Question

Euler’s buckling formula can be expressed as Pcr = π2EI(KL)2 where Pcr is the critical buckling load, E is the column’s Young’s modulus, I is the column’s moment of inertia, and L is the column’s length. Derived using a quantity called effective length, the constant K depends upon the column’s end conditions.

This problem will compare various end conditions of a slender column under compression. The studied column has a length of L = 2 meters, and its square cross-section has a side length of b = 7 centimeters. The material is a grade of steel with E = 200 GPa and σy = 500 MPa.What is the column’s critical buckling load in meganewtons (MN)? (You must provide an answer before moving to the next part.)The column's critical buckling load is MN.

Answer 1

Both end hinge-Pcr= 0.98 MN

Both end fix-Pcr=3.9 MN

Step-by-step explanation:

E= 200 GPa

L=2 m

b=7 cm =70 mm

The critical load given as

`P_(cr)=(\pi^2EI)/(L^2)`

For square section

`I=(b^4)/(12)`

`P_(cr)=(\pi^2E* (b^4)/(12))/(L^2)`

Lets take column is hinge at the both ends :

Now by putting all the values

`P_(cr)=(\pi^2E* (b^4)/(12))/(L'^2)`

L'= L

`P_(cr)=(\pi^2* 200* 1000* (70^4)/(12))/(2000^2)`

Pcr=987371.6 N

Pcr= 0.98 MN

Therefore critical load = 0.98 MN

When both end fixed :

`P_(cr)=(\pi^2E* (b^4)/(12))/(L'^2)`

L' = 0.5 L

`P_(cr)=(\pi^2* 200* 1000* (70^4)/(12))/(1000^2)`

Pcr=3.9 MN

Therefore critical load = 3.9 MN