• posted
Could someone tell me the correct answer to this test question please.
Sorry about the poor picture quality.
Question is "Which girder will take the most weight, A. Equilateral
angles, B. Obtuse angles, C. Acute angles.
Picture is at :-

Thanks,
Phill.
• posted
I would have thought A. In B the struts are very rigid but there is a long gap between them which means the lower beam can buckle. In C there are short gaps so the lower beam is less likely to buckle but the struts are weaker. A seems intuitively the best compromise but I think it might depend on the thickness and strength of the beams.
• posted
:-
dont know what they are supposed to mean ... if they are talking triangles transfering weight above them to floor ... the multitriangle set-up is best ...
all the best.mark
• posted
:-
C Acute angles.
• posted
Puzzling! I'm inclined to say that all 3 take the same load.
Arguments: * the area of the triangles is the same for all 3 solutions * The upper belt is compressed, and all 3 solutions cover the full length of the upper belt * the lower belt is expanded, no buckling * the cross-section of the triangles in the neutral line (where you have the biggest shearing forces) is the same.
The reason why I-beams that are cut, shifted and re-welded (that honey-comb pattern) look more like A) is to reduce the length of cut and to get a beam as high as possible.
I might be corrected.
How many points did I get? Nick
• posted
I think, because it is a girder, that it is supported on the left and right end and takes a load on top of it. Like a bridge.
Nick
• posted
in that case then ...it would be "c" which is the best .because it has the most metal in it .
if it was a case of which one could be most economically made, it would be "b"
if it was a case for which one was best compromise for price and load carrying it would be "a"
wonder what drug exam compiler of questions is on ..he's far removed .if you have only the choice of answering a,b, or c
all the best.mark
• posted
Well, I am a at least not alone. I thought C. because of the fact that the triangles sides are more vertical, so less side stress, but the answer paper says A.
I would really like to get a professional engineers opinion on this one.
Thanks for the input. It is interesting.
Phill.
Nick Mueller wrote:
• posted
from Phill
I wasn't going to offer an opinion because (to me is seemed so simple) but the variety of anwers has amazed me.
Without looking at the diagrams I deduced that the Equilateral had to take the most weight since the virtical and sideways stresses will be equal and be directed down the length of the side of the triangle - putting the material in compression - with the obtuse the vertical stress component will be greater than the sideways component and in the acute the sideways stress will exceed the vertical.
J (nearly failed HND mechanical engineering but was reassessed ;) ) G
• posted
Well, you only "nearly" failed. I haven't even sat the test, so I'll take your word for it.
Phill.
JG wrote: > from Phill > >> Well, I am a at least not alone. I thought C. because of the fact that >> the triangles sides are more vertical, so less side stress, >> but the answer paper says A. > >> I would really like to get a professional engineers opinion on this one. > >> Thanks for the input. It is interesting. > > I wasn't going to offer an opinion because (to me is seemed so simple) > but the variety of anwers has amazed me. > > Without looking at the diagrams I deduced that the Equilateral had to > take the most weight since the virtical and sideways stresses will be > equal and be directed down the length of the side of the triangle - > putting the material in compression - with the obtuse the vertical > stress component will be greater than the sideways component and in the > acute the sideways stress will exceed the vertical. > > J (nearly failed HND mechanical engineering but was reassessed ;) ) G
• posted
Is there significance in the (faint)arrow to the left of each diagram with indecipherable writing attached?
Is it meant to be the direction of load?
if so, then your assumption is wrong, this girder is resisting en loads, in which case the choice of a) is more rational.
Robi
• posted
You might want to rethink that before the examiner hears or he'll be after you for the remedial geometry class I'm afraid.
• posted
PS. Here's a clue. Why is the area of any triangle, symmetrical or otherwise, always 1/2 base x height?
• posted
the exam question is flawed ... if you wanted to show differences
he should of set out out with the same amount of metal spars doing the their thing in each diagram ..
c has a multitude of spars ...despite being all at the wrong angle it will win over the others ...because of the shear amount of metal in it .
so stuffy examiner does not know how to set out questions in my opinion.
you just cant answer that one ...if you have the choice of leaving it and answering another question instead ...i would ..
all the best.mark
• posted
Mark -
...and you would (rightly) fail the test.
Firstly, the "amount of metal" is the same in each case. Dave already tried to point that out to you, but you weren't paying attention. Sum the areas of the triangles in each diagram and you will find they are the same, so the "amount of metal" is the same in each case. My guess is that was entirely deliberate on the part of the examiner.
Secondly, it isn't how big it is, but how you use it. Take a trivial example - a round bar of a given weight of metal per unit length will support a smaller load than a round tube of the same weight per unit length.
So the important issue isn't the amount of metal in the three different designs but the way that metal is distributed.
Regards, Tony
• posted
well the site isn't working now
so i cant check out what you said tony
you're saying same amount of metal used in all three diagrams ...
didn't look like it to me
and cant check now
but will take your word on it
and not argue any longer
i know all the principles
but this just says to me corrugated cardboard.
all the best.mark
• posted
Another way to look at is the load distribution per unit area. If we are assuming that the amount of material in the triangular supports is the same in all cases, and the load is evenly distributed across the top surface, then there will differing apex loadings at the lower surface.
The top surface support is across the base of the triangles and will concentrate the load across 2, 4, or 7 points on the bottom. Therefore there should be a lower force per area with the acute triangles.
As such, perhaps the failure mode would be on the lower beam, rather than the triangulation?
Peter
• posted
I'm a professional engineer, unfortunately not a mechanical one, so I don't really have a clue. :)
I'd suggest that in reality, if the weight was evenly distributed, there would be no difference in their load bearing capability. The bigger issue is the load bearing capability of the ground it is resting on - those pointy bits could do no end of damage!
Sorry, I told you I wasn't a mech. eng.
Mark
• posted
I am, and I don't understand the question.
• posted
So am I and I think the question is appallingly badly set. How is the load applied, how is the girder supported and how is it constructed? As it stands it may as well have read 'Are apples better than monkeys?' or some such sureal gibberish.
Richard MIMechE

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.