Response of single degree of freedom mechanisms to base excitation

Mechanism and Machine Theory 36 (2001) 833±842

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Response of single degree of freedom mechanisms to base excitation Samuel Doughty Naval Surface Warfare Center, Philadelphia Naval Ship Yard, Carderock Divison Code 811, 5001 S. Broad Street, Philadelphia, PA 19112, USA Received 5 May 2000; received in revised form 12 January 2001; accepted 23 February 2001

Abstract Design dynamic analysis for mechanisms is usually done under the assumption that the mechanism is supported on an inertial frame. When the mechanism is mounted on an accelerating base, the previous analysis must be extended to include the dynamic e€ects of the base acceleration. This paper provides the necessary extension for single degree of freedom (SDOF) mechanisms and also presents an example. Ó 2001 Elsevier Science Ltd. All rights reserved.

1. Introduction Countless useful devices fall under the heading of single degree of freedom (SDOF) mechanisms. In the design of such devices, they are usually assumed to be mounted to a steady support, and the dynamic response of the system is determined assuming that the support provides an inertial reference. Often there are questions that arise later regarding the response of the system to base motion. In some cases, the question may be caused by a possible shock to the support. In other cases, the device may be mounted on a rotating member. In either case, the dynamic analysis done during design is insucient, and a more complete analysis, including base motion, is required. A speci®c example is found in the trip release mechanism for a vacuum circuit breaker. The intended action of the system is controlled by forces developed in springs, electromagnetic actuators, gravity, or some other prime mover. None of these forces is changed due to the imposition of support translation, and with the exception of gravity, none is changed due to support rotation. Thus, in general, but subject to the exception noted, the driving forces in the system

E-mail address: [email protected] (S. Doughty). 0094-114X/01/$ - see front matter Ó 2001 Elsevier Science Ltd. All rights reserved. PII: S 0 0 9 4 - 1 1 4 X ( 0 1 ) 0 0 0 2 4 - 6

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S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842

depend only on the relative positions of the mechanism components, and possible other inputs such as current in a coil. The thing that is di€erent when support motion is included is the development of a number of inertial reactions, terms involving the product of an inertia with an acceleration. The purpose of this note is to focus on the development of those additional terms in the equation of motion. 2. Analysis The basic kinematic analysis for the SDOF mechanism will be assumed to have been made with reference to a rectangular Cartesian coordinate system o±xy, assumed to be stationary, but now to be subject to an input motion. To be speci®c, it is assumed that some particular point ®xed in o±xy, denoted as …xRef ; yRef †, is subject to an input motion X …t†; Y …t†, and that the entire o±xy system is subject to an input rotation H…t†, where X ; Y ; and H are measured with respect to an inertial coordinate system O±XY . The system positions and displacements relative to o±xy are shown in Fig. 1(a), and those relative to O±XY are shown in Fig. 1(b). It is assumed that there exists a comprehensive analysis of the mechanism kinematics without any reference to support motion [1,2]. The SDOF is associated with a generalized coordinate q. In particular, it is assumed that the loop closure equations have been used to develop position solutions, and further, that velocity coecients and velocity coecient derivatives have been developed for all secondary variables and for all the centers of mass. The unit vectors i and j will be associated with the inertial coordinate system. It will also be useful to make use of a pair of unit vectors, ex ˆ cos Hi ‡ sin Hj and ey ˆ sin Hi ‡ cos Hj, which are always aligned with the (moving) x- and y-axes. The velocity of the reference point, with respect to an inertial coordinate system, is vRef ˆ X_ i ‡ Y_ j ˆ …X_ cos H ‡ Y_ sin H†ex ‡ … X_ sin H ‡ Y_ cos H†ey :

Fig. 1. (a) Displacements relative to o±xy. (b) Displacements relative to O±XY .

…1†

S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842

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Note that this is a velocity with respect to the inertial system, but it is resolved on the moving coordinates. 2.1. Motion of the jth center of mass Consider the jth body, for which the center of mass, referred to o±xy, is located at ‰xcj …q†; ycj …q†Š. Fig. 1 shows this body in relation to both o±xy and O±XY , including also the location of the center of mass with respect to the body coordinates at …Uj ; Vj †. (The body coordinate system U ±V moves with the body at all times and the center of mass is ®xed in this coordinate system.) The angular orientation of the body with respect to o±xy is given by the angular coordinate Bj …q†, positive in the counter clockwise sense. It is assumed that velocity coecients and velocity coecient derivatives are known for this center of mass: dxcj ; dq dycj Kcyj …q† ˆ ; dq Kcxj …q† ˆ

…2† …3†

Lcxj …q† ˆ

dKcxj d2 xcj ˆ 2 ; dq dq

…4†

Lcyj …q† ˆ

dKcyj d2 ycj ˆ 2 : dq dq

…5†

The velocity of this center of mass with respect to o±xy is _ cxj ex ‡ Kcyj ey †: vRel ˆ x_ cj ex ‡ y_ cj ey ˆ q…K

…6†

When considered relative to the inertial coordinate system, this point also has a velocity due to the rotation of the coordinate system: 



_  xcj xRef ex ‡ ycj yRef ey ˆ H _ xcj xRef ey : _ ycj yRef ex ‡ H vRot ˆ Hk …7† The complete velocity of this center of mass, with respect to the inertial coordinate system, is vcj ˆ vRef ‡ vRel ‡ vRot h
i _ ycj yRef _ cxj ‡ X_ cos H ‡ Y_ sin H H ˆ ex qK h
i _ xcj xRef : _ cyj X_ sin H ‡ Y_ cos H ‡ H ‡ ey qK

…8†

Taking a time derivative of the velocity gives the acceleration, again referred to the inertial coordinates but resolved in the moving system:

_ 2 xcj xRef Š _ cyj H  ycj yRef H qKcxj ‡ q_ 2 Lcxj 2q_ HK acj ˆ ex ‰X cos H ‡ Y sin H ‡ 

_ 2 ycj yRef Š: _ cxj ‡ H  xcj xRef H ‡ ey ‰ X sin H ‡ Y cos H ‡  qKcyj ‡ q_ 2 Lcyj ‡ 2q_ HK …9†

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S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842

2.2. Kinetic energy for the jth body Expressions have been developed for the velocity components of the jth body, referred to the inertial coordinate system:
_ ycj yRef ; _ cxj ‡ X_ cos H ‡ Y_ sin H H vcxj ˆ qK …10†
_ xcj xRef : _ cyj X_ sin H ‡ Y_ cos H ‡ H …11† vcyj ˆ qK In addition to these, it is necessary to recognize that the angular velocity of the body is simply the sum of the relative angular velocity, B_ j and the superimposed rotational velocity of the coordinate _ With these, the kinetic energy of the jth body is simply system, H.  1  2 1  _ ‡ qK _ Bj : …12† Tj ˆ Mj v2cxj ‡ v2cyj ‡ Jj H 2 2 2.3. Contribution of the jth body to the Lagrange equation The contribution of the jth body to the Lagrange equation of motion for the system is now assembled. The ultimate form for the Lagrange equation will be d oT dt oq_

oT ˆ Qq oq

…13†

so the contribution of the jth body to the left-hand side is     d oTj oTj _ cxj ‡ Jj H  ‡ qKBj ‡ q_ 2 LBj KBj _ cxj Kcyj vcyj HK ˆ Mj acxj Kcxj ‡ acyj Kcyj ‡ Hv dt oq_ oq …14† after some simpli®cation. Note that acxj and acyj are the components of the center of mass acceleration resolved on the accelerating coordinate system and extracted from Eq. (9). 2.4. Lagrange equation for the system response At this point, all that is necessary to extend the results developed above to a complete, coupled system is to sum the contributions from all bodies. These summations will be on the left-hand side of the equals sign, leaving the right-hand side unchanged. Thus the Lagrange equation of motion is d oT dt oq_

oT ˆ Qq oq X X X _ ˆ Mj acxj Kcxj ‡ Mj acyj Kcyj ‡ H Mj vcxj Kcyj j

 ‡H

X j

j

Jj KBj ‡  q

X j

Jj KB2j ‡ q_ 2

X j

j

Jj KBj LBj :

_ H

X

Mj vCyj Kcxj

j

…15†

S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842

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The equation of motion is of the form  qI…q† ‡ q_ 2 C…q† ˆ Qq

X CX

Y CY

 H HC

_ 2 CH_ 2 H

_ X_ CH_ X_ H

_ Y_ CH_ Y_ ; H

…16†

where the coecients are identi®ed as follows: Coecient of  q X X X 2 2 Mj Kcxj ‡ Mj Kcyj ‡ Jj KB2j ˆ I…q†; Cq ˆ j

j

…17†

j

where this coecient is recognized as the expected generalized inertia for the SDOF mechanism. Coecient of q_ 2 X X X Cq_ 2 ˆ Mj Kcxj Lcxj ‡ Mj Kcyj Lcyj ‡ Jj KBj LBj ˆ C…q†; j

j

…18†

j

where this term is recognized as the expected centripetal coecient for the SDOF mechanism. Coecient of X CX ˆ cos H Coecient of Y CY ˆ sin H

X

Mj Kcxj

sin H

X

j

Mj Kcyj :

…19†

Mj Kcyj :

…20†

j

X

Mj Kcxj ‡ cos H

X

j

j

 Coecient of H X X X CH ˆ Mj Kcxj ycj ‡ yRef Mj Kcxj ‡ Mj Kcyj xcj j

j

_2 Coecient of H X CH_ 2 ˆ 2 Mj Kcxj xcj j

_ X_ Coecient of H CH_ X_ ˆ cos H _ Y_ Coecient of H CH_ Y_ ˆ sin H

X

2

X

j

Mj Kcyj ycj ‡ 2xRef

j

Mj Kcyj ‡ sin H

j

X j

X j

X

xRef

X

Mj Kcyj ‡

j

Mj Kcxj ‡ 2yRef

X

Jj KBj :

…21†

j

X

Mj Kcyj :

…22†

j

Mj Kcxj :

…23†

Mj Kcxj :

…24†

j

Mj Kcyj

cos H

X j

The term in q_ has been omitted since the coecient is zero. If all base motion is excluded, the equation of motion reduces to

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S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842

 qI…q† ‡ q_ 2 C…q† ˆ Qq

…25†

which is simply Eksergian's equation of motion [1±3], as expected. The term Qq contains the spring forces, electromagnetic actuator forces, and whatever else acts on the mechanism in the non-base excited mode. The imposition of base excitation causes no change to this term. Thus it is evident that the development of the equation of motion, including the base excitation, has simply added several terms to the generalized force driving the motion, but does not change the essential dynamic characteristics I…q† and C…q† at all.

3. Example For an example, consider the simple slider±crank mechanism shown in Fig. 2. The signi®cant aspects of the individual piece parts are shown at the top of the ®gure, and the assembled mechanism is shown below. The crank is pivoted at the point …D; 0†, and the ®xed end of the spring is located at …xs ; ys †, both with respect to the o±xy coordinate system. The necessary kinematic analysis for this mechanism is given in Appendix A. This mechanism might be used in a device such as a circuit breaker where the switching action is accomplished by the rotation of the crank when the actuation force FAct is applied to the slider. A tension spring acts to restrain the mechanism against the applied force in the ordinary operating mode. For the sake of the example, the actuator force FAct will be taken as zero. The reference point is taken at the origin of the x±y coordinate system, and the frame of the mechanism will be assumed to be subject to a displacement:

Fig. 2. Circuit breaker mechanism example.

 XBase …t† ˆ X_ Base …t† ˆ XBase …t† ˆ

 

S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842


pt

A sin2 0;

T

A Tp sin 0; 2A 0;

;

2pt T


p 2 T

839

06t 6T; T < t;


;

cos

…26†

06t 6T; T < t; 2pt T


;

…27†

06t 6T; T < t;

…28†

where T is the period of the input displacement and A is the amplitude of that motion. The center of mass of the slider is assumed to be located along the center line and to move only along the x-axis. The slider is initially held at rest against the support by the preloaded spring, and the initial position is q0 ˆ 0:15 m. The center of mass of the connecting rod is at the center of its length. The center of mass for the crank, body 3, is located by body coordinates …U3 ; V3 †. The spring has coecient K and free length S0 . The speci®c data used for this example includes L2 ˆ 0:2 m L3 ˆ 0:08 m D ˆ 0:28 m K ˆ 350 N=m S0 ˆ 0:25 m A ˆ 0:10 m T ˆ 0:020 s m1 ˆ 0:01 kg; m2 ˆ 0:015 kg; m3 ˆ 0:03 kg;

connecting rod length crank radius position of crank pivot spring rate spring free length base displacement amplitude period of input displacement J1 ˆ 0:0 kg m2 ;

Uc1 ˆ 0:0 m;

Uc2 ˆ 0:01 m;

Vc2 ˆ 0:0 m;

2

Uc3 ˆ 0:12 m;

Vc3 ˆ 0:05 m;

J2 ˆ 0:00005 kg m ; J3 ˆ 0:000144 kg m ;

uS ˆ 0:18 m;

vS ˆ 0:12 m;

xS ˆ 0:04 m;

yS ˆ 0:06 m:

Vc1 ˆ 0:0 m;

2

Without base motion, the equation of motion is given by Eksergian's equation [1±3] I…q† q ‡ C…q†q_ 2 ‡ K…S

S0 †

dS ˆ FAct : dq

…29†

When we include the base motion, from Eq. (16) the equation of motion becomes I…q† q ‡ C… q†q_ 2 ‡ K…S

S0 †

dS ˆ FAct dq

X CX

Y CY

 H HC

_ 2 CH_ 2 H

_ X_ CH_ X_ H

_ Y_ CH_ Y_ : H …30†

The equation of motion is a second-order di€erential equation with variable coecients. There is no known way to solve this in closed form, but a numerical solution is certainly possible. A fourth-order Runge±Kutta that is directly applicable to the second-order form is given by Abramowitz and Stegun [4], and this algorithm has been used as the basis for a simple simulation applied to this example.

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S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842

Fig. 3. Normalized motion curves.

The input motion and results are shown in Fig. 3, where all curves have been normalized to their maximum values. The acceleration of the base, as described above, is shown, beginning with an initially positive acceleration, dropping to zero, and then going negative. During the time that the base acceleration is positive, the slider position is ®xed at its initial value because the slider is against a stop. At t ˆ 0:0056 s, the mechanism begins to move. This is after the base acceleration has changed signs, and the delay is because the generalized acceleration does not become positive until there is sucient negative acceleration to overcome the spring force. If the spring preload is set to zero, this delay is eliminated. The slider position increases monotonically while the velocity rises to a peak value and begins to drop. The slider acceleration, q, is positive for a short period after the mechanism begins to move, but it soon peaks and begins to drop to negative values before the mechanism motion is complete. The resulting crank motion, B3 …t†, is shown also, holding constant at its initial value B3 …0† ˆ 36:593° for t < 0:0056 s, and dropping monotonically thereafter. This example illustrates the application of the general result to a particular case. While there is no particular importance to this example system, it serves to demonstrate the process and to illustrate the sort of results that can be obtained. Appendix A. Kinematic analysis for example The horizontal and vertical loop equations for the example problem are q ‡ L2 cos B2 L2 sin B2

L3 cos B3

L3 sin B3 ˆ 0:

D ˆ 0;

…A:1† …A:2†

S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842

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The velocity coecient equations are obtained by taking a derivative with respect to q, thus 1

L2 KB2 sin B2 ‡ L3 KB3 sin B3 ˆ 0;

L2 KB2 cos B2

…A:3†

L3 KB3 cos B3 ˆ 0;

…A:4†

which may be recast as      KB2 1 L3 sin B3 L2 sin B2 ˆ : KB3 L2 cos B2 L3 cos B3 0

…A:5†

The velocity coecient derivative equations are L2 LB2 sin B2

L2 KB22 cos B2 ‡ L3 LB3 sin B3 ‡ L3 KB23 cos B3 ˆ 0;

…A:6†

L2 LB2 cos B2

L2 KB22

…A:7†

sin B2

L3 LB3 cos B3 ‡

L3 KB23

sin B3 ˆ 0;

which may be recast as     L2 KB22 cos B2 LB2 L2 sin B2 L3 sin B3 ˆ LB3 L2 cos B2 L3 cos B3 L2 KB22 sin B2

 L3 KB23 cos B3 : L3 KB23 sin B3

…A:8†

The bodies are numbered 1 for the slider, 2 for the connecting rod, and 3 for the crank. The positions, velocity coecients, and velocity coecient derivatives for the centers of mass are as follows: xc1 ˆ q

const;

…A:9†

Kcx1 ˆ 1;

…A:10†

Lcx1 ˆ 0;

…A:11†

yc1 ˆ 0;

…A:12†

Kcy1 ˆ 0;

…A:13†

Lcy1 ˆ 0;

…A:14†

1 xc2 ˆ q ‡ L2 cos B2 ; 2

…A:15†

Kcx2 ˆ 1 Lcx2 ˆ

1 L2 KB2 sin B2 ; 2 1 L2 LB2 sin B2 2

1 L2 KB22 cos B2 ; 2

…A:16† …A:17†

1 yc2 ˆ L2 sin B2 ; 2

…A:18†

1 Kcy2 ˆ L2 KB2 cos B2 ; 2

…A:19†

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S. Doughty / Mechanism and Machine Theory 36 (2001) 833±842

1 Lcy2 ˆ L2 LB2 cos B2 2

1 L2 KB22 sin B2 ; 2

…A:20†

xc3 ˆ D ‡ U3 cos B3

V3 sin B3 ;

…A:21†

Kcx3 ˆ

U3 KB3 sin B3

V3 KB3 cos B3 ;

Lcx3 ˆ

U3 LB3 sin B3

U3 KB23 cos B3

…A:22† V3 LB3 cos B3 ‡ V3 KB23 sin B3 ;

…A:23†

yc3 ˆ U3 sin B3 ‡ V3 cos B3 ;

…A:24†

Kcy3 ˆ U3 KB3 cos B3

V3 KB3 sin B3 ;

…A:25†

Lcy3 ˆ U3 LB3 cos B3

U3 KB23 sin B3

V3 LB3 sin B3

V3 KB23 cos B3 :

…A:26†

The length of the spring is S, and the velocity coecient KS ˆ dS=dq will also be required. These are S 2 ˆ …D ‡ US cos B3 KS ˆ

VS sin B3

xS †2 ‡ …US sin B3 ‡ VS cos B3

yS †2 ;

KB3 … D ‡ US cos B3 VS sin B3 xS †…US sin B3 ‡ VS cos B3 † S KB ‡ 3 …US sin B3 ‡ VS cos B3 yS †…US cos B3 VS sin B3 †: S

…A:27†

…A:28†

References [1] B. Paul, Kinematics and Dynamics of Planar Machinery, Prentice-Hall, Englewood Cli€s, NJ, 1979. [2] S. Doughty, Mechanics of Machines, Wiley, New York, 1988. [3] R. Eksergian, Dynamical analysis of machines, in 15 parts, Journal of the Franklin Institute, vols. 209, 210, 211 (1930). [4] M. Abramowitz, I.E. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, Applied Mathematics Series, vol. 55, item 25.5.19, p. 897 (1964, corrected 9th printing, 1970).