Understanding Vibration and Resonance

Thanks to Curiosity Stream for sponsoring
this video. A good understanding of how structures behave
when vibrating is what allows engineers to build rotating machinery, to launch sensitive
instruments into space, and to safely design buildings in seismic areas, to name just a
few of the many applications. Systems like these can be very complex, so
to study their vibrating behaviour engineers usually start by building a simple model that
approximates the dynamics of the system, but is easier to assess. The two most important parameters in any vibrating
system are its mass and its stiffness. A common modelling approach is to lump all
of the different contributions to mass and stiffness together, and represent them using
a point mass with a mass m and a spring with a stiffness k. This is called the lumped parameter modelling
approach. This kind of simplified model might seem quite
abstract, but can actually represent the dynamic behaviour of a lot of real systems quite accurately. The beauty of this simplicity is that we now
have something that we can analyse mathematically.

But first we have to make a few assumptions. We'll assume that the mass can only move up
and down. Since the system behaviour is defined by a
single output, the x coordinate of the mass, this is what's called a single degree-of-freedom
model. We'll also neglect the effects of gravity,
and for now we'll assume that there's no damping, meaning that no energy is lost from the system
as it vibrates, by friction or other means.

No external loads are acting on the system
– the purpose of the model is to understand how the system behaves in free vibration,
or in other words how it will oscillate when it's displaced, and then released. Since we've assumed there's no damping, the
mass will continue to oscillate like this indefinitely. The way the system vibrates is defined by
its equation of motion, which can be determined by applying Newton's second law. The second law states that the sum of the
forces acting on the point mass is equal to the product of its mass and its acceleration,
F=ma. We can figure out the sum of the forces acting
on the mass by drawing a free body diagram.

There is only one force, the force exerted
by the spring, which is equal to the displacement x multiplied by the spring stiffness k. And
so we obtain the equation of motion for the system. This is an ordinary differential equation,
and the solution is a sinusoidal function. t is time, Phi is the phase angle and A is
the amplitude of vibration. We can determine A and Phi by considering
the initial position and velocity of the mass. Let's look at an example where the system
has a mass of 5 kilograms and a spring stiffness of 20 Newtons-per-metre, and vibration is
triggered by applying an upwards velocity of 2 centimetres per second to the mass. Since the displacement x is initially zero,
the phase angle Phi must also be equal to zero. And then we can differentiate the equation
for x to calculate the amplitude of vibration A. An important property that can be calculated
from a mass-spring model is the system's natural frequency, the frequency at which it will
oscillate naturally when in free vibration. It's given by this term in the solution to the equation of
motion, and is denoted using the Greek letter Omega.

It depends only on the mass and the spring
stiffness, so no matter what the initial conditions are a system will always oscillate at the
same frequency. It has units of radians per second, so is
called the angular natural frequency. But it's sometimes more practical to think
of the natural frequency as a number of cycles per second, in which case it's denoted using
the letter f and has units of Hertz. The inverse of the natural frequency is the
period T, which is the duration of each cycle in seconds. Let's compare how two different systems oscillate. Both of these models have the same spring
stiffness, but different masses, and so different natural frequencies. The heavier mass oscillates at a much lower
frequency. A neat demonstration of the natural frequency
is the tuning fork. When the fork is struck it vibrates at its
natural frequency, which is much faster than shown here.

This causes the air molecules to vibrate at
that same frequency, which produces a corresponding tone. By assuming that the prongs behave like cantilever
beams in bending, beam theory can be used to derive a formula for the natural frequency
of the fork. The density, length and cross-section of the
prongs can be calibrated to obtain the desired tone. Of course when a mass oscillates freely it
doesn't do so indefinitely. Energy within the system is dissipated as
heat over time, so the oscillations progressively decrease in magnitude and eventually stop
altogether. This loss of energy is called damping, and
it occurs in all real mechanical systems. There are several different mechanisms that
can contribute to the overall damping of a system.

With structural damping, energy in a vibrating
structure is dissipated due to the relative motion of components at structural joints. And material damping is damping provided by
the material itself, where energy dissipates in a vibrating material due to interactions
occurring at the molecular level. To improve our spring-mass model, we can lump
the damping from all of the different sources together, and represent them by a single dampening
device called a dashpot, which is essentially a plunger that moves through a liquid-filled
cylinder. Whenever the plunger moves a force will act
to oppose its displacement, and the magnitude of this damping force is proportional to the
velocity of the displacement – the faster the plunger moves, the larger the damping
force. C is the viscous damping coefficient – it
defines the total amount of damping in the system. This model of damping is called viscous damping,
because it behaves in a similar way to viscous forces in a fluid, which are proportional
to the fluid velocity. There are other damping models, but viscous
damping is commonly used because of its simplicity.

If we include the dashpot in our spring-mass
model, the equation of motion is the same as for the undamped system, but also includes
the damping force. It's a little more difficult to solve this
equation, and the solution will depend on the amount of damping. If a system is underdamped, it will oscillate,
and the magnitude of each successive oscillation will decrease until it stops. If the damping of the system is increased
significantly, which you can think of as the dashpot being filled with a far more viscous
fluid, any oscillation will be completely suppressed by the damping. This is called an overdamped system. And a critically damped system occurs right
at the limit between these two cases – it has just enough damping to suppress vibration. Each of these cases has a different function
that defines the displacement of the system, obtained by solving the equation of motion. The ratio of the actual damping coefficient
of the system to the damping coefficient that would result in a critically damped response
is the damping ratio. Most engineering systems and structures have
a damping ratio of less than 1, so they’re underdamped.

Of course if we're modelling a real system
we need a way of figuring out which value to use for the damping coefficient. It usually has to be determined experimentally,
and one way of doing this is by measuring the displacement of the system as it oscillates. A parameter called the logarithmic decrement
can be calculated based on this test data, as the natural logarithm of the ratio of any
two successive amplitudes. The damping ratio can be calculated from the
logarithmic decrement, providing an estimate of the overall damping in the system. So far we've looked at free vibration, where
oscillation is only caused by the initial conditions – there are no externally applied
loads. But another scenario is forced vibration,
where oscillation is driven by an external force. This type of loading often occurs in rotating
machinery. A common problem with turbines and motors
occurs when a rotating component is unbalanced, meaning that its mass is unevenly distributed. This introduces a load that has a sinusoidal
component in the vertical direction, and can cause vibration.

Unbalance can occur because components were
poorly fabricated or have been distorted. But eccentric masses are sometimes added to
motors on purpose – intentionally unbalanced motors are how phones and video game controllers
are able to vibrate. We can analyse this type of forced vibration
using the spring and dashpot model, by adding a sinusoidal external load. The resulting equation of motion is similar
to the free vibration case, but is a non-homogeneous differential equation. Its solution is the sum of two functions – a
complementary solution and a particular solution. The complementary solution is the solution
to the homogeneous form of the equation, where the right hand side is equal to zero. This is just the solution to the equation
of motion for an underdamped system in free vibration that we saw earlier.

And the particular solution captures the effect
of the external loads and is given by this expression. R is the ratio of the frequency of the external
force to the natural frequency of the model. Since there's damping in the system, the complementary
solution that represents free vibration will eventually reduce to zero. At this point the motion of the system is
defined by the particular solution only. For this reason the particular solution describes
what's called the steady-state response. It has the same frequency as the forcing function,
but is offset by a certain phase angle, meaning that the response of the system lags the external
force. Something interesting happens to the steady-state
response when the frequency of the forcing load is very close to the natural frequency
of the system. R approaches 1, and so the first term in the
square root is close to zero. And if the system has very little damping
the second term will also be close to zero, which means that the displacement will become
very large.

Let's plot the normalised maximum displacement
against the frequency ratio. For an undamped system the damping ratio is
equal to zero, so when the forcing and natural frequencies match, the displacement becomes
infinite. All systems have some level of damping, so
this is just a theoretical case – here's what the response looks like for different damping
ratios. We can see this effect if we adjust the speed
of the unbalanced motor. As the frequency of the force caused by the
eccentric mass approaches the natural frequency of the system, the displacements become very
large. This happens because when the natural frequency
and forcing frequency are the same, the energy added to the system by the external force
is timed just right so that it increases the amplitude of the displacement with each cycle.

This is called resonance. Resonance can be very dangerous and needs
to be assessed carefully. It's one of the reasons it's so important
to be able to calculate the natural frequency of a system. If the natural frequency of a bridge matches
the frequency of wind loading acting on it, or of pedestrians crossing it, the results
can be catastrophic, particularly if the structure has little damping. Devices called tuned mass dampers are sometimes
installed in buildings and bridges to control the dynamic response at resonant frequencies. With a rotating eccentric mass the external
force acting on the system is a simple sine wave function.

This makes it easy to solve the equation of
motion, since we can obtain a neat closed-form solution. But if the loading is defined by a complicated
function, or is completely arbitrary, which might be the case if it's based on test data,
it might not be possible to solve the equation of motion directly, and numerical integration
methods will have to be used instead. Designing a structure to withstand seismic
events is difficult because the loading caused by an earthquake is random and can't be predicted
ahead of time. And so engineers have to use special probabilistic
techniques like the response spectrum method to design structures to withstand seismic
loads. Single degree of freedom models are really
useful, but sometimes the dynamics of a system are better modelled using multiple degrees
of freedom. Say we want to model the dynamic response
of a three storey building. If we assume that the columns between the
floors are axially rigid but can bend laterally, we can model it as a system with three degrees
of freedom, the x coordinate of each floor. Each of the masses in the model has its own
equation of motion, and if we rewrite this system of equations in matrix form we can
see it has the same familiar form as the equation of motion for an undamped single degree-of-freedom
system.

A single degree-of-freedom system has one
natural frequency and can only vibrate in one way. But since our model has three degrees of freedom
it will also have three natural frequencies, and at each natural frequency the system will
vibrate in a specific way, which is called a mode shape. Exactly how the system vibrates in practice
will depend on the initial conditions that are applied. For the three storey building the three modes
of vibration will look like this. As the number of degrees of freedom increases
it becomes necessary to use numerical techniques like the finite element method to determine
the natural frequencies and associated mode shapes. We've only covered mass-spring models in this
video, which oscillate by translating. But there are other types of vibration too,
like pendulums, which oscillate by rotating. This video is long enough already, but I've
published a short companion video that covers the motion of pendulums over on Nebula, where
we take a look at how the motion of a pendulum differs from a mass-spring system and how
to derive the equation of motion. Nebula is a streaming service built by a group
of educational creators, which means it's completely free of ads, and revenue generated
by the project is distributed directly to the creators.

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and resonance. Thanks for watching!.

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Patterns in raising 1 and -1 to different powers | Pre-Algebra | Khan Academy

Let’s think about exponents with ones and zeroes. So let’s take the number 1 and let’s raise it to the eighth power.

So we’ve already seen that there’s two ways of thinking about this.

You could literally view this as taking eight 1’s and then multiplying them together.

So let’s do that.

So you have one two, three, four five, six, seven, eight 1’s and then you’re going to multiply them together And if you were to do that, you would get well 1 times.

1 is 1 times 1.

It doesn’t matter how many times, you multiply 1 by 1.

You are going to just get 1, You are just going to get 1 And you could imagine I did it eight times.

I multiplied eight 1’s, But even if this was 80 or if this was 800 or if this was 8 million.

If I just multiplied 1, if I had 8 million 1’s and I multiplied them all together, it would still be equal to 1. So 1 to any power is just going to be equal to 1 And you might say, hey what about 1 to the 0 power.

Well, we’ve already said anything to 0 power, except for 0 that’s, where we’re going to it.’s actually up for debate, But anything to the 0 power is going to be equal to 1 And just as a little bit of intuition.

Here you could literally view this as our other definition of exponentiation, which is you start with a 1, and this number says how many times you’re going to multiply that 1 times.

This number So 1 times 1 zero times is just going to be 1, And that was a little bit clearer when we did it like this, where we said 2 to the let’s say, fourth, power is equal to this was the other definition of exponentiation.

We had, which is you start with a 1, and then you multiply it by 2, four times so times, 2 times 2 times 2 times 2, which is equal to let’s see.

This is equal to 16.

So here, if you start with a 1 and then you multiply it by 1 zero times, you’re still going to have that 1 right over there And that’s.

Why anything that’s not 0 to the 1 power is going to be equal to 1? Now let’s try some other interesting scenarios.

Let’s start try some negative numbers.

So let’s take negative 1 And let’s first raise it to the 0 power. So once again, this is just going based on this definition.

This is starting with a 1 and then multiplying it by this number 0 times.

Well, that means we’re just not going to multiply it by this number.

So you’re just going to get a 1.

Let’s try negative 1.

Let’s try negative 1 to the first power.

Well, anything to the first power.

You could view this, and I like going with this definition as opposed to this one right over here.

If we were to make them consistent, if you were to make this definition consistent with this, you would say: hey let’s start with a 1 and then multiply it by 1 eight times, And you’re still going to get a 1 right over here.

But let’s do this with negative 1, So we’re going to start with a 1 and then we’re going to multiply it by negative 1 one time times negative 1, And this is of course going to be equal to negative 1. Now let’s take negative 1 and let’s take it to the second power.

We often say that we are squaring it when we take something to the second power, So negative, 1 to the second power.

Well, we could start with a 1.

We could start with a 1 and then multiply it by negative 1.

Two times multiply it by negative 1 twice And what’s this going to be equal to And once again by our old definition.

You could also just say: hey ignoring this one, because that’s not going to change the value.

We took two negative 1’s and we’re multiplying them Well negative 1 times negative 1 is 1 And I think you see a pattern forming.

Let’s take negative 1 to the third power.

What’s this going to be equal to? Well, by this definition, you start with a 1 and then you multiply it by negative 1, three times so negative 1 times negative 1 times negative 1 Or you could just think of it.

As you’re taking three negative 1’s and you’re multiplying it because this 1 doesn’t change the value, And this is going to be equal to negative 1 times. Negative 1 is positive 1 times negative 1 is negative.

1.

So you see the pattern Negative 1 to the 0 power is 1 Negative.

1 to the first power is negative 1.

Then you multiply it by negative 1.

You’re going to get positive 1.

Then you multiply it by negative 1 again to get negative 1 And the pattern you might be seeing is, if you take negative 1 to an odd power, you’re going to get negative 1 And if you take it to an even power, you’re Going to get 1 because a negative times a negative is going to be the positive And you’re going to have an even number of negatives so that you’re always going to have negative times negatives.

So this right over here.

This is even Even is going to be positive, 1, And then you could see that if you went to negative 1 to the fourth power Negative 1, the fourth power.

Well, you could start with a 1 and then multiply it by negative 1 four times. So a negative 1 times negative 1 times negative 1 times negative 1, which is just going to be equal to positive 1.

So if someone were to ask you, we already established that if someone were to take 1 to the I don’t know 1 millionth power.

This is just going to be equal to 1.

If someone told you let’s take negative 1 and raise it to the 1 millionth power.

Well, 1 million is an even number, so this is still going to be equal to positive 1.

But if you took negative 1 to the 999 999th power, this is an odd number.

So this is going to be equal to negative 1 .

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Understanding the Finite Element Method

Thanks to CuriosityStream for sponsoring this
video. There are a lot of different analytical methods
that engineers can use to solve structural mechanics problems, whether it's to calculate
the deflection of a beam or the stresses in a flat plate. But we often encounter problems that can't
be solved in this way, typically because the geometry, loads or materials are too complex. The finite element method is a powerful
numerical technique that uses computational power to calculate approximate solutions to
these types of problems. It's widely used in all major engineering
industries. It could be used to check that satellite components
will survive the launch conditions, for example. Or to optimise the design of automotive components,
like the lower control arm of this car's suspension system. Finite element analysis software can be used
to analyse a wide range of solid mechanics problems, including static, dynamic, buckling,
and modal analyses.

But it can also be used for fluid flow, heat
transfer, and electromagnetic problems. For this introduction to the finite element
method, we'll focus on how it applies to static linear-elastic stress analysis. Imagine we want to analyse the brackets supporting
this air conditioning unit. The goal of a static stress analysis would
typically be to calculate the stresses, strains and displacements within the bracket. These unknowns are called "field variables". Internal stresses develop within a body in
such a way as to maintain equilibrium over any volume of the body, so we can apply the
concept of equilibrium to calculate the field variables. This is easy to do for a simple beam – we
can use equilibrium to calculate the bending moment and shear force along it, and from
there we can calculate the normal and shear stresses in the beam.

But enforcing equilibrium over a two dimensional
shape like this bracket is difficult, and it becomes even more complicated for a three
dimensional body. The finite element method approaches this
problem by splitting the body into a number of small elements, that are connected together
at nodes. This process is called discretisation, and
the collection of nodes and elements is called the mesh. Discretisation is useful because the equilibrium
requirement now only needs to be satisfied over a finite number of discrete elements,
instead of continuously over the entire body. Several different element shapes can be used. We've used triangular surface elements to
model this bracket. Surface elements are two dimensional elements
that are typically used to model thin surfaces. They can be triangular or quadrilateral. Triangular elements are good for modelling
awkward shapes, although quadrilateral elements tend to perform better. Solid elements are used for three-dimensional
bodies.

And then we have line elements. Choosing the right element for your model
will depend on the specific scenario being analysed, and requires some expertise. In the case of our bracket we could have used
solid elements, or even line elements, depending on how much we wanted to simplify the problem. Even for elements of the same shape, there
are hundreds of different types to choose from that each have different formulations,
and introduce different levels of approximation. A line element can be a bar, for example,
that only carries axial loads, or a beam, that can carry axial, bending, shear and torsional
loads. We can model the bracket using plane stress
surface elements, because the bracket is thin and the loading is all in the same plane.

But that's only one of many surface element
types. These are all first order elements, but we
can also use second order elements, which have additional mid-side nodes and are more
accurate. For stress analysis problems the fundamental
variable we want to calculate is the displacement at each node. If we know how a body displaces when loads
are applied, we'll easily be able to calculate secondary outputs like stress and strain. For each element we can define a vector {u}
that contains all of the possible displacements for the nodes of the element, including rotations.

If we're analysing a two-dimensional case
with beam elements, each node can translate along the X and Y axes and it can rotate about
the Z axis, so the vector {u} will look like this. Each of these displacements is called a degree
of freedom. For the beam element we have 3 degrees of
freedom per node, or 6 in total. For a 3D case that increases to 6 degrees
of freedom per node. A shell element node also has 3 degrees of
freedom in two dimensions, but since the element has 4 nodes, it has 12 degrees of freedom
in total. The nodes of a solid element only have the
3 translational degrees of freedom.

The nodes aren't allowed to rotate and instead
rotation of the element is captured by translation of the nodes. So how can we calculate all of the displacements
at every node in our mesh? For a spring, the relationship between force
and displacement is defined by Hooke's law. The spring stiffness k determines how far
the spring will displace for a given force. In the same way, we can think of the elements
of our mesh as having a certain amount of stiffness, that resists deformation. In this equation {f} is a vector of the nodal
forces and moments, {u} is the vector of the nodal displacements, and [k] is the stiffness
matrix of the element. A 2D beam element has 6 degrees of freedom,
so the displacement vector looks like this.

And the force vector and the stiffness matrix
will look like this. The element stiffness matrix defines how much
each node in the element will displace for a set of forces and moments applied to the
nodes, and so is the key to solving the displacements at every node of our mesh. It's a square matrix – the number of rows
and the number of columns are equal to the number of degrees of freedom of the element. We can figure out what the terms of the stiffness
matrix are by enforcing equilibrium. We'll come back to this later on in the video,
but for a 2D beam the matrix looks like this. We can think of this equation as a system
of linear equations that we can solve to obtain the displacements at the nodes of our mesh. If we apply a lateral displacement to node
2, for example, and all of the other degrees of freedom are fixed, and so are equal to
zero, we can use the stiffness matrix to calculate the forces and moments at both of the nodes. To make the next steps easier to visualise,
let's represent the stiffness matrix in a more abstract form.

This is just one element, but our overall
mesh will be made up of many more elements. Let's look at a simple example where we have
a mesh made up of three 2D beam elements, that we're using to model a cantilever beam. We can assemble the individual stiffness matrices
for all of the elements in our mesh into a huge global stiffness matrix that defines
how the entire structure will displace when loads are applied to it. Like the element stiffness matrix, the global
stiffness matrix is a square matrix and the number of rows and columns is equal to the
total number of degrees of freedom in the model. The element stiffness matrices are assembled
together to form the global stiffness matrix based on how the elements are connected together. Elements 1 and 2 are connected at node 2 for
example. Continuity tells us that since these two elements
are connected at the same node, the displacements for both elements must be the same at the
common node. So when we assemble the global stiffness matrix,
the terms in the element stiffness matrices corresponding to node 2 should be summed for
each degree of freedom.

Element 3 is not connected to node 2, so this
element's stiffness matrix should have no effect on the displacements at node 2. This is what the actual global stiffness matrix
looks like for this model. It has some interesting characteristics. It is said to be sparse, because it contains
a lot of zeros, and banded, because the non-zero terms are grouped around the diagonal. For linear-elastic problems the matrix will
also be symmetric. If we modify the mesh so that the three elements
are connected differently, the global stiffness matrix will change. In this case we have three nodes instead of
four, so the matrix will be smaller, and the fact that elements 1 and 3 are connected is
reflected in the matrix. An important thing to note here is that the
elements are no longer aligned to the same coordinate system, so we have to transform
the stiffness matrix for each element so that it aligns with a global coordinate system.

We can do this by multiplying each element
stiffness matrix by a rotation matrix. Now that we've assembled the global stiffness
matrix, we need to solve this equation to obtain the displacements at each of the nodes. To do this we need to define the external
loads, and the boundary conditions. The boundary conditions are known displacements
at specific nodes, typically because specific degrees of freedom are fixed. In this model, vertical and horizontal translations
are fixed at node 1, and vertical translations are fixed at node 2, so the displacement vector
looks like this. And the force vector will look like this. It includes the applied force and the reaction
forces at the supports. Now we can solve the equation. We could do this by inverting the global stiffness
matrix and solving the displacements from there.

But in practice inverting the matrix isn't
very efficient, particularly because it's a sparse matrix. Commercial solvers mostly use methods that
involve iteratively approximating the displacement vector, like the conjugate gradient method. Once we've solved for the nodal displacements,
we can calculate the strains and then stresses throughout the mesh. A typical finite element mesh could easily
have a hundred thousand degrees of freedom, which would be impossible to solve by hand,
and so applying the finite element method to anything more complicated than a very basic
model requires the use of appropriate software. Now that we have an overall understanding
of the finite element method, let's return to the element stiffness matrix to see how
it's derived.

The matrix shown here is for a 2D beam element,
but it will look very different for different element types. Several different methods can be used to derive
these stiffness matrices, and they are all fundamentally based on the concept of equilibrium. The direct method derives the stiffness matrix
directly from the equilibrium equations that govern the behaviour of the element. The lateral deflection of a beam is governed
by this equation, for example, so we can solve the equation to obtain the stiffness matrix
for a beam element. These governing equations are usually differential
equations. The differential equations and associated
boundary conditions are what we call the "strong" form of the equilibrium problem. But it's only really possible to solve the
strong form for simple elements. For more general cases we can use "weak" forms
that describe the differential equations in integral form, instead of solving the differential
equations directly. These give approximate solutions to the equilibrium
equations, but are easier to solve.

The first of the weak form methods is based
on variational principles. One such principle commonly used for structural
mechanics problems is the Principle of Minimum Potential Energy. It states that the displacement configuration
that satisfies equilibrium conditions is the one that minimises the total potential energy,
where the potential energy is the sum of the strain energy and the potential energy of
the external loads. By applying a mathematical technique called
the calculus of variations to minimise the total potential energy, we can obtain an approximate
solution to the equilibrium equations.

The other weak form method is the Galerkin
method of weighted residuals. In this method the function that satisfies
the differential equation is approximated as the sum of a number of assumed trial functions
that each have unknown coefficients. This approximate solution is substituted into
the differential equation, and an equation for the error, called the residual, is obtained. If we multiply each trial function by the
residual and set the integral of this product to zero, we can calculate the unknown coefficients
that minimise the residual. This gives us an approximate solution to the
differential equation.

This is a more widely applicable approach
than the principle of minimum potential energy, but for stress analysis problems both methods
give the same result. Regardless of which method we use, we end
up with the stiffness matrix for our element. But to apply these methods we need to be able
to describe how displacements and other field variables vary inside the element, instead
of just at the nodes of the element. To overcome this issue, an element needs to
have a defined function that calculates values inside the element by interpolating the values
at the nodes.

The shape function is just an assumption. It's usually chosen to be a polynomial, since
they're relatively simple and sufficiently accurate. And with that we've covered all of the key
aspects of the finite element method. In summary, the first step in the finite element
method is defining the problem, including the relevant material properties, loads and
boundary conditions. Next the body being analysed is split into
a number of small elements connected at nodes, and the element types are chosen. Then a stiffness matrix is defined for each
element, using one of the three methods we covered earlier. The element stiffness matrices are then assembled
into a global stiffness matrix based on element connectivity. This global stiffness matrix defines how the
structure will respond to applied loads, and we can use it along with boundary conditions
to solve for the displacement at each node in the structure.

Once we have displacements we can calculate
stresses, strains and other field variables of interest. Then all that's left to do is post-processing
to obtain the desired results, and validation of the model. A lot of the hard work like calculating the
element stiffness matrices, assembling the global stiffness matrix and solving the model
is done by the software being used. But the engineer is responsible for making
sure that the problem has been properly defined, that the mesh is suitable, and for interpreting
and validating the results. I hope this video has helped you develop a
better understanding of the fundamentals of the finite element method. If you're interested in learning more, you
can check out the extended version of this video, that's available now over on Nebula,
where I spend a few more minutes covering the problem definition, discretisation, post-processing,
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And that's it for this introduction to the
finite element method. Thanks for watching!.

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Different types of shares | Video 1. DSCG1 UE6. Finance. Stocks and Shares.

one ordinary shares the most common type of shares are known as ordinary shares purchasing ordinary shares gives the owner of stake in the company and an entitlement to dividends dividends are the part of the company's profits that are paid out to the shareholders every six months when the company is making a profit most ordinary shares also give the shareholder a right to attend the company's Annual General Meeting and vote on issues relevant to the company's future to preference shares preference shares return a fixed dividend to the investor that is not linked to the company's annual profit result although preference shareholders receive dividends before ordinary shareholders they do not receive the same voting rights as ordinary shareholders three contributing shares contributing shares of those that have not been fully paid for and require further payment in the future dividends are generally paid according to the proportion of the paid up amount for bonus issues a bonus issue is a free issue of new shares to existing shareholders every now and then when a company makes an extraordinary profit or if it is amassed accumulated profits over a period of time it gives its shareholders a present of a bonus issue of shares at no cost receiving a bonus issue does not increase the proportion of a company owned by the shareholder as all shareholders receive the same present in proportion to their ownership of the company it is in effect a cashless dividend five rights issues a rights issue is also an issue of new shares to existing shareholders however these are not free a rights issue occurs when a company needs to raise extra capital and it gives its shareholders our right but not an obligation to purchase extra shares there are two types of rights issues renowned scible and non renowned civil rights pronounceable rights can be traded on the share market if an existing shareholder does not wish to purchase the new shares being offered to them they are therefore of some value to the shareholder a little bonus non renowned scible rights cannot be traded or sold to others so if the shareholder does not take up their right to buy the new shares by a particular date the writer of no value to them so those are the five main types of shares in general when we refer to shares in this and other training modules we are talking about ordinary shares you

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SPOT A DIFFERENT NUMBER 1 #quiz #puzzle

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