Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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}
Question 4 |
Two prismatic beams having the same flexural rigidity of 1000 kN-m^2 are shown in the figures.
If the mid-span deflections of these beams are denoted by \delta_1 \; and \; \delta _2 (as indicated in the figures). The correct option is
If the mid-span deflections of these beams are denoted by \delta_1 \; and \; \delta _2 (as indicated in the figures). The correct option is
\delta _{1}=\delta _{2} | |
\delta _{1} \lt \delta _{2} | |
\delta _{1} \gt \delta _{2} | |
\delta _{1} \gt \gt \delta _{2} |
Question 4 Explanation:
\begin{aligned} \delta_{1} &=\frac{5}{384} \times \frac{w l^{4}}{E I} \\ &=\frac{5}{384} \times \frac{6(4)^{4}}{1000}=20 \times 10^{-3} \mathrm{m}\\ \delta_{2}&=\frac{P l^{3}}{48 E I}=\frac{120 \times 2^{3}}{48 \times 1000}=20 \times 10^{-3} \mathrm{m} \\ \therefore \quad \delta_{1}&=\delta_{2} \end{aligned}\\
Question 5 |
Two beams PQ (fixed at P and with a roller support at Q, as shown in Figure I, which allows vertical movement) and XZ (with a hinge at Y) are shown in the Figures I and II respectively. The spans of PQ and XZ are L and 2L respectively. Both the beams are under the action of uniformly distributed load (W) and have the same flexural stiffness, EI (where, E and I respectively denote modulus of elasticity and moment of inertia about axis of bending). Let the maximum deflection and maximum rotation be \delta _{max1} \; and \; \theta _{max1} respectively, in the case of beam PQ and the corresponding quantities for the beam XZ be \delta _{max2} \; and \; \theta _{max2}, respectively.
Which one of the following relationships is true?
Which one of the following relationships is true?
\delta _{max1}\neq\delta _{max2} \; and \; \theta _{max1}\neq \theta _{max2} | |
\delta _{max1}= \delta _{max2} \; and \; \theta _{max1}\neq \theta _{max2} | |
\delta _{max1}\neq \delta _{max2} \; and \; \theta _{max1}= \theta _{max2} | |
\delta _{max1}= \delta _{max2} \; and \; \theta _{max1}= \theta _{max2} |
Question 5 Explanation:
Deflection in beam xy at y= Deflection in beam
yz at y
\begin{aligned} \Rightarrow \quad \frac{w L^{4}}{8 E I} &-\frac{R L^{3}}{3 E I}=\frac{R L^{3}}{3 E I}+\frac{w L^{4}}{8 E I} \\ & A=0 \end{aligned}
In beam PQ also at support Q , vertical reaction in zero because of roller support.
So, bearn PQ, xy and yz are same
\therefore \quad \theta _{\max 1}=\theta _{\max 2} and \delta_{\max 1 }=\delta_{\max 2}
Question 6 |
A 3 m long simply supported beam of uniform cross section is subjected to a uniformly distributed load of w=20kN/m in the central 1 m as shown in the figure
If the flexural rigidity (EI) of the Beam is 30\times 10^{6}N-m^{2}, the Maximum slope (expressed in Radians ) of the deformed beam is:
If the flexural rigidity (EI) of the Beam is 30\times 10^{6}N-m^{2}, the Maximum slope (expressed in Radians ) of the deformed beam is:
0.681\times10^{-7} | |
0.943\times10^{-7} | |
4.310\times10^{-7} | |
5.910\times10^{-7} |
Question 6 Explanation:
Due to symmetrical loading,
R_{p}=R_{Q}=\frac{w \times 1}{2}=\frac{w}{2}
M_{x}=R_{\rho} x /-\frac{w}{2}(x-1)^{2} /+\frac{W}{2}(x-2)^{2}
According to Macaulay method
E I \frac{d^{2} y}{d x^{2}}=R_{\rho} x /-\frac{w}{2}(x-1)^{2} /+\frac{w}{2}(x-2)^{2}
After integrating once
E I \frac{d y}{d x}=\frac{R_{p} x^{2}}{2}+c_{1} /-\frac{w}{6}(x-1)^{3} /+\frac{w}{6}(x-2)^{3}
Due to symmetrical loading slope will be zero at
mid section (x=1.5 m)
\begin{aligned} \therefore \text{at }x&=15 \mathrm{m}, \frac{d y}{d x}=0 \\ 0 &=\frac{R_{p}(1.5)^{2}}{2}+c_{1}-\frac{w}{6}(1.5-1)^{3} \\ &=\frac{9 R_{p}}{8}+c_{1}-\frac{w}{48}=\frac{9 w}{16}+c_{1}-\frac{w}{48} \\ \therefore \quad c_{1} &=\frac{w}{48}-\frac{9 w}{16}=-\frac{26 w}{48}=-\frac{13 w}{24} \end{aligned}
\therefore \quad Equation of slope
E I \cdot \frac{d y}{d x}=\frac{R_{p} x^{2}}{2}-\frac{13 w}{24} /-\frac{w}{6}(x-1)^{3} /+\frac{w(x-2)^{3}}{6}
The slope will be maximum at the support
\therefore \theta_{\max }(x=0)
=\frac{-13 w}{24 E I}=\frac{-13 \times 20 \times 10^{3}}{24 \times 30 \times 10^{6}}=-0.361 \times 10^{-3} \mathrm{radiar}
Alternatively.
Moment Area Method
\begin{aligned} M_{A} & =P \times x \\ & =10 \times 1=10 \mathrm{kNm} \\ M_{B} & =P_{x}-\frac{W}{2}(x-1)^{2} \\ & =10 \times 1.5-\frac{20 \times 0.5^{2}}{2}=12.5 \mathrm{kNm} \end{aligned}
\theta_{A}-\theta_{C}= Area of M/EI diagram between points
P and B
\begin{aligned} &\theta_{B}-\theta_{A}=A_{l}+A_{II}+A_{III}\\ &=\left(\frac{1}{2} \times 1 \times \frac{10}{E I}\right)+\left(0.5 \times \frac{10}{E I}\right)+\left(\frac{2}{3} \times 0.5 \times \frac{2.5}{E I}\right) \\ &=\frac{5}{E I}+\frac{5}{E I}+\frac{0.833}{E I} \\ \theta_{A} &=-\frac{10.833}{E I}=-\frac{10.833 \times 10^{3}}{30 \times 10^{6}} \\ &=-0.3611 \times 10^{-3} \mathrm{radians} \end{aligned}
Question 7 |
A steel strip of length, L = 200 mm is fixed at end A and rests at B on a vertical spring of stiffness, k = 2 N/mm. The steel strip is 5 mm wide and 10 mm thick. A vertical load, P = 50 N is applied at B, as shown in the figure. Considering E = 200 GPa, the force (in N) developed in the spring is _________.
9 | |
6 | |
5 | |
3 |
Question 7 Explanation:
Deflection of point B= Deflection of spring
\Delta_{B}=\frac{(P-R) L^{3}}{3 E I}=\frac{R}{k}
Where, R= Force in the spring
\begin{aligned} \Rightarrow \frac{(50-R) 200^{3}}{3 \times 200 \times 10^{3} \times \frac{5 \times 10^{3}}{12}}&=\frac{R}{2} \\ 0.064(50-R)&= R \\ 3.2&=R+0.064 R \\ R &=3.0075 \mathrm{N} \end{aligned}
Question 8 |
A simply supported reinforced concrete beam of length 10 m sags while undergoing shrinkage. Assuming a uniform curvature of 0.004 m^{-1} along the span, the maximum deflection (in m) of the beam at mid-span is _______.
0.05 | |
0.005 | |
0.0005 | |
0.004 |
Question 8 Explanation:
\begin{aligned} \text{Radius. }R&=\frac{1}{C}=250 \mathrm{m} \\ O A &=\sqrt{250^{2}-5^{2}} \\ &=249.95 \mathrm{m} \\ \text { Deflection } &=A A^{\prime}=250-249.95 \\ &=0.05 \mathrm{m} \end{aligned}
Question 9 |
Two beams are connected by a linear spring as shown in the following figure. For a load P as shown in the figure, the percentage of the applied load P carried by the spring is __________.
33.33 | |
44.44 | |
66.66 | |
25.00 |
Question 9 Explanation:
\Delta_{\text {spring }}=\Delta_{B}-\Delta_{D}
Compression of spring
\begin{aligned} =\frac{(P-R) L^{3}}{3 E I}-\frac{R L^{3}}{3 E I}&=\frac{R}{K_{S}} \\ \frac{(P-R) L^{3}}{3 E I}-\frac{R L^{3}}{3 E I}&=\frac{R \times\left(2 L^{3}\right)}{3 E I}\\ (P-A)-R &=2 R \\ P &=4 R \\ R &=\frac{P}{4} \end{aligned}
% force carried by spring =25 %
Question 10 |
The tension (in kN) in a 10m long cable, shown in figure, neglecting its self-weight is
120 | |
75 | |
60 | |
45 |
Question 10 Explanation:
\begin{aligned} \Sigma F_{y} &=0 \\ \Rightarrow\quad 2 \text{Tc} 0 s \theta &=120 \\ \text{Here, }\quad \cos \theta &=\frac{4}{5} \\ 2 T \times \frac{4}{5}&=120 \\ \Rightarrow\quad T=\frac{120 \times 5}{2 \times 4}&=75 \mathrm{kN} \end{aligned}
There are 10 questions to complete.