Engineering Entire HSC Course PDF

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Detailed summary of the entire Engineering Studies HSC Course 2021....

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Year 12 Engineering Notes ________________________________________________________ Fundamental Mechanics

________________________________________________________ Simple Machines: A simple machine is a machine that makes work easier to perform by: - Changing the direction of a force - Increasing the magnitude of a force - Increasing the distance or speed of a force - Transferring a force from one place to another These consist of:

Levers: A rigid bar that pivots on a fixed fulcrum/hinge. Effort is applied to produce an output to move the lever. There are three classes of levers:

Mechanical Advantage - MA = Load/Effort = L/E The benefit obtained using simple machines, calculated using the relationship between load (resistance) and effort. It is dimensionless.

Velocity Ratio - VR=Distance Moved by Effort/Distance Moved by Load = DE/DL The ratio of the distance moved by the effort compared to the distance moved by the load. Values of VR greater than 1 indicate there is magnification of effort operating, whilst values less than indicate demagnification of effort operating. It is dimensionless.

Efficiency - η= Mechanical Advantage/Velocity Ratio = MA/VR Measure of the effectiveness, expressed as a percentage, with which the effort supplied is translated to work. It is a measure of the output compared with the input.

Gears: Gears are designed to interlock with other gears. The design prevents slippage, allowing change in speed, torque and rotational direction. When the gears mesh, they will rotate in opposite directions. A gear system consists of:

Driver Gear The first gear in the system, and is the one that has an energy source attached to it. It moves the system. The driver gear transports its energy to the driven gear which is moved in the system.

Driven Gear The gear that has energy transferred to it from the driver gear. It rotates in the opposite direction of the driver gear.

Idler Gear Is inserted between the driver and driven gear to keep the driver gear and driven gear rotating in the same direction. Reducing Gears In this configuration, the driver gear is smaller than the driven gear, meaning with every 2 rotations of the smaller gear, the bigger gear would complete 1 rotation. This reducing gear acts as a force multiplier. Like 1st gear in easy mode on a bicycle. Better acceleration.

Multiplying Gears -

The driver gear is bigger than the driven gear, meaning that with every rotation of the larger gear, the smaller gear would complete 2 rotations. This multiplying gear acts as a speed multiplier. Like 5th gear on hard mode on a bicycle. Better speed.

Gear Ratio VR = (Radius, Diameter, Circumference, #Teeth on the Driven Gear) / (Radius, Diameter, Circumference, #Teeth on the Driving Gear)

Speed/Velocity Ratio To determine how many revolutions per minute a gear experiences, we multiply the input speed by the speed ratio. The input speed is equal to the RPM of the driver/driven gear. The speed ratio is the reciprocal of VR: SR = (Radius, Diameter, Circumference, #Teeth on the Driver Gear) / (Radius, Diameter, Circumference, #Teeth on the Driven Gear) RPM = Input Speed x SR or = Input RPM/VR

Compound Gear Trains A compound gear consists of 2 different sized gears that rotate around the same shaft. 2 or more of these compound gears make up a compound gear train. The RPM is the same for both gears if they are on the same shaft, but vary in teeth.

Pulleys: A pulley consists of one or more wheels with a grooved rim carrying a rope. This rope is pulled making it easier to lift the object. Increasing the number of wheels will increase the ability to lift heavier objects. There are a variety of pulleys, such as:

One-Wheel Pulleys -

The mechanical advantage of this system is 1. Effort = Load

Two-Wheel Pulleys -

Adding an extra wheel will reduce the effort required to lift the load by half. The max is now supported by two portions of the same rope instead of one.

Mechanical Advantage The greater the mechanical advantage, the less effort required. Hence the more pulleys in the system, the less work needed. Ma is # pulleys

Velocity Ratio VR = #rope sections connected to load

Inclined Planes: An inclined plane is a surface that is at an angle to the horizontal

Velocity Ratio When the effort moves along the plane over a distance of L, the load moves through a vertical distance of H. Where H = Lsin() VR = distance that effort moves/distance that the load moves = L/H = L/LsinΘ = 1/sinΘ

Mechanical Advantage MA = Load/Effort

Screws & Screw Jack: A screw is a type of inclined plane with a circular or helical ramp with threads cut into the outside of the ramp. A Screw jack is used to raise and lower the horizontal heavy objects and heavy weights. Due to the pitch of the screw being very small compared to the length of the rod that is used to operate the screw, it has a high velocity ratio, and hence a high mechanical advantage.

Velocity Ratio VR = distance that the effort moves/distance that the load moves = Circumference/Pitch

Hydraulics: Hydraulics operate according to Pascal’s Principle which states that a pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid and the walls of the container.

Formula P1 = F1 / A1 P1 = P2

Pressure = Force / Area

VR = distance that the input piston moves / distance that the output piston moves = (diameter of output piston)^2 / (diameter of input piston)^2 = Area of output piston / Area of input piston

Friction: Friction is the force that acts OPPOSITE to the body’s direction of motion. Static Friction is the force that keeps an object at rest or just at the point of moving. Once it starts moving, the object experiences kinetic friction. The frictional force is determined by Fs = µ𝑁 = coefficient of friction x Normal Force (weight force perpendicular to the surface). The coefficient of friction is dependent upon the nature of the surface that the object is in contact with. Such as material, roughness and the angle of repose. The angle of repose is the angle of the inclined plane at which the object begins to slide down without an acting external force. As the angle of incline increases, the weight component acting down the plane will increase upto a point whereby the weight force is greater than the frictional force, causing the object to slide. Coefficient of Friction = µ= tanΘ = tan (angle of repose) Note: If the angle found using the equation is lower than the angle at which the plane is inclined, then the object will slide down.

Work, Energy & Power: -

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Work is defined as the energy that is transferred by a force. Therefore, for work to be done, the force applied must be able to move an object a certain distance If the force has a component in the same direction as the object's displacement, then we say that positive work has been done. Work done by acceleration forces. If the force has a component in the direction opposite to the object’s displacement (opposing object’s motion) then we say negative work has been done. I.e friction Work is measured in Joules (J) and is given by W = Fs where W = work, F = applied Force (N) and s = distance (m) Energy is defined as the ability to do work Potential energy is the energy held by an object due to its position relative to other objects. Known as stored energy. It is given by potential energy (PE in J) = Mass (m in kg) x acceleration due to gravity (g = 9.8 m/s^2) x height of object (h in m) Kinetic energy is the energy that an object possesses due to its motion. A car travelling down a road is an example of kinetic energy. It is given by kinetic energy (K in J) = ½ x mass (m in kg) x Velocity^2 (v^2 in m/s) A positive work value will suggest that the system has gained mechanical advantage. A negative value will suggest that the system has lost mechanical advantage. A zero work value will occur when the only force doing work is the gravitational force. This usually implies when the rider coasts down a hill and the friction is assumed zero.

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To find the work possessed by a system we can use; W = ∆KE (KEf - KEi) + ∆PE (PEf - PEi) As an object begins to move, the potential energy is transformed into kinetic energy as per the conservation of energy law. I.e AN EXAMPLE OF THIS IS WHEN A BOOK IS RESTING ON A TABLE. RELATIVE TO THE GROUND, THIS BOOK HAS MAXIMUM POTENTIAL ENERGY AND ZERO KINETIC ENERGY. WHEN THE BOOK GETS KNOCKED TO THE GROUND, IT PICKS UP KINETIC ENERGY DUE TO ITS MOTION AND LOSES POTENTIAL ENERGY (MAXIMUM KINETIC ENERGY IS ACHIEVED JUST BEFORE IT HITS THE GROUND AND POTENTIAL ENERGY IS ZERO AS IT HAS BEEN CONVERTED TO KINETIC

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Power is the rate at which work is done and is expressed in watts which is equivalent to Joules per second. P = W/t = KE + PE/t = Fs/t = Fv

Forces: Tension: defined as a pulling or stretching force. Elevator control cables undergo tensile forces when the pilot moves the control column. These cables cannot support a compressive force. Compression: forces push or try to reduce the length of a piece of material. While stationary on the runway aircraft landing gear struts are in compression. Shear: stress is the outcome of sliding one part over another in opposite directions. Shear in an aircraft structure is a stress exerted when two pieces of fastened material try to separate. Rivets and bolts of aircraft experience both shear and tensile stresses.

Bending & Shear Forces: Shear Force - at any point on a beam, may be defined as the algebraic sum of all external forces and its reaction to one side of a beam. The shear force is the reaction at a given point along a beam of the material to being sheared apart by the external forces. Always taken from the left. To calculate Shear Forces, we first find the sum of the vertical forces that exist to the left of the section. As shear force is equal to the sum of all external forces,we make it the subject of the equation. Although downward shear force is positive, we consider downward-external forces to be negative. Bending Moment - at any given point along a beam is equal to the total moment developed at that point by the external force system. If we sum moments at a given point along a beam that is in equilibrium, the value will be zero. Therefore to find bending stresses we have to only find the moment set up in one section of the beam.

Method of Joints: When applying this methods; -

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We must select a joint that has no more than 2 unknown forces Assumptions can be made on whether the member is in tension or compression, although may be altered when calculations have been completed If the force in a member is calculated to be a negative value, then this means that the force is opposite in actions. E.g if determining compression, a negative value would determine stress We usually begin with the reaction joints and work our way around the Truss Draw a free body diagram of all the forces acting on the joint that you are analysing Break up each force into its horizontal and vertical components where necessary Apply resultant f(x)=0 and/or f(y)=0 by determining how many unknown forces there are in each direction. If there are 2 unknown forces in the x direction, then we cannot use the f(x) equation. However, if there is 1 unknown force in the y direction, then we can apply the f(y) equation to solve for the unknown. We then apply f(x) to find remaining force Determine whether the member is in tension of compression using - If a member force points away from the joint it is attached to then we assume the force is in tension. If it points towards the joint, we say it is in compression. https://docs.google.com/presentation/d/1JE2kwQ_YbD5d3l4gpCFlKn_ilM-WN9xl/e dit#slide=id.p14

Method of Sections: Analytically cutting the truss into sections and using the static equilibrium equations to solve -

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Cutting the members exposes the force inside each members A section has a fixed size which allows us to use the moment equation to solve the unknown forces. We can solve up to 3 unknown, due to this, one should choose sections that involve cutting through no more than 3 members If a member points away from the joint that it is attached to then we say that the force is in tension, if it points towards the joint, it is in compression This technique has the advantage of going straight to the member in need of resolution with working individual joints along the way When cutting members remember to cut two or three members, one of which is the member whose force is to be found ONLY CONSIDER EXTERNAL FORCES AND THOSE IN THE MEMBERS

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BEING CUT Make a cut through the members in the question Members that have been cut now become forces Choose either the left or right side of the cut to analyse. Usually we take the side with less forces Either side is correct though Draw a labelled free body diagram with all the necessary information. Assume the force’s sense. Apply the appropriate static equilibrium equation to solve for the unknown forces. If the value is negative, then the correct sense is opposite to the assumed sense. Determine the nature of force by stating either tension (t) or compression (c)

Moments of Rollers: 1. 2. 3. 4. 5. 6. 7. 8.

Take the moments from the fixed point Determine where it is a positive (clockwise) or negative (anti-clockwise) moment We usually do not take it from the horizontal due to no perpendicular distance, until final steps when finding fixed point Force at the fixed point = sum of vertical moments minus the reaction moment (at the roller) Find the reaction force (roller) Force at the fixed point equals to sum of horizontal and vertical forces (including roller which will act as a minus) Split the horizontal and vertical forces in a force diagram, and calculate the total force in horizontal and vertical Make a force triangle, and using trigonometry to determine resultant force on the fixed point and the angle at which this occurs at

The fixed point has a vertical and horizontal component. The roller has only a vertical component as it cancels out the horizontal component.

Measured Properties: Static/Dead Loads- Loads that are derived from the weight of the structure. Dynamic/Live Loads- Loads that include shock/impact/vibration, additional weight (traffic), expansion/contraction, wind, water and ground movement. Compression Force- A type of force that presses inward on an object causing it to be pushed together/condensed.

Tension Force - A type of force that pulls on an object causing it to be stretched.

Torsion Force - The twisting of an object due to an applied force

Bending Force - A perpendicular force that is applied to a certain point on a surface, causing the object to bend

Stress & Strain Stress is the body’s internal reaction to an externally applied force. Tensile and compressive stresses are axial stresses because the external force is applied along the axis of the member.

A shear stress is a reaction to an external force applied at right angles to the axis. Stress is calculated by dividing the external force (or load) by the area. Stress = load/area

σ= L/A

For tensile and compressive stresses, it is always the area that is at right angles to the force. As the force is axial, then the area is perpendicular to the axis. This is commonly called the cross sectional area (CSA). For shear stress, the area is always measured parallel to the applied force. This is known as the shear area, which is the area that needs to break if the component is to fail. Shear stresses act along planes inside the material. These will be parallel to the applied force and the shear force will cause one section to slide over an adjacent section. If the member fails along two separate parallel planes, this is known as double shear. Stress - Pascal (Pa = N/m^2) = Force (N) / (m^2) Strain is the indication to how much a material will deform under a load. It is given by ϵ= e/l. The tension test involves the application of a load to a material sample. From this test, a load-extension graph is produced. A steadily increasing axial tensile load is applied to a small specimen until it breaks. During the test, the applied load is plotted against the extension of the material. A load extension graph will have exactly the same shape as a stress strain diagram. From the load/extension graph, a stress strain diagram can be derived and hence can also be determined, the; - Proportional limit stress - Yield stress - Proof stress - Ultimate tensile strength - Young's Modulus (stiffness) - Breaking point -

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Proportional limit stress is the stress at the end of the straight line section of the stress strain diagram. Also known as the elastic limit. Yield stress is the stress at which a marked increase in strain occurs without a corresponding increase in stress. This is known on the graph by the flattening out of the curve. When this happens, the yield continues after the proportional limit, and the yield stress can only be determined by another method. This offset method is known as the proof stress. Proof stress is the stress necessary to produce a certain amount of strain in the material. Depending on the service, an offset percentage of strain is requested by the engineer. Common values for strain are 0.1% and 0.2%. The offset method involves drawing a line parallel to the straight line section, from the percentage required, until it intersects with the curve. This approximates the yield stress.

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Ultimate Tensile stress (UTS) is the maximum stress a material can withstand before it fails but not necessarily breaks. This is read from the top of the graphed line. Because the material has deformed plastically, it is necessary to compensate for this by applying a factor of safety into design calculations. A factors of safety is a multiplier by which the calculated value is increased The factor of safety is calculated by the ultimate load (strength)/allowable load (stress). In other words it is the minimum breaking strength by the maximum force it will support. The FoS is always greater than 1. Young’s modulus is a measure of the stiffness of the material. This is shown on a stress strain diagram by the slope of the straight line section up to the proportional limit. The steeper the slope, the stiffer the material, the higher the value of Young’s modulus and the smaller the deformation. It is calculated by dividing stress by strain. E = σ/ϵ 1 GPA = 10^9 Pa or 10^3 MPa Toughness can also be determined from the stress strain diagram. It is represented by the area under the graph, from the initial point to the point of fracture. Fracture is indicated by where the graph ends. Toughness is the ability of a material to absorb energy when being deformed and therefore to resist deformation and failure. Breaking point is known as the fracture point. This is where the material breaks or fails under a tensile loading. It is normally less than the ultimate strength, as many materials undergo some stretching before failure. This demonstrates the ductility of the material. Because the material has increased in length, there must be a corresponding decrease in cross sectional area. Because this area has been reduced, a smaller force is necessary to continue to elongate the material

________________________________________________________ Communications

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________________________________________________________ Electronics

________________________________________________________ In Australia, Electricity is produced by; - ...