Euler’s Crippling Load Formula and Example

Euler’s crippling load formula is used to find the buckling load of long columns. The load obtained from this formula is the ultimate load that column can take. In order to find the safe load, divide ultimate load with  the factor of safety (F.O.S)

Euler’s Formula

Mathematically, Euler’s formula can be expressed as;

$$\large P\space =\space \frac{\Pi ^{2}EI}{l^{2}}$$

Where,

E = Modulus of Elasticity of material,

I = Moment of Interia of column section

l = Equivalent/ Effective length of the column

Euler’s Formula Example

Statement

Second story of a residential building has a column, whose diameter is 10 cm and 3 meter long. Both ends of this column are hinged. Find the buckling load that a column can carry

Take Elastic Modulus E = 2x1o^6 Kg /centimeter square

Given Data

Length of column = L = 3 m = 300 cm

Diameter of column = d = 10 cm

Modulus of Elasticity = E = 2×10^6 kg/centimeter square

Required

Buckling Load = P = ?

Solution

In order to find the buckling load, fist find the moment of inertia of that column.

$$\large Moment\space\ of\space\ Inertia\space =\space \frac{\pi d^{4}}{64}$$

Hence,

$$I\space\ =\frac{(3.142)(10)^{4}}{64}$$

Therefore,

$$I\space\ =490.94\space\ cm^4$$

As both ends of column are hinged,

Length of column = L = 300 cm

$$\small Buckling\space\ load =\space\frac{\pi^{2}EI}{l^{2}}=\space\ \frac{(3.142)^2(2\times10^6)(490.94)}{(300)(300)}$$