Euler’s Theory of Column

 Question 1
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 1 Explanation:

For a given material in 0.1 and length,
$\mathrm{Pe} \propto \frac{1}{\alpha^{2}}$
where, $\quad \alpha=$ length fixity coefficient
\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 2
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 2 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 3
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 3 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 4
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 4 Explanation:
Condition for buckling,
(Reaction offered by the fixed support)$\gt$ Buckling
\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 5
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 5 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 6
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 6 Explanation:
$P_{e}=\frac{n \pi^{2} E I_{\min }}{L^{2}}$
n=4 for both end clamped (fixed)
n=1 for both end hinged
$\therefore$ Ratio =4
 Question 7
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 7 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}
 Question 8
The rod PQ of length L with flexural rigidity EI is hinged at both ends. For what minimum force F is it expected to buckle?
 A $\frac{\pi^2EI}{L^2}$ B $\frac{\sqrt{2}\pi^2EI}{L^2}$ C $\frac{\pi^2EI}{\sqrt{2}L^2}$ D $\frac{\pi^2EI}{2L^2}$
GATE ME 2008   Strength of Materials
Question 8 Explanation:

Let the axial comp-load acting on the rod is R
\begin{aligned} R_{H} &=R \cos 45^{\circ} \\ R_{V} &=R \sin 45^{\circ} \\ \text{Hence,}\quad R &=\frac{F}{\cos 45^{\circ}}=\sqrt{2} F \end{aligned}
Condition for buckling, $R>P_{\theta}$
\begin{aligned} \sqrt{2} F &>\frac{\pi^{2} E I_{\min }}{L_{e}^{2}} \quad \text { where } L_{e}=\alpha L \\ F &>\frac{\pi^{2} E I_{\min }}{\sqrt{2} \alpha^{2} L^{2}}\\ \alpha &=1 \quad(\because \text{both ends hinged})\\ \text{Hence }\;F_{\min } &=\frac{\pi^{2} E I_{\min }}{\sqrt{2} L^{2}} \end{aligned}
 Question 9
A pin-ended column of length L, modulus of elasticity E and second moment of the cross-sectional area I is loaded centrically by a compressive load P. the critical buckling load $(P_{cr})$ is given by
 A $P_{cr}= \frac{EI}{\pi ^{2}L^{2}}$ B $P_{cr}= \frac{\pi ^{2} EI}{3L^{2}}$ C $P_{cr}= \frac{\pi EI}{L^{2}}$ D $P_{cr}= \frac{\pi^{2} EI}{L^{2}}$
GATE ME 2006   Strength of Materials
Question 9 Explanation:
$P_{E}=\frac{n \pi^{2} E I}{L^{2}}$
For pin-ended column
n=1
Here n= End fixity coefficient
E = Modulus of elasticity
I = Second moment of area
L = Actual length of column
There are 9 questions to complete.