Design of Machine Elements - Bhandari - Solution Manual PDF

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Summary

The feeder-selector device at one of the workstations of an assembly machine has a
feed rate of 56 components/min and provides a throughput of one part in five. The ideal
cycle time of the assembly machine is 6 sec. The low-level sensor on the feed track is set at
10 components,...

Description

Solutions Manual to DESIGN OF MACHINE ELEMENTS (First Revised Edition)

V. B. Bhandari Formerly Professor & Head of Mechanical Engineering Vishwakarma Institute of Technology, Pune.

McGraw-Hill Education (India) Limited New Delhi

1

CHAPTER 1 1.1 The series factor for R10 series is given by, 10

10 = 1.2589

First number = 1 Second number = 1 (1.2589) = 1.2589 = (1.25) Third number = (1.2589)( 1.2589) = (1.2589 )2 = 1.5848 = (1.6) Fourth number = (1.2589)2(1.2589) = (1.2589 )3 = 1.9951 = (2) Fifth number = (1.2589)3(1.2589) = (1.2589 )4 = 2.5117 = (2.5) Sixth number = (1.2589)4(1.2589) = (1.2589 )5 = 3.1620 = (3.16) Seventh number = (1.2589)5(1.2589) = (1.2589 )6 = 3.9806 = (4) Eighth number = (1.2589)6(1.2589) = (1.2589 )7 = 5.0112 = (5) Ninth number = (1.2589)7(1.2589) = (1.2589 )8 = 6.3086 = (6.3) Tenth number = (1.2589)8(1.2589) = (1.2589 )9 = 7.9418 = (8) Eleventh number = (1.2589)9(1.2589) = (1.2589 )10 = 9.9980 = (10) In above calculations, the rounded numbers are shown in bracket. 1.2 The series factor for R20 series is given by, 20

10 = 1.122

Since every third term of R20 series is selected, the ratio factor ( φ ) is given by, φ = (1.122) 3 = 1.4125

First number = 200 Second number = 200 (1.4125) = 282.5 = (280)

2 2

Third number = 200(1.4125)( 1.4125) = 200 (1.4125) = 399.03 = (400) Fourth number = 200(1.4125)2( 1.4125) = 200 (1.4125)3 = 563.63 = (560) Fifth number = 200(1.4125)3( 1.4125) = 200 (1.4125)4 = 796.13 = (800) Sixth number = 200(1.4125)4( 1.4125) = 200 (1.4125)5 = 1124.53 = (1120) In above calculations, the rounded numbers are shown in bracket. The complete series is given by, 200, 280(282.5), 400(399.03), 560(563.63), 800(796.13), 1120(1124.53), … 1.3 Let us denote the ratio factor as ( φ ). The derived series is based on geometric progression. The power rating of seven models will as follows, (1) 40 (φ)0 ,

(2) 40 (φ) 1 ,

(3) 40 (φ) 2 ,

(5) 40 (φ)4 ,

(6) 40 (φ) 5 ,

(7) 40 (φ)6

4) 40 (φ) 3 ,

The maximum load capacity is 630 kN. Therefore, 40 (φ) = 630 6

or

⎛ 630 ⎞ φ =⎜ ⎟ ⎝ 40 ⎠

1/ 6

= 1.5832

Load capacity of first model = (40) kN Load capacity of second model = 40 (1.5832) = 63.33 = (63) kN Load capacity of third model = 40 (1.5832) 2 = 100.26 = (100) kN Load capacity of fourth model = 40 (1.5832)3 = 158.73 = (160) kN Load capacity of fifth model = 40 (1.5832) 4 = 251.31 = (250) kN Load capacity of sixth model = 40 (1.5832) 5 = 397.87 = (400) kN Load capacity of seventh model = 40 (1.5832)6 = 629.90 = (630) kN 1.4 Let us denote the ratio factor as ( φ ). The derived series is based on geometric progression. The speeds of different steps will as follows,

3

(1) 72 (φ) ,

(2) 72 (φ) ,

(3) 72(φ) ,

(4) 72 (φ) ,

(5) 72 (φ)4 ,

(6) 72 (φ) 5 ,

(7) 72 (φ)6

(8) 72 (φ) 7

(9) 72 (φ) 8

(10) 72 (φ)9

(11) 72 (φ)10

0

1

2

3

The maximum speed is 720 r.p.m. Therefore, 72 (φ) 10 = 720

or

⎛ 720 ⎞ φ =⎜ ⎟ ⎝ 72 ⎠

1 / 10

= (10) 1 / 10 = 1010 = 1.2589

Speed of first step = 72 r.p.m. Speed of second step = 72 (1.2589) = 90.64 = (91) r.p.m. Speed of third step = 72 (1.2589)2 = 114.11 = (114) r.p.m. Speed of fourth step = 72 (1.2589)3 = 143.65 = (144) r.p.m. Speed of fifth step = 72 (1.2589)4 = 180.84 = (181) r.p.m. Speed of sixth step = 72 (1.2589)5 = 227.66 = (228) r.p.m. Speed of seventh step = 72 (1.2589)6 = 286.60 = (287) r.p.m. Speed of eighth step = 72 (1.2589)7 = 360.80 = (361) r.p.m. Speed of ninth step = 72 (1.2589)8 = 454.22 = (454) r.p.m. Speed of tenth step = 72 (1.2589)9 = 571.81 = (572) r.p.m. Speed of eleventh step = 72 (1.2589)10 = 719.85 = (720) r.p.m.

1

CHAPTER 3 3.1

From Tables 3.2 and 3.3b, the tolerances for the small end of connecting rod and bush

are as follows: 15.011 mm 15.000

Connecting rod (inner diameter) (15H6) =

Bush (outer diameter) (15r5) =

15.031 mm 15.023

Maximum interference = 15.031 – 15 = 0.031 mm Minimum interference = 15.023 - 15.011 = 0.012 mm 3.2

From Tables 3.2 and 3.3a,

Limiting dimensions of valve stem (5d8) =

4.970 mm 4.952

Limiting dimensions of guide for valve stem (7H7) =

5.012 mm 5.000

Maximum clearance = 5.012 - 4.952 = 0.06 mm Minimum clearance = 5 - 4.97 = 0.03 mm From Tables 3.2 and 3.3b, Limiting dimensions of valve seat (20s5) =

Limiting dimensions of housing (20H6) =

20.044 mm 20.035

20.013 mm 20.000

Maximum interference = 20.044 – 20 = 0.044mm Minimum interference = 20.035 – 20.013 = 0.022 mm

1

CHAPTER 4 4.1 Rod diameter: σt =

S yt (fs)

=

380 = 152 N / mm2 2.5

⎛π 2⎞ P = ⎜ D ⎟σ t ⎝4 ⎠

∴D =

4 (25 x 103 ) = 14.47 mm (i) π (152)

4P = π σt

Pin diameter: τ=

S sy (fs)

=

0.577 S yt (fs)

⎛π 2⎞ P = 2⎜ d ⎟ τ ⎝4 ⎠

4.2

τ max =

Ssy (fs)

=

=

0.577(380) 2 = 87.7 N / mm 2.5

∴d =

0.5 S yt (fs)

=

2 (25 x 10 3 ) = 13.47 mm π (87.7)

2P = πτ

(ii)

0.5(310) 2 = 62 N / mm 2.5

A = cross sectional area of bolt σt =

12000 A

τ=

and 2

⎛σ ⎞ 2 τ max = ⎜ t ⎟ + (τ) = 2 ⎝ ⎠

⎛ 6000⎞ 62 = ⎜ ⎟ 2 ⎝ A ⎠ π 2 6000 2 d = 62 4

or

6000 A

2

⎛ 12000 ⎞ ⎛ 6000 ⎞ ⎜⎜ ⎟⎟ + ⎜ ⎟ 2 A ⎝ A ⎠ ⎝ ⎠

A=

2

6000 2 62

d = 13.2 mm

(Ans.)

4.3 The maximum force in tie-rod is denoted by P. From Fig.4.71(a), P sin(30) x 2500 = (50x103) x (2000) Diameter of rod:

∴

P = 80 000 N

2

σt =

S yt (fs)

=

250 = 83.3 N / mm 2 3

⎛π 2⎞ P = ⎜ d r ⎟σ t ⎝4 ⎠

or

⎛π ⎞ 80 000 = ⎜ d r2 ⎟83.3 ⎠ ⎝4

dr = 34.96 mm (i)

Diameter of pin: τ=

Ssy (fs)

=

0.5 S yt (fs)

⎞ ⎛π P = 2 ⎜ d 2p ⎟ τ ⎝4 ⎠

4.4

σt =

=

0.5 (250) = 41.67 N / mm2 3 ⎛π 2⎞ ∴ 80 000 = 2 ⎜ d p ⎟(41.67) ⎝4 ⎠

dp = 34.96 mm (ii)

S ut 300 = = 120 N / mm 2 (fs) 2.5

P 15000 ⎛ 3000 ⎞ 2 = =⎜ ⎟ N / mm A ( t )(5t ) ⎝ t 2 ⎠ P e y 15000 (7.5t ) (2.5 t ) ⎛ 27 000 ⎞ 2 = =⎜ ⎟ N / mm ⎡1 I t2 ⎠ ⎝ 3⎤ ⎢ 12 ( t )(5t ) ⎥ ⎣ ⎦

From Eq.(4.24), σt =

P Pe y + A I

t = 15.81 mm 4.5

or

⎛ 3000 ⎞ ⎛ 27 000 ⎞ ⎛ 30 000 ⎞ 120 = ⎜ 2 ⎟ + ⎜ ⎟ =⎜ ⎟ 2 2 ⎠ ⎠ ⎝ t ⎝ t ⎠ ⎝ t

(Ans.)

(σ1 − σ 2 ) = 50 N / mm 2

(σ1 − σ 3 ) = 200 N / mm 2

(Maximum value)

(σ 2 − σ 3 ) = 150 N / mm2

Maximum shear stress theory: Eq.(4.39) ( σ1 − σ 3) =

Syt (fs)

or

(200) =

460 (fs)

(fs) = 2.3

(i)

3

Distortion energy theory: Eq.(4.44) S yt (fs)

4.6

(σ

=

2 1

− σ 1σ 2 + σ 2 2

)

460 = (fs)

(200)2 − (200)(150) + (150) 2

⎛σ x+σ y ⎜ ⎜ 2 ⎝

⎞ ⎛ 100+ 40 ⎞ 2 ⎟ =⎜ ⎟ = 70 N / mm ⎟ ⎝ 2 ⎠ ⎠

⎛ σ x −σ y ⎜ ⎜ 2 ⎝

⎞ ⎛ 100− 40 ⎞ 2 ⎟=⎜ ⎟ = 30 N / mm ⎟ ⎝ 2 ⎠ ⎠

(fs) = 2.55

(ii)

From Eqs. (4.31) and (4.32), ⎛ σx + σ y σ 1 , σ 2 = ⎜⎜ 2 ⎝

⎞ ⎟⎟ ± ⎠

⎛ σx − σ y ⎜⎜ 2 ⎝

σ1 = 155.44 N / mm2

2

⎞ ⎟⎟ + (τ xy ) 2 = 70 ± ⎠

σ 2 = −15.44 N / mm 2

(30) 2 + (80) 2

σ3 = 0

From Eq.(4.34), τ max =

⎛ σx − σy ⎜ ⎜ 2 ⎝

2

⎞ ⎟ + ( τxy ) 2 = (30) 2 + (80) 2 = 85.44 N/mm2 ⎟ ⎠

Maximum normal stress theory: (fs) =

S yt σ1

=

380 = 2.44 155.44

(i)

Maximum shear stress theory: (fs) =

S sy τ max

=

0.5 S yt τ max

=

0.5 (380) = 2.22 85.44

(ii)

Distortion energy theory: Eq.(4.44)

(σ

2 1

)

− σ1 σ2 + σ22 =

[(155.44)

2

]

− (155.44)( −15.44) + ( −15.44) 2 = 163.71 N / mm 2

4

(fs) =

4.7

Syt (163.71)

=

380 = 2.32 163.71

(iii)

Refer to Fig.4.73. R i = 4 d − 0.5 d = (3.5 d) mm

R=4d

R o = 4 d + 0.5 d = (4.5 d) mm π 2 d = (0.7854 d 2 ) mm 2 4

A =

M b = (1x10 3 ) (4 d ) = (4000 d) N − mm

From Eq.(4.60),

RN

( =

Ro +

Ri

)

2

4

( =

(4.5 d) + (3.5 d)

)

2

4

= (3.9843 d) mm

e = R − R N = 4 d − 3.9843 d = (0.0157 d ) mm h i = R N − R i = 3.9843 d − 3.5 d = (0.4843 d) mm

From Eq.(4.56), σ bi =

Mb hi ( 4000 d ) ( 0. 4843 d) ⎛ 44 886 . 51 ⎞ 2 = =⎜ ⎟ N / mm 2 2 A e R i ( 0. 7854 d ) ( 0. 0157 d) (3.5 d) ⎝ d ⎠

Direct tensile stress: σt =

S yt (fs)

=

P 1000 ⎛ 1273 . 24 ⎞ 2 = =⎜ ⎟ N / mm A ( 0. 7854 d 2 ) ⎝ d 2 ⎠

P M b hi + A A e Ri

∴

380 1273.24 44 886.51 = + (4.5) d2 d2

d = 23.38 mm 4.8

Refer to Fig.4.74. Ri = 4 t

bi = 4 t

(Ans.) At section XX, h=6t

R=7t

5

Ro = 10 t

ti = to = t

bo = 4 t

From Eq. (4.64), RN =

[ t (4 t − t ) + t (4 t − t ) + t (6 t )] ⎧ ⎛ 4t+t⎞ ⎛ 10 t − t ⎞ ⎛ 10 t ⎞ ⎫ ⎟⎟ + t log e ⎜⎜ ⎟⎟ + 4 t log e ⎜⎜ ⎨ 4 t log e ⎜⎜ ⎟⎟ ⎬ ⎝ 4t ⎠ ⎝ 4t+t ⎠ ⎝ 10 t − t ⎠ ⎭ ⎩

= (6.3098 t) mm

e = R − R N = (7 − 6.3098) t = (0.6902 t ) mm h i = R N − R i = (6.3098 − 4) t = (2.3098 t ) mm Mb = (100x10 3 )(4 t + R ) = (100x10 3)(4 t + 7 t ) = (11x10 5) t N − mm A = 4 t 2 + 4 t 2 + 4 t 2 = (12 t 2 ) mm 2

From Eq.(4.56), σ bi =

⎛ 9. 2031x10 5 ⎞ M b hi (11x10 5 t ) ( 2. 3098 t ) 2 ⎜ = = ⎟⎟ N / mm A e R i (12 t 2) (0. 6902 t ) ( 4 t ) ⎜⎝ 12 t 2 ⎠

Direct tensile stress: σt =

P 100 x10 3 ⎛ 105 = = ⎜⎜ 2 A (12 t 2 ) ⎝ 12 t

S ut P M b hi = + (fs) A A e R i

t = 26.62 mm

∴

⎞ 2 ⎟⎟ N / mm ⎠

300 105 9.2031x105 = + (2.5) 12 t 2 12 t 2

(Ans.)

4.9 Permissible stresses: σt =

Syt ( fs )

=

300 = 60 N / mm2 5

Refer to Fig.4.1-solu, R=

τ=

Ssy ( fs )

0.5 S yt ( fs )

=

( 7.5 x 10 3 ) x 100 = P x 500

( 7500 )2 + (1500 ) 2 = 7648.53 N

R = p ( d x l ) or

=

0.5 x 300 = 30 N / mm 2 5

or P = 1500 N

From Eq.(4.51),

7648.53 = 10 ( d x 1.5 d )

6

∴ d= 22.58 mm and l= 1.5 d = 1.5 x 22.58 = 33.87 mm

(i)

R 7648.53 = 9.55 N / mm 2 = ⎞ ⎛π 2 ⎞ ⎛π 2⎜ d ⎟ 2 ⎜ ( 22.58 )2 ⎟ ⎝4 ⎠ ⎝4 ⎠

τ=

( ii )

The dimensions of the boss of lever at the fulcrum are as follows, inner diameter = 23 mm, outer diameter = 46 mm, length = 34 mm ( iii ) For the lever, σb =

∴

4.10

Mb y I

d=4b or

b = 16.74 mm

σt =

S yt ( fs )

=

60 =

Mb = ( 7500 x 100 ) N- mm ( 7500 x 100 ) ( 2 b ) ⎡1 3⎤ ⎢⎣12 b ( 4b ) ⎥⎦

d = 4 b = 4 x 16.74 = 66.94 mm

( iv )

200 2 = 50 N / mm 4

Components of force P :P v = P cos ( 30 ) = 5000 cos ( 30 ) = 4330.13 N

Ph = P sin ( 30 ) = 5000 sin ( 30 ) = 2500 N Mb = Ph x 250 + Pv x 125 = 2500 x 250 +4330.13 x 125 = 1166 266.25 N-mm σb =

σt =

∴ 50 =

1749.4 x 10 3 1166 266 .25 x t Mb y = N / mm2 = 3 1 I t ⎡ 3⎤ ⎢12 t (2 t ) ⎥ ⎦ ⎣ PV 4330.13 2165.07 = = N / mm2 A 2 t2 t2

1749.4 x 103 2165.07 + 3 t t2

or

t3 – 43.3 t = 34988

(i)

( ii )

7

The cubic equation is solved by trial and error

4.11

t

t3 – 43.3 t

35

41 359.5

34

37 831.8

33

34 508.1

∴t =33.5 mm

σt =

S ut 400 = = 100 N / mm 2 ( fs ) 4

σt =

P Mby + I A

or

100 =

(Ans.)

At inner fibre, 25 x 10 3 ( 25 x 103 x140 ) t + ( t x 2t ) ⎡1 3⎤ ⎢12 t ( 2 t ) ⎥ ⎦ ⎣

t3 – 125 t = 52 500 The cubic equation is solved by trial and error. t

t3 – 125 t

40

59 000

39

54 444

38.5

52 254

∴

t = 38.5 or 40 mm b = 2 t = 80 mm

(Ans.)

1

CHAPTER 5 5.1

At the hole of 3 mm diameter, σo =

P 20 x10 3 = 60.61 N / mm 2 = ( w −d ) t (25 − 3) 15

⎛ d⎞ ⎛ 3⎞ ⎜ ⎟ = ⎜ ⎟ = 0.12 ⎝ w ⎠ ⎝ 25 ⎠

K t =2.67

From Fig.5.2,

σmax = Kt σo = 2.67 (60.61) = 161.82 N / mm 2

(i)

At the hole of 5 mm diameter, σo =

P 20 x10 3 = 66.67 N / mm 2 = (w −d ) t (25 − 5) 15

⎛ d⎞ ⎛ 5⎞ ⎜ ⎟ = ⎜ ⎟ = 0. 2 ⎝ w ⎠ ⎝ 25 ⎠

K t =2.51

From Fig.5.2,

σmax = Kt σo = 2.51(66.67) = 167.33 N / mm 2

(ii)

At the hole of 10 mm diameter, σo =

P 20 x10 3 = 88.89 N / mm2 = ( w −d ) t (25 − 10) 15

⎛ d ⎞ ⎛ 10 ⎞ = ⎜ ⎟ = ⎜ ⎟ 0.4 ⎝ w ⎠ ⎝ 25 ⎠

K t =2.25

From Fig.5.2,

σ max = K t σo = 2.25 (88.89) = 200 N / mm 2

5.2

D = 0.25d +d +0.25 d = 1.5 d

(iii) ⎛ D⎞ ⎜ ⎟ =1.5 ⎝ d⎠

From Fig.5.5,

(D/d = 1.5 and Kt = 1.5 )

⎛ r⎞ ⎜ ⎟ = 0.17 ⎝ d⎠

d=

r 2 = = 11.76 mm 0.17 0.17

(i)

2 3

σb =

32 Mb 32 (15x10 ) = 93.94 N / mm 2 = 3 3 π d π (11.76)

σmax = Kt σo = 1.5 (93.94) = 140.91 N / mm 2 (fs) =

5.3

Sut 200 = = 1.42 σ max 140.91

(ii) (iii)

By symmetry, the reaction at each bearing is 2500 N. At fillet section, M b = 2500 (25) = 62 500 N − mm

σb =

32 Mb 32 (62 500) = 9.947 N / mm2 = 3 π d π (40) 3

⎛ D ⎞ 60 =1.5 ⎜ ⎟= ⎝ d ⎠ 40

and

From Fig.5.5,

K t = 2.05

2 ⎛r ⎞ = 0.05 ⎜ ⎟= d 40 ⎝ ⎠

σmax = Kt σo = 2.05 (9.947) = 20.39 N / mm 2

5.4

σmax =

(Ans.)

Sut 350 = = 140 N / mm 2 (fs) 2.5

σo =

P 20 x10 3 = = 66.67 N / mm 2 d t (30) (10)

Kt =

σmax 140 = = 2.1 σo 66.67

and

⎛ D ⎞ 45 =1.5 ⎜ ⎟= ⎝ d ⎠ 30

From Fig.5.3,

(D/d = 1.5 and Kt = 2.1 )

⎛ r⎞ ⎜ ⎟ = 0.095 ⎝ d⎠

r = 0.095 d = 0.095 (30) = 2.85 or 3 mm (Ans.)

3

5.5

S'e = 0.5 Sut = 0.5 (600) = 300 N / mm 2

From Fig. 5.24 (Forged shaft and Sut = 600 N / mm2 ), For 25 mm diameter,

K a = 0.45

K b = 0.85

K f = 1+ q (K t − 1) = 1+ 0.84 (2.1− 1) = 1.924 Kd =

1 1 = = 0.52 K f 1.924

Se = Ka Kb Kd Se' = 0.45 (0.85) (0.52) (300) =59.67 N / mm 2

5.6

S'e = 0.5 Sut = 0.5 (660) = 330 N / mm 2

From Fig. 5.24 (Machined surface and Sut = 660 N / mm 2 ), For 40 mm diameter,

K b = 0.85

For 99% reliability,

K c = 0.814

K a = 0.76

K f = 1+ q ( K t − 1) = 1+ 0.90 (1.6 − 1) = 1.54 Kd =

1 1 = = 0.649 K f 1.54

Se = Ka Kb Kc Kd S'e = 0.76 (0.85) (0.814) (0.649) (330) =112.62 N / mm2

5.7

S'e = 0.5 Sut = 0.5 (540) = 270 N / mm 2

From Fig. 5.24 (Machined surface and Sut = 540 N / mm 2 ), Assuming (7.5< d...