Design of Machine Elements Bhandari Solution Manual

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

Third number = 200(1.4125)( 1.4125) = 200 (1.4125) 2 = 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) 6 = 630

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) 0 ,

(2) 72 (φ)1 ,

(3) 72(φ) 2 ,

(4) 72 (φ) 3 ,

(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

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

or

⎛ 720 ⎞ φ =⎜ ⎟ ⎝ 72 ⎠

1 / 10

= (10)1 / 10 = 10 10 = 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: Connecting rod (inner diameter) (15H6) =

Bush (outer diameter) (15r5) =

15.011 mm 15.000

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, 4.970 mm 4.952

Limiting dimensions of valve stem (5d8) =

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 / mm 2 2 .5

⎛π ⎞ P = ⎜ D2 ⎟ σt ⎝4 ⎠

∴D =

4 (25 x 10 3 ) = 14.47 mm (i) π (152)

4P = π σt

Pin diameter: τ=

Ssy (fs)

=

0.577 S yt (fs)

⎛π ⎞ P = 2⎜ d2 ⎟ τ ⎝4 ⎠

4.2

τ max =

S sy (fs)

=

=

0.577(380) = 87.7 N / mm 2 2 .5

∴d =

0.5 S yt (fs)

=

2P = πτ

2 (25 x 10 3 ) = 13.47 mm π (87.7)

(ii)

0.5(310) = 62 N / mm 2 2 .5

A = cross sectional area of bolt σt =

12000 A

τ max

⎛ 12000 ⎞ ⎛ 6000 ⎞ ⎛σ ⎞ ⎟⎟ + ⎜ = ⎜ t ⎟ + (τ) 2 = ⎜⎜ ⎟ ⎝ 2 ⎠ ⎝ 2A ⎠ ⎝ A ⎠

τ=

and 2

2

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

6000 A

or

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

S yt

250 σt = = = 83.3 N / mm 2 (fs) 3 ⎛π ⎞ P = ⎜ d 2r ⎟ σ t ⎝4 ⎠

or

⎛π ⎞ 80 000 = ⎜ d 2r ⎟ 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 / mm 2 3 ⎛π ⎞ ∴ 80 000 = 2 ⎜ d 2p ⎟ (41.67) 4 ⎝ ⎠

dp = 34.96 mm (ii)

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

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

From Eq.(4.24), σt =

P Pey + 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 / mm 2

Maximum shear stress theory: Eq.(4.39) (σ 1 − σ 3 ) =

S yt (fs)

or

(200) =

460 (fs)

(fs) = 2.3

(i)

3

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

4.6

(σ

=

2 1

− σ1 σ 2 + σ 22

)

460 = (fs)

(200) 2 − (200)(150) + (150) 2

⎛ σx + σy ⎜⎜ ⎝ 2

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

⎛ σx − σy ⎜⎜ ⎝ 2

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

(fs) = 2.55

(ii)

From Eqs. (4.31) and (4.32), ⎛ σx + σy σ1 , σ 2 = ⎜⎜ ⎝ 2

⎞ ⎟⎟ ± ⎠

⎛ σx − σy ⎜⎜ 2 ⎝

σ1 = 155.44 N / mm 2

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

⎞ ⎟⎟ + (τ xy ) 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) =

Ssy 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 σ 2 + σ 22 =

[(155.44)

2

]

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

4

S yt

380 (fs) = = = 2.32 (163.71) 163.71

4.7

(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) 4

)

2

= (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), Mb hi ( 4000 d ) (0.4843 d ) ⎛ 44 886.51 ⎞ 2 = =⎜ ⎟ N / mm 2 A e Ri (0.7854 d ) (0.0157 d ) (3.5 d) ⎝ d2 ⎠

σ bi =

Direct tensile stress: σt =

S yt (fs)

P 1000 ⎛ 1273.24 ⎞ 2 = =⎜ ⎟ N / mm 2 A (0.7854 d ) ⎝ d 2 ⎠

=

P Mb 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 M b = (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 ⎞ Mb hi (11x10 5 t ) (2.3098 t ) ⎜ ⎟⎟ N / mm 2 = = A e Ri (12 t 2 ) (0.6902 t ) ( 4 t ) ⎜⎝ 12 t 2 ⎠

Direct tensile stress: σt =

P 100 x10 3 ⎛ 10 5 = = ⎜⎜ 2 A (12 t 2 ) ⎝ 12 t

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

t = 26.62 mm

∴

⎞ ⎟⎟ N / mm 2 ⎠

300 10 5 9.2031x10 5 = + (2.5) 12 t 2 12 t 2

(Ans.)

4.9 Permissible stresses: σt =

S yt ( fs )

=

300 = 60 N / mm 2 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 = 50 N / mm 2 4

Components of force P :Pv = 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 =

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

σt =

PV 4330.13 2165.07 = = N / mm2 A 2 t2 t2

∴ 50 =

1749.4 x 10 3 2165.07 + 3 t t2

or

(i)

( ii )

t3 – 43.3 t = 34988

7

The cubic equation is solved by trial and error

4.11

σt =

t

t3 – 43.3 t

35

41 359.5

34

37 831.8

33

34 508.1

∴ t = 33.5 mm

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

P Mb y σt = + A I

or

(Ans.)

At inner fibre,

25 x 10 3 ( 25 x 10 3 x 140 ) t + 100 = ( 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,

σ max = K t σ o = 2.67 (60.61) = 161.82 N / mm 2

(i)

At the hole of 5 mm diameter, σo =

P 20x10 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,

σ max = K t σ o = 2.51(66.67) = 167.33 N / mm 2

(ii)

At the hole of 10 mm diameter, σo =

P 20x10 3 = = 88.89 N / mm 2 ( w − d ) t (25 − 10) 15

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

K t = 2.25

From Fig.5.2,

σ max = K t σ o = 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

32 M b 32 (15x10 3 ) σb = = = 93.94 N / mm 2 π d3 π (11.76) 3 σ max = K t σ o = 1.5 (93.94) = 140.91 N / mm 2 (fs) =

5.3

S ut 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 M b 32 (62 500) = = 9.947 N / mm 2 3 3 πd π (40)

⎛ D ⎞ 60 = 1.5 ⎜ ⎟= ⎝ d ⎠ 40

and

2 ⎛r⎞ = 0.05 ⎜ ⎟= ⎝ d ⎠ 40

From Fig.5.5,

K t = 2.05

σ max = K t σ o = 2.05 (9.947) = 20.39 N / mm 2

5.4

σ max =

(Ans.)

S ut 350 = = 140 N / mm 2 (fs) 2.5

σo =

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

Kt =

σ max 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 S ut = 0.5 (600) = 300 N / mm 2

From Fig. 5.24 (Forged shaft and S ut = 600 N / mm 2 ),

K a = 0.45

K b = 0.85

For 25 mm diameter,

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

1 1 = = 0.52 K f 1.924

S e = K a K b K d S 'e = 0.45 (0.85) (0.52) (300) = 59.67 N / mm 2

5.6

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

From Fig. 5.24 (Machined surface and S ut = 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

S e = K a K b K c K d S 'e = 0.76 (0.85) (0.814) (0.649) (330) =112.62 N / mm 2

5.7

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

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