Hey guys welcome back to the channel this is Gabriel
Oi. What? What's that? What's that what? Why are you wearing a suit?
I don't know, just felt like it would be nice. No mate, you're an engineer, engineers don't wear suit
Hello young grasshopper. Gabriel here and welcome back to the brazilian engineer in Australia
and in today's video we're gonna go through the design of a timber post
i'm gonna hold your hands and walk you through a worked example and hopefully in the end of
the video you're going to be able to design any timber member subject to compression
all righty
i'm pumped today guys it's saturday morning sun is shining it's a beautiful day out there so i'm
just gonna finish recording this lesson for you my beautiful students and then i'm off for surf
okay so let's suppose you need to design a timber post a column or or a strut in a truss
or stud in a wall framing it doesn't matter they all mean the same thing okay so it's just
a member subject to axial compression
so how do we start that the capacity in compression in the member needs to be greater
than the load obviously okay so we need to work out the design capacity and compression parallel
to the grain so what's parallel to the grain means that the force is in the same direction
that the fibers in the timber so this design capacity like in
any other material it could be steel could be concrete or whatever is essentially given by the
characteristic strength in compression times the cross-sectional area of the column but that's not
all we have some other factors that we need to take into account when designing a member
in compression. Don't freak out with all these k values because they're easy peasy we're going
to go through this numerical example here and i'll explain you all these factors as we go okay
so the example we've got here is to calculate the compression capacity of a 3 meters high
100 by 100 f-17 unseason hardwood post in brisbane okay so we will start with the Phi
factor which is the capacity factor and you can find its value on table 2.1 of as1720.1
so this table is divided in categories and timber materials so these categories
are essentially the importance of the structure you're designing so category one
would be a member that if it fails will only cause a small amount of damage okay so if one
joist fails it won't cause a huge amount of damage for example category two if it fails a significant
portion of the structure may fail as well i would classify bearers and post as a category two
and category three would be a primary member in a hospital for example or a post-disaster structure
so we're going to focus mainly on category two members and you can see that for stress grades mgp
15 or F17 or higher the Phi factor is 0.85 while if you have any grade lower than that phi is 0.7
so essentially this is a standard telling you to use a higher grade for primary members otherwise
only 70 percent of the capacity can be accounted for instead of 85 so let's say i'm designing
a column or a bearer for a house i'm going to try not to use anything lower than mgp15 or f17 so a
Phi factor for our example or worked example is 0.85 next is the duration of load factor k1
which can be found on table 2.3 or even better on table g1 in the appendix g
so timber can withstand impact much more easily from short-term loads than from long-term loads
so you can see that for an effective duration of peak action of 5 seconds k1 is equal to 1
meanwhile for 50 years plus k1 is 0.57 so it's a reduction of almost 50 percent in the capacity
of the timber this table is interesting because when you design for steel aluminum or concrete
you will have a single number for the capacity of the member regardless of the load combination
however with timber with timber different load combinations can result in different member
capacities so load combination 1.35 g could yield results way lower than 1.2 g plus 1.5 q so this is
just something for you to be mindful of okay so when you design a member you might have to check
the capacity for different combinations all right so for our example i'm using k1 equals to 0.57
next is the moisture condition factor k4 which can be found on clause 2.4.2 this value is usually one
sometimes you can take advantage of these higher values on this table for unseasoned
timber so generally these occur when the timber is not exposed to humidity and can partially
dry during the process of construction um so this would be a team a a timber beam
under the roof for for example okay you can you can read the standard
for for more for a more detailed explanation on this so for seasoned timber k4 is usually one as
well unless the beans average moisture content for a 12 month period is expected to exceed 15 percent
then you will have to calculate k4 as per the formulas presented on clause 2.4.2.3
so you just head to the standard have a look at these formulas if this is your case
you might you might have to get in touch with the manufacturer as well to assess if the timber
can dry in um if the moisture content in the timber will be expected to exceed 15 okay
all right i'll give you a quick explanation here so um
this actually means that if you have a chamber a timber member unprotected from rain for example
in a in a patio or you're building on the coast where the timber will be subject to ocean spray or
or an exposed timber in the bathroom okay so you young padawan you come back from school
and then you're dirty and tired and you just want to have that nice and hot shower
so the steam from the shower will increase the moisture in the timber
and that means you need to work out k4 accordingly all right
we will take it for as one for our worked example as we're using unseasoned hardwood
you guys do that you're not sleeping are you hope everybody's following me if you're getting tired
get up a little bit move your body move your head move your shoulders just going to put
a put a quick music here quick music for us quick music oh what happened here yeah
all right come on let's go let's go let's focus um
all right ronnie k6 is a temperature factor it's very easy to work out this one basically k-6
is equal to one for most of australia unless you are in the far north like darwin
or coastal regions of queensland north like cairns townsville mackay
rockhampton um gladstone bundaberg pretty nice place by the way if you haven't been to darwin or
cairns check them out i definitely recommend it's worth a visit eh so just check the map
in as1720 if you are located near these areas and in this place k6 is equal to 0.9
now the last modification factor is k12 which is the stability factor this number allows for
the reduction in capacity of slender members so remember when you study buckling you've got
different modes of failure so k-12 takes this into account work out k12 we're going to start with
g13 which is the effective length factor we're assuming our column will be pink pink therefore
g13 equals to 1 as per table 3.2 and the condition of restrain is what is the yeah
the condition of restrain the restraint is restrained at both ends in position only
i'm assuming the minor and the major axes are restraining the same way
but it's quite common to have different g13 values for the two buckling directions okay so you've got
direction x and direction y and then we've got different g13 values that you need to work out
we could have used g13 at 0.75 which is each end held by two bolts substantially restrained
that means all column must be connected to a very rigid structure so since we don't have
much information about that in our example i'll be conservative and just assume g13 equals one
so we're going to use g13 to find s3 which is the slenderness coefficient for buckling about
the major axis s3 is the lesser of lax divided by d or g13 times l divided by d um lax is the
distance between intermediate lateral restraints in this case it's just the length of the column
so plugging the numbers into these equations we get that s3 is equal to 30.
obviously the calculation for the minor x will be the same because the cross section is square it
could have been different if you if we had if you had lateral restraints in one x and three movement
in the other axis um you know sometimes you you can restrain the column in one direction but it
doesn't restrain in the other direction so you've got different values for different directions
this is not the case here so as for for the minor axis will be 30 as well
continuing our calculations to find k12 we now need to find the material constant
rho c on table 3.3 for F 17 rho c is equal to 1.25
and if rho c times s is equal or less than 10 k 12 is 1. and this is a k12 for a stocky
compression member this column will not fail due to buckling uh this this column would probably
would probably crush the the load would probably crush the column so if rho c times s is between
10 and 20 k 12 is equal to 1.5 minus 0.05 times rho c times s so this is a k 12 for a compression
member at the transition from a stocky member to a slanted member so if rho c times s is equal or
greater than 20 k 12 is equal to 200 divided by rho c times s squared this is a cage well for a
slender compression member and buckling would be the main mode of failure or rho c times s
is equal to 37.5 so 37.5 is much larger than 20. so this tells us that
the column is slender and the capacity is limited substantially by buckling k12 is equal to 0.142
and this means that we will use only approximately 14 percent of the capacity of the timber
due to buckling that's that's very low eh the characteristic
strength in compression parallel to the grain fc prime is found on table h2.1
for F17 is 34 MPa. Ac is the area of the cross section of the column and finally we can work
out the design capacity and compression of this column so plugging all the numbers in we found
that we get Ndc equal to 23.4 kN this column can withstand 23.4 kilonewtons
you see that how much the all these k values can reduce the capacity of a membrane compression
so you got to be very mindful of that a couple of comments before i finalize this video
all these calculations we've been through our strength check although the serviceability check
rarely governs for compression members you might need to do this check afterwards okay as you may
know as well many compression members also have bending actions so a stud for example could be
resisting wind loads a column could have an eccentric load so you will need to design
this member for combined actions okay so we would it would be you
would have to design for compression and then you would have to design for bending
it is important okay so it is important that you assess every scenario before you send your
design out for construction all right so that was it for today um i hope you enjoyed the video
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