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
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 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
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 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: 
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 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.

Leave a Comment

Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.