기계설계 솔루션(기계요소설계 - R.Juvinall - Scitech).pdf

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JUVINALL: Machine DesignFig. 1-1a W-1

JUVINALL: Machine DesignFig. 1-2 W-2


JUVINALL: Machine DesignFig. 1-3b W-3

JUVINALL: Machine DesignFig. 1-3c W-3


R

F

Torq

ue (

N •

mm

)Cam rotation (rad)

(b)(a)

0.10

10

0

Forc

e (N

)

Followerdisplacement (mm)

(c)

10

1

0



n = 1000 rpm

T = 10 N • mm

�

3

Cra

nk t

orqu

e (k

N •

m)

Crank angle (rad)0

Actual torquerequirement

� 2�0

2

4

6

8

10


2�

Cra

nk t

orqu

e (k

N •

m)

Crank angle (rad)

Uniform torquesupplying equal

energy

Average torque

Actual torquerequirement

0 �0

2

4

6

8

10


�

3


0.2d

Rim

Arm

Hub

0.8d d

Vehi

cle

road

load

pow

er (

hp)

Vehicle speed (mph)0 20 40 60 80 100 120

0

40

80

120

160


Eng

ine

outp

ut p

ower

(hp

)

Engine speed (rpm)0 1000 2000 3000 4000 5000

0

40

80

120

160


Spe

cifi

c fu

el c

onsu

mpt

ion

of e

ngin

e (l

b/hp

• h)

Engine output power (hp)20 40 60

1000 rpm

1500 rpm

2000 rpm

2500 rpm

3000 rpm3500 rpm

4000 rpm

80 100 120 140 160 1800.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3



g = 9.81 m/s2

5 mm

30mm

1 N

m

JUVINALL: Machine DesignFig. P1.18 W-15

JUVINALL: Machine DesignFig. P1-23 W-16

a = 5 ft/s2

g = 32.2 ft/s2F

Pulley


V = 5 ft /sm = 5 lb

R = 3 in.


I = 10 A

Outputshaft

t = 2 hrn = 1000 rpmT = 9.5 N • m

V = 110 V +–

0

T max

0.5

Crankshaft rotation, revolutions

Cra

nksh

aft

torq

ue

1.51.0 2.0

Curve A (linear variation)Curve B (half-wave rectified sinusoid)

JUVINALL: Machine DesignFig. P1-42 W-18A


55 mph

2.64 axle ratio


100 in.

V = 60 mph50 in.

CP

3000 lb

25 in.

Fd

Ft

Wr Wf

20 in.CG


Engine power, 96 hp

W = 3000 lb

Fd = 100 lb

Ft = 567.5 lb

Wr = 1618.5 lb Wf = 1381.5 lb

Fi = 467.5 lb


A

X

X

RT

RT

RT

RT – T

RT – T

T

T

RTA

RTA

2 in.

3 in.

2 in.

5 in.

3000 lb.in.

T = 3000 lb.in.(input)RT – T = 5333 lb.in.

Transmissionin low gearR = 2.778RT = 8333 lb.in.

(output)

5333 lb.in.

8333 lb.in.

(b)

(a)

(d)

(c)

1087 lb

2864 lb

1087 lb

1087 lbII

IV

1087 lb

4444 lb

4444 lb

2864 lb

2864 lb

2667 lb

2667 lb

III

III

IV

I

I

II

B

D

A

C



A

AF F

(a) (b)

a

F

F

Fa

SectionAA plane

Fixedsupport

(bending moment)


(a)

Fa

b

F

a

(b)

F

Fb (torque) F (shear force)


F32 = 40 lb

H12

F42

V12

30°

1

1

in.12

2

1 in.


F32 = 40 lb

F42

F42

F12

F32 = 40 lb

0

F12

1

Force polygon for link 2

2

Ft = 60 lb

Fb = 55 lb

Fb (bonecompression force)

0.5 in.

0

Ft (tendonforce)

10 lbpinch

10 lb

3 in.


Force polygon for finger

b2

2L

wb2

2Lb2

4L

FaL

FbL

FbL

–

+

FaL

–

aL

b

F

V

MM

V


M

a +

VV

R2R1

+

+

Positive shear force

Positive bendingmoment

wb (a + b/2 )L

w

wb (a + b/2 )L

+wb2/2L

wb 2

2La

Lb

R2R1

FabL

+

M

(b)(a)

Distributed loadSingle concentrated load

JUVINALL: Machine DesignFig. 2-11 W-29A

2667 lb

2 in.

2864 lb

4444 lb

–2864 lb

–5728 in..lb

Torque

Moment

Shear

Loads

5000 lb.in.

2174 in..lb

2667 lb

IV

III

2 in.5 in.

+ 1580 lb

– 1087 lb

1087 lb

Critical section

C

V

M

T

B

1087 lb

4444 lb

2864 lb

III

IV

B

C

IV


2864 lb

4444 lb

T = 5000 lb.in.

V = 1580 lb

CM = 5728 lb.in.*

* Actually slightly less, dependingupon the width of gear C

d

F

FF

F


m

2b

b

b

d

a

1

2

3

443

2

1

1

4'

4'

6

4'5

5

3

2

2

2 1

PinFork Blade

F

F

F

F4'

2

2

3

7



w lb/ft

Spring in tensionk1 = 10 lb/in.

Spring incompressionk2 = 40 lb/in.


100 lb

100 lb

10�

40�

(a)

(b)


3

1strap

Topstrap

Rightplate

Bottomstrap

1 pitch

1 pitch


Inner rowMiddle row

Outer row Inner rowOuter row

Area

Plate

Inner row

Topstrap

Bottomstrap

Leftplate

Outer row

(a)

(b)

(d)

(e)

( f )

(c)

Strap

Straps

Inner path

Middle path

Outer path Bearing with plate

Shear

Bearing with straps

Plate

2 straps

Bearing with strap

Bearing with plate

Shear

Plate

Strap

1 2

8

9

6

7

5

4

s bp

bs


gWall channel, C

2 in.

A

B

Density = �

Density = �r =

1.25 in.

r =1.25 in.


1500 N 1500 N

1000 mm

45°

45° 45°

45°

B

D

C

A


A

B

125 N

1000 N


10 in.

Motor1 hp

1800 rpmGear box

Directionof rotation

Blower6000 rpm


50 mm

50 mm

Clockwiserotation

Forwardair velocity

A

B

A'

Pump450 rpm

Connectingtube


4:1 ratiogear reducer

C'

B

A

B'

Motor 1.5 kW1800 rpm

These unitsare attached toa fixed support.


C


Engine is attachedto aircraft structure here.

Engine

Reduction gear,ratio = 1.5

Propeller

Rotation

Verticaldriveshaft

Mountingflange

Propellerrotation

Y

X

Y

Z

Z

X


2:1 ratio bevelgears are inside

this housingSuggested notation

for moments appliedto mounting flange

500 mm

Fowarddirection ofboat travel

Mx MyMz

150 mm

600400

330 R330 R

40 R

100 R

800 N160


Bevel gearreducer

Attachesto motor

Attachesto load

600 rpm1800 rpm


100 mm

100 mmA

BC

D

8 in.

4 in.

6 in.

Output

Output shaft

Mountings

6 in. dia. gear

2 in. dia. pinion

Front bearings

Front bearings

Rearbearings

Motor input torque100 lb.ft

100 lb.ft

Reducer assembly

Housing andgear-shaft assembly

details



X

Y

Y

X

B

D

A

C

12 in.

24 in.

Transmission2.0 ratio

Rear drive shaft–1200 rpm

Rear axle(not part offree body)

Front drive shaft–1200 rpm

Left-frontwheel axle

shaft–400 rpm

Engine2400 rpm

100 lb.ft torque

Right-front wheel axleshaft–400 rpm

Mixing paddle

g


Motor

Radialflow


Mass of mixersystem = 50 kg

200 mm

A B

g


B

Mountingwidth = 75 mm

to 150 mm

Fan

Radialair flow

Mass of blowersystem = 15 kg

A

Motor

Direction ofrotation


Y

FC

FA

Z

D

B

A

C

X

45 mm

30 mm

20 mm

Shaft

Gear 224-mm dia.

Gear 150-mm dia.

20°

20°

300200


100 N

A B

30050 N140

200

100 N

A B

12-in.pulley radius


Cable

Cable

100 lb

100 lb

48 in. 12 in.

27 in.

Motorattaches

to this endof shaft

6040

50


Fa = 1000 N

Ft = 2000 NFr = 600 N

A B


Pump shaft iscoupled to this

end of shaft

15050

100

Fa = 100 N

Ft = 1000 NFr = 200 N

A B

8020

40

20

30


Fa = 200 N

200 N

Fr = 400 N

A B

Ft = 1000 N


Fa = 150 N

Ft = 1500 NFr = 500 N

150

120

A B

250 N

100 N

200 300 140

L

Section on A-A

Key


Rb

r

A

A

F

t

t/2

dD

F


t

d

Total gas force = F


a2aa

R

A B

L

F

L

A

L

ttf


P Pd D

d D


E

A

B

C

D


k

k

kk

a

100 N

a

F F


P

t '

t '

Rivet diameter = 10 mm

t

Str

ess

� (

ksi)

Strain � (arbitary nonlinear scale)

Slope = modulus of elasticity, E

F (Fracture)

Sy = 39

Su = 66

Se = 36

C

AB

0.2% offset0

20

40

60


Str

ess

� (

ksi)

Strain � (%)0 10 20 30 40 50 60 70 80 90 100 110

Hot-rolled 1020 steel

Su = 66 ksi

Sy = 39 ksi

Se = 36 ksi

G

AB

DC

F

H

120 130 140 150 160

Area ratio R

Area reduction Ar

0 1.1

0.1 0.2 0.3 0.4 0.5 0.6

1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6

0

20

40

60


True

str

ess

�T (

ksi (

log)

)

True strain �T (% (log))3 4 5 6 7 8 0.1 32 4 5 6 7 8 1.0 32 4 5 6 7 8 10 32 4 5 6 7 8100

10

20

30

40

60

80100

200

Elasticregion

Transitionregion


Plastic strain-strengthening region

� T =

E� T

(E =

30

× 103 ks

i)�T = Sy = 39 ksi

F115

�Tf = 0.92

�T = �0 �T (�0 = 115 ksi, m = 0.22)

m

True

str

ess

�T (

log)

True strain �T (log)

Elastic line (�T = E�T)

Curve II

Curve I (�T = Se ≈ Sy)

Plastic line (�T = �0�T )m

"Ideal" material

Se3

Se1

Se2


Se �f

E


Str

ess

�

0

Strain �

Se

Sy

Su

F

dD

KB,

rati

o S u

/HB

m, strain hardening exponent0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

300

400

500

600

700

800

900

1000


Steel

d = indentation dia.D = ball dia.

0.2

0.3

0.4

0.5

0.6

Dia

mon

d py

ram

id h

ardn

ess

(Vic

kers

)

Bri

nell

hard

ness

Ult

imat

e te

nsile

str

engt

h (k

si)

Rockwell C hardness

Shore hardness

Rockwell B hardness

(0)

72 80 90 100 (110)

(10) 20 30 40 50 60 70950

900

760

740

720

700

680

660640620600580560540520500480460440420400380360340320300280260240220200180160140120100

60

70

80

90

100110120

130

140

150160170180190200210220230240250260270280290300

9590807060504030

85756555464238343228262422

850

800

750

700

650

600

550

500

450

400

350

300

250

200

150

100



Distance fromquenched end

(b)

(a)

Roc

kwel

l C h

ardn

ess

Youn

gs M

odul

us,

E (

GP

a)

Strength S (MPa)0.1 1 10 100 1000 10,000

0.01

0.1

= 10–3

Woods

Polymersfoams

AshLead

W

WC

Si CCermets

Boron

Al203Beryllium

Cast irons

OakPine

PMMA

Porousceramics

1.0

10

100

1000


Engineeringalloys

SE

= CSE

= CS3/2

E

= 10–4SE

= 10–2SE

Min. energystorage perunit volume

Yield beforebuckling

Max energystorage perunit volume

Bucklingbefore yield

Cork

Pu

Silicone

Hardbutyl

Softbutyl

= CS2

E

Elastomers

PP

II tograin

Mel

LDPE

PTFE

Pine

⊥ tograin

Balsa

HDPE

AshOak

Engineeringpolymers

PVC = 0.1SE

BalsaWood

products

Epoxies

PS

Nylons

Polyester

Designguidelines

CFRP

CFRPuniply

GFRP

LaminatesGFRP

Glasses

Sn

Concrete+

Al alloysCommon

rocksBrick etc

Zn alloysCu alloys Ti alloys

Mo alloys

SteelsNi alloys

Diamond

Si3N3Mg0

Be0

GeSilicon

Zr 02

Engineeringceramics

Engineeringcomposites

Mg alloysCement

S1/2

�

Str

engt

h S

(MP

a)

Density � (Mg/m3)0.1 0.3 1

= CS2/3

�= C

3 10 300.1

1

10

Polymersfoams

100

1000

10,000

JUVINALL: Machine DesignFig. 3-12 W-74a

Guide linesfor minimumweight design

S�

= C

Cork

Balsa

Balsa

Woods

Softbutyl Elastomers

Engineeringpolymers

Porousceramics


Engineeringceramics

Engineeringalloys

KFRP

CFRP

Mg0

Diamond

SialonsSi3N4

Al203

Ge

GFRPUniplyKFRPCFRP

Glasses

AshOak

Pine

PineFir

Parallelto grain

Ash

Perpendicularto grain

Fir

Woodproducts

LDPE

PU

PTFE

MelPVC

EpoxiesPolyesters

NylonsPMMA

PS

Silicone

PP

OakHDPE

Leadalloys

Engineeringalloys

W alloys

Mo alloys

Ni alloys

Steels

Castirons

Znalloys

Al alloys

Tialloys

Stone,rock

Cu alloysMgalloys

Pottery

Si C

B

Si

GFRPLaminates

Be

Cementconcrete

CermetsZr02

Str

engt

h at

tem

pera

ture

S (

T)

(MP

a)

Temperature T (C)0

Elastomers

T-IndependentYield strength

Upper limit on strength at temperature


Woods

100 200 300 400 600 800 1000 14000.1

1

10

100

1000

10,000

JUVINALL: Machine DesignFig. 3-13 W-74b

Engineeringalloys

Porousceramics

Engineeringceramics

Range typicalof alloy series

Polymerfoams

SiliconesButylsIce

LDPE

II tograin

Engineeringpolymers

PolyestersHDPE

PP

PC

PVCPF

Mg alloysNylons

⊥ tograin

⊥ tograin

PTFE

Epoxies

PMMAPolymides

Laminates

GFRP

CFRPUniply

KFRP

GFRP

CFRPZn alloys

Al alloys

Ti-alloys

Glasses

Bricketc.

Ni alloysSteels

SiC

Al203

Si3N4

Compression

Mullites

Mg0Zr02

Har

dnes

s (R

ockw

ell C

)

Distance from quenched end (mm)0 10 20 30 40 50

10

20

30

40

50

60

70


P


(c)

�

�

�

�

(a)

Isometric view of tensile linkloaded through a pin at oneend and a nut at the other.

(b)

(d)

� =

Equilibrium of left half showing uniform stress distribution at cutting plane

(e)

View showing "lines of force" through the link

Nut

P

EE

E

Direct view of element EEnlarged view of element E

+

+

DPA

�D2A =

4

P2

P2


3

1

5

(b)(a) P

P

4

3

1

25

6

3

1

3

P

P


2

PP



(a)

T

T

Isometric view

(b)

E

E

E E

Enlarged view ofelement

(c)

Direct view ofelement E

(d)Positive shear

Positive shear

Negativeshear

Negativeshear

Shear signconvention

(a)

T

(b)

2

Torque axis

Zero shear stress existsalong all edges.

Maximum shear stress existsalong this line.

Enlarged view ofelement 2


a

T

b

Line "A"

3

21

Top

Side Front

(a)

�max

(b) (c)

Partial beam in equilibrium

Entire beam in equilibrium

Neutralsurface

Transversecutting plane

Neutralbendingaxis and

centroidalaxis Typical cross sections

M M


Mc

cy

Neutral (bending) surface

(a)

�max

(b) (c)

Partial beam in equilibrium

Entire beam in equilibrium

Neutralsurface

CGCG CG CG

Typical cross sections

Neutral bending axisand centroidal axis

M M


M

cy

Neutral (bending) surface

(a)

Initially straight beam segment

(b)

(c)

Initially curved beam segment

CG

Hyperbolic stress distribution withincreased stress at inner surface

M

e

Typical cross sectionCenter of

initial curvature

Neutral surface


M

Centroidal surface

Neutral surfacedisplaced distance"e " toward inner

surface

co

ci

Centroidal surface

d�

�

b

ci

a d'

c'c

e

ey

d

M M

Neutral surface

Neutral axis

CG

Center of initial curvature

Centroidal axis


c

cro

co y

rrnri

�

rrn

Mc I

Valu

es o

f K

in E

q. 4

.11

: �

= K

Ratio r /c1 2 3 4

Round, elliptical or trapezoidal

Values of Ko for outside fiber as at B

U or T

I or hollow rectangular

5 6 7 8 9 100

0.5

1.0

1.5

2.0

2.5

3.0

3.5


Values of Ki for inside fiber as at A

I or hollow rectangular

Trapezoidal

b

B

B A B

B A

A

B A

b

bA

8

b2

b3b

6A

ABB b

bc

c

c

c

4

Round or elliptical

U or T

r

r


M

M M

M

h

b

h

h2

Centroidal axis

CGr = h

c =

dA = b d�

�

b


(b)(a)

Loaded "curved beam"Unloaded "curved beam"


MN.A.

�

�

M

M + dM

Enlarged view of beam segment

V

V

dA dA

y

My/I (M + dM)y/I

dA

b

y0

y c

x dx

Neutral axis


(a)

Marked and unloaded

(b)

Loaded as a beam

43

N.A.

�av = V/A

�max = V/A


32

�av = V/A

�max = V/A

N.A.


M

V

M

V


M

V

X X

100

80

60

100

+40,000 N

–40,000 N

40,000 N

80,000 N

40,000 N

60

40


(c)

b = 20

dA = 20dy

dA = 60dy

dx

�

40

(b)

b = 20

dA = 60dy

dx

�

10+

(a)

dA = 60dy

dx

10–

b = 60

�

� = 7.61 MPa0

� = 22.83 MPa

� = 32.61 MPa


Oblique viewDirect view

�x

(a)

Marked eraser

(b)

(c)

Enlarged view of element

Oblique viewDirect view

(e)

Element subjected to �max

Loaded eraser

(d)

+�

+�

�max

x

S'

y0

S

Mohr's circle


x

y

y

y

x xS'

S'SS

S'

S

yx

yx

(a)

Marked eraser (for twisting)

(b)

Enlarged element

(c)

+�

+�

�1�2

#1#2

x

y

Mohr's circle


(d)

y

y

y

x

xTT

x#2 #1

0

2000 lb


2 in.

1 in.

3 in. rad.


"B" is at bottom of shaft, opposite "A"

Top of shaft

B

2 in.

A


B

V = 2000 lb

2000 lb

M = 4000 in.lb2000 lb

4000 in.lbLoad diag.

2000 lb

Shear diag.

Moment diag.2000 lb

4000 lb

V

M

T = 6000 lb in.

A

y

y

xy

x

x

yA A x


2 in.

(a)

Isometric view

(b)

Enlarged isometric view

(c)

Direct view

Calculated values: � = 40.8 ksi � = 30.6 ksi

(d)

Isometric view

A

�yx

�yx�yx

�yx

�xy

�x�x

�x

�x

�xy

�xy

�xy

A

x y

�yx�xy

y

y

x A x


�yx = 30.6 ksi

�yx

�xy

�x = 40.8 ksi

Direct view ofelement A

y (0, +30.6)

x (40.8, –30.6)

�max = 37 ksi

�2 = –17 ksi

+�

+�

�xy

34°

56°0

�1 = 57 ksi

y

y

xA

�1 = 57 ksi

�2 = –17 ksi

28°

x


y

y

xA

� = 20 ksi

� = +37 ksi

� = –37 ksi

� = 20 ksi

17°

x


–�


(�2, 0) (�1, 0)

(�y, �yx)

(�x, �xy)

�x + �y

2

�x – �y

2

0

2

+�

–�

+�2�

1

�xy2 +

2

2

�x – �y

�1 + �22

�1 – �22

cos 2�

�1 – �22

sin 2�

0

2


+�

+�

�2

�1

2�

1


z

x

y

A A

(2) (= z)

(a)Original element

(c)1-2 plane

(b)Principal element

(3)

(1) (2)

(1)

(d)1-3 plane

(3)

(1)

(e)2-3 plane

(3)

(2)A


+�

�max = 37

Principal circle

123

(57, 0)(0, 0)(–17, 0)+�


A

A �1 (tangential)

�2 (axial) �3 = 0 (radial)

+�

0 +��2 �1�3

Correct value of �max

Erroneous value of �max obtained if �3 is neglected

r/d0 0.1 0.2 0.3

1.0

1.2

1.4

1.6

1.8

2.0Kt (a)

2.2

2.4

2.6

2.8

3.0


D d

r

M M

McI

32M

�d3�nom = =

D/d = 631.51.11.031.01

r/d0 0.1 0.2 0.3

1.0

1.2

1.4

1.6

1.8

2.0Kt

(b)

2.2

2.4

2.6

D d

rPP

PA

4P

�d2�nom = =

D/d = 21.51.21.051.01

r/d0 0.1 0.2 0.3

1.0

1.2

1.4

1.6

1.8

2.0Kt

(c)

2.2

2.4

2.6

D d

r

T T

TcJ

16T

�d3�nom = =

D/d = 21.21.09

r/d0 0.1 0.2 0.3

1.0

1.2

1.4

1.6

1.8

2.0Kt (a)

2.2

2.4

2.6

2.8

3.0


M M

McI

32M

�d3�nom = =

D/d ≥ 21.11.031.01

0 0.1 0.2 0.31.0

1.2

1.4

1.6

1.8

2.0Kt

(c)

2.2

2.4

2.6

D/d ≥ 21.11.01

r/d

r/d

0 0.1 0.2 0.31.0

1.2

1.4

1.6

1.8

2.0Kt (b)

2.2

2.4

2.6

2.8

3.0

D/d ≥ 21.11.031.01

PP

PA

4P

�d2�nom = =

T T

D d

D d

D d

TcJ

16T

�d3�nom = =

r

r

r

d/D0 0.1 0.2 0.3

1.0

1.2

1.4

1.6

1.8

2.0

Kt

2.2

2.4

2.6

2.8

3.0

MMTT

PPD

d

Axial load:

Bending (in this plane):

�nom = ≈PA

P

(�D2/4) – Dd

�nom = ≈McI

M

(�D3/32) – (dD2/6)

Torsion:

�nom = ≈TcJ

T

(�D3/16) – (dD2/6)


r/h(a)

0 0.05 0.10 0.15 0.20 0.25 0.301.0

1.2

1.4

1.6

1.8

2.0Kt

2.2

2.4

2.6

2.8

3.0


H/h = 621.21.051.01

M

r b

M

McI

6M

bh2�nom = =

r/h(b)

0 0.05 0.10 0.15 0.20 0.25 0.301.0

1.2

1.4

1.6

1.8

2.0Kt

2.2

2.4

2.6

2.8

3.0

H/h = 3

1.52

1.151.05

1.01

PA bh

P�nom = =

PP

hH

r b

hH

r/h(a)

0 0.05 0.10 0.15 0.20 0.25 0.301.0

1.2

1.4

1.6

1.8

2.0Kt

2.2

2.4

2.6

2.8

3.0


H/h = ∞1.51.151.051.01

M

rb

h MH

McA

6M

bh2�nom = =

r/h(b)

0 0.05 0.10 0.15 0.20 0.25 0.301.0

1.2

1.4

1.6

1.8

2.0Kt

2.2

2.4

2.6

2.8

3.0

H/h = ∞1.51.15

1.05

1.01

r

bh H

PA bh

P�nom = =

PP


d/b(a)

0 0.1 0.2 0.3 0.4 0.5 0.6

d/b(b)

0 0.1 0.2 0.3 0.4 0.5 0.6

1.0

7

6

5

4

3

2

1

1.2

1.4

1.6

1.8

2.0Kt

Kt

2.2

2.4

2.6

2.8

3.0

d/h = 0

0.25

0.5

MM

PP

d bh

d bh

McI

6M

(b-d)h2�nom = =

PA (b-d)h

P�nom = =

2.0

1.0

Pinloadedhole

Unloadedhole

W/w1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.01

2

3

4

5

6

7

8

9Kt

10

11

12

13

14

15

16

17

18


t

h

w

P

r

W

PA

�nom = = Pwt

r/w = 0.050r/w = 0.10r/w = 0.20

h/w = 0.5

0.5

0.75

0.5

0.751.0

0.751.0

1.03.0

3.0

3.0

F F

�

�

Sy

Su


F F

(a) Unnotched

Stress

F = ASy

(c)

(e)

� = Sy

Stress

�max = Sy

�av = Sy /2

Stress con. factor = K = 2Same cross-section area = K

F =ASy

2

(d)

Stress

� = Sy

F

( f )

ab

cd

(b) Notched


0

(a) Load causes no yielding

+ Sy

0

(b) Load causes partial yielding

+ Sy

0

(c) Load causes partial yielding

+ Sy

0

(d) Load causes total yielding

Load stress + =

+

+

+

+

+

=

=

=

=

Sy –2Sy –Sy

–Sy – 0

– 0

– 0

– 0

– 0 +

– 0 +

– 0 +

– 0 +

Load removalstress change

Residual stress


F1

F2

M1 M1

F2

F110 mm

10 mm

50 mm

0

–300 MPa

300 mPa

25 mm

Z = bh2

6

Z =

Z = 1.042 × 10–5 m3

(0.025)(0.050)2

6

0 300 396 –396

–238

–96

62

Load stress

+

+

=

=

(a) Given information (see text)

(b)

––

0

Load removal stress change

––

0

Residual stress

––

0–96

62

62

62

–104

–62

–200

Residual stress

+

+

=

=(c)

––

0

Load stress

––

0

Total stress (straight beam)

––

0–96 300

Residual stress

+

+

=

=(d)

––

0 396

238

Load stress

––

0

Total stress (ready to yield)

––

0–96 –204

–122

–300

–60

Residual stress

+

+

=

=(e)

––

0

Load stress

––

0

Total stress (ready to yield)

––


10.000 in.

T = 80°F

P = 0 lb P = 0 lb

10.008 in.

T = 480°F

P = 60,000 lb P = 60,000 lb


60 mm

10 mm dia.

20 mm

P = 400 kN

P


FEDCBA


di = 20 mm

do = 24 mm

�max = 100 mm

T


b

2rT

T

T

T


M M

h

r

b


4 in.200 lb

200lb

1-in.-dia round rod

3 in.


A

P Q

A60

70,000 N

40 80

120


h

X

b

a

c

F


2 in.13

in.16

3in.16

3 in.4

3in.16

1 in.


A

A

12,000 N

30 mm

24 mm

8 mm

5 mm

5 mm

Section AA

30 mm

400 N120-mm-dia.

sheave

Free endof shaft

1002000 N


B

Connected toflexible coupling

and clutch

20-mm-dia. shaft

A

S

T

100


8 in.

in.3

in.

in.

12

12

in.12

38


5

5 5

5 50

40

12 kN

L2

Cement L2


100 mm

250 mm

200 mm

25-mm-dia. roundrod bent into crank

1000 N

1 in.

6-in. dia.

1-in. dia.shaft

Motor

1000-lbbelt tension

3000-lbbelt tension


50-mm dia.

100-mm dia.

30-mm dia.


50 mm

100 mm

50 mm

4000 lb

F

B

A

500 lb

JUVINALL: Machine DesignFig. P4-38 W-134a

3 in.

4 in.

3 in.

2 in.

1000 lb

4 in.

1-in.-dia. shaft

A

B

A



y x

18 45

30


Free surface, �3 = 0

20 ksi

30 ksi


100350

75


5000 N5000 N 30 mm 200 mm

15 mm25 mm

500 mm 250 mm


d = 40 mm d = 40 mm

r = 5 mm

A

RA

B

RB

1000 N

D = 80 mm


5000 N 5000 N50 mm 100 mm

15 mm

25 mm

Cal

cula

ted

elas

tic

stre

ss (

MP

a)

Time0 1 2 3 4 5 6 7 8 9 10 11 12

–200

0

200

400



(a) (d) (e)(b) (c)

Z

Y

Xx

y

z

dy (neg.)

dz (neg.)

dx

x → 0�x = lim dx

xy → 0

�y = limdyy

z → 0�z = lim dz

z

Unloaded elementElement loaded in uniaxialtension in X direction (withdeflections shown exaggerated)



�

dx

�yx

�xy

�xy (shown counterclockwise, hence negative)�yx (shown clockwise, hence positive)

y

xZ X

Y

y → 0absolute value = lim = tan � ≈ �dx

y


+�/2

x

y

0+�

�2

�y

�x

�xy

�1


(a)Single-element gages oriented

to sense horizontal strain

(b)Two-element rosettes oriented

to measure horizontal andvertical strain

(c)Three-element equiangular

rosettes

(d)Three-element rectangular

rosettes


�2

�1

�240

�120

�0+�

+�

120°

240°

�2

�1

�120 �240

�0+�

+�

�2

�1

�240 �120

�0

+�

+�

(a) (b) (c)


17°

240° gage(� = +0.00185)

120° gage (� = +0.0004)

0° gage (� = –0.00075)

�1 = +0.0020

�2 = –0.0010


+� /2

120°

240°0°

34°

+0.00185–0.00075

+0.0004

�2 = –0.001 �1 = +0.0020

+�


90°

�2

�1

�0

�90

(a)

45°

+�

+�

�2

�1

�90

�0

�45�45

(b)

+�

+�

�90

�2

�1

�0

�45

(c)

+�

+�


+2600 �m/m

+450 �m/m

–200 �m/m

(a)

(b)

�0 = +2600

�90 = +450

�45 = –200

�0 = +2600

�90 = +450�2 = –510

�1 = +3560

�45 = –200


29°

119°


+� /2

0°

45°

90°

2600

–510 3560

+�–200

450058°

+�

0+�


�2 = –24

�3 = 0

�1 = 134 �1 = 0.002

(a)

+� /2

+�

�2 = –0.001 �3 = –0.0005

(b)

0

0

–0.04

–1.2

–0.8

–0.4

0

–0.8

–0.4

0

0.4

0.8

0.4

0.8

1.2

, (1

0–6

/mm

)M

, be

ndin

g m

omen

t (k

N•m

m)

V,

shea

r fo

rce

(kN

)

0

4

8

12

0

200

400

–2

–1

0

1

2

–0.08

–0.12

Def

lect

ion

(mm

)R

elat

ive

slop

e (m

illir

adia

ns)


0.093

–0.677

–0.220

0.2700.835

0.838

0.0220.007

0.115

0.126

Tangent point

Parallel to true lineof zero deflection

True line ofzero deflectionLo

cati

on o

f ze

roab

solu

te s

lope

Init

ially

ass

umed

loca

tion

of

zero

slo

pe

0.119

+0.001

–1.109–1.096

A =0.093 mm

A = 0.011

A = 13 × 10–6

A = 11,000

A = 36,000

M EI

A = 3 × 10–6

A =0.457 × 10–3 A =

0.270 × 10–3

A = 0.565 × 10–30.220 × 10–30.419× 10–3

A = 0.120 mm

Abs

olut

e sl

ope

(mra

d)

0.852.67

8.46

220 255325 360

1822

9.80

5.12 5.670.69

1.94 2.47 0.28

Area = 58,000 kN•mm2

2.2

0.7Area =

220 kN•mmArea = 140 kN•mm

Area = 360 kN•mm

–1.8

d = 30

2.2 kN 1.8 kN

10

d = 40

1050 50100 200 200

d = 40

1.5 kN 2.5 kN

d = 50d = 60

0 ∆Deflection

Area = U'

Area = U

Area = dU

dQ


QLoad

Area = dU'


L/2 P/2P/2

x dx

L

P

b

h


200 mm 2500 N2500 NL = 400 mm

P = 5000 N

b = 25 mm

h = 50 mm


L

P

h

bx

c

a

y

Q


F

h

b

V

F

F

P M

F

R

R

�

2R

�

�R – R cos �

(a) (b)

sin

�

–1.0

–0.5

0sin � d� = 1;

(From this, average value = can be determined)

�/2

0

0.5

1.0

2�

2�

2�

4�

2�

Avg value =

cos

�

–1.0

–0.5

0

0.5

1.02�Avg value =

sin

� c

os �

� (radians)0 �/2 � 3�/2 2�

–0.5

0

0.5

(0.707)(0.707) = 0.52�Avg value = 0.5 = 1

� = (Half of avg value shown in sin � plot)

cos2

�

0

0.5

1.0Avg value = 0.5

sin2

�

0

0.5

1.0Avg value = 0.5

∫ sin � d� = 02�

0∫sin � d� = 2;�

0∫

cos � d� = 1;�/2

0∫ cos � d� = 02�

0∫cos � d� = 0;�

0∫

sin2 � d� = (0.5) = ;�/2

0∫ sin2 � d� = �(0.5) =�

0∫sin2 � d� = �

2�

0∫

2�

4�

2�

cos2 � d� = (0.5) = ;�/2

0∫ cos2 � d� = �(0.5) =�

0∫cos2 � d� = �

2�

0∫

2�

�121

sin � cos � d� = = ;�/2

0∫ sin � cos � d� = 0�

0∫sin � cos � d� = 0

2�

0∫



F

h = 0.3 in.

b = 0.2 in.

E = 18 × 106 psiG = 7 × 106 psi

F

R = 2.0 in.

�

2R = 4 in.

(a)

CopperCast ironSteel

0.2 0.3Width, h (in.)

(b)

Def

lect

ion,

� (

in.)

0.4 0.50.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035


Def

lect

ion,

� (i

n.)

Radius, R (in.)1.0 1.5 2.0 2.5 3.0

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035


CopperCast ironSteel

0.06

0.05

0.04

0.03

Def

lect

ion,

� (

in.)

0.02

0.01

0.5

0.4

Thickness, b (in.)

0.3

0.2

0.3

Width, h (in.)

0.2

0.1

JUVINALL: Machine DesignFig. 5-21d W-169

F


3 m

10 m

ya

Point of zerodeflection

1.2 m

500 kg mass

P/2


a x

y

b2

a

3'

2'

1

2

3

b2

Pa/2

M0

P/2

P

P/2

P/2

M0

M0

Pa/2

M0

P/2M0

P/2

M0

(a)

(b)

P


e

L or Le

P

P

P

P

x x

y

y

(b)Column cross section

(a)Two views of column

Axis of least I and � becomesneutral bending axis whenbuckling occurs. With columnformulas, always use I and �with respect to this axis.

Slenderness ration Le /�

10 1000.0001

0.001

0.010

ScrE

0.100


Slenderness ratio Le /�20 40 60 80 100 120 140 1600

20

40

60

80

A

B

C

D

100

120

140

160

180


0

100

200

300

400

500

600

700

800

900

1000

1100

1200

Cri

tica

l uni

t lo

ad S

cr (

ksi)

Cri

tica

l uni

t lo

ad S

cr (

MP

a)

Sy = 689 MPa

Arbitrary valuesfor illustration

Sy = 496 MPa

Euler, E = 203 GPa (steel)

EF

Euler, E = 71 GPa (alum)


L

Le

(b)

L Le

(c)

Le = L

(a)

(Buckledshape showndotted)

L Le

(d)

LLe2

(e)

Le = 0.707L Le = 0.5LLe = L Le = L Le = 2LTheoretical

Le = 0.80L Le = 0.65LLe = L Le = 1.2L Le = 2.1L

MinimumAISCRecommend

Source: From Manual of Steel Construction, 7th ed., American Institute of Steel Construction, Inc., New York, 1970, pp. 5–138.

Slenderness ratio Le /�20 40 60 80 100 120 140 1600

20

40

60

80

A

B

C

D

100

120

140

160

180


0

100

200

300

400

500

600

700

800

900

1000

1100

1200

Cri

tica

l uni

t lo

ad S

cr (

ksi)

Cri

tica

l uni

t lo

ad S

cr (

MP

a)

Sy = 689 MPa

Johnson,E = 71 GPa,Sy = 496 MPa

Johnson, E = 203 GPa, Sy = 689 MPa

Euler, E = 203 GPaTangentpoints

Euler, E = 71 GPa

EF

Sy = 496 MPa


80,000 N 80,000 N

D

1 m

SF = 2.5Sy = 689 MPaE = 203 GPa (steel)


80,000 N 80,000 N

D

200 m

SF = 2.5Sy = 496 MPaE = 71 GPa (aluminum)

Cri

tica

l uni

t lo

ad S

cr (M

Pa)

Cri

tica

l uni

t lo

ad S

cr (

ksi)

Slenderness ratio Le/�20 40 60 80 100 120 140 160 180 2000

100

50

150

200

250

300

350

400

450

500

550

600

10

0

20

30

40

50

60

70

80


ec/�2 = 1.0

ec/�2 = 0

0.6

0.3

0.1

Euler curve

0.05


(a)Wrinkling, or "accordian

buckling" of thin-wall tube

(b)Typical local buckling ofan externally pressurized

thin-wall tube

(c)Wrinkling of thin, unsupportedflanges of a channel section


Beam

Triangle

Quadrilateral

Tetrahedron

Pentahedron

Hexahedron

JUVINALL: Machine DesignFig. 5-34 W-180B

(1)

(3)(6)

(7) (9)

(2) 31

2

5 7

W

4

6

(5) (10)

(4)

(8)

(11)

JUVINALL: Machine DesignFig. 5-35 W-180C

�3x

�3x

3'

3

F (7)3y F (7)

3x

6'

6

F (7)

L 6y

F (7)6x �6x

�6x

JUVINALL: Machine DesignFig. 5-36 W-180D

�

3 3'

F (7)3x

�3x

F (7)

A, E

L

3y

F3

6

�

2 m

22' (Constant A, E)

3 N

2 N

3.464 m

4 m

�2x

�2y

�3x

(3)

(2)

3 3'1

JUVINALL: Machine DesignFig. 5-37 W-180E

F (1)2y F (2)

2y

F (1)2x

F (1)1y

F (3)1y

F (3)1x

F (1)1x

F (2)3y

F (2)3x

F (2)2x

(1)

(2)

(3)

4 m

1

1

3

F (3)3y

F (3)3x

3

2 2

2 m

3.464 m

JUVINALL: Machine DesignFig. 5-38 W-180F

JUVINALL: Machine DesignTable 5-1 W-157

L

P

�� = PL

AE1. Tension or compression

Cross-section area = A

LT

�

�

�

2. Torsion

For solid round bar anddeflection in degrees,

K'a = section property. For solidround section, K' = J = �d4/32.

3. Bending (angular deflection)

I = moment of inertia aboutneutral bending axis

M

M

L

4. Bending (linear deflection)


L

�

5. Cantilever beam loaded at end


LP

� = TLK'G

� = MLEI

� = ML2

2EI

� = PL3

3EI

k = =P�

AEL

K = =T�

K'GL

K = =M�

EIL

k = =M�

2EI

L2

k = =P�

3EI

L3

�° = 584TL

d4G

(d4 – d4)


d

do

2b

2a

a

a

b

a

K' = J = �d4

32

di K' = J = o i�32

– 3.36 1 –K' = ab3

16b4

12a4163

ba

d

t

K' = �dt4

32

K' =

K' = 0.0216a4

K' = 2.69a4

�a3b3

a2 + b2

a

a

K' = 0.1406a4


�120 = +625

�240 = +300

�0 = +950


�0 = –300

�135 = –380

�270 = –200

�90 = –300

�45 = –380

�0 = –200

Gage readings Equivalent rosettes


k = 5 N/mm

A

B

C

100 mm

100 mm

100 mm

F

F

F


200 mm

25 mm-dia. steel

1000 N

100 mm

A


d = 30d = 50 d = 40

4 kN

2 kN100 200 150


a

Z

b

YX

F (used in Problem 5.17)

T (used in Problem 5.18)

Solid round rod ofproperties E, G, A,

I, and J.


w = 200 lb/in.

d

5 in. 15 in. 5 in.

0.75d 0.75d


a

a F

F

b


P

R


F

b

h

2

b2

L


5 kN

S

500 mm

300 mm


2

1 1

2

1 m

0.7 m

1 m0.7 m

Boom

12 mm-dia. tie-rod

6 kN



2w

2c

t

P

P

(a) Center crack

w

t

P

P

(b) Edge crack

c


2w = 6 in.

2c = 1 in.

t = 0.06 in.

7075–T651 Aluminum,Su = 78 ksi, Sy = 70 ksi,

Kic = 60 ksi

P

P

in.


P

P

a2c

2w

t


P

P

a2c

2w = 6 in.

t = 1 in.

�g = 0.73 Sy

a/2c = 0.25

Ti – 6Al – 4V (annealed)titanium plate


+�

�2 = 40 ksi

(�max = 60)

�1 = 80 ksi

(a) Proposed application involving�1 = 80, �2 = –40, �3 = 0

+�

+�

(�max = 50)

�2 = �3 = 0�3 = 0

�1 = 100 ksi

(a) Standard tensile test of proposed material. Tensile

strength, S = 100 ksi

+�


+�

+�

Uniaxialcompression Uniaxial

tension

Principal Mohr circlemust lie within thesebounds to avoid failure

For biaxial stresses (i.e., �3 = 0),�1 and �2 must plot within thisarea to avoid failure

0

+�2

+�10

Suc Sut Suc

Sut

Suc

Sut

(a) Mohr circle plot (b) �1 – �2 plot

+�

+�


Principal Mohrcircle must liewithin thesebounds to avoidfailure

Uniaxial tension

Syt

+�2

+�1Syt

Syt

0




Principal plane

Principal planes

Octahedral plane

+�2

(0, 100)(–100, 100) (100, 100)

(100, 0)(–100, 0)

(–100, –100)(0, –100)

(100, –100)

+�1


(–58, 58)

(58, –58)

(50, –50)

0

Shear diagonal (�1 = –�2)

Distortion energy theory

Normal stress theory

Shear stresstheory

Note: �3 = 0

+�

+�


Principal Mohr circle mustlie within these boundsto avoid failure

SutSuc

+�2

+�1Sut

Suc

Suc

+Sut

0



0


+�2

+�1SutSuc

Sut

Suc

0

Shear diagonal


�2 = –25 ksi

�2 = –25 ksi

�1 = 35 ksi �1 = 35 ksi

Note: �3 = 0

Steel,Sy = 100 ksi


–25

–100

�2 (ksi)

�1 (ksi)35 58 66 100

(58, –58)

0

Normal loadpoint

Limitingpoints

� theory

D.E. theory

� theory

Load line

Shear diagonal

Str

ess

(% o

f ul

tim

ate

stre

ngth

)

Load (% of ultimate load)0 50

SF = 2 based onload, Eq. 6.10

SF = 2 based onstrength, Eq. 6.9

Fracture

100

50

100


Freq

uenc

y p(

x) a

nd p

(y)

Strength (x), and stress (y) (MPa or ksi)0 40 70


x (strength)y (stress)

�y �x

Freq

uenc

y p(

z)

Margin of safety, z, where z = x – y300


z (margin of safety)

�z

Freq

uenc

y p(

x)

Quantity x0 � x1

�1

�1 < �2 < �3

�2

�3

x2


Freq

uenc

y p(

x)

Quantity x

–3� –2� –1� � +1� +2� +3�


0.13% 0.13%2.14% 2.14%

13.60% 13.60%

34.13% 34.13%of totalarea

Inflection point Inflection point

% r

elia

bilit

y(%

of

surv

ivor

s or

% c

umul

ativ

e pr

obab

ility

of

surv

ival

)

Number of standard deviations, k–4 –3 –2 –1 0 +1 +2 +3 +4

0.01

0.050.10.2

0.5

1

2

5

10

20

30

40

50

60

70

80

90

95

98

99

99.899.9

99.99

% o

f fa

ilure

s or

P,

% c

umul

ativ

e pr

obab

ility

of

failu

re

99.99

99.999.8

99

98

95

80

70

60

50

40

30

20

10

5

2

1

0.5

0.20.10.05

0.01


�

–k�

% of failures

Fatigue life, strength, etc.

Extremevalues

k = –4, 99.99683% reliabilityk = –5, 99.9999713% reliabilityk = –6, 99.9999999013% reliability

% ofsurvivors

Freq

uenc

y p(

z)

Torque z (N • m)

(5.22)0


z = x – y

�z

Freq

uenc

y p(

x) a

nd p

(y)

Torque x and y (N • m)

(a)

(b)

0 (14.8) 20.0

One bolt in500 twists off

x (bolt twist-off strength)

y (wrenchtwist-off torque)

�x�y

�x = 1 N • m�y = 1.5 N • m

k�z

One bolt in500 twists off


�1 = 200 MPa

�2 = 100 MPa

�2

�1 �1 = 150 MPa

�2 = –100 MPa

�2

�1 ba


�x �x = 50 MPa

Sy = 500 MPa

�xy = 100 MPa

�xy

k


c

k

m

(b)(a) (c)

k

m

m

S u a

nd S

y (k

si)

Average strain rate (s–1)10–6 10–5 10– 4 10–3 10–2 10–1 1 101 102 1030

10

20

30

40

50

60

70

80

90

100

S y /

S u (%

)E

long

atio

n (%

)

0

10

20

30

40

50

60

70

80

90

100


Yield strength Sy

Ultimate strength Su

Total elongation

Ratio Sy /Su

�

k

(a) (b) (c)

Elastic-strain energy stored

in structure = Fe�ForceFe

W O

hDeflection

Work of falling weight = W (h + �)

Guide rod

h

�

�st

12

JUVINALL: Machine DesignFig. 7-3a-c W-218

W

k

W

d

L /2

L /2

2d


(b)

d

L

(a)

�d2

4Area = A =

L

Bumper ofcross section A;volume = AL

h


W

1 in. × 3 in., I = bh3/12 = 6.46 in.4

2 × 4 white pineE = 106 psi

Mod. of rupture = 6 ksi


30 in.

12 in.

60 in.

100 lb/in. 100 lb/in.58

58

Z = I/c = 3.56 in.3

100 lb

100-mmdia

120-mmdia

20-mmdia


20 mm20 mm

250 mm

Tors

iona

l def

lect

ion

of s

haft

(de

g)

Shaft radius (mm)5.0 7.5 10.0 12.5 15.00

5

10

15

20

25

30

35

40


Sha

ft s

hear

str

ess

(MP

a)

Shaft radius (mm)5.0 7.5 10.0 12.5 15.0

100

200

300

400

500

600

700

AluminumCast ironSteel

AluminumCast ironSteel


Ki = 1.5

Ki = 1.5

d


Ki = 1.5

Ki = 3

Ki = 1.5d

d2

Ki = 3.4

A = 700 mm2


"Very long"(>10d)

"Negligible"

Ki = 3.5

A = 600 mm2

Shank

Head

Ki = 3.4

A = 300 mm2

Ki =1.5

Ki = 3.0

Ki = 1.5

A = 600 mm2

(a) Original design (b) Modified design

d


K = 1.55

K = 4

Dropweight

24 mm dia.

K = 1.4

30 mm dia.


(a)

Axial hole

(b)

"Very long"

K = 2

K = 21-in dia.


0.1-in dia. hole

Steel cableA = 2.5 in.2

E = 12 × 106



K = 5000 N/mmL = 5 m

v = 4 km/hr m = 1400 kg

Rope


(a)

Original design New design

(b)

Thread: Area = 600 mm2

K = 3.9

Thread:Area = A

K = 2.6

Area = AK = 2.6

"Very long"

A= 600 mm2

K = 3.9

K = 1.3

K = 1.3

Platform

Area = 800 mm2

Area = 800 mm2

dia. = 36 mm

K = 2.2

250 mm

(Fracturelocation)

3 mm(negligible)


K = 1.5

Hole dia., d

A = 800 mm2

K = 3.8

(a)

Original design

(b)

Modified design

Assume that holedrilled to this depth,does not significantlychange the K = 3.8factor at the thread.


Small region behaves plastically

Main body behaves elastically

0.300"

C

R

110-Volts AC

Flexiblecoupling


Weights

Revolutioncounter

Test specimen

9 "78

Motor

++

++

Fati

gue

stre

ngth

, or

Pea

k al

tern

atin

g st

ress

S (

ksi (

log)

)

Life N (cycles (log))103 104 105 106 107 108

10

20

30

40

50607080

100


(c) Log-log coordinates

Sn (endurance limit)'

'

'

"Knee" of curve

8:7 ratio

Not broken

7:1ratio

Life N (cycles (log))

(a) Linear coordinates (not used for obvious reasons)

(b) Semilog coordinates

Sn (endurance limit)

Fati

gue

stre

ngth

, or

Pea

k al

tern

atin

g st

ress

S (

ksi)

103 104 105 106 107 1080

10

20

30

40

50

Not broken

Life N (cycles � 106)

Sn (endurance limit)

Fati

gue

stre

ngth

, or

Pea

k al

tern

atin

g st

ress

S (

ksi)

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

Not broken

++++++

++++++

++++

++

++

+++++++++++

++++++++++++

S/S u

(log

)

Life N (cycles(log)) Sn = 0.5 Su(in ksi, Sn � 0.25 � Bhn;

in MPa, Sn � 1.73 � Bhn)

103 1042 4 6 1052 4 6 1062 4 6 1072

Not broken

4 60.4

0.5

0.6

0.8

0.9

1.0

0.7


S = 0.9Su (in ksi, S � 0.45 � Bhn; in MPa, S � 3.10 � Bhn)

Sn

'

'

''

End

uran

ce li

mit

S

n (k

si)

' 'E

ndur

ance

lim

it

Sn

(MP

a)

Hardness (Rockwell C)0 10 20 30 40 50

0.25 Bhn

60

187 229 285

Hardness (HB)

375 477 653

700

20100

SAE 4063SAE 5150SAE 4052SAE 4140

200

300

400

500

600

700

800

900

40

60

80

100

120

140


Actual maximum stressCalculated maximum

stress (Mc /I )


MM

Pea

k al

tern

atin

g be

ndin

g st

ress

S (

ksi (

log)

S (M

Pa

(log

))

Life N (cycles (log))103 104 105 106 107

Wrought

Permanentmold cast

Sand cast

108 10956

78

10

12

14161820

30

35

25

40

50

60

7080

50

75

100

150

200

250

300

400

500


Fati

gue

stre

ngth

at

5 �

10

8 c

ycle

sS

n (k

si)

Sn

(MP

a)

Tensile strength Su (ksi)

Su (MPa)

0 10 20 30

Alloys represented:1100-0, H12, H14, H16, H183003-0, H12, H14, H16, H185052-0, H32, H34, H36, H38

2014-0, T4, and T62024-T3, T36 and T46061-0, T4 and T6

6063-0, T42, T5, T67075-T6

40 50

0 50 100 150 200 250 300 350 400 450 500 550

60 700 0

50

100

150

10

20

30


Sn = 19 ksi

Sn = 0.4Su'

'' '

Pea

k al

tern

atin

g st

ress

S (

ksi (

log)

)

S (

MP

a (l

og))

Life N (cycles (log))105 1062 4 6 8 1072 4 6 8 1082 4 6 85

6

50

75

100

150

200

7

89

10

12

14

161820

25

30

35

40


Sand-cast

Extruded and forged

Rat

io,

(log

)pe

ak a

lter

nati

ng s

tres

s, S

or

S sS u

Life, N (cycles (log))103

Torsion

Axial (no eccentricity)

Bending

Sn = Sn = 0.5Su'

'

'

Sn = 0.9Sn = 0.45Su

Sn = 0.58Sn = 0.29Su

S103 = 0.9SuS103 = 0.75SuS103 = 0.9Sus (≈ 0.72Su)

104 105 106 1070.1

0.3

0.5

1.0


–1.2Reversed bending

Reversed torsion

Reversed bending

DE theory

–0.8

–0.6

–0.4

–0.2

0.2

0.4

0.6

0.8

1.2

–1.2 –0.8 –0.4

Note: Dotted portion issuperfluous for completelyreversed stresses

0 0.4 0.8 1.2

�1Sn


�1Sn

�2Sn

–1.0

1.0

Sur

face

fac

tor

Cs

Tensile strength Su (ksi)60 80 100 120 140 160

Su (GPa)

180 200 220 240 260

120 160 200 240 280 320

Hardness (HB)

360 400 440 480 520

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0


0.4 1.81.61.41.21.00.80.6

Mirror-polished

Fine-ground orcommerciallypolished

Machined or cold-drawn

Hot-rolled

As forged

Corroded in salt water

Corroded intap water


Equalsurfacestresses

(a) d = (0.3" or 7.6 mm)

(b) d > (0.3" or 7.6 mm)

(c) d < (0.3" or 7.6 mm)

0

+

–

�min

�m = mean stress; �a = alternating stress (or stress amplitude)�max = maximum stress; �min = minimum stress�m = (�max + �min)/2�a = (�max – �min)/2

Str

ess


�max

�max

�a

�m

�m �a

�a

�min

0 Sy–Sy

A' A

Sn

Sy

F

E

A"

D

C

G

H

106 ~

Values from S–N curve

105 ~

104 ~

10 3 ~

10 4 ~

10 5 ~10 6

~

103 ~

H'

B

�m (tension)–�m (compression) Su


�a

Max

imum

str

ess

�m

ax (

% o

f S u

)

Minimum stress �min (% of Su)–100 –80 0 20 40 60 80 1000

10

20

30

40

50

60

70

80

90

100


–60 –40 –20

–30

–20

40

50

Mea

n str

ess �

m (%

of S u

)

60

80

70

100

90

–10

10

30

20

60

50

40

Alternating stress �a (%

of Su ) 20

30

10

90

70

80

Su

103-cycle life

104

10610

5

107

3 × 103

3 × 104


–20

–10

Max

imum

str

ess

�m

ax (

ksi)

Minimum stress �min (ksi)–80 –60 0 20 40 60 800

10

20

30

40

50

60

70

80

–20–40

50

40

Alternating stress �a (ksi) 20

30

10

70

60

103-cycle life

104

105

109

106

107

4 × 104

4 × 105

Su

40

50

60

70

80

10

30Mea

n str

ess �

m (k

si)

20


Max

imum

str

ess

�m

ax (

ksi)

Minimum stress �min (ksi)–80 0 20 40 60 800

10

20

30

40

50

60

70

80

–40–60 –20

–30

–10

–20

40

50

60

70

80

10

30Mea

n str

ess �

m (k

si)

20

50

40

Alternating stress �a (ksi) 20

30

10

70

60

103 -cycle life

104

105

109

106 10

7

4 × 104

4 × 105

Sn

–Sn

Sy

Su

0

(a)

(b)

(c)

(d)

(e)

( f )

Cal

cula

ted

fluc

tuat

ing

axia

l str

ess

(ign

orin

g yi

eldi

ng)


�a�m

�a

�a

�m

0

d < 2.0 in.

P(t)P(t)

P(t)

t

Commerciallypolished surface

Sy = 120 ksiSu = 150 ksi


20 = 0.67

A10 3

~

10 4 ~

10 5 ~

10 6 ~

Axial loading stresses in ksi

Axial loading stresses in ksi

1�a

�a

Sy

�m

40

Point O

80

100

112 ksi

92 ksi

61 ksi

75 ksi

120

–120 –100 –80 –60 –40 –20 0 20 40

(used in Sample Problem 8.2)

60 80 100 120 150–�m (compression) +�m (tension) SySy Su


Pea

k al

tern

atin

g st

ress

, S

(log

)


S = 0.75Su = 0.75(150) = 112

'Sn = SnCLCGCS = (0.5 × 150)(1)(0.9)(0.9) = 61

103 104 105 106 107

61

75

92

112

3 2

Cal

cula

ted

nom

inal

str

ess

S

Life N (cycles)103 104 105

Unnotched specimensNotched specimens

106

(a) Unnotched specimen ("u")

(b) Notched specimen ("n") (c) Illustration of fatigue stress concentration factor, Kf

107

Kf = Sn(u)

Sn(u)

Sn(n) Sn(n)


d

d

Notch radius r (in.)0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

Notch radius r (mm)0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

0

0.1

0.2

0.3

q

0.4

0.5

0.6

0.7

0.8

0.9

Use these values with bending and axial loads

Use these values with torsion

SteelSu (ksi) and Bhnas marked

1.0


Aluminum alloy (based on 2024-T6 data)

200 (400 Bhn) 180 (360 Bhn)

120 (240 Bhn)

140 (280 Bhn)

100 (200 Bhn)

80 (160 Bhn)

80 (160 Bhn)

60 (120 Bhn)

60 (120 Bhn)

50 (100 Bhn)

100

200

d

cc'

b

a

b'

280300

0 100 200 300

106 ~ Life105 ~ Life

104 ~ Life

103 ~ Life

400 450�m, tension (MPa) Su

Sy

�a

(MP

a)

0

–300

300

300

600

0(a) (b) (c) (d)

(a) (b') (c') (d')

Calculatedstresses

Actualstresses

Constant-lifefatiguediagram

Cal

cula

ted

notc

h st

ress

(P/A

)Kf (

MP

a)A

ctua

l not

ch s

tres

s(P

/A)K

f + �

resi

dual

(M

Pa)


d'

Commercial ground finishHeat-treated alloy steel,Su = 1.2 GPa, Sy = 1.0 GPa

T = 1000 ± 250 N • m

T = 1000 ± 250 N • m

D/d = 1.2 r/d = 0.05 SF = 2.0


d

r

D

100

150

1200

B'

B

ANB

NB'

NA

(0.58)(0.9)(0.87) = 272 MPa

�max = Ssy106 ~ = ∞ life

Assuming 10 mm < d < 50 mm

2

116

200

300

–200 –100 0

0'

100 200

Sn = SnCLCGCS = '

300 400

Mean torsional stress �m (MPa)

500 600 700 800 900Ssy ≈ 0.58(1000) = 580 Sus ≈ 0.8(1200) = 960

Alt

erna

ting

tor

sion

al s

tres

s� a

(M

Pa)


Alt

erna

ting

ben

ding

str

ess

�ea

(M

Pa)

Mean bending stress �em (MPa)0 100

(15.7, 65.0)"Operating point"

200 300 400 500 600 700 800 9000

100

200

300 "Failure point"

�max = Sy

Sy = 750

Su = 900

106 ~ = ∞ life


900

r = 5 mm rad., machined surface

D = 18 mm (bearing bore)d = 16 mm (shaft dia.)

50 mm

f = 0.6 (between the objectand the disk)

T = 12 N • m (friction torque)

Su = 900 MPa

Sy = 750 MPa

' (1)(0.9)(0.72) = 291 MPa2

Sn = SnCLCGCS =

100 mm

Ft Fn

Str

ess

(ksi

)

(a)Stress-time plot

(b)S-N curve

–80

–40

0

40

80


Rev

erse

d st

ress

S (

ksi (

log)

)

N (cycles (log))103 104 105 106 107

40

50

60

140

120

100

80

Representive 20-second test

1.6

× 1

04

3.8

× 1

04

10

5

0

–100

–200

–300

100

200

300

2~a

3~

(a)

Stress-time plot

2~ 1~

1~

Ben

ding

str

ess

(MP

a)σ a

(M

Pa)

Rev

erse

d st

ress

S (

MP

a)

�m (MPa)

�m – �a plot

(b)

Sy Su

N (cycles)

(c)

(d)

S-N plot

100 200 300 400 500

103 104 105 106 107 108

0

100

200

300

400

100

150

200

250

300

400

500


b c

b d''

c

b'b

a'

a

c'd

d'

c''

d'

b''

d

Representative 6-sec test

c'

a' b'

2.5

× 1

03

3.5

× 1

06

2 ×

10

4Bending stress at

critical notch

Part

Aluminum alloySy = 410 MPaSu = 480 MPa

P(t)

0

Pmax

Pmax

T(d) Strength

(c) Total stress, (a) + (b)(a) Load stress

(b) Residual stress

Com

pres

sive

str

ess

(ksi

)Te

nsile

str

ess

and

stre

ngth

(ks

i)


Axis of specimensymmetry andaxis of load

(1) (2)

P


30 mm

30 mm

PP

P

30 mm

r = 2.5 mm

r = 2.5 mm

35 mm

30 mm


3 in.3 in.

1 in. dia. 1 in. dia.14

2 in. 1 in.2 in.

F lb

1 in. dia.

1R8

1R8

1R16

Flb2

Flb2


24 mm 24 mm20 mm

2-mm rad. 0.8-mm rad.


High-carbonsteel, 490 Bhn,machined finish

1

0.1094 in.

3 in.4

8r = in.Kt = 1.7

F

4 in.


1

0.050 in.

0.191-in. rad.

0.125-in. dia.

2 in.

0

+80

–16Time

Nom

inal

str

ess

(MP

a)

JUVINALL: Machine DesignFig. P8-38 W-268b

60 mm50 mm

1.5 rad. 5-mm rad. 5-mm rad.

50 mm60 mm

JUVINALL: Machine DesignFig. P8-39 W-268c

-in. dia. hole116

1.0-in. dia.1.2-in. dia.

0.1-in. rad.

0.1-in. rad.

Torq

ue

7000 lb • in.

3000 lb • in.

Time

JUVINALL: Machine DesignFig. P8-44 W-268d

Helicalspur gear

Pump

Fillet

25-mm solidround shaft

500 N

750 N2000 N

Bending Kf = 2.0Torsional Kf = 1.5Axial Kf = 1.8

50 mm

250-mm dia.

Forces act at 500-mm dia.

Fx = 0.2625FyFz = 0.3675Fy

C

BAx

Forces act at 375-mm dia. (2)

120 dia. Keyway

(Kf = 1.6 for bend and torsion; 1.0for axial load. Use CS = 1 with these values.)

80 dia.

B

E

C D

A

(1)

Fx = 1.37 kN

Fz = 5.33 kN

Fy = 1.37 kN

Fy

D

y

z

JUVINALL: Machine DesignFig. P8-45 W-268e

550

400

450400

JUVINALL: Machine DesignFig. P8-46 W-268f

Tors

ion

stre

ss (

ksi)

0

–10

–20

–30

10

20

30

30 seconds

JUVINALL: Machine DesignFig. 9-1 W-269a

++

+

+

+

++

+

++

+

Electrolyte

Fe2+ ions in solution

Iron electrode with surplus of electrons

~ ~ ~ ~

~~

~~

~~~

~~ ~

JUVINALL: Machine DesignFig. 9-2 W-269b

Electrolyte

Exposed iron (anode or cathode)

Plating, as tin or zinc (cathode or anode)~ ~~ ~~~ ~

~~~~ ~~ ~ ~ ~

~~ ~

~~

JUVINALL: Machine DesignFig. 9-3 W-269d

Electrolyte

~~

~

~~ ~

~~

~~~~

~~

~

A B

JUVINALL: Machine DesignFig. 9-4 W-269e

Magnesiumanode

Zinc strips between steel spring leaves

Outlet

Inlet

(a) Water tank

(c) Leaf spring

(d) Ship

(b) Underground pipe

Magnesiumanode

Insulated copper wire

Zinc anodes

JUVINALL: Machine DesignFig. 9-5 W-269f

Insulated copper wire

+–

Steel tank

Steel

Rust particles Rust particles

(a)

Rust begins at center of drop

(b)

"Crevice corrosion"

JUVINALL: Machine DesignFig. 9-6 W-269g

Steel Steel

Steel bolt and nut

Nonporous, pliableelectrical insulator


Aluminum plates

JUVINALL: Machine DesignFig. 9-8a W-271a

Salt water

Strongacids

Strongalkalis

Aeratedwater

U-Vradiation

A ExcellentB GoodC PoorD BadOrganic solvents

Ceramics,Glasses

KFRP

Polymers

PTFE, PP Epoxies, PS, PVC

HDPE, LDPE, PolyestersPhenolics

NylonsPMMA

CompositesGFRP

CFRP

A B C D D

Alloys

Alloys

CeramicsGlasses

Lead alloysSteels

Ti-alloys Cu-alloys

Al-alloys C-steels

Cast ironsNi-alloys

Ceramics,Glasses

KFRP

GFRPCFRP Polymers

Many elastomers

PTFE PVCPMMA

Nylons

LDPEEpoxies, HDPEpolyesters, PP,phenolics, PSFilled polymers

All

All

AlloysAllAll

All

Composites

KFRP

GFRPCFRP

Polymers

PSPVC

Mostelastomers

PhenomicsPolyesters

PULDPEHDPE

EpoxiesNylons

PP

PTFE

Composites

Polymers

PTFEEpoxies

LDPE/HDPEPP PS PVC

Nylons

PolyestersPhenolics

PMMAAlloys

Lead alloysNickel alloys

S-steelsCu-alloys

Al-alloys

Cast irons

Low alloysteels

C- steels

Ti-alloys CFRPKFRP

GFRPComposites Ceramics,

Glasses

All

Alloys

GoldLead

PTFEPVC

HDPE

Nylons

LDPE, EpoxiesElastomers

AlloysTi-alloys

Castirons

C-steel

Al-alloys

Ni-alloys

S-steels

PolyestersPhenolics

PSPMMA

PU Composites

CFRPCFRP

KFRPCeramics,Glasses

Glasses

Vitreousceramics

Mg 0 Zr02

Al203Si C

Si3N4Si02

Alloys

Al-alloys Cu-alloysZn-alloys

Ni-alloysSteels

S-steelsCast irons

Ti-alloys

C AB

Polymers Nylons

PMMA ElastomersPhenomics

Polyesters LDPE/HDPEP.V.,PS,PP,PTFEPVC,EpoxyComposites

GFRP

KFRP

CFRPCeramics,Glasses

Si02Glasses

Vitreous ceramics

Si C, Si3N4Al203

Zr02Graphites

Polymers


Mo Cr Co Ni Fe Nb Pt Zr Ti Cu Au Ag Al Zn Mg Cd Sn Pb In

W

Mo

Cr

Two liquid phases

Increasingcompatibility;hence,increasingwear rate

One liquid phase, solidsolubility below 0.1%

Solid solubility between1 and 0.1%

Solid solubility above 1%

Identical metals

Co

Ni

Fe

Nb

Pt

Zr

Ti

Cu

Au

Ag

Al

Zn

Mg

Cd

Sn

Pb

In


Wear coefficient, K10–2 10–3 10–4 10–5 10–6

Unlubed Poorlube

Goodlube Excellent lube

Unlubed Excellent lubePoorlube

Poorlube

Poorlube

Goodlube

Excellentlube

Goodlube

Unlubed

Unlubed Good lube Exc.lube

Unlubed Lubed

2-body 3-bodyHigh abr.concentr.

Low abr.concentr.

Unlubed LubedFretting

Abrasivewear

Adhesivewear

Nonmetal on metal or nonmetal

Incompatible metals

Partly compatible

Compatiblemetals

Identicalmetals


r = 16 mm

F = 20 N

n = 80 rpm

Copper pin80 Vickers

Pin Pin

DiskDisk

Initial profiles Final profiles

t = 2 ht = 0 h

Disk volume lost

Pin volume lost = 2.7 mm3

= 0.65 mm3Steel disk210 Brinell


Contact area

(a)

Two spheres

(b)

Two parallel cylinders

Contact area

x

z

y


a

p

p

p0

R1

R2

z

xL

b

y

y

a

p0

Dis

tanc

e be

low

sur

face

–p0 –0.8p0 –0.4p0 0.4p00Stress

4a

(a)

Two spheres(a is defined in Fig. 9.13a)

3a

2a

a

0


p0 = max. contactpressure

�z

�max

�x = �y

Dis

tanc

e be

low

sur

face

–p0 –0.8p0 –0.4p0 0.4p00Stress

7b

(b)

Two parallel cylinders(b is defined in Fig. 9.13b)

6b

5b

4b

3b

2b

b

0

�z

p0 = max. contactpressure

�y

�max

�x

–0.6p0 –0.4p0 –0.2p0 0 0.2p0

0

0.1p0

0.2p0

0.3p0


�y ≈ –0.1p0

�max ≈ 0.3p0

�x ≈ –0.25p0

�z ≈ –0.7p0

One plane of maximum shear stress

�z ≈ –0.7p0

+�

+�

�y ≈ –0.1p0

�x ≈ –0.25p0

Str

ess

� yz

Distance y from load plane–4b –3b –2b –b 0 b 2b 3b 4b

–0.3p0

–0.2p0

–0.1p0

0

0.1p0

0.2p0

0.3p0

A

A B

F

F

b

B

�yz �yz

�z �z

�y �y


p0 = max contact pressure

0.5b below surface

A B

A B

y


�yt = 2fp0

�yt = 2fp0

�yzt = fp0

�yzt = fp0

�yt = –2fp0

�yt = –2fp0

p0 = maximum contact pressuref = coefficient of friction

Loaded cylinder(resists rotation to

cause some sliding)

Driving cylinder

y y

z

z

Direction of rotation


b b


Hard-bronze bearingalloy spherical seat

10 mm

2000 N

Hardened-steel sphere

10.1 mm

Max

imum

con

tact

str

ess

p 0 (

MP

A)

Sphere radius, R1 (mm)5.00 5.01 5.02 5.03 5.04

50

100

150

200

250


Cast ironCopperSteel

Com

pute

d m

axim

um e

last

ic c

onta

ct s

tres

sp 0

or

�z

(ksi

)

Life N (cycles (log))105 106 107 108 109 1010

100

150

200

300

400

500

600

700800


Spur gears–high-qualitymanufacture, case-hardenedsteel, 60 Rockwell C (630 Bhn)

Roller bearings

Angular-contact ball bearings

Radial ball bearings

Parallel rollers

1

C

Protected ("noble", more cathodic)C

C

C

C

X

C

C

C

C

C

C

C

C

C

C

C

C

CC

JUVINALL: Machine DesignTab. 9-1 W-269c

2

10Key:

11

12

13

14

15

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Gold, platinum, gold-platinum alloysC

C

C

X

X

X

X

X

X

X

C

X

X

X

F

F

X

X

F

C

C

C

C

C

C

X

X

X

C

X

X

X

F

F

X

X

F

C

C

C

C

C

C

X

X

C

X

X

X

F

F

X

X

F

C

C

C

C

C

C

C

C

C

C

C

F

F

C

C

F

C

C

C

C

C

C

C

P

P

X

F

F

P

X

F

C

C

C

C

C

C

X

X

X

F

F

X

X

F

C

C

C

C

C

C

P

X

F

F

P

P

F

C

C

C

C

C

X

X

F

F

X

X

F

C

C

C

C

C

X

F

F

X

X

F

C

C

C

C

C

F

F

X

X

F

C

C

C

C

F

F

P

X

F

C

C

C

F

F

C

P

F

C

C

F

F

X

X

F

C

F

F

C

P

F

F

F

C

C

F

F

F

F

F

F

F

F

C

FF

Rhodium, graphite, palladium

Silver, high-silver alloys

Titanium

Nickel, manel, cobalt, high-nickel andhigh-cobalt alloys

Nickel-copper alloys per QQ-N-281,QQ-N-286, and MIL-N-20184

Steel, AISI 301, 302, 303, 304, 316,321, 347*, A286

Copper, bronze, brass, copper alloys per QQ-C-551,QQ-B-671, MIL-C 20159, silver solder per QQ-5-561

Commercial yellow brass and bronze;QQ-B-611 brass

Leaded brass, naval brass, leaded bronze

Steel, AISI, 431, 440; AM 355; PH steels

Chromium plate, tungsten, molybdenum

Steel, AISI 410, 416, 420

Tin, indium, tin-lead solder

Lead, lead-tin solder

Aluminum , 2024, 2014, 7075

Steel, (except corrosion-resistant types), iron

Aluminum, 1100, 3003, 5052, 6063,6061, 356

Cadmium and zinc plate, galvanized steel,beryllium, cald aluminum

Magnesium

Legend:X – Not compatibleC – CompatibleP – Compatible if not exposed within two miles of a body of salt waterF – Compatible only when finished with at least one coat of primer*Applicable forms: 301, 302, 321, and 347 sheetand plate; 304 and 321 tubing; 302, 303, 316, 321,and 347 bar and forgings; 302 and 347 casting;and 302 and 316 wire. These materials must be finished with at leastone coat of primer.

C

+

Corroded ("active", more cathodic)

–

JUVINALL: Machine DesignProb. 9-1 W-269

RivetsTotal exposed area = 100 cm2

Metal platesTotal exposed area = 1 m2

Chromium-plated steel cap screwsTotal exposed area = 110 cm2

301 Stainless steel platesTotal exposed area = 1.5 m2

Electrolytic environment


Drain plug(steel)

Insert rod(magnesium)

Oil

Crankcase(steel)

JUVINALL: Machine DesignProb. 9-8d W-287

100 N 100 N

Latchclosed

Latchopen

30 cycles/day, every day

Steel, 100 BhnSteel, 300 Bhn

JUVINALL: Machine DesignProb. 9-12 W-287a

30 mm

Locking plateArm

Geneva wheel


100-mmradius


�

Lp

End of thread

�

Lp

End of thread

End of thread

(a)Single thread–right hand

(b)Double thread–left hand

Pitch dia. dp

Crest

p

4

p

8

Root

Axis of thread

30�

p


Major dia. d

Root (or minor) dia. dr

60�

Nut tolerance zone

Basic profile(as shown inFig. 10.2)

Screw

Basic profile

Screw tolerance zone


Nut

p2

p2


2� = 29�

5�45�

0.663p

0.163p

p

p

2

p

2

d

dr

p

22� = 29�

(a) Acme

(c) Square (d) Modified square (e) Buttress

(b) Acme stub

p

dmdr

0.3p

dm dr d

pp

2

p

2

dm dr d

p p

dmdr

� = 5�� = 7�

dm

d

d

a


(c)(b)(a)

Force F

dc

Weight

A

A

L

q


�

�dm

fn

w

nn cos �n

Section A A(normal to thread)

Scale 4:1

�n

�n

q fn

dm


�n

�

h

h

b

A

A

B

B

Section B-B(normal to thread)

Section A-A(through screw axis)

tan �n =

Screw axis

bh

tan � = bh cos �

�

b/cos ��

Eff

icie

ncy,

e (%

)

Helix angle, �0� 10� 20� 30� 40� 50� 60� 70� 80� 90�

0

10

20

30

40

50

60

70

80

90

100


e =cos �n – f tan � cos �n + f cos �

, where

�n = tan–1 (tan 14 � cos �)12

f = 0.01

f = 0.02

f = 0.05

f = 0.10

f = 0.15

f = 0.20

a


Force F

f = 0.12

fc = 0.09

1-in. double-threadAcme screw

dc = 1.5 in.

Weight = 1000 lb

Forceflowlines

Nut

A - shear fracture line for nut thread stripping

B - shear fracture line for bolt thread stripping

B

A

JUVINALL: Machine DesignFig. 10 -11 W-299

dr

didp

d

1

Total = P

Total = P

Clampedmember

2

3

Bolt

t


Clamped member

BoltNut


Pilot surfaceof bolt

Materialbeing

compressed


Motor

Materialbeing

compressed

Spur gears

Ball thrustbearings

Ball thrustbearings

Thrustwashers

(b) Screws in tension (good)(a) Screws in compression (poor)

Thrustwashers

Motor

Flat washer

(a) Screw (b) Bolt and nut (c) Stud and nut (d) Threaded rod and nuts


0.65d

1.5d

d

(a) Hexagon head

(b) Square head

(d) Flat head (f ) Oval head

(h) Hex socket headless setscrew

(j) Round head with Phillips socket

(c) Round head

(e) Fillister head

(g) Hexagon socket head (i) Carriage bolt



(a)Conventional screwdriver

will tighten but notloosen the screw

(Plug in socket)

(b)Special tool required totighten or loosen screw

(5-sided head) ("Spanner head")

(c)Break-away heads

T4

T3


y

Mohr circles

Fi

FiFi

Fi

y x

Fi

Fi

� +��

+�

�max = per max � theorySy

2Stresses with torques applied

Stresses aftertorques are relieved

x'

x (�, –�)

y'

y

T2

T1

Bol

t te

nsio

n

Bolt elongation

Torqued tension,galvanized

(high friction)

Torqued tension,galvanized and

lubricated

Torqued tension,black oxide


Direct tension,black oxide

Direct tension,galvanized


(a)Helical (split) type

(b)Twisted-tooth type

(Teeth may be external,as in this illustration,

or internal.)


(b)(a)

(a)Insert nut (Nylon insert is compressed when

nut seats to provide both locking and sealing.)

(c)Single thread nut (Prongs pinch bolt threadwhen nut is tightened. This type of nut isquickly applied and used for light loads.)


(b)Spring nut (Top of nut pinches

bolt thread when nut is tightened.)

Spring- top nut(Upper part of nut is tapered.

Segments press against bolt threads.)

Starting Fully locked

Nylon-insert nuts(Collar or plug of nylon exerts friction

grip on bolt threads.)Distorted nut (Portion of nut is distortedto provide friction grip on bolt threads.)

(a) (b) (c)



Fe

Fe Fe

(a)Complete joint

(b)Free body without

external load

Fb = Fi Fc = Fi

(c)Free body withexternal load

Fb Fc

(a) (b) (c)

(d)

g


Fe Fb

Fc

Fb

Fb

Fc

Fc

Fc

F b a

nd F

c

Fc = Fi

F b =

F i and

F e

Fb

Softgasket

(Separating force per bolt)0 Fe

Fi

(a) (b) (c)

g


Fe

Fb

Fc

Fb

Fb

Fc

Fb

"O-ring"gasket

(d)

F b a

nd F

c

Fb = Fi

Fc = Fi – Fe

Fc = 0

(Separating force per bolt)0

"Rubber"portion of

bolt

Fe

Fi

Forc

e

External load Fe

0

Fi

Fc

Fe

Fb

Fc = 0

Fb = Fe Fb = Fi +


�Fb

�Fc

C A

Bkbkb + kc

Fekc

kb + kc

Fluctuationsin Fb and Fc

correspondingto fluctuationsin Fe between

0 and C

Fc = Fi –

Conical effectiveclamped volume

(Hexagonal bolthead and nut)

30°


d2

d3

d1

g

d

(a)Bolt bending caused by nonparallelism

of mating surfaces. (Bolt will bendwhen nut is tightened.)

Connectingrod and cap

P

A

a

(b)Bolt bending caused by deflection

of loaded members. (Note tendencyto pivot about A; hence, bending is

reduced if dimension a is increased.)

a

AP



Pillow blockRotating

shaft

Metric(ISO) screw

9 kN

F

F2

F2

F2

F2

(a) (b)

Normal load, carried byfriction forces

Overload, causing shear failure

F


150

150500

400

100

24 kN24 kN

150

D

A

D

E


48

144-kN appliedoverload


F

F

V

F

100

150

CG of bolt groupcross section

200180

4848

144 kN

144 kN (150 mm) =21.6 kN � m

200

180180

Su = 830

Str

ess,

� (M

Pa)

� = 0.0032 @ Sy = 660 on idealized curve

Strain, �0 0.01 0.02 0.03 0.04

Fluctuation in thread root stress – Case 2, Fi = 35.3 kN

Fluctuation in thread root stress – Case 1, Fi = 10 kN

12% elongation @ Su specified for class 8.8

0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.120

100

200

300

400

500

600

700

800

900

JUVINALL: Machine DesignFig. 10-34ab W-323

Sp = 660

Sp = 600

505

Forc

e (k

N)

Bolt tight,machine notyet turned on

38.3

35.3

Case 2 – Fi = 38.3 kN(bolt tightened to full

yield strength)

Bolt yields slightly, with noincrease in load or stress during

first application of Fe

29.3

4

0

10

20

30

40Fb

Machineoperating atnormal load

Machineturned off

Case 1 – Fi = 10 kN

Fc

Fb

Fe

Fc9

13

Time

(a) Fluctuation in Fb and Fc caused by fluctuations in Fe

(b) Idealized (not actual) stress–strain curve for class 8.8 bolt steel

Alt

erna

ting

str

ess,

�a

(MP

a)

Mean stress, �m (MPa)0

After initialtightening

After shut-downfollowing normal operation

(c) Mean stress-alternating stress diagram for plotting thread root stresses

130

77

200 400 600 8000

100

200

300

400


� Life

Su = 830Sy = 660

4

2

1 3

During normaloperation

Operation at overload on vergeof causing eventual fatigue failure

Sn = SnCLCS CG = (1)(1)(0.9) = 373 MPa8302

'

Steel bolt, '' – 13 UNC,

grade 5 with cut threads

Ext

erna

l loa

d F e

Time

(b)Fluctuating separating

force versus time

(a)Simplified model of machine

members bolted together

g = 2''

Fmax

JUVINALL: Machine DesignFig. 10-35a-c W-325, 326a

0 to Fmax

Steelmember

0 to Fmax,fluctuating

external force

Fe

12

�a

(ksi

)

�m (ksi)0

�a = 37

�a = 23

�a = 22.7

20 40 60 80 100 1200

20

40

60

Limiting point for case a

Limiting point for case b

Sn = S 'nCLCGCS = (1)(0.9)(1) = 54 ksi1202

Sy = 120

Sy = 92

�a = 37

�a = 23

�a = 22.7

Limiting point for case a

Limiting point for case b

Sn = SnCLCGCS = (1)(0.9)(1) = 54 ksi'120

2

(c)Fatigue diagram for thread root

Su = 120

Sy = 92

Fe

"O-ring" gasket

250 mm

Aluminum cover plate,E = 70 GPa

Cast-iron cylinder,E = 100 GPa

Class 8.8steel bolt

350 mm

g/2

g/2



a

f = 0.13

fc = 0.10

1-in. double-threadAcme screw

dc = 2.0 in.

Weight = 10,000 lb

Force F


5 in.

1/2 in. Acme threaddc = 5/8 in.

Fe = 0 to 8,000 lbkc = 6kb

Fe

Fe



(1)

1000 N

Springwasher

1000 N

(2)

1000 N1000 N

AA

Spring washer



280 mm

140 mm

230 mm


Forc

e (k

N)

Time

15 kN 15 kN

30 kN

0

20

40

60

80 Fb and Fc

Fe

Light load Light load

Heavy load

Initialtighten


22

22

22

22

22

22

22

22

22

22

22

22

22 222222



Before setting

After setting

Full tubular

Bifurcated (split)

Metal-piercing


(c)Compression

(a)Semitubular

(b)Self-piercing


"Built-up" lightweight structure

Acute corner

Back (blind) sidenot accessible

Back (blind) side not accessible

Blindside upset

Blindside upset

Drive pinblind rivet


Open-endbreak mandrelblind rivet

Blindside upset

Pull-throughblind rivet

Blindside upset

Closed-endbreak mandrelblind rivet

Blindside upset

Mandrel head collapsesas rivet is expanded

and pulled through rivet

Trim or grindmandrelPulling

head

Mandrel breaksafter seating andrivet expansion

Self-pluggingblind rivet

(b) (c)(a)


60�


(a) (c) (d)(b)

45�60�

FF

ht

t = 0.707

h

h

t

h

(a)

A

B

C

D

(b)

(d)Transverse loading

50 mm

(a' )Convex

weld bead

(a" )Concave weld

bead (poorpractice)

(c)Parallel loading

(e)Transverse loading

F

A

B

C

D

50 mm

F


1052 + 202

J

T

802 + 452

JT

690.0t

525.8t

691.9t

131.4t690.0

t80t

674.1t

80t

5600 N • m

20 kN

295.7t

y

G

B A


(a)

(b) (c)

100

–

x–

XX

C

B

Y

G

A

Y

150

300

20 kN

T (80)J

525.8t

105

8020

G

B

T=

T (105)J

=T (20)

J131.4

t=

T (45)J

295.7t

Torsional stresses Torsional plus direct shear stresses

=

B

C

A

B

A

X

X

120

16010 kN

60

70

(a) (b) Stresses on weld AB

121.2t

� =26.3t

� =

124tResultant stress =


D

L /2

L /2

G (CG of total weld group)

G' (CG of this weld segment)

b

a

y

Y'

Y' Y

Y

X'X'

X

t

X


�a

(ksi

)

�m (ksi)190 50 62

13.6


= = 0.5�a�m

3060


(a) Adhesive-bonded metal lap joint (b) Brazed tubing fittings (c) Glued wood joint

Removing thismaterial reduces

stress concentrationat bond edges

F


15 mm

Weld length = 90 mmSy = 400 MPaSF = 3E70 series weld

F

3.0 in.

F

F


E60 series welding rodsSy = 50 ksi (plates)h = 0.375 in.SF = 3

Note: There are two3 in. welds.

100 mm

Note: Each plate has two 75 mmwelds and one 100 mm weld.

60 kN


75mm

55mm

4 in.

3 in.


4000lb

Note: There are two 4 in. welds.

Fixedend

Spline

Spline

Generousradius

Bearing

Bearing

Torsion barportion

(a)Torsion bar with splined ends

(type used in auto suspensions, etc.)

(b)Rod with bent ends serving as torsion bar spring

(type used for auto hood and trunk counterbalancing, etc.)


d


D

�

F

End surfaceground flat (c)

Tension spring

(a)Compression spring

(ends squared and ground)

D

d

F

F

d

FD2

T =

FD2

T =

F

F

F

F

d

(d)Top portion of tensionspring shown as a freebody in equalibrium

D

F

F


(a)Straight torsion bar

(b)Curved torsion bar

Pla

ne m

Pla

ne n

0

0

� =

� �

� �

TcJ

TcJ

TcJ

TcJ

� =

T

b

dc

a

T

T

T

Preferred range,ends ground

Kw

and

Ks

Spring index, C = D/d 2 4 6 8 10 12 14

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

2

4

6

8

10

12

14

16

18


Kw

C a

nd K

sC

Preferred range, ends not ground

Ks

KwC

Kw

KsC

Ks = 1 + (shear correction only, use for static loading)

Kw = + (shear and curvature corrections, use for fatigue loading)

0.5C

0.615C

4C – 14C – 4



Min

imum

ult

imat

e te

nsile

str

engt

h (M

Pa)

Min

imum

ult

imat

e te

nsile

str

engt

h (k

si)

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1

Wire diameter (mm)

Wire diameter (in.)

1.00.10 10.0 100.02 3 4 5 6 7 8 9 1

0

50

100

150

200

250

300

350

400

450

500

0

1000

1500

2000

2500

3000

0.0400.0200.0080.004 0.080 0.200 0.400 0.800


ASTM A229

ASTM A313(302)

ASTM A228 music wire (cold-drawn steel)

ASTM A401 (Cr-Si steel)

ASTM A230(oil-tempered carbon steel)

ASTM A232 (Cr-Va steel)

ASTM A229(oil-tempered carbon steel)

ASTM A227(cold-drawn carbon steel)

ASTM A313(302 stainless steel)

ASTM B159(phosphor bronze)

Inconel alloy X-750 (spring temper)

ASTM A227


(a)Ls = (Nt + 1)d

(c)Ls = (Nt + 1)d

(b)Ls = Nt d

(d)Ls = Ntd

(b)(a)Contouredand plate

(d)(c)


Rat

io,

defl

ecti

on–f

ree

leng

th,

�/L

f

Ratio, free length–mean diameter, Lf /D2 3 4 5 6

Unstable

Stable

7 8 9 10 110.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6

0.65

0.7

0.75


A

A- end plates are constrained parallel (buckling pattern as in Fig. 5.27c)B- one end plate is free to tip (buckling pattern as in Fig. 5.27b )

B

12


Lf

Ls

Fs

D + d � 1.5 in.

(Springfree) (Spring

withmin. load)

(Springwith

max. load)

105 lb

60 lb

(Springsolid)

Cashallowance

� 2.5 in.

in.



Str

ess

S (%

Su)

(log

)

103 104 105 106

30

20

40

50

60

70

800.9Sus � 0.72 Su

0.54 Su

0.395 Su

Su2

CLCSCGSn =

Su2

= (0.58)(1)(1)

= 0.29 Su

� a/S

u

�m /Su

0.1 0.2

(0.38, 0.38)

(0.325,0.325)

(0.265, 0.265)(0.215, 0.215)

0.72

0.3 0.4 0.5 0.6 0.7 0.80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8


103�

104�

105�

106 + �

0.54

0.395

0.29

Region of interest

0

0

�m = 0= 1

�a�m

0� 1

�a�m

P

Static load

0= 0

�a�m

� max

/Su

�min /Su

0.20 0.40 0.60 0.800

0.20

0.40

0.60

0.80


103 �

104 �

105 �

106 + �

P

0.76

0.65

0.53

0.43

Tors

iona

l str

ess

S s,

max

(%

Su)

Life N, (cycles)104

Shot-peened wire

103 105 106 107

30

40

50

60

80

70


Design curves [1]Non-shot-peened wire

Calculated curve (from Fig. 12.14)

Note: For zero-to-max torsionalstress fluctuation

76

65

53

43

� max

(M

Pa)

�min (MPa)0 200 400

689

800

510

600 800

(965, 965)

(862, 862)

(Static load line)

(Load line, slope 600/300for Sample Problem 12.2)

1000

200

400

600

800

1000


Infinite life withoutshot peening

Infinite life withshot peening


Without presetting

Load stress

Load stressplus residualstress

With presetting

–�

0

+�

�resid from presetting


25 mm

600 N

300 N

600 N

Squared andground ends,ASTM A232spring wire

300 N

Key

Cam

n =650rpm Shaft

�� + 25

� max

(M

Pa)

�min (MPa)0 200 400 600 800 1000 1200

200

400

600

800

975

750

540

1000

1200


�max�min

600300

=

D

r2r4 B

dd

F

D

A

F

r1

r3


Bending stress at Sec. A:

� =16FD

�d3

r1r3

Torsional stress at Sec. B:

� =8FD

�d3

r4r2

F


(b)Semi-elliptic

6FL

bh2

L L

� =

12FL3

Ebh3� =

(c)Full-elliptic

2F

2F

6FL

bh2

L L

� =

6FL3

Ebh3� =

2F

6FL

bh2� =

6FL3

Ebh3� =

(a)Quarter-elliptic

(simple cantilever)

L F F F


F

L

x


(a)

t

h

b

w

F

L

(b)t is constant; w varies linearly with L

(c)w is constant; t varies parabolically with L

L

h

b

If � and t areconstant, then x /wmust be constant.

If � and w areconstant, then x /t2

must be constant.

McI

6Fx

wt2� = =

F

b

h

Half of nth leaf

Half of nth leaf

n leaves

Half of 3rd leaf

Half of 3rd leaf

Half of 2nd leaf

Half of 2nd leaf

Main leafb

L L

h

F


F

h

bn

(a) (b)

ClipsShackle (permitssmall fluctuationsin spring length)

Fixed pivot

Fixed pivot


2F

Bolt, Kf = 1.3

�a�m

�a

(MP

a)

�m (MPa)

�a = 525

0

Design overload point

= 0.67

400 800 1200 1600

400

800


� life, bendingSn

Sy Su

F = 1000 to 5000 N

(b)(a)

k = 30 N/mmh = 7 mm

F

6FL3

Ebh3

FL3

Eh3

L = 682

L = 682 L = 682

b = 416

b = 416 b = 333

L = 682


� = = 0.0144 ; F�

Eh3

L3k = = 69.33

6FL3

Ebh3

FL3

Eh3�1 = = 0.0180

F�1

Eh3

L3k1 = = 55.55

Eh3

L3k = 76.30

4FL3

Ebh3

FL3

Eh3�2 = = 0.0482

Eh3

L3k2 = 20.75

(a) Triangular-plate solution to Sample Problem 12.4

(b) Trapezoidal plate solution to Sample Problem 12.5

+ =

k = k1 + k2

8383

D

a

F

d

F


Thickness = h

b


FF

a

Dd

Dh

Fact

ors

for

inne

r su

rfac

e st

ress

con

cent

rati

onK

i, ro

und

and

Ki,

rect

Spring index, C = or

2 4 6 8 10 121.0

1.1

1.2

1.3

1.4

1.5

1.6


Ki, round

Ki, rect

Belleville Wave Slotted Finger Curve Internally slotted(as used in automotive clutches)


In series


In parallel In series-parallel


+ + + +

Storage drum Output drum Storage drum Output drum

(a) Constant-force extension springs

(b) Electric motor brush spring

(c) Two forms of constant spring motors


End attachedto door

Fixedend

Torsion bar

Center of gravityof 60-lb door

Door stop


24 in.

110�

Deflection

Support

Spring

Retainer

Nut

Force

Forc

e

Threaded bolt


Deflection

C

B

A


F = 3.0 kN

DiDo

Do = 45 mm do = 8 mm No = 5

do

di

Di = 25 mm di = 5 mm Ni = 10

F

F = 45 to 90 lbDeflection = 0.5 in.D = 2 in.


FSquared andground end

0

+

–

3600 engine rpm1800 camshaft rpm

"Reversal point"

Valve lift is 0.384 in.(maximum-on "nose" of cam)

"Reversal point"Valve lift is 0.201 in.

Valv

e ac

cele

rati

on


Cam angle

Key

Cam

Shaft

Stationary guide

Oscillating assembly

Roller follower

Spring

Adjusting nut

Cap screw


Support

e


F

Tire

Stationarysupport

Handle

Brakeshoe

Spring

Pin stops

Pivot A35 mm

Stationarysupport


25 mm shaft

Torsion springs

Cable

Cable

110 mm dia.

110 mm dia.



Main bearing

Connecting rod

Connecting rod bearing

Thrust bearing(flanged portion of main bearing)

Main bearing

Crankshaft

Main bearing cap

Connecting rod bearing cap


(a) Hydrodynamic(surface separated)

(b) Mixed film(intermittent local contact)

(c) Boundary (continuousand extensive local contact)


(a)At rest

W

Oil inlet

W

(b)Slow rotation

(boundary lubrication)

W

Bearing

Journal

Minimum filmthickness, h0

(c)Fast rotation

(hydrodynamic lubrication)

Oil flow

W

e

W

Resultant oilfilm force

W

�n/P

f

A


Boundary lubrication

Mixed-film lubrication

Hydrodynamic lubrication

(viscosity × rps ÷ load per unit of projected bearing area)


T

Cross-sectional area, A

where G = shear modulus

(b)

At equilibrium, torque Tproduces elastic displacement,�, across a solid element

F

h h

FhAG

�

� =

Surface velocity, U

where � = absolute viscosity

(c)

At equilibrium, torque Tproduces laminar flowvelocity, U, across a fluidelement

F

FhA�

U =

(a)

Rubber element

Fluid element

Abs

olut

e vi

scos

ity

(mP

a • s

)

Abs

olut

e vi

scos

ity

(�re

yn)

Temperature (°C)

Temperature (°F)

10 20 30 40 50 60 70 80 90 100 110 120 130 140

28026024022020018016014012010080

0.3

0.4

0.5

0.6

0.7

0.91.0

2

3

4

5

10

23

5

102

23

103

2

3

4

5

10

2

3

45

102

2

3

5

103

235

104


SAE 70

20

40

5060


Overflow rim

Oil

Level of liquidin bath

Saybolt viscometer

Kinematic viscosity,58 s at 100°C

Bottom of bath

Orifice

Bath


D

R

L

c

n


5000 N

D = 100 mm

R = 50 mmn = 600 rpm

Oil viscosity,50 mPa • s

c = 0.05 mm

L = 80 mm


Journal

Oil hole

W Partial bronzebearing

Oil level


Rotating journal

Fixed bearing

Lubricant

Lubricant flow

Coordinates:x = tangentialy = radialz = axial

W

dx

dy

(� + dy) dx dz

� dx dz

dx

dyp dy dz

∂�∂y

(p + dx) dy dzdpdx


y

h Lubricant flow

Stationary bearing

Rotating journal

U


1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0.01 0.02 0.04 0.06 0.08 0.1 0.2 0.4 0.6 0.8 1.0 42 6 8 10

Min

imum

film

thi

ckne

ss v

aria

ble,

h 0 c0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Ecc

entr

icit

y ra

tio,

e/c

Bearing characteristic number, S =2R

c�nP

Optimum zone

Min. friction

Max.load

L /D = ∞

1

12

14


1

2

3

45

10

20

30

4050

100

200

0 0.01 0.02 0.04 0.1 0.2 0.4 0.6 0.8 1.0 2 4 6 8 100.08

Coe

ffic

ient

of

fric

tion

var

iabl

e,f

R c

L /D =

∞


c�nP

14

12

1


1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.2

0.3

0.1

0 0.01 0.02 0.04 0.06 0.1 0.2 0.4 0.6 0.81.0 2 4 6 8 10

Min

imum

film

pre

ssur

e ra

tio,

Pp m

ax (

gage

)

L /D = ∞

1


c�nP

12

14


100

90

80

70

60

50

40

20

30

10

0 0.01 0.02 0.04

* Defined in Figure 13.20

0.06 0.1 0.2 0.4 0.6 0.8 1.0 2 4 6 8 10

Pos

itio

n of

min

imum

film

thi

ckne

ss,

� (

deg*

)

L /D = ∞

1


c�nP

12

14


100

90

80

70

60

50

40

30

20

10

0 0.01 0.02 0.04 0.06 0.08 0.1 0.2 0.4 0.6 0.8 1.0 42 6 8

Term

inat

ing

posi

tion

of

film

, �

p 0 (

deg*

)25

20

15

10

5

0

Pos

itio

n of

max

imum

film

pre

ssur

e, �

p max (

deg*

)

L /D = ∞ 1

1

∞

�p0

�pmax


c�nP

12

12

14

14

* Defined in Figure 13.20


6

5

4

3

2

1

0 0.01 0.02 0.04 0.1 0.2 0.4 0.6 0.81.0 2 4 6 8 10

Flow

var

iabl

e,Q

Rcn

LL /D =

L /D =

L /D = 1

L /D = ∞


c�nP

14

12


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.01 0.02 0.04 0.1 0.2 0.4 0.6 1.0 2 4 6 8 10

Flow

rat

io,

Qs

QL /D =

L /D = ∞

1


c�nP

14

12


W

e

n

Oil of viscosity �and flow rate Q

R = D/2

Average film pressure = P =

Film pressure, p

WDL

h0

�p0

�

pmax

�pmax


1000 lb

D = 2.0 in.

R = 1.0 in.n = 3000 rpm

SAE 20 OilTavg = 130°F

c = 0.0015 in.

L = 1.0 in.


Oil hole

Steel backingBearing material

Axial groove

Oil

film

pre

ssur

e


Groovedbearing

Oil inlet hole

Ungrooved bearing

2

Circumferentialgroove

L2L

D


"Rifle drilled" passagein connecting rod*

Circumferential groovein main bearing

PistonPiston pin or wrist pin

Drilled passage in crankshaft

Circumferential groove in rod bearing

Circumferential groove in main bearing

*If omitted, piston pin bearing is splash lubricated.

Oil in Oil in


W = 17 kN

D = 150 mm

R = 75 mmn = 1800 rpm

Force feed,SAE 10 oilTavg = 82°C

c = ? mm

L = ? mm

f = ?

Qs = ?

Power loss = ?

Oil temperature rise = ?

f, co

effi

cien

t of

fri

ctio

n

h 0,

min

imum

oil

film

thi

ckne

ss (

mm

)

Q a

nd Q

s, o

il fl

ow r

ate

(cm

3/s

)

c, radial clearance (mm)

*As defined in Fig. 13.13

0 0.05 0.10

Optimum band*

0.150

0.001

0

0.005

0.010

0.015

0.020

50

100

150

200

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.010


Max

. lo

ad Min

. fr

icti

on

h0

Qs

Q

f


aa > b

bPads

RoRi

Runner


D = 4.0 in.

R = 2.0 in.n = 900 rpm

SAE 10 OilTavg = 150°F

c = 0.002 in.

L = 6.0 in.


JUVINALL: Machine DesignFig. P13-29D W-425

4.5 kN

D = ? in.

R = ? in.n = 660 rpm

SAE ? OilTavg = ?°C

c = ? in.

L = ? L = ?

4


3

(b)

Steps in assembly

21

(a)Relative proportions of bearings

with same bore dimension

(b)Relative proportions of bearings

with same outside diameter


(LL00) (L00)

Lightseries

(200)

Mediumseries

(300)

Mediumseries

(300)

Extra-lightseries

(L00)

Extra-lightseries

Extra-extra-lightseries

(LL00)

Extra-extra-lightseries

Lightseries

(200)

Loadinggrooves

(a) Filling notch (loading groove) type


(c) Double row


(d) Internal self-aligning

JUVINALL: Machine DesignFig. 14-3d W-428

(e) External self-aligning

JUVINALL: Machine DesignFig. 14-3e W-428


Oneshield

Twoshields

One seal Two seals Shieldand seal

Snap ring

Snap ringshield and

seal

Snap ringand twoshields

Snap ringand oneshield

Snap ringand two

seals

Snap ringand one

seal


Addedstabilizing ring

(b) One-direction locating (c) Two-direction locating


Bearing axis

Commonapex

Crowned roller body

Spherical roller head

Roller axis

(d) Idler sheave(unground bearing)

(e) Rod end bearing


JUVINALL: Machine DesignFig. 14-10h W-437

(h) Integral spindle, shown with V-belt pulley


dS dH

r

r

+

+

Per

cent

age

of f

ailu

res

Life1 2 3 4 5 6 7 8 9 10 11 12

2

4

6

8

10

12

14

16

18

20

22

24


Median

Life

adj

ustm

ent

relia

bilit

y fa

ctor

Kr

Reliability r (%)90 91 92 93 94 95 96 97 98 99 100

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0


1800 rpm

Radial bearing


1800 rpm

Angular bearing

Ft = 1.5 kN, Fr = 1.2 kNLight-to-moderate impactEight hours/day operation

1800 rpm

Case a: 90 percent reliability


Case b: 30,000-hour life

No. 211 radial-contact bearing,C = 12.0 kN

Fr = 1.2 kNFt = 1.5 kN

Light-to-moderate impact,Ka = 1.5

21 mm

100 mm

55 mm

1000 rpm

Ka = 1.0 (uniform load)

Fr

No. 207 radial-contact bearing


Rad

ial l

oad

(kN

)

01

1 2 3 1

Time

32

4567

20%100%

50% 30%

Note spacers



F1

L1


F2

2L1

F2

3L1

3500 rpm


No. 204 radial ball bearingFr = 1000 N, Ft = 250 N90% reliabilityLight-moderate shock loadingL = ? hr life

Gear Shaft Bearing



150-mm dia.

Gear120-mm dia.

1.2 kN

20°

B

A

300 mm

100mm

60mm

60mm

Clamping supportsChain sprocket

Rotating shaftChain

JUVINALL: Machine DesignProb. 14-21d W-450

1.75 ft 0.75 ft

600 lb 600 lb

2.25 ft

Right-angle gearing

Parallel gearing




Line of centers

Point of contact

Pitch point

Common normal (to tooth surfacesat point of contact)

Driving gear

P

Driven gear


Involute curves

Base circle

Base pitch, pb


�p

�g

Gearpitchcircle

Pinionpitchcircle

P (pitch point)c

dp

dp


Pinionbasecircle

� (pressure angle)

�p

�g

Gearbasecircle

b

P

a


Base circlePitch circle

Pitch circle

Pinion

a

Pc

fe

dg

b

�

�

Base circle

Gear

rp

rg


rg

Addendum circle

Dedendum

Angle ofapproach

p

Angle ofrecess

Position ofteeth leaving

contact

0g

0p

Pitch circle

Addendum

Addendum

Dedendum circle

Base circlePitch circle

Addendum circle

Angle ofapproach

Dedendum

Angle ofrecess

Pinion (driving)

Positionof teethenteringcontact

Base circleDedendum circle

Gear (driven)

c

b

a

n

n

�

rp


Workingdepth

Wholedepth

Addendum

Dedendum

Clearance

Top

land

Face

Flan

kBo

ttom

land

Circular pitch p

Pitch circle

Filletradius

Dedendumcircle Clearance circle

(mating teeth extendto this circle)

Addendum circle

Tooththickness t

Face

width

b

t0

Width ofspace

18

3632

30288064

4048

16

14

2022

2426

65

412

11

10

78

9


Pinion

Rack

Circular pitch

p


�


Pitch circle

Base circle

�Base circle

Pitch circle Dedendumcircle

Addendumcircle

��


Gear blank rotatesin this direction

Rack cutter reciprocates in a directionperpendicular to this page

Driving gear

Driven gear

Base circle

Base circle

02

01

�2

�1

�

�


Interference is on flankof driver during approach

(This portion of profileis not an involute)

(This portion of profileis not an involute)

Pitch point, P

�

Addendum circlesab

P = 6 teeth/in.� = 20°�p�g


= –3.0

rg

rp

c = 4 in.

Pitch circle(gear)


dg

dp

P

P

Fr

Fr

Pitch circle(pinion)

(Driving pinion rotates clockwise)

F

F

�

�

Ft

Ft

�g

�p

N = 12 teeth (input pinion)P = 3

600 rpm; 25 hp

(a)

(b)


a

b cN = 36 teeth

(idler)N = 28 teeth(output gear)

Resultant force (applied by shaft to gear) = (1313 + 478) 2 = 2533 lb.

45°

Vab = 478 lb

Hab = 1313 lb

Vcb = 1313 lb

20°

20°

b

Hcb = 478 lb

c

a

rf

Fr

Ft

x

F

a

F

Constant-strenghparabola


t

h

b

Lew

is f

orm

fac

tor

Y

Number of teeth N (Rack)12 15 17 20 24 30 35 40 4550 60 80 125 275 ∞

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60


� = 14 °1

2

� = 25°

� = 20°� = 20°, st

ub teeth

Load

Driving and driven gears0

F2

Time

Time

Fa = Fm =

Driving gear Idler gear Driven gear

0

(b) Stress fluctuation

Sn

�m

�a

Su


1 revolution

F

Load

0

Idler gear

40% higher fat. strengthfor o-max loading(driver and driven)

Fat. strengthfor reversed

loading (idler)

Fa = FFm = 0

1 revolution

F

F

�a �m

�a

Geo

met

ry f

acto

r J

Number of teeth N

(b) 25° full-depth teeth

12 15 17 20 24 30 40 50 80 275 ∞0.15

0.20

0.25

Load applied attip of tooth(no sharing)

Load applied athighest point ofsingle-toothcontact(sharing)

851000

502517

0.30

0.35

0.40

0.45

0.50

0.55

0.60


Geo

met

ry f

acto

r J

Number of teeth N

(b) 20° full-depth teeth

12 15 17 20 24 30 40 50 80 275 ∞0.15

0.20

0.25

Load applied attip of tooth(no sharing)

Load applied athighest point ofsingle-toothcontact(sharing)

8550

2535

17

0.30

0.35

0.40

0.45

0.50

0.55

0.60

Number of te

eth in matin

g gear

Number of te

eth in matin

g gear

1000

12545 6035

12545 6035

Velo

city

fac

tor

Kv

Pitch line velocity V (ft/min)* Limited to about 350 Bhn

E D

C

B

A

0 1000 2000 3000 4000 5000 6000 70001

2

3

4

5

0 10 20

Pitch line velocity V (m/s)

30


Hobs, shaping cutters*

High precision, shaved and ground

Highest precision, shaved and ground

Hobs,

form

cutte

rs*

Precision, shaved and ground

Pinion

Gear

Steel, 290 Bhn

(manufacture of pinion and gear correspondsto curve D, Fig. 15.24)

P = 10

20° full-depth teethSteel, 330 Bhn

Np = 18 (teeth)


Electricmotor1720 rpm

Conveyordrive(involvesmoderateshocktorsionalloading)860 rpm

Vgn = Vpn

Vgn = Vpn

Vp = VgVg

VpVgt Vpt = Vgt

Commonnormal

Commonnormal

Gear(driver)

Gear(driver)

Commontangent

(a) General contact position sliding velocity as shown

(b) Teeth in contact at pitch pointno sliding

Pinion(driven)

Pinion(driven)

Commontangent

Slidingvelocity

�


Vpt

Surface fatigue life (cycles)104

CLi

105 106 107 108 109 1010 1011

0.6

0.8

1.0

1.2

1.41.61.82.0



Win = 100 hp

3600 rpm

•

900 rpm

Negligible shock loadingLife: 5 years, 2000 hours/year Full power: 10 percent of time Half power: 90 percent of timeFailure in 5 years: 10 percent likely


Motor(input)

Driven machine(output)

p1

g1

g2

p2

a c

b

(a)With three planets (typical)


P

P

Ring

PlanetArmSun

R

A S

P

(b)With one planet (for analysis only)

P

R

A S

(R = input; A = output; S = fixed member)


R/2

2Ti

2Ti /3R

2Ti /3R

4Ti /3R 4Ti /3R

R + S4

Ti

R P

A

3R

R + S4

2Ti

3R

2Ti

3R4Ti

3RTo

Ti

�i

�o

SR

SR

4Ti

R

4Ti

3R

To =

= = 1 +

= Ti 1 +

(R = input; A = output; S = fixed member)


P

R

V

S

�0

�0

�i

�i

A

R2 R + S

4

V2

VR/2V/2

=

SR�0

�i = 1 +

(R + S)/4


P

P

R

S

90° = 17 teeth (ring)12

Planet willfit here

Planet willfit here 90° = 5 teeth (sun)

Planet willNOT fit here

Planet willNOT fit here


Driven machinecoupled to this

shaft

45 teeth

P = 5, � = 25°15 teeth

45 teeth

25mm

100mm

25mm

1 kW, 1200-rpmmotor coupledto this shaft

c

B

b

A

a

JUVINALL: Machine DesignProb. P15-23 W-482A

A

B

36T

64T

24T, P = 6

18T, P = 9

2''

To drivenmachine

8''

2''

Coupled to 20 lb.in.torque motor


a

a

Motor

Driven machine

32T

24T To drivenmachine

1''

B

A


90°2.0''

16T

1700-rpmmotor

100-lb.in.torque

(b)(a)


3

3

44

2

7

6

78

9

4

1

Low (L)

Neutral (N)

High (H)

3

2

Memberpushes here todisengage pawl

Hub can"overrun"in thisdirection

5

JUVINALL: Machine DesignProb. P15-47 W-485A

P2 P1

P2

S2

P1

S1

Input

Output

Arm

JUVINALL: Machine DesignProb. P15-50 W-485B

P1 P2

Input

S1 (100 teeth)

P2 (102 teeth)

P1 (101 teeth)

S2 (99 teeth)

Output

Arm

Reverse brake bandholds S2 fixed

Low brake bandholds S3 fixed

S3 (21 teeth)

P3 (33 teeth) P1 (27 teeth)

P2 (24 teeth)

S1 (27 teeth)S2 (30 teeth)

OutputInput



(b) Rotated spur gear laminationsapproach a helical gear as laminationsapproach zero thickness.

ppn

pn

pa

�n

p

�� Section NN

(normal plane)

Section RR(in plane of rotation)


Rb

R

b

N

N

c

d

a�

e

Section NN(normal plane)

Section RR(plane of rotation)

d

2 cos2 �


Re =

R

d

N

N

R�

�n

�F

Top of tooth

Pitchcylinder

Spur gear(helical gear with � = 0�)

d


Fa

FrFt

Fr

Ft

Ft

Isometric viewshowing helical

gear forces

�

Top of tooth

Pitchcylinder

Helical gear

Section RR(plane of rotation)

Section NN(normal plane)

Fr

Ft

Ft

�

�

Fa

Fb

Fr

F

Fb

R

N

N

R

Np = 18 (teeth)Pn = 14�n = 20�

(a)

Electric motor hp 1800 rpm

(normalplane)

Input shaft ofdriven machine

600 rpm

� = 30�(right hand)


Fr

FaFt

(b)Isometric view of

motor shaft and pinion

Direction ofrotation

� = 30�(left hand)

12

Geo

met

ry f

acto

r J

Helix angle �0� 5� 10� 15� 20� 25� 30� 35�

0.40

0.30

0.50

0.60

0.70


J-fa

ctor

mul

tipl

ier

Num

ber

of t

eeth

Teet

h in

mat

ing

gear

Helix angle �0� 5� 10� 15� 20� 25� 30� 35�

0.95

0.90

1.00

5001507550

30

12141618203060150500

20

1.05

Developed backcone radius, rbg

rbp�p�p

b

Pitch conelength, L

Dedendum

Dedendum

Addendum

Pinion back cone Pitch cone

angles

Gearpitch cone

dp


Gear pitchdia., dg

Gearbackcone

Pinionpitchcone

Spiralangle

Circular pitch

Face advance

�

Meanradius


b

Ft

Ft

Fa

FrFn

Fn

F

Fr

Note: Fn is normalto the pitch cone.

�

d2

b2

dav

2


�


Geo

met

ry f

acto

r J

Number of teeth in gear for which geometry factor is desired0 10 20 30 40 50 60 70 80 90 100

0.16

0.18

0.20

0.22

0.24

0.26

0.28

0.30

0.32

0.34

0.36

0.38

15

20

40

30

50

80

Teeth in mating gear

60

100

0.40


250 mm

Load

Motor

Rotation

140 mm

A

B

C

D140 mm

65mm


Geo

met

ry f

acto

r J

Number of teeth in gear for which geometry factor is desired0 20 40

100

80

60

50

40

12

30

2520 15

60

Teeth in mating gear

80 1000.16

0.20

0.24

0.28

0.32

0.36

Geo

met

ry f

acto

r I

Number of teeth in pinion NP

0 10 20 30 40 500.05

0.06

0.07

0.08

0.09

0.10

0.11


Ng = 100

Ng = 70

Ng = 50

15 Teeth in gear

90

80

60

4030

2520

Geo

met

ry f

acto

r I

Number of teeth in pinion NP

0 10 20 30 40 500.06

0.08

0.10

0.12

0.14

0.16

0.18


Ng = 100

30 Teeth in gear

1520 25

5040

60

80

�S


�R

A

P

P Fixed arm, A

RP (planet)

R (ring)

S (sun)

S

P


Arm(input member)

Right axleLeft axle

P

P

RS


Worm leadangle, �, and

gear helixangle, �

Centerdistance

c

Keyway

Axial pitch, P

Lead, L

Note: � and � aremeasured on pitchsurfaces.

Wormoutside dia.,

dw, out

Pitchdia., dw

Pitchdia., dg

Facewidth, b

Fwt

Fwr

Fwa

Worm-drivingtorque


Fgt

Fgr

Fga


fFn cos �

Fn sin �n

Fn cos �n

Fn cos �n cos �

Fn

Fn cos �n sin �

�n

Direction ofFga and Fwt

Direction of

Fgt and Fwa

Direction ofFgr and Fwr

(a) Worm driving (as in Fig. 16.20) (b) Gear driving (same direction of rotation)

�Fn

fFn sin ��

fFn cos �

Direction ofFga and Fwt

Direction ofFgr and Fwr

Direction of

Fgt and Fwa

fFn sin �

fFn

Fn cos �n sin �

�n

Fn cos �n

Fn sin �n

Fn cos �n

Fn

Fn cos �n cos �

�


Vs

Vw

Vg Gearrotation

�

Worm

Gear

Worm rotation

Coe

ffic

ient

of

fric

tion

f

Sliding velocity Vs (ft /min)1 2 4 6 10 2 4 6 100 2 4 6 1000 2 4 6 10,0000

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0


Worm: Steel, hardened and ground

Nw = 2, RH, p = in., �n =14 �

c = 5 in.


Gear

60 rpm

Worm

Gear: Bronze

Motor2 hp., 1200 rpm

58

12

Worm rpm, nw

Coe

ffic

ient

C

0 400 800 1200 16000

10

20

30

40

50

60

70

80


ft �

lb

min

� ft

2 �

�F

With fan (as in Fig. 16.26)

Without fan

c ≈ 6 in.


Gear

Worm: hardened steelrpm =1200

Gear: Chill-cast bronze

Speed ratio, 11:1

Worm


p

b


�

Pa

40 teeth 60 teeth

20 teethOutput


24 teeth Input


Motor

Output

50 teeth

20 teeth� = 0.50 rad

left hand

25 teeth� = 0.35 radright hand

A

B 125

100

20050 teeth

Gear

Pinion

1000 rpm

400 rpm

Bevel gears:35 hp

Np = 36Ng = ?

b = 2 in.� = 20�

P = 6


Pinion

1500 rpm

Gear

A B

3 in. 3 in.

Bevel gears: 50 hp

Np = 30Ng = 60b = 3 in.� = 20�

P = 6

Flexiblecoupling tomachine

JUVINALL: Machine DesignFig. P16-23D W-520

Nw = 2Ng = 55

P = ?


c = 8 in. Gear

Worm

Gear material: Chill-cast bronzeWorm material: Hardened steelNw = 3 p = 0.5 in. f = 0.029

Ng = 45 �n = 20�

c = 4.5 in.

b =1.0 in.

1200 rpm



(a) Square key

d

w

ww

2

(b) Flat key

w ≈ d/4 w ≈ d/4; h ≈ 3w/4

d

h

hw

2

(c) Round key

Key usually has drivefit; is often tapered

(d) Kennedy keys

Keys are tapered anddriven tightly; forheavy-duty service

(e) Woodruff key

Widely used in automotive andmachine tool industries

( f ) Gib-head key

Usually tapered, giving tightfit when driven into place;gib head facilitates removal

(g) Feather key

Key is screwed to shaft; hub is free toslide axially – easier sliding is obtained

with two keys spaced 180° apart


d

D

(a) Straight round pin (b) Tapered round pin(c) Split tubular spring pin Grooves are produced by rolling,

and provide spring action toretain pin

(d) Grooved pin


Basic Inverted

E-ring

External rings (fit on shaft)

(a) Conventional type, fitting in grooves

(b) Push-on type – no grooves required

Teeth deflect when installed to "bite in" and resist removal(less positive than conventional type)

Basic

Internal rings (fit in housing)

Inverted

I

I Section II

External ring (fit on shaft)

I

I Section II

Internal ring (fit in housing)

4-spline 6-spline 10-spline

(a) Straight-sided (b) Involute

16-spline



�stGravitationalforce, w

Mass, m

Shaft of spring rate k = w/�st

(a) Single mass


m1

w1 w2w3

m2

�1 �2 �3

m3

w1 w2

w5

m1 m2m3

m4m5

w3w4

(b) Multiple masses


�st

(c) Shaft mass only

50 300

Tracksprockets

Track

5060

T1

A BC

T2

Chain

30°

Chainsprocket100 diameter

Track sprocket250 diameter

Track

Fc (= 2500 N)

FT (= 1000 N)

(a) General arrangement


d

JUVINALL: Machine DesignFig. 17-6b-d W-530

55 Small gap orspring washer

(b) Shaft layout

d

A

T1 T2

BC

S

A T1 S T2

2490Vertical forces

C

B325 2165

50245

Torque

62,500

55 50 60

95,800

125,000

(c) Loading diagrams

VV

MV

T1 S T2

Horizontal forces

B

CA

687.5 1250

937.5

VH

MH

500(Ft /2)

500(Ft /2)

80,30090,600

75,000

�ea

(M

Pa)

�em (MPa)

�ea�em

(d) Fatigue diagram

0 100 200

"Design overload" point

300 400 450

530

500

100= 2.9

200

165

Sn = S' CLCGCS = (0.5)(550)(1)(0.9)(0.78) = 186n

113,700 130,000


d/4

d

(c) Shear failure of a tightly fitted key

d/4

d/8

d

(b) Key tightly fitted at top and bottom(a) Loosely fitted key


Sled-runner keyway Profiled keyway

Fatigue stress concentration factor, Kf*Steel

Annealed(less than 200 Bhn)

Quenched and drawn(over 200 Bhn)

1.3

Bending

1.6

* Base nominal stress on total shaft section.

1.3

Torsion

1.6

1.6

Bending

2.0

1.3

Torsion

1.6


(b) Constant-stress, constant strain shear coupling


(a) Basic shear-type coupling

Bonded rubber element

(c) Tube form shear coupling


(a) Basic Oldham type (b) Modified type



20 in.

Flexible couplingMotor 0.25-in.-dia. shaft


600 mm600 mm

50 kg

25-mm dia.


40 in. 30 in.20 in.

120 lb80 lb

2-in.-dia. shaft

JUVINALL: Machine DesignFig. P17-13a-f W-541,542,543

Driver56 kw

Bearing (2)

ShaftDriven28 kw

Driven28 kw

(b) Gear input shaft

(a) Connecting shaft

(c) Hydroelectric generator shaft

(d) Idler gear shaft

Bearings Shaft

Electricgenerator

rotor

Hydraulicturbine

DriverShaft

Bearing (2)

Driven

Idler

(e) Gear countershaft ( f ) Stationary countershaft

Shaft

Needle bearing (2)

Shaft

DriverDriven

Driver Driven

Coupling (2)Motor

Bearing (2)

Shaft

Driven machine


5 in.

2 in.

Bearing ABearing B

Fr = 450 lb

Fa = 400 lb

Ft = 800 lb

3 in. rad


125 mm

50 mm

B

A

Fr = 2.4 kN

Ft = 4.0 kN

Fa = 1.5 kN

Note: Gear forces actat a 75-mm radius

from shaft axis.

Chain sprocket


(a)

Stationary shaft

Clamping supports

Chain sprocket

(b)

Rotating shaft

Clamping supports

Drivingshaft

Drivenshaft

Provision foraxial movement

Ring elementsubjected to clamping

pressure, p

Friction liningmaterial


dr

rri

ro

Releaselever

To release

To transmission

Housing

Cover

Spring

Flywheel

Frictionplanes

Enginecrankshaft

Pressure plate

Clutch plate(driven disk)

Releasebearing


Oil passage

Input

Oil passage

Piston

Oil chamber(pressurized toengage clutch)

Seals

Bushing

Key

Key

Output

Disks b – driven disks (3 disks, 6 friction surfaces)

Disks a – driving disks (4 disks, 6 friction surfaces)


ri ro

dr

F

Cone

� Cone angle

Spline (sliding fit)

Shifting groove

Spring

Key

Cup

(a) (b)

drsin �

Local pressure = p


r�

ri

ro

Lever

(a)Brake assembly

a A

F

Shoe(block)

O

Drum



(b)Shoe and lever as a free body

(c)Drum as a free body

(d)Lever proportions for a

self-locking brake

F

OOv

Oh

A

r

bAv

fN

fN

N

N

Inertial and/orload torque, T

Ah

c

b

a


6

23

1

8080

All dimensions in millimeters

1

4

6

(rotation)

23

4

400

300

250rad.

45°

300 Shoe width, 80 mmf = 0.20pmax = 0.40 N/mm2

400

120120

(a)

(b)

(c)

( f )

(d) (e)

5

5H25 = 4F

H52 = 4F

V26 = 1.41F

H54 = 4FH34 = 4FH43 = 4F

H62 = 7.07FH26 = 7.07F

V16 = 0.70F

H16 =3.46F

T = 880F

H36 = 10.53FH63 = 10.53F

V63 = 0.2H63

V36 = 2.11F

H13 = 6.53F

V13 = 2.11F

V12 = 2.41F

H12 = 3.07FO12

O16

O13

V62 = 0.2H62

V52 = F

H45 = 4F

V25 = FO25

F

45°

90° 90°

F

A'

�n

�

�


O3

O3 O2

A

B

�'

�


O3

d�

O2

A

dN

f dN

d cos (180° – �)= –d cos �

d sin (1

80° – �)

= d sin �

Drum rotationNote: d = O2O3 b = width of shoe

B

r �1

�2

c

F

�

180° – �


300 rpm

C = 500

200

Springcompressedto force F

Force torelease brake

Cast-iron drumMolded composite

shoe lining ofwidth b = 50

All dimensions in millimeters

(a) Complete brake

(b)Drum and right shoe

F

150

150 150

300

200

150

O2

O3

�1

�15045°

45°�2

d


O2

N

Pivot P

F

fN

�2

r f

r

�2


P

f dN2 + f dN2

r

rf

f dN1

dN2

dN1

dN2

��

f dN2

f dN2

'

'

'

Drum

Rotation

Adjustingcam

Brake lining

Anchor pins

Hydraulic wheel cylinder

Return spring

O2

O3


C

r

d �1

�2

Anchor pin


Forwardrotation

Adjusting camand guide

Brake shoe

Brake liningAnchor pin


Brake drum

Adjusting camand guide


Returnspring

P1

P1

P2

F

P2

a

Band of width = b

Rotation

Cutting plane forfree-body diagrams


�

�

c

r

d�/2d�/2

P + dP P

Rotation

dNd�


P1

s

P2

Band of width = b

Rotation


�

a

F

c

P1

P2

Band width, b = 80 mm

Rotation


� = 270° Friction coefficient, f = 0.20

Maximum lining pressure,pmax = 0.5 MPa

a = 150 mm

F

c = 700 mms = 35 mm

r = 250 mm

Pads of diameter = 60 mm

125 mm

320 mmpmax = 500 kPaf = 0.30


4m/s

1000 kg



Electricmotor

Gearreducer

Friction clutch

T = 6 N • m, 600 rpm

Rotary inertial load,I = 0.7 N • m • s2


Rotation

F = 1500 N

200 mm340 mm

400 mm 500 mm


Rotation

A

F

240 mm

250 mm 320 mm

400 mm

150 mm


500 mm

Spring

300 mm400 mm

350 mm

Woven lining

Cast- irondrum


2

5

3

4

61

18 in.

25 in.

150

5 in.

5 in.

23 in.

18 in.

30 in.

Rotation5


Rotation

F = 300 N

500 mm

�


55

258°

240

F

75

72

370


Wt.

sa

c

270°


�

Motorrotation

(a) Manual adjustment

(b) Pivoted, overhung motor

(c) Weighted idler pulley

Adjustment

Tightside

Pivot

Pivot

Overhang

Idler

Weight

Tightside


0.50 in.

0.31 in.

0.66 in.

0.41 in. 0.53 in.

0.75 in.0.91 in.

A

0.38 in.

0.32 in.3 V

0.62 in.

0.54 in.0.88 in.

5 V

8 V

DE

C

B

(a) Standard sizes A, B, C, D, and E

(b) High-capacity sizes 3V, 5V, and 8V

0.88 in.

1.25 in.1.50 in.

1.0 in.


2�≈ 36°

dNdN2

dN/2sin �

(a) (b)


Tension-carrying cords

Fabric cover

Rubber


p

�r Pitchcircle

(a)

rc

p

(b)

A

B

AB

rc r

Chordal rise, r – rc


Pitch p

(a)

JUVINALL: Machine DesignFig. 19-10 W-584B

A

A

r

r4(D/2)

r3

D

Ti

�i �o

To

Inputshaft

Oil particle

Impeller

Gasket

Case

Turbine

Fluidcirculation

Oil seal

Output shaft

Core or innershroud

r2

r1

�i

Blades

Section AA

Per

cent

rat

ed t

orqu

e

Input speed �i (rpm)0 200 400 600 800 1,000 1,200 1,400 1,600 1,800

0

40

80

120

160

200

240

280

320

360

JUVINALL: Machine DesignFig. 19-11 W-584C

12 8

7

5

20

100

16 14

10

6

4

3

2

Percent slip

Maximumcouplingtorque

Electric motortorque curve

Turbine

Fluidcirculation

Inputshaft Output

shaft

Impeller

Reactor


Turbine

Inputshaft Output

shaft

Impeller

One-wayclutchReactor


Torq

ue r

atio

Speed ratio0 0.2 0.4 0.6 0.8 1.0

0.8

1.2

1.6

2.0

2.4

2.8

3.2

Eff

icie

ncy

(%)

0

20

40

60

80

100


Converter efficiency

Couplingefficiency

Convertertorque ratio

Coupling torque ratio


�

�

�

�

�2

c

�1r2

r1


A

B

C

D


V-belt� = 18°f = 0.20

Belt maximum tension = 1300 NBelt unit weight = 1.75 N/m

r = 100 mmPulley radius� = 170°

n = 4000 rpm


Electric motorn = 1780 rpm

55% rated power

Fluid coupling—performance curves—

Fig. 19.11

Drivenmachine

JUVINALL: Machine DesignFig. U19-1 W-580

Multiple V-belt, � = 18°, size 5VUnit weight = 0.012 lb/in.Power input = 25 hpPmax = P1 = 150 lbf = 0.20

3.7 in. dia.

Drivingpulley

Number of belts = ?

Drivenpulley

n = 1750 rpm

165° angle of wrap

1.44 –2.99

B1 engaged

–4.31Reverse

B3 engaged

1.00 1.00

C1 engaged

1.004

C2 engaged

1.44 1.00

B1 engaged

39%

61%

39%

61%

1.443

C2 engaged

1.00

(b) Power flow block diagram

(c) Gear shift pattern(a) Internal power flow diagram

2.53

C1 engaged

2.532

B2 engaged

1.44 2.53

B1 engaged

3.661

–Neutral

Torqueratio

Tout/Tin

Frontplanetary

train

Fluidcoupling

Rearplanetary

trainGear

No clutches or brakes engaged

B2 engaged

Gear Ratio

Neutral –

1 3.66

2 2.53

3 1.44

4 1

Reverse –4.31

C1 B1 C2 B2 B3

S1, P1, R1 S2, P2, R2 S3, P3, R3

B3

B2B1

C2C1Neutral

S1, P1, R1 S2, P2, R2 S3, P3, R3

B2B1

C2 B3C11

S1, P1, R1 S2, P2, R2 S3, P3, R3

B3

B2B1

C2C12

S1, P1, R1 S2, P2, R2 S3, P3, R3

B3

B2B1

C2C13

S1, P1, R1 S2, P2, R2 S3, P3, R3

B3

B2B1

C2C14

S1, P1, R1 S2, P2, R2 S3, P3, R3

B3

B2B1

C2C1Reverse

Fluidcoupling

Gearinterface

Bearing



Input torqueTi

54-tooth ring

R1

×Ti3

TiP

81TiP

81

TiP

81

TiP

81

TiP

81

TiP

81

TiP

81

TiP

40.5

P1

S1

TiP

40.5

TiP

40.5

TiP

40.5A1

TiP

81

P

2715-tooth planet

12-tooth sun

P

27

P12

P19.5

P12

P

7.5

Torque from B1:

TB1 =

TB1 = 0.44Ti

(3)

TiP

40.5 P19.5

Output torque:

To =

To = 1.44Ti

(3)

Arm

To


TiP

67.5

TiP

67.5

TiP

67.5

Ti3

P

22.5

TiP

Ti

67.5

TiP

67.5

S2

6P

TiP

33.75

TiP

67.5

R2

34.5P

TiP

33.75

TiP

33.75

TiP

33.75A2

P28.5

TiP

33.75 P28.5

Output torque:

To = (3)

To = 2.53Ti

Arm

22.5

45-tooth sun

69-tooth ring

•

P

TiP

67.5 P34.5

TB2 = (3)

TB2 = 1.53Ti

P2

12-tooth planet


TiP

67.5

34.5TiP

675

69TiP

675

TiP

67.5

34.5TiP

67.5

34.5TiP

675

TiP

67.5

TiP

67.5

TiP

67.5TiP

67.5

TiP

Ti

67.5

S2

TiP

33.75

34.5P

10P

34.5

10•

TiP

33.7569TiP

675

A2, A3

P28.5

P18

TiP

33.7569TiP

675P28.5

Output torque:

To = – 3

To = –2.99Ti

P18

Arms

22.5

45-tooth sun

P

P2

12-tooth planet

P3

16-tooth planet

R2, S3

69-tooth ringand 20-tooth sun


34.5TiP

675

34.5TiP

675

34.5TiP

675

34.5TiP

675

P26

Torque from reverse lock, B3

TB3 = (3)

TB3 = 3.99Ti

26 R3

52-tooth ring

P


To

TiP

40.5

TiP

81

S1

24-tooth sun

TiP

81

TiP

81 19.5P

TiP

40.5

TiP

40.5

Output torque:To = 1.00Ti

TiP

81 P12

Torque from C1:

TC1 = 3

TC1 = 0.44Ti

A1

TC1

TC1

Arm

12P

0.0117Ti P

Tf

Tf P

67.5

TC2P

Clutchtorque

TC2

103.5

P28.5

Arm

A222.5

45-tooth sun

P 34.5P

P2

12-toothplanet

69-tooth ring

R2S2

To

P28.5

Output torque:

To = 0.117Ti P (3)

To = 1.00Ti

Tf P

67.5

TC2P

103.5

For equilibrium of P2:

∴ Planet pin force = 0.0117Ti P

=

Tf + TC2 = Ti

Tf = 0.39Ti

TC2 = 0.61Ti


JUVINALL: Machine DesignFig. B-1 W-601

h

h

b

d

b

Rectangle

Triangle

Circle

d

Hollow circle

y

y

di

Rod

Disk

Rectangular prism

Cylinder

y

d

d

b

x

xz

z

y

y

t

JUVINALL: Machine DesignFig. B-2 W-602

L

c

L

a

z

y

d

xz

Hollow cylinder

L

y

do

di

xz

Tempering temperature (°F)

Reduction of area

400

%

ksi

600 800 1000 12000

20

40

60

60

80

100

120

140

160

180

200

220

240

JUVINALL: Machine DesignFig. C-5a W-603

4130

4130 1095

1050

1095

1050

1095

1030, 1040

1095, 4130

1040

1040

1030

1030

1050

1030, 1040, 4130

1050

Elongation

Ultimate strength

Yield strength


Reduction of area

400

%

ksi

600 800 1000 12000

20

40

60

60

80

100

120

140

160

180

200

JUVINALL: Machine DesignFig. C-5b W-604

Elongation

Ultimate strength

Yield strength

1095

1050

1040

1050

1040

1040

1040

1050

1095

10951050

1095


Reduction of area

400

%

ksi

600 800 1000 12000

20

40

60

100

140

120

160

200

180

220

260

240

280

300

JUVINALL: Machine DesignFig. C-5c W-605

Elongation

Ultimate strength

Yield strength

4340

9255

4140

43409255

4340

4140

9255

9255

4140

4140, 4340

Diameter of test specimen (in.)

Diameter of test specimen (mm)

0 1 2 3 450

60

70

80

90

ksi MPa

100

110

120

130

140

150

160

170

180

190 1300

1200

4340

4340

4140

3140

4140

3140

1040

1040

1100

1000

900

800

700

600

500

400

0 200 400 600 800

Su

Sy

1000

JUVINALL: Machine DesignFig. C-6 W-606

–PL

P

P

�

�max

V = P

M = –PL

+

0

0

–

V

M

JUVINALL: Machine DesignFig. D-1(1) W-607

Lx


V = P

M = –Pa

+

0

0

–

V

M

–Pa

P

P

b

�

�max

xa

�max

M = –wL2/2

+

0

0

–

V

M


wLV = wL

wL2

2

w

x

L


0

0

–

V

M

–Mb

–Mb

Mb

P�

�max

xL

+

+

0

0

–V

M


�

Lx

P2

LP

2

P

P/2

–P/2

PL/4

2

+

+

0

0

–

V

M


�

Lx

b

PaL

PbL

P

Pb/L

Pab/L –Pa/L

a

+

+

0

0

– –

V

M


Lx

w

wL

wL /2

wL2

2

2wL2

wL2

+

0

0

–

–

V

M


L

x

a

PL

P

PPba

a

–Pb/a

–Pb

b�

�max

z

+

+

0

0

– –

V

M


a

L

x

M0

M0

Lb

M0

L

M0

L

M0a

L

M0b

L

0

–

–

0

V

M


a a

L

x

M0

M0

x'

b

�max

aM0

M0

aM0

�

–


�

+

+

0

0

–

–

V

M

PL8

Px

L

–

–

–

PL8

PL8

– PL8

PL8

P2

P2

P2

P2

L2


+

+

0

0

–

– –

V

M

Pb2

L3

P

xa b

L

–

–

Pab2

L2Pa2b

L2

(3a + b)

Pb2

L3

2Pa2b2

L3

Pab2

L2

(3a + b)

Pb2

L3 (a + 3b)

Pb2

L3

– Pa2b

L2

(a + 3b)


+

+

0

0

–

–

V

M

xw

�

L

–

–

wL2

12

– wL2

12– wL2

12

wL2

24

wL2

12wL2

wL2

wL2

wL2

JUVINALL: Machine DesignFig. D-4 W-613

107

95

.00

12

2.1

7

14

4.8

1 R

OO

T1

75

.84

PIT

CH

20

6.8

8 O

.D.

25

82

.21

N •

m

10

1.6

85

.00

82

.63

152

276

400

530

646

680

716

1060

A shaft with integral worm, dimensions in millimeters.

RB

RA

8.68 kN

10

98

7654

32

1

1

h hole

s shaft

Classof fit

Bargraph(basichole

system)

2 3 4 5 6 7 8

Loose fit Free fit Medium fit Snug fit Wringing fit Tight fitMediumforce fit

Heavyforce andshrink fit

.0216 (.0025) .0112 (.0013) .0069 (.0008) .0052 (.0006) .0052 (.0006) .0052 (.0006) .0052 (.0006) .0052 (.0006)

.0216 (.0025) .0112 (.0013) .0069 (.0008) .0035 (.0004) .0035 (.0004) .0052 (.0006) .0052 (.0006) .0052 (.0006)

.0073 (.0025)

Note: Numbers in the table are for use with all dimensions in millimeters, except for those in parentheses, which are for use with inches.

.0041 (.0014) .0026 (.0009) 0 (0)

Cb

Cs

Ca

Ci 0 (0) .00025 (.00025) .0005 (.0005) .0010 (.0010)

JUVINALL: Machine DesignFig. E-1 W-614

aa

hh h h s h h h

s ss

sa

ss

JUVINALL: Machine Designp754 b1 W-???

JUVINALL: Machine Designp754b2 W-604

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