steel beam design if you want to be a structural engineer you cannot run from it it's everywhere i
see dead beams in today's video we're gonna design a steel beam together and we're gonna quickly look
at lateral restraints critical flange and a couple of more things that it's important for you to know
when designing a steel beam having said that let's get started a simply supported steel floor beam
spans six meters and supports four meter long joists and a three meters high timber framed wall
select a suitable steel PFC which is a parallel flange channel for the loading
below so we've got a timber floor dead load of 0.5 kPa a timber floor live load of 1.5 kPa
and a timber wall dead load of 0.2 kpa we don't have any roof loads or wind loads in this exercise
so the wall is 3 meters high the joist is four meters long and the main beam we're going to
design is six meters long this is what a PFC looks like in case you don't know
and the beam is simply supported and we need to find the loads in kilonewtons per meter
so 0.5 kpa for wall 0.2 kpa for for floor and 0.2 for wall and 0.5 times half of the joists span
and for the wall is going to be 0.2 times the wall height so we've got a total of 1.6
kilonewtons per meter the live load is 1.5 times half of the joist span
you will have to work out the worst case combinations for your scenario in my case
i will use 1.2 g plus 1.5 q for ultimate limit state and g plus 0.5 0.7 q for the serviceability
limit state so uls is 6.42 kilonewtons per meters which will be assessed for strength and sls
is 3.7 kilonewtons per meter which will be assessed for deflection
so just emphasizing this you will have to use one load combination to work out
the bending moment capacity and the other to work out the maximum deflection of the beam
to calculate the design action we simply use the formula for bending moment which is
moment equals to load times the length squared divided by eight
we've got 28.89 kilo newtons meter and the deflection limit will be the span divided by 250
you can also check live load divided by 360 and if the wall is made of a brittle material
we can do span divided by 500 so since 24 millimeters is a little too much we will limit
the beam to deflect 12 millimeters and this number can vary according to the designer assessment
we will also check the beam for dynamic effects we don't want the floor to bounce when people walk on
it the way we check this is by applying a force of one in kilonewtons in the mid span of the beam
and limiting this deflection to one millimeter
in order to find the moment capacity of the beam
we will consider it as full lateral restraint by the joists and by doing
so we can utilize the design section moment capacity 5 msx found on this on this red book
and i will explain a little bit more about lateral restraint in the end of the video
so the 150 pfc has a section capacitor of 37 kilo newtons meters which is way over then what we need
let's have a look at deflection now because most of the times deflection will govern the design of
a beam so by using this formula for deflection of a simply supported beam and substituting
the delta for 12 millimeters which is the maximum deflection we get a second moment of area required
for the beam that we need that's 26 times 10 to the power of 60. looking at the one steel book
we find that we need a 230 pfc so instead of using a 150 pfc we actually need a 230 pfc
we will do the same process process to the dynamic calculations the deflection formula
is a little different for a point load but the concept here is the same we need to find
the i required for a deflection limit of 1 millimeters and check if the 230 pfc works
so just by looking at the tables again we find that the 230 pfc actually works so we're happy
with this section and let's run with a 230 pfc and this ends our beam design if you want to stick
around for a couple more minutes i will explain you a little bit more about lateral restraint
of flexural members which is very important to know in order to choose the correct tables to
select the bending moment capacity okay so in this example we use the tables for the section
moment capacity phi msx which is simply the huge stress times the effects effective section modulus
okay now if you have a check at 4100 clause 5.3.2.1 you will see that a beam can achieve
full lateral restraint by a continuous lateral restraint to the critical flange
right and this would be the case of a concrete slab on top of a beam for downward loads
or by full or partial restraint provided at sufficient locations along the beam
intermediate lateral restraints as well it's missing here sorry so the segment length that
this segment length that as4100 refers to is the frflr on the table as you can see here and again
um your restrains in the beam needs to be located at this maximum flr distance so you can use
msx for msx if it's not that distance if it's more than that that distance you cannot use phi
msx you cannot use the section moment capacity theoretically in our case the joists
are spaced 450 centers and therefore the bending capacity of the 150 pfc that we chose
is in reality less than 37 kilo newton meters because flr is 422
i simply went ahead with the size because i know it will not change much from 422 to 450 okay
for instance a rafter of a portal frame has the purlins which laterally restrain the top flange
for downward loads and fly brakes which laterally restrain the bottom flange for
upward loads and when you have a fly brace and a purling at the same section
you get a fully restrained section also when i mention upward or downward loading
it is for you to know where the critical flange is and i'll explain what critical flange is
in a second a couple more examples here a steel beam supporting timber joists in plane
so if the joists are closed fitted closed fitted like in this image the joists can be considered to
provide full restraint meanwhile with the joists running on top of the beam the steel beam might
be considered to have lateral restraint at the joist locations where the critical flange is
and as i explained before the case of purlins and fly brace in restraining the flanges
a beam can fail for quite a few reasons okay one of them is yielding you must have studied that at
uni before uh that happens to compact sections and these beams will develop the full cross-section
plastic moment and will not buckle it can also fail for local buckling by local buckling which
happens to non-compact and slender sections it could fail by lateral torsional buckling and
to explain this a bit more so when a beam is loaded one flange goes into compression
which means it is trying to get shorter this flange will tend to buckle out sideways
effectively the flange is behaving like a column and like a column the longer it is the more easily
it will buckle however the tension flange right it could be the the bottom flange if it's a
if there is a downward load the tension flange does not tend to buckle
so the beam must twist as the compression flange buckles
so you can see here if i draw a plane view of a beam it will buckle sideways when loaded
however if you restrain the beam like you restrain a column with the floor diaphragm in the building
the effective length reduces therefore the bending capacity increases okay that's why the joists that
are important to restrain the beam now when we talk about restraining the beam we usually mean
restraining the critical flange of the beam which is the flange that in the absence of any restraint
would deflect the further during buckling so in practice the critical flange is the compression
flange for a simply supported beam and for a cantilever beam it is the tension flange if the
unsupported end's not restrained against lateral movement but if the tension flange or the whole
cross-section is prevented from moving sideways so the compression flange will be the critical flange
okay um when a beam has closely spaced restraints which prevent lateral deflection of the critical
flange then this prevent lateral buckling and the beam can be considered to have
full lateral restraint so that's the rule if you want to use the section moment capacity
for for for the maximum flr that we explained before okay otherwise we have to calculate the
effective length of the segment of a beam and you find these calculations on as4100 clause
5.6.3 and then you're gonna look at these tables for members without
full lateral restraint okay so if you look at these tables here we're going to notice that we're
not working with phi msx anymore now we're using phi mb which accounts for the effective length
that is unrestrained and you can use that for ubs ucs pfcs or any type of beams
so i hope that wasn't too rushed guys i hope you learned something
on if you if you didn't understand something just drop a comment below and leave me some comments
suggesting some videos or whatever show some love all right see you next time