Calculate Euler Buckling Load with Different End Fixity Conditions

[thebest99]

a steel rod, 40mm in diameter and 1.00m long is pinned at each end

calculate the euler bucking load

Answer:

using PE = pie sqaured 210x10cubed kN/mm x 1.25 x 10power 5mm to the 4 divided by 1000mm sqaured

=259.1kN

b) indentify three other possible end fixity conditions for the rod and demonstrate how the Euler buckling load will be effected in each case

i have pinned -fixed, PE=2.04pie sqaured EI divided by L sqaured

fixed -fixed , PE= 4pie sqaured EI dived by L sqaured
and fixed free, PE= quater pie sqaured EI divide by L sqaured

can someone help with question b.

thnak you

 

[PhanthomJay]

What sort of help are you looking for? You've identified 3 other end conditions and noted how it affects the critical buckling load.

 

[thebest99]

believe i have to do the calculations too for the equations

 

[PhanthomJay]

Plug and chug.

 

[DeeNos]

Answer:

In the case of pinned-fixed, the Euler buckling load will be slightly lower than the pinned-pinned case because one end is fixed, providing some stability to the rod. This can be calculated using the formula PE=2.04pie squared EI divided by L squared, where E is the modulus of elasticity, I is the moment of inertia, and L is the length of the rod.

In the case of fixed-fixed, the Euler buckling load will be significantly higher than the pinned-pinned case because both ends are fixed, providing maximum stability to the rod. This can be calculated using the formula PE=4pie squared EI divided by L squared.

In the case of fixed-free, the Euler buckling load will be the highest among all the end fixity conditions because one end is free to move, allowing for maximum deflection and instability. This can be calculated using the formula PE=1/4 pie squared EI divided by L squared.