Question 1 |
A rigid beam AD of length 3a = 6 m is hinged at
frictionless pin joint A and supported by two strings
as shown in the figure. String BC passes over two
small frictionless pulleys of negligible radius. All
the strings are made of the same material and have
equal cross-sectional area. A force F = 9 kN is
applied at C and the resulting stresses in the strings
are within linear elastic limit. The self-weight of the
beam is negligible with respect to the applied load.
Assuming small deflections, the tension developed
in the string at C is _________ kN (round off to 2
decimal places).
2.57 | |
3.65 | |
1.45 | |
4.65 |
Question 1 Explanation:
F.B.D of beam
\begin{aligned} \delta l_1:\delta l_2:\delta l_3&=T_1:T_2:T_3\\ 1:2:3&=T_1:T_2:T_3\\ \text{Let }T_1&=T\\ \therefore \; T_2&=2T,T_3=3T\\ \Sigma M_A&=0\\ \therefore \; -T \times a-2T &\times 2a-3T \times 3a+F \times 2a=0\\ \therefore \;\; F \times 2a &=14 Ta\\ 2F&=14T\\ T&=\frac{F}{7}=\frac{9}{7}kN\\ T_2&=2T=\frac{18}{7}=2.57kN \end{aligned}
\begin{aligned} \delta l_1:\delta l_2:\delta l_3&=T_1:T_2:T_3\\ 1:2:3&=T_1:T_2:T_3\\ \text{Let }T_1&=T\\ \therefore \; T_2&=2T,T_3=3T\\ \Sigma M_A&=0\\ \therefore \; -T \times a-2T &\times 2a-3T \times 3a+F \times 2a=0\\ \therefore \;\; F \times 2a &=14 Ta\\ 2F&=14T\\ T&=\frac{F}{7}=\frac{9}{7}kN\\ T_2&=2T=\frac{18}{7}=2.57kN \end{aligned}
Question 2 |
An L-shaped elastic member ABC with slender
arms AB and BC of uniform cross-section is
clamped at end A and connected to a pin at end C.
The pin remains in continuous contact with and is
constrained to move in a smooth horizontal slot.
The section modulus of the member is same in both
the arms. The end C is subjected to a horizontal
force P and all the deflections are in the plane of the
figure. Given the length AB is 4a and length BC is
a, the magnitude and direction of the normal force
on the pin from the slot, respectively, are
3P/8, and downwards | |
5P/8, and upwards | |
P/4, and downwards | |
3P/4, and upwards |
Question 2 Explanation:
No vertical deflection allowed
\delta _{VC}=\delta _{VB}=0 \text{ (Vertical deflection) }
\begin{aligned} \delta _B&=\frac{ML^2}{2EI}-\frac{NL^2}{3EI}=0\\ \frac{M}{2}&\frac{NL}{3}\\ N&=\frac{3M}{2L}\;\;(\because L=4a)\\ N&=\frac{3Pa}{2 \times 4 a}\\ N&=\frac{3}{8}P \text{ (downward)} \end{aligned}
Question 3 |
A uniform light slender beam AB of section modulus
EI is pinned by a frictionless joint A to the ground
and supported by a light inextensible cable CB to
hang a weight W as shown. If the maximum value
of W to avoid buckling of the beam AB is obtained
as \beta \pi ^2 EI, where \pi is the ratio of circumference to
diameter of a circle, then the value of \beta is
0.0924\; m^{-2} | |
0.0713\; m^{-2} | |
0.1261\; m^{-2} | |
0.1417\; m^{-2} |
Question 3 Explanation:
Draw FBD of AB
\Sigma M_A=0
W \times 2.5=T \sin 30^{\circ} \times 2.5
T=2W
Compressive load acting on AB =T \cos 30^{\circ}=2W \times \frac{\sqrt{3}}{2}=\sqrt{3}W
Buckling happens when \sqrt{3}W=P_{cr}=\frac{\pi ^2 EI}{L_e^2}
\sqrt{3}W=\frac{\pi ^2 EI}{L^2} \;\;(\because L_e=L \text{as both ends hinged})
W=\frac{1 \times \pi ^2 EI}{\sqrt{3} \times (2.5)^2}=0.0924 \pi^2 EI
W0.0924 \pi^2 EI=\beta \pi ^2 EI
\beta =0.0924 m^{-2}
\Sigma M_A=0
W \times 2.5=T \sin 30^{\circ} \times 2.5
T=2W
Compressive load acting on AB =T \cos 30^{\circ}=2W \times \frac{\sqrt{3}}{2}=\sqrt{3}W
Buckling happens when \sqrt{3}W=P_{cr}=\frac{\pi ^2 EI}{L_e^2}
\sqrt{3}W=\frac{\pi ^2 EI}{L^2} \;\;(\because L_e=L \text{as both ends hinged})
W=\frac{1 \times \pi ^2 EI}{\sqrt{3} \times (2.5)^2}=0.0924 \pi^2 EI
W0.0924 \pi^2 EI=\beta \pi ^2 EI
\beta =0.0924 m^{-2}
Question 4 |
A column with one end fixed and one end free has a critical buckling load of 100 N. For the same column, if the free end is replaced with a pinned end then the critical buckling load will be ________N (round off to the nearest integer).
800 | |
450 | |
1020 | |
680 |
Question 4 Explanation:
For a given material in 0.1 and length,
\mathrm{Pe} \propto \frac{1}{\alpha^{2}}
where, \quad \alpha= length fixity coefficient
Pe= Buckling or critical load.
\begin{aligned} \frac{(P e)_{I I}}{(P e)_{I}} &=\left(\frac{\alpha_{I}}{\alpha_{I I}}\right)^{2}=\left(\frac{2 L}{1 / \sqrt{2}}\right)^{2}=8 \\ \left(P_{e}\right)_{I I} &=8\left(P_{e}\right)_{I}=800 \mathrm{~N} \end{aligned}
Question 5 |
A right solid circular cone standing on its base on a horizontal surface is of height H and base radius R. The cone is made of a material with specific weight w and elastic modulus E. The vertical deflection at the mid-height of the cone due to self-weight is given by
\frac{wH^2}{8E} | |
\frac{wH^2}{6E} | |
\frac{wRH}{8E} | |
\frac{wRH}{6E} |
Question 5 Explanation:
\begin{aligned} P_{\mathrm{x}-\mathrm{x}}&=\frac{(-) \mathrm{W} A_{x-x}(x)}{3}\\ \text{Contraction of small strip }&=(\delta l)_{\text {strip }}\\ &=\frac{P_{x-x} d x}{(A E)_{x-x}} \\ &=\frac{w\left(A_{x-x}\right)(x) d x}{3\left(A_{x-x}\right) E}=\frac{w x}{3 E} d x \end{aligned}
Contraction of conical bar at mid height ( i.e. x=\frac{H}{2} )
\begin{aligned} &=\int_{H / 2}^{H}(\delta l)_{\operatorname{stn} p} \\ &=\frac{W}{3 E} \int_{H / 2}^{H} x d x \\ &=\frac{w}{6 E}\left[H^{2}-\frac{H^{2}}{4}\right]=\frac{w H^{2}}{8 E} \end{aligned}
There are 5 questions to complete.