ME 304-DOM EQUATIONS AND IMPORTANT POINTS TO REMEMBER MODULE-1 1. Conditions of static equilibrium 2. Conditions of Equilibrium of Two Forces 3. Conditions of Equilibrium of Three Forces 4. Torque 5. Example: PRINCIPLE OF SUPER POSITION 6. 7. 8. µFR RsinΟ = µR MODULE-2 D’Alembert’s Principle DYNAMIC ANALYSIS OF SLIDER CRANK MECHANISM VERY IMPORTANT FOR PROBLEM SOLVING ON DYNAMIC ANALYSIS OF SLIDER CRANK MECHANISM MODULE-3 1. Turning Moment Diagram Torque on Crank πΉπ × π π ππ2θ π = πΉπ × π = πππ θ + Ο = πΉπ × π π ππθ + πππ Ο 2 π2 − π ππ2 θ Turning Moment Diagram = Plot between T and θ Crank Effort Diagram= Plot between FT and θ 2. Turning moment diagram for a single cylinder, double acting steam engine (TWO STROKE ENGINE) 3. Turning Moment Diagram for a Four Stroke Cycle Internal Combustion Engine –Single Cylinder 4. Turning Moment Diagram for a Multi-cylinder Engine 5. Fluctuation of Energy οΌ The variations of energy above and below the mean resisting torque line are called fluctuations of energy. οΌ The areas BbC, CcD, DdE, etc. represent fluctuations of energy. οΌ Maximum fluctuation of energy, β E = Maximum energy – Minimum energy 6.Coefficient of Fluctuation of Energy (CE) It may be defined as the ratio of the maximum fluctuation of energy to the work done per cycle. 7.Coefficient of Fluctuation of Speed 8.Energy Stored in a Flywheel (Derivation Important -5 Marks Question) 9.Energy Stored in a Flywheel (Derivation Important -5 Marks Question)………… ΔE 10. Maximum Acceleration of Flywheel (αf) Maximum Torque= Tmax Mean Torque = Tmean Fluctuation in Torque, ΔT = Tmax – Tmin We, Know that, ΔT απ = πΌ I = mass moment of Inertia = mk2 where, k= radius of gyration 11. Dimensions of Rim of a Flywheel Centrifugal Force Flywheel for Punching Press MODULE-4 Gyroscopic Couple/Torque Gyroscopic Couple Summary of Direction Active and Gyroscopic Couple for Aeroplane Viewing Direction Propellor Rotation Precision of Aeroplane Active Gyroscopic Couple Reactive Gyroscopic Couple REAR Clockwise Left Clockwise Anti-Clockwise REAR Clockwise Right Anti-Clockwise Clockwise REAR Anti-Clockwise Left Anti-Clockwise Clockwise REAR Anti-Clockwise Right Clockwise Anti-Clockwise FRONT Clockwise Left Anti-Clockwise Clockwise FRONT Clockwise Right Clockwise Anti-Clockwise FRONT Anti-Clockwise Left Clockwise Anti-Clockwise FRONT Anti-Clockwise Right Anti-Clockwise Clockwise Equations Gyroscopic Effect on Ships Terms Used in a Naval Ship οΌ The top and front views of a naval ship are shown in Fig . The fore end of the ship is called bow and the rear end is known as stern or aft. The left hand and right hand sides of the ship, when viewed from the stern are called port and star-board respectively. οΌ We shall now discuss the effect of gyroscopic couple on the naval ship in the following three cases: 1. Steering 2. Pitching, and 3. Rolling On Ship Conditions for Pitching of Ship οΌ A little consideration will show, that the weight of the vehicle (W) will be equally distributed over the four wheels which will act downwards. οΌ The reaction between each wheel and the road surface of the same magnitude will act upwards. οΌ Therefore Road reaction over each wheel, = W/4 = m.g /4 newtons The positive sign is used when the wheels and rotating parts of the engine rotate in the same direction. If the rotating parts of the engine revolves in opposite direction, then negative sign is used. οΌ Due to the gyroscopic couple, vertical reaction on the road surface will be produced. οΌ The reaction will be vertically upwards on the outer wheels and vertically downwards on the inner wheels. οΌ Let the magnitude of this reaction at the two outer or inner wheels be P newtons. Then P × x = C or P = C/x οΌ Vertical reaction at each of the outer or inner wheels, P /2 = C/ 2x οΌ Note: We have discussed above that when rotating parts of the engine rotate in opposite directions, then –ve sign is used, i.e. net gyroscopic couple, C = CW – CE, When C E > CW, then C will be –ve. οΌ Thus the reaction will be vertically downwards on the outer wheels and vertically upwards on the inner wheels. 65 2. Effect of the centrifugal couple οΌ Since the vehicle moves along a curved path, therefore centrifugal force will act outwardly at the centre of gravity of the vehicle. οΌ The effect of this centrifugal force tends to overturn the vehicle. οΌ We know that centrifugal force, οΌ This overturning couple is balanced by vertical reactions, which are vertically upwards on the outer wheels and vertically downwards on the inner wheels. οΌ Let the magnitude of this reaction at the two outer or inner wheels be Q. Then οΌ A little consideration will show that when the vehicle is running at high speeds, PI may be zero or even negative. οΌ This will cause the inner wheels to leave the ground thus tending to overturn the automobile. In order to have the contact between the inner wheels and the ground, the sum of P/2 and Q/2 must be less than W/4. Stability of a Two Wheel Vehicle Taking a Turn οΌ Consider a two wheel vehicle (say a scooter or motor cycle) taking a right turn as shown in Fig m = Mass of the vehicle and its rider in kg W = Weight of the vehicle and its rider in newtons = m.g h = Height of the centre of gravity of the vehicle and rider rw = Radius of the wheels R = Radius of track or curvature, 68 οΌ A little consideration will show that when the wheels move over the curved path, the vehicle is always inclined at an angle θ with the vertical plane as shown in Fig. οΌ This angle is known as angle of heel (θ). οΌ In other words, the axis of spin is inclined to the horizontal at an angle θ, as shown in Fig. οΌ Thus the angular momentum vector Iω due to spin is represented by OA inclined to OX at an angle θ. οΌ But the precession axis is vertical. Therefore the spin vector is resolved along OX. Notes: οΌ When the engine is rotating in the same direction as that of wheels, then the positive sign is used in the above expression and if the engine rotates in opposite direction, then negative sign is used. οΌ The gyroscopic couple will act over the vehicle outwards i.e. in the anticlockwise direction when seen from the front of the vehicle. The tendency of this couple is to overturn the vehicle in outward direction. 71 MODULE-5 Part-I- Free Undamped Vibration 1. General Equation of Vibration: m = mass; x = displacement; s = Stifness of spring π2π₯ = Acceleration= 2 ππ‘ 2. Natural Frequency of Longitudinal Vibration: Axial or Longitudinal Deflection 3. Natural frequency of Transverse Vibration: If Inertia is considered 4. Energy Method: 5. Newton’s Method: 6. Equivalent Stiffness of Spring: PART-II DAMPED FREE VIBRATION Displacement of underdamped vibration PART-III FORCED VIBRATION 4 5. Amplitude at Resonance, A Force Transmitted at Resonance, Ft π΄= 6. Maximum Amplitude of Vibration, Amax = πΉπ‘ π 1 + 4ζ2 MODULE-6 1. Natural frequency of Transverse Vibration: If Inertia is considered 2. Whirling of Shaft: Also, M= Bending Moment y1=Distance from neutral Axis to outer Fiber of beam I = Moment of Inertia 3. Dunkerley’s Equation: Natural frequency of Transverse Vibration 4. Natural frequency of Torsional Vibration: 5. Torsionally Equivalent Shaft: Length of Torsionally Equivalent Shaft, l, Values of static deflection (δ) for the various types of beams and under various load conditions and under various load conditions Equations of Maximum Bending Moment of Beams