Equivalent Length

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

Equivalent Lengths

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 Examples

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:

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:

Correct Answer: D

Solved Example:

48-2-03

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

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:

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

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$^2$

Solution:
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