Welcome back to statics Ege 2013 at Lawrence Tech.
I'm Jim, and as I have hinted about
in a couple previous videos we're going
to go from a loading, to the shear forces,
to the bending moments via integration
without having to cut sections and
things. We'll just doing this pretty much
mathematically. We are going to do two examples, one
would just be a cantilever beam with
the triangular loading on top of it and
in the other one we will add this strut
in here. We're kind of going to use this
airplane is kind of a model and we're assuming
a triangular lift distribution
that's not a very accurate assumption
but this is statics, not aerodynamics
Do I'm going to use a little bit of license
here and simplify the math to make it
easier to follow along -  and also easier
for me to do too. So if we just take this
in look at one wing we have our
simplified lift distribution which I'm
assuming be triangular and I have an
equation for that where this is the
total lift in pounds and b is the maximum
value of lift. And I picked a value of
108 for that because it's
basically just the number that works out
to give me a number - I think the total lift
came out to be 1298 pounds which is a
reasonable number for this particular
airplane operating under straight level
flight.  A is the wing semi span the
length of one wing which is 12 feet
X is a variable which starts at zero
at the wing root. And so we have an
equation for lift as a function of X. With all
the numbers that simplifies to just
simply 108-9*X
So here's a diagram of the forces on our
wing the lift is pointing up.
and it has a triangular distribution. 
There's a reaction force here at the
wing root points down to counteract. That
of course the moment to counteract the
bending moment caused by the lift. We
have the equation for our lift from before
We're saying that the shear force is the
integral of the load as a function X
plus some initial condition, and we're also saying
that the bending moment is the integral
of the shear as a function X and then some
other initial condition there which we
will find. Now what we could do is find
the initial conditions or find the
reaction forces  by solving
our load - you know coming up with the
equivalent load which is going to be
acting somewhere, well it's a triangle so
it's a third of the distance - acting there -
we could find the value of that load and that
would tell us the value of the reaction
force, that load times a distance tells
us the value that moment and it wouldn't
be hard to do because it's a simple
triangle. But I'm going to avoid that we're
just gonna run the math and see what
happens and see if we get the right
answers and if we had a more complicated
distribution that wasn't so easy to just
look up in the back of the book you know
we may not have that option of doing the
simple way so let's run the math. Now I
said that the sheer is the integral of the load
as a function X plus the initial
conditions ok so if I write that out i can say
that the shear load is equal to the
integral of 108-9 X dx
plus an initial condition and the initial
condition comes from the loading at X
equals zero if I do the integration I
get V equals 108 X minus 4.5 X
squared plus initial condition and I
know that the sheer load when I get all
the way out here to the end has to be 0
right because there's no forces beyond
that there's no nothing it's just
everything fades away into the sunset
you know so I know that if I evaluate
this at X equals 12 this should be
equal to 0, right? because at the end my
sheer is equal to 0 so I get 108 times
12-4.5 times 12 squared plus
the initial condition equals zero when I
run that all out I get my initial
condition equals -648 and the units are
pounds. I don't know that I mentioned
that before and if we had done the
triangle calculation we've got same
answer and so that tells us that if this
is 648 downwards
somewhere in here I have some equivalent
force equals 648 but in theory we don't know
where that is quite yet. and the other
thing we should do is be careful
with our signs. we have an upward force
that's going to result in our downward
reaction force that's a negative force
in our up-and-down convention. when we
look at the internal loads if we have a
section of the beam here the downward
force at the left there's some upward
force at the right side and some force
due to the lift... this is our sheer V. this
results in a counter clockwise rotation
here so our sheer force is negative at
the root and then gets less negative as
we go out. that gives us our final
equation for the sheer as a function of
x equals  -648 to give us that
negative number right at the root + 108
X minus 4.5 x squared. and I'll go to the
next slide there are the numbers we have
so far we have the equation for the
loading we integrated that we got this
equation raised here there's the graph
of it. It starts out negative
and ends up at zero and has a positive
slope everywhere so it's decreasing in
magnitude and I would point out that our
linear loading- it's a straight line here-
it's a linear first order equation - 
ended up with a second order sheer force
so given the sheer - what is our moment as
a function of X? once again our moment
is the integral of the sheer + initial
conditions so my moment is equal to the
integral of -648 plus 108 X minus
4.5 x squared dx + my
initial condition whatever that might be.
so if I do the integration i get i get
minus 648 X plus 54 X minus 1.5 X cubed
+ our initial condition and again we know that
at X equals 12 we know that the bending
moment has to be zero because again
there's nothing else at the end of that
being or cantilever. So if I evaluate this
at 12, I get minus 648 times 12 +54 times
12 squared minus 1.5 12 cubed equals zero
+ my initial conditions sorry forgot
that
and when I solve that out I get minus 2592
+ the initial condition equals zero
which gives us an initial condition
equal to 2592 pounds and that gives me
my final M of X - I'm going to put it up here that gives
you my moment as a function of x equals
2592 -648 X  + 54 X squared minus
1.5 X cubed. So having determined our
initial conditions and we found this
2592 number here we look at our beam and
at X equal to 0, I've got a twenty five
ninety-two foot pound moment and it's
acting in that direction because our our
forces are up, right? so I've got a reaction
force which has to be clockwise so
that's a negative 2592 and we can
fill that back in in our other diagrams
thing to look at is if we have a beam segment and
we have our reaction force at the end
and we have our bending moment here in
our section these moments will tend to
bend this up into a concave upwards or a
smiley face shape so we know we have a
positive moment you know at least at the
root and at x = 0 a positive number
here for our initial condition gives us
a positive results were confident there
are signs are just fine there
and finally just as a double check were
added we know that the moment is equal
to the equivalent force if we convert
that triangle to load into a single
equivalent force acting at some
equivalent distance ok so I can re-arrange
that and say that the distance is equal
to the moment we just said a moment at
the root there was 2592 divided by the
force, well the force is that load
we've applied, and I'm not too worried
about signs at the moment I'm looking
for magnitudes, from the other side that
was 648 and if we divide that out that
gives his distance of 4 feet given
this is pound-feet and this is pounds and if
we look at our triangular load
distribution you know this is 12 feet
there, the forces going to act at a
distance one third of the way so it's
going to act at 4 feet and that's
what we got. so you know it's always nice
to confirm our results. and the last
slide on this i've kind of put all the
diagrams together. The top one is our loading
we've got the reaction forces moment and
vertical load at the wing root
calculated we've got a equation for the
sheer force and we've we've got that
graphed out and the same thing for moment
we've got an equation and graphs and that's
pretty much
everything we needed to accomplish given simply this initial equation
the couple dimensions in going through
and integrating which was our plan for
the whole thing. So let's look at a more
complicated example when I throw that
strut on there a lot of things change.
first of all I have a force here which
is generated by the strut and
structurally
having a fixed and like I would with a
cantilever I have a a pin connection
here and there's a pin connection there
as well at
the strut. and that, by the way,
corresponds to reality in my
airplane there is a single bolt right
there at the root that holds the main
spar to the fuselage and there's a
single bolt right there at the
strut that holds the strut to a
bracket attached to the main spar so our
assumption is will have hinged joint so
there's no moments in these joints - that
reflects the reality - gonna have some
force here at the root in the y
direction some force at the root in the
x-direction
we're gonna have some force from the
strut. and the angle between the strut
and the wing is 31 degrees our loading
is the same equation we had before so...
this one is a bit messy and I think it's
worthwhile to find those reaction forces
up front. so we're going to assume that
there's some equivalent force that we
can calculate that we can use to
calculate our reactions I'm gonna go
through this pretty quickly because you
guys should have this down pretty
pat and if there is some point where I go too
fast just stop the tape and  go back and
figure it out. There is going to be a y component and an x
component to that force
the magnitude of that equivalent force F sub L
= one half our base of the triangle which is
twelve times the height of the triangle
which we said was108 and that's equal to
648 pounds which is the answer we
got before. the location of that load L sub X
is one third of the twelve feet,
that's equal to 4 feet which we got
before, so we find a force on the strut we
can do the sum of the moments about our
root = zero,  sum of the moments at the root
equals zero, and that's equal to the 648
pounds equivalent times the 4 feet
minus the force of the strut
F sub sy times five feet and that gives
us a
F sub sy equal to 518.4, from the
geometry we know that f sub s equals the
518.4 over the sine of 31 degrees
and that gives us one thousand and six
pounds in the strut and that is
going to point that direction this
points downward
and we can also determine
that the load in the strut in the x-direction is
equal to 1006 pounds time the
cosine of 31 degrees and that's equal to
862.3 pounds to the left and since that's
the only force with an x component it's
pointing to the left, 862.3 we can, by
inspection,  determine that this is 862.3
pounds pointing to the right
the sum of the forces Y equals 0, I've got
my lift of 648,  I've got the y
component of the force in the strut equal 518
- that doesn't look like a 5 ... 518.4
downwards minus F at the root
in the y direction and this gives us our 
Fry equal to 129.6 pounds
downward ok so those are reaction forces
129.6 is that one we've got
all of the forces X&Y  and we can
now do our analysis. for the normal
forces, again by inspection, we've got
that minus 862.3 and its
negative because it's in compression by
our our conventions for internal force
between the root and the strut and
beyond the strut it's just zero. so I
won't spend any more time on that
Let's work out the rest of our stuff. as
before the change in shear is a
function of the integral of the load and we get  delta sheer is an integral of load plus
we'll have some initial conditions. we
determined the lift is that equation
which will substitute into the load here the
forces at the root we had minus 129.6
in the y direction and minus 518.4 in the y direction
downward at the strut  location:
X equals five,  so we're kinda gonna have
to look at this maybe in two pieces
because there's something funny
happening in the middle. but if I do the...
we just start from the left and I do the
integral of my lift or load I get the
sheer I get my sheer V of x equals the integral of
108 - 9X dx plus our initial
conditions and that's equal to 108 X minus 4.5 X squared
plus our initial conditions. There is nothing
new here, same stuff here. I know at the root
X equals 0 my shear force from the
results from the reaction had x equals
zero shear force is equals minus 129.6
so I know that if I substitute 0 into
this equation I get my initial
conditions there and I get my shear
force V of X equals -129.6 + 108x
minus 4.5 x squared and that applies
when we're to the left of the strut
at X less than five now when we get to
that strut, there is, in theory, a point
source shear applied, actually it's applied
over small distance, but we'll assume for
convenience that it's a point force. so
at X equal to 5, I'm adding a minus 518.4
pounds downward. ok that''s this right here, so I'm going to have to add
that. that gives me a step change in my
sheer force that's going to be added as
you get beyond X equals 5. at X greater
than or equal to 5,  we're going to add that
518 to the negative 518 plus our
existing initial value here: minus 129.6
gives us a -648 and the
rest of the  equations is the same + 108 X - 4.5 X
squared and that's X greater than or equal to 5
so there is our shear; it's got that step
in it like we'd expect. and if I go to the
next slide we see the results here is
-129.6 this component here is
-518.4 and so have my
518 point four pounds just continuity in
the shared diagram I start out a that 
-129.6 right there
the shape of the curve in between is the
same as a shape in the previous example
of course we got that step in it. and
here is my equation written out neatly
for those who have troubles my
handwriting. Next we gotta find the bending
moment, right? so again the moment is the
integral of the shear plus our initial
conditions or we could say that the
change in the moment is the integral
this year over some specific - between two
specific points. We also know that at X = 0
lets go back a couple slides, 
actually one size enough. At this point
here there's no moment there; we said
that was a pin joint. Our moment at
X =0 is 0.Our  moment at X = 0 is 0, so that's
going to give us our initial condition of 0
at that point. So if we do the integration, and
I'm just going to skip the intermediate
steps I'm gonna have my moment equals
minus 129 X + 54 X squared minus 1.5 X
cubed plus our initial conditions: this is for X
less than five, and you've seen this
equation before. And we know that at
the moment at x equals 0 our moments
equals 0, so that initial conditions equals
0 and I can just take that away. Beyond 
that for X greater than or equal to 5;
we have the initial condition, whatever
it is, minus 648 X + 54 X squared
minus 1.5 X cubed For X greater or equal to 5. And what we
don't know is our initial condition; and
we can't look at X equals zero. We could
look at X equals 12 and say well at
X = 12 this whole equation has to be
equal to 0 or we can look at look at
what happens at five feet.  If I 
take my first equation, the top equation,
and evaluated at X equal to 5 I get 129 * 5
+ 54* 5 squared
minus 1.5 * 5 cuebed; and that gives
me 517.5 pound-feet moment at
X =five feet. and I know that
at X greater than five, less than
five, right around five, it's going
to be that value on both sides of that.
The application of force at a point does
not cause a step change in the magnitude
of the moment, it causes a discontinuity
in slope, but the magnitude
on either side of the that loading is
going to be the same. And if I
had applied a couple at that point, then I
get a step-change moment, but we don't have
any couples this problem. So I know that
if I take my second equation: that the
moment equals that 517.5, that's going
to be equal to my initial condition for
a second question,  minus 648 X plus 
648 * 5 + 54 times
times five squared minus 1.5 times 5
cubed. And run all that through, I get
my initial condition is equal to
2595.
And I'll write out the final equation on
the next slide, cuz I put it there anyhow
and rather than re-write it, you can read it.
So here are our equations for the moment,
here based those were based on
integrating the equations from the shear
and finding those initial conditions. And
we had kind of funny initial conditions
here at this point right here in the
middle, but you can see that we start
with the moment 0 we end with a moment
of zero, we had a pin at the left and we
just had the team ending with no forces
and no reactions on the right. There is
the discontinuity in a shared which
results in discontinuity in the slope.
We also note that where the sheer
crosses zero that's a local minimum in
our moment as we expect - the slope is equal to 0.
We have a maximum. And I think that's pretty much
all I need to say about this. There is one more
slide I wanted to touch on, and this is the
two systems compared. The top one is our
cantilever and the bottom one had the
strut. Obviously the shapes of these
curves have some changes but the things
we should probably notice here; for our
cantilever we had a minus 648,
was our maximum shear load, the maximum
shear load with the strut was 
297.3, so we reduced the maximum shear in
our thing. The moment here for our
cantilever was 2592 was a maximum value,
with the strut we only had 517.5 that's
almost an 80% reduction in the maximum
bending moment. So you can see why
putting a strut in this kind of
structure as attractive from
a structural perspective if you're
trying to keep weight down to support
support a particular load. In both
these cases we're supporting the same
airplane the same aerodynamic forces. But
we've got a lot lower maximum load here
when we put their strut in. So I think
that's everything I needed to show in
this example, hope you found it useful
and I'll see you next time. Bye.
