Column Analysis

  • 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 Examples

Solved Example:

48-1-01

Buckling of column means:

Solution:
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:

Solution:
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:

Correct Answer: A

Solved Example:

48-1-04

The slenderness ratio is the ratio of:

Correct Answer: B

Solved Example:

48-1-05

Pure Buckling occurs in a:

Correct Answer: C

Solved Example:

48-1-06

Pure Buckling uses the equation of:

Correct Answer: B

Solved Example:

48-1-07

The buckling load for a given material depends on:

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)

48.1-08



Solution:

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