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JUVINALL: Machine Design Fig. 1-1a W-1

JUVINALL: Machine Design Fig. 1-2 W-2


JUVINALL: Machine Design Fig. 1-3b W-3

JUVINALL: Machine Design Fig. 1-3c W-3

F

R


0

(a)

1 Force (N)

Torque (N • mm)

10

0

0.1

0

Cam rotation (rad)

1 Follower displacement (mm)

(b)

(c)


0

n = 1000 rpm

T = 10 N • mm


10

Actual torque requirement

Crank torque (kN • m)

8

6

4

2 ␲ 3 0 0

␲ Crank angle (rad)


2␲

10

Crank torque (kN • m)

Actual torque requirement 8 Average torque 6

4 2 0 0

␲ 3

Uniform torque supplying equal energy

␲ Crank angle (rad)

2␲


Hub

Rim

0.8d d

Arm 0.2d


Vehicle road load power (hp)

160

120

80

40

0 0

20

40

60 80 Vehicle speed (mph)


100

120

Engine output power (hp)

160

120

80

40

0 0

1000

2000 3000 Engine speed (rpm)


4000

5000

1.3 Specific fuel consumption of engine (lb/hp • h)

1000 rpm 1.2 1.1

1500 rpm

1.0 2000 rpm 0.9

2500 rpm 3000 rpm 3500 rpm

0.8

4000 rpm

0.7 0.6 0.5 0.4 20

40

60

80 100 120 Engine output power (hp)


140

160

180


5 mm 1N

30 mm

g = 9.81 m /s 2 m

JUVINALL: Machine Design Fig. P1.18 W-15

a = 5 ft/s2

F

g = 32.2 ft/s2

JUVINALL: Machine Design Fig. P1-23 W-16

Pulley R = 3 in.

V = 5 ft /s m = 5 lb


I = 10 A V = 110 V

Output shaft

+ –

n = 1000 rpm T = 9.5 N • m

t = 2 hr


Crankshaft torque

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

0

0.5 1.0 1.5 Crankshaft rotation, revolutions

JUVINALL: Machine Design Fig. P1-42 W-18A

2.0

55 mph

2.64 axle ratio


3000 lb

50 in.

V = 60 mph CP

CG

20 in.

25 in.

Ft

Wr

100 in.


Fd

Wf

W = 3000 lb Engine power, 96 hp Fi = 467.5 lb

Fd = 100 lb

Ft = 567.5 lb Wr = 1618.5 lb


Wf = 1381.5 lb

RT A

X

RT – T T

RT – T

T A

RT RT RT

X


RT A

T = 3000 lb.in. (input)

RT – T = 5333 lb.in.

Transmission in low gear R = 2.778

RT = 8333 lb.in. (output) 3 in.

5333 lb.in. (a)

I 3000 lb.in.

2 in. 5 in. 2 in.

1087 lb

I

1087 lb II

A 1087 lb

D

(b)

IV

II

2667 lb 2864 lb

2667 lb 8333 lb.in.

2864 lb 4444 lb

III (d)

B

4444 lb 1087 lb

2864 lb

C

(c) IV


III

F

F

F

A a F Fa A (a)

(b)


a b F Section AA plane Fixed support

(a) F Fa (bending moment)

Fb (torque) F (shear force)

(b)


F32 = 40 lb

1

2

1 in. 2

1 H12 V12 1 in.

30°

F42


F32 = 40 lb 0

2 1 F32 = 40 lb F12

F42 F12 F42

Force polygon for link 2


0 Fb (bone 0.5 in. compression force)

3 in.

Ft (tendon force)

10 lb 10 lb pinch

Ft = 60 lb

Fb = 55 lb Force polygon for finger


V

+

V

Positive shear force M

+

M

Positive bending moment

w

F R1 Fb L

V

R1

R2 a

b L

Fa L

R2

wb 2 2L

a

b L

+wb2/2L

Fb + L

wb (a + b/2 ) L

b2 2L

V Fa – L

+

2 wb2 a+ b 2L 4L

Fab L

M

M (a)

(b)

Single concentrated load

Distributed load


–

wb (a + b/2 ) L

Critical section

2667 lb

2864 lb 2667 lb

III

C

B 4444 lb 1087 lb 2864 lb

III

Loads IV

C

B

4444 lb 2 in.

IV

1087 lb 5 in.

2 in.

+ 1580 lb Shear

V – 1087 lb –2864 lb 2174 in..lb

Moment

M

–5728 in..lb 5000 lb.in.

Torque

T

JUVINALL: Machine Design Fig. 2-11 W-29A

2864 lb 4444 lb T = 5000 lb.in.

IV M = 5728 lb.in.* C V = 1580 lb


* Actually slightly less, depending upon the width of gear C

b F

2b

d

d

F

b

F

m

a


F

Fork

Pin

Blade 4'

5 2

6

F

F

4' 5

1

2

1

4' 2 2

2 3

3

4

F

4

F

4' 3

3

1

1 2

2

7


w lb/ft


Spring in tension k1 = 10 lb/in. 10␦ 100 lb 100 lb

40␦ Spring in compression k2 = 40 lb/in.

(a)


(b)


Outer row Middle row

1

pi tc h

Inner row

Bottom strap

Left plate

Top Right strap plate

Top strap

Bottom strap

(a)

Outer row

Inner row Outer row

Inner row

1

pi tc h

Area

(c) (b)

bp Plate

s Plate

Strap

Bearing with plate Shear

bs

Bearing with strap Strap

(d)

(e)

1 strap

2 straps

3 Straps

d le p

r pa

6 7

th

th

a th

r pa

4

In n e

M id

O u te

5

Bearing with straps 8 Shear 9 Bearing with plate

Plate 1

2 (f)


Wall channel, C

g

2 in. r= 1.25 in. A Density = ␳

r= 1.25 in. B Density = ␳


1500 N

1500 N B 45°

C

45° 1000 mm

A

45°

D

45°


A

1000 N B

125 N


10 in.

Direction of rotation Motor 1 hp 1800 rpm Gear box


Blower 6000 rpm

50 mm 50 mm

A

B Forward air velocity Clockwise rotation


4:1 ratio gear reducer

Motor 1.5 kW 1800 rpm

Direction of rotation Connecting tube C Pump 450 rpm C' A A'

B'

B


These units are attached to a fixed support.

Engine

Engine is attached to aircraft structure here. Reduction gear, ratio = 1.5 Propeller


Vertical drive shaft

Z Rotation X

Mounting flange Y Z

Foward direction of boat travel

500 mm

X

Mx

Propeller rotation

Mz

My

Y 2:1 ratio bevel gears are inside this housing

150 mm


Suggested notation for moments applied to mounting flange

800 N 160 40 R

330 R

330 R 100 R

400

600


100 mm D A

C

100 mm B

1800 rpm

Bevel gear reducer

600 rpm

Attaches to motor

Attaches to load


4 in.

8 in.

Output shaft

Rear bearings

Front bearings

Mountings

6 in.

Housing and gear-shaft assembly details

100 lb.ft

Reducer assembly

Output

6 in. dia. gear

Motor input torque 100 lb.ft 2 in. dia. pinion Front bearings


Right-front wheel axle shaft–400 rpm

Engine 2400 rpm 100 lb.ft torque

24 in.

Y

A X

D

Left-front wheel axle shaft–400 rpm

12 in. C

B

Transmission 2.0 ratio

Y Front drive shaft–1200 rpm Rear drive shaft–1200 rpm

Rear axle (not part of free body) JUVINALL: Machine Design Fig. P2-14 W-51

X

Motor g Mass of mixer system = 50 kg A

B

Direction of rotation

Radial flow Mixing paddle 200 mm


Mounting width = 75 mm to 150 mm Radial air flow

g

Fan

B Direction of rotation A

Mass of blower system = 15 kg

Motor


Y 45 mm

D

30 mm

20 mm

FC

20°

Z B

Gear 1 50-mm dia.

A Shaft

FA C 20°

X Gear 2 24-mm dia.


100 N A

100 N B

200

300

A

B 200


300

140

50 N

12 in.

48 in.

Cable 27 in. 100 lb

Cable

12-in. pulley radius 100 lb JUVINALL: Machine Design Fig. P2-19 W-54

Fr = 600 N

Ft = 2000 N Fa = 1000 N

50 A

B

40

Motor attaches to this end of shaft

60


Fr = 200 N

Ft = 1000 N Fa = 100 N

100 A

B

50

Pump shaft is coupled to this end of shaft

150


Fr = 400 N

Ft = 1000 N

200 N

Fa = 200 N 40

30

A

B

20

80


20

Fr = 500 N Ft = 1500 N Fa = 150 N 150 A

B

200

300

120

140

100 N

250 N


A L F

t/2

b

R

Key

t

r

Section on A-A A


F

t

d D


Total gas force = F

a

2a

a d


f

t

t B

A

d D

R

L

P

L

L P

d D F

A


A

B E

C

D


k

k

a

a

k

k 100 N


P

t' t

F t'

Rivet diameter = 10 mm


F

C Su = 66

Stress ␴ (ksi)

60

A 40

B

F (Fracture) Sy = 39 Se = 36

20 Slope = modulus of elasticity, E

0

0.2% offset Strain ⑀ (arbitary nonlinear scale)


Su = 66 ksi

Stress ␴ (ksi)

60

C

D

Sy = 39 ksi

40 B A

F

Se = 36 ksi Hot-rolled 1020 steel

20

0 0

0

G 10 20

H 30 40

50

60

70 80 90 100 110 120 130 140 150 160 Strain ⑀ (%)

1.1 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 Area ratio R

0.1

0.2

0.3

0.4 0.5 Area reduction Ar


0.6

Transition region

Elastic region

Plastic strain-strengthening region

200

True stress ␴T (ksi (log))

⑀Tf = 0.92 115

100 80 60 40 30 20 ␴T

10 3

=

T E⑀

(E

=

30

×

) si 3 k 0 1

4 5 6 7 8 0.1

15 m (␴ 0 = 1

␴T = Sy = 39 ksi

2

3

ks

0.2 i, m =

F

2)

⑀T ␴T = ␴0

4 5 6 7 8 1.0 2 3 4 5 6 7 8 10 True strain ⑀T (% (log))


2

3

4 5 6 7 8 100

True stress ␴T (log)

"Ideal" material Curve I ( ␴T = Se ≈ Sy ) Se2 Se1 Se3

Plastic line ( ␴T = ␴0⑀Tm ) Curve II

Elastic line ( ␴T = E⑀T ) True strain ⑀T (log)


Su

Stress ␴

F Sy Se

0

Se E

⑀f Strain ⑀


1000

900

KB, ratio Su /HB

800 d D

2 0. 3 0. 4 0. 5 0. 6 0.

700

600

500 d = indentation dia. D = ball dia. 400 Steel 300

0

0.1

0.2 0.3 0.4 0.5 m, strain hardening exponent


0.6

0.7

200 250 300

Diamond pyramid hardness (Vickers)

350 400 450 500 550 600 650 700

680 750 700 720

800

740 850 760 900 950

72

(0)

80

(10)

20 30 40 Rockwell C hardness

90 100 Rockwell B hardness

(110)


50

60

70

60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300

Ultimate tensile strength (ksi)

150

100 120 140 160 180 200 220 240 260 280 300 320 340 360 380 400 420 440 460 480 500 520 540 560 580 600 620 640 660

Brinell hardness

100

Shore hardness 30 40 50 60 70 80 90 22 24 26 28 32 34 38 42 46 55 65 75 85 95

Rockwell C hardness

Distance from quenched end (b)

(a)


1000 Diamond WC

Cermets W

Engineering alloys

Beryllium

Mg0 Steels Ni alloys

Cast irons

100

Cu alloys

Min. energy storage per unit volume Yield before buckling

Al alloys Common Cement rocks Mg alloys +

Sn

Lead

Ash Oak Pine

10 Mel

II to grain Epoxies

CFRP uniply Ti alloys GFRP

Brick etc

CFRP Laminates GFRP

Concrete

Youngs Modulus, E (GPa)

Zn alloys

Si C

Boron

Mo alloys

Glasses

Si3N3 Al203

Zr 02 Silicon Ge

Be0

Engineering ceramics

Engineering composites

Porous ceramics PMMA

PS

S = 0.1 E

Balsa

S = 10–4 E

PVC Nylons Wood products Polyester PP Ash Oak

Woods

1.0

Pine ⊥ to grain

0.1

PTFE

Balsa

S = 10–3 E

S = 10–2 E

Design guide lines

HDPE

Engineering polymers

LDPE

S =C E S 3/2 =C E

Polymers foams Hard butyl

Pu

Silicone

1

Buckling before yield

Elastomers S2 =C E

Cork

0.01 0.1

Max energy storage per unit volume

Soft butyl 10

100 Strength S (MPa)


1000

10,000

10,000 Engineering ceramics

Diamond Si3N4 Sialons Al203

Si C

Zr02

B Mg0

Glasses

CFRP GFRP Uniply KFRP CFRP Be

1000 Engineering composites

Si

Ge

Steels Pottery

Al alloys Mg alloys

Strength S (MPa)

Parallel to grain

PS

Balsa Ash Oak Pine Fir Perpendicular to grain LDPE

Woods 10

Mel PVC Epoxies Polyesters HDPE PTFE

Soft butyl

Ni alloys Cu alloys

Lead alloys

Cement concrete

Porous ceramics

PU

Engineering polymers

Silicone

Balsa

Cast irons

Engineering alloys

PP

Wood products

Mo alloys

Stone, Zn rock alloys

Nylons PMMA

Ash Oak Pine Fir

W alloys

Ti alloys

GFRP Laminates KFRP

100

Cermets

Guide lines for minimum weight design

Elastomers

Polymers foams 1

Engineering alloys

Cork

S =C ␳ S 1/2 =C ␳

S 2/3 =C ␳ 0.1 0.1

0.3

3

1 Density ␳ (Mg/m3)

JUVINALL: Machine Design Fig. 3-12 W-74a

10

30

10,000 Engineering ceramics

Porous ceramics Engineering alloys Engineering composites

SiC

1000

CFRP Uniply KFRP

Zr02

Glasses

Strength at temperature S (T) (MPa)

Si3N4

Tialloys

Mg alloys

Nylons PC

Polymides

PMMA

PP PVC

Woods

Compression

Brick etc.

GFRP

⊥ to grain

Mullites Ni alloys

Steels

CFRP Laminates

Zn alloys

⊥ to grain

Al203

GFRP

Al alloys

100

Mg0

Epoxies

PF

HDPE Polyesters

II to grain

PTFE LDPE Engineering polymers

10

Butyls

Ice

Silicones

Elastomers 1 T-Independent Yield strength

Polymer foams

Upper limit on strength at temperature

Range typical of alloy series

0.1

0

100

200

300 400 Temperature T (C)

JUVINALL: Machine Design Fig. 3-13 W-74b

600

800

1000

1400

Hardness (Rockwell C)

70 60 50 40 30 20 10 0

10

20 30 40 Distance from quenched end (mm)


50

P

E

␴ P 2

␴

E

P 2 (a) Isometric view of tensile link loaded through a pin at one end and a nut at the other.

P

+

␴

␴

E

(b)

(c)

Enlarged view of element E

Direct view of element E

␴=

D

A=

P A

(d) Equilibrium of left half showing uniform stress distribution at cutting plane

+

Nut (e) View showing "lines of force" through the link


␲D2 4

P

3 3

1

4

5

1 2

5 6

(a)

P


(b)

2 3

3

P P 1


P

P


T E E (b) Enlarged view of element

T (a)

Positive shear

Isometric view E

Negative shear

E

(c)

Positive shear (d)

Direct view of element E

Shear sign convention


Negative shear

T Torque axis Line "A" 1

2 2

3

p To b

Sid

T a

Maximum shear stress exists along this line. (a) Zero shear stress exists along all edges.


F ro

e

nt

(b) Enlarged view of element 2

Neutral (bending) surface M

M (a) Entire beam in equilibrium ␴max c

y

Neutral surface

c M

Transverse cutting plane (b) Partial beam in equilibrium

Neutral bending axis and centroidal axis


(c) Typical cross sections

Neutral (bending) surface M

M (a) Entire beam in equilibrium Neutral bending axis and centroidal axis

␴max c

CG

CG

CG

y

M (b)

Neutral surface

Partial beam in equilibrium

(c) Typical cross sections


CG

Hyperbolic stress distribution with increased stress at inner surface

Centroidal surface

CG co

M Neutral surface

(a) Initially straight beam segment

e ci

M

(b) Typical cross section

Neutral surface displaced distance "e " toward inner surface

Center of initial curvature (c) Initially curved beam segment JUVINALL: Machine Design Fig. 4-9 W-84

d␾ b

c

Centroidal surface c' Neutral surface CG

co

y

c

ci

ro

a

d'

c

d

␳

␾ M

e

y

e

ri

rn

r

r M

Center of initial curvature


rn

Centroidal axis Neutral axis

b 8

r

b 4

c

3.5 A

U or T

B

b

A

B c 3.0 Round or elliptical Values of K in Eq. 4.11: ␴ = K

Mc I

B

A B

A c

2.5 b 2

Trapezoidal

2.0 I or hollow rectangular

b 3 B

b 6 B

1.5

B

A

b

c A b

A r

Values of Ki for inside fiber as at A

1.0

Values of Ko for outside fiber as at B

I or hollow rectangular

0.5

U or T Round, elliptical or trapezoidal 0 1

2

3

4

5

6 Ratio r / c


7

8

9

10

b

M

M

Centroidal axis

b

dA = b d␳

h

h

M


M

c=h 2

r=h CG

␳

(a)

(b)

Unloaded "curved beam"

Loaded "curved beam"


dA

dA

b y

␶

c

y0

Neutral axis

x

dA (M + dM)y/I

My/I

dx

V V

␶

y M M + dM N.A. M

Enlarged view of beam segment


(a)

(b)

Marked and unloaded

Loaded as a beam


␶av = V/A ␶max =

␶av = V/A

4 V/A 3

N.A.

␶max = N.A.


3 V/A 2

V

V

M

M


80,000 N

80

60 X

X 40

40,000 N

60

100

40,000 N 100

+40,000 N V –40,000 N

M


dx

dx

dx

dA = 60dy 10–

dA = 60dy

dA = 60dy

10+

40 b = 60 ␶

␶ b = 20

dA = 20dy

␶

b = 20 (a)

(b)


(c)

␶ = 32.61 MPa

0

␶ = 22.83 MPa ␶ = 7.61 MPa


+␶ ␶max yx

S

(a)

0

Marked eraser

x

y

+␴

S' yx (d) Mohr's circle

(b) Loaded eraser

y y x

x

␴x

x

y Direct view

Oblique view

S

S' S S S'

S'

Direct view

Oblique view

(c)

(e)

Enlarged view of element

Element subjected to ␶max


+␶ y y x T

T

#2

+␴

0

#1 x

(a)

(c) Mohr's circle

Marked eraser (for twisting)

␴1

␴2 y x

#2 #1

x y (b)

(d)

Enlarged element


2 in. 3 in. rad.

1 in.

2000 lb


2 in. A Top of shaft B "B" is at bottom of shaft, opposite "A"


V = 2000 lb

A B

2000 lb

M = 4000 in.lb

4000 in.lb

T = 6000 lb in. Load diag.

2000 lb

2000 lb

V

Shear diag. 2000 lb

M

Moment diag. 4000 lb


xy

yx y 2 in. x x

(a) Isometric view

yx y

x A

A

x

y

xy

yx

x

A

x xy

A

xy x

x

y yx

y

x xy

yx

(b)

(c)

(d)

Enlarged isometric view

Direct view

Isometric view

Calculated values: = 40.8 ksi = 30.6 ksi


+␶ y (0, +30.6)

␶max = 37 ksi 34° ␶yx = 30.6 ksi ␶xy

y ␴1 = 57 ksi ␴2 = –17 ksi

0

+␴

x ␶xy

56°

A

x

y ␶yx

x (40.8, –30.6)

Direct view of element A


␴x = 40.8 ksi

28° ␴1 = 57 ksi y x

A

x

y ␴2 = –17 ksi


␴ = 20 ksi 17°

␶ = +37 ksi y x

A

x

␶ = –37 ksi

y


␴ = 20 ksi

+␶ ␴x + ␴y

␴x – ␴y 2

2 (␴x, ␶xy) 2 –␴

1

2␾

+␴

0

(␴1, 0)

(␴2, 0) (␴y, ␶yx)

␶xy2 +

␴x – ␴y

–␶ JUVINALL: Machine Design Fig. 4-30 W-106

2

2

+␶ ␴1 + ␴2

␴1 – ␴2

2

2

cos 2␾

␴1 – ␴2

2

2␾

0

2 1

␴2 ␴1


sin 2␾ +␴

y

(3) (1)

(2)

(3)

(3)

x A

A

z

(1)

A

(1)

(2)

(2) (= z) (a) Original element

(b) Principal element

(c) 1-2 plane


(d) 1-3 plane

(e) 2-3 plane

+␶ ␶max = 37 Principal circle

3 (–17, 0)

2

1

(0, 0)

(57, 0)


+␴

+␶

Correct value of ␶max Erroneous value of ␶max obtained if ␴3 is neglected

␴1 (tangential)

A

0 ␴ 3 A ␴2 (axial)

␴3 = 0 (radial)


␴2

␴1

+␴

3.0 r 2.8 M

M

d

D

2.6 ␴nom = Mc = 32M I ␲d 3

2.4 2.2 Kt

(a)

2.0 1.8 1.6 1.4 1.2 1.0

0

0.1

0.2

0.3

D/d = 6 3 1.5 1.1 1.03 1.01

r/d 2.6

r P

P d

D

2.4 2.2

␴nom = P = 4P A ␲d 2

2.0 Kt

(b)

1.8 1.6

D/d = 2 1.5 1.2 1.05 1.01

1.4 1.2 1.0

0

0.1

0.2

0.3

r/d 2.6

r T

2.4 2.2

d

D

T

␶nom = Tc = 16T J ␲d 3

2.0 Kt

(c)

1.8 1.6 1.4 D/d = 2 1.2 1.09

1.2 1.0

0

0.1

0.2 r/d


0.3

3.0

r

2.8 M

M

d

D

2.6 ␴nom = Mc = 32M I ␲d 3

2.4 2.2

(a)

Kt 2.0 1.8 1.6 D/d ≥ 2 1.1 1.03 1.01

1.4 1.2 1.0

0

0.1

0.2

0.3

r/d 3.0

r

2.8

P

P d

D 2.6

␴nom = P = 4P A ␲d 2

2.4 2.2

(b)

Kt 2.0 1.8 1.6 D/d ≥ 2 1.1 1.03 1.01

1.4 1.2 1.0

0

0.1

0.2

0.3

r/d r

2.6 T

2.4 2.2

d

D

T

␶nom = Tc = 16T J ␲d 3

2.0 Kt

(c)

1.8 1.6 1.4 D/d ≥ 2 1.1 1.01

1.2 1.0

0

0.1

0.2 r/d


0.3

3.0 T

M

M P

2.8

P

T

D

2.6

d Axial load:

2.4

P ␴nom = P ≈ A (␲D2/4) – Dd

2.2 Kt 2.0

Bending (in this plane):

1.8

M ␴nom = Mc ≈ I (␲D3/32) – (dD2/6)

1.6 Torsion: T ␶nom = Tc ≈ J (␲D3/16) – (dD2/6)

1.4 1.2 1.0

0

0.1

0.2

0.3

d/D


3.0 2.8 M

2.6

H

M

h

2.4

r

2.2

␴nom = Mc = 6M I bh2

Kt 2.0

b

1.8 1.6

H/h = 6 2 1.2 1.05 1.01

1.4 1.2 1.0

0

0.05

0.10

0.15 r/h (a)

0.20

0.25

H

h

0.30

3.0 2.8

P

2.6

P

r

2.4

␴nom

2.2

b = P = P A bh

Kt 2.0 1.8

H/h = 3 2 1.5

1.6

1.15 1.05

1.4 1.2 1.0

1.01 0

0.05

0.10

0.15 r/h (b)

0.20


0.25

0.30

3.0 2.8 M 2.6

M

H

h r

2.4

b ␴nom = Mc = 6M A bh2

2.2 Kt 2.0 1.8

H/h = ∞ 1.5 1.15 1.05 1.01

1.6 1.4 1.2 1.0

0

0.05

0.10

0.15 r/h (a)

0.20

0.25

0.30

3.0 2.8

P

P H

h

2.6

b

2.4

r

2.2

␴nom = P = P A bh

Kt 2.0

H/h = ∞ 1.5 1.15

1.8 1.6

1.05

1.4

1.01

1.2 1.0

0

0.05

0.10

0.15 r/h (b)

0.20


0.25

0.30

3.0 M

2.8

h

d

b

M

2.6 ␴nom = Mc = 6M I (b-d)h2

d/h = 0

2.4 2.2

0.25 Kt 2.0

0.5

1.8

1.0

1.6

2.0

1.4 1.2 1.0

0

0.1

0.2

0.3 d/b (a)

0.4

0.5

0.6

7 Pin loaded hole

6

P

h

d

b

P

␴nom = P = P A (b-d)h

5 Kt 4 Unloaded hole 3

2

1 0

0.1

0.2

0.3 d/b (b)

0.4


0.5

0.6

18

P

17 16

t

w

␴nom = P = P A wt r/w = 0.050 r/w = 0.10 r/w = 0.20

h/w = 0.5

15 r 14

h W

13 12 0.5

11 10 Kt

0.75

9

0.5

8 0.75 1.0

7 6 5

0.75 1.0

4

1.0 3.0

3

3.0 3.0

2 1 1.5 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 3.0 W/w


Same cross-section area = K

F

F

Stress con. factor = K = 2

F

F

(a) Unnotched

(b) Notched ␴av = Sy /2 ␴max = Sy

␴ = Sy Stress F=

F = ASy

ASy

Stress

2

(d)

(c)

␴ = Sy

Su Sy

Stress

␴

b

F

d

⑀ (e)

(f )


a c

0

+

–Sy

Sy

–

–0+

0

+

=

(a) Load causes no yielding 0

+

–

Sy

–0+

0

+

=

(b) Load causes partial yielding 0

+

–

Sy

–0+

0

+

=

(c) Load causes partial yielding 0

+

–2Sy

Sy

–

+

–Sy

0

–0+

=

(d) Load causes total yielding

Load stress

+

Load removal stress change


=

Residual stress

2 Z = bh 6 (0.025)(0.050)2 Z= 6 Z = 1.042 × 10–5 m3

300 mPa 0 F1

10 mm

F2

50 mm F2 F1

10 mm M1

M1

–300 MPa

25 mm

(a) Given information (see text) 0

300 396

–

(b)

–

–396

+

+

Load stress

0

62

–238 –

62

Residual stress

+

–

+

=

0

–

Residual stress

+

+

–96 0

0

–

Residual stress

+

+

–

Total stress (straight beam) 0

=

=

Load stress 0

–122 –

–

396 238 –

–

–204 62

=

Load stress

62

(e)

Residual stress –200

–

–96 0

–

–

–62 –

–

–

=

–104 0

–

(d)

=

–

Load removal stress change

–96 0

(c)

–96 0

–

300

–

Total stress (ready to yield) –300

0 –60

–

Load stress


=

=

–

–

Total stress (ready to yield)

P = 0 lb

10.000 in.

10.008 in.

T = 80°F

T = 480°F

P = 0 lb

P = 60,000 lb


P = 60,000 lb

10 mm dia.

P

60 mm P = 400 kN

20 mm


A

B

C D

E


F

di = 20 mm

T ␶max = 100 mm do = 24 mm


T

T

T T

2r b


b h

r

M

M


200 lb 3 in. 1-in.-dia round rod

4 in. 200 lb


A 40

80

A 60 P

Q

120

70,000 N


b

h c X a


3 in. 4

F 2 in.

3 in. 16

1 in.


3 in. 16

13 in. 16

24 mm

A

5 mm 30 mm

8 mm 5 mm

30 mm

A

Section AA

12,000 N


100

B Free end of shaft

T

20-mmdia. shaft

100

S

2000 N A

400 N

120-mm-dia. sheave

Connected to flexible coupling and clutch JUVINALL: Machine Design Fig. P4-27 W-131

1

3 2 in.

1 2 in.

8 in.

3 in. 8 1 2 in.


12 kN Cement

5

L 2

5

50

5

5

40


L 2

200 mm

25-mm-dia. round rod bent into crank 250 mm

100 mm 1000 N


1-in. dia. shaft

3000-lb belt tension

Motor 6-in. dia. 1000-lb belt tension 1 in.


F B

100-mm dia. 50-mm dia.

30-mm dia. A

50 mm 4000 lb 100 mm

50 mm


B 1-in.-dia. shaft

2 in. A

1000 lb 3 in.

4 in. 500 lb 4 in.

3 in.

JUVINALL: Machine Design Fig. P4-38 W-134a

A


30 y x 18

45


20 ksi Free surface, ␴3 = 0 30 ksi


75

350 100


5000 N

200 mm

30 mm

15 mm 25 mm


5000 N

1000 N r = 5 mm d = 40 mm

d = 40 mm

A

B D = 80 mm

500 mm

250 mm

RA

RB


5000 N

50 mm

100 mm

5000 N

15 mm 25 mm JUVINALL: Machine Design Fig. P4-55 W-138

Calculated elastic stress (MPa)

400

200

0

–200

0

1

2

3

4

5

6 Time

7

8


9

10

11

12

(a)

(b)

(c)


(d)

(e)

Y

Unloaded element Element loaded in uniaxial tension in X direction (with deflections shown exaggerated)

z dy (neg.)

y dx

Z

x X dz (neg.)

dx ⑀x = lim x x→0

dy ⑀y = lim y y→0

dz ⑀z = lim z z→0


Y

dx ␶yx ␪ y ␶xy Z

X

x

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

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


+␥/2

⑀y y ␥xy

⑀1

⑀2 ⑀x

+⑀ 0

x


(a) Single-element gages oriented to sense horizontal strain

(b) Two-element rosettes oriented to measure horizontal and vertical strain

(d) Three-element rectangular rosettes

(c) Three-element equiangular rosettes


⑀2

⑀120

⑀2

⑀120

⑀2

⑀240

120°

+␣ 240°

+␣ ⑀0

+␣

⑀1 +␣

⑀0

+␣ ⑀0

⑀1

⑀1

+␣

⑀240

⑀240 (a)

(b)


⑀120 (c)

⑀1 = +0.0020 240° gage (⑀ = +0.00185)

17° 0° gage (⑀ = –0.00075) ⑀2 = –0.0010 120° gage (⑀ = +0.0004)


+␥ /2 120°

⑀2 = – 0.001

–0.00075

+0.00185

⑀1 = +0.0020 +⑀

34° +0.0004 240° 0°


⑀90

⑀0

⑀2 ⑀45

⑀45

⑀0

⑀2 +␣

90°

⑀2 +␣

+␣ 45°

+␣ ⑀90

⑀0

⑀90

+␣ ⑀1

⑀1

⑀1

+␣ ⑀45 (a)

(b)


(c)

+2600 ␮m/m

+450 ␮m/m –200 ␮m/m

⑀90 = +450 ⑀45 = –200

(a)

⑀0 = +2600 (b)


⑀1 = +3560

⑀2 = –510

⑀90 = +450 ⑀45 = –200

29° 119° ⑀0 = +2600


+␥ /2 0° 45°

58°

–510

3560

0 450 –200

+⑀ 2600

90°


+␥ /2

+␶

⑀2 = –0.001

␴2 = –24

⑀3 = –0.0005

␴3 = 0 +␴

0

+⑀

0

⑀1 = 0.002

␴1 = 134

(a)

(b)


2.5 kN

1.5 kN 200

100 50

10

10

50 d = 50

d = 60

d = 40

d = 30

200

d = 40 1.8 kN

V, shear force (kN)

2.2 kN 2.2

2 1

Area = 220 kN•mm

0.7 Area = 140 kN•mm

0

Area = 360 kN•mm

M, bending moment (kN•mm)

–1 –2

–1.8

400

255

220 200

Area = 58,000 kN•mm2

22

18

A = 11,000

4

A= 0.457 × 10–3 5.12

0.85 2.67

5.67

2.47 1.94 0.220 × 10–3

0.419 × 10–3

A = 0.565 × 10–3

Initially assumed location of zero slope

A = 13 × 10–6

1.2 0.8 0.4

A= 0.270 × 10–3 0.69 0.28

A = 3 × 10–6

0.838 0.835 0.270 A = 0.120 mm

0.8 0.4

A = 0.011 0

–0.8 –1.2

0.007 –0.4 –0.220

–0.677 –1.096 –1.109

Location of zero absolute slope

–0.4

0

0.022 A= 0.093 mm

–0.8

True line of zero deflection

0

+0.001

–0.04 –0.08 –0.12

0.093 0.115 0.126

0.119

Tangent point


Parallel to true line of zero deflection

Absolute slope (mrad)

M , (10–6/mm) EI

9.80

8.46 8

0

Relative slope (milliradians)

A = 36,000

0

12

Deflection (mm)

360

325

Area = dU'

dQ

Area = U' Load

Area = dU

Q Area = U

0 Deflection


∆

P x

dx h

P/2

L

L/2


P/2

b

P = 5000 N

h = 50 mm 2500 N

L = 400 mm

200 mm


2500 N

b = 25 mm

x

c

b L h y a

P

Q


R

F

F

R

␦

P F

h

M

␪

␪

F

V 2R

b

(a)

(b)


R – R cos ␪

1.0 2 Avg value =

sin

0.5

∫

0

/2 0

sin d = 1;

∫

0

sin

d = 2;

∫

2 0

sin d = 0

2 can be determined) (From this, average value =

–0.5

–1.0 1.0 2 Avg value =

cos

0.5

∫

0

/2 0

cos d = 1;

∫

0

cos d = 0;

∫

2 0

cos d = 0

–0.5

–1.0 1.0

∫ ∫

sin2

Avg value = 0.5 0.5

0 1.0

∫ ∫

cos2

Avg value = 0.5 0.5

0

/2 0 2 0

/2

0

(0.5) = ; 2 4

∫

0

∫

0

sin2 d = (0.5) =

2

cos2 d = (0.5) =

2

sin2 d =

0 2

sin2 d =

cos2 d =

(0.5) = ; 2 4

cos2 d =

(0.707)(0.707) = 0.5 2 1 Avg value = 0.5 = = (Half of avg value shown in sin plot) sin cos

0.5

∫ ∫

0

–0.5

0

/2

3/2 (radians)

/2 0 2 0

sin cos d =

1 1 = ; 2 2

sin cos d = 0

2


∫

0

sin cos d = 0

2 R=

.0

in.

E = 18 × 106 psi G = 7 × 106 psi F

F ␦ h = 0.3 in. b = 0.2 in. 2R = 4 in. (a)


0.035 Copper Cast iron Steel

Deflection, ␦ (in.)

0.030 0.025 0.020 0.015 0.010 0.005 0.000 0.2

0.4

0.3 Width, h (in.) (b)


0.5

0.035

Deflection, (in.)

0.030 0.025

Copper Cast iron Steel

0.020 0.015 0.010 0.005 0.000 1.0

1.5

2.0 Radius, R (in.)


2.5

3.0

Deflection, ␦ (in.)

0.06 0.05 0.04 0.03 0.02 0.01 0.2

0. 1 0.3 Wid th

,h

(in.

0. 2

0.4

)

0. 3 0.5

JUVINALL: Machine Design Fig. 5-21d W-169

,b ess ckn Thi

) (in.

1.2 m

500 kg mass

10 m Point of zero deflection F

a y 3m


3

Pa/2

a

M0 2

P/2

P 1

y b 2

M0

P/2

3' Pa/2 2' a

M0

b 2

P

M0

x P/2 M0

(a)

P/2 M0

P/2


(b)

P/2

P P

y x

x y

Axis of least I and ␳ becomes neutral bending axis when buckling occurs. With column formulas, always use I and ␳ with respect to this axis.

L or Le e

(b) Column cross section

P P (a) Two views of column


0.100

0.010

Scr E

0.001

0.0001

10

100

Slenderness ration Le /␳


180

1200 1100

160

1000

800 700 600 500 400 300

Critical unit load Scr (ksi)

Critical unit load Scr (MPa)

140 900

120

Sy = 689 MPa A

Arbitrary values for illustration

C

100 Sy = 496 MPa 80

B

D

60 Euler, E = 203 GPa (steel)

40

200 20 100 0

Euler, E = 71 GPa (alum) 0

20

40

60 80 100 120 Slenderness ratio Le /␳


140

E F 160

(Buckled shape shown dotted)

Le Le = L

L

Le

L

Le

L

Le 2

L

(a)

(b)

(c)

(d)

(e)

Theoretical

Le = L

Le = 0.707L

Le = 0.5L

Le = L

Le = 2L

Minimum AISC Recommend

Le = L

Le = 0.80L

Le = 0.65L

Le = 1.2L

Le = 2.1L

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


180

1200 1100

160

1000

800 700 600 500 400 300



140 900 120 100 80

0

Sy = 496 MPa B D

Sy = 689 MPa Johnson, E = 203 GPa, Sy = 689 MPa

60 40

200 100

C

A

Johnson, E = 71 GPa, Sy = 496 MPa

Tangent points

Euler, E = 203 GPa Euler, E = 71 GPa

20 E F 0

20

40

60 80 100 Slenderness ratio Le /␳


120

140

160

D 80,000 N

80,000 N 1m SF = 2.5 Sy = 689 MPa E = 203 GPa (steel)


D

80,000 N

80,000 N 200 m

SF = 2.5 Sy = 496 MPa E = 71 GPa (aluminum)


600 550

80

500 70

Euler curve

400

60

ec/␳ 2 = 0

350

0.05

50

0.1 300 40 250

0.3

200

0.6

30

150

ec/␳ 2 = 1.0

20

100 10 50

0

20

40

0 60 80 100 120 140 160 180 200 Slenderness ratio Le /␳




450

(a) Wrinkling, or "accordian buckling" of thin-wall tube

(b) Typical local buckling of an externally pressurized thin-wall tube


(c) Wrinkling of thin, unsupported flanges of a channel section

Beam

Tetrahedron

Triangle

Pentahedron

Quadrilateral

Hexahedron


1

(2)

3

(5)

5

(7) (3)

(10)

7

(9) (11)

(6) 6

(1) (8) 4 (4) 2

JUVINALL: Machine Design Fig. 5-34 W-180B

W

␦3x 3' F (7) 3y

F (7) 3x

␦3x

3

L

F (7) 6y 6' F (7) 6x 6 ␦6x

JUVINALL: Machine Design Fig. 5-35 W-180C

␦6x

F3

F (7) 3x 3

F (7) 3y

␦3x

3'

␦

L A, E

␪ 6

JUVINALL: Machine Design Fig. 5-36 W-180D

3N 2N

2

␦2y

2'

(Constant A, E) 3.464 m

2m

␦2x (2)

1

4m

(3)

JUVINALL: Machine Design Fig. 5-37 W-180E

3

␦3x 3'

F (1) 2y

F (2) 2y F (1) 2x

2

F (2) 2x 2

(1)

3.464 m 2m

F (1) 1y

F (2) 3y

(2) F (1) 1x

F (2) 3x

1

3

(3) F 1y

F (3) 1x

F (3) 3y F (3) 3x

(3) 1

4m

JUVINALL: Machine Design Fig. 5-38 W-180F

3

1.

Tension or compression

␦=

␦

L

PL AE

k=

P AE = ␦ L

K=

T K'G = ␪ L

P Cross-section area = A

2.

Torsion

␪

L T K' a = section property. For solid round section, K' = J = ␲d 4/32.

3.

Bending (angular deflection)

TL K'G For solid round bar and deflection in degrees, 584TL ␪° = d 4G ␪=

␪=

ML EI

K=

M EI = ␪ L

␦=

ML2 2EI

k=

M 2EI = ␦ L2

␦=

PL3 3EI

k=

P 3EI = ␦ L3

L ␪ M

I = moment of inertia about neutral bending axis

4.

Bending (linear deflection) ␦

L

M


5.

Cantilever beam loaded at end L

P

␦


JUVINALL: Machine Design Table 5-1 W-157

di

d

4 K' = J = ␲d 32

do

K' = J = ␲ (d 4 – d 4i ) 32 o

t 4 K' = ␲dt 32

d

2b

3 3 K' = ␲a b 2 a + b2

2a

K' = 0.0216a4 a

K' = 2.69a4 a

b

3 4 K' = ab 16 – 3.36 b 1 – b 16 3 a 12a4

a a

K' = 0.1406a4

a


⑀120 = +625

⑀0 = +950

⑀240 = +300


⑀135 = –380 ⑀0 = –300

⑀90 = –300

⑀45 = –380 ⑀270 = –200 Gage readings

⑀0 = –200 Equivalent rosettes


100 mm A F 100 mm

k = 5 N/mm B

F 100 mm C F


200 mm

100 mm A

25 mm-dia. steel 1000 N


4 kN d = 30

d = 50

d = 40

2 kN 100

200


150

a Z Solid round rod of properties E, G, A, I, and J.

b X

Y T (used in Problem 5.18)

F (used in Problem 5.17)


w = 200 lb/in. 0.75d

0.75d d

5 in.

15 in.


5 in.

a

F

a

F

b


R

P


b 2 b 2

L F

h


500 mm

5 kN

S

300 mm


2

1

1

2


12 mm-dia. tie-rod

1m 0.7 m Boom

0.7 m 1m


6 kN

P

P

t t 2c c 2w

w

P

P

(a) Center crack

(b) Edge crack


P

7075–T651 Aluminum, Su = 78 ksi, Sy = 70 ksi, Kic = 60 ksi

2c

t = 0.06 in. =1

2w

i n.

=6

in.

P


in.

P

a

2c 2w

P


t

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

a

2c

2w = 6 ␴g = 0.73 Sy

in. P

a/2c = 0.25


t = 1 in.

+␶

+␶

(␶max = 60)

(␶max = 50)

␴2 = 40 ksi ␴1 = 80 ksi

␴2 = ␴3 = 0

␴3 = 0 +␴

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

␴1 = 100 ksi +␴

(a) Standard tensile test of proposed material. Tensile strength, S = 100 ksi


Uniaxial compression

+␴2

Uniaxial tension

Sut

+␶

Sut

Suc 0

+␴

Suc

Sut 0

Principal Mohr circle must lie within these bounds to avoid failure Suc For biaxial stresses (i.e., ␴3 = 0), ␴1 and ␴2 must plot within this area to avoid failure (b) ␴1 – ␴2 plot

(a) Mohr circle plot


+␴1

+␴2 Syt +␶

Uniaxial tension

Principal Mohr circle must lie within these bounds to avoid failure

Syt

Syt

+␴

0

For biaxial stresses (i.e., ␴3 = 0), ␴1 and ␴2 must plot within this area to avoid failure (b) ␴1 – ␴2 plot



+␴1

Principal plane

Octahedral plane

Principal planes


Shear diagonal (␴1 = –␴2) +␴2 (0, 100)

(100, 100)

(–100, 100) Shear stress theory

Distortion energy theory

(–58, 58)

(–100, 0)

(100, 0) 0

(50, –50)

+␴1

Normal stress theory (58, –58) Note: ␴3 = 0

(–100, –100)

(100, –100) (0, –100)


+␴2 +Sut +␶

Suc

Sut 0

+␴

Suc

Sut 0

Principal Mohr circle must lie within these bounds to avoid failure For biaxial stresses (i.e., ␴3 = 0), ␴1 and ␴2 must plot within this area to avoid failure

Suc

(b) ␴1 – ␴2 plot



+␴1

+␴2 Sut

Suc

Sut 0

+␴1 Shear diagonal

Suc


␴2 = –25 ksi Note: ␴3 = 0

␴1 = 35 ksi

␴1 = 35 ksi

Steel, Sy = 100 ksi ␴2 = –25 ksi


0

35

58

66

100

␴1 (ksi)

Normal load point –25 ␶ theory D.E. theory ␴ theory (58, –58)

Load line

–100

Shear diagonal

␴2 (ksi)


Limiting points

Stress (% of ultimate strength)

100 Fracture

SF = 2 based on strength, Eq. 6.9

50 SF = 2 based on load, Eq. 6.10

0

50 100 Load (% of ultimate load)


Frequency p(x) and p(y) 0

y (stress)

x (strength)

␮y

␮x

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


Frequency p(z)

z (margin of safety) ␮z 0

30 Margin of safety, z, where z = x – y


␴1 Frequency p(x)

␴1 < ␴2 < ␴3

0

␴2 ␴3

␮ Quantity x


x1 x2

Frequency p(x)

Inflection point

Inflection point

34.13%

34.13% of total area

2.14%

2.14%

0.13%

0.13% 13.60%

–3␴

–2␴

–1␴

13.60% ␮

+1␴

Quantity x


+2␴

+3␴

99.99

0.01

99.9 99.8

0.05 0.1 0.2

1

98

2

95

5

90

10

80

20

70

30

60

40

50

50

40

60

30

70

20

80

% of failures –k␴

10 5

95

% of survivors

2

98

␮

1

99

Fatigue life, strength, etc.

0.5 0.2 0.1 0.05 0.01

Extreme values

–4

–3

99.8 99.9

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

–2

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


+3

+4

99.99

% of failures or P, % cumulative probability of failure

% reliability (% of survivors or % cumulative probability of survival)

0.5 99

One bolt in 500 twists off

Frequency p(x) and p(y)

␴x = 1 N • m ␴y = 1.5 N • m

0

y (wrench twist-off torque)

x (bolt twistoff strength)

␮y

␮x

(14.8)

20.0

Torque x and y (N • m) (a)

k␴z Frequency p(z)

One bolt in 500 twists off z=x–y

␮z 0

(5.22) Torque z (N • m) (b)


␴2 = 100 MPa

␴1

a

␴2

␴2 = –100 MPa

␴1 = 200 MPa

␴1

b

␴2


␴1 = 150 MPa

Sy = 500 MPa

␶xy = 100 MPa ␴x

␴x = 50 MPa ␶xy


c

m

m

m

k

k

(a)

(b)


k

(c)

100

100

80

80

70 60

60 50

50

Ultimate strength Su

40

Total elongation

30 20

40 30

Yield strength Sy

20

10

10

0 10–6 10–5 10– 4 10–3 10–2 10–1 1 Average strain rate (s–1)


101

102

0 103

Elongation (%)

Su and Sy (ksi)

70

Ratio Sy /Su

Sy / Su (%)

90

90

W Guide rod h

Elastic-strain energy stored 1

in structure = 2 Fe ␦

Force Fe W

␦ k

Work of falling weight = W (h + ␦) k

W

O ␦st

h Deflection

␦ (a)

(c)

(b)

JUVINALL: Machine Design Fig. 7-3a-c W-218

2d

L/2

␲d2 Area = A = 4 L

L/2 d

d

(b)

(a)


W

h

Bumper of cross section A; volume = AL L


30 in. 100 lb

2 × 4 white pine E = 106 psi Mod. of rupture = 6 ksi

100 lb/in.

5

12 in.

5

1 8 in. × 3 8 in., I = bh3/12 = 6.46 in.4 Z = I/c = 3.56 in.3 60 in.


100 lb/in.

250 mm 20-mm dia 120-mm dia

100-mm dia

20 mm

20 mm


700 Aluminum Cast iron Steel

Shaft shear stress (MPa)

600 500 400 300 200 100 5.0

7.5

10.0 Shaft radius (mm)

12.5

15.0

Torsional deflection of shaft (deg)

40 35

Aluminum Cast iron Steel

30 25 20 15 10 5 0 5.0

7.5

10.0 Shaft radius (mm)


12.5

15.0

K i = 1.5

d

K i = 1.5


K i = 1.5

Ki = 3 d 2 d K i = 1.5


Head Ki = 3.4 Ki = 3.4 A = 700 mm2

Ki =1.5 A = 300 mm2

"Very long" (>10d)

Shank d Ki = 1.5 Ki = 3.5 A = 600 mm2

Ki = 3.0 A = 600 mm2

"Negligible" (a) Original design


(b) Modified design

K = 1.55 Drop weight K =4

24 mm dia.

30 mm dia.

K = 1.4


Axial hole

(a)

(b)


K =2

1-in dia. 0.1-in dia. hole

"Very long"

K =2


Steel cable A = 2.5 in.2 E = 12 × 106


v = 4 km/hr m = 1400 kg

K = 5000 N/mm L=5m Rope


Thread: Area = A K = 2.6

Thread: Area = 600 mm2 K = 3.9

K = 1.3

Area = 800 mm2

Area = 800 mm2

"Very long"

K = 1.3

A = 600 mm2 K = 3.9

Area = A K = 2.6

Platform

Original design

New design

(a)

(b)


Hole dia., d

K = 2.2

K = 1.5

250 mm

Assume that hole drilled to this depth, does not significantly change the K = 3.8 factor at the thread.

dia. = 36 mm (Fracture location) A = 800 mm2 K = 3.8

3 mm (negligible) Original design

Modified design

(a)

(b)


Small region behaves plastically Main body behaves elastically


Test specimen

Flexible coupling

9 8" R 7

Motor 0.300" Revolution counter C Weights 110-Volts AC


Fatigue strength, or Peak alternating stress S (ksi)

50 + + + + + + 40 + + + + + + + 30 +

Not broken +++

20 Sn' (endurance limit) 10 0 0

10

20

30

40 50 60 70 80 Life N (cycles ⫻ 106)

90 100

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

Fatigue strength, or Peak alternating stress S (ksi)

50

+ +

40

+

+

++ + ++

30

Not broken + ++ + +

+ +

20 Sn' (endurance limit) 10 0 103

104

105

106 107 Life N (cycles (log))

108

Fatigue strength, or Peak alternating stress S (ksi (log))

(b) Semilog coordinates

100 80 70 60 50

8:7 ratio +

40 30 20

+ + "Knee" of curve Not broken + + + + + + + + + + 7:1 ratio Sn' (endurance limit)

10 103

104

105

106 Life N (cycles (log))

(c) Log-log coordinates


107

108

S = 0.9Su (in ksi, S ⬇ 0.45

Bhn; in MPa, S ⬇ 3.10 Bhn)

1.0 0.9

S/Su (log)

0.8 0.7

Not broken

0.6 0.5 Sn' 0.4 103

2

4

6

104

2

4 6 105 2 Life N (cycles(log))

4


6

106

2

4

6

107

Sn' = 0.5 Su (in ksi, Sn' ⬇ 0.25 Bhn; in MPa, Sn' ⬇ 1.73 Bhn)

140

187

229

Hardness (HB) 375 285

477

653

900 120

Endurance limit S 'n (ksi)

100

700 600

80

500 60 400 0.25 Bhn 300

40 SAE SAE SAE SAE

20

0 0

10

20

30 40 50 Hardness (Rockwell C)


4063 5150 4052 4140

60

200 100

70

Endurance limit S 'n (MPa)

800

Actual maximum stress Calculated maximum stress (Mc /I )

M

M


500

60

400

50 300

40 35

250

30

200

25

Wrought

20 18 16 14

Permanent mold cast

150

100

12 75

10 8 7 6 5 103

Sand cast 50

104

105


107


108

109

S (MPa (log))

Peak alternating bending stress S (ksi (log)

80 70

2014-0, T4, and T6 2024-T3, T36 and T4 6061-0, T4 and T6

6063-0, T42, T5, T6 7075-T6

30

150

20 S n' = 19 ksi

100

S n' = 0.4S u

10

50 0

0 0

0

10

50

20

100

150

30

200

40 50 Tensile strength Su (ksi) 250

300 Su (MPa)

350


60

400

70

450

500

550

S n' (MPa)

Fatigue strength at 5 108 cycles S n' (ksi)

Alloys represented: 1100-0, H12, H14, H16, H18 3003-0, H12, H14, H16, H18 5052-0, H32, H34, H36, H38

40 35 200

25 Extruded and forged

20 18 16

150

100

14

Sand-cast

12 10 9 8

75

7

50

6 5 105

2

4

6 8 106

2 4 6 8 107 Life N (cycles (log))


2

4

6 8 108

S (MPa (log))

Peak alternating stress S (ksi (log))

30

Sn = Sn' = 0.5Su 0.5 Bending Axial (no eccentricity)

0.3

Torsion

Sn = 0.9Sn' = 0.45Su Sn = 0.58Sn' = 0.29Su

Ratio,

peak alternating stress, S or Ss (log) Su

1.0

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

0.1

103

104

105 Life, N (cycles (log))


106

107

␴2 Sn 1.2 Note: Dotted portion is superfluous for completely 1.0 reversed stresses 0.8 0.6 0.4 Reversed bending

0.2 ␴1 Sn

–1.2

–0.8

–0.4

0 – 0.2

0.4

0.8

1.2

␴1 Sn

DE theory

– 0.4

Reversed torsion

– 0.6 – 0.8 –1.0

Reversed bending –1.2


Hardness (HB) 120 160 200 240 280 320 360 400 440 480 520 1.0 Mirror-polished 0.9 Fine-ground or commercially 0.8 polished

Mac hine d or

Surface factor Cs

0.7 0.6

cold -dra wn

0.5 Hot -rol led

0.4 0.3

As f orge d

0.2 0.1

Corroded in tap water Corroded in salt water

0 60

0.4

80

0.6

100 120 140 160 180 200 220 240 260 Tensile strength Su (ksi)

0.8

1.0

1.2

1.4

Su (GPa)


1.6

1.8

Equal surface stresses

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

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

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


␴a ␴max

␴a Stress

␴max + 0 –

␴m ␴m

␴a

␴min

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


103 ~

F

Values from S–N curve

Sy A" 104 ~ E D

105 ~

10 3

106 ~ H'

10 4

H

Sn

~

~

C 10 6

␴a

10 5

~

~ G

A' – Sy

A – ␴m (compression)

0


␴m (tension)

Sy

B Su

90 4

7

10

40

20

10

10

20

20 ) 30 f S u o

30

–40

–20

0

–1 0

–60

20 40 60 80 Minimum stress ␴min (% of Su)

–2 0

–80

–3 0

0 –100

30 M ea 40 n st re 5 ss 0 ␴ m (% 60 of S

50

u)

70

10 3× 5 10 6 10

% 40 a( ␴ ss 50 stre g in 60 nat r te Al

60

10

80

4

70

70

10

80

3 le life 10 -cyc 3 0 3×1

80

Maximum stress ␴max (% of Su)

90

90

10 0

Su

100


100

80

80

60 50

si 40 ss ␴

5

st

re

10

3 M 0 ea n

6

20

10 7 10 9 10 10

10

4×

10

–20 0

0

20 40 60 Minimum stress ␴min (ksi)

–1

– 40

0

– 60

–2

0 – 80

m

(k

5

10

20

20

)

4

10

i) 30 (ks ␴a

Maximum stress ␴max (ksi)

4×

ss

30

4

10

40 stre g in at 50 ern t Al

40

60

50

life 3 cycle 10 -

70

60

70

Su 70


80

70 60 4

5 (k 0 si ) 10

10

0

– 20 –1 0

– 40

20 40 60 Minimum stress ␴min (ksi)

–2 0

– 60

–3 0

0 – 80

3 M 0 ea n st 40 re ss ␴

5

10 7 4× 6 10 10 9 10

20

20

m

5

10

20

30

10

10

4×

i) 30 (ks ␴a

Maximum stress ␴max (ksi)

4

10

s 40 tres s g

40

fe le li

in 50 nat r te Al

50

3 - cyc 10

60

60

70

70

80

80


80

Calculated fluctuating axial stress (ignoring yielding)

Su ␴a

Sy Sn ␴a

(e)

␴a ␴m

(d)

0

(c) (b)

– Sn

(a)


␴m

(f)

d < 2.0 in.

Commercially polished surface

P(t)

P(t)

P(t) 0 Sy = 120 ksi Su = 150 ksi


t

Peak alternating stress, S (log)

Axial loading stresses in ksi

S = 0.75Su = 0.75(150) = 112 112

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

92 75 61

103

104


120

106

107

Sy

112 ksi Axial loading stresses in ksi

100

92 ksi

10 4

80

75 ksi

10 5

61 ksi

A

10 3

~

~

10 6

␴a

~

~

3

2

40 1 ␴a ␴m = 0.67 (used in Sample Problem 8.2)

20 Point O –120 Sy

–100

– 80 – 60 – 40 – ␴m (compression)

– 20

0

20


40

60 80 100 +␴m (tension)

120 Sy

150 Su

(a) Unnotched specimen ("u")

d

(b) Notched specimen ("n")

Calculated nominal stress S

d

U nn

103

104

o tc h

ed sp e c im ens Not che ds pec ime ns

105 Life N (cycles)

106

Sn(u) S (u) Kf = n Sn(n)

Sn(n)

107

(c) Illustration of fatigue stress concentration factor, Kf


Use these values with bending and axial loads Use these values with torsion 1.0 0.9 0.8 0.7 0.6 q 0.5

Steel Su (ksi) and Bhn as marked

180 (360 Bhn) Bhn) (400 200 0 Bhn) 120 (24 Bhn) 0 8 2 0 Bhn) ( 80 (16 140 hn) n) B 0 20 Bh 60 (1 (20 ) 100 n h 0B ) (16 Bhn 80 (120 hn) B 60 (100 Aluminum alloy (based on 2024-T6 data) 50

0.4 0.3 0.2 0.1 0 0 0

0.02

0.04

0.06 0.08 0.10 Notch radius r (in.)

0.12

0.14

0.16

0.5

1.0

1.5 2.0 2.5 Notch radius r (mm)

3.0

3.5

4.0


Calculated notch stress (P/A)Kf (MPa)

600

Calculated stresses

300

0

(a)

(b)

(c)

(d)

Actual notch stress (P/A)Kf + ␴residual (MPa)

Sy 300

Actual stresses

0

– 300

300 280

(a)

(b')

(c')

(d')

d' d c'

␴a (MPa)

c 200

b' a

106 ~ Life 105 ~ Life 104

~ Life 103

100

0

100

Constantlife fatigue diagram

b

200 300 ␴m, tension (MPa)


~ Life

400

450 Su

r

d

D

T = 1000 ± 250 N • m

T = 1000 ± 250 N • m D/d = 1.2 r/d = 0.05 SF = 2.0

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


Alternating torsional stress ␶a (MPa)

Assuming 10 mm < d < 50 mm

Sn = Sn' CLCGCS =

(0.58)(0.9)(0.87) = 272 MPa ␶max = Ssy

300 10 6 ~

200

=∞

100

–100

life

B'

150 116 N'B

B A NA

0' – 200

1200 2

0

100

200

NB 300 400 500 600 Ssy ≈ 0.58(1000) = 580

Mean torsional stress ␶m (MPa)


700 800 900 Sus ≈ 0.8(1200) = 960

50 mm Ft

Fn r = 5 mm rad., machined surface

100 mm

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

Alternating bending stress ␴ea (MPa)

f = 0.6 (between the object and the disk) T = 12 N • m (friction torque)

Sn = Sn' CLCGCS =

Sy = 750 MPa

900 2

(1)(0.9)(0.72) = 291 MPa

"Failure point"

300

10 6

200

100

0

Su = 900 MPa

~=∞

␴max = Sy life

Sy = 750 Su = 900

(15.7, 65.0) "Operating point" 0

100

200

300 400 500 600 700 Mean bending stress ␴em (MPa)


800

900

0 – 40 – 80 Representive 20-second test

80 60 50 40 103

(a) Stress-time plot

104

105

40

100

3.8 × 104

Stress (ksi)

80

120

1.6 × 104

Reversed stress S (ksi (log))

140

105 N (cycles (log)) (b) S-N curve


106

107

Representative 6-sec test 500

0 –100 2~

– 200

3~

a

– 300

1~ d

2~ c

b

c'

300 250 200 150

100 103 104 d'' c''

1~ b

b'

105

106

N (cycles)

(a)

(c)

Stress-time plot

S-N plot

107 b''

d'

400 d

Bending stress at critical notch

c' 300

σa (MPa)

a'

3.5 × 106

100

d'

400

2 × 104

200

2.5 × 103

Reversed stress S (MPa)

Bending stress (MPa)

300

P(t) Part c

200

Aluminum alloy Sy = 410 MPa Su = 480 MPa

b' b a' 100

(d)

a

0

100

200 300 ␴m (MPa)

400 Sy

500 Su

(b) ␴m – ␴a plot


108

T Tensile stress and strength (ksi)

(d) Strength

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

Compressive stress (ksi)

(b) Residual stress 0

Axis of specimen symmetry and axis of load

Pmax


P r = 2.5 mm P 35 mm

30 mm 30 mm 30 mm 30 mm P

r = 2.5 mm

P (1)

(2)


F lb 3 in.

3 in. 2 in. 2 in.

1 in.

1

1 in. dia.

1 16

F lb 2

R

1 4 in. dia.

1 in. dia.

1 8

R


1 8

R F lb 2

2-mm rad. 24 mm

0.8-mm rad.

20 mm


24 mm

r = 81 in. Kt = 1.7

4 in.

0.1094 in. F High-carbon steel, 490 Bhn, machined finish 3 4


in.

0.050 in. 1 2

in. 0.191-in. rad.

0.125-in. dia.


60 mm

5-mm rad.

50 mm

5-mm rad.

60 mm

50 mm

+80

Nominal stress (MPa)

1.5 rad.

0 –16 Time

JUVINALL: Machine Design Fig. P8-38 W-268b

1.2-in. dia.

7000 lb • in.

1.0-in. dia. Torque

0.1-in. rad.

1 16

-in. dia. hole

3000 lb • in.

0.1-in. rad. Time

JUVINALL: Machine Design Fig. P8-39 W-268c

25-mm solid round shaft Fillet 2000 N

750 N

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

500 N

Pump

Helical spur gear

50 25 0-m m

dia .

JUVINALL: Machine Design Fig. P8-44 W-268d

mm

y x

B

A

Forces act at 500-mm dia. C z

A

B 550 C

400

D

D 400

450

Fy = 1.37 kN Fz = 5.33 kN Fz = 0.3675Fy

Fx = 0.2625Fy

120 dia.

Fy

(1)

Fx = 1.37 kN Forces act at 375-mm dia.

E

Keyway

80 dia.

(Kf = 1.6 for bend and torsion; 1.0 for axial load. Use CS = 1 with these values.) (2)

JUVINALL: Machine Design Fig. P8-45 W-268e

Torsion stress (ksi)

30 20 10 0 –10 – 20 – 30 30 seconds

JUVINALL: Machine Design Fig. P8-46 W-268f

Iron electrode with surplus of electrons

~ +~ + ~ + ~ + + ~ ~ + + ~ ~ ~ +~ ~ + ~ + + ~ ~

Fe2+ ions in solution

Electrolyte

JUVINALL: Machine Design Fig. 9-1 W-269a

~ ~ ~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~

Plating, as tin or zinc (cathode or anode) Exposed iron (anode or cathode)

Electrolyte

JUVINALL: Machine Design Fig. 9-2 W-269b

A B

~ ~ ~ ~ ~

~ ~ ~ ~

~ ~ ~

~

~

~

Electrolyte

JUVINALL: Machine Design Fig. 9-3 W-269d

Outlet

Inlet

Magnesium anode Insulated copper wire

Magnesium anode

(b) Underground pipe

(a) Water tank

Zinc anodes

Zinc strips between steel spring leaves (c) Leaf spring

JUVINALL: Machine Design Fig. 9-4 W-269e

(d) Ship

–

+

Insulated copper wire

Steel tank

JUVINALL: Machine Design Fig. 9-5 W-269f

Rust particles

Rust particles

Steel

Steel

Steel

(a)

(b)

Rust begins at center of drop

"Crevice corrosion"

JUVINALL: Machine Design Fig. 9-6 W-269g

Steel bolt and nut

Aluminum plates

Nonporous, pliable electrical insulator


Salt water

All

Ceramics, Glasses

PTFE Epoxies LDPE/HDPE PP PS PVC Nylons

Lead alloys Nickel alloys S-steels Cu-alloys Ti-alloys

Polymers Polyesters Phenolics PMMA

Alloys

Composites GFRP CFRP PTFE, PP Epoxies, PS, PVC Nylons Polymers PMMA HDPE, LDPE, Polyesters Phenolics Lead alloys Alloys Steels CuAlTi-alloys alloys alloys Ni-alloys Cast irons

A

B

KFRP CFRP

Gold Lead Alloys Ti-alloys S-steels Ni-alloys

Al-alloys Cast irons

KFRP

Aerated water

All

GFRP

Composites Ceramics, Glasses

PTFE PVC

Al-alloys

HDPE LDPE, Epoxies Elastomers

Alloys Low alloy steels

C-steel Cast irons

Nylons Polyesters CFRP Phenolics CFRP PS PU Composites PMMA KFRP

Csteels

Mg 0 Zr02 Vitreous ceramics

C-steels

C

D

D

C

B

Al-alloys Many elastomers PTFE PVC PMMA

Ceramics, Glasses KFRP CFRP GFRP

U-V radiation

Glasses

Ceramics, Al203 Glasses Si C

All

Composites

Strong acids

Polymers

Alloys

Nylons

Polymers

PS PVC Most elastomers

LDPE Epoxies, HDPE polyesters, PP, phenolics, PS Filled polymers

Composites GFRP CFRP

All

Alloys All

A Ni-alloys Steels S-steels Cast irons Ti-alloys

PMMA Elastomers Phenomics Si02 Polymers Nylons GFRP Glasses Polyesters LDPE/HDPE Vitreous P.V.,PS,PP,PTFE KFRP ceramics Composites PVC,Epoxy

Polymers KFRP

Cu-alloys Zn-alloys

Si3N4 Si02

Phenomics Polyesters PU LDPE HDPE Epoxies Nylons PP

Ceramics Glasses

Ceramics, Glasses

Strong alkalis

CFRP

Si C, Si3N4 Al203 Zr02 Graphites All

Alloys PTFE

Organic solvents

JUVINALL: Machine Design Fig. 9-8a W-271a

A B C D

Excellent Good Poor Bad


Mo Cr Co Ni Fe Nb Pt

Zr

Ti Cu Au Ag Al Zn Mg Cd Sn Pb In

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

Two liquid phases One liquid phase, solid solubility below 0.1% Solid solubility between 1 and 0.1% Solid solubility above 1% Identical metals

W


Increasing compatibility; hence, increasing wear rate

Identical metals Compatible metals Adhesive wear

Poor lube

Unlubed

Unlubed

Good lube Good lube

Poor lube

Unlubed

Incompatible metals

2-body

High abr. 3-body concentr.

10–2

Excellent lube

Poor lube

Good lube

Lubed

Low abr. concentr. Unlubed

Fretting

Good lube

Unlubed

Nonmetal on metal or nonmetal Abrasive wear

Excellent lube

Poor lube

Unlubed

Partly compatible

Excellent lube

10–3 10–4 10–5 Wear coefficient, K


Lubed

10–6

Exc. lube

F = 20 N

Copper pin 80 Vickers

Pin

Pin

Steel disk 210 Brinell r = 16 mm

Pin volume lost = 2.7 mm3

Disk

Disk volume lost = 0.65 mm3

Disk t=0h

Initial profiles n = 80 rpm


t=2h Final profiles

z z

Contact area Contact area

R2

p0 p R1 a

p

a y x

x

p0

y

L b y

(a)

(b)

Two spheres

Two parallel cylinders


– p0 – 0.8p0 0

Stress – 0.4p0

0

0.4p0

0

␴x = ␴y

Stress – 0.4p0

b

a

0

0.4p0

␴y ␴x

2b

2a

␶max 3a p0 = max. contact pressure 4a (a) Two spheres (a is defined in Fig. 9.13a)

Distance below surface

Distance below surface

␴z

– p0 – 0.8p0

␴z

3b ␶max 4b

5b

6b p0 = max. contact pressure 7b (b) Two parallel cylinders (b is defined in Fig. 9.13b)


␶max ≈ 0.3p0

␴z ≈ – 0.7p0 0.3p0

␴y ≈ – 0.1p0

+␶

0.2p0 0.1p0 +␴

0 ␴z ≈ – 0.7p0

␴y ≈ – 0.1p0

␴x ≈ – 0.25p0 ␴x ≈ – 0.25p0 One plane of maximum shear stress

– 0.6p0

– 0.4p0


– 0.2p0

0

0.2p0

F

b ␴z

A

y ␴z

B

␶yz

␶yz ␴y

A

B

F

0.3p0 0.5b below surface 0.2p0

Stress ␶yz

0.1p0 0 – 0.1p0 p0 = max contact pressure – 0.2p0 – 0.3p0 – 4b

– 3b

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


3b

4b

␴y

z

Driving cylinder Direction of rotation

␴yzt = fp0

␴yt = – 2fp0

␴yt = 2fp0

y

y ␴yt = 2fp0

␴yt = – 2fp0

␶yzt = fp0 b

b

Loaded cylinder (resists rotation to cause some sliding)

Direction of rotation

z

p0 = maximum contact pressure f = coefficient of friction


2000 N 10.1 mm Hardenedsteel sphere

Hard-bronze bearing alloy spherical seat


10 mm

Maximum contact stress p0 (MPA)

250 Cast iron Copper Steel

200

150

100

50 5.00

5.01

5.02 Sphere radius, R1 (mm)


5.03

5.04

Computed maximum elastic contact stress p0 or ␴z (ksi)

800 700 600

Pa ral Rad lel ial rol bal ler l s bea Ang ring ula r-co s nta ct b all Ro bea lle ring rb s ea rin gs

500 400 300

200 150

100 105

Spur gears–high-quality manufacture, case-hardened steel, 60 Rockwell C (630 Bhn) 106

107 108 Life N (cycles (log))


109

1010

C Key:

+

10 C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

X

Protected ("noble", more cathodic)

11

1 2

13

3

14 C

4

15 Legend: X – Not compatible C – Compatible P – Compatible if not exposed within two miles of a body of salt water F – Compatible only when finished with at least one coat of primer *Applicable forms: 301, 302, 321, and 347 sheet and 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 least one coat of primer.

6 7

15 16 17 18 19 20

C

C

C

C

C

C

C

C

X

C

C

C

C

X

C

C

C

C

C

X

C

C

C

C

C

C

X

C

C

C

C

C

C

X

X

C

C

C

C

C

C

X

X

X

Silver, high-silver alloys

Titanium

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,

8 QQ-B-671, MIL-C 20159, silver solder per QQ-5-561 9 10 11

12 13

Rhodium, graphite, palladium

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

5

14

Gold, platinum, gold-platinum alloys C

12

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

Leaded brass, naval brass, leaded bronze C

C

C

C

C

C

C

X

X

X

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

X

P

C

X

X

X

C

C

C

C

C

X

P

X

P

C

X

X

X

C

C

C

C

C

X

X

X

X

X

C

X

X

X

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

C

C

X

C

P

X

X

X

P

X

P

C

X

X

X

C

F

F

C

P

X

P

X

X

X

X

P

X

X

C

X

X

X

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

F

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

F

Magnesium

Corroded ("active", more cathodic) –

JUVINALL: Machine Design Tab. 9-1 W-269c

Rivets Total exposed area = 100 cm2 Metal plates Total exposed area = 1 m2

JUVINALL: Machine Design Prob. 9-1 W-269

Chromium-plated steel cap screws Total exposed area = 110 cm2 301 Stainless steel plates Total exposed area = 1.5 m2

Electrolytic environment


Drain plug (steel)

Insert rod (magnesium) Oil

Crankcase (steel)

JUVINALL: Machine Design Prob. 9-8d W-287

30 cycles/day, every day 100 N

100 N Steel, 100 Bhn Steel, 300 Bhn

30 mm Latch closed

Latch open

JUVINALL: Machine Design Prob. 9-12 W-287a

Locking plate Arm Geneva wheel

100-mm radius


End of thread

L p

␭

End of thread

L p

End of thread

(a) Single thread–right hand

␭

(b) Double thread–left hand


p p 8

Crest

60⬚ 30⬚

p 4

Root Root (or minor) dia. dr

Major dia. d Pitch dia. dp

Axis of thread


Nut tolerance zone Basic profile (as shown in Fig. 10.2) Nut

Basic profile Screw

Screw tolerance zone


p

p 2

p 2

2␣ = 29⬚

p 2␣ = 29⬚

p 2

0.3p

d

d dr

dm

dr

(a) Acme

dr

(c) Square

d

␣ = 5⬚

p

p

p 2 p 2

dm

(b) Acme stub

p

p 2

dm

0.163p ␣ = 7⬚

5⬚

dm

dr

p 2

d

(d) Modified square


45⬚ 0.663p

dr

dm

(e) Buttress

d

Weight a dc

Force F

(a)

(b)


(c)

w A ␣n

fn

dm q ␭ L q

n cos ␣n

␲dm fn

A Scale 4:1


␣n

n

Section A A (normal to thread)

h ␣n

b tan ␣n = h

b

Screw axis

Section B-B (normal to thread) A h

␭

B

␣ b/cos ␭ ␣ tan ␣ =

A

B

Section A-A (through screw axis)


b h cos ␭

100

f=

90

1 0.0 f=

80

2 0.0 f=

5 0.0 f=

70

0 0.1

Efficiency, e (%)

f= 60

5 0.1

f=

0 0.2

50

40

30 e= 20

cos ␣n – f tan ␭ , where cos ␣n + f cos ␭ 1

␣n = tan–1 (tan 14 2 ⬚ cos ␭)

10 0 0⬚

10⬚

20⬚

30⬚

40⬚ 50⬚ Helix angle, ␭

60⬚


70⬚

80⬚

90⬚

Weight = 1000 lb a dc = 1.5 in.

fc = 0.09 Force F f = 0.12 1-in. double-thread Acme screw


Total = P

Force flow lines

Clamped member

Total = P

Bolt 1 A - shear fracture line for nut thread stripping

2 t

B - shear fracture line for bolt thread stripping

3

A

B dr di dp d

JUVINALL: Machine Design Fig. 10 -11 W-299

Nut

Clamped member

Bolt

Nut


Pilot surface of bolt


Motor

Spur gears Motor

Ball thrust bearings

Material being compressed

Material being compressed

Thrust washers

Thrust washers Ball thrust bearings

(a) Screws in compression (poor)

(b) Screws in tension (good)


Flat washer

(a) Screw

(b) Bolt and nut

(c) Stud and nut


(d) Threaded rod and nuts

0.65d d 1.5d (b) Square head

(c) Round head

(e) Fillister head

(f ) Oval head

(a) Hexagon head

(d) Flat head

(g) Hexagon socket head

(h) Hex socket headless setscrew (i) Carriage bolt

(j) Round head with Phillips socket


(Plug in socket) (a) Conventional screwdriver will tighten but not loosen the screw

(5-sided head)

("Spanner head")

(b) Special tool required to tighten or loosen screw


(c) Break-away heads

T4 T3 Fi ␶

Sy 2

per max ␶ theory

Stresses with torques applied Fi

Fi

T1

␶max =

+␶

y x

Fi T2

␴

Fi

y

y

Stresses after torques are relieved

y'

x'

Fi

+␴

x (␴, –␶)

Mohr circles


Direct tension, galvanized

Direct tension, black oxide

Bolt tension

Torqued tension, galvanized and lubricated

Torqued tension, galvanized (high friction)

Torqued tension, black oxide

Bolt elongation


(a) Helical (split) type

(b) Twisted-tooth type (Teeth may be external, as in this illustration, or internal.)


(a)

(b)


(a) Insert nut (Nylon insert is compressed when nut seats to provide both locking and sealing.)

(b) Spring nut (Top of nut pinches bolt thread when nut is tightened.)


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

Starting

Fully locked

Spring- top nut (Upper part of nut is tapered. Segments press against bolt threads.)

Nylon-insert nuts (Collar or plug of nylon exerts friction grip on bolt threads.)

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

(a)

(b)

(c)


Fe

Fb = Fi

Fc = Fi

Fe (a) Complete joint

Fb

Fe (b) Free body without external load


(c) Free body with external load

Fc

g Fe

Fc

Fb (a)

(b)

Fi

Fc = Fi

Fb

Fc

Soft gasket

b =

Fb

F

Fc Fc

Fb and Fc

F

i a n

d

F

e

Fb

0

Fe (Separating force per bolt)

(c) (d)


"Rubber" portion of bolt

Fb

Fi

Fc

g Fb

Fc

Fb and Fc

Fb

Fb = Fi

Fc = Fi – Fe

Fe Fc = 0

Fb

"O-ring" gasket

0 (a)

(b)

(c)


Fe (Separating force per bolt) (d)

Force

Fb = Fi +

kb F kb + kc e

Fb = Fe

B

Fb

⌬ Fb

Fc

⌬ Fc

Fi

kc Fc = Fi – F kb + kc e

C

A

Fc = 0

0 External load Fe


Fluctuations in Fb and Fc corresponding to fluctuations in Fe between 0 and C

Conical effective clamped volume

d g

30°

d1

d2 d3


(Hexagonal bolt head and nut)

P

Connecting rod and cap a

a A

A P

(a) Bolt bending caused by nonparallelism of mating surfaces. (Bolt will bend when nut is tightened.)

(b) Bolt bending caused by deflection of loaded members. (Note tendency to pivot about A; hence, bending is reduced if dimension a is increased.)


Rotating shaft

9 kN Pillow block Metric (ISO) screw


F 2

F 2 F

F 2

F F 2

(a)

(b)

Normal load, carried by friction forces

Overload, causing shear failure


150 24 kN

500

24 kN D

400

150

D

150 E

100 A


144-kN applied overload F 150

F

144 kN 180

180 100

48 CG of bolt group cross section 144 kN (150 mm) = 21.6 kN m

200

F

200 180

48


48

V

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

40

Bolt yields slightly, with no increase in load or stress during first application of Fe

38.3 Fb

Force (kN)

35.3

Fc

30 29.3

20

Case 1 – Fi = 10 kN 13 Fb

10

9

Fc Fe

4 0 Bolt tight, machine not yet turned on

Machine operating at normal load

Machine turned off

Time (a) Fluctuation in Fb and Fc caused by fluctuations in Fe 900 Su = 830 800 Sp = 660 700 Sp = 600

Stress, (MPa)

600 500

Fluctuation in thread root stress – Case 2, Fi = 35.3 kN 505

400

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

300 12% elongation @ Su specified for class 8.8 200 = 0.0032 @ Sy = 660 on idealized curve

100 0 0

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 Strain, (b) Idealized (not actual) stress–strain curve for class 8.8 bolt steel

JUVINALL: Machine Design Fig. 10-34ab W-323

Alternating stress, ␴a (MPa)

400

300

Sn = Sn' CLCS CG = 830 (1)(1)(0.9) = 373 MPa 2 ⬁ Life

Operation at overload on verge of causing eventual fatigue failure 4

200 130 100 0 0

During normal operation

2

Sy = 660

77

200 400 Mean stress, ␴m (MPa)

Su = 830

600

800 1 After initial 3 tightening After shut-down following normal operation

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


Steel bolt, 1 '' – 13 UNC, 2

Fmax

0 to Fmax

External load Fe

grade 5 with cut threads 0 to Fmax, fluctuating external force

g = 2''

Fe

Steel member

Time (b) Fluctuating separating force versus time

(a) Simplified model of machine members bolted together

SSnn == SS'n' CLCG CS ==

␴a (ksi)

60

40

120 120 (1)(0.9)(1) (1)(0.9)(1) == 54 54 ksi ksi 22

Limiting point for case a Limiting point for case a

␴a = 37

Limiting point for case b Limiting point for case b

␴a = 23 20

␴a = 22.7 Suy = 120 Sy = 92

0 0

20

40

60 ␴m (ksi)

80

(c) Fatigue diagram for thread root

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

100

120

Aluminum cover plate, E = 70 GPa

Class 8.8 steel bolt g/2 g/2

Fe "O-ring" gasket

250 mm 350 mm


Cast-iron cylinder, E = 100 GPa

Weight = 10,000 lb a dc = 2.0 in.

fc = 0.10 Force F f = 0.13 1-in. double-thread Acme screw


1/2 in. Acme thread dc = 5/8 in.

5 in.


Fe

Fe = 0 to 8,000 lb kc = 6kb

Fe


A

Spring washer

A

1000 N 1000 N

1000 N

1000 N

Spring washer

(1)

(2)



230 mm

140 mm 280 mm


80

Fb and Fc

Force (kN)

60

Heavy load 40 Light load

30 kN

Light load Fe

20

15 kN

15 kN

Initial tighten 0 Time


22

22

22

22

22

22

22

22

22

22

22

22

22

22

22

22



Before setting After setting


Full tubular

Bifurcated (split)

Metal-piercing (a) Semitubular

(b) Self-piercing


(c) Compression

Back (blind) side not accessible

Back (blind) side not accessible

Acute corner "Built-up" lightweight structure


Mandrel breaks after seating and rivet expansion Trim or grind mandrel

Pulling head Self-plugging blind rivet Blind side upset

Mandrel head collapses as rivet is expanded and pulled through rivet

Pull-through blind rivet Blind side upset

Closed-end break mandrel blind rivet Blind side upset

Open-end break mandrel blind rivet Blind side upset

Drive pin blind rivet

Blind side upset


(a)

(b)


(c)

60⬚

60⬚

(a)

(b)

(c)


45⬚

(d)

t = 0.707

t

h

h

t

h

h

(a" ) Concave weld bead (poor practice)

(a' ) Convex weld bead (a)

(b)

F

F A

A D

B

D

B

C

C

50 mm

mm 50

F

F

(c) Parallel loading

(d) Transverse loading


(e) Transverse loading

Y 300 100 B

X

A

X

–y G x– 150 20 kN

C Y

20

(a) 295.7 T (45) = t J

80

B

T G

A

B G

802 + 452 J

525.8 t

20 kN T

T (80) 525.8 = J t

80 t

5600 N m •

105 T

1052 + 202 J

T (105) 690.0 = t J

691.9 t

T (20) 131.4 = J t

690.0 t

80 t

131.4 t

(b)

(c)

Torsional stresses

Torsional plus direct shear stresses


295.7 t

674.1 t

70

A X

A

B 60

B D

=

X 120

26.3 t

=

121.2 t

124 Resultant stress = t

C 160 10 kN

(a)

(b) Stresses on weld AB


Y' Y

t

G (CG of total weld group) L /2

X

X

y

b

X'

X' G ' (CG of this weld segment) L /2

a Y'

Y


␴a 30 ␴m = 60 = 0.5

␴a (ksi)

13.6

0

19

50 ␴m (ksi)


62

Removing this material reduces stress concentration at bond edges

(a) Adhesive-bonded metal lap joint

(b) Brazed tubing fittings


(c) Glued wood joint

F

F

15 mm


Weld length = 90 mm Sy = 400 MPa SF = 3 E70 series weld

F

3.0 in.

E60 series welding rods Sy = 50 ksi (plates) h = 0.375 in. SF = 3 Note: There are two 3 in. welds.

F JUVINALL: Machine Design Fig. P11-07 W-349

60 kN 100 mm 75 mm 55 mm

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


4 in.

4000 lb

3 in.

Note: There are two 4 in. welds.


Fixed end

Bearing Torsion bar portion

Spline Generous radius Bearing

d Spline

(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.)


FF

F

␭

F

F d T= D

FD 2

d

F

T= D

F

F

D F d End surface ground flat

(c) Tension spring

(a) Compression spring (ends squared and ground) JUVINALL: Machine Design Fig. 12-2 W-353

(d) Top portion of tension spring shown as a free body in equalibrium

FD 2

=

Tc J

T

0

(a) Straight torsion bar

c

a T

Plane m

n T

= Tc J

Plane

0

d

⬎

b

Tc ⬎ J

(b) Curved torsion bar


Tc J

T

18

1.8

Ks = 1 + 1.7

0.5 (shear correction only, C use for static loading)

16

1.6

14

1.5

12

Kw and Ks

KwC KsC

1.4

10

8

1.3

Kw

1.2

Ks

1.1

6

Preferred range, ends ground

4

Preferred range, ends not ground 1.0 2

4

6 8 10 Spring index, C = D/d


12

2 14

KwC and KsC

Kw = 4C – 1 + 0.615 (shear and curvature C 4C – 4 corrections, use for fatigue loading)



Wire diameter (in.) 0.004

0.008

0.020

0.040

0.080

0.200

0.400

0.800

450 3000

2500

350 ASTM A313 (302)

ASTM A228 music wire (cold-drawn steel) 300

2000

ASTM A401 (Cr-Si steel)

ASTM A229 ASTM A227

250

ASTM A230 (oil-tempered carbon steel)

1500

ASTM A232 (Cr-Va steel) 200

Inconel alloy X-750 (spring temper)

ASTM A229 (oil-tempered carbon steel) 150

1000

ASTM A227 (cold-drawn carbon steel)

ASTM B159 (phosphor bronze)

100

ASTM A313 (302 stainless steel) 500

50

0 1 0.10

2

3

4

5 6 7 8 91 1.0

2

3

4

5 6 7 8 91 10.0 Wire diameter (mm)


2

3

4

0 5 6 7 891 100.0

Minimum ultimate tensile strength (ksi)

Minimum ultimate tensile strength (MPa)

400

(a) Ls = (Nt + 1)d

(b) Ls = N t d

(c) Ls = ( Nt + 1)d


(d) Ls = N t d

(a) Contoured and plate

(b)

(c)


(d)

0.75 0.7 0.65 Ratio, deflection–free length, ␦/L f

0.6 0.55

Unstable

0.5 0.45 0.4 0.35 A

0.3 B

0.25 0.2 Stable

0.15 0.1 0.05 2

3

4 5 6 7 8 9 10 Ratio, free length–mean diameter, L f /D

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 )


11

60 lb

D+d ⬍ 1.5 in.

105 lb 1 in. 2

(Spring free)

Lf

(Spring with ⬍ 2.5 in. min. load)

(Spring with max. load)


Cash allowance

Fs

(Spring solid)

Ls

80 70

0.9Sus ⬇ 0.72 Su

Stress S (% Su)(log)

60 0.54 Su 50 0.395 Su

40

Sn = Su CLCS CG 2 Su = (0.58)(1)(1) 2 = 0.29 Su

30

20 103

104



106

0.8

␶m = 0

0.72

0

␶a ␶m = 1

0.7

␶a /Su

0.6

0.54

0.3 0.2

Region of interest

103⬃

0.5 0.4

0

␶a ␶m ⬍ 1

104⬃

0.395

(0.38, 0.38) 0.29

105⬃

106 + ⬃

0

(0.325, 0.325)

P Static load

0.1

0

␶a ␶m = 0

(0.265, 0.265) (0.215, 0.215) 0.1

0.2

0.3

0.4

0.5

0.6

0.7 ␶m /Su

0.8


0

0.80

0.76 0.65

103 ⬃ 104 ⬃

0.53

0.60

␶max /Su

0.43

0.40

105 ⬃ 106 + ⬃

P

0.20

0

0.20

0.40 ␶min /Su

0.60


0.80

Torsional stress Ss, max (% Su)

80

76

Calculated curve (from Fig. 12.14) 65

70

Note: For zero-to-max torsional stress fluctuation

60

53

50

43

Shot-peened wire 40 Design curves [1] Non-shot-peened wire 30 103

104

105 Life N, (cycles)


106

107

1000

(965, 965)

800 800

Infinite life with shot peening (862, 862)

689

Infinite life without shot peening

␶max (MPa)

510 600

(Static load line) 400

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

200

0

200

600 400 ␶min (MPa)

800


1000

Load stress Load stress plus residual stress +␶

Without presetting With presetting

0 –␶

␶resid from presetting


600 N

300 N

Squared and ground ends, ASTM A232 spring wire

25 mm 600 N 300 N

Cam

n= 650 rpm

Shaft Key

+ 25


1200

1000

␶max (MPa)

800

975

750

600 540 400 ␶max 600 = ␶min 300 200

0

200

400

600 ␶min (MPa)

800


1000

1200

F

F

d

r1

d

A r3

r2 r4

B

D D Bending stress at Sec. A: 16FD r1 ␴= r3 ␲d3

Torsional stress at Sec. B: 8FD r4 ␶= ␲d3 r2


F


2F L

F

F

L

L

F

L

L

2F = =

6FL bh2 6FL3 Ebh3

(a) Quarter-elliptic (simple cantilever)

2F = =

6FL

=

bh2 6FL Ebh

3

3

(b) Semi-elliptic


=

6FL bh2 12FL3 Ebh3

(c) Full-elliptic

=

Mc 6Fx = I wt 2

If and t are constant, then x /w must be constant.

L

If and w are constant, then x /t 2 must be constant.

L b

b

x b

F F w

h

F h

h L

t

(a)

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


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

F

L

F

L

h

h n leaves Half of nth leaf Half of 3rd leaf Half of 2nd leaf b

Main leaf Half of 2nd leaf Half of 3rd leaf

b n

Half of nth leaf

(a)

(b)


Clips Fixed pivot

Shackle (permits small fluctuations in spring length)

Fixed pivot


800 F = 1000 to 5000 N 2F Bolt, Kf = 1.3

Sn ␴a (MPa)

F

␴a ␴m = 0.67

life, bending

Design overload point

␴a = 525 400

h = 7 mm k = 30 N/mm

Sy 0

(a)

400

800 ␴m (MPa) (b)


1200

Su 1600

=

b = 416

6FL3 Ebh3

= 0.0144

FL3 Eh3

;

k=

F Eh3 = 69.33 L3

L = 682 (a) Triangular-plate solution to Sample Problem 12.4

L = 682

L = 682 L = 682 83

b = 416

Eh

3

L3

+

b = 333

1 =

k = k1 + k2 k = 76.30

=

6FL3 Ebh

3

= 0.0180

FL3 Eh

3

83

2 =

4FL3 Ebh

3

= 0.0482

3

k1 =

F Eh = 55.55 1 L3

(b) Trapezoidal plate solution to Sample Problem 12.5


k2 = 20.75

Eh

3

L3

FL3 Eh3

F

F d

a D


Thickness = h

F

a


F

b

Factors for inner surface stress concentration Ki, round and Ki, rect

1.6

1.5

1.4

1.3

1.2

Ki, round Ki, rect

1.1

1.0 2

4

6

8

Spring index, C =

10 D D or d h


12

Belleville

Wave

Slotted

Finger


Curve

Internally slotted (as used in automotive clutches)

In series

In parallel


In series-parallel


(b) Electric motor brush spring (a) Constant-force extension springs

+

+

+

+

Storage drum

Output drum

Storage drum

Output drum

(c) Two forms of constant spring motors


Door stop

Center of gravity of 60-lb door 24 in.

110⬚ Torsion bar

End attached to door Fixed end


Force

Threaded bolt Nut B Support Force

C A

Spring Deflection Deflection Retainer


F = 3.0 kN di do

Do = 45 mm do = 8 mm No = 5

Di Do


Di = 25 mm di = 5 mm Ni = 10

F

F = 45 to 90 lb Deflection = 0.5 in. D = 2 in.

F

Squared and ground end


Valve acceleration

3600 engine rpm 1800 camshaft rpm

+ 0

Cam angle

– "Reversal point" "Reversal point" Valve lift is 0.201 in.

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


Adjusting nut Cap screw

Spring

Support

Roller follower

Stationary guide Oscillating assembly

Cam

Key

Shaft


F e


Stationary support

Handle

Tire 35 mm Pivot A

Brake shoe

Spring

Stationary support

Pin stops


110 mm dia.

25 mm shaft Cable 110 mm dia.

Torsion springs

Cable


Connecting rod Connecting rod bearing Thrust bearing (flanged portion of main bearing)

Main bearing

Main bearing Crankshaft

Main bearing cap Connecting rod bearing cap


(a) Hydrodynamic (surface separated)

(b) Mixed film (intermittent local contact)


(c) Boundary (continuous and extensive local contact)

Oil inlet

Bearing Journal Oil flow

W

W

W

e

W (a) At rest

Minimum film thickness, h0 W (b) Slow rotation (boundary lubrication)


Resultant oil film force W

(c) Fast rotation (hydrodynamic lubrication)

Boundary lubrication

f Mixed-film lubrication Hydrodynamic lubrication

A ␮n/P (viscosity × rps ÷ load per unit of projected bearing area)


T

Rubber element Fluid element

Cross-sectional area, A

(a)

␦

Surface velocity, U

F

F h

h

Fh AG where G = shear modulus

Fh A␮ where ␮ = absolute viscosity

(b)

(c)

At equilibrium, torque T produces elastic displacement, ␦, across a solid element

At equilibrium, torque T produces laminar flow velocity, U, across a fluid element

␦=

U=


Temperature (°F) 80

100

120

140

160

180 200

220 240 260 280

104

103

5 3 2

3 2

103

102

5 5

3 2

3 2

102

5 4 3

5

40

4 3

20

2

2 10 1.0 0.9 5

0.7 0.6

4

0.5 3 0.4

2 10

0.3 20

30

40

50

60 70 80 Temperature (°C)

90


100 110 120 130 140

Absolute viscosity (␮reyn)

Absolute viscosity (mPa • s)

10 SA E 60 70 50

Overflow rim

Level of liquid in bath Saybolt viscometer

Oil

Bath

Kinematic viscosity, 58 s at 100°C

Bottom of bath Orifice


n

L

R

c

D


5000 N Oil viscosity, 50 mPa • s D = 100 mm n = 600 rpm

R = 50 mm

c = 0.05 mm L = 80 mm


Oil hole W

Partial bronze bearing

Oil level

Journal


Fixed bearing

Lubricant Rotating journal

W

(␶ + ∂␶ dy) dx dz ∂y

Lubricant flow

dx dy

Coordinates: x = tangential y = radial z = axial

p dy dz


dy dx ␶ dx dz

(p +

dp dx) dy dz dx

Rotating journal

U

h

Lubricant flow

y

Stationary bearing


0

1.0

0.1

0.9 L /D = ∞

0.2

1

0.7

1 2

Max.load

0.6

0.3 0.4

Min. friction

1 4

0.5

0.5

0.4

0.6

0.3

0.7

0.2

0.8

0.1

0

0.9

Optimum zone 0.01

0.02

0.04 0.06 0.08 0.1

0.2

0.4

Bearing characteristic number, S =


0.6 0.8 1.0 R c

2

␮n P

2

4

6

1.0 8 10

Eccentricity ratio, e/c

Minimum film thickness variable,

h0 c

0.8

200

100

Coefficient of friction variable,

R f c

50 40 30 20 L /D =

1 4

10 1 2

5 4 3

1

2

∞

1

0

0.01

0.02

0.04

0.08 0.1

0.2

0.4

0.6 0.8 1.0



R c

2

␮n P

2

4

6

8 10

1.0

P pmax (gage)

0.8

Minimum film pressure ratio,

0.9

0.6

L /D = ∞

0.7

1

0.5 0.4

1 2

0.3

1 4

0.2 0.1 0

0.01

0.02

0.04 0.06

0.1

0.2

0.4 0.6 0.81.0



R c

2

␮n P

2

4

6

8 10

100

Position of minimum film thickness, ␾ (deg*)

90 80 L /D = ∞

70 60

1

50 40

1 2

30

1 4

20 10 0

0.01

0.02

0.04 0.06

0.1

0.2

0.4 0.6 0.8 1.0

Bearing characteristic number, S = * Defined in Figure 13.20


R c

2

␮n P

2

4

6

8 10

100

25

L /D = ∞

1 2

1

20

max

70 60

1 4

15

50 1 2

40

10

30 1 5

20 ␪p

0

10

␪p

∞

0.04 0.06 0.08 0.1

0.2

max

0

0.01

0.02

0.4

Bearing characteristic number, S = * Defined in Figure 13.20


0.6 0.8 1.0 R c

2

␮n P

2

4

6

8

0

Position of maximum film pressure, ␪p

Terminating position of film, ␪p (deg*) 0

80

(deg*)

1 4

90

6

L /D =

1 4

L /D =

1 2

5 L /D = 1

Flow variable,

Q RcnL

4

L /D = ∞

3

2

1

0

0.01

0.02

0.04

0.1

0.2

0.4 0.6 0.81.0



R c

2

␮n P

2

4

6

8 10

1.0 L /D = 0.9

1 4

0.8 1 2

Flow ratio,

Qs Q

0.7 0.6 0.5 1

0.4 0.3 0.2

L /D = ∞

0.1 0

0.01

0.02

0.04

0.1

0.2

0.4 0.6



R c

1.0 2

␮n P

2

4

6

8 10

W

n

Oil of viscosity ␮ and flow rate Q

e

R = D/2 h0

␪p

␾

0

␪p

max

pmax

Film pressure, p Average film pressure = P = W DL


1000 lb SAE 20 Oil Tavg = 130°F D = 2.0 in. n = 3000 rpm

R = 1.0 in.

c = 0.0015 in. L = 1.0 in.


Steel backing Bearing material Oil hole Axial groove


Ungrooved bearing Oil film pressure

Grooved bearing

Oil inlet hole

L 2 Circumferential groove

L 2

D


Piston Piston pin or wrist pin "Rifle drilled" passage in connecting rod* Oil in

Oil in

Circumferential groove in main bearing

Drilled passage in crankshaft Circumferential groove in rod bearing Circumferential groove in main bearing *If omitted, piston pin bearing is splash lubricated.


W = 17 kN Force feed, SAE 10 oil Tavg = 82°C D = 150 mm n = 1800 rpm

R = 75 mm

c = ? mm L = ? mm f=?

Power loss = ?

Qs = ?

Oil temperature rise = ?


0.006 0.005 0.004 0.003

Max. load

h0, minimum oil film thickness (mm)

f, coefficient of friction

0.007

0.015

0.010

200

Q 150

100

Qs f

0.005

50

0.002 0.001 0

Optimum band* 0 0

0.05 0.10 c, radial clearance (mm)

*As defined in Fig. 13.13


0.15

Q and Qs, oil flow rate (cm3/s)

h0

0.009 0.008

Min. friction

0.020

0.010

Runner

Ro Ri

a>b a

b

Pads


SAE 10 Oil Tavg = 150°F D = 4.0 in. n = 900 rpm

R = 2.0 in.

c = 0.002 in. L = 6.0 in.



4.5 kN SAE ? Oil Tavg = ?°C D = ? in. n = 660 rpm

R = ? in.

c = ? in. L=?

L=?

JUVINALL: Machine Design Fig. P13-29D W-425

1

2

3 (b) Steps in assembly


4

Extraextralight series

Extralight series

Light series

Medium series

(LL00) (L00)

(200)

(300)

(a) Relative proportions of bearings with same bore dimension

Extraextralight series

Extralight series

Light series

Medium series

(LL00)

(L00)

(200)

(300)

(b) Relative proportions of bearings with same outside diameter


Loading grooves

(a) Filling notch (loading groove) type


(c) Double row


(d) Internal self-aligning

JUVINALL: Machine Design Fig. 14-3d W-428

(e) External self-aligning

JUVINALL: Machine Design Fig. 14-3e W-428

One shield

Snap ring and one shield

Two shields

One seal

Snap ring and two shields

Two seals

Snap ring and one seal


Shield and seal

Snap ring and two seals

Snap ring

Snap ring shield and seal

Added stabilizing ring (b) One-direction locating

(c) Two-direction locating


Spherical roller head Crowned roller body

Ro ller axi s Bearing axis Common apex


(d) Idler sheave (unground bearing)

(e) Rod end bearing


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

JUVINALL: Machine Design Fig. 14-10h W-437

+ r dS

+ r


dH

24 22

Percentage of failures

20 18 16 14 12 10 8 6 Median

4 2 1

2

3

4

5

6 7 Life

8

9


10 11 12

1.0

Life adjustment reliability factor Kr

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

90

91

92

93 94 95 96 Reliability r (%)

97


98

99 100

Radial bearing

1800 rpm

Angular bearing

1800 rpm

Ft = 1.5 kN, Fr = 1.2 kN Light-to-moderate impact Eight hours/day operation


Case a: 90 percent reliability

Case b: 30,000-hour life No. 211 radial-contact bearing, C = 12.0 kN

1800 rpm

Fr = 1.2 kN Ft = 1.5 kN

100 mm 55 mm

Light-to-moderate impact, Ka = 1.5

21 mm


Radial load (kN)

7 6 5 4 3 2 1 0

Fr 1000 rpm

1

50%

2

30% 100%

3

No. 207 radialcontact bearing Ka = 1.0 (uniform load)

1

20%

Time


Note spacers



F1

F2

F2

L1

2L1

3L1


3500 rpm

No. 204 radial ball bearing Fr = 1000 N, Ft = 250 N 90% reliability Light-moderate shock loading L = ? hr life


Gear

Shaft


Bearing

A 150-mm dia. 1.2 kN 20°

60 mm

B 300 mm 60 mm

100 mm

Gear 120-mm dia.


Chain sprocket Clamping supports

2.25 ft

Rotating shaft

1.75 ft

Chain

0.75 ft

600 lb

JUVINALL: Machine Design Prob. 14-21d W-450

600 lb

Right-angle gearing

Parallel gearing



Driven gear

Line of centers

Point of contact Pitch point P

Driving gear

Common normal (to tooth surfaces at point of contact)


Involute curves Base pitch, pb Base circle


␻p

Pinion pitch circle

dp

c P (pitch point)

dp


␻g

Gear pitch circle

␻p Pinion base circle

a

␾ (pressure angle)

P b

Gear base circle


␻g

Pinion

Base circle Pitch circle

rp ␾ a

d g c

P

Pitch circle

f b

Base circle

e

␾

rg

Gear


Pinion (driving)

Dedendum circle rp

Base circle Pitch circle 0p

Addendum circle Angle of approach

Angle of recess

Dedendum Addendum Position of teeth entering contact Addendum

Position of teeth leaving contact

n a p

c ␾ b

Angle of approach Dedendum

Angle of recess

n

Addendum circle Pitch circle Base circle Dedendum circle rg

Gear (driven)

0g


To p

la nd

Fa ce

wi dt h

b

t0

Fa ce

Addendum ci rcle

Dedendum

Tooth thickness t

Width of space

Pitch

circle

la nd

Whole depth

Fl an k

Working depth

Circular pitch p

Bo tto m

Addendum

Clearance Fillet radius

Dedendum circle Clearance circle (mating teeth extend to this circle)


6

7

18

20

28

5

80

24 10

26

12

11


14

30

48

32

22

8

16

36

64

9

40

4

Rack

␾ p Circular pitch

Pinion


Pitch circle ␾

␻

Dedendum circle

Base circle Base circle

␻

Pitch circle


Addendum circle

Gear blank rotates in this direction

Rack cutter reciprocates in a direction perpendicular to this page


01

␻1

Driven gear

␾

Base circle (This portion of profile is not an involute)

Pitch point, P

␾

b

Addendum circles

a

Interference is on flank of driver during approach Base circle (This portion of profile is not an involute)

␾

Driving gear ␻2

02


rp P = 6 teeth/in. ␾ = 20° ␻p ␻ = –3.0 g

c = 4 in. rg


(Driving pinion rotates clockwise) ␻p Pitch circle (pinion)

dp

Ft

P ␾ Fr

F Fr

F ␾

P

dg ␻g


Ft

Pitch circle (gear)

b N = 36 teeth (idler)

c N = 28 teeth (output gear) (a)

a N = 12 teeth (input pinion) P=3

600 rpm; 25 hp

Resultant force (applied by shaft to gear) = (1313 + 478)

2 = 2533 lb.

20°

Hcb = 478 lb

45°

(b) c

b

20° a

Hab = 1313 lb Vab = 478 lb


Vcb = 1313 lb

F

Fr

Ft F h b rf a Constantstrengh parabola

x t


0.60 0.55

␾=

Lewis form factor Y

0.50 ␾

0.45

25

°

th tee ub t s 0°, =2

␾=

20

°

0.40 0.35

␾=

14

1° 2

0.30 0.25 0.20 0.15 12

15

17

20 24 30 Number of teeth N

35 40 4550 60 80 125 275 ∞ (Rack)


Load

1 revolution Fa = Fm = F 2

F

Time Driving and driven gears

Load

0

Fa = F Fm = 0

F

Driving gear

Idler gear

0 Time F

1 revolution Idler gear

Fat. strength for reversed loading (idler)

40% higher fat. strength for o-max loading (driver and driven) ␴a ␴a

␴m

Sn

␴a

0

␴m (b) Stress fluctuation


Su

Driven gear

0.60 1000

0.55 r gea ting a in m th tee f o er mb Nu

0.50 Geometry factor J

85 50 35 25 17

0.45 0.40

Load applied at highest point of single-tooth contact (sharing)

0.35 Load applied at tip of tooth (no sharing)

0.30 0.25 0.20 0.15 12

35 15

17

20 24 30 Number of teeth N

45

60

40 50

125 275 ∞

80

(b) 20° full-depth teeth 1000

0.60 0.55

Geometry factor J

0.50 er mb Nu

0.45

ear gg atin m h in eet of t

Load applied at 85 highest point of 50 single-tooth 25 contact 17 (sharing)

Load applied at tip of tooth (no sharing)

0.40 0.35 0.30 0.25 0.20 0.15 12

35 15

17

20 24 30 Number of teeth N (b) 25° full-depth teeth


45 40 50

60

125 80

275 ∞

Pitch line velocity V (m/s) 0

10

20

30

5 D

cu

tte

rs

*

4

or

m

3

,f

s,

Hob

C

nd and grou , shaved Precision d shaved and groun High precision,

2

1

pin sha

*

tters

g cu

bs Ho

Velocity factor Kv

E

B A

d Highest precision, shaved and groun

0

1000

2000

3000 4000 5000 Pitch line velocity V (ft/min)

* Limited to about 350 Bhn JUVINALL: Machine Design Fig. 15-24 W-474

6000

7000

Electric motor 1720 rpm

Conveyor drive (involves moderate shock torsional loading) 860 rpm

Pinion P = 10 Np = 18 (teeth) 20° full-depth teeth Steel, 330 Bhn

Gear Steel, 290 Bhn (manufacture of pinion and gear corresponds to curve D, Fig. 15.24)


Gear (driver)

Common normal

Gear (driver)

Common tangent

Vpt Sliding velocity Vp

Vgt

Common normal

Vpt = Vgt Vp = Vg ␾

Vg Vgn = Vpn

Vgn = Vpn

Common tangent Pinion (driven)

Pinion (driven)

(a) General contact position sliding velocity as shown

(b) Teeth in contact at pitch point no sliding


2.0 1.8 1.6 1.4 1.2 CLi 1.0 0.8 0.6 104

105

106

107 108 Surface fatigue life (cycles)


109

1010

1011

•

Win = 100 hp 3600 rpm

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

900 rpm


p1 Motor (input)

g2

a

c

Driven machine (output)

b g1

p2


Ring R

R Planet Arm Sun

P

A

P

P

A

S

S

P

(a) With three planets (typical)

(b) With one planet (for analysis only)


2Ti /3R

Ti 2Ti

R

3R

4Ti /3R P

R/2 2Ti /3R

2Ti 3R

4Ti R + S 4 R S = Ti 1 + R

To =

4Ti /3R R+S 4

4Ti

A

2Ti 3R

4Ti 3R

3R To Ti

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


=

␻i ␻o

=1+ S R

␻i V

P R 2

V 2

R+S 4

␻0 A S

R

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


␻i ␻0 =

V R/2 V/2 (R + S)/4

␻i S ␻0 = 1 + R

1

90° = 17 2 teeth (ring)

R Planet will fit here P

S Planet will NOT fit here

Planet will NOT fit here

Planet will fit here P


90° = 5 teeth (sun)

45 teeth c P = 5, ␾ = 25° 15 teeth

Driven machine coupled to this shaft

1 kW, 1200-rpm motor coupled to this shaft B a A b 25 mm

45 teeth

100 mm

25 mm


To driven machine

36T B 24T, P = 6 64T 2'' A

2''

8''

18T, P = 9

Coupled to 20 lb.in. torque motor

JUVINALL: Machine Design Prob. P15-23 W-482A

Motor

Driven machine

a a


1700-rpm motor 100-lb.in. torque

16T

32T 2.0''

1'' 90°

A B


24T

To driven machine

4 5

1

9

3

2 8

Low (L)

7

Neutral (N) High (H)

6 7

4 Hub 4 can "overrun" in this direction

3 Member 2 pushes here to disengage pawl 3

(a)

(b)


P2

P1 Arm

Output

S1 Input

S2 P2

P1

JUVINALL: Machine Design Prob. P15-47 W-485A

P1 (101 teeth) P2 (102 teeth) Arm S2 (99 teeth) Output

Input S1 (100 teeth)

P1

P2

JUVINALL: Machine Design Prob. P15-50 W-485B

P3 (33 teeth)

P1 (27 teeth) P2 (24 teeth)

S1 (27 teeth) S2 (30 teeth)

S3 (21 teeth) Input

Output

Reverse brake band holds S2 fixed

Low brake band holds S3 fixed


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


pn

b

d

Section NN (normal plane) b

␾

␺

N

p

R

R

␾n

a

pn

␺

pa

e c

p N


Section RR (in plane of rotation)

N

R

␺ R

N

Re =

Section NN (normal plane)

d Section RR (plane of rotation)


d 2 cos2 ␺

Spur gear (helical gear with ␺ = 0⬚)

Helical gear Ft

Pitch cylinder

Fr

Pitch cylinder Fa

Ft

Ft

␾

F

␾

Fr

Fr

Section RR (plane of rotation) N

␺

Ft

Isometric view showing helical gear forces

Ft ␺ R Top of tooth

N

Fb

R Fa

Top of tooth

d Fr F ␾n Fb

Section NN (normal plane)


Fr Ft

Fa

Input shaft of driven machine 600 rpm

= 30⬚ (left hand)

Direction of rotation Np = 18 (teeth) Pn = 14 (normal n = 20⬚ plane)

Electric motor hp 1800 rpm

1 2

= 30⬚ (right hand) (a)

(b) Isometric view of motor shaft and pinion


0.70

500 150 60 30 20 18 16 14 12

0.50

0.40

Number of teeth

Geometry factor J

0.60

0.30 0⬚

5⬚

10⬚

15⬚ 20⬚ Helix angle ␺

25⬚

30⬚

35⬚

500 150 75 50

1.00

30 20

0.95

0.90 0⬚

5⬚

10⬚

15⬚ 20⬚ Helix angle ␺


25⬚

30⬚

35⬚

Teeth in mating gear

J-factor multiplier

1.05

Pitch cone angles

Pitch cone length, L ␥p b

Pinion back cone

Pinion pitch ␥p cone

r bp

dp

Dedendum

Gear pitch cone

Addendum

Dedendum

Gear pitch dia., dg Gear back cone

Developed back cone radius, r bg


Circular pitch Face advance

Mean radius

␺


Spiral angle

Fn F b

Ft

b 2

␾ Fr

Fr

Fn Fa

Ft d 2 dav 2

␥


Note: Fn is normal to the pitch cone.

0.40 100

0.38 0.36 80

Geometry factor J

0.34 0.32

Teeth in mating gear

0.30

60

50

0.28

40

0.26

30

0.24

20

0.22 15

0.20 0.18 0.16 0

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


65 mm 140 mm

Load D

B

Rotation

Motor

A

C

250 mm

140 mm


0.36 100 Geometry factor J

0.32 80 60

0.28

50 0.24 40 0.20

30 Teeth in mating gear 12 15 60 80 20 40 Number of teeth in gear for which geometry factor is desired

25 20

0.16 0


100

0.11

Ng = 100 90

0.10 80

Geometry factor I

0.09

Ng = 70 60

0.08

Ng = 50

0.07

40 20

0.06 15 0.05 0

10

25

30

Teeth in gear

20 30 Number of teeth in pinion NP


40

50

0.18

0.16

Geometry factor I

Ng = 100 0.14

80

0.12

60

0.10

50 40 0.08 30 0.06 0

15 10

20

25

20 30 Number of teeth in pinion NP


Teeth in gear 40

50

P

P (planet) R R (ring) S (sun)

S

␻R

␻S A

P Fixed arm, A P


Arm (input member)

P S

R

Left axle

Right axle

P


Lead, L Axial pitch, P Worm outside dia., dw, out

Worm lead angle, ␭, and gear helix angle, ␺

Pitch dia., dw

Center distance c Pitch dia., dg Keyway

Note: ␭ and ␺ are measured on pitch surfaces.

Face width, b


Worm-driving torque

Fwt

Fwa Fwr Fgr

Fgt

Fga


f Fn cos f Fn cos

Fn cos n sin Fn

f Fn sin

Direction of

Fn cos n cos

Fgt and Fwa

f Fn

f Fn sin

Fn cos n cos

Direction of Fgt and Fwa

Fn cos n sin

Direction of Fga and Fwt n

Fn

Fn sin n

Fn cos n

Direction of Fga and Fwt

Fn sin n

Direction of Fgr and Fwr

Fn cos n

Fn n Direction of Fgr and Fwr

(a) Worm driving (as in Fig. 16.20)

Fn cos n

(b) Gear driving (same direction of rotation)


Vs ␭

Vg

Gear rotation

Vw Worm Worm rotation Gear


0.14

Coefficient of friction f

0.12

0.10

0.08

0.06

0.04

0.02

0 0 1

2

4 6

10

2

4 6 100 2 4 6 1000 Sliding velocity Vs (ft /min)


2

4 6 10,000

Worm: Steel, hardened and ground Nw = 2, RH, p =

5 8

Motor 2 hp., 1200 rpm

1

in., ␾n =14 2 ⬚

Worm Gear

60 rpm Gear: Bronze


c = 5 in.

80

Coefficient C

ft lb

min ft2 F

70 With fan (as in Fig. 16.26)

60 50 40

Without fan 30 20 10 0 0

400

800 Worm rpm, nw


1200

1600

Worm: hardened steel rpm =1200

Worm Gear

c ≈ 6 in. Gear: Chill-cast bronze

Speed ratio, 11:1



␺ Pa b

p


24 teeth

40 teeth

Output

20 teeth


Input

60 teeth

100 50 teeth 200 125

25 teeth ␺ = 0.35 rad right hand

B Motor

A 20 teeth ␺ = 0.50 rad left hand 50 teeth Output


Pinion

Gear 1000 rpm

400 rpm


Bevel gears: 35 hp Np = 36 Ng = ? b = 2 in. ␾ = 20⬚ P=6

1500 rpm

Pinion A

B

Flexible coupling to machine

Gear 3 in.

3 in.

JUVINALL: Machine Design Fig. P16-23D W-520

Bevel gears: 50 hp Np = 30 Ng = 60 b = 3 in. ␾ = 20⬚ P=6

Worm Gear

c = 8 in.

Nw = 2 Ng = 55 P=?


1200 rpm

Gear material: Chill-cast bronze Worm material: Hardened steel Nw = 3 p = 0.5 in. f = 0.029 Ng = 45 ␾n = 20⬚

c = 4.5 in.

b= 1.0 in.


w 2

h 2

w

h w

w

d

d

w ≈ d/4

w ≈ d/4;

(a) Square key

Key usually has drive fit; is often tapered

h ≈ 3w/4

(b) Flat key

(c) Round key

Keys are tapered and driven tightly; for heavy-duty service

Widely used in automotive and machine tool industries

(d) Kennedy keys

(e) Woodruff key

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

Key is screwed to shaft; hub is free to slide axially – easier sliding is obtained with two keys spaced 180° apart

( f ) Gib-head key

(g) Feather key


d

D (a) Straight round pin

(b) Tapered round pin (c) Split tubular spring pin

Grooves are produced by rolling, and provide spring action to retain pin (d) Grooved pin


Basic

Inverted

Basic

Inverted

Internal rings (fit in housing) E-ring External rings (fit on shaft) (a) Conventional type, fitting in grooves

I

I

I

Section II

External ring (fit on shaft)

I

Section II

Internal ring (fit in housing)

(b) Push-on type – no grooves required Teeth deflect when installed to "bite in" and resist removal (less positive than conventional type)


4-spline

6-spline

10-spline

16-spline

(a) Straight-sided


(b) Involute

(a) Single mass Mass, m

␦st

Gravitational force, w

Shaft of spring rate k = w/␦st


(b) Multiple masses ␦1

m1 w1 m1

w1

␦2

␦3

m2

m3

w3

w2 m2

m3

m4

w2

w3

w4

m5


w5

(c) Shaft mass only

␦st


Fc (= 2500 N) 30° 50

300

50

Chain

60

FT (= 1000 N) Chain sprocket 100 diameter

Track d T1 A

Track sprockets

T2 B

Track

C

Track sprocket 250 diameter (a) General arrangement


A

B T2

T1

C

S

d

55

Small gap or spring washer

(b) Shaft layout

Vertical forces A

T1

2490

S

T2

C

T1

B 325

Horizontal forces

687.5

A 2165

50

55

245

50

S 500 (Ft /2)

T2

B C

500 (Ft /2) 937.5

60

VV

1250

VH

113,700 95,800

130,000 MH

MV 80,300 125,000 62,500 Torque (c) Loading diagrams

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

␴ea (MPa)

200

"Design overload" point

165

100

␴ea ␴em = 2.9

0

100

530 200

75,000 90,600

300 ␴em (MPa)

400

(d) Fatigue diagram

JUVINALL: Machine Design Fig. 17-6b-d W-530

450 500

d/8 d/4 d

(a) Loosely fitted key

(b) Key tightly fitted at top and bottom

d/4

d

(c) Shear failure of a tightly fitted key


Sled-runner keyway

Steel

Profiled keyway

Fatigue stress concentration factor, K f * Bending

Torsion

Bending

Torsion

Annealed (less than 200 Bhn)

1.3

1.3

1.6

1.3

Quenched and drawn (over 200 Bhn)

1.6

1.6

2.0

1.6

* Base nominal stress on total shaft section.



Bonded rubber element

(a) Basic shear-type coupling

(b) Constant-stress, constant strain shear coupling

(c) Tube form shear coupling


(a) Basic Oldham type

(b) Modified type



Flexible coupling Motor

0.25-in.-dia. shaft

20 in.


50 kg

25-mm dia.

600 mm

600 mm


120 lb 80 lb 2-in.-dia. shaft

20 in.

40 in.


30 in.

Driven machine Motor

Coupling (2) Shaft

Bearing (2)

(a) Connecting shaft

Electric generator rotor Driver 56 kw

Bearings Shaft Driven 28 kw Bearing (2)

Hydraulic turbine

Driven 28 kw

(c) Hydroelectric generator shaft

(b) Gear input shaft

Bearing (2) Driver

Shaft

Idler

Driven

(d) Idler gear shaft

Needle bearing (2) Shaft Driver

Driven

Shaft

Driver (e) Gear countershaft

Shaft

Driven ( f ) Stationary countershaft

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

Fr = 450 lb Fa = 400 lb

3 in. rad Bearing A Ft = 800 lb

5 in. 2 in.


Bearing B

Fr = 2.4 kN Ft = 4.0 kN A

B

Fa = 1.5 kN

Note: Gear forces act at a 75-mm radius from shaft axis.

125 mm

50 mm


Chain sprocket

Chain sprocket Clamping supports

Clamping supports

Rotating shaft

Stationary shaft

(a)

(b)


Friction lining material

Driving shaft

dr Ring element subjected to clamping pressure, p

Driven shaft

ri ro

Provision for axial movement


r

Flywheel

Clutch plate (driven disk) Pressure plate

Friction planes

Spring Cover

Engine crankshaft

Release bearing

Housing

To transmission

To release Release lever


Disks a – driving disks (4 disks, 6 friction surfaces) Disks b – driven disks (3 disks, 6 friction surfaces) Seals Oil chamber (pressurized to engage clutch) Piston

Bushing

Oil passage

Output

Key

ri ro Input

Key

Oil passage


␣ Cone angle

Cone Key

dr sin ␣

Spline (sliding fit) dr

r

ro

ri

␣ F

Spring Cup

Shifting groove

Local pressure = p (b)

(a)


c

Av

F

F

b

Lever

Ah a

Shoe (block)

A N

Direction of rotation r

A

fN

(b) Shoe and lever as a free body

O Drum

b (a) Brake assembly

a

N fN Inertial and/or load torque, T

Oh Ov

O

(d) Lever proportions for a self-locking brake

(c) Drum as a free body


120 120

F

400

4 5 3

2

400

Shoe width, 80 mm f = 0.20 pmax = 0.40 N/mm2

300

250 rad.

6

45°

45°

(a) 90°

90° 1 300 F

H45 = 4F H25 = 4F

5

1 O25

80

V25 = F

80

(b)

All dimensions in millimeters

(c) H43 = 4F

H34 = 4F

H54 = 4F 4

H52 = 4F

3 V16 = 0.70F

V63 = 0.2H63 H63 = 10.53F

H36 = 10.53F

6 V36 = 2.11F

H26 = 7.07F O16 T = 880F

(rotation) H13 = 6.53F O13

2

V26 = 1.41F

H16 = 3.46F

(f)

V13 = 2.11F (d)

H62 = 7.07F

V62 = 0.2H62 V12 = 2.41F

H12 = 3.07F (e)


V52 = F

O12

A'

A ␦n

␤

␣ O3'

␣

␪ O3

B


O2

c F

0 (18 ␪ in d s sin =d

°–

␪)

dN A

fd

d␪

␪ ␪2

N

O2

␪1

r O3

180° – ␪ Drum rotation

B

Note: d = O2O3 b = width of shoe

d cos (180° – ␪) = – d cos ␪


F

Force to release brake 300 Spring compressed to force F 150 300 rpm

150

200

150

Cast-iron drum Molded composite shoe lining of width b = 50

All dimensions in millimeters

␪2

O3

C = 500

150

45°

␾ 45° ␪1

200

O2

d 150

(a) Complete brake (b) Drum and right shoe


F

N Pivot P fN

␪ 2

␪ rf 2

r


O2

f dN2 + f dN2' P f dN1 f dN2' ⑀

⑀ f dN2

dN2

rf

dN1 dN2'

r


Hydraulic wheel cylinder Drum

Return spring

Rotation r

O3

C ␪2

Adjusting cam d

␪1

Brake lining O2 Anchor pins


Brake lining

Anchor pin Hydraulic wheel cylinder

Brake drum

Brake shoe Adjusting cam and guide Return spring

Forward rotation

Adjusting cam and guide Hydraulic wheel cylinder Anchor pin


␪

Band of width = b

␾ Rotation

Cutting plane for free-body diagrams

r

F a

P1 P1

P2 P2 c


d␪/2

d␪/2

P + dP

P d␪ dN

Rotation


Band of width = b Rotation ␾

P2

F

s P1 a c


Band width, b = 80 mm Rotation ␾ = 270°

Friction coefficient, f = 0.20 Maximum lining pressure, pmax = 0.5 MPa r = 250 mm P2

P1 s = 35 mm

a = 150 mm c = 700 mm


F

Pads of diameter = 60 mm

125 mm

320 mm pmax = 500 kPa f = 0.30


4m/s

1000 kg


Friction clutch

Electric motor

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

Gear reducer

T = 6 N • m, 600 rpm


F = 1500 N 400 mm

500 mm

Rotation

200 mm 340 mm


F 250 mm

320 mm

A

240 mm

Rotation 150 mm

400 mm


500 mm Spring

400 mm 300 mm

350 mm


Woven lining 23 in. 4 6 5 in.

5 in.

Cast- iron drum

3

30 in.

5 Rotation

5

25 in.

18 in.

150

18 in.

2


1

Rotation ␾

500 mm

F = 300 N


75

258° 240

72 55

370

F


270°

a s

Wt.

c


Tight side Pivot

Motor rotation

Overhang (b) Pivoted, overhung motor

␾ Adjustment Pivot

Idler

Weight Tight side (a) Manual adjustment (c) Weighted idler pulley


A

0.88 in.

0.66 in.

0.50 in.

C

0.41 in.

0.31 in.

0.53 in.

B 1.50 in. 1.25 in. E

D

0.91 in.

0.75 in.

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

1.0 in. 0.62 in. 0.38 in.

8V 0.54 in.

3 V 0.32 in. 5V

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


0.88 in.

dN/2 sin ␤

dN 2

dN

2␤ ≈ 36°

(a)

(b)


Rubber Fabric cover Tensioncarrying cords


p

p

␪r

Chordal rise, r – rc A

A

B

B Pitch circle

rc

(a)

rc

(b)


r

Pitch p

(a)


A

Gasket Case

Core or inner shroud

␻i

Impeller Turbine

Oil particle

Fluid circulation

r

Input shaft

r2 r4 (D/2)

Oil seal

r3

Output shaft D

r1

To Ti ␻i

␻o

Blades

A

Section AA

JUVINALL: Machine Design Fig. 19-10 W-584B

360

320 100 280 Percent slip 20 16 14 12

Percent rated torque

240

8

10

7

200

6

Electric motor torque curve

160

5 120

4

Maximum coupling torque

80

3 2

40

0

0

200

400

600

800 1,000 1,200 Input speed ␻i (rpm)

JUVINALL: Machine Design Fig. 19-11 W-584C

1,400

1,600

1,800

Impeller Turbine

Fluid circulation

Input shaft

Output shaft Reactor


Turbine

Impeller

Reactor

Input shaft

Output shaft


One-way clutch

3.2

100

2.8 Converter efficiency

Torque ratio

Coupling efficiency 60

2.0

40

1.6 Converter torque ratio

20

1.2 Coupling torque ratio 0.8

0

0.2

0.4 0.6 Speed ratio


0.8

0 1.0

Efficiency (%)

80

2.4

␣ ␣ ␾1

r2

r1

␾2

␣ ␣ c


C

B A D


V-belt ␤ = 18° f = 0.20 n = 4000 rpm

r = 100 mm Pulley radius ␾ = 170°

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


Electric motor n = 1780 rpm 55% rated power

Driven machine

Fluid coupling — performance curves — Fig. 19.11


3.7 in. dia. Multiple V-belt, ␤ = 18°, size 5V Unit weight = 0.012 lb/in. Power input = 25 hp Pmax = P1 = 150 lb f = 0.20

Driving pulley n = 1750 rpm 165° angle of wrap

Driven pulley Number of belts = ?

JUVINALL: Machine Design Fig. U19-1 W-580

Fluid coupling

B1 C1

B2 C2

Gear interface

Gear

Torque ratio Tout /Tin

Neutral

–

1

3.66

B3

Neutral

Front planetary train

Fluid coupling

Rear planetary train

Bearing

S1, P1, R1

B1 C1

B2 C2

S2, P2, R2 S3, P3, R3

B1 C1

B2 C2

1.44

2.53

B1 engaged

B2 engaged

1.00

2.53

C1 engaged

B2 engaged

B3

1

S1, P1, R1

No clutches or brakes engaged

S2, P2, R2 S3, P3, R3

2

2.53

39% B3

3

2

1.44

1.44

1.00

61%

B1 engaged S1, P1, R1

39%

S2, P2, R2 S3, P3, R3 4 B1

C1

B2 C2

1.00

Reverse

B1

B2 C2

C2 engaged

1.44

–2.99

B1 engaged

B3 engaged

B3

(b) Power flow block diagram

Gear

Ratio

Neutral

–

1

3.66

2

2.53

3

1.44

4

1

Reverse

–4.31

C1

B1

4

S1, P1, R1

S2, P2, R2 S3, P3, R3

B1 C1

B2 C2

B3

Reverse

S1, P1, R1

1.00

61%

C1 engaged

–4.31

S2, P2, R2 S3, P3, R3

C1

1.00

B3

3

S1, P1, R1

C2 engaged

S2, P2, R2 S3, P3, R3

(a) Internal power flow diagram

(c) Gear shift pattern


C2

B2

B3

54-tooth ring P Ti × 3 27 Input torque Ti

15-tooth planet Ti P 81

Ti P 40.5

7.5 P 27 P

R1

P1

12-tooth sun

Ti P 81 12 P

Ti P 81

S1

Ti P 81 Ti P 81

Torque from B1: T P 12 (3) TB1 = i 81 P

Ti P 81 Ti P 81

TB1 = 0.44Ti Arm

Ti P 40.5 Output torque: Ti P 19.5 To = (3) 40.5 P To = 1.44Ti

19.5 P To

Ti P 40.5


A1

Ti P 40.5

Ti P 67.5

Ti P 67.5

12-tooth planet Ti P P2 33.75 Ti P

TB2 =

Ti P 34.5 (3) P 67.5

TB2 = 1.53 Ti

67.5 6 P

Ti

R2 S2

Ti P 67.5

22.5 P

45-tooth sun

Ti 3

•

P 22.5

Ti P 67.5

34.5 P Ti P 67.5 69-tooth ring

Output torque: Ti P 28.5 To = (3) P 33.75

Ti P 33.75

To = 2.53 Ti

28.5 P A2

Ti P 33.75

Ti P 33.75 Arm


Ti P 67.5

12-tooth planet Ti P P2 33.75

Ti P 67.5

34.5Ti P

Ti

R2, S3

675 S2 Ti P

34.5 •

67.5 Ti P 67.5

10

10 P

22.5 P Ti P 67.5

Ti P 67.5

34.5Ti P 67.5

45-tooth sun

34.5 P

Ti P 67.5

P3 69Ti P

34.5Ti P

675

675

69-tooth ring and 20-tooth sun

16-tooth planet Ti P 33.75

69Ti P 675

28.5 P

18 P

Output torque: Ti P 69Ti P 18 28.5 To = – P P 33.75 675 To = –2.99Ti

A2, A3

Arms


3

Torque from reverse lock, B3 34.5Ti P 26 TB3 = (3) P 675 TB3 = 3.99Ti

34.5Ti P 675 R3

26 P

34.5Ti P 675 34.5Ti P 675

52-tooth ring


Torque from C1: T P 12 TC1 = i 3 81 P TC1 = 0.44Ti Ti P 40.5

24-tooth sun Ti P 81 12 P Ti P 81

Output torque: To = 1.00Ti

TC1

A1

TC1 To

19.5 P

S1 Ti P 81 Ti P 40.5

Arm


Ti P 40.5

For equilibrium of P2: Tf P T P = C2 Tf = 0.39Ti 67.5 103.5 TC2 = 0.61Ti Tf + TC2 = Ti

TC2P 103.5 Tf P

P2

67.5

Clutch torque TC2

0.0117Ti P

28.5 (3) P

To = 1.00Ti R2

S2

22.5 P

Output torque: To = 0.117Ti P

12-tooth planet Tf

∴ Planet pin force = 0.0117Ti P

28.5 P A2

To 34.5 P

45-tooth sun

Arm 69-tooth ring


h y b Rectangle

h y b Triangle

d

Circle

d

di

Hollow circle

JUVINALL: Machine Design Fig. B-1 W-601

y

d L

z

x Rod y t d

x

z Disk y

b

c

z

a Rectangular prism y

d z

L x Cylinder y

di do z

L x

Hollow cylinder

JUVINALL: Machine Design Fig. B-2 W-602

240 4130

220

Ultimate strength

200

Yield strength

1095

4130 180

160

1050

140

1095

ksi 1040 120

1030 1050

100

1040

80

1030

60

60

1030, 1040, 4130 Reduction of area

1050 1095

40

Elongation

% 20

1030, 1040 1095, 4130

0 400

1050 600

800 1000 Tempering temperature (°F)

JUVINALL: Machine Design Fig. C-5a W-603

1200

200

180

1095 Ultimate strength

160 Yield strength

1050 140 ksi 120

1095 1040 1050

100 1040 80

60

60

1040 Reduction of area

1050 40 1095

Elongation

% 20

1040 1095

0 400

1050 600


JUVINALL: Machine Design Fig. C-5b W-604

1200

300

280 9255 Ultimate strength

260 9255

4140, 4340 Yield strength

240 4140 220

4340

ksi 200 180

160

140

120

100 60

Reduction of area 4140

40 4340

%

Elongation 9255 9255

20 4140 0 400

4340 600


JUVINALL: Machine Design Fig. C-5c W-605

1200

Diameter of test specimen (mm) 0

200

400

600

800

1000

190

1300

Su Sy

180

1200 170

4340

160

1100 4340

150 1000 140 4140 900

130 3140

ksi

MPa

120 800 110 4140 700

100 3140 90

1040

600

80 500 70

60

50

1040

0

1

2 3 Diameter of test specimen (in.)

JUVINALL: Machine Design Fig. C-6 W-606

400

4

P

L x –PL

␦max P

V

+

V=P

␦

0 0

M –

M = –PL

JUVINALL: Machine Design Fig. D-1(1) W-607

P a

b

x –Pa

␦max

P

V M

+

␦ V=P

0 0 –

M = –Pa


L w

wL2 2 V

␦max

x wL +

V = wL

0 0 M –

M = –wL2/2


x

–Mb

␦max P

V M

L

␦

Mb

0 0 –

–Mb


P

␦ P 2 + V

P 2

L

0 – +

M

x P/2

L 2

–P/2 PL/4

0


a Pb L + V

x

P

␦

b

Pb/L

0 – +

M

Pa L

L

Pab/L

–Pa/L

0


w wL 2

x L

+ V

wL 2

wL /2

0 –

wL2 2

– wL 2

+ M

0


Pb a

L z x ␦ a

V

+ 0 –

M

0 –

P ␦max

PL a b

P

–Pb/a

–Pb


a

M0

x M0 L

b L M0 L

+ V

0 M0a L

+ M

M0 L

0 –

–

M0b L


M0 a

L x' x

␦ a

V

␦max M0

0 –

M

M0 a b M – a0

0 – M0


–

PL 8 P 2 V

P 2

+ 0 –

0 –

L 2

␦

PL 8

+ M

L P

x

–

PL 8


PL 8 P 2

–

P 2

–

PL 8

a P 2 – Pab L2

V

M

L b

x

Pa2b L2

2

Pb (3a + b) L3 Pb2 + (3a + b) L3 0 – 2Pa2b2 + L3 0 –

Pab2 – L2

Pb2 (a + 3b) L3 –

Pb2 (a + 3b) L3

–

Pa2b L2


L x –

wL2 12 wL 2 V

w

wL 2

+ 0 –

M

␦

wL2 24

+

wL2 12 wL 2

–

wL 2

–

wL2 12

0 – –

wL2 12


RB 82.63

85.00

101.6

95.00

RA

2582.21 N • m

122.17

144.81 ROOT 175.84 PITCH 206.88 O.D.

8.68 kN

2

1

3 107

4

5

6

7

10

8

152

9 276 400 530 646 680 716 1060 A shaft with integral worm, dimensions in millimeters.

JUVINALL: Machine Design Fig. D-4 W-613

Class of fit

1

2

3

4

5

6

7

s

h hole h Bar graph (basic hole system)

a

8

h s

h a

a

s

h

s

h

s h

h

s s

s shaft Heavy force and shrink fit

Loose fit

Free fit

Medium fit

Snug fit

Wringing fit

Tight fit

Medium force fit

Cb

.0216 (.0025)

.0112 (.0013)

.0069 (.0008)

.0052 (.0006)

.0052 (.0006)

.0052 (.0006)

.0052 (.0006)

.0052 (.0006)

Cs

.0216 (.0025)

.0112 (.0013)

.0069 (.0008)

.0035 (.0004)

.0035 (.0004)

.0052 (.0006)

.0052 (.0006)

.0052 (.0006)

Ca

.0073 (.0025)

.0041 (.0014)

.0026 (.0009)

.00025 (.00025) .0005 (.0005)

.0010 (.0010)

Ci

0

(0) 0

(0)

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

JUVINALL: Machine Design Fig. E-1 W-614

JUVINALL: Machine Design p754 b1 W-???

JUVINALL: Machine Design p754b2 W-604

Diseño de maquinas juvinall.pdf

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