As regards to the co-efficient of Friction calculation, between the Mass and surface. Isn't the formula F=umg, where u=Co-eff of friction, m=Mass and g=gravitational effect?
So with F=umg, then u=F/mg
Giving u=80/6x9.81 = 1.359 as the frictional co-efficient?
Hi Adam
Looks like you have arrived at the correct ESR figure (118.74). Basically the minimum E.S.R. for which buckling occurs, is the one when the critical stress σc equal to the yield stress of the material σy. For a mild steel with E = 200 GPa and σc = σy = 240 MPa, the slenderness ratio is...
You are on the right track by using Eulers equation and the question is asking what load would the failure occur?
Remember Eulers Equation isn't just Fc=σ*A it is also pi^2*EI / Le^2.
Also remember that the column is fixed at both ends
Thanks Steamking.
Looks like I missed the column design notes!
So Le=L/2
L=Le x 2
L=2.97m x 2 = 5.94m, which is the minimum length of the column at which buckling is likely to occur.
Hi All
I have just looked at this problem and just wanted to confirm my answer, which is different to everyone else!
The question stated the diameters of 80mm and 60mm with E = 200GNm^2 and σ = 140MNm^2
To calculate the minimum length of the column at which buckling is likely to occur
ESR =...
Hi Steamking
Thanks for the reply. I have had a look through my notes and realize that the formula for calculating the stress uses the second moment of area with the neutral axis, so Ixx=bd^3/12.
Again thanks for the confirmation
Hi
Am new to the forum, but as regards to the original question of the second moment of area I.
Should this be the second moment of area for a rectangle Ixx = bd^3/3?
The formula being used is for second moments of area about an axis passing through the centroid, Ixx=bd^3/12?