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 Ullah Author

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.

Comments

    mudasar

    (January 29, 2018 - 5:01 pm)

    Sir please send details

    Sahil khan

    (January 31, 2018 - 3:02 pm)

    Can i get a video about this topic (Find safe compressive strength of column)
    Please tell me where i can get

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