Design of Axial Compression Member Video Lecture - Civil Engineering (CE)

we have already covered the design of
members experiencing axial tension
loading now it's time to learn how to
design members experiencing axial
compression loading and for axial
compression loading things are a little
bit more complicated
please note we'll limit ourselves to
materials where the yield stress in
tension can be assumed to be equal to
the yield stress in compression which is
the case for steel now as well as yield
failure we must also consider buckling
failure because a member experiencing
axial compression loading has one of two
possible failure modes
it will either buckle if it's long
enough or it will yield and the greatest
load of member can carry is of course
equal to its yield stress times its
cross-sectional area and we commonly
refer to this load as the squash load
you've previously seen this video
showing the shape of the buckling curve
as a function of the effective length of
the member but what you now need to know
is when the the length of the member
reduces the buckling load actually tends
to infinity and this is not possible the
upper limit as I previously mentioned is
equal to the squash load these two
relationships has drawn combined to form
a failure envelope shown here in blue
and any member with a load and length
that plots below the curve is deemed
safe will not fail and any member with a
point that plots up above a failure
envelope here would be unsafe let's now
do an example the question posed here is
is member a be here safe to carry the
loading shown the points ABC are all
pins and that is telling us that if we
are required to do a buckling check our
effective length is equal to our true
length and it's also telling us we can
do a method of joints analysis to find
the forcing member a B which is of
course what we first require this
question is a typical exam type question
so pay close attention before we do
start on
I'd like to point out this type of
arrangement of members can be found in
cranes you can see here this is a
similar triangle system and also in
structures up here okay so what we've
been asked to do here is to carry out an
ultimate limit State design that is a
design against failure we'll assume this
load of 150 kilonewtons is already an
ultimate limit state load that is a load
that is already been multiplied by the
relevant load factors of 1.2 dead load
and 1.5 live load assuming it comprises
both dead and live load components and
unless you're being told otherwise in
this subject you can always assume the
load is a limit state load already now
in order to solve for member for say B
one would do a method of joints analysis
considering as our free body diagram the
loop has drawn here now redraw in this
system we know all Freebody diagrams of
stable structures are in equilibrium we
draw our unknown forces away from the
joint and as we have two unknown forces
we can use a to translational equations
of equilibrium to solve for those forces
I won't do that calculation here now
I'll leave you to do that in your own
time when you do soft with a forcing
member a B you should get a value of 86
point six kilonewton that force will be
negative which indicates the member is
in compression and because it is a
compressive force we need to check
member a B will not fail by either
yielding or Buckley so let's now check
our member against a yield failure first
and in this check we have to insure our
squash load is greater than our member
load so our squash load equal to our
yield stress times our cross sectional
area is greater than our load of eighty
six point six kilonewtons we've been
told our yield stress is three hundred
mega Pascal's now as our section has
dimensions
of 100 by 100 by 3 mil 3 mil indicating
the walls of the section of 3 mil thick
so that's now inner dimension in here
would be 94 mil so that's when we
calculate the cross-sectional area here
we would calculate an solid box outer
area and take away and we never area now
I'm currently working in mega Pascal's
and millimeters so I'll just convert
this figure here to Newtons which is the
corresponding force and doing this
calculation we get a value for our
squash load of 349 Newton's which is
greater than our applied load of 68,000
Newtons and thus our member will not
fail by yielding moving on now to do our
buckling check to ensure buckling
failure does not occur our Euler
buckling load has to be greater than our
member load I'm in belowed being our 86
and once again our working Newton's so
80 6.6 by 10 to the 3 Newtons so
plugging into our formula now we have
been given the young's modulus above
it's equal to 200,000 so that's telling
us the memories made out of steel and we
now need to calculate our second moment
of area so going back to a previous
slide and you can see here it's equal to
the outer dimensions of the cross
sectional area to the power of 4 on 12
take away the inner dimensions so coming
back here out over you will be out power
to dimensions of 100 to the power 4
minus our inner dimensions which we said
we're 94 millimeters
now we are working here in mega Pascal's
and millimeters and so that's when we
come to our length but we need to also
present that in terms of millimeters
because I'm to ends append our effective
length is just equal to our true length
that value has to be greater than now
eighty six point six kilonewtons and
plugging in we get a value here of
56,000 which is not greater than 86,000
so that's what we're saying here is our
member will fail you're not safe to
carry the load relating this now to the
failure curve we found our squashed load
was equal to 349 kilonewtons and at a
length of eight meters our buckle load
was equal to fifty six point three
kilonewtons and our actual point plotted
up here at eight meters it had a value
of 86 point six kilonewtons and thus it
did not lie under the curve which
resulted in failure with a member that
brings us to the end of our videos an
axial member design I hope you've
learned Lots and I will talk to you
again soon