WEEKS 2-3 Dynamics of Machinery References Theory of Machines and Mechanisms, J.J. Uicker, G.R.Pennock ve J.E. Shigley, 2003 Makine Dinamiği, Prof. Dr. Eres SÖYLEMEZ, 2013 Uygulamalı Makine Dinamiği, Jeremy Hirschhorn, Çeviri: Prof.Dr. Mustafa SABUNCU, 2014 Prof.Dr.Hasan ÖZTÜRK 1
Example:This slider-crank mechanism is in static equilibrium in the shown configuration. A known force F acts on the slider block in the direction shown. An unknown torque acts on the crank. Our objective is to determine the magnitude and the direction of this torque in order to keep the system in static equilibrium. 2 3 4 F 32y F 32x F 23x F 34x F 34y F 12x F 23y F 43y T 12 F 14 =F 14y T 14 F F 12x 32x 12 y 32 y F 12y + F = 0 + F = 0 af + bf + T = 0 32x 32 y 12 (Sum of moments about O ) 2 F23x + F43x = 0 F + F = 0 cf 23 y 43 y + df = 0 43x 43 y F 43x (Sum of moments about A ) Prof.Dr.Hasan ÖZTÜRK F + F = 0 14 34x F + F = 0 T 34 y 14 = 0 (Sum of moments about B ) 2
Numerical values for the link lengths are L 2 = 2 m and L 3 = 4 m. From the figures we extract the following measurements: a = 1.8 m, b = 1 m, c = 2 m, d = 3.6 m. Assume the applied force is given to be F = 10 N in the negative direction. F F F F T 12x 12 y 32x 32 y 12 = 10N = 5.56N = 10N = 5.56 = 23.55N F F F T 43x 43 y 14 14 = 10N = 5.56N = 5.56N = 0 Simplified FBD method: The connecting rod of this mechanism is a two-force member. The reaction forces at A and B must be equal but in opposite directions. These reaction forces are named F 2 3 and F 43, and given arbitrary directions T 2 2 A F 32 F 23 A F 12x 3 F 12y F 34 B 4 F B F 43 1 F 14 Prof.Dr.Hasan ÖZTÜRK 3
Graphical: F = 0 F = 0 x Fy = 0 F + F + F = 14 34 0 F = F 23 43 F = F 32 12 F = 0 Prof.Dr.Hasan ÖZTÜRK T M = 0 T hf. = 0 O2 2 32 = hf. 2 32 4
Coulomb Friction: Coulomb friction can be included between two contacting surfaces in a static force analysis. Given the static coefficient of friction, μ (s), the friction force can be described as the product of the coefficient of friction and the reaction force normal to the contacting surfaces. The friction force must act in the opposite direction of the tendency of any motion. V B F friction =µ F 14 F F + F = 0 + F = 0 12x 32x 12 y 32 y af + bf + T = 0 32x 32 y 12 (Sum of moments about O ) 2 F23x + F43x = 0 F + F = 0 cf 23 y 43 y + df = 0 43x 43 y (Sum of moments about A ) F + F + F = 34 x fric. F + F = 0 34 y 14 0 Prof.Dr.Hasan ÖZTÜRK 5
Example: An external force of 10 N is acting horizontally on the rocker link, 30 mm from the point D. Find the amount of torque to be applied to the crank AB to keep the mechanism in static equilibrium. [ME 302 DYNAMICS OF MACHINERY, Prof. Dr. Sadettin KAPUCU] θ = 60, b = 20.02, φ = 89.86 0 0 0 Prof.Dr.Hasan ÖZTÜRK 6
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Graphical method: F 14 & F 34 are measured directly from the scaled force polygon. 10 N stands for 50 mm Prof.Dr.Hasan ÖZTÜRK 9
Example: Onto link 6 of the mechanism given, a 100 N vertical force acting. Calculate the amount of the torque required on the crank AB to keep the mechanism in static equilibrium, using the graphical approach. [ME 302 DYNAMICS OF MACHINERY, Prof. Dr. Sadettin KAPUCU] Prof.Dr.Hasan ÖZTÜRK 10
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GEAR KINEMATICS The kinematic function of gears is to transfer rotational motion from one shaft to another. Since these shafts may be parallel, perpendicular, or at any other angle with respect to each other, gears designed for any of these cases take different forms and have different names: spur, helical, bevel, worm, etc. Vt1 = Vt2 pitch diameter ω 1.r 1= ( ) ω2.r 2 ω 1.D 1 = ( ) ω2.d2 N 1.D 1 = ( )N 2.D 2, N : rpm 2πN ω=,rad / sn 60 N. m.z = ( )N. m.z, D = m.z ( ) ( ) 1 1 1 2 2 2 *****m = m N.z = ( )N.z 1 2 1 1 2 2 number of teeth Kinematics of meshing gears. a = a t1 t2 α.r = ( ) α.r 1 1 2 2 tangential acceleration module Prof.Dr.Hasan ÖZTÜRK 13
Spur Gear Force Analysis (Static): The reaction forces between the teeth occur along the pressure line AB, tipped by the pressure angle, tipped by the pressure angle φ from the common tangent to the pitch circles. F F = F = radial force component = F = transmitted force component r r 32 23 t t 32 23 F r t 32 32 ( ) = F tan φ In applications involving gears, the power transmitted and the shaft speeds are often specified. remembering that power is the product of force times velocity or torque times angular velocity, we can find the relation between power and the transmitted force. Using the symbol P to denote power, we obtain 14
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Spur Gear Force Analysis (Dynamic): The reaction forces between the teeth occur along the pressure line AB, tipped by the pressure angle, tipped by the pressure angle φ from the common tangent to the pitch circles. F F = F = radial force component = F = transmitted force component r r 32 23 t t 32 23 F = F tan φ r t 32 32 ( ) Prof.Dr.Hasan ÖZTÜRK 17
Example (Midterm 1-2015): Find all the pin (joint) forces and the external torque (T 5 ) that must be applied to gear 5 of the mechanism by using the analytical method. Neglecting friction. Prof.Dr.Hasan ÖZTÜRK 18
A 3 M O3 = 0 Prof..Dr.Hasan ÖZTÜRK 19
M O4 = 0 Prof.Dr.Hasan ÖZTÜRK 20
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Helical Gears F a : axial, F r : radial F t : tangential φ t : transverse pressure angle ψ: helix angle Prof.Dr.Hasan ÖZTÜRK 22
Straight Bevel Gears Prof.Dr.Hasan ÖZTÜRK 23
Dynamic Force Analysis 24
Sometimes it is convenient to arrange these mass moments of inertia and mass products of inertia into a symmetric square array or matrix format called the inertia tensor of the body: 25
Dynamic Force Analysis D'Alembertls principle: The vector sum of all external forces and inertia forces acting upon a system of rigid bodies is zero. The vector sum of all external moments and inertia torques acting upon a system of rigid bodies is also separately zero. i F + F = 0 or F ma G = 0 inertia force F i has the same line of action of a G but is in opposite direction i M + T = 0 or M I α = 0 G inertia torque T i is in opposite sense of the angular acceleration a Prof.Dr.Hasan ÖZTÜRK 26
Slider Crank Mechanism external force and torque, F 4 and T 2 All frictions are neglected except for the friction at joint 14 Prof.Dr.Hasan ÖZTÜRK 27
EXAMPLE We use the four-bar linkage of the below Figure. The required data, based on a complete kinematic analysis, are illustrated in the Figure and in the legend. At the crank angle shown and assuming that gravity and friction effects are negligible, determine all the constraint forces and the driving torque required to produce the acceleration conditions specifıed. Prof.Dr.Hasan ÖZTÜRK 28
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We start with the following kinematic information. Next we calculate the inertia forces and inertia torques. Because the solution is analytical, we do not need to calculate offset distances nor do we replace the inertia torques by couples. The six equations are 30
Considering the free-body diagram of link 4 alone, we formulate the summation of moments about point 04: Also, considering the free-body diagram of link 3 alone, we formulate the summation of moments about point A: Prof.Dr.Hasan ÖZTÜRK 31
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PRINCIPLE OF SUPERPOSITION Linear systems are those in which effect is proportional to cause. This means that the response or output of a linear system is directly proportional to the drive or input to the system. An example of a linear system is a spring, where the deflection (output) is directly proportional to the force (input) exerted on the spring. The principle of superposition may be used to solve problems involving linear systems by considering each of the inputs to the system separately. If the system is linear, the responses to each of these inputs can be summed or superposed on each other to determine the total response of the system. Thus, the principle of superposition states that for a linear system the individual responses to several disturbances, or driving functions, can be superposed on each other to obtain the total response of the system. Prof.Dr.Hasan ÖZTÜRK 38
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PLANAR ROTATION ABOUT A FIXED CENTER F = ma G For fixed-axis rotation, it is generally useful to apply a moment equation directly about the rotation axis O. Application of the parallel-axis theorem for mass moments of inertia Prof.Dr.Hasan ÖZTÜRK 48
Center of percussion P P P P Prof.Dr.Hasan ÖZTÜRK 49
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SHAKING FORCES AND MOMENTS Of special interest to the designer are the forces transmitted to the frame or foundation of a machine owing to the inertia of the moving links. When these forces vary in magnitude or direction, they tend to shake or vibrate the machine (and the frame); consequently, such effects are called shaking forces and shaking moments. If we consider some machine, say a four-bar linkage for example, with links 2,3, and 4 as the moving members and link I as the frame, then taking the entire group of moving parts as a system, not including the frame, and draw a free-body diagram, we can immediately write This makes sense because if we consider a free-body diagram ofthe entiremachine including the frame, all other applied and constraint forces have equal and opposite reaction forces and these cancel within the free-body system. Only the inertia forces, having no reactions, are ultimately extemal to the system and remain unbalanced. These are not balanced by reaction forces and produce unbalanced shaking effects between the frame and whatever bench or other surface on which it is mounted. These are the forces that require that the machine be fastened down to prevent it from moving. Prof.Dr.Hasan ÖZTÜRK 52
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Force Analysis using the method of Virtual Work 1. If a rigid body is in equilibrium under the action of external forces, the total work done by these forces is zero for a small displacement of the body. 2. Work: W = F dx, W = T dθ 1. With F, x, T, q, vectors and W a scalar. 2. To indicate that we are dealing with infinitesimal displacements (virtual displacements), use the notation: 3. Now apply the virtual work definition: δw δw = F δx δw = Fi δxi + T i j j = T δθ δθ j = 0 Prof.Dr.Hasan ÖZTÜRK 54
Virtual Work (cont.) 4. If we divide the virtual work by a small time step, we get: F v T 0 i i i j j j 5. These are all external torques and forces on the body, and include inertial forces and gravity. Rewrite, to clearly show this as: i i F v+ T ω+ F v + T ω= external,friction,weight external,friction G 0 Prof.Dr.Hasan ÖZTÜRK 55
Example (2015-Midterm 1). Neglect the gravity force for the mechanism shown in the Figure. Links 2 and 3 are uniform. E and D points are the centroids of the triangles. All frictions are neglected. a) Find all the pin (joint) forces and the external torque that must be applied to link 2 of the mechanism by using the analytical method. b) Find the external torque that must be applied to link 2 of the mechanism by using the Virtual Work method. Prof.Dr.Hasan ÖZTÜRK 56
ANSWER- a): Prof.Dr.Hasan ÖZTÜRK 57
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( F 43 = F34 ) Prof.Dr.Hasan ÖZTÜRK 59
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ANSWER- b) Virtual Work method : Prof.Dr.Hasan ÖZTÜRK 61
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Example (2010-Midterm 1): For the mechanism shown in the figure, Links 2 and 3 are uniform. All frictions and mass of link 2 are neglected. Link 4 can rotate about its axis. Find all the pin (joint) forces and the external torque that must be applied to link 2 of the mechanism by using the grapical method. g α 3 =545 r/s 2 Prof.Dr.Hasan ÖZTÜRK 66
Free Body Diagram Prof.Dr.Hasan ÖZTÜRK 67
F + F + F = 43 23 0 F =13.37 N O T 2 F 12 0.1 cm F 12 O F = F 23 12 T 2 F 23 1 cm: 5 N F 43 F43 = 4.2cm 5 = 21N A F 32 A F 32 F =13.37 N F 43 F 23 F23 = 3.28cm 5 = 16.4 N 0.1 MO = 0 T2 + F32 = 0 100 0.1 T2 = 16.4 N = 0.0164 Nm 100 Prof.Dr.Hasan ÖZTÜRK 68
Example (2011-Final): For the mechanism shown in the figure, Links 2 and 3 are uniform. All frictions (except for 3-4) and mass of link 2 are neglected. Link 4 can rotate about its axis. Find all the pin (joint) forces and the external torque that must be applied to link 2 of the mechanism by using the grapical method. g α 3 =545 r/s 2 V 34 µ=0.5 Prof.Dr.Hasan ÖZTÜRK 69
tanφ = µ φ = 26.56 0 F 23 A B F R 43 F 23 26.56 0 A 22 0 G 3 58.8 0 F 43 α 3 i F = ma 3 G3 = 12.4 N 2 ag3 = 24.8 m/ s G 3 B F fric 13.37 N C d 0 0.227 Nm = d 13.74 cos(53.2 ) d = 0.0275m d = 27.5 mm i T3 = IG3α 3 = 0.227 Nm m 3 g=5 N C Prof.Dr.Hasan ÖZTÜRK 70
A F 23 F =13.37 N F 23 B F R 43 26.56 0 F R 43 G 3 58.8 0 22 0 d F =13.37 N F 23 13.37 N C A F R 43 V 34 B 26.56 0 F R 43 G 3 F 43 F fric F fric C Prof.Dr.Hasan ÖZTÜRK 71
F R 43 A 1 cm: 5 N F =13.37 N G 3 B 58.8 0 26.56 0 F 23 F =13.37 N F R 43 F 23 13.37 N 22 0 C d F R 43 F + F + F = R 43 23 0 F 23 R F 43 = 4.3cm 5 = 21.5 N F23 = 4.7cm 5 = 23.5 N F 12 O T 2 F 12 O T 2 1.5 cm 1.5 M = 0 T2 F32 = 0 100 O A A 1.5 23.5 0.3525 T2 = N = Nm 100 F 32 F 32 Prof.Dr.Hasan ÖZTÜRK 72