Learning Objectives:

• To understand stability and buckling phenomena for a slender member under an axial compressive force.

• Develop an appreciation of the phenomenon of buckling and the various types of structure instabilities.

• Understand the use of buckling formulas in the analysis and design of structure.

• Distinguish between long column and short column.

• Understand buckling phenomenon and its effect on load carrying capacity.

• Understand Rankine’s formula.

• Solve simple problems and find load carrying capacity of columns.

Buckling:

Buckling can be defined as the sudden large deformation of structure due to a slight increase of an existing load under which the structure had exhibited little, if any, deformation before the load was increased.

Case I: Concentrically loaded long Columns:

Euler’s Formula:

\[P_{cr} = \frac{\pi ^{2}EI}{(Kl)^{2}}\] where,
\(\large P_{cr}\) = critical axial loading
l = unbraced column length
K = effective length factor to account for end supports
\(\dfrac{l}{r}\) = slenderness ratio for the column

Solved Example: 48-1-01

Buckling of column means:

A. Lateral deflection

B. Axial deflection

C. Torsional deflection

D. None of the listed

Buckling is characterised by lateral deflection but it is different from lateral deflection as there is sudden lateral deflection in buckling unlike lateral deflection where there is gradual deflection.

Correct Answer: D

Solved Example: 48-1-02

Short column and long column are classified on the basis of:

A. Slenderness ratio

B. Diameter

C. Length

D. None of the listed

Slenderness ratio takes into consideration length and radius of gyration and thus is preferred.

Correct Answer: A

Solved Example: 48-1-03

Buckling of a column occurs under:

A. Axial load

B. Transverse load

C. Direct load

D. None of the listed

Correct Answer: A

Solved Example: 48-1-04

The slenderness ratio is the ratio of:

A. Area of column to least radius of gyration

B. Length of column to least radius of gyration

C. Least radius of gyration to area of column

D. Least radius of gyration to length of column

Correct Answer: B

Solved Example: 48-1-05

Pure Buckling occurs in a:

A. Short column

B. Medium Column

C. Long column

D. None of the listed

Correct Answer: C

Solved Example: 48-1-06

Pure Buckling uses the equation of:

A. Rankin-Gordon

B. Euler

C. Stiffness

D. None of the listed

Correct Answer: B

Solved Example: 48-1-07

The buckling load for a given material depends on:

A. Slenderness ratio and area of cross-section

B. Poisson’s ratio and modulus of elasticity

C. Slenderness ratio and modulus of elasticity

D. Slenderness ratio, area of cross-section and modulus of elasticity

Correct Answer: D

Solved Example: 48-1-08

An elevated cylindrical water storage tank is shown in the figure. The tank has inner diameter of 1.5m. It is supported on a solid steel circular column of diamter 75 mm and total height (L) of 4 m. Take water density = 1000 $kg/m^3$ and acceleration due to gravity = $10 m/s^2$. If elastic modulus (E) of steel is 200 GPa, ignoring self-weight of the tank, for the supporting steel column to remain unbuckled, the maximum depth (h) of the water permissible (in m, round off to one decimal place) is: (GATE Civil 2021)

A. 2.1

B. 2.4

C. 2.7

D. 3.0

Given: \[L = 4 m, \rho = 1000 kg/m^3, g = 10 m/s^2, \\ E = 200 GPa, D_i = 1.5 m, d = 75 mm\] \[\begin{aligned} \dfrac{\pi^2 EI}{L_e^2} &= \dfrac{\pi}{4} D_i^2 \times h \rho g\\ \dfrac{\pi^2 E\times \dfrac{\pi}{64}d^4}{(2L)^2} &= \dfrac{\pi}{4} D_i^2 \times h \rho g\\ h &= \dfrac{4 \pi^2Ed^4}{4 \times 64L^2 \times D_i^2\times \rho g}\\ h &= \dfrac{4 \pi^2 \times 200 \times 10^9 \times (0.075)^4}{4 \times 64 \times \times 4^2 \times 1.5^2 \times 10^4}\\ h &= 2.7\ m\end{aligned}\]

Correct Answer: C

Learning Objectives:

• Understand concept of equivalent length and for columns with different end conditions.

When column has both ends hinged, its entire length gets buckled to form a bow. Therefore equivalent length of a column with both ends hinged is ‘L’. It is written as \(L_e\).
For other end conditions, equivalent lengths forming a bow are:

• Both ends hinged \(L_e\)= L,

• One end fixed and other free \(L_e\)= 2L

• One end fixed other hinged \(L_e\)= \(\dfrac{L}{\sqrt{2}}\)

• Both ends fixed \(L_e\)= \(\dfrac{L}{2}\)

Solved Example: 48-2-01

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. 4

B. 8

C. 1

D. 2

Correct Answer: A

Solved Example: 48-2-02

A vertical column has two moments of inertia (i.e. $I_{xx}$ and $I_{yy}$ ). The column will tend to buckle in the direction of the:

A. Axis of load

B. Perpendicular to the axis of load

C. Maximum moment of inertia

D. Minimum moment of inertia

Correct Answer: D

Solved Example: 48-2-03

A long column with fixed ends can carry load as compared to both ends hinged:

A. 4 times

B. 8 times

C. 16 times

D. None

Correct Answer: A

Solved Example: 48-2-04

Two steel columns P (length L and yield strength $f_y$ = 250 MPa) and Q (length 2L and yield strength $f_y$ = 500 MPa) have the same cross-sections and end-conditions. The ratio of buckling load of column P to that of column Q is:

A. 0.5

B. 1.0

C. 2.0

D. 4.0

\[P=\dfrac {\pi ^{2}EI}{\left( l_{eff}\right) ^{2}}\] \[\dfrac {P_{p}}{P_{q}}=\dfrac {\dfrac {\pi ^{2}EI}{L^{2}}}{\dfrac {\pi ^{2}EI}{\left( 2L\right) ^{2}}}=4\]

Correct Answer: D

Solved Example: 48-2-05

Find the Euler critical axial load for a hollow cylindrical cast iron column 200mm external diameter and 25 mm thick, if it is 6 m long and hinged at both ends. Take E = 1.2 $\times$ 10$^6$ N/mm$^2$

A. 13.31 $\times$ 10$^6$ N

B. 17.66 $\times$ 10$^6$ N

C. 19.40 $\times$ 10$^6$ N

D. 21.94 $\times$ 10$^6$ N

For columns hinged at both ends K = 1.0 \begin{align*} P_{cr}&= \dfrac{\pi^2EI}{(Kl)^2}\\ &= \dfrac{\pi^2 (1.2 \times 10^6 \times 10^6) (\dfrac{\pi}{64} \times (200^4 - 150^4) \times 10^{-12})}{6^2}\\ &=17.66 \times 10^6\ N \end{align*}

Correct Answer: B