Dynamic force analysis slider crank mechanism

Dynamics of Machinery

1. DYNAMICS FORCE ANALYSIS Presented By: Mr. N V S G SASIKIRAN Assistant Professor Mechanical Dept. GIET (A) Lecture Details: DOM : Unit 3 DOM/Mechanical, V th Semester.

2. OBJECTIVES ● ● To learn about Dynamic Force Anlaysis and various techniques involved. Study the concept of Inertia Torque. ● U n der s t a n d t h e D y nam i c A n a l y si s o f F ou r - L i n k & S li d e r - C r a n k Mechanisms. ● Emphasise the Forces acting on Reciprocating parts of an Engine.

3. DYNAMIC FORCE ANALYSIS Definition: When the inertia forces are considered in the analysis of the mechanism, the analysis is known as dynamic force analysis. Now applying D’Alembert principle one may reduce a dynamic system into an equivalent static system and use the techniques used in static force analysis to study the system. D’Alembert Principle The principle of virtual work states that the sum of the incremental virtual works done by all external forces F i acting in conjunction with virtual displacements δS i of the point on which the associated force is acting is zero. i.e, δ W = ∑ F i · δ S i = 0.

4. DYNAMICS OF RIGID LINK IN PLANE MOTION In many situations, it is necessary to determine the forces to be applied on a mechanism to keep it in Equilibrium or to accelerate it. Both are part of the same general problem. However, we treat them separately here. First we look at how the force is specified. K = 1 m v 2 = 1 m x ˙ 2 , P = 1 k x 2 2 2 2 However, the Dynamic force analysis of planar mechanisms is different. When the mechanism is at a position with no velocity, and the forces on the mechanism do not cause any acceleration, the mechanism is said to be in equilibrium in that position. The various techniques to carry on the dynamic analysis of planar mechanisms are described in upcoming slides.

5. FORCE ANALYSIS TECHNIQUES Method 1 ­ Superposition: Given a mechanism with known position, velocity, and acceleration conditions, derive Newtons equations for dynamic equilibrium. These equations are linear in the forces and therefore Superposition principles can be applied: Inertial and applied forces on each link can be considered individually and then superposed to determine their combined effect. This approach is good for building intuition and solving by hand but can be very long. Method 2 – Matrix Method: All inertial and applied forces are considered at once. The dynamic equations become coupled in the unknown forces and are solved using linear algebra techniques. This approach is the best for computer application. Method 3 – Energy Method: Here, only forces that do work on the mechanism are considered. An equation of conservation of energy is written that results in 1 scalar equation with 1 unknown. This is the easiest method if only the input force is required.

6. IN E R T I A T OR Q UE In plane motions involving accelearations, the inertia force acts on a body through its centre of mass. But if it is acted upon by forces such that their resultant does not pass through the center of mass, a couple also acts on the body. In graphical methods, it is possible to replace the inertia force and inertia couple by an equivalent offset inertia force from the center of mass. The perpendicular displacement h of the force from the centre of mass is such that the torque so produced is equal to inertia couple acting on the body. i . e. , h is taken in such a way that the force produces a moment about the centre of mass, which is opposite in sense to the angular accelaration α . T i = C i F i × h = C i C i − I g α m k 2 α k 2 α = f h = F = − mf = m f i g g g

7. DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS Consider the motion of a crank and connecting rod of a reciprocating steam engine as shown in Figure. Let OC be the crank and PC the connecting rod. Let the crank rotates with angular velocity of ω rad/s and the crank turns through an angle θ from the inner dead centre (I.D.C). Let x be the displacement of a reciprocating body P from I.D.C. after time t seconds, during which the crank has turned through an angle θ . Let l = Length of connecting rod between the centres, r = R adi us o f cr an k or cr ank p i n c i r c le, φ = Inclination of connecting rod to the line of stroke PO, and n = Ratio of length of connecting rod to the radius of crank = l/r.

8. ; Velocity of the Piston: From the geometry, x = P′P = OP′ − OP = ( P′C′ + C′O ) − ( PQ + QO ) = ( l + r ) − ( l cos φ + r cos θ ) = r [(1 − cos θ ) + n (1 − cos φ )] From triangles CPQ and CQO , CQ = l sin φ = r sin θ or l/r = sin θ /sin φ n = sin θ /sin φ or sin φ = sin θ / n On expanding, we get i.e., DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

9. Substituting the value of (1 – cos φ ) in above equation, we have Differentiating w.r.t θ , Therefore, Velocity of P with respect to O or velocity of the piston P, Substituting the value of dx/d θ , we have DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

10. Acceleration of the piston: Since the acceleration is the rate of change of velocity, therefore acceleration of the piston P, Differentiating velocity equation with respect to θ , p p d v d v p d θ dv p dt d θ dt d θ a = = × = × ω S u b s t i t u t i ng t he v a l ue of dv p i n t h e a b ove e q uat i on , w e have d θ P a = ω . r ( cos θ + cos 2 θ n 2 ) × ω = ω . r ( cos θ + c o s 2 θ n ) DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

11. Velocity of the connecting rod: From the geometry of the figure, we find that CQ = l sin φ = r sin θ i.e, Differentiating both sides with respect to time t, Since the angular velocity of the connecting rod PC is same as the angular velocity of point P with respect to C and is equal to d φ /dt, therefore angular velocity of the connecting rod We know that, i.e, ω P = DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

12. Acceleration of the connecting rod: Angular acceleration of the connecting rod PC, P C α = Angular acceleration of P with respect to C = We know that, Now, differentiating velocity equation, we get d ( ω ) P C dt dt d θ dt d θ d ω P C = d ω P C × d θ = d ω P C × ω DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

13. DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

14. Problem 2: The crank and connecting rod of a steam engine are 0.3 m and 1.5 m in length. The crank rotates at 180 r.p.m. clockwise. Determine the velocity and acceleration of the piston when the crank is at 40 degrees from the inner dead centre position. Also determine the position of the crank for zero acceleration of the piston. Given : r = 0.3; l = 1.5 m ; N = 180 r.p.m. or ω = π × 180/60 = 18.85 rad/s; θ = 40 ° DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

15. Solu t ion: DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

16. Solution (contd.): Position of the crank for zero acceleration of the piston is determined as, Let θ 1 = Position of the crank from the inner dead centre for zero acceleration of piston. We know that acceleration of piston, i.e, DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

17. Problem 3: In a slider crank mechanism, the length of the crank and connecting rod are 150 mm and 600 mm respectively. The crank position is 60 ° from inner dead centre. The crank shaft speed is 450 r.p.m. (clockwise). Using analytical method, determine: Velocity and acceleration of the slider, and Angular velocity and angular acceleration of the connecting rod. Given : r = 150 mm = 0.15 m ; l = 600 mm = 0.6 m ; θ = 60 ° ; N = 400 r.p.m or ω = π × 450/60 = 47.13 rad/s DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

18. Solution: Velocity and acceleration of the slider We know that ratio of the length of connecting rod and crank, n = l / r = 0.6 / 0.15 = 4 DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

19. Solution: Angular velocity and angular acceleration of the connecting rod DYNAMIC ANALYSIS OF SLIDER- CRANK MECHANISMS

20. FORCES ON THE RECIPROCATING PARTS OF AN ENGINE o r The various forces acting on the reciprocating parts of a horizontal engine are shown in Figure. The expressions for these forces are: Piston Effort: F P = N et l o ad o n pi s t o n ∓ I n e rti a f o r ce = F L ∓ F I − R F Where R F = Frictional resistance. The –ve sign is used when the piston is accelerated, and +ve sign is used when the piston is retarded. Force acting on connecting rod: Q F = F P c o s ϕ

21. Thrust on the sides of the cylinder walls or normal reaction on the guide bars: Crank­pin effort and thrust on crank shaft bearings: Crank­pin effort Bearings thrust Crank­effort: T FORCES ON THE RECIPROCATING PARTS OF AN ENGINE

22. ASSIGNMENT QUESTIONS State and explain D’Alembert’s principle. Derive an expression for the angular acceleration of the connecting rod of a reciprocating engine. What do you mean by dynamical equivalent system? Explain. In a four­bar link mechanism ABCD, the link AB revolves with an angular velocity of 10rad/s and angular acceleration of 25 rad/s 2 at the instant when it makes an angle of 45 o with AD, the fixed link. The lengths of the links are AB = CD = 800mm, BC = 1000mm & AD = 1500mm. The mass of the links is 4kg/m. Determine the torque required to overcome the inertial forces, neglecting the gravitational effects. Assume all links to be of uniform cross­sections.

23. “ Theory of Machines,” S.S Ratan, Mc. Graw Hill Publ. “ Theory of machines,” Khurmi, S.Chand. Concepts borrowed from: