Design Data Book

Machine

Design

Data Book

Faculty In Charge Head of Department

Machine Design Data Book

POWER SCREWS-


Sr.No. Table Equation

1 Power Required to rotate the Screw

2. . ........

60. . ...... ( . )

N TPower watt

N r p m T Torque N m

p=

- -

2Torque Required to rotate the

Screw (T)

1) Considering Screw Only-

.2

dT P=

2) Considering Screw and Collar –

1. . .2

dT P W Rm= +

1 Coefficient of friction for collarm -W- Load on the Screw

4

Tangential Force Required at

the circumference of the screw

(P)

1) To Raise or Lift the Load,-

( ) ...( )1 .

tan tanP W tan W N

tan tan

a fa fa f

È ˘+= + = Í ˙-Î ˚

2) To Lower the Load:-

( ) ...( )1 .

tan tanP W tan W N

tan tan

f af af a

È ˘-= - = Í ˙+Î ˚

5 For Acme Thread

( ) 11

1

1

...( )1 .

tancos

14.5o

tan tanP W tan W N

tan tan

Semi angleof Acmethread

a fa fa f

mfb

b

È ˘+= + = Í ˙-Î ˚

=

= - =

6 Helix angle

nptan

da

p=

n- no of starts of the thread

p-pitch of the Screw

d- Mean diameter of the screw

7 Friction Angle tanm f=


8 Mean Diameter of the Screw (d)

002 2 2

cc

d d p pd d d

+= = - = -

Where- do – Nominal or Outside Diameter

p- Pitch of the Screw

dc- Core or Root Diameter

9 Coefficient of Friction for Screw

tanm a=

10 Mean radius of the Collar

1 2

2

R RR

+=

R1 – Outer Radius of CollarR2 – Inner Radius of Collar

11

12Effort Required at the end of

Lever arm of length l, (P1)

1.T P l= ……T – Total Torque

1. ........2

DT P D Diameter of HandWheel= -

13Speed of the Screw per

revolution

( / min)......( . . )

( )

velocity mmN r p m

Pitch mm=

14 Efficiency of the Screw

1) Considering Screw Only-

.tan .2

.2

dW

dP

ah =

2) Considering Screw and Collar –

1

.tan .2

. . .2

dW

dP W R

ah

m=

+

HELICAL SPRINGS-



1 Mean Diameter of the Spring Coil (D)

Using Spring Index C = D/d

D = C. d

Outer Diameter of Spring Coil (Do) = D + d

Inner Diameter of spring Coil (Di) = D - d

2 Diameter of the Spring wire (d) 1) When Wahl’s Factor (K) is Given –

3

8.

4 1 0.615

4 4

WDK

dC

KC C

tp

=

-= +

-2) When load is variable (W1 And W2),

then for Maximum load-

3max. . .

2 16

DT W d

p t= =

Note:- Standard size of wire is taken from table below----



3 Number of turns of Coil (n)

38. .

.

W C n

G dd =

d- Deflection of Spring

C- Spring Index = D/d

n- no. of turns of coil

G- Modulus of Rigidity

d- diameter of Spring Wire

1) If the Load variation is given Wmin to Wmax

Then W in above Equation is W= (Wmax- Wmin)

For Squared and Ground Ends;,

n’ = n + 2

4 Free length of spring (LF) 'max max0.15FL n d d d= + +

1) If the d is Load variation given for (Wmax – Wmin) then

dmax is calculated for Maximum load (Wmax)

5 Pitch of the Coil (p) ' 1

Free LengthPitch

n=

-

6 Energy stored in the Spring (U) 1.

2U W d=

7Helical springs subjected to

fatigue Loading

3

max min

3

max min

2.1

.

8. .........

.1

. 1 ......2 2

8. ........

.4 1 0.615

.. ......4 4 2

m v v

y e

mm s

s m

vv

v

F S

W DK

dW W

K WC

W DK

dW WC

K WC C

t t tt t

tp

tp

-= +

=

+Ê ˆ= + =Á ˜Ë ¯

=

--Ê ˆ= + =Á ˜-Ë ¯


LEAF SPRINGS-


1 Effective Length of spring (2L)

12 2L L l= -

2L1 – Length of Span or Overall length of Spring

l- width of the Band

2 Total Number of Leaves (n)

n = nF + nG

nF – No. of Full length leavesnG- No. of Graduated Leaves

3 Central Load acting on the spring 2W

4When no. of Springs are given, then

Load on each spring (2W)2W = (Total Load/No. of Springs)

5Depth to Width Ratio

nt

b

n- Number of leaves, t – Thickness of plate

and b – Width of the plate

6 Final stresses in springs

When Leaves are Equally stressed When the leaves are not Initially stressed

i) 2

6. .,

. .

W LStress

n b ts = 2

18. .,

. (2 3 )G F

W LStress

b t n ns =

+


ii)3

3

6. .,

. . .

W LDeflection

n E b td = 3

12. .,

. . (2 3 )G F

W LDeflection

E b t n nd =

+

7

Calculation Initial Gap between full length and graduated leaves

3

3

2. .

. . .

W LC

n E b t=

8Load Exerted on the Band after the

spring is assembled (Wb)2. . .

.(2 3 )F G

bG F

n n WW

n n n=

+

Knuckle Joint

A knuckle joint is used to connect two rods which are under the action of tensile loads. However,

if the joint is guided, the rods may support a compressive load. A knuckle joint may be readily

disconnected for adjustments or repairs. Its use may be found in the link of a cycle chain, tie rod

joint for roof truss, valve rod joint with eccentric rod, pump rod joint, tension link in bridge

structure and lever and rod connections of various types.

Figure:- Knuckle joint


DESIGN PROCEDURE


1 Diameter of the Rod (d)

2.4 tP dp s=

P- Load Transmitted,

st - Tensile Stress

2 Diameter of Knuckle Pin (d1) d1 = d

3 Outer diameter of eye (d2) d2 = 2 d

4 Diameter of knuckle pin Head and Collar (d3) d3 = 1.5 d

5 Thickness of single eye or rod end (t) t = 1.25 d

6 Thickness of Fork (t1) t1 = 0.75 d

7 Thickness of Pin head (t2) t2 = 0.5 d

8 Failure of knuckle pin in shear (t) 212. .

4P d

p t=

9 Failure of the single eye or rod end in tension (st) 2 1( ). . tP d d ts= -

10 Failure of the single eye or rod end in shearing (t) 2 1( ). .P d d tt= -

11 Failure of the single eye or rod end in crushing (sc) 1( ). . cP d ts=

12 Failure of the forked end in tension (st) 2 1 1( ).2 . tP d d t s= -

13 Failure of the forked end in shear (t) 2 1 1( ).2 .P d d t t= -

14 Failure of the forked end in crushing (sc) 1 1( ).2 . cP d t s=

If the induced stresses are less than the given design stresses, therefore the joint is safe


S.No Table Equation

1 Stress

sec

P

AP Forceor load acting onaboby

A Cross tional Areaof thebody

s =

==

2 Strain

l

ll Changeinlengthof body

l original lengthof the body

de

d

=

==

3Young Modulus or Modulus of

Elasticity (E)

. ......

. ......

. .

E E

P l P lE l

A l A E

s a ess ee

dd

= =

= =

4 Shear Stress 2

2

sin , .4

, 2. .4

P

A

For gle Shear A d

For Double Shear A d

t

p

p

=

=

=

5Shear Modulus or Modulus of

Rigidity

C

Shear Strain

C Modulusof Rigidity

t a ft f

f=

==

6 Factor of Safety

int.

.

Yeild Po stressF S

Working or Design stress

Ultimate stressF S

Working or Design stress

=

=

7 Impact stress2. . .

1 1.i

W h AE

A W ls

È ˘= + +Í ˙

Î ˚

8 Torsional shear Stress


9For Solid Shaft of

Diameter d

10For Hollow Shaft of

Diameter d

11Power transmitted by the

shaft

12Bending Stress in

straight Beam

13Bending Stress in

straight Beam



14 Bending stress in Curved Beams

e = R - Rn, Yi = Rn-Ri, Yo = Ro-Rn,

15 Resultant Bending Stress

Ri t bi

Ro t bo

s s ss s s

= += +


16 Axial Direct Stress... .( )t

WA Areaof Cross SectionGiven see AboveTable

As = -

17

Maximum Bending stress at the inside

fibre

18Maximum Bending

stress at the outside fibre


RIVETED JOINTS:-

S.No Table Equation

1Tearing Resistanceof

Rivet (Pt)

( ). .

, ,

,

t t

t

P p d t

p pitchof rievt d diameter of rivet hole

t Thickness of plate PermissibleTensile stress

s

s

= -

- -- -

2Shearing resistance of a

Rivet (Ps)

2. . .4

. , . ,

,

sP n x d

n no of rievt x no of shear

d diameter of rivet hole Permissible shear stress

p t

t

=

- -- -

3Crushing resistance of a

Rivet (Pc)

. . .

. , ,

,

c c

c

P n d t

n no of rievt t thicknessof rivet plate

d diameter of rivet hole PermissibleCrushing stress

s

s

=

- -- -

4 Efficiency of the Rivet

WELDED JOINTS:-


S.No Table Equation

1 Strength of the plate

. ......

. ....

si

t

t

P A kN

A b t band t widthand thicknessof a plate

Permis ble tensile stress

s

s

== --

2Strength of a single

Transverse fillet weld

0.707. . . t

t

P s l

s sizeof weld Thicknesof plate

l Lengthof weld

PermissibleTensile Stress

s

s

=- =--

3Strength of a Double

Transverse fillet weld

1.414. . . t

t

P s l


l Lengthof weld

PermissibleTensile Stress

s

s

=- =--

4Strength of a Parallel

fillet weld

1.414. . .P s l


l Lengthof weld

PermissibleShear Stress

t

t

=- =--

5

Strength of single

Transverse and Parallel

Fillet weld

1 2

1 2

0.707. . . 1.414. . .tP s l s l


l and l Lengthof Transverseand Parallel fillet weld

Permissible Shear Stress

s t

t

= +

- =-

-

6 For Starting and Stopping of weld run 12.5 is added to the lengths


7 For Fatigue Loading(max) (max)......t

tt t

t

K K

K Stress concentration Factor

s ts t= =

-

8

POWER SCREWS:-

STRESSES IN POWER SCREWS-

1 Stresses in Power Screws

1) Direct Compressive Stress.

2....... .4

( )

. .

c c cc

c o

o

WA d

A

d d p

d Nomimal dia or outsidedia

p pitchof the screw

ps = =

= -

-

-

2 Shear Stresses in Screws3

16.

.

( . )c

T

d

T Torque N mm

tp

=

-

3Maximum shear Stress in

Screws 2 2max

14

2 ct s t= +

4 Bearing Pressure on Screw

. . .

.

....

2 2

b

WP

n d th

n No of threads inengagement withnutp

h height of nut p pichof screw

pitch pt thicknessof screw

p=

- =

- -

- = =

Design Data Book

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