MACHINE DESIGN I
MACHINE DESIGN I ME 4133
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This 9 page Class Notes was uploaded by Dolly Koepp on Tuesday October 13, 2015. The Class Notes belongs to ME 4133 at Louisiana State University taught by W. Wang in Fall. Since its upload, it has received 23 views. For similar materials see /class/222916/me-4133-louisiana-state-university in Mechanical Engineering at Louisiana State University.
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Date Created: 10/13/15
ME 4133 295 WNWMOM VECTOR LOOP VELOCITY AND ACCELERATION ANALYSIS Prerequisites Before proceeding with the Velocityacceleration analysis two tasks must be completed 1 The mobility DOF must be determined Each DOF requires one input for position velocity and acceleration analysis 2 The displacement analysis must be completed so that the position variables for each link are known Given the vector loop for a mechanism with the position parameters resolved and input velocities and accelerations de ned what steps are necessary to solve for the unknown velocities and acclerations of every link in the mechanism or any point on any link As an example consider the vector loop for the fourbar mechanism shown on the next page The loop closure equation or vector loop constraint equation for this loop is given by a gt gt gt ZR R1RZR3R4 0 The magnitudes and directions of all these components are known as a result of the displacement analysis However neither of the unknowns 93 or 94 are explicit functions of time ie 03 f 2 so that the desired velocities cannot be obtained by explicit time differentiation However the loop closure equation de nes a multivariable vector function 62 93 94 of the input and unknowns 92 63 04 23 5 29quotquotZ H3563 Rlei9 R4ei quot 0 This function can be used to develop additional vector loop constraint equations for determining the unknown velocities and accelerations in the mechanism 33 ME 4133 295 WNWMOM Veloci Anal sis Im licit Time Differentiation The multivariable vector function is a function of the input link kinematics 02602052 and the unknown link kinematics 93a3oc3 and 94604064 Although time does not appear explicitly in the function all of the kinematic variables are implicit functions of time Im licit differentiation of the function With respect to time should yield expressions for the link velocities The total derivative of 02 03 94 with respect to time is 32 d9 92 d6 32 d9 26 35 2 3 4 gtampar9339 3 as dt as dt as dt 291 e A Wquot 30 This is a vector equation 2 scalar equations which depends only on the two unknown scalar velocities 63 and 94 and the input Rearranging into matrix form 9Z 92 g s as 2 as 994 399 The form of this equation is d gt Es9936420 9G 939 B J T r g i 9 where b avector functlon of the known 1nputs a I C and loop geometry 819k 1 t J the Jacobian matrix X a vector of the unknown velocities This is similar to the matrix expression developed for the iterative displacement solution and J is the same Jacobian matrix which is characteristic of the mechanism vector loop The vector on the lefthand side 3 contains functions of the input velocity and loop geometry Expanding the equation into real and imaginary parts with the top row of the matrix containing the real part 819293 94 and the bottom row the imaginary part 2 92 03 94 then substituting for the partial derivatives x Vile 381 881 381 73 Z 755 35 892 2 893 864 R2 sin 92 6392 RS sin 63 4 sin 94 9393 64 Fx 2 cos 62 92 1 cos 93 R4 cos 94 ME 4133 295 WNWMGM The solution for the unknown velocities can be obtained using standard linear algebra techniques such as LU Decomposition if the Jacobian is nonsingular J i O In the twodimensional case Cramer39s rule is an effective approach but it is inef cient for larger problems The resulting velocities are ACTUAL exact within the limits of the computer M solutions No iteration is required 7 The disadmh is that the input velocity appears explicitly in the equation inj For each value of in ut velocity evaluated the entire equation must be solved again whigh can be 50th tedious and computationa ly expensivef W Veloci Analysis Kinematic Coefficients There is an alternative approach to evaluating the derivatives Instead of differentiating with respect to time differentiate the constraint equation with respect to the input variable 92 a function of time d a 32 d0 32 d9 92 d6 geae o 4 3 4 d9 2 3 4 86 d9 86 d0 89 d9 We rigIa as 3 9 d9 99 d9 In matrix form de 862 90 a9 51 Ldez The equation is in the same form as our previous result from time differentiation except that the velocities are not explicitly present The matrix is still the Jacobian but the left hand vector contains a partial derivative with respect to the input and the vector of unknowns contains total derivatives with respect to the unknown displacement The total derivatives in the unknown vector are de ned as kinematic or in uence coef cients with respect to the input In your text Hall uses the following notation to distinguish kinematic coef cients for rotational and translational variables of angle d input d displacement dinput gth 62 The input referred to in the above equation may be either rotational or translational Substituting and expanding the vector equation into real and imaginary parts ME 4133 295 WNWMOM as 881 861 9 92 8 93 597 992 W3 994 F2 sin 62 4 3 sin 63 Fi 4 sin 04 h3 RZ cos 62 2 R3 cos 93 I4 cos 94 iihj The resulting kinematic coef cients shown below do not depend on the input velocity so they only need to be calculated once for each position of the linkage R2 sin 02 Ff4 sin 64 ngin 03 R2 sin 92 R2 cos 92 H4 cos 94 R3 cos 63 R2 cos 92 Ff3 sin 63 R4 sin 94 1 3sin93 H4sin 94 R3 cos 63 R4 cos 64 R3 cos 63 R4 cos 64 h 3 h4 The true unknowns are still the link velocities and the kinematic coef cients are related to the velocities by d wk 2 9k a z an dt d92 dt Why bother with kinematic coef cients instead of velocities directly There are several reasons hk92 hsz k 34 Kinematic coef cients are a function of GEOMETRY only 39 Actual velocities are simply scalar multiples of the kinematic coefficients Kinematic coef cients can be quickly determined using instant centers IC39s For 1 DOF systems all velocities are related to the one input wan73 ME 4133 295 WNWMCM Mechanism motion can be characterized independent of a particular input velocity value Once the kinematic coef cients are determined a change in input does not require a complete new analysis just a rescaling Velocig of an Arbitra Point Up to this point only link and joint velocities have been considered Most of the time the point of interest on a machine will not be at a joint or on the vector loop for the velocity analysis If the position solution is known and the geometric relationship between the unknown point and the linkage is known solving for the velocity of an arbitrary point is a straightforward extension of the methods developed so far Consider the example linkage shown on the following page with an arbitrary point on link 3 Create a new vector loop containing point C and write the vector 100p equation a gt gt gt 0 2 ER FICHZH5 Rearrangng and substituting gt a i9 3 RC R2R5 Rze 1 529 5 X0RevaH2v3ost92 31500895 lt Ewes plexwagj 02 yem ostin62Hssin95 RASL 9L frgpgl h 0 DL 3 at X The compnents of velocity at C are just the time derivatives of the displacements or ww 2 Q Ve7e Ob was a 9148quot L quot r b V Now the total derivatlve w1th respect to tlme o the new multlvariable function can be ME 4133 295 WNWMCM V dXC 02 X0 dt dB2 dt X dy dy d9 y z Qz C szymz 0 dt dQZ dt 5 a Evaluating the kinematic coef cients from the components of RC 9 dx he 9er 916 C39 K fxc 261 90 RZSIn62 R5h3sm93a5 10 o 5 OJ d 2 VLZMWIAOJB WM 0 ny y chosezH5h3cos93a5 Mg The angle 95 has been replaced by the geometric relation 93 15 where as is a constant The kinematic coef cient h3 and 63 are known from the original vector loop solution and the other parameters are known from the mechanism geometry so the components of the velocity of any point on any link may now be determined Acceleration Analysis Implicit Time Differentiation The development of the equations for acceleration is similar to that for velocity In that case the starting point was recognition that the loop closure equation was a multivariable gt function e 92 3 94 Since the time derivative of that function is also a function of the a input and unknown variables it is also a multivariable vector function 8 92 93 64 and de ned by V e d e e 926394B EM ML 5 woo 9263640 evaluated d a d2 a d a a e E5 629394W89293640 Z 3 926024 863 w3a 94 04 zi wa wi m 2 5 dt a9 a 993 3 dt as 4 I 1 L gt M 019 88 L8 i 03 3 86 0529 93053f364 14 b 2 2quot 2 29 c9 8 2 8 E 2 amp e 2 03w kpz 392 2392 3 as 4 a ry gt gt gt 9398 88aa aa 8a 0 gig3 J9 as 2 39 3 394 quot 2L6 7 M e gt3 29 VLQ 7972 6 ME 4 133 295 WN WMCM R2ei za 1335036032 R4ei9 a iRZe gzaz iR3e39 3053 iR4e 94a4 a The coef cients of all of the terms in the equation are all known functions of the loop geometry The rst four terms also depend on known quantities the results of the velocity analysis and the input link acceleration Only the last two terms are functions of the unknown accelerations of links 3 and 4 Rearranging the equation the result is the farmliarmatnx form g 6 L5 922 2 a2 2 322 2 a 92 a2 a3 0M L021sz agt a w a 2 as 2 992 3 as 4 as 2 39 394 a The righthand side of the equation is the product of the Jacobian the same Jacobian as found in the displacement and velocity analysis previously and the unknown acclerations The left hand side contains a function of known quantities As with the velocity solution of this equation will yield an ACTUAL solution for the unknown ac or each new value of the input acceleratlon and velocity the equations must be solved again Substituting the exponential form for the terms in and expanding into real and imaginary parts the equations for the 4bar are Film2 cos 6 Fi3a32 cos 9 341 cos 94 R2052 sin 62 4 sin 93 Fi 4 Sin 94 0 32a sin 92 Ra2 sin 03 Hp sin 94 H2052 cos 6 R3 COS 03 R4 COS 64 04 Acceleration Analysis Kinematic Coefficients Kinematic coef cients provided a solution to the problem of solving the velocity equation for each new value of the input velocity It is natural to see if a similar result can be obtained for acceleration In the same manner as for the time derivative de ne the multivariable e mction h 9 i x anges 6 M New H83 rt v ca d f d6 9 d9 f d6 i9 a 79 quot 8 2 8 3 g 4 8 600 999 0 2 3 4 d928 2 3 4 aezde Bade 864d92 i gigantic 9 a A 8 h I 3i 9L as as 3 994 4 249 Find the total derivative of 2192 63 94 with respect to the input displacement 92 ME 4133 295 WNWMGM 0i e d L6 W9 de ade a de 88 bd64 39 d6 as de 39 d9 Q 5299 3337 96 26 30 993 4 as 4 r 79 363 559293940 2 1151 Egregbihpk evaluated gt 2 g d293 iced261 j 39 d9 as d9 29quot gt gt gt a 928 h39 h39 394 as 3 39 4 95 9390 Wu z a 7 a42 3 63398 929364 9932 ha i a 7 82 3 948 94 94 94a e 4 In the partial derivatives with respect to 93 and 64 only the second partial derivatives carry through because they are the only remaining terms which are functions of those variables This is made possible by the use of absolute angles in de ning the vector loop once the rst partial with respect to 93 and 94 was taken only those terms retain any function of those respective variables If relative angles had been used so that 94 was a function of 62 and 93 for example the equation would be much more complicated Substituting the partial derivatives into the total derivative yields the following equation 92 a 8 s 8 s h39ath28ZZh2 T 99 893 3 864 4 80 3 99 4 Rearranging and collecting functions of known quantities on the lefthand side of the equation F54 er saw v gt a l 7 l t 943W 8482 a 88 ha ravage 304 8932 3 9642 4 393 394 h lg 5 Lg A t as k The result is similar to all of the other equations obtained so far The righthand side again contains the matrix product of the Jacobian and a vector of unknown quantities The left hand side contains only known quantities which are functions of the loop geometry There is no explicit reference to the input angular acceleration but there is a difference from the equation resulting from the time differentiation In that equation the input acceleration appeared as a product with a rst partial derivative but there is no rst partial derivative term on the lefthand side of this equation This suggests that the d7 d0 993940 ME 4133 295 WNWMOM unknown accelerations are not just scalar multiples of the second order kinematic coef cients h and h like the unknown velocities are for the rst order kinematic coef cients h3 and 74 The relationship between the second order kinematic coef cients h and h and the unknown accelerations can be shown by differentiating one of the unknown velocities with respect to time 03sz m3 dh3w2 3 dt dt dh da WSW dh3 d92 a7 h d92 dt 2 3 12 hgco h3ot2 So the unknown accelerations are the sum of scalar multiples of the rst and second order kinematic coef cients not just scalar multiples of the second order coef cients The rst order partial with respect to the input missing from the second order kinematic coef cient equation appears in this sum as the product with the input acceleration Substituting the derivatives of the 4bar loop closure equation into the kinematic coef cient form for acceleration yields H2 cos 92 H3h32 cos 93 R4hj cos 94 FI3 sin 03 F 4 sin 94 7 Hz sin 92 F3h32 sin 63 94h sin 94 H3 cos 93 R4 cos 94 hi