Materi Khurmi Power Screw

7252019 Materi Khurmi Power Screw

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CONTENTS

624 n A Textbook of Machine Design

C

HAP

TER 17Power Screws

1 Introdction Tes of Screw Threads

sed for Power Screws$ Mtie Threadsamp Tore Reired to Raise

(oad b Sare ThreadedScrews

) Tore Reired to (ower(oad b Sare ThreadedScrews

E+cienc of Sare Threaded Screws

Maxi-- E+cienc of Sare Threaded Screws

E+cienc s Heix Ange0 erhaing and Sef2ocking Screws

13 E+cienc of Sef (ockingScrews

11 Coe+cient of 4riction1 Ac-e or Trae5oida

Threads1$ Stresses in Power Screws1amp Design of Screw 6ack1) Di7erentia and Co-ond

Screws

11 IntrodctionThe power screws (also known as translation screws)

are used to convert rotary motion into translatory motion

For example in the case of the lead screw of lathe the rotarymotion is available but the tool has to be advanced in the

direction of the cut against the cutting resistance of the

material In case of screw jack a small force applied in the

horizontal plane is used to raise or lower a large load Power

screws are also used in vices testing machines presses

etc

In most of the power screws the nut has axial motion

against the resisting axial force while the screw rotates in

its bearings In some screws the screw rotates and moves

axially against the resisting force while the nut is stationary

and in others the nut rotates while the screw moves axially

with no rotation

624

CONTENTS

Power Screws



625

1 Tes of Screw Threads sed for Power Screws

Following are the three types of screw threads mostly used for power screws

1 Square thread A square thread as shown in Fig 171 (a) is adapted for the transmission of

power in either direction This thread results in maximum efficiency and minimum radial or bursting

Fig 171 Types of power screws

pressure on the nut It is difficult to cut with taps and dies It is usually cut on a lathe with a single

point tool and it can not be easily compensated for wear The

square threads are employed in screw jacks presses and

clamping devices The standard dimensions for square threads

according to IS 4694 ndash 1968 (Reaffirmed 1996) are shownin Table 171 to 173

2 Acme or trapezoidal thread An acme or trapezoidal

thread as shown in Fig 171 (b) is a modification of square

thread The slight slope given to its sides lowers the efficiency

slightly than square thread and it also introduce some burstingpressure on the nut but increases its area in shear It is usedwhere a split nut is required and where provision is made to

take up wear as in the lead screw of a lathe Wear may be

taken up by means of an adjustable split nut An acme thread

may be cut by means of dies and hence it is more easily

manufactured than square thread The standard dimensions

for acme or trapezoidal threads are shown in Table 174

(Page 630)

3 Buttress thread A buttress thread as shown in Fig

171 (c) is used when large forces act along the screw axis in

one direction only This thread combines the higher efficiency

of square thread and the ease of cutting and the adaptability toScrew 8acks

a split nut of acme thread It is stronger than other threads because of greater thickness at the base of the thread The buttress thread has limited use for power transmission It is employed as the thread for

light jack screws and vices

Tabe 11 9asic di-ensions for sare threads in -- 4ine series according

to IS lt amp0amp = 10 Rea+r-ed 100


Power Screws n 627

Nominal

diameter

(d )1

Major diameter Minor

diameter

(d )

Pitch

( p )

Depth of Area of

core2

( A ) mmcBolt

(d)

Nut

(D)

Bolt

(h)

Nut

(H)

10

12

10

12

105

125

8

10

2 1 125 503

785

d1

d D dc

p h H Ac

14

16

18

20

14

16

18

20

145

165

185

205

12

14

16

18

2 1 125

113

154

201

254

22

24

26

28

30

32

(34)

36

(38)

40

42

44

(46)

48

50

52

55

(58)

60

(62)

22

24

26

28

30

32

34

36

38

40

42

44

46

48

50

52

55

58

60

62

225

245

265

285

305

325

345

365

385

405

425

445

465

485

505

525

555

585

605

625

19

21

23

25

27

29

31

33

35

37

39

41

43

45

47

49

52

55

57

59

3 15 175

284

346

415

491

573

661

755

855

962

1075

1195

1320

1452

1590

1735

1886

2124

2376

2552

2734

65

(68)

70

(72)

75

(78)

80

(82)

(85)

(88)

90

(92)

95

(98)

65

68

70

72

75

78

80

82

85

88

90

92

95

98

655

685

705

725

755

785

805

825

855

885

905

925

955

985

61

64

66

68

71

74

76

78

81

84

86

88

91

94

4 2 225

2922

3217

3421

3632

3959

4301

4536

4778

5153

5542

5809

6082

6504

6960



Note Diameter within brackets are of second preference

Tabe 1 9asic di-ensions for sare threads in -- gtor-a

iesaccording to IS lt amp0amp = 10 Rea+r-ed 100


Power Screws n 629


Tabe 1$ 9asic di-ensions for sare threads in -- Coarse series according

toIS lt amp0amp = 10 Rea+r-ed 100


d1

d D dc

p h H Ac

100

(105)

110

100

105

110

1005

1055

1105

96

101

106

4 2 225

7238

8012

8825

(115)

120

(125)

130

(135)

140

(145)

150

(155)

160

(165)

170

(175)

115

120

125

130

135

140

145

150

155

160

165

170

175

1155

1205

1255

1305

1355

1405

1455

1505

1555

1605

1655

1705

1755

109

114

119

124

129

134

139

144

149

154

159

164

169

6 3 325

9331

10207

11 122

12 076

13 070

14 103

15 175

16 286

17437

18 627

19 856

21124

22 432

Nominal

diameter

(d )1


diameter

(d )c

Pitch

( p)

Depth of Area of

core2

( A ) mmcBolt

(d)

Nut

(D)

Bolt

(h)

Nut

(H)

22

24

26

28

22

24

26

28

225

245

265

285

17

19

21

23

5 25 275

227

284

346

415

30

32

(34)

36

30

32

34

36

305

325

345

365

24

26

28

30

6 3 325

452

531

616

707

(38)

40

(42)

44

38

40

42

44

385

405

425

445

31

33

35

37

7 35 375

755

855

962

1075

d1

d D dc

p h H Ac

(46)

48

50

52

46

48

50

52

465

485

505

525

38

40

42

44

8 4 425

1134

1257

1385

1521

55

(58)

(60)

(62)

55

58

60

62

555

585

605

625

46

49

51

53

9 45 525

1662

1886

2043

2206

65

(68)

70

(72)

75

(78)

80

(82)

65

68

70

72

75

78

80

82

655

685

705

725

755

785

805

825

55

58

60

62

65

68

70

72

10 5 525

2376

2642

2827

3019

3318

3632

3848

4072

85

(88)

90(92)

95

(98)

100

(105)

110

85

88

9092

95

98

100

105

110

855

885

855925

955

985

1005

1055

1105

73

76

7880

83

86

88

93

98

12 6 625

4185

4536

47785027

5411

5809

6082

6793

7543

(115)

120

(125)

130

(135)

140

(145)

115

120

125

130

135

140

145

116

121

126

131

136

141

146

101

106

111

116

121

126

131

14 7 75

8012

882

9677

10 568

11 499

12 469

13 478

150

(155)

160

150

155

160

151

156

161

134

139

144

16 8 85

14 103

15 175

16 286

Nominal

diameter

(d )1


diameter

(d )c

Pitch

( p)

Depth of Area of

core2

( A ) mmcBolt

(d)

Nut

(D)

Bolt

(h)

Nut

(H)

22

24

26

28

22

24

26

28

225

245

265

285

14

16

18

20

8 4 425

164

204

254

314

30

32

(34)

36

(38)

30

32

34

36

38

305

325

345

365

385

20

22

24

26

28

10 5 525

314

380

452

531

616

40

(42)

44

(46)

48

50

52

40

42

44

46

48

50

52

405

425

445

465

485

505

525

28

30

32

34

36

38

40

12 6 625

616

707

804

908

1018

1134

1257

55

(58)

60

(62)

55

58

60

62

56

59

61

63

41

44

46

48

14 7 725

1320

1521

1662

1810

65

(68)

70

(72)

75

(78)

80

(82)

65

68

70

72

75

78

80

82

66

69

71

73

76

79

81

83

49

52

54

56

59

62

64

66

16 8 85

1886

2124

2290

2463

2734

3019

3217

3421

d1

d D dc

p h H Ac

(165)

170

(175)

165

170

175

166

171

176

149

154

159

16 8 85

17 437

18 627

19 856



Note Diameters within brackets are of second preference

Tabe 1amp 9asic di-ensions for trae5oidaAc-e threads

Power Screws n 631


1$ Mtie Threads

The power screws with multiple threads such as double triple etc are employed when it is

desired to secure a large lead with fine threads or high efficiency Such type of threads are usually

found in high speed actuators

1amp Tore Reired to Raise (oad b Sare Threaded Screws

The torque required to raise a load by means of square threaded screw may be determined by

considering a screw jack as shown in Fig 172 (a) The load to be raised or lowered is placed on the

head of the square threaded rod which is rotated by the application of an effort at the end of lever for

lifting or lowering the load

d1

d D dc

p h H Ac

85

(88)

90

(92)

95

(96)

85

88

90

92

95

96

86

89

91

93

96

99

67

70

72

74

77

80

18 9 95

3526

3848

4072

4301

4657

5027

100

(105)

110

100

105

110

101

106

111

80

85

90

20 10 105

5027

5675

6362

(115)

120(125)

130

115

120125

130

116

121126

131

93

98103

108

22 11 115

6793

75438332

9161

(135)

140

(145)

150

(155)

135

140

145

150

155

136

141

146

151

156

111

116

121

126

131

24 12 125

9667

10 568

11 499

12 469

13 478

160

(165)

170

(175)

160

165

170

175

161

166

171

176

132

137

142

147

28 14 145

13 635

14 741

15 837

16 972

Nominal or major

dia-

Minor or core dia-

meter (d ) mmc

Pitch

( p ) mm

Area of core2

( A ) mmc

10

12

65

85

3 33

57

14

16

18

20

95

115

135

155

4

71

105

143

189

22

24

26

28

165

185

205

225

5

214

269

330

389

30

32

34

36

235

255

275

295

6

434

511

594

683

eads

d dc

p Ac

38

40

42

44

305

325

345

365

7

731

830

935

1046

46

48

50

52

375

395

415

435

8

1104

1225

1353

1486

55

58

60

62

455

485

505

525

9

1626

1847

2003

2165

65

68

70

72

75

78

80

82

545

575

595

615

645

675

695

715

10

2333

2597

2781

2971

3267

3578

3794

4015

85

88

90

92

95

98

100

105

110

115

725

755

775

795

825

855

875

925

975

100

12

4128

4477

4717

4964

5346

5741

6013

6720

7466

7854

120

125

130

135

140

145

150

105

110

115

120

125

130

133

14

8659

9503

10 387

11 310

12 272

13 273

13 893

155

160

165

170

175

138

143

148

153

158

16

14 957

16 061

17 203

18 385

19 607



Fig 172A little consideration will show that if one complete turn of a screw thread be imagined to be

unwound from the body of the screw and developed it will form an inclined plane as shown in

Fig 173 (a)

Fig 173

Let p = Pitch of the screw

d = Mean diameter of the screw

lang = Helix angle

Power Screws n 633

P = Effort applied at the circumference of the screw to lift the load

W = Load to be lifted and

prop = Coefficient of friction between the screw and nut

= tan where is the friction angle

From the geometry of the Fig 173 (a) we find that

tan lang = p d

Since the principle on which a screw jack works is similar to that of an inclined plane thereforethe force applied on the circumference of a screw jack may be considered to be horizontal as shown

in Fig 173 (b)

Since the load is being lifted therefore the force of friction (F = propRN ) will act downwards Allthe forces acting on the body are shown in Fig 173 (b)

Resolving the forces along the plane

P cos lang = W sin lang + F = W sin lang + propRN

and resolving the forces perpendicular to the plane

RN = P sin lang + W cos langSubstituting this value of RN in equation (i ) we have

P cos lang = W sin lang + prop (P sin lang + W cos lang)

= W sin lang + prop P sin lang + propW cos lang

(i )

(ii )

or

or

P cos lang ndash prop P sin lang = W sin lang + propW cos langP (cos lang ndash prop sin lang) = W (sin lang + prop cos lang)

or

(sin lang + prop cos lang)

(cos lang minus prop sin lang)Substituting the value of prop = tan in the above equation we get

sin lang + tan cos lang

cos lang minus tan sin langMultiplying the numerator and denominator

by cos we have

sin lang cos + sin



c

o

s

langP = W sdot

cos lang cos minus sin lang sin

Screw 8ack

sin (lang + )

cos (lang + )there4 Torque required to overcome friction between the screw and nut

d d

2 2

When the axial load is taken up by a thrust collar as shown in Fig 172 (b) so that the load does

not rotate with the screw then the torque required to overcome friction at the collar

where

2 =

R1 and R2 =

R =

prop1 =

3 (Assuming uniform pressure conditions)

R 1 2 (Assuming uniform wear conditions)

2Outside and inside radii of collar

R 1 + R 2Mean radius of collar = and

2Coefficient of friction for the collar


there4 Total torque required to overcome friction (ie to rotate the screw)

= 1 + 2

If an effort P1 is applied at the end of a lever of arm length l then the total torque required toovercome friction must be equal to the torque applied at the end of lever ie

d

2

Notes 1 When the nominal diameter (do) and the core diameter (dc) of the screw is given then

Mean diameter of screw d = = do minus = dc +2 2 2

2 Since the mechanical advantage is the ratio of the load lifted (W ) to the effort applied (P1) at the end of

the lever therefore mechanical advantage

W W sdot 2 l

MA = P P sdot d 2 2l sdot l or P

W sdot 2 l 2 l

= W tan (lang + ) d d tan (lang + )

1) Tore Reired to (ower (oad b

Sare Threaded ScrewsA little consideration will show that when the load

is being lowered the force of friction (F = propRN) willact upwards All the forces acting on the body are shown

in Fig 174


P cos lang = F ndash W sin lang= prop RN ndash W sin lang


RN = W cos lang ndash P sin langSubstituting this value of RN in equation (i ) we have

P cos lang = prop (W cos lang ndash P sin lang) ndash W sin lang= prop W cos lang ndash prop P sin lang ndash W sin lang

Fig 174

(i )

(ii )

or

or

P cos lang + prop P sin lang = propW cos lang ndash W sin langP (cos lang + prop sin lang) = W (prop cos lang ndash sin lang)

(prop cos lang minus sin lang)(cos lang + prop sin lang)

Substituting the value of prop = tan in the above equation we have

sdot W = propprop1 1 W R

+ R

sdot prop1 sdot W 2 2 (R1 ) minus (R2 )

2 (R1 )3 minus (R2 )3

1 = P sdot = W tan (lang + )

= W sdot = W tan (lang + )

P = W sdot

there4 P = W sdot



(tan

cos

sin lang )

(cos lang + tan sin lang)Multiplying the numerator and denominator by cos we have

P = W sdot (sin cos lang minus cos sin lang)(cos cos lang + sin sin lang)sin (minus lang )

cos (minus lang )

The nominal diameter of a screw thread is also known as outside diameter or major diameter

The core diameter of a screw thread is also known as inner diameter or root diameter or minor diameter

Power Screws n 635

there4Torque required to overcome friction between the screw and nut

d d

2 2

Note When lang gt then P = W tan (lang ndash )

1 E+cienc of Sare Threaded Screws

The efficiency of square threaded screws may be defined as the ratio between the ideal effort

(ie the effort required to move the load neglecting friction) to the actual effort ( ie the effort re-

quired to move the load taking friction into account)

We have seen in Art 174 that the effort applied at the circumference of the screw to lift the

load is

P = W tan (lang + ) (i )

where W = Load to be lifted

lang = Helix angle

= Angle of friction and

prop = Coefficient of friction between the screw and nut = tan If there would have been no friction between the screw and the nut then will be equal to zero

The value of effort P0 necessary to raise the load will then be given by the equation

P0 = W tan lang [Substituting = 0 in equation (i )]

there4Efficiency = Ideal effort

Actual effort

P W tan lang tan langP W tan (lang + ) tan (lang + )

This shows that the efficiency of a screw jack is independent of the load raised

In the above expression for efficiency only the screw friction is considered However if the

screw friction and collar friction is taken into account then

=

=

Torque required to move the load neglecting friction

Torque required to move the load including screw and collar friction

= P sdot d 2 + prop1W R

Note The efficiency may also be defined as the ratio of mechanical advantage to the velocity ratio

We know that mechanical advantage

MA =W W sdot 2l W sdot 2l 2l

P P sdot d W tan (lang + ) d d tan (lang + )(Refer Art 174)

and velocity ratio R=Distance moved by the effort (P1) in one revolution

Distance moved by the load (W ) in one revolution

=2 l

p

2 l 2 l

tan lang sdot d d tan lang ( tan lang = p d)

there4 Efficiency =M A 2l d tan lang R d tan (lang + ) 2 l

tan langtan (lang + )

1 Maxi-- E+cienc of a Sare Threaded Screw

We have seen in Art 176 that the efficiency of a square threaded screw

=tan lang sin lang cos lang sin lang sdot cos (lang + )

tan (lang + ) sin (lang + ) cos (lang + ) cos lang sdot sin (lang + )(i )

= P sdot = P1 sdot l

do + dc p p

= P sdot = P1 1 = d P sdot d

1

=

P = W sdot

P = W sdot

= W sdot = W tan (minus lang )

1 = P sdot = W tan (minus lang )

= = =0

0 P0 sdot d 2

= = =1

= =

= sdot =

= =




Multiplying the numerator and denominator by 2 we have

=2 sin lang sdot cos (lang + ) sin (2lang + ) minus sin 2 cos lang sdot sin (lang + ) sin (2 lang + ) + sin

(ii )

2sin A cos B = sin ( A + B) + sin ( A minus B)

The efficiency given by equation (ii ) will be maximum when sin (2lang + ) is maximum ie when

sin (2lang + ) = 1 or when 2lang + = 90deg

there4 2lang = 90deg ndash or lang = 45deg ndash 2Substituting the value of 2lang in equation (ii ) we have maximum efficiency

ma =sin (90deg minus + ) minus sin

sin (90deg minus + ) + sin

sin 90deg minus sin

sin 90deg + sin

1 minus sin

1 + sin Example 171 A ertical $cre ith $inample $tart $uare thread$ of () mm mean diameter

and

2( mm pitch i$ rai$ed aampain$t a load of ) +N b mean$ of a hand heel the bo$$ of hich i$threaded to act a$ a nut he aial load i$ ta+en up b a thru$t collar hich $upport$ the heelbo$$and ha$ a mean diameter of ) mm he coecient of friction i$ )( for the $cre and )0 forthecollar 1f the tanampential force applied b each hand to the heel i$ )) N nd $uitable diameter of thehand heel

Solution Given d = 50 mm p = 125 mm W = 10 kN = 10 times 103N D = 60 mm or

R = 30 mm prop = tan = 015 prop1 = 018 P1 = 100 N

We know that tan lang = p

d

125

sdot 50and the tangential force required at the circumference of the screw

tan lang + tan

1 minus tan lang tan 3 008 + 015

1 minus 008 sdot 015 We also know that the total torque required to turn the hand wheel

d 50 3

2 2= 58 200 + 54 000 = 112 200 N-mm (i )Let D1 = Diameter of the hand wheel in mmWe know that the torque applied to the handwheel

D1 D1

2 2(ii )

Equating equations (i ) and (ii )

D1 = 112 200 100 = 1122 mm = 1122 m Ans

Example 172 An electric motor drien poer $cre moe$ a nut in a hori3ontal planeaampain$t

a force of 4( +N at a $peed of 5)) mm 6 min he $cre ha$ a $inample $uare thread of mm pitchon

a major diameter of 7) mm he coecient of friction at $cre thread$ i$ ) 8$timate poer ofthemotor

Solution Given W = 75 kN = 75 times 103 N = 300 mmmin p = 6 mm do = 40 mm

prop = tan = 01

=

2 cos A sin B = sin ( A + B) minus sin ( A minus B)

= = 008

P = W tan (lang + ) = W

= 10 sdot 10 = 2328 N

= P sdot + prop1 W R = 2328 sdot + 018 sdot 10 sdot 10 sdot 30 N-mm

sdot = 2 sdot 100 sdot = 100 D = 2 p1 1 N-mm



now t at mean iameter of t e screw

d = do ndash p 2 = 40 ndash 6 2 = 37 mm

Power Screws n 637

and tan lang = p

d

6

sdot 37

We know that tangential force required at the circumference of the screw

tan lang + tan 1 minus tan lang tan

00516 + 01 3

1 minus 00516 sdot 01and torque required to operate the screw

= P sdot d 37 3

2 2

Since the screw moves in a nut at a speed of 300 mm min and the pitch of the screw is 6 mm

therefore speed of the screw in revolutions per minute (rpm)

and angular speed

Speed in mm min 300

N = Pitch in mm 6 = 2 N 60 = 2 times 50 60 = 524 rad s

there4 Power of the motor = = 21145 times 524 = 1108 W = 1108 kW Ans

Example 173 he cutter of a broachinamp machine i$ pulled b $uare threaded $cre of ((

mm

eternal diameter and ) mm pitch he operatinamp nut ta+e$ the aial load of 7)) N on a 9at$urfaceof ) mm and ) mm internal and eternal diameter$ re$pectiel 1f the coecient of friction i$)(for all contact $urface$ on the nut determine the poer reuired to rotate the operatinamp nut henthecuttinamp $peed i$ m6min Al$o nd the ecienc of the $cre

Solution Given do = 55 mm p = 10 mm = 001 m W = 400 N D1 = 60 mm or

R1 = 30 mm D2 = 90 mm or R2 = 45 mm prop = tan = prop1 = 015 Cutting speed = 6 m min

Power required to operate the nut

We know that the mean diameter of the screw


p 10

d sdot 50and force required at the circumference of the screw


00637

+ 015

1 minus 00637 sdot 015We know that mean radius of the flat surface

R =2 2

d 50

2 2= 4410 N-mm = 441 N-m

We know that speed of the screw

N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N

= 1143 sdot 103 sdot = 21145 sdot 10 N-mm = 21145 N-m

= = 50 rpm

= = 00637there4 tan lang =


= 400 = 864 N

= = 375 mmR1 + R2 30 + 45

there4Total torque required

= P sdot + prop1 W R = 864 sdot + 015 sdot 400 sdot 375 N-mm

= = 600 rpm



A Textbook of Machine Design

and angular speed = 2 N 60 = 2 times 600 60 = 6284 rad s

there4Power required to operate the nut

= = 441 times 6284 = 277 W = 0277 kW Ans

Efciency o the screw

We know that the efficiency of the screw

= 0 W tan lang sdot d 2 400 sdot 00637 sdot 50 2 4410

= 0144 or 144 Ans

Example 174 A ertical to $tart $uare threaded $cre of a )) mm mean diameter and

2) mm pitch $upport$ a ertical load of 0 +N he aial thru$t on the $cre i$ ta+en b a collar bearinamp of 2() mm out$ide diameter and )) mm in$ide diameter Find the force reuired at theendof a leer hich i$ 7)) mm lonamp in order to lift and loer the load he coecient of friction fortheertical $cre and nut i$ )( and that for collar bearinamp i$ )2)

Solution Given d = 100 mm p = 20 mm W = 18 kN = 18 times 103N D2 = 250 mm

or R2 = 125 mm D1 = 100 mm or R1 = 50 mm l = 400 mm prop = tan = 015 prop1 = 020

Force required at the end o lever Let P = Force required at the end of lever

Since the screw is a two start square threaded screw therefore lead of the screw

= 2 p = 2 times 20 = 40 mm

We know that tan lang =Lead 40

d sdot 100

1 For raisin the load


tan lang + tan

1 minus tan lang tan

3 0127 + 015 1 minus 0127 sdot 015and mean radius of the collar

R = = = 875 mm2 2

there4Total torque required at the end of lever

= P sdot + prop1 WR

100 3

2

We know that torque required at the end of lever ( )

569 150 = P1 times l = P1 times 400 or P1 = 569 150400 = 1423 N Ans2 For lowerin the load


tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639

and the total torque required the end of lever

d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125

= 5083 sdot + 020 sdot 18 sdot 10 sdot 875 = 569 150 N-mm

P = W tan (minus lang ) = W 1 + tan tan lang

=18 sdot 10 = 4063 N



2

100 3

2


335 315 = P1 times l = P1 times 400 or P1 = 335 315 400 = 8383 N Ans

Example 175 The mean diameter of the $uare threaded $cre hainamp pitch of ) mm i$

() mm A load of 2) +N i$ lifted throuamph a di$tance of 4) mm Find the or+ done in liftinamp theloadand the ecienc of the $cre hen

he load rotate$ ith the $cre and

2 he load re$t$ on the loo$e head hich doe$ not rotate ith the $cre

he eternal and internal diameter of the bearinamp $urface of the loo$e head are ) mm and

) mm

re$pectiel he coecient of friction for the $cre and the bearinamp $urface ma be ta+en a$))0

Solution Given p = 10 mm d = 50 mm W = 20 kN = 20 times 103 N D2 = 60 mm or

R2 = 30 mm D1 = 10 mm or R1 = 5 mm prop = tan = prop1 = 008


d

10

sdot 50

there4Force required at the circumference of the screw to lift the load

tan lang + tan


minus sdot and torque required to overcome friction at the screw

= P times d 2 = 2890 times 50 2 = 72 250 N-mm = 7225 N-m

Since the load is lifted through a vertical distance of 170 mm and the distance moved by the

screw in one rotation is 10 mm (equal to pitch) therefore number of rotations made by the screw

N = 170 10 = 17

1 hen the load rotates with the screw

We know that workdone in lifting the load

= times 2 N = 7225 times 2 times 17 = 7718 N-m Ans

and efficiency of the screw

tan lang tan lang (1 minus tan lang tan ) = tan (lang + ) tan lang + tan

00637 (1 minus 00637 sdot 008)=

2 hen the load does not rotate with the screw

We know that mean radius of the bearing surface

R1 + R2 5 + 30R =

2 2and torque required to overcome friction at the screw and the collar

d

2

Ans


= 2890 sdot50

2

+ 008 sdot 20 sdot 103 sdot 175 = 100 250 N-mm

= 10025 N-m

there4Workdone by the torque in lifting the load

= times 2 N = 10025 times 2 times 17 = 10 710 N-m Ans

We know that torque required to lift the load neglecting friction

0 = P0 times d 2 = W tan lang times d 2 ( Po = W tan lang)= 20 times 103 times 00637 times 50 2 = 31 850 N-mm = 3185 N-m

= P sdot + prop1 W R

= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637





there4Efficiency of the screw

= 0

3185

10025

1 E+cienc s Heix Ange

We have seen in Art 176 that the efficiency of a square threaded screw depends upon the helix

angle lang and the friction angle The variation of efficiency of a square threaded screw for raising the

load with the helix angle lang is shown in Fig 175 We see that the efficiency of a square threaded screw

increases rapidly upto helix angle of 20deg after which the increase in efficiency is slow The efficiencyis maximum for helix angle between 40 to 45deg

Fig 175 Graph between efficiency and helix angle

When the helix angle further increases say 70deg the efficiency drops This is due to the fact that

the normal thread force becomes large and thus the force of friction and the work of friction becomes

large as compared with the useful work This results in low efficiency

10 er Haing and Sef (ocking Screws

We have seen in Art 175 that the effort required at the circumference of the screw to lower the

load is

P = W tan ( ndash lang)and the torque required to lower the load

d d

2 2

In the above expression if lt lang then torque required to lower the load will be neative Inother words the load will start moving downward without the application of any torque Such a

condition is known as over haulin o screws If however gt lang the torque required to lower the load

will be positive indicating that an effort is applied to lower the load Such a screw is known as

Power Screws n 641

Mechanica ower screw drier

sel locin screw In other words a screw will be self locking if the friction angle is greater than

helix angle or coefficient of friction is greater than tangent of helix angle ie prop or tan gt tan lang

= = 0318 or 318 Ans

= P sdot = W tan (minus lang )



113 E+cienc of Sef (ocking Screws

We know that the efficiency of screw

=tan (lang + )

and for self locking screws ge lang or lang le

there4Efficiency for self locking screws

le tan tan tan (+ ) tan 2

tan (1 minus tan 2 )2 tan

1 tan2 2 2

2 tan

From this expression we see that efficiency of self locking screws is less than

efficiency is more than 50 then the screw is said to be overhauling

Note It can be proved as follows

Let


h = Distance through which the load is lifted

there4

-13Output = Wh

Output W hand Input =

there4Work lost in overcoming friction

or 50 If the

= Input minus Output =W h

1

For self locking

1

there4 minus 1 le 1 or le 1

2or 50


111 Coe+cient of 4riction

The coefficient of friction depends upon various factors like material of screw and nut work-

manship in cutting screw quality of lubrication unit bearing pressure and the rubbing speeds The

value of coefficient of friction does not vary much with different combination of material load or

rubbing speed except under starting conditions The coefficient of friction with good lubrication and

average workmanship may be assumed between 010 and 015 The various values for coefficient of friction for steel screw and cast iron or bronze nut under different conditions are shown in the follow-

ing table

Tabe 1) Coe+cient of friction nder ferent conditions

If the thrust collars are used the values of coefficient of friction may be taken as shown in the

following table

Tabe 1 Coe+cient of friction when thrst coars are sed

tan

le le le minus

tan 2=1 minus tan 2

=

minus W h = W h minus 1

W h le W hminus 1



11 Ac-e or Trae5oida Threads

We know that the normal reaction in case of a squarethreaded screw is

RN = W cos langwhere

lang is the helix angle

But in case of Acme or trapezoidal thread the normal re-

action between the screw and nut is increased because the axial

component of this normal reaction must be equal to the axial

load (W )

Consider an Acme or trapezoidal thread as shown in

Fig 176

Let 2983214 = Angle of the Acme thread and

983214 = Semi-angle of the threadFig 176 Acme or trapezonidal threads

The material of screw is usually steel and the nut is made of cast iron gun metal phosphor bronze in order

to keep the wear to a mininum

For Acme threads 2 983214 = 29deg and for trapezoidal threads 2 983214 = 30deg

Power Screws n 643

there4 RN = W cos 983214

W and frictional force

where prop cos 983214 = prop1 known as virtual coefcient o riction

Notes 1 When coefficient of friction prop1 =

thread

prop

cos 983214is considered then the Acme thread is equivalent to a square

2 All equations of square threaded screw also hold good for Acme threads In case of Acme threads prop1

(ie tan 1) may be substituted in place of prop (ie tan ) Thus for Acme threads

P = W tan (lang + 1)where 1 = Virtual friction angle and tan 1 = prop1

Example 176 he lead $cre of a lathe ha$ Acme thread$ of () mm out$ide diameter and0 mm pitch he $cre mu$t eert an aial pre$$ure of 2()) N in order to drie the tool carriaampehethru$t i$ carried on a collar ) mm out$ide diameter and (( mm in$ide diameter and the lead$crerotate$ at 5) rpm Determine (a) the poer reuired to drie the $cre and (b) the ecienc ofthelead $cre A$$ume a coecient of friction of )( for the $cre and )2 for the collar

Solution Given do = 50 mm p = 8 mm W = 2500 N D1 = 110 mm or R1 = 55 mm

D2 = 55 mm or R2 = 275 mm N = 30 rpm prop = tan = 015 prop2 = 012

(a) Power required to drive the screwWe know that mean diameter of the screw


p 8

d sdot 46

Since the angle for Acme threads is 2983214 = 29deg or 983214 = 145deg therefore virtual coefficient of friction

prop 015 015

We know that the force required to overcome friction at the screw

ltNo

=ondition Aeraampe coecient of

lttartinamp Runninamp

1

2

3

High grade materials and workmanship

and best running conditions

Average quality of materials and workmanship

and average running conditions

Poor workmanship or very slow and in frequent motion

with indifferent lubrication or newly machined surface

014

018

021

010

013

015

ltNo Material$ Aeraampe coecient of

lttartinamp Runninamp1

2

3

4

Soft steel on cast iron

Hardened steel on cast iron

Soft steel on bronze

Hardened steel on bronze

017

015

010

008

012

009

008

006

di7er



tan lang + tan 1

1 minus tan lang tan 1

0055 + 0155

and torque required to overcome friction at the screw

1 = P times d 2 = 530 times 46 2 = 12 190 N-mm

We know that mean radius of collar

R =R1 + R2 55 + 275

2 2

Assuming uniform wear the torque required to overcome friction at collars

2 = prop2 W R = 012 times 2500 times 4125 = 12 375 N-mm

there4Total torque required to overcome friction

= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m


We know that power required to drive the screw

sdot 2 N 24565 sdot 2 sdot 30

60 60Ans

( = 2N 60)

() Efciency o the lead screwWe know that the torque required to drive the screw with no friction

d 46

2 2

there4Efficiency of the lead screw

= o

3163

24565Ans

11$ Stresses in Power Screws

A power screw must have adequate strength to withstand axial load and the applied torque

Following types of stresses are induced in the screw

1 $irect tensile or compressive stress due to an aial load The direct stress due to the axial

load may be determined by dividing the axial load (W ) by the minimum cross-sectional area of the

screw ( Ac) ie area corresponding to minor or core diameter (dc )

there4Direct stress (tensile or compressive)

W

= Ac

This is only applicable when the axial load is compressive and the unsupported length of thescrew between the load and the nut is short But when the screw is axially loaded in compression and

the unsupported length of the screw between the load and the nut is too great then the design must be

based on column theory assuming suitable end conditions In such cases the cross-sectional area

corresponding to core diameter may be obtained by using Rankine-Gordon formula or JB Johnsonrsquos

formula According to this

W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N

P = W tan (lang + 1) = W

prop1 = tan 1 = cos 983214= cos 145deg = 09681 = 0155

= = 0055tan lang =there4

cos 983214= prop1W F = prop R N = prop sdot



Crit

ical

load

Yiel

d

st

re

ss

Length of screw

Least radius of gyration

End-fixity coefficient

Modulus of elasticity and

Stress induced due to load W

Note In actual practice the core diameter is first obtained by considering the screw under simple compression

and then checked for critical load or buckling load for stability of the screw

2 amporsional shear stress Since the screw is subjected to a twisting moment therefore torsional

shear stress is induced This is obtained by considering the minimum cross-section of the screw We

know that torque transmitted by the screw

=

16

sdot (dc )3

Power Screws n 645

or shear stress induced

16

(d c )3

When the screw is subjected to both direct stress and torsional shear stress then the design mustbe based on maximum shear stress theory according to which maximum shear stress on the minor

diameter section

ma =1

2(Bt or Bc 2 amp 2

It may be noted that when the unsupported length of the screw is short then failure will takeplace when the maximum shear stress is equal to the shear yield strength of the material In this case

shear yield strength

= ma times Factor of safety

3 Shear stress due to aial load The threads of the

screw at the core or root diameter and the threads of the nut

at the major diameter may shear due to the axial load

Assuming that the load is uniformly distributed over the

threads in contact we have

Shear stress for screw

W

n dc t and shear stress for nut

where

W

n do t W = Axial load on the screw

n = Number of threads in engagement

4riction between the threads of screw

and nt as i-or tant roe indeter-ining the e+cienc and ockingroerties of a screw

dc = Core or root diameter of the screw

do = Outside or major diameter of nut or screw and

t = Thickness or width of thread

4 Bearin pressure In order to reduce wear of the screw and nut the bearing pressure on the

thread surfaces must be within limits In the design of power screws the bearing pressure depends

upon the materials of the screw and nut relative velocity between the nut and screw and the nature of

lubrication Assuming that the load is uniformly distributed over the threads in contact the bearing

pressure on the threads is given by

where

W W pb =

)2 minus (d 2

4d = Mean diameter of screw

t = Thickness or width of screw = p 2 and

n = Number of threads in contact with the nut

=Height of the nut

Pitch of threads=

h

p

Therefore from the above expression the height of nut or the length of thread engagement of

the screw and nut may be obtained

= = = = 77 W = 0077 kW

o = W tan lang sdot = 2500 sdot 0055 sdot = 3163 N-mm = 3163 N-m

= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +



The following table shows some limiting values of bearing pressures

We know that = sdot = d sdot = d t 4 2 2 2


Tabe 1 (i-iting aes of bearing essres

Example 177 A poer $cre hainamp double $tart $uare thread$ of 2( mm nominaldiameter

and ( mm pitch i$ acted upon b an aial load of ) +N he outer and inner diameter$ of $crecollar are () mm and 2) mm re$pectiel he coecient of thread friction and collar friction ma be a$$umed a$ )2 and )( re$pectiel he $cre rotate$ at 2 rpm A$$uminamp uniform ear condition at the collar and alloable thread bearinamp pre$$ure of (0 N6mm2 nd the toruereuired to rotate the $cre 2 the $tre$$ in the $cre and 5 the number of thread$ of nut inenampaampement ith $cre

Solution Given do = 25 mm p = 5 mm W = 10 kN = 10 times 103 N D1 = 50 mm or

R1 = 25 mm D2 = 20 mm or R2 = 10 mm prop = tan = 02 prop1 = 015 N = 12 rpm pb = 58 Nmm2

1 amporque required to rotate the screw

We know that mean diameter of the screw


Since the screw is a double start square threaded screw therefore lead of the screw

= 2 p = 2 times 5 = 10 mm

there4 tan lang =Lead 10

d sdot 225We know that tangential force required at the circumference of the screw

tan lang + tan


01414 + 02

1 minus 01414 sdot 02

and mean radius of the screw collar

R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n

(do ) 2 minus (dc ) 2 do + dc do minus dc p

Application of

$cre

Material ltafe bearinamp

pre$$ure

2

Rubbinamp $peed

at

thread pitchltcre Nut

1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175



otal torque required to rotate the screw

Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans

2 Stress in the screw

We know that the inner diameter or core diameter of the screw

dc = do ndash p = 25 ndash 5 = 20 mm

there4 Corresponding cross-sectional area of the screw

Ac =

4 4(20)2 = 3142 mm2

We know that direct stress

and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3

We know that maximum shear stress in the screw

ma =1

22 1

2(3183)2 + 4 (4186)2

448 Nmm2 = 448 MPa Ans

3 umer o threads o nut in enaement with screw

Let n = Number of threads of nut in engagement with screw and

t = Thickness of threads = p 2 = 5 2 = 25 mm

We know that bearing pressure on the threads ( pb)

58 =W

d sdot t sdot n

3

= = sdot 225 sdot 25 sdot n n

there4 n = 566 58 = 976 say 10 Ans

Example 17 he $cre of a $haft $traiamphtener eert$ a load of 5) +N a$ $hon in Fiamp

44

he $cre i$ $uare threaded of out$ide diameter 4( mm and mm pitch Determine

Force reuired at the rim of a 5)) mm diameter hand heel a$$uminamp the coecient of

friction for the thread$ a$ )2

2 Maimum compre$$ie $tre$$ in the $cre bearinamp pre$$ure on the thread$ and maimum

$hear $tre$$ in thread$ and5 8cienc of the $traiamphtner

Solution Given W = 30 kN = 30 times 103 N do = 75 mm p = 6 mm D = 300 mm

prop = tan = 012

1 Force required at the rim o handwheel

Let P1 = Force required at the rim of handwheel




Mean diameter of the screw

10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =

+ prop1 W R = 3513 sdot + 015 sdot 10 sdot 103 sdot 175 N-mm



d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265

there4 Torque required to overcome friction at the threads

= P sdot d2

= W tan (lang + )2

tan lang + tan d 1 minus tan lang tan 2

00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm

We know that the torque required at the rim of handwheel ( )

All dimensions in mm

Fig 177

there4

D 300

2 2

P1 = 158 728 150 = 1058 N Ans2 (aimum compressive stress in the screw

We know that maximum compressive stress in the screw

W W 30 sdot 103

Ac 2

4 4Bearin pressure on the threads

We know that number of threads in contact with the nut

Ans

n =Height of nut 150

Pitch of threads 6

and thickness of threads t = p 2 = 6 2 = 3 mm

We know that bearing pressure on the threads

pb =W

d t n

30 sdot 103

sdot 72 sdot 3 sdot 25

(aimum shear stress in the threads

We know that shear stress in the threads

=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans

The mean diameter of the screw (d ) is also given by

d do p 6 2 = 75 ndash 6 2 = 72 mm

there4 Maximum shear stress in the threads

Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2

= 47 Nmm2 = 47 MPa Ans

3 Efciency o the straihtener

We know that the torque required with no friction

d 3 72

2 2

there4Efficiency of the straightener

=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans

Example 17 A $luice ampate eiamphinamp 0 +N i$ rai$ed and loered b mean$ of $uarethreaded

$cre$ a$ $hon in Fiamp 40 he frictional re$i$tance induced b ater pre$$ure aampain$t theampatehen it i$ in it$ loe$t po$ition i$ 7))) N

he out$ide diameter of the $cre i$ ) mm and pitch i$ ) mm he out$ide and in$ide

diameter

of a$her i$ () mm and () mm re$pectiel he coecient of friction beteen the $cre and nut i$ ) and for the

a$her and $eat i$ )2 Find he maimum force to be eerted at the end$ of

the leer rai$inamp and loerinamp the ampate2 8cienc of thearranampement and 5 Number of thread$ and heiampht of nutfor an alloable bearinamp pre$$ure of 4 N6mm2

Solution Given W 1 = 18 kN = 18 000 N

F = 4000 N do = 60 mm p = 10 mm D1 = 150 mm or R1

= 75 mm D2 = 50 mm or R2 = 25 mm prop = tan = 01 prop1 = 012 pb = 7 Nmm2

1 (aimum orce to e eerted at the ends o lever Let P1 = Maximum force exerted at

each end of the lever 1 m

(1000 mm) long



We know that inner diameter or core diameter of the

screw

dc = do ndash p = 60 ndash 10 = 50 mmMean diameter of the screw

Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55

(a) For raisin the ate

Since the frictional resistance acts in the opposite direction to the motion of screw therefore for

raising the gate the frictional resistance ( F ) will act downwards

there4Total load acting on the screw

W = W 1 + F = 18 000 + 4000 = 22 000 N


d d tan lang + tan d2 2 1 minus tan lang tan 2


Mean radius of washer

0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2

there4Torque required to overcome friction at the washer

2 = prop1 W R = 012 times 22 000 times 50 = 132 000 N-mm

and total torque required to overcome friction = 1 + 2 = 96 148 + 132 000 = 228 148 N-mm

We know that the torque required at the end of lever ( )

there4228 148 = 2P1 times Length of lever = 2P1 times 1000 = 2000 P1

P1 = 228 148 2000 = 1411 N Ans() For lowerin the ate

Since the gate is being lowered therefore the frictional resistance (F ) will act upwards


W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N

We know that torque required to overcome friction at the screw

= W tan (minus lang )2 2 1 + tan tan lang 2 01 minus 0058 55

1 + 01 sdot 0058 2and torque required to overcome friction at the washer



= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans

2 Efciency o the arranement

We know that the torque required for raising the load with no friction

d 55

2 2

there4Efficiency of the arrangement

= 0

35 090

228 148Ans

3 umer o threads and heiht o nut

Let n = Number of threads in contact with the nut

0 = W tan lang sdot = 30 sdot 10 sdot 00265 sdot = 28 620 N-mm

= = 018 or 18

do + dc 60 + 50

= = 0058

1 = P sdot = W tan (lang + ) = W



h = Height of nut = n times p and

t = Thickness of thread = p 2 = 10 2 = 5 mm

We know that the bearing pressure ( pb)

and

there4

W 22000 25467 =

n = 2546 7 = 364 say 4 threads Ans

h = n times p = 4 times 10 = 40 mm Ans

Power Screws n 651

Example 171 he $cre a$ $hon in Fiamp 4 i$ operated b a torue applied to theloer

end he nut i$ loaded and preented from turninamp b ampuide$ A$$ume friction in the ball bearinamptobe neampliampible he $cre i$ a triple $tart trape3oidal thread he out$idediameter of the $cre i$ 70 mm and pitch i$ 0 mm he coecient of

friction of the thread$ i$ )( Find gtoad hich can be rai$ed b a torue of 7) N-m

2 Whether the $cre i$ oerhaulinamp and

5 Aeraampe bearinamp pre$$ure beteen the $cre and nut thread

$urface

Solution Given do = 48 mm p = 8 mm prop = tan = 015 = 40 N-m = 40 000 N-mm

1 )oad which can e raised

Let W = Load which can be raised



Since the screw is a triple start therefore lead of the screw

= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44

and virtual coefficient of frictionprop 015 015

983214 deg ( For trapezoidal threads 2 983214 = 30deg)

We know that the torque required to raise the load

there4

= P sdot = W tan (lang + 1 ) = W

= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns

2 hether the screw is overhaulin

We know that torque required to lower the load

d

2

We have discussed in Art 179 that if 1 is less than lang then the torque required to lower the loadwill be neative ie the load will start moving downward without the application of any torque Sucha condition is known as overhauling of screws

In the present case tan 1 = 0155 and tan lang = 0174 Since 1 is less than lang therefore the screw

is overhauling Ans3 Averae earin pressure etween the screw and nut thread suraces

We know that height of the nut

h = n times p = 50 mm

there4Number of threads in contact

n = h p = 50 8 = 625

and thickness of thread t = p 2 = 8 2 = 4 mm

(Given)

sdot 55 sdot 5 sdot n ndt n= =

= = 0154 or 154

0 = W tan lang sdot = 22 000 sdot 0058 sdot = 35 090 N-mm

= 14 000 = 16 077 N-mm

= W d d tan minus tan lang d

1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =

d d tan lang + tan 1 d

2 2 1 minus tan lang tan 1 2

0174 + 0155 4440 000 = W

= W tan ( 1 minus lang)




We know that the average bearing pressure

pb = W dt n

5380 sdot 44 sdot 4 sdot 625

Example 1711 A =-clamp a$ $hon in Fiamp 4) ha$trape3oidal thread$ of 2 mm out$ide diameter and 2 mm pitch hecoecient of friction for $cre thread$ i$ )2 and for the collar i$)2( he mean radiu$ of the collar i$ mm 1f the force eerted b the operator at the end of the handle i$ 0) N nd he lenampth of handle 2 he maimum $hear $tre$$ in the bod of the $cre andhere doe$ thi$ ei$t and 5 he bearinamp pre$$ure on the thread$

Solution Given do = 12 mm p = 2 mm prop = tan = 012

prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle

Let l = Length of handle


d = do ndash p 2 = 12 ndash 2 2 = 11 mm Fig 171

there4 tan lang = p

d

2

sdot 11

Since the angle for trapezoidal threads is 2983214 = 30deg or 983214 = 15deg therefore virtual coefficient of

friction

prop 012 012

cos 983214 cos 15deg 09659

We know that the torque required to overcome friction at the screw

d d

2 2

tan lang + tan 1 d

1 minus tan lang tan 1 2

= 4033 N-mm

1 minus 0058 sdot 0124 2Assuming uniform wear the torque required to overcome friction at the collar

2 = prop2 W R = 025 times 4000 times 6 = 6000 N-mm

there4Total torque required at the end of handle

= 1 + 2 = 4033 + 6000 = 10 033 N-mm

We know that the torque required at the end of handle ( )

10 033 = P1 times l = 80 times l or l = 10 033 80 = 1254 mm Ans

2 (aimum shear stress in the ody o the screw

Consider two sections A-A and B-B The section A-A just above the nut is subjected to torqueand bending The section B-B just below the nut is subjected to collar friction torque and direct

compressive load Thus both the sections must be checked for maximum shear stress

+onsiderin section AA

We know that the core diameter of the screw


and torque transmitted at A-A

=

16sdot (dc )3

Power Screws n 653there4 S h ear stress =

= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124

1 = P sdot = W tan (lang + 1 ) = W

0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2

Bending moment at A-A

M = P1 times 150 = 80 times 150 = 12 000 N-mm

= (d

32

32 M 32 sdot 12 0003 3

We know that the maximum shear stress

ma =

1 2 2 14

2 2

2 2

= 7965 MPa

+onsiderin section BB

Since the section B-B is subjected to collar friction torque ( 2) therefore the shear stress

16 2 16 sdot 60003 3

and direct compressive stress

W

Ac

there4Maximum shear stress

4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa

From above we see that the maximum shear stress is 7965 MPa and occurs at section

A-A Ans

3 Bearin pressure on the threads




n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000


Example 1712 A poer tran$mi$$ion $cre of a $cre pre$$ i$ reuired to tran$mit

maimumload of )) +N and rotate$ at ) rpm rape3oidal thread$ are a$ under

he $cre thread friction coecient i$ )2 orue reuired for collar friction and journalbearinamp i$ about ) of the torue to drie the load con$iderinamp $cre friction Determine $cre

dimen$ion$ and it$ ecienc Al$o determine motor poer reuired to drie the $cre Maimum permi$$ible compre$$ie $tre$$ in $cre i$ )) MPa


Solution Given W 100 kN 100 C 103 N

N 60 rpm prop 012 ⌠ c 100 MPa 100 Nmm2

$imensions o the screw

Let

We know that the direct compressive stress (⌠ c)100 =

Nominal dia 7) () ) 4)

=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3

= = 1222 Nmm 2there4Bending stress ⌠ b = (dc ) (10)

(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3

Since the core area is 1000 mm2 therefore we

shall use the following dimensions for the screw (for

core area 1353 mm2)

Nominal diameter do = 50 mm

Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans

Efciency o the screw p 8

d sdot 46

and virtual coefficient of friction

This screw ress was -ade in 1$) andinstaed in the Segoia Mint to strike a newseries of coer coins which began in 1 This ress is resent on disa in the Aca5arcaste of Segoia

prop1 = tan 1 =prop

cos 983214

012

cos 15deg012

09659= 0124 ( For trapezoidal threads 2983214 = 30deg)

there4 Force required at the circumference of the screw

tan lang + tan 1 tan tan

3 0055 + 0124

1 minus 0055 sdot 0124 and the torque required to drive the load

1 = P times d 2 = 18 023 times 46 2 = 414 530 N-mm

We know that the torque required for collar friction

2 = 10 1 = 01 times 414 530 = 41 453 N-mm

there4 Total torque required

= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2

there4 Efficiency of the screw

= 0

126 500

455 983

Power Screws n 655

Power required to drive the screw

We know that the power required to drive the screw


60 60= 2865 kW Ans

Example 1713 A ertical to $tart $uare threaded $cre of )) mm mean diameter and

2)

mm pitch $upport$ a ertical load of 0 +N he nut of the $cre i$ tted in the hub of a ampear

heelhainamp 0) teeth hich me$he$ ith a pinion of 2) teeth he mechanical ecienc of the pinionandampear heel drie i$ ) percent he aial thru$t on the $cre i$ ta+en b a collar bearinamp 2() mmout$ide diameter and )) mm in$ide diameter A$$uminamp uniform pre$$ure condition$ nd mini-mum diameter of pinion $haft and heiampht of nut hen coecient of friction for the ertical $creandnut i$ )( and that for the collar bearinamp i$ )2) he permi$$ible $hear $tre$$ in the $haft

Ac = Core area of threads

=

Ac = 100 sdot 10 100 = 1000 mm2

= = 0055We know that tan lang =

P = W tan (lang + 1 ) = W 1 minus lang 1

= 100 sdot 10 = 18 023 N

0 = W tan lang sdot = 100sdot 10 sdot 0055 sdot = 126 500 N-mm

= = 0278 or 278 Ans



materiali$ ( MPa and alloable bearinamp pre$$ure i$ 7 N6mm2

Solution Given d = 100 mm p = 20 mm W = 18 kN = 18 times 103N No of teeth on gear wheel

= 80 No of teeth on pinion = 20 m = 90 = 09 D1 = 250 mm or R1 = 125 mm D2 = 100 mm or

R2 = 50 mm prop = tan = 015 prop1 = 020 = 56 MPa = 56 Nmm2 pb = 14 Nmm2

(inimum diameter o pinion shat

Let D = Minimum diameter of pinion shaft


= 2 p = 2 times 20 = 40 mm


d sdot 100

and torque required to overcome friction at the screw and nut


3

1 minus 0127 sdot 015 2= 25416 N-m

We know that for uniform pressure conditions torque required to overcome friction at the

collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m

Since the nut of the screw is fixed in the hub of a gear wheel therefore the total torque required

at the gear wheel

= 1 + 2 = 25416 + 33429 = 58845 N-mAlso the gear wheel having 80 teeth meshes with pinion having 20 teeth and the torque is

proportional to the number of teeth therefore torque required at the pinion shaft

=

80 80Since the mechanical efficiency of the pinion and gear wheel is 90 therefore net torque required

at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm


We know that the torque required at the pinion shaft ( p)

there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut

n = Number of threads in contact and

t = Thickness or width of thread = p 2 = 20 2 = 10 mm


there4and height of nut

3

14 = = =

dt n sdot 100 sdot 10 sdot n nn = 573 14 = 409 say 5 threads


Example 1714 A $cre pre$$ i$ to eert a force of 7) +N he un$upported lenampth of the

$cre

i$ 7)) mm Nominal diameter of $cre i$ () mm he $cre ha$ $uare thread$ ith pitch eualto )mm he material of the $cre and nut are medium carbon $teel and ca$t iron re$pectiel For

= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm

(R1 )3 minus (R2 )3 sdot prop1 W

sdot 020 sdot 18 sdot 103 N-mm

= 58845 sdot = 14711 N-m sdot 20 20



the$teel u$ed ta+e ultimate cru$hinamp $tre$$ a$ 52) MPa ield $tre$$ in ten$ion or compre$$ion a$ 2))MPa and that in $hear a$ 2) MPa Alloable $hear $tre$$ for ca$t iron i$ 2) MPa and alloablebearinamp pre$$ure beteen $cre and nut i$ 2 N6mm2 ounampE$ modulu$ for $teel 2) +N6mm2Determine the factor of $afet of $cre aampain$t failure Find the dimen$ion$ of the nut What i$theecienc of the arranampement a+e coecient of friction beteen $teel and ca$t iron a$ )5

Solution Given W = 40 kN = 40 times 103 N gt = 400 mm = 04 m do = 50 mm p = 10 mm

⌠ cu = 320 MPa = 320 Nmm2 ⌠ = 200 MPa = 200 Nmm2 = 120 MPa = 120 Nmm2

c = 20 MPa = 20 Nmm2 pb = 12 Nmm2 8 = 210 kNmm2 = 210 times 103 Nmm2 prop = tan = 013We know that the inner diameter or core diameter of the screw

dc = do ndash p = 50 ndash 10 = 40 mmand core area of the screw

4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45

there4Torque required to move the screw

= P sdot =W tan (lang + ) = W

3

1 minus 007 sdot 013 2We know that torque transmitted by the screw ( )

1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2

According to maximum shear stress theory we have

Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety

We know that factor of safety

=

ma

120

215

Now considering the screw as a column assuming one end fixed and other end free According

to JB Johnsons formula critical load2

sdot⌠ 4= 2 8 G

For one end fixed and other end free = = 025

there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)

there4Factor of safety =W cr

W

212 700

40 sdot 103

We shall take larger value of the factor of safety

there4Factor of safety = 558 say 6

$imensions o the nut

Ans

sdot sdot D3 = sdot sdot D3 = 11 D3

18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =

there4Direct compressive stress on the screw due to axial load

= = 318 Nmm 2

do + dc 50 + 40

= = 007and tan lang =

d d tan lang + tan d

2 2 1 minus tan lang tan 2

= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568



Let n = Number of threads in contact with nut and

h = Height of nut = p times n

Assume that the load is uniformly distributed over the threads in contact

We know that the bearing pressure ( pb)3

= =n

4 4

and

there4 n = 566 12 = 47 say 5 threads

h = p times n = 10 times 5 = 50 mm Ans

Ans

Now let us check for the shear stress induced in the nut which is of cast iron We know that

W 40 sdot 103

ndo t sdot 5 sdot 50 sdot 5( t = p 2 = 10 2 = 5 mm)

This value is less than the given value of c = 20 MPa hence the nut is safe

Efciency o the arranement

We know that torque required to move the screw with no friction

d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103


11amp Design of Screw ack

A bottle screw jack for lifting loads is shown in Fig 1711 The various parts of the screw jack are as follows

1 Screwed spindle having square threaded screws

2 Nut and collar for nut

3 Head at the top of the screwed spindle for handle

4 Cup at the top of head for the load and

5 Body of the screw jack

In order to design a screw jack for a load W the following procedure may be adopted

1 First of all find the core diameter (dc) by considering that the screw is under pure compressionie

(d 2

The standard proportions of the square threaded screw are fixed from Table 171

(⌠ c )2 + 4 =

= = 558

⌠ gtW cr = Ac 1 minus

W cr = 1257 sdot 200 1 minus4 sdot 025 sdot sdot 210 sdot 103

= = 53

40 sdot 10 566W

12 = (do c )2 n n)2 minus (d (50)2 minus (40)2

= = 102 N mm2 = 102 MPanut =

0 = W tan lang sdot = 40 sdot 10 sdot 007 sdot = 63 sdot 10 N-mm

= = 0347 or 347 Ans



Fig 1711 Screw jack

2 Find the torque ( 1) required to rotate the screw and find the shear stress () due to thistorque

We know that the torque required to lift the load

d d

2 2where P = Effort required at the circumference of the screw and


there4Shear stress due to torque 1

16 1

(d c )3

Also find direct compressive stress (⌠ c) due to axial load ieW

4

3 Find the principal stresses as follows

Maximum principal stress (tensile or compressive)

1 ⌠ c(ma ) =

and maximum shear stress

Power Screws n 659

ma = 12

(⌠ c )2 + 4 2

These stresses should be less than the permissible stresses

4 Find the height of nut (h) considering the bearing pressure on the nut We know that the

bearing pressure on the nut

pb =

4

W

2 2

where

there4 Height of nut

where

n = Number of threads in contact with screwed spindle

h = n times p

p = Pitch of threads

5 Check the stressess in the screw and nut as follows

($cre) =

(nut ) =

W

ndct W

ndot where t = Thickness of screw = p 2

6 Find inner diameter (D1) outer diameter (D2) and thickness (t 1) of the nut collar

The inner diameter (D1) is found by considering the tearing strength of the nut We knowthat

W =4

2 2

The outer diameter (D2) is found by considering the crushing strength of the nut collar We

know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

The thickness (t 1) of the nut collar is found by considering the shearing strength of the nutcollar We know that

W = D1t 17 Fix the dimensions for the diameter of head (D3) on the top of the screw and for the cup

6ac

W = ⌠ c c c c )sdot A =⌠ sdot




Take D3 = 175 do The seat for the cup is made equal to the diameter of head and it is

chamfered at the top The cup is fitted with a pin of diameter D4 = D3 4 approximately Thispin remains a loose fit in the cup


Find the torque required ( 2) to overcome friction at the top of screw We know that

2 2

R 3 + R 4

(Assuming uniform pressure conditions)

(Assuming uniform wear conditions)

where R3 = Radius of head and

R4 = Radius of pin

1

Now the total torque to which the handle will be subjected is given by

= 1 + 2

Assuming that a person can apply a force of 300 ndash 400 N intermittently the length of handlerequired

= 300

The length of handle may be fixed by giving some allowance for gripping

The diameter of handle (D) may be obtained by considering bending effects We know that

bending moment

M =

32sdot ⌠ b sdot D3 ( ⌠ b = ⌠ t or ⌠ c)

11

12

The height of head (H) is usually taken as twice the diameter of handle ie H = 2D

Now check the screw for buckling load

Effective length or unsupported length of the screw

gt = Lift of screw + Height of nut

We know that buckling or critical load

2

where ⌠ = Yield stress

= = End fixity coefficient The screw is considered to be a strut with lower

end fixed and load end free For one end

fixed and the other end free = = 025

+ = Radius of gyration = 025 dc

The buckling load as obtained by the above expression must behigher than the load at which the screw is designed

13 Fix the dimensions for the body of the screw jack

14 Find efficiency of the screw jack

Example 1715 A $cre jac+ i$ to lift a load of 0) +N throuamph a

heiampht of 7)) mm he ela$tic $trenampth of $cre material in ten$ion andcompre$$ion i$ 2)) MPa and in $hear 2) MPa he material for nut i$ pho$phor-bron3e for hich the ela$tic limit ma be ta+en a$ )) MPa inten$ion ) MPa in compre$$ion and 0) MPa in $hear he bearinamp pre$$ure beteen the nut and the $cre i$ not to eceed 0 N6mm2 De$iampnand dra the $cre jac+ he de$iampn $hould include the de$iampn of $cre 2 nut 5 handle and cupand 7 bod

Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n

(D1 ) minus (do ) ⌠ t

4 = 2 8 +

⌠ gt W cr = Ac ⌠ 1 minus

= prop1 W = prop 1 W R

(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661

Solution Given W = 80 kN = 80 times 103N H1 = 400 mm = 04 m ⌠ et = ⌠ ec = 200 MPa

= 200 Nmm2 e = 120 MPa = 120 Nmm2 ⌠ et (nut ) = 100 MPa = 100 Nmm2 ⌠ ec (nut ) = 90 MPa

= 90 Nmm2 e(nut ) = 80 MPa = 80Nmm2 pb = 18 Nmm2

The various parts of a screw jack are designed as discussed below

$esin o screw or spindle

Let dc = Core diameter of the screw

Since the screw is under compression therefore load (W )

80 times 103 = (dc ) 2 sdot = (dc ) 2 = 7855 (dc ) 2

F lt 4 2

(Taking factor of safety Flt = 2)

there4 (dc )2 = 80 times 103 7855 = 10185 or dc = 32 mmFor square threads of normal series the following dimensions of the screw are selected from

Table 172

Core diameter dc = 38 mm AnsNominal or outside diameter of spindle

do = 46 mm AnsPitch of threads p = 8 mm Ans

Now let us check for principal stressesWe know that the mean diameter of screw

d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42

Assuming coefficient of friction between screw and nut

prop = tan = 014

there4Torque required to rotate the screw in the nut



3

1 minus 00606 sdot 014 2Now compressive stress due to axial load

⌠ c =W W 80 sdot 103

Ac 2 4 4

and shear stress due to the torque

16 1 16 sdot 340 sdot 103

=

there4Maximum principal stress (tensile or compressive)

⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2

From Table 172 we see that next higher value of 32 mm for the core diameter is 33 mm By taking dc = 33

mm gives higher principal stresses than the permissible values So core diameter is chosen as 38 mm


The given value of ⌠ c is equal to ieF lt 2

= 100 Nmm 2

We know that maximum shear stress

⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2

The given value of is equal to ieF lt 2

2

Since these maximum stresses are within limits therefore design of screw for spindle is safe

2 $esin or nut

Let n = Number of threads in contact with the screwed spindle


t = Thickness of screw = p 2 = 8 2 = 4 mmAssume that the load is distributed uniformly over the cross-sectional area of nut

We know that the bearing pressure ( pb )


3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n

Now let us check the stresses induced in the screw and nut

We know that shear stress in the screw

($cre) =

W

ndc t

80 sdot 103

sdot 10 sdot 38 sdot 4 ( t = p 2 = 4 mm)

and shear stress in the nut

(nut ) =

W

ndot

80 sdot 103

sdot 10 sdot 46 sdot 4Since these stresses are within permissible limit therefore design for nut is safe

Let D1 = Outer diameter of nut

D2 = Outside diameter for nut collar and

t 1 = Thickness of nut collar

First of all considering the tearing strength of nut we have

W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or

there4(D1)2 ndash 2116 = 80 times 103 393 = 2036

(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans

Now considering the crushing of the collar of the nut we have

Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or

there4 (D2)2 ndash 4225 = 80times

103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans

Considering the shearing of the collar of the nut we have

W = D1 times t 1 times

80 times 103 = sdot 65 sdot t i sdot80

2= 8170 t 1

e(nut )

there4 t 1 = 80 times 103 8170 = 98 say 10 mm Ans

= 393 ( D1 )2 minus (46)2 ( D1 )2 minus 2116 ⌠ t = et ( nut )

= = 1384 Nmm2

= = 1615 Nmm2

(do )2 minus (dc )2 n (46)2 minus (38)2 n18 =

W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200



3 $esin or handle and cup

The diameter of the head (D3) on the top of the screwed rod is usually taken as 175 times the

outside diameter of the screw (do)

there4 D3 = 175 do = 175 times 46 = 805 say 82 mm Ans

The head is provided with two holes at the right angles to receive the handle for rotating the

screw The seat for the cup is made equal to the diameter of head ie 82 mm and it is given chamfer

at the top The cup prevents the load from rotating The cup is fitted to the head with a pin of diameter

D4 = 20 mm The pin remains loose fit in the cup Other dimensions for the cup may be taken asfollows

Height of cup = 50 mm Ans

Thickness of cup = 10 mm Ans

Diameter at the top of cup = 160 mm Ans

Now let us find out the torque required ( 2) to overcome friction at the top of the screw

Assuming uniform pressure conditions we have

2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2

(Assuming prop1 = prop)

3

(41)2 minus (10)2

there4Total torque to which the handle is subjected

= 1 + 2 = 340 times 103 + 321 times 103 = 661 times 103 N-mm

Assuming that a force of 300 N is applied by a person intermittently therefore length of handle

required

= 661 times 103 300 = 2203 mm

Allowing some length for gripping we shall take the length of handle as 2250 mm


A little consideration will show that an excessive force applied at the end of lever will causebending Considering bending effect the maximum bending moment on the handle

M = Force applied times Length of lever

= 300 times 2250 = 675 times 103 N-mm

Let D = Diameter of the handle

Assuming that the material of the handle is same as that of screw therefore taking bending

stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2

We know that the bending moment (M)

675 times 103 =

32 32

there4 D3 = 675 times 103 982 = 6874 times 103 or D = 4096 say 42 mm Ans

The height of head (H) is taken as 2D

there4 H = 2 D = 2 times 42 = 84 mm Ans

Now let us check the screw for buckling load

We know that the effective length for the buckling of screw

gt = Lift of screw + Height of nut = H1 + h 2

= 400 + 80 2 = 440 mm

When the screw reaches the maximum lift it can be regarded as a strut whose lower end is fixed

and the load end is free We know that critical load

(D2 ) minus (65) = 353 (D2 )2 minus 4225 ⌠ c =F lt

=F lt

2 = 3 sdot prop1 W (R3 ) minus (R4 )

82 20 3

sdot 014 sdot 80 sdot 103 2 2 2 minus

82 20 minus

( 41)3 minus (10)3 = 747 sdot 103 = 321 sdot 10 N-mm



2 For one end fixed and other end free = = 025

Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2

(Taking ⌠ = ⌠ et )

= 226 852 (1 ndash 0207) = 179 894 N

Since the critical load is more than the load at which the screw is designed (ie 80 times 103 N)

therefore there is no chance of the screw to buckle

4 $esin o ody The various dimensions of the body may be fixed as follows

Diameter of the body at the top

D5 = 15 D2 = 15 times 82 = 123 mm Ans

Thickness of the body

t 3 = 025 do = 025 times 46 = 115 say 12 mm Ans

Inside diameter at the bottom

D6 = 225 D2 = 225 times 82 = 185 mm Ans

Outer diameter at the bottom

D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665

Thickness of base t 2 = 2 t 1 = 2 times 10 = 20 mm Ans

Height of the body = Max lift + Height of nut + 100 mm extra= 400 + 80 + 100 = 580 mm Ans

The body is made tapered in order to achieve stability of jack

Let us now find out the efficiency of the screw jack We know that the torque required to rotate

the screw with no friction

d 3 42

2 2

there4Efficiency of the screw jack

= 0

101808

661 sdot 103

Example 1716 A toggle jack as shown in Fig 1712 is to be designed for lifting a load of

4 kN When the jack is in the top position the distance between the centre lines of nuts is and

in the botto position this distance is 21 he eight links of the jack are s$etrical and 11

long he link pins in the base are set apart he links screw and pins are ade fro ild

steel for which the perissible stresses are 1 ampa in tension and ampa in shear he bearing

pressure on the pins is liited to 2 N(2

Assue the pitch of the s)uare threads as and the coefficient of friction between threads

as 2

sdot ⌠ b sdot D3 = sdot 100 sdot D3 = 982 D3

W cr = Ac sdot ⌠ 1 minus⌠ gt

4= 2 8 +

440 200(38) 200 1 minus

4 sdot 025 sdot sdot 210 sdot 103 95



Fig 1712

Solution Given W = 4 kN = 4000 N l = 110 mm ⌠ t = 100 MPa = 100 Nmm2 = 50 MPa

= 50 Nmm2 pb = 20 Nmm2 p = 6 mm prop = tan = 020

The toggle jack may be designed as discussed below

1 $esin o square threaded screw

A little consideration will show that the maximum load on the square threaded screw occurs

when the jack is in the bottom position The position of the link =D in the bottom position is shownin Fig 1713 (a)

Let be the angle of inclination of the link =D with the horizontal


(a) (b)

Fig 1713

From the geometry of the figure we find that

cos = 105 minus 15

110= 08112 or = 351deg

Each nut carries half the total load on the jack and due to this the link =D is subjected to tensionwhile the square threaded screw is under pull as shown in Fig 1713 (b) The magnitude of the pull on

the square threaded screw is given by

F =W W

2 tan 2 tan 351deg4000

2 sdot 07028Since a similar pull acts on the other nut therefore

total tensile pull on the square threaded rodW 1 = 2F = 2 times 2846 = 5692 N


We know that load on the screw (W 1)

5692 = (dc t (dc )2 100 The rotationa seed of the ead screwreatie to the sinde seed can bead8sted -ana b adding andre-oing gears to and fro- the gear

there4 (dc)2 = 5692 7855 = 725 or dc = 85 say 10 mmSince the screw is also subjected to torsional shear stress therefore to account for this let us

adopt

dc = 14 mm Ans

there4Nominal or outer diameter of the screwdo = dc + p = 14 + 6 = 20 mm Ans

and mean diameter of the screw

d = do ndash p 2 = 20 ndash 6 2 = 17 mmLet us now check for principal stresses We know that

tan lang = p

d

6

sdot 17 (where lang is the helix angle)

We know that effort required to rotate the screw


= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020

there4Torque required to rotate the screw

and shear stress in the screw due to torque

=

We know that direct tensile stress in the screw

there4Maximum principal (tensile) stress

W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2

Since the maximum stresses are within safe limits therefore the design of square threaded

screw is satisfactory

2 $esin o nut

Let n = Number of threads in contact with the screw (ie square threaded rod)

Assuming that the load W 1 is distributed uniformly over the cross-sectional area of the nut

therefore bearing pressure between the threads ( pb )

there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n

In order to have good stability and also to prevent rocking of the screw in the nut we shall

provide n = 4 threads in the nut The thickness of the nut

t = n times p = 4 times 6 = 24 mm Ans

The width of the nut (b) is taken as 15 do

there4 b = 15 do = 15 times 20 = 30 mm AnsTo control the movement of the nuts beyond 210 mm (the maximum distance between the

centre lines of nuts) rings of 8 mm thickness are fitted on the screw with the help of set screws

there4Length of screwed portion of the screw

= 210 + t + 2 times Thickness of rings

= 210 + 24 + 2 times 8 = 250 mm Ans

The central length (about 25 mm) of screwed rod is kept equal to core diameter of the screw ie

14 mm Since the toggle jack is operated by means of spanners on both sides of the square threaded

rod therefore the ends of the rod may be reduced to 10 mm square and 15 mm long

there4Toal length of the screw

= 250 + 2 times 15 = 280 mm Ans

Assuming that a force of 150 N is applied by each person at each end of the rod therefore length

of the spanner required

15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =

= = (do ) minus (dc ) n2 2 2 2 (20) minus (14) n

= = 5162 mm




We shall take the length of the spanner as 200 mm in order to facilitate the operation and evena single person can operate it

3 $esin o pins in the nuts

Let d1 = Diameter of pins in the nuts

Since the pins are in double shear therefore load on the pins (F )

2846 = 2 sdot 4 4

2

there4 (d1)2 = 2846 7855 = 3623 or d1 = 602 say 8 mm Ans

The diameter of pin head is taken as 15 d1 (ie 12 mm) and thickness 4 mm The pins in the nutsare kept in position by separate rings 4 mm thick and 15 mm split pins passing through the rings and

pins

4 $esin o lins

Due to the load the links may buckle in two planes at right angles to each other For buckling in

the vertical plane (ie in the plane of the links) the links are considered as hinged at both ends and for

buckling in a plane perpendicular to the vertical plane it is considered as fixed at both ends We know

that load on the link

= F 2 = 2846 2 = 1423 N

Assuming a factor of safety = 5 the links must be designed for a buckling load of

W cr = 1423 times 5 = 7115 N

Let t 1 = Thickness of the link and

b1 = Width of the link

Assuming that the width of the link is three times the thickness of the link ie b1 = 3 t 1 therefore

cross-sectional area of the link A = t 1 times 3t 1 = 3(t 1)2

and moment of inertia of the cross-section of the link

1 =1 1

12 12

We know that the radius of gyration

+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1

Since for buckling of the link in the vertical plane the ends are considered as hinged therefore

equivalent length of the link

gt = l = 110 mm

and Rankinersquos constant a =1

7500

According to Rankinersquos formula buckling load (W cr )

7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2

Power Screws n 669or

71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1

there4 (t 1)2 =237 plusmn (237)2 + 4 sdot 51 237 + 277

2 2

or

andt 1 = 507 say 6 mm

b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)

Now let us consider the buckling of the link in a plane perpendicular to the vertical plane

Moment of inertia of the cross-section of the link

1 =

1 1

12 12

and cross-sectional area of the link

A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2

there4Radius of gryration

+ =1

A

025 (t l )4

3 (t 1 )2

Since for buckling of the link in a plane perpendicular to the vertical plane the ends are considered

as fixed therefore

Equivalent length of the link

gt = l 2 = 110 2 = 55 mm

Again according to Rankines formula buckling load

W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2

Substituting the value of t 1 = 6 mm we have

300 sdot 62

48

62

Since this buckling load is more than the calculated value (ie 7115 N) therefore the link is safefor buckling in a plane perpendicular to the vertical plane

there4 We may take t 1 = 6 mm and b1 = 18 mm Ans

11) Di7erentia and Co-ond ScrewsThere are certain cases in which a very slow movement of the screw is required whereas in other

cases a very fast movement of the screw is needed The slow movement of the screw may be obtained

by using a small pitch of the threads but it results in weak threads The fast movement of the screw

may be obtained by using multiple-start threads but this method requires expensive machning and

the loss of self-locking property In order to overcome these difficulties differential or compound

screws as discussed below are used

1 $ierential screw When a slow movement or fine adjustment is desired in precision

equipments then a differential screw is used It consists of two threads of the same hand (ie right

handed or left handed) but of different pitches wound on the same cylinder or different cylinders as

shown in Fig 1714 It may be noted that when the threads are wound on the same cylinder then two


nuts are employed as shown in Fig 1714 (a) and when the threads are wound on different cylindersthen only one nut is employed as shown in Fig 1714 (b)

4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714

In this case each revolution of the screw causes the nuts to move towards or away from each

other by a distance equal to the difference of the pitches

Let p1 = Pitch of the upper screw

d1 = Mean diameter of the upper screw

lang1 = Helix angle of the upper screw andprop1 = Coefficient of friction between the upper screw and the upper nut

= tan 1 where 1 is the friction angle

p2 d2 lang2 and prop2 = Corresponding values for the lower screw

We know that torque required to overcome friction at the upper screw

d tan lang1 + tan 1 d2

Similarly torque required to overcome friction at the lower screw

+ 2 1 minus tan lang 2 tan 2 2

there4Total torque required to overcome friction at the thread surfaces

(i )

(ii )

= P1 times l = 1 ndash 2

When there is no friction between the thread surfaces then prop1 = tan 1 = 0 and prop2 = tan 2 = 0Substituting these values in the above expressions we have

there4 1E = W tan lang1 sdotd 1

2

andd 2

2

there4Total torque required when there is no friction

0 = 1E ndash 2E

= W tan lang1 sdot d1 d2

2 2

Power Screws n 671

sdot minus sdot2 d1 2 d2 2

W

2 ( p1 minus p2 )

p1 p2

1 2 We know that efficiency of the differential screw

= 0

2 +ompound screw When a fast movement is desired then a compound screw is employed It

consists of two threads of opposite hands (ie one right handed and the other left handed) wound on

the same cylinder or different cylinders as shown in Fig 1715 (a) and (b) respectively

In this case each revolution of the screw causes the nuts to move towards one another equal to

the sum of the pitches of the threads Usually the pitch of both the threads are made equal


+ 2 1 minus tan lang1 tan 1 2

1 = W tan (lang1 + 1 ) 1 = W 1 1 minus tan lang1 tan 1 2

2 = W tan (lang 2 2 ) 2 = W 2

d tan lang 2 + tan 2 d

2E = W tan lang 2 sdot

minus W tan lang 2 sdot



Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W

tan lang 2 + tan 2 d 2

1 minus tan lang 2 tan 2 2(ii )


= P1 times l = 1 + 2


1E = W tan lang1 sdot


d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2

2 We know that efficiency of the compound screw

0

=

Example 1717 A dierential $cre jac+ i$to be

made a$ $hon in Fiamp 4 Neither $crerotate$ heout$ide $cre diameter i$ () mm he $crethread$ areof $uare form $inample $tart and the coecient ofthreadfriction i$ )(

Determine 8cienc of the $cre jac+

2 gtoadthat can be lifted if the $hear $tre$$ in the bodof the$cre i$ limited to 20 MPa

Solution Given do = 50 mm prop = tan = 015

p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )

Fig 1716 Differential screw

1 Efciency o the screw jac

We know that the mean diameter of the upper screw

d1 = do ndash p1 2 = 50 ndash 16 2 = 42 mm

and mean diameter of the lower screw


there4 tan lang1 = p1

d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44

Let W = Load that can be lifted in N


= W =

p1 d1 p2 d

tan lang1 = d and tan lang2 = d

1 = W tan (lang1 1 ) 1 = W 1

d tan lang1 + tan 1 d



d tan lang1 + tan d

2 1 minus tan lang1 tan 2

01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)

d 2 tan lang 2 minus tan d 2

00868 minus 015 44= W


= minus 137 W N-mm

= 1 ndash 2 = 58 W ndash (ndash 137 W ) = 717 W N-mm

We know that the torque required when there is no friction

0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W

2 )oad that can e lited

Since the upper screw is subjected to a larger torque therefore the load to be lifted (W ) will be

calculated on the basis of larger torque ( 1)We know that core diameter of the upper screw

dc1 = do ndash p1 = 50 ndash 16 = 34 mm

= W tan lang1 sdot + W tan lang2 sdot

= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2

= W 1 + tan lang 2 tan 2

1 + 00868 sdot 015 2

( p1 2 ) =minus p



Since the screw is subjected to direct compressive stress due to load W and shear stress due to

torque 1 thereforeDirect compressive stress

W W W W

Ac1 2 908

4 4

Nmm 23 3

We know that maximum shear stress (ma )

28 =1 2 1

2 2

2

908 1331

2

=1 1

2 21863 sdot 10minus3 W

there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES

1 In a hand vice the screw has double start square threads of 24 mm outside diameter If the lever is 200

mm long and the maximum force that can be applied at the end of lever is 250 N find the force with

which the job is held in the jaws of the vice Assume a coefficient of friction of 012 Ans 17 42 N

2 A square threaded bolt of mean diameter 24 mm and pitch 5 mm is tightened by screwing a nut whose

mean diameter of bearing surface is 50 mm If the coefficient of friction for the nut and bolt is 01 and

for the nut and bearing surfaces 016 find the force required at the end of a spanner 05 m long when

the load on the bolt is 10 kN3 The spindle of a screw jack has a single

start square thread with an outside

diameter of 45 mm and a pitch of 10

mm The spindle moves in a fixed nut

The load is carried on a swivel head

but is not free to rotate The bearing

surface of the swivel head has a mean

diameter of 60 mm The coefficient of

friction between the nut and screw is

012 and that between the swivel head

and the spindle is 010 Calculate the

load which can be raised by efforts of

100 N each applied at the end of two

levers each of effective length of 350

mm Also determine the efficiency of

the lifting arrangement

Ans 45 N 227+

Ans 12 N

(ead screw sorted b coar bearing


4 The cross bar of a planner weighing 12 kN is raised and lowered by means of two square threadedscrews of 38 mm outside diameter and 7 mm pitch The screw is made of steel and a bronze nut of

38 mm thick A steel collar has 75 mm outside diameter and 38 mm inside diameter The coefficient of

friction at the threads is assumed as 011 and at the collar 013 Find the force required at a radius of

100 mm to raise and lower the load Ans 425 N 267 N

5 The lead screw of a lathe has square threads of 24 mm outside diameter and 5 mm pitch In order to

drive the tool carriage the screw exerts an axial pressure of 25 kN Find the efficiency of the screw

and the power required to drive the screw if it is to rotate at 30 rpm Neglect bearing friction

Assume coefficient of friction of screw threads as 012 Ans 3776+ 1655

6 The lead screw of a lathe has Acme threads of 60 mm outside diameter and 8 mm pitch It supplies

drive to a tool carriage which needs an axial force of 2000 N A collar bearing with inner and outer

radius as 30 mm and 60 mm respectively is provided The coefficient of friction for the screw threadsis 012 and for the collar it is 010 Find the torque required to drive the screw and the efficiency of the

screw Ans 15 N-m 136+

7 A cross bar of a planer weighing 9 kN is raised and lowered by means of two square threaded screws

of 40 mm outside diameter and 6 mm pitch The screw is made of steel and nut of phosphor bronze

having 42 mm height A steel collar bearing with 30 mm mean radius takes the axial thrust The

coefficient of friction at the threads and at the collar may be assumed as 014 and 010 respectively

Find the force required at a radius of 120 mm of a handwheel to raise and lower the load Find also the

shear stress in the nut material and the bearing pressure on the threads

Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2

= =and shear stress =16 1 16 sdot 58 W W

(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4

1213 sdot 10minus6 W 2 + 2258 sdot 10minus6 W 2 =



A machine slide weighing 3000 N is elevated by a double start acme threaded screw at the rate of

840 mmmin If the coefficient of friction be 012 calculate the power to drive the slide The end of

the screw is carried on a thrust collar of 32 mm inside diameter and 58 mm outside diameter The pitch

of the screw thread is 6 mm and outside diameter of the screw is 40 mm If the screw is of steel is it

strong enough to sustain the load Draw a neat sketch of the system Ans 165

A sluice valve used in water pipe lines consists of a gate raised by the spindle which is rotated by the

hand wheel The spindle has single start square threads The nominal diameter of the spindle is 36 mm

and the pitch is 6 mm The friction collar has inner and outer diameters of 32 mm and 50 mm respectively

The coefficient of friction at the threads and the collar are 012 and 018 respectively The weight of

the gate is 75 kN and the frictional resistance to open the valve due to water pressure is 275 kN

Using uniform wear theory determine 1 torque required to raise the gate and 2 overall efficiency

Ans 1365 N-m 71+1 A vertical square threads screw of a 70 mm mean diameter and 10 mm pitch supports a vertical load

of 50 kN It passes through the boss of a spur gear wheel of 70 teeth which acts as a nut In order toraise the load the spur gear wheel is turned by means of a pinion having 20 teeth The mechanical

efficiency of pinion and gear wheel drive is 90 The axial thrust on the screw is taken up by a collar

bearing having a mean radius of 100 mm The coefficient of friction for the screw and nut is 015 and

that for collar bearing is 012 Find

(a) Torque to be applied to the pinion shaft

(b) Maximum principal and shear stresses in the screw and

(c) Height of nut if the bearing pressure is limited to 12 Nmm2

Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm

11 A single start square threaded screw is to be designed for a C-clamp The axial load on the screw may

be assumed to be 10 kN A thrust pad is attached at the end of the screw whose mean diameter may be

taken as 30 mm The coefficient of friction for the screw threads and for the thrust pads is 012 and

008 respectively The allowable tensile strength of the screw is 60 MPa and the allowable bearing

pressure is 12 Nmm2 Design the screw and nut The square threads are as under

Ans d c 17 mm n 1 h 3 mm

Power Screws n 675

12 Design a screw jack for lifting a load of 50 kN through a height of 04 m The screw is made of steel

and nut of bronze Sketch the front sectional view The following allowable stresses may be assumed

For steel Compressive stress = 80 MPa Shear stress = 45 MPa

For bronze Tensile stress = 40 MPa Bearing stress = 15 MPa

Shear stress = 25 MPa

The coefficient of friction between the steel and bronze pair is 012 The dimensions of the swivel

base may be assumed proportionately The screw should have square threads Design the screw nut

and handle The handle is made of steel having bending stress 150 MPa (allowable)13 A screw jack carries a load of 22 kN Assuming the coefficient of friction between screw and nut as

015 design the screw and nut Neglect collar friction and column action The permissible compressive

and shear stresses in the screw should not exceed 42 MPa and 28 MPa respectively The shear stress

in the nut should not exceed 21 MPa The bearing pressure on the nut is 14 Nmm2 Also determine

the effort required at the handle of 200 mm length in order to raise and lower the load What will be

the efficiency of screw Ans d c 3 mm h 36 mm 31 N 166 N 276+

14

15

Design and draw a screw jack for lifting a safe load of 150 kN through a maximum lift of 350 mm

The elastic strength of the material of the screw may be taken as 240 MPa in compression and 160

MPa in shear The nut is to be made of phosphor bronze for which the elastic strengths in tension

compression and shear are respectively 130 115 and 100 MPa Bearing pressure between the threads

of the screw and the nut may be taken as 18 Nmm2 Safe crushing stress for the material of the body

is 100 MPa Coefficient of friction for the screw as well as collar may be taken as 015

Design a toggle jack to lift a load of 5 kN The jack is to be so designed that the distance between the

centre lines of nuts varies from 50 to 220 mm The eight links are symmetrical and 120 mm long The

link pins in the base are set 30 mm apart The links screw and pins are made from mild steel for which

the stresses are 90 MPa in tension and 50 MPa in shear The bearing pressure on the pin is 20 Nmm2

Assume the coefficient of friction between screw and nut as 015 and pitch of the square threaded

screw as 6 mm

Ans d c 1 mm d o 22 mm d 1 mm n 4 t 24 mm 33 mm d 1 1

mm

Nominal diameter mm 16 18 20 22

Core diameter mm 13 15 17 19

Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS

1 Discuss the various types of power threads Give atleast two practical applications for each type

Discuss their relative advantages and disadvantages

2 Why are square threads preferable to V-threads for power transmission

3 How does the helix angle influence on

the efficiency of square threaded screw

4 What do you understand by overhauling

of screw

5 What is self locking property of threads

and where it is necessary

6 Show that the efficiency of self lockingscrews is less than 50 percent

7 In the design of power screws on what

factors does the thread bearing pressure

depend Explain

Why is a separate nut preferable to an

integral nut with the body of a screw jack

Differentiate between differential screw

and compound screwScrew 8ack biding2bock sste-


Eamp8E 9E ESSNS

1 Which of the following screw thread is adopted for power transmission in either direction

(a) Acme threads (b) Square threads

(c) Buttress threads (d) Multiple threads

2 Multiple threads are used to secure(a) low efficiency (b) high efficiency

(c) high load lifting capacity (d) high mechanical advantage

3 Screws used for power transmission should have

(a) low efficiency (b) high efficiency

(c) very fine threads (d) strong teeth

4 If lang denotes the lead angle and the angle of friction then the efficiency of the screw is written as

(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )

5 A screw jack has square threads and the lead angle of the thread is lang The screw jack will be self-

locking when the coefficient of friction (prop) is

(a)

(c)

prop gt tan langprop = cot lang

(b)

(d)

prop = sin langprop = cosec lang

6 To ensure self locking in a screw jack it is essential that the helix angle is

(a) larger than friction angle (b) smaller than friction angle

(c) equal to friction angle (d) such as to give maximum efficiency in lifting

7 A screw is said to be self locking screw if its efficiency is

(a) less than 50 (b) more than 50

(c) equal to 50 (d) none of these

A screw is said to be over hauling screw if its efficiency is



While designing a screw in a screw jack against buckling failure the end conditions for the screw are

taken as

(a) both ends fixed (b) both ends hinged

(c) one end fixed and other end hinged (d) one end fixed and other end free

1 The load cup a screw jack is made separate from the head of the spindle to

(a) enhance the load carrying capacity of the jack



reduce the effort needed for

ng the working load

(c) reduce the value of frictional torque required to be countered for lifting the load

(d) prevent the rotation of load being lifted

ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



625

1 Tes of Screw Threads sed for Power Screws

Following are the three types of screw threads mostly used for power screws

1 Square thread A square thread as shown in Fig 171 (a) is adapted for the transmission of

power in either direction This thread results in maximum efficiency and minimum radial or bursting

Fig 171 Types of power screws

pressure on the nut It is difficult to cut with taps and dies It is usually cut on a lathe with a single

point tool and it can not be easily compensated for wear The

square threads are employed in screw jacks presses and

clamping devices The standard dimensions for square threads

according to IS 4694 ndash 1968 (Reaffirmed 1996) are shownin Table 171 to 173

2 Acme or trapezoidal thread An acme or trapezoidal

thread as shown in Fig 171 (b) is a modification of square

thread The slight slope given to its sides lowers the efficiency

slightly than square thread and it also introduce some burstingpressure on the nut but increases its area in shear It is usedwhere a split nut is required and where provision is made to

take up wear as in the lead screw of a lathe Wear may be

taken up by means of an adjustable split nut An acme thread

may be cut by means of dies and hence it is more easily

manufactured than square thread The standard dimensions

for acme or trapezoidal threads are shown in Table 174

(Page 630)

3 Buttress thread A buttress thread as shown in Fig

171 (c) is used when large forces act along the screw axis in

one direction only This thread combines the higher efficiency

of square thread and the ease of cutting and the adaptability toScrew 8acks

a split nut of acme thread It is stronger than other threads because of greater thickness at the base of the thread The buttress thread has limited use for power transmission It is employed as the thread for

light jack screws and vices

Tabe 11 9asic di-ensions for sare threads in -- 4ine series according

to IS lt amp0amp = 10 Rea+r-ed 100


Power Screws n 627

Nominal

diameter

(d )1


diameter

(d )

Pitch

( p )

Depth of Area of

core2

( A ) mmcBolt

(d)

Nut

(D)

Bolt

(h)

Nut

(H)

10

12

10

12

105

125

8

10

2 1 125 503

785

d1

d D dc

p h H Ac

14

16

18

20

14

16

18

20

145

165

185

205

12

14

16

18

2 1 125

113

154

201

254

22

24

26

28

30

32

(34)

36

(38)

40

42

44

(46)

48

50

52

55

(58)

60

(62)

22

24

26

28

30

32

34

36

38

40

42

44

46

48

50

52

55

58

60

62

225

245

265

285

305

325

345

365

385

405

425

445

465

485

505

525

555

585

605

625

19

21

23

25

27

29

31

33

35

37

39

41

43

45

47

49

52

55

57

59

3 15 175

284

346

415

491

573

661

755

855

962

1075

1195

1320

1452

1590

1735

1886

2124

2376

2552

2734

65

(68)

70

(72)

75

(78)

80

(82)

(85)

(88)

90

(92)

95

(98)

65

68

70

72

75

78

80

82

85

88

90

92

95

98

655

685

705

725

755

785

805

825

855

885

905

925

955

985

61

64

66

68

71

74

76

78

81

84

86

88

91

94

4 2 225

2922

3217

3421

3632

3959

4301

4536

4778

5153

5542

5809

6082

6504

6960







Power Screws n 629





d1

d D dc

p h H Ac

100

(105)

110

100

105

110

1005

1055

1105

96

101

106

4 2 225

7238

8012

8825

(115)

120

(125)

130

(135)

140

(145)

150

(155)

160

(165)

170

(175)

115

120

125

130

135

140

145

150

155

160

165

170

175

1155

1205

1255

1305

1355

1405

1455

1505

1555

1605

1655

1705

1755

109

114

119

124

129

134

139

144

149

154

159

164

169

6 3 325

9331

10207

11 122

12 076

13 070

14 103

15 175

16 286

17437

18 627

19 856

21124

22 432

Nominal

diameter

(d )1


diameter

(d )c

Pitch

( p)

Depth of Area of

core2

( A ) mmcBolt

(d)

Nut

(D)

Bolt

(h)

Nut

(H)

22

24

26

28

22

24

26

28

225

245

265

285

17

19

21

23

5 25 275

227

284

346

415

30

32

(34)

36

30

32

34

36

305

325

345

365

24

26

28

30

6 3 325

452

531

616

707

(38)

40

(42)

44

38

40

42

44

385

405

425

445

31

33

35

37

7 35 375

755

855

962

1075

d1

d D dc

p h H Ac

(46)

48

50

52

46

48

50

52

465

485

505

525

38

40

42

44

8 4 425

1134

1257

1385

1521

55

(58)

(60)

(62)

55

58

60

62

555

585

605

625

46

49

51

53

9 45 525

1662

1886

2043

2206

65

(68)

70

(72)

75

(78)

80

(82)

65

68

70

72

75

78

80

82

655

685

705

725

755

785

805

825

55

58

60

62

65

68

70

72

10 5 525

2376

2642

2827

3019

3318

3632

3848

4072

85

(88)

90(92)

95

(98)

100

(105)

110

85

88

9092

95

98

100

105

110

855

885

855925

955

985

1005

1055

1105

73

76

7880

83

86

88

93

98

12 6 625

4185

4536

47785027

5411

5809

6082

6793

7543

(115)

120

(125)

130

(135)

140

(145)

115

120

125

130

135

140

145

116

121

126

131

136

141

146

101

106

111

116

121

126

131

14 7 75

8012

882

9677

10 568

11 499

12 469

13 478

150

(155)

160

150

155

160

151

156

161

134

139

144

16 8 85

14 103

15 175

16 286

Nominal

diameter

(d )1


diameter

(d )c

Pitch

( p)

Depth of Area of

core2

( A ) mmcBolt

(d)

Nut

(D)

Bolt

(h)

Nut

(H)

22

24

26

28

22

24

26

28

225

245

265

285

14

16

18

20

8 4 425

164

204

254

314

30

32

(34)

36

(38)

30

32

34

36

38

305

325

345

365

385

20

22

24

26

28

10 5 525

314

380

452

531

616

40

(42)

44

(46)

48

50

52

40

42

44

46

48

50

52

405

425

445

465

485

505

525

28

30

32

34

36

38

40

12 6 625

616

707

804

908

1018

1134

1257

55

(58)

60

(62)

55

58

60

62

56

59

61

63

41

44

46

48

14 7 725

1320

1521

1662

1810

65

(68)

70

(72)

75

(78)

80

(82)

65

68

70

72

75

78

80

82

66

69

71

73

76

79

81

83

49

52

54

56

59

62

64

66

16 8 85

1886

2124

2290

2463

2734

3019

3217

3421

d1

d D dc

p h H Ac

(165)

170

(175)

165

170

175

166

171

176

149

154

159

16 8 85

17 437

18 627

19 856





Power Screws n 631


1$ Mtie Threads









d1

d D dc

p h H Ac

85

(88)

90

(92)

95

(96)

85

88

90

92

95

96

86

89

91

93

96

99

67

70

72

74

77

80

18 9 95

3526

3848

4072

4301

4657

5027

100

(105)

110

100

105

110

101

106

111

80

85

90

20 10 105

5027

5675

6362

(115)

120(125)

130

115

120125

130

116

121126

131

93

98103

108

22 11 115

6793

75438332

9161

(135)

140

(145)

150

(155)

135

140

145

150

155

136

141

146

151

156

111

116

121

126

131

24 12 125

9667

10 568

11 499

12 469

13 478

160

(165)

170

(175)

160

165

170

175

161

166

171

176

132

137

142

147

28 14 145

13 635

14 741

15 837

16 972

Nominal or major

dia-

Minor or core dia-

meter (d ) mmc

Pitch

( p ) mm

Area of core2

( A ) mmc

10

12

65

85

3 33

57

14

16

18

20

95

115

135

155

4

71

105

143

189

22

24

26

28

165

185

205

225

5

214

269

330

389

30

32

34

36

235

255

275

295

6

434

511

594

683

eads

d dc

p Ac

38

40

42

44

305

325

345

365

7

731

830

935

1046

46

48

50

52

375

395

415

435

8

1104

1225

1353

1486

55

58

60

62

455

485

505

525

9

1626

1847

2003

2165

65

68

70

72

75

78

80

82

545

575

595

615

645

675

695

715

10

2333

2597

2781

2971

3267

3578

3794

4015

85

88

90

92

95

98

100

105

110

115

725

755

775

795

825

855

875

925

975

100

12

4128

4477

4717

4964

5346

5741

6013

6720

7466

7854

120

125

130

135

140

145

150

105

110

115

120

125

130

133

14

8659

9503

10 387

11 310

12 272

13 273

13 893

155

160

165

170

175

138

143

148

153

158

16

14 957

16 061

17 203

18 385

19 607





Fig 173 (a)

Fig 173



lang = Helix angle

Power Screws n 633






tan lang = p d


in Fig 173 (b)








(i )

(ii )

or

or


or





by cos we have

sin lang cos + sin



c

o

s

langP = W sdot


Screw 8ack

sin (lang + )


d d

2 2



where

2 =

R1 and R2 =

R =

prop1 =








= 1 + 2


d

2





W W sdot 2 l


W sdot 2 l 2 l





in Fig 174






Fig 174

(i )

(ii )

or

or





+ R


2 (R1 )3 minus (R2 )3



P = W sdot

there4 P = W sdot



(tan

cos

sin lang )



cos (minus lang )



Power Screws n 635


d d

2 2







load is



lang = Helix angle






Actual effort





=

=










=2 l

p

2 l 2 l









do + dc p p


1

=

P = W sdot

P = W sdot



= = =0

0 P0 sdot d 2

= = =1

= =

= sdot =

= =






(ii )







sin 90deg minus sin

sin 90deg + sin

1 minus sin


and





d

125


tan lang + tan



d 50 3


D1 D1

2 2(ii )


D1 = 112 200 100 = 1122 mm = 1122 m Ans





prop = tan = 01

=


= = 008


= 10 sdot 10 = 2328 N







Power Screws n 637

and tan lang = p

d

6

sdot 37



00516 + 01 3


= P sdot d 37 3

2 2



and angular speed

Speed in mm min 300




mm







p 10



00637

+ 015


R =2 2

d 50

2 2= 4410 N-mm = 441 N-m


N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N


= = 50 rpm



= 400 = 864 N

= = 375 mmR1 + R2 30 + 45



= = 600 rpm






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST







Power Screws n 629





d1

d D dc

p h H Ac

100

(105)

110

100

105

110

1005

1055

1105

96

101

106

4 2 225

7238

8012

8825

(115)

120

(125)

130

(135)

140

(145)

150

(155)

160

(165)

170

(175)

115

120

125

130

135

140

145

150

155

160

165

170

175

1155

1205

1255

1305

1355

1405

1455

1505

1555

1605

1655

1705

1755

109

114

119

124

129

134

139

144

149

154

159

164

169

6 3 325

9331

10207

11 122

12 076

13 070

14 103

15 175

16 286

17437

18 627

19 856

21124

22 432

Nominal

diameter

(d )1


diameter

(d )c

Pitch

( p)

Depth of Area of

core2

( A ) mmcBolt

(d)

Nut

(D)

Bolt

(h)

Nut

(H)

22

24

26

28

22

24

26

28

225

245

265

285

17

19

21

23

5 25 275

227

284

346

415

30

32

(34)

36

30

32

34

36

305

325

345

365

24

26

28

30

6 3 325

452

531

616

707

(38)

40

(42)

44

38

40

42

44

385

405

425

445

31

33

35

37

7 35 375

755

855

962

1075

d1

d D dc

p h H Ac

(46)

48

50

52

46

48

50

52

465

485

505

525

38

40

42

44

8 4 425

1134

1257

1385

1521

55

(58)

(60)

(62)

55

58

60

62

555

585

605

625

46

49

51

53

9 45 525

1662

1886

2043

2206

65

(68)

70

(72)

75

(78)

80

(82)

65

68

70

72

75

78

80

82

655

685

705

725

755

785

805

825

55

58

60

62

65

68

70

72

10 5 525

2376

2642

2827

3019

3318

3632

3848

4072

85

(88)

90(92)

95

(98)

100

(105)

110

85

88

9092

95

98

100

105

110

855

885

855925

955

985

1005

1055

1105

73

76

7880

83

86

88

93

98

12 6 625

4185

4536

47785027

5411

5809

6082

6793

7543

(115)

120

(125)

130

(135)

140

(145)

115

120

125

130

135

140

145

116

121

126

131

136

141

146

101

106

111

116

121

126

131

14 7 75

8012

882

9677

10 568

11 499

12 469

13 478

150

(155)

160

150

155

160

151

156

161

134

139

144

16 8 85

14 103

15 175

16 286

Nominal

diameter

(d )1


diameter

(d )c

Pitch

( p)

Depth of Area of

core2

( A ) mmcBolt

(d)

Nut

(D)

Bolt

(h)

Nut

(H)

22

24

26

28

22

24

26

28

225

245

265

285

14

16

18

20

8 4 425

164

204

254

314

30

32

(34)

36

(38)

30

32

34

36

38

305

325

345

365

385

20

22

24

26

28

10 5 525

314

380

452

531

616

40

(42)

44

(46)

48

50

52

40

42

44

46

48

50

52

405

425

445

465

485

505

525

28

30

32

34

36

38

40

12 6 625

616

707

804

908

1018

1134

1257

55

(58)

60

(62)

55

58

60

62

56

59

61

63

41

44

46

48

14 7 725

1320

1521

1662

1810

65

(68)

70

(72)

75

(78)

80

(82)

65

68

70

72

75

78

80

82

66

69

71

73

76

79

81

83

49

52

54

56

59

62

64

66

16 8 85

1886

2124

2290

2463

2734

3019

3217

3421

d1

d D dc

p h H Ac

(165)

170

(175)

165

170

175

166

171

176

149

154

159

16 8 85

17 437

18 627

19 856





Power Screws n 631


1$ Mtie Threads









d1

d D dc

p h H Ac

85

(88)

90

(92)

95

(96)

85

88

90

92

95

96

86

89

91

93

96

99

67

70

72

74

77

80

18 9 95

3526

3848

4072

4301

4657

5027

100

(105)

110

100

105

110

101

106

111

80

85

90

20 10 105

5027

5675

6362

(115)

120(125)

130

115

120125

130

116

121126

131

93

98103

108

22 11 115

6793

75438332

9161

(135)

140

(145)

150

(155)

135

140

145

150

155

136

141

146

151

156

111

116

121

126

131

24 12 125

9667

10 568

11 499

12 469

13 478

160

(165)

170

(175)

160

165

170

175

161

166

171

176

132

137

142

147

28 14 145

13 635

14 741

15 837

16 972

Nominal or major

dia-

Minor or core dia-

meter (d ) mmc

Pitch

( p ) mm

Area of core2

( A ) mmc

10

12

65

85

3 33

57

14

16

18

20

95

115

135

155

4

71

105

143

189

22

24

26

28

165

185

205

225

5

214

269

330

389

30

32

34

36

235

255

275

295

6

434

511

594

683

eads

d dc

p Ac

38

40

42

44

305

325

345

365

7

731

830

935

1046

46

48

50

52

375

395

415

435

8

1104

1225

1353

1486

55

58

60

62

455

485

505

525

9

1626

1847

2003

2165

65

68

70

72

75

78

80

82

545

575

595

615

645

675

695

715

10

2333

2597

2781

2971

3267

3578

3794

4015

85

88

90

92

95

98

100

105

110

115

725

755

775

795

825

855

875

925

975

100

12

4128

4477

4717

4964

5346

5741

6013

6720

7466

7854

120

125

130

135

140

145

150

105

110

115

120

125

130

133

14

8659

9503

10 387

11 310

12 272

13 273

13 893

155

160

165

170

175

138

143

148

153

158

16

14 957

16 061

17 203

18 385

19 607





Fig 173 (a)

Fig 173



lang = Helix angle

Power Screws n 633






tan lang = p d


in Fig 173 (b)








(i )

(ii )

or

or


or





by cos we have

sin lang cos + sin



c

o

s

langP = W sdot


Screw 8ack

sin (lang + )


d d

2 2



where

2 =

R1 and R2 =

R =

prop1 =








= 1 + 2


d

2





W W sdot 2 l


W sdot 2 l 2 l





in Fig 174






Fig 174

(i )

(ii )

or

or





+ R


2 (R1 )3 minus (R2 )3



P = W sdot

there4 P = W sdot



(tan

cos

sin lang )



cos (minus lang )



Power Screws n 635


d d

2 2







load is



lang = Helix angle






Actual effort





=

=










=2 l

p

2 l 2 l









do + dc p p


1

=

P = W sdot

P = W sdot



= = =0

0 P0 sdot d 2

= = =1

= =

= sdot =

= =






(ii )







sin 90deg minus sin

sin 90deg + sin

1 minus sin


and





d

125


tan lang + tan



d 50 3


D1 D1

2 2(ii )


D1 = 112 200 100 = 1122 mm = 1122 m Ans





prop = tan = 01

=


= = 008


= 10 sdot 10 = 2328 N







Power Screws n 637

and tan lang = p

d

6

sdot 37



00516 + 01 3


= P sdot d 37 3

2 2



and angular speed

Speed in mm min 300




mm







p 10



00637

+ 015


R =2 2

d 50

2 2= 4410 N-mm = 441 N-m


N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N


= = 50 rpm



= 400 = 864 N

= = 375 mmR1 + R2 30 + 45



= = 600 rpm






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST





Power Screws n 631


1$ Mtie Threads









d1

d D dc

p h H Ac

85

(88)

90

(92)

95

(96)

85

88

90

92

95

96

86

89

91

93

96

99

67

70

72

74

77

80

18 9 95

3526

3848

4072

4301

4657

5027

100

(105)

110

100

105

110

101

106

111

80

85

90

20 10 105

5027

5675

6362

(115)

120(125)

130

115

120125

130

116

121126

131

93

98103

108

22 11 115

6793

75438332

9161

(135)

140

(145)

150

(155)

135

140

145

150

155

136

141

146

151

156

111

116

121

126

131

24 12 125

9667

10 568

11 499

12 469

13 478

160

(165)

170

(175)

160

165

170

175

161

166

171

176

132

137

142

147

28 14 145

13 635

14 741

15 837

16 972

Nominal or major

dia-

Minor or core dia-

meter (d ) mmc

Pitch

( p ) mm

Area of core2

( A ) mmc

10

12

65

85

3 33

57

14

16

18

20

95

115

135

155

4

71

105

143

189

22

24

26

28

165

185

205

225

5

214

269

330

389

30

32

34

36

235

255

275

295

6

434

511

594

683

eads

d dc

p Ac

38

40

42

44

305

325

345

365

7

731

830

935

1046

46

48

50

52

375

395

415

435

8

1104

1225

1353

1486

55

58

60

62

455

485

505

525

9

1626

1847

2003

2165

65

68

70

72

75

78

80

82

545

575

595

615

645

675

695

715

10

2333

2597

2781

2971

3267

3578

3794

4015

85

88

90

92

95

98

100

105

110

115

725

755

775

795

825

855

875

925

975

100

12

4128

4477

4717

4964

5346

5741

6013

6720

7466

7854

120

125

130

135

140

145

150

105

110

115

120

125

130

133

14

8659

9503

10 387

11 310

12 272

13 273

13 893

155

160

165

170

175

138

143

148

153

158

16

14 957

16 061

17 203

18 385

19 607





Fig 173 (a)

Fig 173



lang = Helix angle

Power Screws n 633






tan lang = p d


in Fig 173 (b)








(i )

(ii )

or

or


or





by cos we have

sin lang cos + sin



c

o

s

langP = W sdot


Screw 8ack

sin (lang + )


d d

2 2



where

2 =

R1 and R2 =

R =

prop1 =








= 1 + 2


d

2





W W sdot 2 l


W sdot 2 l 2 l





in Fig 174






Fig 174

(i )

(ii )

or

or





+ R


2 (R1 )3 minus (R2 )3



P = W sdot

there4 P = W sdot



(tan

cos

sin lang )



cos (minus lang )



Power Screws n 635


d d

2 2







load is



lang = Helix angle






Actual effort





=

=










=2 l

p

2 l 2 l









do + dc p p


1

=

P = W sdot

P = W sdot



= = =0

0 P0 sdot d 2

= = =1

= =

= sdot =

= =






(ii )







sin 90deg minus sin

sin 90deg + sin

1 minus sin


and





d

125


tan lang + tan



d 50 3


D1 D1

2 2(ii )


D1 = 112 200 100 = 1122 mm = 1122 m Ans





prop = tan = 01

=


= = 008


= 10 sdot 10 = 2328 N







Power Screws n 637

and tan lang = p

d

6

sdot 37



00516 + 01 3


= P sdot d 37 3

2 2



and angular speed

Speed in mm min 300




mm







p 10



00637

+ 015


R =2 2

d 50

2 2= 4410 N-mm = 441 N-m


N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N


= = 50 rpm



= 400 = 864 N

= = 375 mmR1 + R2 30 + 45



= = 600 rpm






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST





Fig 173 (a)

Fig 173



lang = Helix angle

Power Screws n 633






tan lang = p d


in Fig 173 (b)








(i )

(ii )

or

or


or





by cos we have

sin lang cos + sin



c

o

s

langP = W sdot


Screw 8ack

sin (lang + )


d d

2 2



where

2 =

R1 and R2 =

R =

prop1 =








= 1 + 2


d

2





W W sdot 2 l


W sdot 2 l 2 l





in Fig 174






Fig 174

(i )

(ii )

or

or





+ R


2 (R1 )3 minus (R2 )3



P = W sdot

there4 P = W sdot



(tan

cos

sin lang )



cos (minus lang )



Power Screws n 635


d d

2 2







load is



lang = Helix angle






Actual effort





=

=










=2 l

p

2 l 2 l









do + dc p p


1

=

P = W sdot

P = W sdot



= = =0

0 P0 sdot d 2

= = =1

= =

= sdot =

= =






(ii )







sin 90deg minus sin

sin 90deg + sin

1 minus sin


and





d

125


tan lang + tan



d 50 3


D1 D1

2 2(ii )


D1 = 112 200 100 = 1122 mm = 1122 m Ans





prop = tan = 01

=


= = 008


= 10 sdot 10 = 2328 N







Power Screws n 637

and tan lang = p

d

6

sdot 37



00516 + 01 3


= P sdot d 37 3

2 2



and angular speed

Speed in mm min 300




mm







p 10



00637

+ 015


R =2 2

d 50

2 2= 4410 N-mm = 441 N-m


N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N


= = 50 rpm



= 400 = 864 N

= = 375 mmR1 + R2 30 + 45



= = 600 rpm






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



c

o

s

langP = W sdot


Screw 8ack

sin (lang + )


d d

2 2



where

2 =

R1 and R2 =

R =

prop1 =








= 1 + 2


d

2





W W sdot 2 l


W sdot 2 l 2 l





in Fig 174






Fig 174

(i )

(ii )

or

or





+ R


2 (R1 )3 minus (R2 )3



P = W sdot

there4 P = W sdot



(tan

cos

sin lang )



cos (minus lang )



Power Screws n 635


d d

2 2







load is



lang = Helix angle






Actual effort





=

=










=2 l

p

2 l 2 l









do + dc p p


1

=

P = W sdot

P = W sdot



= = =0

0 P0 sdot d 2

= = =1

= =

= sdot =

= =






(ii )







sin 90deg minus sin

sin 90deg + sin

1 minus sin


and





d

125


tan lang + tan



d 50 3


D1 D1

2 2(ii )


D1 = 112 200 100 = 1122 mm = 1122 m Ans





prop = tan = 01

=


= = 008


= 10 sdot 10 = 2328 N







Power Screws n 637

and tan lang = p

d

6

sdot 37



00516 + 01 3


= P sdot d 37 3

2 2



and angular speed

Speed in mm min 300




mm







p 10



00637

+ 015


R =2 2

d 50

2 2= 4410 N-mm = 441 N-m


N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N


= = 50 rpm



= 400 = 864 N

= = 375 mmR1 + R2 30 + 45



= = 600 rpm






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



(tan

cos

sin lang )



cos (minus lang )



Power Screws n 635


d d

2 2







load is



lang = Helix angle






Actual effort





=

=










=2 l

p

2 l 2 l









do + dc p p


1

=

P = W sdot

P = W sdot



= = =0

0 P0 sdot d 2

= = =1

= =

= sdot =

= =






(ii )







sin 90deg minus sin

sin 90deg + sin

1 minus sin


and





d

125


tan lang + tan



d 50 3


D1 D1

2 2(ii )


D1 = 112 200 100 = 1122 mm = 1122 m Ans





prop = tan = 01

=


= = 008


= 10 sdot 10 = 2328 N







Power Screws n 637

and tan lang = p

d

6

sdot 37



00516 + 01 3


= P sdot d 37 3

2 2



and angular speed

Speed in mm min 300




mm







p 10



00637

+ 015


R =2 2

d 50

2 2= 4410 N-mm = 441 N-m


N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N


= = 50 rpm



= 400 = 864 N

= = 375 mmR1 + R2 30 + 45



= = 600 rpm






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST






(ii )







sin 90deg minus sin

sin 90deg + sin

1 minus sin


and





d

125


tan lang + tan



d 50 3


D1 D1

2 2(ii )


D1 = 112 200 100 = 1122 mm = 1122 m Ans





prop = tan = 01

=


= = 008


= 10 sdot 10 = 2328 N







Power Screws n 637

and tan lang = p

d

6

sdot 37



00516 + 01 3


= P sdot d 37 3

2 2



and angular speed

Speed in mm min 300




mm







p 10



00637

+ 015


R =2 2

d 50

2 2= 4410 N-mm = 441 N-m


N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N


= = 50 rpm



= 400 = 864 N

= = 375 mmR1 + R2 30 + 45



= = 600 rpm






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST





Power Screws n 637

and tan lang = p

d

6

sdot 37



00516 + 01 3


= P sdot d 37 3

2 2



and angular speed

Speed in mm min 300




mm







p 10



00637

+ 015


R =2 2

d 50

2 2= 4410 N-mm = 441 N-m


N =Cutting speed 6

Pitch 001

638

= = 00516


= 75 sdot 103 = 1143 sdot 10 N


= = 50 rpm



= 400 = 864 N

= = 375 mmR1 + R2 30 + 45



= = 600 rpm






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST






= = 441 times 6284 = 277 W = 0277 kW Ans




= 0144 or 144 Ans







= 2 p = 2 times 20 = 40 mm


d sdot 100



tan lang + tan



R = = = 875 mm2 2


= P sdot + prop1 WR

100 3

2




tan minus tan lang

3 015 minus 0127 1 + 015 sdot 0127

Power Screws n 639


d

= =

= = 0127


=18 sdot 10 = 5083 N

R1 + R2 50 + 125



=18 sdot 10 = 4063 N



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



2

100 3

2








) mm





d

10

sdot 50


tan lang + tan






N = 170 10 = 17






00637 (1 minus 00637 sdot 008)=



R1 + R2 5 + 30R =


d

2

Ans


= 2890 sdot50

2


= 10025 N-m






= = 175 mm

00637 + 008= 0441 or 441

=

= 20 sdot 10 1 00673 008 = 2890 N


= = 00637






= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST




= 0

3185

10025












load is


d d

2 2




Power Screws n 641




= = 0318 or 318 Ans






=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST





=tan (lang + )





1 tan2 2 2

2 tan




Let



there4

-13Output = Wh



or 50 If the


1

For self locking

1


2or 50








ing table



following table


tan

le le le minus

tan 2=1 minus tan 2

=


W h le W hminus 1










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST










load (W )


Fig 176






Power Screws n 643





thread

prop










p 8

d sdot 46


prop 015 015


ltNo



1

2

3







014

018

021

010

013

015



2

3

4





017

015

010

008

012

009

008

006

di7er



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



tan lang + tan 1


0055 + 0155




R =R1 + R2 55 + 275

2 2




= 1 + 2 = 12 190 + 12 375 = 24 565 N-mm = 24565 N-m




60 60Ans

( = 2N 60)


d 46

2 2


= o

3163

24565Ans








W

= Ac






W cr =

Ac sdot⌠ 1 minus

2

there4 ⌠ c =W Ac

1 2

where W cr

⌠ gt

=

8

⌠

c

= = 4125 mm

= 2500 1 minus 0055 sdot 0155 = 530 N







Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



Crit

ical

load

Yiel

d

st

re

ss

Length of screw










=

16

sdot (dc )3

Power Screws n 645


16

(d c )3


diameter section

ma =1

2(Bt or Bc 2 amp 2










W


where

W













where

W W pb =

)2 minus (d 2




=Height of the nut

Pitch of threads=

h

p



= = = = 77 W = 0077 kW


= = 013 or 13

⌠ gt

4 = 2 8 +

1 minus

⌠ gt 4 = 8 +















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST















= 2 p = 2 times 5 = 10 mm



tan lang + tan


01414 + 02



R =R1 + R2 25 + 10

2 2

=

($cre) =

(nut ) =

=

d t n (dc c ) n


Application of

$cre


pre$$ure

2

Rubbinamp $peed

at


1 Hand press

2 Screw jack

3 Hoisting screw

4 Lead screw

Steel

Steel

Steel

Steel

Steel

Steel

Bronze

Cast iron

Bronze

Cast iron

Bronze

Bronze

175 - 245

126 ndash 175

112 ndash 175

42 ndash 70

56 ndash 98

105 ndash 17

Low speed well

lubricated

Low speed

lt 24 m min

Low speed

lt 3 m min

Medium speed

6 ndash 12 m min

Medium speed

6 ndash 12 m min

High speed

gt 15 m min

ressres

= = 01414


= 10 sdot 103 = 3513 N

= = 175




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST




Power Screws n 647

= P sdot d 225

2 2

= 65 771 N-mm = 65771 N-m Ans





Ac =

4 4(20)2 = 3142 mm2


and shear stress

⌠ c =

=

W 10 sdot 103

Ac 3142

16 16

sdot 65 771

3 3


ma =1

22 1

2(3183)2 + 4 (4186)2






58 =W

d sdot t sdot n

3


there4 n = 566 58 = 976 say 10 Ans


44







prop = tan = 012







10 sdot 10 566

(⌠ c )2 + 4 =

(dc ) (20)= = 4186 Nmm 2

= = 3183 Nmm2

(dc )2 =




d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



d =do + dc 75 + 69

2 2

= 72 mm

and tan lang = p

d

6

sdot 72

= 00265


= P sdot d2

= W tan (lang + )2


00265 + 012 72

1 minus 00265 sdot 012 2

= 158 728 N-mm



Fig 177

there4

D 300

2 2



W W 30 sdot 103

Ac 2



Ans


Pitch of threads 6



pb =W

d t n

30 sdot 103




=16

(dc )

16 sdot 158 728

(69)3

= 246 Nmm 2 Ans


d do p 6 2 = 75 ndash 6 2 = 72 mm


Power Screws n 649

ma =

1

2(

⌠ c )2

+ 4

2 =

1

2(802)2 + 4 (246)2




d 3 72

2 2


=

d

= W

= 30 sdot 103

158 728 = P1 sdot = P1 sdot = 150 P1

= = = 802 Nmm2 = 802 MPa⌠ c =(dc ) (69)2

= = 25 threads

= = 177 Nmm2 Ans



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



28 620

158 728Ans




diameter









(1000 mm) long




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST




screw


Fig 17

d = = = 55 mm2 2

and tan lang = p

d

10

sdot 55





W = W 1 + F = 18 000 + 4000 = 22 000 N





0058 + 01 55

1 minus 0058 sdot 01 2= 96 148 N-mm

R1 + R2 75 + 25R =

2 2









W = W 1 ndash F = 18 000 ndash 4000 = 14 000 N






= 1 + 2 = 16 077 + 84 000 = 100 077 N-mm


100 077 = 2P1 times 1000 = 2000 P1 or P1 = 100 0772000 = 5004 N Ans



d 55

2 2


= 0

35 090

228 148Ans




= = 018 or 18

do + dc 60 + 50

= = 0058







and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST






and

there4

W 22000 25467 =



Power Screws n 651






$urface







= 3 p = 3 times 8 = 24 mm Fig 17


d sdot 44




there4


= 7436 W 1 minus 0174 sdot 0155 2

W = 40 000 7436 = 5380 NAns



d

2







n = h p = 50 8 = 625


(Given)


= = 0154 or 154


= 14 000 = 16 077 N-mm


1 = P sdot

= = 50 mm

= 22 000

= = 0174

prop1 = tan 1 = cos cos 15 09659 = 0155= =



0174 + 0155 4440 000 = W






pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST





pb = W dt n




prop2 = 025 R = 6 mm P1 = 80 N W = 4 kN = 4000 N

1 )enth o handle




there4 tan lang = p

d

2

sdot 11


friction

prop 012 012

cos 983214 cos 15deg 09659


d d

2 2

tan lang + tan 1 d


= 4033 N-mm




= 1 + 2 = 4033 + 6000 = 10 033 N-mm










=

16sdot (dc )3


= = 156 Nmm2 Ans

= = 0058

prop1 = tan 1 = = = = 0124


0058 + 0124 11= 4000



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



(dc ) 1 6 sdot 10 033

sdot 103

= 511 Nmm2



= (d

32

32 M 32 sdot 12 0003 3


ma =

1 2 2 14

2 2

2 2

= 7965 MPa



16 2 16 sdot 60003 3


W

Ac


4W 4 sdot 4000

2 2

ma =1 2 1

2 2

(51)2 + (306)2 = 3983 Nmm2 = 3983 MPa


A-A Ans





n = h p = 25 2 = 125



(Given)

pb =W

dt n

4000










Let



=ore dia mm 52( 7( ()( ((

Mean dia mm 5( 7 ((( (2

=ore area mm 05) 5(5 2))5 240

Pitch mm 4 0 )

sdot ⌠ b c )3


(⌠ b ) + = (1222) + = 7965 Nmm 24 (511)

= = 306 Nmm2

(dc ) = sdot 10

⌠ c = = = = 51 Nmm 2

(dc ) sdot 10

(⌠ c )2 + 4 =

= = 926 Nmm2 Ans



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



00 sdot 103 Ac Ac

or 3



core area 1353 mm2)


Core diameter

Mean diameter

Pitch

dc = 415 mm

d = 46 mm

p = 8 mm Ans


d sdot 46



prop1 = tan 1 =prop

cos 983214

012

cos 15deg012




3 0055 + 0124




2 = 10 1 = 01 times 414 530 = 41 453 N-mm


= 1 + 2 = 414 530 + 41 453 = 455 983 N-mm = 455983 N-m


d 3 462 2


= 0

126 500

455 983

Power Screws n 655




60 60= 2865 kW Ans


2)




=

Ac = 100 sdot 10 100 = 1000 mm2



= 100 sdot 10 = 18 023 N


= = 0278 or 278 Ans










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST










= 2 p = 2 times 20 = 40 mm


d sdot 100



3

1 minus 0127 sdot 015 2= 25416 N-m


collar bearing

2 =

=

2

3

2

3

(R1 )2 minus (R2 )2 (125)3 minus (50)3

(125)2 minus (50)2 = 334 290 N-mm = 33429 N-m


at the gear wheel



=


at the pinion shaft

p = 14711 sdot 10080

= 16346 N minus m = 163 460 N-mm



there4-eiht onut

Let

163 460 = 56

16 16D3 = 163 460 11 = 14 860 or D = 246 say 25 mm Ans

h = Height of nut





3

14 = = =




$cre


= sdot = = = 2865 W

= = 0127


= 18 sdot 10 0127 + 015 100 = 254 160 N-mm



= 58845 sdot = 14711 N-m sdot 20 20








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST








4 4

⌠ c =W

Ac

40 sdot 103

1257


d = = = 45 mm2 2

p 10

d sdot 45



3


1816 times 103 =

16 16

there4 = 1816 times 103 12 568 = 1445 Nmm2


Power Screws n 657

ma =1

22

1

2(318)2 + 4 (1445)2 = 215 Nmm2

Factor o saety


=

ma

120

215



sdot⌠ 4= 2 8 G


there4

200

2

400 10

2 N

= 212 700 N

( + = dc 4 = 40 4 = 10 mm)


W

212 700

40 sdot 103




Ans


18 sdot 10 573W

(dc )2 = (40)2 = 1257 mm 2 Ac =


= = 318 Nmm 2

do + dc 50 + 40




= 40 sdot 10 3 007 + 013 45 = sdot 10 N-mm1816

sdot (d c )3 = sdot (40)3 = 12 568







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST







= =n

4 4

and



Ans


W 40 sdot 103





d 3 45 3

2 2


= 0

63 sdot 103

1816 sdot 103











(d 2


(⌠ c )2 + 4 =

= = 558



= = 53

40 sdot 10 566W


= = 102 N mm2 = 102 MPanut =


= = 0347 or 347 Ans



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



Fig 1711 Screw jack



d d




16 1

(d c )3


4



1 ⌠ c(ma ) =


Power Screws n 659

ma = 12

(⌠ c )2 + 4 2




pb =

4

W

2 2

where


where


h = n times p



($cre) =

(nut ) =

W

ndct W




W =4

2 2


know that

W = ( D2 )2 minus ( D1 ) 2 ⌠ c



6ac









2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST







2 2

R 3 + R 4




R4 = Radius of pin

1


= 1 + 2


= 300



bending moment

M =


11

12






2











Screw 8ack

Power Screws

=

⌠ c =(dc )2

⌠ c c )2 + 4 2 + (⌠ 2

(do ) minus (dc ) n


4 = 2 8 +



(R3 ) minus (R4 ) 2 (R3 )3 minus (R4 )3

2 = 3 sdot prop1 W



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



661









F lt 4 2



Table 172




d =do + dc 46 + 38

2 2

and tan lang = p

d

8

sdot 42


prop = tan = 014




3


⌠ c =W W 80 sdot 103

Ac 2 4 4


16 1 16 sdot 340 sdot 103

=


⌠ c(ma ) =

1 1 7053 + (7053)2 + 4 (3155)2

= [7053 + 9463] = 8258 Nmm 2





= 100 Nmm 2


⌠ ec 200

= = 42 mm

= = 00606


= 80 sdot 10 3 00606 + 014 42 = 340 sdot 10 N-mm

= = = 7053 Nmm 2

(dc ) (38)2

= = 3155 Nmm2

(dc )3 3 (38)

(⌠ c )2 + 4 2 =⌠ c +2 2



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



ma =2 2 1

2 2

= sdot 9463 = 47315 Nmm 2


2


2 $esin or nut






3

= =

4 4n = 1516 18 = 84 say 10 threads Ans


n



($cre) =

W

ndc t

80 sdot 103



(nut ) =

W

ndot

80 sdot 103






W = ( D1 )2 minus (do )2 ⌠ t

80 times 103 =4 2 F lt

or


(D1)2 = 2036 + 2116 = 4152 or D1 = 65 mm Ans


Power Screws n 663

W = ( D2 )2 minus ( D1 ) 2 ⌠ c

80 times 103 =2 2 90

4 2

⌠ ec(nut )

or


103 353 = 2266(D2)2 = 2266 + 4225 = 6491 or D2 = 806 say 82 mm Ans




2= 8170 t 1

e(nut )



= = 1384 Nmm2

= = 1615 Nmm2


W 80 sdot 10 1516

= 60 Nmm e 120

(7053)2 + 4 (3155)2 (⌠ c ) + 4 =1

⌠ ec 200
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST
















2 (R3 )3 minus (R4 )3 2 2

=

2

3

3

2

2 2


3

(41)2 minus (10)2




required

= 661 times 103 300 = 2203 mm





= 300 times 2250 = 675 times 103 N-mm



stress ⌠ b = ⌠ t = ⌠ et 2 = 100 Nmm2


675 times 103 =

32 32







= 400 + 80 2 = 440 mm




=F lt


82 20 3


82 20 minus





Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST




Also + = 025 dc = 025 times 38 = 95 mm

there4 W cr =

4

2

2

2


= 226 852 (1 ndash 0207) = 179 894 N





D5 = 15 D2 = 15 times 82 = 123 mm Ans




D6 = 225 D2 = 225 times 82 = 185 mm Ans


D7 = 175 D6 = 175 times 185 = 320 mm Ans

Power Screws n 665






d 3 42

2 2


= 0

101808

661 sdot 103








as 2



4= 2 8 +

440 200(38) 200 1 minus




Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



Fig 1712









(a) (b)

Fig 1713


cos = 105 minus 15

110= 08112 or = 351deg



F =W W








adopt

dc = 14 mm Ans




tan lang = p

d

6




= = 0154 or 154 Ans




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST




01123 + 020 1 minus 01123 sdot 020



=



W 1 W 1 56922 2

Power Screws n 667

⌠ t (ma ) =2

2 2 2 2(37) 2 + 4 (287) 2

= 185 + 341 = 526 Nmm2


ma =1 2 1

2 2(37)2 + 4 (287)2 = 341 Nmm2



2 $esin o nut




there4

W 1 5692 35520 =

4 4

n = 355 20 = 1776

n









= 210 + 24 + 2 times 8 = 250 mm Ans





= 250 + 2 times 15 = 280 mm Ans



15487

2 sdot 150 300

668

= 5692 = 1822 N

P = W 1 tan (lang + ) = W 1

= = 01123

)2 ⌠ =

= = 2846 N

=

= P sdot = 1822 sdot = 15 487 N-mm

= = 287 Nmm2

(dc ) (14)

= = = 37 Nmm2⌠ t = 07855 (dc ) 07855 (14)(dc )

⌠t 1 3 7 1+ ( ⌠t ) 2 + 4 = +

(⌠ t )2 + 4 =


= = 5162 mm








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST








2846 = 2 sdot 4 4

2



pins

4 $esin o lins





= F 2 = 2846 2 = 1423 N


W cr = 1423 times 5 = 7115 N






1 =1 1

12 12


+ =1

A= 225 (t 1 )4

3(t 1)2

= 0866 t 1



gt = l = 110 mm


7500


7115 =⌠ c sdot A

gt +

1 +

100 sdot 3(t 1 )2

1 110 7500 0866 t 1

300 (t 1 )2

215

(t 1 )2


71

15

300

(d1 )2 = 2 sdot (d1 ) 50 = 7855 (d1 )2

sdot t 1 1 1 1 1t ( ) 4(b )3 = sdot t (3t )3 = 225

1 + a 1 +



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



(t 1)4

(t 1)2 + 215

(t 1) 1


2 2

or


b1 = 3 t 1 = 3 times 6 = 18 mm

(Taking + ve sign)



1 =

1 1

12 12


A = t 1b1 = t 1 times 3 t 1 = 3 (t 1)2


+ =1

A

025 (t l )4

3 (t 1 )2


as fixed therefore


gt = l 2 = 110 2 = 55 mm


W cr = A

gt +

1 +

100 sdot 3(t 1 ) 2

1 55

7500 029 t 1

300 (t 1 ) 2

48

(t 1 ) 2


300 sdot 62

48

62















4 ndash 237 (t )2 ndash 51 = 0

= = 257

sdot b1 (t 1)3 = sdot 3t 1 (t 1)3 = 025 (t 1)4

= = 029 t 1

1 + a 1 +

= 9532 NW cr =1 +

Di7er



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



Fig 1714













(i )

(ii )




2

andd 2

2


0 = 1E ndash 2E


2 2

Power Screws n 671


W

2 ( p1 minus p2 )

p1 p2


= 0









2 = W tan (lang 2 2 ) 2 = W 2






Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



Fig 1715


(i )

2 = W tan (lang 2 + 2 )d 2

2= W




= P1 times l = 1 + 2




d 1

2

d 2

2



0 = 1E + 2E

d2 d2

2 2

sdot + sdot 2

d1

2 d2


0

=






p1 = 16 mm p2 = 12 mm ma = 28 MPa = 28 Nmm2

W

2 ( p1 + p2 )








d1

16

sdot 42

and tan lang2 = p2

d2

12

sdot 44



= W =

p1 d1 p2 d


1 = W tan (lang1 1 ) 1 = W 1




d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



d tan lang1 + tan d


01212 + 015 42= W = 58 W N-mm


2

=

W

t

a

n

(

lang

2

minus

)


00868 minus 015 44= W


= minus 137 W N-mm



0 =

W W

2 2 (16 minus 12) = 0636 W N-mm


Power Screws n 673

= 0

0636 W

717 W






= W = p1 d 1 p2 d

= = 01212

= = 00868

1 = W tan (lang1 + ) 1 = W 1

1 minus 01212 sdot 015 2


1 + 00868 sdot 015 2

( p1 2 ) =minus p





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST





W W W W

Ac1 2 908

4 4

Nmm 23 3


28 =1 2 1

2 2

2

908 1331

2

=1 1


there4 W =28 sdot 2

1863 sdot 10minus3

= 30 060 N = 3006 kN Ans

E$EampSES























Ans 45 N 227+

Ans 12 N





















Ans 45 N 346 N 17 a 14 N0mm2

= = 00887 or 887 Ans

= = =⌠ c = Nmm 2

(dc1 ) (34)2


(dc1 ) (34) 1331

(⌠ c )2 + 4 = W W + 4























Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST






















Ans 26 N-m 266 N0mm2 1 N0mm2 4 mm







Power Screws n 675














14

15












screw as 6 mm


mm



Pitch mm 3 3 3 3



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



t 1 7 mm 1 21 mm

ESSNS







of screw






depend Explain






Eamp8E 9E ESSNS










(a)

(c)

tan (lang minus )

tan lang

tan (lang + )

tan lang

(b)

(d)

tan lang

tan (lang minus )

tan lang

tan (lang + )



(a)

(c)


(b)

(d)












taken as








ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST




ng the working load



ANSES

1 (b)

6 (b)

2 (b)

7 (a)

3 (b)

(b)

4 (d)

(d)

5 (a)

1 (d)



GO To FIRST



GO To FIRST

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Materi Khurmi Power Screw