today i'm going to be going over buckling analysis and the number one mistake i see
people doing out there and since the majority of engineering officers nowadays have
access to a structural analysis software i'll quickly demonstrate a column buckling analysis
in autodesk robot and spacegass and we can go through the results together based on AS4100
i'm Gabriel i'm a qualified civil engineer in Queensland specializing in structural engineering
and i make engineering related videos to hopefully help you to be a better engineer. To begin with
let's draw a simple bar in autodesk robot with one end fix and the other end free
next i'll change this bar to a 360 UB by 44.7 section
i'm going to apply a 100 kilonewtons axial force to the top node
and you can switch on the display of loads by clicking on these two buttons at the bottom
some programs may require you to add intermediate nodes
along the length of the member that's why i'm dividing this column in five segments
in the analysis parameters tab i'm going to set up a
buckling analysis with three modes of buckling hit okay calculations
in the results tab you can check the deformation for the three different modes of buckling
mode 1 mode 2 mode 3
and as you can see the first mode will be about the weakest axis of the column which according
to the local axis is the z axis represented by the red arrow the critical loads can be
seen in the results advanced tab activate all the columns but we're only interested
in the critical coefficient and critical force so write down this number 161.97 kilonewtons
and i'll show you what this value represents
let's jump to mathcad you might be familiar with this equation
it is the euler's formula which gives us the maximum axial load on the column just
before it begins to buckle and this is pi squared times the modulus of elasticity for the material
times the moment of inertia divided by effective length square
so let's see if we can get the same results that the software gave us we can check the material
and section properties by right clicking the bar and selecting object properties the weakest axis
in this case is the z axis therefore i'll take i z i z we can also use the one steel table to find
the section properties whichever you prefer okay so the modulus of elasticity i'm going to use is
200 000 mpa moment of inertia 8.1 times 10 to the power of 6 millimeters 4 effective length factor
of 2 and the length of the bar 5 meters plugging the numbers in the equation we get a result
of approximately 160 kilonewtons which is pretty much the critical force calculated by the software
uh just a heads up robot used effective length factor of 2 for fixed free column in fact to
be more precise it's 1.99 so if we change it here to 1.9 we get a 161 kilonewtons and
AS 4100 would suggest ke equal to 2.2 as you can see in this table in the clause 4.6
so i'm going to quickly demonstrate the same procedure in space gas and hopefully get the
same results the procedure is pretty much the same i'm going to draw the column assign supports
assign loads divide the bar into segments run a buckling analysis and generate a text report
so i've got my 360 ub by 44.7 fixed in one end
i'm applying the same 100 kilonewtons point load at the top node
just going to apply a self weight as
well
um let's subdivide this bar into five equal segments
so just go to member twos subdivide and five equally spaced segments
all right so i'm gonna run a static analysis
and i'm going to run a buckling analysis as well
i'm going to choose three modes of buckling
and hit ok
so we get a load factor of 1.582
so if you generate a tax report you can check the critical load which is pretty much 1.582
times 100 okay so let's do this gonna go to next okay
so just scroll down until you find pcr
and we can see here for the first mode pcr is 161 kilonewtons as well
now here is where people make the mistake of assuming the column can take a load of
161 kilonewtons before it buckles however 161 kilonewtons is a reference buckling value and
not the capacity this calculation is for an ideal column therefore you still have to account for
second order effects like geometrical imperfections and residual stresses so in
other words an ideal column is perfectly straight made of homogeneous material and upon which the
load is applied through the centroid of the cross section so what you should do next with the result
i'm using as4100 but whichever code you're using it should be pretty similar go to as4100
page 93 clause 6.3.4 nominal capacity of a member of varying cross section
so in case you're asking yourself if this clause applies to a member of
constant cross-section as well the answer is yes and i'll show you that in a moment
um yeah so cos 6.3.4 yields same results as clause 6.3.3 which is nominal capacity of a member of
constant cross section so let's jump back to mathcad i've got here the calculation
process for the first clause and right next to it the calculation for the second clause
all the information you will need about the 360 ub 44.7 you can find only one steel tables so the
gross area is 5720 millimeters squared the huge stress 300 mpa the form factor is 0.93 and the
radius of gyration 37.6 millimeters and the column is five meters long or five thousand millimeters
for both procedures we have to determine the section capacity Ns um which is 1.596
times 10 cubed um the let's adopt effective length factor k e equal to 2.2 as per clause 4.6.3.2 and
in the first design process we find the modified member slanderness through this formula on
page 19. um i'll show you now and in the second design process the formula is this one on page 93
this one here in which we need to work out first the euler's buckling load as i've done here
so you can see already that the modified member slenderness is quite similar for both equations
hence the member section constant and the member slenderness reduction factor
will also be the same and you will find these values on page 91.
we can find the value of member section constant alpha b for kf less than one
which is zero for hot rolled u b and with the value of lambda and n over b we are able to find
alpha c which is the memory slanderness reduction factor okay so if we scroll
down here is zero and all lambda n is 310 approximately 310 that gives us a 0.077
and finally we plug these numbers in the memory capacity equation
and voila we've got the same results for both both design processes in conclusion our buckling
capacity is actually 110 kilonewtons and not 161 kilonewtons as we got from the software and
if you change ke to 2.0 you'll get a buckling capacity of around 135 kilonewtons but it's still
less than 161. so in the next column buckling video we're going to learn how to use the buckling
analysis to estimate the critical buckling load of a frame it's going to be an exercise
a little bit more complex than this example and i hope you like the video and see you next time