i beam design theory

huangyhg

i beam design theory
hypothetical situation:
2 pieces of aluminum rod 1/2" dia.
one piece is machined to create the largest possible(given the 1/2" dia.) i-beam.
is the resultant i-beam
     a) the same strength as the original rod?
     b) weaker than the original rod?
     c) stronger than the original?
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this sounds almost like a 'homework' problem.  nonetheless, when you are refering to strength do you mean some allowble stress or are you refering to load carrying capacity.  if you mean strength as some allowable stress (sy, su), then you have a trick question...strength will not change with cross sectional area (assuming the machining process does not alter the material properties).
if you are looking add load carrying capacity in simple bending, then the cross section with the largest moment of inertia will be 'stronger' but will have the same material propeties (machining process aside).
how are you loading these beams and how will they be used (what purpose)..this plays an important role in understanding your question. with such small beams, the machining process will have a tremendous impact on the strength.
rg88
this is not a "homework question" per se.
i am not an engineer and this is a home project to "stiffen" the forearm of a rifle stock.
the metal insert will be epoxied into an injection moulded plastic stock.
the discussion that evolved among our group of hobbyists related to whether the 1/2" dia. rod was more prone to bending, given a force applied perpendicular to the axis, than the same piece of rod machined to resemble a small "i beam".  this assumes that the load is applied to the "i beam" down through the depth of the web, of course.
i realize this is probably elemantary engineering to those trained in the field.  could you suggest some texts that expain such basic concepts to hobbyists?
thanks,
rooster
rooster1717,
rooster1717,
sorry for the 'homework' issue...we don't like to see to lazy students.  when i suspect that, i try to lead them down the path to making an informed decision on their own.  
being a firearms hobbiest myself, i understand what you are trying to do.  the round section will be heavier and stiffer than a machined i-beam section.  if you can accept the additional, although nominal increase in weight, go for the round section.  machining can cause the aluminum to 'work-harden' and/or reduce the strength through the surface finish, but in your instance i don't think it would be a big issue.  
defining your conditions for a thorough analysis is quite difficult, but it most nearly matches a cantilevered beam.  given that, the higher moment of inetia will produce a stiffer beam.  the round section has the higher moment of inertia in your case.
have you thought about fiber reinforecement in polyesther?  or maybe a square cross-section bar instead of round.  
a good source is to find a book on solid mechanics or basic machine design at your local library.  because of number of books on the subject, try an author search for roarke  or timoshenko.  maybe a machinery's handbook. but to be honest, for your application, a rectangular section will give you the best weight vs. stiffness without all of the hard work.  fyi this isn't always the case in other engineering applications.
hi rooster1717
having read the earlier replies i would agree that the rod
would be stiffer then an "i" beam made from the same rod.
the stiffness of a beam loaded vertically through the web as you put it increases by a cube law as the depth of the beam or web increases, so take a simple rectangle say 1" wide by 2" deep and another rectangle 1" wide 4" deep therefore the 4" deep beam would have 8 times the stiffness than that of the 2" deep beam eg:-
       (2")^3=8
       (4")^3=64
again as mentioned earlier the "i" beam as much greater stiffness when loaded vertically through the web, in comparison with the same beam loaded horizontally through the web. the rod on the other hand will take the same load in any direction because it is symmetrical although it carries the penalty of extra mass.
my suggestion would be not to machine an "i" beam but to put a hole through the rod and make it a tube, if you do this you can have a lightweight reinforcement which can handle equal loads in any direction and save yourself a lot of machining.
whilst i appreciate your not an engineer if you bear with me i'll calculate an example.
the "i" value of a beam whether circular,rectangular or square is known as the "second moment of area" this is a property of the shape of the beam.take your 1/2" dia rod of
aluminium its density is 0.097lb/in^3 therefore its weight per foot would be about 0.229lb.
now if we calculate the "second moment of area" for this solid rod the formula would be:-
    second moment of area= 3.142 x (rod dia)^4/(64)
     this would =0.00307 in"^4
look upon this figure as the stiffness of the beam made from solid rod.
now if we drill a 3/8" hole through the 1/2" rod the "second moment of area" becomes:-
second moment of area=3.142 x (rod o.d^4-rod i.d^4)/(64)
where the o.d. = 1/2" and i.d.=3/8"
this would now = 0.00209in"^4
therefore %loss in second moment of area
                                      
                         =1-(.00209/.00307)x100=31.6%
however your new weight for the tube would be 0.09999lbs
reduction in weight would be 56.33%.
a tube of this size would be stiffer than any "i" beam machined from the same size rod and a lot easier to make, even easier if you can use standard tube instead.
note as you don't give any load figures it is hard to determine exactly what size beam you should use but working with that rod this will be your best option "turn it to tube" if you cannot carry the excess weight.
hope this helps
desertfox

desertfox, it appears as thought my thread got deleted, this is hello2u, we discussed tubing properties the other day, can you please email me at the addy i gave you?
hi hello2u
yes your thread got deleted before i could get your address
down could you post it again please.