67071_24.pdf

CHAPTER

24 M I S C E L L A N E O U S M A C H I N E E L E M E N T S 2,3

24.1 C R A N K S H A F T S 2'3

SYMBOLS

A b c

d de do d,. E F

F~

comb Fic

Fr Fo G h

i ,=¢ do

I Di

K = m Do

K~

l

Mb

area of cross section, m 2 (in 2) width of crank cheek, m (in) distance from the neutral axis of section to outer fiber, m (in) diameter (also suffixes), m (in) equivalent diameter, m (in) diameter of crankpin, m (in) diameter of main bearing, m (in) modulus of elasticity, GPa (psi) force acting on the piston due to steam or gas pressure

corrected for inertia effects of the piston and other reciprocating parts, kN (lbf)

the component of force F acting along the axis of connecting rod, kN (lbf)

combined force, kN (lbf) magnitude of inertia force due to the weight of connecting rod

itself, kN (lbf) total radial force acting on the crankpin, kN (lbf) total tangential force acting on the crankpin, kN (lbf) modulus of rigidity, GPa (psi) thickness of cheek or web (also with suffixes), m (in)

ratio of length to diameter of crank

moment of inertia, m 4, cm 4 (in 4)

ratio of inner to outer diameter of a hollow shaft

numerical combined shock and fatigue factor to be applied to the computed bending moment

numerical combined shock and fatigue factor to be applied to the computed twisting moment

length (also with suffixes), m (in) equivalent length, m (in) bending moment, N m (lbf in)

24.1

24.2 CHAPTER TWENTY-FOUR

m t

P F

Z O"

T

twisting moment, N m (lbf in) allowable pressure, MPa (psi) radius, throw of crankshaft, m (in) section modulus, m 3, cm 3 (in 3) normal stress (also with suffixes), MPa (psi) shear stress, MPa (psi)

SUFFIXES

b bending c compressive comb combined e elastic m main max maximum r radial ra resultant in arm rh resultant in hub t torque s shaking 0 tangential

Other factors in performance or special aspects which are included from time to time in this chapter and are applicable only in their immediate context are not given at this stage.

Particular Formula

FORCE ANALYSIS (Fig. 24-1)

The radial component of force F,. acting along the axis of connecting rod (Fig. 24- l)

The tangential component of force F,. acting along the axis of connecting rod (Fig. 24-1)

The radial component of force Fit (Fig. 24-1)

The tangential component of force Fit (Fig. 24-1)

The total radial force acting on the crank

Fcl - - F t c o s ( 0 + ~ ) - - _ sin 0)2 cos(0 + 4)

(24-1)

F F c 2 = F " s i n ( O + ~ ) = V / 1 - ( s i n 0 ) 2

sin(O + ~b)

(24-2)

Fi,., = 2 Fit cos 3' (24-3)

where 3' = angle between the force Fic and the radial component of Fic

Fic2 __ 2 Fi c sin 3' (24-4)

Fr = Ficl 4- Fcl (24-5)

Fr = 2 Fic cos 7 ± F~ cos(0 + ,¢,) (24-6)

MISCELLANEOUS MACHINE ELEMENTS 24.3

Particular Formula

The total tangential force acting on the crank

The resultant force on the crankpin

F 0 : Fic 2 -Jr- Fc2

= 2 Fic sin 7 + Fc sin(0 + ~b)

F~om~= V/F~r + F~

(24-7)

(24-8)

_ _ F r = Fic 1 - Fc l c

( 0 + ~ I " ~ c tF i ~ / ~ ~ <~t.~-4--N/ h/= / ~ F s , / ~ ~ [ (b) Shak ing force d iag ram

(a)

FIGURE 24-1 (a) Forces acting on crankshaft. (b) Vector sum of F and Fr.

Fcomb I I I Neutra l axis of a rm / I*hl~l

w --~ ' - b 2 = 1 . 3 5 d

I o = 1.25 d o C1 =

/ Ic~.N~ 1[ i ' 1 d~ 1.75a " . x ~ , - x , x -v - -J - r_ -~ - - -~ . - -x d l = 2 d o

I-~ _~v~,~4 i1 ~, , ~ . - D 2 Arm > ~ ~ ~,,,k,,~ l . '~-"+Or, h, = 1.4 do

. . . . d2

~ - h 2 - ~ Im

FIGURE 24-2 Overhung built-up crank.

SIDE CRANK

Crankpin

The maximum bending moment on the crankpin (Fig. 24-2)

The crankpin diameter with respect to the bending moment

The diameter of crankpin from the consideration of bearing pressure

From Eqs. (24-10) and (24-11) neglecting t /2 and eliminating lo the equation for crankpin diameter

Empirical relation to determine the length of crankpin

Mb(max) -- fcomb X Jr- -~ (24-9)

-- Fcomb X 1

lo where l = ~ + c2 = distance from centroidal axis

to the application of load (Fig. 24-2), m (in)

3/321Fcomb do = V ~ (24-10)

/

where ab = allowable bending stress, MPa (psi)

f comb do = - - (24-11) lop

6Fzc°mb (24-12) do= 1 ,2 7rpcrb

lo=i 'do (24-13)

where i' lo = ~ - 1.25 to 1.5


Particular Formula

Another relation for the crankpin length/diameter ratio

Another relation for the crankpin diameter

H O L L O W C R A N K P I N

The crankpin length/diameter ratio

The crankpin outside diameter

i, lO=doo = ~ 0 . ~ b (24-14)

Fcomb do = V l-~ (24-15)

lo ~/0.2a(1 - K 4)

l: o: p

where K = m Oi Do

(24-16)

/Fcomb D o = g t-~ (24-17)

Crank arm

CRANK ON HEAD-END DEAD-CENTER POSITION

When the crank is on the head-end dead-center position, the section X X (Fig. 24-2) of the arm is subjected to bending moment

The direct compressive stress due to the load Fco,,,b (i.e., more specifically by its component F,.)

The resultant stress in the crank arm at X X

CRANK ON CRANK-END DEAD-CENTER POSITION

The direct tensile stress in the plane of the hub of crankshaft section passing through the shaft center due to load Fcom6 (Fig. 24-2)

The bending stress in the section due to bending moment Fcom6 x a

M b = Fcomb × 1 (24-18)

Fc°mb (24-19) °c -- A

Fcomb gbC Crra :-" ~ at- T (24-20)

where

A - area of cross section of the arm at XX, m 2 (in 2)

c = distance from the neutral axis of section to outer fiber of arm, m (in)

I = moment of inertia of the section, cm 4 (in 4)

Fcomb (24-21) ~' = h2(d2 - a )

Fcomb X a O'b = Z

where Z = section modulus, cm 3 (in 3)

(24-22)


Particular Formula

The resultant stress in the plane of the hub of crankshaft section passing through the shaft center

CRANK PERPENDICULAR TO THE CONNECTING ROD

The bending moment in the plane of rotation of the crank

The bending stress

The torsional moment

The shear stress

The maximum normal stress for crank made of cast iron

The maximum shear stress for the crank made of steel

DIMENSION OF CRANKSHAFT MAIN BEARING (Fig. 24-2b)

The shaking force on the main bearing from F and Fr (Fig. 24' lb)

The diameter of main bearing taking into consideration the bearing pressure on the projected area of the crankshaft

The bending movement on the crankshaft

The torque on the crankshaft

The diameter of crankshaft taking into consideration indirectly the fatigue and shock factors

O" r ~--- O" t ~ O" b (24-23)

M b -- Fcomb X l (24-24)

MbCl (24-25) o. b -- Zb

M t = Fcomb x r 1 (24-26)

MtCl (24-27) Zt

Crmax = 1 [Ob + V/if2 + 4T 2 ] (24-28)

"/-max --- 1 V/O.2 _+_ 47.2 (24-29)

F~ = vector sum of F and Fr (24-30)

dm = Fs (24-31) tmp

where

lm length of bearing, m (in)

p al lowable bearing pressure, MPa (psi)

Mb -- Fcomb × ll (24-32)

lo lm h = -~+ h2 + 7

where

h2 = hub length, m (in)

lo length of crankpin, m (in)

lm = length of bearing on crankshaft, m (in)

M t = Fcomb X r (24-33)

where r = throw of the crank, m (in)

/ 16 dm = V -~e { K b M b -+- w/(KbMb)2 q- (KtMt) 2 }

(24-34)


Particular Formula

The length of main bearing m - - - m

F,

~ P (24-35)

P R O P O R T I O N S OF C R A N K S H A F T S

For proport ions of crankshaft Refer to Figs. 24-2 to 24-10.

1.4d0~ ~,~ ~o~_.

• .

P

1 . l d ,

FIGURE 24-3 Overhung built-up crank.

t ph*l

FIGURE 24-5 Disk crank.

w =-H -

,e

0.8 to ,1.1do . . . . .

- a ? - - ~ I- - - ~ 0 . 5

2

to 0.9d o o_t._ ~ ~ ~ ."

i g-

to 2.5d-~ ~ J '-1.1 to 1.2d'

FIGURE 24-4 Overhung forged crank.

I

]-I Center o.~jl, m bearing ~-h =

I v

12ram c

Fcomb

t

~' ,

: I O -~

Center of bea~ng

I

12mm~ ~-Im L - - ~ - I ~

i S ~ " ~ ' x

I

j j ~

N N N N

b d

FIGURE 24-6 Center crank (American Bureau of Shipping method).

s S

_ / # - 'k k. _j •

s s

Y .I b -I

FIGURE 24-7 Equivalent length of crankshaft. FIGURE 24-8 Center hollow crank.


Particular Formula

0.06 to 0.1

,9-0.58B -~

R = 0.35BM-X B = Cylinder b o r ~ ~ . ~

111 b,.

d

_i 0.53---~

31B + 9.5 to 0.6B ~ - - 0.825B + 9---~,.

All dimensions in mm

FIGURE 24-9 Empirical proportion for center crank.

0 58d 3 25d • - - ~ •

I ~ \~ ' ,~~\~ ' , , \ \ \ \~ , \ \ \ \~ ]

I

CENTER CRANK (Fig. 24-6)

Crankpin

The maximum bending moment treating the crankpin as a simple beam with concentrated load at the center

Fcomb(to + h + tin) Mbc= 4

where

lo = length of crankpin, m (in)

lm = length of main bearing, m (in)

h - thickness of cheek, m (in)

(24-36)


Particular Formula

The diameter of the crankpin based on maximum bending moment Mbc

The diameter of crankpin based on bearing pressure between pin and the bearing

3/32Mbc

where ab = design stress, MPa (psi)

(24-37)

d o -- Fc°m-------~b (24-38) lop

Dimensions of main bearing

The maximum bending moment treating the center crank as a simple beam with load concentrated at the center

The twisting moment

The diameter of crankshaft at main bearing taking into consideration the fatigue and shock factors

The diameter of the crankshaft based on bearing pressure

American Bureau of Shipping formulas for center crank

The thickness h of the cheeks or webs (Fig. 24-6)

The diameter of crankpins and journals (Fig. 24-6)

The thickness h and the width b of crank cheeks must satisfy the conditions

Fcomb X l e (24-39) Mbb-- 4

where le = equivalent length of crankshaft, m (in)

Mt = Fcomb X r (24-40)

dm-- ?/16 {KbMbbq-V/(KbMbb)2-~-(KtMt) 2 } v TrO'e

(24-41)

dm = F~ (24-42) tmp

h = 0.4d to 0.6d (24-43)

3/Dpc (24-44) d= a V -~b

where

a = coefficient from Table 24-1A

D = diameter of cylinder bore, m (in)

p = maximum gas pressure, MPa (psi)

c = distance over the crank web plus 25 mm (1.0 in) (Fig. 24-6)

ab = allowable fiber stress, MPa (psi)

bh 2 >_ 0.4d 3 (24-45a)

b2h > d 3 (24-45b)


Particular Formula

EQUIVALENT SHAFTS

A portion of a shaft length I and diameter d can be replaced by a portion of length le and diameter de

The length he equivalent to crank web

The equivalent length crankshaft le of Fig. 24-7 varies between

The equivalent length of commercial crankshaft for solid journal and crankpin according to Carter (Fig. 24-8)

The equivalent length of commercial crankshaft for hollow journal and crankpin according to Carter (Fig. 24-8)

The equivalent length of crankshaft for solid journal and crankpin according to Wilson (Fig. 24-8)

The equivalent length of crankshaft for hollow journal and crankpin according to Wilson (Fig. 24-8)

4 le = 1(~)4 (24-46) rC

he = - - (24-47) B

where

C = ~27r~G = torsional rigidity of the crankpin

B = ~ hb3E = flexural rigidity of the web

0.95/< le < 1.10/ (24-48)

Le = ~ ( e + 0 . 8 a ~ 0.75bD 4 1.5r) D----a---+ + --ac 3 (24-49)

Le = ~ ( e + 0.8a_a)) D4-------~c0"75b 1.5r) Da----- ~ + + ~ac 3 (24-50)

Le = ~ ( e + O'4DJD~ + b + O.4DCD~ + r - O.2(Dj + D~) 3

(24-51)

{e + 0.4Dj t e --" 4 ~ "D3 =-~j J " - ] - ~

b+O.4Dc + r - 0 . 2 ( D j + Dc)'~

ac 3 ) (24-52)

EMPIRICAL PROPORTIONS

For empirical proportions of side crank, built-up crank, and hollow crankshafts

The film thickness in bearing should not be less than the values given here for satisfactory operating condition:

Main bearings

Big-end bearings

The oil flow rate through conventional central circumferential grooved bearings

Refer to Figs. 24-2 to 24-10.

h = 0.0025 mm (0.0001 in) to 0.0042 mm (0.0017 in)

h = 0.002 mm (0.00008 in) to 0.004 mm (0.00015 in)

(24-52a)

(24-52b)

Q kpc 3 d = - - ~ ~ (1 + 1.5e 2) (24-52c)

2 4 . 1 0 CHAPTER TWENTY-FOUR

Particular Formula

For oil flow rate in medium and large diesel engines at 0.35 M P a 0.5 psi

The velocity of oil in ducts on the delivery side of the p u m p

The velocity of oil in ducts on the suction side of the p u m p

The delivery pressure in modern high-duty engines

Fo r housing tolerances

where

Q = oil flow rate, m3/s (gal/min)

k = a constant - -0 .0327 SI units

= 4.86 x 104 US Cus tomary System Units

p = oil feed pressure, Pa (psi)

c = D - d = diametra l clearance, m (in)

r / = absolute viscosity (dynamic viscosity), Pa s (cP)

d = bearing bore, m (in)

L = land width, m (in)

e = at t i tude or eccentricity ratio

Refer to Table 24-1B.

v = 1.8 to 3 .0m/s (6 to 10 ft/s) (24-52d)

v = 1.2 m/ s (4 ft /s) (24-52e)

p = 0.28 to 0.42 M P a (40-60 psi)

Pmax = 0.56 MPa (80 psi)

Refer to Table 24-1C.

(24-52f)

(24-52g)

T A B L E 24-1A Coefficient a in the American Bureau of Shipping formula [Eq. (24-44)]

Number of cylinder Ratio of stroke to distance over crank webs = I/c

Type Four-stroke Two-stroke 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4

Explosion engines l, 2, 4 1, 2 I. 17 I. 17 1.17 1.17 1.17 I. 17 I. 17 3, 5, 6 3 1.17 1.17 1.17 1.17 1.19 1.20 1.22 8 8 1.17 1.19 1.21 1.23 1.25 1.28 1.30 10, 11, 12 5, 6 1.18 1.20 1.23 1.25 1.28 1.31 1.33

Air-injection 1, 2, 4 1, 2 1.17 1.19 1.22 1.25 1.28 1.31 1.34 diesel engines 3, 5, 6 1.19 1.22 1.25 1.28 1.32 1.35 1.38

8 3 1.20 1.24 1.27 1.30 1.33 1.37 1.40 12 4 1.22 1.25 1.29 1.32 1.36 1.39 1.42 16 5, 6 1.25 1.29 1.33 1.36 1.40 1.44 1.47

8

1.17 1.24 1.32 1.35 1.36 1.41 1.43 1.45 1.50


TABLE 24-1B Oil flow rate in medium and large diesel engines at 0.35 MPa

Oil flow rate

Different parts of engine liters/min/kW liters/min/hp (gal/h/hp)

Bed plate gallery to mains 0.536 0.4 (5) with piston cooling

Mains to big end (with 0.362 0.27 (3.5) piston cooling) Big ends to pistons (with 0.201 0.15 (2) oil cooling) Total flow of oil with un- 0.335 0.25 (3) cooled pistons

TABLE 24-1C Housing tolerances

Parts Tolerances

Waviness of the surface Run-out of thrust faces Surface finish

Journals

Gudgeon pins

Housing bores

Alignment of adjacent housing

The fine grinding or honing

~O.O001d ~O.O003d

0.2-0.25 ~tm R a (8-10 # in clearance)

O. 1-0.16 ~tm Ra (4-6 # in clearance)

O. 75-1.6 ~tm R a (30-60 # in clearance)

<1 in 10,000 to 1 in 12,000

0.025-0.05 mm (0.001-0.002 in)

TABLE 24-1D Properties of some steel-backed crankshaft plain bearing materials

Lining materials

Lining or G u i d a n c e Recommended Nominal overlay Relative peak loading journal composition, thickness, fatigue limits, ~sl, hardness, per cent mm strength MPa V.P.N.

Tin-based white metal Sn 87, Sb 9 Over 0.1 1.0 12-14 160 Cu 4, Pb 0.35 max Up to 0.1 1.3 14-17 160

Tin-based white metal Sn 89, Sb 7.5 No overlay 1.1 12-15 160 with cadmium Cu 3, Cd 1

Copper-lead, overlaid Cu 70, Pb 30 0.2 1.4 15-17 160 with cast white metal

Sintered copper-lead, Cu 70 0.05 1.8 21-23 a 230 overlay plated with Bb 30 0.025 2.4 28-3 la 280 lead-tin

Cast copper-lead, overlay Cu 76, Pb 24 0.025 2.4 31 a plated with lead-tin or lead-indium

Sintered lead-bronze, Cu 74, Pb 22 0.025 2.4 28-31 a 400 overlay plated with Sn 4 lead-tin

Tin-aluminum A1 60, Sn 40 No overlay 1.8 21-23 230

Tin-aluminum A1 80, Sn 20 No overlay 3.3 28-35 230

Tin-aluminum, overlay A1 92, Sn 6 0.025 2.4 28-31 a 400 plated with lead-tin Cu 1, Ni 1

a Limit set by overlay fatigue in the case of medium/large diesel engines. Suggested limits are for big-end applications in medium/large diesel engines and are not to be applied to compressors. Maximum design loadings for main bearings will generally be 20% lower. (Courtesy: Extracted from M. J. Neale, ed., Tribology Handbook, Section A11, Newnes-Butterworth, London, 1973)


24.2 C U R V E D B E A M 2'3

S Y M B O L S

C i - - C 1 - - e

C O - - C 2 + e

/4(= a) e

E F G h I k K l

Mb rc

ri

ro

rn y o"

semimajor axis of ellipse, m area of cross section, m 2 width of beam, m semiminor axis of ellipse, m distance from the centroidal axis to the inner surface of curved

beam, m distance from the centroidal axis to the outer surface of curved

beam, m distance from the neutral axis to inner surface of curved

beam, m distance from the neutral axis to outer surface of curved

beam, m diameter of curved beam of circular cross section, m distance from centroidal axis to neutral axis of the section, m modulus of elasticity, GPa load, kN modulus of rigidity, GPa depth of beam, m moment of inertia, m 4, c m 4

stress factor (also with suffixes) constant length of straight section between the semicircular ends of chain

link, m pure number to be determined for each particular shape of the

cross section by performing the integration applied bending moment (also with suffixes), N m radius of centroidal axis, m inner radius of curved beam, radius of curvature, m outer radius of curved beam, m radius of neutral axis, m deflection, m normal stress (also with suffixes), MPa

Please note: The US Customary System units can be used in place of the above SI Units.

S U F F I X E S

b bending i inner h horizontal o outer n neutral max maximum r resultant or combined v vertical x x direction y y direction


Particular Formula

GENERAL

Pure bending

The general equation for the bending stress in a fiber at a distance y from the neutral axis (Figs. 15-11 and 15-12)

The maximum compressive stress due to bending at the outer fiber (Fig. 24-12)

The maximum tensile stress due to bending at the inner fiber (Fig. 24-12)

y ) Orb = =t: --A--ee r n + y

MbCo (24-54) O-bo - - A e r o

M b C i (24-55) O'bi -'- Aer i

Stress due to direct load

The direct stress due to load F F \ ),

dA ~///~//,//~,

Mb b

i 0 2 ,

FIGURE 24-11 Bending stress in curved beam. FIGURE 24-12 Analysis of stresses in curved beam.

24 .14 CHAPTER TWENTY-FOUR

Particular Formula

Combined stress due to load F and bending

The general expression for combined stress

The combined stress in the outer fiber

The combined stress in the inner fiber

F o r values of radius to neutral axis for curved beams

(7 r =-A ±-~e rn + y

O'r o ~ - F MbCo A Aero

F M b C i

O'ri - - - A q - A er-----[

Refer to Table 24-2.

(24-57)

(24-58)

(24-59)

TABLE 24-2 Values of radius to neutral axis for curved beams

Type Section Radius of neutral surface, r,,

• r . ~ 0

ro ;I

e~ , 1" rn =

l: r o ;i ('-9

e r n =

=

0

ro

,- r O

r c rn L

0

0

~-C4

( ~ o -}- V/~') 2 rn = 4

h r n = - -

'n(r~i )

lh(bi+bo) rn = biro - bori ln['ro'~, •

bo) h \ r i /

i i

If bo = 0, this section reduces to a triangle

r n = b i I n ri -+- ai ro - a o + b 2 In + b o In

ri ri + a i r o - - a o

If ao = 0, this section reduces to a l section; r, is the same for a box section in dotted lines with each side panel ½ b2 thick

(24-60)

(24-61)

(24-62)

(24-63)

MISCELLANEOUS MACHINE ELEMENTS 24 .15

Particular Formula

APPROXIMATE EMPIRICAL EQUATION FOR CURVED BEAMS

An approximate empirical equation for the maximum stress in the inner fiber

The stress at inner radius for a curved beam of rectangular cross section

The stress at inner radius of circular cross section

The stress at inner radius of elliptical sections according to Bach a

O'i -- g b +7""-- - J - - to

where

K = 1.05 for circular and elliptical sections

= 0.5 for all other sections

b = maximum width of the section, m (in)

Mb in N m (lbf ft)

on"6Mb ( h ) ai = ~ 1 + 0 . 2 5 - f i

( cr i - - 71.d 3 1 + 0 . 3 - /"i

( a) ai= "na 2b l + 0 . 3 - ri

(24-64)

(24-65)

(24-66)

(24-67)

STRESSES IN RINGS (Fig. 24-13a)

Maximum moment for a circular ring at the point of application of the load, A, Fig. 24-13a

Another max imum moment b for a circular ring at a point B 90 ° away from the point of application of load

Direct stress for the ring at point B 90 ° away from the point of application of load

The general expression for bending moment at any cross section DD at an angle 0 with the horizontal (Fig. 24-13b)

The stress due to direct load F at any cross section DD at an angle 0 with the horizontal

m b ( m a x ) -4- Fr = - - = ~0318Fr__._____ /

71" (24-68)

where - ve sign refers to tensile load, + ve sign refers to comprehensive load

m b ( m a x ) - - - + - 0 . 1 8 2 F r ( 2 4 - 6 9 )

where - ve sign refers to comprehensive load, + ve sign refers to tensile load

F cr = 2A (24-70)

M ; = M ~ - 1 F r ( 1 - cos 0) (24-71)

F sin 0 o = 2 ~ (24-72)

a Courtesy: Bach, Maschinenelemente, 12th ed., p. 43. b Moments which tend to decrease the initial curve of the bar are taken as positive.


Particular Formula

F

A B A D B

B

MMA~~__~ ~ MB

(a)

FIGURE 24-13 Bending moments in a ring. FIGURE 24-14 Bending moments in a link.

The combined stress at any cross section

DEFLECTION

The increase in the vertical diameter of the ring (Fig. 24-13a)

The decrease in the horizontal diameter of the ring (Fig. 24-13a)

1 F Mb ( Y ) (24-73) ar = -2 -A s i n O - t- -~e rn + Y

Fr 3 Yv = 0 . 1 4 9 ~ (24-74)

E1z

Fr 3 Yh = 0 . 1 3 7 ~ (24-75)

Elz

LINK (Fig. 24-14)

The moment, MbA, at the point of application of load (Fig. 24-14)

The moment, Mb, , at the section 90 ° away from the point of application of load (Fig. 24-14)

CRANE HOOK OF CIRCULAR SECTION (Fig. 24-15)

The combined stress in any fiber of a crane hook sub- ject to a load F

The maximum combined stress

Fr(2r + l) MbA = ~ _t S/~ (24-76)

F r ( Z r - 7rr) (24-77) MbB -- 2(Trr + l)

where l - length of straight section between the semicircular ends.

F Mb( y ) Fy (24-78) Cr r = -A i -~e r . + y = A m ( r x - y )

O'r(max) -- A ~ =--A ki (24-79)


Particular Formula

The minimum combined stress

For crane hook of trapezoidal section

Crr(min) = A 2-~ro = Ak° (24-80)

where k i and ko are stress factors which depend on H/2re; ki is the critical one which varies from 13.5 to 15.4 as ratio H/2re changes from 0.6 to 0.4

Refer to Fig. 24-16 and Table 24-3.

I Undercut

FIGURE 24-15 Hook of circular section.

/I" UNDFRCUT

l H K ~,x~~ .~, Z M

FIGURE 24-16 Crane hook of standard trapezoidal section.

¢',1

°~

om

om

g~

¢',I

G"

G"

G"

,..,,

G" ¢

'4

C

r..,

C

C

r:.

,---~ ,---~

,--~ ,-'-~

,---~ ,--~

,.-- ,---~

,--~ ,--

t"-I ,--~

t"4

,.-.~ ¢

'4

¢'4

,... ,.,

,-~

,.~

,.~

,.~

,-.~ ,..~

,.,

¢-4 ,-~

t',,,I

t--4 t-.,l

t',-4 t'-4

t'-4 ¢¢'~

t'-4

,-. ,.,

~ ,..

,-- t"4

,--

t"4

,-- t"4

¢

'4

¢'4

¢

'4

t"~

t"4

¢'4 ,---

¢'4 ,---

~~

~-

~~

~~

~~

~~

-~

~

..

..

~

~~

=~

-~

~~

.

..

.

=~

--. t--.1

~ ~

-.i ~,,1

¢-.i ~

-,i ~

-i e-,i

e~

r-,i

e-.l ,.-

~-,.1

e-,i e-,l

1--.i ~

-,,1 ~

.1

e~

~,1

¢

~

e~

r---

t--4

II

z~

ff

,.. ,.-.,

-~

24.18


24.3 C O N N E C T I N G A N D C O U P L I N G R O D 2'3

S Y M B O L S 2,3

A a

b d dl de d~ E F

F~

Fi

F~r g

h k l it le

Mb n

n l

1 Flt - - _

f

P Pl v

w

W Z

ct

area of cross section, m 2 (in 2) Rankine's constant width, m (in) diameter, m (in) core diameter of bolt, m (in) crankpin diameter, m (in) gudgeon pin diameter, m (in) modulus of elasticity, GPa (psi) force acting on the piston due to steam or gas pressure

corrected for inertia effects of the piston and other reciprocating parts, kN (lbf)

the component of F acting along the axis of connecting rod, kN (lbf)

inertia force, kN (lbf) inertia force due to reciprocating masses, kN (lbf) crippling or critical force, kN (lbf) acceleration due to gravity, 9.8066 m/s 2 9806.6 mm/s 2 (32.2 ft/s 2) depth of rectangular or other sections, m (in) radius of gyration, m (in) length of connecting rod, m (in) length of crankpin, m (in) equivalent length, m (in) length of gudgeon pin, m (in) bending moment, N m (lbf in) speed of crank, rpm safety factor

ratio of connecting rod length to radius of crank

allowable pressure, MPa (psi) load due to gas or steam pressure on the piston, MPa (psi) velocity of crank, m/s (fps) specific weight of material of connecting road, kN/m 3 (lbf/in 3) weight of the reciprocating masses, kN (lbf) section modulus, m 3, cm 3 (in 3) angular speed of crank, rad/s angle between the crank and the center line of connecting rod,

deg angle between the crank and the center line of the cylinder

measured from the head-end dead-center position, deg angle between the center line of piston and the connecting rod,

deg normal stress (also with suffixes), MPa (psi)


Particular Formula

The velocity 2 7rrn 60

where r in m

(24-81)

DESIGN OF CONNECTING ROD (Fig. 24-17)

Gas load

Load due to gas or steam pressure on the piston ~ a ~

F.=--£-pe (24-82)

Inertia load due to reciprocating motion

Inertia due to reciprocating parts and piston cos 0) Fir---- ~ - cos0-k- n' (24-83a)

Wrn 2 i" Fir = 0.01095 [ COS 0 +

g \

cos 20 F/t

(24-83b)

The maximum value of Fir occurs when 0 = 0 ° or when the crank is at the head-end dead center Flit(max) -- 0.01095

(1) Wrn2 1 + g

(24-84)

At the crank-end dead center, when 0 = 180 °, Fir attains the maximum negative value, acting in opposite direction

The combined force on the piston

F2ir(max) = - 0 . 0 1 0 9 5

F= Fg + Fir

Wrn2g ( 1 - ~ (24-85)

(24-86)

The component of F acting along the axis of connecting rod

The stress induced due to column action on account of load Fc acting along the axis of connecting rod

f c - - _ sin 0 )2

(24-87)

(a) As per Rankine's formula a l = ~ - l + a ~ (24-88a)

Fc

FIGURE 24-17 Forces acting on a connecting rod.


P a r t i c u l a r F o r m u l a

(b) As per Ritter 's formula

(c) As per Johnson 's parabolic formula

inertia load due to connecting rod

The magnitude of inertia force (Fig. 24-17) due to the weight of the rod itself, not including the ends

The maximum bending moment produced by the inertia force Fic is at a distance (2 /3) / f rom wrist pin

The maximum bending stress developed in the rod due to inertia force Fic

The crank angle (0) at which the maximum bending moment occurs according to B. B. Low

The relation between the moment of inertia in the xx and yy planes in order to have same resistance in either plane

E ,24-8 b ~,1=~- 1 + ~ ;

F~ al = (24-88c)

All O'Y (~e)2J 4n~ -2 E

where

le = equivalent length, m (in)

k = radius of gyration, m (in)

n = end-condit ion coefficient (Table 2-4)

a -- constant obtained from Table 2-3

Awv21 . Fic -- 2gr sin a (24-89a)

WF2 when a = 90 ° (24-89b) Fic = 2gr

where W = Awl = weight of the rod itself, not including the ends, kN (lbf)

• 2Ficl 2Wv21 sin a (24-90a) Mb(max)-- 9----~ = 9v/3 x 2gr

W v 2 l Mb(max) 9x/3gr when a = 90 ° (24-90b)

mb(max) W v 2 l O'b(max) - - Z - - 9x/~grZ sin a (24-91a)

W v 2 l °'b(max) -- 9x/~grZ when a = 90 ° (24-91b)

3500 0 = 90 ° - ( n ' + 7.82) 2 (24-92)

Iyy = ¼ Ixx (24-93)

o r

G = ¼G (24-94)


Particular Formula

DESIGN OF SMALL AND BIG ENDS

The diameter of crankpin at the big end F (24-95) de : tcp

where

p = allowable bearing pressure based on projected area, MPa (psi)

= 4.9 to 10.3 MPa (700 to 1500 psi), and

lc = 1.25 to 1.5 de

The diameter of the gudgeon pin at the small end F (24-96) dg = lgp

where

p = 10.3 to 13.73 MPa (1.2 to 2.0kpsi), and

L = 1.5 to 2

d

DESIGN OF BOLTS FOR BIG-END CAP

The diameter of bolts used for fixing the big-end cap

The expression for checking load for measuring peripheral length of each thin-walled half-bearing according to J. M. Conway Jones a

The expression for total minimum nip, n

2Flir(max) di = V ~Td (24-97)

where Flir(max) is obtained from Eq. (24-84)

aa = design stress of bolt material, MPa (psi)

Lhh SI (24-97a) W,. = 6000--D--

where

W, = checking load, N

L = axial length of bearing, mm

hb = wall thickness of bearing, mm

D - diameter of housing, mm

n = 44 × 10 -6 D2 hb or 0 .12mm SI (24-97b)

whichever is larger

a In M. J. Neale, ed., Tribology Handbook, Section A20, Newnes-Butterworth, London, 1973.


Particular Formula

Note: The "nip" or "crush" is the amoun t by which the total peripheral length of both halves of bearing under no load exceeds the peripheral length of the housing of the bearing.

The compressive load on each bearing joint face to compress nip a

F ~ h~ b . . . _

, l r ~

b b' . . . - ~ -b'

. a ' ( a ) , a '

~ ' ~ ~ W{ \ \" W k ~ " ,a . , .

b -b' b-( .. -b'

a !

(0)

FIGURE 24-18 Forces acting on a coupling rod.

The bolt load required on each side of bearing to compress nip for extremely rigid housing

ELhslm W = 7r(D - hsl)106 SI (24-97c)

o r

W = Lhstcry × 10 -6 SI (24-97d)

whichever is smaller

where

D = housing diameter, m m (in)

h~t = steel thickness + ½ lining thickness, m m (in)

m = sum of m a x i m u m circumferential nip on both halves of bearing, m m (in)

W = compressive load on each bearing joint face, N (lbf)

E -- modulus of elasticity of material of backing, Pa (psi)

= 2 1 0 G P a (30.45 Mpsi) for steel

L = bearing axial length, m m (in)

ay = yield stress of steel backing, Pa (psi)

= 350 M P a (50 kpsi) for white-metal-l ined bearing

= 300 to 4 0 0 M P a (43.5 to 58kpsi) for bearing with copper-based lining

= 600 M P a (87 kpsi) for bearing with a luminum- based lining

Wb = 1.3 W (24-97e)

The bolt load required on each side of bearing to compress nip for normal housing with bolts very close to back of bearing

The ratio of connecting rod length (/) to crank radius (r)

Wb = 2 W (24-97f)

, 1 n = - = 3.4 to 4.4 single-acting engines

r (24-98)

= 4.6 to 5.4 for double-acting engines

= 6.0 or more for steam locomotive engines

= 5 to 7 for s tat ionary steam engines

= 3.2 to 4 for in ternal-combust ion engines

= 1.5 to 2 for aero engines



Particular Formula

DESIGN OF COUPLING ROD (Fig. 24-18)

The centrifugal force due to the weight of the rod

The bending component of centrifugal force

The maximum bending moment due to the uniformly distributed load of Feb

The axial component of the centrifugal force

For some of the common cross sections of connecting rods

For forces acting on a coupling rod

For proport ions of ends of round and H-section connecting rod

For proport ions and empirical relations of steam engine common strap end

WV 2 Fc = hbl (24-99)

gr

W v 2 Fc = - - (24-100)

gr

wv2hbl Wv 2 Fcb = ~ cos a = ~ cos a (24-101)

gr gr

wv2 hbl 2 Wv2 l Mb(max)- 8g-----"~-- 8g------~ (24-102)

wv 2 Wv 2 sin a Fca = hbls ina = ~ (24-103)

gr gr

Refer to Fig. 24-19.




Y

bt

Y (b) (c) B = cylinder bore

' y

x /,0"36 to 0.4B l~OCk washer

• ( lS0P:~re -- . r ~ . ~ o.02B . Groove in bush

. . . . . . . . . 00 to " 4 l I [oo4 to olo4B

~_ b_4t .~1_.[" o

• to

I - - ! - - - - [ ~ . ~ 0.26 to 0.33B

p--

.,E

FIGURE 24-19 Connecting rod sections. FIGURE 24-20 Two typical end designs for round and H-section connecting rods.


Taper of cot ter 1 in 16 _ , ~ , ~

b__) 1 -, G

Y A = 0.9d E = 1.15d + 5.5 H = 0.3d + 1.5 B = 0.5d + 10 F = 0.37d + 3 C = 0.3d + 1.5

b = d G = 0.1d + 4 t = 0.2d + 1.5 D = 0.35d + 3

All d imens ions in mm

FIGURE 24-21 Steam engine common strap end.


24.4 PISTON AND PISTON RINGS

S Y M B O L S 2'3

D Dr E F

Fo

hi, h2, h3 H i,=t L Mb n

Pb P F

R, Ri, R h th tr r~ re W

O"

0 TABLE 24-4

area of cross section of piston head, m 2 (in 2) width of face of piston, m (in) diameter of bore, m (in) heat-conduction factor, kJ/mZ/m/h/K

(Btu/in 2/in/h/°F) higher heat value of fuel used, kJ/kg (Btu/lb) nominal diameter of piston ring, m (in) diameter of piston rod, m (in) diameter of gudgeon pin, m (in) diameter of bore (cylinder), m (in) root diameter of the piston ring groove, m (in) modulus of elasticity, GPa (psi) force, kN (lbf) diametral load on the piston ring to close the gap which is less

than 2.45 N (0.55 lbf) tangential load on the piston ring to close the gap which is less

than 2.45 N (0.55 lbf) thickness (also with subscripts), m (in)

radial thickness of piston ring, m (in) thickness as shown in Fig. 24-24, m (in) heat flowing through the head, kJ/h (Btu/h) length/diameter ratio length of gudgeon pin, m (in) length of piston, m (in) bending moment, N m (lbf in) safety factor brake horsepower (bhp) pressure, MPa (psi) radius, m (in) radius as shown in Fig. 24-22b, m (in) thickness of head, m (in) thickness under ring groove, m (in) temperature at center of head, °C (°F) temperature at edge of head, °C (°F) weight of fuel used, kg/bhp/h axial width of ring, m (in) stress (also with subscripts), MPa (psi) angle (Fig. 24-23), deg

Dimensions of cast-iron piston up to 430 mm diameter (Fig. 22-24) (all dimensions in cm)

Diameter of Diameter of cylinder, D piston rod, d dt b a h ht h2 h3

15.0 2.8 2.5 7.5 1.2 1.2 1.4 20.0 3.4 3.1 8.1 1.4 1.2 1.6 25.0 4.0 3.7 9.0 1.6 1.4 1.7 30.0 5.0 4.7 10.0 1.7 1.6 1.7 35.0 5.6 5.0 11.2 1.9 1.6 1.9 40.0 5.9 5.6 11.8 1.9 1.7 2.2

0.8 0.95 0.95 1.2 1.2 1.2

0.8 0.8 0.95 0.95 0.95 0.95

3.1 3.4 4.0 4.6 5.3 5.8

2.5 3.1 3.7 4.7 5.0 5.6


Particular Formula STEAM E N G I N E P I S T O N S

Piston rods

The diameter of piston rod

The diameter of piston rod according to Molesworth

(b)

FIGURE 24-22 Plate piston.

T--W/////////////////~/~

I,t dal

D

--I

--IV

--g

FIGURE 24-24 Cast-iron piston of diameter < 400 mm (16.0 in).

D P-- d = ~ (24-104)

where

p = unbalanced pressure or difference between the steam inlet pressure and the exhaust, MPa (psi) O " u

aa = - = allowable stress, MPa (psi) n

N o t e : o a is based on a safety factor of 10 for double- acting engines and 8 for single-acting engines. (The diameter of piston rod is usually taken as 1 to ~ the diameter of the piston.)

d = 0.0044Dx/- fi for cast-iron pistons (24-105)

d = 0.00338Dv/- fi for steel pistons (24-106)

. m

FIGURE 24-23 Conical plate piston.

24.28 C H A P T E R T W E N T Y - F O U R

Particular Formula

PROPORTIONS FOR PRELIMINARY L A Y O U T F O R P L A T E P I S T O N S

Box type (Figs. 24-22a and 24-24)

Width of face

Thickness of walls and ribs for low pressure

b - 0.3 to 0.5D

h = (2R + 50cm) (0.003v/fi + 0.0275 cm)

or

2R h = ~ + 10 mm (0.40 in)

(24-107)

(24-108)

(24-109)

The thickness of walls and ribs for high pressure

For dimensions of conical plate piston

2R h = ~ + 10mm (0.40in)


(24-110)

For dimensions of cast-iron piston of <_400mm diameter


Disk type (Fig. 24-22b)

Width of face b = 0.3 to 0.5D (24-111)

Thickness of walls and ribs for low pressure h = (2R + 12.5cm)(0.0096x/fi+0.057cm ) (24-112)

The hub thickness h I = 0.45d (24-113)

The hub diameter Dh = 2Rh = 1.6 x the piston diameter (24-114)

Width of piston rings w = 0.03D to 0.06D (24-115)

Thickness of piston rings

For dimensions of cast-iron piston

h = 0.025D to 0.03D (24-116)


S T R E S S E S

(a) Distributed load over the plate inside the outer cylindrical wall (i.e., the area 7rR 2)

(1) Stress at the outer edge (Fig. 24-22b) 4R~ Ri) 3p R2 _ 3R2 + ln~-£ ffl =4-~-~ 2 2

R i - R h (24-117)

(2) Stress at the inner edge (Fig. 24-22b) 3p (R 2 + R2h - 4R2R ] Ri ) cr 2 = ~ R 2 _ R 2 ln2 ~-£h (24-118)


Particular Formula

(b) Load on the outer wall, pTr(R 2 - R 2) distributed around the edge of the plate

(1) Stress at the outer edge (Fig. 24-22b)

(2) Stress at the inner edge (Fig. 24-22b)

(3) The sum of the stresses at the outer edge

(4) The sum of the stresses at the inner edge

3p(R2 - R2) ( l _ 2R2 Ri ) (24-119) 0-3 = 2h 2 R 2 - R--------~ In

3p(RZ - R2i) ( 1 - 2R2i Ri ) (24-120) 0"4 : 2h 2 R 2 - R'-------~ In Rhh

0-0 --- 0-1 + 0"3 (24-121)

O" i - - 0" 2 -]-- 0" 4 (24-122)

(Note: 0-0 or 0-i should not be greater than the permissible stress of the material. A safety factor, n, of 8 can be used.)

Dished or conical type (Fig. 24-23)

An empirical formula for the thickness of conical piston (Fig. 24-23)

The height of boss

The diameter of boss

The thickness hi measured on the center line

For calculating hub diameter, width of piston rings, and thickness of piston rings

h - 0.288v/pD/0-sinO SI (24-123a)

where p and 0- in MPa, and D and h in m

h = 9.12v/pD/0-sin O Customary Metric (24-123b)

where p and 0- in kgf/mm 2, D and h in mm

h = i.825v/pD/0-sinO USCS (24-123c)

where p and 0- in psi, D and h in in

H = 1.1K (24-124)

Dh = 1.7K for small pistons (24-125a)

Dh = 1.SK for large pistons and light engines

(24-125b)

hi = Kc (24-126)

where

c = 1 to 0.75 depending on the angle of inclination 0 (Refer to Table 24.5.)

0 = varies from 6 ° to 35 °

K = 1 to 4.5 for varying pressure and diameter

Also refer to Table 24-6 for values of K.

Refer to Eqs. (24-114) to (24-116).

2 4 . 3 0 CHAPTER TWENTY-FOUR

Particular Formula

PISTONS FOR INTERNAL-COMBUSTION ENGINES

Trunk piston (Fig. 24-25)

The head thickness of t runk pistons (Fig. 24-25a) ~ /3 PD 2 th = V i6--aa (24-127)

where

cr = 39 MPa (5.8 kpsi) for close-grained cast iron

= 56.4 MPa (8.2kpsi) for semisteel or aluminum alloy

= 83.4 MPa (12.0 kpsi) for forged steel

C O M M O N L Y USED EMPIRICAL F O R M U L A S IN THE DESIGN OF TRUNK PISTONS FOR AUTOMOTIVE-TYPE ENGINES

Thickness of head (Fig. 24-25a) th = 0.032D + 1.5 mm SI (24-128)

th = 0.00D + 0.06in USCS (24-128a)

The head thickness for heat flow

'

~~.'.'.~lllllllllllH.. ~) t r t, I

F" (a) .Dr ~-i

N /.J i ~ ~(c)

(b)

HD 2

th = 0.16c(Tc - Te)A

H = SI (24-129a)

0.194¢(Tc- Te )

where

7",.- Te = 205°C (400°F) and Tc = 698K, 425°C (800°F) for cast-iron piston

A T = T c - Te = 55°C (130°F) and Tc = 533K, 260°C (500°F) for a luminum piston

c = 2.2 for cast iron

c = 7.7 for aluminum

HD 2 H th = 16cATA = 12.5cA-------T USCS (24-129b)

FIGURE 24-25 Trunk piston for small internal-combustion engine. (a) piston laid out for heat transfer; (b) piston modified for structural efficiency; (c and d) alternate pin designs.


Particular Formula

Thickness of wall under the ring (Fig. 24-25a and b)

The thickness of wall under the ring groove

The heat flow through the head

The root diameter of ring grooves, allowing for ring clearance

Length L of piston

For chemical composition and properties of aluminum alloy piston

Gudgeon pin

The diameter of gudgeon pin

The length/diameter ratio of gudgeon pin

For gudgeon pin allowable oval deformation

For empirical relations and proportions of pistons

t r = thickness of head = th (24-130)

tr - - l ( O r + v / D Z - 4 D t h ) (24-131)

H - K C w P b (24-132)

where

W = weight of fuel used, kJ/kW/h (lbf/bhp/h)

K = constant representing that part of heat supplied to the engine which is absorbed by the piston

= 0.05 (approx.)

P b -- brake horsepower per cylinder

D r = D - (2w + 0.006D + 0.02 in) USCS

D r - - D - (2w + 0.006D + 0.5 mm)

at the compression rings

D r = D - (2w + 0.006D + 1.5 mm)

D r = D - (2w + 0.00D + 0.06 in)

at the oil grooves

where D r and D in mm (in)

SI (24-132a)

SI (24-132b)

USCS

L = D to 1.5D (24-133)

Refer to Table 24-10B.

f-} dr = V/@p (24-134)

where

F = maximum gas pressure corrected for inertia effect of the piston and other reciprocating parts, kN (lbf)

p = working bearing pressure

= 9.81 MPa (1.42 kpsi) to 14.7 MPa (2.13 kpsi)

lg = 1.5 to 2 (24-135)

Refer to Fig. 24-28c.

Refer to Figs. 24-26 to 24-28a.

24.32 CH APTER T W E N T Y - F O U R

]I

I B I j

0. 0.04B

~ .~ca l c learance

- - .9.~B

B = cy l inder bore

FIGURE 24-26 Proportions of a typical alloy piston.

' - B (Nomina l ) ~, I-

'

, i

, ~

*0"2B*I • 1

B

O t t ~ h.- t'~l O " - " - j . - m

¢0 ¢~4¢0

0

O.05B

0.03 to

O.04B

FIGURE 24-27 Proportions of an iron piston.

Fig. 24-27 m m

(in) m m

(in) mm

(in) m m

(in) m m

(in)

Cylinder bore

Crown thickness

Clearance

Clearance

Clearance

152.5 (6) 16 (5/8) 0.760 (0.03) 0.125 (0.005) 0.125 (0.005)

203.2 (8) 19 (3/4) 0.900 (0.035) 0.225 (0.008) 0.150 (0.006)

254 (lO) 32 (1¼) 1.145 (0.045) 0.230 (0.009) 0.180 (0.007)

305 (12) 41.5 (15 )

1.525 (0.06) 0.255 (0.01) 0.200 (0.008)

Piston weight = 40.715B3 N (approx.) SI where B = cylinder Piston weight - 0.15B 3 lbf USCS where B in in

406.5 (16) 47.5 (17 )

2.030 (0.08) 0.255 (0.01) 0.230 (0.009)

bore, m


/ r~ b 2.2B

. :~, B.less 1000

o ml~ I 0 ; , B I 0 ~ 0 4 5 ~

o 0.045B 4000 o_ .i,v i . , "- Relief

in w a y / l l ~ i I ' ~ / of pin ~,,~//~, ~ ~ 0.036B _~os~es~ ~ ~o~ ~Loo~~

_. lOOOj~ "~lA~Bless 4000

FIGURE 24-28(a) Iron piston for small engines (B = cylinder bore).

700

E 650 z

6 600 oo

550

500

(.9 . 450

z h" 400

~ 350

~ 300

250

200

i k k k k k k . ~.~...~' ........~ I

i °1

- ~ to % d 2 _ GAS LOAD

0

X e,i

100 .=_ c~

9O -~

ao ~ 6

zo ~_

LU

6o ~

z 50 ~.

4o ~ 121

30 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450

GUDGEON PIN / + )

FIGURE 24-28(b) Fatigue stress in gudgeon pins for various pin and piston geometries. (M. J. Neale, Tribology Handbook, Butterworth-Heinemann, 1973.)


Particular Formula

60

50

E ::L

o U_ W o 3O

2O

10

CYLINDER BORE DIAMETER (NOMINAL), in 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

I I I I I I I I I I I I I I I I

P IN OVAL DEFORMATION ~ " ( a -b )

t= 1- '~' \L 8J rn d (D2 P'~ A

(ORt= ~ k - ~ ' / in)

= L =

I tt i

1 b

D DIA. _ l

= p MAX. PRESSURE-I MN/m 2 (Ibf/in 2)

!i'l, I I I I I I I

100 150 200 250 300 350 400

CYLINDER BORE DIAMETER (NOMINAL), mm

2.6 - 2.5

- 2.4

- 2 . 3 -r" O

- 2 . 2 z

-2 .1 z <

- 2 . 0 ,, - 1 . 9 0

cO - 1 . 8 -r-

I - - 1 . 7 cl

Z - 1 . 6 <

1.5 D O - 1 . 4 -r

- 1 . 3 ~ Z

- 1 . 2 0 m

-1 .1 ,~ - 1 . 0 :~ n,' -0 .9 0

LL

- 0 . 8 m r"t - 0.7

- 0.6 - 0.5

FIGURE 24-28(c) Gudgeon pin allowable oval deformation. (M. J. Neale, Tribology Handbook, Butterworth-Heinemann, 1973.)

For fatigue stress in gudgeon pins Refer to Fig. 24-28b.

For empirical proportions and values of cylinder cover, cylinder liner, and valves


Piston rings

Width of rings

For land width or axial width of piston ring (w) required for various groove depths (g) and maximum cylinder pressure, Pmax.

D w = 2-0 for concentric rings (24-136a)

D w = ~ opposite the joint of eccentric rings

(24-136b)

D w = ~ at the joint of eccentric rings (24-136c)

Refer to Fig. 24-28d.


0.200

16.0

15.0

14.0

13.0

12.0

11.0

E E ~- lo.o

9.0

8.0

Groove depth g, in

0.250 0.300 0.350 0.400 i i i i

ALUMINIUM PISTON LAND WIDTH W

TO DETERMINE LAND WIDTHS FOR PISTONS IN CAST IRON OR STEEL USE THE FOLLOWING CONSTANTS

CAST IRON W FROM GRAPH x 0.700

STEEL W FROM GRAPH x0.5

0.450 0.500 i i

2400 2200

7o L i - I

6.0 I I I w

I I

5.0

4.0

3.0 i i '0 50 6'0 7'0 8'0 go ~00 ~1 12'0 13'0 Groove depth, mm

2000

1800

1600

1400

1200

1000

MAX. CYLINDER PRESSURE, Ibf/in 2-

- 0.650

0.600

0.550

0.500

0.450

t - -

0.400 ~-

0.350

0.300

0.250

0.200

- 0.150

FIGURE 24-28(d) The land width required for various groove depths and maximum cylinder pressures. (M. J. Neale, Tribology Handbook, Butterworth-Heinemann, 1973.)

J ~--Sf l, I

= -~!ll-- ~cl

Free ring - " "

~~.~ Compressed ring

/ 8 c = circumferential clearance gap

(~f = free piston ring gap h = radial depth or wall thickness

of piston ring w = axial width of piston ring d = nominal diameter of piston ring r = radius of neutral axis of ring

--I. I--j_ I i I F/'/~~

FIGURE 24-28(e) Nomenclature of piston ring and tangential force, Fo.

FIGURE 24-28(f) Typical variable and constant contact pressure distribution around piston rings for four-stroke engines. (Courtesy: Piston Ring Manual, Goetze AG, D- 5093 Burscheid, Germany, August 1986.)

FIGURE 24-28(g) Diametrically opposite force (Fd) applied on piston ring.

FIGURE 24-29 Tangentially applied force, Fo, on a piston ring.

Fd

Gap clearance ,,.._1 ... ,

" - I ' ~

Fo I

Variable (~)

Constant

pc I

24.36


Particular Formula

The modulus of elasticity of piston ring as per Indian Standards

The bending momentproduced at any cross section of the ring by the pressure uniformly distributed over the outer surface of the ring at an angle 4~ measured from the center line of the gap of the ring (Figs. 24-28e and 24-28f)

The bending moment (Mb) in Eq. (24-138a) in terms of tangential force, Fo

The uniform contact pressure of the piston ring on the wall

The radial distance from a point in piston ring to obtain a uniform pressure distribution (Fig. 24-28e) according to R. Munro a

5.37 ~ - 1 F E =

w6

when the ring is diametrically loaded

(24-137a)

(d )3 14.14 ~ - 1 Fo

E = w6

when the ring is tangentially loaded

where

(24-137b)

E = modulus of elasticity, MPa (psi)

6 = difference between free gap and gap after apply- ing the load, mm (in)

M b - 2 p w r 2 sin 2 ~ (24-138a) 2

Mb = pwr2(1 + cos 4~) (24-138b)

where

r = radius of neutral axis, mm (in)

p = pressure at the neutral axis of the piston ring, Pa (psi)

Mb = For(1 + cos q~) (24-138c)

En6f (24-138d) P = 7.OYd(d/h 1) 3

where

d = external piston ring diameter

h = radial depth or wall thickness of piston ring

En = nominal modulus of elasticity of material of the ring

6f = free ring gap

ro = r + v + dv (24-138e)



Particular Formula

~.065 B ~ ' 0 9 to . 12B

tt') u5 +

~D 0


FIGURE 24-30 Proportion for four-stroke cover, 100- to 450-mm bore (B = cylinder bore).

where

f r 4 v = ~ . / ( l - cos 4~ + ½~b sin ~b) (24-138f)

dv = ~ J (~ _ 1 ~b cos ~b - 1 sin 4~)

x (3 sin ¢ + ¢ cos ¢) ]

where

F = (mean wall pressure x ring axial width)

r = radius of neutra l axis, when the ring is in place inside the cylinder (Fig. 24-28e)

If)

0"1 i Clear on dia. ~

1.25 to 1.35B 1.18 to 1.2B-

~ 0.12B

0.03B

0.08 0.12

B +0.0012

~-0.07to 0.08B

H,[

• 18D to .21 D

.25 to .30D

ii ~ ~ - 18 to 221bf/sq in of

valve area

7.5

Ob

7R 5B

8000 Clear on dia

/o

• B+0.0125 '


Lowest ring on BDC

FIGURE 24-31 Empirical rules for average practice in liner design (B = cylinder bore).

~ . I _ ~ \ g 3 0 °to45 °

D

FIGURE 24-32 Valve seated directly in the cylinder head.


P a r t i c u l a r F o r m u l a

The relation between the ratio of fitting stress oft to nominal modulus of elasticity (E,) in terms of h, d, and by

The relation between the ratio of working stress (O-w) to nominal modulus of elasticity (En) in terms of h, d, and by

The relation between the ratio of the sum of (oft + Ow) to nominal modulus of elasticity (En) in terms of d and h

For preferred number of piston rings

For properties of typical piston ring materials

4) = angle measured from bottom of the vertical line passing through the center of the gap of the ring as shown in Fig. 24-28e

I = moment of inertia of the ring

oft = 4(8h - by+ 0.00 4 (24-138g) En 37rh(d/h - 1) 2

where aft = opening stress when fitting the piston ring onto the piston

cr__Zw = 4 (b f - 0.00 4 (24-138h) En 37rh(d/h - 1) 2

where cr w = working stress when the piston ring is in the cylinder

oft + Crw 32 m u

En 3~(d/h - ])2 (24-138i)

Equation (24-138i) is independent of by.

Refer to Table 24-10C.

Refer to Table 24-10D.

Hardened r--11D--.1 Two bosses thus

0.27D 1 3 off,,,2, m

Cooling 4 ~ " water

connection N . , Two ribs i" "\ "" "J. thus • ~~-l[(rev°lved '~. / section) ii i' Coreplug!

i ' i 0.16D- . . - - . , - - " ,

! ./ ~ . . . . I

! . . . . - . . - .3 • Copper

~---- D joint ~-------- C -----------~ FIGURE 24-33 Valve with removable cage.


Particular Formula

The circumferential clearance (tS~) or gap between ends of ring

An expression for pressure acting on ring from Eqs. (24-138b) and (24-138c)

The pressure in the radial outward direction against the cylinder

For variable and constant radial contact pressure distribution of piston ring

The diametral load which acts at 90 ° to the gap required to close the ring to its nominal diameter, d (Fig. 24-28g)

The maximum bending stress at any cross section which makes an angle 4~ measured from the center line of the gap of the ring

The maximum bending stress which occurs at 0 = 7r, i.e., at the cross section opposite to the gap of the ring

The bending stress present in the ring of rectangular cross section in terms of free gap (6f) of the ring, when it is in place in the cylinder

The bending stress present in the ring of rectangular cross section in terms of tangential force, Fo (Fig. 24- 29)

The bending stress present in the case of slotted oil control ring of rectangular cross section in terms of free ring gap, 6f

6~ = do~pT (24-138j)

where

c~p = coefficient of expansion of piston ring material

T = operating temperature

d = cylinder diameter

ro p = - - (24-139a)

/'w

2Fo (24-139b) P = dw

Refer to Fig. 24-28f

Fd = 2.05F0 for modulus of elasticity E < 150 GPa

(24-140a)

Fa = 2.15Fo for modulus of elasticity E > 150 GPa

(24-140b)

Fd ~, 2.21F0 (24-140c)

12pr2 (q~) (24-141) ~r h = h 2 sin2

12pr 2 tTma x ~- h2ax

Eh ab = 0.4246j ( d - h) 2

where ab and E in N/mm 2

h, d, and 6f in mm

6 ( d - h) O'b = wh 2 Fo (24-142b)

where ab in N/mm2; Fo in N; d, h, and w in mm

(24-142)

Sl (24-142a)

O'bso = 0.424 6fElcolm (a-h)2I,,~

where Obso in N/mm 2 and

(24-142c)

Ius - moment of inertia of the unslotted cross-section ring, mm 4

Iu,+Is Ira= 2


Particular Formula

The bending stress present in the case of slotted oil control ring of rectangular cross section in terms of tangential load, Fo

The tangential load or force required for opening of a rectangular cross-section piston ring a

The piston ring parameter (k) in terms of tangential load Fo for rectangular cross-section rings

The tangential load or force required for opening of rectangular cross-section slotted oil control rings

The piston ring parameter (k) in terms of free ring gap (~Sf) for rectangular cross-section slotted oil rings for use in Eq. (24-142g)

The piston ring parameter (k) in terms of the constant pressure (p) for rectangular cross-section rings also for use in Eqs. (24-142f) and (24-142g)

The radial thickness of the ring at a section which makes an angle ~b measured from the center line of the gap of the ring

Is = moment of inertia of the slotted cross-section ring, mm 4

lco = twice the diameter between center of gravity and outside diameter, mm

6 ( d - h)leolm Crbso = wh3i s Fo (24-142d)

where ores o in N/mm2; Fp in N; lco, d, h, and w in mm; Im and Is in mm ~

hE °'°'max - - d - h (1"26e~ - 1.84k + 0.025)

SI (24-142e)

where

d + h 1 o r = d _ h

k- - piston ring parameter from Eq. (24-142f) and (24-142h)

3 ( d - h) 2 Fo k = wh----------g~ -~ (24-142f)

lci E °'0,max T ~-- d - h

where

(1 .26e7- - 1 . 8 4 k + 0.025) Im

(24-142g)

d + h 1 SI e T = d _ h

lei = twice the distance between center of gravity and inside diameter, mm

k = piston ring parameter from Eq. (24-142e) and (24-142f)

k --- 2 ~f (24-142h) 37r d - h

3 p d ( a - h ) ~ k = ~ ~ h-----g---- (24-142i)

3/24pr 4 dp (24-142j) h = V E~5 sin22

a Goetze AG, Piston Ring Manual, 3rd ed., Burscheid, Germany, 1987.


Particular Formula

The m a x i m u m thickness of the ring which occur oppos i te the gap of the ring (i.e., at 4~ = 70

F o r p is ton ring d imensional deviat ion, hardness, and m i n i m u m wall pressure

F o r cylinder bore d iameter

~ /24pr 4 hmax : V E6

Refer to Tables 24-7 to 24-9.


(24-142k)

T A B L E 24-5 Values of c for various inclinations of coned pistons

Inclination ranges, Cone 0, deg

0-6 1 Slightly 6-18 0.85-0.95 Medium 18-28 0.75-0.85 Strong 28-35 0.65-0.75

T A B L E 24-6 Values of coefficient K for pistons (admissible pressures, kgf ]mm 2 absolute)

Diameter of Pressure, kgf/mm 2

cylinder, mm 0.01 to 0.02 0.02 to 0.04 0.04 to 0.06 0.06 to 0.08 0.08 to 0.10 0.10 to 0.12 0.12 to 0.14 0.14 to 0.16

380-575 1.000 1.125 1.375 1.500 575-775 1.375 1.500 1.750 2.000 775-975 1.500 1.750 2.000 2.500 975-1175 1.750 2.000 2.375 3.000

1175-1375 2.000 2.250 2.750 3.125 1375-1575 2.375 2.500 3.125 3.500 1575-1775 2.500 3.000 3.375 4.000 1775-1975 2.750 3.125 3.500 4.125 1975-2150 3.000 3.375 3.750 4.375 2150-2350 3.000 3.500 4.000 2350-2550 3.125 3.500 2550-2750 3.375 3.750

1.750 2.000 2.500 2.750 3.125 3.500 3.500 4.000 3.750 4.125 4.000 4.375 4.375

2.125 2.500 3.000 3.375 3.750 4.000 4.500

Key: 1 kgf/mm 2 = 1.42247 kpsi; 1 kpsi = 6.894757 MPa.


TABLE 24-7 Recommended hardness for piston rings of IC engines

Nominal diameter, d, mm Hardness HRD

<100 95-107 100-200 93-105 > 200 90-102

TABLE 24-8 Minimum wall pressure for piston rings of IC engines

Compression rings Oil rings

MPa kgf/cm 2 MPa kgf]cm 2

Petrol a 0.059 0.60 0.137 1.40 Diesel 0.013 1.05 0.196 2.00

a G a s o l i n e .

TABLE 24-9 Permissible deviation on the dimensions of piston rings of IC engines

Dimensions Deviations, mm

Axial width, b -0.010 -0.022

Radial thickness <80 mm ring diameter >80mm with <175 mm ring diameter 175 mm ring diameter

Parallelism of sides--40% of tolerance on axial width

+0.08 4-0.12 +0.15

TABLE 24-10A Preferred cylinder bore diameters for internal-combustion (IC) engines (all dimensions in mm)

30 (62) 95 125 (152.4) (188) (241.3) 32 65 98 (127) 155 190 (245) 34 (68) (98.4) (128) (158) (190.5) (250)

(35) 70 100 (128.2) (158.8) (192) (254) 36 (72) (101.6) 130 160 195 (255) 38 (73) (102) (132) (162) (196.8) 266 40 74 (103.2) (133.4) 165 198 (265) 42 (76) (104.8) 135 (165.1) 200 270 44 (78) 105 (138) (168) (205) (273) 46 (79.4) 108 (139.7) 170 (209.6) (275) 48 80 110 140 (171.4) 210 280 50 82 (111.1) (142) (172) (215) (285) 52 85 112 142.9 175 (215.9) 290 54 87.3 (114.3) (145) (177.8) 220 (292.1) 56 88 115 (146) (178) (225) (295)

(57) (88.9) (118) (148) 180 (228.6) (298.4) 58 90 120 (149.9) (182) 230 300

(59) (91.4) (120.6) 150 (184.2) (235) (305) 60 (92) (122) (152) 185 240 103

315 (317.5) 320

(325) 330

(335) 340

(343) (345) 350

<

omm

Im

e~

0 ,,!

0

om

n e~

k,m

o I,m

o I,m

J omm

o m

-m

e~

0

omm

_e~

• ~. e~

~D

om

e~

Jopu

!sm0H

tz%

~ t~

t~

o.

o. o.

o.

tz,~ tz%

tz-~

~ o

o o.

o.

('xl ('xl

('~l ~

c~ c~

c~ c~

• "

c~c~

~ o

o

~ ~

.

c~ c~

c~ o

o. o.

o.

t~

oo

o

o

oo

o

o

oo

~ ¢xl

vh

tz~

o

o

oo

o

o

t~

¢'xl

~ o

o

oo

o

o

.9

~~

~.'~ .. ~

~~

~

~ o

c~ c~

~ c~

~

o ~.~

¢~ ~

24

.44


TABLE 24-10C Preferred number of piston rings

Differential pressure

Minimum number of rings

Std. atm. 0-9 10-14 15-24 25-29 30-49 50-99 100-200 MPa 0-0.88 0.98-1.37 1.47-2.35 2.45-2.85 2.94-4.80 4.90-9.71 9.81-19.61 psi 0-128 142-199 213-341 355-412 426-696 710-1406 1422-2844

2 3 4 5 6 7 8

Source: M. J. Neale, Tribology Handbook, Butterworth-Heinemann, London, 1973; reproduced with permission.

TABLE 24-10D Properties of typical piston ring materials

Typical Nominal modulus of coefficient of

elasticity, E. Brinell Bulk expansion, c~ hardness density, Wear

Material MPa kpsi GPa Mpsi number, HB g]cm 3 x 106/°C rating

Tensile strength, cr t

Metallic: Gray irons 230-310 33.4-45.0 83-124 12.1-18.0 210/310 Good Carbide malleable irons 400-580 58.0-84.1 140-160 20.3-23.2 250/320 Excellent Malleable and/or nodular 540-820 78.3-119.0 155-165 22.5-24.0 200/440 Poor

irons Sintered irons

Nonmetallic:

Carbon-filled PTFE 10.3 1.49

Graphite/MoS2-filled 19.6 2.85 PTFE

Resin-bonded PTFE 29,4 4.27 Carbon 43.4 6.30 Resin-bonded carbon 19.6 2.85 Glass-filled PTFE 16.7 2.42 Bronze-filled PTFE 12.8 1.85 Resin-bonded fabric 110.8 16.07

250-390 36.5-56.6 120 17.4 130/150 Good

2.05 55 2.20 115

1.75 1.8 1.9 2.26 3.90 1.36

a Material is anisotropic. Source: M. J. Neale, Tribology Handbook, Butterworth-Heinemann, London, 1973, extracted with permission.

30 43 20 80

118 22.5/87.5 a


T A B L E 24-10E Piston rings and piston ring elements

Designation Grade

Mechanical properties

Tensile strength, Modulus of Hardness O'st , MPa elasticity, E, MPa Main application

Steel GOE61 GOE62

GOE 64 GOE65A

GOE65B

Cr steel, 17% Cr min 380-450 HV 30 1200* approx. 230 000 approx. Compression rings Cr-Si steel 500-600 HV 30 1900* approx. 210 000 approx. Coil spring loaded

rings Cr-Si steel 450-550 HV 30 1700" approx. 210000 approx. Compression rings Cr steel, 11% Cr min, 300-400 HV 30 1300* approx. 210 000 approx. Compression rings,

high C nitrided Cr steel, 11% Cr rain, 270-420 HV 30 1300* approx. 220 000 approx. Coil spring loaded

low C tings and segments, nitrided

Cast Iron GOE 12 Unalloyed non heat- 94-106HRB 350min 85000 typical Compression and oil

treated gray cast iron control rings GOE 13 Unalloyed non heat- 97-108 HRB 420 min 95000-125000 Compression and oil

treated gray cast iron control rings GOE 32 Alloyed heat-treated 109-116 HRB 650 min 130000-160000 Compression rings

gray cast iron with carbides

GOE 44 Malleable cast iron 102-111 HRB 800 min 150 000 min Compression rings GOE 52 Spheroidal graphite cast 104-112 HRB 1300 min 150000 min Compression rings

iron GOE 56 Spheroidal graphite cast 40-46 HRC 1300 min 150 000 min Compression rings

iron

Source: Goetze Federal Mogul Burscheid GmbH, Piston Ring Manual, 4th ed., January 1995, Burscheid, Germany, reproduced with permission


24.5 DESIGN OF SPEED REDUCTION GEARS AND VARIABLE-SPEED DRIVES

S Y M B O L S 2'3

An

A £ Aw

b dl

42

dmax

amin

dmax D-- dmin D1

D2

e Fmax

h

hn

hc

h a

hi i

knl -- Fmax Fm

L m

Mts n

nl , /72 q

center distance, m (in) number of pinions or planetary pinion (Fig. 24-36) center distance (also with subscripts) (Fig. 24-36) area of reduction gear housing, m 2 (in 2) noncooled, i.e., ribbed, surface of housing of reduction gear

drive, m 2 (in 2) cooled surface of reduction gear drive, m 2 (in 2) surface area of contact of teeth when one-fourth of all teeth of

wheel in wave-type reduction gears are engaged, m 2 (in 2) width of rim, m (in) diameter of pinion, m (in) diameter of rigid immovable rim with internal teeth of

wave-type reduction gears, m (in) diameter of gear, m (in) diameter of flexible movable wheel rim with external teeth of

wave-type reduction gear, m (in) maximum diameter of the circumference of the belt

arrangement on the• V-belt of a variable-speed drive, m (in) minimum diameter of the circumference of the belt

arrangement on the V-belt of a variable-speed drive, m (in)

velocity control range for a V-belt drive

velocity control range for a V-belt drive with only one adjustable pulley

velocity control range for a V-belt drive with two adjustable pulleys

working height of a V-groove of the pulley, m (in) maximum load acting on the pinion, kN (lbf) mean load acting on thepinion, kN (lbf) height of tooth, m (in) coefficient of heat transfer, W/m 2 K (Btu/ftZh °F) coefficient of heat transfer of noncooled surface, W/m 2 K (Btu/

ft2ti °R) coefficient of heat transfer of cooled surface, W/m 2 K (Btu/ft 2 h

°R) addendum of tooth, m (in) dedendum of tooth, m (in) transmission or speed ratio

nonuniform load distribution factor

distance between the axes of the pinions (Fig. 24-36d) module, m (in) torque acting on smaller wheel, N m (lbf in) speed, rpm speeds of pinion and gear, respectively, rpm a whole number heat generated, W (Btu/h)


rmax

rmin

tl ta Z1, Z2

Z3

Zs

031 , 03 2

A

OL

O'ca

maximum radius of the circumference of the belt arrangement on the V-belt of a variable-speed drive, m (in)

minimum radius of the circumference of the belt arrangement on the V-belt of a variable-speed drive, m (in)

temperature of lubricant, °C (°F) ambient temperature, °C (°F) number of teeth on sun pinion and planetary pinion of epicyclic

gear transmission, respectively, Fig. 24-36 number of teeth on pinion and gear, respectively number of teeth on ring gear 3 (Fig. 24-36a) number of teeth on smaller wheel angular speed of pinion and gear, respectively, rad/s deformation, m (in) clearance between the pinions which should be at least 1 mm

(in) half-cone angle of V-belt, deg allowable compressive stress, MPa (psi)

Particular Formula

Transmission or speed ratio for single reduction gear

For different types of gear reduction drives

&l n l d2 z2 i ~ .__ ~ _ ~ . . ~

a;2 n2 dl Zl


(24-143)

-y

X ~ B e l t

E "1o

.

(a) V-belt at top position (b) V-belt at bottom position FIGURE 24-34 Dimension of V-belt variable-speed drive.


1. Single-reduction spur gear 5. Single-reduction bevel gear

H S, ~ _ L ~ -)

2. Double-reduction coaxial gear

3. Double-reduction spur and helical gear

7 Single-reduction worm gear with worm underneath

- I ~ L

(a) I i'/12 ' l /( i I / ~ _ L ~

4. Triple-reduction spur

6. Double-reduction bevel spur gear

11. Double-reduction worm

8. Single-reduction worm gear

9. Single-reduction worm gear with worm arranged sideways

L

10. Single-reduction worm gear with worm arranged vertically

gear

L -I

and helical gear

(a) h ' , ,2 ' /~ll '51t6 ' l / ( I I

H-,- I I L

(b) h',,2 '/(,, ~,m ',,/(, I / I i t ,14 , i \ l I I

with worm on top 12. Combination wormspur

helical reduction gear

H = High Speed I = Intermediate Speed L = Low Speed

FIGURE 24-35 Schematic diagrams of various types of spur, helical, herringbone, bevel, and worm reduction gears.


Particular Formula

P L A N E T A R Y R E D U C T I O N G E A R S

F i r s t c o n d i t i o n ~ m a t i n g

The sum of the radii of the addendum circles of the mat ing pinions in planetary reduction gears should be smaller than the distance between their axes (Fig. 24-36d) so that the top of the pinions should not touch each other

71" L = 2A1, 2 s i n - = z 2 m -+- 2m(1 + ~) + A (24-144)

a

where

a = number of pinions

A = clearance between the pinions, which should be at least 1 m m

Al,2 = center distance as shown in Fig. 24-36

3 3' 3 3' 1 2 3

2 . ~ 2 2 2'

i = 8 i = 15 i = 20 to 100

FIGURE 24-36 Planetary reduction gears.

S e c o n d c o n d i t i o n - - c o a x i a l i t y

The center distance of each pair of wheels should be equal (Fig. 24-36)

The relationship between teeth in corrected or uncorrected gears (Fig. 24-36a)

The relationship between teeth in corrected or uncorrected gears (Fig. 24-36c) to ratify two conditions

(i) First condit ion

(ii) Second condit ion

A I 2 = A 2 3 ; A 1 2 = A 2 3 = A 2 , y

Z 1 --[-Z 2 - - Z3 - - Z 2

o r

z l + 2z2 = z 3

Refer to Eq. (24-146).

mz(z3 - z2) = mlz(Z13 - zl2)

o r

! Z 3 - - -7 2 - - 2" 3 - - Z i since m2 = m~

(24-145)

(24-146a)

(24-146b)

(24-147a)

(24-147b)

M I S C E L L A N E O U S M A C H I N E E L E M E N T S 24.51

Particular Formula

Third condition--coincidence

The condition for the teeth and spaces of the meshed gears should coincide when the pinions are arranged uniformly over the circumference

The moment acting on smaller wheel

Z 1 -~-Z 3 ~ = q (24-148) a

where q is a whole number

M t l k n l Zs Mrs = (24-149)

a z 1

where

Z s - - Z 1 or Zs = z2 i f z 1 > z 2

k,t = 2 maximum value

= 1.4 to 1.6 for gears of 7th degree of accuracy

- 1.1 to 1.2 when floating central wheels are used to equalize the load

C O N D I T I O N S O F P R O P E R A S S E M B L Y O F

P L A N E T A R Y G E A R T R A N S M I S S I O N

Two planetaries

Both the driving pinion (sun pinion) and the planetaries may have either an even or an odd number of teeth.

Three planetaries

If Zl (number of teeth on sun pinion) is divisible by 3, then z2 (number of teeth on planetary pinion) must also be divisible by 3.

If z 2 - - 1 is divisible by 3, then Z 2 -~- 1 must be divisible by 3.

If Zl + 1 is divisible by 3, then z 2 - - 1 must be divisible by 3.

Four planetaries

If Zl is even, then z2 must be even.

If Zl is odd, then z2 must be odd.


Particular Formula

W A V E - T Y P E R E D U C T I O N G E A R S

Transmission or gear ratio

The necessary deformation

The condition for obtaining the module for the drive

The module of the drive from Eq. (24-152)

i - z2 = d2 (24-150) Z 1 - - Z 2 d l - d 2

For a double-wave drive, zl - z2 - - 2.

6 - - - d 1 - d 2 ~- d2 ( 2 4 - 1 5 1 ) l

d 1 - d 2 = (Zl - z2)m = 6 (24-152)

6 m = ~ = 0 . 5 6 (24-153)

Z 1 - - Z 2

The tooth height h = 6 (24-154)

The tooth addendum h a = 0.446 (24-155)

The tooth dedendum

The rim width

The total surface area of contact of teeth when one- fourth of all teeth of wheel are engaged

The torque transmitted

VARIABLE-SPEED DRIVES (Figs. 24-34 and 24-37, and Table 24-12)

For schematic arrangements of various variable- speed drives

The velocity control range for V-belt drive with only one adjustable pulley

The relation between dmax and dmi n of V-belt drive

h i = 0.566 (24-156)

b = 0.1d2 to 0.2d2 (24-157)

Aw = 0.5h × 0.25z2b (24-158)

The velocity control range for V-belt drive from Eqs. (24-160) and (24-161)

M t = 0 .5d2Awo ' ca " O .06d26bz2 f f ca ( 2 4 - 1 5 9 )

where O'ca - - 29.5 MPa (4.28 kpsi) for hardened steel wheels

Refer to Figs. 24-34 and 24-37.

dmax ( 2 4 - 1 6 0 ) D1 - dmin

dma x - - dmi n + 2 ( e - h )

dma x - - dmi n + b c o t a - 2 h

(24-161a)

(24-161b)

= ~ 2e 2h D dmax= 1-~ (24-162a) dmin dmin dmin

b 2h D = 1 + - S i nam c o t a -- nam--i~ (24-162b)


Particular Formula

The velocity control range for V-belt drive when two pulleys are adjustable

The total range of velocity control of variable-speed drive of two adjustable pulleys of V-belt drive

The working height of the V-groove of the pulley

The width of standard V-belt

D 2 -- D 1 (24-163)

D = D12 (24-164)

b e > ~ cot c~ (24-165)

b ~ 1.8h (24-166)

1. Frontal friction

H 2. Bevel friction

6. V-belt

i ' L

3. Toroidal friction

4. Ball

5. Disk

I TTTT, n ,-i I I I I I I I I I ' " * I - - ,-WWUo..,-

7. Chain

8. Combination

H

FIGURE 24-37 Variable-speed drives.


Particular Formula

The larger ratio of width to height of specially profiled broad V-belts

The total velocity control range for adjustable pulleys of V-belt drive

b ~_ 2 to 3 (24-167)

D = D 4 (24-168)

D I S S I P A T I O N O F H E A T I N R E D U C T I O N G E A R D R I V E S

The area of housing required for dissipating heat generated in a closed-type reduction gear drive operating in an oil bath at stable thermal equilibrium condition

The thermal equilibrium condition of reduction gear drive which has a housing of noncooled surface (ribbed surface) and cooled surface (cooled by blow- ing of air by fan)

The expression for coefficient of heat transfer of the housing or reduction gear drive blown over by air

The velocity of air which depends on impeller velocity

For minimum weight equations for gear systems

For total weight equations for gear systems

For K factors for preliminary estimate of spur and helical gear size

For comparison of five gear systems

A = ~ (24-169) h( t 1 - ta)

where h = coefficient of heat transfer, which varies from 8.75 to 17.5W/m2 K (1.54 to 3.1Btu/ ft 2 h °R)

<_ (h,,An + hcAc) W (Btu/h) (24-170)

he = 12x/~ W/m 2 K (Btu/ft 2 h °R)

where v - velocity of air, m/s (ft/min)

v ~ 0.005hi m/s (ft/min)

n i = impeller speed, rpm




(24-171)



T A B L E 24-11 Transmission ratio (t), efficiency (r/), and allowable transmitted power (Pat) for reduction gears

Type of reduction gear Fig. no. i rl Pat, kW

Single- and triple-spur and helical reduction gear

Single-spur reduction gear

24-35, 10-60 serial nos. 3a, 4a 24-35, <8-10 serial no. 1

Single worm 108 Helical worm 100 Harmonic drive 100 Planetary reduction gear 24-36a 8 0.97-0.99

24-36b 15 0.97-0.99 24-36c 20-100

Wave-type toothed reduction gear 100 0.75-0.85

1000 100

TABLE 24-12 Velocity control range (D), efficiency (r/), and allowable power transmitted (Pal) for variable-speed drives

Serial no. in Particular Type of drive Fig. 24-37 D Pal, kW

Frontal friction Single 1 3-4 Twin type 8-10

Bevel friction Single 2 3-4 Double 4-10 Self-locking ring 16

Toroidal friction 3 4-6 Ball 4 10-12 Disk drives 5 <3

4-5 V-belt drives Solid disk 6 1.3-

1.7 Grooved disk 2

Chain drives First type of drive 7 6 Second type of drive 7-10

20

0.7-0.8 10 0.95 20

0.8-0.9

0.8-0.9

8OO <300

50

30 75


MINIMUM AND TOTAL WEIGHT EQUATION FOR GEAR SYSTEMS

The following symbols are used in Tables 24-13 to 24-16: a = number of branches in an epicyclic gear: C = ( 2 M t / K ) , m3; d = pitch diameter, m (in); i = gear speed rati0~ io = overall ratio; is = dp /ds = Zp/Zs = speed ratio of planet gear to sun gear; j = number of idlers; K = a factor from Table 24-15; M t - - input torque, N m (lbf in ) ; (io + 1 ) / i o = {o.

TABLE 2 4 - 1 3

Minimum weight equations for gear systems TABLE 24-14 Total weight equations for gear systems

Particular Equation Particular Equation

Simple t ra in (offset) 2i 3 + i 2 -- 1 Offset wi th idler 2i3 + i 2 .2 = t o + l

.2

Offset wi th two idlers 2i3 _+_ i 2 _ _ to + 1 2

Offset wi th j idlers 2i 3 -+- i 2 =

2i 2 D o u b l e - r e d u c t i o n 2i 3 + m :!

lo

D o u b l e - r e d u c t i o n , d o u b l e 2i 3 -+ b r a n c h

D o u b l e - r e d u c t i o n , four

b r a n c h

D o u b l e - r e d u c t i o n , j

b r a n c h e s

P l a n e t a r y ( theore t ica l )

.2 l o + l

.2 t o + l

:! lo

.2 2i 2 to + 1

" 2i'o lo

.2 2i 2 to + 1 2i3 +-'7-, =

to 4{o

2i 3 + .2 2i 2 t o + 1

" jilo lo

• 2 0.4(i o - - 1 ) 2 -+- 1 2i~ + Is =

a

• 2 0.4i2o + 1 2i~ + l s = a

Star ( theore t ica l )

Offset

Offset wi th

idler

Offset wi th

two idlers

D o u b l e -

r educ t ion

D o u b l e -

r educ t ion , d o u b l e

b r a n c h

D o u b l e -

r educ t ion ,

four-

b r a n c h

P l a n e t a r y

Star

1 i2 E ( b d 2 / C ) = 1 + - + i + l

.2 1 i 2 lo .2 E ( b d 2 / C ) = l + - + i + + - + t o l l

• 2 i 2 1 1 i 2 to o

E ( b d2 / C ) = -~ + ~i + i + +~ii + - ~

• 2 .2 1 i 2 to to ..2 E ( b d 2 / C ) = 1 + - + 2 i + + - + + l l o -2i l °

1 1 i 2 i 2 .2 .2 E ( b d2 / C ) = _~ + ~ii + 2 i + + _ + t o ~ii + __ 5to

I 1 i 2 i 2 .2 .2 E ( b aa / C ) = _~ + ~ + 2 i + + _ + t o to ~ + -41°

E ( hd2 / C~ 1 1 .2 = _ + m + % + t,. + - - \ . . ~ 1 V / a at s

0.4(i o - 1) 2 +

a

0.4(i o - 1) 2

ai,

E ( b d 2 / C ) 1 1 .2 0-4i2o 0.4i2

a a t s a t s a


TABLE 24-15 K factors for preliminary estimate of spur and helical gear size

Hardness HB; Pitch line Particular pinion gear velocity, m/s kgf]mm 2

K factor

MN/m 2 MPa

Motor driving compressor 225-180 >20.5 0.036 Engine driving compressor 225-180 >20.5 0.032-0.050

575-575 0.155-0.320 Turbine driving generator 225-180 >20.5 0.066-0.077

575-575 0.280-0.56 Industrial drives 575-575 5.1 0.350-0.703

350-300 5.1 0.246-0.316 210-180 5.1 0.120-0.176 575-575 1 5 . 3 0.334-0.527 300-300 1 5 . 3 0.193-0.264 210-I80 15.3 0.088-0.141

Large industrial gears 225-180 5.1 max 0.056-0.070 such as hoists, kilns, and mills 260-210 0.091-0.120

Aircraft, single pair 60Rc-6ORc 51 0.703 (at take off) Aircraft, planetary 60Rc-6ORc 15.3-51 0.492 (at take off) Automotive transmission 60Rc-6ORc 1.055 Small commercial 350; phenolic <5.1 0.52

vehicles laminated nylon 0.035 Small gadgets 200; zinc alloy < 5.1 0.018

die casting 200; brass or A1 <2.55 0.018 Brass or A1 <2.55 0.016

TABLE 24-16 A comparison of five gear systems (all systems producing 0.746 kW at 18 rpm)

0.353 0.353 0.314-0.49 0.314-0.49 1.52-3.14 1.52-3.14 0.65-0.76 0.65-0.76 2.746-5.50 2.746-5.50 3.434-6.89 3.434-6.89 2.234-3.100 2.234-3.10 1.177-1.726 1.177-1.726 3.277-5.170 3.277-5.170 1.893-2.589 1.893-2.589 0.873-0.138 0.873-0.138 0.550-0.687 0.550-0.687

0.893-1.177 0.893-1.177 6.89 6.89 4.82 4.82 10.35 10.35 5.10 5.10 0.343 0.343 0.176 0.176

0.176 0.176 0.157 0.157

Parameter Epicyclic Herringbone Single w o r m Helical worm Harmonic drive

Speed ratio 97.4 96.2 108 100 100 Safety factor 3 2 2 2 36 Height, mm 330 356 580 406 152 Length, mm 381 508 483 432 152 Width, mm 330 254 356 254 152 Cubic volume, m 3 0.0410 0.0458 0.1000 0.0442 0.003 Weight, kgf 111.60 127.00 104.33 93.00 13.61 Efficiency, ~7% 85 85 40 78 82 Number of gears 13 4 2 4 2 Number of bearings 17 6 6 6 2 Tooth-sliding velocity, m/s 12.75 12.75 7.65 12.75 0.143 Pitch line velocity, m/s 7.65 7.65 7.65 7.65 0.092 Tooth contact pressure, kgf/mm 2 35 35 3.5 35 0.425 Tooth contact pressure, GPa 0.343 0.343 0.034 0.343 0.0042 Tooth in contact, % 7 5 2 3 50 Tooth contact Line Line Line Line Surface Quiet operation No Yes Yes Yes Yes Balanced forces Yes No No No Yes


24.6 FRICTION GEARING

S Y M B O L S 2,3

b dl

F Fa fr

FR

F, h i n t

n

P pl

Vm R ct

P # #l

031, 032

61, 62

center distance, m (in) dimensions as shown in Fig. 24-42 gear face width, m (in) diameter of smaller wheel, m (in) diameter of larger wheel, m (in) pressure on wheels, kN (lbf) thrust, kN (lbf) radial force on the grooved spur wheel for each groove, kN

(lbf) normal reaction between two bevel friction gears (Fig. 24-40),

kN (lbf) tangential force, kN (lbf) depth of groove, m (in) number of grooves, m (in) speed, rps speed, rpm power transmitted, kW (hp) permissible pressure, kN/m (lbf/in) mean circumferential velocity, m/s (ft/min) cone distance, m (in) (Fig. 24-40) half the included angle of the groove, deg ranges from 12 ° to 18 °

(should not exceed 20 ° ) angle of friction, deg coefficient of friction between wheels coefficient of friction between shaft of wheel and bearings angular speeds of smaller and larger wheels, respectively, rad/s cone center angles of smaller and larger wheels, respectively,

deg

Particular Formula

SPUR FRICTION GEARS

Plain spur friction wheels (Fig. 24-38)

The radial pressure on the wheels

The tangential force due to radial pressure F

The power transmitted

F = bp t (24-172)

1:, = ubp' (24-173)

p = Ft Vm SI 1000

where P in kW, Ft in N, and V m in m/s

(24-174a)

p = Ft 12m USCS 33,000

(24-174b)

where P in hp, Ft in lbf, and F m in ft/min

Particular

F t

. . . . . . F = b p '

Formula


FIGURE 24-38 Plain spur friction gears.

The gear face width

P = Ftl:m 75 Customary Metric (24-174c)

where P in hpm, Ft in kgf, and Vm in m/s

1000P b - 7r#p, dn---------- ~ SI (24-175a)

where P in kW, p' in N/m, n' in rps, and b and d in m

102 × 103p b = :r#p,dn~ Customary Metric (24-175b)

where P in kW, p' in kgf/mm, n' in rps, and b and d in mm

33,000P b - 7 r # p ' ~ USCS (24-175c)

where P in hp, p~ in lbf/in, n in rpm, and b in in and d i n f t

126,000P b = #p'---------d~n USCS (24-175d)

where b and d in in and p' in lbf/in, n in rpm, and P in hp

Grooved spur friction wheel (Fig. 24-39)

The radial force on the wheel for each groove

The total tangential force

The power transmitted

Fr = 2p'h(tan c~ + #) (24-176)

Ft = 2 # i p ' h sec c~ (24-177)

p = 27r i#hp tdl n ~ S| (24-178a)

1000 cos c~

where P in kW, p' in N/m, n' in rps, and h and dl in mm


Particular Formula

[=r

A

d2

dl

FIGURE 24-39 Grooved spur friction gears.

p = i#hp'dln USCS (24-178b) 63,000cos a

where P in hp, p' in lbf/in, n' in rps, and h and dl in in

p = 27ri#hp'dln Customary Metric (24-178c) 4500 × 103 cos c~

where P in hpm, p' in kgf/mm, n in rpm, and h and dl in mm

The empirical relation for the depth of the groove h = 0.006dl + 4 mm (0.15 in) (24-179)

The recommended value for the mean circumferential velocity

1~ m ~ 6 + 0.08dl

w h e r e i ' m in m/s and dl in m

SI (24-180a)

I' m ~ 1200 + 4dl USCS (24-180b)

where v in ft/min

BEVEL F R I C T I O N G E A R S (Fig. 24-40)

Starting

The reaction is inclined from the normal by an angle of friction p

~la F'2a (24-181) F'R = sin(61 + p) = cos(61 + p)


Particular Formula

The tangential force transmitted

The least axial thrust on the small wheel

The least axial thrust on the big wheel

Running The reaction in this case is designated by FR <_ bp' (where p' is the permissible unit pressure)

I

I • I o11

I~ d ~1

m o ) 2

A F2 a

Fla

FIGURE 24-40 Bevel friction gears.

Ft = #bur cos p = 1000P

Vm SI (24-182a)

~t = #/JR COS P = 75P

Vm

Customary Metric (24-182b)

b l a - - 1000P(sin~l + ~ c o s ~ l ) Sl (24-183a) #Vm

-~/;~'a = 33,000P(sin 61 n t- # COS 61 ) USCS (24- 1 83b)

#Vm

~2a : 1000P(c°stS1 -- #s ina i ) Sl (24-184a) lZVm

bZa = 33,000P(cos 61 - # sin 61) USCS (24-184b) #Vm

Fla f2a (24-185) FR = sin t~ 1 --" COS 61


Particular

The tangential force transmitted

The least axial thrust on the small wheel

Ft = # F R = ~ 1000P

Vm

Ft = # F R = ~ 33,000P

Vm

F1 a --- 1000P sin ~51

#Vm looo [ , 1 ~ Vm V / 4 -.]- d~2

F1 a --- 33,000 P sin 61

#Vm

33 ' 000 I'(dl~!_.~_ d~2 1 ~V m

F2a -" 1000Pcos ~51

#Vm

]-Z l~ m "t- d~2

F2a = 33,000P cos /51

~Vm

l vm

The least axial thrust on the big wheel

1000P

Formula

SI (24-189a)

9550P SI (24-189b)

where M t in N m, P in kW, and n in rpm

159P Sl M t = n I

where Mt in N m, P in kW, and n' in rps

(24-189c)

DISK F R I C T I O N GEARS (Fig. 24-41)

The torque on the driving shaft

m t -'-

SI (24-186a)

USCS (24-186b)

SI (24-187a)

USCS (24-187b)

SI (24-188a)

USCS (24-188b)

where M t in N m, P in kW, and ~v in rad/s


Particular Formula

Driven spur

f ' "1 "~ ~ '

Driving isc

FIGURE 24-41 Variable-speed disk friction gearing.

The tangential force acting on the driven wheel for the minimum speed at minimum diameter of driving disk

The tangential force acting on the driven wheel for the maximum speed at maximum diameter of driving disk

716,000P Mt = ~ Customary Metric (24-189d)

n

where Mt in kgf mm, P in hpm, and n in rpm

63,000P Mt = ~ USCS (24-189e)

H

where M t in lbf in, P in hp, and n in rpm

1000P Ftl--- 7r-'l n--------T" SI (24-190a)

where Ftl in N, P in kW, dl in m, and n ~ in rps

33,000P = ~ USCS (24-190b) Ftl :rdi n

where Ftl in lbf, P in hp, dl in ft, and n in rpm

102 × 103p Ftl = 7rdln' Customary Metric (24-190c)

where Ftl in kgf, P in kW, d 1 in mm, and n ~ in rps

1000P Ft2 = 7rdzn' SI (24-191a)


Particular Formula

The minimum thrust to be applied to the disk for the minimum speed

The maximum thrust to be applied to the disk for maximum speed

The axial thrust required to shift the driven wheel underload

The efficiency

The minimum force available on the chain sprocket at minimum speed of driven wheel

The maximum force available on the chain sprocket at maximum speed of driven wheel

33,000P Ft2 = USCS (24-191b)

7rd2n

where d2 in ft

102 × 103p Ft2 -- 7 r d 2 n ' Customary Metric (24-191c)

Fal = Ft---Ll = 1000_____P_P SI (24-192a) # #Trdl n'

Fal = 33,000_______~P USCS (24-192b) #~rdl n

where Fal = bp' (b = face width of driven cylindrical wheel) and d2 in ft

Ft2 1000P SI (24-193a) F 2 a " - - ~ - - -

I,Z #Trdzn'

Fza = 33,000_______~P USCS (24-193b) #zrdzn

where d2 in ft

Fza = 126,000P USCS (24-193c) #nd2

where d2 in in and Fza in lbf, n in rpm, and P in hp

Fa = F, (# + #') (24-194)

where J is the coefficient of friction between the shaft of driven wheel and its bearings

d 77 = d + b (24-195)

where r/varies from 0.6 at low speeds when d = dl to 0.8 at high speeds, when d = d2

Flcs -- r/Ftld (24-196) d3

where

d = diameter of driven wheel, m (in)

d3 = diameter of chain sprocket, m (in)

Fzcs = ~TFtz~d (24-197) 't3


Particular Formula

BEARING LOADS OF FRICTION GEARING (Fig. 24-42, Table 24-17)

Driven shaft

The horizontal force on bearing A due to the tangential force Ft

The vertical force on bearing A due to thrust Fa and the force on the chain sprocket Fcs

The resultant load on bearing A

The horizontal force on bearing B due to the tangential force Ft

The vertical force on bearing B due to the thrust Fa and the force on the chain sprocket Fcs

The resultant force on bearing B

FhA = (L + e)F t

( e + L + c )

FVA -- (L + e)Fa + eFes

( e + L + c )

FR. = V /G + r~.

FhB --- cF~

(e + L + c)

FVB = c& + (c + L ) G

( e + L + c )

FR~ = v/F]~ + F2v.

(24-198)

(24-199)

(24-200)

(24-201)

(24-202)

(24-203)

Driving shaft N

Bearing \ j

I I

.lq k~j/l~..1"-- Bearing

~ Driving ]~ disc

I

wheel Driven r-~ E/-Chain sprocket A . . . . shaft "~ t I I ~ e a r i n g

cha'n I L e

m

FIGURE 24-42 Bearing loads of disk friction gearing.


Particular Formula

Driving shaft

The horizontal force due to thrust F a on bearing D

The horizontal force due to the tangential force Ft on the bearing D

The resultant force on the bearing D

The horizontal force due to thrust Fa on the bearing C

The horizontal force due to the tangential force Ft on the bearing C

The resultant force on the bearing C

dlFt (24-204) FhDa: ~a

where dl and dE denote the minimum and maximum diameters of driving disk

FhD t -- bEt (24-205) a

FRD = v/F2Da + F2hDt (24-206)

Fhca = d l F t (24-207) 2a

Fhct = (a + b)F t (24-208)

FRc -- V/Fhca + F2hct (24-209)

TABLE 24-17 Design data for friction gearing

Allowable pressure, p'

Material of driver kN/m lbf/in

Coefficient of friction/z with cast iron

Allowable pressure, p'

Material of driver kN/m Ibf/in

Coefficient Coefficient of of friction friction/~ with p with cast iron aluminum

Cast iron 530 3000 0.15 Cork composition 8.9 50 0.21 Paper 26.5 150 0.15 Rubber 17.7 100 0.20 Wood 26.5 150 0.15

Leather Leather fiber Straw fiber Sulfite fiber Tarred fiber

26.5 150 0.09 0.13 42.2 240 0.18 0.18 26.5 150 0.15 0.16 24.5 140 0.20 0.19 44.1 250 0.28 0.28


24.7 M E C H A N I C S OF VEHICLES

S Y M B O L S 2,3

B c

C Ot Dw EI

Esh

F Fmax

G h i k l { ll 12 13 14 15 16 17 18 m Mt M. mti mto

~/ItF

n /7 t

ni no P F

rmi Fmo

center distance, m (in) a constant in Eq. (24-216b) frontal projected area of vehicle, m 2 (ft 2) face width of gear, m (in) a constant in Eq. (24-216b) width of bearing, m (in) distance between adjacent rotating parts, m (in) constant (also with suffixes) maximum diameter of torus, m (in) diameter of wheel, m (in) flow loss in each member of hydraulic torque converter, N m

(lbf in) shock loss in each member of hydraulic torque converter, N m

(lbf in) driving force at the tire, kN (lbf) maximum permissible load on the pitch circle of any particular

pair of gears, kN (lbf) gradient thickness of housing, m (in) gear ratio (total) a constant distance between support bearings on a shaft in gearbox, m (in) distance between bearings of overhanging shaft, m (in) distance of rotating part from the bearing, m (in) distance of bearing from the wall, m (in) cap height from bolt to end, m (in) distance of rotating parts from the bearing cap, m (in) width of boss of rotating parts, m (in) distance of coupling to cap, m (in) distance between gear and shaft, m (in) distance of rotating parts from inner wall of housing, m (in) module, m (in) output torque of the engine, N m (lbf in) torque at the tire surface, N m (lbf in) the input torque, N m (lbf in) the reaction to the output torque, which is opposite in direction

to output torque, N m (lbf in) the torque that must be applied to transmission housing to

balance the moments of internal friction, oil churning, etc., N m (lbf in)

the torque reaction of the transmission housing due to the gear reduction in transmission, N m (lbf in)

speed, rpm speed, rps speed of driving shaft, rpm speed of driven shaft, rpm power, kW (hp) radius of the driving wheel, m (in) mean radius of inflow to the runner, m (in) mean radius of outflow from the runner, m (in)


R a

Rr R" Rg

Rt t

tj V

V vj Vsh W

Z

OL

A

air resistance, kN (lbf) rolling resistance, kN (lbf) road resistance, kN/tf (lbf/ton) gradient resistance, kN (lbf) total resistance, kN (lbf) tonne, t tonne force, tf velocity, m/s (ft/min) speed of vehicle, km/h (ft/s) velocity of fluid relative to the vane, m/s (ft/min) shock velocity, m/s (ft/min) weight of the vehicle, kN (Tonf) number of teeth angle of inclination of road, deg angle of repose, deg minimum clearance between gears and inner wall of housing, m

(in) transmission efficiency

SUFFIXES

1 pinion 2 gear b brake t tonne max maximum min minimum

Other factors in performance or in special aspects which are included from time to time in this section and, being applicable only in their immediate context, are not given at this stage.

Particular Formula

CALCULATION OF POWER

Torque M t _ _ 1000_____~P

FIGURE 24-43 Forces on the vehicle moving up the gradient.

SI (24-210a)

where P in kW, ,~ in rad/s, and Mt in N m

9550P M t = SI (24-210b)

where P in kW, n in rpm, and M t in N m

63,000P M t = ~ USCS (24-210c)

where P in hp, n in rpm, and M t in lbf in


Particular Formula

Torque at the tire surface

The driving force at the tire

Tractive factor

Force required to pull the vehicle of weight W up the slope (Fig: 24-43)

M t t - - T]Mt

where

r /= 0.90 at top gear

r /= 0.80 at other gears

F = ~TMt r

g . fir = looow

(24-211)

(24-212)

SI (24-213a)

m t t

fir = 2240 W USCS (24-213b)

1000Wsin(a + ¢) SI (24-214a) Rg = cos¢

2240Wsin(a + ¢) USCS (24-214b) Rg = COS

Gradient G = W (24-215) R.

The air resistance

For values of air resistance at different speeds of vehicle

The rolling resistance

For rolling or road resistance R'r for various road surfaces

The general formula for total resistance or tractive resistance (Fig. 24-44)

Another formula for total resistance

R a - - k A V 2 (24-216a)

where k = constant obtained from Table 24-18


Rr- - ( a + b V ) W

where

a - constant varies from 15 to 600

b - constant varies from 0.1 to 3.5


(24-216b)

R t - - k A V 2 -Jr- W S'9(OL; + ¢)

COS

+ (a + b V)W (24-217a)

Rt = Ra + Rg + Rr

1000 ) = W R'r + --'G- + kA V2 (24-217b)

where k and R'r are obtained from Tables 24-19 and 24-20 where R'r in N/tf, W in tf, A in m 2, V inm/s


Particular Formula

Tractive effort at the tire surface

The speed of the vehicle

2

J - - R O L L I N G R E S I S T A N C E R r = ( a+bv ) ,W m

FIGURE 24-44 Various resistances on the moving vehicle.

R t : - R a-k-Rg-+-R r

= W R ' r + ~ + k A V 2 USCS (24-217c)

where R'r in lbf/t, W = weight of vehicle, tonf

A = projected frontal area of vehicle, ft 2

V = speed of vehicle, ft/s

Ftr= irlMt (24-218) r

where i = gear ratio obtained from Table 24-21

V = 0.00297 nD---~w USCS (24-219a) i

where V in mph (miles per hour), Dw in in, and n in rpm

V = 0 . 0 5 2 ~ nDw

SI (24-219b)

where V in m/s, Dw in m, and n in rpm

p O.O02VMtt

DW SI (24-220a)

where V in m/s, Mtt in N m, Dw in m, and P in kW

p _.._ O.O0163VMtt

Dw USCS (24-220b)

where V in mph, Mtt in lbf ft, Dw in in, and P in hp [ ' G R A D I E N T R E S I S T A N C E R . . . . ,

5.5VM n

DW Customary Metric (24-220c)

where V in km/h, Mtt in kgfm, Dw in mm, and P in kW

T R A N S M I S S I O N G E A R B O X (Fig. 24-45)

The equation for center distance between main and countershafts for the case of three-speed passenger car

a = 0 . 5 ~ t

where a in in and Mt in lbf ft

a = 0.0106 v'Mt

where a in m and M t in N m

USCS (24-221a)

SI (24-221b)


Particular Formula

i A 4

Clutch

- - C

. . . . t Z

m

The distance between support bearings of shaft

The maximum permissible load at the pitch circle of any pair of gears

The face width of gear tooth

The expression for center distance for th e case of four- speed truck transmission

The distance between support of bearings of shaft


FIGURE 24-45 A typical four-speed gearbox.

0 to 0 0 18

where l in m and Mt in N m

SI (24-222a)

1--1.2 ~//Mt to 1.5 ~ t

where I in in and Mt in lbf ft

USCS (24-222b)

Fma x =- clbm = Clb (24-223) Pm

where c 1 - - - c o n s t a n t obtained from Table 24-22

b - - Fma-------~x - - F m a x P n

mcl Cl

a = 0.017

where a in m and M t in N m

(24-224)

SI (24-225a)

a = 0 . 8 ~ t

where a in in and M t in lbf ft

1= 0.0.254 ~ t to 0.0318 ~ t

where I in m, and Mt in N m

l = 1.2 ~ t to 1.5 ~ t

where l in in and Mt in lbf ft

b = Fmax = FmaxP---------~n mcl c1

For values of cl, refer to Table 24-22.

USCS (24-225b)

SI (24-226a)

USCS (24-226b)

(24-227)


Part icu lar F o r m u l a

The expression for center distance for the case of five-speed and reverse truck transmission

The distance between support of bearings of shaft


The expression for center distance for a farm tractor transmission

Effective face width of gear tooth

The efficiency of transmission

Distance of rotating parts from the inner wall of housing

Distance between adjacent rotating parts

Minimum clearance between gears and inner wall of housing

Distance between bearings of overhanging shaft

Distance of bearing from the wall

Cap height from bolt end

Distance of rotating parts from the bearing cap

a = 0.0170 ~ t SI (24-228a)

where a in m and Mt in N m

a = 0.8 ~ t USCS (24-228b)

where a in in and Mt in lbf ft

l = 0.0254 ~ t to 0.0318 ~ t SI (24-229a)

where l in m and Mt in N m

l = 1 . 2 ~ t to 1.5 ~ t USCS (24-229b)

where l in in and Mt in lbf ft

b = Fma-----~x -- FmaxP~n (24-230) mcl Cl

For values of c~, refer to Table 24-22.

a = 0.021 ~ t SI (24-231a)

where a in m and Mt in N m

a = ~ t USCS (24-231b)

where a in in and M t in lbf ft

FmaxV Sl (24-232a) b = 28 x 105m

where b in m, Fmax in N, m in m, and v in m/s

b = FmaxvPn USCS (24-232b) 8,000,000

where b in in, Fmax in lbf, Pn in in -1 , and v in ft/min

noMto Mto = (24-233)

'17 : n i M t i i r [ M t o _ ( M t r _ M t f ) ]

where ir = reduction ratio of transmission = (ni/no)

18 = 10 to 15 mm or more for high-power and heavy- duty operation

/8 = 0.4 to 0.6 in

c = 10 to 15 mm (0.4 to 0.6 in)

A > 1.2h B

where h = thickness of housing

1' = 1.2d to 3d

where d = diameter of shaft

12 = 5 to l Omm (0.2 to 0.4in)

USCS (24-234)

(24-235)

(24-236)

(24-237)

(24-238)

/3 = depends on the design by empirical formula

14 -- 15 to 20mm (0.6 to 0.8 in) (24-239)

MISCELLANEOUS MACHINE ELEMENTS 2 4 . 7 3

Particular Formula

Width of boss of rotating part

Distance of coupling to cap

Distance between gear and shaft

Distance of rotating part from the bearing

For planetary gear transmission

For detail design equations of spur, helical, bevel, crossed-helical and worm gears

ls-- 1.2d to 1.5d

(depends on the type of coupling)

/7 _> 20 mm (0.8 in)

B ls t~ = ~ + t3 + 14 -+-~

Refer to Chapter 24, Section 24.5.

Refer to Chapter 23.

(24-240)

(24-241)

(24-242)

HYDRAULIC COUPLING (Fig. 24-46)

Torque transmitted by the coupling

Percent slip between primary and secondary speeds

T J . _

~ FIGURE 24-46 Hydraulic coupling.

The mean radius of the inner passage (Fig. 24-46)

The mean radius of the outer passage (Fig. 24-46)

The expression for number of times the fluid circu- lates through the torus in one second

The torque capacity of hydraulic coupling at a given slip

Mt __ ksn 2 W(r2o 2 - rmi) (24-243)

where k = coefficient = 1.42 x 10 -7 (approx.)

(np - ns) s = ~ x 100 (24-244)

np

where np and ns are primary and secondary speeds of impeller, respectively, rpm

2 (r 3 _ r] rmi r~ r~j

2 rmo = ~ (~4 - ~ ) 13,000Mr

il = ,, w ( r ~ o 2 Fmi )

Mt = KnZD~

where

K = coefficient varying from 0.166 x 108 to 0.244 x 108 SI

= 1.56 to 2.28 USCS

D t = diameter of torus, m (ft)

Mt = torque capacity, N m (lbf ft); n in rpm

(24-245)

(24-246)

(24-247)

(24-248)


Particular Formula

H Y D R O D Y N A M I C T O R Q U E C O N V E R T E R

(F ig . 2 4 - 4 7 )

The equation for input torque

Runner Impeller

Mti = Kn2D~ (24-249)

where

K = coefficient depending on design

n i -- speed of input shaft, rpm

D t - - a n y linear dimension such as maximum diameter of impeller

FIGURE 24-47 Hydrodynamic torque converter.

The equation for the input power

The expression for flow loss or friction loss in each member of the torque converter under any particular operating conditions in energy unit per kilogram of fluid circulated

The expression for shock loss per kg fluid circulated in the impeller of a torque converter

The maximum inside diameter of torus

P - Cn~D~ (24-250)

where C = coefficient depending on design

cfvZf (24-251) Ef= 2g

where

(7].= coefficient whose value depends mainly on the Reynolds number and the relative smoothness of the metallic surface

= 0.445 to 0.890 SI (where EU in N m and VU in m/s)

= 0.328 to 0.656 USCS (where VU in ft/s and EU in lbf ft)

c~, Vs~ E~ = ~ (24-252)

2g

where C,, = coefficient

3 / Dt = 0.00135C {¢/Mt/n '2 SI

where Dt in m, Mt in N m, and n ~ in rps

(24-253a)

/Mt D, = 0 . 0 0 1 6 8 C v ~ USCS (24-253b)

where Dt in in, Mt in lbf in, and n in rpm

C - coefficient = 14 for a ratio of minimum inside diameter to maximum diameter of torus of one-third

n = speed in hundreds of rpm


Particular Formula

T R A C T I V E E F F O R T C U R V E S F O R C A R S , T R U C K S , A N D C I T Y B U S E S

For finding the diameter of tire of vehicles for a particular wheel speed

For tractive effort of a passenger car

For tractive effort of trucks, tractors, and city buses




850

800

750

700

650

600 E L,-

550

• -= 500

450

400

350

300

\

25050 60 70 80 90 100 110

Tyre diameter, cm

FIGURE 24-48 Wheel speed vs. tire diameter of vehicles.

-.,,.

120 130


200

co 100

~ 50 L

f

I

f

500 1000 1500 2000 Gross vehicle weight, kgf

2500

F I G U R E 24-49 Tractive effort curve for passenger cars (l kgf = 9.8066 N = 2.2046 lbf).

950[ ]

9001

850

800

750

700

650

600

"-" 550 o}

~_ soo

• -> 450

400

350

300

250

200

150

100 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000

Gross vehicle weight, kgf

F I G U R E 24-50 Tractive effort curve for trucks, tractors, and city buses.

MISCELLANEOUS MACHINE ELEMENTS 2 4 . 7 7

TABLE 24-18 Values of k for use in Eq. (24-216) and (24-217)

Particular k in USCSU a k in s i

Average automobile of 0.0017 0.20 modern design

Streamlined racing car 0.0006 0.07 Truck or omnibus 0.0024 0.28

a US Customary System units.

TABLE 24-19A Air resistance a

Speed of Velocity of vehicle, wind, V, mph ft/s 0.0024V 2 0.0017V 2 0.0006V 2

i0 14.67 0.516 0.366 20 29.35 2.060 1.460 0.516 30 44.00 4.650 3.300 1.160 40 58.60 8.240 5.830 2.060 50 73.30 12.900 9.130 3.220 60 88.00 18.600 13.160 4.650 90 132.00 29.650 10.450

150 220.00 29.000

a Values given in this table are in US Customary System units.

TABLE 24-19B Air resistance in SI units

Speed of Velocity of vehicle wind km/h ~ V, m/h 0.28V 2 0.20V 2 0.07V 2

10 20 30 4O 50 60 70 80 90

100

2.78 5.56 8.34

11.i2 13.90 16.68 19.46 22.21 25.02 27.80

2.17 1.55 0.54 8.68 6.20 2.17

19.50 13.92 4.88 34.72 24.80 8.68 54.25 38.80 13.56 78.00 57.60 19.50

106.40 76.00 26.60 139.00 99.20 34,75 176.00 125.60 44.00 217.00 155.00 54.25

TABLE 24-20 Road resistance, R'r

Surface N/tf

Solid

lbf/ton % N/tf

Pneumatic

lbf/ton %

Polished marble Concrete Asphalt Stone

Good quality Poor quality

Vitrified bricks Good macadam (metal road)

Good Fair Rough Clay Sand

29.3 62.4 67.4

7!.8 153.2 85.0

146.2 220.0 307.0 438.4

1314.0 i

12.12 14.25 15.40

16.40 35.00 19.45

33.50 50.00 70.00

100.00 300.00

0.541 0.636 0.687

0.732 1.562 0.866

1.491 2.240 3.130 4.470

13.400

35.3 41.5

448.2

477.6 102.0 56.7

97.9 186.4 389.3 389.3 874.8

8.08 9.50

10.25

10.92 23.30 12.95

22.20 33.20 46.60 66.60

200.00

0.360 0.423 0.457

0.487 1.040 0.578

0.998 1.900 2.100 3.970 8.920

24.78

¢',1

=

.,..~

0 ~O

,.C

o.~

o~

°~

¢',I

<--

×

% )<

! ×

! ×

X

= 0 o

~

°! 0

Z

t~

t"4

¢'4

to%

to%

t'-4

¢'-4

¢-4

~0

[-

t"4 t"4

r-- r-:

tr5

to%

tr~

~.

r--

oo

,-- t"4

¢'4 ¢'4

Lr~

r--- t -~


24.8 INTERMITTENT-MOTION MECHANISMS

S Y M B O L S 2,3

a

b

d

e l , e2

f . f.r F, h

m

Mb M t

! 11

11

P

P F

S1

$2

Z

Z

Oz

# = tan p = p = tan -1 # =

b @ = - - m

O"

O" b

7"

distance of the pawl pivot point, m (in) face width of ratchet tooth, m (in) diameter (also with suffixes), m (in) hub diameter, m (in) dimensions as shown in Fig. 24-52, m (in) normal force through O, Figs. 24-51 and 24-52, kN (lbf) peripheral force normal to the tooth of ratchet, kN (lbf) tangential force at diameter, d, kN (lbf) tooth height or distance from the critical section to the line of

action of the load Fnr, m (in) module, m (in) bending moment, N m (lbf in) twisting moment, N m (lbf in) speed, rps speed, rpm tooth pitch, m (in) linear unit pressure, N/m (lbf/ft) radius, m (in) dimension as shown in Fig. 24-53 breadth of tooth land (Fig. 24-53), m (in) thickness of tooth at base (Fig. 24-52), m (in) number of teeth on ratchet wheel section modulus, m 3 (crn 3) (in 3) pressure angle or angle of the pawl force, deg angle at pawl (= 90 ° - a°), deg coefficient of friction friction angle, deg ratchet tooth angle or pitch angle, deg

varies from 1.5 to 3

stress, MPa (psi) bending stress, MPa (psi) shear stress, MPa (psi)

Particular Formula

P A W L A N D R A T C H E T

The ratchet tooth angle (Fig. 24-51)

The ratchet diameter (Fig. 24-51)

27r

Z , r a d

360

Z , deg

d2 = m z - p z 7"f

(24-254a)

(24-254b)

(24-255)

Particular Formula

b>F,, - r n

The face width of ratchet tooth

The allowable unit pressure or force

The tangential force

The normal force through O (Fig. 24-51)

The normal force through O (Fig. 24-52)

r=Fn b

2Mt El= d

Ft F n = ~

COS o~

F.=Ft

Allowable bending stress (O'b)

M b 6Fth The bending stress orb-- Z - - ~ <- O'ba

Module


Number of teeth z - 6 to 30

m > 6 (mostly from 10 to 20)

The ratio of h/m

For ratchet wheel definitions and dimensions

The ratio of s2/m

The tooth height

h/m = O.6 to 1


s2/m = 0.6 to 0.9

h = 5 to 15 for toothed ratchet

~l~Fnr ~ PFnr Fnr = ~

(24-256)

(24-257)

(24-258)

(24-259a)

(24-259b)

(24-260)

(24-261)

(24-262)

(24-263)

(24-264)

(24-265)


FIGURE 24-51 Ratchet wheel with radial tooth flanks and pawl.

FIGURE 24-52 Ratchet wheel with non-radial tooth flanks and pawl.


Particular Formula

FIGURE 24-53 Definitions and dimensions of ratchet wheel.

For external ratchet

For internal ratchet

The ratio of a/d (internal ratchet)

The module

The bending moment on pawl

The bending stress

The diameter of pawl pin

TABLE 24-23

F* O" b

Material kN/m lbf]in MPa kpsi

Cast iron Steel or

cast steel Hardened

steel

49-98 280-560 19.5-29.5 2.85-4.27 98-196 560-1120 39-68.5 5.69-10.0

196-392 1120-2240 58.8-98 8.54-14.23

a = 14°to 17 °

a = 17 ° to 30 °

a/d = 0.35 to 0.43

~ M t m = 2 V z ~~b a

mbl -- Fne2

6Mbl Fn ~b = b~ +-~l <- ~a

I F n b ) where th = thickness of hub on pawl

(24-266)

(24-267)

(24-268)

(24-269)

(24-270)

(24-271)

(24-272)


24.9 GENEVA M E C H A N I S M

S Y M B O L S 2'3

r l a = .

sin q5 F1

F2

f2(max)

f#(max)

f i (max)

z - 2

z

J

k Mlt M2t Mzti M2,~ n

n

P r l

r2 ! r2

ra2

rp r2

e r - - - - r l

t ti tr 1;

z

oL

OZ 2a

o~ m

r l ,-)/ - - _ a

center distance, m (in)

the component of force acting on the crank or the driving shaft due to the torque, Ml,, kN (lbf) (Fig. 24-57)

the component of force acting on the driven Geneva wheel shaft due to the torque M2,, kN (lbf) (Fig. 24-57)

maximum force (pressure) at the point of contact between the roller pin and slotted Geneva wheel, kN (lbf)

the component of maximum friction force at the point of contact due to the friction torque M2w, on the driven Geneva wheel shaft, kN (lbf)

the component of maximum inertia force at the point of contact due to the inertia torque on the driven Geneva wheel shaft, kN (lbf)

gear ratio

polar moment of inertia of all the masses of parts attached to Geneva wheel shaft, m 4 (in 4)

the working time coefficient of the Geneva wheel total torque on the driver or crank, N m (lbf in) total torque on the driven or Geneva wheel, N m (lbf in) inertia torque on the Geneva wheel, N m (lbf in) friction or resistance torque on Geneva wheel, N m (lbf in) speed, rps speed, rpm power, kW (hp) radius to center of driving pin, m (in) radius of Geneva wheel, m (in) distance of center of semicircular end of slot from the center of

Geneva wheel, m (in) outside radius of Geneva wheel, which includes correction for

finite pin diameter, m (in) pin radius, m (in)

radius ratio

total time required for a full revolution of the driver or crank, s time required for indexing Geneva wheel, s time during which Geneva wheel is at rest, s velocity, m/s number of slots on the Geneva wheel crank angle or angle of driver at any instant, deg (Fig. 24-54) angular acceleration, m/s 2 (ft/s:) angular acceleration of Geneva wheel, m/s 2 (ft/s 2) angular position of the crank or driver radius at which the

product wc~2~ is maximum, deg angle of the driven wheel or Geneva wheel at any instant, deg

(Fig. 24-54)

the ratio of the driver radius to center distance

efficiency of Geneva mechanism


360

~b= 2z

27rn

6O 031, 032

locking angle of driver or crank, rad or deg ratio of time of motion of Geneva wheel to time for one

revolution of driver or crank

semi-indexing or Geneva wheel angle, or half the angle subtended by an adjacent slot, deg (Fig. 24-54)

crank or driver angle, deg (Fig. 24-54)

angular velocity of driver or crank (assumed constant), rad/s

angular velocities of driver or crank and Geneva wheel, respectively, rad/s

a

FIGURE 24-54 Design of Geneva mechanism.

Particular Formula

The angular velocity (constant) of driver or crank

Gear ratio

The semi-indexing angle or Geneva wheel angle or half the angle subtended by two adjacent slots

The angle through which the Geneva wheel rotates

EXTERNAL GENEVA WHEEL

The angle of rotation of driver through which the Geneva wheel is at rest or angle of locking action (Fig. 24-55)

The crank or driver angle

271-n 031 = 60 (24-273)

angle moved by crank or driver during rotation i =

angle moved by Geneva wheel during rotation

z - 2 i = ~ (24-274)

Z

360 7r @ = ~ or - (24-275)

Z

2@ 360 27r = or (24-276)

Z Z

A = 2 ( 7 r - ~b) = 7r + 2@ = 7r (z + 2) (24-277) Z

7r 7 r ( z - 2) (24-278) ~b= ~ - ~ b = 2------~----


Particular Formula

o31

FIGURE 24-55 External Geneva mechanism.

D I S P L A C E M E N T

The center distance (Fig. 24-55)

The radius ratio

The ratio of crank radius to center distance

The relat ion between crank angle and Geneva wheel angle

r l a = .

s i n q5

Rr r2 = - - = cot 4~ r l

r l 7 = - = sin ~b = sin 7r a z

7 s i n a ) fl = tan-1 1 - 7 c ~ a

(24-279)

(24-280)

(24-2Sl)

(24-282)

V E L O C I T Y

The angular velocity of the Geneva wheel

The m a x i m u m angular velocity of Geneva wheel at angle a = 0

d3 ~,(cos a - 7) CO2-- d t - 1 - 2 7 c o s a + 7 2 C O l

sin(Tr/z)(cos a - sin 7r/z) CO2 =

1 - 2 sin(Tr/z)cos a + sin 2 7r/z

( ~ ) - -Y CO2(max) - - ~ max - - 1 - 7 ~ C O l

sin /(1 sin )l _ _ _ CO 1 Z Z

COl

(24-283a)

(24-283b)

(24-283c)


Particular Formula

A C C E L E R A T I O N

The angular acceleration, aoQa , of Geneva wheel

Fo r angular velocity and angular accelerat ion curves for three-slot external Geneva wheel with driver velocity, ~1 = 1 rad/s

The m a x i m u m angular acceleration of Geneva wheel which occurs at a - a(max)

The angular acceleration of Geneva wheel at start and finish of indexing

Total t ime required for a full revolut ion of the crank or driver

30

20

10

O~2a 0012 "~

//I - 3 0 ° - 2 0 ° - 1 0 ° 0

G2a 0)12

, I

k 10 ° '2~,~30 °'

~ ~t~- 0112

V

0 t°-2 0} 1

l o

20

30

d 2/3 _ (73 - "y) sin a ~o~ (24-284a) a0/.2 a = dt 2 _ (1 + ,),2 _ 27 cos 002

a sin(u/z) cos 2 (u/z) sin a O~2a - - --1-- 2

1 - 2 sin(Tr/z) cos a + sin 2 (Tr/z) (24-284b)


COS O~(max ) - - --/'i; + V / ~ 2 + 2

where l(1) sin(Tr/z) cos 3 (Tr/z)

(OQa)i,f--" i [1 - 2 sin2(Tr/z) + sin2(Tr/z)]

= +wE tan ¢ = +w2 tan 7r/z

(24-284c)

= 24-28

60 = - - (24-286) t

n

FIGURE 24-56 Angular velocity and angular acceleration k

curves for three-slot external Geneva wheel.

a O/2a is the symbol used for angular acceleration of Geneva wheel; a is the crank or driver angle at any given instant.


Particular Formula

The ratio of t i to t

The ratio of tr to t

The sum of angles of (¢ + ~)

The time required for indexing Geneva wheel, in seconds

The time during which Geneva wheel is at rest, in seconds

The working time coefficient of Geneva wheel

Ratio of time of motion of Geneva wheel to time for one revolution of crank or driver

The required speed of the driver shaft or crankshaft

SHOCK OR JERK

The jerk or shock, J2, on Geneva wheel

The jerk or shock at a = 0

The jerk or shock at start, i.e.,/3 = 05

The length of the slot (Fig. 24-54)

The condition to be satisfied by diameter on which the driver or crank is mounted

The condition to be satisfied by the diameter on which Geneva wheel is mounted

t i 2 ~ ~b

t 27r 7r

tr = 2(7r- ~ ) _ 1 t 27r

~ + ~ b = 9 0 °

z - 2 t i = - - ~ z t = ~

z+2(6o) t r ~ ~

z ~

k = t i _ = z - 2

tr : : + 2

v = - - ~ <~

z + 2 ( 6 0 )

z ~r

where n in rpm

z - - 2

2z

z + 2 7r 2z

z_2 (60) z

7(7--1)[2")' cOs2 °~ + (1 + 72) COSO~ -- 47] J2 =

(1 + 7 2 - 27cosa) 3

39)~ _ ' r ( 'Y+ 1 ) ~

(J2).=0 = -d7 = 0 - ( 7 - l) 3

(J2)3=~ = ( d3/3 372 (1 a2-r,+r -a--a(sin -z + cos ')

dl < 2 a 3 = 2 ( a - r z ) = 2 a 1 - c o s - Z

or

- - < 2 1 - c o s - = 4 s i n 2 - a z 2z

d 2 < 2 ( 1 s i n T r ) = 4 s i n Z ( T r 7c) a z 4 ~zz

(24-287)

(24-288)

(24-289)

(24-290)

(24-291)

(24-292)

(24-293)

(24-294)

+~ (24-295)

(24-296)

(24-297)

(24-298)

(24-299)

(24-300)

(24-301)


Particular Formula

T O R Q U E A C T I N G O N S H A F T S O F

G E N E V A W H E E L A N D DRIVER

The total torque acting on Geneva wheel shaft

The torque on the shaft of crank or driver

The efficiency of Geneva mechanism

I N S T A N T A N E O U S P O W E R

The instantaneous power on the crank or driving shaft

Calculation of average power

The average torque Mti(av) for complete cycle

The average torque for first half-cycle

M 2 t -- M2tl z _qt_ M2t i -- M2t# _qt_ joe2 a

It is assumed that M2tu is constant.

M1 t = M 2 t 602 1__ = (M2t # -+- J°Qa ) 60 2 1 601 ~ 601 T]

~7 - 0.80 to 0.90 when Geneva wheel shaft is mounted on journal bearings

~7 = 0.95 when drive shaft is mounted on rolling contact bearings

~7 = 0.75 when the diameter of bearing surface is larger than the outside diameter of Geneva wheel

(24-302)

(24-303)

(24-304a)

(24-304b)

(24-304c)

e = Mt60

1000 SI (24-305a)

where P in kW, Mt in N m, and co in rad/s

m t60 P = 102 x 10 3 Customary Metric (24-305b)

where P in kW, M t in kgf mm, and 60 in rad/s

Mt60 P = 75 x 10 ------------~ Customary Metric (24-305c)

where P in h p m , M t in kgf mm, and co in rad/s

Mtn USCS (24-305d) P = 63,000

where P in hp, Mt in lbf in, and n in rpm

Mti(av) = 0 (24-306)

Mt(av) = Mu(av) 2[ )2 ]1 = ~ ' 601 (24-307) z - 2 Mzt# nt- - ~ 1 7 -~

where J = polar m o m e n t of inertia, m 4, c m 4 (in 4)


Particular Formula

The average power required on the crank or driving shaft

Calculation of maximum power

The maximum torque on the driven shaft of Geneva wheel

The maximum torque on the driving shaft of the crank

F2 ' •

F1 ,

FIGURE 24-57 Forces acting on Geneva wheel.

The maximum power required on the shaft of the crank or driver

Pav - - Mt(av-'-----~) c° SI (24-308a) 1000

where Pay in kW, Mt(av ) in N m, and co in rad/s

Mt(av)co Customary Metric (24-308b) Pay = 75 x 103

where Pay in hpm , Mt(av ) in kgf mm, and co in rad/s

Pa~ -- Mt(av)'-'-'~n U S E S (24-308c) 63,000

where Pav in hp, Mt(av ) in lbf in, and n in rpm

M 2 t ( m a x ) - - M 2 t # -Jr- M2ti(max)

where M2t # is constant

J°12a(max)(27rn) 2 Mzti(max) = J°lza(max) - co2 " ~

(24-309)

1 co2(max)

Mlt(max ) = M2tla rl COl

J 1 + - - -- (O12aco2)ma x

co1 77 (24-310a)

[( 1 Mlt(max) ~ M2tl~ 1 - ' 7 ~7

'72(1 - - ' 7 2 ) ( C 0 S OZm - '7) sin ol m ) +

( 1 - 2'7 cos Ol m + '72)3

X Jco~ 1177 (24-310b)

where c~ = o~ m at which Mtl is maximum

Mlt(max)co SI (24-31 la) P1 (max) = 1000

where el(max) in kW, Mlt(max ) in N m, and co in rad/s

M l / ( m a x ) c o Customary Metric (24-31 lb) P1 (max) - - 102 x 103

where el(max) in kW, Ml/(max ) in kgf mm, and co in rad/s


Particular Formula

FORCES AT THE POINT OF CONTACT (Fig. 24-57)

The maximum force at the point of contact between the roller pin and slotted Geneva wheel

The component of maximum friction force at the point of contact due to the friction torque M 2 t , on the driven Geneva wheel shaft

For maximum values of F2i

For design data for external Geneva mechanism

Mlt(max)n USCS (24-31 lc) Pl(max)-- 63,000

where Pl(max) in hp, Mlt(max ) in lbf in, and n in rpm

M 2 t F2(max) = ~ (24-312a)

r2

F2(max) = F#(max) -~- Fi(max)

where

r2 = ? a 2 - 2ar l cos a + rl 2

_ _ r l 9 1 _ 27cos a + ,),2 7

F2#(max ) - - M 2 t # M 2 t # "),

r2(min) r 1 1 -- 7

where (1) r2(min) -- a - - r 1 -- 1 - ~ rl


Refer to Table 24-25A.

(24-312b)

(24-313)

INTERNAL GENEVA WHEEL

The time required for indexing Geneva wheel, s

The time during which Geneva wheel is at rest, s

The t i / t ratio

The tr/t ratio

The working time coefficient of Geneva wheel

The relationship between crank or driver angle a and Geneva wheel angle/3

z+2(60) t i - - _

z

t r -- z (60)z t i z + 2 m t 2z

t r Z ~ 2 u

t 2z

z + 2 k = > 1

z - 2

7 sin a ) / 3 = tan-1 1 +7. /c~sa

(24-314)

(24-315)

(24-316)

(24-317)

(24-318)

(24-319)

24 . 90 CHAPTER TWENTY-FOUR

Particular Formula

The angular velocity of Geneva wheel

The m a x i m u m angular velocity of Geneva wheel

The angular acceleration, c~2a, of Geneva wheel

F o r values of Ol2a at s tart and finish of indexing

Fo r curves of angular velocity and angular accelera- t ion of internal Geneva wheel

The contac t forces between the slotted wheel and the pin on the driving crank of the internal Geneva wheel are calculated in a m a n n e r similar to tha t for the external Geneva wheel

Materials

C h r o m i u m steel 15 Cr 65 case-hardened to Rc 58 to 65 is used for the roller pin on the driver or crank.

C h r o m i u m steel 40 Cr 1 hardened and tempered to Rc 45 to 55 is used for the sides of slotted Geneva wheel.

TABLE 24-24 Maximum F2i values

z 3 4 5 6 8

F2i(max)/(Jrl2~ \ r , ]

1.966 0.126 0.0318 0.0131 0.00424

d/3 ( , 7 ( c o s o L + , 7 ) ) 032 --- "-~ = 1 + 2,7 cos ct + ,72 031

,7 --- 0j 1 032(max) 1 + ,7

d2/~ -{- , 7 ( 1 - - ,7),2) sin c~

OZ2a "-- dt 2 - (1 + 2"7 cos c~ + ,72)2 032

Use Eq. (24-285) of external Geneva wheel.


(24-320)

(24-321)

(24-322)

0.8 1.00

0.6 ~IG2a 0)2 0)12 ~ . . . . . . , . j r 0)1 0)2

0)~T 0.4

0 ~ - 35 ° - 9 0 ° - 4 5 °o~ 0

\

45 ° ~ t 90 ° 135 °

~2a j 0)12

0.75

0.50

0.25

~2a 0)12

FIGURE 24-58 Angular velocity and angular acceleration for four-slot internal Geneva wheel.

TABLE 24-25A Design data for external Geneva mechanism

Z q~ ~b i rl /a r2/a Rr r[/a A OQa( initiai)

U W2(max) ¢~ -- - - ~ ~(max) Jmax J ~ =o

3 60 ° 30 ° 0.5 0.886 0.500 0.577 0.134 4 45 ° 45 ° 1 0.707 0.707 1.000 0.293 6 30 ° 60 ° 2 0.500 0.866 1.732 0.500 8 22°30 ' 67°30 t 3 0.383 0.924 2.414 0.617

10 18 ° 72 ° 4 0.309 0.951 3.078 0.690

300 ° 270 ° 240 ° 225 ° 216 °

0.167 6.46 1.732 4o46 , 31.44 -672 0.250 2.41 1.000 1 l°24 ' 5.41 -48 0.333 1.00 0.577 22°54 / 1.35 - 6 0.375 0.620 0.414 31°38 / 0.699 -2.25 0.400 0.447 0.325 38°30 ' 0.465 -1.24


24.10 U N I V E R S A L J O I N T

S Y M B O L S 2,3

d K~

l L Mt Mtd n

l n

P Pd

0

031 , 032

diameter, m (in) ~hn~k fnntnr

correction factor to be applied to torque to be transmitted correction factor to be applied to power to be transmitted length (also with subscripts), m (in) life, h torque to be transmitted by universal joint, N m (lbf in) design torque, N m (lbf in) speed, rpm speed, rps power to be transmitted by universal joint, kW (hp) design power, kW (hp) angle between two intersecting shafts 1 and 2, deg angle of rotation of the driver shaft 1, deg angle of rotation of the driven shaft 2, deg angular velocities of driver and driven shafts respectively, rad/s

Particular Formula

SINGLE UNIVERSAL JOINT (Figs. 24-59 and 24-61a)

The relation between 0, ~, and fl tan q5 =

tan 0

cos fl

The relation between the angular velocities of driving &2 shaft 1 or driver (aJ1) to the driven shaft 2 or the a~l follower (a~2)

cos fl 1 - sin 2/3 sin 2 0

(24-323)

(24-324)

d2

IP

FIGURE 24-59 A single universal joint.


Particular Formula

The max imum value of CO2/COl

The min imum value of 032/031

The angular acceleration of the driven shaft 2, if col is constant

The value of 0 for which the angular acceleration of the driven shaft is max imum

The power transmit ted by universal joint

The design torque of universal joint

The design power of universal joint

For calculation of torque and power t ransmit ted by universal joint for various angles of inclination fl

For design data of universal joint

(co~-~l) _ cosfl _ ~ 1 (24-325) max - 1 - sin 2 fl - cos fl

when sin 0 = + 1, i.e., 0 = 90 °, 270 °, or 7r/2 or 37r/2, etc.

( ~-~11 )min -- COS fl

when sin = 0, i.e., 0 = 0, 7r, 27r, etc.

(24-326)

d2~5 _ dj_22 = cos flsin 2 f ls in20 co2 (24-327) dt 2 - dt (1 - sin 2 0sin 2 fl)2

COS 20(max ) = t~- V/~ 2 + 2 (24-328)

where ~ = (2 - sin 2 f l ) / 2 sin 2 fl

The angular acceleration of driven shaft is max imum when 0 is approximately equal to 45 °, 135 °, etc., when the arms of cross are inclined at 45 ° to the plane containing the axes of the two shafts.

P = Mtco/lO00 SI (24-329a)

where P in kW, Mt in N m, and co in rad/s

P = M t n / 6 3 , 0 0 0 USCS (24-329b)

where P in hp, M t in lbf in, and n in rpm

Mid = MtKsKct (24-330)

P - (24-331)

P d - K c N


Refer to Tables 24-25B and 24-25C.

D O U B L E U N I V E R S A L JOINT (Figs. 24-60 and 24-61b)

The angular velocities ratio for a double universal joint which will produce a uniform velocity ratio at all times between the input and output ends

co_~l = 1 (24-332)

CO2

1

1

" ~ )

3 I

Intermediate shaft

(a)

(b)

~ (driven)

• A

- - - - ~ L 2 ~ 0)2 (driven)

Yoke 1 or fork 1

~1 ~ .... r , , ~ , ~ _ ~ 7 , ~ Yoke 2 or fork 2

3 2

I"

(a) 13 FI 3 ~ 2

FIGURE 24-60 Double universal joints.

0°

, . J / . - ' ~ " \ 7 % . \

,,<7 ( \ 1 " i ' , , , , , • V / / / / A ~¢ ", ~ i /]i~ T

~ ~ L - ~ ~ " I F ~- +, / I " v ~ .~' I 'W// /A ,I,

D x,,~o T I _JI o.~ x 4~o , . . , L1 L3 "!-" v , - ~ ._ ~

< L 1 ---! , 2 I~-L~ ~. L 4 . ~,I

(Refer to Table 25.25B)

(b) Double universal joint d i = 10 to 50

D x 45 ° 7-

1 - y L1 ! < i" 3

(a) Single universal joint d i F 6 to 50

tD

FIGURE 24-61 Dimensions of universal joints.

,0.5 x 45 °

-,,7 ~ L,=

24.93

2 4 . 9 4 C H A P T E R T W E N T Y - F O U R

2.5 :o 2.2 g 2 , "5 1.8 ~ 1.6 r-

• ~ 1.4

~ 1.2 L

1

/ f

J

/ /

J f

. j , -

_ /

10 15 20 25 30 35 40 Angle of inclination (13), deg

45

FIGURE 24-62 Angle between two intersecting shafts vs. correction factor (K~t).

1.0 \

z 0.9 \ ' \

,,,' 0.8 g 0.7 ~

o 0.6

~- 0.5 o

0.4 . . . . . . . . - - ~ k • 0 . 3 L U "--'-- L

o 0.2 0 I 0.1 I

0 '

10 15 20 25 30 35 40 45

Angle of inclination (13), deg

FIGURE 24-63 Angle between two intersecting shafts vs. correction factor (KEN).

8 E ~ 16x25 - - ,

g 20x32 "

4 .12x25 16x25

3 , v ,

2

.10x20- , 1 12×20

"~ 0"8 ~l~d 2 o 0.6

0-5 0.4

0.3

100 - 50x901 \ _m ! 981 "0 80 40x75! \ N - I " 784"5

: : :

60 \ ,, i : 588.4 '- \ • 490.3

50 32x63 -~ " \ ' \ i 392.3 40 40x63 " \ " ' J " " 294.2 30 " I', "

' ' ~" " i 196.1 20 .25x50 ,~ ~ \ 18kgfm .32x50. - ~ . . . . . - I ~ - - " ~ " I . . . . . . . 176.5Nm

15 ,20x40 \ \ m ~, N!

10 25x40 \ X : ~ ;m " \ \ \ i I, i_ ~ i 98.10

0.2

0.1

~m m m m mlmk~ m m m

~mmmm:,mm mm~mnm~m ,:qm,-immmm~ ml,,liaimm ,Immellaai mp.mm~mll .qi imm im l i l i l n i l l l l J i l l l l l l l J h ~ l mmmmm.mi~mmk~l mmii,mmm¢~mm mmmm~mnm )mmi[mmmma m,~wmmmmmml mmwnlmmim mmmm mmm mmmnl,~mm mmmelmp,| mnmllnmw immmlmmm

1 2 3 5 1 0 2 0

" ~,'i~, 78.45 "4. ,, 58"84 E

I" \ 49.03 Z • 39.23

"~" 29.42 • \ ~"

"J 19.61 =

~ l \ 7.85 a

i'~,, 5"88 ~ ~, 4.90

~, \ 3.92

I I I \ 2 . 9 4

1.96 m

098 | | | ,

30 100 300 1000 300010000x104

Speed x life (nL), rpm h

FIGURE 24-64 Design curves for single universal joint with needle bearings for/3 = l0 °


kW 10 H / / / I / X J l l .Jr 1,4" I,

8 ~ ' x z L 4 ~ 4 I / X V I V I I.

_ V V I X I M 5-7 kW t,,,"1"/~" ~,,~ t"

4

3

/ / / / " / / . / / d l ' J t .m . / f ~' 1'111/'I I I X I / I I l J r l v v ' I / l J F I I , 4 1 1 A L I . ~ j ( ~ IU ' I I 1 1 1 1 / I I I X / I / V I / I Y ' ~ IX, I ' A" I Iz r , , r / v i i i I / I I X I 1 , 4 , 4 A / I / J . . , K T - X I I J q l 1 , 4 . ' I ' V I 1 , 4 . ~ 4 / CO 11114 1,4I I.,44,4~,#'~XN"IHIII,4"dXI~I.44zIA/ I'~ ':14 1- I - ~ / ' ~ ::l'ff b ' l ~ ~ I/VYJ/J4. ~

1 0.8

-00.6 0.5

i,.. 0-4

o 0.3 o . ¢..

.~ 0.2 go

a

0.1 0-08

0.06 0.05 0.04

0.03

0"02

0-01 rpm

10 20 30 40 60 100 200 3004006001000 Speed, (n), rpm

FIGURE 24-65(a) Design curves for single universal joint with plain bearings for/3 = 10 °.

T A B L E 24-25B Dimensions of universal joint (Fig. 24-61)

di do L3 L4 H7 k l l L2 +1 +1

Maximum allowable rotational play

Test torque

N m kgf m

Angular rotational play at an angle of inclination of 0 deg in minutes

Tolerance on coaxiality of the two bores

6 9 34 8 16 11 40

10 15 52 74 20 13 48

12 18 62 88 25 15 56 86

16 22 74 104 32 19 68

20 25 86 124 40 23 82 128

25 32 108 156 50 29 105 160

32 40 132 188 63 36 130 198

40 50 166 238 75 44 160 245

50 90 54 190 290

0.5

0.196 0.02 45

0.392 0.04 40

0.981 0.10 32

0.667 0.17 28

3.334

0.06

0.34 25 0.09

5.296 0.54 20

14.710 1.5 18 0.12

21.575 2.2 16 27.458 2.8 14 0.15

~ - - - - ITI -

--¢4d

FIGURE 24-65(b) Taper pin joint. The length of the taper pin should conform to diameter do in Table 24-25B.

T A B L E 24-25C Dimensions of taper pin a (Fig. 24-65b)

a/ a, m

6 2 4.5 8 3 5

10 4 6 12 5 7.5 16 6 9 20 8 11 25 10 15 32 12 18 40 14 22 50 16 27

a The shear stress of taper pin = r = 158.5 to 247.5 MPa (16 to 25 kgf/mm2).

U S E OF C U R V E S IN FIGS. 24-62 TO 24-65

Worked example 1

A single universal jo int has to t ransmi t a torque of 10 k g f m at 1500rpm. The angle between intersecting shafts is 25 ° . The joint is subjected to a minor shock. The shock factor (K~.) is 1.5. Design a universal jo in t with needle bearings for a life o f 800 h.

S O L U T I O N F r o m Fig. 24-62 correct ion factor for /3 = 25 ° is K,., = 1.2. Design torque = Mid = MtK, .Ka = 10 x 1.5 x 1.2 - 1 8 k g f m (176 .5Nm) . Speed × l i f e : nL = 1500 x 800 = 120 x 104 r p m h . F r o m Fig. 24-64 for M n t - 1 8 k g f m ( 1 7 6 . 5 N m ) and nL = 120 x 104 rpm h, the size of a single universal jo in t is (di x do)40 x 7 5 m m .

W o r k e d e x a m p l e 2

Design a single universal joint with plain bearings to t ransmi t 2 kW power at 325 rpm. The angle between two intersecting shafts is 27.5 ° .

S O L U T I O N F r o m Fig. 24-63 correct ion factor for / 3 = 27.5 ° is KeN = 0 . 3 5 . Design p o w e r - - P d - - ( P / K c N ) = ( 2 / 0 . 3 5 ) - 5 .7kW. F r o m Fig. 24-65a the size of a single universal joint for Pd = 5 .7kW and s p e e d = n = 3 2 5 r p m is ( d i x d o ) 4 0 x 7 5 m m . The permissible to rque for this size of joint (Fig. 24-65a) is 1 7 . 5 k g f m ( 1 7 1 . 5 N m ) .

24.96


24.11 U N S Y M M E T R I C A L BENDING AND TORSION OF NONCIRCULAR CROSS-SECTION MACHINE ELEMENTS

S Y M B O L S 2,3

a semimajor axis of elliptical section, m (in) width of rectangular section, m (in) (in k)

A area of cross section, m 2 (in) b semi-minor axis of elliptical section, m (in)

height of rectangular section, m (in) c distance of the plane from neutral axis, m (in)

thickness of narrow rectangular cross section (Fig. 24-68) e the distance from a point in the shear center S (Table 24-26) E Young's modulus, GPa (MPsi) G modulus of rigidity, GPa (MPsi) I moment of inertia, area (also with suffixes), m 4 (cm 4) (in4~ I,, Iv moment of inertia of cross-sectional area, respectively, m

(cm 4) (in 4) ark polar moment of inertia, m 4 (cm 4) (in 4) kl, k 2 constants from Table 24-28 for use in Eqs. (24-343) and (24-

344) L length, m (in) Mb bending moment, N m (lbf ft) Mt twisting moment, N m (lbf ft) Mb~ = Mb cos 0 bending moment about the U principal centroidal axis or any

axis parallel thereto Mbv = Mb sin 0 bending moment about the V principal centroidal axis or any

axis parallel thereto .~ q = ~-t shear flow Q the first moment of the section, m 4 (cm 4) (in 4) S the length of the center of the ring section of the thin tube, m

(in) t width of cross section at the plane in which it is desired to find

the shear stress, m (in) thickness of the wall of the thin-walled section, m

u, v coordinates of any point in the section with reference to principal centroidal axes

V shear force on the cross section, kN (lbf) Vy resultant shear force acting at the shear center, kN (lbf) x the distance of the section considered from the fixed end (Fig.

24-73) x, y coordinates in x and y directions crb bending stress (also with suffixes), MPa (psi) ~- shear stress (also with suffixes), MPa (psi) ~5 variable thickness of thin tube wall (Fig. 24-70), m (in) 0 angle measured from the V principal centroidal axis, deg q5 angle of twist, deg


Particular Formula

SHEAR CENTER

The shear stress at any point in transverse plane or section of a member

The flexural stress in a thin-walled open section

Shear flow

For the equations for locating the shear centers of various thin open sections

VQ '7" - . - - u

It

Mbc ab = I

VQ q : r t = ~

I


(24-333)

(24-334)

(24-335)

UNSYMMETRICAL BENDING

The flexural stress in case of sections subjected to unsymmetrical bending

Flexural modulus for any cross section on which the stress is desired

(M b cos 0) v ab= lu +

MbuV mbvU = -~ lu Iv

(Mb sin O)u

Z = lulv/(vlvcosO + ulusinO)

(24-336)

(24-337)

TORSION

Solid sections

ELLIPTICAL CROSS SECTION

Shear stress acting in the x direction on the xz plane (Fig. 24-66)

Shear stress acting in the y direction on the yz plane (Fig. 24-66)

Maximum shear stress on the periphery at the extremities of the minor axis (Fig. 24-66 and Table 24-27)

Minimum shear stress on the periphery at the extremities of the major axis

Angle of twist (Fig. 24-66)

2Mty Txz 7tab 3

2Mtx ryz = 7ra3 b

2Mt Tmax 7tab 2

2Mt 7"min 7ra 2 b

¢ Mt (a2 + b2) L

-- --G-- 7ra 3b ' ' T -

(24-338)

(24-339)

(24-340)

(24-341)

(24-342)


Particular Formula

1RECTANGULAR CROSS SECTION

The maximum shear stress at point A on the boundary, close to the center (Fig. 24-67 and Table 24-27)

Angle of twist (Table 24-27)

m t "r A = k lab 2 (24-343)

where kl depends on ratio a / b (Refer to Table 24-28.)

M t L

= k2ab 3 G

where k2 depends on ratio a / b (Refer to Table 24-28.)

(24-344)

Mt

/ ,

FIGURE 24-66

X

",91----- a ~

-C-

FIGURE 24-67

L %'max

a

~ X

N A R R O W R E C T A N G U L A R C R O S S S E C T I O N S (F ig . 2 4 - 6 8 )

Equation for twisting moment (Fig. 24-68)

Equation for angle of twist

The maximum shear stress

Mt -- 1 Gc~c3b

3Mt

= Gc3b

3Mt 7-ma x --- bc 2

(24-345)

(24-346)

(24-347)

24.100 CHAPTER T W E N T Y - F O U R

TABLE 24-26 Location of shear center for various cross sections

Section Location of shear center Section Location of shear center

x .

tf

IIio I x

s - 1 ; r c =']'~ x

Vy b 6 q-'m

T

I' a ,

Vy

3b2tf e - - - ~ htw + 6btf

e=b]

lbl 4 hi 2 }

l h 2 b I 2 h , ( 2hi) 1 + g ~--~-! + ~-~-i - -~ - 2 1 - ~h-~2

wherebl = b - t h 2 = h - t

1 b I 4 ht 2 }

e = b I 1 h2 bl ( 2hl'~ l + ~ g + g + 2 ~ . I + N j

where b, = b - t/2 h2 =h+t

m e - - m

n

m = 12 + 67r + 6 + 12-~-

+ 37r(hl ) 2 ( ~ ) 3b -- - 4 a a

n=3rr+12( b+h')-a +4(~- )2(3+- -h l )a

C is at the centroid of triangle e = 0.47a for narrow triangle ((t > 12 °) approx.

I~b2 ,J- bl -~

A -r;N¢ - - ×

-t-~-

3(b~ - b~) e~- - (tw/tf)h + 6(b, + b2)

for b 2 < bt

X

.L. -~ t2 ~-

e = ( btlh~ ) tlh~ + t2h32

X

e=bl b-i ~ -

+3 ~ - 3~(+ 1

L = Length of ~l[C Dotted line

-x 2A e

h + L(th/th---------~

where A = area Y e 4 Vy


TABLE 24-26 (Cont.) Location of shear center for various cross sections

Section Location of shear center Section Location of shear center

s

t

Y

ex~l.i _h L ey'l qJl Iq ,f 4 - - ; ~ L i x T~ ~vy~

Y

~ r - - - b ~

Y

e = 2a[(Tr - ~) cos ~b + s ine] [(Tr- ¢) + sin ~bcos ¢]

7r 4a F o r ¢ = 2 ' e =--Tr

b ht3w ex = ~ ( ht3 + tfb3 )

= h ey -~ ( ht3w ht3w + t:b~ )

e = g ( t f + b ) 1 + t 3 b l

) I / .2

Vy Y

b~ht el = c - 6(iyi _ I]y )

x [3Ixy(h - c) +/~(2b, - 3d)]

b~ht e2 - d +

6(Iylx - I]y)

x [Ixy(2b, - 3d) + 3 Iy (h - c)]

where

h 2 + 2blh b 2 + b 2 c = ~ d = ~

2h(bl + b2) ' 2h(bl q- b2)

Ix, Iy - m o m e n t of inert ia of section a b o u t x and y axes, respectively

Ixy - p roduc t of m o m e n t of inert ia

( l + 3v~ J xta dx

e=~ l+v : i7.~

2 4 . 1 0 2 CHAPTER TWENTY-FOUR

TABLE 24-27 Approximate formulas for torsional shearing stress and angle of twist for various cross sections

Cross section Shearing stress, Ibf/in 2 (N/m 2 or MPa)

Angle of twist per unit length 05, tad/in (rad/m)

~ 2 b ~ . . _ 2M t re = 7tab 2

C _ 2M___L

Ab

~ ; 20Mr 7 = b3

~ - - ~ mt % = klab 2

7. = 2rrr2------ ~

3Mt 7. = 2rrr~2

fi

8 C Mt

~ 2 a ~l 4~ 7. = 27rab~5 2b

81~ C M t V / / / / / / / / / / ~ 7.c =

D _ ~ q ~"A "~ 2 a b 6 , 8 ~t ° ~l lllllllllJf~ i M ,

.~ a v, 7"D = 2ab5

75 - - -

75 =

Mt(a 2 + b 2)

GTra3 b 3

47r2 jMt A4G

46.2Mt Gb 4

g t = k2ab 3 G

g t ~b - 2,n_r3 ~SG

3Mt q5 = 2rrr~3 G

(~) - - Mtv/2(a 2 + b 2)

47ra 2 b 2 ~G

¢1 - - - Mt(a6 + b61)

26/~1 a 2 b 2 G

A = area of cross section

TABLE 24-28 Variation of kl and k 2 with the ratio a/b for use in Eqs. (24-343) and (24-344)

a/b 1 1.2 1.5 2.0 2.5 3.0 4.0 5.0 10.0

kl 0.208 0.219 0.231 0.246 0.258 0.267 0.282 0.391 0.312 0.333 k2 0.141 0.166 0.196 0.229 0.249 0.263 0.281 0.291 0.312 0.333


Particular Formula

COMPOSITE SECTIONS

Cross Sections Composed of Narrow Rectangles

Equation for torque of a narrow rectangular cross section

Equation for torque of a narrow rectangular section ( b / c =

Equation for torque for a cross-section composed of several narrow rectangles (Table 24-27)

Angle of twist for a cross section composed of several narrow rectangles

Maximum shear stress

For approximate formulas for torsional shearing stress and angle of twist for various cross sections

For variation of stress-concentration factor Ko with ratio r/c for structural angle (Fig. 24-69)

k2 bc3 Gd~ (24-348) M r = L

6~ Mt = - ~ ~bc 3 (24-349)

6¢ m t -- --~ ~kzbc 3 (24-350)

3MtL (24-351) 0 = G~bc 3

3Mtcmax for k2 "/-max : ~bc------W- k l l - 1 (24-352)

where Cma x - - maximum thickness of the narrow section



f f

o t Section on AA

" ~ ~X

Section on BB

FIGURE 24-68


Particular Formula

TABLE 24-29 StreSS concentration factors for structural angle, K~ (Fig. 24-69)

r/c 0.125 0 . 2 5 0 0 .500 0.750 1.000 K~ 2.550 2 . 2 5 0 2.000 1.875 1.800 ~¢ r - r

FIGURE 24-69

H O L L O W T H I N - W A L L E D T U B E S (Fig . 24-70)

The equation for the twisting moment

The angle of twist

By membrane analogy the value of ~g 7" ds

The equation for the shear stress

The difference in level between DC and ,4B of membrane

The equation for twisting moment of thin webbed tubes (or box beams) (Fig. 24-71)

The equations for shear stress

S

io B FIGURE 24-70

m t - - q(2A) = 2A67" (24-353)

where A = area enclosed by the median line of the tubular section

M t S (24-354) dp =4A2G 6

~ 7" ds = 2G~A (24-355)

7" = ~ 6 (24-356)

h = 7"6 (24-357)

Mt -- 2(Alhl + Azhz) (24-358a)

M t - 2(A1617"1 + A2627"2) (24-358b)

¢53S2AI + ~zS3(AI + A2) 7"1 -- Mt R (24-359a)

63SIA2 -t-61S3(A1 + A2) 7"2 = Mt R (24-359b)

61S3Al - t52SIA2 7"3 - Mt R (24-359c)

where

R = 216,63S2 A2 + 6263S1A2 + 6162S3(A, + A2) 2]

(24-360)


Particular Formula

S 2 ~ S l ,51

(~2

D E B C hi

,51 A

FIGURE 24-71

7w '-'T ~ ~ " -""~ Mt t2 ×

y

FIGURE 24-72

BENDING STRESSES CAUSED BY TORSION

Torsion of I-beam having one section restrained from warping

The lateral bending moment in the flanges of an I- beam subjected to twisting moment at one end, the other end being fixed, Fig. 24-72

The maximum bending moment for long beam

Twisting moment at any section, distance x from the fixed end

The angle of twist per unit length

The total angle of twist at the free end

Maximum bending moment

Mb = Mt s i n h ( L - x ) / k -£-k cosh(L/k)

Mtk L Mb(max) = h as tan h ~ = 1

where k = (h/2)[EI/JG] 1/2

Mtx = Mt [ 1 - c°sh(L - x ) / k

co_s_h! - l ~ . - - j ~ 1 - cos h(L/k) J

-J-d,M' ( L) q5 = L - k t a n h ~

Mt L Mb(max) -- --h-- k tan h

(24-361)

(24-362)

(24-363)

(24-364)

(24-365)

(24-366)


Particular Formula

The angle of twist at free end if l / k> 2.5

The maximum bending moment if l / k> 2.5

The bending stress

For beams subjected to torsion

TRANSVERSE LOAD ON BEAM OF CHANNEL SECTION NOT THROUGH SHEAR CENTER (Fig. 24-73)

M, (L- k)

Mt Mb(max) = --~-- k

Mb(max)b ab = 2If

where M6(max) obtained from Table 24-30


(24-367)

(24-368)

(24-369)

Mh

~.__~/~ - N _ 6Fel

- ht b ' - - ~ i ~ _ _ _ b - ~ l - - -

. . . .

L__

_ Mt.

I_., b ,~l I -~ "-I

(c)

F I G U R E 24-73

v

"1••

•

Shear centre


Particular Formula

The direct stress

The bending stress

The maximum longitudinal stress (Fig. 24-73b)

For geometrical properties, weight, and nominal dimensions of beams, channels, T-bars, and equal and unequal angles

M h crt= i 2

where M = Fe

(24-370)

VIb/2 6Fel (24-371) crb = ~ = htb---g

M h 6Fel cr = ~ + htb---- ~ (24-372)

Refer to Tables 24-31 and 24-32 and Figs. 24-74 to 24-79.


TABLE 24-30 Formulas for maximum lateral bending moment and angle of twist of beams subjected to torsion a

Type of loading and support Maximum lateral bending moment in flange, lbf in (N m) Angle of twist of beam of length L, ~b rad

M t = w L e .....

I "

~ L v I

M t = wLe , (

i I

i~ L

-M t-

L I ~ L 2 ~

Mtk L Mb(max) = --if-- tan h - - 2k

Mtk _..__

h L

if ~ > 2.5

Mtk ( L 2 k ) Mb(max) -- - - ~ cot h ~-~ -

Mtk h

if ~k is large

Mtk ( L k) Mh(max) = ~ c o t h ~ - -

Mtk _._.

h

L i f ~ is large

Mtk sin h ~ sin h Mh(max) -- h L

sinh-- k

Mtk 2h

L~ L 2 if--~- and --k-- > 2

error is small

0 = ~ - ~ L - 2 k t a n h

= Mt ( L - 2k) b JG

L if ~-~ > 2.5

Mt (L L) 0 = ~ ~ - - ktanh~-~

)b 2JC ~ - k

L if ~-~ > 2.5

~ = j - ~ - ~ - k t anh

MI L k) JG ('2 -

L if ~-~ > 2.5

IMI(L ¢~ =-~ 7-6 ~ - k L ) (approx.) tan h~-~

IMt(L )b 2JG ~ - k

if ~k > 2.5

a Formulas given in Table 24-30 can also be used for Z and channel sections. b Error is small for the conditions L/2k > 2.5 and L/4k > 2.5.

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REFERENCES

1. Lingaiah, K., and B. R. Narayana Iyengar, Machine Design Data Handbook, fps system, Engineering College Co-operative Society, Bangalore, India, 1962.

2. Lingaiah, K., and B. R. Narayana Iyengar, Machine Design Book Handbook, Volume I. (SI and Customary Metric Units), Suma Publishers, Bangalore, India, 1986.

3. Lingaiah, K., Machine Design Data Handbook, Volume II (S.I. and Customary Metric Units), Suma Publishers, Bangalore, India, 1986.

4. Lingaiah, K., Machine Design Data Handbook, McGraw-Hill Publishing Company, New York, 1994. 5. Maleev, V. L., and J. B. Hartman, Machine Design, International Text book Company, Scranton,

Pennsylvania, 1954. 6. Black, P. H., and O. E. Adams, Jr., Machine Design, McGraw-Hill Book Company Inc., New York, 1968. 7. Neale, M. J., Tribology Handbook, Newnes-Butterworth, London, 1973. 8. Bach, Maschinenelemente, 12th ed., P-43. 9. Heldt, P. M. Torque Converters for Transmissions, Chiltan Company, Philadelphia, 1955.

10. Newton, K., and W. Steeds, The Motor Vehicle, Iliffe and Sons Ltd, London, 1950. 11. Steeds, W., Mechanics of Road Vehicle, Iliffe and Sons, Ltd., London, 1960. 12. Arkhangelsky, V., et al., Motor Vehicles Engines, Mir Publisher, Moscow, 1971. 13. Heldt, P. M., High Speed Combustion Engines, 6th ed., Chilton Company, Philadelphia, 1955. 14. Thimoshenko, S., and J. N. Goodier, Theory of Elasticity, McGraw-Hill Book Company, Inc., and

Kogakkusha Company Ltd., Tokyo, 1951. 15. Timoshenko, S., and S. Woinowsky-Krieger, Theory of Plates and Shells, McGraw-Hill Book Company, Inc.,

New York, 1959. 16. Seely, F. B., and J. O. Smith, Advanced Mechanics of Materials, John Wiley and Sons, Inc., 2nd ed., 1959. 17. Timoshenko, S., and J. M. Gere, Mechanics of Materials, Von Nostrand Reinhold Company, New York,

1972. 18. Goetze, A. G. Piston Ring Manual, 3rd ed., Burscheid, Germany, 1987. 19. Bureau of Indian Standards, New Delhi, India.

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