Euler’s Theory of Column

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).

A
2.57
B
3.65
C
1.45
D
4.65
GATE ME 2022 SET-2   Strength of Materials
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}
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

A
3P/8, and downwards
B
5P/8, and upwards
C
P/4, and downwards
D
3P/4, and upwards
GATE ME 2022 SET-1   Strength of Materials
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

A
0.0924\; m^{-2}
B
0.0713\; m^{-2}
C
0.1261\; m^{-2}
D
0.1417\; m^{-2}
GATE ME 2022 SET-1   Strength of Materials
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}
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).
A
800
B
450
C
1020
D
680
GATE ME 2021 SET-2   Strength of Materials
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
A
\frac{wH^2}{8E}
B
\frac{wH^2}{6E}
C
\frac{wRH}{8E}
D
\frac{wRH}{6E}
GATE ME 2021 SET-1   Strength of Materials
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}
Question 6
The truss shown in the figure has four members of length l and flexural rigidity EI, and one member of length l\sqrt{2} and flexural rigidity 4EI. The truss is loaded by a pair of forces of magnitude P, as shown in the figure.

The smallest value of P, at which any of the truss members will buckle is
A
\frac{\sqrt{2}\pi ^2 EI}{l^2}
B
\frac{\pi ^2 EI}{l^2}
C
\frac{2\pi ^2 EI}{l^2}
D
\frac{\pi ^2 EI}{2l^2}
GATE ME 2020 SET-1   Strength of Materials
Question 6 Explanation: 


Since buckling will happen due to compression, and it occurs in diagonal only
R_{BD}\geq \frac{\pi ^2 EI}{L^2}
P\geq \frac{\pi ^2 (4EI)}{(\sqrt{2}L)^2}
P\geq \frac{2\pi ^2 EI}{L^2}
Question 7
An initially stress-free massless elastic beam of length L and circular cross-section with diameter d (d \lt \! \lt l) is held fixed between two walls as shown. The beam material has Young's modulus E and coefficient of thermal expansion \alpha .

If the beam is slowly and uniformly heated, the temperature rise required to cause the beam to buckle is proportional to
A
d
B
d^{2}
C
d^{3}
D
d^{4}
GATE ME 2017 SET-1   Strength of Materials
Question 7 Explanation: 
Condition for buckling,
(Reaction offered by the fixed support)\gt Buckling
load
\begin{aligned} i.e. R&>P e\\ \sigma_{\mathrm{Th}} A>& \frac{\pi^{2} E I_{\mathrm{min}}}{L_{e}^{2}} \\ \alpha(\Delta T) E A>& \frac{\pi^{2} E I_{\min }}{\left(L_{\theta}\right)^{2}}\\ \text{for cicular cross-section}\\ \alpha(\Delta T) E \frac{\pi}{4} d^{2} &>\frac{\pi^{2} E}{\left(L_{e}\right)^{2}} \frac{\pi}{64} d^{4} \\ \Delta T &>\frac{\pi^{2} d^{2}}{16 \alpha\left(L_{\theta}\right)^{2}} \\ \text{Hence,}\quad \Delta T & \propto d^{2} \end{aligned}
Question 8
Consider a steel (Young's modulus E = 200 GPa) column hinged on both sides. Its height is 1.0 m and cross-section is 10 mm x 20 mm. The lowest Euler critical buckling load (in N) is __________
A
3289.868N
B
1254.3256N
C
1289.6985N
D
7845.9685N
GATE ME 2015 SET-1   Strength of Materials
Question 8 Explanation: 
\begin{aligned} F &=\frac{\pi^{2} E I}{L_{e}^{2}}=\frac{\pi^{2} \times 200 \times 10^{3} \times 20 \times 10^{3}}{12 \times(1000)^{3}} \\ &=3289.868 \mathrm{N} \end{aligned}
Question 9
For a long slender column of uniform cross section, the ratio of critical buckling load for the case with both ends clamped to the case with both ends hinged is
A
1
B
2
C
4
D
8
GATE ME 2012   Strength of Materials
Question 9 Explanation: 
Euler's buckling load,
P_{e}=\frac{n \pi^{2} E I_{\min }}{L^{2}}
n=4 for both end clamped (fixed)
n=1 for both end hinged
\therefore Ratio =4
Question 10
A column has a rectangular cross-section of 10mm x 20mm and a length of 1m. The slenderness ratio of the column is close to
A
200
B
346
C
477
D
1000
GATE ME 2011   Strength of Materials
Question 10 Explanation: 
Assuming both ends of column to be hinged. Given:
\begin{aligned} \qquad L_{\mathrm{eq}}&=1 \mathrm{m} \\ \text { Slenderness ratio }&=\frac{\mathrm{L}_{\mathrm{eq}}}{r_{\min }} \\ r_{\min }= \sqrt{\frac{I_{\min }}{A}}&=\frac{\sqrt{\frac{1}{12} \times(10)^{3} \times 20 \times 10^{-12}} }{10 \times 20 \times 10^{-6}}\\ &=2.88 \times 10^{-3}\\ \text{Slenderness }&=\frac{1}{2.88 \times 10^{-3}}=347.22 \approx 346 \end{aligned}
There are 10 questions to complete.

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