Solutions Manual Theory Of Machine And Mechanisms 4th edition by Uicker , Pennock, Shigley PDF

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Theory of Machines and Mechanisms, 4e

Uicker et al.

Solutions Manual to accompany

THEORY OF MACHINES AND MECHANISMS Fourth Edition International Version

John J. Uicker, Jr. Professor Emeritus of Mechanical Engineering University of Wisconsin – Madison

Gordon R. Pennock Associate Professor of Mechanical Engineering Purdue University

Joseph E. Shigley Late Professor Emeritus of Mechanical Engineering The University of Michigan

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Theory of Machines and Mechanisms, 4e

Uicker et al.

PART 1

KINEMATICS AND MECHANISMS

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Theory of Machines and Mechanisms, 4e

Uicker et al.

Chapter 1

The World of Mechanisms 1.1

Sketch at least six different examples of the use of a planar four-bar linkage in practice. They can be found in the workshop, in domestic appliances, on vehicles, on agricultural machines, and so on. Since the variety is unbounded no standard solutions are shown here.

1.2

The link lengths of a planar four-bar linkage are 0.2, 0.4, 0.6 and 0.6 m. Assemble the links in all possible combinations and sketch the four inversions of each. Do these linkages satisfy Grashof's law? Describe each inversion by name, for example, a crankrocker mechanism or a drag-link mechanism.

s  0.2, l  0.6, p  0.4, q  0.6 ; since 0.2  0.6  0.4  0.6 .

1.3

these

linkages

all

satisfy

Grashof’s

law Ans.

Drag-link mechanism

Drag-link mechanism

Ans.

Crank-rocker mechanism

Crank-rocker mechanism

Ans.

Crank-rocker mechanism Ans. Double-rocker mechanism A crank-rocker linkage has a 250 mm frame, a 62.5 mm crank, a 225 mm coupler, and a 187.5 mm rocker. Draw the linkage and find the maximum and minimum values of the transmission angle. Locate both toggle positions and record the corresponding crank angles and transmission angles.

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Theory of Machines and Mechanisms, 4e

4.46

Uicker et al.

The planar four-bar linkage illustrated in Fig. P4.46 has link dimensions RO4O2  50 mm ,

RAO2  20 mm , RBA  63 mm , and RBO4  30 mm . For the position indicated, link 2 is 30 counterclockwise from the ground link O2O4 and the angular velocity and angular

acceleration of the coupler link AB are 3  5 rad/s ccw and 3  20 rad/s2 cw , respectively. For the instantaneous motion of the coupler link AB show: (a) the velocity pole I, the pole tangent T, and the pole normal N; (b) the inflection circle and the Bresse circle; (c) the acceleration center of the coupler link AB. Then determine; (d) the radius of curvature of the path of coupler point C where RCB  25 mm , (e) the magnitude and direction of the velocity of coupler point C, (f) the magnitude and direction of the angular velocity of link 2, (g) the magnitude and direction of the velocity of the pole, (h) the magnitude and direction of the acceleration of coupler point C, and (i) the magnitude and direction of the acceleration of the velocity pole. (a) The pole I is coincident with the instant center I13 shown in the figure below. The instant center I 24 and the collineation axis are as shown in the figure. From Bobillier's theorem, the angle from the collineation axis to the first ray (say link 2) is measured as 84° cw. This is equal to the angle from the second ray (link 4) to the pole tangent T; that is, 84° cw. Therefore, the pole tangent T is as shown in the figure and the pole normal N, which is perpendicular to the pole tangent T, is also shown. (b) The inflection point J A for point A on link 3 can be obtained from the Euler-Savary equation; that is,

13.8 mm   9.5 mm R2  AI  20 mm RAOA 2

RAJ A

The location of the inflection point J A is shown in the figure. Similarly, the inflection point J B for point B on link 3 can be obtained from the EulerSavary equation; that is,

 56.2 mm   105.3 mm R2  BI  30 mm RBOB 2

RBJ B

The location of the inflection point J B is shown on the figure. Knowing the pole normal and the two inflection points, the inflection circle can be drawn. The inflection circle for the motion of link 3 with respect to 1, the inflection pole J, and the center of the inflection circle (denoted as point O) are shown on the figure. Note that the pole normal N points from the pole I toward the inflection pole J and the pole tangent T is 90 clockwise from the pole normal. The diameter of the inflection circle for the motion 3/1 is measured as RJI  49.2 mm Ans. The diameter of the Bresse circle is

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Theory of Machines and Mechanisms, 4e

9.20

Uicker et al.

In the clock mechanism illustrated in the Fig. P9.20, a pendulum on shaft A drives an anchor (see Fig. 1.12c). The pendulum period is such that one tooth of the 30T escapement wheel on shaft B is released every 2 s, causing shaft B to rotate once every minute. In the figure, note that the second (to the right) 64T gear is pivoted loosely on shaft D and is connected by a tubular shaft to the hour hand. (a) Show that the train values are such that the minute hand rotates once every hour and that the hour hand rotates once every 12 hours. (b) How many turns does the drum on shaft F make every day?

(a)

 B  1.0 rev/min N B NC 8 teeth 8 teeth 1   NC N D 60 teeth 64 teeth 60  B  1 60 rev/min D   DB tD  60 min/rev

   DB

N D N E  1  28 teeth 8 teeth 1    N E N H  60  42 teeth 64 teeth 720 720 min/rev  B  1 720 rev/min tH   12 hr/rev H   HB 60 min/hr

Ans.

   DB   HB

(b)

N D  1  8 teeth 1    N F  60  96 teeth 720  B  1 720 rev/min  60 min/hr  24 hr/day   2 rev/day F   FB

Ans.

   DB   FB

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Ans.

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Theory of Machines and Mechanisms, 4e

Uicker et al.

13.35 The horizontal link 2 is subjected to the load F  150 kN at C as illustrated in Fig. P13.35. The link is supported by the solid circular aluminum link 3. The lengths of the links are L2  RCA  5 m, RBA  3 m, and L3  RBD  3 m. The end D of link 3 is fixed in the ground and the opposite end B is pinned to link 2 (that is, the effective length of the link is LEFF  0.5 L3 ). For aluminum, the yield strength is Syc = 370 MPa and the modulus of elasticity is E = 207 GPa. Determine the diameter d of the solid circular cross-section of link 3 to ensure that the static factor of safety is N = 2.5.

The cross-sectional area and the area moment of inertia of link 3, respectively, are A = π d2/ 4 and I = π d 4 / 64 Therefore, the radius of gyration of the link is

I  d 4 64 d   A d2 4 4 Using the effective length LEFF  0.5 L3 , the slenderness ratio of the link is k

Sr  LEFF k  0.5  3 m  d / 4   6 m d

(1)

The slenderness ratio at the point of tangency is

 Sr  D  

2E Syc   2  207 109 Pa 370 106 Pa  105.09

(2)

Taking moments about A gives 3 m P  5 m F cos60 where P is the compressive load acting at B on link 3. Solving this equation, the compressive load is  5 m 150 000 N 0.5  125 000 N P 3m The factor of safety guarding against buckling of link 3 is defined as N  Pcr P Substituting N = 2.5, the critical unit load is Pcr  2.5 125 000 N   312 500 N Using the Euler column formula, the critical unit load can be written as Pcr   2 E A Sr 2

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Theory of Machines and Mechanisms, 4e

12.6

Uicker et al.

For the gantry robot of Problem 12.3 in the position described, find the instantaneous velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 43.75 mm, if the actuators have (constant) velocities of 1  2  0 , 3  40 mm/s , and 4  20 rad/s . 0 0 D1  Q1   0  0

0 0  1  0 0 0 0 0 1 D3  T13Q3T13   0 0  0 0 0 0 0 0

0 0 0 0

0 0 0 0

0 1 0  0

0 0 1 D2  T12Q2T12   0  0

0 0 0 0

0 0 1 D4  T14Q4T14   1  0

0 0 0 0

1 0  0  0 1 450  0 0  0 180   0 0 

0 0 0 0

3  2  D22  0

2  1  D11  0 0 0 4  3  D33   0  0

0 0 0 0

0 0  0 40  0 0   0 0 

 0  0 5  4  D44   500   0

0 500 9000  40  0 0 0 0 3600   0 0 0 

 2  1  D11  1D1  D11 1  0

3   2  D22  2 D2  D22 2  0

 4  3  D33  3 D3  D33 3  0

5   4  D44  4 D4  D44 4  0

0    40 mm/s   R5  5 R1     0   0  

0  0  R5   5  55  R1    0    0 

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Ans.

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Theory of Machines and Mechanisms, 4e

Uicker et al.

Chapter 18

Cam Dynamics 18.1

In Fig. P18.1a, the mass m is constrained to move only in the vertical direction. The circular cam has an eccentricity of 50 mm, a speed of 25 rad/s, and the weight of the mass is 26.7 N. Neglecting friction, find the angle    t at the instant the cam jumps.

 F  W  F  my my  mg  F y   2 y0 cos t y  y0 1  cos t  F  m 2 y0 cos t  mg Jump begins when F = 0; that is, when m 2 y0 cos t  mg  0 9650 mm/s 2 mg cos t      0.308 8 2 m 2 y0  25 rad/s   50 mm 

  t  cos1  0.308 8  107.99

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Ans....