### Equivalent Length

• 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:

### 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:

### Solved Example:

#### 48-2-03

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

### 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:

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

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$^2$