Contour Equations - Kinematics and Dynamics of Machines | MECH 2120, Study notes of Mechanical Engineering

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Contents

• 5 Contour Equations
  • 5.1 Contour Velocity Equations
  • 5.2 Contour Acceleration Equations
  • 5.3 Independent Contour Equations
  • 5.4 Example
  • 5.5 Problems

5 Contour Equations

This section aims at providing an algebraic method to compute the veloci- ties and accelerations of any closed kinematic chain. The classical method for obtaining the velocities and accelerations involves the computation of the derivative with respect to time of the position vectors. The method of con- tour equations avoids this task and utilizes only algebraic equations [4, 55]. Using this approach, a numerical implementation is much more efficient. The method described here can be applied to planar and spatial mechanisms. Two rigid links (j) and (k) are connected by a joint (kinematic pair) at A, Fig. 5.1. The point Aj of the rigid body (j) is guided along a path prescribed in the body (k). The points Aj belonging to body (j) and the Ak belonging to body (k) are coincident at the instant of motion under consideration. The following relation exists between the velocity vAj of the point Aj and the velocity vAk of the point Ak

vAj = vAk + vrAjk , (5.1)

where vrAjk = vrAj Ak indicates the velocity of Aj as seen by an observer at Ak attached to body k or the relative velocity of Aj with respect to Ak, allowed at the joint A. The direction of vrAjk is obviously tangent to the path

prescribed in body (k). From Eq. (5.1) the accelerations of Aj and Ak are expressed as

aAj = aAk + arAjk + acAjk , (5.2)

where acAjk = acAj Ak is known as the Coriolis acceleration and is given by

acAjk = 2 ωk × vrAjk , (5.3)

where ωk is the angular velocity of the body (k). Equations (5.1) and (5.2) are useful even for coincident points belonging to two links that may not be directly connected. A graphical representation of Eq. (5.1) is shown in Fig. 5.1(b) for a rotating slider joint. Figure 5.2 shows a monocontour closed kinematic chain with n rigid links. The joint Ai, i = 0, 1 , 2 , ..., n is the connection between the links (i) and (i − 1). The last link n is connected with the first link 0 of the chain. For the closed kinematic chain, a path is chosen from link 0 to link n. At the joint Ai there are two instantaneously coincident points: 1) the point Ai,i

belonging to link (i), Ai,i ∈ (i), and 2) the point Ai,i− 1 belonging to body (i − 1), Ai,i− 1 ∈ (i − 1).

5.1 Contour Velocity Equations

The absolute angular velocity, ωi = ωi, 0 , of the rigid body (i), or the angular velocity of the rigid body (i) with respect to the ‘fixed” reference frame Oxyz is

ωi = ωi− 1 + ωi,i− 1 , (5.4)

where ωi− 1 = ωi− 1 , 0 is the absolute angular velocity of the rigid body (i − 1) (or the angular velocity of the rigid body (i − 1) with respect to the ‘fixed” reference frame Oxyz) and ωi,i− 1 is the relative angular velocity of the rigid body (i) with respect to the rigid body (i − 1). For the n link closed kinematic chain the following expressions are ob- tained for the angular velocities

ω 1 = ω 0 + ω 1 , 0 ω 2 = ω 1 + ω 2 , 1 ................................. ωi = ωi− 1 + ωi,i− 1 ................................. ω 0 = ωn + ω 0 ,n. (5.5)

Summing the expressions given in Eq. (5.5), the following relation is obtained

ω 1 , 0 + ω 2 , 1 + ... + ω 0 ,n = 0 , (5.6)

which may be rewritten as

∑

(i)

ωi,i− 1 = 0. (5.7)

Equation (5.7) represents the first vectorial equation for the angular velocities of a simple closed kinematic chain. The following relation exists between the velocity vAi,i of the point Ai,i and the velocity vAi,i− 1 of the point Ai,i− 1

vAi,i = vAi,i− 1 + vrAi,i− 1 , (5.8)

where vrAi,i− 1 = vrAi,iAi,i− 1 is the relative velocity of Ai,i on link (i) with respect to Ai,i− 1 on link (i − 1). Using the velocity relation for two particles on the rigid body (i) the following relation exists

vAi+1,i = vAi,i + ωi × rAiAi+1 , (5.9)

where ωi is the absolute angular velocity of the link (i) in the reference frame Oxyz, and rAiAi+1 is the distance vector from Ai to Ai+1. Using Eqs. (5.8) and (5.9) the velocity of the point Ai+1,i ∈ (i + 1) is written as

vAi+1,i = vAi,i− 1 + ωi × rAiAi+1 + vrAi,i− 1. (5.10)

For the n link closed kinematic chain the following expressions are obtained

vA 3 , 2 = vA 2 , 1 + ω 2 × rA 2 A 3 + vrA 2 , 1 vA 4 , 3 = vA 3 , 2 + ω 3 × rA 3 A 4 + vrA 3 , 2 ............................................................ vAi+1,i = vAi,i− 1 + ωi × rAiAi+1 + vrAi,i− 1 ............................................................ vA 1 , 0 = vA 0 ,n + ω 0 × rA 0 A 1 + vrA 0 ,n vA 2 , 1 = vA 1 , 0 + ω 1 × rA 1 A 2 + vrA 1 , 0. (5.11)

Summing the relations in Eq. (5.11) [ ω 1 × rA 1 A 2 + ω 2 × rA 2 A 3 + ... + ωi × rAiAi+1 + ... + ω 0 × rA 0 A 1

]

[ vrA 2 , 1 + vrA 3 , 2 + ... + vrAi,i− 1 + ... + vrA 0 ,n + vrA 1 , 0

] = 0. (5.12)

Because the reference system Oxyz is considered “fixed”, the vector rAi− 1 Ai is written in terms of the position vectors of the points Ai− 1 and Ai

rAi− 1 Ai = rAi − rAi− 1 , (5.13)

where rAi = rOAi and rAi− 1 = rOAi− 1. Equation (5.12) becomes

[rA 1 × (ω 1 − ω 0 ) + rA 2 × (ω 2 − ω 1 ) + ... + rA 0 × (ω 0 − ωn)] + [ vAr 1 , 0 + vrA 2 , 1 + ... + vAri,i− 1 + ... + vAr 0 ,n

] = 0. (5.14)

Using Eq. (5.5), Eq. (5.14) becomes

[rA 1 × ω 1 , 0 + rA 2 × ω 2 , 1 + ... + rA 0 × ω 0 ,n] + [ vAr 1 , 0 + vrA 2 , 1 + ... + vAr 0 ,n

] = 0. (5.15)

The previous equation is written as

∑

(i)

rAi × ωi,i− 1 +

∑

(i)

vrAi,i− 1 = 0. (5.16)

Equation (5.16) represents the second vectorial equation for the angular ve- locities of a simple closed kinematic chain. Equations ∑

(i)

ωi,i− 1 = 0 and

∑

(i)

rAi × ωi,i− 1 +

∑

(i)

vrAi,i− 1 = 0 , (5.17)

represent the velocity equations for a simple closed kinematic chain.

5.2 Contour Acceleration Equations

The absolute angular acceleration, αi = αi, 0 , of the rigid body (i) (or the angular acceleration of the rigid body (i) with respect to the ‘fixed” reference frame Oxyz) is

αi = αi− 1 + αi,i− 1 + ωi × ωi,i− 1 , (5.18)

where αi− 1 = αi− 1 , 0 is the absolute angular acceleration of the rigid body (i−1) (or the angular acceleration of the rigid body (i−1) with respect to the ‘fixed” reference frame Oxyz) and αi,i− 1 is the relative angular acceleration of the rigid body (i) with respect to the rigid body (i − 1). For the n link closed kinematic chain the following expressions are ob- tained for the angular accelerations

α 2 = α 1 + α 2 , 1 + ω 2 × ω 2 , 1 α 3 = α 2 + α 3 , 2 + ω 3 × ω 3 , 2 .............................................. αi = αi− 1 + αi,i− 1 + ωi × ωi,i− 1 .............................................. α 1 = α 0 + α 1 , 0 + ω 1 × ω 1 , 0. (5.19)

Summing all the expressions in Eq. (5.19)

α 2 , 1 + α 3 , 2 + ... + α 1 , 0 + ω 2 × ω 2 , 1 + ... + ω 1 × ω 1 , 0 = 0. (5.20)

Equation (5.20) is rewritten as

∑

(i)

αi,i− 1 +

∑

(i)

ωi × ωi,i− 1 = 0. (5.21)

Equation (5.21) represents the first vectorial equation for the angular accel- erations of a simple closed kinematic chain. Using the acceleration distributions of the relative motion of two rigid bodies (i) and (i − 1)

aAi,i = aAi,i− 1 + arAi,i− 1 + acAi,i− 1 , (5.22)

where aAi,i and aAi,i− 1 are the linear accelerations of the points Ai,i and Ai,i− 1 , arAi,i− 1 = arAi,iAi,i− 1 is the relative acceleration between Ai,i on link (i) and Ai,i− 1 on link (i − 1). Finally, acAi,i− 1 is the Coriolis acceleration defined as

acAi,i− 1 = 2 ωi− 1 × vrAi,i− 1. (5.23)

Using the acceleration distribution relations for two particles on a rigid body

aAi+1,i = aAi,i + αi × rAiAi+1 + ωi × (ωi × rAiAi+1 ), (5.24)

where αi is the angular acceleration of the link (i). From Eqs. (5.22) and (5.24)

aAi+1,i = aAi,i− 1 + arAi,i− 1 + acAi,i− 1 + αi × rAiAi+1 + ωi × (ωi × rAiAi+1 ). (5.25)

Writing similar equations for all the links of the kinematic chain, the following relations are obtained

aA 3 , 2 = aA 2 , 1 + arA 2 , 1 + acA 2 , 1 + α 2 × rA 2 A 3 + ω 2 × (ω 2 × rA 2 A 3 ), aA 4 , 3 = aA 3 , 2 + arA 3 , 2 + acA 3 , 2 + α 3 × rA 3 A 4 + ω 3 × (ω 3 × rA 3 A 4 ),

..................................................................... aA 1 , 0 = aA 0 ,n + arA 0 ,n + acA 0 ,n + α 0 × rA 0 A 1 + ω 0 × (ω 0 × rA 0 A 1 ), aA 2 , 1 = aA 1 , 0 + arA 1 , 0 + acA 1 , 0 + α 1 × rA 1 A 2 + ω 1 × (ω 1 × rA 1 A 2 ). (5.26)

Summing the expressions in Eq. (5.26) [ arA 1 , 0 + arA 2 , 1 + ... + arA 0 ,n

]

[ acA 1 , 0 + acA 2 , 1 + ... + acA 0 ,n

]

[α 1 × rA 1 A 2 + α 2 × rA 2 A 3 + ... + α 0 × rA 0 A 1 ] + ω 1 × (ω 1 × rA 1 A 2 ) + ω 2 × (ω 2 × rA 2 A 3 ) + ...+ ω 0 × (ω 0 × rA 0 A 1 ) = 0. (5.27)

Using the relation rAi− 1 Ai = rAi − rAi− 1 in Eq. (5.27)

[ arA 1 , 0 + arA 2 , 1 + ... + arA 0 ,n

]

[ acA 1 , 0 + acA 2 , 1 + ... + acA 0 ,n

]

[rA 1 × (α 1 , 0 + ω 1 × ω 1 , 0 ) + ... + rA 0 × (α 0 ,n + ω 0 × ω 0 ,n)] + ω 1 × (ω 1 × rA 1 A 2 ) + ω 2 × (ω 2 × rA 2 A 3 ) + ...+ ω 0 × (ω 0 × rA 0 A 1 ) = 0. (5.28)

Equation (5.28) is rewritten as

∑

(i)

arAi,i− 1 +

∑

(i)

acAi,i− 1 +

∑

(i)

rAi × (αi,i− 1 + ωi × ωi,i− 1 )+

∑

(i)

ωi × (ωi × rAiAi+1 ) = 0. (5.29)

Equation (5.29) represents the second vectorial equation for the angular ac- celerations of a simple closed kinematic chain. Equations

∑

(i)

αi,i− 1 +

∑

(i)

ωi × ωi,i− 1 = 0 and

∑

(i)

rAi × (αi,i− 1 + ωi × ωi,i− 1 ) +

∑

(i)

arAi,i− 1 +

∑

(i)

acAi,i− 1 +

∑

(i)

ωi × (ωi × rAiAi+1 ) = 0. (5.30)

are the acceleration equations for the case of a simple closed kinematic chain. Remarks

1. For a closed kinematic chain in planar motion, simplified relations are obtained because

ωi × (ωi × rAiAi+1 ) = −ω i^2 rAiAi+1 and ωi × ωi,i− 1 = 0. (5.31)

Equations

∑

(i)

αi,i− 1 = 0 and

∑

(i)

rAi × αi,i− 1 +

∑

(i)

arAi,i− 1 +

∑

(i)

acAi,i− 1 − ω^2 i rAiAi+1 = 0. (5.32)

represent the acceleration equations for a simple closed kinematic chain in planar motion.

2. The Coriolis acceleration, given by the expression

acAi,i− 1 = 2ωi− 1 × vAri,i− 1 (5.33)

vanishes when ωi− 1 = 0 , or vrAi,i− 1 = 0 , or when ωi− 1 is parallel to vrAi,i− 1.

5.3 Independent Contour Equations

A diagram is used to represent to a mechanism in the following way: the numbered links are the nodes of the diagram and are represented by circles, and the joints are represented by lines which connect the nodes. Figure 5.3 shows the diagram which represents a planar mechanism. The maximum number of independent contours is given by

N = c − n or nc = N = c − p + 1, (5.34)

where c is the number of joints, n is the number of moving links, and p is the number of links. The equations for velocities and accelerations is written for any closed contour of the mechanism. However, it is best to write the contour equations only for the independent loops of the diagram representing the mechanism.

Step 1. Determine the position analysis of the mechanism.

Step 2. Draw a diagram representing the mechanism and select the inde- pendent contours. Determine a path for each contour.

Step 3. For each closed loop write the contour velocity relations, Eq. (5.17), and contour acceleration relations, Eq. (5.30). For a closed kinematic chain in planar motion the following equations will be used ∑

(i)

ωi,i− 1 = 0 ,

∑

(i)

rAi × ωi,i− 1 +

∑

(i)

vrAi,i− 1 = 0. (5.35)

∑

(i)

αi,i− 1 = 0 ,

∑

(i)

rAi × αi,i− 1 +

∑

(i)

arAi,i− 1 +

∑

(i)

acAi,i− 1 − ω i^2 rAiAi+1 = 0. (5.36)

Step 4. Project on a cartesian reference system the velocity and accel- eration equations. Linear algebraic equations are obtained where the unknowns are

• the components of the relative angular velocities ωj,j− 1 ;
• the components of the relative angular accelerations αj,j− 1 ;
• the components of the relative linear velocities vrAj,j− 1 ;
• the components of the relative linear accelerations arAj,j− 1.

Solve the algebraic system of equations and determine the unknown kinematic parameters.

Step 5. Determine the absolute angular velocities ωj and the absolute angular accelerations αj. Compute the velocities and accelerations of the characteristic points and joints.

In the following examples, the contour method is applied to determine the velocities and accelerations distribution for several planar mechanisms. The following notation will be used: ωij is the relative angular velocity vector of the link i with respect to the link j. When the link j is the ground (denoted as link 0), then ωi = ωi 0 also denotes the absolute angular velocity vector of the link i. The magnitude of ωij is ωij i.e. |ωij | = ωij. vrAij is the relative linear velocity of the point Ai on link i with respect to the point Aj on link j. The point Ai belonging to link j and the point Aj belonging to link j are coincident at the instant of motion under considera- tion. αij is the relative angular acceleration vector of the link i with respect to the rigid body j. When the link j is the ground, then αi = αi 0 also denotes the absolute angular acceleration vector of the rigid body i. arAij is the relative linear acceleration vector of Ai on link i with respect to Aj on link j. acAij is the Coriolis acceleration of Ai with respect to Aj. rBC denotes a vector from the joint B to the joint C. xB , yB , zB denote the coordinates of the point B with respect to the fixed reference frame. vB denotes the linear velocity vector of the point B with respect to the fixed reference frame.

aB denotes the linear acceleration vector of the point B with respect to the fixed reference frame.

5.4 Example

The planar mechanism considered in this example is depicted in Fig. 5.4(a). The following data are given: AC=0.100 m, BC=0.300 m, BD=0.900 m, and La=0.100 m. The angle of the driver element (link AB) with the horizontal axis is φ = 45◦. A Cartesian reference frame with the origin at A (xA = yA = 0) is selected. The coordinates of joint C are xC = AC, yC = 0. The coordinates of joint B are xB = 0.256 m, yB = 0.256 m. The coordinates of joint D are xD = 1.142 m, yD = 0.100 m. The position vectors rAB , rAC and rAD are defined as follows

rAB = xB ı + yB  = 0. 256 ı + 0. 256 , rAC = xC ı + yC  = 0. 100 ı, rAD = xD ı + yD  = 1. 142 ı + 0. 100 .

The angular velocity of the driver link is n 1 = 100 rpm, or

ω 10 = ω 1 = n

π 30

= 100

π 30

rad/s = 10.472 rad/s.

The mechanism has six links and seven full joints. Using Eq. (5.34), the number of independent loops is given by

nc = l − p + 1 = 7 − 6 + 1 = 2.

This mechanism has two independent contours. The first contour I contains the links 0, 1, 2 and 3, while the second contour II contains the links 0, 3, 4 and 5. The diagram representing the mechanism is given in Fig.5.4(b). Clockwise paths are chosen for each closed loop I and II.

First contour According to Fig. 5.5, the first contour has

• rotational joint R between the links 0 and 1 (joint A);
• translational joint T between the links 1 and 2 (joint BT);
• rotational joint R between the links 2 and 3 (joint BR);
• rotational joint R between the links 3 and 0 (joint C). For the velocity analysis, the following equations are written using Eq. (5.35)

ω 10 + ω 32 + ω 03 = 0 , rAB × ω 32 + rAC × ω 03 + vrB 21 = 0 , (5.37)

where ω 10 = ω 10 k = 10. 47 k rad/s, ω 32 = ω 32 k, and ω 03 = ω 03 k. The relative velocity of B 2 on link 2 with respect to B 1 on link 1, vBr 21 , has ı and  components

vrB 21 = vBr 21 x ı + vrB 21 y  = vrB 21 cos φ ı + vrB 21 sin φ ,

where vBr 21 is the magnitude of the vector vrB 21 i.e. |vrB+21| = vrB 21. The sign of the unknown relative velocities is selected as positive as shown in Figs. 5.4(a) and 5.5(b). Then the numerical computation will give the correct orientation of the unknown vectors. The unknowns in Eq. (5.37) are ω 32 , ω 03 , and vBr 21. Equation (5.37) becomes

ω ∣ 10 k + ω 32 k + ω 03 k = 0 , ∣∣ ∣∣ ∣∣

ı  k xB yB 0 0 0 ω 32

∣∣ ∣∣ ∣∣ ∣

+

∣∣ ∣∣ ∣∣ ∣

ı  k xC yC 0 0 0 ω 03

∣∣ ∣∣ ∣∣ ∣

+

vBr 21 cos φ ı + vBr 21 sin φ  = 0. (5.38)

Equation (5.38) is projected onto the “fixed” reference frame Oxyz

ω 10 + ω 32 + ω 03 = 0, yB ω 32 + yC ω 03 + vBr 21 cos φ = 0, −xB ω 32 − xC ω 03 + vrB 21 sin φ = 0, (5.39)

or numerically

10 .472 + ω 32 + ω 03 = 0,

1. 256 ω 32 + vrB 21 cos 45◦^ = 0, − 0. 256 ω 32 − 0. 100 ω 03 + vrB 21 sin 45◦^ = 0. (5.40)

Equation (5.40) represents a system of three equations with three unknowns: ω 32 , ω 03 , and vBr 21. Solving the algebraic equations, the following numerical values are obtained ω 32 = 2.539 rad/s, ω 03 = − 13 .011 rad/s, and vrB 21 = − 0 .920 m/s. The absolute angular velocity of the link 3 is

ω 30 = −ω 03 = 13. 011 k rad/s. (5.41)

The velocity of the point B 2 = B 3 is computed with the expression of velocity field of two points (B 3 and C) on the same rigid body (link 3)

vB 2 = vB 3 = vC + ω 30 × rCB =

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 ω 30 xB − xC yB − yC 0

∣∣ ∣∣ ∣∣ ∣

=

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 13. 011

1. 256 − 0. 100 0. 256 0

∣∣ ∣∣ ∣∣ ∣

= − 3. 333 ı + 2. 032  m/s,

where vC = 0 because the joint C is grounded. The link 2 and the driver link 1 have the same angular velocity

ω 10 = ω 20 = ω 30 + ω 23 = 13. 011 k − 2. 539 k = 10. 472 k rad/s.

The velocity of the point B 1 on link 1 is calculated with the expression of velocity field of two points (B 1 and A) on the same rigid body (link 1)

vB 1 = vA + ω 10 × rAB = ω 10 × rAB =

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 ω 10 xB yB 0

∣∣ ∣∣ ∣∣ ∣

=

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 10. 472

1. 256 0. 256 0

∣∣ ∣∣ ∣∣ ∣

= − 2. 682 ı + 2. 682  m/s.

Another way of calculating the velocity of the point B 2 = B 3 is with the help of velocity field of two points (B 1 and B 2 ) not situated on the same rigid body (B 1 is on link 1 and B 2 is on link 2)

vB 2 = vB 1 + vrB 21 ,

where vrB 21 = vrB 21 cos φ ı + vrB 21 sin φ  = − 0. 651 ı − 0. 651  m/s.

For the acceleration analysis, the following equations are written using Eq. (5.36)

α 10 + α 32 + α 03 = 0 , rAB × α 32 + rAC × α 03 + aB 21 + acB 21 − ω^210 rAB − ω^230 rBC = 0 ,(5.42)

where α 10 = ˙ω 10 k = 0 , α 32 = α 32 k, and α 03 = α 03 k. The relative acceleration of B 2 on link 2 with respect to B 1 on link 1, arB 21 , has ı and  components

arB 21 = arB 21 x ı + arB 21 y  = arB 21 cos φ ı + arB 21 sin φ .

The sign of the unknown relative accelerations is selected positive and then the numerical computation will give the correct orientation of the unknown acceleration vectors. The expression for the Coriolis acceleration is

acB 21 = 2 ω 10 × vrB 21 = 2 ω 20 × vrB 21 =

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 ω 10 vBr 21 cos φ vrB 21 sin φ 0

∣∣ ∣∣ ∣∣ ∣

= − 2 vrB 21 ω 10 sin φ ı + 2vrB 21 ω 10 cos φ  =

−2(− 0 .920)(10.472) sin 45◦^ ı + 2(− 0 .920)(10.472) cos 45◦^  =

13. 629 ı − 13. 629  m/s^2.

The unknowns in Eq. (5.42) are α 32 , α 03 , and arB 21. Equation (5.42) becomes

α 32 k + α 03 k = 0 , ∣∣ ∣∣ ∣∣ ∣

ı  k xB yB 0 0 0 α 32

∣∣ ∣∣ ∣∣ ∣

+

∣∣ ∣∣ ∣∣ ∣

ı  k xC yC 0 0 0 α 03

∣∣ ∣∣ ∣∣ ∣

• arB 21 cos φ ı + arB 21 sin φ  +

acB 21 − ω^210 (xB ı + yB ) − ω^230 [(xC − xB ) ı + (yC − yB ) ] = 0.

The previous equations are projected onto the “fixed” reference frame Oxyz

α 32 + α 03 = 0, yB α 32 + yC α 03 + arB 21 cos φ − 2 vrB 21 ω 10 sin φ − ω^210 xB − ω^230 (xC − xB ) = 0, −xB α 32 − xC α 03 + arB 21 sin φ + 2vrB 21 ω 10 cos φ − ω 102 yB − ω^230 (yC − yB ) = 0,

or numerically

α 32 + α 03 = 0,

1. 256 α 32 + arB 21 cos 45◦^ + 13. 626 − (10.472)^2 (0.256) − (13.011)^2 (0. 100 − 0 .256) = 0, − 0. 256 α 32 − 0. 100 α 03 + arB 21 sin 45◦^ − 13. 626 − (10.472)^2 (0.256) − (13.011)^2 (0 − 0 .256) = 0. (5.43)

Equation (5.43) represents a system of three equations with three unknowns: α 32 , α 03 , and arB 21. Solving the algebraic equations, the following numerical

values are obtained α 32 = − 25 .032 rad/s^2 , α 03 = 25.032 rad/s^2 , and arB 21 =

− 7 .865 m/s^2. The absolute angular acceleration of the link 3 is

α 30 = −α 03 = − 25. 032 k rad/s^2.

The velocity of the point B 2 = B 3 is computed with the expression of velocity field of two points (B 3 and C) on the same rigid body (link 3)

vB 2 = vB 3 = vC + ω 30 × rCB =

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 ω 30 xB − xC yB − yC 0

∣∣ ∣∣ ∣∣ ∣

=

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 13. 011

1. 256 − 0. 100 0. 256 0

∣∣ ∣∣ ∣∣ ∣

= − 3. 333 ı + 2. 032  m/s,

where vC = 0 because the joint C is grounded. The link 2 and the driver link 1 have the same angular velocity

ω 10 = ω 20 = ω 30 + ω 23 = 13. 011 k − 2. 539 k = 10. 472 k rad/s.

The velocity of the point B 1 on link 1 is calculated with the expression of velocity field of two points (B 1 and A) on the same rigid body (link 1)

vB 1 = vA + ω 10 × rAB = ω 10 × rAB =

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 ω 10 xB yB 0

∣∣ ∣∣ ∣∣ ∣

=

∣∣ ∣∣ ∣∣ ∣

ı  k 0 0 10. 472

1. 256 0. 256 0

∣∣ ∣∣ ∣∣ ∣

= − 2. 682 ı + 2. 682  m/s.

Another way of calculating the velocity of the point B 2 = B 3 is with the help of velocity field of two points (B 1 and B 2 ) not situated on the same rigid body (B 1 is on link 1 and B 2 is on link 2)

vB 2 = vB 1 + vrB 21 ,

where vrB 21 = vrB 21 cos φ ı + vrB 21 sin φ  = − 0. 651 ı − 0. 651  m/s. The angular acceleration of the link 3 is

α 30 = −α 03 = α 32 = − 25. 032 k rad/s^2.

The absolute linear acceleration of the point B 3 is computed as follows

aB 3 = aC + α 30 × rCB − ω 302 rCB = − 20. 026 ı − 47. 277  m/s^2.

Second contour analysis According to Fig. 5.6, the second contour is described as

• rotational joint R between the links 0 and 3 (joint C);
• rotational joint R between the links 3 and 4 (joint B) ;
• rotational joint R between the links 4 and 5 (joint DR);
• translational joint T between the links 5 and 0 (joint DT). For the velocity analysis, the following equations are written

ω 30 + ω 43 + ω 54 = 0 , rAC × ω 30 + rAB × ω 43 + rAD × ω 54 + vrD 05 = 0. (5.44)

The relative linear velocity vDr 05 has only one component, along the x axis

vrD 05 = vrD 05 ı.

The unknown parameters in Eq. (5.44) are ω 43 , ω 54 and vrD 05. The following numerical values are obtained ω 43 = − 15 .304 rad/s, ω 54 = 2.292 rad/s, and vrD 05 = 3.691 m/s. The angular velocity of the link BD is

ω 40 = ω 30 + ω 43 = −ω 54 = − 2. 292 k rad/s.

The absolute linear velocity of the point D 4 = D 5 is computed as follows

vD 4 = vD 5 = vB 4 + ω 40 × rBD = −vrD 05 = − 3. 691 ı m/s,

where vB 4 = vB 3. For the acceleration analysis the following equations exist

α 30 + α 43 + α 54 = 0 , arD 05 + acD 05 + rAC × α 30 + rAB × α 43 + rAD × α 54 − ω^230 rCB − ω^240 rBD = 0. (5.45)

Because the slider 5 does not rotate (ω 50 = 0 ), the Coriolis acceleration is

acD 05 = 2ω 50 × vrD 05 = 0.

The unknowns in Eq. (5.45) are α 43 , α 54 and arD 05. The following numerical

results are obtained α 43 = 77.446 rad/s^2 , α 54 = − 52 .414 rad/s^2 , and arD 05 =

16 .499 m/s^2. The absolute angular acceleration of the link BD is

α 40 = α 30 + α 43 = α 45 = 52. 414 k rad/s^2 ,

and the linear acceleration of the point D 4 = D 5 is

aD 4 = aD 5 = aB 4 + α 40 × rBD − ω 402 rBD = − 16. 499 ı m/s^2 ,

where aB 4 = aB 3.

5.5 Problems

5.1 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 200 rpm. Find the velocities and the accelerations of the four-bar mechanism using the contour equations method for the case when the angle of the driver link AB with the horizontal is φ = 60◦.

5.2 The angular speed of the driver link 1, of the mechanism shown in Fig. 4.9, is ω = ω 1 = 20 rad/s. The distance from the link 3 to the horizontal axis Ax is a = 55 mm. Using the contour equations find the velocity and the acceleration of the point C on the link 3 for φ = 30◦.

5.3 The slider crank mechanism shown in Fig. 4.10 has the dimensions AB = 0.4 m and BC = 1 m. The driver link 1 rotates with a constant angular speed of n = 160 rpm. Find the velocity and acceleration of the slider 3 using the contour equations when the angle of the driver link with the horizontal is φ = 30◦.

5.4 The planar mechanism considered is shown in Fig. 3.19. The fol- lowing data are given: 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 angular speed of the driver link 1 is constant and has the value 180 rpm. Using the contour equations method find the velocities and the accel- erations of the mechanism for φ = φ 1 = 30◦.

5.5 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 with a constant angular speed of n = 1200 rpm. For φ = 145◦^ find the veloci- ties and the accelerations of the mechanism with the contour equations.

5.6 The mechanism shown in Fig. 3.21 has the following dimensions: AB = 80 mm, AD = 250 mm, BC = 180 mm, CE = 60 mm, EF = 200 mm, and a = 170 mm. The constant angular speed of the driver link 1 is n = 400 rpm. Find the velocities and the accelerations of the mechanism using the contour equations when the angle of the driver link 1 with the horizontal is φ = φ 1 = 300◦.

5.7 The dimensions for the mechanism shown in Fig. 3.22 are: AB = 150 mm, BD = 400 mm, BC = 140 mm, CD = 400 mm, DE = 250 mm, CF = 500 mm, AE = 380 mm, and b = 100 mm. The constant angular speed of the driver link 1 is n = 40 rpm. Find the velocities and the accelerations of the mechanism for φ = φ 1 = 210◦. Use the contour equations method.

5.8 The mechanism in Fig. 3.23 has the dimensions: AB = 200 mm, AC = 100 mm, BD = 400 mm, DE = 550 mm, EF = 300 mm, La = 500 mm, and Lb = 100 mm. Using the contour equations method find the velocities and the accelerations of the mechanism if the constant angular speed of the driver link 1 is n = 70 rpm and for φ = φ 1 = 210◦.

5.9 The dimensions for the mechanism shown in Fig. 3.24 are: AB = 150 mm, BC = 400 mm, AD = 360 mm, CD = 210 mm, DE = 130 mm, EF = 400 mm, and La = 40 mm. The constant angular speed of the driver link 1 is n = 250 rpm. Utilizing the contour equations find the velocities and the accelerations of the mechanism for φ = φ 1 = 30◦.

5.10 The mechanism in Fig. 3.25 has the dimensions: AB = 250 mm, AC = 800 mm, BD = 1200 mm, La = 180 mm, and Lb = 300 mm. The driver link 1 rotates with a constant angular speed of n = 50 rpm. Find the velocities and the accelerations of the mechanism for φ = φ 1 = 210◦. Use the contour equations method.

5.11 Figure 3.26 shows a mechanism with the following dimensions: AB = 120 mm, BD = 400 mm, and La = 150 mm. The constant angular speed of the driver link 1 is n = 600 rpm. Find the velocities and the accelerations of the mechanism, using the contour equations method, when the angle of the driver link 1 with the horizontal is φ = 210◦.

5.12 The mechanism in Fig. 3.27 has the dimensions: AB = 200 mm, AC = 500 mm, BD = 800 mm, DE = 400 mm, EF = 270 mm, La = 70 mm, and Lb = 300 mm. The constant angular speed of the driver link 1 is n = 40 rpm. Utilizing the contour equations find the velocities and the accelerations of the mechanism for φ = φ 1 = 300◦.

5.13 Figure 3.28 shows a mechanism with the following dimensions: AB = 200 mm, BC = 750 mm, CD = DE = 300 mm, EF = 500 mm, La = 750 mm, and Lb = Lc = 250 mm. Find the velocities and the

accelerations of the mechanism, using the contour equations method, if the constant angular speed of the driver link 1 is n = 1100 rpm and for φ = φ 1 = 120◦.

5.14 Figure 3.29 shows a mechanism with the following dimensions: AB = 120 mm, BC = 550 mm, CE = 180 mm, CD = 350 mm, EF = 300 mm, La = 320 mm, Lb = 480 mm, and Lc = 600 mm. The constant angular speed of the driver link 1 is n = 100 rpm. Find the velocities and the accelerations of the mechanism, using the contour equations, for φ = φ 1 = 30◦.

5.15 Figure 3.30 shows a mechanism with the following dimensions: AB = 180 mm, BC = 520 mm, CF = 470 mm, CD = 165 mm, DE = 540 mm, La = 630 mm, Lb = 360 mm, and Lc = 210 mm. The constant angular speed of the driver link 1 is n = 70 rpm. Use the contour equations to calculate the velocities and the accelerations of the mechanism when the angle of the driver link 1 with the horizontal is φ = 210◦.

5.16 Figure 3.31 shows a mechanism with the following dimensions: AB = 60 mm, BC = 150 mm, AD = 70 mm, and BE = 170 mm. The constant angular speed of the driver link 1 is n = 300 rpm. Find the velocities and the accelerations of the mechanism, using the contour equations, if the angle of the driver link 1 with the horizontal is φ = 210 ◦.

5.17 The dimensions of the mechanism shown in Fig. 3.32 are: AB = 90 mm, BC = 240 mm, BE = 400 mm, CE = 600 mm, CD = 220 mm, EF = 900 mm, La = 250 mm, Lb = 150 mm, and Lc = 100 mm. The constant angular speed of the driver link 1 is n = 50 rpm. Employing the contour equations find the velocities and the accelerations of the mechanism for φ = φ 1 = 210◦.

5.18 The dimensions of the mechanism shown in Fig. 3.33 are: AB = 180 mm, AC = 300 mm, CD = 400 mm, DE = 200 mm, and La = 360 mm. The constant angular speed of the driver link 1 is n = 90 rpm. Use the contour equations to calculate the velocities and the accelerations of the mechanism when the angle of the driver link 1 with the horizontal is φ = 30◦.

??.19 The dimensions of the mechanism shown in Fig. 3.34 are: AB = 80 mm, AC = 40 mm, CD = 100 mm, and DE = 300 mm. The constant angu- lar speed of the driver link 1 is n = 60 rpm. Use the contour equations to calculate the velocities and the accelerations of the mechanism for φ = φ 1 = 210◦.

5.20 The dimensions of the mechanism shown in Fig. 3.35 are: AB = 200 mm, AC = 350 mm, and CD = 600 mm. For the distance b select a suitable value. The constant angular speed of the driver link 1 is n = 90 rpm. Find the velocities and the accelerations of the mecha- nism for φ = φ 1 = 120◦.

5.21 The dimensions of the mechanism shown in Fig. 3.36 are: AB = 80 mm, AC = 60 mm, and CD = 70 mm. The constant angular speed of the driver link 1 is n = 220 rpm. Find the velocities and the accelerations of the mechanism, using the contour equations, for φ = φ 1 = 240◦.

??.22 The dimensions of the mechanism shown in Fig. 3.37 are: AB = 150 mm, AC = 420 mm, BD = La = 650 mm, and DE = 350 mm. The constant angular speed of the driver link 1 is n = n 1 = 650 rpm. Find the velocities and the accelerations of the mechanism, using the contour equations, for φ = φ 1 = 240◦.

5.23 The dimensions of the mechanism shown in Fig. 3.38 are: AB = 200 mm, AD = 500 mm, and BC = 250 mm. The constant angular speed of the driver link 1 is n = 160 rpm. Use the contour equations to calculate the velocities and the accelerations of the mechanism for φ = φ 1 = 240◦. Select a suitable value for the distance a.

5.24 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. The constant angular speed of the driver link 1 is n = 2600 rpm. Using the contour equations find the velocities and the accelerations of the mechanism for the given input angle φ = φ 1 =210◦.

5.25 The mechanism in Fig. 3.12 has the dimensions: AB = 0.03 m, BC = 0 .05 m, CD = 0.08 m, AE = 0.07 m, and La = 0.025 m. The constant angular speed of the driver link 1 is n = 90 rpm. Employing the contour equations find the velocities and the accelerations of the mechanism for the given input angle φ = φ 1 = π/3.

5.26 The mechanism in Fig. 3.15 has the dimensions: AC = 0.200 m, BC = 0 .300 m, BD = 1.000 m, and La = 0.050 m. The constant angular speed of the driver link 1 is n = 1500 rpm. Use the contour equations to calculate the velocities and the accelerations of the mechanism for φ = 330◦^.

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