### 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.

### 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.

### 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.

### Solved Example:

#### 48-1-03

Buckling of a column occurs under:

### Solved Example:

#### 48-1-04

The slenderness ratio is the ratio of:

### Solved Example:

#### 48-1-05

Pure Buckling occurs in a:

### Solved Example:

#### 48-1-06

Pure Buckling uses the equation of:

### Solved Example:

#### 48-1-07

The buckling load for a given material depends on:

### 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) 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}