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[music playing]

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Welcome back to DIY Project #2,
in which we‘re designing,

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\hbuilding, and testing a
cardboard tower capable of

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supporting some very demanding
\h\hgravity and lateral loads.

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Last lecture, we defined the
problem, formulated a design

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\h\h\hconcept, and initiated the
detailed design by modeling and

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analyzing our structural system,
called a braced frame, to

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calculate this set of internal
member forces, which will form

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the basis for our final design.

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\hOur principal design challenge
in this lecture is to design the

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members, such that the strength
of each one exceeds its internal

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force by a safe margin.

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\hAs such, the next step in our
process must be to figure out a

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way to predict the strength of
\h\hstructural members made of

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cardboard.

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If our tower were going to be
\hmade of steel, concrete, or

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commercial lumber, we could take
advantage of published design

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\h\h\hcodes which incorporate
scientifically-based equations

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that can be used to predict
member strengths with great

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confidence.

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\hBut, alas, there are no design
codes for cardboard structures,

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so we‘ll need to develop our own
program of DIY experimentation

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to predict cardboard member
\h\h\h\h\h\h\h\h\hstrengths.

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Let‘s begin by delving more
deeply into this concept of

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strength.

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\hIf I subject this structural
member to gradually increasing

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tension, it eventually breaks in
two, a failure mode called

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rupture.

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\hThe applied force at which
rupture occurs is called the

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tensile strength of the member.

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\hFor example, if I was pulling
with five pounds at the instant

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before the member failed, then
\hits tensile strength is five

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pounds.

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\h\hIn general, tensile strength
depends on only two factors: the

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\h\h\htype of material and the
member‘s cross-sectional area,

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\h\hwhich corresponds to the
rectangular area highlighted

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here on this model.

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Now, note that when I subject an
identical member to compression,

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rather than tension, it fails in
a fundamentally different way.

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\h\hWith relatively little
compressive load, it fails

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by kicking out sideways.

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\h\hThis failure mode, called
buckling, is much more complex

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than tensile rupture.

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\h\hThe buckling strength of a
member depends not only on the

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\h\h\h\h\htype of material and
cross-sectional area, but also

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the cross-section shape and,
most importantly, the member

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length.

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Let‘s explore these influences
\hby doing some more buckling

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experiments.

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Now here are three structural
members: a solid square bar, a

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solid flat bar that‘s much wider
than it is thick,

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and a hollow square tube.

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They‘re all three feet long and
all made of the same type of

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wood.

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Furthermore, they all use
exactly the same amount of

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material.

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\hWe can prove this by weighing
each one on my digital scale 32

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grams, 32 grams, and 32 grams.

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Now let‘s use this testing
apparatus to determine the

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compressive strength of each
\h\h\h\h\h\h\h\h\h\hmember.

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We‘ll start with the flat bar.

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\hI‘ll load it into the machine,
and now when I apply load to the

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\h\htop of the machine, a single
one-pound weight, is sufficient

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to cause buckling, indicating a
compressive strength of this

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member is actually less than one
pound.

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Now let‘s try the square
\h\h\h\h\h\h\h\hsection.

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It‘s in position, and now I‘ll
\h\h\h\h\h\h\happly one pound.

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No problem, carries load just
\h\h\h\h\h\h\h\h\h\h\h\hfine.

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What about two?

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Three, four, five, six, seven,
\height, nine, ten, and on 11

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pounds, we can just barely see
\hthe onset of buckling here.

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I won‘t add the 12.

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11 pounds!

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\h\h\hClearly, the compressive
strength of this solid square

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member is significantly greater.

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Now, let‘s try the hollow box
\h\h\h\h\h\h\h\h\h\h\h\hshape.

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When we load the tube, well, we
run out of weights long before

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even approaching failure.

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\h\hIndeed, the theoretical
compressive strength of this

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\h\hmember works out to be 86
pounds, a load I wouldn‘t even

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be able to apply safely with
\h\h\h\h\h\h\h\hthis device.

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\h\h\h\hRemember that the only
difference between these three

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members we just tested is the
\hshape of the cross-section;

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otherwise, they use the same
\h\h\h\hamount of material.

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Clearly a very important factor
in compressive strength.

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\h\hIs it any wonder why hollow
tubes like these are used in so

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many different structural
\h\h\h\h\h\happlications?

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So, what about length?

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\hTo answer this question, let‘s
test a member that has the same

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flat, rectangular cross-section
as our first specimen, the one

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\h\h\hthat failed at less than a
pound, but only half the length.

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The longer one buckled with the
application of 1 pound, in fact

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\h\hit had buckled at a load
somewhat less than 1 pound.

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And now, I‘ll load a column with
an identical cross-section, but

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half the length, with one pound,
two, three, four, five, six, and

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finally with seven pounds, once
again we see just beginning to

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occur, the onset of buckling for
this member.

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Substantially stronger.

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Evidently, reducing the length
\hof a compression member can

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\hsignificantly increase its
strength, particularly for a

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slender column like this one.

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Furthermore, the beneficial
\h\h\heffect of shortening a

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compression member can be
achieved without actually

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shortening.

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Let‘s put our original 3-foot
\hmember back in the machine, 

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but now I‘m going to add this
lateral brace at mid-height of

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the column.

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And even though the brace only
\h\h\hprevents the member from

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displacing laterally at this one
point, the member now needs one,

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two, three, four, five, six, and
again, seven pounds to cause

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buckling.

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Its compressive strength has
been substantially increased

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because the brace constrains it
to buckle in an S-shape, rather

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than in a single curve as
\h\h\h\h\h\h\h\h\hbefore.

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Evidently, adding intermediate
\h\h\hbraces to a long column

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increases compressive strength
\h\h\hby reducing the column‘s

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effective length to the distance
between brace-points.

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And this is why, in the concept
design for our structure, we

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subdivided each column into four
segments by adding intermediate

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beams.

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Before we move on, I should add
that if we had a machine capable

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of loading these specimens to
failure in tension, we‘d find

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\hthat all four have a tensile
strength of approximately 800

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pounds!

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That‘s nearly ten times the
compressive strength of the

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hollow tube, and over 100 times
that of the solid square bar.

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\hWe can conclude that tension
members are almost always more

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structurally efficient than
\h\hcompression members, an

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\hobservation that will also
influence our final design.

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At this point, we can combine
\h\hour newfound knowledge of

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tensile and compressive strength
with the structural analysis

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\h\hresults from our previous
lecture to make some important

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design decisions.

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First, we know that the columns
will experience compressive

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forces in excess of 40 pounds,
\hso we‘ll prefabricate these

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members by forming corrugated
\hcardboard into hollow square

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cross-sections, like this, for
optimum compressive strength.

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\h\hSecond, the beams will
experience a much smaller

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compressive force of only five
\h\h\h\hpounds, so we can use

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significantly smaller tubes for
these members.

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And third, the diagonals will
experience a tensile force of

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seven pounds, which we can carry
quite safely with strips of

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file-folder material, like this.

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\h\hBut what size should we make
these members, such that they‘ll

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be sufficiently strong to carry
their respective internal

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forces?

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\h\h\hClearly we need to do some
strength tests on cardboard, but

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here we encounter a significant
challenge.

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As you‘ll see shortly, testing a
member like this one to ensure

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that it can carry 40 pounds
safely will actually require

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compressive loading somewhere in
excess of 80 pounds.

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\h\hAnd this testing device is
wholly inadequate for the task

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because there‘s simply no way to
balance 80 pounds up on that

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loading platform on top of the
\h\h\h\h\h\h\h\h\h\h\hmachine.

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Not to worry!

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\h\hThis is really just another
engineering design challenge, so

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\h\h\hlet‘s solve it as any good
engineer would, by figuring out

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a clever way to do more with
\h\h\h\h\h\h\h\h\h\h\hless.

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\h\hWhat we need is a means of
multiplying force such that we

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put in, say, 10-15 pounds, and
we get out a force five or six

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times larger.

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In engineering terms, what we
need is mechanical advantage.

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There are many ways to achieve
\h\hmechanical advantage, but

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perhaps the simplest is the
\h\h\h\h\h\h\h\h\h\h\hlever.

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\hAnd here is the lever-based
machine we‘ll be using to test

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the strength of cardboard
structural members in both

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tension and compression.

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It consists of a fixed base, a
\h\h\hfulcrum, which uses this

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\hhalf-inch bolt as a pivot, and
the lever-arm itself, loaded by

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\hmeans of a bucket hanging from
its far end and with provisions

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for placing a test specimen on
this side for tension testing,

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\h\hand over on this side of the
fulcrum for compression testing.

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\hThe fulcrum itself has three
sets of holes, so we can test

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specimens with lengths from six
inches up to 12 inches.

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\h\h\h\hAnd this post keeps the
lever-arm from drifting sideways

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\h\h\h\h\hunder load while also
incorporating a set of holes for

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a safety pin, here, that can be
used to support the lever-arm

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temporarily, as it is right now.

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If you‘d like to try some DIY
materials testing of your own,

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I‘ve provided drawings for this
machine in your Course Guide.

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Let‘s see how it works by
\hrunning a tension test.

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Here‘s our first test specimen,
a piece of file-folder material

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that‘s exactly 3/4” wide here in
the center, but flares out to a

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width of 2” at both ends.

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\hThis characteristic shape,
called a “dog bone,” is very

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commonly used in real-world
materials testing, as these

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00:12:20,340 --> 00:12:23,043
metal test specimens attest.

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\h\h\h\hThe dog bone shape is
advantageous because it causes

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00:12:26,746 --> 00:12:30,984
the specimen to fail here, in
\hthe measured center section,

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00:12:31,017 --> 00:12:33,286
where we‘re interested in
\h\h\h\h\hacquiring data.

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Were it not for these wide ends,
the clamps would cause premature

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\hfailure at these locations and
our data would be compromised as

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00:12:41,227 --> 00:12:43,229
a result.

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00:12:43,263 --> 00:12:46,900
Before I mount the specimen,
note that I‘ve balanced the

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00:12:46,933 --> 00:12:51,171
\h\hlever-arm on its fulcrum by
hanging these weights out on the

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00:12:51,204 --> 00:12:52,672
short end.

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00:12:52,705 --> 00:12:56,175
This procedure ensures that the
weight of the arm itself doesn‘t

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00:12:56,209 --> 00:12:59,412
contribute to the load on the
\h\h\h\h\h\h\h\h\h\hspecimen.

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00:12:59,445 --> 00:13:02,248
\h\h\hNow, we‘ll hang this empty
bucket from the opposite end of

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00:13:02,282 --> 00:13:12,325
\hthe arm, clamp the specimen in
position, and remove the safety

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00:13:12,325 --> 00:13:17,664
\hthe arm, clamp the specimen in
position, and remove the safety

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00:13:17,697 --> 00:13:22,302
pin.

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00:13:22,335 --> 00:13:36,482
To initiate the test, slowly add
sand to the bucket until the

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specimen ruptures.

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\hIt‘s a good clean break, so
we‘ll now weigh the bucket and

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and the result is 2.5 pounds.

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00:13:56,469 --> 00:13:57,937
But what does this number mean?

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00:13:57,970 --> 00:14:04,076
When a force of 2.5 pounds was
\hpulling down here, how much

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00:14:04,110 --> 00:14:09,649
tension was being applied to the
specimen here?

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00:14:09,682 --> 00:14:13,586
To answer this question, we need
to introduce a new engineering

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00:14:13,619 --> 00:14:18,257
mechanics concept, the moment,
\hdefined as the tendency of a

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00:14:18,291 --> 00:14:22,295
force to cause rotation about a
point or axis.

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00:14:22,328 --> 00:14:26,032
\hThis is another concept we‘ll
return to repeatedly throughout

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00:14:26,065 --> 00:14:28,167
the course, so please learn it
\h\h\h\h\h\h\h\h\h\h\h\hwell!

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00:14:28,201 --> 00:14:33,506
Mathematically, the moment of a
force about a point is equal to

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00:14:33,539 --> 00:14:37,376
\h\hthe magnitude of that force
times the perpendicular distance

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00:14:37,410 --> 00:14:39,612
from the force to the point.

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00:14:39,645 --> 00:14:46,018
\hIn this diagram, the moment of
Force F about Point A is F x d,

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00:14:46,052 --> 00:14:50,290
and its direction is clockwise.

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00:14:50,323 --> 00:14:53,526
\hLast lecture we learned that,
when working in two dimensions,

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00:14:53,559 --> 00:14:57,063
the mathematical condition for
equilibrium is that the sum of

222
00:14:57,096 --> 00:15:01,033
all forces acting on a body, in
both the x and y directions,

223
00:15:01,067 --> 00:15:02,502
must equal zero.

224
00:15:02,535 --> 00:15:06,539
\h\hBut there‘s actually a third
equilibrium condition, which we

225
00:15:06,572 --> 00:15:10,743
can now state: That the moments
of all forces acting on the body

226
00:15:10,777 --> 00:15:13,713
must also add to zero.

227
00:15:13,746 --> 00:15:18,618
Therefore, if we call the weight
of our bucket W, and the tension

228
00:15:18,651 --> 00:15:22,788
in our test specimen T, then the
Principle of Equilibrium

229
00:15:22,822 --> 00:15:26,959
\hrequires that the clockwise
moment of T about the fulcrum

230
00:15:26,993 --> 00:15:31,965
must equal the counterclockwise
moment of W about the fulcrum.

231
00:15:31,998 --> 00:15:35,802
Each of these moments is the
\hassociated force times its

232
00:15:35,835 --> 00:15:37,603
distance from the fulcrum.

233
00:15:37,637 --> 00:15:43,076
Therefore, TL1=WL2.

234
00:15:43,109 --> 00:15:46,279
And if we divide both sides of
\hthis equation by L1, we find

235
00:15:46,312 --> 00:15:51,083
that the tension force is equal
to the ratio L2-over-L1 times

236
00:15:51,117 --> 00:15:52,285
the weight.

237
00:15:52,318 --> 00:15:57,757
For this machine, L1 is 8” and
\h\h\hL2 is 48”, so the ratio

238
00:15:57,790 --> 00:16:02,495
L2-over-L1 is 48 ÷ 8 = 6.

239
00:16:02,528 --> 00:16:06,599
This number is the mechanical
advantage of the lever, and it

240
00:16:06,632 --> 00:16:09,502
tells us that for every pound
applied to the long end of the

241
00:16:09,535 --> 00:16:14,273
lever, 6 pounds are applied to
\h\h\h\h\h\hthe test specimen.

242
00:16:14,307 --> 00:16:17,477
And by the way, this machine is
configured such that its

243
00:16:17,510 --> 00:16:21,681
\h\h\hmechanical advantage for
compression testing is also 6.

244
00:16:21,714 --> 00:16:25,685
Therefore, for the test we just
completed, the tension in the

245
00:16:25,718 --> 00:16:31,023
test specimen at the instant of
failure was 2.5 pounds times the

246
00:16:31,057 --> 00:16:35,695
mechanical advantage, 6, equals
15 pounds.

247
00:16:35,728 --> 00:16:39,298
\h\hPrior to this lecture, I ran
several iterations of this test

248
00:16:39,332 --> 00:16:45,071
using 1/2”, 3/4”, and 1”-wide
\h\h\h\h\h\h\h\h\h\hspecimens.

249
00:16:45,104 --> 00:16:47,940
\hAnd the results are shown on
this graph of tensile strength

250
00:16:47,974 --> 00:16:50,577
vs. width.

251
00:16:50,610 --> 00:16:54,114
This is very encouraging because
it shows that the tensile

252
00:16:54,146 --> 00:16:57,349
strength of file-folder
\hcardboard varies very

253
00:16:57,383 --> 00:17:00,820
predictably as a linear function
of member width.

254
00:17:00,853 --> 00:17:04,357
Given this predictability, we
can confidently use this graph

255
00:17:04,390 --> 00:17:07,426
\has a tool for designing the
diagonal members of our braced

256
00:17:07,460 --> 00:17:09,595
frame.

257
00:17:09,629 --> 00:17:12,465
\h\h\hThe general mathematical
requirement for the design of

258
00:17:12,498 --> 00:17:17,403
\hany structural member is shown
here: The actual condition must

259
00:17:17,436 --> 00:17:21,206
\h\hbe less than or equal to the
failure condition divided by the

260
00:17:21,240 --> 00:17:23,142
factor of safety.

261
00:17:23,175 --> 00:17:25,511
\h\h\h\hThe actual and failure
conditions in this expression

262
00:17:25,545 --> 00:17:28,481
\hcan be defined in a variety of
different ways, depending on the

263
00:17:28,514 --> 00:17:32,485
situation, but for this project,
the actual condition will be

264
00:17:32,518 --> 00:17:36,255
internal force, and the failure
condition will be member

265
00:17:36,289 --> 00:17:38,491
strength.

266
00:17:38,524 --> 00:17:42,828
So, rearranging this expression
algebraically, we arrive at an

267
00:17:42,862 --> 00:17:47,233
\hinequality that will serve as
the design basis for all of the

268
00:17:47,266 --> 00:17:51,203
\h\hmembers in our tower, member
strength must be greater than or

269
00:17:51,237 --> 00:17:55,742
equal to internal force times
\h\h\h\h\h\hfactor of safety.

270
00:17:55,775 --> 00:17:58,745
In this expression, the factor
of safety is simply a number,

271
00:17:58,778 --> 00:18:02,382
\halways greater than 1, which
provides a margin of error to

272
00:18:02,415 --> 00:18:05,451
account for the many sources of
uncertainty inherent in

273
00:18:05,484 --> 00:18:10,889
structural design: unanticipated
loads, natural variation in

274
00:18:10,923 --> 00:18:14,960
material properties, fabrication
errors, and so forth.

275
00:18:14,994 --> 00:18:18,998
In engineering practice, factors
of safety are normally specified

276
00:18:19,031 --> 00:18:24,036
in structural design codes, and
typically range between 1.5 and

277
00:18:24,070 --> 00:18:26,773
2, with the actual number
\h\h\h\hdetermined by such

278
00:18:26,806 --> 00:18:31,077
considerations as the type of
\hstructural member, material,

279
00:18:31,110 --> 00:18:34,413
type of loading, and the
consequences of failure.

280
00:18:34,447 --> 00:18:39,986
\h\hFor our project, we‘ll use a
factor of safety of 1.75, which

281
00:18:40,019 --> 00:18:43,623
\h\h\h\h\hgives us a reasonably
comfortable 75 percent margin of

282
00:18:43,656 --> 00:18:45,558
error.

283
00:18:45,591 --> 00:18:46,759
Okay, let‘s do some design!

284
00:18:46,792 --> 00:18:50,262
\h\hRecall from the structural
analysis of our tower that the

285
00:18:50,296 --> 00:18:55,168
internal force in all diagonals
in the structure is 6.6 pounds.

286
00:18:55,201 --> 00:19:00,106
Substituting this force into our
design basis and multiplying by

287
00:19:00,139 --> 00:19:05,111
the safety factor, 1.75, we get
a required tensile strength of

288
00:19:05,144 --> 00:19:07,780
11.6 pounds.

289
00:19:07,813 --> 00:19:11,150
\h\h\h\hNow we can return to our
experimentally-derived graph of

290
00:19:11,183 --> 00:19:15,254
tensile strength versus member
width, start with the required

291
00:19:15,287 --> 00:19:20,192
strength of 11.6 pounds, and
\h\hread across and down to

292
00:19:20,226 --> 00:19:23,229
\hdetermine the minimum member
width required to attain this

293
00:19:23,262 --> 00:19:24,597
much strength.

294
00:19:24,630 --> 00:19:29,635
The result is 0.6”, which we‘ll
round up to 5/8”, just for ease

295
00:19:29,669 --> 00:19:31,371
of measurement.

296
00:19:31,404 --> 00:19:35,742
Now let‘s apply this same
\hprocess to the columns.

297
00:19:35,775 --> 00:19:39,112
\hIn our structural analysis, we
calculated a range of different

298
00:19:39,145 --> 00:19:43,249
internal forces in the various
\hcolumn segments, but for the

299
00:19:43,282 --> 00:19:46,819
sake of simplicity, it‘s quite
reasonable to use the absolute

300
00:19:46,852 --> 00:19:51,757
largest force, 43 pounds, as the
basis for selecting a single

301
00:19:51,791 --> 00:19:55,928
member size that will be used
\h\h\hfor all of our columns.

302
00:19:55,961 --> 00:20:00,065
When we substitute this force
and the safety factor of 1.75

303
00:20:00,099 --> 00:20:04,470
\h\h\hinto the design basis, the
result is a required compressive

304
00:20:04,503 --> 00:20:09,675
strength of just under 75
\h\h\h\h\h\h\h\h\hpounds.

305
00:20:09,709 --> 00:20:12,779
Our next challenge, then, is to
determine what size

306
00:20:12,812 --> 00:20:16,316
corrugated-cardboard tube will
provide a compressive strength

307
00:20:16,348 --> 00:20:18,517
of at least 75 pounds.

308
00:20:18,551 --> 00:20:21,154
The only practical way to meet
\hthis challenge is through a

309
00:20:21,187 --> 00:20:24,190
series of trial-and-error
\h\h\h\h\h\h\hexperiments.

310
00:20:24,223 --> 00:20:26,292
We‘ll begin with this test
\h\h\h\h\h\h\h\hspecimen.

311
00:20:28,461 --> 00:20:33,633
This one, a hollow square tube
\hthat I fabricated by forming

312
00:20:33,666 --> 00:20:38,905
\h\hcardboard around this
1/2”-square aluminum core.

313
00:20:38,938 --> 00:20:42,008
We‘ll see how this was done
\h\h\h\h\h\h\h\h\h\hshortly.

314
00:20:42,041 --> 00:20:45,311
Its nine-inch length corresponds
to the lengths of all columns in

315
00:20:45,344 --> 00:20:46,912
our braced frame.

316
00:20:46,946 --> 00:20:50,583
Now, if we placed this specimen,
without modification, into the

317
00:20:50,616 --> 00:20:53,519
\h\h\h\htesting machine at its
designated location and loaded

318
00:20:53,552 --> 00:20:56,522
it, the ends of the tube would
\h\h\h\halmost certainly crush

319
00:20:56,555 --> 00:21:00,392
\h\h\h\h\h\h\hprematurely due to
concentrations of internal force

320
00:21:00,426 --> 00:21:04,030
here, at the interface between
\hthe specimen and the testing

321
00:21:04,063 --> 00:21:05,331
machine.

322
00:21:05,364 --> 00:21:09,568
To prevent this problem from
occurring, I have reinforced

323
00:21:09,602 --> 00:21:13,272
both ends of an identical test
\hspecimen with a few wraps of

324
00:21:13,305 --> 00:21:19,378
duct tape, and we‘ll also insert
these hardwood plugs into the

325
00:21:19,411 --> 00:21:21,413
ends.

326
00:21:22,648 --> 00:21:27,019
Now, once we put the specimen in
place, each end-cap makes

327
00:21:27,052 --> 00:21:31,590
\hcontact with the machine at a
single point, so rotation of the

328
00:21:31,624 --> 00:21:35,228
lever-arm during testing is less
likely to cause localized

329
00:21:35,261 --> 00:21:38,397
crushing at the ends of the
\h\h\h\h\h\h\h\h\hspecimen.

330
00:21:38,430 --> 00:21:41,566
With everything set, I‘ll
\hrelocate the safety pin

331
00:21:41,600 --> 00:21:45,337
\h\hdownward, so the bucket will
only fall a few inches when the

332
00:21:45,371 --> 00:21:46,606
specimen fails.

333
00:21:46,639 --> 00:21:51,043
\hI know that this test is going
to take a lot more load than the

334
00:21:51,076 --> 00:21:54,179
\hprevious one we did, so I‘m
going to begin the actual load

335
00:21:54,213 --> 00:22:01,520
process by adding five of these
iron weights to the bucket.

336
00:22:01,554 --> 00:22:03,222
And now I‘ll continue the test
by adding sand slowly as I did

337
00:22:05,524 --> 00:22:08,227
before, until the specimen
\h\h\h\h\h\h\h\h\h\hfails.

338
00:22:21,874 --> 00:22:28,214
\h\h\hOnce again, I‘ll weigh the
bucket, and that weight is 10.5

339
00:22:28,247 --> 00:22:29,482
pounds.

340
00:22:29,515 --> 00:22:32,885
And after multiplying this by
our mechanical advantage of 6,

341
00:22:32,918 --> 00:22:35,621
we get a compressive strength of
63 pounds.

342
00:22:35,654 --> 00:22:40,292
\hNot bad, but somewhat short of
the 75 pounds we need for a safe

343
00:22:40,326 --> 00:22:42,328
design.

344
00:22:45,130 --> 00:22:48,166
Well, if at first you don‘t
succeed, try a larger size!

345
00:22:48,200 --> 00:22:56,508
\hThis one, this test specimen,
was formed around a 3/4”-square

346
00:22:56,542 --> 00:23:01,180
metal core, which now produces
\h\h\han outside dimension of

347
00:23:01,213 --> 00:23:04,116
approximately 1” square.

348
00:23:04,149 --> 00:23:07,719
\hAs you see, I‘ve already taped
the ends so we can very quickly

349
00:23:07,753 --> 00:23:10,356
\hmount the specimen in the
testing machine and run our

350
00:23:10,389 --> 00:23:14,193
test. And there‘s a failure.

351
00:23:37,516 --> 00:23:40,953
\h\hThis time when I weigh the
bucket, I get 14 pounds, which

352
00:23:40,986 --> 00:23:43,422
\hcorresponds to a very
satisfactory compressive

353
00:23:43,455 --> 00:23:45,424
strength of 84 pounds.

354
00:23:45,457 --> 00:23:49,695
\hI should add, however, that
tests of compressive strength

355
00:23:49,728 --> 00:23:54,166
are notoriously variable, far
more so than tensile tests, so

356
00:23:54,199 --> 00:23:57,569
it would be unwise to make any
\hdefinitive design decisions

357
00:23:57,603 --> 00:24:00,039
based on a single test.

358
00:24:00,072 --> 00:24:03,943
\h\hFortunately, I ran multiple
iterations of this test prior to

359
00:24:03,976 --> 00:24:08,314
this lecture and got reasonably
consistent results, all above 75

360
00:24:08,347 --> 00:24:12,418
\hpounds, so we can be confident
that this member size will work

361
00:24:12,451 --> 00:24:14,820
for our design.

362
00:24:14,853 --> 00:24:18,323
\h\h\hBefore continuing with the
design process, we need to pause

363
00:24:18,357 --> 00:24:22,161
and think ahead, to envision the
configuration of the actual

364
00:24:22,194 --> 00:24:26,365
three-dimensional structural
system we‘re going to build.

365
00:24:26,398 --> 00:24:29,568
\h\h\h\h\hWe‘ll be using this
corrugated-cardboard tube for

366
00:24:29,601 --> 00:24:34,739
\hour columns, this strip of
file-folder material for our

367
00:24:34,773 --> 00:24:38,277
\h\h\h\h\h\h\hdiagonals, and a
yet-to-be-determined tube for

368
00:24:38,310 --> 00:24:39,578
our beams.

369
00:24:39,611 --> 00:24:43,081
But how are we going to connect
them all together in a way that

370
00:24:43,115 --> 00:24:47,386
transmits internal forces safely
from one member to the next?

371
00:24:47,419 --> 00:24:50,689
To answer to this question,
let‘s consult with the real

372
00:24:50,723 --> 00:24:53,926
world of structural engineering,
where we‘ll find this: the

373
00:24:53,959 --> 00:24:57,863
gusset-plate connection, one of
the most common connection types

374
00:24:57,896 --> 00:25:00,031
used in modern steel-frame
\h\h\h\h\h\h\hstructures.

375
00:25:00,065 --> 00:25:05,604
Its key elements are the gusset
plates, here, which ties all of

376
00:25:05,637 --> 00:25:07,672
the other members together.

377
00:25:07,706 --> 00:25:10,275
This drawing shows how we‘ll
\h\h\himplement gusset-plate

378
00:25:10,309 --> 00:25:13,179
connections in our very own
\h\h\h\h\h\hcardboard frame.

379
00:25:13,212 --> 00:25:16,148
Note three key characteristics
\h\h\h\h\h\hof this connection

380
00:25:16,181 --> 00:25:18,750
configuration.

381
00:25:18,784 --> 00:25:21,987
First, to ensure the structural
integrity of our main

382
00:25:22,021 --> 00:25:25,458
\h\hload-carrying frames, each
connection is composed of two

383
00:25:25,491 --> 00:25:28,861
gusset plates, one on either
\h\h\h\hside of the column.

384
00:25:28,894 --> 00:25:32,865
Second, to maintain the symmetry
of the frame, the diagonals are

385
00:25:32,898 --> 00:25:37,102
\hconfigured in parallel pairs,
with one attached to each gusset

386
00:25:37,136 --> 00:25:38,404
plate.

387
00:25:38,437 --> 00:25:42,274
Thus, while our design actually
calls for one 5/8”-wide strip of

388
00:25:42,307 --> 00:25:46,945
\hfile-folder cardboard for each
diagonal, we‘ll actually use two

389
00:25:46,979 --> 00:25:51,283
5/16” strips to provide the
required symmetry while also

390
00:25:51,316 --> 00:25:54,319
maintaining the required
\h\h\h\h\h\h\hstrength.

391
00:25:54,353 --> 00:25:58,490
And third, to incorporate a beam
into this connection, the beam

392
00:25:58,524 --> 00:26:02,328
width must be the same as the
column width, so the beam will

393
00:26:02,361 --> 00:26:05,364
fit snugly between the two
\h\h\h\h\h\hgusset plates.

394
00:26:05,397 --> 00:26:07,599
So let‘s design the beams!

395
00:26:07,633 --> 00:26:10,903
Based on our structural analysis
results, these members need to

396
00:26:10,936 --> 00:26:14,640
carry an internal force of only
five pounds in compression.

397
00:26:14,673 --> 00:26:17,943
\hSo after accounting for our
factor of safety, the required

398
00:26:17,976 --> 00:26:20,645
compressive strength is only
\h\h\h\h\habout nine pounds.

399
00:26:20,679 --> 00:26:25,484
\hIt would be extremely wasteful
to use the same 1”-square tubes

400
00:26:25,517 --> 00:26:28,020
we‘re using for the columns,
\h\hwhich have a compressive

401
00:26:28,053 --> 00:26:31,423
strength eight times higher than
this requirement.

402
00:26:31,457 --> 00:26:38,030
Instead, we‘ll use these smaller
rectangular cross-sections,

403
00:26:38,063 --> 00:26:45,337
which was formed around a 1/4” x
3/4” core, so its width matches

404
00:26:45,370 --> 00:26:47,138
that of the column.

405
00:26:47,172 --> 00:26:50,175
\h\h\hI‘ve already tested its
compressive strength, and the

406
00:26:50,209 --> 00:26:54,313
result is 25 pounds, more than
adequate for the requirement.

407
00:26:57,749 --> 00:27:00,652
\hHaving selected the required
sizes for our three principal

408
00:27:00,686 --> 00:27:04,690
\h\hstructural members, columns,
beams, and diagonals, we‘re now

409
00:27:04,723 --> 00:27:08,527
ready to finalize the design and
communicate it with a 3D

410
00:27:08,560 --> 00:27:10,896
computer model in SketchUp.

411
00:27:10,929 --> 00:27:13,665
There are two important aspects
of this design that we haven‘t

412
00:27:13,699 --> 00:27:15,367
yet discussed.

413
00:27:15,400 --> 00:27:19,905
First, note that we‘ve added a
second set of diagonals within

414
00:27:19,938 --> 00:27:23,942
each panel of the main frames,
such that the paired diagonals

415
00:27:23,976 --> 00:27:27,012
now form an X.

416
00:27:27,045 --> 00:27:30,348
\h\hTo understand the purpose of
these additional members, let‘s

417
00:27:30,382 --> 00:27:34,519
revisit the simple braced-frame
model we used last lecture, but

418
00:27:34,553 --> 00:27:37,856
now I‘ve replaced the wooden
\hdiagonal with one made of

419
00:27:37,890 --> 00:27:39,925
file-folder cardboard.

420
00:27:39,958 --> 00:27:42,394
\hRecall that the principal
purpose of this member is to

421
00:27:42,427 --> 00:27:47,132
carry lateral loads, like this
\h\hand like wind applied to a

422
00:27:47,166 --> 00:27:48,901
real-world structure.

423
00:27:48,934 --> 00:27:51,537
And when the wind blows in this
direction, the diagonal works

424
00:27:51,570 --> 00:27:53,238
quite nicely.

425
00:27:53,272 --> 00:27:55,474
But when the wind blows the
\h\h\h\h\h\h\h\h\hother way?

426
00:27:55,507 --> 00:27:58,310
Here the diagonal is compressed,
and it buckles with the

427
00:27:58,343 --> 00:28:00,712
slightest application of load.

428
00:28:00,746 --> 00:28:04,583
\hSadly, we don‘t get to choose
which way the wind blows, so our

429
00:28:04,616 --> 00:28:07,786
structure must carry load both
\h\h\h\h\h\h\h\h\h\h\h\hways.

430
00:28:07,819 --> 00:28:11,923
We could address this challenge
by redesigning the diagonal as a

431
00:28:11,957 --> 00:28:16,094
\h\htube, so it can carry both
tension and compression, but a

432
00:28:16,128 --> 00:28:19,898
\h\h\hsimpler and more elegant
solution is to add this second

433
00:28:19,932 --> 00:28:25,171
diagonal, like this.

434
00:28:25,204 --> 00:28:29,542
\h\hNow, whichever way the wind
blows, one of the two diagonals

435
00:28:29,575 --> 00:28:34,680
carries the load in tension,
while the other goes slack.

436
00:28:34,713 --> 00:28:38,150
\h\h\h\hThis system of crossed
tension-only diagonals, called

437
00:28:38,183 --> 00:28:41,753
X-bracing, is quite common in
real-world structures, as you

438
00:28:41,787 --> 00:28:44,189
can see in this example.

439
00:28:44,223 --> 00:28:47,226
\hThe second aspect of our final
design that warrants additional

440
00:28:47,259 --> 00:28:52,331
discussion is the use both beams
and diagonals to connect the two

441
00:28:52,364 --> 00:28:54,333
frames together.

442
00:28:54,366 --> 00:28:58,170
\hThe principal purpose of these
members is to provide stability

443
00:28:58,203 --> 00:29:01,339
perpendicular to the plane of
\h\h\h\h\h\h\hthe main frames.

444
00:29:01,373 --> 00:29:04,076
If we didn‘t use diagonals here,
the entire frame would be

445
00:29:04,109 --> 00:29:08,013
susceptible to toppling sideways
under gravity loads, just like a

446
00:29:08,046 --> 00:29:09,514
house of cards.

447
00:29:09,548 --> 00:29:13,118
To facilitate construction, I‘ve
translated this 3D computer

448
00:29:13,151 --> 00:29:15,954
model into a set of full-size
drawings, which are available

449
00:29:15,988 --> 00:29:18,190
in pdf format in your Course
\h\h\h\h\h\h\h\h\h\h\hGuide.

450
00:29:18,223 --> 00:29:20,892
You can have these printed at a
commercial printing

451
00:29:20,926 --> 00:29:23,629
establishment for just a few
\h\h\h\h\h\h\h\h\h\h\hbucks.

452
00:29:23,662 --> 00:29:26,932
So, armed with our drawings,
we‘re ready to head for the

453
00:29:26,965 --> 00:29:28,800
workshop.

454
00:29:28,834 --> 00:29:31,103
\h\h\h\h\h\h\hAs a source of
corrugated-cardboard for the

455
00:29:31,136 --> 00:29:36,608
\hproject, I used an 18” moving
box, so each of the 36” columns

456
00:29:36,642 --> 00:29:40,679
could be fabricated in just two
18” halves.

457
00:29:40,712 --> 00:29:44,149
\h\hThe other required supplies
include legal-size file-folders,

458
00:29:44,182 --> 00:29:48,486
\h\hwood glue, wax paper, sewing
pins, a wooden base measuring at

459
00:29:48,520 --> 00:29:53,025
\hleast 14” square, the metal or
wooden cores around which we‘ll

460
00:29:53,058 --> 00:29:56,395
form our tubular compression
\h\hmembers, and this simple

461
00:29:56,428 --> 00:29:59,698
U-shaped jig, made from scrap
\h\h\h\h\h\h\h\h\h\h\hlumber.

462
00:29:59,731 --> 00:30:03,068
Let‘s begin by prefabricating
\h\h\h\hthe eight 3/4”-square

463
00:30:03,101 --> 00:30:06,571
corrugated-cardboard tubes we‘ll
be using for the columns of our

464
00:30:06,605 --> 00:30:07,840
tower.

465
00:30:07,873 --> 00:30:11,009
And, as I noted earlier, this
procedure also applies to the

466
00:30:11,043 --> 00:30:14,046
test specimens we just used for
our compressive strength

467
00:30:14,079 --> 00:30:15,581
experiments.

468
00:30:15,614 --> 00:30:19,184
For each tube, start by cutting
out a cardboard rectangle

469
00:30:19,217 --> 00:30:23,188
measuring 18” by 5”, with the
\hcorrugations running in the

470
00:30:23,221 --> 00:30:25,390
longer direction.

471
00:30:25,424 --> 00:30:28,694
\hNow measure and mark the
locations of the four fold

472
00:30:28,727 --> 00:30:31,997
\hlines, using the dimensions
provided in your Course Guide,

473
00:30:32,030 --> 00:30:35,934
and use a ball-point pen to
crease each fold line, like

474
00:30:35,967 --> 00:30:39,337
this.

475
00:30:39,371 --> 00:30:43,508
\h\hNext, carefully fold the
rectangle inward along each

476
00:30:43,542 --> 00:30:47,546
\h\h\hcrease, wrap the cardboard
around the 3/4”-square aluminum

477
00:30:47,579 --> 00:30:52,250
core, and slide it into the
\h\h\h\h\h\h\hU-shaped jig.

478
00:30:52,284 --> 00:30:56,355
\hRun a bead of glue along the
inside of the outer flap, then

479
00:30:56,388 --> 00:31:00,359
close the overlap, place another
piece of wood on top, and clamp

480
00:31:00,392 --> 00:31:04,129
it in position until the glue
\h\h\h\h\h\h\h\h\h\h\h\hsets.

481
00:31:04,162 --> 00:31:07,332
After about five minutes, remove
the cardboard and core from the

482
00:31:07,366 --> 00:31:12,104
\hjig, slide out the core, and
you‘ve got a completed column!

483
00:31:12,137 --> 00:31:15,340
That‘s two; only six more to go.

484
00:31:15,374 --> 00:31:18,744
\h\h\hNext, the 16 beams are
prefabricated in exactly the

485
00:31:18,777 --> 00:31:23,782
same way, except they use a
smaller 1/4” x 3/4” piece of

486
00:31:23,815 --> 00:31:28,687
wood or metal for the core, and
they‘re only 8 3/4” long.

487
00:31:28,720 --> 00:31:35,093
\hTo make the 48 diagonals, draw
parallel lines 5/16” apart on a

488
00:31:35,127 --> 00:31:39,965
legal-size file-folder material,
and then slice along each line

489
00:31:39,998 --> 00:31:44,736
carefully with a hobby knife or
razor blade and a straightedge.

490
00:31:44,770 --> 00:31:48,440
\hTo fabricate the gusset plates
and connecting plates, download

491
00:31:48,473 --> 00:31:52,644
\hand print the templates, use
spray adhesive to attach these

492
00:31:52,677 --> 00:31:57,015
\h\h\h\h\htemplates to sheets of
cardboard, and then cut ´em out

493
00:31:57,048 --> 00:31:59,183
with a knife or scissors.

494
00:31:59,217 --> 00:32:03,822
\h\h\h\h\hNow glue one of these
rectangular connecting plates to

495
00:32:03,855 --> 00:32:09,761
each end of each beam, as shown
here, and set them aside to dry.

496
00:32:09,795 --> 00:32:10,996
Okay.

497
00:32:11,029 --> 00:32:13,665
\h\h\hWith all of our structural
components prefabricated, we‘re

498
00:32:13,698 --> 00:32:16,568
now ready to put it all
\h\h\h\h\h\h\htogether.

499
00:32:16,601 --> 00:32:20,472
\h\h\hWe‘ll construct the frames
directly on this building-board,

500
00:32:20,505 --> 00:32:23,742
a sheet of particle board with a
thin layer of cork glued to its

501
00:32:23,775 --> 00:32:25,477
upper surface.

502
00:32:25,510 --> 00:32:28,947
Place the full-size plans onto
the board, followed by a sheet

503
00:32:28,980 --> 00:32:31,716
of wax paper, to ensure that we
don‘t accidentally glue any

504
00:32:31,750 --> 00:32:34,319
cardboard components to the
\h\h\h\h\h\h\h\h\h\h\hplan.

505
00:32:34,352 --> 00:32:38,122
Next, pin the gusset plates in
their proper positions on both

506
00:32:38,156 --> 00:32:40,158
frame drawings.

507
00:32:41,560 --> 00:32:45,497
Then glue the diagonals to the
gussets, using a pin to anchor

508
00:32:45,530 --> 00:32:48,033
each end of each diagonal.

509
00:32:48,066 --> 00:32:51,770
On one of the two frames, glue
\h\hthe columns to the gusset

510
00:32:51,803 --> 00:32:55,974
plates, with the overlapped side
of each tube facing inward.

511
00:32:56,007 --> 00:33:00,411
And glue the beams to both the
\h\h\h\h\hgussets and columns.

512
00:33:00,445 --> 00:33:02,847
Place a weight over each joint
\hto hold everything together

513
00:33:02,881 --> 00:33:05,851
while the glue sets.

514
00:33:05,884 --> 00:33:10,355
Finally, remove this frame from
the board, apply glue at the

515
00:33:10,388 --> 00:33:15,326
gusset-plate locations, flip it
around, position it on the

516
00:33:15,360 --> 00:33:19,097
adjacent subassembly of gusset
\h\hplates and diagonals, and

517
00:33:19,130 --> 00:33:21,132
replace the weights.

518
00:33:22,801 --> 00:33:26,471
Once the glue is completely dry,
remove this completed frame from

519
00:33:26,505 --> 00:33:30,542
the board and build another one
just like it.

520
00:33:30,575 --> 00:33:34,813
Now, using the same procedure,
build the two subassemblies of

521
00:33:34,846 --> 00:33:39,284
gusset plates, transverse beams,
and diagonals over the same set

522
00:33:39,317 --> 00:33:42,020
of plans.

523
00:33:42,053 --> 00:33:45,223
Once they‘re dry, glue the two
main frames onto one of these

524
00:33:45,257 --> 00:33:50,529
\hsubassemblies, and then glue
this entire assembly onto the

525
00:33:50,562 --> 00:33:52,564
other one.

526
00:33:53,798 --> 00:33:57,035
\h\hAt this point, our awesome
cardboard tower is essentially

527
00:33:57,068 --> 00:33:59,871
complete, and now we‘re ready
\h\h\h\h\h\h\h\h\hfor testing!

528
00:33:59,905 --> 00:34:03,709
Our base is clamped to the
table, the structure is in

529
00:34:03,742 --> 00:34:07,746
position, and I‘ve threaded this
loop of nylon string through the

530
00:34:07,779 --> 00:34:11,049
\huppermost gusset plates, to
facilitate the application of

531
00:34:11,082 --> 00:34:15,720
lateral load with my luggage
\h\h\h\h\h\h\h\h\h\h\hscale.

532
00:34:15,754 --> 00:34:22,461
\h\hBut first, I‘ll apply the
gravity load, and I‘ll do that

533
00:34:22,494 --> 00:34:24,496
as quickly as possible.

534
00:34:28,600 --> 00:34:32,437
\hOkay, the gravity load is in
place, and its currently being

535
00:34:32,470 --> 00:34:34,539
carried entirely by the columns.

536
00:34:34,573 --> 00:34:38,343
But now, as I add the lateral
\h\hload, you‘ll see that the

537
00:34:38,376 --> 00:34:43,047
tension diagonals and beams are
also coming into play, while the

538
00:34:43,081 --> 00:34:45,517
compressive force in the
downwind columns is also

539
00:34:45,550 --> 00:34:49,087
increasing significantly.

540
00:34:49,120 --> 00:34:53,891
I‘ve now reached 10 pounds, and
so we can breathe a sigh of

541
00:34:53,925 --> 00:34:58,396
\hrelief and proclaim our first
structural engineering project a

542
00:34:58,430 --> 00:35:00,699
success!

543
00:35:00,732 --> 00:35:03,935
To wrap up, let‘s take a moment
to review all we‘ve done to

544
00:35:03,969 --> 00:35:05,971
implement this project.

545
00:35:06,004 --> 00:35:09,741
We created a concept design for
a braced frame, performed a

546
00:35:09,774 --> 00:35:13,144
structural analysis to determine
the internal forces in the frame

547
00:35:13,178 --> 00:35:16,715
members, performed experiments
\hto determine the strength of

548
00:35:16,748 --> 00:35:20,218
\h\hcardboard members in both
tension and compression, used

549
00:35:20,251 --> 00:35:24,188
\h\hour analysis results and
experimental data to select

550
00:35:24,222 --> 00:35:28,626
member sizes capable of carrying
their respective internal forces

551
00:35:28,660 --> 00:35:33,665
safely, and built this structure
using a very unconventional

552
00:35:33,698 --> 00:35:37,235
construction material, yet
\h\hreplicating real-world

553
00:35:37,268 --> 00:35:40,972
structural details with
reasonable authenticity.

554
00:35:41,006 --> 00:35:44,543
\h\hNext lecture, we‘ll use this
same general approach to design

555
00:35:44,576 --> 00:35:48,480
\ha very different sort of
structure, a beam bridge!