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,
P = Buckling Load,
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
Now Buckling load,
\(\small Buckling\space\ load =\space\frac{\pi^{2}EI}{l^{2}}=\space\ \frac{(3.142)^2(2\times10^6)(490.94)}{(300)(300)}\)Buckling load = P = 107702.56 Kg
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