Bending of Beams  Easy Approach 

Link to: Bending of Beams  Full Theory 

Getting Acquainted  
Calculating the bending of beams was
a timehonored centerpiece in the early teaching of "technical
mechanics", something every budding engineer and scientist had to go
through as a kind of initiation rite. Many didn't make it. Mastering
beambending is just not something done on the back of an envelope, while
having a beer or two. You really need to apply yourself to pass those exams.
Let's first look at the enormity of the task and then cut it down to pieces we can handle. The figure below shows what it is all about under rather simple circumstances. We have what the illuminati call a "beam", solidly fixed at one end. It just as well could be a sword seen edgeon. Some forces act on our beam; as a result it bends. Let's look at the major points connected to this one by one: 



First, let's define a bit more precisely what I mean
with the word "beam".
"Beam" shall be the generic name for any longish object with some
crosssection that is the same everywhere along its length l.
Moreover, the length l must be considerably larger than the
dimensions of the crosssection. Since the crosssection of most swords changes with the length, those swords are not proper beams but at best approximate beams. 

Second, let's define "bending". If you bend something, in particular a beam, you deform it mechanically. As far as the word bending is concerned here, we restrict ourselves to purely elastic deformations anywhere in the beam. Take off the forces acting on the beam above, and it must be perfectly straight again.  
Third, we
look at the forces that can bend a
beam. We allow all kinds of forces acting anywhere on the beam  as longs as
they act in the zdirection only. Otherwise we have two bending beam problems  bending in the
zdirection and in the
ydirection perpendicular to the drawing plane. Applying forces
in the xdirection would mean a tensile (or compressive) test, so
we don't allow them either. Of course, the forces acting on real beams or
swords do not care about those restrictions but I'm looking at simple beam
bending here. Forces could be point forces, like the force your sword experiences when it is hit with the blade of another sword, or distributed forces, as you get them when you put, for example, a weight on your beam as shown above. Our beam is supposed to bend but not to move otherwise. It will neither rotate nor is it going anywhere, and that means that the sum of all forces and moments must be zero by definition. That is obviously not the case in the picture above. Well, we forgot: 

Fourth, the
boundary conditions.
In the picture above, the beam is fixed solidly in some wall. Mathematically
that means the deflection
z(x), the amount of bending at the coordinate
x or simply the bending line, is zero for x=0. Expressed mathematically, we have as boundary condition: For the mathematically challenged: The derivative dz/dx gives the slope of the bending curve z(x), which must be zero (i.e. it is parallel to the xaxis) next to the wall These boundary conditions can only be met if forces and moments act on the beam at x=0 that exactly compensate the forces and moments of the external forces. That is quite nice because it means we don't have to worry about that. Simply speaking, if you immobilize one point of your sword somehow (by putting it into a vice) it can't move or rotate by definition, whatever forces act on it somewhere else. 

Fifth, and
not to forget, the beam can have some arbitrary
crosssection, including
hollow ones. The three examples shown are already rather special since they are highly symmetric. This is not required, nevertheless we will only consider crosssections with some symmetry from now on, in order not to boggle the mind too much. 

Now let's consider the task at hand.
We want to calculate Even if you are mathematically so challenged that you had to go into banking, you will appreciate that this cannot be an easy task. How do you express the crosssection No 2 above (a sword with a fuller) in equations? What would the boundary conditions be if you try hold the sword in a fixed position with your hands instead of putting it into a vice? If it is hit by another sword you cannot possibly keep the position of hand and sword completely unchanged  in contrast to a (very strong) vice. 

Well, the task is certainly not easy  but also not quite as difficult as it may appear. Let's do in in two steps and modules. First, in this module, I give a rather simple description of a special situation, which I then will use to derive a few general rules about bending beams. In a second module I will give a more general treatment.  
Guessing a Special Solution  
The good news are that I'm not a mathematician. If I would be one, I couldn't have drawn the picture above because it already contains a kind of solution to the problem at hand since it shows (by guessing) how the beam bends. Mathematicians only solve problems without any prior idea of the result and that means without cute little pictures. That's why they often don't actually solve real problems but only discuss in great detail if the problem might have a solution, many solutions  or possibly none.  
All of us
nonmathematicians, however, just know that
a beam bends pretty much as shown above for the given situation. We know even
more. We know that if we look at a small part of the beam, we can, at least to
a good approximation, assume that the bending can locally be described as being circular. The upper and lower surface of the beam then can be drawn by a compass using two radii that differ by the thickness d. Compared to the length of the undeformed beam, the upper part is now a bit longer, the lower part a bit shorter. We have looked at this before and there can be no doubt that we always have a neutral axis inside the beam, the length of which has not changed. In our simple example we actually have a neutral surface but for more complex crosssections sticking to a neutral axis (sometimes also called neutral line or centroid) is better since it is more general. This neutral axis (and here the whole neutral surface), also no doubt, will be in the middle of the rectangular beam as drawn below. 



Now, in a thought process, we
"cut out" the yellow part of the bend beam. To keep it in the shape
it had when it still was an integral part of the beam, we must attach forces.
The obvious thing to note is that only a distribution of forces as drawn in
blue on the right hand side, and the same thing mirrored at the other end (not
shown) can produce a deformation as shown. We can always replace the distribution of the blue
forces by two point forces F
by some smart averaging, and the result will be what is shown in
brown on the lefthand end of our beam part. Now we see that the total effect of the forces is to induce a torsional moment or a torque T with an guessed magnitude of T=½Fd since the distance between the force pair at a guess is d/2, or half the beam thickness as drawn. If you actually calculate the whole thing, the distance would be 2d/3. 

So far I'm just describing the
figure, generating first equations as I go along. If you keep looking at the figure, you will realize that there is more to see that can be described in simple equations. In particular, the maximum strain right at the top is e^{t}_{max}=d/2r; it is tensile. The maximum strain right at the bottom is e^{b}_{max}=–d/2r; it is compressive. 

Why? The picture shows it all. The
length of the utmost part is Equally of course, the strain decreases linearly from the maximum value on the outside to zero at the neutral plane as one moves inside the beam. 

On average, we guess, the upper or lower part thus experiences an average strain of e_{av}=±d/4r.  
As long as we only consider elastic deformations, strain e and stress s are related: s=Y · e; with Y=Young's modulus.  
The average stress going with our average strain is the average force F divided by half of the
cross sectional area bd. In other words: That gives us for the force 

It is now easy to realize that in the end it is the torque T=F · d/2 that causes the bending. So let's relate the torque T to the radius of curvature r. With the equation above we get as the final result:  


Deducing a Few Essentials  
We have a nice and simple equation.
We did a lot of guessing so we cannot expect that this equation is absolutely
correct. However, we didn't do anything outrageously wrong so it shouldn't be
completely wrong either but a halfway decent approximation. But what does the
equation mean? First, I'm going to tell you what it definitely means, despite the guessing. Then I'm going to discuss the shortcomings of this simple approach and how we can remedy this.. 

The essential deductions we can make are:




The figure illustrates a number of
things. First of all, it makes clear that for the point force shown, the local
bending is strongest around the clamping of the beam because there you have the
highest torque and thus smallest radius of curvature. The radius of curvature
thus changes smoothly along the length of the beam up to the location of the
force. To the right of the force there is no bending at all in this case
(meaning infinite radius of curvature). Note that we have not yet calculated the deflection curve but just seen how it
results from local curvatures. Without illustrating it directly, we can also infer a few other points: 

If we wouldn't have had a
point force but some distributed force, we still can calculate the torque
at any point by summing up or integrating over the force distribution. How that
is done is not important at this point, all we need to know is that it can be
done. If we would apply a second point force somewhere to the right of the one shown above, we simply sum up the moments of the forces. 

Isn't it amazing, how much one can deduce about a complex problem without actually solving it "from first principles"? Before I proceed to go beyond guessing, let's look at the problems, oversights and shortcomings of my "guessing" approach.  
The first point to make is that little cubes inside our
bend beam experience only uniaxial normal
stresses. Any
small cube
inside the beam is just getting a little longer in xdirection;
this is shown
here in a slightly different context. The strain increases linearly with
the distance to the neutral surface. This is an approximation. Real beams would experience lateral contraction in the upper half and a corresponding widening in the lower. The crosssectional shape would change from rectangular to trapezoidal. Moreover, the external forces also induce some shear components in the stress, which we neglected. It is clear that looking into that ain't going to be easy, so let me give you the good news right away: We don't need to do this. While treating beam bending with only uniaxial normal stresses is certainly not quite correct, it is good enough for pretty much all practical cases. 

The second point is more serious and will open a large
can of worms. In our guessing approach we have assumed a rectangular
crosssection, defined by the thickness d times the width or
"broadness" b. We have guessed that the neutral plane
is right in the middle. If we rewrite the final equation for the radius of
curvature as above, we can express it as
It should thus not be a big surprise that the correct value of that factor resulting from proper calculations is actually: (bd^{3} / 12). Well, that's just too bad. Nevertheless, the guessing approach wasn't all that bad and certainly worth doing for the simple rectangular crosssection. 

The real problem is that for
arbitrary crosssections, the guessing
approach doesn't work at all anymore for
deducing the value of the factor in front of the Y / T
that takes care of the beam geometry. While one could still make a good guess
about the position of the neutral surface for the
crosssections given above, you
would have a hard time for, e.g., triangular crosssections and many others.
Worse, you couldn't guess the averages easily either for obvious reasons. So you see the problem: 



Well, our guessing result gives a
clear hint: We can keep the general relation r µ (Y / T). All that changes is that
the proportionality constant (bd^{3}/16) in our
guess work) must be calculated from the real crosssection of the beam. That is actually not
very difficult to do (provided you know a bit about integrating in two
dimensions). The result is simply a number that is called the area moment of inertia I_{A}. If there ever was a misnomer, this is it. The area moment of inertia has nothing whatsoever to do with what is called "moment of inertia" for good reasons when you calculate how bodies rotate around some axis. The only connection between the two is that if you write down the integrals for calculating those moments, the mathematical expressions are similar. 

So all we need to do is to calculate the area moment of inertia I for the crosssection used and to rewrite our basic equation as  



That requires to calculate the position of the neutral axis or centroidfirst. It then will define the coordinate system since we center it on the neutral axis.  
Look at the rectangular crosssection in the picture below to get this: The red circle denotes the position of the neutral axis (it runs perpendicular to the picture plane in the xdirections, of course). Since bending about the zaxis (by applying a force in ydirection) will produce results different form bending about the yaxis (by applying a force in zdirection), we need to calculate twoarea moments of inertia, I_{y} and I_{y}, to take account of that.  
How the centroid and
area moments I_{A} are defined, and how they are
calculated, is the subject of another module. All we need to know here is that
if somebody has calculated it in the past, we don't have to do it ourselves. We
just can use the numbers generated. Here are a few examples; the centroid position is drawn as a red dot. 



The red dot denotes the neutral line. For the same Young's modulus Y, the area moments now give the resistance to bending in y or z direction (I_{z} or I_{y}, respectively. The inde denotes the axis around which the bending occurs). What you can appreciate right away is that a lightweight tube with a somewhat larger radius than a heavy solid circular rod can have the same area moment of inertia and thus resistance to bending. That's why you always have hollow beams (or the and double shaped ones) in any structural construction.  
Sword crosssections tend to be more complicated than the ones given above, but with the elliptical one we could at least make an educated guess for swords. That allows us to estimate the strain in the outermost layers as a function of bending and that gives us a last important insight:  
As soon as that strain e (or the stress related by s/e=Y) reaches the limit for either plastic deformation or fracture, it is all over. Since we must assume that somewhere along the blade there might be a "weak" spot, it is all over somewhat earlier than what we would calculate.  
Finally, I just tell you, that what we learned here can also be carried over to the "buckling of bemas" problem, or the effect that your sword might bend when you try to pierce straight into something. That will be dealt with in a special module  
Link to: Bending of Beams  Full Theory 

Bending of Beams  Full Theory
Science of Uniaxial Deformation
12.2. Static Sword Properties; 12.2.1 Sword Bending
12.2.2 Sword Bending: Deformation and Fracture
12.3 Dynamic Properties; 12.3.1 Moving a Stick
© H. Föll (Iron, Steel and Swords script)