In this Equation I will find the Safe Compressive Load for a Column using Euler’s formula

**Example no 1**

A start 3 cm long is rectangle section 12 x 20 cm size. Its one end is fixed and other pinned. Fine the safer compperisve load using Euler’s formula allowing a factor of safety of 3.

Take = E = 210 x 10^{6} KN/m^{2}

**Given data**

Actual length of strut = L = 3m = 3 x 100 = 300 cm

Breadth of strut = b = 12 cm

Depth of strut = d = 20 cm

Factor of safety = 3

Modl of elasticity = 210 x 10^{6} KN/m^{2}

210 x 10^{6} / 100 x 100 = 21000 KN/cm^{2}

**Required**

Safe compressive load = p_{safe } = ?

**Solution**** **

For one end fixed and other pinned equivalent length is:

L = L/ = 300/ = 212.12 cm

Least moment of inertia = I = Iyy = db^{3}/12 = 20 x 12^{3}/12

Crippling load = p = π^{2} E I/l^{2}

π^{2 }x 21000 x 2880 / (212.13)^{2} = 597,068,478.7/(212.13)^{2}

p= 13268.14 KN

**safe load ** = p_{safe} = crippling load /factor of safety

13268.14 / 3 = **4422.71 KN**

**Also Read: Equation How to Find Deflection in cantilever Beam**

## Comments

## mudasar

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