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 <50 mm),
⎛D⎞
⎜ ⎟ = 1.5
⎝d⎠
From Fig.5.5,
and
⎛r⎞
⎜ ⎟ = 0.1
⎝d⎠
K t =1.72
K b = 0.85
K a = 0.78
4
K f = 1 + q (K t − 1) = 1 + 0.9 (1.72 −1) = 1.648
Kd =
1
1
=
= 0.61
K f 1.648
S e = K a K b K d S 'e = 0.78 (0.85) (0.61) (270) = 109.20 N / mm 2
5.8
σb =
Se
109.20
=
= 54.6 N / mm 2
(fs)
2
σb =
32 M b
or
π d3
32 (5x10 3 )(100)
πd3
54.6 =
d = 45.35 mm
(Ans.)
S 'e = 0.5 S ut = 0.5 (620) = 310 N / mm 2
From Fig. 5.24 (Ground surface),
K a = 0.89
Assuming (7.5< d <50 mm),
K b = 0.85
For 90% reliability,
K c = 0.897
S e = K a K b K c S 'e = 0.89 (0.85) (0.897) (310) = 210.36 N / mm 2
Sse = 0.577 S e = 0.577 (210.36) = 121.38 N / mm 2
τa =
Sse
121.38
=
= 60.69 N / mm 2
(fs)
2
(M t ) max = 400 N − m
(M t ) a =
τa =
(M t ) min = − 200 N − m
1
[(M t ) max − (M t ) min ]= 1 [(400) − (−200) ] = 300 N − m
2
2
16 (M t ) a
or
π d3
60.69 =
16 (300 x10 3 )
πd3
d = 29.31 mm (Ans.)
5.9
τ xym
35 + 0
=
= 17.5 N / mm 2
2
τ xya
30 + (−15)
= 7.5 N / mm 2
2
σ xm =
35 − 0
=
= 17.5 N / mm 2
2
σ xa =
30 − (−15)
= 22.5 N / mm 2
2
From Eqs. (5.50) and (5.51),
σ m = (σ xm ) 2 + 3 (τ xym ) 2 = (7.5) 2 + 3 (17.5) 2 = 31.22 N / mm 2
σ a = (σ xa ) 2 + 3 (τ xya ) 2 = (22.5) 2 + 3 (17.5) 2 = 37.75 N / mm 2
tan θ =
σa
37.75
=
= 1.2
σ m 31.22
or
θ = 50.4 0
The modified Goodman diagram is shown in Fig.5.1-solu.
X is the point of intersection of following two lines,
S
Sm
+ a =1
540 200
5.10
and
Sa
= 1.2
Sm
∴
S a = 152.83 N / mm 2 and
(fs) =
S a 152.83
=
= 4.05
σa
37.75
S m = 127.36 N / mm 2
(Ans.)
τ xym = 70 N / mm 2
τ xya = 35 N / mm 2
σ xm = 60 N / mm 2
σ xa = 80 N / mm 2
From Eqs. (5.50) and (5.51),
σ m = (σ xm ) 2 + 3 (τ xym ) 2 = (60) 2 + 3 (70) 2 = 135.28 N / mm 2
σ a = (σ xa ) 2 + 3 (τ xya ) 2 = (80) 2 + 3 (35) 2 = 100.37 N / mm 2
tan θ =
σa
100.37
=
= 0.742
σ m 135.28
or
θ = 36.57 0
5
6
The modified Goodman diagram is similar to Fig.5.1-solu.
X is the point of intersection of following two lines,
S
Sm
+ a =1
700 220
and
Sa
= 0.742
Sm
∴
S a = 154.54 N / mm 2 and
(fs) =
S a 154.54
=
= 1.54
σ a 100.37
S m = 208.28 N / mm 2
(Ans.)
1
CHAPTER 6
6.1
l = 2p = 2 x 6 = 12 mm
dm = d – 0.5p = 30 – 0.5 x 6 = 27 mm
tan α =
l
12
=
= 0.1415
πdm
π (27)
tan φ = µ = 0.1
or
α = 8.052 0
or φ = 5.7110
For square threads, Eq.(6.10)
η=
tan α
0.1415
=
= 0.5776
tan ( φ + α ) tan ( 5.711 + 8.052 )
or η = 57.76 %
(i)
For Acme threads,
µ sec θ =
η=
µ
0.1
=
= 0.10329
cos θ cos (14.5 )
From Eq. (6.16),
tan α (1 − µ sec θ tan α ) 0.1415 (1 − 0.10329 x 0.1415 )
=
= 0.5696
( µ sec θ + tan α )
( 0.10329 + 0.1415 )
∴ η = 56.96%
6.2
( ii )
dm = d – 0.5p = 36 – 0.5 x 6 = 33 mm
l = p = 6 mm
tan α =
l
6
=
πdm
π (33)
tan φ = µ = 0.15
Mt =
or
α = 3.312 0
or
φ = 8.5310
W dm
(10 x 10 3 ) (33)
tan ( φ + α ) =
tan ( 8.531 + 3.312 ) = 34 599.55 N-mm
2
2
( M t )c =
µ c W ( D o + D i ) ( 0.2 ) (10 x 10 3 ) ( 50 + 30 )
= 40 000 N-mm
=
4
4
( M t ) t = 34 599.55 + 40 000 = 74 599.55 N-mm or 74.6 N – m
ηo =
Wl
(10 x 10 3 ) (6 )
=
= 0.128
2 π ( Mt )t
2 π ( 74 599.55 )
or 12.8 %
(i)
( ii )
2
6.3
l = 2p = 2 x 9 = 18 mm
dm = d – 0.5p = 60 – 0.5 x 9 = 55.5 mm
tan α =
l
18
=
= 0.1032
πdm
π (55.5)
θ = 14.5 0
µ sec θ =
Raising load :
or
α = 5.894 0
0.15
= 0.1549
cos (14.5 )
From Eq.(6.13),
W dm
( µ sec θ + tan α )
x
2
(1 − µ sec θ tan α )
Mt =
=
( 5 x 10 3 ) (55.5 )
( 0.1549 + 0.1032 )
x
2
(1 − 0.1549 x 0.1032 )
= 36 393.14 N-mm or 36.39 N-m
Lowering load :
Mt =
(i)
From Eq.(6.15),
W dm
( µ sec θ − tan α )
x
2
(1 + µ sec θ tan α )
( 5 x 10 3 ) (55.5 )
( 0.1549 − 0.1032 )
=
x
2
(1 + 0.1549 x 0.1032 )
= 7060.51 N-mm or 7.06 N-m
( ii )
From Eq.(6.16),
η=
tan α (1 − µ sec θ tan α ) 0.1032 (1 − 0.1549 x 0.1032 )
= 0.3935
=
( µ sec θ + tan α )
( 0.1549 + 0.1032 )
∴ η = 39.35 %
6.4
d c = d − p = 60 − 9 = 51 mm
z=
( iii )
From Eq.(6.23),
4W
4 ( 50 x 10 3 )
=
= 6.37 or 7 threads
π S b ( d 2 − d c2 ) π (10 ) (60 2 − 512 )
length of nut = 7 x 9 = 63 mm
t = p/2 = 9/2 = 4.5 mm
(i)
From Eq.(6.22),
3
50 x 10 3
W
τn =
=
= 8.42 N / mm 2
π d t z π ( 60 ) ( 4.5 ) ( 7 )
6.5
(ii)
l = 3p = 3 x 8 = 24 mm
dm = d – 0.5p = 50 – 0.5 x 8 = 46 mm
l
24
=
= 0.166
πdm
π (46)
tan α =
tan φ = µ = 0.12
α>φ
Mt =
or
α = 9.429 0
or
φ = 6.843 0
The screw is not self locking.
W dm
( 7500 ) (46)
tan ( φ + α ) =
tan (6.843 + 9.429 ) = 50 351.04 N-mm
2
2
d c = d − p = 50 − 8 = 42 mm
σc =
τ=
W
7500
=
= 5.41 N / mm 2
π
π
⎛
⎛
2⎞
2⎞
⎜ d c ⎟ ⎜ (42 ) ⎟
⎝4
⎠ ⎝4
⎠
16 ( M t ) t 16 ( 50 351.04 )
=
= 3.46 N / mm 2
3
3
π dc
π ( 42 )
⎛ σc ⎞
⎜ ⎟ + τ2 =
⎝ 2 ⎠
2
τ max =
z=
2
⎛ 5.41 ⎞
2
2
⎜
⎟ + (3.46 ) = 4.39 N / mm
⎝ 2 ⎠
48
= 6 threads
8
(i)
t = p/2 = 8/2 = 4 mm
From Eqs.(6.21) to (6.23),
τs =
W
7500
=
=2.37 N / mm 2
π d c t z π (42) (4 ) (6 )
τn =
W
7500
=
= 1.99 N / mm 2
π d t z π (50) (4 ) (6 )
Sb =
4W
4 ( 7500 )
=
= 2.16 N / mm 2
2
2
2
2
π z ( d − d c ) π ( 6 ) (50 − 42 )
( ii )
( iii )
4
6.7 dm = d – 0.5p = 20 – 0.5 x 5 = 17.5 mm
tan α =
l
5
=
= 0.0909
πdm
π (17.5)
tan φ = µ = 0.15
Mt =
or
α = 5.197 0
or
φ = 8.5310
W dm
W (17.5)
tan ( φ + α ) =
tan ( 8.531 + 5.197 ) = 2.1376 W N − mm
2
2
( M t )c =
µc W ( DO + Di )
= µ c W rm = 0.15 W (8) =1.2 W N − mm
4
( M t ) t = M t + ( M t ) c = ( 2.1376 + 1.2 ) W = 3.3376 W
∴ 3.3376 W = 50 x 150
or
W = 2247.12 N
1
CHAPTER 7
7.1
σt =
σt =
7.2
S yt
( fs )
=
400
= 80 N / mm 2
5
P
⎛π 2⎞
⎜ dc ⎟
⎠
⎝4
80 =
or
( 7.5 x 10 3 )
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
∴ d c = 10.93 mm
d=
dc
10.93
=
= 13.66 mm
0.8
0.8
P=
π 2
π
D p = ( 250 ) 2 (1.5 ) = 73 631.08 N
4
4
P=
π 2
dc n σt
4
or
or
14 mm
73 631.08 =
(Ans.)
Also,
π 2
d c (12 ) (30 )
4
∴ d c = 16.14 mm
d=
dc
16.14
=
= 19.21 mm or 20 mm
0.84
0.84
(i)
t=
π d π (400 )
=
= 104.72 mm
n
12
(ii)
5 d = 5( 20 ) = 100 mm
10 d = 10( 20 ) = 200 mm
7.3
τ=
Ssy
( fs )
=
∴ 5 d < t < 10 d
0.5 ( 380 )
= 95 N / mm 2
2
P1' = P2' = P3' = P4' =
3000
= 750 N
4
e = 200 + 50 = 250 N
(o.k.)
Refer to Fig.7.1-solu.
2
r1 = r2 = r3 = r4 =
50 2 + 50 2 = 70.71 mm
( P e ) r1
3000 x 250 x 70.71
=
= 2651.68 N
2
4r
4 x 70.712
P1" =
Force on bolt 2 or 4 is maximum. As shown in Fig.7.1-solu, the angle
between vectors P2' and P2" is 450
P2 =
( P2' + P2" cos 45 ) 2 + ( P2" sin 45 ) 2
[ ( 750 + 2651.68 cos 45 )
=
2
+ ( 2651.68 sin 45 ) 2
]
= 3225.9 N
d=
7.4
τ=
4 ( 3225.9 )
= 6.58 mm
π ( 95 )
Ssy
( fs )
0.5 ( 400 )
= 66.67 N / mm 2
3
=
P1' = P2' = P3' =
P1" = P3" =
P
5000
=
= 1666.67 N
3
3
(Ans.)
Refer to Fig.7.2-solu.
e = 250 mm
( P x e ) r1
(5000 x 250 ) ( 75 )
=
= 8333.33 N
2
2
( r1 + r3 )
( 75 2 + 75 2 )
As shown in Fig.7.2-solu, the vectors P1' and P1" are perpendicular to
each other.
P1 = P3 =
d=
4 P1
πτ
( 8333.33 ) 2 + (1666.67 ) 2 = 8498.36 N
=
4 ( 8498.36 )
= 12.74 mm
π (66.67 )
(Ans.)
3
7.5
Two bolts at A are denoted by 1 and two bolts at B by 2.
P1' = P2' =
τ=
P1' ⎛ 1250 ⎞
=⎜
⎟ N / mm
A ⎝ A ⎠
P1" =
σt =
P 5000
=
= 1250 N
4
4
From Eq.(7.10),
P e l1
5000 (250 ) (375 )
=
= 1602.56 N
2
2
2 ( l1 + l 2 )
2 ( 375 2 + 75 2 )
P1" ⎛ 1602.56 ⎞
=⎜
⎟ N / mm ( ii )
A ⎝
A ⎠
σ max =
=
σt
+
2
⎛ 1602.56 ⎞
⎛ σt ⎞
⎟⎟ +
⎜
⎟ + τ 2 = ⎜⎜
⎝ 2 ⎠
⎝ 2A ⎠
2
2286.05
A
π 2
d c = 30.08
4
d=
7.6
(i)
∴
⎛ 1602.56 ⎞
⎜⎜
⎟⎟
⎝ 2A ⎠
S yt
2286.05
380
=
=
A
( fs )
5
2
⎛ 1250 ⎞
+ ⎜
⎟
⎝ A ⎠
2
∴ A = 30.08 mm 2
dc = 6.19 mm
dc
6.19
=
= 7.74 mm or 8 mm
0 .8
0 .8
(Ans.)
Two bolts at A are denoted by 1 and two bolts at B by 2. The direct
tensile force in each bolt is given by,
P1' = P2' =
P1" =
P 10 000
=
= 2500 N
4
4
From Eq.(7.10),
P e l1
10 000 (550 ) (325 )
=
= 8033.71 N
2
2
2 ( l1 + l 2 )
2 ( 325 2 + 75 2 )
P2 = P1' + P1" = 2500 + 8033.71 = 10 533.71 mm
4
S yt
σt =
7.7
7.8
( fs )
=
400
= 66.67 N / mm 2
6
Also,
σt =
P2
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
d =
dc
14.18
=
= 17.73 or 18 mm
0 .8
0 .8
66.67 =
10 533.71
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
dc = 14.18 mm
(Ans.)
Two bolts at A are denoted by 1 and two bolts at B by 2.
P2 =
P l l2
10 000 (300 ) (225 )
=
= 6000 N
2
2
2 ( l1 + l 2 )
2 ( 75 2 + 225 2 )
σt =
S yt
( fs )
=
400
= 80 N / mm 2
5
Also,
σt =
P2
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
d =
dc
9.77
=
= 12.22 or 13 mm
0 .8
0. 8
σt =
S yt
a=
( fs )
=
80 =
6000
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
dc = 9.77 mm
(Ans.)
380
= 76 N / mm 2
5
750
= 375 mm
2
b=
l = r – a = 750 – 375 = 375 mm
⎡
⎛ 180 ⎞⎤
2 P l ⎢ a + b cos ⎜
⎟⎥
⎝ n ⎠⎦
⎣
P1 =
4 2 a 2 + b2
[
]
600
= 300 mm
2
From Eq.(7.18),
Therefore,
5
⎡
⎛ 180 ⎞⎤
2 ( 50 x 10 3 ) ( 375 ) ⎢ 375 + 300 cos ⎜
⎟⎥
⎝ 4 ⎠⎦
⎣
P1 =
4 2 x 375 2 + 300 2
[
]
σt =
P1
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
d =
dc
15.76
=
= 19.7 or 20 mm
0. 8
0 .8
76 =
14 826.57
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
= 14826.57 N
dc = 15.76 mm
(i)
From Eq.(7.16),
P1 =
7.9
2Pl [a+b
4 2 a 2 + b2
[
]
]
=
2 ( 50 x 10 3 ) (375 ) (375 + 300 )
= 17 045.45 N
4 ( 2 x 375 2 + 300 2 )
σt =
P1
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
d =
dc
16.9
=
= 21.12 or 22 mm
0 .8
0. 8
76 =
17 045.45
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
dc = 16.9 mm
( ii )
It can be proved that the maximum principal stress (σ max ) in the tie rods
is 1.207 times of direct tensile stress (σ t ) .
σ max = 1.207 σ t =
dc = 9.54 mm
1.207 P
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
d =
∴
380 1.207 ( 4500 )
=
5
⎛π 2⎞
⎜ dc ⎟
⎝4
⎠
dc
9.54
=
= 11.92 mm
0 .8
0 .8
(Ans.)
6
7.10
k 'c = 2 k 'b
From Eqs. (7.23) and (7.24),
⎡ k'
⎤
⎡
⎤ P ⎛ 7.5x10 3 ⎞
k'
⎟N
∆P = P ⎢ ' b ' ⎥ = P ⎢ ' b ' ⎥ = = ⎜⎜
3 ⎟⎠
⎣kb + kc ⎦
⎣kb + 2kb ⎦ 3 ⎝
Pb = Pi + ∆P = 10 000 +
A=
7500
= 12 500 N
3
Pb (fs) 12 500 (3)
=
= 93.75 mm 2
S yt
400
(Ans.)
From Table7.1, the standard size of bolt is M16.
7.11
S 'e = 0.5 S ut = 0.5 (630) = 315 N / mm 2
Kc = 1
For 50% reliability,
Kd =
1 1
=
Kf 3
S e = K c K d S 'e = 1 (1 / 3) (315) = 105 N / mm 2
k 'c = 3 k 'b
From Eqs. (7.23) and (7.24),
⎡ k'
⎤
⎡
⎤ P
k'
∆P = P ⎢ ' b ' ⎥ = P ⎢ ' b ' ⎥ = = o.25 P
⎣kb + kc ⎦
⎣kb + 3kb ⎦ 4
Pb = Pi + ∆P = Pi + 0.25 P
(Pb ) max = Pi + 0.25 P = 4500 + 0.25 (5000) = 5750 N
(Pb ) min = Pi + 0.25 P = 4500 + 0.25 (0) = 4500 N
(Pb ) m =
1
[ 5750 + 4500] = 5125 N
2
(Pb ) a =
1
[ 5750 − 4500] = 625 N
2
From Eq.(7.30),
7
Sa =
S ut − (Pi / A) 630 − (4500 / 36.6)
=
= 72.44 N / mm 2
1 + (S ut / S e )
1 + (630 / 105)
Sa
(P )
= b a
(fs)
A
(fs) = 4.24
or
72.44 625
=
(fs)
36.6
(Ans.)
1
CHAPTER 8
8.1
From Eq.(8.7),
P = 1.414 h l τ
∴ 35 x 10 3 = 1.414 (10 ) l (75 )
or l = 33 mm
Adding 15 mm length for starting and stopping of the weld run,
8.2
l = 33 + 15 = 48 mm
(Ans.)
P = w t σ t = 80 (10 ) (100 ) = 80 000 N
(i)
The strength of the transverse fillet weld is denoted by P1.
From Eq.(8.9),
P1 = 0.707 h l σ t = 0.707 (10 ) ( 80 ) (100 ) = 56 560 N
(ii)
The strength of double parallel fillet weld is denoted by P2.
P2 = 1.414 h l τ = 1.414 ( 10 ) l ( 70 ) = ( 989.8 ) l
N
(iii)
The strength of the welded joint is equal to the strength of the plate.
80 000 = 56 560 + (989.8) l
8.3
or l = 23.68 mm (Ans.)
120 x 10 3
= 120 mm
l1 + l 2 =
1 x 10 3
The
resisting
forces
at
(i)
welds
l1 and l 2
are
P1 and P2 respectively.
P1 x 100 = P2 x 50
∴2 l 1 = l 2
or
(1000 x l1 ) x 100 = (1000 x l 2 ) x 50
(ii)
denoted
by
2
From (i) and (ii),
8.4
l1 = 40 mm and l 2 = 80 mm
(Ans.)
A = 2(50 t ) = 100 t mm 2
τ1 =
P 10 000 ⎛ 100 ⎞
2
=
=⎜
⎟ N / mm
A
100 t
⎝ t ⎠
e = 150 + 25=175 mm
M = P x e = 10 000 x 175 = 175 x 10 4 N − mm
Refer to Fig.8.1-solu.
25 2 + 25 2 = 35.36 mm
r = distance of farthest point =
From Eq.(8.25),
⎞
⎛ l2
J 1 = J 2 = A 1 ⎜⎜ 1 + r12 ⎟⎟ = 50 t
⎠
⎝ 12
⎞
⎛ 50 2
⎜⎜
+ 25 2 ⎟⎟ = 41 667 t mm 4
⎠
⎝ 12
J = J 1 + J 2 = 2 J 1 = 83 333 t mm 4
τ2 =
M r (175 x 10 4 ) (35.36 ) ⎛ 742.56 ⎞
2
=
=⎜
⎟ N / mm
J
( 83 333 ) t
t ⎠
⎝
2
τ max =
816.34
= 95
t
8.5
2
100 ⎞
816.34
⎛ 742.56
⎛ 742.56
⎞
cos 45 +
sin 45 ⎟ =
N / mm 2
⎜
⎟ +⎜
t ⎠
t
⎝ t
⎝ t
⎠
t = 8.59 mm
(Ans.)
There are two horizontal welds W1 and W2 and one vertical weld W3.
From Fig.8.2-solu,
y = 100 mm
Taking moment about vertical
weld and treating the weld as line,
( 100 + 200 + 100 ) x = 100 x 50 + 100 x 50 + 200 x 0
x = 25 mm
3
The areas of welds are as follows,
A1 = 100 t
A2 = 100 t
A3 = 200t
A = A1 + A2 + A3 = (400 t) mm2
τ1 =
P 50 000
⎛ 125 ⎞
2
=
= ⎜
⎟ N / mm
A
400 t
⎝ t ⎠
(i)
A is the farthest point from the centre of gravity G. Its distance r is given
by,
r=
100 2 + (100 − 25 ) 2 = 125 mm
∴ θ = 53.130
Also
tan θ =
100
75
φ = 90 − θ = 90 − 53.13 = 36.87 0
The secondary shear stress τ 2 is inclined at 36.870 with horizontal.
e = ( 100 – x ) + 200 = ( 10 –25 ) + 200 = 275 mm
G G1 = G G 2 =
25 2 + 100 2 = 103.08 mm or r1 = r2 = 103.08mm
r3 = G G 3 = x = 25 mm
⎞
⎛ l2
J 1 = J 2 = A 1 ⎜⎜ 1 + r12 ⎟⎟ = 100 t
⎠
⎝ 12
⎛ 100 2
⎞
⎜⎜
+ 103.08 2 ⎟⎟ = 1145 881.97 t mm 4
⎝ 12
⎠
⎛ l2
⎞
⎛ 200 2
⎞
J 3 = A 3 ⎜⎜ 3 + r32 ⎟⎟ = 200 t ⎜⎜
+ 25 2 ⎟⎟ = 791 666.67 t mm 4
⎝ 12
⎠
⎝ 12
⎠
J = 2 J1 + J3 = ( 3 083 430.61) t mm4
τ2 =
M r ( 50 000 ) (275 ) (125) ⎛ 557.41 ⎞
=
=⎜
⎟
J
( 3 083 430.61 ) t
t ⎠
⎝
Refer to Fig.8.3-solu,
N/mm2
(ii)
4
Vertical component of τ 2 =
334.45
⎛ 557.41 ⎞
N / mm 2
⎟ sin ( 36.87 ) =
t
⎝ t ⎠
= τ 2 sin φ = ⎜
Horizontal component of τ 2 =
445.93
⎛ 557.41 ⎞
N / mm 2
⎟ cos ( 36.87 ) =
t
⎝ t ⎠
= τ 2 cos φ = ⎜
Resultant shear stress –
2
⎛ 445.93 ⎞
⎛ 334.45 125 ⎞
+
⎟
⎜
⎟ +⎜
t
t ⎠
⎝ t ⎠
⎝
τ=
640.27
= 70
t
h=
8.6
2
⎛ 640.27 ⎞
2
=⎜
⎟ N / mm
t
⎝
⎠
t = 9.15 mm
t
9.15
=
= 12.94 mm
0.707 0.707
(Ans.)
A = 2 ( 50 t ) = ( 100 t ) mm 2
τ1 =
P 2500 ⎛ 25 ⎞
2
=
=⎜
⎟ N / mm
A 100 t ⎝ t ⎠
⎛ t ( 50 ) 3 ⎞
⎟⎟ = ( 20 833.33 t ) mm 4
I = 2 ⎜⎜
12
⎝
⎠
σb =
M b y ( 2500 x 150 ) ( 25 ) ⎛ 450 ⎞
2
=
=⎜
⎟ N / mm
I
20 833.33 t
t
⎝
⎠
τ=
⎛ 450 ⎞ ⎛ 25 ⎞
226.38
⎛ σb ⎞
⎟⎟ + ⎜
N / mm 2
⎜ ⎟ + τ12 = ⎜⎜
⎟ =
t
⎝ 2 ⎠
⎝ 2t ⎠ ⎝ t ⎠
2
226.38
= 50
t
2
t = 4.53 mm
2
5
t
4.53
h=
=
= 6.4 mm
0.707 0.707
8.7
From Eq.(8.32),
τ=
8.8
(Ans.)
Mt
2π t r2
∴ 70 =
3000 x 10 3
2π t ( 37.5 ) 2
h=
t
4.85
=
= 6.86 mm
0.707 0.707
τ1 =
P
P
5000
63.66
=
=
=
N / mm 2
A π d t π ( 25 ) t
t
t = 4.85 mm
(Ans.)
I xx = π t r 3 = π t (12.5 ) 3 mm 4
σb =
M b y ( 5000 x 100 ) (1 2.5 ) ⎛ 1018.59 ⎞
2
=
=⎜
⎟ N / mm
I
t
π t (12.5 ) 3
⎝
⎠
τ=
⎛ 1018.59 ⎞ ⎛ 63.66 ⎞
513.26
⎛ σb ⎞
⎟⎟ + ⎜
N / mm 2
⎜ ⎟ + τ12 = ⎜⎜
⎟ =
2
2
t
t
t
⎠
⎝ ⎠
⎝
⎠ ⎝
2
2
513.26
= 95
t
h=
2
t = 5.4 mm
t
5.4
=
= 7.64 mm
0.707 0.707
(Ans.)
1
CHAPTER 9
9.7
τ=
Ssy
=
( fs )
0.5 S yt
( fs )
=
0.5 (380)
= 47.5 N / mm 2
4
For motor shaft,
Mt =
τ=
60 x 10 6 ( kW )
60 x 10 6 (10) (1.5)
x (1.5) =
= 99 471.84 N − mm
2πn
2 π (1440)
16 M t
π d3
∴ 47.5 =
16 ( 99 471.84 )
π d3
∴d = 22.01 mm
(i)
For pump shaft,
Mt =
9.8
60 x 10 6 ( kW )
60 x 10 6 (10) (1.5)
x (1.5) =
= 298 415.52 N − mm
2πn
2 π (480)
τ=
16 M t
τ=
Ssy
πd
( fs )
Mt =
3
=
∴ 47.5 =
0.5 S yt
( fs )
=
16 ( 298 415.52 )
π d3
∴d = 31.75 mm
0.5 (300)
= 50 N / mm 2
3
60 x 10 6 ( kW )
60 x 10 6 (12.5 )
=
= 397 887.36 N − mm
2πn
2 π (300 )
Mt = ( P1 – P2 ) R
(ii)
Also,
∴ 397 887.36 = ( P1 – P2 ) 225
∴( P1 – P2 ) = 1768.39 N
(a)
and
⎛ P1
⎜⎜
⎝ P2
⎞
⎟⎟ = 2
⎠
(b)
From Eqs. (a) and (b),
P1
=
3536.78 N and P2 = 1768.39 N ∴(P1 + P2 + W) = 5605.17 N
From Fig. 9.1-solu (b),
( P1 + P2 + W ) x 200 = Rb x 750
∴ Rb = 1494.71 N
2
Mb = Rb x 550 = 1494.71 x 550 = 822 091.6 N-mm
τ
50 =
9.9
=
max
16
π d3
16
π d3
( M )2 + ( M )2
b
t
( 822 091.6 ) 2 + ( 397 887.36 ) 2
From Eq. (9.8), σ1 =
=
16
[M b +
πd3
∴ d = 45.31 mm
(Ans.)
(M b ) 2 + (M t ) 2 ]
[
16
(1250 x 10 3 ) +
π (40) 3
(1250 x 10 3 ) 2 + ( 250 x 10 3 ) 2
= 200.91 N/mm2
From Eq. (9.10),
τ max =
=
16
[ ( Mb )2 + ( M t )2 ]
3
πd
16
π (40) 3
[
(1250 x 10 3 ) 2 + ( 250 x 10 3 ) 2
]
= 101.44 N/mm2
According to maximum principal stress theory,
( fs) =
S yt
σ1
=
580
= 2.89
200.91
(i)
According to maximum shear stress theory,
Ssy = 0.5 Syt = 0.5 (580) = 290 N/mm2
( fs) =
Ssy
τ max
=
290
= 2.86
101.44
( ii )
]
3
Ssy
0.5 S yt
0.5 (380)
τ=
=
=
= 47.5 N / mm 2
( fs )
( fs )
4
9.10
Mt =
d
C = i = 0. 8
do
60 x 10 6 (kW )
60 x 10 6 (50)
=
= 795 774.72 N − mm
2πn
2 π (600)
From Eq. (9.19),
τ=
16 M t
π d 3o (1 − C 4 )
∴ 47.5 =
16 ( 795 774.72 )
π d 3o (1 − 0.8 4 )
∴do = 52.48 mm
di = C do = 0.8 (52.48) = 41.98 mm
9.11
(Ans.)
0.30 Syt = 0.30 (480) = 144 N/mm2
0.18 Sut = 0.18 (620) = 111.6 N/mm2 (minimum)
The gears are keyed to the shaft.
τ max = 0.75 (111.6 ) = 83.7 N / mm 2
Refer to Fig.9.2-solu. The resultant bending and turning moments are given
by,
( M b ) max =
(69334 ) 2 + (114400 ) 2 = 133 770.56 N − mm
M t = (1144 ) x 125 = 143 000 N − mm
d3 =
=
16
π τ max
16
π ( 87.3 )
From Eq. (9.15),
(k b M ) 2 + (k t M ) 2
b
t
( 2 x 133 770.56 ) 2 + (1.5x143 000 ) 2
∴ d = 27.15 mm
(Ans.)
4
9.12
From Eq. (9.13),
θ=
584 M t l
or
G d4
0.25 =
584 (143 000 ) (1000 )
( 79 300 ) d 4
d = 45.3 mm
9.13
(Ans.)
The deflections at gear A and gear B are calculated by the principle of
superimposition.
Vertical Plane:
(i) Deflection due to a force of 208 N [Fig. 9.3-solu (a), Eq. (27) of Table
9.4]:
a = 200 mm
(δ A ) 1 =
b = 400 mm
P b x ( x 2 + b2 − l2 )
6EIl
=
l = 600 mm
x = 200 mm
(208) (400) (200) ( 200 2 + 400 2 − 600 2 )
6 E I (600)
=−
7.3955 x 10 8
mm
EI
(i)
From Eq. (28) of Table 9.4,
(δ B ) 1 =
P a (l − x ) ( x 2 + a 2 − 2 l x )
6EIl
=
x = 400 mm
(208) (200) (600 − 400) ( 400 2 + 200 2 − 2 x 600 x 400)
6 E I (600)
=−
6.4711 x 10 8
mm
EI
(ii)
5
(ii) Deflection due to a force of 416 N [Fig. 9.3-solu (b), Eq. (27) of Table
9.4]:
a = 400 mm
b = 200 mm
P b x ( x 2 + b2 − l2 )
6EIl
(δ A ) 2 =
=
12.942 x 10 8
mm
EI
P b x ( x 2 + b2 − l2 )
6EIl
=
x = 200 mm
(416) (200) (200) ( 200 2 + 200 2 − 600 2 )
6 E I (600)
=−
(δ B ) 2 =
l = 600 mm
(iii)
x = 400 mm
(416) (200) (400) ( 400 2 + 200 2 − 600 2 )
6 E I (600)
=−
14.791 x 10 8
mm
EI
(iv)
Horizontal Plane:
(iii) Deflection due to a force of 572 N [Fig. 9.3-solu (c), Eq. (27) of Table
9.4]:
a = 200 mm
(δ A ) 1 =
b = 400 mm
P b x ( x 2 + b2 − l2 )
6EIl
=
l = 600 mm
x = 200 mm
(572) (400) (200) ( 200 2 + 400 2 − 600 2 )
6 E I (600)
=−
20.338 x 10 8
mm
EI
From Eq. (28) of Table 9.4,
(i)
6
P a (l − x ) ( x 2 + a 2 − 2 l x )
(δ B ) 1 =
6EIl
=
x = 400 mm
(572) (200) (600 − 400) ( 400 2 + 200 2 − 2 x 600 x 400)
6 E I (600)
=−
17.796 x 10 8
mm
EI
(ii)
(ii) Deflection due to a force of 1144 N [Fig. 9.3-solu (d), Eq. (27) of Table
9.4]:
a = 400 mm
b = 200 mm
P b x ( x 2 + b2 − l2 )
6EIl
(δ A ) 2 =
=
=
x = 200 mm
(−1144) (200) (200) ( 200 2 + 200 2 − 600 2 )
6 E I (600)
=+
(δ B ) 2 =
l = 600 mm
35.591 x 10 8
mm
EI
P b x ( x 2 + b2 − l2 )
6EIl
(iii)
x = 400 mm
(−1144) (200) (400) ( 400 2 + 200 2 − 600 2 )
6 E I (600)
=+
40.676 x 10 8
mm
EI
Vertical deflection at A is denoted by (δ A ) v .
( δ A ) v = ( δ A )1 + ( δ A ) 2
− 12.942 x 10 8
7.3955 x 10 8
=−
+
EI
EI
(iv)
7
20.3375 x 10 8
=−
EI
Vertical deflection at B is denoted by (δ B ) v .
( δ B ) v = ( δ B )1 + ( δ B ) 2
=−
− 14.791 x 10 8
6.4711 x 10 8
+
EI
EI
=−
21.2621 x 10 8
EI
Horizontal deflection at A is denoted by (δ A ) h .
( δ A ) h = ( δ A )1 + ( δ A ) 2
=−
20.338 x 10 8
35.591 x 10 8
+
EI
EI
=+
15.253 x 10 8
EI
Horizontal deflection at B is denoted by (δ B ) h .
( δ B ) h = ( δ B )1 + ( δ B ) 2
=−
17.796 x 10 8
40.676 x 10 8
+
EI
EI
=+
22.88 x 10 8
EI
The radial deflection at gear A and B consists of two components-vertical
deflection (δ ) v and horizontal component (δ ) h . The radial deflections are given
by,
δA =
[(δ A ) v ] 2 + [(δ A ) h ] 2
8
⎛ 20.3375x10
⎜⎜
EI
⎝
=
=
δB =
25.4218 x 10 8
EI
8
2
⎛ 15.253x10
⎞
⎟⎟ + ⎜⎜
EI
⎝
⎠
8
mm
[(δ B ) v ] 2 + [(δ B ) h ] 2
2
⎛ 22.88x10 8 ⎞
⎛ 21.2621x10 8 ⎞
⎟⎟
⎟⎟ + ⎜⎜
⎜⎜
EI
EI
⎠
⎝
⎠
⎝
=
=
δ max =
⎞
⎟⎟
⎠
2
31.2341 x 10 8
EI
31.2341x10 8
EI
2
mm
(0.01x10) =
31.2341x10 8
⎛ π d4
(207 000)⎜⎜
⎝ 64
d = 41.87 mm
9.14
⎞
⎟⎟
⎠
(Ans.)
0.30 Syt = 0.30 (380) = 114 N/mm2
0.18 Sut = 0.18 (600) = 108 N/mm2 (minimum)
τ max = 0.75 (108 ) = 81 N / mm 2
Refer to Fig.9.4-solu (b). The resultant bending moments are given by,
( M b )a =
(136 709 ) 2 + ( 33 333.33 ) 2 = 140 714.11 N − mm
( M b )b =
(89 871) 2 + (100 000 ) 2 =134 449.98 N − mm
M t = (1500 − 500 ) x 150 = 150 000 N − mm
d3 =
=
16
π τ max
16
π ( 81)
(k b M ) 2 + (k t M ) 2
b
t
(1.5 x 140 714.11) 2 + (150 000 ) 2
9
∴ d = 25.35 mm
θ=
584 M t l
or
G d4
(Ans.)
0.5 =
584 (150 000 ) ( 200 )
( 79 300 ) d 4
d = 25.78 mm
9.15
σc =
τ=
S yc
( fs )
Ssy
=
( fs )
Mt =
=
(Ans.)
300
= 107.14 N / mm 2
2.8
0.577 S yt
( fs )
=
0.577 (300)
= 61.82 N / mm 2
2.8
60 x 10 6 (kW )
60 x 10 6 (25)
=
= 795 774.72 N − mm
2πn
2 π ( 300 )
From Eq.(9.27),
l =
2Mt
2 (795 774.72 )
=
= 29.26 mm
τd b
( 61.82 ) ( 40 ) ( 22 )
(i)
From Eq.(9.28),
l =
4Mt
4 (795 774.72)
=
= 53.05 mm
σc d h
(107.14 ) ( 40 ) (14 )
∴
9.16
l = 53.05 mm
σc =
τ=
S yc
( fs )
Ssy
( fs )
b=h=
=
=
(Ans.)
380
= 126.67 N / mm 2
3
0.577 S yt
( fs )
=
(ii)
0.577 (380)
= 73.09 N / mm 2
3
d 50
=
= 12.5 mm
4
4
10
60 x 10 6 (kW )
60 x 10 6 (10)
=
= 477 464.83 N − mm
Mt =
2πn
2 π ( 200 )
From Eq.(9.27),
l =
2Mt
2 ( 477 464.83 )
=
= 20.9 mm
τd b
( 73.09 ) ( 50 ) ( 12.5 )
(i)
From Eq.(9.28),
l =
4Mt
4 ( 477 464.83)
=
= 24.12 mm
σc d h
(126.67 ) ( 50 ) (12.5 )
∴
9.17
size of key = 12.5x12.5x25 mm
σc =
τ=
(ii)
S yc
( fs )
Ssy
( fs )
Mt =
=
=
(Ans.)
230
= 76.67 N / mm 2
3
0.5 S yt
( fs )
=
0.5 (230)
= 38.33 N / mm 2
3
60 x 10 6 (kW )
60 x 10 6 (15)
=
= 397 887.36 N − mm
2πn
2 π ( 360 )
From Eq.(9.27),
l =
2Mt
2 (397 887.36 )
=
= 32.95 mm
τd b
( 38.33 ) ( 45 ) (14 )
(i)
From Eq.(9.28),
l =
4Mt
4 (397 887.36)
=
= 51.26 mm
σc d h
( 76.67 ) ( 45 ) ( 9 )
(ii)
11
9.18
9.19
From Eq. (9.32),
Mt =
1
1
p m l n (D 2 − d 2 ) = (6.5) (50) (8) (40 2 − 36 2 ) = 98 800 N − mm
8
8
kW =
2 π n Mt
2 π (700) (98 800)
=
= 7.24 kW
6
60 x10
60 x10 6
τ=
Ssy
( fs )
=
0.577 S yt
( fs )
=
(Ans.)
0.577 (380)
= 87.7 N / mm 2
2.5
60 x 10 6 (45)
60 x 10 6 (kW )
x 2.25 = 671 434.92 N − mm
x 2.25 =
2 π (1440 )
2πn
Mt =
From Eq. (9.41),
d2 =
9.20
8 Mt
8 (671 434.92)
=
π D N τ π (150) (8) (87.7)
d = 4.03 mm
(Ans.)
From Eq. (9.42),
2 (R 3o − R 3i ) 2 (80 3 − 47.5 3 )
Rf =
=
= 65.13 mm
3 (R o2 − R i2 ) 3 (80 2 − 47.5 2 )
From Eq. (9.43),
M t = µ Pi N R f = 0.15 (10 x10 3 )(6)(65.13) = 586 170 N − mm
kW =
9.21
τ=
2 π n Mt
2 π (100) (586 170)
=
= 6.14 kW
6
60 x10
60 x10 6
Ssy
( fs )
Mt =
=
0.5 S yt
( fs )
=
(Ans.)
0.5 (400)
= 40 N / mm 2
5
60 x 10 6 (kW )
60 x 10 6 (7.5)
x 1.5 =
x 1.5 = 149 207.76 N − mm
2πn
2 π ( 720 )
12
Diameter of shaft:
τ=
16 M t
πd
or 40 =
3
16 (149 207.76 )
π d3
∴d = 26.68 mm or 30 mm
(i)
Diameter of bolts:
For 30 mm diameter shaft,
N=3
(ii)
D = 3d = 3 x 30 = 90 mm
From Eq. (9.41),
8 Mt
8 (149 207.76 )
=
π D N τ π (90) (3) (40)
d 12 =
d1 = 5.93 or 6 mm
9.22
(iii)
Permissible stresses for shafts and keys :
τ=
Ssy
=
( fs )
S yc
σc =
( fs )
0.5 S yt
( fs )
=
S yt
( fs )
=
0.5 (240)
= 40 N / mm 2
3
=
240
= 80 N / mm 2
3
Permissible stress for pins :
S yt
σt =
( fs )
=
τ = 35 N / mm 2
240
= 80 N / mm 2
3
Shaft diameter:
Mt =
τ=
60 x 10 6 (kW )
60 x 10 6 (5)
=
= 66 314.56 N − mm
2πn
2 π ( 720 )
16 M t
πd
3
40 =
16 ( 66 314.56 )
πd3
13
∴d = 20.36 mm or 22 mm
(i)
Dimensions of key:
b=h=
d 22
=
= 5.5 or 6 mm
4
4
lh = 1.5d = 33 mm
Size of key: 6 x 6 x 33 mm
(ii)
τ =
2Mt
2 (66 314.56)
=
= 30.45 N / mm 2
dbl
( 22 ) ( 6 ) ( 33 )
From Eq.(9.28), σ c =
4Mt
4 (66 314.56)
=
= 60.90 N / mm 2
dhl
( 22 ) ( 6 ) (33 )
From Eq.(9.27),
∴ τ < 40 N / mm 2
and
Dimensions of bushes:
σ c < 80 N / mm 2
N=4
D = 4d = 4 x 22 = 88 mm
From Eq. (9.50),
D 2b =
2Mt
2 (66314.56)
=
DN
(88) (4)
Db = 19.41 or 20 mm
lb = Db = 20 mm
(iv)
Diameter of pins : P =
2 Mt
2 (66 314.56)
=
= 376.79 N
DN
(88) ( 4)
⎛l
⎞
⎛ 20
⎞
+ 5 ⎟ = 5651.81 N − mm
M b = P ⎜ b + 5 ⎟ = 376.79 ⎜
2
2
⎝
⎠
⎝
⎠
σb =
32 M b
π d 13
or
d1 = 8.96 or 9 mm
80 =
32 (5651.81)
π d 13
(iii)
14
9.23
The deflections at gear A and gear B are calculated by the principle of
superimposition.
Vertical Plane:
(i) Deflection due to a force of 200 N [Fig. 9.5-solu (a), Eq. (27) of Table
9.4]:
a = 300 mm
(δ A ) A =
=
b = 700 mm
P b x ( x 2 + b2 − l2 )
6EIl
l = 1000 mm
x = 300 mm
(200) (700) (300) ( 300 2 + 700 2 − 1000 2 )
6 E I (1000)
=−
0.294 x 1010
mm
EI
(i)
From Eq. (28) of Table 9.4,
(δ B ) A =
=
P a (l − x ) ( x 2 + a 2 − 2 l x )
6EIl
x = 600 mm
(200) (300) (1000 − 600) ( 600 2 + 300 2 − 2 x1000x 600)
6 E I (1000)
=−
0.3 x 1010
mm
EI
(ii)
(ii) Deflection due to a force of 3500 N [Fig. 9.5-solu (b), Eq. (27) of Table
9.4]:
a = 600 mm
(δ A ) B =
b = 400 mm
P b x ( x 2 + b2 − l2 )
6EIl
l = 1000 mm
x = 300 mm
15
(3500) (400) (300) ( 300 2 + 400 2 − 1000 2 )
=
6 E I (1000)
=−
(δ B ) B =
=
5.25 x 1010
mm
EI
(iii)
P b x ( x 2 + b2 − l2 )
6EIl
x = 600 mm
(3500) (400) (600) ( 600 2 + 400 2 − 1000 2 )
6 E I (1000)
=−
6.72 x 1010
mm
EI
(iv)
Vertical deflection at A is denoted by (δ A ) .
( δA ) = ( δA )A + ( δA )B
=−
0.294 x 1010
− 5.25 x 1010
+
EI
EI
=−
5.544 x 1010
EI
=−
5.544 x1010
⎛ π (50) 4
207 000 ⎜⎜
⎝ 64
⎞
⎟⎟
⎠
= 0.873 mm = (0.873x10-3) m
Vertical deflection at B is denoted by (δ B ) .
( δB ) = ( δB )A + ( δB )B
=−
0.3 x 1010
− 6.72 x 1010
+
EI
EI
=−
7.02 x 1010
EI
16
7.02x1010
=−
⎛ π (50) 4
207 000 ⎜⎜
⎝ 64
⎞
⎟⎟
⎠
= 1.105 mm = (1.105x10-3) m
From Eq. (9.57),
ωn =
g ( W1 δ1 + W2 δ 2 )
( W1 δ12 + W2 δ 22 )
[
]
]
=
(9.81) (200)(0.873)(10 −3 ) + (3500)(1.105)(10 −3 )
(200)(0.873) 2 (10 −3 ) 2 + (3500)(1.105) 2 (10 −3 ) 2
=
9.81(4042.1)(10) −3
4426.01(10) −6
[
= 94.65 rad/s
=
(1 revolution = 2 π radians)
94.65(60)
= 903.86 r.p.m.
2π
(Ans.)
1
CHAPTER 10
10.1
The permissible shear stress is given by,
τ = 0.5 Sut = 0.5(1000) = 500 N/mm2
From Eq. (10.7),
K=
4C − 1 0.615 4 (6) − 1 0.615
+
=
+
= 1.2525
4C − 4
C
4(6) − 4
6
From Eq. (10.13),
⎛ 8PC ⎞
τ = K⎜ 2 ⎟
⎝ πd ⎠
or
⎧ 8(500)(6) ⎫
500 = (1.2525)⎨
⎬
2
⎩ πd
⎭
∴d= 4.37 or 5 mm
(1)
D = Cd = 6(5) = 30 mm
(2)
From Eq. (10.8),
δ=
8PD 3 N
Gd 4
or 20 =
8(500)(30) 3 N
(81370)(5) 4
∴N= 9.42 or 10 coils
(3)
The spring has square and ground ends. The number of inactive coils is 2.
Therefore,
Nt = N+2 = 10+2 = 12 coils
(4)
The actual deflection of the spring is given by,
δ=
8PD 3 N 8(500)(30) 3 (10)
=
= 21.24 mm
Gd 4
(81370)(5) 4
solid length of spring = Nt d = 12(5) = 60 mm
There is a gap of 1 mm between consecutive coils when the spring is
subjected to the maximum force. The total number of coils is 12.
The total axial gap between the coils is (12-1) x 1 = 11 mm.
2
free length
= solid length + total axial gap + δ
= 60 + 11 + 21.24
= 92.24
pitch of coil =
10.2
(5)
free length 92.24
=
= 8.39 mm
( N t − 1) (12 − 1)
(6)
There are six springs in parallel. The force acting on each spring is given
by,
P=
1500
=250 N
6
The permissible shear stress for the spring wire is given by,
τ = 0.5 Sut = 0.5 (1200) = 600 N/mm2
K=
4C − 1 0.615
4x 6−1 0.615
+
+
=
= 1.2525
4C − 4
C
4x 6−4
6
From Eq. (10.13),
⎛ 8PC ⎞
⎟
⎝ πd 2 ⎠
τ= K ⎜
or
⎧ 8(250)(6) ⎫
⎬
2
⎩ πd
⎭
600 = (1.2525) ⎨
∴ d = 2.82 or 3 mm
(1)
D = C d = 6 (3) = 18 mm
(2)
From Eq. (10.8),
δ =
8 PD 3 N
Gd 4
or
∴ N = 5.65 or 6 coils
10 =
8(250)(18) 3 N
(81370)(3) 4
(3)
The springs have square and ground ends. The number of inactive coils is
2. Therefore,
3
Nt = N + 2 = 6+2 = 8
(4)
The actual deflection of the spring is given by,
δ =
8 PD 3 N 8(250)(18) 3 (6)
=
= 10.62 mm
Gd 4
(81370)(3) 4
solid length = Ntd = 8(3) = 24 mm
(5)
There is a gap of 1 mm between the adjacent coils when the spring is
subjected to the maximum force of 250N. The total number of coils is 8.
Therefore, the total axial gap is (8-1) x1 = 7 mm.
free length = solid length + total axial gap + δ
= 24 + 7 + 10.62
= 41.62
(6)
The required stiffness of the spring is given by,
k=
P 250
=
=25 N / mm
δ 10
(7)
The actual stiffness of spring is given by,
k=
10.3
Gd 4 (81370)(3) 4
=
= 23.54 N / mm
8D 3 N 8(18) 3 (6)
(8)
The length of scale of the pointer is 75 mm. In order words, the spring
deflection is 75 mm when the force is 500 N.
The permissible shear stress for spring wire is given by,
τ = 0.5 Sut = 0.5 (1400) = 700 N/mm2
K=
4C − 1 0.615
4 x 6 − 1 0.615
+
+
= 1.2525
=
4C − 4
C
4 x6−4
6
4
From Eq. (10.13),
⎛ 8PC ⎞
τ = K ⎜ 2⎟
⎝ πd ⎠
or
⎧ 8(500 )(6 )⎫
700 = 1.2525 ⎨
⎬
2
⎩ πd
⎭
∴ d = 3.7 or 4 mm
(i)
D = Cd = 6 (4) = 24 mm
(ii)
From Eq. (10.8),
8 PD 3 N
δ=
Gd 4
8(500 )(24 ) N
3
or
75 =
(81370)(4)4
∴ N = 28.25 or 29 coils
(iii)
The required spring rate is given by,
k=
P 500
=
= 6.67 N / mm
δ 75
(iv)
The actual spring rate is given by,
(81370)(4) = 6.5 N / mm
Gd 4
=
3
8D N 8(24 )3 (29 )
4
k=
10.4
(v)
The kinetic energy of the moving wagon is absorbed by the springs. The
kinetic energy of the wagon is given by,
1
1
2
KE = mv 2 = (1000 )(2 ) =2000 J or N-m =(2000x103 ) N-mm
2
2
(a)
Suppose P is the maximum force acting on each spring and causing it to
compress by 150 mm. The strain energy absorbed by two springs is given
by,
5
⎡1
⎤
⎡1
⎤
E = 2 ⎢ P δ ⎥ = 2 ⎢ P (150 )⎥ = (150P) N-mm
⎣2
⎦
⎣2
⎦
(b)
The strain energy absorbed by two springs is equal to the kinetic energy of
the wagon. Therefore,
(150 P) = 2000 x 103
P = 13 333.33 N
(i)
The permissible shear stress for spring wire is given by,
τ = 0.5 (1500) = 750 N / mm2
K=
4C − 1 0.615
4(6) − 1 0.615
+
= 1.2525
+
=
4C − 4
C
4(6) − 4
6
From Eq. (10.13),
⎛ 8PC ⎞
τ = K ⎜ 2⎟
⎝ πd ⎠
or
⎧ 8(13 333.33)(6) ⎫
⎬
πd 2
⎩
⎭
750 = (1.2525) ⎨
∴ d = 18.44 or 19 mm
(ii)
D = Cd = 6 (19) = 114 mm
(iii)
From Eq, (10.8),
δ =
8 PD 3 N
Gd 4
or
∴ N = 10.07 or 11 coils
150 =
8(13 333.33)(114) 3 N
(81370)(19) 4
(iv)
6
10.5
The permissible shear stress for spring wire is given by,
τ = 0.5 Sut = 0.5 (1000) = 500 N/mm2
K=
4C − 1 0.615
4x8−1 0.615
+
= 1.184
+
=
4C − 4
C
4x8−4
8
From Eq. (10.13),
⎛ 8PC ⎞
τ = K ⎜ 2⎟
⎝ πd ⎠
or
⎧ 8(1000 )(8) ⎫
500 = 1.184 ⎨
⎬
2
⎩ πd
⎭
∴ d = 6.95 or 7 mm
(1)
D = C d = 8 (7) = 56 mm
(2)
From Eq. (10.8),
8 PD 3 N
δ=
Gd 4
8(1000 − 500 )(56 ) N
3
or
25 =
(81370)(7 )4
∴ N = 6.95 or 7 coils
(3)
The spring has square and ground ends. The number of inactive coils is 2.
Therefore,
Nt = N + 2 = 7+2 = 9
(4)
solid length = Ntd = 9(7) = 63 mm
(5)
The actual deflection of the spring is given by,
δ =
8 PD 3 N 8(1000 )(56) 3 (7)
=
= 50.34 mm
Gd 4
(81370)(7) 4
Gap = (9 – 1)(2) = 16 mm
free length = solid length + total axial gap + δ
= 63 + 16 + 50.34
7
= 129.34 mm
(6)
The required spring rate is given by,
k=
P 1000 − 500
=
= 20 N / mm
25
δ
(7)
The actual spring rate is given by,
(81370)(7 ) = 19.87 N / mm
Gd 4
=
3
3
8D N
8(56 ) (7 )
4
k=
10.6
P1 =
π
(40)2 (1.2)= 1507.96 N
4
(8)
δ1 = 20 mm
δ 2 = 20 +12 = 32 mm
P2 δ 2
=
P1
δ1
∴
P2
32
=
1507.96 20
P2 = 2412.74 N
The permissible shear stress for the spring wire is given by,
τ = 0.5 Sut = 0.5 (1400) = 700 N/mm2
K =
4C − 1 0 . 615 4(6) − 1 0.615
+
+
=
= 1.2525
4C − 4
C
4(6) − 4
6
From Eq. (10.13),
⎛ 8PC ⎞
τ = K⎜ 2 ⎟
⎝ πd ⎠
or
⎧ 8(2412.74)(6)⎫
⎬
πd 2
⎭
⎩
700 = 1.2525 ⎨
∴d = 8.12 or 9 mm
(i)
D= Cd = 6(9) = 54 mm
(ii)
From Eq. (10.8),
δ=
8 PD 3 N
Gd 4
8(1507.96 )(54 ) N
3
or
20 =
(81370)(9)4
8
∴ N = 5.62 or 6 coils
10.7
(iii)
60x10 6 (kW ) 60x10 6 (25)
=
= 238 732.41 N − mm
Mt =
2πn
2 π (1000)
There are two pairs of contacting surfaces and the torque transmitted by
each pair is (238 732.41/2) or 119 366.2 N-mm. Assuming uniform-wear
theory (Chapter 11), the total normal force Pt required to transmit the torque
is given by,
Pt =
4 (M t ) f
4 (119 366.2)
=
= 3589.96 N
µ (D + d ) 0.35 (2 x190)
Since there are six springs, the force exerted by each spring is
P=
3589.96
= 598.33 N
6
From Eq. (10.7),
K=
4C − 1 0.615 4 (6) − 1 0.615
+
=
+
= 1.2525
4C − 4
C
4(6) − 4
6
From Eq. (10.13),
⎛ 8PC ⎞
τ = K⎜ 2 ⎟
⎝ πd ⎠
or
τ =
or
⎡ 8 (598.33) (6) ⎤
τ = (1.2525) ⎢
⎥
πd2
⎣
⎦
11450.12
N / mm 2
2
d
(a)
The permissible shear stress is denoted by τ d in order to differentiate it
from induced stress τ . It is given by,
τ d = 0.5 Sut
Equations (a) and (b) are solved by the trial and error method.
(b)
9
From Table 10.1 (Grade-2),
Sut
τ d = 0.5 S ut
2.5
1640
820
1832.02
3.0
1570
785
1272.24
3.6
1510
755
883.50
4.0
1480
740
715.63
d (mm)
Therefore,
10.8
d = 4 mm
τ=
11450.12
d2
(Ans.)
⎛π ⎞
⎛π
⎞
Pmax = ⎜ d 2 ⎟ p max = ⎜ (25) 2 ⎟ (1) = 490.87 N
4
4
⎠
⎝
⎠
⎝
Assume
D = 15 mm
Trial 1
d = 3.6 mm
S ut = 1400 N / mm 2
From Table 10.2,
τ d = 0.5 Sut = 0.5(1400) = 700 N/mm2
C=
K=
D 15
=
= 4.167
d 3.6
4C − 1 0.615 4 (4.167) − 1 0.615
+
=
+
= 1.3844
4C − 4
C
4(4.167) − 4 4.167
⎛ 8PD ⎞
τ = K⎜ 3 ⎟
⎝ πd ⎠
or
Therefore,
τ < τd
The design is safe.
⎡ 8 (490.87) (15) ⎤
2
τ = (1.3844) ⎢
⎥ = 556.35 N / mm
3
⎣ π (3.6)
⎦
10
Trial 2
d = 3 mm
S ut = 1430 N / mm 2
From Table 10.2,
2
τ d = 0.5 Sut = 0.5(1430) = 715 N/mm
C=
K=
D 15
=
=5
d
3
4C − 1 0.615 4 (5) − 1 0.615
+
=
+
= 1.3105
4C − 4
C
4(5) − 4
5
⎛ 8PD ⎞
τ = K⎜ 3 ⎟
⎝ πd ⎠
or
⎡ 8 (490.87) (15) ⎤
2
τ = (1.3105) ⎢
⎥ = 910.06 N / mm
3
π (3)
⎣
⎦
Therefore,
τ > τd
The design is not safe.
From Trial No. 1 and 2,
d = 3.6 mm and
D = 15 mm
(i), (ii)
⎛π ⎞
⎛π
⎞
Pmin = ⎜ d 2 ⎟ p min = ⎜ (25) 2 ⎟ (0.25) = 122.72 N
⎠
⎝4
⎠
⎝4
k=
Pmax − Pmin
490.87 − 122.72
=
= 61.36 N / mm
lift
6
Gd 4
k=
8D 3 N
N = 8.25 or 9
61.36 =
(iii)
(81370)(3.6)4
3
8(15) N
(iv)
11
10.9
C=5
K=
From Eq. (10.7) and (10.5),
4C − 1 0.615 4 (5) − 1 0.615
+
=
+
= 1.311
4C − 4
C
4(5) − 4
5
Ks = 1 +
0.5
0.5
=1+
= 1.1
C
5
Pm =
1
1
( Pmax + Pmin ) = (300 + 50) = 175 N
2
2
Pa =
1
1
( Pmax − Pmin ) = (300 − 50) = 125 N
2
2
From Eq. (10.18) and (10.19),
⎛ 8P D ⎞
⎛ 8 (175)(25) ⎞
⎟⎟ = 98.04 N / mm 2
τ m = K s ⎜⎜ m 3 ⎟⎟ = (1.1) ⎜⎜
3
(
5
)
d
π
π
⎠
⎠
⎝
⎝
⎛ 8 (125)(25) ⎞
⎛ 8P D ⎞
⎟⎟ = 83.46 N / mm 2
τ a = K ⎜⎜ a 3 ⎟⎟ = (1.311) ⎜⎜
3
⎝ π (5)
⎠
⎝ πd ⎠
From. Eq. (10.21), the relationships for oil-hardened and tempered steel
wire are as follows,
2
Sse' = 0.22 Sut = 0.22(1440) = 316.8 N/mm
Ssy = 0.45 Sut = 0.45(1440) = 648 N/mm2
1 '
1
Sse = (316.8) =158.4 N / mm 2
2
2
From Eq. (10.22),
τa
⎛ Ssy ⎞
⎜⎜
⎟⎟ − τ m
⎝ fs ⎠
1 '
Sse
2
=
1
Ssy − Sse'
2
158.4
83.46
=
648 − 158.4
⎛ 648 ⎞
⎟ − 98.04
⎜
⎝ fs ⎠
12
(fs)= 1.82
(Ans.)
10.10 Suppose suffix ‘i’ and ‘o’ refer to inner and outer spring respectively.
Di = 64 mm
di = 8 mm
Ni = 10 coils
Do = 80 mm
do = 10 mm
No = 15 coils
Stiffness of springs:
From Eq. (10.9),
ko =
G d o4
(81370) (10) 4
=
= 13.24 N / mm
8 D 3o N o
8 (80) 3 (15)
ki =
G d i4
(81370) (8) 4
=
= 15.89 N / mm
8 D 3i N i
8 (64) 3 (10)
k = k o + k i = 13.24 + 15.89 = 29.13 N / mm
(i)
(ii)
10.11 From Eq. (10.32),
σb =
12 M
b t2
or
750 =
12 (1000)
(10 t ) t 2
t = 1.17 or 1.2 mm
(i)
b = 10 t = 12 mm
(ii)
θ = 2.5 revolutions = 2.5 (2 π) = 5 π
radians
From Eq. (10.33),
θ=
∴
12 M l
E b t3
or
5π=
12 (1000) l
(207 000) (12) (1.2) 3
l = 5618.68 mm or 5.62 m
(iii)
13
10.12 2 L = 1x103 mm
L = 500 mm
From Eq. (10.41),
σb =
6PL
n bt2
or
(350) =
6 P (500)
(8) (50) (7.5) 2
P = 2625 N
2 P = 5250 N
10.13 2 P = 10x103 N
2 L = 1x103 mm
(Ans.)
P = 5000 N
L = 500 mm
From Eq. (10.41),
σb =
6PL
n bt2
or
(350) =
t = 9.26 or 10 mm
6 (5000) (500)
(10) (50) t 2
(i)
From Eq. (10.38),
δ=
12 P L3
12 (5 000) (500) 3
=
E b t 3 (3 n f + 2 n g ) (207 000) (50) (10) 3 (3 x 2 + 2 x 8)
δ = 32.94 mm
10.14 2 P = 30x103 N
2 L = 1x103 mm
σb =
S yt
(fs)
=
(ii)
P = 15 000 N
L = 500 mm
1500
= 750 N / mm 2
2
From Eq. (10.41),
14
6PL
σb =
n bt2
or
6 (15000) (500)
(750) =
(10) b t 2
b t2 = 6000
Let
b = 32 mm
Cross-section = 32x14 mm
t2 =
6000
32
or
t = 13.69 or 14 mm
(i)
From Eq. (10.38),
δ=
12 P L3
12 (15 000) (500) 3
=
E b t 3 (3 n f + 2 n g ) (207 000) (32) (14) 3 (3 x 2 + 2 x 8)
δ = 56.27 mm
(ii)
1
CHAPTER 11
11.1
Uniform wear theory:
From Eq. (11.8),
Mt =
µP
(0.25) (15 x10 3 )
(D+d) =
(250 + 125) = 351 562.5 N-mm
4
4
kW =
2πn Mt
2 π (500) (351 562.5)
=
= 18.41
6
60 x 10
60 x 10 6
(i)
Uniform pressure theory: From Eq. (11.5),
11.2
Mt =
µ P ( D3 − d 3 )
( 0.25 ) (15x10 3 ) ( 250 3 − 125 3 )
=
= 364 583.33 N-mm
3 ( D2 − d 2 )
3
( 250 2 −125 2 )
kW =
2πn Mt
2 π (500) (364 583.33)
= 19.09
=
6
60 x 10
60 x 10 6
(ii)
There are two pairs of contacting surfaces. Therefore, the torque transmitted by one
pair of contacting surfaces is given by,
Mt =
531x10 3
= 265 500 N − mm
2
Mt =
πµ pa d 2
(D − d 2 )
8
From Eq. (11.7),
(265 500 ) =
π ( 0.3 ) ( 0.3 ) d
(270 2 − d 2 )
8
d (270 2 − d 2 ) = 7 512 113.314
The above equation is solved by trial and error method. It is a cubic equation.
d
d ( 2702 – d2)
d = 167 mm
Check:
166
167
168
7 527 104
7 516 837
7 505 568
(i)
2
πµ pa d 2
π ( 0.3 ) ( 0.3 ) (167)
(270 2 −167 2 )
Mt =
(D − d 2 ) =
8
8
= 265 660.95 N-mm
(> 265 500 N-mm) (O.K.)
From Eq. (11.8),
P=
11.3
4 Mt
4 (265 500)
=
= 8100.69 N
µ (D + d) 0.3 (270 + 167)
(ii)
There are two steel plates and one bronze plate. The total number of plates is 3.
number of disks = z +1 = 3
or
z=2
From Eq. (11.6),
P=
π pa d
π (0.4) (200)
(250 − 200 ) = 6283.19 N
(D − d ) =
2
2
(i)
From Eq. (11.10),
11.4
Mt =
µPz
(0.1) (6283.19) (2)
(D+d) =
(250 + 200) = 141 371.78 N-mm
4
4
kW =
2πn Mt
2 π (720) (141 371.78)
= 10.66
=
6
60 x 10
60 x 10 6
From Eq. (11.6),
P=
π pa d
π (0.5) (100)
(200 − 100 ) = 7853.98 N
(D − d ) =
2
2
60 x10 6 (kW ) 60 x10 6 (15)
Mt =
=
= 99 471.84 N − mm
2πn
2 π (1440)
From Eq. (11.8),
(ii)
3
4 Mt
4 (99 471.84)
z=
=
= 1.13 or 2
µ P (D + d ) 0.15 (7853.98) (200 + 100)
Number of plates = z +1 = 2 +1 = 3
Two steel plates and one bronze plate
11.5
D+d
= 300
2
(Ans.)
D + d = 600 mm
(a)
From Eq.(11.19),
b=
D−d
2 sin α
100 =
D−d
2 sin (12.5 0 )
D – d = 43.29 mm (b)
From (a) and (b),
D = 321.64 mm
d = 278.36 mm
From Eq. (11.6),
P=
π pa d
π (0.07) (278.36)
(D − d ) =
(321.64 − 278.36 ) = 1324.68 N (i)
2
2
From Eq. (11.18),
11.6
Mt =
µ P (D + d ) (0.2) (1324.68) (321.64 + 278.36)
=
= 183 609.64 N − mm
4 sin α
4 sin(12.5 o )
kW =
2πn Mt
2 π (500) (183 609.64)
=
= 9.61
6
60 x 10
60 x 10 6
ω2 =
2πn2
2 π (720)
=
= 75.40 rad / s
60
60
ω1 = 0.75 ω1 = 0.75 (75.40) = 56.55 rad / s
Mt =
60 x 10 6 ( kW )
60 x 10 6 (18.5 )
=
= 245 363.87 N-mm
2πn2
2 π (720)
(ii)
4
From Eq. (11.23),
m=
1000 M t
1000 ( 245 363.87 )
=
2
2
µ rg rd z (ω 2 − ω1 )
0.25 (140) (165) (4) (75.4 2 − 56.55 2 )
= 4.27 kg
(Ans.)
1
CHAPTER 12
12.1
v1 = 95 km / hr =
deceleration =
95 x 10 3
m / s = 26.39 m / s
60 x 60
( v1 − v 2 )
t
6=
and
v2 = 0
(26.39 − 0)
t
t = 4.4 s
v ave =
(i)
1
1
v1 = (26.39) m / s
2
2
distance = v ave t =
1
(26.39) (4.4) = 58.06 m
2
(ii)
KE of the vehicle = Ev
1
1 ⎛ 13 . 5 x10 3
⎜
m ( v 12 − v 22 ) =
2
2 ⎜⎝
9 . 81
Ev =
(E t ) per brake =
3
Et ⎛ 1 ⎞
⎛ 1 ⎞ ⎛ 13.5x10
= ⎜ ⎟(1.1) ⎜ ⎟ ⎜⎜
4
⎝ 2 ⎠ ⎝ 9.81
⎝4⎠
= 131 779 J
ωave =
⎞
⎟⎟ ( 26 . 39 ) 2
⎠
E t = 1.1 E v
⎞
⎟⎟ ( 26.39 ) 2
⎠
(iii)
v ave
26.39
=
rad / s
r
2 (0.375)
θ = ω ave t =
26.39
(4.4) = 154.82 rad
2 (0.375)
Mt =
E t 131 779
=
= 851.18 N − m
θ
154.82
(iv)
∆t =
Et
131 779
=
= 28.65 0
m c 10 (460)
(v)
2
12.2
From Eq. (12.12),
h=
12.3
4 R sin θ
4 (150) sin (45 0 )
=
= 165 mm
2 θ + sin 2 θ
⎛ 2 x 45 π ⎞
0
⎜
⎟ + sin(90 )
⎝ 180 ⎠
(Ans.)
From Eq. (12.6),
N=
Mt
(15 x 10 3 )
=
= 333.33 N
µR
0.3 (150)
The free-body diagram of forces for clockwise rotation of the drum is shown in Fig.
12.1-solu. Taking moments about the hinge-pin,
µ N (60) + P (650) − N (200) = 0
or
0.3(333.33)(60) + P(650) – 333.33(200) = 0
∴ P = 93.33 N
N=lwp
(i)
333.33 = w2 (1)
w = l = 18.26 mm
(ii)
R X = µ N = 0.3 (333.33) = 100 N
R Y = N − P = 333.33 − 93.33 = 240 N
R=
(100) 2 + (240) 2 = 260 N
(iii)
⎛2 π n⎞
⎡ 2 π (50) ⎤
⎟r=⎢
⎥ (0.15) = 0.7854 m/s
⎝ 60 ⎠
⎣ 60 ⎦
Initial velocity = v1 = ω r = ⎜
Final velocity = 0
v ave =
1
1
v1 = (0.7854) = 0.3927 m/s
2
2
Rate of heat generated = µ N v ave = 0.3 (333.33) (0.3927)
= 39.27 N-m/s or W
(iv)
3
12.4
R = 150 mm
h = 125 mm
C = 177 mm
θ1 = 0
θ 2 = 90 0
φ max = 90 0
w = 50 mm
pmax = 0.8 N/mm2
µ = 0.4
sin φ max = 1
From Eq. (12.19),
Mf =
=
µ p max R w [4R (cos θ1 − cos θ 2 ) − h (cos 2θ1 − cos 2θ 2 )]
4 sin φ max
[
0.4 (0.8)(150)(50) 4(150)(1 − cos 90 0 ) − 125(1 − cos 180 0 )
4(1)
]
= 210 000 N-mm
From Eq. (12.20),
Mn =
p max R w h [ 2(θ 2 − θ1 ) − (sin 2θ 2 − sin 2θ1 )]
4 sin φ max
⎡ ⎛ 90π ⎞
0 ⎤
0.8 (150)(50)(125) ⎢2 ⎜
⎟ − sin(180 )⎥
⎣ ⎝ 180 ⎠
⎦
=
4(1)
= 589 048.62 N-mm
From Eq. (12.22),
P=
`
Mn − Mf
589 048.62 − 210 000
=
= 2141.52 N
C
177
(i)
From Eq. (12.21), the torque (Mt)R for the right hand shoe is given by
(M t ) R =
=
µ R 2 p max w (cos θ1 − cos θ 2 )
sin φ max
0.4(150) 2 (0.8)(50)(1 − cos 90 0 )
1
= 360 000 N-mm
4
The maximum intensity of pressure for the left hand shoe is unknown. For identical
shoes, it can be seen from the expressions of Mn and Mf, that both are proportional to
( p max ). For left hand shoe, the maximum intensity of pressure is taken as ( p 'max ).
Therefore, for the left hand shoe,
M 'f =
210 000 p 'max (210 000) p 'max
=
= 262 500 p 'max
p max
(0.8)
Similarly,
M 'n =
589 048.62 p 'max (589 048.62) p 'max
=
= 736 310.78 p 'max
p max
(0.8)
For the left-hand shoe,
P=
M 'n + M 'f
C
∴
2141.52 =
or
( 262 500 + 736 310.78 ) p 'max
177
p 'max = 0.3795 N / mm 2
Since the shoes are identical,
⎛ p'
( M t ) L = 360 000 ⎜⎜ max
⎝ p max
⎞
0.3795 ⎞
⎟⎟ = 360 000 ⎛⎜
⎟ = 170 775 N − mm
⎝ 0.8 ⎠
⎠
The total torque-absorbing capacity of the brake is given by,
M t = 360 000 + 170 775 = 530 775 N-mm
M t = 530.78 N-m
12.5
(ii)
From Eq. (12.27),
P1 = R w p max = 250 (60) (0.25) = 3750 N
e µθ = e
⎛ 225 ⎞
0. 4 ⎜
⎟π
⎝ 180 ⎠
= 4.81
P1
= 4.81
P2
(i)
5
P
3750
= 779.63 N
P2 = 1 =
4.81 4.81
(i)
P x 750 = P2 x 250 = 779.63(250)
P = 259.88 N
(ii)
M t = ( P1 − P2 ) R = (3750 − 779.63) (0.25) = 742.59 N − m (iii)
1
CHAPTER 13
13.1
v=
πd n
π (300) (720)
=
= 11.31 m / s
60 (1000)
60 (1000)
(P1 − P2 ) =
1000 (kW ) 1000 (35)
=
= 3094.61 N
v
11.31
P1 = P2 + 3094.61
(a)
m v2 = 2(11.31)2 = 255.83
⎛ D− d ⎞
⎟⎟ = 180 − 2 sin −1
α s = 180 − 2 sin −1 ⎜⎜
2
C
⎝
⎠
ef α = e
⎛ 162.75 ⎞
( 0.35 ) ⎜
⎟π
⎝ 180 ⎠
⎛ 900 − 300 ⎞
⎜⎜
⎟⎟ = 162.750
2
x
2000
⎝
⎠
= 2.70
From Eq. (13.6),
P1 − m v 2
= ef α
P2 − m v 2
or
P1 − 255.83
= 2.70
P2 − 255.83
P1 – 2.70 P2 + 434.91 = 0
(b)
From (a) and (b),
P1 = 5170.80 N
and
P2 = 2076.19 N
(i)
From Eq. (13.3),
L=2C+
π( D + d ) ( D − d )2
+
2
4C
= 2 (2000) +
π ( 900 + 300 ) (900 − 300 ) 2
+
2
4 (2000)
= 5929.96 mm
(ii)
13.2
2
πd n
π (250) (1000)
v=
=
= 13.09 m / s
60 (1000)
60 (1000)
(P1 − P2 ) =
1000 (kW ) 1000 (7.5)
=
= 572.96 N
v
13.09
P1 = P2 + 572.96
(a)
m v2 = 0.55(13.09)2 = 94.24
⎛ D+ d ⎞
⎟⎟ = 180 + 2 sin −1
α s = 180 + 2 sin −1 ⎜⎜
2
C
⎝
⎠
ef α = e
⎛ 208.96 ⎞
( 0.3 ) ⎜
⎟π
⎝ 180 ⎠
⎛ 500 + 250 ⎞
⎜⎜
⎟⎟ = 208.960
2
x
1500
⎝
⎠
= 2.99
From Eq. (13.6),
P1 − m v 2
= ef α
P2 − m v 2
or
P1 − 94.24
= 2.99
P2 − 94.24
P1 – 2.99 P2 + 187.54 = 0
(b)
From (a) and (b),
P1 = 955.12 N
and
P2 = 382.16 N
(i)
From Eq. (13.5),
L=2C+
π ( D + d ) ( D + d )2
+
2
4C
= 2 (1500) +
π ( 500 + 250 ) (500 + 250 ) 2
+
2
4 (1500)
= 4271.85 mm
13.3
d=
(ii)
60 (1000) v
60 (1000) (13)
=
= 248.28 mm (or 250 mm )
πn
π (1000)
d = 250 mm
(i)
3
250 (1000)
D=
= 500 mm
(500)
(i)
The correct belt velocity is given by,
v=
πd n
π ( 250 ) (1000)
=
= 13.09 m / s
60 (1000)
60 (1000)
From Eq. (13.5),
L=2C+
π ( D + d ) ( D + d )2
+
2
4C
= 2 (1500) +
π ( 500 + 250 ) ( 500 + 250 ) 2
+
2
4 (1500)
= 4271.85 mm
(ii)
⎛ D+ d ⎞
⎟⎟ = 180 + 2 sin −1
α s = 180 + 2 sin −1 ⎜⎜
2
C
⎝
⎠
e
fα
=e
⎛ 208.96 ⎞
( 0.3) ⎜
⎟π
⎝ 180 ⎠
⎛ 500 + 250 ⎞
⎜⎜
⎟⎟ = 208.960
2
x
1500
⎝
⎠
= 2.99
⎛ b ⎞⎛ 6 ⎞
-3
⎟ ⎜ ⎟ = (5.7x10 ) b
⎝ 10 ⎠ ⎝ 10 ⎠
m = (0.95x10-3 )(100) ⎜
m v 2 = (5.7 x10 −3 ) b (13.09) 2 = (0.977 b)
From Eq. (13.6),
P1 − m v 2
= ef α
P2 − m v 2
or
P1 − 0.977 b
= 2.99
P2 − 0.977 b
P1 – 2.99 P2 + 1.944 b = 0
(a)
From Eq. (13.8),
P1 − P2 =
1000(kW ) 1000(7.5)
=
= 572.96
v
13.09
P1 = σ A = 1.75(6 b) = (10.5 b)
(b)
(c)
4
From (a), (b) and (c),
b = 90.4 mm
(ii)
P1 = 949.2 N
13.4
Assume
d=
P2 = 376.24 N
(iii)
v = 18 m/s
60 (1000) v
60 (1000) (18)
= 238.73 mm (or 250 mm )
=
πn
π (1440)
d = 250 mm
D=
(i)
250 (1440)
= 1000 mm
(360)
(i)
The correct belt velocity is given by,
v=
∴
πd n
π ( 250 ) (1440)
= 18.85 m / s
=
60 (1000)
60 (1000)
17.8 < v < 22.9 m/s
(o.k.)
Maximum power = 10 (load factor) = 10 (1.2) = 12 kW
⎛ D− d ⎞
⎟⎟ = 180 − 2 sin −1
α s = 180 − 2 sin −1 ⎜⎜
2
C
⎠
⎝
⎛ 1000 − 250 ⎞
⎟⎟ = 158.390
⎜⎜
2
x
2000
⎠
⎝
Assuming linear interpolation of values in Table 13.2,
Fd = 1.08 +
(1.13 − 1.08)
(160 − 158.39) = 1.088
(160 −150)
Corrected power = 1.088(12) = 13.056 kW
⎛ 18.85 ⎞
⎟ = 0.04379
⎝ 5.08 ⎠
Corrected belt rating = (0.0118) ⎜
⎛ 13.056 ⎞
⎟ = 298.15
⎝ 0.04379 ⎠
(width x number of plies) = ⎜
5
⎛ 298.15 ⎞
w= ⎜
⎟ = 74.54 or 76 mm (standard size)
⎝ 4 ⎠
4 ply
Belt specification = 4 ply x 76 mm wide belting
(ii)
From Eq. (13.3),
L=2C+
π( D + d ) ( D − d )2
+
2
4C
= 2 (2000) +
π (1000 + 250 ) (1000 − 250 ) 2
+
2
4 (2000)
= 6033.81 or 6035 mm
13.5
Assume
d=
(iii)
v = 18 m/s
60 (1000) v
60 (1000) (18)
= 429.72 mm (or 450 mm standard size)
=
πn
π (800)
d = 450 mm
D=
(i)
450 (800)
= 900 mm
(400)
(i)
The correct belt velocity is given by,
v=
∴
πd n
π ( 450 ) ( 800)
= 18.85 m / s
=
60 (1000)
60 (1000)
17.8 < v < 22.9 m/s
(o.k.)
Maximum power = 30 (load factor) = 30 (1.3) = 39 kW
⎛ D− d ⎞
⎟⎟ = 180 − 2 sin −1
α s = 180 − 2 sin −1 ⎜⎜
⎝ 2C ⎠
⎛ 900 − 450 ⎞
⎟⎟ = 171.40
⎜⎜
⎝ 2 x 3000 ⎠
Assuming linear interpolation of values in Table 13.2,
Fd = 1 +
(1.04 − 1)
(180 − 171.4) = 1.034
(180 − 170)
6
Corrected power = 1.034(39) = 40.33 kW
⎛ 18.85 ⎞
⎟ = 0.0545
⎝ 5.08 ⎠
Corrected belt rating = (0.0147) ⎜
⎛ 40.33 ⎞
⎟ = 740
⎝ 0.0545 ⎠
(width x number of plies) = ⎜
⎛ 740 ⎞
⎟ = 185 mm
⎝ 4 ⎠
4 ply
w= ⎜
5 ply
w= ⎜
⎛ 740 ⎞
⎟ = 148 or 152 mm (standard size)
⎝ 5 ⎠
Belt specification = 5 ply x 152 mm wide belting
(ii)
From Eq. (13.3),
L=2C+
π( D + d ) ( D − d )2
+
2
4C
= 2 (3000) +
π ( 900 + 450 ) (900 − 450 ) 2
+
2
4 (3000)
= 8137.45 or 8138 mm
13.6
(iii)
In this application, an induction motor is driving a compressor of 20 kW capacity for
15 hr per day. From Table 13.15, the correction factor according to service (Fa) is 1.2.
Therefore,
Design power = Fa (transmitted power) = 1.2(20) = 24 kW
Plot a point with coordinates 24 kW and 1440 r.p.m. speed in Fig. 13.24. It is
observed that the point is located in the region of C – section belt. Therefore, for this
7
application the cross-section of V-belt is C. From Table13.12, the minimum pitch
diameter for the smaller pulley of C section is 315 mm.
Speed ratio =
1440
=3
480
d = 315 mm and
D = 3(315) = 945 mm
(ii)
From Eq. (13.3),
L=2C+
π( D + d ) ( D − d )2
+
2
4C
= 2 (1200) +
π (945 + 315 ) (945 − 315 ) 2
+
2
4 (1200)
= 4461.89 mm
From Table 13.14, the preferred pitch length for C- section belt is 4060 or 4600 mm.
It is assumed that the pitch length of the belt is 4600 mm.
Belt specification = C section 4600 mm pitch length V belt
(i)
Substituting this value of pitch length in Eq.(13.3),
L = 2C+
π( D + d ) ( D − d )2
+
2
4C
4600 = 2 C +
π (945 + 315) (945 − 315 ) 2
+
2
4C
Simplifying the above expression,
C 2 − 1310.4 C + 49 612.5 = 0
C=
1310.4 ± 1310.4 2 − 4 ( 49 612.5)
2
= 1271.38 mm
The correct center distance is 1271.38 mm.
From Table 13.21 (C-section and 4600 mm pitch length),
(iii)
8
Fc = 1.05
From Eq. (13.1),
⎛ D− d ⎞
⎟⎟ = 180 − 2 sin −1
α s = 180 − 2 sin −1 ⎜⎜
⎝ 2C ⎠
⎛ 945 − 315 ⎞
⎜⎜
⎟⎟ = 151.310
⎝ 2 x 1271.38 ⎠
From Table 13.22, Fd is approximately 0.93.
From Table 13.18, ( 1440 r.p.m., 315 mm pulley, C-section) ( speed ratio = 3)
Pr = 14.76 + 1.27 = 16.03 kW
From Eq.(13.18),
13.7
Number of belts =
P x Fa
20 (1.2)
=
Pr x Fc x Fd
16.03 (1.05) (0.93)
Number of belts
= 1.53 or 2 belts
(iv)
In this application, an induction motor is driving a centrifugal pump for a service of
24 hr per day. From Table 13.15, the correction factor according to service (Fa) is 1.3.
Therefore,
Design power = Fa (transmitted power) = 1.3(15) = 19.5 kW
Plot a point with coordinates 19.5 kW and 1440 r.p.m. speed in Fig. 13.24. It is
observed that the point is located on the border of B and C – section belts. We will
select B cross-section V-belt. From Table13.12, the minimum pitch diameter for the
smaller pulley of B section is 200 mm.
Speed ratio =
1440
=4
360
d = 200 mm and
From Eq. (13.3),
D = 4(200) = 800 mm
(ii)
9
π( D + d ) ( D − d )2
+
L=2C+
2
4C
= 2 (1000) +
π (800 + 200 ) (800 − 200 ) 2
+
2
4 (1000)
= 3660.80 mm
From Table 13.14, the preferred pitch length for B- section belt is 3600 or 4060 mm.
It is assumed that the pitch length of the belt is 3600 mm.
Belt specification = B section 3600 mm pitch length V belt
(i)
Substituting this value of pitch length in Eq.(13.3),
L = 2C+
π( D + d ) ( D − d )2
+
2
4C
3600 = 2 C +
π (800 + 200) (800 − 200 ) 2
+
2
4C
Simplifying the above expression,
C 2 − 1014.6 C + 45 000 = 0
C=
1014.6 ± 1014.6 2 − 4 ( 45 000)
2
= 968.12 mm
The correct center distance is 968.12 mm.
(iii)
From Table 13.21 (B-section and 3600 mm pitch length),
Fc = 1.1
From Eq. (13.1),
⎛ D− d ⎞
⎟⎟ = 180 − 2 sin −1
α s = 180 − 2 sin −1 ⎜⎜
2
C
⎝
⎠
⎛ 800 − 200 ⎞
⎜⎜
⎟⎟ = 143.90 0
2
x
968
.
12
⎝
⎠
From Table 13.22, Fd is approximately 0.90.
From Table 13.17, ( 1440 r.p.m., 200 mm pulley, B-section) ( speed ratio = 4)
10
Pr = 5.90 + 0.46 = 6.36 kW
From Eq.(13.18),
13.8
Number of belts =
P x Fa
15 (1.3)
=
Pr x Fc x Fd
6.36 (1.1) (0.90)
Number of belts
= 3.097 or 3 belts
e=
(P1 a + P2 b) 1500(75) + 500(250)
=
= 242.1 mm
W
100 x9.81
(iv)
(Ans.)
1
CHAPTER 14
14.1
From Table 14.1, the pitch dimension (p) of I.S.O. 10B chain is 15.875 mm. When
both shafts rotate at same speed, the number of teeth on driving and driven sprockets
is same.
z2 = z1 = 19 teeth
p = 15.875 mm
a = 550 mm
From Eq. (14.6),
2
⎛ a ⎞ ⎛ z +z ⎞ ⎛ z −z ⎞ ⎛ p ⎞
L n = 2 ⎜⎜ ⎟⎟ + ⎜ 1 2 ⎟ + ⎜⎜ 2 1 ⎟⎟ x ⎜ ⎟
⎝p⎠ ⎝ 2 ⎠ ⎝ 2 π ⎠ ⎝a ⎠
⎛ 550 ⎞ ⎛ 19 + 19 ⎞
=2⎜
⎟ +0
⎟+⎜
⎝ 15.875 ⎠ ⎝ 2 ⎠
= 88.29 or 88 links
(i)
From Eq. (14.7),
⎧
p⎪ ⎡
⎛ z +z
a = ⎨ ⎢ Ln − ⎜ 1 2
4⎪ ⎣
⎝ 2
⎩
Substituting,
a=
⎞⎤
⎟⎥ +
⎠⎦
⎡
⎛ z1 + z 2
⎢ Ln − ⎜ 2
⎝
⎣
2 ⎫
2
⎡ z 2 − z1 ⎤ ⎪
⎞⎤
⎟⎥ − 8 ⎢
⎥ ⎬
⎠⎦
⎣ 2π ⎦ ⎪
⎭
z2 =z1
15.875 ⎧ ⎡
p⎧ ⎡
⎛ z1 + z 2 ⎞⎤ ⎫⎪
⎛ 19 + 19 ⎞⎤
⎟⎥ ⎬ =
⎟⎥
⎨ ⎢ Ln − ⎜
⎨ ⎢ 88 − ⎜
2 ⎩⎣
2⎩ ⎣
⎝ 2 ⎠⎦
⎝ 2 ⎠⎦ ⎪⎭
= 547.69 mm
14.2
⎫⎪
⎬
⎪⎭
(ii)
Refer to Table 14.2. The kW rating of simple chain 08B at a speed of 200 r.p.m. is
1.18 kW.
For smooth operation without shock, (Table 14.3)
Ks = 1
2
From Table 14.4 (single strand),
K1 = 1
From Table 14.5 (19 teeth),
K2 = 1.11
From Eq. (14.8),
(kW to be transmitted) = (kW rating of chain)
= 1.31
From Table 14.1,
K1 K 2
(1) (1.11)
= (1.18)
Ks
1
(i)
(chain 08B)
p =12.7 mm
v=
zpn
(19) (12.7) (200)
=
= 0.8 m / s
3
60x10
60x10 3
(ii)
P1 =
1000 kW 1000 (1.31)
=
= 1637.5 N
v
0.8
(iii)
From Table 14.1, the breaking load for simple 08B chain is 17 800 N.
(fs) =
14.3
17 800
= 10.87
1637.5
Table 14.1
(chain 06B)
(iv)
p = 9.525 mm
From Eq. (14.2),
D1 =
D2 =
p
⎛ 180 ⎞
⎟⎟
sin ⎜⎜
⎝ z1 ⎠
p
⎛ 180 ⎞
⎟⎟
sin ⎜⎜
⎝ z2 ⎠
=
=
9.525
= 63.91 mm
⎛ 180 ⎞
sin ⎜
⎟
⎝ 21 ⎠
9.525
= 106.26 mm
⎛ 180 ⎞
sin ⎜
⎟
⎝ 35 ⎠
(i)
(i)
3
v=
zpn
(21) (9.525) (500)
=
= 1.667 m / s
3
60x10
60x10 3
(ii)
P1 =
1000 kW 1000 (1)
= 599.88 N
=
v
1.667
(iii)
⎛ D2 ⎞
⎛ 106.26 ⎞
3
⎟ = (599.88) ⎜
⎟ = 31.87x10 N-mm
2
2
⎝
⎠
⎝
⎠
Torque = P1 ⎜
= 31.87 N-m
14.4
Assume
(iv)
z1 = 17
(i)
⎛ 1400 ⎞
⎛ 1400 ⎞
z 2 = z1 ⎜
⎟ = (17) ⎜
⎟ = 68
⎝ 350 ⎠
⎝ 350 ⎠
(i)
For moderate shock with electric motor, (Table 14.3)
Ks = 1.3
From Table 14.4 (two strands),
K1 = 1.7
From Table 14.5 (17 teeth),
K2 = 1
From Eq. (14.8),
kW rating of chain =
( kW to be transmitted ) x K s
(15) (1.3)
=
= 11.47 kW
K1 x K 2
(1.7) (1)
Refer to Table 14.2. The required kW rating is 11.47 kW at 1400 r.p.m. Therefore,
chain No.10 B (kW rating = 11.67) is suitable for the above application.
Recommended chain = 10B
From Table 14.1
p = 15.875 mm
(ii)
4
From Eq. (14.2),
D1 =
D2 =
p
⎛ 180 ⎞
⎟⎟
sin ⎜⎜
⎝ z1 ⎠
p
⎛ 180 ⎞
⎟⎟
sin ⎜⎜
⎝ z2 ⎠
=
=
15.875
= 86.39 mm
⎛ 180 ⎞
sin ⎜
⎟
⎝ 17 ⎠
(iii)
15.875
= 343.74 mm
⎛ 180 ⎞
sin ⎜
⎟
⎝ 68 ⎠
(iii)
a = 40 p = 40 (15.875) = 635 mm
From Eq. (14.6),
2
⎛ a ⎞ ⎛ z +z ⎞ ⎛ z −z ⎞ ⎛ p ⎞
L n = 2 ⎜⎜ ⎟⎟ + ⎜ 1 2 ⎟ + ⎜⎜ 2 1 ⎟⎟ x ⎜ ⎟
⎝p⎠ ⎝ 2 ⎠ ⎝ 2 π ⎠ ⎝a ⎠
2
⎛ 635 ⎞ ⎛ 17 + 68 ⎞ ⎛ 68 − 17 ⎞ ⎛ 15.875 ⎞
⎟⎟ x ⎜
=2⎜
⎟ + ⎜⎜
⎟+⎜
⎟
⎝ 15.875 ⎠ ⎝ 2 ⎠ ⎝ 2 π ⎠ ⎝ 635 ⎠
= 124.15 or 124 links
(iv)
From Eq. (14.7),
a=
=
⎧
p⎪ ⎡
⎛ z1 + z 2 ⎞⎤
⎟⎥ +
⎨ ⎢ Ln − ⎜
4⎪ ⎣
⎝ 2 ⎠⎦
⎩
2 ⎫
2
⎡ z 2 − z1 ⎤ ⎪
⎡
⎛ z 1 + z 2 ⎞⎤
−
−
L
8
⎢ 2π ⎥ ⎬
⎢ n ⎜ 2 ⎟⎥
⎝
⎠⎦
⎣
⎣
⎦ ⎪
⎭
⎧
15.875 ⎪ ⎡
⎛ 17 + 68 ⎞⎤
⎟⎥ +
⎨ ⎢ 124 − ⎜
4 ⎪⎣
⎝ 2 ⎠⎦
⎩
= 633.81 mm
2
⎡
⎛ 17 + 68 ⎞⎤
⎢ 124 − ⎜ 2 ⎟⎥ − 8
⎠⎦
⎝
⎣
(v)
2 ⎫
⎡ 68 −17 ⎤ ⎪
⎢ 2π ⎥ ⎬
⎣
⎦ ⎪
⎭
1
CHAPTER 15
15.1
From Eq. (15.6),
p
L10
3
⎛ 22.8 ⎞
⎛ C⎞
=⎜ ⎟ = ⎜
⎟ = 11.85 million revolutions
⎝ 10 ⎠
⎝ P⎠
L10 h =
L10 x 10 6 (11.85) x 10 6
=
= 136.23 hr
60 n
60 (1450)
L 50 h = 5 L10 h = 5 (136.23) = 681.17 hr
15.2
(ii)
(iii)
⎛D ⎞
⎛ 40 ⎞
M t = µ Fr ⎜ i ⎟ = 0.0012 (25x10 3 ) ⎜ ⎟ = 600 N-mm
⎝ 2 ⎠
⎝ 2 ⎠
(kW ) f =
15.3
(i)
2 π n M t 2 π (1440) (600)
=
= 0.09
60 x10 6
60 x10 6
(Ans.)
From Eq. (15.2),
P = X Fr + Y Fa = 0.56(2500)+1.6(1000) = 3000 N
p
⎛ C⎞
⎛ 7350 ⎞
L10 = ⎜ ⎟ = ⎜
⎟
⎝ P⎠
⎝ 3000 ⎠
L10 h
15.4
3
million revolutions
L x 10 6 ⎛ 7350 ⎞
= 10
=⎜
⎟
60 n
⎝ 3000 ⎠
3
10 6
= 340.42 hr
60 (720)
Consider the work cycle of one minute duration.
(Ans.)
2
Element
P
Element time
Speed
Revolutions N
No.
(N)
(sec.)
(r.p.m.)
in element time
1
3000
18
720
216
2
7000
30
1440
720
3
5000
12
900
180
60
total
1116
Total
Average speed of rotation = 1116 r.p.m.
(i)
From Eq. (15.13),
Pe =
=
3
⎡ N 1 P13 + N 2 P23 + N 3 P33 ⎤
⎢
⎥
N1 + N 2 + N 3
⎣
⎦
3
⎡ 216(3000) 3 + 720(7 000) 3 + 180(5000) 3 ⎤
⎢
⎥
1116
⎣
⎦
= 6271.57 N
(ii)
According to the load-life relationship,
3
L10
L10 h =
15.5
Pe =
=
3
⎛ C⎞
⎛ 16.6 x10 3 ⎞
⎟⎟ = 18.54 million rev.
= ⎜⎜ ⎟⎟ = ⎜⎜
⎝ 6271.57 ⎠
⎝ Pe ⎠
L10 x 10 6 18.54 x 10 6
=
= 276.94 hr
60 n
60 (1116)
3
⎡ N 1 P13 + N 2 P23 ⎤
⎢
⎥
⎣ N1 + N 2 ⎦
3
⎡ 5(2500) 3 + 10(1500) 3 ⎤
⎢
⎥
5 + 10
⎣
⎦
= 1953.8 N
(iii)
3
C = Pe (L10 )1 / 3 = 1953.8 (20)1 / 3 = 5303.43 N
15.6
L 95 =
(Ans.)
60 n L 95 h
60 ( 720 )(10 000)
=
= 432 million rev
6
10
10 6
From Eq. (15.17),
⎡
⎛ 1 ⎞⎤
⎟⎟ ⎥
⎢ log e ⎜⎜
R
⎛ L 95 ⎞
⎝ 95 ⎠ ⎥
⎜⎜
⎟⎟ = ⎢
⎢
L
⎛
1 ⎞⎥
⎝ 10 ⎠
⎟⎟ ⎥
⎢ log e ⎜⎜
⎝ R 90 ⎠ ⎦⎥
⎣⎢
L10 =
15.7
1 / 1.17
=
⎡
⎛ 1 ⎞⎤
⎢ log e ⎜ 0.95 ⎟ ⎥
⎝
⎠⎥
⎢
⎛ 1 ⎞⎥
⎢
⎢ log e ⎜⎝ 0.90 ⎟⎠ ⎥
⎦
⎣
1 / 1.17
= 0.5405
L 95
432
=
= 799.26 million revolutions
0.5405 0.5405
C = P (L10 )1 / 3 = 3000 (799.26)1 / 3 = 27 840.94 N
(i)
R s = (R ) N = (0.95) 4 = 0.8145
(ii)
R s = (R ) N
0.82 = (R ) 4
or
81.45%
R = 0.95
From Example 15.6,
⎛ L 95
⎜⎜
⎝ L10
⎞
⎟⎟ = 0.5405
⎠
L10 =
or
L 95
5
=
= 9.25 million revolutions
0.5405 0.5405
C = P (L10 )1 / 3 = 2500 (9.25)1 / 3 = 5247.92 N
(Ans.)
1
CHAPTER 16
16.1
From Eq. (16.10),
⎡
⎤
⎡
⎤
⎢
2
2 ⎥
2
2 ⎥
⎢
π Pi ⎢ R o − R i ⎥ π (5) 200 − 125
⎢
⎥ = 407 317.71 N
W=
=
2 ⎢
2 ⎢
⎛ Ro ⎞⎥
⎛ 200 ⎞ ⎥
⎟⎟ ⎥
⎢ log e ⎜⎜
⎢ log e ⎜⎝ 125 ⎟⎠ ⎥
⎣
⎦
⎝ R i ⎠ ⎦⎥
⎣⎢
W = 407.32 kN
(i)
From Eq. (16.9),
π Pi h 3o
Q=
⎛R
6 µ log e ⎜⎜ o
⎝ Ri
⎞
⎟⎟
⎠
=
π (5) (0.15) 3
3
= 626 642.63 mm /s
⎛ 200 ⎞
6 (30 x 10 −9 ) log e ⎜
⎟
⎝ 125 ⎠
= 626 642.63 (10-3) (10-3) 60 litre/min
= 37.598 or 37.6 litre/min
(ii)
From Eq. (16.11),
(kW)p = Q(Pi – Po) (10-6)=(626 642.63)[5 – 0] (10-6)
= 3.13
(iii)
From Eq. (16.12),
⎛
1
(kW ) f = ⎜⎜
6
⎝ 58.05 x 10
⎛
1
= ⎜⎜
6
⎝ 58.05 x 10
⎞ µ n 2 (R o4 − R i4 )
⎟⎟
ho
⎠
[
⎞ (30 x 10 −9 ) (720) 2 (200) 4 − (125) 4
⎟⎟
(0.15)
⎠
= 2.42
]
(iv)
The total energy loss is given by
(kW)t = (kW)p + (kW)f = 3.13 + 2.42 = 5.55 kW or kJ/s
2
Q = 626 642.63 mm3/s
= 626 642.63(10-3) c.c./s
= 626 642.63(10-3) (0.86) gm/s
= 626 642.63(10-3) (0.86) (10-3) kg/s
= 0.5389 kg/s
H = m Cp ∆ t
or
5.55 = 0.5389 (1.75) ∆ t
∆ t = 5.880C
16.2
(v)
W = (500x50)(Pi) + (2x500x225)(0.5 Pi) = 137 500 Pi
Pi =
W
500 x10 3
=
= 3.64 N/mm2 or MPa
137 500 137 500
(i)
From Eq. (16.7),
Q1 =
3.64 (500)(0.2) 3
∆p b h 3
=
= (10.785x103) mm3/s
−9
12 µ l
12 (500 x 10 )(225)
Q = 2 Q1(10-6)(60) litres/min
= 2 (10.785x103)(10-6)(60)
= 1.29 l/min
(ii)
3
16.3
300
W=
= 75 kN
4
For each pad
From Eq. (16.10),
⎛R ⎞
⎛ 100 ⎞
2 W log e ⎜⎜ o ⎟⎟
2 (75 x 10 3 ) log e ⎜
⎟
R
⎝ i⎠ =
⎝ 25 ⎠
Pi =
π (R o2 − R i2 )
π (100 2 − 25 2 )
= 7.06 N/mm2 or MPa
(i)
From Eq. (16.4)
⎡
180 ⎤
180 ⎤
⎡
= 54.28 cSt
z k = ⎢0.22 t −
= ⎢0.22 (250) −
⎥
t ⎦
(250) ⎥⎦
⎣
⎣
z = ρ z k = 0.88(54.28) = 47.77 cP
µ=
z
= (47.77) (10-9) N-s/mm2
9
10
From Eq.(16.9),
Q=
π Pi h 3o
⎛R
6 µ log e ⎜⎜ o
⎝ Ri
⎞
⎟⎟
⎠
=
π (7.06) (0.1) 3
⎛ 100 ⎞
6 (47.77) (10 −9 ) log e ⎜
⎟
⎝ 25 ⎠
= 55820.36 mm3/s
Qt = 4 Q
= 4 (55820.36) mm3/s
= 4 (55820.36) (10-6) (60) l/min
= 13.4 l/min
16.4
(ii)
Consider the flow of lubricant through a slot of length R dθ and thickness ho as shown
in Fig.16.1-solu.
Q=
∆p b h 3
12 µ l
(a)
∆ p = dp
b = 2 π R sin θ
l = R dθ
h = ho
Substituting above values in Eq. (a),
dp = −
6 µ Q dθ
π h 3o sin θ
Intrgrating,
6 µQ
⎡ ⎛ θ ⎞⎤
log ⎢ tan⎜ ⎟⎥ + C
3
π ho
⎣ ⎝ 2 ⎠⎦
p=−
(b)
The first boundary condition is,
p=0
C=
when
θ = φ1
⎡ ⎛ φ ⎞⎤
6 µQ
log ⎢ tan ⎜ 1 ⎟⎥
3
π ho
⎣ ⎝ 2 ⎠⎦
(c)
From (b) and (c),
⎧ ⎛ φ1 ⎞ ⎫
⎪ tan⎜ 2 ⎟ ⎪
6 µQ
⎪
⎪
p=
log e ⎨ ⎝ ⎠ ⎬
3
π ho
⎪ tan ⎛⎜ θ ⎞⎟ ⎪
⎝ 2 ⎠ ⎭⎪
⎩⎪
(d)
The second boundary condition is,
p = Pi
when
θ = φ2
Substituting above condition in Eq. (d),
⎧ ⎛ φ1 ⎞ ⎫
⎪ tan⎜ 2 ⎟ ⎪
6 µQ
⎪
⎪
log e ⎨ ⎝ ⎠ ⎬
Pi =
3
φ
π ho
⎪ tan ⎛⎜ 2 ⎞⎟ ⎪
⎪⎩ ⎝ 2 ⎠ ⎪⎭
Q=
π Pi h 3o
⎡ tan(φ1 / 2) ⎤
6 µ log e ⎢
⎥
⎣ tan(φ 2 / 2) ⎦
or
(i)
4
5
The load carrying capacity is given by,
φ1
W = π (R sin φ 2 ) 2 Pi + ∫ (2 π R sin θ) R dθ (p cos θ)
φ2
Substituting value of p from Eq. (d),
W = π Pi R 2 sin 2 φ 2 +
6 µQR2
h 3o
φ1
⎡ tan(φ1 / 2) ⎤
∫ log⎢⎣ tan(θ / 2) ⎥⎦ sin 2θ dθ
φ2
Define I as,
⎡ tan(φ / 2) ⎤
1
I = ∫ log ⎢⎣ tan(θ / 2) ⎥⎦ sin 2θ dθ = ∫ u dv
where,
⎡ tan(φ / 2) ⎤
1
u = log ⎢
tan(
θ
/ 2) ⎥⎦
⎣
∴ du =
v=−
dv = sin 2θ dθ
and
tan(φ1 / 2)
1
tan(θ / 2)
1
x
x sec 2 (θ / 2) x = −
2
tan(φ1 / 2) [ − tan (θ / 2)]
2
sin θ
cos 2θ
2
Since,
I = ∫ u dv = u v − ∫ v du
=−
⎡ tan(φ1 / 2) ⎤ 1
1
cos 2θ log ⎢
⎥−
2
⎣ tan(θ / 2) ⎦ 2
=−
⎡ tan(φ1 / 2) ⎤ 1 ⎡ dθ
1
⎤
cos 2θ log ⎢
−
− 2∫ sin θ dθ⎥
⎥
∫
⎢
2
⎦
⎣ tan(θ / 2) ⎦ 2 ⎣ sin θ
= −
⎡ tan(φ1 / 2) ⎤ 1 ⎡
⎤
1
⎛θ⎞
cos 2θ log ⎢
− ⎢ log tan⎜ ⎟ + 2 cos θ⎥
⎥
2
⎝2⎠
⎦
⎣ tan(θ / 2) ⎦ 2 ⎣
Substituting limits,
∫
cos 2θ
dθ
sin θ
(e)
6
φ1
⎡ 1
⎤
⎡ tan(φ1 / 2) ⎤ 1
⎛θ⎞
− log tan⎜ ⎟ − cos θ⎥
cos 2θ log ⎢
⎥
⎝2⎠
⎣ tan(θ / 2) ⎦ 2
⎣ 2
⎦ φ2
I = ⎢−
=+
⎡ tan(φ1 / 2) ⎤
⎡ tan(φ1 / 2) ⎤ 1
1
cos 2φ 2 log ⎢
⎥ − (cos φ1 − cos φ 2 )
⎥ − log ⎢
2
⎣ tan(φ 2 / 2) ⎦
⎣ tan(φ 2 / 2) ⎦ 2
⎡ tan(φ1 / 2) ⎤ 2
⎥ sin φ 2 − (cos φ1 − cos φ 2 )
⎣ tan(φ 2 / 2) ⎦
= − log e ⎢
Substituting above expression in Eq.(e),
W=
16.5
p=
π Pi R 2 (cos φ 2 − cos φ1 )
⎡ tan(φ1 / 2) ⎤
log e ⎢
⎥
⎣ tan(φ 2 / 2) ⎦
(ii)
W
50 x10 3
=
= 2.222 N / mm 2
l d 150 x150
ho
0.03
=
= 0 .2
c
0.15
From Table 16.1,
l 150
=
=1
d 150
S = 0.0446
2
2
2.222
⎛c⎞ p
⎛ 0.15 ⎞
ns = S ⎜ ⎟
= 49.55 rev/s
= (0.0446) ⎜
⎟
−9
⎝r⎠ µ
⎝ 75 ⎠ (8x10 )
n = 60 (49.55) = 2973 r.p.m.
16.6
p=
(Ans.)
W
50 x10 3
=
= 5 N / mm 2
l d 100 x100
−9
⎛ 50 ⎞ (16x10 )(1440 / 60)
⎛ r ⎞ µ ns
=⎜
= 0.0133
S=⎜ ⎟
⎟
(5)
p
⎝c⎠
⎝ 0.12 ⎠
2
2
The values of dimensionless performance parameters are obtained by linear
interpolation from Table (16.1).
For (l/d =1)
7
(0.1 − 0.03)
⎛ ho ⎞
(0.0133 − 0.00474) = 0.07262
⎜ ⎟ = 0.03 +
(0.0188 − 0.00474)
⎝ c ⎠
h o = 0.07262 c = 0.07262 (0.12) = 0.008714 mm or 0.0087 mm
(i)
(1.05 − 0.514)
⎛r⎞
(0.0133 − 0.00474) = 0.84
⎜ ⎟ f = 0.514 +
(0.0188 − 0.00474)
⎝c⎠
⎛ 0.12 ⎞
⎛c⎞
−3
f = 0.84⎜ ⎟ = 0.84 ⎜
⎟ = 2.016 x10
⎝r⎠
⎝ 50 ⎠
( kW ) f =
16.7
(ii)
2 π n s f W r 2 π (1440 / 60) ( 2.016 x10 −3 ) (50 x10 3 ) (50)
=
= 0.76
10 6
10 6
(iii)
l=d
p=
W
ld
(2.5) =
25x10 3
l2
l = d = 100 mm
(i)
Refer to Chapter-3 for values of tolerances. The hole and shaft limits for H7e7
running fit are as follows:
Hole limits: (100.00) and (100.035) mm
Shaft limits: (100 - 0.072) and (100 - 0.107) mm
If the manufacturing processes are centered, the average diameter of the bearing and
journal will be 100.0175 and 99.9105 mm respectively.
⎛1⎞
c = ⎜ ⎟ (100.0175 − 99.9105) = 0.0535 mm
⎝2⎠
−9
⎛ 50 ⎞ (20x10 )(900 / 60)
⎛ r ⎞ µ ns
=⎜
= 0.1048
S=⎜ ⎟
⎟
(2.5)
p
⎝ 0.0535 ⎠
⎝c⎠
2
2
The values of dimensionless performance parameters are obtained by linear
interpolation from Table (16.1).
For (l/d =1)
8
(0.1048 − 0.0446)
⎛ ho ⎞
(0.4 − 0.2) = 0.3576
⎜ ⎟ = 0.2 +
(0.121 − 0.0446)
⎝ c ⎠
h o = 0.3576 c = 0.3576 (0.0535) = 0.0191 mm
(ii)
⎛ Q ⎞
(0.1048 − 0.0446)
⎟⎟ = 4.62 −
⎜⎜
(4.62 − 4.33) = 4.391
(0.121 − 0.0446)
⎝ r c ns l ⎠
Q = 4.391 r c ns l = 4.391(50)(0.0535)(15)(100)
= 17618.89 mm3/s
= 17618.89 (60x10-6) litres/min
= 1.057 litres/min
16.8
ho = 5
(iii)
[∑ surface roughness]= 5 (0.8 + 0.4) = 6 µm
c=
1
(50.02 − 49.93) = 0.045 mm
2
p=
W
8000
=
= 3.2 N / mm 2
l d 50x 50
(i)
−9
⎛ 25 ⎞ (12x10 )(1440 / 60)
⎛ r ⎞ µ ns
=⎜
= 0.02777
S=⎜ ⎟
⎟
(3.2)
p
⎝ 0.045 ⎠
⎝c⎠
2
2
The values of dimensionless performance parameters are obtained by linear
interpolation from Table (16.1).
For (l/d =1)
(0.0278 − 0.0188)
⎛ ho ⎞
(0.2 − 0.1) = 0.1349
⎜ ⎟ = 0.1 +
(0.0446 − 0.0188)
⎝ c ⎠
h o = 0.1349 c = 0.1349 (0.045) = 0.00607 mm or 6.07 µm
(0.0278 − 0.0188)
⎛r⎞
(1.7 − 1.05) = 1.277
⎜ ⎟ f = 1.05 +
(0.0446 − 0.0188)
⎝c⎠
r f = 1.277 c = 1.277 (0.045) = 0.05747
(ii)
9
2 π n s f W r 2 π (1440 / 60) (0.05747 ) (8000)
( kW ) f =
=
= 0.069
10 6
10 6
(iii)
1
CHAPTER 17
17.1
zp = 20
zg = 100
m = 6 mm
From Eq. (17.5),
a=
m (z p + z g )
2
=
6 (20 + 100)
= 360 mm
2
(i)
d 'p = m zp = 6(20) = 120 mm
d 'g = m zg = 6(100) = 600 mm
(ii)
addendum (ha) = m = 6 mm
dedendum (hf ) = 1.25 m = 1.25(6) = 7.5 mm
(iii)
tooth thickness = 1.5708 m = 1.5708(6) = 9.4248 mm
clearance (c) = 0.25 m = 0.25(6) = 1.5 mm
i=
17.2
zg
zp
=
100
=5
20
zp = 25
i=
np
ng
(vi)
np = 1200 r.p.m.
=
(iv)
ng = 200 r.p.m.
m = 4 mm
1200
=6
200
z g = i z p = 6 (25) =150
From Eq. (17.5),
a=
17.3
m (z p + z g )
2
=
a = 495 mm
From Eq. (17.5),
4 (25 + 150)
= 350 mm
2
i = 4.5
m = 6 mm
(Ans.)
m (z p + z g )
a=
495 =
2
z p + z g = 165
zg
zp
17.4
2
(a)
= 4.5
zp = 30
2
6 (z p + z g )
(b)
Solving (a) and (b),
zg = 135
(Ans)
d 1' = m z1 = 4 (20) = 80 mm
d 3' = m z 3 = 4 (30) = 120 mm
Forces between gears 1 and 2:
( M t )1 =
Pt =
60 x 10 6 (kW )
60 x 10 6 (5)
=
= 66 314.56 N − mm
2πn
2 π (720)
2 (M t )1 2 (66 314.56)
=
= 1657.86 N
80
d 1'
Pr = Pt tan α = 1657.86 tan (20) = 603.41 N
(i)
Forces between gears 3 and 4:
(M t ) 3 =
Pt =
z2
⎛ 50 ⎞
(M t )1 = ⎜ ⎟ (66314.56) = 165 786.4
z1
⎝ 20 ⎠
2 (M t ) 3 2 (165 786.4)
=
= 2763.11 N
120
d 3'
Pr = Pt tan α = 2763.10 tan (20) = 1005.69 N
(ii)
3
17.5
Torques:
(M t ) A =
60 x 10 6 (kW )
60 x 10 6 (10)
=
= 66 314.56 N − mm
2πn
2 π (1440)
(M t ) B =
z2
⎛ 100 ⎞
(M t ) A = ⎜
⎟ (66314.56) = 331 572.8 N − mm
z1
⎝ 20 ⎠
(M t ) C =
z4
⎛ 150 ⎞
(M t ) B = ⎜
⎟ (331 572.8) = 1 989 436.79 N − mm
z3
⎝ 25 ⎠
(M t ) D =
z6
⎛ 150 ⎞
(M t ) C = ⎜
⎟ (1 989 436.8) = 11 936 620.73 N − mm
z5
⎝ 25 ⎠
Gear tooth forces:
Forces between gears 1 and 2:
d 1' = m z1 = 5 (20) = 100 mm
Pt =
2 (M t ) A
2 (66 314.56)
=
= 1326.29 N
'
100
d1
Pr = Pt tan α = 1326.29 tan (20) = 482.73 N
Forces between gears 3 and 4:
d 3' = m z 3 = 6 (25) = 150 mm
Pt =
2 (M t ) B 2 (331 572.8)
=
= 4420.97 N
150
d 3'
Pr = Pt tan α = 4420.97 tan (20) = 1609.1 N
Forces between gears 5 and 6:
d 5' = m z 5 = 6 (25) = 150 mm
Pt =
2 (M t ) C
2 (1 989 436.79)
=
= 26 525.82 N
'
150
d5
(i)
4
Pr = Pt tan α = 26 525.82 tan (20) = 9654.61 N
(ii)
Reactions at B1 and B2 :
Free body diagram of forces is shown in Fig.17.1-solu. It is observed that tangential
components on gears 2 and 3 are in same direction. The radial components on gear 2
and 3 are in opposite directions. Refer to Fig. 17.2-solu (a).
Taking moments about bearing B1,
(B 2 ) h x 350 = 4420.97 x 100 + 1326.29 x 250
(B 2 ) h = 2210.48 N
Taking moments about bearing B2 ,
(B1 ) h x 350 = 1326.29 x 100 + 4420.97 x 250
(B1 ) h = 3536.78 N
Refer to Fig. 17.2-solu (b).
Taking moments about bearing B1,
(B 2 ) v x 350 + 482.73 x 250 = 1609.1 x 100
(B 2 ) v = 114.94 N
Taking moments about bearing B2 ,
(B1 ) v x 350 + 482.73 x 100 = 1609.1 x 250
(B1 ) v = 1011.43 N
Resultant reactions:
B1 =
3536.78 2 + 1011.43 2 = 3678.56 N
B2 =
2210.48 2 + 114.94 2 = 2213.47 N
(iii)
Reactions at C1 and C2 :
Free body diagram of forces is shown in Fig.17.3-solu. It is observed that tangential
components on gears 4 and 5 are in same direction. The radial components on gear 4
and 5 are in opposite directions. Refer to Fig. 17.4-solu (a).
Taking moments about bearing C1,
5
(C 2 ) h x 350 = 4420.97 x 250 + 26525.82 x 100
(C 2 ) h = 10736.64 N
Taking moments about bearing C2 ,
(C1 ) h x 350 = 26525.82 x 250 + 4420.97 x 100
(C1 ) h = 20210.15 N
Refer to Fig. 17.4-solu (b).
Taking moments about bearing C1,
(C 2 ) v x 350 + 1609.1 x 250 = 9654.61 x 100
(C 2 ) v = 1609.1 N
Taking moments about bearing C2 ,
(C1 ) v x 350 + 1609.1 x 100 = 9654.61 x 250
(C1 ) v = 6436.41 N
Resultant reactions:
17.6
C1 =
20210.15 2 + 6436.412 = 21210.32 N
C2 =
10736.64 2 + 1609.12 = 10856.55 N
Torques:
( M t )1 =
60 x 10 6 (kW )
60 x 10 6 (50)
=
= 1 591 549.43 N − mm
2πn
2 π (300)
(M t ) 2 =
z2
⎛ 60 ⎞
(M t )1 = ⎜ ⎟ (1 591 549.4) = 3 183 098.86 N − mm
z1
⎝ 30 ⎠
Gear tooth forces:
Forces between gears 1 and 2:
d 1' = m z1 = 8 (30) = 240 mm
Pt =
2 (M t )1 2 (1 591 549.43)
=
= 13 262.91 N
240
d 1'
(iv)
6
Pr = Pt tan α = 13 262.91 tan (20) = 4827.31 N
Forces between gears 3 and 4:
d 3' = m z 3 = 8 (25) = 200 mm
Pt =
2 (M t ) 2
2 (3 183 098.83)
=
= 31 830.99 N
'
200
d3
Pr = Pt tan α = 31 830.99 tan (20) = 11 585.53 N
(i)
Reactions at B1 and B2 :
Free body diagram of forces is shown in Fig.17.5-solu.
Refer to Fig. 17.6-solu (a).
Taking moments about bearing B1,
(B 2 ) v x 625 + 31830.99 x 150 = 4827.31 x 500
(B 2 ) v = − 3777.59 N
Taking moments about bearing B2 ,
(B1 ) v x 625 = 4827.31 x 125 + 31830.99 x 775
(B1 ) v = 40435.89 N
Refer to Fig. 17.6-solu (b).
Taking moments about bearing B1,
(B 2 ) h x 625 + 11585.53 x 150 = 13262.91 x 500
(B 2 ) h = 7829.80 N
Taking moments about bearing B2 ,
(B1 ) h x 625 = 13262.91 x 125 + 11585.53 x 775
(B1 ) h = 17018.64 N
Resultant reactions:
B1 =
40435.89 2 + 17018.64 2 = 43 871.35 N
B2 =
3777.59 2 + 7829.80 2 = 8693.44 N
(ii)
7
17.7
From Table17.3 Lewis form factor Y for 24 teeth is 0.337
From Eq.(17.16),
⎛ 600 ⎞
S b = m b σ b Y = 3 (30) ⎜
⎟ (0.337) = 6066 N
⎝ 3 ⎠
(i)
d 'p = m z p = 3 (24) = 72 mm
v=
π d 'p n p
60 x 10
3
=
π (72) (1200)
= 4.52 m / s
60 x 10 3
3
3
=
= 0.3987
3 + v 3 + 4.52
Cv =
(ii)
S b = Peff (fs)
Cs
Pt (fs)
Cv
Sb =
6066 =
(1.5)
Pt (1.5)
(0.3987)
Pt = 1074.90 N
⎛ d 'p
M t = Pt ⎜
⎜ 2
⎝
17.8
⎞
⎟ = 1074.90 ⎛⎜ 72 ⎞⎟ N − mm
⎟
⎝ 2 ⎠
⎠
kW =
2 π n M t 2 π (1200) (1074.90) (72 / 2)
=
= 4.86
60 x10 6
60 x10 6
Mt =
60 x 10 6 (kW ) 60 x 10 6 (5)
=
= 95 492.97 N − mm
2 π np
2 π (500)
Pt =
v=
2 Mt
2 (95 492.97)
=
= 1909.86 N
'
100
dp
π d 'p n p
60 x 10
3
=
π (100) (500)
= 2.618 m / s
60 x 10 3
(iii)
8
3
3
Cv =
=
= 0.534
3 + v 3 + 2.618
Peff =
Cs
1.5 (1909.86)
Pt =
= 5364.78 N
Cv
0.534
It is assumed that gear is weaker than pinion and Lewis form factor for gear is
approximately 0.42.
(
)
⎛ 300 ⎞
2
S b = m b σ b Y = m (10 m) ⎜
⎟ (0.42) = 420 m N
3
⎠
⎝
S b = Peff (fs)
420 m2 = 5364.78 (1.5)
m = 4.38 mm
The first preference value of module is 5 mm.
m = 5 mm
zp =
d 'p
m
=
(i)
100
= 20
5
zg =
d 'g
m
=
300
= 60
5
(ii)
Check for design:
Fro Table 17.3,
Yp = 0.32 (for 20 teeth)
Yg = 0.421 (for 60 teeth)
(σ b Y) p = 0.32(200) = 64
(σ b Y) g = 0.421(100) = 42.1
Therefore, gear is weaker than pinion.
17.9
Mt =
60 x 10 6 (kW ) 60 x 10 6 (7.5)
=
= 71 619.72 N − mm
2 π np
2 π (1000)
d 'p = m z p = 4 (25) = 100 mm
9
2 Mt
2 (71 619.72)
Pt =
=
= 1432.39 N
'
100
dp
v=
π d 'p n p
60 x 10
3
=
π (100) (1000)
= 5.236 m / s
60 x 10 3
Cv =
3
3
=
= 0.3643
3 + v 3 + 5.236
Peff =
Cs
2 (1432.39)
Pt =
= 7863.79 N
Cv
0.3643
(i)
From Table17.3 Lewis form factor Y for 25 teeth is 0.34
Peff = S b = m b σ b Y
7863.79 = 4 (45) σ b (0.34)
σ b = 128.49 N/mm2
17.10 Q =
2 zg
zg + zp
=
(ii)
2 (60)
= 1.4118
60 + 25
2
2
⎛ BHN ⎞
⎛ 220 ⎞
⎟ = 0.16 ⎜
⎟ = 0.7744
⎝ 100 ⎠
⎝ 100 ⎠
K = 0.16 ⎜
d 'p = m z p = 5(25) = 125 mm
S w = b Q d 'p K = 45(1.4118)(125)(0.7744) = 6149.8 N
v=
π d 'p n p
60 x 10
3
=
π (125) (500)
= 3.2725 m / s
60 x 10 3
Cv =
3
3
=
= 0.4783
3 + v 3 + 3.2725
Sw =
Cs
Pt (fs)
Cv
6149.8 =
1.75
Pt (2)
0.4783
(i)
10
Pt = 840.41 N
⎛ d 'p
M t = Pt ⎜
⎜ 2
⎝
kW =
7.11
(ii)
⎞
⎟ = 840.41⎛⎜ 125 ⎞⎟ = 52525.88 N − mm
⎟
⎝ 2 ⎠
⎠
2 π n M t 2 π (500) (52525.88)
=
= 2.75
60 x10 6
60 x10 6
(iii)
Beam strength:
From Table17.3 Lewis form factor Y for 24 teeth is 0.337
⎛ 450 ⎞
S b = m b σ b Y = 6 (60) ⎜
⎟ (0.337) = 18 198 N
⎝ 3 ⎠
(i)
Wear strength:
Q=
2 zg
zg + zp
=
2 (48)
= 1.333
(48 + 24)
2
2
⎛ BHN ⎞
⎛ 250 ⎞
K = 0.16 ⎜
⎟ = 0.16 ⎜
⎟ =1
⎝ 100 ⎠
⎝ 100 ⎠
d 'p = m z p = 6(24) = 144 mm
S w = b Q d 'p K = 60(1.333)(144)(1) = 11 517.12 N
v=
π d 'p n p
60 x 10
Cv =
3
=
π (144) (1000)
= 7.5398 m / s
60 x 10 3
3
3
=
= 0.2846
3 + v 3 + 7.5398
Sw < Sb
Sw =
Cs
Pt (fs)
Cv
11 517.12 =
1.5
Pt (2)
0.2846
(ii)
11
Pt = 1092.59 N
⎛ d 'p
M t = Pt ⎜
⎜ 2
⎝
kW =
⎞
⎟ = 1092.59 ⎛⎜ 144 ⎞⎟ = 78 666.48 N − mm
⎟
⎝ 2 ⎠
⎠
2 π n M t 2 π (1000) (78 666.48)
=
= 8.24
60 x10 6
60 x10 6
(iii)
17.12 zp = 18
⎛ 720 ⎞
zg = i zp = ⎜
⎟ (18) = 90
⎝ 144 ⎠
Cv =
(i)
3
3
3
=
=
3+v 3+5 8
From Table17.3 Lewis form factor Y for 18 teeth is 0.308
⎡
⎢
60 x 10 6
m = ⎢
⎢
π
⎢
⎣
⎡
⎢
60 x 10 6
= ⎢
⎢
π
⎢
⎣
⎧
⎫⎤
⎪
⎪⎥
( kW ) C s (fs)
⎪
⎪⎥
⎨
⎬
⎪ z n C ⎛⎜ b ⎞⎟ ⎛⎜ S ut ⎞⎟ Y ⎪⎥
⎪⎩ p p v ⎝ m ⎠ ⎝ 3 ⎠ ⎪⎭⎥⎦
1/ 3
⎧
⎫⎤
⎪⎪
⎪⎪⎥
( 5 ) (1.25) (2)
⎨
⎬⎥
3
410
⎥
⎞
⎛
⎞
⎛
⎪ (18) (720) ⎜ ⎟ (10) ⎜
⎟ (0.308) ⎪⎥
⎪⎩
⎪⎭⎦
⎝ 3 ⎠
⎝8⎠
1/ 3
= 4.89 mm
The first preference value of the module is 5 mm.
m = 5 mm
(ii)
d 'p = m z p = 5(18) = 90 mm
d 'g = m z g = 5(90) = 450 mm
b = 10 m = 10 (5) = 50 mm
(iii)
12
Static load:
Mt =
60 x 10 6 (kW ) 60 x 10 6 (5)
=
= 66 314.56 N − mm
2 π np
2 π (720)
2 Mt
2 (66 314.56)
=
= 1473.66 N
'
90
dp
Pt =
Dynamic load:
For Grade 6,
e = 8 + 0.63 φ
For pinion,
φ = m + 0.25 d 'p = 5 + 0.25 90
ep = 8 + 0.63 φ = 12.644 µm
For gear,
φ = m + 0.25 d 'g = 5 + 0.25
450
ep = 8 + 0.63 φ = 14.491 µm
∴ e = ep + eg = 12.644 + 14.491 = 27.135 µm or (27.135 x 10 ) mm
-3
From Table 17.7, the value of deformation factor C is 11 400 N/mm2.
v=
π d 'p n p
60 x 10
3
=
π (90) (720)
= 3.3929 m / s
60 x 10 3
From Eq. (17.26),
Pd =
21 v ( C e b + Pt )
21 v +
( C e b + Pt )
=
[
21 (3.3929) 11 400 (27.135 x 10 −3 ) (50) + 1473.66
21 (3.3929) +
[11 400 (27.135 x 10
−3
]
) (50) + 1473.66
= 5993 N
Effective load:
Peff = ( C s P t + Pd ) = 1.25(1473.66) + 5993 = 7835.08 N
Beam strength:
⎛ 410 ⎞
S b = m b σ b Y = 5 (50) ⎜
⎟ (0.308) = 10 523.33 N
⎝ 3 ⎠
(fs) =
Sb
10 523.33
=
= 1.34
Peff
7835.08
(iv)
Surface hardness:
S w = (fs) Peff = 2 (7835.08) = 15670.16 N
Q=
2 zg
zg + zp
=
2 (90)
= 1.667
(90 + 18)
S w = b Q d 'p K
⎛ BHN ⎞
15670.16 = 50(1.667)(90) (0.16) ⎜
⎟
⎝ 100 ⎠
BHN = 361.33 or 370
2
(v)
13
]
1
CHAPTER 18
18.1
From Eq.(18.7),
a=
m n (z 1 + z 2 )
2 cos ψ
or
165 =
4 (25 + 50)
2 cos ψ
ψ = 24.62 0
18.2
dp =
zp mn
dg =
zg mn
a=
cos ψ
cos ψ
20 (5)
= 103.53 mm
cos (15 0 )
=
60 (5)
= 310.58 mm
cos (15 0 )
(i)
pa =
(ii)
mn
3
=
= 3.26 mm
cos ψ cos (230 )
tan α =
tan α n tan (20 0 )
=
cos ψ cos (23 0 )
(i)
or
α = 21.57 0
πm
π (3.26)
p
=
=
= 24.13 mm
tan ψ
tan ψ tan (23 0 )
(M t ) =
d=
=
m n (z1 + z 2 ) 5 (20 + 60)
=
= 207.06 mm
2 cos ψ
2 cos 15 0
18.3 m =
18.4
(Ans)
60 x 10 6 (kW ) 60 x 10 6 (7.5)
=
= 35 809.86 N − mm
2πn
2 π (2000)
z mn
18 (6)
=
= 117.33 mm
cos ψ cos (23 0 )
(ii)
(iii)
2
2 (M t )
2 (35 809.86)
=
= 610.41 N
Pt =
d
117.33
Pa = Pt tan ψ = 610.41 tan (230 ) = 259.10 N
⎡ tan (20 0 ) ⎤
⎡ tan α n ⎤
=
Pr = Pt ⎢
610
.
41
= 241.36 N
⎢
⎥
0 ⎥
⎣ cos ψ ⎦
⎣ cos (23 ) ⎦
18.5
a=
m n (z1 + z 2 )
2 cos ψ
285 =
5 (35 + 70)
2 cos ψ
ψ = 22.92 0
(i)
Beam strength:
Since both gears are made of the same material, the pinion is weaker than the gear.
z 'p =
zp
cos ψ
3
=
35
= 44.79
cos (22.92 0 )
3
From Table17.3,
Y = 0.389 +
σb =
(0.399 − 0.389) (44.79 − 40 )
= 0.3986
(45 − 40)
Su
600
=
= 200 N / mm 2
3
3
Sb = mn b σ b Y = 5(50)(200)(0.3986) = 19 930 N
Wear strength:
Q=
dp =
2 zg
zg + zp
zp mn
cos ψ
=
2 (70)
= 1.333
70 + 35
=
35(5)
= 190 mm
cos (22.92 0 )
(ii)
3
2
2
⎛ BHN ⎞
⎛ 300 ⎞
2
K = 0.16 ⎜
⎟ = 0.16 ⎜
⎟ = 1.44 N / mm
⎝ 100 ⎠
⎝ 100 ⎠
b Q dp K
Sw =
=
cos ψ
2
50 (1.333) (190)(1.44)
= 21495.64 N
cos 2 (22.92 0 )
(iii)
Static load:
Mt =
60 x 10 6 (kW ) 60 x 10 6 (20)
=
= 265 258.24 N − mm
2 π np
2 π (720)
2 Mt
Pt =
2 (265 258.24)
= 2792.19 N
190
=
dp
(iv)
Dynamic load:
From Table 17.8, the error for Grade 6 is given by,
e = 8 + 0.63 φ
φ = mn + 0.25 d
where
For pinion,
φ = mn + 0.25
d p = 5 + 0.25 190
ep = 8 + 0.63 (5 + 0.25 190 ) = 13.32 µm
For gear,
dg =
zg mn
=
cos ψ
70(5)
= 380 mm
cos (22.92 0 )
φ = mn + 0.25
d g = 5 + 0.25
380
eg = 8 + 0.63 (5 + 0.25 380 ) = 14.22 µm
e = ep + eg = 13.32 + 14.22 = 27.54 µm
Also, C = 11400 N/mm2
v=
π dp np
60 x 10
3
=
or (27.54x10-3 ) mm
b = 50 mm
π (190) (720)
= 7.163 m / s
60 x 10 3
Pt = 2792.19 N
4
From Eq. (18.21),
Pd =
21 v ( C e b cos 2 ψ + Pt ) cos ψ
=
21 v + ( C e b cos 2 ψ + Pt )
[ 11400 (27.54 x 10 ) (50) cos (22.92) + 2792.19] cos (22.92)
21(7.163) + [ 11400 (27.54 x 10 ) (50) cos (22.92) + 2792.19]
21 (7.163)
−3
2
−3
= 8047.29 N
2
(v)
From Eq. (18.22),
Peff = ( C s P t + Pd ) = 1.25(2792.19) + 8047.29 = 11 537.53 N (vi)
For bending,
(fs) =
Sb
19 930
=
= 1.73
Peff 11 537.53
(vii)
For pitting,
(fs) =
Sw
21 495.64
=
= 1.86
Peff
11 537.53
(viii)
1
CHAPTER 19
19.1
D p = m z p = 4 (30) = 120 mm
D g = m z g = 4 (48) = 192 mm
tan γ =
zp
tan Γ =
zg
zg
zp
=
A0
(i)
=
30
48
γ = 32 0
=
48
30
Γ = 58 0
⎛ Dp
⎜⎜
⎝ 2
2
⎞
⎛D ⎞
⎟⎟ + ⎜⎜ g ⎟⎟
⎠
⎝ 2 ⎠
2
(ii)
2
=
⎛ 120 ⎞
⎛ 192 ⎞
⎜
⎟ + ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
2
A 0 = 113.21 mm
19.2
Mt =
(iii)
60 x 10 6 (kW ) 60 x 10 6 (5)
=
= 59 683.1 N − mm
2 π np
2 π (800)
tan γ =
zp
zg
=
1
= 0.5
2
or
γ = 26.565 0
⎡ Dp
b sin γ ⎤ ⎡ 80 40 sin (26.565) ⎤
rm = ⎢
−
−
⎥=
⎥ = 31.056 mm
2 ⎦ ⎢⎣ 2
2
⎦
⎣ 2
Pt =
19.3
M t 59 683.1
=
= 1921.79 N
rm
31.056
(i)
Pr = Pt tan α cos γ = 1921.79 tan (20) cos (26.565) = 625.63 N
(ii)
Pa = Pt tan α sin γ = 1921.79 tan (20) sin (26.565) = 312.81 N
(iii)
Since the same material is used for both pinion and gear, the pinion is weaker than
the gear.
2
D p = m z p = 6 (30) = 180 mm
D g = m z g = 6 (45) = 270 mm
⎛ Dp
⎜⎜
⎝ 2
Ao =
zp
tan γ =
z 'p =
cos γ
2
30
45
=
zg
zp
2
⎛D ⎞
⎞
⎟⎟ + ⎜⎜ g ⎟⎟ =
⎝ 2 ⎠
⎠
=
2
2
⎛ 180 ⎞
⎛ 270 ⎞
⎜
⎟ +⎜
⎟ = 162.25 mm
⎝ 2 ⎠
⎝ 2 ⎠
γ = 33.69 0
or
30
= 36.06
cos (33.69)
From Table 17.3,
Y = 0.373 +
σb =
(0.380 − 0.373) (36.06 − 35)
= 0.3767
(37 − 35)
S ut
600
=
= 200 N / mm 2
3
3
⎡
b ⎤
S b = m b σ b Y ⎢1 −
⎥ = 6 (50) ( 200) (0.3767)
⎣ Ao ⎦
Sb = 15 636.82 N
19.4
50 ⎤
⎡
⎢⎣1 − 162.25 ⎥⎦
(Ans)
Since the same material is used for both pinion and gear, the pinion is weaker than
the gear.
D p = m z p = 6 (24) = 1 44 mm
D g = m z g = 6 (48) = 288 mm
Ao =
⎛ Dp
⎜⎜
⎝ 2
2
⎞
⎛D
⎟⎟ + ⎜⎜ g
⎠
⎝ 2
2
⎞
⎟⎟ =
⎠
2
2
⎛ 144 ⎞
⎛ 288 ⎞
⎜
⎟ +⎜
⎟ = 161 mm
⎝ 2 ⎠
⎝ 2 ⎠
3
zp
24
tan γ =
=
= 0.5
zg
48
z 'p =
zp
cos γ
=
γ = 26.57 0
or
24
= 26.83
cos (26.57)
From Table 17.3,
Y = 0.344 +
(0.348 − 0.344) (26.83 − 26)
= 0.3473
(27 − 26)
⎡
b ⎤
⎛ 220 ⎞
S b = m b σ b Y ⎢1 −
⎟ (0.3473)
⎥ = 6 (50) ⎜
⎝ 3 ⎠
⎣ Ao ⎦
⎡ 50 ⎤
⎢⎣1 − 161 ⎥⎦
Sb = 5267.74 N
v=
π Dp n p
60 x 10
3
=
(i)
π (144) (300)
= 2.262 m / s
60 x 10 3
For generated teeth,
5 .6
Cv =
5 .6 +
v
=
5 .6
5 .6 +
2.262
= 0.7883
S b = Peff (fs)
Sb =
Cs
Pt (fs)
Cv
5267.74 =
(1.5)
Pt (2)
(0.7883)
Pt = 1384.19 N
(ii)
⎛ Dp ⎞
144 ⎞
⎟⎟ = 1384.19 ⎛⎜
M t = Pt ⎜⎜
⎟ = 99 661.68 N − mm
⎝ 2 ⎠
⎝ 2 ⎠
kW =
2 π n M t 2 π (300) (99 661.68)
=
= 3.13
60 x10 6
60 x10 6
(iii)
4
19.5
Ep = Eg = 114 000 N/mm2
⎡ 1
1 ⎤
σ c2 sin α cos α ⎢
+
⎥
E g ⎥⎦
⎢⎣ E p
K=
1.4
⎡ 1
1 ⎤
(90) 2 sin (20) cos (20) ⎢
+
⎥
⎣114 000 114 000 ⎦
=
1.4
= 0.0326 N/mm2
D p = m z p = 6 (30) = 180 mm
D g = m z g = 6 (40) = 240 mm
tan γ =
zp
Q=
2 zg
zg
=
30
= 0.75
40
z g + z p tan γ
Sw =
=
0.75 b Q D p K
cos γ
γ = 36.87 0
or
2 (40)
= 1.28
40 + 30 tan(36.87)
=
0.75 (50) (1.28) (180) (0.0326)
cos (36.87)
Sw = 352.08 N
19.6
(Ans)
D p = m z p = 3 (40) = 120 mm
tan γ =
zp
zg
=
40
65
or
γ = 31.610
5
2 zg
2 (65)
Q=
=
= 1.45
z g + z p tan γ
65 + 40 tan (31.61)
2
2
⎛ BHN ⎞
⎛ 300 ⎞
2
K = 0.16 ⎜
⎟ = 0.16 ⎜
⎟ = 1.44 N / mm
⎝ 100 ⎠
⎝ 100 ⎠
Sw =
0.75 b Q D p K
cos γ
=
0.75 (35) (1.45) (120) (1.44)
cos (31.61)
Sw = 7723.02 N
19.7
(Ans)
Beam strength:
Since the same material is used for both pinion and gear, the pinion is weaker than the
gear.
D p = m z p = 6(30) = 180 mm
D g = m z g = 6(45) = 270 mm
⎛ Dp
⎜⎜
⎝ 2
Ao =
tan γ =
z 'p =
zp
=
zg
zp
cos γ
2
2
⎞
⎛D ⎞
⎟⎟ + ⎜⎜ g ⎟⎟ =
⎠
⎝ 2 ⎠
=
30
45
2
2
⎛ 180 ⎞
⎛ 270 ⎞
⎟ = 162.25 mm
⎜
⎟ +⎜
⎝ 2 ⎠
⎝ 2 ⎠
or
γ = 33.69 0
30
= 36.06
cos (33.69)
From Table 17.3,
Y = 0.373 +
(0.38 − 0.373) (36.06 − 35)
= 0.3767
(37 − 35)
6
⎡
b ⎤
⎛ 570 ⎞
S b = m b σ b Y ⎢1 −
⎟ (0.3767)
⎥ = 6 (50) ⎜
⎝ 3 ⎠
⎣ Ao ⎦
50 ⎤
⎡
⎢⎣1 − 162.25 ⎥⎦
Sb = 14 854.98 N
Wear strength:
Q=
2 zg
z g + z p tan γ
=
2 (45)
= 1.3846
45 + 30 tan (33.69)
2
2
⎛ BHN ⎞
⎛ 350 ⎞
2
K = 0.16 ⎜
⎟ = 0.16 ⎜
⎟ = 1.96 N / mm
⎝ 100 ⎠
⎝ 100 ⎠
0.75 b Q D p K
Sw =
cos γ
=
0.75 (50) (1.3846) (180) (1.96)
cos (33.69)
Sw = 22 015.79 N
Static load:
Mt =
Pt =
60 x 10 6 (kW )
60 x 10 6 (16.5)
=
= 315 126.79 N − mm
2 π np
2 π (500)
2 Mt
2 (315 126.79)
=
= 3501.41 N
Dp
180
Dynamic load:
e = 20 µm = 20 x10 −3 mm
v=
π Dp n p
60 x 10
3
=
π (180) (500)
= 4.7124 m / s
60 x 10 3
Also, C = 11400 N/mm2
From Eq. (19.21),
b = 50 mm
Pt = 3501.41 N
7
Pd =
=
21 v ( C e b + Pt )
21v + ( C e b + Pt )
[
21 (4.7124) 11400 (20 x10 −3 )(50) + 3501.41)
21(4.7124) +
[ 11400 (20x10
−3
]
)(50) + 3501.41)
]
Pd = 6671.66 N
Effective load:
From Eq. (19.22),
Peff = C s Pt + Pd = 1.5(3501.41) + 6671.66
Peff = 11 923.78 N
Factor of safety:
For bending consideration,
(fs ) =
Sb
14 854.98
=
= 1.25
Peff
11 923.78
(i)
For wear consideration,
(fs ) =
Sw
22 015.79
=
= 1.85
Peff
11923.78
(ii)
1
CHAPTER 20
20.1
For the given pair,
z1 = 2
z2 = 54 teeth
q = 10
m = 5 mm
From Eq. (20.9) and (20.10),
a=
1
1
m ( q + z 2 ) = (5) (10 + 54) = 160 mm
2
2
(i)
i=
z 2 54
=
= 27
z1
2
(ii)
Dimensions of worm:
d1 = q m = 10(5) = 50 mm
d a1 = m (q + 2 ) = 5 (10 + 2) = 60 mm
tan γ =
z1
2
=
= 0.2
q 10
γ = 11.310
or
[
]
d f 1 = m (q + 2 − 4.4 cos γ ) = 5 10 + 2 − 4.4 cos(11.310 ) = 38.427 mm
px = π m = π (5) = 15 .708 mm
Dimensions of worm wheel:
d2 = m z2 = 5 (54) = 270 mm
[
]
d a 2 = m (z 2 + 4 cos γ − 2) = 5 54 + 4 cos (11.310 ) − 2 = 279.612 mm
[
]
d f 2 = m (z 2 − 2 − 0.4 cos γ ) = 5 54 − 2 − 0.4 cos (11.310 ) = 258.039 mm
20.2
z1 = 2
z2 = 52 teeth
q = 10
d1 = q m = 10(4) = 40 mm
tan γ =
z1
2
=
= 0.2
q 10
or
γ = 11.310
m = 4 mm
2
60 x 10 6 (kW ) 60 x 10 6 (10)
Mt =
=
= 132 629.12 N − mm
2 π n1
2 π (720)
From Eq. (20.29),
(P1 ) t =
2 Mt
2 (132 629.12)
=
= 6631.46 N
d1
40
From Eqs. (20.30),
(P1 ) a = (P1 ) t x
( cos α cos γ − µ sin γ )
(cos α sin γ + µ cos γ )
= 6631.46 x
[ cos (20) cos(11.31) − 0.04 sin (11.31)] = 27 105.78 N
[(cos (20) sin(11.31) + 0.04 cos(11.31)]
From Eq. (20.31),
(P1 ) r = (P1 ) t x
sin α
(cos α sin γ + µ cos γ )
= 6631.46 x
sin (20)
= 10 147.47 N
[cos (20) sin (11.31) + 0.04 cos (11.31)]
The force components acting on the worm wheel:
20.3
(P2)t = (P1)a = 27105.78 N
(i)
(P2)a = (P1)t = 6631.46 N
(ii)
(P2)r = (P1)r = 10 147.47 N
(iii)
z1 = 1
z2 = 52 teeth
q = 10
d1 = q m = 10(8) = 80 mm
tan γ =
z1
1
=
= 0.1
q 10
From Eq. (20.33),
or
γ = 5.710
m = 8 mm
3
π d1 n 1
π (80) (1000)
vs =
=
= 4.21 m / s
60 000 cos γ 60 000 cos (5.71)
From Fig. 20.9, the coefficient of friction is 0.027.
(i)
From Eq. (20.34),
η=
[cos (20) − 0.027 tan (5.71)] = 0.7745
( cos α − µ tan γ )
=
[cos (20) + 0.027 cot (5.71)]
(cos α + µ cot γ )
η = 77.45%
20.4
z1 = 1
i=
(ii)
z2 = 40 teeth
q = 10
m = 4 mm
z 2 40
=
= 40
z1
1
n1 = 1000 r.p.m.
n2 =
1000 1000
=
= 25 r.p.m.
i
40
d2 = m z2 = 4(40) = 160 mm
tan γ =
z1
1
=
= 0.1
q 10
or
γ = 5.710
From Eq. (20.20),
F = 2 m (q + 1) = 2 (4) (10 + 1) = 26.533 mm
From Eq. (20.13) and (20.14),
c = 0.2 m cos γ = 0.2(4) cos (5.71) = 0.796 mm
d a1 = m (q + 2 ) = 4 (10 + 2) = 48 mm
From Eq. (20.21),
4
⎡
⎤
F
l r = ( d a1 + 2 c) sin −1 ⎢
⎥
⎣ ( d a 1 + 2 c) ⎦
⎡
⎤
26.533
= ( 48 + 2 x 0.796) sin −1 ⎢
⎥ = 27.997
⎣ (48 + 2 x 0.796) ⎦
mm
For case-hardened carbon steel 14C4, (Table 20.2),
Sb1 = 28.2
For sand cast phosphor-bronze,
Sb2 = 5.0
From Fig. 20.14,
Xb1 = 0.27
for
n1 = 1000 r.p.m.
Xb2 = 0.51
for
n2 = 25 r.p.m.
From Eqs. (20.35) and (20.36),
(M t )1 = 17.65 X b1 S b1 m l r d 2 cos γ
= 17.65 (0.27) (28.2)(4)(27.997) (160) cos (5.71)
= 2 396 011.05 N-mm
(M t ) 2 = 17.65 X b 2 S b 2 m l r d 2 cos γ
= 17.65 (0.51) (5) (4) (27.997) (160) cos (5.71)
= 802 446.57 N-mm
The lower value of the torque on the worm wheel is 802 446.57 N-mm.
kW =
2 π n 2 (M t ) 2 π (25) (802 446.57 )
=
= 2 .1
60 x 10 6
60 x 10 6
(Ans)
5
20.5
For the given pair of worm gears,
m = 4 mm
d2 = 160 mm
q = 10
z1 = 1
For (q = 10) and (z1 = 1), the zone factor Yz from Table 20.4 is given by
Yz =1.143
For case-hardened carbon steel 14C4 (Table 20.3),
Sc1 = 4.93
For sand cast phosphor-bronze,
Sc2 = 1.06
d1 = q m = 10 (4) = 40 mm
From Eq. (20.33),
vs =
π d1 n 1
π (40) (1000)
=
= 2.1 m / s
60 000 cos γ 60 000 cos (5.71)
For vs = 2.1 m/s
and
n1 = 1000 r.p.m. (Fig.20.15),
and
n1 = 25 r.p.m.
Xc1 = 0.145
For vs = 2.1 m/s
Xc2 = 0.35
From Eqs. (20.38) and (20.39),
(M t ) 3 = 18.64 X c1 S c1 Yz (d 2 )1.8 m
= 18.64 (0.145) (4.93) (1.143) (160)1.8 (4)
= 565 175.1 N-mm
(M t ) 4 = 18.64 X c 2 S c 2 Yz (d 2 )1.8 m
= 18.64 (0.35) (1.06) (1.143) (160)1.8 (4)
= 293 320.22 N-mm
(lower value)
6
2 π n 2 (M t ) 2 π (25) (293 320.22)
kW =
=
= 0.77
60 x 10 6
60 x 10 6
20.6
(Ans)
From Fig. 20.9, the coefficient of friction is 0.035 for ( vs = 2.1 m/s)
From Eq. (20.34),
η=
( cos α − µ tan γ )
=
(cos α + µ cot γ )
[cos (20) − 0.035 tan(5.71)]
[cos (20) + 0.035 cot(5.71)]
= 0.7259
From Eq. (20.40),
kW =
k (t − t o ) A
25 (45) (0.25)
=
= 1.03
1000 (1 − η ) 1000 (1 − 0.7259 )
(Ans)
1
CHAPTER 21
21.1
U o = 3000 N − m
ω=
K = 0.9
Cs = 0.2
2 π n 2 π (200)
=
= 20.944 rad / s
60
60
From Eq. (21.14),
Ir =
Uo K
(3000) (0.9)
=
= 30.776 kg − m 2
2
2
ω C s (20.944) (0.2)
From Eq. (21.15),
mr =
Ir
30.776
=
= 123.1 kg
2
R
(0.5) 2
⎛ b ⎞⎛ t ⎞
mr = 2 π R ⎜
⎟⎜
⎟ρ
⎝ 1000 ⎠ ⎝ 1000 ⎠
⎛ t ⎞⎛ t ⎞
123.1 = 2 π (0.5) ⎜
⎟⎜
⎟ (7100)
⎝ 1000 ⎠ ⎝ 1000 ⎠
t = 74.29 or 75 mm
21.2
b = t = 75 mm
(Ans)
The turning moment diagram is shown in Fig. 21.1-solu. It is assumed that the energy
stored in the flywheel is U at point A. Therefore,
Energy at B = U - 30
Energy at C = U - 30 + 400 = U + 370
Energy at D = U + 370 - 270 = U + 100
Energy at E = U + 100 + 330 = U + 430 (maximum)
Energy at F = U + 430 – 310 = U + 120
2
Energy at G = U + 120 + 230 = U + 350
Energy at H = U + 350 - 380 = U – 30
(minimum)
Energy at I = U – 30 + 270 = U + 240
Energy at J = U + 240 – 240 = U
The maximum and minimum energy occurs at points E and H. The angular velocity of
the flywheel will be maximum at point E and minimum at point H.
U o = U E − U H = ( U + 430) − ( U − 30) = 460 mm 2
⎛ 2π ⎞
U o = 460 (1250) ⎜
⎟ N − m or J
⎝ 180 ⎠
= 20 071.29 N-m
ω=
2 π n 2 π (240)
=
= (8 π) rad / s
60
60
From Eq. 21.14,
Ir =
U o K (20 071.29) (0.9)
=
= 1429.91 kg − m 2
2
2
ω Cs
(8 π) (0.02)
From Eq. 21.16, the mean radius R of the rim is given by,
R<
30
ω
or
Therefore,
R<
30
(8 π)
R = 1.15 m
From Eq. (21.15),
mr =
Ir
1429.91
=
= 1081.22 kg
2
R
(1.15) 2
⎛ b ⎞⎛ t ⎞
mr = 2 π R ⎜
⎟⎜
⎟ρ
⎝ 1000 ⎠ ⎝ 1000 ⎠
or
R < 1.19 m
(i)
3
⎛ 1.5 t ⎞ ⎛ t ⎞
1081.22 = 2 π (1.15) ⎜
⎟⎜
⎟ (7100)
⎝ 1000 ⎠ ⎝ 1000 ⎠
t = 118.53 or 120 mm
b = 1.5(118.53) = 177.8 or 180 mm
Dimensions of cross-section = 120x180 mm
21.3
(ii)
From Eq. (21.21),
m = b t ρ = (200)(100)(7200 x 10 −9 ) = 0.144 kg / mm
A = b t = (200)(100) = 20 000 mm2
For four spokes,
2α=
π
2
From Eq. (21.19) (four spokes),
⎡ 72 960 R 2
20 000 ⎤
A ⎤ ⎡ 72 960 (1) 2
0
.
643
C=⎢
+
+
+ 0.643 +
⎥=⎢
⎥
2
2
6500 ⎦
A 1 ⎦ ⎣ (100)
t
⎣
= 11.016
⎛2 π n⎞
⎛ 2 π (720) ⎞
⎟R = ⎜
⎟ (1) = 75.398 m / s
60
⎝ 60 ⎠
⎝
⎠
v = ωR = ⎜
⎛ 1000 m v 2
⎜⎜
bt
⎝
⎞ (1000)(0.144)(75.398) 2
⎟⎟ =
= 40.931
(20 000)
⎠
From Eq. (21.18), the stresses in the rim are given by:
At φ = 45 0 ,
σt =
cos φ
2 (1000) R
(1000) m v 2 ⎡
±
⎢1 −
3 C sin α
Ct
bt
⎣
⎛ 1 cos φ
⎜⎜ −
⎝ α sin α
⎡
cos (45)
2 (1000) (1)
= (40.931) ⎢1 −
±
3 (11.016) sin (45)
(11.016) (100)
⎣
⎞⎤
⎟⎟ ⎥
⎠⎦
⎛ 4 cos (45) ⎞ ⎤
⎟⎟ ⎥
⎜⎜ −
⎝ π sin (45) ⎠ ⎦
4
= 59.997 or 60 N/mm2
(using positive sign)
(i)
At φ = 0 0 ,
⎡
2 (1000) (1)
1
σ t = (40.931) ⎢1 −
±
3 (11.016) sin (45)
(11.016) (100)
⎣
= 49.66 N/mm2
⎛4
1
⎜⎜ −
⎝ π sin (45)
(using negative sign)
From Eq. 21.17, the stress in the spoke is given by,
σt =
2 ⎡ (1000) m v 2 ⎤ 2 ⎡ (1000) (0.144) (75.398) 2 ⎤
⎢
⎥= ⎢
⎥
(11.016) (6500)
3 ⎣ C A1
⎦ 3 ⎣
⎦
= 7.62 N/mm2
(ii)
⎞⎤
⎟⎟ ⎥
⎠⎦
1
CHAPTER 22
22.1
σt =
t=
22.2
S yt
(fs)
=
230
= 92 N / mm 2
2 .5
Pi D i (10) (200)
=
= 10.87 mm
2 σt
2 (92)
σt =
S yt
(fs)
=
(Ans)
200
= 80 N / mm 2
2 .5
Thickness of cylindrical wall:
t=
Pi D i (3) (500)
=
= 9.38 mm
2 σt
2 (80)
(i)
Thickness of hemispherical ends:
t=
22.3
Pi D i (3) (500)
=
= 4.69 mm
4 σt
4 (80)
(ii)
Assumptions: (i) The cylinder is made of ductile material.
(ii) The cylinder is closed at two ends.
When the cylinder is subjected to internal pressure, the three principal stresses are as
follows,
σt = +
[Eqs. (22.9), (22.8) and (22.12)]
Pi ( D o2 + D i2 )
( D o2 − D i2 )
σ r = − Pi
σl = +
Pi D i2
( D o2 − D i2 )
Rearranging the given equation,
(a)
(b)
(c)
2
σ=
S yt
=
(fs)
[
1
( σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + ( σ 3 − σ 1 ) 2
2
]
σ 2 = ( σ 2t + σ 2r + σ i2 ) − ( σ t σ r + σ r σ l + σ l σ t )
we get,
(d)
Substituting (a), (b) and (c) in above expression,
σ=
3 Pi D o2
(D o2 − D i2 )
Rearranging the terms,
D o2
σ
=
2
Di
σ − 3 Pi
σ
Do
=
Di
or
σ − 3 Pi
Substituting D o = (D i + 2 t ) in above expression,
t=
22.4
Di
2
⎡
⎤
σ
− 1⎥
⎢
⎢⎣ ( σ − 3 Pi )
⎥⎦
where σ =
S yt
(fs)
(Ans)
Assumptions: (i) The cylinder is made of ductile material.
(ii) The cylinder is open at two ends.
When the cylinder is subjected to internal pressure, the maximum stresses at the inner
surface of the cylinder are given by,
σt = +
Pi ( D o2 + D i2 )
( D o2 − D i2 )
(a)
σ r = − Pi
(b)
The maximum shear stress is given by,
τ=
1
1
(σ t − σ r ) =
2
2
[Eqs. (22.9) and (22.8)]
⎡ Pi ( D o2 + D i2 )
⎤
+ Pi ⎥
⎢
2
2
⎣ ( Do − Di )
⎦
3
Pi D o2
τ =
(D o2 − D i2 )
or
D o2 − D i2 Pi
=
τ
D o2
Rearranging the terms,
Do
=
Di
τ
τ − Pi
Substituting D o = (D i + 2 t ) in above expression,
t=
22.5
⎤
Di ⎡
τ
− 1⎥
⎢
2 ⎣ ( τ − Pi )
⎦
where τ =
Ssy
(fs)
=
0.5 S yt
(fs)
(Ans)
Since the cylinder material is brittle, maximum principal stress theory is applicable.
Pi ( D o2 + D i2 ) 15 ( 240 2 + 200 2 )
σt =
=
= 83.18 N / mm 2
2
2
2
2
( Do − Di )
( 240 − 200 )
(fs) =
S ut
300
=
= 3.61
σt
83.18
(i)
It is seen from expression of ( σ t ) that when pressure is raised by 50%, ( σ t ) also
increases by 50%.
(fs) =
22.6
S ut
300
=
= 2 .4
1.5 σ t 1.5 (83.18)
(ii)
From Eq. (22.7),
⎡ D o2
⎤
Pi D i2
50 (20) 2
σt =
+ 1⎥ =
⎢
( D o2 − D i2 ) ⎣ 4 r 2
( 40 2 − 20 2 )
⎦
⎡ 400 ⎤
σ t = 16.667 ⎢ 2 + 1⎥
⎣ r
⎦
(a)
⎡ 40 2
⎤
⎢ 2 + 1⎥
⎣4 r
⎦
4
From Eq. (22.6),
σr =
⎡ D o2
⎤
⎤
Pi D i2
50 (20) 2 ⎡ 40 2
=
−
1
⎢
⎥
⎢ 2 − 1⎥
2
2
2
2
2
( Do − Di ) ⎣ 4 r
⎦
⎦ ( 40 − 20 ) ⎣ 4 r
⎡ 400 ⎤
σ r =16.667 ⎢ 2 −1⎥
⎣ r
⎦
or
(b)
Substituting value of r from 0 to 20 mm,
R (mm) 10
22.7
12
14
16
18
20
(σ t ) ( N / mm 2 )
83.34
62.96
50.68
42.71
37.24
33.33
(σ r ) ( N / mm 2 )
50
29.63
17.35
9.38
3.91
0
D1 = 200 mm
D2 = 300 mm
δ = 0.25 mm
E = 207x103 N/mm2
D3 = 400 mm
From Eq. (22.22),
δ =
⎤
2 D 22 ( D 32 − D12 )
P D2 ⎡
⎢
2
2
2
2 ⎥
E ⎣ ( D 3 − D 2 ) ( D 2 − D1 ) ⎦
0.25 =
⎤
P (300) ⎡
2 (300) 2 ( 400 2 − 200 2 )
⎢
3
2
2
2
2 ⎥
(207 x 10 ) ⎣ ( 400 − 300 ) ( 300 − 200 ) ⎦
P = 27.95 MPa
(i)
The maximum principal stress is tangential stress at the inner surface of jacket.
From Eq. (22.9),
σt =
P (D 32 + D 22 ) 27.95 (400 2 + 300 2 )
=
(D 32 − D 22 )
(400 2 − 300 2 )
= 99.82 N/mm2
(ii)
5
22.8
kb
= 1 .5
kc
k b = 1.5 k c
Pl = 10 000 N
⎡ kb ⎤
1.5 k c
1.5
=
= 0.6
⎢
⎥=
2.5
⎣ k b + k c ⎦ 1.5 k c + k c
( Pi ) per bolt =
π
⎛1⎞
(300) 2 (1) ⎜ ⎟ = 5890.49 N
4
⎝ 12 ⎠
The resultant load on the bolt is given by Eqs. (22.26) and (22.27).
⎡ kb
P = Pl + Pi ⎢
⎣ kb + kc
22.9
⎤
⎥ = 10 000 + 5890.49 (0.6) = 13 534.29 N
⎦
Pi = 1.05(1.5) = 1.575 MPa or N/mm2
σt =
S yt
1.5
=
255
= 170 N / mm 2
1.5
Thickness of cylindrical shell:
From Eq. (22.29),
t=
Pi D i
(1.575) (1650)
+ CA =
+2
2 σ t η − Pi
2 (170) (0.8) − 1.575
t = 11.61 mm
Thickness of torispherical end:
t=
(i)
[Eq. (22.33)]
0.885 Pi L
0.885 (1.575) (1300)
+ CA =
+2
σ t η − 0.1 Pi
(170) (0.8) − 0.1 (1.575)
t = 15.34 mm
(ii)
6
22.10
Syt
210
σt =
=
= 140 N / mm 2
1.5 1.5
tr =
Pi D i
2 σ t η − Pi
=
(0.75) ( 2000)
= 6.32 mm
2 (140) (0.85) − 0.75
d = d i + 2 (CA ) = 300 + 2 (2) = 304 mm
A = d t r = 304 (6.32) = 1921.28 mm 2
trn =
Pi d i
2 σ t η − Pi
=
(a)
(0.75) (300)
= 0.95 mm
2 (140) (0.85) − 0.75
The limiting dimension X is the higher of the following two values:
X = d = 304 mm
⎡d
⎤
X = ⎢ i + t + t n − 3 CA ⎥ = (150+10+10–6) = 164 mm
2
⎣
⎦
Therefore,
X = 304 mm
h 1 = 2.5 ( t − CA ) = 2.5 (10 − 2) = 20 mm
h 2 = 15 mm
The areas available for reinforcement within the above limits are as follows:
A 1 = ( 2 X − d ) ( t − t r − CA ) = [ 2 (304) − 304 ] (10 − 6.32 − 2) = 510.72 mm 2
A 2 = 2 h 1 ( t n − t r n − CA ) = 2(20)(10 – 0.95 - 2) = 282 mm2
A 3 = 2 h 2 ( t n − 2 CA ) = 2 (15) (10 – 2x2) = 180 mm2
∴
(Al +A2+A3) = 972.72 mm2
From (a) and (b),
(b)
7
A > (Al +A2+A3)
Therefore a reinforcing pad is necessary. The area of the reinforcing pad A4 is given
by
A4 = A - (Al +A2+A3) = 1921.28 – 972.72 = 948.56 mm2
The thickness of the reinforcing pad is 10 mm. Therefore the width of the pad is given
by,
w=
948.56
= 94.86 or 95 mm
10
Dimensions of pad:
inner diameter of the pad = 300 + 20 = 320 mm.
(i)
outer diameter of pad = 320 + 95 = 415 mm
(ii)
1
CHAPTER 23
23.1
Let us assume that the number of wire ropes is z. The force acting on each wire rope
comprises the following factors;
(i)
the weight of the hoist with material to be raised;
(ii)
the weight of the wire rope, and
(iii)
the force due to acceleration of the material and the wire rope.
The weight of the hoist with material raised by each wire rope is given by,
⎛ 10 x10 3 ⎞
⎜⎜
⎟
z ⎟⎠
⎝
N
(i)
The mass of 100 m long wire rope is 34.6 kg. Since the height is 3 m, the weight of
the wire is given by,
⎛ 3 ⎞
34.6 ⎜
⎟ (9.81)
⎝ 100 ⎠
or
10.18 N
(ii)
⎡⎛ 10 x10 3 ⎞ ⎛ 1 ⎞⎤
⎟⎟ ⎜ ⎟⎥ and that of each
⎣⎝ 9.81 ⎠ ⎝ z ⎠⎦
The mass of the material raised by each wire rope is ⎢⎜⎜
⎡
⎛ 3 ⎞⎤
⎝
⎠⎦
wire rope is ⎢34.6 ⎜
⎟ . The force due to acceleration (i.e. mass x acceleration) is
100 ⎥
⎣
given by,
⎡⎛ 10 x10 3 ⎞ ⎛ 1 ⎞
⎛ 3 ⎞⎤
⎟⎟ ⎜ ⎟ + 34.6 ⎜
⎟⎥ (1 )
⎢⎜⎜
⎝ 100 ⎠⎦
⎣⎝ 9.81 ⎠ ⎝ z ⎠
From Table 23.4, the breaking strength of the wire rope is 54 kN. The factor of safety
is 10.
2
⎡10 x10 3
10x10
54 000
⎛ 3 ⎞⎤
=
+ 34.6 ⎜
+ 10.18 + ⎢
⎟⎥
10
z
⎝ 100 ⎠⎦
⎣ 9.81 z
3
z = 2.045 or 3 wire ropes
23.2
(Ans)
From Eq. 23.3 and Table 23.7,
Pb =
A Er dw
(0.40 d 2r )(83 000)(0.063 d r )
=
= 46.48 d 2r = 46.48(10) 2 = 4648 N
D
(45 d r )
The total force acting on the wire rope consists of three factors discussed in the
previous example, plus the bending load. The total force is given by,
⎡10 x10 3
⎤
10 x10 3
+
10
.
18
+
+ 1.038⎥ + 4648 or
⎢
9.81 (3)
⎣ 3
⎦
(fs) =
23.3
54 000
= 6.48
8332.34
(Ans)
p = 0.0015 S ut = 0.0015 (1770) = 2.655 N / mm 2
From Eq. (23.4),
P=
23.4
8332.34 N
p=
2P
dr D
p d r D 2.655 (10) (450)
=
= 5973.75 N
2
2
The maximum force on the piston rod is given by,
P=
π 2
π
D p=
(200) 2 (1) N
4
4
The factor of safety is 5. Therefore,
⎤
⎡π
Pc r = 5 P = 5 ⎢ (200) 2 (1) ⎥ = 157 079.63 N
⎦
⎣4
For circular cross-section,
(Ans)
3
⎛d⎞
k=⎜ ⎟
⎝4⎠
⎛ l ⎞ ⎛ 1000 ⎞ ⎛ 4000 ⎞
⎜ ⎟=⎜
⎟=⎜
⎟
⎝k⎠ ⎝ d/4 ⎠ ⎝ d ⎠
At this stage, it is not clear whether, one should use Euler's or Johnson's equation.
Using Johnson's equation as a first trial,
⎡
S yt ⎛ l ⎞ 2 ⎤
Pc r =S yt A ⎢ 1−
⎜ ⎟ ⎥
2
⎢⎣ 4 n π E ⎝ k ⎠ ⎥⎦
2
⎡
⎛ 4000 ⎞
380 ⎜
⎟
⎢
⎛ π d2 ⎞ ⎢
d ⎠
⎝
⎜
⎟
157 079.63 = (380) ⎜
2
⎟ 1−
⎝ 4 ⎠ ⎢ 4 ( 2 ) π (207 000)
⎢
⎢⎣
⎤
⎥
⎥
⎥
⎥
⎥⎦
d = 29.97 mm
(Ans)
Check for design:
⎛ l ⎞ 4000 4000
=
= 133.47
⎜ ⎟=
d
29.97
⎝k⎠
(i)
The boundary line between Euler's and Johnson's equations is given by
S yt
2
=
n π2 E
(l / k )2
or
⎛l⎞
⎜ ⎟ = 146.65
⎝k⎠
380 ( 2 ) π 2 (207 000)
=
2
(l / k )2
(ii)
In this example, the slenderness ratio (133.47) is less than the boundary value of
(146.65). Therefore the rod is treated as a short column and Johnson's equation used
in the first trial is justified.
4
23.5
d c = d − p = 30 − 6 = 24 mm
⎛ d ⎞ 24
k=⎜ c⎟=
= 6 mm
4
⎝ 4 ⎠
S yt
2
=
n π2 E
or
(l / k )2
⎛ l ⎞ ⎛ 500 ⎞
⎜ ⎟=⎜
⎟ = 83.33
⎝k⎠ ⎝ 6 ⎠
(i)
380 ( 0.25 ) π 2 (207 000)
=
2
(l / k )2
⎛l⎞
⎜ ⎟ = 51.85
⎝k⎠
(ii)
In this example, the slenderness ratio (83.33) is more than the boundary value of
(51.85). Therefore the screw is treated as a long column and Euler’s equation is
applicable.
nπ EA
2
Pcr =
(fs) =
23.6
(l / k )2
=
⎛ π (24) 2
( 0.25 ) π 2 (207 000) ⎜⎜
⎝ 4
2
(83.33)
Pcr 33 275.13
=
= 3.33
P
10 000
⎛d⎞ 6
k = ⎜ ⎟ = = 1.5 mm
⎝4⎠ 4
S yt
2
=
⎞
⎟⎟
⎠ = 33 275.13 N
n π2 E
(l / k )2
⎛l⎞
⎜ ⎟ = 101.07
⎝k⎠
or
(Ans)
⎛ l ⎞ ⎛ 300 ⎞
⎜ ⎟=⎜
⎟ = 200 (i)
⎝ k ⎠ ⎝ 1.5 ⎠
400 (1 ) π 2 (207 000)
=
2
(l / k )2
(ii)
5
In this example, the slenderness ratio (200) is more than the boundary value of
(101.7). Therefore the screw is treated as a long column and Euler’s equation is
applicable.
Pcr =
P=
n π2 E A
(l / k )2
=
⎛ π (6) 2
(1 ) π 2 (207 000) ⎜⎜
⎝ 4
2
(200)
Pcr
1444.12
=
= 412.6 N
(fs)
3.5
⎞
⎟⎟
⎠ = 1444.12 N
(Ans)
1
CHAPTER 24
24.1
x1 = 10 – 0.025 = 9.975 mm
x2 = 10 + 0.025 = 10 .025 mm
Z1 =
9.975 − 10.02
= − 4.5
0.01
Z2 =
10.025 − 10.02
= 0.5
0.01
From Table 24.6, the area below normal curve from Z = 0 to Z = 0.5 is
0.1915. Since Z1 is more than (-3.9), the area below normal curve in
negative half is 0.5. Refer to Fig.24.1-solu.
Shaded area = 0.5 – 0.1915 = 0.3085
% of rejection = 30.85 %
24.2
(Ans)
x1 = 25 + 0.15 = 25.15 mm
x2 = 25 - 0.15 = 24.85 mm
Z1 =
25.15 − 25
= + 1.5
0.1
Z2 =
24.85 − 25
= −1.5
0.1
From Table 24.6, the area below normal curve from Z = 0 to Z = 1.5 is
0.4332. Refer to Fig.24.2-solu.
% of rejected forgings = (1 – 2x0.4332)x100 = 13.36 %
(Ans)
2
24.3
Five bolts are rejected out of a sample of 100 bolts. For 5% rejection, the
area below normal curve in positive half is given by,
A=
1
(1 − 0.05) = 0.475
2
From Table 24.6, the corresponding values of Z1 and Z2 are +1.96 and –
1.96.
x 1 = µ + σˆ Z1 = 10.5 + 0.02x1.96 = 10.5 + 0.0392 mm
x 2 = µ − σˆ Z 2 = 10.5 - 0.02x1.96 = 10.5 - 0.0392 mm
Tolerances = 10.5 ± 0.0392 mm
24.4
(Ans)
There are two populations – population of bearing dimension denoted by
letter B and that of journal dimension denoted by letter J.
The limiting dimensions for bearing 20H7 are (Table 3.2)
20.021
mm or
20.000
20.0105 ± 0.0105 mm
The design tolerance and natural tolerance are equal. Therefore,
µ B = 20.0105 mm
and
σˆ B =
0.0105
= 0.0035 mm
3
The limiting dimensions for journal 20e8 are [Table 3.3(a)]
19.960
mm or
19.927
Therefore,
19.9435 ± 0.0165 mm
3
µ J = 19.9435 mm
σˆ J =
and
0.0165
= 0.0055 mm
3
The population for clearance is denoted by the letter C. It is obtained by
subtracting the population of journal from the population of bearing. From
Eq. (24.12),
µ C = µ B − µ J = 20.0105 – 19.9435 = 0.067 mm
From Eq. (24.14),
σ̂ c =
( σˆ B ) 2 + ( σˆ J ) 2
=
( 0.0035 ) 2 + ( 0.0055 ) 2 = 0.00652 mm
For higher limit of clearance,
Z1 =
0.08 − 0.067
= + 1.99
0.00652
For lower limit of clearance,
Z1 =
0.05 − 0.067
= − 2.61
0.00652
From Table 24.6, the areas below the normal curve from Z = 0 to Z = 1.99
and Z = 0 to Z = 2.61 are 0.4767 and 0.4955 respectively.
% of rejected assemblies = [ 1 – (0.4767 + 0.4955)]x100
= 2.78%
24.5
For the population of component A,
µ A = 10.00 mm
σˆ A =
0.6
= 0.2 mm
3
(Ans)
4
The standard deviation and the mean of the assembly dimension (a) are
denoted by ( σ̂ a ) and ( µ a ) respectively.
µa = µA + µB
σˆ a =
40 = 10 + µ B
or
µ B = 30 mm
0.9
= 0.3 mm
3
σ̂ a =
( σˆ A ) 2 + ( σˆ B ) 2
σ̂ B =
( σˆ a ) 2 − ( σˆ A ) 2
( 0.3 ) 2 − ( 0.2) 2 = 0.2236 mm
=
3 σ̂ B = 3 (0.2236) = 0.67 mm
Dimensions of component B = 30 ± 0.67 mm
24.6
(Ans)
The standard deviation and the mean of the assembly dimension (a) are
denoted by ( σ̂ a ) and ( µ a ) respectively.
µ a = µ A + µ B + µ C = 20 + 10 +15 = 45 mm
σˆ a =
σ̂ a =
0.9
= 0.3 mm
3
( σˆ A ) 2 + ( σˆ B ) 2 + ( σˆ C ) 2
0.3 = 3 σ̂
or
= 3( σˆ ) 2
σ̂ = 0.173 mm
Tolerance for individual components = ± 3 σˆ
= ± 0.519 or ± 0.52 mm
(Ans)
5
24.7
The interference population is denoted by letter I.
µ I = µ A − µ B = 75 – 75.125 = - 0.125 mm
σ̂ I =
( σˆ A ) 2 + ( σˆ B ) 2 =
(0.025) 2 + (0.0375) 2 = 0.045 mm
When interference is zero (I = 0),
Z=
0 − ( − 0.125)
= + 2.78
0.045
Therefore, interference occurs when (Z > 2.78). From Table 24.6, the area
below normal curve from Z = 0 to Z = 2.78 is 0.4973.
Probability of interference fit = (0.5 – 0.4973) x 100 = 0.27%
24.8
(Ans)
w denotes the population of width. For this population,
µ w = 50 mm
and
σˆ w = 0.5 mm
t denotes the population of thickness of the cross-section. For this
population,
µ t = 25 mm
and
σˆ t = 0.3 mm
A denotes the third population of the area of cross-section. It is obtained by
multiplication of the width population by the population of thickness. From
Table 24.7,
µ A = µ w µ t = 50 (25) = 1250 mm 2
σˆ A =
µ 2w (σˆ t ) 2 + µ 2t (σˆ w ) 2 + (σˆ w ) 2 (σˆ t ) 2
(i)
6
=
(50) 2 (0.3) 2 + (25) 2 (0.5) 2 + (0.5 ) 2 (0.3) 2
= 19.53 mm2
(ii)
The population of cross-sectional area has a mean of 1250 mm2 and a
standard deviation of 19.53 mm2. We can predict the mean and the standard
deviation of the population of cross-sectional area. However, this
population is not exactly normally distributed random variable.
24.9
S denotes the population of strength. For this population,
µ S = 300 N / mm 2
σˆ S = 10 N / mm 2
and
The population of stress is denoted by σ . For this population,
µ σ =150 N / mm 2
σˆ σ = 20 N / mm 2
and
A third population of factor of safety is denoted by fs. It is obtained by
dividing the strength population by the population of stress. From Table
24.7,
µ fs =
µ S 300
=
=2
µ σ 150
1 ⎡ µ S2 σˆ σ2 + µ σ2 σˆ S2 ⎤
σˆ fs =
⎢
⎥
µσ ⎣
µ σ2 + σˆ σ2
⎦
(i)
1/ 2
1 ⎡ 300 2 x 20 2 + 150 2 x 10 2 ⎤
=
⎥
⎢
150 ⎣
150 2 + 2 0 2
⎦
1/ 2
= 0.27
(ii)
The population of factor of safety has a mean of 2 and a standard deviation
of 0.27. We can predict the mean and the standard deviation of factor of
7
safety population. However, the population is not normally distributed
random variable.
24.10 µ =15 000 hr
and
σˆ = 1000 hr
From Eq.24.7,
Z=
X − µ 16500 − 15 000
=
= 1.5
σˆ
1000
From Table 24.6, area below normal curve from Z = 0 to Z = 1.5 is 0.4332.
The area below the normal curve from Z = - ∞ to Z = 1.5 represents the
probability of bearings that may fail within first 16 500 hr. This area is
equal to (0.5 + 0.4332) or 0.9332.
Therefore, 93.32% bearings may fail within first 16 500 hr.
(Ans)
24.11 Half the area of the curve is located on either side of the mean load of 5 kN.
The area below normal curve from Z = 0 to Z = + ∞ indicates the
probability that the load will be more than 5 kN. This area is 0.5. Therefore,
we can conclude that the probability of randomly selected load being more
than 5 kN is 0.5 or 50 %.
When randomly selected load is 6 kN,
Z=
X−µ 6−5
=
= +2
σˆ
0.5
(i)
X = 6 kN
8
The area below normal curve from Z = 0 to Z = + 2 indicates the
probability that the load selected at random will be between 5 to 6 kN.
From Table 24.6, the area below normal curve from Z = 0 to Z = 2 is
0.4772. Therefore, the probability that a randomly selected load will be
between 5 to 6 kN is 47.72 %.
(ii)
24.12 S denotes the population of yield strength. For this population,
µ S = 380 N / mm 2
σˆ S = 50 N / mm 2
and
The population of bending stress is denoted by σ .
µ σ = 250 N / mm 2
σˆ σ = 25 N / mm 2
and
A third population of margin of safety is denoted by m. It is obtained by
subtracting bending stress population from the population of yield strength.
Therefore,
µ m = µ S − µ σ = 380 − 250 =130 N / mm 2
σˆ m =
( σˆ S ) 2 + ( σˆ σ ) 2 =
(50) 2 + (25) 2 = 55.90 N / mm 2
When (m = 0), the standard variable Z0 is given by
Z0 =
m −µm
0 − 130
=
= − 2.33
σˆ m
55.90
Conclusions:
(i) When Z = - 2.33, the value of ‘m’ is zero or there is no margin of safety.
9
(ii) The area below the normal curve from Z = − ∞ to Z = - 2.33 indicates
the region of unreliability. In this region, ‘m’ is negative or the bending
stress is more than the yield strength.
(iii) The area below the normal curve from Z = − 2 . 33 to Z = + ∞ indicates
the region of reliability. In this region, ‘m’ is positive or the bending stress
is less than the yield strength.
From Table 24.6, the area below normal curve from Z = 0 to Z = +2.33 is
0.4901.
The total area below normal curve from Z = - 2.33 to Z = + ∞ consists of
two parts namely, the area from Z = - 2.33 to Z = 0 and area from Z = 0 to
Z = + ∞ . Therefore, total area is equal to (0.4901+0.5) or 0.9901.
Reliability of beam = 0.9901 or 99.01%
Average factor of safety =
µS
380
=
=1.52
µσ
250
(i)
(ii)
Minimum yield strength = µ S − 3 σˆ S = 380 − 3 (50) = 230 N / mm 2
Maximum bending stress = µ σ + 3 σˆ σ = 250 + 3 (25) = 325 N / mm 2
Minimum factor of safety =
230
= 0.71
325
(iii)