Calculating Euler Buckling of a Steel Rod: End Fixities & Failure Loads

AI Thread Summary
The discussion focuses on calculating the Euler buckling load for a steel rod with specific dimensions and end conditions. The initial calculation provided yields a buckling load of 259.1 kN using the formula PE = π²EI/L. Participants discuss the significance of the effective length factor (K) based on different end fixity conditions, noting that pinned ends yield K = 1.0, while fixed ends result in K = 0.50. Additionally, the conversation highlights the discrepancies between theoretical Euler loads and actual failure loads, attributing these differences to geometric and material imperfections. The complexities of experimentally determining the buckling load are also acknowledged, emphasizing the challenges in achieving accurate results.
thebest99
Messages
32
Reaction score
0
a steel rod, 40mm in diameter and 1.00m long, is pinned at each end
i) calculate the euler buckling for the rod
ii) identify three other possible end fixity conditions for the rod and demonstrate how euler buckling load would be affected in each case
iii) explain the relation between the Euler Buckling load and the true failure load of a real strut

my attempt at i)

pie squared EI/L

π (40mm) ²

PE= π ² 210 x 10 cubed kN/mm²squared x 1.25 x 10 to the power of 5 mm power 4

= 259.1kN

Can some one check this thank you

And help me with question ii and iii
 
Physics news on Phys.org
thebest99 said:
a steel rod, 40mm in diameter and 1.00m long, is pinned at each end
i) calculate the euler buckling for the rod


pie squared EI/L

π (40mm) ²

Have you checked what the units would be for "pie squared EI/L"?
What is the meaning of "n"?
 
F=pi^2 EI/(KL)^2}

where

F = maximum or critical force (vertical load on column),
E = modulus of elasticity,
I = area moment of inertia,
L = unsupported length of column,
K = column effective length factor, whose value depends on the conditions of end support of the column, as follows.

For both ends pinned (hinged, free to rotate), K = 1.0.
For both ends fixed, K = 0.50.
For one end fixed and the other end pinned, K = 0.699...
For one end fixed and the other end free to move laterally, K = 2.0.

KL is the effective length of the column.

from http://en.wikipedia.org/wiki/Buckling
 
sorry that n is meant to be pie
 
"π (40mm) ²" gives you 4 x the cross-sectional area. is that what you meant?
For part iii, the main reasons for the differences between Euler loads and actual buckling loads is said to be a combination of geometric and material imperfections. But my personal view is that the buckling load is quite difficult to obtain experimentally. My understanding is that if a loaded strut is practically straight, and under load, you perturb it sideways, then buckling load has been reached if it remains in the perturbed position. When I have done experiments on timber struts, that definition never quite satisfied me, as it was hard to determine the accuracy with which it was obtained. The struts were never quite so well-behaved.
 
thank you pongo this is what i worked out
 

Attachments

Back
Top