removed, the mid-span deflection ____
reduces to \Delta_{0} / 2 | |
reduces to a value less than \Delta_{0} / 2 | |
reduces to a value greater than \Delta_{0} / 2 | |
remains unchanged at \Delta_{0} |
Question 2 Explanation:

As Case-I and II are mirror image of each other,.
Hence, \quad \Delta_{\mathrm{C}, 1}=\Delta_{\mathrm{C}, \mathrm{II}}
Also, \quad \Delta_{\mathrm{C}}=\Delta_{\mathrm{C}, 1}+\Delta_{\mathrm{C}, 1}
\Rightarrow \quad \Delta_{\mathrm{C}, \mathrm{I}}=\Delta_{\mathrm{C}, \mathrm{II}}=\frac{\Delta_{\mathrm{C}}}{2}
Question 3 |
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}
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Question 3 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 4 |
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 ______________

The curvature of the beam at the mid-span (in units, in integer) will be ______________
1 | |
2 | |
3 | |
4 |
Question 4 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 5 |
A cantilever beam PQ of uniform flexural rigidity (EI) is subjected to a concentrated
moment M at R as shown in the figure.

The deflection at the free end Q is

The deflection at the free end Q is
\frac{ML^2}{6EI} | |
\frac{ML^2}{4EI} | |
\frac{3ML^2}{8EI} | |
\frac{3ML^2}{4EI} |
Question 5 Explanation:

\begin{aligned} \delta _c&=\delta _1=CC_1+C_1C_2\\ &=BB_1+C_1C_2\\ &=\frac{M\left ( \frac{L}{2} \right )^2}{2EI}+\frac{M\left ( \frac{L}{2} \right )}{EI}\left ( \frac{L}{2} \right )\\ \delta _c&=\frac{3}{8}\frac{ML^2}{EI} \end{aligned}
There are 5 questions to complete.