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I.6 Dynamic Force Analysis 0

Contents

6 Dynamic Force Analysis 1

6.1 Equation of Motion for the Mass Center . . . . . . . . . . . . 16.2 Angular Momentum Principle for a System of Particles . . . . 46.3 Equations of Motion for General Plane Motion . . . . . . . . . 66.4 DAlemberts principle . . . . . . . . . . . . . . . . . . . . . . 96.5 Free-Body Diagrams . . . . . . . . . . . . . . . . . . . . . . . 116.6 Joint Forces Analysis Using Individual Links . . . . . . . . . . 116.7 Joint Force Analysis Using Contour Method . . . . . . . . . . 136.8 Joint Force Analysis Using Dyads . . . . . . . . . . . . . . . . 196.9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

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6 Dynamic Force Analysis

For a kinematic chain it is important to know how forces and moments aretransmitted from the input to the output, so that the links can be properlydesignated. The friction effects are assumed to be negligible in the forceanalysis presented here.

6.1 Equation of Motion for the Mass Center

Consider a system of N particles. A particle is an object whose shape andgeometrical dimensions are not significant to the investigation of its motion.An arbitrary collection of matter with total mass m can be divided into N

particles, the ith particle having mass, mi, Fig. 6.1

m =N

i=1

mi.

A rigid body can be considered as a collection of particles in which thenumber of particles approaches infinity and in which the distance betweenany two points remains constant. As N approaches infinity, each particleis treated as a differential mass element, mi dm, and the summation isreplaced by integration over the body

m =

body

dm.

The position of the mass center of a collection of particles is defined by

rC =1

m

Ni=1

miri, (6.1)

where ri is the position vector from the origin O to the ith particle. AsN the summation is replaced by integration over the body

rC = 1m

body

r dm, (6.2)

where r is the vector from the origin O to differential element dm.

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The time derivatives of Eq. (6.1) gives

Ni=1

mid2ridt2

= md2rCdt2

= maC, (6.3)

where aC is the acceleration of the mass center. The acceleration of themass center can be related to the external forces acting on the system. Thisrelationship is obtained by applying Newtons laws to each of the individualparticles in the system. Any such particle is acted on by two types of forces.One type is exerted by other particles that are also part of the system. Suchforces are called internal forces (internal to the system). Additionally aparticle can be acted on by a force that is exerted by a particle or object notincluded in the system. Such a force is known as an external force (externalto the system). Let fij be the internal force exerted on the jth particle bythe ith particle. Newtons third law (action and reaction) states that the jthparticle exerts a force on the ith particle of equal magnitude, and oppositedirection, and collinear with the force exerted by the ith particle on the jthparticle, Fig. 6.1

fji = fij, j = i.

Newtons second law for the ith particle must include all of the internal forcesexerted by all of the other particles in the system on the ith particle, plusthe sum of any external forces exerted by particles, objects outside of thesystem on the ith particle

j

fji + Fexti = mi

d2ridt2

, j = i, (6.4)

where Fexti is the external force on the ith particle. Equation (6.4) is writ-ten for each particle in the collection of particles. Summing the resultingequations over all of the particles from i = 1 to N the following relation isobtained

i

j

fji +

i

Fexti = maC, j = i. (6.5)

The sum of the internal forces includes pairs of equal and opposite forces.The sum of any such pair must be zero. The sum of all of the internal forceson the collection of particles is zero (Newtons third law)

i

j

fji = 0, j = i.

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The term

iFexti is the sum of the external forces on the collection of particles

i

Fexti = F.

One can conclude that the sum of the external forces acting on a closedsystem equals the product of the mass and the acceleration of the masscenter

m aC = F. (6.6)

Considering Fig. 6.2 for a rigid body and introducing the distance q inEq. (6.2) gives

rC =1

m

body

r dm =1

m

body

(rC + q) dm = rC +1

m

body

q dm. (6.7)

It results

1

m

body

q dm = 0, (6.8)

that is the weighed average of the displacement vector about the mass centeris zero. The equation of motion for the differential element dm is

a dm = dF,

where dF is the total force acting on the differential element. For the entirebody

body

a dm =

body

dF = F, (6.9)

where F is the resultant of all forces. This resultant contains contributionsonly from the external forces, as the internal forces cancel each other. Intro-ducing Eq (6.7) into Eq. (6.9) the Newtons second law for a rigid body is

obtained m aC = F

The derivation of the equations of motion is valid for the general motion ofa rigid body. These equations are equally applicable to planar and three-dimensional motions.

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Resolving the sum of the external forces into cartesian rectangular com-

ponents

F = Fx + Fy + Fz k,

and the position vector of the mass center

rC = xC(t) + yC(t) + zC(t) k,

the Newtons second law for the rigid body is

mrC = F, (6.10)

or

mxC = Fx, myC = Fy, mzC = Fz. (6.11)

6.2 Angular Momentum Principle for a System of Par-

ticles

An arbitrary system with the mass m can be divided into N particlesP1, P2,...,PN. The position vector of the ith particle relative to an originpoint, O, is ri = rOPi and the mass of the ith particle is mi, Fig. 6.3. The

position of the mass center, C, of the system is rC =

Ni=1

miri/m. The position

of the the particle Pi of the system relative to O is

ri = rC + rCPi. (6.12)

Multiplying Eq. (6.12) by mi, summing from 1 to N, the following relationis obtained

Ni=1

mirCPi = 0. (6.13)

The total angular momentum of the system about its mass center C, is thesum of the angular momenta of the particles about C

HC =N

i=1

rCPi mivi, (6.14)

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where vi =dri

dt

is the velocity of the particle Pi.

The total angular momentum of the system about O is the sum of the angularmomenta of the particles

HO =N

i=1

ri mivi =N

i=1

(rC + rCPi)mivi = rCmvC + HC, (6.15)

or the total angular momentum about O is the sum of the angular momentumabout O due to the velocity vC of the mass center of the system and the totalangular momentum about the mass center, Fig. 6.4.

Newtons second law for the ith particle is

j

fji + Fexti = mi

dvidt

, j = i,

and the cross product with the position vector ri, and sum from i = 1 to Ngives

i

j

ri fji +

i

ri Fexti =

i

ri d

dt(mivi), j = i. (6.16)

The first term on the left side of Eq. (6.16) is the sum of the moments aboutO due to internal forces, and

ri fji + ri fij = ri (fji + fij) = 0, j = i.

The term vanishes because the internal forces between each pair of particlesare equal, opposite, and directed along the straight line between the twoparticles, Fig. 6.1. The second term on the left side of Eq. (6.16)

i

ri Fexti =

MO,

represents the sum of the moments about O due to external forces and cou-ples. The term on the right side of Eq. (6.16) is

i

ri d

dt(mivi) =

i

d

dt(ri mivi) vi mivi

=

dHOdt

, (6.17)

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which represents the rate of change of the total angular momentum of the

system about the point O.Equation. (6.16) is rewritten as

dHOdt

=

MO. (6.18)

The rate of change of the angular momentum about O equals the sum of themoments about O due to external forces and couples.Using Eqs. (6.15) and (6.18) the following result is obtained

MO =

d

dt(rCmvC + HC) = rCmaC +

dHCdt

, (6.19)

where aC is the acceleration of the mass center.With the relation

MO =

MC + rC F =

MC + rCmaC,

Eq. (6.19) becomes

dHCdt

=

MC. (6.20)

The rate of change of the angular momentum about the mass center equals

the sum of the moments about the mass center.

6.3 Equations of Motion for General Plane Motion

An arbitrary rigid body with the mass m can be divided into N particlesPi, i = 1, 2,...,N. The position vector of the Pi particle is ri = OPi and themass of the particle is mi. Figure 6.5(a) represents the rigid body movingwith general planar motion in the (X, Y) plane. The origin of the cartesianreference frame is O. The mass center C, of the rigid body is located in theplane of the motion, C (X, Y).

Let dO = OZ be the axis through the fixed origin point O that is per-pendicular to the plane of the motion of the rigid body (X, Y), dO (X, Y).Let dC = Czz be the parallel axis through the mass center C, dC||dO. Therigid body has a general planar motion and the angular velocity vector is = k. The unit vector of the dC = Czz axis is k.

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The velocity of the Pi particle relative to the mass center is

dRidt

= kRi,

where Ri = rCPi. The sum of the moments about O due to external forcesand couples is

MO =

dHOdt

=d

dt[(rCmvC) + HC], (6.21)

where

HC = i

[Ri mi(kRi)],

is the angular momentum about dC. The magnitude of the angular momen-tum about dC is

HC = HC k =

i

[Ri mi(kRi)] k =

i

mi [(Ri k)Ri)] k =

i

mi [(Ri k) (Ri k)] =

i

mi |Ri k|2 =

i

mir2i , (6.22)

where the term |kRi| = ri is the perpendicular distance from dC to the Piparticle. The identity

(a b) c = a (b c).

has been used.The summation

i

mir2i is replaced by integration over the body

r2 dm

and is defined as mass moment of inertia ICz z of the body about the z-axisthrough C

ICz z = imir

2i .

The mass moment of inertia ICz z is a constant property of the body and is ameasure of the rotational inertia or resistance to change in angular velocitydue to the radial distribution of the rigid body mass around z-axis throughC.

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Equation (6.22) defines the angular momentum of the rigid body about

dC (z-axis through C)HC = ICz z or HC = ICz z k = ICz z .

Substituting this expression into Eq. (6.21) gives

MO =

d

dt[(rCmvC) + ICz z] = (rCmaC) + ICz z. (6.23)

The rotational equation of motion for the rigid body is

ICz z =

MC or ICz z k =

MC k. (6.24)

For general planar motion the angular acceleration is

= = k, (6.25)

where the angle describes the position, or orientation, of the rigid bodyabout a fixed axis.

If the rigid body is a plate moving in the plane of motion (X, Y) themass moment of inertia of the rigid body about z-axis through C becomesthe polar mass moment of inertia of the rigid body about C, ICz z = IC. Forthis case the Eqs. (6.24) gives

IC =

MC. (6.26)

A special application of the above equation is for rotation about a fixedpoint. Consider the special case when the rigid body rotates about the fixedpoint O as shown in Fig. 6.5(b). It follows that the acceleration of the masscenter is expressed as

aC = rC 2rC. (6.27)

The relation between the sum of the moments of the external forces aboutthe fixed point O and the product ICz z is given by Eq. (6.23)

MO = rCmaC + ICz z. (6.28)

Equations (6.28) and Eq. (6.27) giveMO = rCm ( rC 2rC) + ICz z =

m rC ( rC) + ICz z =

m [(rC rC) (rC )rC] + ICz z =

m r2C+ ICz z = (m r2C + ICz z). (6.29)

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According to parallel-axis theorem

IOzz = m r2C + ICz z,

where IOzz denotes the mass moment of inertia of the rigid body about z-axisthrough O. Therefore for the special case of rotation about a fixed point Oone can use the formula

IOzz =

MO. (6.30)

The general equations of motion for a rigid body in plane motion are(Fig. 6.6)

F = m aC or F = mrC, (6.31)MC = ICz z , (6.32)

or using the cartesian components

mxC =

Fx, myC =

Fy, ICz z =

MC. (6.33)

Equations (6.31) and (6.32) are interpreted in two ways

1. The forces and moments are known and the equations are solved forthe motion of the rigid body (direct dynamics).

2. The motion of the RB is known and the equations are solved for the

force and moments (inverse dynamics).

The dynamic force analysis in this chapter is based on the known motion ofthe mechanism.

6.4 DAlemberts principle

Newtons second law can be writen as

F + (maC) = 0, or F + Fin = 0, (6.34)

where the term Fin = maC is the inertia force. Newtons second law can

be regarded it as an equilibrium equation.Equation (6.23) relates the total moment about a fixed point O to the

acceleration of the mass center and the angular accelerationMO = (rCmaC) + ICz z,

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or

MO + [rC (maC)] + (ICz z) = 0. (6.35)

The term Min = ICz z is the inertia moment. The sum of the momentsabout any point, including the moment due to the inertial force ma actingat mass center and the inertial moment, equals zero.

The equations of motion for a rigid body are analogous to the equationsfor static equilibrium:The sum of the forces equals zero and the sum of the moments about anypoint equals zero when the inertial forces and moments are taking into ac-count This is called DAlemberts principle.

The dynamic force analysis is expressed in a form similar to static forceanalysis

R =

F + Fin = 0, (6.36)

TC =

MC + Min = 0, (6.37)

where

F is the vector sum of all external forces (resultant of external force),and

MC is the sum of all external moments about the center of mass C

(resultant external moment).For a rigid body in plane motion in the xy plane,

aC = xC + yC , = k,

with all external forces in that plane, Eqs. (6.36) and (6.37) become

Rx =

Fx + Fin x =

Fx + (m xC) = 0, (6.38)

Ry =

Fy + Fin y =

Fy + (m yC) = 0, (6.39)TC =

MC + Min =

MC + (IC ) = 0. (6.40)

With dAlemberts principle the moment summation can be about any arbi-trary point P

TP =

MP + Min + rP C Fin = 0, (6.41)

where

MP is the sum of all external moments about P, Min is the inertia moment,

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Fin is the inertia force, and

rP C is a vector from P to C.The dynamic analysis problem is reduced to a static force and moment bal-ance problem where the inertia forces and moments are treated in the sameway as external forces and moments.

6.5 Free-Body Diagrams

A free-body diagram is a drawing of a part of a complete system, isolated inorder to determine the forces acting on that rigid body.

The following force convention is defined: Fij represents the force exertedby link i on link j.

Figure 6.7 shows various free-body diagrams that are considered in the anal-ysis of a crank slider mechanism Fig. 6.7(a).In Fig. 6.7(b), the free body consists of the three moving links isolated fromthe frame 0. The forces acting on the system include a driving moment M,external driven force F, and the forces transmitted from the frame at jointA, F01, and at joint C, F03. Figure 6.7(c) is a free-body diagram of the twolinks 1 and 2. Figure 6.7(d) is a free-body diagram of a single link.The force analysis can be accomplished by examining individual links orsubsystem of links. In this way the joint forces between links as well as therequired input force or moment for a given output load are computed.

6.6 Joint Forces Analysis Using Individual Links

Figure 6.8(a) is a schematic diagram of a crank slider mechanism comprisedof a crank 1, a connecting rod 2, and a slider 3. The center of mass of link 1is C1, the center of mass of link 2 is C2 and the center of mass of slider 3 is C.The mass of the crank is m1, the mass of the connecting rod is m2, and themass of the slider is m3. The moment of inertia of link i is ICi, i = 1, 2, 3.The gravitational force is Gi = mi g, i = 1, 2, 3, where g=9.81 m/s

2 isthe acceleration of gravity.For a given value of the crank angle and a known driven force Fext the

joint reactions and the drive moment M on the crank are computed usingfree-body diagrams of the individual links.

Figures 6.8(b), (c), and (d) show free-body diagrams of the crank 1,the connecting rod 2, and the slider 3. For each moving link the dynamicequilibrium equations are applied.

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For the slider 3 the vector sum of the all the forces (external forces

Fext, gravitational force G3, inertia forces Fin3, joint forces F23, F03) is zero,Fig. 6.8(d) F(3) = F23 + Fin3 + G3 + Fext + F03 = 0.

Projecting this force onto x and y axes givesF(3) = F23x + (m3 xC) + Fext = 0, (6.42)F(3) = F23y m3 g + F03y = 0. (6.43)

For the connecting rod 2, Fig. 6.8(c), two vectorial equations can be written

F(2) = F32 + Fin 2 + G2 + F12 = 0,M

(2)B = (rC rB) F32 + (rC2 rB) (Fin 2 + G2) + Min 2 = 0,

or F(2) = F32x + (m2 xC2) + F12x = 0, (6.44)F(2) = F32y + (m2 yC2) m2 g + F12y = 0, (6.45)

k

xC xB yC yB 0F32x F32y 0

+

k

xC2 xB yC2 yB 0m2 xC2 m2 yC2 m2 g 0

IC2 2 k = 0. (6.46)

For the crank 1, Fig. 6.8(b), there are two vectorial equationsF(1) = F21 + Fin 1 + G1 + F01 = 0,M

(1)A = rB F21 + rC1 (Fin1 + G1) + Min 1 + M = 0,

or F(1) = F21x + (m1 xC1) + F01x = 0, (6.47)

F(1) = F21y + (m1 yC1)m1 g + F01y = 0, (6.48) k

xB yB 0F21x F21y 0

+

k

xC1 yC1 0m1 xC1 m1 yC1 m1 g 0

IC1 1 k + Mk = 0, (6.49)

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where M = |M| is the magnitude of the input moment on the crank.

The eight scalar unknowns F03y, F23x = F32x, F23y = F32y, F12x = F21x, F12y =F21y, F01x, F01y, and M are computed from the set of eight equations (6.42),(6.43), (6.44), (6.45), (6.46), (6.47), (6.48) and (6.49).

6.7 Joint Force Analysis Using Contour Method

An analytical method to compute joint forces that can be applied for bothplanar and spatial mechanisms will be presented. The method is based on thedecoupling of a closed kinematic chain and writing the dynamic equilibriumequations. The kinematic links are loaded with external forces and inertiaforces and moments.

A general monocontour closed kinematic chain is considered in Fig. 6.9.The joint force between the links i 1 and i (joint Ai) will be determined.When these two links i 1 and i are separated, Fig. 6.9(b), the joint forcesFi1,i and Fi,i1 are introduced and

Fi1,i + Fi,i1 = 0. (6.50)

Table 6.1 shows the joint forces for several joints. The following notationshave been used: M is the moment with respect to the axis , and F isthe projection of the force vector F onto the axis .

It is helpful to mentally disconnect the two links (i 1) and i, which

create the joint Ai, from the rest of the mechanism. The joint at Ai will bereplaced by the joint forces Fi1,i and Fi,i1. The closed kinematic chain hasbeen transformed into two open kinematic chains, and two paths I and IIare associated. The two paths start from Ai.

For the path I (counterclockwise), starting at Ai and following I the firstjoint encountered is Ai1. For the link i1 left behind, dynamic equilibriumequations are written according to the type of the joint at Ai1. Followingthe same path I, the next joint encountered is Ai2. For the subsystem (i 1and i 2) equilibrium conditions corresponding to the type of the joint atAi2 can be specified, and so on. A similar analysis is performed for thepath II of the open kinematic chain. The number of equilibrium equations

written is equal to the number of unknown scalars introduced by the joint Ai(joint forces at this joint). For a joint, the number of equilibrium conditionsis equal to the number of relative mobilities of the joint.

The five moving link (j = 1, 2, 3, 4, 5) mechanism shown in Fig. 6.10(a)has the center of mass locations designated by Cj(xCj , yCj , 0). The following

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analysis will consider the relationships of the inertia forces Fin j, the inertia

moments Min j, the gravitational force Gj , the driven force, Fext, to thejoint reactions Fij and the drive moment M on the crank 1.To simplify the notation the total vector force at Cj is written as Fj =Fin j + Gj and the inertia moment of link j is written as Mj = Min j.The diagram representing the mechanism is depicted in Fig. 6.10(b) and hastwo contours 0-1-2-3-0 and 0-3-4-5-0.

Remark. The joint at C represents a ramification point for the mech-anism and the diagram, and the dynamic force analysis will start with this

joint. The force computation starts with the contour 0-3-4-5-0 because thedriven load Fext on link 5 is given.

I. Contour 0-3-4-5-0Reaction F34

The rotation joint at C (or CR, where the subscript R means rotation),between 3 and 4, is replaced with the unknown reaction, Fig. 6.11,

F34 = F43 = F34x + F34y .

If the path I is followed, Fig. 6.11(a), for the rotation joint at E (ER) amoment equation is written

M(4)E = (rC rE) F32 + (rC4 rD) F4 + M4 = 0,

or k

xC xE yC yE 0F34x F34y 0

+

k

xC4 xE yC4 yE 0F4x F4y 0

+M4k = 0. (6.51)

Continuing on path I, the next joint is the translational joint at D (DT).The projection of all the forces, that act on 4 and 5, onto the sliding direction (x axis) should be zero.

F(4&5) = F(4&5) = (F34 + F4 + F5 + Fext) =F34x + F4x + F5x + Fext = 0. (6.52)

The system of Eqs. (6.51) and (6.52) is solved and the two unknowns F34xand F34y are obtained.

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Reaction F45

The rotation joint at E (ER), between 4 and 5, is replaced with the unknownreaction, Fig. 6.12,F45 = F54 = F45x + F45y .

If the path I is traced, Fig. 6.12(a), for the pin joint at C (CR) a momentequation is written

M

(4)C = (rE rC) F54 + (rC4 rC) F4 + M4 = 0,

or

k

xE xC yE yC 0F45x F45y 0

+ k

xC4 xC yC4 yC 0F4x F4y 0

+M4k = 0. (6.53)

For the path II the slider joint at E (ET) is encountered. The projection ofall the forces, that act on 5, onto the sliding direction (x axis) should bezero.

F(5) =

F(5) = (F45 + F5 + Fext) =

F45x + F5x + Fext = 0. (6.54)

The unknown force components F45x and F45y are calculated from Eqs. (6.53)and (6.54).

Reaction F05The slider joint at E (ET), between 0 and 5, is replaced with the unknownreaction, Fig. 6.13,

F05 = F05y.

The reaction joint introduced by the translational joint is perpendicular onthe sliding direction F05 . The application point P of the of the forceF05 is unknown.

If the path I is followed, as in Fig. 6.13(a), for the pin joint at E (ER) amoment equation is written for link 5

M

(5)E = (rP rE) F05 = 0,

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or

x F05y = 0 x = 0. (6.55)

The application point is at E (P E).Continuing on path I, the next joint is the pin joint C (CR).

M(4&5)C = (rE rC) (F05 + F5 + Fext) + (rC4 rC) F4 + M4 = 0,

or

k

xE xC yE yC 0F5x + Fext F05y 0

+

k


+M4k = 0. (6.56)The joint reaction force F05y is computed from the above Eq. (6.56).

II. Contour 0-1-2-3-0

For this contour the joint force F43 = F34 at the ramification point C,is considered as a known external force.

Reaction F03The pin joint DR, between 0 and 3, is replaced with the unknown reactionforce, Fig. 6.14

F03 = F03x + F03y .If the path I is followed, Fig. 6.14(a), a moment equation is written for thepin joint CR for the link 3

M(3)C = (rD rC) F03 + (rC3 rC) F3 + M3 = 0,

or k

xD xC yD yC 0F03x F03y 0

+

k


+M3k = 0. (6.57)

Continuing on path I the next joint is the pin joint BR and a momentequation is written for links 3 and 2

M(3&2)B = (rD rB) F03 + (rC3 rB) F3 + M3 + (rC rB) F43

+(rC2 rB) F2 + M2 = 0,

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or

k

xD xB yD yB 0F03x F03y 0

+

kxC3 xB yC3 yB 0

F3x F3y 0

+M3k +

k


+

k

xC2 xB yC2 yB 0F2x F2y 0

+ M2k = 0. (6.58)

The two components F03x and F03y of the joint force are obtained fromEqs. (6.57) and (6.58).

Reaction F23The pin joint CR, between 2 and 3, is replaced with the unknown reactionforce, Fig. 6.15,

F23 = F23x + F23y .

If the path I is followed, as in Fig. 6.15(a), a moment equation is written forthe pin joint DR for the link 3

M

(3)D = (rC rD) (F23 + F43)(rC3 rD) F3 + M3 = 0,

or k

xC xD yC yD 0F23x + F43x F23y + F43y 0

+

k

xC3 xD yC3 yD 0F3x F3y 0

+M3k = 0. (6.59)

For the path II the first joint encountered is the pin joint BR and a momentequation is written for link 2

M(2)B = (rC rB) (F23) + (rC2 rB) F2 + M2 = 0,

or k


+

k


+M2k = 0. (6.60)

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The two force components F23x and F23y of the joint force are obtained from

Eqs. (6.59) and (6.60).

Reaction F12The pin joint BR, between 1 and 2, is replaced with the unknown reactionforce, Fig. 6.16

F12 = F12x + F12y .

If the path I is followed, as in Fig. 6.16(a), a moment equation is written forthe pin joint CR for the link 2

M

(2)C = (rB rC) F12 + (rC2 rC) F2 + M2 = 0,

or k

xB xC yB yC 0F12x F12y 0

+

k


+M2k = 0. (6.61)

Continuing on path I the next joint encountered is the pin joint DR and amoment equation is written for links 2 and 3

M(2&3)D = (rB rD) F12 + (rC2 rD) F2 + M2 +

(rC rD) F43 + (rC3 rD) F3 + M3 = 0,

or k

xB xD yB yD 0F12x F12y 0

+

k


+ M2k +

k

xC xD yC yD 0F43x F43y 0

+

k


+M3k = 0. (6.62)

The two components F12x and F12y of the joint force are computed fromEqs. (6.61) and (6.62).

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Reaction F01 and driver moment M

The pin joint AR, between 0 and 1, is replaced with the unknown reactionforce, Fig. 6.17F01 = F01x + F01y .

The unknown driver moment is M = Mk. If the path I is followed,Fig. 6.17(a), a moment equation is written for the pin joint BR for the link 1

M

(1)B = (rA rB) F01 + (rC1 rB) F1 + M1 + M = 0,

or

k

xA xB yA yB 0F01x F01y 0

+ k


+M1 k + Mk = 0. (6.63)

Continuing on path I, the next joint encountered is the pin joint CR and amoment equation is written for links 1 and 2

M

(1&2)C = (rA rC) F01 + (rC1 rC) F1 + M1 + M +

(rC2 rC) F2 + M2 = 0. (6.64)

Equation (6.64) is the vector sum of the moments about DR of all forces and

moments that act on links 1, 2, and 3.M

(1&2&3)D = (rA rD) F01 + (rC1 rD) F1 + M1 + M +

(rC2 rD) F2 + M2 + (rC rD) F43 + (rC3 rD) F3

+M3 = 0. (6.65)

The components F01x, F01y and M are computed from Eqs. (6.63), (6.64),and (6.65).

6.8 Joint Force Analysis Using Dyads

RRR dyad

Figure 6.18 shows a RRR dyad with two links 2 and 3, and three pin joints

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B, C, and D. The unknowns are the joint reaction forces

F12 = F12x + F12y,

F43 = F43x + F43y,

F23 = F32 = F23x + F23y. (6.66)

The inertia forces and external forces Fj = Fj + Fj, inertia moments andexternal moments Mj = Mj k, (j=2,3) are given.

To determine F12 and F43 the following equations are written sum of all forces on links 2 and 3 is zero

F

(2&3)

= F12 + F2 + F3 + F43 = 0,or

F(2&3) = F12x + F2x + F3x + F43x = 0,F(2&3) = F12y + F2y + F3y + F43y = 0. (6.67)

sum of moments of all forces and moments on link 2 about C is zeroM

(2)C = (rB rC) F12 + (rC2 rC) F2 + M2 = 0. (6.68)

sum of moments of all forces and moments on link 3 about C is zeroM

(3)C = (rD rC) F43 + (rC3 rC) F3 + M3 = 0. (6.69)

The components F12x, F12y, F43x, and F43y are calculated from Eqs. (6.67), (6.68),and (6.69).The reaction force F32 = F23 is computed from the sum of all forces on thelink 2

F(2) = F12 + F2 + F32 = 0,

or

F(2)

= F12x + F2x + F32x = 0,F(2) = F12y + F2y + F32y = 0. (6.70)

RRT dyad

Figure 6.19 shows a RRT dyad with the unknown joint reaction forces F12,

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F43, and F23 = F32. The joint reaction force F43 is perpendicular to the

sliding direction F43 orF43 = (F43x + F43y) (cos + sin ) = 0. (6.71)

In order to determine F12 and F43 the following equations are written sum of all the forces on links 2 and 3 is zero

F(2&3) = F12 + F2 + F3 + F43 = 0,

or F(2&3) = F12x + F2x + F3x + F43x = 0,

F(2&3) = F12y + F2y + F3y + F43y = 0. (6.72) sum of moments of all the forces and the moments on link 2 about C

is zero M

(2)C = (rB rC) F12 + (rC2 rC) F2 + M2 = 0. (6.73)

The components F12x, F12y, F43x, and F43y are calculated from Eqs. (6.71), (6.72),and (6.73).The reaction force components F32x and F32y are computed from the sum ofall the forces on the link 2

F(2) = F12 + F2 + F32 = 0,or

F(2) = F12x + F2x + F32x = 0,F(2) = F12y + F2y + F32y = 0. (6.74)

RTR dyad

The unknown joint reaction forces F12 and F43 are calculated from the rela-tions, Fig. 6.20

sum of all the forces on links 2 and 3 is zero

F(2&3) = F12 + F2 + F3 + F43 = 0,or

F(2&3) = F12x + F2x + F3x + F43x = 0,F(2&3) = F12y + F2y + F3y + F43y = 0. (6.75)

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sum of the moments of all the forces and moments on link 2 and 3 about

B is zero M

(2&3)B = (rD rB) F43 + (rC3 rB) F3 + M3 +

(rC2 rB) F2 + M2 = 0. (6.76)

sum of all the forces on link 2 projected onto the sliding direction = cos + sin is zero

F(2) = (F12 + F2) (cos + sin ) = 0. (6.77)

The components F12x, F12y, F43x, and F43y are calculated from Eqs. (6.75), (6.76),

and (6.77).The force components F32x and F32y are computed from the sum of all theforces on link 2

F(2) = F12 + F2 + F32 = 0,

or

F(2) = F12x + F2x + F32x = 0,F(2) = F12y + F2y + F32y = 0. (6.78)

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6.9 Examples

Example 6.1. The R-RTR mechanism shown in Fig. 6.21(a) has thedimensions: AB = 0.14 m, AC = 0.06 m, and CF = 0.2 m. The driver link

1 makes an angle = 1 =

3rad with the horizontal and rotates with a

constant speed of n = n1 = 30 rpm. The position vectors of the points A,B, C, and F are

rA = 0 + 0 m,

rB = rC2 = xB + yB = 0.07 + 0.121 m,

rC = xC + yC = 0 + 0.06 m,

rF = xF + yF = 0.150 + 0.191 m,

where the mass center of the slider 2 is at B (B = C2). The position vectorsof the mass centers of the links 1 and 3 are

rC1 = xC1 + yC1 =xB2

+yB2

= 0.035 + 0.06 m,

rC3 = xC3 + yC3 =xC + xF

2 +

yC + yF2

= 0.075 + 0.125 m.

The total forces and moments at Cj , j = 1, 2, 3 are Fj = Fin j + Gj andMj = Min j, where Fin j is the inertia force, Mj is the inertia moment, andGj = mj g is the gravity force with gravity acceleration g = 9.81 m/s

2.

F1 = 0.381 + 0.437 N, M1 = 0k N m,

F2 = 0.545 + 0.160 N, M2 = 0.001k N m,

F3 = 3.302 0.539 N, M3 = 0.046k N m.

The external moment on link 3 is M3ext = 1000k Nm. Determine themoment M required for dynamic equilibrium and the joint forces for themechanism using the free body diagrams of the individual links.

SolutionFor each link two vectorial equations are written

Fj + Fin j = 0 and

MCj + Min j = 0, (6.79)

where

Fj is the vector sum of all external forces (resultant of externalforce) on link j, and

MCj is the sum of all external moments on link j

about the mass center Cj.

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The force analysis will start with link 3 because the moment M3ext is

known.Link 3For the free body diagram of link 3 shown in Fig. 6.21(b), Eq. (6.79) gives

F03 + Fin 3 + G3 + F23 = 0,

rC3C F03 + rC3Q F23 + Min 3 + M3ext = 0,

or

F03 + F3 + F23 = 0,

rC3C F03 + rC3Q F23 + M3 + M3ext = 0, (6.80)

where the unknowns are

F03 = F03x + F03y, F23 = F23x + F23y,

and the position vector rQ = xQ + yQ of the application point of the jointforce F23. Numerically Eq. (6.80) becomes

3.302 + F03x + F23x = 0, (6.81)

0.539 + F03y + F23y = 0, (6.82)

1000.05 + 0.065F03x 0.075F03y + 0.125F23x 0.075F23y +

F23yxQ F23xyQ = 0. (6.83)

The application point Q of the joint force F23 is on the line BC

yB yCxB xC

=yQ yCxQ xC

or 0.874yQ 0.06

xQ= 0. (6.84)

The joint force F23 is perpendicular to the sliding direction BC

F23 rBC = 0 or 0.07F23x 0.061F23y = 0. (6.85)

There are five scalar equations, Eqs. (6.81)-(6.85), and six unknownsF03x, F03y, F23x, F23y, xQ, yQ. The force analysis will continue with link 2.

Link 2Figure 6.21(c) shows the free body diagram of link 2 and Eq. (6.79) gives

F12 + Fin2 + G2 + F32 = 0,

rBQ F32 + Min 2 = 0,

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or

F12 + F2 F23 = 0,

rBQ (F23) + M2 = 0,

where the new unknown is introduced (the reaction of link 1 on link 2)

F12 = F12x + F12y.

Numerically the previous equations becomes

0.545 + F12x F23x = 0, (6.86)

0.160 + F12y F23y = 0, (6.87)0.001 0.121 F23x + 0.07 F23y xQ F23y + yQ F23x = 0. (6.88)

Now there is a system of eight scalar equations, Eqs. (6.81)-(6.88), eightunknowns, and the solution is

F03 = F03x + F03y = 7078.41 8093.7 N,

F23 = F23x + F23y = 7081.72 + 8094.24 N,

F12 = F12x + F12y = 7082.26 + 8094.08 N,

rQ = xQ + yQ = 0.069 + 0.121 m.

Link 1

Figure 6.21(d) shows the free body diagram of link 1. The sum of all theforces for the driver link 1 gives

F21 + Fin1 + G1 + F01 = 0, or F12 + F1 + F01 = 0.

The reaction of the ground 0 on the link 1 is

F01 = F12 F1 = 7082.26 + 8094.08 (0.381 + 0.437)

= 7082.64 + 8094.52 N.

The sum of the moments about the mass center C1 for link 1 gives theequilibrium moment

rC1B F21 + rC1A F01 + M = 0,

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or

M = rC1B F12 rC1A F01

= 712.632 k + 712.671 k = 1425.303 k N m.

Example 6.2. Calculate the moment M required for dynamic equilib-rium and the joint forces for the mechanism shown in Fig. 6.22 using the

contour method. The position of the crank angle is =

4rad. The dimen-

sions are AC=0.10 m, BC=0.30 m, BD=0.90 m, and La=0.10 m, and theexternal force on slider 5 is Fext=100 N. The angular speed of the crank 1 is

n1=100 rpm, or 1=100 30 rad/s. The center of mass locations of the links

j = 1, 2,..., 5 (with the masses mj ) are designated by Cj (xCj , yCj , 0). Theposition vectors of the joints and the centers of mass are

rA = 0 + 0 m,

rC1 = 0.212 + 0.212 m,

rB = rC2 = 0.256 + 0.256 m,

rC3 = 0.178 + 0.128 m,

rC = 0.100 + 0.000 m,

rC4 = 0.699

+ 0.178

m,

rD = rC5 = 1.142 + 0.100 m.

The total forces and moments at Cj are Fj = Fin j + Gj and Mj = Min j ,where Fin j is the inertia force, Mj is the inertia moment, and Gj = mj gis the gravity force with gravity acceleration g = 9.81 m/s2.

F1 = 5.514 + 3.189 N,

F2 = 0.781 + 1.843 N,

F3 = 1.202 + 1.660 N,

F4 = 6.466 + 4.896 N,

F5 = 0.643 0.382 N,

M1 = M2 = M5 = 0 k N m,

M3 = 0.023k N m,

M4 = 1.274k N m.

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Solution

The diagram representing the mechanism is shown in Fig. 6.22(b) andhas two contours 0-1-2-3-0 and 0-3-4-5-0.I. Contour 0-3-4-5-0

The joint at B represents a ramification point, and the dynamic forceanalysis will start with this joint.

Reaction F34The rotation joint at BR between 3 and 4, is replaced with the unknownreaction, Fig. 6.23,

F34 = F43 = F34x + F34y .

If the path I is followed, Fig. 6.23(a), a moment equation is written for therotation joint DR

M(4)D = (rB rD) F34 + (rC4 rD) F4 + M4 = 0. (6.89)

Continuing on path I the next joint is the slider joint DT and a force equationis written. The projection of all the forces, that act on 4 and 5, onto thesliding direction x is zero

F(4&5) = (F34 + F4 + F5 + Fext) =

F34x + F4x + F5x + Fext = 0. (6.90)

Solving the system of Eqs. (6.89) and (6.90)

F34x = 107.110 N and F34y = 14.415 N.

Reaction F45The pin joint at DR between 4 and 7, is replaced with the reaction force,Fig. 6.24,

F45 = F54 = F45x + F45y .

For the path I, shown Fig. 6.24(a), a moment equation about BR is writtenfor the link 4

M

(4)B = (rD rB) F54 + (rC4 rB) F4 + M4 = 0. (6.91)

For the path II an equation for the forces projected onto the sliding directionof the joint DT is written for the link 5

F(5) = (F45 + F5 + Fext) =

F45x + F5x + Fext = 0. (6.92)

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The joint force F45 is obtained from the system of Eqs. (6.91) and (6.92)

F45x = 100.643 N and F45y = 19.310 N.

Reaction F05The reaction force F05 is perpendicular to the sliding direction of joint DT,Fig. 6.25

F05 = F05y.

The application point of the unknown reaction force F05 is computed from amoment equation about DR, for the link 5 (path I), Fig. 6.25(a)

M

(5)D = (rP rD) F05 = 0, (6.93)

or

x F05y = 0 x = 0. (6.94)

The application point of the reaction force F05 is at D (P D). Themagnitude of the reaction force F05y is obtained from a moment equationabout BR, for the links 5 and 4, (path I)

M(5&4)B = (rD rB) (F05 + F5 + Fext) +

(rC4 rB) F4 + M4 = 0. (6.95)

Solving the above equation

F05y = 18.928 N.

II. Contour 0-1-2-3-0

The reaction force F43 = 107.110 14.415 N, is considered as an externalforce for this contour at B.

Reaction F23The rotation joint at BR between 2 and 3, is replaced with the unknownreaction force, Fig. 6.26,

F23 = F32 = F23x + F23y .If the path I is followed, as in Fig. 6.26(a), a moment equation is written forthe pin joint CR for the link 3

M(3)C = (rB rC) (F23 + F43) + (rC3 rC) F3 + M3 = 0. (6.96)

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For the path II an equation for the forces projected in the direction the

sliding direction of the joint BT, is written for the link 2F(2) = (F32 + F2) (cos + sin ) = 0. (6.97)

The joint force F23 is calculated from the Eqs. (6.96) and (6.97)

F23x = 71.155 N and F23y = 73.397 N.

Reaction F03For the joint reaction force F03 at CR, there is only path I. For the pin jointBR one moment equation is written for link 3, Fig. 6.27,

M(3)B = (rC rB) F03 + (rC3 rB) F3 + M3 = 0. (6.98)

A force equation is written for the links 3 and 2 for the slider joint BT

F(3&2) = (F03 + F3 + F43 + F2) (cos + sin ) = 0. (6.99)

The components of the unknown force are obtained by solving the system ofEqs. (6.98) and (6.99)

F03x = 37.156 N and F03y = 60.643 N.

Reaction F12The slider joint at BT between 1 and 2, is replaced with the reaction force,Fig. 6.28

F12 = F21 = F12x + F12y .

The reaction force F12 is perpendicular to the sliding direction

F12 = (F12x + F12y) (cos + sin ) =

F12x cos + F12y sin = 0. (6.100)

The point of application of force F12 is determined from the equation (path

I) M(2)B = (rQ rB) F12 = 0, (6.101)or

x F12 = 0 x = 0, (6.102)

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and the force F12 acts at B.

Continuing on path I, a moment equation is written for links 2 and 3 withrespect to the pin joint CRM

(2&3)C = (rB rC) (F12 + F2 + F43) +

(rC3 rC) F3 + M3 = 0. (6.103)

The two components of the joint force F12 are computed from Eqs. (6.100)and (6.103)

F12x = 71.936 N and F12y = 71.936 N.

Reaction F01 and equilibrium moment MThe pin joint AR, between 0 and 1, is replaced with the unknown reaction,

Fig. 6.29,F01 = F01x + F01y .

The unknown equilibrium moment is M = Mk. If the path I is followed, asin Fig. 6.29(a), for the slider joint BT a force equation is written for link 1

F(1) = (F01 + F1) (cos + sin ) = 0. (6.104)

Continuing on path I the next joint encountered is the pin joint BR and amoment equation is written for links 1 and 2

M(1&2)B = rB F01 + (rC1 rB) F1 + M = 0. (6.105)

Equation (6.105) is the vector sum of the moments about CR of all forcesand moments that act on links 1, 2, and 3.

M(1&2&3)C = rC F01 + (rC1 rC) F1 + M +

(rB rC) (F2 + F43) + M3 + (rC3 rC) F3 = 0. (6.106)

From Eqs. (6.104), (6.105), and (6.106) the components F01x, F01y and Mare computed

F01x = 77.451 N, F01y = 68.747 N, and M = 37.347 N m.

Example 6.3. For the R-TRR-RRT mechanism in Example 6.2 calculatethe moment M required for dynamic equilibrium of the mechanism and the

joint forces using the dyad method.

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Solution

BR DR DT dyadFigure 6.30(a) shows the last dyad BR DR DT with the unknown joint reac-tions F34, F05, and F45 = F54. The joint reaction F05 is perpendicular tothe sliding direction F05 = or

F05 = F05y. (6.107)

The following equations are written to determine F34 and F05 sum of all the forces on links 4 and 5 is zero

F(4&5) = F34 + F4 + F5 + Fext + F05 = 0,

or

F(2&3) = F43x + F4x + F5x + Fext = 0,F(2&3) = F43y + F4y + F5y + F05y = 0. (6.108)

sum of moments of all the forces and moments on link 4 about DR iszero

M

(4)D = (rB rD) F43 + (rC4 rD) F4 + M4 = 0. (6.109)

From Eqs. (6.108) and (6.109) the unknown components are calculated

F34x = 107.110 N, F34y = 14.415 N, and F05y = 18.928 N.

The reaction components F54x and F54y are computed from the sum ofall the forces on link 4, Fig. 6.30(b)

F(4) = F34 + F4 + F54 = 0,

or

F(4) = F34x + F4x + F54x = 0,

F(4) = F34y + F5y + F54y = 0, (6.110)

andF54x = 100.643 N and F54y = 19.310 N.

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BT BR CR dyad

Figure 6.31(a) shows the first dyad BT BR CR with the unknown joint reactionforces F12, F03, and F23 = F32. The joint reaction force F12 is perpendicularto the sliding direction F12 or

F12 = (F12x + F12y) (cos + sin ) = 0. (6.111)

The following equations are written in order to determine the forces F12 andF03

sum of all forces on links 2 and 3 is zeroF(2&3) = F12 + F2 + F3 + F43 + F03 = 0,

or F(2&3) = F12x + F2x + F3x + F43x + F03x = 0,F(2&3) = F12y + F2y + F3y + F43y + F03x = 0. (6.112)

sum of moments of all the forces and the moments on link 3 about BRis zero

M(3)B = (rC rB) F03 + (rC3 rB) F3 + M3 = 0. (6.113)

From Eqs. (6.111), (6.112), and (6.113) the following components are ob-

tained F12x = 71.936 N and F12y = 71.936 N,

F03x = 37.156 N and F03y = 60.643 N.

The reaction components F23x and F23y are computed from the sum of allthe forces on link 3, Fig. 6.31(b)

F(3) = F23 + F3 + F43 + F03 = 0,

or

F(3) = F23x + F3x + F43x + F03x = 0,F(2) = F23y + F3y + F43y + F03y = 0, (6.114)

and solving the equations

F23x = 71.155 N and F23y = 73.397 N.

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Driver link

A force equation for the driver can be written to determine the joint reactionF01, Fig. 6.32 F(1) = F01 + F1 + F21 = 0,

or

F(1) = F01x + F1x + F21x = 0,F(1) = F01y + F1y + F21y = 0, (6.115)

Solving the above equations gives

F01x = 77.451 N and F01y = 68.747 N.

Sum of the moments about AR for link 1 gives the equilibrium momentM

(1)A = rB F21 + rC1 F1 + M = 0, (6.116)

and M=37.347 Nm.

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6.10 Problems

6.1 Figure 6.33 shows a uniform rod of mass m and length L. The rod isfree to swing in a vertical plane. The rod is connnected to the groundby a pin joint at the distance D from one end of the rod. The rod makesan angle (t) with the horizontal. The local acceleration of gravity isg. 1. Find the differential equation or equations describing the motionof the rod. 2. Determine are the axial and shear components of theforce exerted by the pin on the rod as the rod swings by any arbitraryposition? 3. The rod is released from rest in the horizontal position,and then the initial value of the angular velocity is zero. Find the initialangular acceleration and the initial pin force components.

6.2 The four-bar mechanism shown in Fig. 3.10(a) has the dimensions:AB = 80 mm, BC = 210 mm, CD = 120 mm, and AD = 190 mm.The driver link AB rotates with a constant angular speed of 2400 rpm.The links are homogeneous rectangular prisms made of steel with thewidth h = 0.010 m and the depth d = 0.001 m. The external momentapplied on the link CD is opposed to the motion of the link and hasthe value |Mext| = 600 Nm. The density of the material is Steel =8000 kg/m3 and the gravitational acceleration is g = 9.807 m/s2. Findthe equilibrium moment on link AB and the joint forces for = 120

using: 1. free-body diagram of individual links; 2. contour method;

and 3. dyads.

6.3 The slider crank mechanism shown in Fig. 4.10 has the dimensionsAB = 0.4 m and BC = 1 m. The driver link 1 rotates with a constantangular speed of n = 1600 rpm. The links 1 and 2 have rectangu-lar shape made of steel with the width h = 0.010 m and the depthd = 0.001 m. The steel slider 3 has the width wSlider = 0.050 m, theheight hSlider = 0.020 m, and the depth d = 0.001 m. The external forceapplied on the slider 3 is opposed to the motion of the slider and has thevalue |Fext| = 800 N. The density of the material is Steel = 8000 kg/m

3

and the gravitational acceleration is g = 9.807 m/s2. Find the equi-

librium moment on the driver link 1 and the joint forces for = 30

using: 1. free-body diagram of individual links; 2. contour method;and 3. dyads.

6.4 The planar mechanism considered is shown in Fig. 3.19 and has the fol-

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lowing data: AB=0.150 m, BC=0.400 m, CD=0.370 m, CE=0.230 m,

EF=CE, La=0.300 m, Lb=0.450 m, and Lc=CD. The constant angu-lar speed of the driver link 1 is 1800 rpm. The links 1, 2, 3, and 4 arehomogeneous rectangular prisms with the width h = 0.010 m and thedepth d = 0.001 m. The slider 5 has the width wSlider = 0.050 m, theheight hSlider = 0.020 m, and the depth d = 0.001 m. The external forceapplied on the slider 5 is opposed to the motion of the slider and has thevalue |Fext| = 500 N. The density of the material is Steel = 8000 kg/m

3

and the gravitational acceleration is g = 9.807 m/s2. Find the equilib-rium moment on the driver link 1 and the joint forces for = 1 = 60.

6.5 The R-RRR-RTT mechanism is shown in Fig. 3.20. The following data

are given: AB=0.080 m, BC=0.350 m, CE=0.200 m, CD=0.150 m,La=0.200 m, Lb=0.350 m, Lc=0.040 m. The driver link 1 rotates witha constant angular speed of n = 2200 rpm. The links 1, 2, 3, and 5are homogeneous rectangular prisms made of aluminum with the widthh = 0.010 m and the depth d = 0.001 m. The aluminum slider 4 hasthe width wSlider = 0.050 m, the height hSlider = 0.020 m, and thedepth d = 0.001 m. The external force applied on 5 is opposed to themotion of the link and has the value |Fext| = 1000 N. The density ofthe material is Al = 2.8 Mg/m

3 and the gravitational acceleration isg = 9.807 m/s2. For = 145 find the equilibrium moment on thedriver link 1 and the joint forces. Select suitable dimensions for the

link 5.

6.6 The mechanism shown in Fig. 3.21 has the following dimensions: AB =100 mm, AD = 350 mm, BC = 240 mm, CE = 70 mm, EF = 300 mm,and a = 240 mm. The constant angular speed of the driver link 1 isn = 1400 rpm. The links 1 and 4 are homogeneous rectangular prismswith the width h = 0.010 m and the depth d = 0.001 m. The link 2 hasthe width h = 0.010 m and the depth d = 0.001 m. The sliders 3 and5 have the width wSlider = 0.050 m, the height hSlider = 0.020 m, andthe depth d = 0.001 m. The external force applied on 5 is opposed tothe motion of the link and has the value |Fext| = 1200 N. The densityof the material is Iron = 7.2 Mg/m3 and the gravitational accelerationis g = 9.807 m/s2. Find the equilibrium moment on the driver link 1and the joint forces for = 1 = 30.. Select a suitable dimension forthe link 2.

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6.7 The dimensions for the mechanism shown in Fig. 3.22 are: AB =

60 mm, BD = 160 mm, BC = 55 mm, CD = 150 mm, DE = 100 mm,CF = 250 mm, AE = 150 mm, and b = 40 mm. The constant angularspeed of the driver link 1 is n = 1400 rpm. The links 1, 3, and 4 arehomogeneous rectangular prisms with the width h = 0.010 m and thedepth d = 0.001 m. The slider 5 has the width wSlider = 0.050 m, theheight hSlider = 0.020 m, and the depth d = 0.001 m. The plate 2has the width h = 0.010 m and the depth d = 0.001 m. The externalforce applied on 5 is opposed to the motion of the link and has the value|Fext| = 1500 N. The density of the material is Bronze = 8.7 Mg/m

3 andthe gravitational acceleration is g = 9.807 m/s2. Find the equilibriummoment on the driver link 1 and the joint forces for = 1 = 60.

6.8 The mechanism in Fig. 3.23 has the dimensions: AB = 110 mm, AC =55 mm, BD = 220 mm, DE = 300 mm, EF = 175 mm, La = 275 mm,and Lb = 65 mm. The links 1, 2, 4, and 5 are homogeneous rectangularprisms with the width h = 0.010 m and the depth d = 0.001 m. Theslider 3 has the width wSlider = 0.050 m, the height hSlider = 0.020 m,and the depth d = 0.001 m. The constant angular speed of the driverlink 1 is n = 2400 rpm. The external moment on 5 is opposed to the

motion of the link Mext = |Mext|5

|5|where |Mext| = 600 Nm. The

density of the material is Steel = 8000 kg/m3 and the gravitational

acceleration is g = 9.807 m/s2

. Find the equilibrium moment on thedriver link 1 and the joint forces for = 1 = 150.

6.9 The dimensions for the mechanism shown in Fig. 3.24 are: AB =250 mm, BC = 650 mm, AD = 600 mm, CD = 350 mm, DE =200 mm, EF = 600 mm, and La = 100 mm. The constant angularspeed of the driver link 1 is n = 2500 rpm. The links 1, 2, 3, and 4are homogeneous rectangular prisms with the width h = 0.010 m andthe depth d = 0.001 m. The slider 5 has the width wSlider = 0.050 m,the height hSlider = 0.020 m, and the depth d = 0.001 m. The external

force on 5 is opposed to the motion of the link Fext = |Fext|vF

|vF|where

|Fext| = 1600 N. The density of the material is Steel = 8000 kg/m3 andthe gravitational acceleration is g = 9.807 m/s2. Find the equilibriummoment on the driver link 1 and the joint forces for = 1 = 60.

6.10 The mechanism in Fig. 3.25 has the dimensions: AB = 50 mm, AC =

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160 mm, BD = 250 mm, La = 30 mm, and Lb = 60 mm. The driver

link 1 rotates with a constant angular speed of n = 1500 rpm. Thelinks 1, 2, and 5 are homogeneous rectangular prisms with the widthh = 0.010 m and the depth d = 0.001 m. The sliders 3 and 4 have thewidth wSlider = 0.050 m, the height hSlider = 0.020 m, and the depthd = 0.001 m. The external moment on 5 is opposed to the motion of

the link Mext = |Mext|5

|5|where |Mext| = 900 Nm. The density of

the material is Steel = 8000 kg/m3 and the gravitational acceleration

is g = 9.807 m/s2. Find the equilibrium moment on the driver link 1and the joint forces for = 1 = 130. Select a suitable dimension forthe link 5.

6.11 Figure 3.26 shows a mechanism with the following dimensions: AB =150 mm, BD = 500 mm, and La = 180 mm. The constant angu-lar speed of the driver link 1 is n = 1600 rpm. The links 1, 2, and4 are homogeneous rectangular prisms with the width h = 0.010 mand the depth d = 0.001 m. The sliders 3 and 5 have the widthwSlider = 0.050 m, the height hSlider = 0.020 m, and the depth d =0.001 m. The external force on 5 is opposed to the motion of the

link Fext = |Fext|vD

|vD|where |Fext| = 2000 N. The density of the

material is Steel = 8000 kg/m3 and the gravitational acceleration is

g= 9

.807 m/s2. Find the equilibrium moment on the driver link 1 and

the joint forces for = 210. Select a suitable dimension for the link 4.

6.12 The mechanism in Fig. 3.27 has the dimensions: AB = 20 mm, AC =50 mm, BD = 150 mm, DE = 40 mm, EF = 27 mm, La = 7 mm,and Lb = 30 mm. The constant angular speed of the driver link 1 isn = 1400 rpm. The links 1, 2, 4, and 5 are homogeneous rectangularprisms with the width h = 0.010 m and the depth d = 0.001 m. Theslider 3 has the width wSlider = 0.050 m, the height hSlider = 0.020 m,and the depth d = 0.001 m. The external moment on 5 is opposed to

the motion of the link Mext = |Mext|5

|5|where |Mext| = 1500 Nm.

The density of the material is Steel = 8000 kg/m3 and the gravitationalacceleration is g = 9.807 m/s2. Find the equilibrium moment on thedriver link 1 and the joint forces for = 1 = 120.

6.13 Figure 3.28 shows a mechanism with the following dimensions: AB =

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250 mm, BC = 940 mm, CD = DE = 380 mm, EF = 700 mm,

La = 930 mm, and Lb = Lc = 310 mm. The driver link 1 rotates witha constant angular speed of n = 1500 rpm. The links 1, 2, 3, and 4are homogeneous rectangular prisms with the width h = 0.010 m andthe depth d = 0.001 m. The slider 5 has the width wSlider = 0.050 m,the height hSlider = 0.020 m, and the depth d = 0.001 m. The external

force on 5 is opposed to the motion of the link Fext = |Fext|vF

|vF|where

|Fext| = 2000 N. The density of the material is Steel = 8000 kg/m3 and

the gravitational acceleration is g = 9.807 m/s2. Find the equilibriummoment on the driver link 1 and the joint forces or = 1 = 120.

6.14 Figure 3.29 shows a mechanism with the following dimensions: AB =200 mm, BC = 900 mm, CE = 300 mm, CD = 600 mm, EF =600 mm, La = 500 mm, Lb = 800 mm, and Lc = 1100 mm. Theconstant angular speed of the driver link 1 is n = 1000 rpm. Thelinks 1, 2, 3, and 4 are homogeneous rectangular prisms with the widthh = 0.010 m and the depth d = 0.001 m. The slider 5 has the widthwSlider = 0.050 m, the height hSlider = 0.020 m, and the depth d =0.001 m. The external force on 5 is opposed to the motion of the

link Fext = |Fext|vF

|vF|where |Fext| = 3000 N. The density of the


g= 9

.807 m/s2. Find the equilibrium moment on the driver link 1 and

the joint forces for = 1 = 150.

6.15 Figure 3.30 shows a mechanism with the following dimensions: AB =200 mm, BC = 540 mm, CF = 520 mm, CD = 190 mm, DE =600 mm, La = 700 mm, Lb = 400 mm, and Lc = 240 mm. The constantangular speed of the driver link 1 is n = 1200 rpm. The links 1, 2, 3, and4 are homogeneous rectangular prisms with the width h = 0.010 m andthe depth d = 0.001 m. The slider 5 has the width wSlider = 0.050 m,the height hSlider = 0.020 m, and the depth d = 0.001 m. The external

force on 5 is opposed to the motion of the link Fext = |Fext|vE

|vE|where

|Fext| = 900 N. The density of the material is Steel = 8000 kg/m3 andthe gravitational acceleration is g = 9.807 m/s2. For = 30 find theequilibrium moment on the driver link 1 and the joint forces.

6.16 Figure 3.31 shows a mechanism with the following dimensions: AB =

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80 mm, BC = 200 mm, AD = 90 mm, and BE = 220 mm. The

constant angular speed of the driver link 1 is n = 1300 rpm. Thelinks 1, 2, and 4 are homogeneous rectangular prisms with the widthh = 0.010 m and the depth d = 0.001 m. The sliders 3 and 5 havethe width wSlider = 0.050 m, the height hSlider = 0.020 m, and thedepth d = 0.001 m. The external force on 3 is opposed to the motion

of the link Fext = |Fext|vC

|vC|where |Fext| = 1900 N. The density of


is g = 9.807 m/s2. Find the equilibrium moment on the driver link 1and the joint forces for = 60..

6.17 The dimensions of the mechanism shown in Fig. 3.32 are: AB = 80 mm,BC = 150 mm, BE = 300 mm, CE = 450 mm, CD = 170 mm,EF = 600 mm, La = 200 mm, Lb = 150 mm, and Lc = 50 mm.The constant angular speed of the driver link 1 is n = 1500 rpm.The links 1, 3, and 4 are homogeneous rectangular prisms with thewidth h = 0.010 m and the depth d = 0.001 m. The slider 5 hasthe width wSlider = 0.050 m, the height hSlider = 0.020 m, and thedepth d = 0.001 m. The plate 2 has the width h = 0.010 m and thedepth d = 0.001 m. The external force applied on 5 is opposed to themotion of the link and has the value |Fext| = 2000 N. The density ofthe material is Bronze = 8.7 Mg/m

3 and the gravitational acceleration

is g = 9.807 m/s2

. Find the equilibrium moment on the driver link 1and the joint forces for = 1 = 210.

6.18 The dimensions of the mechanism shown in Fig. 3.33 are: AB =140 mm, AC = 200 mm, CD = 350 mm, DE = 180 mm, andLa = 300 mm. The constant angular speed of the driver link 1 isn = 900 rpm. The links 1, 3, and 4 are homogeneous rectangular prismswith the width h = 0.010 m and the depth d = 0.001 m. The sliders 2and 5 have the width wSlider = 0.050 m, the height hSlider = 0.020 m,and the depth d = 0.001 m. The external force on 5 is opposed to

the motion of the link Fext = |Fext|vE

|vE|where |Fext| = 1000 N. The

density of the material is Steel = 8000 kg/m3 and the gravitationalacceleration is g = 9.807 m/s2. Find the equilibrium moment on thedriver link 1 and the joint forces for = 60.

6.19 The dimensions of the mechanism shown in Fig. 3.34 are: AB =

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250 mm, AC = 100 mm, CD = 280 mm, and DE = 800 mm. The

constant angular speed of the driver link 1 is n = 1600 rpm. Thelinks 1, 3, and 4 are homogeneous rectangular prisms with the widthh = 0.010 m and the depth d = 0.001 m. The sliders 2 and 5 havethe width wSlider = 0.050 m, the height hSlider = 0.020 m, and thedepth d = 0.001 m. The external force on 5 is opposed to the motion

of the link Fext = |Fext|vE

|vE|where |Fext| = 900 N. The density of


is g = 9.807 m/s2. For = 1 = 210 find the equilibrium moment onthe driver link 1 and the joint forces.

6.20 The dimensions of the mechanism shown in Fig. 3.35 are: AB =100 mm, AC = 200 mm, and CD = 350 mm. The constant angu-lar speed of the driver link 1 is n = 900 rpm. The links 1, 3, and5 are homogeneous rectangular prisms with the width h = 0.010 mand the depth d = 0.001 m. The sliders 2 and 4 have the widthwSlider = 0.050 m, the height hSlider = 0.020 m, and the depth d =0.001 m. The external force on 5 is opposed to the motion of the

link Fext = |Fext|vG

|vG|where |Fext| = 2500 N. The density of the


g = 9.807 m/s2. Find the equilibrium moment on the driver link 1 and

the joint reaction forces for = 1 = 45

. Select suitable dimensionsfor the link 5 and the distance b.

6.21 The dimensions of the mechanism shown in Fig. 3.36 are: AB =140 mm, AC = 60 mm, and CD = 140 mm. The constant angularspeed of the driver link 1 is n = 2200 rpm. The links 1, 3, and 5 arehomogeneous rectangular prisms with the width h = 0.010 m and thedepth d = 0.001 m. The sliders 2 and 4 has the width wSlider = 0.050 m,the height hSlider = 0.020 m, and the depth d = 0.001 m. The external

moment on 5 is opposed to the motion of the link Mext = |Mext|5

|5|where |Mext| = 1500 Nm. The density of the material is Steel =

8000 kg/m3 and the gravitational acceleration is g = 9.807 m/s2. Findthe equilibrium moment on the driver link 1 and the joint forces for = 1 = 60. Select suitable lengths for the link 3 and 5.

6.22 The dimensions of the mechanism shown in Fig. 3.37 are: AB =

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110 mm, AC = 260 mm, BD = La = 400 mm, and DE = 270 mm.

The constant angular speed of the driver link 1 is n = n1 = 1000 rpm.The links 1, 2, and 4 are homogeneous rectangular prisms with thewidth h = 0.010 m and the depth d = 0.001 m. The sliders 3 and 5have the width wSlider = 0.050 m, the height hSlider = 0.020 m, and thedepth d = 0.001 m. The external force on 5 is opposed to the motion




is g = 9.807 m/s2. Find the equilibrium moment on the driver link 1and the joint forces for = 1 = 45.

6.23 The dimensions of the mechanism shown in Fig. 3.38 are: AB =180 mm, AD = 450 mm, and BC = 200 mm. The constant angu-lar speed of the driver link 1 is n = 1600 rpm. The links 1, 2, and5 are homogeneous rectangular prisms with the width h = 0.010 mand the depth d = 0.001 m. The sliders 3 and 4 have the widthwSlider = 0.050 m, the height hSlider = 0.020 m, and the depth d =0.001 m. The external force on 5 is opposed to the motion of the

link Fext = |Fext|vG

|vG|where |Fext| = 1500 N. The density of the


g = 9.807 m/s2. Find the equilibrium moment on the driver link 1 andthe joint forces. Select suitable lengths for the link 5 for

=

1= 135.

6.24 The mechanism in Fig. 3.11(a) has the dimensions: AB = 0.20 m,AD = 0.40 m, CD = 0.70 m, CE = 0.30 m, and yE = 0.35 m. Theconstant angular speed of the driver link 1 is n = 2600 rpm. Thelinks 1, 3, and 4 are homogeneous rectangular prisms with the widthh = 0.010 m and the depth d = 0.001 m. The sliders 2 and 5 havethe width wSlider = 0.050 m, the height hSlider = 0.020 m, and thedepth d = 0.001 m. The external force on 5 is opposed to the motion




is g = 9.807 m/s2. Find the equilibrium moment on the driver link 1and the joint forces for = 1 = 30.

6.25 The mechanism in Fig. 3.12 has the dimensions: AB = 0.04 m, BC =0.07 m, CD = 0.12 m, AE = 0.10 m, and La = 0.035 m. The constant

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angular speed of the driver link 1 is n = 900 rpm. The links 1, 2, and

4 are homogeneous rectangular prisms with the width h = 0.010 mand the depth d = 0.001 m. The sliders 3 and 5 have the widthwSlider = 0.050 m, the height hSlider = 0.020 m, and the depth d =0.001 m. The external force on 5 is opposed to the motion of the




g = 9.807 m/s2. Find the equilibrium moment on the driver link 1 andthe joint forces for = 1 = 60.

6.26 The mechanism in Fig. 3.15 has the dimensions: AC = 0.080 m,BC = 0.150 m, BD = 0.400 m, and La = 0.020 m. The constantangular speed of the driver link 1 is n = 1500 rpm. The links 1, 3,and 4 are homogeneous rectangular prisms with the width h = 0.010 mand the depth d = 0.001 m. The sliders 2 and 5 have the widthwSlider = 0.050 m, the height hSlider = 0.020 m, and the depth d =0.001 m. The external force on 5 is opposed to the motion of the




g = 9.807 m/s2. Find the equilibrium moment on the driver link 1 andthe joint forces for = 1 = 60. Select a suitable length for the link

1.

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References

[1] P. Antonescu, Mechanisms, Printech, Bucharest, 2003.

[2] P. Appell, Traite de Mecanique Rationnel le, Gautier-Villars, Paris, 1941.

[3] I.I. Artobolevski, Mechanisms in Modern Engineering Design, MIR,Moscow, 1977.

[4] M. Atanasiu, Mecanica, EDP, Bucharest, 1973.

[5] H. Baruh, Analytical Dynamics, WCB/McGraw-Hill, Boston, 1999.

[6] A. Bedford and W. Fowler, Dynamics, Addison Wesley, Menlo Park, 1999.

[7] A. Bedford and W. Fowler, Statics, Addison Wesley, Menlo Park, 1999.

[8] M.I. Buculei, Mechanisms, University of Craiova Press, Craiova, 1976.

[9] M.I. Buculei, D. Bagnaru, G. Nanu, D.B. Marghitu, Analysis of Mecha-nisms with Bars, Scrisul romanesc, Craiova, 1986.

[10] A.G. Erdman, and G.N. Sandor, Mechanisms Design, Prentice-Hall, Up-per Saddle River, 1984.

[11] A. Ertas and J.C. Jones, The Engineering Design Process, John Wiley

& Sons, New York, 1996.

[12] F. Freudenstein, An Application of Boolean Algebra to the Motion ofEpicyclic Drives, Transaction of the ASME, Journal of Engineering forIndustry, pp.176-182, 1971.

[13] J.H. Ginsberg, Advanced Engineering Dynamics, Cambridge UniversityPress, Cambridge, 1995.

[14] D.T. Greenwood, Principles of Dynamics, Prentice-Hall, EnglewoodCliffs, 1998.

[15] A.S. Hall, Jr., A.R. Holowenko, and H.G. Laughlin, Theory and problemsof machine design, McGraw-Hill, New York, 1961.

[16] R.C. Hibbeler, Engineering Mechanics - Statics and Dynamics, Prentice-Hall, Upper Saddle River, New Jersey, 1995.

7/29/2019 CI_6_Forces

45/50

[17] R.C. Juvinall and K.M. Marshek, Fundamentals of Machine Component

Design, John Wiley & Sons, New York, 1983.[18] T.R. Kane, Analytical Elements of Mechanics, Vol. 1, Academic Press,

New York, 1959.

[19] T.R. Kane, Analytical Elements of Mechanics, Vol. 2, Academic Press,New York, 1961.

[20] T.R. Kane and D.A. Levinson, The Use of Kanes Dynamical Equtionsin Robotics, MIT International Journal of Robotics Research, No. 3, pp.3-21, 1983.

[21] T.R. Kane, P.W. Likins, and D.A. Levinson, Spacecraft Dynamics,McGraw-Hill, New York, 1983.

[22] T.R. Kane and D.A. Levinson, Dynamics, McGraw-Hill, New York,1985.

[23] J.T. Kimbrell, Kinematics Analysis and Synthesis, McGraw-Hill, NewYork, 1991.

[24] R. Maeder, Programming in Mathematica, AddisonWesley PublishingCompany, Redwood City, California, 1990.

[25] N.H. Madsen, Statics and Dynamics , class notes,www.eng.auburn.edu/users/nmadsen/, 2004.

[26] N.I. Manolescu, F. Kovacs, and A. Oranescu, The Theory of Mechanismsand Machines, EDP, Bucharest, 1972.

[27] D.B. Marghitu, Mechanical Engineers Handbook, Academic Press, SanDiego, California, 2001.

[28] D.B. Marghitu and M.J. Crocker, Analytical Elements of Mechanisms,Cambridge University Press, Cambridge, 2001.

[29] D.B. Marghitu and E.D. Stoenescu, Kinematics and Dy-namics of Machines and Machine Design , class notes,www.eng.auburn.edu/users/marghitu/, 2004.

7/29/2019 CI_6_Forces

46/50

[30] J.L. Meriam and L.G. Kraige, Engineering Mechanics: Dynamics, John

Wiley & Sons, New York, 1997.[31] D.J. McGill and W.W. King, Engineering Mechanics: Statics and an

Introduction to Dynamics, PWS Publishing Company, Boston, 1995.

[32] R.L. Mott, Machine elements in mechanical design, Prentice Hall, UpperSaddle River, New Jersey, 1999.

[33] D.H. Myszka, Machines and Mechanisms, Prentice-Hall, Upper SaddleRiver, New Jersey, 1999.

[34] R.L. Norton, Machine Design, Prentice-Hall, Upper Saddle River, New

Jersey, 1996.

[35] R.L. Norton, Design of Machinery, McGraw-Hill, New York, 1999.

[36] W.C. Orthwein, Machine Component Design, West Publishing Com-pany, St. Paul, 1990.

[37] L.A. Pars, A treatise on analytical dynamics, Wiley, New York, 1965.

[38] R.M. Pehan, Dynamics of Machinery, McGraw-Hill, New York, 1967.

[39] I. Popescu, Mechanisms, University of Craiova Press, Craiova, 1990.

[40] I. Popescu and C. Ungureanu, Structural Synthesis and Kinematics ofMechanisms with Bars, Universitaria Press, Craiova, 2000.

[41] I. Popescu and D.B. Marghitu, Dyad Classification for Mechanisms,World Conference on Integrated Design and Process Technology, Austin,Texas, December 3-5, 2003.

[42] I. Popescu, E.D. Stoenescu, and D.B. Marghitu, Analysis of SpatialKinematic Chains Using the System Groups, 8th International Congresson Sound and Vibration, St. Petersburg, Russia, July 5-8, 2004.

[43] M. Radoi and E. Deciu, Mecanica, EDP, Bucharest, 1981.[44] F. Reuleaux, The Kinematics of Machinery, Dover, New York, 1963.

[45] C.A. Rubin, The Student Edition of Working Model, AddisonWesleyPublishing Company, Reading, Massachusetts, 1995.

7/29/2019 CI_6_Forces

47/50

[46] I.H. Shames, Engineering Mechanics - Statics and Dynamics, Prentice-

Hall, Upper Saddle River, New Jersey, 1997.[47] J.E. Shigley and C.R. Mischke, Mechanical Engineering Design,

McGraw-Hill, New York, 1989.

[48] J.E. Shigley and J.J. Uicker, Theory of Machines and Mechanisms,McGraw-Hill, New York, 1995.

[49] R.W. Soutas-Little and D.J. Inman, Engineering Mechanics: Statics andDynamics, Prentice-Hall, Upper Saddle River, New Jersey, 1999.

[50] A. Stan and M. Grumarescu, Mechanics Problems, EDP, Bucharest,

1973.

[51] A. Stoenescu, A. Ripianu, and M. Atanasiu, Theoretical MechanicsProblems, EDP, Bucharest, 1965.

[52] A. Stoenescu and G. Silas, Theoretical Mechanics, ET, Bucharest, 1957.

[53] E.D. Stoenescu, Dynamics of Linkage Systems with Impact and Clear-ance, Ph.D. Dissertation, Mechanical Engineering, Auburn University,2005.

[54] L. W. Tsai, Mechanism Design: Enumeration of Kinematic Structures

According to Function, CRC Press, Boca Raton, Florida, 2001.

[55] R. Voinea, D. Voiculescu, and V. Ceausu, Mecanica, EDP, Bucharest,1983.

[56] K.J. Waldron and G.L. Kinzel, Kinematics, Dynamics, and Design ofMachinery, John Wiley&Sons, New York, 1999.

[57] C.E. Wilson and J.P. Sadler, Kinematics and Dynamics of Machinery,Harper Collins College Publishers, 1991.

[58] C.W. Wilson, Computer integrated macine design, Prentice Hall, Inc.,

Upper Saddle River, New Jersey, 1997.

[59] S. Wolfram, Mathematica, Wolfram Media/Cambridge University Press,Cambridge, 1999.

7/29/2019 CI_6_Forces

48/50

[60] * * * ,The theory of mechanisms and machines (Teoria mehanizmov i

masin), Vassaia scola, Minsc, 1970.[61] * * * ,Working Model 2D, Users Manual, Knowledge Revolution, San

Mateo, California, 1996.

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Figure captions

Figure 6.1 Rigid body as a collection of particlesFigure 6.2 Rigid body with differential element dmFigure 6.3 System of particlesFigure 6.4 Angular momentum about the mass center for a system of

particlesFigure 6.5 (a) Rigid body in general plane motion; (b) rotation about a

fixed pointFigure 6.6 Rigid body in plane motionFigure 6.7 Free-body diagrams for a crank slider mechanismFigure 6.8 (a) Crank slider mechanism; free-body diagrams: (b) crank 1,

(c) connecting, and (d) slider 3Figure 6.9 (a) Monocontour closed kinematic chain, (b) joint at Ai re-

placed by the joint forces Fi1,i and Fi,i1: Fi1,i + Fi,i1 = 0Figure 6.10 (a) Mechanism, and (b) diagram representing the mechanismFigure 6.11 Joint force F34 (a) calculation diagram, and (b) force diagramFigure 6.12 Joint force F45 (a) calculation diagram, and (b) force diagramFigure 6.13 Joint force F05 (a) calculation diagram, and (b) force diagramFigure 6.14 Joint force F03 (a) calculation diagram, and (b) force diagramFigure 6.15 Joint force F23 (a) calculation diagram, and (b) force diagramFigure 6.16 Joint force F12 (a) calculation diagram, and (b) force diagram

Figure 6.17 Joint forceF01 (a) calculation diagram, and (b) force diagramFigure 6.18 Joint forces for RRR dyad

Figure 6.19 Joint forces for RRT dyadFigure 6.20 Joint forces for RTR dyadFigure 6.21 Joint forces for R-RTR mechanism (Example 6.1)Figure 6.22 (a) Mechanism, and (b) diagram representing the mechanism

with two contoursFigure 6.23 Joint force F34 (a) calculation diagram, and (b) force diagramFigure 6.24 Joint force F45 (a) calculation diagram, and (b) force diagramFigure 6.25 Joint force F05 (a) calculation diagram, and (b) force diagramFigure 6.26 Joint force F23 (a) calculation diagram, and (b) force diagram

Figure 6.27 Joint force F03 (a) calculation diagram, and (b) force diagramFigure 6.28 Joint force F12 (a) calculation diagram, and (b) force diagramFigure 6.29 Joint force F01 (a) calculation diagram, and (b) force diagramFigure 6.30 Joint reactions for the dyad BR DR DTFigure 6.31 Joint reactions for the dyad BT BR CR

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Figure 6.32 Joint reactions for the driver link

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