## Find Safe Compressive Load for a Column Using Euler’s Formula

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 106 KN/m2

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 106 KN/m2

210 x 106  /  100 x 100     = 21000 KN/cm2

Required

Safe compressive load  = psafe  =  ?

Solution

For one end fixed and other pinned equivalent length  is:

L =  L/    =  300/       =  212.12 cm

Least moment of inertia  = I  =  Iyy  =  db3/12    =  20 x 123/12

Crippling  load  =  p  =  π2 E I/l2

π2  x 21000 x 2880 /  (212.13)2    =   597,068,478.7/(212.13)2

p= 13268.14 KN

safe load  = psafe =  crippling load /factor of safety

13268.14 / 3 =  4422.71 KN

Also Read: Equation How to Find Deflection in cantilever Beam

### Engr Sami UllahAuthor

Founder & Admin of civilengineeringinfo.com I am Civil Engineer working as a Site Engineer. With a good subject knowledge in civil Engineering. And I have started the blog to share my valuable information with civil engineer students. Civil Global is the Civil Engineering learning Home around all over the world.