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).
800 | |
450 | |
1020 | |
680 |
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
\frac{wH^2}{8E} | |
\frac{wH^2}{6E} | |
\frac{wRH}{8E} | |
\frac{wRH}{6E} |
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
The smallest value of P, at which any of the truss members will buckle is
\frac{\sqrt{2}\pi ^2 EI}{l^2} | |
\frac{\pi ^2 EI}{l^2} | |
\frac{2\pi ^2 EI}{l^2} | |
\frac{\pi ^2 EI}{2l^2} |
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
If the beam is slowly and uniformly heated, the temperature rise required to cause the beam to buckle is proportional to
d | |
d^{2} | |
d^{3} | |
d^{4} |
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}
(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 __________
3289.868N | |
1254.3256N | |
1289.6985N | |
7845.9685N |
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
1 | |
2 | |
4 | |
8 |
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
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
200 | |
346 | |
477 | |
1000 |
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}
\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?
\frac{\pi^2EI}{L^2} | |
\frac{\sqrt{2}\pi^2EI}{L^2} | |
\frac{\pi^2EI}{\sqrt{2}L^2} | |
\frac{\pi^2EI}{2L^2} |
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
P_{cr}= \frac{EI}{\pi ^{2}L^{2}} | |
P_{cr}= \frac{\pi ^{2} EI}{3L^{2}} | |
P_{cr}= \frac{\pi EI}{L^{2}} | |
P_{cr}= \frac{\pi^{2} EI}{L^{2}} |
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
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.