Normally force are ditributed along the structure, but can be simplified into a point

This is when the resultant force of a particle is zero. This means that the formula is Fr=0
Meaning that the Fx=0 and Fy=0

The resultant force being 0 means that the particle is not experiencing any acceleration. This could mean that the object is not moving or that he object is moving at a constant velocity.

A free body diagram is used to see cleary where are all the forces coming from and how they are interacting with the particle.

When a particle is only moving in one direction (let’s say along s-axis), it is said to be in rectilinear motion

With kinematics, there are four main variables that we will be working with. Velocity, displacement, time and acceleration.
The relations between these quantities are described as time derivatives

If our motion is not in just one direction but two directions (let’s say x and y), then our motion is said to be curvilinear

In these cases, both our velocity and acceleration become Pythagorean products
of their x- and y-components, so their magnitudes are:

Direction angles can be calculated from x- and y-components using inverse
tangent:

If our situation is different in such a way that the curvilinear nature is forced because of our particle is moving along a path that is known beforehand, then it is usually simpler to perform our calculations in a plane which consists of normal and tangential components (n and t)

Here at represents the acceleration caused by change in the magnitude of the velocity and an represents the acceleration caused by change in the direction of the velocity

Tangential and normal components of the acceleration can be easily calculated
using the following formulas

Tangential acceleration is the time derivative of velocity

Normal acceleration is velocity squared
and divided by radius of curvature ρ

If our path is circular, radius of curvature is constant (= radius of circle)
If our path can be expressed as y = f(x), the radius of curvature is

Third possible coordinate system that we can use is the system of cylindrical coordinates (or, in plane, polar coordinates r and θ )

This is useful in applications where our particle moves at certain angular velocity, The radius of the path of our particle changes respect to time

In polar coordinates, our velocity equation takes the form

• Likewise, our acceleration equation takes the form

Acceleration is linked to the resultant force acting on the particle as well as the mass
of the particle via Newton’s 2nd law

Mass is also present in Newton’s 4th law – the law of gravitation. If we consider a particle with mass m that is falling near surface of Earth (mass 𝑚2),
we can write that the particle and Earth act upon each other via gravitation, so

If an object has mass, it also has inertia, which means that the object will resist changes in its state of movement. Accelerations and decelerations are often measured in g’s

All calculations must be performed in inertial coordinate system. i.e. a coordinate system which has an origin that does not rotate or have any acceleration. Unfortunately, a coordinate system like this has not (yet) been discovered

In “traditional” engineering tasks we can use an inertial coordinate system which
is fixed on the surface of Earth

If a particle is subjected to force F and the particle undergoes a finite displacement along its path from s1 to s2 , the work done by force F is the product of force component in the direction of the path and displacement

we must remember, that our forces usually act on tangential and normal directions – which are not the same as θ and r

Hence, we will introduce a new angle ψ which means the angle between the
extension of the radial coordinate r and tangent of the path curve

If our force component in the direction of movement is constant, we can just multiply: U = F∆s

If the displacement is in the same direction as the force, work U is positive, and if it's opposite then U is negative.

The force of a spring is kx, and by performing the integration we get:
But becuase the force that is exerted on the particle is in the opposite direction, the final formula is negative

If we consider a particle moving along its path in an inertial coordinate system, the direction of movement is naturally tangential direction we can use the kinematic equation and intigrate to find the work done. Which results in:

This principle can be applied not only to a single particle but also a system of particles – with following limitations:

Power is the time-rate of doing work – so, time derivative of work:. 𝑷 = 𝑭v
The concept of power is often needed when we need to select a suitable motor for
certain applications

s it is essential to differentiate between power input and power
output. Between these two we can form the relationship, This relationship is called mechanical efficiency of the machine

Forces are said to be conservative if the work done by them is independent of the
path followed by the particle

On the other hand, for example friction is a non-conservative force.
If our system only consists of forces which are conservative, we can write the
principle of work and energy into the following form:

T = Kinetic energy, V = potential energy (gravetational and spring)

Linear Impulse and Momentum: If we consider the already familiar Newton’s 2nd law and remember that
acceleration is the time derivative of velocity, we get
F=ma=mv dv/dt
If we multiply this by dt and put in bounds for integrals, we get

This equation comes in handy when we have problems which are dealing with force, velocity and time. The equation can be applied separately in x- and y-directions, also be applied to a system of particles.

When thinking about dynamics processes, we can sometimes simplify our calculation by neglecting forces which have small impulses

Forces which produce a very small impulse are called
non-impulsive forces.
Forces which are large (in comparison to other forces) but act for a small time period are called impulsive forces

CENTRAL DIRECT IMPACT: The collision of two objects during a very short time is called an impact. The tangential plane of the contact surfaces is called plane of contact. The normal drawn to this plane at the contact point is called line of impact

If the mass centers of both objects lie on the line of impact, the impact in
question is a central impact
If the velocity vectors of contact points of both objects are parallel to the line of
impact, the impact in question is direct impact

In the beginning of the collision, objects exert deformation impulses integral P dt to
each other.
For a very short instant of maximum deformation, the objects will move with a
common velocity v

After this instant, the objects exert restitution impulses integral R dt to each other. As a result the velocity of the two objects changes

Because both deformation and restitution impulses are internal and cancel each
other out, the momentum of our system of two particles is conserved, so

If initial velocities are known, this leaves us with two unknowns. Therefore we need another equation in order to solve for final velocities. The second equation we can get by considering both particles separately at both
phases

The ratio of impulses is called coefficient of restitution:

By performing the same analysis for particle B, we get:

Now, the common velocity v (at the instant of maximum deformation) is only valid for a very short time and it’s not of our interest, so we eliminate that from equations for e

If we fix y-axis along contact plane and x-axis along line of impact, we can see that the impulses only affect in x-direction. The y-components of velocities remain unchanged.
Applying the coefficient of restitution and conservation of momentum for the x-components of velocities.

It can be mathematically derived that the resultant moment about point O is equal to the time derivative of angular momentum. This means that the change in angular momentum of a
particle is caused by a moment impulse!

This moment impulse is known as the angular impulse. The net torque equals the rate of change of the angular momentum! The net torque acting over a time interval is the angular impulse.

UNDAMPED FREE VIBRATION: When the motion is maintained by gravitational or elastic restoring forces (usually
spring force), vibration is called free vibration.
If there are no frictional or other similar forces that would resist the motion, the vibrations can continue indefinitely, the vibration is undamped

one degree of freedom (DOF): 1 DOF = one particle with mass m that can only move in 1 direction

Acceleration is the 2nd time derivative of displacement, so:
x'' + 𝑘/m = 0
This is usually written in form

This equation is a homogenous differential equation (DE) of 2nd order with constant coefficients. Later in math courses you will (hopefully) learn why, but at this point, let’s
just believe that the solution for it is

The solution can also be written in the form:

Here p is called circular frequency or natural frequency (often denoted by 𝜔 instead of p, but you’ll see a bit later why I favor p).
C is the amplitude (= maximum displacement) of our particle

The time period required to complete one cycle is:

The damping force is proportional to the velocity of our particle, so our Newton’s 2 nd law –equation takes the form:
−𝑘𝑥 − 𝑐𝑣 = 𝑚a
Here c is the coefficient of viscous damping (unit kg/s).

If we move everything to the left again and remember that velocity is the 1st time derivative of displacement, we get the differential equation.

Again, we end up with a homogenous differential equation of 2nd order with constant coefficients. Now when there exists a damping term, the behavior of the system can be divided to three different cases – depending on the coefficient of damping

The critical coefficient of damping cc can be calculated from the equation:

For an overdamped system, the solution is:

For a critically damped system, the values for 𝜆1 and 𝜆2 are equal and therefore the solution simplifies to form:

Generally speaking, these 2 cases don’t actually represent vibration; in these cases the damping is so “strong” that the displacement x will reduce to zero without any oscillation.
Critically damped system will do this faster than an overdamped system

In here, constant pd is called the damped natural frequency of the system:

The ratio c/cc is often referred to as damping factor. The time period of damped vibration can be written as:

This is a nonhomogenous differential equation of 2nd order with constant coefficients. The solution is a linear combination of these two:

+

If we take a closer look at the formula of particular solution:

we notice that if values of angular frequency ω and natural frequency p are really close to each other, the denominator will become very small, and therefore the amplitude will become very large

Now our differential equation takes the form:

Again, the solution for this differential equation is a linear combination of homogenous solution and particular solution:

The particular solution takes the form:

Amplitude C’ and phase difference 𝜙′ are given by following equations:

RIGID BODY OR PARTICLE: Particles are small and only have mass; their size is considered negligible, so
they have no “dimensions”. Rigid body, on the other hand, has dimensions or in other words Acting points of forces become significant (not always on mass center)!

The equations for equilibrium are 2/3 the same as for particles:

Structures can be attached to their surroundings via several different support types (pin joints, fixed joints etc.) It is extremely important to know which support reactions will these support types exert on our structure

Different types of support reactions:

Usually our structures have as many unknown support reactions as we are able to solve using the equations of equilibrium (so, maximum of 3 in 2D cases). In reality our structures may have more supports and hence more support
reactions

A truss = structure composed of slender members joined
together at their end points. In calculation, joints of the truss are considered as pin joints

If the truss is welded or the joints are strengthened by gusset
plates, this does not reflect reality 100%.
Still, if the truss members are long compared to their other dimensions (mathematically: 𝐿 ≫ (ℎ, 𝑤, 𝑑)), this simplification provides a viable ground for analysis

If all loadings are applied at the joints, the truss can be
analyzed as a simple truss. As a result of the pin joint simplification, the truss members will only transmit normal forces – i.e. forces in axial direction

Simple trusses consist of triangles. Most practical uses feature planar trusses, which are joined together by purlins or
floor beams in order to form a 3D-structure

Elementary statics gives us tools for evaluating the following things from zeroredundancy trusses: Support reaction and member forces

If the truss has a degree of redundancy greater than zero, we will need additional
information about sizes & possible materials of members

Tools for deformation analysis of trusses will be learned in further courses

The truss member forces can be solved using the method of joints. In this method, we analyze each joint of the truss separately as a particle that
must be in equilibrium, so we apply

If there are n joints in the truss, we can form 2n equations. Draw a FBD of the whole truss and solve the support reactions – or at least write the equations of equilibrium for it (the truss itself is a rigid body!)

Solve the unknown forces (if possible) and move on to the next joint. Always draw unknown forces as tension forces – i.e. away from the joint! When done this way, you can see directly from the results whether the member is in tension (positive result) or compression (negative result)

It is rather common that our truss has members which don’t carry primary load. Therefore, the force in these members is considered to be zero.

These members are added to the structure in order to: Increase stability, Increase redundancy, Provide support in case of secondary loading or if the loading changes.

Two general rules can be used in order to spot zero-force members:

If only two members form a truss joint and no external load or support reaction is applied to the joint, the members must be zero-force members

This is based on the principle that if the whole body is in equilibrium, also any part of it must be in equilibrium.
Hence, we can “cut” any part of the truss and inspect it as a rigid body that must be in equilibrium

When we move on to investigating beams, limiting our analysis only to normal forces will not be sufficient, because they are capable of carrying also shear force and moment. In general case, these three are called internal forces

These internal forces are not constant but take different values along the length axis of the beam. In design, it is essential to be able to find out the maximum values of internal
forces as well as their distribution

Internal forces can be solved at any point of our structure using the method of sections – just like we did with the trusses

Now when we are dealing with shear and moment, it is important to remember the sign convention rule for these forces:

• Shear force: if we use the left part of the cut, positive V is down
• Moment: left part of the cut = counter-clockwise
• For moment, also the “smiley face rule” can be used

If the question is just “calculate the internal forces at point X”, we can perform the cut at this point, draw FBD and simply solve our three unknowns using the three equations of equilibrium

BUT, if the question is to solve internal forces in the whole structure, we need to perform a much longer analysis in order to get the shear and moment equations (so, V and M as functions of length coordinate) and diagrams

In the latter case, we need to specify a length coordinate x starting from one end
of the beam and perform cuts after every discontinuity in the beam

As a result of the previous process, we get the equations (or actually functions, but “equations” is a more commonly used term) for shear and moment

From these, we can recognize the most critical parts of our structure and
calculate the maximum (as well as minimum) values for V and M

eg:

Using a FBD of the whole beam, we can solve the support reactions:
𝐹𝑦 = 0 ⇒ 𝐴𝑦 + 𝐶𝑦 − 15 − 25 = 0

𝑀𝐴 = 0, −80 − 15 ∙ 5 − 25 ∙ 7.5 + 𝐶𝑦 ∙ 10 = 0
10𝐶𝑦 = 80 + 75 + 187.5 ⇒ 𝑪𝒚 = 𝟑𝟒. 𝟐𝟓 𝒌N
𝐴𝑦 + 34.25 − 15 − 25 = 0 ⇒ 𝑨𝒚 = 𝟓. 𝟕𝟓 𝒌N

Now, let’s cut the beam first between A and B. We set a length coordinate x1 starting from left end of the beam. By applying the equations of equilibrium, we get functions for V and M:
0 ≤ x1 ≤ 5 m

𝐹𝑦 = 0 ⇒ 5.75 − 𝑉 = 0 ⇒ 𝑽 = 𝟓. 𝟕𝟓 𝒌N
𝑀 = 0 ⇒ 𝑀 − 5.75𝑥1 − 80 = 0
𝑴 = 𝟓. 𝟕𝟓𝒙𝟏 + 𝟖𝟎 𝒌𝑵

Then, let’s section the beam between B and C. We set a new coordinate x2 starting from left end of the beam. In a similar fashion, we get:
5 m ≤ x2 ≤ 10 m

𝐹𝑦 = 0 ⇒ 5.75 − 15 − 5 𝑥2 − 5 − 𝑉 = 0 ⇒ 𝑽 = (𝟏𝟓. 𝟕𝟓 − 𝟓𝒙𝟐) 𝒌N
𝑀 = 0 ⇒ 𝑀 − 5.75𝑥2 + 15 ∙ (𝑥2 − 5) + 5(𝑥2 − 5)∙(x2-5)/2 - 80 = 0
⇒ 𝑴 = −𝟐. 𝟓𝒙𝟐^𝟐 + 𝟏𝟓. 𝟕𝟓𝒙𝟐 + 𝟗𝟐. 𝟓 𝒌𝑵

Now we can draw the shear and moment diagrams and also calculate maximum values by substitution:

Usually, we are not that interested about the equations but more interested in the general shape of the diagrams and maximum values. The before mentioned relations can be used in order to sketch the shear and moment diagrams in a much easier fashion. Maximum values can be calculated using integration

TRANSLATION: We may use either a fixed coordinate system or a translating coordinate system in our calculations. In vector form, the position of point B can be given as

The time derivatives of this give us velocity and acceleration. If the rigid body is only in translation, vector 𝒓ത𝑩/𝑨 is constant (both in magnitude as well as in direction). • Therefore, taking the derivatives will yield us

ROTAT: ION ABOUT A FIXED AXIS: When a body is rotating about a fixed axis, any point P of the body travels a circular path subject to the fixed axis.

If we draw a perpendicular line from the fixed axis to point P, we can specify the angular motion of this line:

• Angular position is defined by angle θ
• Angular displacement is defined by 𝑑θ
• Angular velocity is the time derivative of angle
• Angular acceleration is the 2nd time derivative of angle

The angular motion of the line can be linked to point P. This is done by multiplication with radius r

• If we eliminate dt from previous differential equations, we get a 3rd equation:

If the angular acceleration is constant (𝛼 = 𝛼𝑐) , integration of these 3 gives us

When converted to motion of point P, we must also remember that the total acceleration is the Pythagorean sum of tangential and normal components:

GENERAL PLANE MOTION (VELOCITY): General plane motion of a rigid body is a combination of translation and rotation. In order to assess these separately, we must establish a relative-motion analysis

If we think back to the position vector again. If we took the derivative of this, in pure translation was constant hence disappeared in derivation. Now when our body is also rotating, that means that the direction of 𝒓ത𝑩/𝑨 is
changing, so its derivative will not be zero

Therefore, now our velocity will take the form

V B/A is the relative circular velocity of B subject to A

So, how to find out the relative velocity 𝒗𝑩/𝑨?. Think that momentarily the body is “pinned” at A. Since the motion is circular, the magnitude can be calculated by

The direction of 𝒗𝑩/𝑨 is perpendicular to 𝒓𝑩/�

In planar case, this vector equation breaks down to x- and y-components. As a result, we get magnitudes for velocity components:

GENERAL PLANE MOTION (ACCELERATION): Since in general plane motion there is both translation and rotation, we must remember this when calculating acceleration

In rotation, the total acceleration is caused by tangential and normal components. Therefore, in general plane motion the acceleration will be

Here (𝒂𝑩/𝑨) 𝒕 is the relative tangential acceleration. Direction is perpendicular to 𝒓𝑩/�

Likewise, (𝒂𝑩/𝑨) 𝒏 is the relative normal acceleration. Direction is from B to A.

MOMENT OF INERTIA: Mass m is the measure for a body’s resistance to acceleration. Likewise, there exists a measure for body’s resistance to angular acceleration; this measure is called moment of inertia or second moment of mass and its symbol is I

Mathematically speaking, this moment is calculated as an integral

Considering that 𝑑𝑚 = 𝜌 𝑑𝑉, this can also be presented in form
It is common that density 𝜌 is constant, so it can be factored out of the integral

We could naturally calculate moments of inertia for different-shaped bodies using integration, but in order to simplify calculations, there are formulas for the most commonly used ones

KINETIC EQUATIONS IN TRANSLATION: The body has acceleration in x- and y-directions, but no rotation. Therefore the equations of motion become. Note: aG
refers to the acceleration of mass center

KINETIC EQUATIONS IN TRANSLATION: No rotation, but the acceleration consists of tangential and normal components. Therefore the equations of motion become
Note: Here 𝜌 is radius of curvature, not density!

KINETIC EQUATIONS IN ROTATION ABOT A FIXED AXIS: If the rotation is accelerating or decelerating at angular acceleration α, the moment sum equals to this times moment of inertia. Hence, the equations of motion become
: Here 𝑟𝐺 refers to distance from O to G

KINETIC EQUATIONS IN GENERAL PLANE MOTION: In case of general plane motion, our equations will be

If the body is in GPM, it means that it rotates and translates at the same time. Hence, we often have to deal with unknown friction and support forces. Because we only have 3 equations with usually 3 unknown accelerations, this means that we need additional equations in order to solve the problem

In this case following tricks are used:
➢ Choose your coordinate system (x,y) in such a way that the other coordinate axis is in the direction of movement and the other is perpendicular (this will make the perpendicular acceleration component zero)

➢ If the movement is rolling with no slipping, use fourth equation that relates straight-line acceleration and angular acceleration:

➢ If the movement has slipping, use fourth equation that relates friction and support forces:

We can solve this kind of problem even if we don’t know whether the body will slip or not. First assume there is no slipping and use 𝑎𝐺 = 𝛼𝑟 as the 4th equation in order to
solve the group of equations

After this, perform a check:
If 𝐹𝜇 < 𝜇𝑘𝑁, the assumption stands and the solution is valid
If 𝐹𝜇 > 𝜇𝑘𝑁, the assumption was wrong (slipping occurs); change the 4th equation and rework the solution

APPLYING THE ROTATIONAL EQUATION ABOUT POINT P: It is not compulsory to write the rotational equation subject to mass center G; it is also possible to write this equation about any point P in our body

This is done by fixing the origin of our coordinate system to P and using a kinetic diagram and calculating the “kinetic moments” of acceleration components

In rotation about fixed axis, the normal component of acceleration is directed towards the point of rotation O; so, if we take this moment sum subject to point O, the equation takes the form

Applying the rotational equation about some other point than G is not necessarily needed, but in some cases, it leads in much easier calculations

KINETIC ENERGY OF A RIGID BODY::

In pure translation (rectilinear or curvilinear), kinetic energy can be calculated the same way as for particles.
Subscript “G” is added to velocity to accentuate that this is the velocity of the mass center

If our rigid body is in pure rotation around its mass center, the velocity of its mass center is zero. However, the body still has kinetic energy due to rotation. The kinetic energy of this rotation can be calculated analogously

If our rigid body is in general plane motion, then its kinetic energy consists of translational and rotational energy. Due to the nature of energy, we can just calculate the total kinetic energy as the
sum of these two:

Because calculation of IO can be troublesome, in general I recommend using the GPM form

WORK OF FORCE AND COUPLE MOMENT:
Work done by a force can be calculated as an integral of F over s

Work done by a couple moment can be calculated as an integral of M over theta

If positive work is done to the system, the total energy of the system increases!

PRINCIPLE OF WORK AND ENERGY: Same principle of work and energy that we already used for particles is also valid
for rigid bodies. We must only remember that for a rigid body the kinetic energy is a sum of translational and rotational energy

Good method: draw the rigid body in the beginning and in the end in order to see
the changes throughout the movement process

CONSERVATION OF ENERGY: Revision: if the work done by the force is independent of the path, the force issaid to be conservative.
If only conservative forces do work on our body, we can write the previous equation in a simpler form using kinetic (T) and potential (V) energies: