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Capter tr

STRESSES IN MACHINE PARTS

Loatl :

It is defined as any exterual force acting upolt a rnachin part.

The following are three types of loads

tr- Dead or steady load. I It does not change in magnitude and direction]

2^ \-we or varying load, [continually changes , and

3- Suddenly applied or shock load

Stress :

When some extemal syetem of forces or loads appiy cir a body, the

intenial forces (equal and opposite) are set irp at various sections of the

body which resist the extemal forces. This intemal force per unit area zt

any section of,the body is known as unit stress or sirnply a stre;s

P6--

A

".,vhr-:l r:

o : unil. stress (stress)

F: Force or noad acting on the body

A: Cross - sectional area of the body

Un!!s ;

o N/mrn2 or N/ln2

- 1-

_t

A

Strain

Newton N

tnm2

When a system of force

defonnation. This defonnatioil

srniply a strain

t-{).

a:strain E( :changeinlength

f : Original leugth of the bocly .

Tensile stress frrp,d strui*t

or load act on a body, it undergi:e$ solne

per unit length is k*ow-n as tmit st"'aitn or

E(. = t. {.6r

P [---l ' P[- __-L* fi-- I P*e-*f- F*'*.s-I! [= i- l*-*L I i F;-l-_*--__l

Tensile stress 0r

Fig. (1)

Wiren a body is subjected to two equal and opposite axial pull p

(also calied tensile load) as shown in Fig 1; then ttrre stless incltrced at any

section of the body is trmown as tensile stress. dite to the tensjle load. there

rvill be an increase in lengtir of the body. The ratio of the icr*ase in lengtli

to the original length is lalown as teneile straiil

P Axial tensile force N

A Cross sectional area mm2

6t Increase in length't-"-=-*::

I enstle strarll " (. originatr length

3-

when a body is subjected to two equal and opposite forces, acting

tangentially across the resisting section, as a result of which the body teldsto shear offthe section, then the stress induced is called shear stress. The

colrespollding strain is latown as shear strain and it is measured by the

angular defonnation accompanying the shear stress

t = shear stress : tangential force

Resisting Area

considering a body consisting of two plates comeceted by a rivet as

shown in Fig 2. In this case the tangential force P tends to shear offtherivet. Therefore, the shear stress on the rivet cross-section is

PPL_

A Ld,4

d: diameter of the rivet (A:cross sectional area of the rivet)

when the tangential force has to be resisted by one section of the body,

then the body is said to be in single shear. lf the tangential force has to be

resisted by two sections of a body, it is said to be in double shear.

In case of tension or compresscion, the area involved is at right angle to

the extemal force, wheares in case or'shear, the area involved in parallel to

the extemal force.

Younq's modulus or modulus of elasticifv

Hook's law states that when a matenial is loaded within elastic lirnit, the

stress is proportional to strain,

t_t

o- e EP

E is a constant of proportionality known as young's modttlus of elesticity.

It is usually expressed in kg/crn2 - N/rnrn2r$'

shear modulus or modulus of rigiditv

The shear stress is proportional to shear straitt within the elastic lirnit

T shear stress\J

-:A shear strain

known as shear tnodr"rlus or modtrlus of

rigidity ?

Table (1) rnodulus of elasticity and rigidity

Bearing stress

A localised compressive stress at the area of contact between two

mernbers is lanowl as bearing stress or cruslting stress. The beariug stress

Material E N/ mm' GN/ mrnz

Steel

Wrought Iron

Cast Iont

Copper

Brass

Tirnber

2 2.2X rcs

1.9 --- 2.0 x 105

1.0 --- 1.6 X 105

0.9 --- 1.1 X105

0.8 --- 0.9 x 105

0.1 x 105

0.8--1.0x105

0.8 --- 0.9 x 105.

0.4 --- 0.5 x l0t

0.3 --- 0.5 x 105

0.3 --- 0.5 x105

0.1 x105

5-

is taken into acconnt iu design of riveted joints, cottorjoilts. l<nuckle

joints, etc.

Consider a pin and an eye loaded as shown in Fig (3) a

Fig (3) cblThe distribution of bearing stress will not be , but it will be

accordi[g to the shape of the surfaces in contact and the physical

properties of the two tnaterials. The distribution of stress will be sirnilar to

that shown in Fig (b)

since the actual distribution of bearing stress is difficuilt to detennine,

therefore, the bearing stress is usually calculated by dividing the ioad to

tlre projected bearing area of the pin

let

P: load acting on the pin

L : Length of the pin in contact

d: diameter of the pin

Bearing or crushing stress.

o r' = orr"u,ing =

Iu case of riveted joints subjected to

stress.

Lxda load P , tire bearing or cnrshing

Po^:" d.t.n

d: diameter of the rive

t : thicloese of the plate '

tu : number of rivets in crushing

Exarnple (1) : A pull of 80 KN is transrnitted frorn a barXto the bar Y

through a pin as shown in Fig (a). If the maximun pennissible tensile

stress in the bars is 100 NArun2 and the maximus shear stress in the pin is

80 NAnrn2 . Find the diarneter of bars and of the pin

Solution

Given tensile load P : 80 KN

P:80 x 103 N

tenseile stress :100 N/rnrn2

Diameter of the bars

Let Du: Diarneter of the bars'

. .A1,: Aro? of the bars: *",1' _80x103 =100relation o, =fr o .il

4

using the

shear stress

= 3Ztntn

Dlarneler-qf-tbe-@

P 80x 103L_

shear stress

2Ao z*L o?ar4

:g0x 80x103 -z*L o?

4U

7

go KM+l

= 25.23mm#toaot1g 2c^........'_aa':Fl)zd,P.tom

1,, rco c^ _ _-l

Example (2) : A rod 100 crn long and 2crn x 2cm cross section is

subjected to a pull of 1000kg . If the modulus of elasticity of tiie rnaterial

is 2x106 kg / cmz detennine the elongation of the rod.

Solution

length of the rod L : 100cm

cross sectional area of the rod A:2x2: 4cifpull P: l000kg,

Modulus of Elacticity E : 2x106 kg I cnz

6L Elongation of the rod

o P.(E:

.'. 6( =

-=-€ Ax 6(,

P.( 1000x 100= 0.0125 cmA.E 4x2x106

P

pxAg.l&-€)- Two plates 16 rnrn thick are joined by a doulbe riveted lap

joint as shomn in Fig (5) . The rivets ar 2.5 cm diameter. Finci the cr:irsliing

stress induced between the plates and the rivet , if the maxirnun tensile

lood on the joint is 4800kg

sloution

Thickness of the plates t: 16 run: 1.6 cm

Diarneter of the.rivets d:25lmrr : 2.5cm

Maximun teusile load P :4800 kg

(rcr: Crushing stress induced between the plates and the rivets

8-

4800 * 6A0kg I cm2d.t.n 2.5x1.6x2

Example (4) : Two steel plates 10 crn wide and L25 cn thick Fig (5) are

to be joined by double transverse fillet weld. The maximum tensile stress

is not to exceed 700 L<glcn2 .

Find the length of the weld for static and dynamic loading

slo-lrtion

b: 10 cm

ot: 700 kg / cm2

Maxirnum lood which the plate cau carry P

P:Areaxstress:txbxot

p : (1 .25 x 10) x 700 : 8750 kg

Length of the weld for static loadirrg: I

t : size of the weld : Thickress of plate : 1.25 ctn

Tensile strength of the joint for double fillet : ot

Jip

,t:1.25cm

8750

Zto o, Ji x1.25 x 700

Por= rr$

...1 - = 7.07Cnr

Length of the weld for dynarnic loading

- stress conceutration factor for dyrarnic loading aud lbr

transverse fillet welds is 1.5

700o.'--' 1.5

o,=-!, '.t=--- n9- =10.6 cnr-a., J2x1.25x465Ll'x'.'

-

& fsl'25 &n ff

------1,--__c; A *---{--*.."**'rii :.. .'-...''.*fi-:-..**-*-'i' "*--- *;:" *,::.----L

-- W

[r ol0 C*E

H-')fr

'*-i:.:-,

Exarnple (5)

- A20 run steel is theaded through a brass sleeve 100 mm long,

24 rntnbore and 32 run outside diarneter. A nut and washer put on and

the nr.rt tightened up until the brass sleeve has shortened by 0.05 mrn.

Calculate the extension in the bolt

where E for steel:200000 N/mrn2

E for brass : 80000 N/run2

M 0.05= 0.0005strain ilr sleeve s = L i00

SITESS Otr-----D- slrain €

Ebott:

:. o: E.€ s brassSleeue

stress: 80000 X 0.0005 :40 N/rrun2

stress on sleel/e :40 N/ mrn2

cross sectional area of sleeve : f,{tr' -24')

A: 352 rnm2

The cornpressiotl load carried by sleeve L: o . A

loadL:40X352=14080N

This will be equal to the tension in the bolt

llence stress u: the bolt : o bolt - L - 14089

A Loo\'4'

o bort:44.8N/rnrn2

-obot, -44'8 =2ooooo€ bot, € bo,-strain

sl/'ess

.'. strain€uo,, =44.8

10-

At_ , * 44.8

L 'boh 2ooooo

Extension of bolt

Exarnple (6)

A 2a rnrn bolt 160 rnrn long carries a laod of 20 kN . Calculaie the

extenesiou in the bolt if E :200000 N/uun2 .

. 44.8 100x448.". AL = Lx.--= =0.0224mnt200000 200000

:0.0224 rnrn

Ioucl 20000

Ar.ctr ,, (20\,4'

P:200000 Nd:20 rnrn

63.7 N lmnr

SolLrtion

stress rn th bolt :

.Extensiott llxtstrarn :

Original length 160

Young's rnodulus P- ,//'e's'r' -,ttt'afit a

63.7 Ext.'..\lroilt t:

-=-

2 x l0' i60. 160 x63 7Extensron = 0051 ntnt

2 x l0-'

63.7=2x105

rr; cti on .il'ucFli

.I?ssio n

Luad in4-

hG*pru

oF

TlilII

FIovl

t-Ie*si

F

A"siut S\"ea'r

- 1l -

Exarnple (7):

A load F of 5 KN is applied to the tensile member shown in the

Figure and is carried at the joint by a single rivet. The angle of the joint is

30o to the axis of the load.

Calculate the tensile and shear stresses it a20 mrn diarneter rivet .

Solution

The axis of the rivet is at 30" to the line of action of the load P. Dl=5Ku

Diarneter of the rivet d:20 rnrn f>/sA rea of tlre rivet A, = tU' = [tzo)' = 314.2 mmz \

/p)f Srrl_BOo

8nDirect pull on the rivet: coutpouetlt of P alolg axis of rivet .

Pr : 5 cos 30o : 5X 0.866 :4.33 KN

Tlrerefore direct stress or: *= n#W= 13.8 I(N /m2

ot: 13.8 KN/rn2

Shear force on rivet equals cornponent of P transverse to rivet .

i.e. along joint face: P,

shear force P, : P sin 30 : 5 X sin 30 = 2'5 KN

Tlrrrs sltearstressott rivet S,: + = -- ?5--r=7.95 KN/m2A, 314.2 x 10-o

11 this case, where the stress is predorninantly tensile rather than shear,

the rivet would be replased by a bolt.

- t2-

Objeets tirat are assembled must be held togetlier with sorne type of

fa.steirer or by a fastening procedure' There are two rnajor classifications of

fasteners,pennanenta,rrdternporary.Pennanentfasterrersareusedwhen

parts will not be disassernbled. Temporary fasteners are used when the

parts will be disassernbled at some ftittre time'

Perrnanent fastening method inclufu:

* weldirrg t Brg,rty, 5 bayli"S , Nuil ;n2, R;r./-;"o(}

and Gluig.-

TzAuW,! gru fdsJ,A !Pe$ ry c I ttfui- 't lr^.,- ^'^ -l A'

'iSci'ews ,- t)alt-; , i-"J= tit'{r' Ir/rD,* tlA I f- _.; ; a'-c; c \*t 'v I I rt >1o,*ror;tr-ljl ll"! uleJl uUl c,li 'J:Cl JJI \:'li cJtrljllj,^.,

Jw. -,i'vr-L'*

;,rt t{'t,M ri -ffi1 *fi r5lr -,'t''tt M Ls'- ):J

M60 x:r ,rot' , i*.i Jii il' .u"'' r"' (tst) '*' q"- ').j.Rr" *Jl!;:rC -")' P'' R Fl.,-'! ir;r Jll

r.48 x 3 {rnrn} i.5L}l '-1--i;-.rir=;lr 'ut T-Jt- !j J,J

s)o x r0 (,rnr) ;,]Jt , i..; -.t-11 ;'lr ,r-;rr s (Burir€sst (.,t-';"tt *-) t;5 -l{I . ,-1, -<!r -l--.ll

";* Jl

- ^'1:q--ildtilJl x (mmt f/ :5'il /';Jr na ' -

-'-* -+l '

: C",s:; :,:'r'ii I _jlj i"ij"" _--- [i i -------rTl_:--

I I E;"ttt tScrews l)tluts - - -i

i ..o,ai'lr:

i" - '-li,!

Tolerance class

clnss of fit-rnetria: system

J1,Jl ,JiJI(;r!l)

9 Lorl

u"UlL

+ljjl(;Lrr)l)

Ltiirtil-}- "11

m

U u"*cdl5

,bt

alrlle

p)lSLILI

,wl,EIJI

;it iNWl

M3 2,35 2 2,1 2,4 5,5 6,1 7 3,2 0,5

MI, 3,09 1!?(

3,2 3,2 7 8,1 9 4,i- !:9-

I

t,5M5 J,96 I I ] 0,4 t! 5,3

v!M8

!:7- -6,3-7_

8,05

!,255

7_

I

1,8 5 t0 I t,5 6,1

6,5. *1,19,9

6t5---

9,5

r1_. !?_

,9

t 6,2 t/ 8,1 2

2,5

-l--l

3 --

1

Mto 19,6 | 0,-5_

t3-tsMt2 9,72 21.9

M11 I t,l !_t 0,5

I 1,5 llt3

))

,;-25,1 2B

Mt6

iiat3lt1:75

t 6,75

t 3,5 t7

i:l5

l7 t6

27 31,2 31 t9

M20 30 31,6 .?6 4

st:rndard metric f:rstetter callout \3

P

THREADS

l* _P._HEXAGON BOLTS , & SCRE\

r-r- -

5:s--lla(-i---'rl-T;r--l--o:c+-ci6-tEtr.t

EH;

l--t-'t-lHIt-i-l-t._

IoltGrnotiv. ticodsI

]g!--o"o"'

Fig 203 ISO METRIC SCREW THREAD FORMMETRIC THREAD IS SIMILAR TO THE AMERICAN STANDARD

Fig2M

NUT BOLT

EXTERNAL THREAD

Nominalsize

Pitch 0l'thrcad

Z'r-t"TI,ir.:

[)iurrc(er o[ Irrrrlhrerdcd i \Vidtlr across

sharrk I ltats

I

I )iarrretrr ol I I lril\\arhcr l.rcr I llear

..t_

-t. is.o7 I ,

().5.1 I :

7..s1 I .

_._-L___l :.irJ I r

D B

M3 rl.5 10..1ioJ llL(l

3.(x) 5.5()

7]lil.-8L0 -

lu00--L:t1r0-

M4 4.00

M5 0.tt ).50 5.00

M6 I.o 0.75 6.00M8 r .15 t.0 ti.(x)Ml0 !s r t5 10.(x) I7.00 t6.4r{ I tMr2 t.75 I al l?.00 19.00 trJ,l? I rM16 2.1) r 6.00 I 22.00 tt.rx I ruM20 2.5 5 20.00 24.00 :,r. rti---l-lM24 3.0 2{r 24.00 36.00 | 5.ur)Ml0 3.5 2.0 30.00 46.00 44.95 I tqM16 4.0 36.00 55.0() :rt*_[,t-

DEPTH OF THRhAD

Unified National Thread

SINGLETHREAD

--l f- errcu

_-[

DOUBLE

.THREAD

TRIPLE..'TIIREAD

Single and Multriple Threadst)

L4

rx)

xri)l)

l)( )

5()

oiirr.:.{,(,

0t)

00

INTE RNAL THTEAO

Fig2M

oLV .rlr.:4. -.,G"+

(rr,t zo .;- -*y1) a l,;jt}l-l ,rL :JrL -rt"

(r,,r ro ;> a&'r))*fJ! :l,Jl onl, ,- J..,

5.8 j 4.8 t"Jtill L,/lJ.- k1ijM10x40DrNs63-s.A rJic t,,L;ru -,rrr-, ?)1 JU

('r kt D? ('n, ('0,

1,6 3 3,9 11 10 6,4 4.8 M 62 4 5 . 14,5 r 't 3 8,4 6,5 M 82,5 5 6 , 18 16 tO,S 8,2 M103 6 22 13 9,8 M124 8 29 17 13,5 M165 10 36 21 16,9 M 20

50...855...1060... I 280... 20

1 00.. .251 00.. . 30

0,3 1,6o,4 2,1o,5 2,60,5 30,5 40.5 5

21d

3

4,5

100...35120...401 60... 50200. . . 60240...70300. ..80

lorxosl ;L;$ t^iL (,]i,) rU *;. ̂ - dr (t; {DINrrl ;Li-[JJ U.t -4lJ JjJl Lli ]i - dr (,

100... ?0 l EoJ 55..,25 120 I 16 I l2J to1 IrJlrl)'f.t!l 6.rrJl (t

('r ,:t t^,)l c) il#l --,/t

L "l- 'l*300...220 J 200...90 J 80 J 75...40 3 35 : Jbt)l 6js (r 10e 3 88 j 5.6 LJGJI a"\i

SW,

M16x60 DIN93B-t6 ,r" !-lE tL-" 7; , Jr:"DIN938, br:d ':)9-Jl J' eJa 3 J-)l *:cDIN 939. br- 1,25 cJ : 5:1"; )t ):.u 3 L )l ,*DrN83s, bt-2d ,!y-* th-ri 4 :-St ;,

9l+-l JU*LJU.ll o,l3,

26 2,2 3,5 M 1030 2,5 4 M1238 3 5 M1646 3,5 6 M2054 4,5 7 M2466 5 8 M30

8.8, 10.9

81 swr t kr

120...30120.,.3515C...45180...50250. . . 60

9,4 I11,7 101 6,3 1419.8 1722,1 1S

2630364654

6 10 16 M107 12 18 M129 16 24 M16

'r 'l 20 30 M20'l 3.5 24 36 M 24

+ffi(DrN 2440) St SS L-J9L ;r;*L

,-ljJl .:)-e

GTw40iJ:t-t(tR

dl(I dr {'NW

ar

d, (rhread) J!Vt"vi'1"

18,6324,1230,29

20,9626,4433,25

't 621 ,6)a ')

Rh,,R%"R1"

18188402822 20 10 45 3430 25 12 so 4A

"1719

. (otr'r zo+o ""-LJl, i\(ttr ^*;- l;l ) Jl rl3

\" t - L

. ,-l (.rtr") fu-.;t = ruw' ( \'

dJ/ll r*lr.l ,r\J;) tmm) ;rJ-LJl F = a, (r

l.!l = d . :,st-.t.s J-9.I1 :-,rL = thread .Jyl ( t-Li*lell fJt ,l-;-

)i = a,. JIX €rLLt

L5

(.^.r) ;)t

Whitworth Screw for pipes Table l0 ISO Metric screw threads-fine and coars€ series

S.ridr d,i gtrded pitches'Notind

tiil(uar)

'li,Pdnltl

1.61.822.22.5

33.544.55

5.567rt9

IOltt2l4l515l7l82022

24252627?a

303?333536

3B39404245

4850525556

5860626465

68

fJ.50,r'-)()

I0.00,rjo

i 4.o0

t5 5017 5019.50

21.00

-24.00

za.Eo

29.50

32'o0

35.00

:,i oo40.50

43.00

47.0O

uiso

54.50

'+62.0O

I

1.2't

t.25t.5

1.5

1.51.51.5

2

2

2

3

3

t3

3

3

4-

t:'jo8.75

10.50rTo

14.50

16.5018.5020.50

22.@+r

2S.oO:28.00

3r.oo

33.o0

36.OO

riocl42.@

45.00

49.OO

52.o0

56.OO

6ooo

t.25I.45r.60r.75?952.502.903.303154.20

5 O'J

tt.u)b15775

. Tolerance ClassDesignation I----]lM5 x .5-5h 6h

lrrronoour'o*^r,or-f T T Tl TTNoMrNALstz€J I Ii llp,rcx+, liil?rr!r D_r4!,E_rrE roLEnANcE

{ ;:l:lll::F;$?,rrl-_j I I

c,,L,s, o,AMErE,, roLE HANcE { ISi: lliil8E f8i?,t"r.,

_--j J/ WHITWORTH

For example :

Screwcallout'R +- 25.4P= Z mm(Pitch)

r = 0.137. P

H = 0.960. P

THREAD I

Standard rletltc lastener callout

()nBe

0.350.350.40.450.45

0.50.6o.7o.75o.8

2

2.52.52.5

3

;3.5

4-

4

oJ4.5

5

.;

.;a6

;!.1+i rs;rJl +

z

D)a->t

-P

Uil, Urlrl irrUtNominaldiameter

Inner diam)inch

dr

,fi;t ,-;:t

l,t. t rd

?8'19

I914

l414

14

1l1tllll

11

It1l1t11

l1'tr

lltlll1l

0,91

r,341.34

1.8 1

1,81

r,8lt,8 t2,31

2,31

2,31

2.31

2.31

2,31

2,31

2,31

2,31

2,31

2.31

2,31

2.31

.?,312.31

8,5 7

1 t,4514,9518.6320,5924,122t,a830,2938,9544,85

50.79

50,DO

62,76

72,23

78,58

84,939l ,0397,a1

I 03,73l ro,o8122,78135.48

9,73

13,16

16,66

20.96

.21,s-r_;26,4430,20 .

33.2541.9t

.47,81 .53,7s

59.62

65.71

75.1 I81,5487.89..93,9_8

r00,33.106,68r 13,03

125,74'138,44

-RtV*#i*-Rt)s

i'fr'*_pisru-R2-&2 WR 2.'/1

-R 23/4

t-'rrt!ffa: fr.a04.R.5

27

_LO

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i : -1'

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ir .l . i i,,li

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r':.,..-nr:.r!! _:,rl:rt'i''.U )l(r, !ii,n,-

I i.lii CF ai-,Ll- -,; ll il i,lrj

(.)'/f!i lli r\i

F!+, ii:-?l $iechiti: ;ireirur'

i .1!1ii ', !..: !i : rr i

-/'r-l: iri.r.-l.-- -l -'-.

r :.-: r- .l ...": i

i:Lr\1.!lr:i\ir

':i i:[\'1:ra

ii.:.1,,---- : ---,,-- j'

l

liil,l_

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I - -., i L. ,!___-- -iI i l, i:

;;-'i"r;-ii;i:lrriiriiii;,ill,,r',llrl

L]<:I=rr "r-+--o.l,1liiril ilrta:d - 0iit:i ilr.lliuc ilHit iiil t:ii r.litiiI

:: ri

lt , :l._;" -j:;l hijll

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t..,rjl,lar, rl '.J

!

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: :-':';'].1.-; i -.'l ,

:i

Power Screws

power screws are usedfor :-

- transrnittitg rnotion (rnotion devices)

- linear actuators that transfonn rotary motion into linear motiotl

Typical applications for power screws ore :

1- automobile jacks

2- lead screw for lathes

3- screw type Presses

4- C clamps'

5- Valve stems

I HREADS

RIGHT TIAND

souaRd THREAD

SINGUE START

BE_SA"UA

Id

o

qulIG

o -rr+tro'lFo6=O

2 START R H Sq,UARE THREAD

- thread helix angle

- coefficient of sliding friction

r1 ball screw 90 % and higher'

POWER SCREW THREAD FORIVIS

The tliread forms used for power screws are showt

1- Acme tluead.

2- Stub Acme screw threads.

3- 60" - stub Acme threads.

4- rnodified square tlueads.

5- buttress threads.

Sonrc Defintions

Fitch P

The axial distance along

an element of the Pitch cYlinder

rneasured from one tluead to an adjacent thread'

Lead I-

The axial distace a nut

will dvance for one revolution of the screw

For a srew with a single thread,

the iead is eqriai to tlte pitch L - P

For a multiPle tlueaded screw

theleadisequa.ltotlteproductofthenrunberof'

threads n and the Pitch (P) L : n P

P

-T

souARE t! *'

T rr,

) -r-, -- I

C- .ai) BUrrHEss

, or-!'5 I ,

i-_-- PrTCH -*1

I -^. r

I \--" --l i

v9

o -s-r)l /.r__

L n.ptanc/ =-trdm ndm

L: lead

n: nurnber of tlueads

p: pitch

drr : mean screw diarneter

r*"* q__1

+<t'-

'- _-r=-j

Toroue Eouotion For Powr Screwes

raised or lowered by

rotating the nut which is

supported by a tlrust collar

camot rotate and lnove

freely up or down without

any frictional resistance

from the waltr

- Holes for wrench

- Thrust strrf-ace or

collar

- Tlrust collar

Screw

r.i : collar inner raditrs

r*c : lnean collor raduis :

=lor.o: collar outer raduis p= ?;bcl+

Model of a power suew used as screw iuck /l ,

Tdn ,rT^ , I

0 : thread angle L = Leal = n, p zr d^Fn : resultant force, nonnal to the tluead surface

tt. = rzut*ler of flrds

Figure 1 5-1 0 Reactive forcc diagram on a thread due to rarsinu I load ,i

a,t i Poran

2A

J^ , nLect'rl strc'D o/"o"'ohn

Exarnple I

An acrne thread automobile screw jack is rnade of cold rolled 1045

steetr haviilg diarneter of 30 rnrn (i%"). The tliread is outforgeneral

purpose application , but the worktnanship is of poor quality ' when used

to raise au autotnobile the base of the screw is supported by a steel collar

rravi,g a mean dimeter of 39 mm (1 y;'). The nut that does the lifti,g is

made of the salne material as the screw'

To rais a22 70kg car to change a tire; What wolld be the reqlired torque

and the screw efficrencY '

2210[Load one each tire ; = 567'5 kg ]

Solrttiort

w 2274w:221Akg 4 =

2

For a 30 rnrn diarneter screw we obtaitt frorn the table of basic

dirnensio[s of AGME GeneralPurpose thread series'

Nominal Size (1 %") (30 rrun)

Threads per inch 5 P

Basic height of tluead 0'1000"

Basic Major diarneter 1 ' 125" '

Helix angle at basic pitch diarneter ct : 3o 33'

- Frorn the table : the coefficient of friction for a dry screw and nut made

of steel and having poor worktnauship is fs : fc : 0'0'25'

- The coefficient of friction for starting is

fs' : fc' : 1.33 x 0.25 :0'33

2274w/tire : 7,d,,," = 1.5"= 38 mm

2L

for an Actne tluead 0 : 14" 30' , a: 3o 33'

tal1 en: Cos tl, tan 0

tan 0n: Cos 3" 33' . tan 14" 30'

tan 0n : (0.9981) (0.2586): 0.2581'

T'lrerefore 0,,: 14o 24.43'

Knowing the basic thread heiglrt h:0'1000": (2.54 nun)

Drnajor : The basic rnajor diarneter: 30 rnm(lt/'")

dm : D,rrajo. '2 xh/Z

:30 - 2.5

:27 .5 lnlrtn

drn :2.7 5 cm

The starting re'ittired torque with fs ': fc' : 0'33

d .W I f' + Co"; 0 ,, tart a1 , d,,," f " 'W

,t, _ ilt 1..) .. l-|.. .tR- 2 lcoso,,-/',tana) 2

'[n:652 kg'Cnt

The runnilg torqtte requirecl to raise the car is based upon the

coefflicie[t of frictio[ fs:0.25 therefore , ttsiug the above calculatio[ with

the only charrge being the friction coeffcient

we find that The rttnniug torque Tn :511 kg'crn

It is required a29.3%higher torque to start the screw in motion than

it does to keep the screw in motion '

If we ass*me a reasonable crank arm len$h 457 mrn, the operator is

required to exert a starting force of Tn :662: Ru' F

: 45.7 xF

.r-tR -

't)

662F:==14.5k945.7

the f,orce is rather large and the screw jack will have to be lubricated

or a longer crank ann . The srew efficency is deterrnined as follows

d,,,.tafiG.ry=

,, I .f , :-c-o-' 9-, t* z1 *,J,,," 1 "" ''lrru: - .f, tane )

2.75xtan3"33'ry=

9?I:929qs(0{9ry10.968s - 0.25(0.0620)

=8.97 o/o

This screw efficiency is relatively poor. It can be rnarkedly irnproved

by reducing the friction coefficient , increasing the helix angle , or by

rnaking both clianges.

Exarmple (9)

A screw - operated arbor press has a sqllare - tlueaded screw 50 rnm

dia 5mm pitch single start. If the coeflicient of friction at the threads is

0.10, what load rnay be applied by the press when an effort of 200 N is

applied at the end of a handle 200 rnrn long attached to the screw.

+ 3.8 x 0.25

Solutlon

Coeffrcient of friction : 0.l0

Friction angle 0:5'431

Since tan $: f :0.10

do:50mln, P: 5mrn

dtn: do-2xWZ:

h=a-drn: 50 -512:47.5 run

Helix angle o,

""" tan o: 'ftd,,

Jo

tan cr

lead

tan cr

mean circumference of the thread

: pitch: 5 mm ( single start)'Fo= 2oo 1r1

_5r x47.5

: 0.0335 .'. cx,- 7.92"

Tlre rnean raduis of the screw ,,,,: # = 23.75rnn2

So tlrat a f,orce of 200N at 200 trun radius will be equivalent to a force of

200mm x 200 N

23.75 at 23.75 nun radius: 1684 ].{

This is the force F up the incluid plane

L=tan(a+A\w

.". \tr/ = F

tan(u+A)1 684

W_ = 12600 Ntan(1.92 + 5.72)

Example (10)

A rnachine slide weighingZ55 kg (2500 N) is elevated by a 2- start

acrne thread (29") thread angle) 40 run dia.,4 mrn pitch . If the coefficient

of friction is 0.12 calculate '.

a) The torque necessary to raise the slide .

b) The torque necessary to lower it .

^A1.+

2@ nt

If the end of the screw is carried on a thrust collar, 32 mm inside

and 56 nun outside dimeter'

Soltttion.

, -d,.Wl f ,*(:g!9utrygl-d,,,'f"'Wta=-T-Wre,lr^")* z

h=P =2mm2

W: 2500 N

2 start aclne tlread

0 : 14'/r"

do:40 rnrn

pitch r: : 4mtn

t:fr:0.12dn, : do-2X 4^ = +o -2 = 38 mnr

2

tan 0,,: Cos cr ' tan 0

Ltana =

-rdm

lead l-2P=2x4--8mnt8

tana _ _-= 0.067lTxJ6

.". d - 3.834'

;.tan7,,=coS 3'834 x tan 14'5"

= 0.998 x 0.2586

: 0.258

Cos 0,, : 0.9683

,..0,,

dn,.

= 14.47'

- 56 +32 = 44mnt

2

2

=40-2=3I w,ttt

&llar

z5

38x25oo Io.t z * 0.9683"r0.067 Iu, a^ I

-

I I/ t\ 2 L0.9683-0.12x0.067 )

44x0.12x2500

JW M 16 i)fst 37- 2

: 47500 [o,rz+o.oo+ql + 660010.e683 - 0.0081

Tn: 9146 + 6600 : l5746N.mm : 15746 N.mm : 15.746 N.m.

b) the torque required to lower the load TL

.T - d,,.Wl .f, - (.',,s 0,,tuta1* d,,..f ,.*'t' 2 lCo.t?,,-/,tanal 2

= 47500[ o'tz - o'oo+q

l+6o0o10.9683 - 0.0081

:47soo[0'0551-]+6600L0.e763 )

= 9280 N.mm : 9.28 N.m.

Example (11)

A trrubuckle has right-and - left hand square tlueadsof l0run

pitch, lneall diarr :ter of 40 rnrn,

p : 0.16 the tumbuckle is used to tigh a wire rope as showen in the Fig. If

the tension in the rope is constant at 12 KN find the tuming moment

required.

Solution

For each tluead angle of friction <p

tan <p : Ir: cofficient of friction

a :tan-ro.16 9.1o e:9.1o

Helix augle tan 0 : P

ftd n,

j^r M 16 Lr- -d3s' Jlr rra to L,l-" *J3fstt2-2

j^r M 16 i3r, J) JU*" il s gtt!

*lB

ZO (,tu) ,: u-, Tirn L""LI e

d,.,r:40 mm, P:l0run

0 : ta*-l { : ta1-' -q, = tan-r 0.0796= 4.55o

ftd n, tt 40

Torqr"re to overcome ftiction on each thread : i * ' tan (q +0 )

Torque T::W.D.tan (q +0 )

: i.*(12 x lo3 ) - # x tan(9'1"+4'ss')

= 58.28 N.M

Total torque for two threads: 2X58'28: 116'56 N'M

Example (12)

ApowcrScrewlias6Sqtlaretlrreadper25,4uun,doubletlueads

and a rnajor diarneter of 25.4 mrn is to be used in a power driven press as

shown in the schematic representatioir ' If fs : tb : b'08 ' dc : 32 rnm'

The load to be raised per each screw F = 6'80 K'N'

a- Find the pitch , thread depth , tluead width' lneall diameter' minor

diameter and lead.

b - Find the torque required to raise the load'

c- Find the torque required to lower the load'

d- Find the over all efficiency '

Sottltion

a) Since N:6pitclr P :25 "416 : 4.233 mn

The thread depth and width are the same and equal to half the pitch

2i

(h) the tlrreacl depth : width: plZ: 4.23312:2.1165 rnrn

dtn : do - 2 xhl?: do - h : rnean diarneter

dtn : 25.4 - 2.1165 :23.2735 nm

dr : roor dirnater: di : do - 2h

: 25 .4 - 2 x 2.1 165 : 2l .167 rnrn

lead : number of tlueads x pitch : n x p:2 x 4.233: 8.466 mrn

(b) The torque required to tum the screw against the load is :

d,.Wl f , * Cos 0,,tuta1 , d,,".f".Wt h

-

-r

-

t-T--x 2 lCos0,,-.f-tana) 2

,l,,.Wl J', - Cos 0,,tarra1 , d,,,".f ".Wrt --r- ra-' 2 lCos?,, + ./ -tuta ) 2

lf the collar friction is negligible fc : o

d..Wl f,*Cos 0,,t*olt^= z lcrte,, 7r^"1

.n ,l,,.Wl f,-Cos 0,,tanal| | :

-r

------1-- t' 2 lCos?,,+Jrtata)

Co,s 0", - f "tan a

" - Co,r e, + .f ,Clot a

tan 0n: Cos cr . tan 0

fr: fc : 0.08 L : 8.466 rnrn

drnc : 32 rnrn dtn : 23.2735 nm

cr : Helix angle , tatt o: u

trd.

W:6.80 KN

tan cr : 0.1 158 g : 6.6o cot u : 8.6356

23

Cos cr : A.9934

tau0n: Cos cr. tan0:0.993 xtau0 0:0:0 0r:0

, - rl *.w | /', + Co, o,,tata) * d *". f".wrR- 2 lCos?,,-.f"tata) 2

T,. -23.2735 * U ^[O.OS +Cos o.tznt6.6'l 32x0'08x6'8z {mf ,

r -.7s, .[ o'os + t'o't t ss

l+8.ZOa =24.34 KN .ntnt^R [i - o.os x 0.1 l58J

Tlre torque required to raise the load : 24 '34 KN'uun

(c) The torque required to lower the load (T1)

23.2735 x 6.8[ 0.08 - C'os0.tan 6.6 | *

gZxO'gSx0'a

2 I CosO+0.08 tan6.6 | 2

T -0.0358 I=l "'"""" l+8.704lr.ooe264 )

= 79.13 x (-0.03515) + 8.704

: -2.791 +8.7A4

T, : 5'923 KN'mm

The tninus siEr in the first tenn indicates that the screw alone is not

self locking aud would rotate due to the actiol of the load except for the

fact that collar friction is present and must be overcome too '

(D) The overall efficincY r"1

Cos0,-f,tana ^,"1= u" * fs : 0.08'' Cos0,+f,Cat a

0n:0 Cos0n:1 , tanu:0.1158 catcr:8'6356

a

ry=I - 0.08 "r 0.1 158

i + 0.08 -r8.6355

I - 0.009264 0.990736

I + 0.69085 1.69085

: 0.5859

Example (13) :

Ascrew jack carries a load of 400 kg. It has a square thread single

strat screw of 20 rnm pitch and 50 lrun lnean diarneter. The coefficient of

friction between screw and nut is A.22. Catculate the torque to raise the

load and the efficiency of the screw. What is the torque required to lower

the load?

Solution:

Helix angle 0 of the thread

tan 0 :ry ( Single tlread.)rD

p: pitch: 20 rln D: lnean diarneter: 50 rnrn

n: nnrnber of threads: tr

^ 1x20tanu:-nx50

tan 0 :0.1275 .'. 0: 7" 16

The weight of 400 kg, W:400 X 9.8:3920 N.

Tlre angle of ftiction Q is ginen by tan $ : n:4.22

hence O: l2o 24

Torque to raise load : )tantan(g + 0)2"

3A

_ L.- - X3920 X0.05 tan ( 12o 24 + 7' 16)

:1x 3920 x 0.05 X tan 19" 40 : 35 N.rn2

Efficrerrcy ,t= L- o'r27'5

tan(q + 0) tan19o40

n - a)275 = 0.357o135.7 percent' 03574

Torque required to lower the ioad: 1 Wp tan (<p - 0 )2

:1x 3g2o xo.o5 tan ( 12' 24 - 7" 16)2

: 1x 3g2o x o.o5 tan 5o 82

:88N.rn

Example (14)

A double - start squre-tluead screw drives the cutter of a machine

tool against an axial load of 450 N. The external diameter of the screw is

52.5 rnrn and the pitch is 5rnrn.

If the coefficient of friction for the thread is 0.15 find the torque

required to rotate the screw, If the cutting speed is 100 uun/s find the

power required at the operating nut of the screw '

Sohttion:

W : 450N

Coefficient of friction P:0-15

Mean thread diameter d,,, :do - ,d,,, : 52'5 - 2'5 :50 rrun

3L

F{elix angle 0 tan-r *

g = tan-' 2 x5

= tan-r 0.064trx50

coefficient of friction p " g: tan -1

tan (0+q) tan ( 3.64o+8.53"):0.2757

. tan0+tanfor tan(0 + tol =

1- tan O.tang

= 3.64

P: tan"i

0. 15 : lt : tan(p

0.15:8.53

0.15 + 0.064

l-0.15x0.064 = 0.216

hence speed ofnut

and o : 2n n: 2nX

power : T.ro :2.43

Torque T::.W.D. tan (0+q),2

]-x+so -so v1s.216;2 1000

: 2.43 N.rn

The operating nut rnoves forward a distance equal to the lead when

the screw rnak;s one revoltttion

-crttting sPeed

= 1oo

= lo ret,l seclead

l0 : 62.8 rad / sec

X62.8: 152 ro

{poor} )ZAllernatire screw pee$s rrrsngements,

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\

lr

HALF ELEVATION HALF SECTIONALEL EVAT ION

II.JCH ES0r23t+r+"fr]+r_-1_I

SCALE = IoI

-----]- - - - - -I

II

:0. -!ot.(r'i itrt{ t o Iieour.ult 9

?rr))

DESIGI\ OF SHAF'TS

Ashaft is a device that supports pulleys, sprookets gears, eranks,

levers and other attachements and often transmits power between them .

The shaft itself is mounted on bearings so that it can hrrn freely in the

housing. It rnay be subiected to transverce loads to, torque or lnore likly to

botli .

Design for strengtlt :

Designers inust be able to calcglate stresses, often caused by a

cornbination of loads at varions points on the axle or shaft, and be able to

appraise these stresses by means of a suitable theory of failure.

- Muterials foraxles snd slrafts:

One or two of the following eiglrt factors are likely to predominate

in choosing a shaft or axle rnaterial

1- Rigidity or stiffrress ( The ablility to resiste change of fonn)'

2- Strength ( the ability to resist loads

3- Wear resistance 4-corrosion resistance .

5- Weight 6- cost ( the expense of production),

7- Size and availability 8- Machinability .

Strengtlt :

If operating stresses are light a - low-corbon steel will be

satisfactory.

39

Where greater toughness, shock resistance . and strength are needed

heat treated alloy steel are sensible choices.

Design for strength: the Basics

The common axle or shuft designers deql witlt :-

1- Rotates. 2- Is supported by two bearings.

3- Is subjected to steady bending loads ( or none at all )

4- carries a steady torque or no torque at all .

5- May be ssubjected to a steady axial load

6- Is Circular in Cross section either salid or hollow.

7- Is rnade of duetile metal usually steel .

Exumple (15)

Draw a working drawing for the following shaft tacking into

account the following rernarks:-

1- a is a centering hole.

2-b: is a pipe thread Rl length 25 do:33 di : 30

3- c is a pafi with hexagonal cross section e : 53, S W:46

4- d is a shaft diameter d: 50

5- e is a fine tnetric thread M 45 X 1.5 , dr : 43rnrn

6- F is A rnetric thread M30

If the power transrnitted is established for any sectiou of the shaft,

the twisting rnornent M1 for that section cau be calculated frorn the usual

relationships:

Ho: o'w -2ztN'M' N r.p.rn if : Horse Power^ 33000 33000

SEcrtoM n-a

w(sm)50-2 - Y

1

t,oA

-r_1

X-X

*lIli1?

II

- -a-I

I

til],i:--=.1H

I

o\e25

B110

/.

-

tJU50--.-

--

215

4U,5HAFT

St

Shaft Design theory 6.r-oe)l.l f4*^,.atJl qJtr

lcircuiar cross section ) .s-F'lr eJji 3!;rl-c ,#LSJ i; 3a :3*tJl

Gears .ryjll; Pulleys Pulll .ljj.e;r clj'+l C- ':lJrjill L)-lii d rJ:jJl*J

. kl l+-ll"Jl Lr-,HiUl"r belts JJ#JIJ Chains -rl"jE+ll-l

shaft :-9,rtJl '-Uil +.,t-i^ bore diameter ,r-ls'lr JA{ elj)+Yl 'te'l .lA-l

y etj.-)l orr -;j$ .&JlJl o\ 6iJ *S-il| IJJ{*J r39olJl *Jt ,i!-tr,rSl diarneter

.motion isFl: Power ;J$il d6Jre '' q

How is the size of a shaft detennired? f .:;*l.Jl -,2J.! r'-a i ir<

.-r. cp-r ,-Jl-l Loads .Jl--Yl jl Forces cr*tsll ir q':"Jl;r^cyl uibJ'ar : YJI

,+iJ Torsiou cl-jiJYl3 Bending eti.-rYl3 Cotnpression E:ll3 Tetlsion r'$ll

tJsJjlS+Jl c-rl r:-iill gljirlit ";\1, uSJ-i.E:'Jf ;''i-c'3q^ll "lAi3U Jl:ficYI (-+ r=-h'i

cI;EJYI JA +lJl ,jl^Jl tJ" ,il3 tnechanical power traustnistiou

T. R.K

.Fig (2) ,J6,1 C-.-r^ t^S rnajor type of loading

kg/crn2 , 1b/in2 N/mm2

kg. ctn, lb.in N.rn.

J r.-iiil pjc [una, incha

s,=

S, : shearing stress r'-iilt rt-6,"1

T : Torqne cl_5ilYl

1: polar motnent of inertia : !-D* 'r+ill

K: stress corlcentration factor (*l'r=-e oJ+) ':hl'e+)l SJJ &l--

J : *D' for solid shaft dr^-^ll :-rU where D :3'^1-Jl j'!3L

3 : a1n* - o1) for hollow shaft dr+Jl r-rtJ!32'

4L

D: Outer Diameter, Dt : Irurer Diameter

T.R.K T.D,Kusuuu

.r 2.J. zor--Du32

16.7-. K.S for solid slraftsr),,. lf . l)t

.c - l'6'D'K lbr hollow shaft"s^ E(Do - Di)"

r_9*L-,Jl Or*o t,.l ,rJr J-"1+ elJilYf i*+ r3*hJl "'r

J*,S a:c &r.=r..5.J1 ,riill :ta;l

- exact loading g++-Jr ,Ji*"-ill raaqAl dJSt &lp Jl ;-it:Y! type of material

Ternperature ';Jl!l i+-rr: - corrosioll dstijl - shocks dll-or'-Jt

&t-*^ JL:cYl "/ ri-H Ol "'il

dli clJrJj i'!+ J^lJ'Jl ora ;gr!i3 LJ'"-ll U^3

. t Jl Y g,o r3r-: "+-l

a+JJ-tl1 e4.ill c.r^ ellv 6rc O 4-l t++ ;a$ sUt

rj,l3 JIJI eL;jYl .rt eil rjt-^al-9 4+tJl cllJtLii^ll cJ-S lTtYl .,i riij '{rSlJ

. i.Loc,Yl i.oU r.tLi*^Jl UUI ,-Jo '-r:."'*Jl ','!' ll ,lrli"'l3J# g\"1 j^1-'

cournoll cold - drawn steel shafting rnaterial is trsed

cLliil i+r,*lr1l i.ucYl JI}II -rl-,.o1 irnperical fonntrla Ll"'Jl ij'rtJl oiA prr1"'i

elsi-Yl il{i*J iL**t i'!:L'^ ,raj i.,;':ill

'=ffiu"n ( 1'":25.4 rnrn)

i,..-dq ::^d 4+ C3* J.tI .Fl D '':+*=

miuitnutn required shaft diarneter (inch)

F{P: systeln horse power

n: r.p.m. of that shaft

,jr L^Jsll ;lD t'lijl LAIir:

T'. D. K

,.1L ^-ll+ e\!ll alyill ;;riill

i#. / iill! -:3*tJl io-*

,.lfu {iti pulleys }*U-t J^.1 .,ll 6:'ocYl iJl-* d t^l

ilrlJl ora

4z

inch

J.*=Jt JA +,tJl (",rA L^rb l-ol J+Jut ,J^-J cl 6r'os)l'J tl+l': i4$ *l 53'5 d+F

+;:Nl iJlrJl plrii".l .s^.-6 ,*jll ir^gYl il\S u+ *JJi3 r3*1Jl os^^ JIE #l *J'

5s HP/ ) l-"Xn*,, *-1, u'r. J-s3 (stress conceutratiol factor (K) drhl'e+Jl -FJ^: &t" t^i

a,- -F.5l Jl ds-i ,i^:I3 irLc ir.ocYl gl '''*- Shaft material variatiOl 9r"oll tlr'-"

elli keyways *+lJ- t,JL+j - holes +rS *& grr:=: er+> etli3iYl p'ro {.JtS'il

retaining rilg cr,"iSll c1,$ti.r-rb. - ttrrned - down sections t'El J}riJ aJ'3;i'

:3*GJl ,-i1--:l ,rJl .s.ri'i:* Cltll d d'l *ii!l orl Llsj j:,s .,ir Jt grooves

: l, r-i

JIJ; :j*LJl Lu^ ,J!ai ,Jl .Ety ;r ':fu il"*Cl Jlljll t,F JF clli t'lc cJu^s-r

. clltl-(+Yl -6F &\-'^ JUgYI ,..,1 r=J* c-tl "'il +'.,ls a'rlc l'E-l 21 '

:3*LJl ar-;r- ,r-i :.Jl: 93 Torque T el:ilYl rt' J€3 alrL',-oll d)" ,.,+ul J"-l'll t"i

yLSlU+ ,i ) ffi ..,1r-.--Jl-.' uJdi^ll 6;:Jill l--br.lJ shaft speed r'p'm' ij"sr /L'il n

. LJ'. # O:s:;tto "*, ( k'wd'l-r

rl3xJl L3LIL .J:l+ C,.6rb+l CS^,.J 6+ill Ji--ill e-ro *+ J t-i

.ilLJl .J!-X ,#t * ,-,Jq .r:ll !-r!': t-rUl JE! rrq;l gSru' iltJl iJr\'^Jl j-;3

43

calculation of Stresses on Square and Rectangular Keys

ylo"e- Area L Cr*si r,,.a

=o.Sf 1 O

L : Length of the key < 1.5 D

Shearing Stress in the key is t =

2MtWhere F :

F

b.L.

T: shear stress M.Pu.

Mt : applied torque N.mm

b : Width of key mm

L: Length of key mm

D: Shaft diameter

2i['4t' b.L.D

Tallowable: 550 Kglcmz

L=1.5D

Crushing stress (compressive stress 6c on the sides of the key (and on thekeyway walls) is:

44

4MtM.Pa.6c:

0.5 t.L. t.L.D

Example

Design a keY connectionworking strengths are as follows

for a steel shaft and a cast iron hub. The

If the shaft transmits 125 kW at 150 r.p.m. Calculations may be based on

pure torsion and a factor of safety 2. [1 M.P a = 145 P.s.i = 10 kg/crn2]

Solution

! 55 P 9550 Pkw 9550 xl?\\1-= =-=

-'=7958 N.m.,-^t n flr.'... I50

shaft in torsion Tn,** : 16 Mt 16Mt

ft'T.u*nD3

Mt : 7g58N.m. :7958x 103 N.mm , rw: 200 Mpa :200 N/mm2

r* - 2oo - loo N/mm2T-*=f== z:=74 mm

Use a standard size D:75 mm.

The coresponding key is obtained from the following table

z}l4.t - -T*L**-^I.S.

SHAFTM.Pa

KEYMpa

HubM.Pa

Working shear stress rw 200 t6s

Working crushing stress o* 400 330 550

16x 7958x 103

Shaft size D Key size b mm Shaft size D Key size b mm

t2-Ls a 50-60 t4

ts -20 4 60 -70 16

20 -30 6 70-80 18

30-40 8 80-90 20

40-50 10 90 - 100 24

KIEIL Shearing r.u" =b.L.D

45

2}r4.t 2x795}x 103= 143 mmb.D"t-* 18x 75 xl65l2f.s.

crushing L: . 4M' - 4x7958x103 =l43mma b.D.t-* 18 x 75 x33012f.s,.

Coupling Design

Couplings: are used to connect the shaft of a driving machine tothe shaft of a driven machine. This affords a pennanent connection.

Classification: of couplings can be made on the basis of rigid orflexible designs.

(A) Rigid Couplings:

I/kestrated by a flange coupling, compression coupling or tapered-sleeve. Coupling. This type of coupling is suitable for low speeds,accurately aligned shafts.

(B) FlexibleCouplings:

Illustrated by the Falk flexible coupling, old ham coupling geartype of flexible coupling, roller or silent coupling etc.

Flexible coupling are used:

(a) To take care of a small amount of unidirectional misalignment.

(b) To provide for "end float" that is axial movement of shaft.

(.) To alleviate shock by providing hansfer of power through springsor to absorb some of the vibration in the coupling.

Couplings may be classified also as to use, specified by the relationof axes of the connected shafts:

(1) Axes of the shafts are collinear.

(2) Axes of the shafts intersect (A universaljointof themanytypesavailable may be used).

(3) Axes of the shafts are parallel but not collinear. (A coupling of theddham type night be used with its sliding member).

46

aoZlo-

=oOtrJ(,ZJu-

93 F99-.: L O

-:q !6EI tr: o/-.rl9

6qo d.--

ail ==-{ 4. -. *i o

o >t &o $d : tt 1dP-

=-J,*E; i3J,Ja c

-^LvEo E ce b€'EoP F"9Ld > Q 7!: ;; - ts- Y o- -_

^; E C ^Lc c 5;= j " o

E g_ o u o E ol; i c.e5.9!:2.f;-6*pIEdo XY+P€ E.5€

zJ

oUE

U)

d

o(r)

eo"\+Js)sU

_o

UJ(,ZJLLIC

o-VIC

sEE r:E sEf;!=:;

ilf,iiffg*iei=i;:E,eil{tE;;;,X;"is EE:rlitsa; es=

€do

o€o'il-6:6g

otr.qEidooE

caoU

?oo.;)tgoo

.,:4 13

d9

=-s,5.A

65a=

ot'of,oti.orB4VAe'9-Boaq&ooL-Og 3&

OJu)(l'JlJ|A-Oaiinto il,oirto

e.o\U

.q,or

\'a)

\J).(\

ib€).c*codVi

x&Ei9eO

bsxo}iOdvcrE

ii 6EE9oEd9:ai: 10o:Qttr

+t

I

€$$$ i$sss E

I ie iii iS rE

$$ *cgs

t

l-ffio!z

___ __l__\-l'ttl ivluJldhc)ll,X\i sr lr- d

l--,2 iiLAI l<;l\ rr\i\ I +.4ri- lrl)SiuJ i {tra I

VIC

€!:

6l-o;,,,,g) ;Jf * * dtl) i.J' r:,li *;11!

(J!i!l '!) lUl .,tr;rJt ;1rl -L; !

a'rr fJa-/^\,\E4- NltlatstY\zay dF

gV).:tyl<qrLt)

- i,1i ._.i"...

,/ '\- -l.rl .-L; \

q$.)tfl

6J,ot)l erly'l

i*"1.*'"1*,,ari,ll rrli ;i.1Lill

*d.il,r.tJl rclrj iniJl ri,Li

\J'

o,^sJl

ffiffiffif& J*

,ia.iiJr & ;:;, J*tt *t *w ** a;ijr qjln ,turt!

tr/t x,Ll" (*-y-'l"rq Jr!'r!

,jrl" J* J* ,r+

;iJl ilriJl }o ;a.-h

i"bLh.lt .tjtyt JKji

.tr 6;i$t h,.rJt eotg;rt'Jl .["-J1

l1

,J.u-- {it.latl ut$-{\J

L)f e-t)!

4h 4tr,.o\,-i, .-:l il,4i .tl-/l)

i+:bj ;b;1 oli ;;y i,ilt!

d-)\ J*;)J aIU; i.:+ i.;1t!

'ru05 .'Janr crl.: a.;'tr

413

5i)y Ll,:r ily -n1\9

co"rTERED JOiNTS

{7k)

t)=24d D,=[75dd,= t.2lc{ dr=,1'5cl

G = c =-'75cil, * t''l d. t * "J! c{

t, = '/,5c1P " 'Trh

cl?fi

c1E*s-_coIrE-ts-CONNI EXION FOR

srw-gN-8,

CO-TTERED JOIN.i.

Fo-8 Jts)rt€

clrorance ft t 'i

egfjEnEgsTAt'l pAI!.p

lh ickness.t

}(")

tr)I

49

1q!,

FLANGED RIGID COUPLING

Flanged Flexible Coupling (with 6 bolts)

1,2- C.l Flan GE GG - 253 Screw belt DIN 267 - M^lz x 40 - 564 Nut DIN 267 -M 12 - 5

50

2

I n,2s/\/ ,'

,//

\\I

tltt

ffiKey-tx V DtN 6886-BBx7"5o iKeY

'-La r DlN93r,-M10 NUt6)-z) Washer---Bush

r-

--." Pin.

Flexible Coupling .

Schemobild f iir den Zusommenbou

?1

5L

C)

rr.1

,-C)

+<

M

t.Eo-Lt,Gd

I

bo.a

,9()

EEL:AC\

LA?

'()6;{trcdu)M tsE4t

B|.1 .9UU

cn6AdJ iJ <F'i o .3kaili9

IJi H c.)

gtrsX E E"i, EZr- ,.'E g o,E 3,t.* EEY,= EE ?.;ti t'-,.EE E?, E

ETgi.EEB EE iiorZ >?o.o =Q YT ;;iE:AE EA E69 EsEE'iq.E o-H €9i E Ij6=.3*'= t', =!d 6E=EEtp y? i"o'i==.i-y- E -: P.E6dEo^d,.C-E.*

e: g.t z X; ! i"= EoSoEElo:E q>Eoo#; zF::=ag ;g3E:=g Px';=2E: E:s=3;I E;Ei;;; Ei;:;EE 3 ';s.!E E8*EE;; H;€i:iE pE;*Ba# E€E;:gE e:iHr;i gi=;a;

$ AE:'IE=e;i1or-$* E i1: iiT=s€ i = E = A i:IESo o- E

GOOF)-o>a

3:oIir

6

o

tcFUJ)J

I

oOo-UE:6 ?',' Z'-o9tr >i= !

u!e- E .Ya(Eq

Ndq?. 939*;-t- l.oLEOo Y-{-E E r P.;sE.! t'f E a8! !P1;A1c-;-'io-:zoE "i6€,Ad-C

'-V d 9 E c$ _ Lr !e q,N '" ar O

"-'o ; 9{ ;o }'c-Yc?ooru -:d j:niif' :-EIs Bsi i e5 E.-L-=Ctro<-(!uouoo- = n-C o CrdX-](g:.evoJ = .r 3! g

1J,r

-- ').J€ -La - cJY =

- x s.->s E -U rE A'tstd6';.

--a

tr,)

Exarnple (L7):

Asolid coupling transmits 100 KW at2 revlsthrough eight equally

spaced bolts . If the bolts are 72 run diarneter and are on a pitch circle of

n50 mrn diarneter calcnlate the average shear stress in each bolt .

solution.

W.T ZrN T' 1000 1000

Torque T in N.rn

Thtrs the torqtre t - l000to pov'er - 1000 x 100t.r/ = 7950 N'm

2rN 2n x2

T'otal slrear load at raduis of 75uun is F, : ?*= 106000 N' 075

sinoe balts are ductile it rnay be assumed that this load is equally

distributed arnong the eight bolts therefore load per bolt : &

P": lo6ooo =B25ON"8

area of bolt : A6

At,,:Ld=n =L621= = 113.1 mml" 4 4'

slrear stress sr: 4 = ""9- . = 118 xlou N /m2Ab 113.1 x 10-6

53

Fig . 11-8 Shows a knuckle joint rnade from SAE 1020 steel . Assurning

a1 allowable tensile stress of 1 5.000psi, allowable shear stress of 7500 psi

and au axial load of 5000 lb detennine

(1) An algebraic expression for each of the following .

(a) Tensile stress at section A-A.

(b) Searing stress in the Pin .

(c) Bearing stress betweeu the pin and rod .

(d) Bearing stress betweeu the piu aud yoke .

(e)Tensile stress across hole in the rod.

(f) Tensile stress across in the hole in the yoke.

(f) Bending stress in the Pin.

(h) Tearout of rod bY the Pin.

(I) Tearout of the yoke by the pin.

Ans"(a)s,:4F lndz

(b) S, :ZElndz(c) S,: F/d.

(2) The dimensions of the various parts.

Ans. D: 0.652 in ., d: 0.652 in., c:0.66. in.,

e:0.505 in

b:0.33 in., rn:I.306 in.,

@) sc = F/zdb

(e)e= F/lc@'41 -

tf) q = illzu<n--allJ(t) sb = 4Fc/nd"

th1 s" = F/Zce

til i = rtl<zul<z"l)

54

Keys ;g*lsill

;Jriill Jji: rr}+i3 -:3raLJl O+ :rSj ;r-.^Jl ,/q;ibd'l*4 iJ! I Key 3,5a1Jl

a#i ir,li ;*,S e3q System torque rL).,ll elJ':Jyl e* r,Jlt 3o *'!-o ,r:Jl: isqi^ll

"r<J dil: r;l;3Jl Jl dYjiYl i--luL 'r-lc, L'.tl Jll.:Jl di^-.,.-, Jts,iltJ c,a3^ 9 t^s

. ;;ljJl clj=)13 r3xtll cx KFI:;J'rill sliil +b+t el**-l

:L"l ;aLtc JLFoYI .,s -riYl ':':,. J-llE Jl+'i'l rieJ

type ;;4uJt 1s (\) size 6r,t!il1 (t )

.r:* +:-,fl! :l: standard size key .,-'W ..,"UL :i J.r+l- JsiI $ sl^'ll ;LniYl e

JlLJl f p JA b e1r= b: t/o d rlxlJl F Yo

iraj ;3qLJl .i.9t 5r:3 ,.,i:! elll :+ i ;gf-tJt 313:Ylj crLULll cj-rlt+r *y I L"S

low horse power 6J+-i--il ol-,rill .Jj,j ut= u# +s: il "iiiJl i;-r!ilJ ellr Ja drJl

Ci O- SLJI ,- :-i L'o'i a.+d" dl^.," Y ;-ellJl j* Ol- high r'p'm tJL*il alcll3

. ,,r,.,.- aJ 6y;ilt !f.c ;r1 Y ,.ri= 15 rnrn ;r ;;'"'l uC: lt+ lJci-i oC -'-i*L'iil

. 4J' ,",il..r.Jl .#6*"1l i;+r ,lrf dr JSI ':'"rLi'oll Jl\ill ,Jrf OA Ol "'r+

Y-l

]-,) 4i)i cl3jlYl eF 6rL+j J' d"'J,Jll ii$+ll c-rLc!13 qlUlt {.rt;"rill AIL' '{.s!

\-oi

L4 Al #ldl d)- Ll,rl ;r-c clL-:a3 .. -\|Sll el3ilYl hA JsX .+-Ml ,JlJl l-r1""'= g'o

-: .,A -l dS'liJt g,oy't Ja

Flat J..,o -\ Sqtlare e;* -i

pratt and whitrreY L55.lJ 6l;'r -tGib - head *i,, -Y

Woodrtrfr (";r.^s) dJrrr -o

'-5

2

i

TI

It-l tr+l*_ r__l .l w l-_:

Standa,d q!J,e key

la)

r--_-__--l n;i-.----l -l "l-.1

Srand.rd llat kcy

{b}

I

r____-r D_ll-.--.---l !"Li

Eila(Sin tey

\.,n(\us (vpcs oI sha[r kcr,s

l-r..11!#.---------]Fwl[]Fa;ll

l-"1 l.--.----l

9ib head

ld)

(r)

FcarM, key ingeried in

;(rz-

t!:-:=), €'r :'==:T::==lp:1 )

r r5. 1x.7. Joirrilwith Ihrrrr ck,,urc. (a) trunsvcrse pin, (b) woorlrullkcy, (c) Lr,,t ilr lci,rhcr ^,:y (piIxrt,:l r.],. t,t; slrrlr,rliIt!rhcr kcy, (c) splincd pronf".,ld ,".i"i.J','*n protrtc, (B) K pr(,rrtc.

{'t t"''lr,

I'-- r ' ["'-ll- : :..1t.- ...+-. .r,

IItIl . :r'., ,)--t |' I

l. rr1

."iy-_=l__.1.-.7 t-..-.-i-____l \

'l \c^.8.1-'r,.',.:{&jt.,/. .. i '-

T---l rT - I/--< i r'*{:+-.J::[:-l \

Ilir: '. l ,-..r Y--.i. j ....1..l"rl

'.1Fig. l&'8, prc-srrcsscdjoint wirh fon(d) J.vcn ,rp, r"y '*-#

"r'*iiirli,i'::';'"1r"),round-t,-pdrkcv(cro*il},1|rj.:,:"1::lr.,,rr,,rr r.r.y,r.rs,,,,a i.,,,t, rj.,

di'j,'.1,,i'"1lil;:,ii,? f"flx:".'l.; li,ti.li; ;,1''lili'il'iJJili,,.,,,.,, ^",

slr'(lt \tlth lluh sr:atirr,' (r) r'rrb (,n o rr{ slr,rrl; (b) on a slcppcd 6hd,l. r lr.: rr,(, r, cilr\ r rprcssr.r rrrii, liei rva1.,!tepl io thc shafts showrr un ilrc nIht ir r.r,,ria"r,t,f1 t"r"

55

ReclanEular key

Fig.10.20 Examples ol keys

Fig.1}.?z A spur Eear fixed to a shait with a Woocjrufi key

Fig.1O.24 A sptine on a shaft

Fig.10.25 Sptined hote in which a sptined shatl ,itsqounC key

Stope 'l irr 100---

Fiq.10 23 [JrawinQ sr;:es for keys anC keyways

H SPLI NE

Fig.1O.27 Third Anglo projection showing BS convention tor a sptine

Hectangiliar key

aWoodrutl key

Fiq.10.26 Front view ol a splined shafl

A pulley fixed to a r:haft with a rec.angulai key

57

JJru,.ll e$Type of Key

cya.llg r3,rtrJl3 .t.XL3..ll irrg ef;-r ;.JAssembly Showing Key'

Shaft and HUB

Citi..alJ*ll

Specilication

Square

&-)4

..yA 1-14,_r 1. 1 4+y -r:r:L=

6mm square key , 28mm LG

fr" J_1t: 1^1 ? El* e;r JIl.6mm square tapered key , 30 LG.

Flat

eL,-

J3J.1 3Y x l*. o, \ AA C.!..,,^ JJ'tLj.

5x3mm falt keY 25 LG ..r o

5 x 3mm flat tapered keY,

25mm LG.

gib head

.,*l-r -r-.:{L'

.l"' cilt: pnA .rlt Ci-)' Jrl^

8mm square gib-head keY,50mm

Pratt and whiney

l'l,L:i o s_.:l u

\ o C-, A:-l u.rl;,r -,2311:.

No, 15 pratt and whitneY keY

Woodruff

(";*s) G;":je

\I i . dt (Gr::) ol ;-5a1-

No. 1210 woodruffkeY

;#lFl A,jlJl gl3iYl

KEY AND KEYWAYS

1. KEYS Ai\D KEYWAYS

B.S. 4235: part 1: 1972 is the first of aseriesfortneterickeysand

keyways and relates to the following types :

- square parallel

- rectallgular parallel

- square taper , gib head and Plain

- rectangular taper , gib head and plain

lt,-))

In the case of the square and rectangular parallel keys and keyways

three classes of fit are specified as follows :

Frqe,,witere the hub is required to slide over the key when in use,

f,{ormal , where the key is to be inserted in the keyway with rninimuln

fitting , as is required for tnass production assembly'

Close ,where €m accllrate fit of key is required'

Purullel Kevs are used in the transmission of uridirectional torques

not involvilg heavy starting loads, ald where axial lnovelnent of the lnrb

member is required frorn tirne to tirne '

The square key (Fig. 3.1 and Table 3.1) is used generally for shafts

up to and including22rnrn diameter. This fonn of key will provid a greater

resistance to crushing than the rectangular sectioned key ' The cross

sectiol is particularly suitable when the sides ofthekeyaretobeheld

lightly in the keyway , since the crushing effect is then gleater than if the

key was held more sectrely. Assuning shearing and crushing resistances

equal and taking .

b : whidth of key waY

I : lenght of keY

h: thickness of keY

1 : pennissible shear stress

o: pemissible cnrshing stress

shearing resistance : crttshing resistance

tbl: o (h/2) I

also , assuming o :2r and substittrting b : h givilg the proportions for a

sqllare sectioned key .

Such a key is often used when sliding is to take place between the

shaft and the hub number Keys allowing relative axial tnovement are

referred to as featrrer keys. parallel keys are ,onnafly side fitti,g with top

59

clearance , and are usnally retained in shaft lnore securely than the hub.

TXre ends rnay be square or rounded . The rectangular key (Fig. 3,2 and

Table 3.3 , 3.4) is in commor use for shafts greater than22 mrn diarneter.

Taper Keys are used in the tralstnission of heary unidirectional ,

reversing or vibratory torques and where removal of the key may be

necessary . The basic taper of these keys and their keyways in the hub is 1

in 100. Taper keys are of rectangrrlar or square section (Fig. 3.3, 3.4 and

Tables 3.5 , 3.6) . The former are for general use, while the latter are

ernployed rnainly with sirafts up to and including22 mrn diameter. These

keys can be either plain or provided with a gib head (Figs. 3.5, 3.6).

Generally the keys are top fitting Taper keys camot be used in

applications requiring a sliding hub tnember.

Woodruff keys are for use with lightly loaded tapered parts. The

shape of the key and its keyway allows easy adjustment to the hub taper

(Fig.3.7). Sec. B.S.46:part 1 : I958 forfilrlherdetalis.

Table (3.1) Keyways for square parallel Keys

(All dimensions in rnillirneters )

Shtft Key Keyrvny

Nominal Scctionrvidth x

thicliness

width ft) Denth Radius

diameter (d) Tole r:rnce for class of fit shdt (t,) Hub (tr)

Over ToNorrr

free Normal Close Nom Tol Nom Tol Max Min

(bxh) Sh:rft(H9)

Hub(Dr0)

Shaft(N 9)

Hub1.IS 9)

Shaft andHub (P 9)

1Z

l1

l1

22

5x5

6r6

5

6

+ 0.{lJ0

0

+0.078

+o.oro

0

+0.010

+0.015

-0.015

-0.0t2

-0.0d2

J

3.5

2.3

2.t

0.25

0.25

0.16

0.16

br#

,</)

Thickness (h)

Tol (h e)Tol (h e)

Table 3.3 Keyways for rectangular Parallel Keys

(A11 dirnensions in millirneters)

Table (3"2) Square Parallel KeYs

(All dirnensious in rnillirneter)

Table 3.4 Rectangular parallel Keys

(all dimensions in rnillirnetres)

Shaft Kcy Keyrvay

Nominal Sectionrvidth x

thicl<ness(bxh)

width (b) Depth Radius

dinmetcr (d) Tolerance for class of lit Shrrft (tr) Hub (t,)

Over To free Normal Close Nom Tol Nom Tol Max Min

Noru Shnft/lr0\

Hub/D10\

Shaft(H 9)

Hub(JS 9)

Shaft ondIIub (P 9)

30

J8

{{

50

58

30

J8

{{

50

58

8x7

l0x8

8

t0

+ 0.0J6

{t

+0,098

+0.0{0

0

-0.0t6

+0.018

-0.018

-0.015

-0.0s1

.l

5

5

5.5

6

7

9.7

0

J,3

J.J

J.3

3,8

.t.3

{.{

+{t.2

0

0.25

0.,10

0.,10

0.{0

0.40

0.10

0.16

0.t5

0.25

0.25

0.25

0.25

l2 x8

l{ x,

16xl0

t8 x l1

t2

l{

16

l8

+ 0.0{3

0

+t).120

{{).050

0

,0.0JJ

+0.0215

-0.til5

-0.018

-0,061

Rtnge of

lengths (l)

Thickness (lt)I,Wdth (b)

Tol (h e)Tol (h 9)

0

-0.090

7

8

8

9

10

11

,6L

b.* -&-=*1 oPr**

Table 3.5. Key Ways For RectangulerlKeys

(All diarnensions in millirnetres) Fig 3.2

Shaft Key Keyway

Nomral Diameter (d) Scctlon

width xThickness

Width (tr)

Shall and

Hub

Depth Radiu

(3)

Shaft (t I ) Hub (ta)

C)ver To (Bxh) Nom Tol

(DXl0)

Nom Tol Norn Tol Mu Min

22

3(,r

l3

l-l

-i0

85

3(l

38

.t-t

-i0

58

-i(i

8X7

t0x8

l2x8

t4x9

l6xr0

lSxl I

IIr)

+0.098

+0.040

4

5

5

!.5

6

1

tt.2

0

2.{

J1

?.rr

2.t)

3.;+

3.1

+Q.2

()

0.25

0..10

0..10

0.:t0

i).1{)

0.40

0. l6

0.2s

[r.25

0.25

0.?5

0.25

6

I

+().12()

+0.0i

Secllon ot dr.r andol l.fro, ln hub

Fis 3.3

Table 3.6 Rectangular TAP& Keys

(all dirnensious in rnillirnetres)

Srclbo ol do.P ndof trytoy in hub

Fis 3.4

;t-tr:=-L-,im{otot \ fatm B

Eosic laprt I to lO0

width (b) 7'ltickness (h) Chan{er (s)Range of

lengths (l)

Gib

head

(ltt)

Radius

(t)

NoEm I Tol (h 9) Nom Tol (he) Min Max From To Nom Nont

aU

t00

-0.0367

8

8

9

10

11

0

-0 090

0.2s0.40

0.400400.400.40

0.400.60

0.600.600.600.60

18

22

28

36

45

50

90110

140

160

180

200

11

t212

14

16

18

1.5

i.51.5

1.5

3.2

3.2

t2t4t618

0

- 0.043

Sosic lopd I in lC0

ot

e\dl .',i Square keys a-+-r^ll J#tdl )l."i al''rr.,Yl .Jr JllJl Jtr1il"l;ir-U

.6.usYl d,UH u's *l:Jl :rly 6^

^+JL!;J*.i.4 Llt.."^ .-r_;J-t^Jl C3,S; 1*.:rr ar:.i.,,I Flat keyS aJ,,*ll ;gldl

jLJlYt ua-+,rj srnall hub diarneter i+ir- lt3 elt D),-a alr{1 *l ijL- Lrg t^S

d;';.o efu .5-+,rtlt ji*-e d,- 4-.,ti +-Jp leL+: ry^[Jl r-iJ 1i o u# .1*t- .,LSa .,,ILr.,

Woodruff ;J,^i J31L-i LlUl orA c+ r+ a.rli dthl, ( ;, ^'iiJl Jl e$ u^ Jll.JI4-It ,.Jlt^ll elylYl fj' +t.,'- .:'i:l d,E thsi-Yl .-,r'- ;3,rLJl gr tsil .r";3 key

3r,Jl :+t 4;6 i;^al e:liJ lj+=6 ajJl hn crLS l5!r Torque to be transrnitted T

. 'tt ,1^.o': ,Je d"..--ll 4+JEJI c.rl e.tijl

dsiilll rtjFJtl e-re +t^ (\ )

Torque T - 7tr624 HP

n

6302s HP't-n

where HP: systeln horsepower

11 : r.p.ln

63025, 7 1620 constarlts

Actuai force

kg. crn

lb.rn

torclue

The force acting on the key JJ+tilI "1"

6j3"ll o3iJl til,". (Y)

P:1=R Raduis of shaft

L"S ;3',1-=Jl ,Jlt *Jr fU,ij! bt_t Hl Ltr-ni53 ,.Jilsrl d,HS_hJl dr# Cs, p i3ill ora3

f-Jlt{ e,*J. Jo

shear area A (cm2, in2; ,.;lr LL,"*

A:txLt - JltJl ,,t ^,., key thickness

L: minirnnm required key length

a)O)

t

Detennine minimurn key length J3'!l.i.JI ,.lsL +t^ (r

Ss: Shear stress

..PP"'- A txL

tlre lerrgth of the key Z = ,a P

" Srxl

Ss:8000 P.S.lg ,:,,,rtj.o u-I :lasl rqor Ss : 550 kglcm2

,"Jrt O15 lil3 j4+ldl C-J+il e-L-l 4qtJl {4:!f_l t o;r! Ul..-l J-at*^ ,.-r;j,,r ,',.r*

JILi c.1.* Y+ Jll.= y r:c fq d.:ll i;;t ;r,,.- d:f cl" J,Sl ,.J-J-t^Jl -;3,rL-Jl

. J*l-f

Design Application +sLie JJitA Jt,,rilJ JJ.t lr 'l!! +L*. crlp s**a,4j dua

Shaft selection U*tc Jt#it

L-* JJ+ J 7.50 HPIAJITL 6J$ Ji:J,,Jc -,;rtiill ,-,r cll r3,oUl jJ.I ,,,.,,-i

550 kg/crn' ,o .r^'Jl cJr*J a+ 69"..*Jr ",-!ill

.rta.=r r,,l uh-p rrl 1000 r.p.m

Solution

rljEJ)l p;c

71620 HP 61620 x 7 5torque I - , =-1000 =537.15 Kg.Crn

7' : 537.75 kg. crn

I anzs HP I|

-

lb.in ILNJ.ti;! L'-lill .rla;l e*L-l & UJlil r3*Ul _,/"!,*,r,,,-r

7-st "s

,"ri '

,. ^ l--) -

LJ4.Ell :rC i I

D,=167n .s,

S" = 550 kg / cmz

167

fiD'

l16T.l-i/" s

D:1.707 Crn (0'672") .g:,L:ll r-5*LJl :J.E

7.5 F{P iJ$J 550 r.p.m olsl :r,l r3,o\*Jl },! +f,.,,=l i+L'Jl ilrtJl s-5rrt^I pr:'r"'l 1i1

Imperical Fonnule e+L,Jl il.rtJl ,lrrl"'Lr rj,alJl fl5 +t"'=

= i/i091 = 1.03"

D:l.03 x2.54 :2.615 ctn

:yxL--ll :1" OY l:Li-r., i,.]l ,lJr6ll d E3J':L J#l dl)'"LJl dS -lStYl d ril JEJ

_1- +J +4 + il r-=g ilu q-q$l ;J-".cYl djLuL cJ;++ r:+-l^ * a-a3a 1'03"

(f Yo,l x \,'1): L-3a 'l "1 c+ 3'i':'ooYl ct-tii" jjr+

Key selection JJrBI *$il

j .5 F{P t-,o;l::L;-rt! ,iji: *lc l;r\! square key e} J}LiJ ':"-UaJl t}tliJl "'"'11

08.00.."JJj-r:r!-102.00.."jJ,!JJ^Lo0#100r.p'rnia;r,l

Torque =63025 HP

-63025 x75 = 4726 lb.in

a

The force P = R 1.0"

J-a !+Li^]l C+/l JlEll ,Jl ' ri 2" ";lrL:-rLlJ Jlg $l:, JiilJJl cj:l:+,lrri''rLr

0.50" square key

Square keY W:0.5" , H :0'5"

if the allowable shear stress

nT 4726

t : 8000 p.s.i

P P 4726:-L- A Lxt Lx0.5

100

= 4726lb.

= 8000

4 tzo

8000 x 0.5: tr.25"

minimutn lengirt

65

l-, -

= ./16 x )-j /. t) =rJ4.g14

1 nx550

MttRlc (MttU,!lfTrRS)

6

B

10

t2

17

22

l(.)

IrJ

44

50

l2loIri

5{}

" - "G2 -*2WHE RI T2

SQUARE A}.ID RECTANGULAR KEYSDWDW

60-70 16 I 80-90 I 2070-80 l8 I 90-100 I 24

L_______j

r (MNl = 2w

PRATT and WHITNEY KEYS

L(Min)=2W

,/-/. LALJILt<"y r,. ,1.1

Hr

.J,8

5.1

l.i)4.tr

SPECIFICATION (SQU & REC.)

- 6 SQUARE KEY 40 LG.

- 6 SQUARE TAPERED KEY 40 LG.

. 4 X 6 FLAT KEY 25 LG.

- 4X 6 FLAT TAPERED KEY, 25LG.

SPECIFICATION (GIB - HEAD)

- 9 SQUARE GIB - HEAD KEY 50 LG.

SPECIFICATION (WOODRUFF KEY)

- NO. 205 WOODRUFF KEY

-iii, *'.,'ir, ri"t, i loi,,rnirral rJt|l -., 2ljlsitc 64, K,.y

-lt-xtt i t I ( r l-)

-._-t_l__!-__xt!7 I r5 | t, | .tu

^r.17 I tt I t, I .18

^ 15., I l.) I ,'l I {,.1

. t)7 r., i :t i .l.u

^ls') l5 lo.. I o.rliL

^ l,,r i ,, .1 " 1 t.,,

a r',.r I Is ] r,.r i t,ta l{,r I rr i ,,, /.6i-lrl j l.j : ',1 (,.1

^ rr.r I L5 i /,., . 7.tt

li.ll.-'i li ')7 I ')lx 15 -l I L5 Il l lt).(J

, -,rr,, | -'o r lr..' ; ll.').-,, i ri 1 ",| ',,. .,, r I I ' . ,l rr,',

---t,-I r.,,,

I

Key I -_No-IA

--l--l0.l I l .r,iur I ,.,]u5 2 .lqo+ I .r .r+os l r,

I

Itltr I I ,5o5 I '1.{l

5{rt, I rrr5a7 I 4.oal, I +rr

I607 | ''i61)r] I 4.it6()() I -1.rJ

t\t)7 i ,, .

1'',', Ll":.1

-r'I

I

I

I

I

l

l

l

I

I

i

WOODRUFF KEYS

r;rp!r{ r:4a -\ w (FLAr)-1 kl- , t

,; "_\_4J )+ c

ffilu,f.* F-- L---r -l ShaftL) i arnc ie r

Square and Flat gib-head keys

tvlITRIC (,\llLtlMETERS)

12-1416-22

34-35

*t) _=---__llti'lt_

wtlct)--__._-=r.-+l.l 2.qSj3.24.ts 3.2 6.4 5

6.4 5 I 6.4

8 6.4 l0 8

r0 6.4 11 10

nr016lit6 ll 20 16

2{l ll 22 ', 20

3.24.8b_4

B

10

1.1

r620

METRIC (MILLI,\ITI T

s.r..,lll!" _ I

}: CDfI

--l

3.2 6..{ 5.4 I I

4.8 r0 t s.+ |

6.4 ll 8.b 8 t, {814rolol

10 tB 1.2 12 |

1l 21 ts 16 I

lb :7 l,) 20 {20 .r;i ll 1_l

--a

4,5

'6.4

8

1t13

ll,

J(:)+t

46-58UJ_7072-82

2.4 3.62.1 3.6.1 6

464 6.6 8.46 8.4576.4 10

7 'B

5.4 l06.4 ]0

81210 14

t2 2016 24

t.a2.4

I

4

6

5

6.4

56,4

6.4

5

8

r012

16

1)

35

3850s0

75

75

2022

21262B

3032

34

, ,'-1=\ r i

t / i-]-tArI NUi!l--;J

Pratl and Whitney keys.

15

1{.1'l

1a

ai

-\.' +.1 -'l

'. Il],a

1{,"1j^

'{-

iJ''-l,1)

).{

::

i ElNdt -)

I ]E{3-f,=-{oji5

\ --l to-.^

6a

trXrt!)x

clo6l

5j4

- t6cEoo'l)T6-o-4

-571

3dJ

'tr

iNo ffiffi;Frh?ffi;

i*PIH;

x-o

7:T3 \]i {_) .l--'r )

X

<f, roxx<f rJ)

(o r\xx(oo

co6xxON

o)oxx\i' (o

r(!

xxcoorN

b:'j*1\9 -;tr

C)()++roc.i co

rOioo++rr)co+

NC{ctd++r.r) ro

NNOO++LO

to to

N C.l

dd++lrr.

lg 'r'1, 1

io++o{\

eNoo++N<'oi oi

NNia++t =t-NN

NNoi++o{NcO

NNic>++{. o)-cD cO

t:l1+

.4 1:), ..,!

ri-:l11Er6

4rrt

lo' xl!'

e"\

-1, 1 Q" ;AEEOEEI

:!\

4-Jl--j

}.--.i

.1

u),

1 I-'t -o\

N cf) $ rr)

gE'. (O f-oi cr)

f\ tocoro(o

LO LO

ro (o r\rr) roo r-'ol

l(o (o5 t\ -ltq4 V l.-

(oLON@-@-\o)Nlf)

FT

roNf-(Of\Looi ro coFrr

N f\ Cr)

\Lo-qto@rrr(!

l'- OttQ s C)cr(orrF

co(oo)Frr

(oo)NrrN

I co ct)€l I- Nbt ..1 ro@coNC9U)

lJ)ooo,ttrlto

ro ro C)\f,lr)f-

+c]

$.r+o

\+o

Noi+o

,:7JI

COMPRESSION COIL

APPLIES ENERGY WHEN SOUEEZED

Iilt

SPRINGS

"EPft/I/G.S

A manufactuing company often uses a standard spring in

assernbling proclucts. Occasionally, however, a spring must be designed by

a drafter for a sPeciai job'

Figure tr7-1 shows several types of springs'

Study each part of the coil spring in figure 17-2'

Turn or coil is one cornplete tum about the center axis.

Total coils are the total ngrnber of ttuts or coils, starting from one poiut to

the exact Point on the next coil'

Active coils, in a compression spring , are usually the total coils rninus the

two end ooils. Figure (17 -2) has four active coils '

Free lensth is the overall lenght of the spring with l1o load ou it '

onacotnpressionspringitismeasrrredasnotedirrfigurelT.2.

On a1 exten;iol1 sprilg. the measttretnent is takel frorn the ilside of the

hooks or ends.

Loaded lenehh is tlie overall len$h of the spring with a given load

applied to it.

solid lenqth, ol1 a compression spring, is the length of the spring

whel it is completely compressed with each coil closed upon the next'

,6\ J+\\v/ u\-- ,rtRoa FLAT

APPLIES ENERGY IN A

CIRCULAR DIRECTIONWHEN LOAD IS PLACED TORSION COII.

APPLIES ENERGY WHEN

END IS MOVED IN ACIRCULAR DIRECTION

APPLIES ENERGY

WHEN LOAD IS PLACED

ABOVE OR BELOWAPPLIES ENERGY WHEN PULLED

Fig. 17-1 TYPes of sPrings 68

1 renr rur

EXTENSIC]N COIL

l, figure (17-2), if the wire diarneter is , 500 inch (12.7), the overall

solid ienglh is 1,5 inches (38)

r#ire diameter is the diarnter of the wire used to rnake the spring'

outside dimater (c D) is the diarneter of the outside of the coil spriug '

Inside diameter (I.D ) is the diarnter of the inside of the coil sprirrg'

Mean diameter is the theoretical diarneter of the spring tneasttred to the

center of the wire diarnter. This diameter is used to draw the spring' To

find the mean diamter , subtract the wire diarneter frorn the outside

diameter.

Direction of a sprilg describes whether the spring is wognd left - hand or

riglrt-hand. Figrrre l7 -2 is a left - hand spring'

.]TANDARDDRATFTINGPROCEDURES

some computres will not pay a drafter to draw a collventional

represelltation of a spring figlre 77 -6 because of the tirne and cost

invloved

A short cut tnethod to draw a spring is shown in figUre 17'7 This

represelts the sprilg in figgre 17-6 but takes less time to draw' Even itt

semi colventional sprilg i1 figure 17-6 but takes less tirne to draw Eveu

in semi conventional spring represeutatious ( schernaties ) all dimensions

and notes tnust be added.

Windins Ditection

Springs can be wound left -hand

or right - hand . figrrre 17-3 if the coil

6{)

Perspective View Section Symbol Specification

BP

e@/--\

ffiffi=sffietrtrfffiffiEffiffirl,HffiH

@@

\tr/

@@@

vwL t)

Compressionspring

Tension spring

oo(ooE

withrectangularcross-section

with roundcross-section

G'oco()

single cup

multiple cup

untensioned

tensioned(with case)

(E

without eyeswithout cgllar

with eyeswith collar

filoJ

#

E'Ico.otr"9a,an0)Lo"Eoo

E')

L

a(E

CLa

r\-\

lltlL+t1l1I

_\/

-+ -.-

ONE TUBNWI RE DIA

TE ENDS OF SPRINGF REE LENGTH

Fig. 17-Z Coil spring, detail drawing

Fiq. 17'3 Determiningdirection of winding

END

{LIMITED, POORSTABILITY)

OUND OPEN END

Fig, 17-4 ComPression sPring ends (IMPROVED STAEILITY)

RIGHT-HAND COILS

rflBfln

N1R.H.

CLOSED END{t\,4ORE STABILITY)

GROUND CLOSED END.(MAXIMUM STABILITY)

W@ W(Fig. 17.6 conventionar spring drawing Fig.'r7-7 schematic spring drawing

LEFT-HAND COILS

7L

winding direction is not called out

on a drawing . it will be rnanufactnred

with a right - hand winding .

Compression Spring trnds :

Compression springs have either open ends, closed ends, ground

open ends, or ground closed ends, figure 17-4. Note that the ground

springs are made from plain open or closed springs. The plain open -

ended spring is the most econornical to manufacture, though its use is

lirnited. Springs with ground closed ends are the most stable and can be

grourd back for even more support.

DRAWING A COMPRBSSION SPRING :

The following

compression spring .

tused

Overall length 100mm

drawing procedure call be used to lay out a

As an example , a spring with these specification is

2 x total coils + 1

2 x total coils - 1

2 xtotal coils -l

L.H. winding were size 6 run

8 Total coils (6 active) 1 ll2 O.D. Plain open ends 1" I.D.

Step 1. Lightly lay out the overall length ald mean diarneter of the spring.

Divide the length into even spaces. The numer of even spaces will

depend on the total number of coils and the type of end required

For plain open ends

For plain closed ends .

For ground open ends.

For ground closed ends. 2 x total coils -1

lz

ttrre spring in this example therefore has 17 even spaces

(2X8:16+1:17).Step 2 " Lightly draw circles at the top and bottorn of each space to

represent a cross section of,each coil "

ffSp_!-. Lightly draw the coil winding direction.

Slgp-!. Draw the style of end required . A plain end is shown here see

firgure 5-4 for other end styles.

Slgg-J Using correct line weight . finish the drawing. Add all dimerision

and required notations.

EXTENSION SPR.ING ENDS

I'-igure I7-5 ilJustrates a few of the lnany types of ends use,j cn

ex.tensiorr springs . A loop completes a tum on itself while a hook is open .

Ari extension spring has tight windings aud is usually wourd with right -

hand windings.

Tlre spring in this example therefore has I7 even spaces (.2 x 8 : 16)+ 1 :17) .

Slgtlightly draw cicles at the top and bottom of each space to represent

a ci:oss section of each coil"

f,4eg_j . Lightly draw the coil winding direction "

ffigpA I)raw tire style of end required. A plain open end is sirowri irere.

See frgure 5-4 fbr other end styles .

o9_d.E-,1_ 51 [Jsing correct line weight, finish the drawing add di Elirn:,:s,io*s

and r*quired irotations.

D6''ArWING AN EXTENSION SPR.ING :

-/aI)

On a separate sheet of paper, draw a ftlll loop over center extensiop

spring following the steps given . Specifications :

Cverail lengttrr approx. 4.75" (120 rnm)

FulX loop . over center

Wire size 188 (5)

R.H winding (standard )

1 .62s" (41) O.D.

Extension soring stvle

Step I : Make a rough sketch "

Step 2 : Draw the end view (O D / I.D) and each end loop at the required

length .

Step 3 : Very careftilly measure off the wire size springs and, with a vicle

ternplate . lightly draw the approximate number of coils. Startiug

at the top frorn the right end. project to the left 112 wire size.

Step 4 Line up the drafting machine or triangle on the edge of the wire

diarnters (at X and Y) and lock on this angle Lightly project up

frorn wire diameters.

Step 5 : Draw the wire diarneters hightly . From the wire size (A), project

straight down. Adjust the end loop in . Note : Study the right end

and be slrre you urderstand how it is drawn (numbers 1,2,3 and

4).

STANDARD DRAFTING PROCEDURES :

A short cnt method to draw a spring is shown in figrre 17-7. This

represeuts the spring in figure 17-6 ,but takes less tirne to draw . Even in

7L

o 1)

DRhWING ,4, COMPRflSSION SFIttrF,tG :

,, ,jr

-- MEAi.,J

DIA

i lQ +,saOVEB.qLL LENGTH _---

LEFT.HAND WINDING

RIGHT.HAND WINDiNG

NOTE END

_t,"-_t

_'.1

l"WIRE OIA

OPEN COIL HELICAL SPRING (2 COILS)

At-L OII',IENSIONS IN MILLIMETRES

7')

NOTE TAGENT POINTS

)-)-. l-

(-l-\- i-+\' ,,

\,._r_t_L

t]4+-/ \+/

DRAWING AN EXTENSION SPRING :

.TOP HALF OVEH I/2 WIRE SIZE

SHORT LOOP.OVER CENTER

LONG HOOK-OVER CENTER

I

CENTER

i: l.r t. L L,IOP - P,T S I L) t

V HOOK.OVER CENT[ T,I

Fig. i7-5 Types of endg on extension springs

-,/---PHS.NTOM LTNES

FULL LOOP-OVER

SECTION VIEW OF A LARGE SPRING

Fig. 17-9 Sectioning smali and large springs

semioonventional spring representations ( scheinatics) all dimensiorrs and

notes must be added.

Another method of drawing loug springs rapiclly is shown in fiplre

n 7-E.

Figure i7-9 shows a section vieUof a srnall spring (top) and a large spring

(bottorn) . tsoth are right-hand springs but, since the back halfofthe

springs are showr, they appear left-hand in section '

PDT=P.R------I

The torsioual shearing stress due to the

Torque T

I.t".n

'l- rl.rL --

J

in torsion

L_

D,iP.- x -22 =nDu32Tl 14

32

8P.DL -

---------:-nd'

the direct shearing stress

PP4P=-"P A frs) ndl

-- LtA+

The rnaxirnuln combined shear stress is

- -*

)- 4tl- t , tp

therefore

8PD 4pr_--Lr- nd'' nd2

Dif C:1 : spring indexcl

,F

hlire in Sh-r";r

tr :d-rJl *!sYl u'-ill rk-)l

[}I

Frc-e- t"* Jt"1*^

'/ -.,

rD)80.D oo \a)

-:-

--

{,r- nd3 ' n.C X d2

8p D 4P.Dnd' n'C x d3

+ -8p.D [,*-Llr,=-iT L'-2Cj

work: ]'r'aL

r.t

lsis\l'.-!q"

EFFect I uP C$rtulure

crij i.r.:.i-oji d'L.sTiJ static ioauing ;.-d#cl"'ii i'l"-)'i ::; # n3+t i'j;L*}i ';"r-r

Jrt.r:;il1 r-rli cjl..^=)tl # Y LpSIJ low cyclic loacling a;.i*'l^11 crl;-e:ll r:c

.urJl

coefficient of concentration factor & ts;i-'iJl "'?:l

illsJl "igl r.:3""' ll':\a3Yl

.91=Jl *Lil rJ^ a.ii cJS &lr ut' .r-rr:rl h-rUbl d sE CIJ

C+ &lJ &l-- (# J^LJI l-r-r3 curvature at tlte inside of each coil'

(s*,tJJyl :La;yl !# 9-)..r.] d.rlr-:.3 d,-,",*l &L-- 3aj Wahl COrreCtiCln faCtOr

: i,r3Yl ihl-^lt+ aio y.'"r Ll^tJl lraj T tldX

4C-l 0.(rl5r(--+---4C _4 C

XZ c* e'"*'iji qjal3 ;i'o"t-r-c ai'L;l^ui t*' O-tii -it;,o,i :''4:"-;i a'l;i** e+; d:-j3

BPD 1,/

stress fbr cyclic loa-Cs r: =;, ' lt

jJrLo.o ;-a rg:\in-,| g5o.:-9 Deflectiorl fii*.IJl .;L$ 4-';Ell L.'L*YI 3'*'Jt aJ'lL'*

. a.iLLll

,i"= .1$1. rlr'; 3r*.3 iiiitiailly UrlOaded qlr'\l .o-{ ih=.Sh i-}*'=' -,hP c'L' tr^;'-}i l)}-i

j;^fi, .-i3-.9'$l c.rF J^sjf ,$U .:" L'*t; exl,emal tblce cs+-tLi'

q* cj-l+^ll &;itt a.l-:L*.o *r:S.i-l 'i.he s':::':i'tr;, de;Lr:';i; 0

Ffl--rld[ /l i

}dtIL---s(::

el3jJl a-r di\ -JJ q -itl**,.gLrtll 's[-- gtr cD-'Yl $lc el3ilYl fJt JJ"E '''-i L.rl.9

3'-{ 4{ dl-.iiYl i-il-t o s3- illJl o.:a ,rr3 twist

1

Strairi erlergy: i "f '0)-. i.l-1.+^ll r-F,jlll sgjt ,5 dl*iiYl iStl l'*'t3

V/ork: strain energy

nl1p.O = a.T.e))

P'D(T'1l) P'D P'D ('

p.6=T.0: Z [Cl,l = , 2 G]

/o\' (.6_o[;] crwhere T:torsional shear stress t Mpo' Ky'Crn2 ' Psi

T r G'ei = ; - -= T : Torque N.m . Kg. Crn. ,b. inch

J - Polar monent of Inertia : #.0'

r - radius of the wire

G : Modulus of rigidity Mp" , Kg/Cr# ' Psi

0 : ,\rgle of twist in radians on a length L slopeaugle

fi*yle oP -Thisb

F.(5('=-E.ATCr:--t..,

- OJ

T:P.92

d.rl.--.rr '\ . ;* dsl O-r-iHl U:rl A'+i"'- iJrtJl ora3

deg

oll =-€

Tc;--Y

T'()".'0=t.G

l; .(D--L,-

A.6(G..1'r--I-

r

L:n.D.u

0

5=Pfql' (-\z) G.I-d"32

8. P"D' .n Pa =ffis K : springrate : stiffness =

6

G .d4stiffiress K: g O, ,,

i1 : iiiiiriiJer of ticiive c';ils

G .d4h-It - 8 D'K.

eri(g rnR[.J /?)

Anelical coit:rpressicn sprirrg has the followilg flimelsigris :

Spi:itli dia:neter Ii : 40 rnm

y.rli'i; ..iiameler C: 5 tn-tn

act,vr, coils tt : iA.Sg". riL'lg gi,i;;i ipj[i';

i-t : li.' .3;," , 0'3 ir'ipa

E:2A7 x. 103 iVlpa

I pa : N/hn2

I Mpa: N / rnrn2 = 0.1 kg/ Pr# = 10 kg/c#

and a maximum load Pn',n*: 66 kg

what are the basic torrsional stress t and the cornbined stressos rr and tz for

static and cYclic loading ?

v.ihat rlistalce cor:s the sprilg deflect to resist the 66 kg lc'ad ?

wliat is the, sPrirrg rate ?

The soltrtion

]I

I ) Ilasic torsional stress, = 8 P?

7r d'

8PD. =;;- P:66 kg D:40 nrrn d: 5 rnm

8,66x40r - .-ri-- ^'- = 54kg Imrn2

n x (5)3

2) Static loading

'l-he rrraxirnum cornbined shear stress

sPD[ tl!: __ll+_lrd'L 2CJ

D .i0r ------Bci 5

t rlt = 541 I -r l= 54 [t + O.OOZS]I z x s.j

r = -i7.175 kg / lnpr2

3) Cyclic loading

t; l'. t C==8

i+ c-r o.6rsl1.,=l + I

p4 ('-4 c l[ +^8-r o6rs-l liz-l 0.6lsl

h*=r.- i-l--------l'*-l.ix8-4 8 )*32-q' 8 J

: t. 107 + 0.076875 : l.tg3SZS

tz..'5,+ x i.i33875:63.93 kghun2

8PD3n4) Detlction 6:

C dr

^ 8PD3n ,a.h-- P:66 Kg- G.d.r D:40 mrn

^ Ex66x(40)3x85-0.8 x ro;lz- n: 8 coils

d:5 ulmG: 0.8 x 106 kglcnf

E: 54.07 run

5)Stiffiiess K=+:g =L222 kg/rnm 1'2

6s4

i..,S..,"QrJt_l d..,S-6U-Yl dt^sYl ,''-'i LJIJI ++;jIJl d:.,!Ul e+^.-l

Helical springs rmder static and dynarnic loads

Jf^iJl+ Jir ll qgt+Jl JEoYI .,-i Uril lil

J^-+ eJsl Y L,1l"gjll $c ij+=Yl i[tll ,",.r- .-X!-,6i.-l JJ]J.,* d.illk] d3-l- .t;*: .5q

csUl .,s .tU .'^rYl Jl ;jljSll ?L*s $c Jt=l:cYl "t' .i-F Y"e j^= g;l

Torque T : P.R

.eljl ciLil JJ.. d drJ-i-dl L,*ill rJ3 ql^*lf9 P ;1S lii

cs+Jlill-, (JliltlJ !--'j.lt :Jiill ' i' I : dUll -,;J*{ ei; .er R

L:2 n R. Na

Working deflection

.SLJI ,'tt 't 1it,.ijl sJj.tll *A L e++

6W *J;,;u !ti...b.ll *--sr 6:q PW - csUt +1^O Ot ;S+ qlr-ti J" .r*dl 6t U-;-ial lrl

J'^=.ei+ dt^s. J+.e -:-r: .lL$ Fl d+ll a,, d-,rll rn Solid defleCtiOn dJeJl 'LLL;jYl

d.h"^Jl gti:;Ylj

6s: hf- hs

R +rR.R mean coil radius ?

L

L. P. R2E: o,

5. -6*o..

_R

P J^=ll Jh$il rr e,Ull .rU! uo, -tti"5Yl : 6

Jlr-ii:ll -tt: r'il

,5Ul ,s**,\t g'li:;Yl :liS

Class allowance rc ci-rU,Ll 1.r,,i-oi!l r-1.o.,* ,"11'-,,

.:lj*+Lijl ?L^,J 20%,.Jl-r LJi( I r-!"[,fill 7- 1.,n"'','1i'''i r 1-,'

j.r_*Jl -tti*:j)l c &+ J5 "g$t Li*'-i", r-iJd zY. ."lh!,! ahJl ;&"" a.ili:,io.i1l ".Io pl.$:i.,l;',

e+-J C 4il e;l3 L,:t-9 (",Ul .rf' i:,1.:lo:F.sUr ..5]c ,ll-.fr iir\t} i 6S *^;Yt IL"*:iYl -rq:

. drljljln)l Jl:il d.,=:i -)!a.ill 31 6Lr. ll JJtl i *L.st U .r,Ji :;"rJi ;r;g]aE C+'4 & li'q

qrsiU sSlFAt gutltg c-.'.a^tt.=ht3*1$t

working and solid deflection of helical sprirg

.?

Shear Stress .rill lk:l

r_F jlc ,,, .r.'^i .r-! shear stress due to torsion cl;ElYl rJ^ Artjll ,-:!ill .rta5l

*l_5ri.ll p;J

15 PRTG)=

"r1t

"siL='l.I i,

",r11-r dlli-. .,jYl Ji^t-^lt rLa-Yl l:-l

4C - i 0.615+-4C-4 0

5 .r" .JE Y Spring index C:

Gr3jJsll dl*ll L*3JL dllrJ

(gbtt .-rh) sri+ll u,,t+ii"i ,r^^"t g;'Jl C

.,jl+ll dX- *,1A! D

.rql -,;.l,! , a' I fr

t6 PRI qC - I 0.615.l---l a- I,- r,lrl+C_q' C _j

Desigr for static load iiS.jE*Yt dL"-X er^.ill

S"Working stress = + : , Li.iSll rl-c-l

",lS lil

t{

Shear yield point u:iill ..,/ ge':;tt :l€il Srvp ''ts-

oull &t" : N

o < -2:!. ! :\ pYt nte 5 dp r.iir Y (C: spring index ) .rUr ob oi ";"i

lilTrDJ

D *sUl dt ;J"! ,-L'^ ,S+. i.-,,lj^ A-;r, r-y ,i R .r!lt ;J'l c'L'-: i'i;"^1 3

G'o' = = /V ilL.ill ci\-illl :lo ','lsl

64 K.R'

squared and gfould .J:t -t eJ-"" 6l=rll ,-11S l5l j

N1:Nn+l

N1.D: h, cr^.-^il 1E51Yl sl j

c,.,-^il -tti;iYl c.rLi.sUl drLiII o"lii 6t"*" ZY' tf;q J

6.=t.z+ , hr= h,,+6"-\f, /\

.5r+ll ; rt= P = D*$ cU ,' +*r' OA Y l-r:cN.,

l4

0-^ +J+..sul ,.,# cl-j-jYl j#L ,-,1_l

(JjJjJJt

2R

D

Exarnple (1)

It is requred to desigu a set of helial cornpression springs to suppord

a load P: 300 Kg/spring, the natural Freauancy of the system is not to

exceed 100 c.P.m.

The selected material is a high carben spring steel with a

Shear yiled poi,t :4600 kg/Crn2 aud a shear modulus G:7'7X105

KglCrnz If we use a clash allowauce of 20% with a factor clf safety N:

1.5

The spring index C: 5

iire ooii ellos wLil be squareci anci gottuo '

The Solutior: :

Hlical ComPressiorr SPring

Load: 300 -K-g / SPring

F,, 100 C p.tn.

.- tr

Spr':rlg 'rr-i':r '^-''3 : ' :::

i-{-lgli' O:'i'.'-i;! :;-j.'ir'ir' 'il'; -'l

'vifinl

Srt,l, : 4:00 ll-i;,if-)irl2

G:'l .7 X 105 i(1y' Cnr)

Factor c,f safetY N: 1'5

Clash aiiowance :Zirh

To find the coil Diamter 2R and the wire diameter D

matetialrnean coil Dianreter I)wire diarneter Du

il:nei d.iatnete;r llr

i-ength ol'vrira t-'

effeciive ]riun:be:c;j'':ttii i'i,'[ctal Ntlrnhe;'iii L]t'l;; ;di

Workirig clefiertion tsclicl rietJ.ectiori $,

solirl height 1"t,,

stiffiress k

pitch P

scof 4c'-l 0.6l5l.

- ----l-t +

--

|r-,rDtlqr- q " lY .,,, Slrctt sli'esr Yielcl Point

., _ , ,t.,t' =

___*_i -- /{ Faclor of sati:ry

, .,0-uo9 = .i ioo l.'-g / ,llr-' 15

g^/rx3CCi,l.(,-l 0615-]-Jt00= --r1rr-\ q . :;-_a" 7_]t:,i;'! {lictmeler I) = 1'3'otttttt

t,

To lind the mean coil Diarneter 2R

2R 2Rtt. 7_ 6-_D 13.6

2R : 2X40.8 rnm 2R: 81.6 mm

Outer Diarneter Do : 2 R + D

: 8l.6 +13.6

Inner Diarneter Di : 2R -D

81,6 -13.6

68.0 run

To calculate the stiflness of the spring K

Na -*

G.DO

64 k.R3

7.7x105x(136)r""'* 64x33.54x(4.08)l

Na: 18 coil

Total Number of coil N, : )r{u +l: 18+2 : 20 coils

The solid height h,: N1 .D

1r,:29 X 13 .6:21 .2 Cm

ET znF, -^,. ^-^ t ^^^ w: P : 3oo Kg

" \ w 60 Fn: 1oo c.p.rn.

k = (27r x F,, ), , 3oo

= 33.54Ks / Cnt60 981

The spring Rate: stiffiress:33.54 Kg/Cm

Deflection K:L d=+=+5 k 33.s4

To find the number of active coil of a spring Na

r6

Working Deflection : 6u, d Solid deflection

1,.- p1

E.:1 .2 +=r2d-1" - a1- k -l'=r)

6*: 1.2X 3oo = 10.73 Cm" 33.54

Free length : 11, * $*

:27 .2 + I0.73 :37 .9 Cn

Working deflection 6,.,

^ P i000tjo,: - = =8.94Kg/(',nt" k 3354

Length of wire L:2n R . Nt

:n(2R).Nt: zr ( 81.6 )20 mm

PitchP:P1 d'N.,

Y:13.6*107'3 o,r,18

:6u

The solid height ofa spring.

Working and solid deflection of a helical spring'

LEFT HAND

SPRING

CIRCULAR SECTION

TI'I

r)k:'6

=-=>=

=-=*-=l-

Frec Heighr

I

c)

lrro

9

a1

6t4p

2,t

o

Hclical compression spring withsquared and ground ends.

I-t

@

I

II

I

[-- R -l37

(fi) Sheets 1 and 2 (cAlsHAFT puMp) are two sheets ofqEqu*= showing parts of the camshaft pump *" gi""" .The body easting is bored to take ball races which recieve thecamshaft . This is retained in place by the eover plate .Ttre two ra''* are hercr in eontaet wittr the .* f*", by springs.The two inlet and ouilet varves are stainless steer balls .Draw an assembly drawing in first angle the following :_1- A sectional elevation taken parailel to the axis of ihe

camshalt .

2- Projeet a seetionar rre*.'ation to right of the erevation on thelltte throueht the eentre Iine of the first carn .

,_l@

lt

i

o@

a

=oc(

FTIJ)Jl&

I

bPZT5F

J

=i o

2t! ul

gE

atrJJoI!O

[;;I lrrJl-i---191.l-; IUeL-Jfl x

tosl;

{

_l"i

I

.t

-{I

_L__I

-.-1I--Ja

:('tfooU(L

--i.r---tE:

1...l r-fol I I

-t 'f-F-'?l.l LnlQl I

-llt

l

I-r.1

lei

-rtti.l'l_l

Irl-1

-t1l6t

_-l

85Jo-

uIU

I

3XIo€f{

}I

6I

g\o

o.o(ox3xofl

Ju.luJFot,

t

FrJu,t^lF6U

EeE

cIcoe6

I

f-*-G{--rl

'tffi.l

}[

35xo

ei\.].-nJLI]Jqi

L-J tlert

ISOMETRICS CAMsHAFT

9O

W.tdl*Welding is used in place of bolts, Screws, rivets and other types of

fasteners . It is also used to fabricate parts that were fonnerly cast or

forged. Welding is used to a considerable extent in the erection of frames

for structural steel buildings , ships and other large structures.

Fusion Welding

ln fusion welding a welding rod is rnelted and cornbiend with the

rnetal parts that are to be fastened together . The parts wiil be

peflnanenetly joined after cooling The process can be done using torches

or high electric power.

Types of Typical Joints

syrnbols

Figrre l4-2 indicates the syrnbols for each type of weld.

Figure 14-3 gives an exarnple of each type of weld .

TYPE OF WELDS

9L

(1)

IJACK OR

BACKING

WELD

(2)

IIILLET

WELD

(3)

i,LUG

wEi.t)

(4)

SQUAITE

WELD

(5)

VWELD

(6)

BEVEL

WELD

(7)

U

WELD

(8)

J

WELD

A N lt V [, ti

Types of welded joints

BACKING WELD A

Figure 14-3 gives an example of each type of weld

(1) BUTT JOINT USING A BACK OR (2) coRNER JOINT TISING A FILLET

(3) I-AP JOINT USING APLUG WELD USING A SQUARE

wmo||

-llll-x

HOLES ARE DRILLED HROUGH ONE

PIECE NEFORE WELDING Note : SPace betrveen Parts

(6) BUTT JOINT US-ryG A BEVEL WELD

V

(7) BUII'JOINT

VUSING A U WELD (S) BUTT JOINT USING A J WELD

Figure 14.3 Exarnples of welding

I(4) BUTT JOINT

notc : All vertical sides of tlllet svnrpols n.tttst be

drotvn on the left.

joints

(5) BUTT JOINT USING A V WELD

Placing welcl syrnblos

n rf. f . Wheu the weld symbol is placed below the refemce iine

figrrre 14-4 (b) the weld appears on the same side as the arrowhead '

Rule 2, When the weld syrnbol is placed above the refemce line

(A)' the weld appears on the oposite side of tlre arrowlread,

Rule 3 . Whel the weld symbol is placed above a1d below the

refemce line (A)aud (B), the weld appears on both sides of the arro1vhead'

A SOLID FLAG INDICATES

T.IHAT THIS WELD IS A FIELD

WELD.

A CIRCLE FMRE MEANS TO

WELD ALL AROLTND

POINTS TO WHERE WELD IS

TO BE IVLADE

FIGURE 14-4 Placing Standrad Welding Syrnbols'

Figrrre 14-5 shown exatnples of rules 1,2. attd 3 using a fillet weld syrnboi'

Usually rnaterial over . 125 inch (3) thick reqtrires a gToove ( sqrrale' V'

beveled,tJ'orJ)Usirrgtlrebasicweldsyrnbol.poillttlreanowhead

toward the part that has the groove" figure 14-6'

Refennce Line Netations

There are various notations plased on or aror-nd the refemce line '

Figure 14-7 lists a few of the rnore widely trsed staldard lotations ' Each

tells the welder exactly how trre drafter wants the part(s) welded , The

25,,x.375(6x10)sizenotationfileallstheweldisapproxiarnetly

25-x.3,T5andisweldedthewholelerrghoftlrepart.

I

I

MEANS.

I

I

MEANS..J=BULE 1

Fig. 14'5 Welding

ence line rules

MEAi'ls .

RULE 3(ele(.

* A field weld syrnbol indicates that the weld must be made at the worksite and not in the welding shop .

MEANS.. , MEANS .

THIS

MEANS MEANS.

zl.-\

v\- t/Y- -r-7rHrs... \ rHrs... \ rHrs._. ,/

MEANS .

TH IS

M EANS

confonning to a specific delineation.

Fig 14-6 Grooves in welding syrnbols

g,l

SIZE OF WE[,D INDICATES TYPE OF WEI-D

(5 x 10)

R.EFERENCE LiNE

fIG l4-7 Ref'erence Line - size uotations ' -- - . ^-,, I\tr;tglgrlvv Lrrrw - rr!!, rrvr4lrvrre THIS.. . N4EANS

In Figure l4-8. two rnore notations are added . The first means the

length of each weld . and the secottd lnealls the distarice ltorn cellter to

celeter of eacl1 weld or pitch pitch ref-ers to the distatlce, center to center

of each weld " The notatioll lnealls that the weld is to be 2 inches long

with a center-tocenter distauce of 4 inches. The notations would indicate

rnillirnetres if the rnetric systern is used '

SIZE OF W'ELD

LENGTH OF EACH WELD

Fig 14-8 Refemce line pitch trotatiott

R.rtrr",r.. WAalry

oP Passi't;4

Resistalce welding is the processfarl electrid cttn'etlt throtrgh a sport

wilere the parts are to be joiled. Syrnbols for resistance weldirlg are showtl

rri figtre 14-9 . ,.i

r.37 5

w

?9

\-o854 B(Jtt5 '-685

4 H(]LES

+60L 's-

-'rTL

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'l62a

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ROUNDS AND fiLL€TS R 3

g/ roazfl'

44.5

;=t--T,

I16a

o za '! 05-

t

R28

Figure;;;;;il;,shaft support.

y' "orr,,,oa

aou ,0,,

of a cast shaft support with a welded - steel

r520 I_r_d

WE LOING SYMEOLS

WELO I

I{ELO ALL.AROUND ON ONE PLANE

NEAR SIDE

NOT€: wELDING SYMEOL REFERS TO N€AR SIDE

WHEN THE FAR SIDE IS IDENTICAL

TO THE NEAR SIDE 'I"HE WELDING

SHOWN FOR THE NEAR SIDE SHALL

BE DUPLICATED ON TI{E FAR SIDE

WELDS 2 AND -s INVOLVES SYI\4METRY

ABOUT AXIS X.X.

Fig 1 1-3-2 Application of fillet welds

:.:

APPLIES TO WELDS 3

AND 4 WHICH ARE SYMMETRICAL

ABOUTAXEXY-Y

r12

FARISTDE

;";'" f fl _|49'_1,,

il":;,lr lT l?:::^| \ ll /r w€LD5

ARBOwT / lrl uaRROwAHHUw i r/ lll uaBEowsrDE ,I lll , srDF*ero:L-!l---i weros

NEAR SIOE

FAN SIDE

a

L l

fbr shaft support.

l*- a oo -*]Fig. 11-3-3 Welded'steel shaft support'

(A) Casted Elements (B)

,,Welded Elements

-tl

(c)

rdl

Comparison of Casted Elements with Welded Elements

6323 HOLES 1

6X3KEYWAY

L-,,,Fig. 11-6-A Swing bracket.

(D)

(D)

DIMENSIONS tN MTLI.IMETRES

StrengthDesign Rigidity Design

507o ol 37o otFull. Full. Full-

_. ll"r" strength sttength strengthThickness Weld Weld Wei-a(t) =0.75T =0.38T =0.2Sr

5

5

6

6

8'10

10tt

12

12

14

Fig. 11.3.f Rule-of-thumb filtet.weld sizes

where the strength of the welmatches the plate.

5

5

5

5

5

6

6

6

II8

B

10'10

5

5

5

5

8'10

l0'I I12

14

16

16

22252528

6I

10

11

12

14

t620222528323538

Fig. 11-6-C Connecting bracket"

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t'-"i Pie (1) Weided pedestal bearing

11, i,'.

GEARINGInvolutes

If a straight line roll without sliding upoll a circle fig I the locus of,

any point on the straight line is an involute of the circle .

To Draw the Involute - Apply a straight line drawn ou traciug -

paper over the base circle ( fig -2) : let Q be the generating point and A the

point of' contact, Usiug a pricker rnark the position of Q Then transfer the

pricker to A and allow the line to tnm about A until it overlaps the circle

slightly ( dotted line) cutting it at B , Transfer the pricker to B and revolve

the line until it is now tangemtial to the circle ( cliain line ) The point has

rlow rnoved to Q. Mark Q in the sarre way. Plot other points arrd join

them with a fiar cnrve .

Toothecl Gearing - Two plain wheels A and B fig 3 are in cotrtact

aud revolve abont parallel axes one transrnitting tnotion to tlie other by

friction at the rubbing surfaces to prevent slipping at P when power is to

be transirntted , grooves rnay be cut in the surfaces aud projecting strips

added between the grooves fonning the gear teeth showu in fig 4 . The

irnaginary circles in Fig 4 corresponding to A and B are called the pitch

circles of the gear wheels and P is called the pitch poin$the pitch of the

teeth is defined on the following page the height of a tooth above of the

teeth is defined on the fbllowing page. Tlie height of the tooth above the

pitcli surf-ace is called the addendurr$the depth below the dedendurn fig 5

Tre differerce between the addendum of a gear and tire dedendurn of its

rnating gear is termed the clearance.

The profiles of the teeth will be correct when the motion transrnitted

is tire sarle as that given by the plain w"heels in rolling coutact ; i,re.when

the angular velocity ratio is constant For this it is necessary and

rol:rlUIF-l

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Specifrcation S i rnple representation StarrdardRepresentation *-

Spur Gears

and

Helical Gears

II ITUtrTT Tf+++U lJ-| ,ti ,.-! ,.r,<.J-i-c riii"

B[+Kfr-

RttSptr Gear ,

Rack and

Helical with

Rack

€:+vz)

ruWN

Bevel Gears

Spiral Gears

,Z+>\ .F:FL/a'\ ffi\V/ LW

ffi'@Wonn

and

Wonn Gears

GEAR FORMS

,.'-\...

(

ffiI

GEA R FOR MS

scHEMATlc 103

l----L WORM REDUCTION GEAR

Fig . I w, 2W, 3W, 4W . Worm Reductan Gear Parts .The Parts of a worm reductri-

on gear are ginen . The worm wheel and Shaft rum in buehes set in the gear housing .

The warm runs in budhes set in the Casing and bearing support . A thrust race fits on

the shorter end of the worm shaft .The wornw heel has twelue teeth .The worn tooth

section is a rack section .

Draw : a - A sectional elevation tacken on the centerline of the worn with all the

parts assembied.

b - Project a Full Plan, From the elevation .

c - Project a Full end elevation which Shows the bearing Suppart .

Add : - a list of Parts and tilles .

I,l.-(r-I

I I40

rnl

lsa i

Q o.o

r-rB. r.Wo.oz /V"r

Dimcnsions in mm li enO ANGLE

DRG NO. I OF 4 t3 BEARING SUPT I STEEL

6 THRUST RACE I STEEL SET t2 BOLTS 22 M6x l.O xlS 69

5 rrvilM I STEEL ll END PLA'fi1 I STEE.L

4 BUSH A& B 2 BRONZE to END PLATE I STEEL

3 SHAFT & SCRE\! I MS 9 COVER PLATE I MS

2 WORMWHEEL I cl 8 BUSH BRONZE

CASING I ct 7 BUSH I BRONZE

ITEM NAME NO. REMARKS ITEM NAME NO. REMARKS

WORM RE DUCT ION G EAR PARTS104

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REFERENCES

l- Cecil Jensen / Jay Helsel :

Engineering Drawing and Design

Mc . Graw Hill Book Company

2- E.FRENCH and C}L{RLES J

Fturdamentals of Engineering Drawing .

3- T.A.Tholnas.

Teclurical Illustration

Mc Graw Hill Book Cornpany .

4- Firth and Vander Willigen

Engineering Drawing Teclinology

5- T.H.Hewitt

Adva,ced Geo'retricar a,d E,gr,eeri,g Drawi,g6- W.ABBOTT

Machine Drawing and DesiEr

Blackie & Son Limited, Glasgow.

7- Engineering and Mechanical Drawing

8- BOGOLYTIBOV rmd VOINOV

Engineering Drawing

MIR Publishers- Moscow.

9- Engineering Workshop Drawing

10- DEANE LENT

MACHINE DRAWING

LONTGMANS, GREEN & CO LTD.

11. R.K. DHAWAN,

Machine Drawing,

S. Chand & Company Ltd.

12. Bertoline, Wiebe, Miller and Nasman

Technical Graphics Comrnunication,

IRWIN GRAPHICS SERIES.

13. Giesecke, Mitchell/Spencer, Hill, Dygdon, Novak and lockhart

Technical Drawing, Tenth Edition

PRENTICE HALL.

14. K.R. HART,

Engineering Drawing.

The English Universities Press Ltd.

110

Figure 23 shows the dctails ol a rvornt antl ttheel gcar. Thc twO slrafts ate supl>orted

by thc ball bcarings al. the places indicated.Draw the following views oI tlie assembly I'ull sizel

(a) A front sectional elevation on the section planc XX'(b) An e nd sectional elevation on tlie section planc YY.Illclrlcn cdges rrecd not bc shown as dottcd lirrcs;!n ci,tlter vicw.

Esfiarrate anv nrissiuc dirnclrsions.En\d'allowed - 2 ho"urs.

GEARBOX CASE

2 -69

BEARING4 OFF

UIiA'EN,SITY COLL'GE

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lcn

tz 6() 60 t?

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35 AA6 35

WHEEL SHAFT

5C I rp lt3

UNIVERSITY OF LONDONFIRST ANGLE PACJECTIG!

RM & WHEEL GEAROIMENSIONS ARE IN MM

WORM AND SHAFT

Vievn(n)

Tlic,bearings are asscnrbled on the worm and shafti at positions A. The distance fronrthi: qentr'e oI the worm to the furthest face of the slceve is 44 mrn which is also the distancelrdm the centre ol the gearbox case to the inside lace olthe bearing housings. This willposition a bcaring llrrsh witlr the inside face of the bearing housing and, for correctasSembly, the. other bearing should be aspembled in a simillr position. Either side oIthe gearbox case may be chosen lor the shalt extension.

I ilttt ,l,)'lhc bearings qr6asscrnbletl i-rn the rvheel shalt at positions A and a similar case to that[or vie w (a)citn bc nradc [or positioning the wheel shaft bearirrgs in thcir housings.'[hatis, llte ccntre of thc 28 ninr dian.reter portion ol the wheel shaft is 2(, mrn frorn thefurthest I'ace ol'the integral collar. which coincides with the distance lrom the centreof:tlre gearbox case to the inside face oIthe berrring housings. Sinrilarlv, either side ofthe case ma1,bc chose n lor the shalt e,rtension.''f{e soltrtion on page 153 illustrates the bearings pictonally to help thE stucjent toco nr prehencl the role lu lfilled by a bearing. Howcver, the student is rerninded t ha t BS 308includes a conventional representation for bearings and it is this reprcsentation thatshould nornrally be used.

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Uzo (f)

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aul(,JL!LrlI=od

to=

Figure 25 shows thc dctails ol'a shaft brrrckcl.

f)raw lhc follorvirrg vicws of thc assenrbly, iull sjzc:(l) A frt>rrt ssctional clcvation on tlrc plane All.

"(b) Onc lralf of tlre plan vicw.: ,,lii[dcrr cdgcs..ccd nor be shown as dotted lines. In view (a), sh.rv, part of a 3&mm

. Ir-tftatt'.to.which is ftrstcrred a mirre gcar witfrlg spitabre gib head riey. The sccond

$rrrlshould be shorvn but thc shaft and key ntli bi iimitred.-In view (b) the gcars may be onritted conrptetely. I

, fime allowed -2 hours.

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V tew (u)

tn. uro.r.., which are in the form oI a split rube, are used to facilit^te assembly and

nlainteRance. I]y rc,noving thc cap un.j.b,asses the shaft and bevcl (or nritre) gear

asscrnt;ly carr [.c,l,rnr."'ir?o'r"iir,Jui-'witt.iritrvinq the shaft axially Thc othcr gcar

runs i. a lrusS *,hic5 i.,,r'r.-af "r..*bled in the gi;en views olth-eshaft bracket' Refer-

enbc to Ils 308 should b. rrracle for the conventlonal method of drarving bevel gears'

chosen to sccure the cnp to the lrratlct' fhis will

lrirlr..t (lue to the cap being retnovetl frequcntly

the line representing the edge oI the 10' sloptng

View (b)

A Jtud ittt<1 hcxagonal nut has bccll

I)rcvcnt <larnage to the thread in thc

f,rr,:taintcnancc, etc The length oflirce in plan vicvr, tcrminating at Qclevatiort be drarvn lightlY"

N---Y-

I

at r) will ..."t"ii"t" thit a portion of the grven end

EW

,ALsHAFrs AND xev ovrrreo)

--

-T

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.'\,'.r\\\\\\\\''.\

SECTION AB

UNNERSITY '..LLEGE

UNIVERSITY OF LONDON DATE I JUNE la!7 PAtr ONE8.s.. (ENG).

SHAFT BRACKETTHIRO ANGLE PROJECII'!TI

25

r':.,rrre 27 slrorvs dliril:i of a lrrillir:g jig. Tlre jig is uscd to lrolcl u special boii (not sliorvr:),lrrrinF a rrraclrining opcratioir in rvhich a square hcad is millcd on tltc end oIthe bolt,

''-l-irp conrporrcnt to [)c machirrccl is gripped in thc collet rvhich is firnrly hclti in the borly

,r!'rlrc.jig. 'l lrc borly is sccrrrctl to thc base by nrcans of llrc clrrrnping ring but is frce torolatc urlo onc oilour possilrlc positions in which it can be lockcd by a pin (not ihown)rvlriclr eirgagcs.in niatirrg holcs drilled in th.c body and.clamping ring.

.-flre loilowing vicws are rcquired fuil size:(lr), A scctionil Iront elcvation ol the assembled.c6mponents,(b) An outside plan view of the assembly with the cap removed.l.io dotted lines are requircd.,1.11 nrissing dimensions and snrall raclii are to b,e asscssed by the candidatc.

"A fr!rv lr:ading dimensions nray be inselted on the views drawn.Time allowed - 2j hours.

o:;+F

1{@f.T

L,J,L-,:_1:,-jtt

CLAMPII'IG RINGrTEu 4

CAP SCREWITEM 5 4I]EOUIREi)

MIO _ 6H

3,,;l

' ;:"-:';-:i !

COLLEl'ITEM 6

lo* 69 -''lt'-l.i

j::

Iil1+-

KEY! r'ir.r 7

I

Mt2 -6!-l

'l'L:'iriiFr llll'lGl]'l:M 0

+RA5tf PLATE

I] EM I

iliilrll

ld-(r)n !3l-

1.r-5

UNIVERSITY OF LONDON

AREIN MM \

/ ictr (tr)-[lrc

5pcci;rl bolt wlric]r is to bc nrirclrined is gripped in the collet by the'wcclgc'cfl'ccrol tirc collct's outcr sttr[ace r.vhcn forced dorvnrvards by the cap. Tirc collct is 1li.u.nt.,lIlorn rotrrting by the snrall key. Thc lour I0 mnr diameter holes in the bodi,, togetherrvith tlre singlc l0 nrnr diameter holc in the clanrping ring, are for locating the s'pecialbolt in each of the four positions necessary ior nrichlniirg its square heacl.-

View'(b)

Tlrc given plan view olthe base plate can be reproduced together with plan views of thecolle t, tgper ring, body and clamping ring. The cap screws do not proirude through tothe top surface oI the clamping ring and as a conscquence tapped holes are drain intheir [our positions.

SECTION XX

UNIVERSITY OF LONDONFRST AN6LE P'IOJECTSIMILLING JIG

The details olla uiliversal coupli[g are shown in t:igure 29.Two similar coupling plates

"nJ t"o similar forks r,r" r.iuirld to lorm the conlplete coupling. The two lorks are

[.f a't^fiif,r their ends bctweeri thc t\Yo couJ:ling plates, which are joitrted together by

rucatts o['lbur l2 ntnr diir bolts illl(l nuts.'Do

not drarv the sep{ratc pilrts as.l,o*",,, brt produc!, full size, the following views

of the assembled coupling:6i 4; outside elevatioi, showing the axis of the two $hBfts in a horizontal position,' I r,nd conforming to the gener;l position indicated',in lhe key diagram slrotvn in

' ihe ligurc.tUt a" eif elevarion. looking in tlrc dirccliorr intliiatcd by tlre arrow E."l'his vicw

I is to be placed to the right of view (a)'(c) A half sictional plan. The planc of thc section is 1o contain thc axis X - X of t he' '.

shafts. The view required ii that which app':ars above the axis of the shafts, and

''iS:to be placed under view (a). ' I

Insert two dimensions only in each view. Draw a title block 100 lnm by 60.mm inthe bottom right hand corner of your drawin-g.arrd_insert the title antl scale.

.No hiridenlart iincs are rcquiie<i in any oithe vicws. Cniy one boi! and nui is to be

shown, and this must beicorrectly projected in vi,:ws (a) and (b).

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A pictorral vicr, rvlrich has bccn partially sectioncd, has been included rvith tlre rcquircdsoiLrtion to irclp visrialise the arrangement of tne ;'rsernbled parts.

'

l'tt't 1rt)

The givcn vrovs oftlrc iirrk are arrangcd in sirnilar positions to those oithe leit hand forkin tlrc rcqtrircrl viervs of thc asscmblcr! c<lupling.

'I'lrc ril,lrr lrirntl [ork is lsr.:rnblc,l in a position disposed at 90o to the lcft hand fork;rnd borh clirrnpcd b,r'thc r..r'r'r coupling plates, For the construction of the curvc ofintcrscction bctwecn 'l'and 'rn', view (c) must first be drawn and a proccdure adoptcdirs lor itrtcrsccting cylindcrs.

I/icy.(b)"l'he solution rctquircs that a viov on arrow E is positioned to the righl of vierv (a) rvhrchis (hc casc for first anglc projection.

shadcd lines have bcen shown to indicite. thc junction of the housinlls for thc firrknivot pins a.n.cl. thc pareDt plittcs. Note that tirc arc across thc kcywal,p rept.esents theI'irrc of thc hitlrlcn fork.

li icy. (L')

Srrrrilirrly. rlrc positi.rr lbr vicw (c) <icnrandcd in tlre question rcsulrs rn first anglc pro-lcctirtt't. 'I ltc ctlnstrLrclion of thc curvc 'qr' is as for peipendicular intcrsccting cy-Jit)dcrs.

It cotrltJ bc argucd lhitt tlrc pivot pin ol t[e flrk is a parr that rlrlrving io,i.rentioni.::1,1,1,,:ir!:

is nor secrronecl but it is rnorc lrkely lhat rhe exanriners inrendJd tlre rappcdlrolq;l;gc shttwn irr scction. -,ii11.

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The det.ils ol'a clanrpi,g rlevicc,arc sho*n in first unglc projcction in Figure 32.

.The. locking piece labclled 'vce jarv' siicles with. its lorver horizonral lace in contacrwith the upper face 'ff' of the base plate and with irs other faces in contact with rhesurlaces'r.',.'s'and't'ofthe strap. The right hand end ofthis sriding piece is kept incontact.wirh rhe cam rever by means of ihe.spring, and engaqemen"tir

"ti.ioi-uyoperating the handle ofthe cim lever lhrough an rnd" ofripio*;n,,"r"iy S0 l.g**,thus inducing rhe vee-jaw ro move from right t-o iert ,.iJii rtl u'.I"-piri.li5'ilr.*.-r.1*,rs enecreo Dy returning the cam rever to its original po(ition, and ilrowing thd-siring toI;:oO,jnJ:"

jr* and canr tever in conracr Uyixe.rinia force on tfre fa,:L oiiie'iiop

___?9.,1o,91"y 1{r. separarc pails.as shown, bur produce.full size, and in lirsr angleprll:"Io_11!1," following views of the assembied "tarping

o"vrce:(p, A secrronar erevation' corrcsponding to the vcrticaiprane ER shown in the orarr' ys of rhe base pla re a nd-rh.i vee jui. rt. ii.* -rrq ;tffi i; fr;i ;;:"r"fi,;;Hffi ;' t,", illthc direction oflhe arrows.

ifl';fiffrd'[':i"'uff,'ii.:,;1"l,l;],Tf il,:"'direition

orthe arrow P rhis view

'(9).fl half end elcvation assee.n. 1vh.9n rooking in the dircction of the arrow H. The, " .;view is roshow rhat narr *rricn ries io iiie T"ii oi';,; *riii"r

""iii.r p;,;;; "' "":The vee- iaw is to be drawn in rhe ct,i,rp.ir p*irl""-*t i.i "n.r.rponds

wirh rhe dime,.siogr.s,givbn on the dni"wins.'f'lo,hiddcn part linds uic recuirid iu *ny of rhc views, arrd o,ry one of rhe 12 rnnrdiatnotci futcd bolts is tb be snowrr.{nserl lwo dimensions onlv in cach ofrhe.vicws (a) and (bl Drlw l title block 100 lnnrbr 60,mnr in rhe bor.om rigtrt ha,rd coi,i;;;;;;;i;;;)in* pup., irnd inscrr rhe rirle

;.Timelhllowr:d - lI hours.

H

BASE PLATE

SLOT Ftr UFFENsmN6 AND SO(InExso oF troP-

(44

t-.r-u?l--_-+-t 32 rEH zlt,lA , TPRKNG TENGIH {l:8ffiIri*i +(rab!6l6qbaL_-lgL:Kp|

c cou-3 or wne. SPRTNG

I

I

I19

THE ENDS OF THE SPRING mffi

'l'hc ckarrtincr's dctailed descriptiorr of the position and function ol each part in tlrerussenr bly assists consitlcrably"

View (o)

Altlrough this is thccrrablb tlrc usscnrbly

I

V ie*, (b)-lhe vce jarv is required in t[e clamping position, i.e', lvhen it is in.its most forward

rrirsition. Tlris rvill occtrr rvhen that part;ithecam face which is furthest from the canr

ccntrc is in corttitct rvitlr tlrc crr<.1 of tlic vee jaw. I\*ote that the cluaratlce hole in the Strap'

rvhcro thc bolt is ornitted, rvill sho\\, t\\'o conccntric circles -the inrrcr one be ing onc

ol Ilrc circlcs [or tlrc ta1;pcd lrolc irl tlte lrasc.

i'11',, { t )

-I'hc slopc olline'lnr'can be determined by transferring the length olthe hidcien detail

iinq rn vicw (b).

SECTION E F

first vicrv ilske(l for, it may hclp ilview (b) is drarvn first This willlrr br hrrilt up gr adrrally.

120

Figure.34 shows the details ofa non-return yalrc.I Draw full size, using first angle projection, thc lollowing views of the assernbled

valve in the closed position:(@) lAscctionalelcvation,theplaneofthcsectiontobealong,an<Iinthcdirectionof,.Ltll.b),.4 sec.tional plan projected from view A, the plane of ihe section to be along CC.' 'Hidden lines need not be shown, but include on ihe drawing a parts list, title uni ,ny

,othiriitandard data.

Ai$''embly instructions: ''.

'Thc valvc (itcrn 3) and.vulvc stop (itcnr 2) arc irsscrnblctl t<l tlrc body (ircnr l)at yi:j,^4,j::q*tivelv.'rhe glan<ttrush (irem 5) with rhe spindle (itcnr 6) purring thiougtrll,lr. l]tt.q to.the body atz. Afrcr rhe insertio, o[suirable packing, itre glani litem-a)l$ scrqwed into tlte gland bush and adjusted to l)re\ent anl,fluid lea*kage riu tt,. ipintllc.Time alloyed - 2j hours.

l{!

ALL tsliL-EI lrr.BE TAKEN A5 J

III,rl TO I

iIiIi

T-,l--

SECTION C C

-laulr.)l

PARI ONEb.s.. (ENG) I

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T--lcIJrTI J.ol

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4 uotes Q 14

SECTION D D

NON - RETIJRN VA'L\,/E

A- u24,5-6g

U-S.T, KUMA

@ sPrNDL.rj

l/i.1,,' ,,,,'['lrc scc(i<lrr plapc I]l] p16virle s l sirrrilar scction of tlte vertical cylindrical Portitln ol th,e

ltorf y to tltirt oisccli()tl ltllrrlt: I)D.'fhc portion of thc bodl'behind thc sccti()n pllrlrc llli*rl/irrcir,.lc outsidc viovs oI tlre gland bush, gland ahd spindle'

Itit:x'(lt)-[-irc lorvcr r>[ tlrc trvo givcn scctional views of thecarr bc rcproduccd, rvith its corrcct oricntatioll, forlorlrl thc component parts o[ a stufling box.

Viex, (,:)

botly is sectioned on Plane CC andvicw (b). Items 4, 5, 6 and 7 together

Tlris has been arlde d as an alternative to producing a parts Iist.

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