Report

Mechanics of Machines Dr. Mohammad Kilani Class 4 Velocity Analysis DERIVATIVE OF A ROTATING VECTOR Derivative of a Rotating Unit Vector A unit vector in the θ direction, uθ, is a vector of unity magnitude and an angle θ with the x- axis. It is written as: uθ cosi sin j If uθ rotates, it angle θ changes with time, the time derivative of uθ is found by applying the standard differentiation rules on the expression of uθ above duθ d d sin i cosj dt dt dt duθ d sin i cosj dt dt sinθ j uθ θ cosθ i Derivative of a Rotating Unit Vector Noting that sin 2 cos cos 2 sin uθ cosi sin j duθ d sin i cosj dt dt the time derivative of the vector uθ is duθ d sin i cosj dt dt duθ d cos 2i sin 2 j dt dt duθ d u 2 dt dt sinθ j uθ θ cosθ i Derivative of a Rotating Unit Vector The derivative of a unit vector whose angle with the x-axis is θ, (θ changes in time), is a vector whose angle with the x-axis is θ+π/2 and whose magnitude is dθ/dt. If we define a vector ω as a vector in the k direction of magnitude ω = dθ/dt, then d k dt duθ d u dt dt uθ cosi sin j duθ d u dt dt 2 duθ /dt uθ θ + π/2 θ ω 2 ω uθ Derivative of a Rotating Vector A vector rθ = r uθ , is a general vector of magnitude r pointing in the θ direction. The time derivative of rθ is found by applying the normal differentiation rules on the expression for rθ rθ ruθ drθ dt drθ dt drθ dt drθ dt dr uθ dt dr uθ dt dr uθ dt dr uθ dt duθ r dt d r u dt r ω uθ ω rθ duθ /dt uθ θ + π/2 2 θ Derivative of a Rotating Vector Given a vector rθ = ruθ which rotates relative to the reference coordinates, the time derivative of rθ has two components; a component in the direction of uθ and a component normal to uθ in the direction of uθ+π/2 The magnitude of the component of drθ /dt in the direction of uθ is equal to dr/dt; that is the time derivative of the length of rθ. The magnitude of the component of drθ /dt in the direction of direction of uθ+π/2 is equal to ωr ω x rθ (dr/dt) uθ θ + π/2 r θ θ VELOCITY ANALYSIS OF FOUR BAR MECHANISMS Derivative of the Loop Closure Equation for a Four Bar Kinematic Chain The loop closure equation of a 4-bar kinematic chain is written as r2 r3 r1 r4 r2 r3 r1 r4 0 d r2 r3 r1 r4 0 dt d r2u2 r3u3 r1u1 r4u4 0 dt r22u2 2 r2u2 r33u3 2 r3u3 r11u1 2 r1u1 r44u4 2 r4u4 0 When all the links in the chain are of constant lengths, the equation above reduces to r u 2 2 2 2 r u 3 3 3 2 r u 1 1 1 2 r u ω2 r2 ω3 r3 ω1 r1 ω4 r4 0 4 4 4 2 0 Derivative of the Loop Closure Equation for a Four Bar Mechanism For a four bar mechanism with link 1 fixed we have dθ1/dt = ω1 = 0. r u 2 2 2 2 r u 3 3 3 2 r u 4 4 4 2 The vector equation above contains two scalar equations and can be solved for two unknowns. Knowledge of dθ2/dt allows the calculation of dθ3/dt and dθ4/dt . r22 sin 2 r33 sin 3 r44 sin 4 r22 cos 2 r33 cos3 r44 cos 4 Derivative of the Loop Closure Equation for a Four Bar Mechanism r u 2 2 2 2 r u 3 3 3 2 r u 4 4 4 2 Eliminate dθ3/dt by carrying out a dot product with uθ3 on both sides of the equation r22 cos 2 2 3 r44 cos 4 2 3 r sin r sin 2 2 2 4 4 3 4 4 4 3 r2 sin 2 3 r2 sin 2 3 2 2 r4 sin 4 3 r4 sin 4 3 Alternatively, eliminate dθ4/dt by carrying out a dot product with uθ4 on both sides of the equation r22 cos 2 2 4 r33 cos3 2 4 0 r sin r sin 2 2 2 3 3 4 3 3 3 4 r2 sin 2 4 r sin 2 4 r sin 2 4 2 2 2 2 2 r3 sin 3 4 r3 sin 3 4 r3 sin 4 3 Angular Velocity Ratio and Mechanical Advantage The angular velocity ratio mV is defined as the output angular velocity divided by the input angular velocity. For a four bar mechanism with link 2 as the input and link 4 as the output this is expressed as out 4 mV in 2 The efficiency of a four bar linkage is defined as the output power over the input power, Pout Toutout Pin Tinin Assuming 100% efficiency, which is normally approached by four bar mechanisms, we have Tout in 1 mT Tinin out mV Angular Velocity Ratio and Mechanical Advantage The mechanical advantage is defined as the ratio between the output force to the input force mA Fout Tout rout r mT in Fin Tin rin rout VELOCITY ANALYSIS OF SLIDER-CRANK MECHANISMS Velocity Analysis of a Slider-Crank Mechanism Design parameters: r2, r3, r4, θ1. r3 Position analysis parameters: r1, θ2, θ3 Velocity analysis parameter. Find dr1/dt, dθ2/dt , dθ3/dt r2 r3 r1 r4 r22u2 2 r33u3 2 r1u1 To eliminate ω3 dot product both sides by uθ3 r22 cos2 2 3 r1 cos1 3 r22 sin2 3 r1 cos1 3 r1 r2 sin3 2 2 cos3 1 r4 r2 rp r1 To eliminate dr1/dt dot product both sides by u(θ1+π/2) r22 cos2 1 r33 cos3 1 3 r2 cos2 1 2 r3 cos3 1 Velocity Analysis of an Inverted Slider-Crank Mechanism Given r1 , r2, θ1, θ2 , ω2 Position analysis: Find r3, θ3 Velocity analysis: Find dr3/dt, dθ3/dt r2 r1 r3 r2u2 r1u1 r3u3 r22u2 2 r3u3 r33u3 2 To eliminate ω3 and find dr1/dt dot product both sides by uθ3 r22 cos2 2 3 r3 r22 sin3 2 r3 To eliminate dr1/dt and find ω3 dot product both sides by u(θ3+π/2) r22 cos2 3 r33 r cos3 2 3 2 2 r3 r2 θ2 r1 r3 θ1 θ3 Velocity Analysis of an Inverted Slider-Crank Mechanism Given r1 , r2, r4 , θ1, θ2 , ω2 Position analysis: Find r3, θ3 , θ4 Velocity analysis: Find dr3/dt, dθ3/dt , dθ4/dt r2 r3 r1 r4 r2u2 r3u 3 r1u1 r4u4 r22u2 2 r3u 3 r33u3 r u 2 4 4 4 with 4 3 2 r22u2 2 r3u 3 r33u3 r u 2 4 4 3 2 r3 r2 r1 r4 Velocity Analysis of an Inverted Slider-Crank Mechanism r22u2 r u r u 2 3 3 3 3 3 Dot bothsidesby u3 2 r u 2 4 4 3 r22 cos 2 3 r33 0 r2 cos 2 3 2 r3 Dot bothsidesby u 3 3 r22 cos 2 2 3 r3 r44 r22 sin 2 3 r3 r44 But 4 3 r2 cos 2 3 2 r3 r22 sin 2 3 r3 r4 r2 cos 2 3 2 r3 r cos 2 3 r3 r4 2 r2 sin 2 3 2 r3 r3 r2 r1 r4 Example r5 b a LoopI r2u2 a b u3 r1u1 r4u4 r2 LoopII r2u2 au3 r5u5 r1u1 su6 s r4 r1 LoopClosureEquation(1) a b u3 r1u1 r4u4 r2u2 Dot bothsidesof the equationby itself a b 2 r12 r22 r42 2r1r4 cos4 1 2r1r4 cos2 1 2r2r4 cos2 4 a b 2 r12 r22 r42 2r1r4 cos4 2r1r4 cos2 2r2 r4 cos2 cos4 sin 2 sin 4 A cos 4 B sin 4 C 0 C At 2 2 Bt C A 0 t1,2 B Alsofromloopclosurerquation(2) a b cos3 r1 cos1 r4 cos4 r2 cos2 2 2 r2 B 2r2 r4 sin 1 3 1,2 tan1 r1 sin 1 r4 cos4 r2 cos2 r1 cos1 r4 cos 4 r2 cos 2 C r12 r22 r42 2r1r2 cos 2 a b 2 t tan 4 2 b a A 2r1r4 2r2 r4 cos 2 a b sin 3 r1 sin 1 r4 cos4 r2 cos2 r5 B C A , 4 1, 2 2 tan1 t1, 2 AC 2 s r4 r1 Example r5 LoopClosureEquation(2) r5u5 r1u1 su6 r2u2 au3 a Dot bothsidesof the equationby itself r2 r5 r12 s 2 r22 a 2 2r1s cos6 1 2r1r2 cos 2 1 2r1a cos3 1 2 2r2 s cos6 2 2as cos3 1 2r2 s cos6 2 2r2a cos 2 3 b s r4 r1 Let1 0, 6 2 r5 r12 s 2 r22 a 2 2r1r2 cos 2 2r1a cos3 2 2r2 s cos6 2 2as cos3 1 2r2 s cos6 2 2r2a cos 2 3 a b2 r12 r22 r42 2r1r4 cos4 2r1r4 cos2 C At 2 2 Bt C A 0 t1, 2 B B 2 C 2 A2 , 4 1, 2 2 tan1 t1, 2 AC r5 Alsofromloopclosurerquation(2) 3 1,2 tan1 r1 sin 1 r4 cos4 r2 cos2 b a a bcos3 r1 cos1 r4 cos4 r2 cos2 a bsin 3 r1 sin 1 r4 cos4 r2 cos2 r1 cos1 r4 cos 4 r2 cos 2 r2 A 2r1r4 2r2 r4 cos 2 B 2r2 r4 sin 1 C r12 r22 r42 2r1r2 cos 2 a b 2 t tan 4 2 s r4 r1 HW#3 6-17, 6-18 (b+c), 6-21 (b), 6-44. METHOD OF INSTANT CENTERS Relative Velocities Between Two Points on a Rigid Body Given any two points A and B that lie on a rigid body, let the line AB be a line passing through A and B, then the components of the velocity of A and the velocity of B on the line AB must be equal. The velocity of point B relative to A must be normal to the line AB A B vA vB vB A A B Instant Center Relative to the Ground If line AA’ is drawn normal to the direction of vA, then the velocity of any point on the rigid body that falls on line AC must be perpendicular to the line AA’ In a similar manner, if line BB’ is drawn perpendicular to the direction of vB, then the velocity of any point on the rigid body that falls on this line must be perpendicular to the line BB’ If point I is the intersection of lines AA’ and BB’, then the velocity of point I must be perpendicular to both AA’ and BB’. This can only happen when the velocity of point I is zero. A B I B’ A’ v A v I k rA I v A k rA I v A rA I v A rA I vB rB I Instant Center Relative to the Ground Point I determined as before is called the instant center of zero velocity of the rigid body with respect to the ground. The direction of the velocity of point A is normal to the line IA. The direction of the velocity of line B is normal to the line IB. The direction of the velocity of any other point C on the same body must be normal to the line IC. The magnitude of the velocity of any point on the link is proportional to its distance from the point I. A B C I B’ A’ v A vB v A IA IB IC Velocity of a Point on a Link by Instant Center Method Consider two points A and B on a rigid link. Let vA and vB be the velocities of points A and B at a given instant. If vA is known in magnitude and direction and vB in direction only, then the magnitude of vB may be determined by the instantaneous centre method. Draw AI and BI perpendiculars to the directions vA and vB respectively. Let these lines intersect at I, which is known as instantaneous centre or virtual centre of the link. The complete rigid link rotates about the centre I at the given instant. The relations shown may be used to determine the magnitude of the velocity of point B. v A v I k rA I v A k rA I v A rA I v A rA I vB rB I Instant Center Between Two Rigid Bodies In the foregoing discussion, the velocities of points A and B were assumed to be reference to a coordinate system attached to the ground. The resulting center is called an instant center relative to the ground. The velocities of points A and B could as well be taken relative to coordinate system attached to another body. The resulting point in this case will have a zero velocity with respect to that body. In other words, point I will have the same velocity for both bodies. A B I B’ A’ Instant Center Between Two Rigid Bodies A instant center between two bodies is a: A point on both bodies A point at which the two bodies have no relative velocity. A point about which one body may be considered to rotate around the other body at a given instant. A B I B’ A’ Instant Center Between Two Rigid Bodies When two links are connected to one another by a revolute joint, the center of the connecting joint is an instant center for the two links. When two links are not connected, an instant center between the two links will also exist and can be determined if the velocities of both links is known. A A’ Instant Center Between Two Rigid Bodies The number of instant centers in a constrained kinematic chain is equal to the number of possible combinations of two links. The number of pairs of links or the number of instantaneous centers is the number of combinations of L links taken two at a time. Mathematically, number of instant centers is, LL 1 N 2 Types of Instantaneous Centers The instant centers for a mechanism are of the following three types : 1. Primary (Permanent) instant centers. They can be fixed or moving 2. Secondary Instant centers (Not permanent) C n n 1 2 Instant Centers of a Four Bar Mechanism D Consider a four bar mechanism ABCD as shown. The number of instant centers (N) in a four bar mechanism is given by N 44 1 12 6 2 2 The instant centers I12 and I14 are fixed instant centers as they remain in the same place for all configurations of the mechanism. The instant centers I23 and I34 are permanent instant centers as they move when the mechanism moves, but the joints are of permanent nature. The instantaneous centres I13 and I24 are neither fixed nor permanent as they vary with the configuration of the mechanism. 3 2 C A 4 1 (Ground) B n n 1 C 2 Location of Instant Centers When the two links are connected by a pin joint (or pivot joint), the instant center lies on the center of the pin as. Such an instant center is of permanent nature. If one of the links is fixed, the instant center will be of fixed type. When the two links have a pure rolling contact (without slipping), the instantaneous centre lies on their point of contact, as this point will have the same velocity on both links. When the two links have a sliding contact, the instant center lies at the center of curvature of the path of contact. This points lies at the common normal at the point of contact. Kennedy (or Three Centers in Line) Theorem Kennedy’s theorem states that any three bodies in plane motion will have exactly three instant centers, and the three centers will lie on the same straight line. Note that this rule does not require that the three bodies be connected in any way. We can use this rule, in conjunction with the linear graph, to find the remaining lCs which are not obvious from inspection. Example: Instant Centers of a Four Bar Mechanism Draw a circle with all links numbered around the circumference Locate as many ICs as possible by inspection. All pin joints will be permanent ICs . Connect the links numbered on the circle to create a linear graph and record those found. Identify a link combination for which the IC has not been found, and draw a dotted line connecting those two link numbers. Identify two triangles on the graph which each contains the dotted line and whose two other sides are solid lines representing the ICs the already found. Use Kennedy’s theorem to locate the needed IC. Example: Instant Centers of a Four Bar Mechanism Example: Instant Centers of a Slider-Crank Mechanism Example: Instant Centers of a Slider-Crank Mechanism Example: Instant Centers of a Cam and Follower: Common Normal Method Example: Instant Centers of a Cam and Follower: Effective Link Method Velocity Analysis with Instant Centers Once the instant centers (ICs) of a linkage with respect to the ground link have been found, they can be used for a very rapid graphic analysis for that link. Note that some of the ICs may be very far removed from the links. For example, if links 2 and 4 are nearly parallel, their extended lines will intersect at a point far away and not be practically available for velocity analysis. Velocity Analysis with Instant Centers From the definition of the instant center, both links sharing the instant center will have identical velocity at that point. Instant center I13 involves the coupler (link 3) which is in general plane motion, and the ground link which is stationary. All points on the ground link have zero velocity in the global coordinate system, which is embedded in link 1. Therefore, I13 must have zero velocity at this instant, and it can be considered to be an instantaneous "fixed pivot" about which link 3 is in pure rotation with respect to link 1. A moment later, I13 will move to a new location and link 3 will be "pivoting" about a new instant center. Velocity Analysis with Instant Centers If ω2 is known for the mechanism shown, the magnitude of the velocity of point A can be computed as vA = ω2 O2A Its direction and sense can be determined by inspection. Note that point A is also instant center and it has the same velocity as part of link 2 and as part of link 3. Since link 3 is effectively pivoting about I13 at this instant, the angular velocity ω3 can be found by ω3 = vA /AI13 . Once ω3 is known, the magnitude of vB can also be found from vB = ω3 AI13 Once vB is known, ω4 can also be found from ω4 = vB /BO4 = vB /BO4 . Finally, vC (or the velocity of any other point on the coupler) can be found from vC = ω3 CI13 3 vA AI13 vB 3BI13