[music playing]
Welcome back to DIY Project #2, in which we‘re designing,
\hbuilding, and testing a cardboard tower capable of
supporting some very demanding \h\hgravity and lateral loads.
Last lecture, we defined the problem, formulated a design
\h\h\hconcept, and initiated the detailed design by modeling and
analyzing our structural system, called a braced frame, to
calculate this set of internal member forces, which will form
the basis for our final design.
\hOur principal design challenge in this lecture is to design the
members, such that the strength of each one exceeds its internal
force by a safe margin.
\hAs such, the next step in our process must be to figure out a
way to predict the strength of \h\hstructural members made of
cardboard.
If our tower were going to be \hmade of steel, concrete, or
commercial lumber, we could take advantage of published design
\h\h\hcodes which incorporate scientifically-based equations
that can be used to predict member strengths with great
confidence.
\hBut, alas, there are no design codes for cardboard structures,
so we‘ll need to develop our own program of DIY experimentation
to predict cardboard member \h\h\h\h\h\h\h\h\hstrengths.
Let‘s begin by delving more deeply into this concept of
strength.
\hIf I subject this structural member to gradually increasing
tension, it eventually breaks in two, a failure mode called
rupture.
\hThe applied force at which rupture occurs is called the
tensile strength of the member.
\hFor example, if I was pulling with five pounds at the instant
before the member failed, then \hits tensile strength is five
pounds.
\h\hIn general, tensile strength depends on only two factors: the
\h\h\htype of material and the member‘s cross-sectional area,
\h\hwhich corresponds to the rectangular area highlighted
here on this model.
Now, note that when I subject an identical member to compression,
rather than tension, it fails in a fundamentally different way.
\h\hWith relatively little compressive load, it fails
by kicking out sideways.
\h\hThis failure mode, called buckling, is much more complex
than tensile rupture.
\h\hThe buckling strength of a member depends not only on the
\h\h\h\h\htype of material and cross-sectional area, but also
the cross-section shape and, most importantly, the member
length.
Let‘s explore these influences \hby doing some more buckling
experiments.
Now here are three structural members: a solid square bar, a
solid flat bar that‘s much wider than it is thick,
and a hollow square tube.
They‘re all three feet long and all made of the same type of
wood.
Furthermore, they all use exactly the same amount of
material.
\hWe can prove this by weighing each one on my digital scale 32
grams, 32 grams, and 32 grams.
Now let‘s use this testing apparatus to determine the
compressive strength of each \h\h\h\h\h\h\h\h\h\hmember.
We‘ll start with the flat bar.
\hI‘ll load it into the machine, and now when I apply load to the
\h\htop of the machine, a single one-pound weight, is sufficient
to cause buckling, indicating a compressive strength of this
member is actually less than one pound.
Now let‘s try the square \h\h\h\h\h\h\h\hsection.
It‘s in position, and now I‘ll \h\h\h\h\h\h\happly one pound.
No problem, carries load just \h\h\h\h\h\h\h\h\h\h\h\hfine.
What about two?
Three, four, five, six, seven, \height, nine, ten, and on 11
pounds, we can just barely see \hthe onset of buckling here.
I won‘t add the 12.
11 pounds!
\h\h\hClearly, the compressive strength of this solid square
member is significantly greater.
Now, let‘s try the hollow box \h\h\h\h\h\h\h\h\h\h\h\hshape.
When we load the tube, well, we run out of weights long before
even approaching failure.
\h\hIndeed, the theoretical compressive strength of this
\h\hmember works out to be 86 pounds, a load I wouldn‘t even
be able to apply safely with \h\h\h\h\h\h\h\hthis device.
\h\h\h\hRemember that the only difference between these three
members we just tested is the \hshape of the cross-section;
otherwise, they use the same \h\h\h\hamount of material.
Clearly a very important factor in compressive strength.
\h\hIs it any wonder why hollow tubes like these are used in so
many different structural \h\h\h\h\h\happlications?
So, what about length?
\hTo answer this question, let‘s test a member that has the same
flat, rectangular cross-section as our first specimen, the one
\h\h\hthat failed at less than a pound, but only half the length.
The longer one buckled with the application of 1 pound, in fact
\h\hit had buckled at a load somewhat less than 1 pound.
And now, I‘ll load a column with an identical cross-section, but
half the length, with one pound, two, three, four, five, six, and
finally with seven pounds, once again we see just beginning to
occur, the onset of buckling for this member.
Substantially stronger.
Evidently, reducing the length \hof a compression member can
\hsignificantly increase its strength, particularly for a
slender column like this one.
Furthermore, the beneficial \h\h\heffect of shortening a
compression member can be achieved without actually
shortening.
Let‘s put our original 3-foot \hmember back in the machine,
but now I‘m going to add this lateral brace at mid-height of
the column.
And even though the brace only \h\h\hprevents the member from
displacing laterally at this one point, the member now needs one,
two, three, four, five, six, and again, seven pounds to cause
buckling.
Its compressive strength has been substantially increased
because the brace constrains it to buckle in an S-shape, rather
than in a single curve as \h\h\h\h\h\h\h\h\hbefore.
Evidently, adding intermediate \h\h\hbraces to a long column
increases compressive strength \h\h\hby reducing the column‘s
effective length to the distance between brace-points.
And this is why, in the concept design for our structure, we
subdivided each column into four segments by adding intermediate
beams.
Before we move on, I should add that if we had a machine capable
of loading these specimens to failure in tension, we‘d find
\hthat all four have a tensile strength of approximately 800
pounds!
That‘s nearly ten times the compressive strength of the
hollow tube, and over 100 times that of the solid square bar.
\hWe can conclude that tension members are almost always more
structurally efficient than \h\hcompression members, an
\hobservation that will also influence our final design.
At this point, we can combine \h\hour newfound knowledge of
tensile and compressive strength with the structural analysis
\h\hresults from our previous lecture to make some important
design decisions.
First, we know that the columns will experience compressive
forces in excess of 40 pounds, \hso we‘ll prefabricate these
members by forming corrugated \hcardboard into hollow square
cross-sections, like this, for optimum compressive strength.
\h\hSecond, the beams will experience a much smaller
compressive force of only five \h\h\h\hpounds, so we can use
significantly smaller tubes for these members.
And third, the diagonals will experience a tensile force of
seven pounds, which we can carry quite safely with strips of
file-folder material, like this.
\h\hBut what size should we make these members, such that they‘ll
be sufficiently strong to carry their respective internal
forces?
\h\h\hClearly we need to do some strength tests on cardboard, but
here we encounter a significant challenge.
As you‘ll see shortly, testing a member like this one to ensure
that it can carry 40 pounds safely will actually require
compressive loading somewhere in excess of 80 pounds.
\h\hAnd this testing device is wholly inadequate for the task
because there‘s simply no way to balance 80 pounds up on that
loading platform on top of the \h\h\h\h\h\h\h\h\h\h\hmachine.
Not to worry!
\h\hThis is really just another engineering design challenge, so
\h\h\hlet‘s solve it as any good engineer would, by figuring out
a clever way to do more with \h\h\h\h\h\h\h\h\h\h\hless.
\h\hWhat we need is a means of multiplying force such that we
put in, say, 10-15 pounds, and we get out a force five or six
times larger.
In engineering terms, what we need is mechanical advantage.
There are many ways to achieve \h\hmechanical advantage, but
perhaps the simplest is the \h\h\h\h\h\h\h\h\h\h\hlever.
\hAnd here is the lever-based machine we‘ll be using to test
the strength of cardboard structural members in both
tension and compression.
It consists of a fixed base, a \h\h\hfulcrum, which uses this
\hhalf-inch bolt as a pivot, and the lever-arm itself, loaded by
\hmeans of a bucket hanging from its far end and with provisions
for placing a test specimen on this side for tension testing,
\h\hand over on this side of the fulcrum for compression testing.
\hThe fulcrum itself has three sets of holes, so we can test
specimens with lengths from six inches up to 12 inches.
\h\h\h\hAnd this post keeps the lever-arm from drifting sideways
\h\h\h\h\hunder load while also incorporating a set of holes for
a safety pin, here, that can be used to support the lever-arm
temporarily, as it is right now.
If you‘d like to try some DIY materials testing of your own,
I‘ve provided drawings for this machine in your Course Guide.
Let‘s see how it works by \hrunning a tension test.
Here‘s our first test specimen, a piece of file-folder material
that‘s exactly 3/4” wide here in the center, but flares out to a
width of 2” at both ends.
\hThis characteristic shape, called a “dog bone,” is very
commonly used in real-world materials testing, as these
metal test specimens attest.
\h\h\h\hThe dog bone shape is advantageous because it causes
the specimen to fail here, in \hthe measured center section,
where we‘re interested in \h\h\h\h\hacquiring data.
Were it not for these wide ends, the clamps would cause premature
\hfailure at these locations and our data would be compromised as
a result.
Before I mount the specimen, note that I‘ve balanced the
\h\hlever-arm on its fulcrum by hanging these weights out on the
short end.
This procedure ensures that the weight of the arm itself doesn‘t
contribute to the load on the \h\h\h\h\h\h\h\h\h\hspecimen.
\h\h\hNow, we‘ll hang this empty bucket from the opposite end of
\hthe arm, clamp the specimen in position, and remove the safety
pin.
To initiate the test, slowly add sand to the bucket until the
specimen ruptures.
\hIt‘s a good clean break, so we‘ll now weigh the bucket and
and the result is 2.5 pounds.
But what does this number mean?
When a force of 2.5 pounds was \hpulling down here, how much
tension was being applied to the specimen here?
To answer this question, we need to introduce a new engineering
mechanics concept, the moment, \hdefined as the tendency of a
force to cause rotation about a point or axis.
\hThis is another concept we‘ll return to repeatedly throughout
the course, so please learn it \h\h\h\h\h\h\h\h\h\h\h\hwell!
Mathematically, the moment of a force about a point is equal to
\h\hthe magnitude of that force times the perpendicular distance
from the force to the point.
\hIn this diagram, the moment of Force F about Point A is F x d,
and its direction is clockwise.
\hLast lecture we learned that, when working in two dimensions,
the mathematical condition for equilibrium is that the sum of
all forces acting on a body, in both the x and y directions,
must equal zero.
\h\hBut there‘s actually a third equilibrium condition, which we
can now state: That the moments of all forces acting on the body
must also add to zero.
Therefore, if we call the weight of our bucket W, and the tension
in our test specimen T, then the Principle of Equilibrium
\hrequires that the clockwise moment of T about the fulcrum
must equal the counterclockwise moment of W about the fulcrum.
Each of these moments is the \hassociated force times its
distance from the fulcrum.
Therefore, TL1=WL2.
And if we divide both sides of \hthis equation by L1, we find
that the tension force is equal to the ratio L2-over-L1 times
the weight.
For this machine, L1 is 8” and \h\h\hL2 is 48”, so the ratio
L2-over-L1 is 48 ÷ 8 = 6.
This number is the mechanical advantage of the lever, and it
tells us that for every pound applied to the long end of the
lever, 6 pounds are applied to \h\h\h\h\h\hthe test specimen.
And by the way, this machine is configured such that its
\h\h\hmechanical advantage for compression testing is also 6.
Therefore, for the test we just completed, the tension in the
test specimen at the instant of failure was 2.5 pounds times the
mechanical advantage, 6, equals 15 pounds.
\h\hPrior to this lecture, I ran several iterations of this test
using 1/2”, 3/4”, and 1”-wide \h\h\h\h\h\h\h\h\h\hspecimens.
\hAnd the results are shown on this graph of tensile strength
vs. width.
This is very encouraging because it shows that the tensile
strength of file-folder \hcardboard varies very
predictably as a linear function of member width.
Given this predictability, we can confidently use this graph
\has a tool for designing the diagonal members of our braced
frame.
\h\h\hThe general mathematical requirement for the design of
\hany structural member is shown here: The actual condition must
\h\hbe less than or equal to the failure condition divided by the
factor of safety.
\h\h\h\hThe actual and failure conditions in this expression
\hcan be defined in a variety of different ways, depending on the
situation, but for this project, the actual condition will be
internal force, and the failure condition will be member
strength.
So, rearranging this expression algebraically, we arrive at an
\hinequality that will serve as the design basis for all of the
\h\hmembers in our tower, member strength must be greater than or
equal to internal force times \h\h\h\h\h\hfactor of safety.
In this expression, the factor of safety is simply a number,
\halways greater than 1, which provides a margin of error to
account for the many sources of uncertainty inherent in
structural design: unanticipated loads, natural variation in
material properties, fabrication errors, and so forth.
In engineering practice, factors of safety are normally specified
in structural design codes, and typically range between 1.5 and
2, with the actual number \h\h\h\hdetermined by such
considerations as the type of \hstructural member, material,
type of loading, and the consequences of failure.
\h\hFor our project, we‘ll use a factor of safety of 1.75, which
\h\h\h\h\hgives us a reasonably comfortable 75 percent margin of
error.
Okay, let‘s do some design!
\h\hRecall from the structural analysis of our tower that the
internal force in all diagonals in the structure is 6.6 pounds.
Substituting this force into our design basis and multiplying by
the safety factor, 1.75, we get a required tensile strength of
11.6 pounds.
\h\h\h\hNow we can return to our experimentally-derived graph of
tensile strength versus member width, start with the required
strength of 11.6 pounds, and \h\hread across and down to
\hdetermine the minimum member width required to attain this
much strength.
The result is 0.6”, which we‘ll round up to 5/8”, just for ease
of measurement.
Now let‘s apply this same \hprocess to the columns.
\hIn our structural analysis, we calculated a range of different
internal forces in the various \hcolumn segments, but for the
sake of simplicity, it‘s quite reasonable to use the absolute
largest force, 43 pounds, as the basis for selecting a single
member size that will be used \h\h\hfor all of our columns.
When we substitute this force and the safety factor of 1.75
\h\h\hinto the design basis, the result is a required compressive
strength of just under 75 \h\h\h\h\h\h\h\h\hpounds.
Our next challenge, then, is to determine what size
corrugated-cardboard tube will provide a compressive strength
of at least 75 pounds.
The only practical way to meet \hthis challenge is through a
series of trial-and-error \h\h\h\h\h\h\hexperiments.
We‘ll begin with this test \h\h\h\h\h\h\h\hspecimen.
This one, a hollow square tube \hthat I fabricated by forming
\h\hcardboard around this 1/2”-square aluminum core.
We‘ll see how this was done \h\h\h\h\h\h\h\h\h\hshortly.
Its nine-inch length corresponds to the lengths of all columns in
our braced frame.
Now, if we placed this specimen, without modification, into the
\h\h\h\htesting machine at its designated location and loaded
it, the ends of the tube would \h\h\h\halmost certainly crush
\h\h\h\h\h\h\hprematurely due to concentrations of internal force
here, at the interface between \hthe specimen and the testing
machine.
To prevent this problem from occurring, I have reinforced
both ends of an identical test \hspecimen with a few wraps of
duct tape, and we‘ll also insert these hardwood plugs into the
ends.
Now, once we put the specimen in place, each end-cap makes
\hcontact with the machine at a single point, so rotation of the
lever-arm during testing is less likely to cause localized
crushing at the ends of the \h\h\h\h\h\h\h\h\hspecimen.
With everything set, I‘ll \hrelocate the safety pin
\h\hdownward, so the bucket will only fall a few inches when the
specimen fails.
\hI know that this test is going to take a lot more load than the
\hprevious one we did, so I‘m going to begin the actual load
process by adding five of these iron weights to the bucket.
And now I‘ll continue the test by adding sand slowly as I did
before, until the specimen \h\h\h\h\h\h\h\h\h\hfails.
\h\h\hOnce again, I‘ll weigh the bucket, and that weight is 10.5
pounds.
And after multiplying this by our mechanical advantage of 6,
we get a compressive strength of 63 pounds.
\hNot bad, but somewhat short of the 75 pounds we need for a safe
design.
Well, if at first you don‘t succeed, try a larger size!
\hThis one, this test specimen, was formed around a 3/4”-square
metal core, which now produces \h\h\han outside dimension of
approximately 1” square.
\hAs you see, I‘ve already taped the ends so we can very quickly
\hmount the specimen in the testing machine and run our
test. And there‘s a failure.
\h\hThis time when I weigh the bucket, I get 14 pounds, which
\hcorresponds to a very satisfactory compressive
strength of 84 pounds.
\hI should add, however, that tests of compressive strength
are notoriously variable, far more so than tensile tests, so
it would be unwise to make any \hdefinitive design decisions
based on a single test.
\h\hFortunately, I ran multiple iterations of this test prior to
this lecture and got reasonably consistent results, all above 75
\hpounds, so we can be confident that this member size will work
for our design.
\h\h\hBefore continuing with the design process, we need to pause
and think ahead, to envision the configuration of the actual
three-dimensional structural system we‘re going to build.
\h\h\h\h\hWe‘ll be using this corrugated-cardboard tube for
\hour columns, this strip of file-folder material for our
\h\h\h\h\h\h\hdiagonals, and a yet-to-be-determined tube for
our beams.
But how are we going to connect them all together in a way that
transmits internal forces safely from one member to the next?
To answer to this question, let‘s consult with the real
world of structural engineering, where we‘ll find this: the
gusset-plate connection, one of the most common connection types
used in modern steel-frame \h\h\h\h\h\h\hstructures.
Its key elements are the gusset plates, here, which ties all of
the other members together.
This drawing shows how we‘ll \h\h\himplement gusset-plate
connections in our very own \h\h\h\h\h\hcardboard frame.
Note three key characteristics \h\h\h\h\h\hof this connection
configuration.
First, to ensure the structural integrity of our main
\h\hload-carrying frames, each connection is composed of two
gusset plates, one on either \h\h\h\hside of the column.
Second, to maintain the symmetry of the frame, the diagonals are
\hconfigured in parallel pairs, with one attached to each gusset
plate.
Thus, while our design actually calls for one 5/8”-wide strip of
\hfile-folder cardboard for each diagonal, we‘ll actually use two
5/16” strips to provide the required symmetry while also
maintaining the required \h\h\h\h\h\h\hstrength.
And third, to incorporate a beam into this connection, the beam
width must be the same as the column width, so the beam will
fit snugly between the two \h\h\h\h\h\hgusset plates.
So let‘s design the beams!
Based on our structural analysis results, these members need to
carry an internal force of only five pounds in compression.
\hSo after accounting for our factor of safety, the required
compressive strength is only \h\h\h\h\habout nine pounds.
\hIt would be extremely wasteful to use the same 1”-square tubes
we‘re using for the columns, \h\hwhich have a compressive
strength eight times higher than this requirement.
Instead, we‘ll use these smaller rectangular cross-sections,
which was formed around a 1/4” x 3/4” core, so its width matches
that of the column.
\h\h\hI‘ve already tested its compressive strength, and the
result is 25 pounds, more than adequate for the requirement.
\hHaving selected the required sizes for our three principal
\h\hstructural members, columns, beams, and diagonals, we‘re now
ready to finalize the design and communicate it with a 3D
computer model in SketchUp.
There are two important aspects of this design that we haven‘t
yet discussed.
First, note that we‘ve added a second set of diagonals within
each panel of the main frames, such that the paired diagonals
now form an X.
\h\hTo understand the purpose of these additional members, let‘s
revisit the simple braced-frame model we used last lecture, but
now I‘ve replaced the wooden \hdiagonal with one made of
file-folder cardboard.
\hRecall that the principal purpose of this member is to
carry lateral loads, like this \h\hand like wind applied to a
real-world structure.
And when the wind blows in this direction, the diagonal works
quite nicely.
But when the wind blows the \h\h\h\h\h\h\h\h\hother way?
Here the diagonal is compressed, and it buckles with the
slightest application of load.
\hSadly, we don‘t get to choose which way the wind blows, so our
structure must carry load both \h\h\h\h\h\h\h\h\h\h\h\hways.
We could address this challenge by redesigning the diagonal as a
\h\htube, so it can carry both tension and compression, but a
\h\h\hsimpler and more elegant solution is to add this second
diagonal, like this.
\h\hNow, whichever way the wind blows, one of the two diagonals
carries the load in tension, while the other goes slack.
\h\h\h\hThis system of crossed tension-only diagonals, called
X-bracing, is quite common in real-world structures, as you
can see in this example.
\hThe second aspect of our final design that warrants additional
discussion is the use both beams and diagonals to connect the two
frames together.
\hThe principal purpose of these members is to provide stability
perpendicular to the plane of \h\h\h\h\h\h\hthe main frames.
If we didn‘t use diagonals here, the entire frame would be
susceptible to toppling sideways under gravity loads, just like a
house of cards.
To facilitate construction, I‘ve translated this 3D computer
model into a set of full-size drawings, which are available
in pdf format in your Course \h\h\h\h\h\h\h\h\h\h\hGuide.
You can have these printed at a commercial printing
establishment for just a few \h\h\h\h\h\h\h\h\h\h\hbucks.
So, armed with our drawings, we‘re ready to head for the
workshop.
\h\h\h\h\h\h\hAs a source of corrugated-cardboard for the
\hproject, I used an 18” moving box, so each of the 36” columns
could be fabricated in just two 18” halves.
\h\hThe other required supplies include legal-size file-folders,
\h\hwood glue, wax paper, sewing pins, a wooden base measuring at
\hleast 14” square, the metal or wooden cores around which we‘ll
form our tubular compression \h\hmembers, and this simple
U-shaped jig, made from scrap \h\h\h\h\h\h\h\h\h\h\hlumber.
Let‘s begin by prefabricating \h\h\h\hthe eight 3/4”-square
corrugated-cardboard tubes we‘ll be using for the columns of our
tower.
And, as I noted earlier, this procedure also applies to the
test specimens we just used for our compressive strength
experiments.
For each tube, start by cutting out a cardboard rectangle
measuring 18” by 5”, with the \hcorrugations running in the
longer direction.
\hNow measure and mark the locations of the four fold
\hlines, using the dimensions provided in your Course Guide,
and use a ball-point pen to crease each fold line, like
this.
\h\hNext, carefully fold the rectangle inward along each
\h\h\hcrease, wrap the cardboard around the 3/4”-square aluminum
core, and slide it into the \h\h\h\h\h\h\hU-shaped jig.
\hRun a bead of glue along the inside of the outer flap, then
close the overlap, place another piece of wood on top, and clamp
it in position until the glue \h\h\h\h\h\h\h\h\h\h\h\hsets.
After about five minutes, remove the cardboard and core from the
\hjig, slide out the core, and you‘ve got a completed column!
That‘s two; only six more to go.
\h\h\hNext, the 16 beams are prefabricated in exactly the
same way, except they use a smaller 1/4” x 3/4” piece of
wood or metal for the core, and they‘re only 8 3/4” long.
\hTo make the 48 diagonals, draw parallel lines 5/16” apart on a
legal-size file-folder material, and then slice along each line
carefully with a hobby knife or razor blade and a straightedge.
\hTo fabricate the gusset plates and connecting plates, download
\hand print the templates, use spray adhesive to attach these
\h\h\h\h\htemplates to sheets of cardboard, and then cut ´em out
with a knife or scissors.
\h\h\h\h\hNow glue one of these rectangular connecting plates to
each end of each beam, as shown here, and set them aside to dry.
Okay.
\h\h\hWith all of our structural components prefabricated, we‘re
now ready to put it all \h\h\h\h\h\h\htogether.
\h\h\hWe‘ll construct the frames directly on this building-board,
a sheet of particle board with a thin layer of cork glued to its
upper surface.
Place the full-size plans onto the board, followed by a sheet
of wax paper, to ensure that we don‘t accidentally glue any
cardboard components to the \h\h\h\h\h\h\h\h\h\h\hplan.
Next, pin the gusset plates in their proper positions on both
frame drawings.
Then glue the diagonals to the gussets, using a pin to anchor
each end of each diagonal.
On one of the two frames, glue \h\hthe columns to the gusset
plates, with the overlapped side of each tube facing inward.
And glue the beams to both the \h\h\h\h\hgussets and columns.
Place a weight over each joint \hto hold everything together
while the glue sets.
Finally, remove this frame from the board, apply glue at the
gusset-plate locations, flip it around, position it on the
adjacent subassembly of gusset \h\hplates and diagonals, and
replace the weights.
Once the glue is completely dry, remove this completed frame from
the board and build another one just like it.
Now, using the same procedure, build the two subassemblies of
gusset plates, transverse beams, and diagonals over the same set
of plans.
Once they‘re dry, glue the two main frames onto one of these
\hsubassemblies, and then glue this entire assembly onto the
other one.
\h\hAt this point, our awesome cardboard tower is essentially
complete, and now we‘re ready \h\h\h\h\h\h\h\h\hfor testing!
Our base is clamped to the table, the structure is in
position, and I‘ve threaded this loop of nylon string through the
\huppermost gusset plates, to facilitate the application of
lateral load with my luggage \h\h\h\h\h\h\h\h\h\h\hscale.
\h\hBut first, I‘ll apply the gravity load, and I‘ll do that
as quickly as possible.
\hOkay, the gravity load is in place, and its currently being
carried entirely by the columns.
But now, as I add the lateral \h\hload, you‘ll see that the
tension diagonals and beams are also coming into play, while the
compressive force in the downwind columns is also
increasing significantly.
I‘ve now reached 10 pounds, and so we can breathe a sigh of
\hrelief and proclaim our first structural engineering project a
success!
To wrap up, let‘s take a moment to review all we‘ve done to
implement this project.
We created a concept design for a braced frame, performed a
structural analysis to determine the internal forces in the frame
members, performed experiments \hto determine the strength of
\h\hcardboard members in both tension and compression, used
\h\hour analysis results and experimental data to select
member sizes capable of carrying their respective internal forces
safely, and built this structure using a very unconventional
construction material, yet \h\hreplicating real-world
structural details with reasonable authenticity.
\h\hNext lecture, we‘ll use this same general approach to design
\ha very different sort of structure, a beam bridge!