This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Muddy Card Responses Lecture M10
Why is the answer 5? (with reference to the last pair of PRS questions) I don’t understand
how there can be no displacement. Admittedly the idea of the pins sitting on rigid supports
is an idealization (nothing can be perfectly rigid), but given that the supports are rigid, the
bar is prevented from extending or contracting axially.
Are there any other constitutive laws for bars besides: d = FL
+ aDTL ? We will meet
AE some other constitutive laws later this term, and next term. However, this one is the most
useful for the bars in trusses.
Could you put the concept questions on the web?. Certainly. † What is superposition? I like colored chalk. I am glad you like the chalk  I hope that it
clarifies things. Superposition refers to the idea that we can break down complicated loading
and or support conditions to a linear combination of simpler ones. This is due to the fact that
the structures are linear and elastic.
Displacement diagrams are just greatly scaled ways of demonstrating how the joints move,
is that correct?. Yes.
Do they really move like that or is it just a theoretical representation, or do they really
move like that but on a much smaller scale? The second. Trusses such as the one you have
tested in the truss lab are quite stiff, so the deflections are small. The displacements are
measured in mm while the bar lengths are 0.51 m. This means that the deflections and
displacements are very small. However, a truss that conforms to the pinjointed model, will
really deform in this way.
Superposition. I don’t quite understand what you were talking about, but I imagine you
will cover it further later on. We will indeed meet superposition on several occasions in the
future. I brought it up today because some people find it easier to think about statically
determinate structures by superimposing statically determinate cases.
Equation (1) (for the restrained thermal expansion problem) in the last part said
SFx =0, FBFC=0, FBFC=Fbar. . Either you or I wrote this down incorrectly. It should have been
FB=FC=Fbar FB and FC are the reactions at B and C, Fbar is the internal force in the bar.
Superposition not clear. Can you please do an example? We will cover this in recitation
next week. It is not a core concept for this part of Unified M&S, but we will meet it again
when we see beams, and it will be more important then.
Although I got it for the single bar, I am generally confused on how to apply
superposition. See comments above.
One small thing: Superposition seems like a very inefficient way to solve a problem
unless you’re a computer. Not necessarily, it allows us to break complicated problems down
into simpler ones, which we are generally more accurate at solving. In addition, some people see problems in terms of superposition of simpler loading cases and it provides them with
insight and intuition to think of the problems this way.
When you use superposition such that: = + Why don’t you include the pins here (roller and pin on the RH supports). I should have
done, see the figure above.
What is the definition of a truss?. See lecture M5. Pin jointed trusses have straight bars
joined by hinges (pins). The diagram you drew is of a beam, which we will discuss in detail
next term.
I didn’t understand the last PRS question, concept question 4. This was the case of the bar
restrained from expanding when a temperature was applied.
This was an example of applying the three great principles in parallel.
1) equilibrium: SFx =0, FBFC=0, FB= FC=Fbar 2) d = FL
+ aDTL
AE 3) compatibility: d=0, i.e the deflection of the bar must be compatible with its ends being
fixed.
Inserting the bar force from (1) into (2) and setting d=0 from 3, we obtain F = aAEDTL †
If in the truss there are two different types of joints (drawing indicates one end of a truss
like the truss lab truss on rollers, the other on pins), is this truss † symmetric? (assume
still
lengths are equal). This is a good question. Technically it is not exactly symmetric.
However if it is only loaded vertically and symmetrically it will behave as though it is
symmetric – as you will have found in analyzing this truss in M7. If you are ever in doubt,
assume that it is not symmetric.
Is the rotational part for solving deflections only approximated as a 90° angle? Is there a
better way? Actually this approximation is extraordinarily good. You could assume that it
the rotation results in a displacement along a circular arc, but you would find that this is
quite tedious to calculate, and would provide no additional accuracy. Not clear why the displacements cancel each other out on the last PRS? They didn’t. The
ends of the bar are fixed, so there is no displacement.
You may be referring to my illustration of how to solve this by superposition. In which case,
if the right hand end of the bar was not fixed, the bar would extend by an amount,
d = aDTL . If I was solving this by superposition, then I would need to apply a force that
would result in an equal and opposite displacement, so that the sum of the two
displacements would be zero. Thus the force I would need to apply would be
F = aAEDTL , which is as I calculated by setting up the three great principles in parallel. †
† Could you give an example of how to do this problem (the last PRS) by superposition.?
You said you could see what F is needed to make sure that d=0, but this can you elaborate
on this more? Sounds like I will need to start tomorrow’s lecture with this.
I didn’t understand the last concept question. Does the bar tend to warp when it feels a
compressive force? Is there no apparent change in its geometry? If the bar is only capable
of carrying axial loads (as we have been assuming up to this point) then it will not bend (or
warp, or buckle). We will reconsider this assumption next term when we learn about
bending and buckling.
In the first PRS question, why is there always a right angle if q can vary? a a This is because the displacement of the end of a bar in a pin jointed truss is always made up
of two components: the extension/contraction along the bar and rotation about the joint at
the other end of the bar. Since the extensions and displacements are small, the rotation can
be represented the tangent to the circular arc (which would be the exact solution to the
displacement) which is a line perpendicular to the original direction of the bar.
Why does Unified make us learn 3 subjects at a time and have tests 10 days after last
lectures? You should make Unified more similar to regular classes. The scheduling is a bit
constraining, so it is more or less inevitable that three classes will be taught together at some
point. Regarding having tests 10 days after the last lectures I am not sure what the complaint
is. Is this too quick or too long? I would personally rather have a longer time between the
lectures and the test – to provide more time for the concepts to sink in. I sense that you may
feel the opposite. I hope that we can always keep Unified from becoming a “regular” class!
15 cards with no mud, or positive comments about the lecture. Thanks! ...
View
Full
Document
This note was uploaded on 02/20/2012 for the course AERO 16.02 taught by Professor Charlescoleman during the Winter '12 term at MIT.
 Winter '12
 CharlesColeman
 Aeronautics, Astronautics

Click to edit the document details