If we apply a load to a beam, it will deform
by bending. This generates internal stresses, which can
be represented by a shear force acting in the vertical direction, and a bending moment. The shear force is the resultant of vertical
shear stresses, which act parallel to the cross-section, and the bending moment is the
resultant of normal stresses, called bending stresses, which act perpendicular to the cross-section. It's important to have a good understanding
of these stresses because any design or analysis of a beam will involve calculating them. Let's look at bending stresses first. To keep things simple we'll consider a case
of pure bending. A section of a beam is said to be in a state
of pure bending when the shear force along it is equal to zero, and so there is a constant
bending moment along its length, like there is for this beam loaded by two moments. We also have a case of pure bending over the
middle section of this beam, where the bending moment is constant. Let's look at how a beam deflects when it
has a constant bending moment along its length. If we imagine the beam as a collection of
very small fibres, as the beam deflects the fibres at the top of the beam get shorter,
meaning that they are in compression. And those at the bottom of the beam get longer,
so they are in tension. Somewhere between the top and the bottom of
the cross-section there will be a surface containing fibres which stay the exact same
length. This is called the neutral surface. It passes through the centroid of the cross-section. When looking at the beam in two dimensions,
we refer to it as the neutral axis. Let's try and quantify the bending stresses
that develop within the beam to resist these applied moments. First let's calculate the strains in the beam. This can be done quite easily just by considering
the geometry of the deformation. Let's watch how a fibre at the neutral axis
between points A and B and a fibre between points C and D located at a distance Y from
the neutral axis deform. Since this is a case of pure bending, we can
see that the fibres bend into a perfectly circular arc. We'll call the centre of the circle O.
Before any deformation, the fibres are all the same length. After the deformation, the length of the neutral
axis has stayed the same, but the length of the fibre between points C and D has increased. If theta is the angle of the arc, and R is
the radius of the arc to the neutral axis, we can calculate the length of the arc between
A and B, like this. And we can calculate the length of the arc
between C and D in the same way. Strain is defined as the change in length
divided by the original length, and so we can derive an equation for bending strain
at any distance Y from the neutral axis. We defined the distance Y as being positive
downwards, and so this equation will give us a positive strain for the bottom of the
cross-section, which is in tension. Sometimes you'll see this equation written
with a minus sign, but that's because Y was defined as being positive upwards. If we assume that stresses remain within the
elastic region of the stress-strain curve, we can then apply Hooke's law for uniaxial
stress to calculate the bending stresses. This gives us the equation for bending stress
as a function of the radius of curvature R of the deformation. But what we're really interested in is how
the bending moment M affects the bending stress. If we make an imaginary cut through the beam
we can expose the internal bending stresses, represented here as a few discrete forces. The resultant moment of these internal forces
must be equal to the bending moment M, and so we can calculate M by integration, like
this. Now we can plug in the equation for bending
stress we just derived. When rearranged into this form, we can notice
that the integral on the right is the definition of the area moment of inertia. This parameter, which I've covered in detail
in a separate video, defines the resistance of a cross-section to bending due to its shape,
and is denoted using the letter I. We can combine this equation for the bending
moment with the bending stress equation to obtain what is known as the flexure formula. So what does it tell us? Bending stress increases linearly as the bending
moment and the distance from the neutral axis increase. And it decreases as the area moment of inertia
increases. The maximum stress occurs at the fibres furthest
from the neutral axis. The term I over Y-max depends only on the
geometry of the cross-section, and so it is called the section modulus, and is denoted
using the letter S. You will often see the section modulus listed for a range of common
beam cross-sections in reference texts. The I-beam is a commonly used cross-section
because it has a large area moment of inertia, which results in lower stresses. Here's how the bending stresses are distributed
over an I-beam cross-section. They are zero at the neutral axis, and reach
a maximum at the outside surfaces of the flanges. For a T-section the neutral axis is shifted
upwards, and so the bending stress distribution looks like this. So we've established how to calculate the
bending stresses, which are normal stresses, for a case of pure bending. Most of the time we won't have pure bending
as there will also be a shear force acting on the beam cross-section, like the beam we
saw at the start of the video. It turns out that the presence of a shear
force doesn't normally significantly affect the bending stresses, and so luckily we can
consider the flexure formula we derived earlier for pure bending to be valid for a more general
case of bending. The shear force V is the resultant of shear
stresses which act vertically, parallel to the cross-section. We denote the shear stresses using the Greek
letter Tau. To maintain equilibrium, these vertical shear
stresses have complementary horizontal shear stresses, which act between horizontal layers
of the beam. One way to visualise these horizontal stresses
is to consider a beam made up of several planks of wood. When a load is applied, there is a tendency
for the planks to slide relative to one another. Now let's glue the planks together. When the load is applied the planks cannot
slide, and so horizontal stresses develop between them. If these shear stresses are larger than the
shear strength of the glue bond, the glue will fail. These horizontal shear stresses don't exist
if we apply a moment instead of a force, because that gives us a state of pure bending. And so there is no
tendency for the planks to slide relative to one another. The presence of these horizontal shear stresses
explains why wooden beams sometimes fail by splitting longitudinally. This failure usually occurs close to the neutral
axis, for reasons which will soon be obvious. So how can we calculate the shear stresses? We can calculate the average shear stress
acting on the cross-section as the shear force V divided by the cross-sectional area. But the shear stresses aren't distributed
uniformly across the beam cross-section. The shear stress has to be zero at the free
surfaces at the top and bottom of the beam. So the average shear stress isn't very useful,
since it doesn't tell us the maximum shear stress. Instead, we can use this equation to calculate
the shear stresses acting on the cross-section. I won't cover the derivation of the equation
here, but it's based on considering equilibrium of stresses acting on a small element within
the beam. The equation assumes that the shear stress
is constant across the width B of the cross-section, so Tau is a function of the distance along
the beam, X, and the distance above the neutral axis, Y. V is the shear force acting on the cross-section,
which varies with the distance along the beam. B is the width of the cross-section. It can vary with the distance y from the neutral
axis, but in this case the cross-section is rectangular, so b is constant. I is the area moment of inertia, which is
a constant value calculated based on the shape of the cross-section. And Q is the first moment of area for the
portion of the cross-section above the location we want to calculate the shear stress for. So it varies with the distance Y above or
below the neutral axis. It is equal to the product of the area above
the location of interest and the distance between the centroid of that area and the
neutral axis. If the location of interest is below the neutral
axis, we consider the area below the axis, instead of the area above it. To calculate the first moment of area at this
line which is at a distance Y from the neutral axis, we multiply the area of the blue rectangle
above the line by the distance from the neutral axis to the centroid. Doing this calculation gives us an equation
for Q as a function of the distance Y from the neutral axis for a rectangular cross-section. And so we can obtain an equation which describes
how the shear stress varies with distance from the neutral axis. The Y term is squared, and so the shear stress
varies parabolically over the height of the cross-section, with the maximum shear stress
occurring at the neutral axis. This is opposite to the bending stress, which
is zero at the neutral axis, and explains why the horizontal shear failure of a wooden
beam we saw earlier occurs close to the neutral axis. By setting Y to zero in this equation, we
obtain an equation for the maximum shear stress in rectangular cross-sections. It is equal to 1.5 times the average shear
stress across the entire cross-section. The derivation of this equation for shear
stress makes a few assumptions, so we need to be careful with how we apply it. First, it assumes that the shear stresses
are constant across the width of the cross-section. For rectangular cross-sections this is a reasonable
assumption if the rectangle is thin. But for cross-sections like this one, the
shear stresses can vary significantly over the width, and so in these cases the equation
can really only give us the average shear stress across the width of the cross-section. It can't tell us what the maximum shear stress
will be. Another assumption this equation makes is
that the shear stresses are aligned with the Y axis. Shear stresses act tangentially at a free
surface, so for a circular cross-section, for example, we can't strictly use this equation
to get the distribution of shear stresses across the height of the cross-section. But we can still use it to estimate the shear
stresses at the neutral axis, because the shear stresses there are aligned with the
Y axis. The equation for shear stress at the neutral
axis in a circular cross-section is similar to the equation for a rectangular section,
where we have the average shear stress V over A, multiplied by a constant. The constant is 4 over 3 for a circular cross-section
and 3 over 2 for a rectangular one. We can also use the shear stress equation
for thin-walled sections like this I beam, although things are a bit more complicated. Because the vertical shear stresses at the
surfaces shown in red must be zero, and because the flanges are very wide, the vertical shear
stress in the flanges is very small. This means that the vertical shear stress
is distributed like this. The web mostly carries the shear force, and
the flanges mostly carry the bending moment, as we saw earlier. You can see that the shear stresses are distributed
quite evenly over the height of the web. This is because the flanges contribute significantly
to the first moment of area Q when calculating the shear stresses in the web, but they don't
carry much of the vertical shear force. Since the web is thin, the shear stresses
are also distributed evenly across its width. Because of this, we can easily calculate the
approximate shear stress in the web, like this. More detailed analysis reveals that there
are shear stresses in the flanges, but they are acting mainly in the horizontal direction. The horizontal stresses on both sides of the
flanges cancel each other out, so the net shear force is still just a vertical force. We can figure out the direction of the horizontal
shear stresses based on the direction of the vertical shear stresses by imagining that
the stresses are flowing through the cross-section. That's it for this review of bending and shear
stresses in beams. If you enjoyed the video and would like to
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