Chemical Engineering - Apllied Mechanics

Exercise n2 on kinematics and dynamics

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APPLIED MECHANICS 1. KINEMATIC AND DYNAMIC ANALYSIS OF A PLANAR MECHANISM The mechanical system shown in Figures 1 and 2 is analysed in this case study. This system, that is arranged in a vertical plane, is composed of an electric induction motor, a gearbox, a gear wheel, a gear rack, a slotted-link, a coupling rod and a slider. The gear wheel is hinged at point O to a fixed frame (in a real system the wheel is rigidly mounted on a shaft supported on journal bearings: in this simplified model the hinge O substitutes the mechanical components mentioned above allowing the wheel to rotate about the axis passing through its centre). The teeth of the gear wheel engage with those of a horizontal gear rack that is mounted on two frictionless supports, H and K. A pin is rigidly connected to the gear rack at point A. This pin is forced to move inside the rectilinear guide of the slotted link BC that is connected to ground by a hinged joint at the lower end B. The pin joint located at the upper end C of this link connects the slotted-link to a coupling rod, CD. In the end, the rod CD is connected to a slider, at point D, by means of a pin joint. The slider translates on an inclined plane. The barycentre of the slotted link is G 1, while that of the slider is G 2. The mass and mass moment of inertia of the slotted link, evaluated about the barycentric axis, are m 1 and J G1, respectively. The mass of the slider is m 2 while that of the gear rack is m 3. The mass of the coupling rod CD is assumed to be negligible. The mass moment of inertia of the rotor of the induction motor is J d. The pin A slides on the internal surface of the guide of the slotted link: the corresponding kinetic friction coefficient is 1kf . The kinetic friction coefficient between the=slider and the supporting plane is= 2kf . The=efficiency coefficient and transmission ratio of the gearϑset=mounted just after the electric motor (that is= before the gear rack and gear wheel)=are 1η =and 1τ , respectively. The efficiency coefficient of the mechanical transmission composed of the gear wheel and=gear rack is= 2η .=The pitch diameter of this gear wheel is denoted R (Fig.=1). The angle of the inclined plane that supports the slider is α. An external resistant force, F, is applied to the slider at point E. The current velocity Av =and acceleration= Aa =of the gear rack=are=assumed to be known.=The unknown=torque= ' d M =illustrated in Figure=1 is the drive torque transmitted to the gear wheel by the electric motor and gear transmission θ 1. = All the geometrical parameters of the system are known (distances, radii, angles, etc.).= You are asked to evaluate:= = ϑ evaluate the angular velocity and angular acceleration of the electric motor; - evaluate the velocity and acceleration of the slider; - evaluate the drive torque provided by the electric motor; - evaluate the reaction forces at point B; - evaluate the internal forces transmitted to the slider by the coupling rod CD; - evaluate the global power lost by the friction forces; - write the expression of the kinetic energy. Figure 1 Configuration of the system considered for the present study Figure 2 Electric motor, gear transmission T 1, gear wheel and gear rack transmission (T 2) 1.1 KINEMATIC ANALYSIS The kinematic analysis described below has been carried out considering the system shown in Figure 1. The global system is composed of three main sub-systems. The sub-system n.1 consists of the gear rack and the slotted link BC. The sub-system n.2 consists of the “offset slider-crank mechanism” B-C-D. A third sub- system is composed of the electric motor, the gear transmission T 1 and the gear wheel of the transmission T 2 (Figure 2). 1.1.1 Kinematics of the sub-system n.1 Figure 1.1 shows a kinematic model of the sub-system n.1. The vector 4 44i eϑ ρρ = , whose origin is a fixed point, gives the absolute position of point=A, that is the pin mounted on the gear rack, which is forced to move inside the slottedϑlink BC.= = = = Figure 1.1 Kinematic model of the subϑsystem n.1= = Besides, the vector 3 33 ieϑ ρρ = =gives=the relative=position of point A with respect to a rotating reference= frame,=x-y, having origin at point B. This reference frame rotates about an axis passing through point B with the same angular velocity and angular acceleration of the slotted-link. In accordance with the kinematic model shown in Figure 1.1 it is possible to write: 12 34 12 3 4 ii i iee eeϑϑ ϑ ϑ ρρ ρρ ++= = (1.1)= Let us assume that the distance 4ρ , as well as the velocity v A and the acceleration a A, are known, at the current time. The two vectors 1 11 ieϑ ρρ = =and 2 22 ieϑ ρρ = =are constant. Therefore, the=two unknowns= of=eq.(1.1)=are:= 3ϑ =and 3ρ .=These unknowns can be easily determined by solving the nonϑlinear eq.(1.1).= By=differentiating eq. (1.1) with respect to time we obtain:= ( ) 3 34 2 3 334 i iiee e ϑπ ϑϑ ρ ρϑ ρ + +=   = (1.2)= The absolute velocity of point A is given by: 4 A4 ieϑ ρ= v  . θherefore, the two unknowns of this 2D= vector equation are:= 3ρ =and 3ϑ.=These unknowns can be easily evaluated.=θhe vector 3 33 ieϑ ρρ =  = expresses the relative velocity of point A with respect to the rotating reference frame with origin at point B. The term= 3ϑ=is the angular velocity of the slottedϑlink (and then of the rotating=reference frame).= By differentiating eq. (1.2) with respect to time we obtain:= ( ) ( ) ( ) ( ) 333 3 34 222 2 3 333333334 iii i iiee e e e e ϑπϑπϑπϑ π ϑϑ ρ ρϑ ρϑ ρϑ ρϑ ρ +++ + + + + +=    = (1.3)= The absolute acceleration of point A is given by: 4 A4 ieϑ ρ= a  . θherefore, the two unknowns of this 2D= vector equation are:= 3ρ =and 3ϑ .=Also in this case, these unknowns can be easily evaluated.=The vector= 3 33i eϑ ρρ =   = expresses the relative acceleration of point A with respect to the rotating reference frame. The= term= 3ϑ =is the angular acceleration of the slottedϑlink (and then of the rotating reference frame).= = Now, the absolute velocity and absolute acceleration of the barycentre G 1=of the=slottedϑlink=can be= evaluated.=It is possible to write:= ( ) 9 19GO ieϑ ρ −= = ( ) ( ) 93 1 22 G 9993 iiee ϑπϑπ ρϑ ρϑ ++ == v  = ( ) ( ) ( ) ( ) 9 93 3 1 22 22 G 9 9999 393 i ii ie ee e ϑπϑπϑπϑπ ρϑ ρϑ ρϑ ρϑ + ++ + = += + a  = (1.4)= = An alternative method can be used to solve this kinematic problem. Let us consider a rotating reference frame with origin at point B (Figure 1). The x-y axes of this frame rotate with the same angular velocity and angular acceleration of the slotted-link BC. According to this reference frame, the absolute velocity of point A, Av , can be expressed as:= A rd= + v vv = (1.5)= The absolute trajectory of point A is a horizontal straight line, that is the trajectory of the gearϑrack. As= point A is forced to move inside the rectilinear guide of the slottedϑlink, hinged at point B, the relative trajectory of point A, with respect to the considered rotating reference frame, is the straight line passing= through points B and C. In the end, the drag trajectory is obtained by fixing point A to the mobile reference frame and by assigning an infinitesimal rotation to this frame. In this motion, the distance AB is constant, therefore point A described a drag trajectory that coincides with a circle with centre B and radius AB. Then, vector Av =is horizontal, vector rv =is parallel to the segment AB, while vector= dv =is orthogonal to= the segment AB. The eq.(1.5) can be rewritten as:= The expression of this vector is given by:= ( ) A rd r 3 AB =+=+ ∧ − v vv v ω = (1.6)= where 3ω =is the angular velocity of the=slottedϑlink BC. The two unknowns of eq.(1.6) are the magnitude of the relative velocity rv =and that of the drag velocity dv . This last velocity vector is given by:= ( ) d3 AB =∧− v ω = (1.7)= Therefore, according to eq.(1.7) it is possible to evaluate the angular velocity 3ω . Vector= rv =coincides with= vector= 3 3 ieϑ ρ =considered in the previous method used to solve the kinematic problem,=while vector= dv = coincides with vector ( ) 3 2 33 ie ϑπ ρϑ +  .= In accordance with the same mobile reference frame it is possible to write:= A r d rt dn dt Cor.=+= + + + aaaaa aa = (1.8)= That is, the absolute acceleration of point A can be=expressed as:= ( ) ( ) A rt 333 r AB AB 2 2 = −ω − + ∧ − + ∧ aav ωω = (1.9)= Vector= rta =is the tangential component of the relative acceleration vector ra . This vector is parallel to= the segment AB. Conversely, as the relative trajectory of point A is rectilinear, the=vector rna =ϑ=the component of the relative acceleration= ra , normal to the relative trajectory, is null. The component of the= acceleration= da , normal to the=drag trajectory, is parallel to the segment AB. This vector is directed from A to B, the centre of curvature of this circular trajectory. The component of the acceleration da , tangential to the drag trajectory, is orthogonal to the segment AB. The Coriolis’ acceleration Cor.a =is orthogonal to the= relative velocity rv , that is orthogonal to the segment AB.= The two unknowns of eq.(1.9) are the magnitude of the relative acceleration= ra =and that of the tangential= component of the drag acceleration dta . This last acceleration vector is given by:= ( ) dt 3 AB =∧− a ω = (1.10)= Therefore, according to eq.(1.10) it is possible to evaluate the angular=acceleration= 3ω . Vector= ra =coincides with vector 3 3 ieϑ ρ =considered in the previous method used to solve the kinematic problem, while vector dta =coincides with vector ( ) 3 2 33 ie ϑπ ρϑ +  .= 1.1.2 Kinematics of the sub-system n.2 Figure 1.2 shows a kinematic model of the sub-system n.2. Figure 1.2 Kinematic model of the sub-system n.2 In accordance with the kinematic model shown in Figure 1.2 it is possible to write: 75 68 75 68 ii iiee eeϑϑ ϑϑ ρ ρ ρρ ++ = = (1.11)= θhe vector 8 88 ieϑ ρρ = =gives=the=absolute position of point D (the slider). In fact, the origin of this vector is fixed. The distances= 5ρ , 6ρ =and 7ρ , as well as the attitude angles= 7ϑ =and 8ϑ =are constant. Moreover it is:= 53ϑϑ = 53ϑϑ =  53ϑϑ =   = ( 1.12)= The two unknowns of=eq.(1.12)=are:= 6ϑ =and 8ρ .=By differentiating eq.(1.5) with respect to time we obtain:= () () 36 8 22 53668 ii i e ee ϑπϑπ ϑ ρϑ ρϑ ρ ++ +=   = (1.13)= The two unknowns of=eq.(1.13)=are:= 6ϑ=and 8ρ.=The term= 6ϑ=is the angular velocity with which the= coupler rod CD rotoϑtranslates, while the term= 8ρ=is the absolute velocity of the slider.= By differentiating eq.(1.13) with respect to time we obtain:= ( ) ( ) ( ) ( ) 3 36 6 8 22 22 5 35 36 6668 i ii i i e e e ee ϑπϑ πϑπϑ π ϑ ρ ϑρϑ ρ ϑρϑ ρ + ++ + ++ + =   = (1.14)= The two unknowns of=eq.(1.14)=are:= 6ϑ =and 8ρ .=The term= 6ϑ =is the angular acceleration with=which the coupler rod CD rotoϑtranslates, while the term= 8ρ =is the absolute acceleration of the slider. As the slider only= translates, the absolute acceleration of the barycentre G 2=coincides with the acceleration of point D.= = Now,=the kinematics of the subϑsystem n.2, that is the offset sliderϑcrank mechanism BϑCϑD, can be solved considering a translating reference frame with origin at point C. The trajectory of the origin of this mobile frame is a circle with centre B and radius BC. It is important to underline that the axes x-y of this translating reference frame do not rotate. The rectilinear absolute trajectory of point D is known: it is a straight line passing through point D, parallel to the inclined supporting plane of the slider. Therefore, the absolute velocity of point D can be expressed as: ( ) ( ) ( ) D rd 6C 63 DCDC CB =+= ∧−+= ∧−+∧− v vvv ω ωω = (1.15)= The coupler rod CD of the sliderϑcrank mechanism rotoϑtranslates. Then, the relative trajectory of point= D, with respect to the translating reference=frame that has been considered, is a circle with centre C and radius CD. As a result, vector= rv =is orthogonal to the segment CD.= The two unknowns of eq.(1.15) are the magnitude of the relative velocity= rv =and that of the absolute= velocity= αv =of the slider. After having evaluated the velocity= rv =it is possible to obtain the angular velocity 6ω =with which the coupler rod CD rotoϑtranslates.= The absolute acceleration of point D can be expressed as:= α r d rn rn dn dt=+= + + + aaaaaa a = (1.16)= This equation can be rewritten as:= ( ) ( ) ( ) ( ) D6 33 DC DC CB CB 226 =−ω −+∧ − −ω −+∧− a ωω = (1.17)= Vector= rna =is parallel to the segment DC (normal to the relative circular trajectory), directed from D to= the centre of curvature C. Vector rta =is orthogonal to the segment=DC (tangent to the relative circular= trajectory).=Vector dna =is parallel to the segment=CB=(normal to the drag circular trajectory), directed from C to the centre of curvature B. Vector dta =is orthogonal to the segment CB (tangent to the drag circular trajectory). After having evaluated the acceleration rta =it is possible to obtain the angular acceleration 6ω = with which the coupler rod CD rotoϑtranslates.= Now, all the kinematic parameters that are necessary to solve the dynamics of the global system have= been evaluated.=In fact,=the mass of the coupler rod CD has been assumed to be negligible.= = The pitch diameter of the gear wheel that moves the gear rack is denoted R. Therefore, the angular= velocity and angular acceleration of the wheel are:= 1A R =v ω and 1A R =a ω = (1.18)= The transmission ratio of the gearϑset is 1τ . Therefore, the angular velocity and angular acceleration of drive shaft (motor shaft) are:= d 11 τ = ωω and = d 11 τ = ωω = (1.19)= = Here below, some results of a numerical solution of the kinematics of this system are shown. The input data are summarized in Table 1. Table 1 Input data INPUT DATA 1ρ == 0.332=m= 7ρ = 0.458 m= 7θ == 285�= 2ρ == 0.250=m= 1θ == 270�= 9ρ = 0.234 m= 5ρ == 1.065 m= 2θ == 4θ = 0�= 4ρ == 0.050 m/s= 6ρ == 0.723 m= 3θ == 63�= 4ρ == 0.002 m/s 2= = = Map of position vectors A B 1 2 3 4 = = Figure 1.3 Map of the position vectors of the subϑsystem n.1= = = Map of velo city ve c to rs vr vd vA = Map of acc e le ratio n ve cto rs adt adn art aCor. aA Figure 1.4 Map of the velocity vectors of the sub -system n.1 Figure 1.5 Map of the acceleration vectors of the sub -system n.1 = Map of position vectors B C D 7 5 6 8 = = Figure 1.6 Map of the position vectors of the subϑsystem n.2= = = Map of velo city ve c to rs vr vd vD = Map of acc e le ratio n ve cto rs adt adn art arn aD Figure 1.7 Map of the velocity vectors of the sub -system n.2 Fig ure 1.8 Map of the acceleration vectors of the sub -system n.2 = 1.2 Dynamic analysis The movement of the system is shown in Figure 1 is assured by an electric motor (Figure 2) whose drive shaft is connected to a first gear transmission, T 1, having an efficiency coefficient 1η =and a transmission= ratio 1τ . A further gear wheel is rigidly mounted on the output shaft of this transmission. The teeth of this rear wheel engage with those of a gear rack. The pitch=diameter of this gear wheel is denoted R (Fig.=1).= First, it is possible to write a power balance for the global system. The power associated with the unknown drive torque, , is:= d dd τ = ×M ω = (1.20)= The power of all the forces and torques that enters the first gear transmission T 1=is:= 1de ddi d ddddd W+ J= × × = ×− ×MC M ω ω ω ωω = (1.21)= that is the power of the drive torque plus the power associated with the inertia torque due to the mass= moment of inertia of the rotor of the induction motor.= Then, the power lost in the first transmission is given by:= ( )[ ] 1 lost 1 ddddd W1 J η =−− × − × M  ω ωω (1.22) The power available at the output shaft of the transmission T 1 is: 11out 1 e WW η = = (1.23)= The drive torque transmitted to the output shaft of the transmission T 1, and then to the gear wheel that engages with the gear rack, is given by:= 11 'd d out 1 e =W τ η ×= M ω = (1.24)= The power lost in the mechanical transmission , composed of the gear wheel and the gear rack, is:= ( ) 21 lost2 out W 1W η =−− = (1.25)= The power lost by the friction forces in the slottedϑlink and in the contact between the slider and the= supporting plane is:= 3 lostA r 1 α W=−−T v Tv = (1.26)= The power associated with the resistant forces (unless the friction forces) can be expressed as:= 1212 r E1G 2 G A1G 2 G W=++ =++ ××××××Fv P v P v Fv P v P v = (1.27)= 3 lostA r 1 α W=−−T v Tv = (1.28)= The power associated with the inertia forces and torques is:= 11 2 21 i 3A A 1G G 2 G G d d d G 3 3 W=J J mm m− ×− × − × − ×− × av a v a v ωω ωω = (1.29)= The power balance of the global system is:= 1 23 d lost lost lost r i τ +W +W +W +W +W =0 = (1.30)= The unknowns of this equation are: d M , θ A=and θ 1.= Now, it is possible to analyse the subϑsystem n.4, shown in Figure=1.9, that consists of the isolated= slottedϑlink. The two vectors V B and H B, are the reactions forces transmitted to the link by the hinged joint B. The two vectors V C and H C are the internal forces transmitted by coupling rod CD to the slotted-link.. In the end, the force N A is the normal force transmitted to the link by the pin mounted on the gear rack. The force T A is the friction force caused by the sliding motion between the pin A and the slotted-link. Figure 10 shows the sub-system n.5 that consists of only the slider. The two vectors V D and H D are the internal forces transmitted to the slider by the coupling rod CD. The force N 1 is the normal force transmitted to the slider by the supporting plane. The force T 1 is the friction force caused by the sliding motion between slider and supporting plane. In the end, as no rotation can occur between slider and supporting plane, the removal on the ground needs to take into account a further reaction that consists of a moment M 1. Figure 11 shows the sub-system n.6 that consists of only the coupling rod CD, whose mass has been assumed to be negligible. The only forces exerted to this link are: V B, H B, V C and H C. Figure 1.9 Sub -system n.4 Figure 1.10 Sub -system n.5 = = = = Figure 1.11 Subϑsystem n.6= = = For each of the subϑsystems 4, 5 and 6, it is possible to write a balance of the forces acting in the horizontal and vertical direction,=respectively, and a balance of moments evaluated about a pole. Therefore, for each subϑsystem it is possible to write three (3) equations of motion. Owing to this,=nine (9) uncoupled equations can be written. The following two additional equations, which take into account the relationship between the friction force and the corresponding normal force (normal to the surface of the relative motion) can be written: 1 AATN kf = = and 2 11TN kf = = (1.31)= The power balance, eq.(1.30), is the twelfth equation of the problem solution. The twelve unknowns= contained in these equations are listed below:= =M d, H B, V B, H C, V C, H α, V α, N A, θ A, N 1, θ 1, M 1= = This set of equations can be easily solved by implementing the equations into a computer code. This= allows one to evaluate the drive torque M d=and all the external and internal reactions forces and moments.= = = = =