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

Answer:

Both end hinge-Pcr= 0.98 MN

Both end fix-Pcr=3.9 MN

Explanation:

E= 200 GPa

L=2 m

b=7 cm =70 mm

The critical load given as

$P_{cr}=\dfrac{\pi^2EI}{L^2}$

For square section

$I=\dfrac{b^4}{12}$

$P_{cr}=\dfrac{\pi^2E\times \dfrac{b^4}{12}}{L^2}$

Lets take column is hinge at the both ends :

Now by putting all the values

$P_{cr}=\dfrac{\pi^2E\times \dfrac{b^4}{12}}{L'^2}$

L'= L

$P_{cr}=\dfrac{\pi^2\times 200\times 1000\times \dfrac{70^4}{12}}{2000^2}$

Pcr=987371.6 N

Pcr= 0.98 MN

Therefore critical load = 0.98 MN

When both end fixed :

$P_{cr}=\dfrac{\pi^2E\times \dfrac{b^4}{12}}{L'^2}$

L' = 0.5 L

$P_{cr}=\dfrac{\pi^2\times 200\times 1000\times \dfrac{70^4}{12}}{1000^2}$

Pcr=3.9 MN

Therefore critical load = 3.9 MN