Solution Manual for Design of Machine Elements – V.B. Bhandari – Complete Step-by-Step Solutions

Solutions3Manual3to3 DESI
GN3 OF3 MACHINE3 ELEMENTS
(First 3Revised 3Edition)




V.3B.3Bhandari3Fo
rmerly
Professor 3 &3 Head3of3 Mechanical3Engineering 3Vishw
akarma3 Institute3 of3 Technology,3 Pune.




McGraw-
Hill3Education 3(India)3Limited3 New3 Delhi

, 1
CHAPTER3 1

1.1 The3 series3 factor3 for3 R103 series3 is3 given3 by,

3103=1.2589
10



First3 number3 =3 1

Second3 number3 =3 13(1.2589)3=3 1.25893 =3 (1.25)

Third3number3=3(1.2589)(31.2589)3=3 (1.25893)323 =31.58483=3(1.6)

Fourth3 number3=3 (1.2589)2(1.2589)3=3 (1.25893) =31.99513 =3(2)
3




Fifth3 number3 =3 (1.2589)3(1.2589)3=3 (1.25893)343 =3 2.51173=3 (2.5)

Sixth3 number3 =3 (1.2589)4(1.2589)3=3 (1.25893)53 =3 3.16203 =3 (3.16)

Seventh3 number3=3 (1.2589)5(1.2589)3 =3 (1.25893)363 =33.98063 =3 (4)

Eighth3number3 =3(1.2589)6(1.2589)3=3 (1.25893)373 =35.01123 =3(5)

Ninth3 number3 =3 (1.2589)7(1.2589)3=3 (1.25893) =36.30863 =3(6.3)
8




Tenth3 number3 =3 (1.2589)8(1.2589)3 =3 (1.25893)93 =3 7.94183 =3 (8)

Eleventh3 number3=3 (1.2589)9(1.2589)3=3 (1.25893)1 =39.99803 =3(10)
0



In3 above3 calculations,3 the3 rounded3 numbers3 are3 shown3 in3 bracket.

1.2

The3 series3 factor3 for3 R203 series3 is3 given3 by,

3103=1.122
20



Since3every3third3term3of3 R203 series3 is3 selected,3the3 ratio3 factor3 (33)3is3 given3 by,
3 =3 (1.122)33 =31.4125

First3 number3=3200
Second3 number3 =3 2003(1.4125)3=3 282.53 =3 (280)

, 2
Third3 number3=3 200(1.4125)(3 1.4125)3 =3 2003(1.4125)23 =3 399.033 =3(400)

Fourth3 number3=3 200(1.4125)2(3 1.4125)3 =3 2003(1.4125)33 =3 563.633=3 (560)

Fifth3 number3 =3 200(1.4125)3(3 1.4125)3=3 2003(1.4125)43 =3 796.133 =3 (800)
Sixth3 number3 =3 200(1.4125)4(3 1.4125)3=3 2003(1.4125)53 =3 1124.533 =3(1120)
In3 above3 calculations,3 the3 rounded3 numbers3 are3 shown3 in3 bracket.3 The3 complete3 series3 is3

given3by,

200,3 280(282.5),3 400(399.03),3 560(563.63),3 800(796.13),3 1120(1124.53),3 …

1.3

Let3 us3 denote3 the3 ratio3 factor3 as3 (33).3 The3 derived3 series3 is3 based3 on3 geometric3 progression.

3The3power3rating3of3seven3models3will3as3follows,

(1)3403 ()03, (2)3 403 ()13, (3)3 403 ()23, 4)3403 ()33,

(5)3403()43, (6)3403()53, (7)3403()6

The3 maximum3 load3capacity3 is3 6303 kN.3 Therefore,
1/36
3 6303
403 ()63=3 630 or 3 =   =31.5832
 403 

Load3capacity3 of3 first3 model3 =3 (40)3 kN
Load3capacity3of3 second3 model3 =3403(1.5832)3=363.333=3(63)3kN

Load3 capacity3 of3 third3 model3=3 403(1.5832)23=3 100.263 =3 (100)3 kN

Load3 capacity3 of3 fourth3model3 =3 403(1.5832)33=3 158.733=3 (160)3 kN

Load3 capacity3 of3 fifth3 model3=3 403(1.5832)43=3 251.313 =3 (250)3 kN

Load3 capacity3 of3 sixth3 model3=3 403(1.5832)53=3 397.873=3 (400)3 kN

Load3 capacity3 of3 seventh3 model3=3 403(1.5832)63=3 629.903=3 (630)3 kN

1.4

Let3 us3 denote3 the3 ratio3 factor3 as3 (33).3 The3 derived3 series3 is3 based3 on3 geometric3 progression.

3The3speeds3of3different3steps3will3as3follows,

, 3
(1)3723 ()03, (2)3 723 ()13, (3)3 72()23, (4)3723 ()33,

(5)3723()43, (6)3723()53, (7)3723()6 (8)3 723()7

(9) 723()8 (10) 723()9 (11) 723()10

The3 maximum3 speed3 is3 7203 r.p.m.3 Therefore,
1 3/310
3 7203
723 ()103=3 720 or 3 =   =3 (10)1/103 =103103 =31.2589
 723 

Speed3 of3 first3 step3 =3723r.p.m.
Speed3of3 second3 step3=3723(1.2589)3=3 90.643 =3(91)3 r.p.m.

Speed3 of3 third3 step3=3 723(1.2589)23=3 114.113 =3 (114)3 r.p.m.

Speed3 of3 fourth3 step3 =3 723(1.2589)33=3 143.653=3 (144)3 r.p.m.

Speed3 of3 fifth3 step3=3 723(1.2589)43=3 180.843 =3 (181)3 r.p.m.

Speed3 of3 sixth3 step3 =3 723(1.2589)53=3 227.663=3 (228)3 r.p.m.

Speed3 of3 seventh3 step3 =3 723(1.2589)63=3 286.603=3 (287)3 r.p.m.

Speed3of3 eighth3 step3 =3 723(1.2589)73=3 360.803=3 (361)3 r.p.m.
Speed3of3 ninth3 step3=3 723(1.2589)83=3 454.223=3 (454)3 r.p.m.
Speed3 of3tenth3 step3 =3723(1.2589)93=3 571.813 =3 (572)3 r.p.m.

Speed3of3eleventh3 step3 =3 723(1.2589)103=3 719.853=3 (720)3 r.p.m.