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Engineering Mechanics - Coggle Diagram

- - - - Normally force are ditributed along the structure, but can be simplified into a point
        
        To do this we use Integrals
        𝑭𝑹 = Integral(𝒘(𝒙)) 𝒅𝒙
    - - These can be broken down into X and Y components. Then use trig to find the resultant
  - - - Right hand rules
      - If the force is parraller to the line of action then the moments is 0
        M=Fd
        M=F*0
        M=0
  - - - 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.
        
        useful applications
        
        These analysis often requie us to look at tension on string and forces on spring. So it good to also know Hook's law, F=kx
        
        This can be used to analyze knots, pulleys, connecting rings, and any small detail where all forces acting on it can be described as
        acting on its center
        
        Opposed to this, larger objects (beams etc) can not be represented as particles. This kind of objects are called rigid bodies
    - - 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:
        
        FREE-BODY DIAGRAMS: In order to apply these equations, we must first find out the forces and moments
        acting on our rigid body AND their locations
        
        SUPPORT TYPES AND SUPPORT REACTIONS: Usually we know the external forces (and moments) acting on our rigid body. In most cases, equilibrium equations are used in order to find out the support reactions of our structure
        
        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:
        
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    - - 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
        
        These rods can either be in tension or compression
        
        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
        
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        If the truss has a degree of redundancy greater than zero, we will need additional
        information about sizes & possible materials of members
        
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        Tools for deformation analysis of trusses will be learned in further courses
        
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        METHOD OF JOINTS
        
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        ZERO-FORCE MEMBERS
        
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        THE METHOD OF SECTIONS
        
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    - - 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
        
        METHOD OF SECTIONS
        
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        SHEAR AND MOMENT EQUATIONS AND DIAGRAMS
        
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        GRAPHICAL METHOD
        
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  - - - 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
        
        v = dS/dt
        a = dv/dt = d2S/dt^2
        a ds = v dv
      - 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:
        
        The most common application for this would be for projectiles but the (x,y)-component method works well for cases, where the curvilinear nature of motion is a product of things happening in both directions – i.e. the motion must not be curvilinear by definition
        
        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
        
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        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
        
        Very common in machine design:
        Slider-connecting rod –mechanisms
        Robot arm movements
        
        In polar coordinates, our velocity equation takes the form
        
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        Note:we often use dots to imply time (t) derivatives
      - Force and acceleration
        
        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
        
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        Friction
        
        For better accuracy, it is possible to separate static (𝜇𝑠) and dynamic (𝜇𝑘) COF. Becaurful with N as it doesn't always equal to the mass of the object.
        
        x, y coordinates
        
        In X,Y-coordinates, we can divide Newton’s 2nd law to x- and y-components and hence get two equations. Sometimes it is clever to set your coordinate system “tilted”
        
        N, T coordinates
        
        If our particle is moving along a known curved path, it is often wiser to write the
        equations in N,T-coordinates
        
        Cylindrical coordinates
        
        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
      - Work and Energy
        
        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
        
        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
        
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        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
    - - Motion of rigid bodies can be divided (in 2D) to three cases:
        
        Translation (rectilinear or curvilinear)
        
        Rotation about a fixed axis
        
        General plane motion (GPM)
        
        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
        
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        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/�
        
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        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.
        
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      - Force and acceleration
        
        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
        
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        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
        
        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
        
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        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:
        
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        ➢ If the movement has slipping, use fourth equation that relates friction and support forces:
        
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      - Work and Energy
        
        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:
        
        Note: if the rigid body is in pure rotation about a fixed axis passing through point O, then the kinetic energy can be expressed as
        
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        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!
        
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        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:
  - - - 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
        
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        CENTRAL OBLIQUE IMPACT: If the velocities of colliding particles are not aligned with the line of impact, the
        impact is called oblique
        
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    - - Here d is the perpendicular distance from O to the line
        of action of mv
        
        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.
  - - - one degree of freedom (DOF): 1 DOF = one particle with mass m that can only move in 1 direction
        
        In undamped free vibration, there is only one force acting on
        the direction of movement: the restoring force F
        −𝑘𝑥 = 𝑚a
        If we move ma to the left and divide by mass, we get:
        𝑎 + 𝑘/m = 0
        
        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
        
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    - - 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:
        
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    - - The equation of motion will now be:
        𝐹0 sin 𝜔𝑡 − 𝑘𝑥 = 𝑚a
        Rearranging gives us:
        𝑚𝑥''+ 𝑘𝑥 = 𝐹0 sin 𝜔t
        
        This is a nonhomogenous differential equation of 2nd order with constant coefficients. The solution is a linear combination of these two:
        
        +
        
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    - - 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:
        
        The free vibration part xc will gradually fade away due to damping, so ultimately the particular solution is the part that will remain (and hence, most interesting). This xp is therefore called “steady-state solution”
- - - - ➢ Draw FBD of the particle
        ➢ Mark all the forces acting on the particle
        ➢ Divide forces to components (now there are three: x, y and z)
        ➢ Set sum of components equal to zero in all directions
        
        We end up with three equations, where we can solve three unknowns. We need at least 2 angles in order to specify the direction
        Usually we give three:
        ➢ α = angle between force vector and x-axis
        ➢ β = angle between force vector and y-axis
        ➢ γ = angle between force vector and z-axis
        
        Calculation of these angles can be troublesome in scalar form. In three-dimensional cases, it is often wisest to make the switch from scalar equations to vectors. Denote each force by a vector, whose i-, j- and k-components represent x-, y- and
        z-components of the force (correspondingly)
        
        If we know the direction vector v of the force, we can construct the force vector easily by first calculating the unit vector uv and then write
        
        The Bolded letter are vector and the unbolded ones are scalar values
        
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  - - - This was completely okay in 2D applications, but now when we move on to 3- dimensional structures, a vector definition makes a lot more sense. Moment of a force F about point O is the cross product of vectors r and F
        
        Vector r is a position vector that goes from point O to any point which lies on the line of action of F
        F is the forece of the vector
        
        Notice: the i-component of the moment is not affected by Fx , the j-component of the moment is not affected by Fy etc. This is natural, because they produce no moment in the said direction; they are parallel.
        
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        • The magnitude of the cross product equals the area of the parallelogram, which has original vectors as sides.
        
        The direction of the cross product we can get using the right hand rule
        
        • A much quicker way is to can calculate the cross product using matrix notation as a determinant
        
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        MOMENT OF A COUPLE USING CROSS PRODUCT: Moment of a couple can be calculated using the cross product as a resultant of
        𝑴𝑶 = 𝒓𝑨 × (−𝑭) + 𝒓𝑩 × F
        
        Anyhow, if we think that the minus sign belongs to rA
        , this can be written as
        𝑴𝑶 = (𝒓𝑩 − 𝒓𝑨) × 𝑭
        Using the triangle rule of vector addition, we get
        𝑴𝑶 = 𝒓 × F
        
        Example:
        
        Replace the two cples acting on the pipe coloumn by a resultatn couple moment.
        
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        MOMENT OF A FORCE ABOUT COORDINATE AXIS: If we calculate the moment of a force about a specific point, we will get a moment vector whose direction is perpendicular to the plane containing the force and the moment arm. This moment is not always the one that we want to use in calculations; we may
        have to break it into parts
        
        Different components of the moment have different effects on our structure:
        ➢ x-component of the moment is trying to turn our structure around x-axis
        ➢ y-component of the moment is trying to turn our structure around y-axis
        ➢ z-component of the moment is trying to turn our structure around z-axis
        The structure usually has different “ability” to fight against the moment
        depending on its direction
        
        If we calculate the moment subject to point O, we will get a moment vector which acts on direction Ob; the magnitude of this moment has no clear physical meaning
        
        The moment about y-axis does; it tries to turn the pipe around y-axis and hence creates torsion.
        The moment about x-axis tries to turn the pipe around x-axis and hence creates bending
        
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        MOMENT OF A FORCE ABOUT AN ARBITRARY AXIS: What if the component that we are interested in is not in the direction of a coordinate axis – i.e. what if the pipe in the previous example would stick out of point O to some arbitrary direction, say 𝒂 = 𝑎𝑥𝒊 + 𝑎𝑦𝒋 + 𝑎𝑧𝒌 (instead of 𝒋)? In this case we need to calculate the projection of the moment vector 𝑴𝑶 on to
        vector 𝒂 – let’s call it 𝑴a or actually we just need the magnitude Ma
        
        Luckily, this can be calculated easily using the dot product:
        Here 𝒖𝒂 is the unit vector in the direction of 𝒂
        
        TRIPLE SCALAR PRODUCT: Using rules of mathematics, we can simplify our calculation by merging the dot
        product into our cross product that we need in calculating the 𝑴o
        
        Hence, we only need to calculate one 3x3-determinant and we get the magnitude of our moment subject to axis in the direction of 𝒂
        
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  - - - This means that we have a maximum of six support reactions or in other words six unknows. Normally not all six are present, In order to help in deciding whiich support reactions some osupport will develop think about the suport in the following way
        
        If the support reaction stop translation in some direction, then it willl develop a support forece in said direction
        If support restricts rotation subject to some axes then it will develop a support moment about said axis
        
        Note: Many time we acn ingnor the couple moments of single supports, if there are other support s which are porperly aligned
        
        Single connections theoretically are able to produce support couple moment, they are not very "good at it"
        Moment carrying capacity is limited
        Stress coaused by the mommnet often harms operations and shortens lifespan (especially bearings)
        
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        CONSTRAINTS FOR A RIGID BODY: Sometimes there may occur problems, where the structure has more supports
        than necessary in order to maintain equilibrium. These “additional” supports are called redundant supports and the number of them is often referred to as degree of redundancy
        
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  - - - Just in the 2D cases, these simplification don't align 100 % wiht reality, but they provide a good enough initial version for design
        
        We can use the same two methods that we used in 2D. Method of joint and Method of sections.
        
        If all of the truss must be determined for the method of joints. Start from a joint where there is one force you know and only three members. Sometime you might need to find the support forces before this. Then work your way up to the next joint that has 1-3 unknowns.
        
        Don't forget about symmetry and how they affect the direction of the vectors as well as how the change the final equation of equilibrium
        
        If you onlt need some of the foreces, it's often possible to ge tinto these quicker by using the method of sections.
        
        Use this as a road map for the method of joints:
        1) Chose a joint(preferably with max 3 unknown member forces, if possible) and draw FBD of the join(assume unknown forcess in tension)
        2) Find out direction of vectors for the members
        3) Write all forces in vector form
        4) Sum of all force acting on the joint must be zero, so write the vector equation for force equilibirum
        
        5) Group your vector components and convert the vector equation to three scalar equations
        6) Solve the scalar equations and mark the solved forces somewhere
        7) Go back to 1) and repeat until all forces have been solved
        
        Another of doing this using momentum instead of equilibirum:
        1) Find a suitable section that goes through 6 unknown member forces at maximum
        2) Draw a FBD of one half of the section (you may choose which one)
        3) Find out direction vectors for the unknown members & write forces in vector form
        4) Sum of moments subject to any point must be zero, so choose a suitable point for moment equilibrium and write it in vector form. A joint that has 3 unknonw forces would be best
        5) Find out required moment arm vectors and calculate cross products
        6) Solve the moment equilibrium equation (breakdown to 3 scalar eqs etc.)
        7) Sum of all forces acting on the section must be zero, so write the force equilibrium
        8) Solve the remaining unknown forces (again breakdown to 3 scalar eqs)
        
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  - - - In 2D internal loadings are normal force, shear force and moment. All of these internal loadings can be calculated at the point of interest by the following proccedure:
        
        1) Cute the structure at this point and draw a FBD
        2) In FBD, mark all the external forces and at the cut point the internal loadings N, V, M
        3) The "Cutaway" must be in equilibrium, so formulate equations of equilibrium
        4) Solve unknown internal loadings
        
        The only thing you have to take into account is the third dimesion, which means tha there can be up to 3 internal forces and 3 internal moments
        
        Fx is the internal normal for in this example
        Fx = N
        Fy and Fz both act as shearing forces, so the total internal shear force is
        
        My and Mz try to bend the structure, so together they produce the internal bending moment
        
        Mx tries to twist the structure, so it is the internal torsional moment
        
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  - - - There are two types of friction: fluid friction and dry friction
        ➢ In fluid friction, contacting surfaces are separated by a film of fluid (which can be gas or liquid)
        ➢ In dry friction, contacting surfaces are in direct contact
        In this course, we will only analyze problems involving dry friction
        
        CHARACTERISTICS OF DRY FRICTION: Dry friction is caused by irregularities on the surfaces of our contacting bodies.
        The actual distribution of normal forces is as shown in picture,
        because the resultant forces in these contacting points get smaller as we proceed
        
        Therefore, when we replace the distributed normal force by its resultant force N, it will not act in the same point as the weight W, but at distance x towards the direction of motion
        
        Same applies to resultant friction force F too, but this doesn’t make a differencebecause F acts along the plane
        
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  - - - If we now introduce a virtual differenetially small displacement, the resulting work will also be virtual and differentially small:
        
        The reason why this is called virtual is because the displacement doesn't actually exist.
        
        Sign rule:
        ➢ If F (or M) and displacement are in the same direction, work is positive
        ➢ If F (or M) and displacement are in opposing directions, work is negative
        
        PRINCIPLE OF VISTUAL WORK:
        If a body is in equilibrium, the sum of the virtual work done by all forces and moments acting on the body is zero.
        
        Idea: even if the displacement would happen, the forces and moments would maintain the body in equilibrium.
        Because the displacement is small, it doesn’t change the forces/moments from the original case.
        
        Example:A beam loaded at center by a point force. If a small virtual displacement y is applied at A, the only forces that do work are P and Ay (point B remains in place, so Bx and By do no work )
        
        Total work done must be zero, so
        
        Because d(theta) is not 0, it must be that
        
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    - - Now when we're working with systems that start from rest and will be in equilibrium in the end, their kinetic energies T1 and T2 will be zero. The amoutn of potential enrgy is alawys relative to the chosen datum, therefore we can write the total potential energy of the system as function V(s).
        
        Eg:
        
        If a differential small work is done to the system, it results in differentially small change to s and therfor ethe work is
        
        The total differential work is
        
        If we apply a small virtual displacement, it will result in virtual change in V:
        
        If in this equation 3 we subtiture 2 and thne 1 we get
        
        On the other hand, principle of virtual work requires that 𝛿𝑈 = 0, so
        
        (𝛿𝑠 ≠ 0)
        
        The result means that a frictionless system of rigid bodies is in equilibrium, when the first derivative of its potential function is zero.
        
        So the steps are the following:
        
        Sketch the system in its initial state (and if it help, also in it finals state)
        
        Establish a datum through a fixed point
        
        Express gravitational potential energies as a function of distance from datum
        
        Express potential energies of springs as function of thier stretch
        
        Link distances to streches orvice versa in such a way that all potential energies are expressed using only one coordinate
        
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    - - Equilibrium has three types of stability:
        
        Neutral (if displaced, will move but remain in equilibrium)
        
        Unstable (if displaced, will no longer be in equilibrium)
        
        Stable equilibrium (if displaced, will return to equilibirum)
        
        We can find out the equilibirum type simply by taking the 2nd derivative of our potential energy function (V''(s)). Then subtituting the value that results in equilirbium to it and checking the sign.
        
        Remember if the value is positive, its concave up
        and if the value is negative its concave donw.
        
        Eg: If the spring AD has a stifness of 18kN/m and it is unstreached when when 𝜃 = 60°, determine the angle 𝜃𝑒𝑞 when the system is in equilibrium. The load has a mass of 1500 kg and its mass center is in G. Investigate also the stability of the equilibrium position. Ignore the mass of links AB and CD.
        
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