Lectures-Part II [PDF]

STRESS

Critical section

P σ t ,c , s = A My σb = I Ty τ= J (a) Normal, tensile (b) normal, compressive; (c) shear; (d) bending; (e) torsion; (f) combined 3/17/2015

Elementary equations. No discontinuity in cross-section. Holes, shoulders, keyways, etc.

1

A B

High concentration of elements are required to estimate stress level.

Finite element model to calculate stresses 3/17/2015

2

Stress Concentration Axial Load on Plate with Hole

σ avg

P = (b − d ) h

Plate with cross-sectional plane

Stress concentration factor

σ max Kt = σ avg Half of plate with stress distribution. 3/17/2015

3

Geometric discontinuity increases the stress. Stress concentration is a highly localized effect. 3/17/2015

4

EX: A 50mm wide and 5mm high rectangular plate has a 5mm diameter central hole. Allowable stress is 300 MPa. Find the max. tensile force that can be applied. Ans: d/b = 0.1; Kt=2.7 A = (50-5)×5 P = 25 kN

Stress concentration factor for rectangular plate with central hole.

3/17/2015

5

EX: Assume H=45mm, h=25mm, and fillet radius r=5mm. Find stress concentration factor. Ans: ~1.8

Stress concentration factor under axial load for rectangular plate with fillet 3/17/2015

6

Stress concentration factor under axial load for rectangular plate with groove 3/17/2015

7

Stress concentration factor under axial load for round bar with fillet 3/17/2015

Gap between lines decrease with increase in r/d ratio. 8

Stress concentration factor for round bar with groove 3/17/2015

9

Ex: Assuming 80 MPa as allowable strength of plate material, determine the plate thickness † Maximum stress near fillet ⎛ 5000 ⎞ 300 ⎟⎟ = b ⎝ 30 b ⎠

σ fillet = 1.8⎜⎜

† Maximum stress near hole σ hole

Kt=1.8

Kt=2.1

⎛ 5000 ⎞ 700 ⎟⎟ = = 2.1⎜⎜ b ⎝ (30 − 15)b ⎠

† Allowable 3/17/2015

σ allowable = 80

b=8.75 mm

10

EX: Assume H=45mm, h=25mm, and fillet radius r=5mm. Find stress concentration factor. Ans: ~1.5

Stress concentration factor under bending for rectangular plate with fillet 3/17/2015

11

Concentration factor for thick plate with central hole is higher compared to thin plate with same size hole.

Stress concentration factor under bending for rectangular plate with central hole 3/17/2015

12

Stress concentration factor under bending for rectangular plate with groove 3/17/2015

Decrease in Kt for r/h > 0.25 is negligible. 13

Stress concentration factor under bending for round bar with fillet 3/17/2015

14

Stress concentration factor under bending for round bar with groove 3/17/2015

15

Ex:

Assuming 100MPa as allowable stress, determine the shaft dia, d. † Due to symmetry, reaction force at each bearing = 1250 N. † Stress concentration will occur at the fillet. † Kt=1.6 σ max = 1.6 σ avg = 3/17/2015

51.2 (1250 × 350)

π (d )

3

σ avg = = 100

32 M 32 (1250 × 350) = 3 πd π (d )3

Diameter d=41.5 mm 16

Stress concentration under torsion loading is relatively low.

Stress concentration factor under torsion for round bar with fillet 3/17/2015

17

Stress concentration factor under torsion for round bar with groove 3/17/2015

18

Notch Sensitivity

q=

K f −1 Kt −1

† “Metals can accommodate stress concentration by deforming & redistributing load more evenly”. † Some materials are not fully sensitive to the presence of geometrical irregularities (notch) and hence for those materials a reduced value of Kt can be used. Notch sensitivity

parameter q = 0 means stress concentration (Kf ) factor = 1; and q=1 means Kf = Kt.

3/17/2015

19

Notch Sensitivity

1 q = notch senstivity = 1+ a

† In case of loading/unloading …. Materials show better sensitivity (lesser q)….. K f −1 q= Kt −1 1≤ K ≤ K f

t

† If notch sensitivity data is not available, it is conservative to use Kt in fatigue calculations. 3/17/2015

20

r

Neuber’s constant for steels

a

Sut (ksi) 50

.13

55

.118

60

.108

70

.093

80

.08

90

.07

100

.062

110

.055

120

.049

130

.044

140

.039

160

.031

180

.024

200

.018

220

.013

240

.009

(in0.5)

Kf is commonly used for fatigue loading cases.

Neuber’s constant for Annealed Aluminum 1 ksi = 6.895 MPa

Sut (ksi)

a

15

.341

20

.264

25

.217

30

.18

35

.152

40

.126

45

.111

(in0.5)

Notch senstivity for steel step bar (having fillet radius = 0.25 inch and ultimate strength = 100 ksi) 1 1 q= = = 0.89 a 1 + 0.062 / .25 1+ r Cast irons have a very low q (≅ 0.2) value. For such materials 'Fracture Mechanics' techniques

Material selection for a plate having central hole and is subjected to Tensile force EX: A 50mm wide (b) and h mm high rectangular plate has a 5mm diameter central hole. Length of plate is equivalent to 100mm. Select a lightest but strong material which bear tensile force P = 25 kN. Ans: Mass = ρ × (50-5)× h × 100*10-9 ; A = (50-5)× h

P 25000 1500 d/b = 0.1; Kt=2.7; σ = Kt MPa = 2.7 = (b − d ) h (50 − 5) h h 1500 -6 or, M = 4.500 *10 ρ

ρ or, M = 6750 *10 σ

σ

−6

3/17/2015

⎛ρ⎞ ⎛ M ⎞ log10 ⎜ ⎟ = log10 ⎜ ⎟ ⎝σ ⎠ ⎝ 0.00675 ⎠

22

⎛ρ⎞ ⎛ M ⎞ log10 ⎜ ⎟ = log10 ⎜ ⎟ log ⎛⎜ M ⎞⎟ = log (ρ ) − log (σ ) 10 10 10 ⎝σ ⎠ ⎝ 0.00675 ⎠ 0 . 00675 ⎠ ⎝ ⎛ 0.00675 ⎞ 3/17/2015 log10 (σ ) = log10 (ρ ) + log10 ⎜ ⎟ M < 0.025 kg=> -0.57 23 ⎝ M ⎠

1500 σ= = 1.89e3 ⇒ h = 0.8 mm h 3/17/2015

24

Avg. density = 1645 kg/m3 Mass = 0.0059 kg

Mass reduction ????

Commonly available. Economic.

Stress concentration ??? 3/17/2015

25

“Factor of Safety”

Is design of element/assembly safe? FoS < 1 ?

† FOS is a ratio of two quantities that have same units: „ „ „ „ „

Strength/stress Critical load/applied load Load to fail part/expected service load Maximum cycles/applied cycles Maximum safe speed/operating speed.

NOTE: FOS is deterministic. Rational assessment of the risks associated with a particular design. Data are statistical and there is a need to use Probabilistic approach. 3/17/2015

26

Random variables: Friction, Load, Strength, Dimensions, environment, ….

Variation in Material Strength (MPa) Material Range (AISI, rolled) 1080

865 - 975

1095

Mean

St. Deviation

920

18.33

865 - 1070

967.5

34.17

1030

495 - 610

522.5

19.17

1040

565 - 690

627.5

20.83

1050

650 - 800

725.0

25.00

1060

725 - 900

812.5

29.17

3/17/2015

27

Cumulative probability!!

Probability density function

σs =

∑

f (S ) =

⎛ ⎞⎞ ⎛ s ⎜ s i − ⎜⎜ ∑ i ⎟⎟ ⎟⎟ ⎜ N ⎠⎠ ⎝ ⎝ N −1

1

σ s 2π

e

2

1 ⎛ S −S − ⎜⎜ 2⎝ σ s

Sigma, two sigma, three sigma…. ⎞ ⎟⎟ ⎠

2

coefficient of variation (COV).. 0.25 unacceptable

+∞

∫ f (S ) dS = 1

−∞

3/17/2015

Mean value

worst value

28

Probabilistic value

Probability density function Ex: Measured ultimate tensile strength data of nine specimen are: 433 MPa, 444, 454, 457, 470, 476, 481, 493, and 510 MPa. Find the values of mean and std. dev. Assuming normal distribution find the probability density function.

3/17/2015

σs =

∑

⎛ ⎞⎞ ⎛ s ⎜ s i − ⎜⎜ ∑ i ⎟⎟ ⎟⎟ ⎜ N ⎠⎠ ⎝ ⎝ N −1

2

μ s = 468 . 67 MPa σ s = 24 . 34 MPa f (S

)=

−

1 24 . 34

2π

e

1 ⎛ S − 468 . 67 ⎞ ⎜ ⎟ 2⎝ 24 . 34 ⎠

+∞

∫ f (S ) dS

=1

−∞

29

2

4.59,4.34,4.5796,4.50, 4.582,4.5847……………4.5948

EX. NOMINAL SHAFT DIA. NUMBER OF SPECIMEN

4.5mm 34 4.58mm 0.0097

μd σd

1 ⎛ di − μd ⎞ ⎟ σ d ⎟⎠

− ⎜⎜ 1 2 f (d ) = e ⎝ σ d 2π

3/17/2015

⎛ ⎛ ∑ di ⎞ ⎞ ⎜ ⎟⎟ ⎟⎟ − d ∑ ⎜ i ⎜⎜ N ⎝ ⎠⎠ ⎝ σd = N −1

2

6 4.5294 0.0987

NOTE: Variation in stress level occurs due to variation in geometric dimensions. 30

Prob: A steel bar is subjected to compressive load. Statistics of load are (6500, 420) N. Statistics of area are (0.64, 0.06) m2. Estimate the statistics of stress.

P ∴σ = A Standard

σσ

Ref: Probabilistic Mechanical Design, Edward B. Haugen, 1980. deviation

⎡⎛ ∂ σ ⎞ 2 = ⎢⎜ ⎟ σ ∂ P ⎠ ⎢⎣ ⎝

of stress ⎤ ⎛ ∂σ ⎞ 2 +⎜ ⎟ σ A⎥ ∂ A ⎝ ⎠ ⎥⎦ 2

2 P

1/ 2

Ans: (10156, 1156.4) Pa.

3/17/2015

31

FoS vs Statistical Design † Deterministic. Lack of information. Conceptual design † Based on availability of data. Find the mean and standard-deviation values of dependent variables. Actual design. Q = Sy −σ Q = Sy −σ

or

μQ = μ S − μσ y

σ Q = σ s2 + σ σ2 y

3/17/2015

32

Ex : Strength and Stress of a tensile bar are : S y = (270, 32 ) MPa

&

σ = (184,15) MPa f (Q ) =

R = 1-0.0075 ???? −2.43 1 −z2 2 Pf = ∫ e dz −∞ 2π

1

σ Q 2π

e

1 ⎛ Q − μQ − ⎜ 2 ⎜⎝ σ Q

Let normal variable Q − μQ Z =

σQ

3/17/2015 Ref: Probabilistic Mechanical Design, Edward B. Haugen, 1980.

33

⎞ ⎟ ⎟ ⎠

2

Ref: Probabilistic Mechanical Design, Edward B. Haugen, 1980. ALGEBRAIC FUNCTIONS

Q=C Q = Cx Q=C+x

MEAN

C Cμx C + μx

STD. DEVIATION

0 Cσ x

σx

Q = x± y

μx ± μ y

σ x2 + σ y2

Q = xy Q=x y

μx μ y

μ y2σ x 2 + μ x2σ y 2

μx μ y

μ y2σ x 2 + μ x2σ y 2 μ y2

Q =1 x 3/17/2015

1 μx

σ x μ x2 34

Example: Tie-rod forged steel component. ± DIA.=(30, 1.2) mm STRENGTH 300 MPa to 540 MPa LOAD 30000 ± 9600 N ?? FACTOR OF SAFETY a. BASED ON MEAN VALUE P = 30000 N ;..... A = S = 420 MPa P = 42.4328 MPa A ∴ SF = 9.9 ∴

π 4

(30)2 = 707 mm 2

b. BASED ON Worst Case AMIN = PMAX

π

(26.4)2 = 547.4 mm 2

4 = 39600 N

S MIN = 300 MPa

APPLIED STRESS

39600 = 72.342 MPa 547.4 300 SFMIN = ≈ 4.15 72.342

=

c. NORMAL DISTRIBUTION

(P ,σ ) = (30000; 3200)N (S , σ ) = (420; 40) MPa (d , σ ) = (30; 1.2)mm

A=

P

π 4

(30)2 = 707 mm 2

S

P ∴σ = = 42.4328 MPa A Q = 420 − 42.4328 = 377.5672 MPa

d

Standard

deviation

⎡⎛ ∂σ ⎞ = ⎢⎜ ⎟ σ ∂ P ⎠ ⎢⎣ ⎝ 2

σσ

[

of stress ⎤ ⎛ ∂σ ⎞ 2 +⎜ ⎟ σd⎥ ∂ d ⎝ ⎠ ⎥⎦ 2

2 P

σ σ = 4 . 5262 + (− 2 . 83 * 1 . 2 ) 2

]

2 1/ 2

σ Q = 40 2 + 5.66 2 = 40.4 Z=

0 − 377.5672 = −9.3457 40.4

1/ 2

= 5 . 66

σσ

2 ⎡ σ P2 ⎤ ⎛ 8P ⎞ 2 σd⎥ = ⎢ 2 + ⎜− 3 ⎟ π A d ⎠ ⎝ ⎢⎣ ⎥⎦

1/ 2

Ex: Consider a structural member with S(μ s = 40 , 6 ) subjected to a static load that develops a stress σ( μ σ = 30 , 8 ). Find the SAFETY of member. Mean value approach, FOS = 40/30. 100% SAFE Worst case approach, FOS = 22/54. 40% SAFE

Sy −σ ≥ 0 Q≥0 μQ = 40 − 30 = 10 μQ =10,σQ =10 3/17/2015

μσ = 30 μ s = 40 σσ = 8 σ s = 6

σ Q = 6 2 + 82 = 10

38

μQ = 40 − 30 = 10 σ Q = 6 2 + 82 = 10

Normal variable Z = at Q = 0

Z =

x−μ

σ

=

Q −Q sQ

0 − 10 = −1 10

Reducing standard deviation !!

In the present case Probability of failure is 0.1587 & reliability is .8413. Selecting stronger material (mean value of strength = 50 units)!!!!.. CALCULATE Reliability

Z-Table provides probability of failure

3/17/2015

40

Useful 3/17/2015

41

Ex: A round 1018 steel rod having yield strength (540, 40) MPa is subjected to tensile load (220, 18) kN. Determine the diameter of rod reliability of 0.999 (z = -3.09).

Given μs = 540 MPa ; σ s = 40 MPa 220000 18000 μσ = MPa ; σ σ = MPa 2 2 π /4d π /4d Z =

Q − μQ

σQ

;R=

1 2π

+∞

∫e

Z0

μQ where Z 0 = − σQ

1 − Z2 2

dZ

880000 μQ = 540 − π d2 ⎛ 72000 ⎞ 2 ⎟ σ Q = 40 + ⎜⎜ 2 ⎟ ⎝ πd ⎠

2

Assuming st. dev. Of d is zero

2

⎛ 72000 ⎞ 880000 ⎜ ⎟ 3.09 40 + ⎜ = 540 − 2 ⎟ 2 π π d d ⎝ ⎠ 2

3/17/2015

d = 26 mm 42

Example: Stress developed in a machine element is given by:

(

σ = P / 4kd 3

)(

4 L12 + 3L22

)

Given P = (1500, 50) N, Strength = (129, 3) MPa, L1=(150, 3) mm, L2=(100, 2) mm. Assume std. dev. of d is 1.5% mean value of d. k = 0.003811. Determine distribution of d if the maximum probability of machine-element-failure is 0.001 2

n ⎛ ∂φ ⎞ ⎟⎟ σ xi2 Standard deviation of a complex function is expressed by : σ φ = ∑i =1 ⎜⎜ ⎝ ∂xi ⎠ μ 2 2 2 ⎡⎛ ∂σ ⎞ 2 ⎤ ⎞ ⎞ ⎛ ⎛ σ σ σ ∂ ∂ ∂ ⎛ ⎞ 2 2 2 2 ⎟⎟ σ L2 ⎥ ⎟⎟ σ L1 + ⎜⎜ σ σ = ⎢⎜ ⎟ σP +⎜ ⎟ σ d + ⎜⎜ ⎝ ∂d ⎠ ⎢⎣⎝ ∂P ⎠ ⎥⎦ ⎝ ∂L2 ⎠ ⎝ ∂L1 ⎠

1/ 2

Statistically independent

2 2 2 ⎡⎛ 22724 ⎞ 2 ⎤ ⎞ ⎛ ⎞ ⎛ 2 e 13635 4 170430 ⎛ 85216 ⎞ 2 2 2 ⎟⎟ (0.003) + ⎜ ⎟⎟ 0.015 μ d + ⎜⎜ ⎟⎟ (50 ) + ⎜⎜ σ σ = ⎢⎜⎜ ⎟ (0.002 ) ⎥ 3 4 3 3 d μ μ μ ⎝ ⎠ ⎢⎣⎝ ⎥⎦ d d d ⎠ ⎝ ⎠ ⎝ ⎠ 1 σ σ = 3 [1.291e12 + 41830 + 261420 + 29047]1/ 2

(

)

μd

σσ =

1136200

μ d3

43

1/ 2

(

μσ = (μ P / 4kμ d3 ) 4μ L21 + 3μ L22 μσ =

)

34087000

μ d3

Z = −3.09 =

(

0 − 129e6 − 34087000 μd3 ⎡ ⎛ 1136200 ⎞ 2 ⎟⎟ ⎢3e6 + ⎜⎜ 3 ⎢⎣ ⎝ μd ⎠ 2

2

⎤ ⎥ ⎥⎦

1

)

2

⎛ 1136.2 ⎞ ⎛ 11031⎞ (3000) + ⎜⎜ 3 ⎟⎟ = ⎜⎜ 41748 − 3 ⎟⎟ μd ⎠ ⎝ μd ⎠ ⎝ μd = 0.06686 m 2

σ d = 0.001 m

Calculating FOS = Strength/stress Æ FOS =129/114=1.13 2

Probabilistic Approach to Design † Conventional design follows “Deterministic Approach” by disregarding statistical nature of „ Material properties „ Component dimensions „ Externally applied load

† Deterministic design considers uncertainties by „ Applying “Factor of safety” „ Considering absolute worst case

† There is a growing trend toward using a probabilistic approach to better quantify uncertainty and thereby increase reliability

Failure of Machine Element † There are only two ways in which an element fails: „ Obsolescence „ Loss of function

† Element losses its utility due to: „ Change in important dimension due to wear. „ Change in dimension due to yielding (distortion) „ Breakage (fracture). „ Jamming (friction) 3/17/2015

46

Yielding (distortion) Wear

Fracture

3/17/2015

Jamming

47

Design of Components before Minor II † † † †

Shafts Keys Couplings Bearings

These components are employed to separate rotating/sliding elements from stationary/ relatively-stationary elements. 3/17/2015

48

Bearing Classification

AntiFriction !! Misnomer

Rolling Element Bearings ¾ “Rotation is always easier than linear motion”. Low friction & moderate lubricant requirements are two important advantages of rolling bearings.

¾ If you can buy it, don’t make it! ¾ Bearing selection….

Bearing Terminology Four main components ?

Roller Needle roller

Cylindrical Roller Bearing Higher coefficient of friction because of small diameter rollers and rubbing action against each other

= ( Z / 4.37).Wmax

Fr

ball

Fr

roller

= ( Z / 4.06).Wmax

Outer ring or Cup Cage

¾ Radial (Load) Bearing (s) Inner ring or Cone

¾Thrust (Load) Bearing (s) ¾ Combined (Load) Bearing (s)

Ball

Cylindrical roller

Spherical roller

Angular contact ball

Ball

Tapered roller

Cylindrical roller

Spherical roller

Rolling Element Bearings Load Direction

Misalignment Capacity

Radial Axial

Both

High

Deep groove ball

y

y

Cylindrical Roller

y

Some types

Needle

y

Taper Roller

y

Self Aligning Ball

y

y

y

Self Aligning Spherical Roller

y

y

y

Med

Low

y y y

y

Angular contact ball

y

Thrust ball/roller

y

Equivalent load: P = V X Fr + Y Fa

y

y

y

y y

Equivalent Dynamic Load Equivalent load: P = V X Fr + Y Fa

Assuming rotation of inner ring P = Fr when Fa / Fr ≤ e P = X Fr + Y Fa

when Fa / Fr > e

V

Rotation factor

X

Radial factor

Fr Applied radial load Y

Thrust factor

Fa Applied thrust load e is a dimensionless ratio, indicating axial load lower than a certain limit does not affect total load

Value of e depends on arrangement & static load capacity (CO ) of bearing 3/17/2015

56

http://www.skf.com/group/products/bearings-units-housings/ball-bearings/deepgroove-ball-bearings/single-row-deep-groove-ball-bearings/single-row/index.html

Basic Dynamic Load Rating: C Radial load (thrust load for thrust bearings) which a group of identical bearings with stationary outer rings can theoretically endure one million revolutions of inner ring. Static Load Rating: C0 Radial load causing permanent deflection greater than 0.01% of ball dia.

Bearing type

Inner ring

Single row

Double row

Rotating

Stationary

Fa/VFr > e

Fa/VFr ≤ e

Fa/VFr > e

e

Deep groove ball bearing

Fa/C0

V

V

X

Y

X

Y

X

Y

.014 .028 .056 .084 .11 .17 .28 .42 .56

1

1.2

0.56

2.30 1.99 1.71 1.55 1.45 1.31 1.15 1.04 1.00

1

0

0.56

2.30 1.99 1.71 1.55 1.45 1.31 1.15 1.04 1.00

.19 .22 .26 .28 .3 .34 .38 .42 .44

Angular contact ball bearing

20 25 30 35 40

1

1.2

.43 .41 .39 .37 .35

1.0 .87 .76 .66 .57

1

1.09 .92 .78 .66 .55

.70 .67 .63 .60 .57

1.63 1.44 1.24 1.07 .93

.57 .68 .80 .95 1.14

1

1

.4

.4 cotα

1

.42 cotα

.65

.65 cotα

1.5 tanα

Self aligning ball bearing

Important points † Bearing lubrication: Oil/Grease –Permissible speed † Cage- to avoid sliding & collision „ Full complement bearing …. Slow speed rigidity

† Bearing selection…. For given shaft dia † Rotating ring (interference)… non-rotating (transition). Rotation factor !!! † Bearing life …….. Dynamic load capacity… Applied load (3 to 10%). † Equivalent load: X & Y factors depend upon bearing type. 3/17/2015

59

Bearing Life

3/17/2015

60

Lundberg Palmgren Approach In ideal case, bearings fail by surface-fatigue. Dynamic load rating (catalogue C0 reading) is the load which 90% (reliability=0.9) of a group of identical bearings will sustain to the minimum of 106 cycles.

Failure Factor probability (%) a1

(C ) 10

6

a=3

for ball bearings

a

10 a= 3

= P L = P L2 = P L3 a 1 1

a 2

a 3

for roller bearings

⎛C ⇒ Bearing life in hours = ⎜⎜ ⎝ P

a

10 5 4 3 2 1

⎞ 1000,000 ⎟⎟ ⎠ 60 Speed

1 0.62 0.53 0.44 0.33 0.21

Example: Radial load = 2 224 N, Speed = 1500 rpm Desired life= 8 hours/day, 5 day/weeks for 5 years, Shock factor = 1.5. For shaft dia of 25 mm.

C C C

> 2224*1.5*(10400*1500*60/106)1/a > 32, 633 N for BALL BEARINGS > 25, 978 N for ROLLER BEARINGS

http://www.skf.com/portal/skf/home/products?newlink=first&lang=en

3/17/2015

In order of increasing outside bearing diameter 63

Deep Groove Ball Bearing

RSH Æ Sheet steel reinforced contact seal of acrylonitrile-butadiene rubber (NBR) on one side of the bearing. L stand for low friction.

3/17/2015

64

Suffix

61804

3/17/2015

61804-2Z

61804-2RS1

65

Deep Groove Ball Bearing (DGBB): Both rings possess

Seals: Often made of elastic

rubber. Bearings sealed on both sides are grease filled and in – deep grooves. Bearing can normal working conditions the support high radial forces as well as some axial forces. There grease filling lasts the entire service life of the bearings. are single-row & double row DGBB. Widely used in industry.

Cage/Separator: Ensures uniform spacing and prevents mutual contact of rolling elements.

Shield: Profiles sheet steel discs pressed into the grooves of outer ring and forming gaptype seals with the inner-ring shoulders.

Angular Contact Ball Bearing (ACBB): Raceways are so

arranged that forces are transmitted from one raceway to other under certain contactangle between line of action of the force & radial planes. Due to CA, ACBB are better suited to sustain high axial loads than DGBB. 66

3/17/2015

67

Designation –

International Organization for Standardization

™ Each rolling bearing is designed by a code that clearly indicates construction, dimensions, tolerances and bearing clearance.

Instrument Ball 3/17/2015 bearing

Multiply by 5 to get bore in mm d500 mm… 511/530 (d=530mm)

00 = 10mm 01= 12mm 02 = 15mm 03 = 17mm

68

0 Double row angular contact ball bearings 1 Self-aligning ball bearings 2 Spherical roller bearings 3 Taper roller bearings 4 Double row deep groove ball bearings 5 Thrust ball bearings 6 Single row deep groove ball bearings 7 Single row angular contact ball bearings 8 Cylindrical roller thrust bearings HK needle roller bearings with open ends K Needle roller and cage thrust assemblies N Cylindrical roller bearings A second and sometimes a third letter are used to identify the configuration of the flanges, e.g. NJ, NU, NUP; double or multi-row cylindrical roller bearing designations always start with NN. QJ Four-point contact ball bearings 3/17/2015

69

Example: Assume radial and axial loads on a bearing are 7500N and 4500N respectively. Rotating shaft dia = 70 mm. Select suitable single row deep groove ball bearing. Bearing type

Deep Fa/C0 groove .014 ball .028 bearing .056 .084 .11 .17 .28 .42 .56

Inner ring

Single row

Rotating

Fa/VFr > e

V

X

Y

1

0.56

2.30 1.99 1.71 1.55 1.45 1.31 1.15 1.04 1.00

0.0662 0.1452

e

.19 .22 .26 .28 .3 .34 .38 .42 .44

Fa/Fr = 0.6; Fa/C0=4500/31000 Æ X = 0.56, Y= 1.37; P=10365 3/17/2015

Fa/C0=4500/68000 Æ X = 0.56, Y= 1.65; P=11625

70

0.03 C ≤ P ≤ 0.1C Life consideration

(C ) 10

6

a=3

for ball bearings

a

10 a= 3

= P1a L1 = P2a L2 = P3a L3

for roller bearings

⎛C ⇒ Bearing life in hours = ⎜⎜ ⎝ P

a

6014 10365 39700 937

⎞ 1000,000 ⎟⎟ ⎠ 60 Speed

6314 11625 111000 14509

Example: Assume radial and axial loads on a bearing are 7500N and 4500N respectively. Shaft dia = 70 mm. Select a deep groove ball bearing. Consider shaft rotates at 1000 rpm and expected

bearing life = 3000 hours 3/17/2015

71

Deep Groove Ball Bearing

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72

Deep Groove Ball Bearing

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73

Deep Groove Ball Bearing

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74

Pressed brass cage

Deep Groove Ball Bearing

3/17/2015

E indicates reinforced ball set. TN9 indicates injection molded snap type cage of glass fibre reinforced polyamide

75

Deep Groove Ball Bearing

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76

Deep Groove Ball Bearing

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77

Deep Groove Ball Bearing

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78

3/17/2015

79

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80

–

single row angular contact ball bearings

– –

double row (O, X, Tandem) angular contact ball bearings four-point contact ball bearings In SKF catalogue contact angle α =40° is designated by suffix B. Similarly contact angles of 25° and 30° are designated with suffixes AC and A respectively.

Tandem arrangement

Back to back

Face to face

Bearing type

Single row

α (°) Angular contact ball bearing Self aligning ball bearing

20 25 30 35 40

Double row

Fa/VFr > e Fa/VFr ≤ e Fa/VFr > e X Y X Y X Y

e

.43 .41 .39 .37 .35

1.0 .87 .76 .66 .57

1

1.09 .92 .78 .66 .55

.70 .67 .63 .60 .57

1.63 1.44 1.24 1.07 .93

.57 .68 .80 .95 1.14

.4

.4 cotα

1

.42 cotα

.65

.65 cotα

1.5 tanα

Example: A radial load of 3000N combined with thrust load of 2500N is to be carried on a 6214 ball bearing for shaft rotating at 1000 rpm. Determine equivalent radial load to be used for calculating fatigue life. Compare life of 6214 bearing with that for a 7214 (nominal contact angle 30°). † Step 1: C0 for 6214 is 45kN and 7214 is 60 kN. C for 6214 is 63.7 kN and 7214 is 71.5 kN † Step 2: Fa/VFr > e Bearing type Deep groove ball bearing Angular contact ball bearing

Single row, Fa/VFr > e

e

Fa/C0

X

Y

.056

0.56

1.71

.26

30

.39

.76

0.8

Fr

X

Y

Fa

Deep groove ball bearing

3000

0.56

1.71

2500

Angular contact ball bearing

3000

0.39

0.76

2500

• Step 3: Radial load for 6214 bearing is 5955N & for 7214 bearing radial load is 3070.

Step 4: Life for 6214 will be 7192 hours and for 7214, life=124,420 Hours 3

⎛ C ⎞ 1000,000 Bearing life in hours = ⎜ ⎟ ⎝ P ⎠ 60 Speed

Appropriate selection of bearing….

Equivalent load under Variable Loading † Bearing operates at 1000 rpm, applied load of 500 N for 100 hours, then bearing operates at 1200 rpm, 250 N for 250 hours…. „ In such situation it is advisable to find an equivalent load using

1

⎛ P L + P L2 + P L3 + ... ⎞ a ⎟⎟ P = ⎜⎜ L1 + L2 + L3 + ... ⎝ ⎠ a = 3 for ball bearings 10 a= for roller bearings 3 L1, L2 , L3 ,... Number of rotations a 1 1

a 2

a 3

IF L = expected life, then

(

)

⎛ P L + P L2 + P L3 + ... L ⎞ ⎟⎟ P = ⎜⎜ (L1 + L2 + L3 + ...) L ⎠ ⎝

(

a 1 1

a 2

a 3

)

P = P f + P f 2 + P f + ... a 1 1

a 2

a 3 3

1

a

1

a

Question 1: A single row cylindrical roller bearing N 205 ECP is subjected to pure radial load of 2800 N and rotational speed = 1500 rpm. Assuming bearing is subjected to light shocks. Estimate the bearing life for reliability = 0.99.

Question 2a: Select a suitable deep groove ball bearing for a shaft of 30 mm dia rotating at 2000 rpm. Bearing needs to support a radial load of 2000 N and axial load of 400 N. Minimum desirable bearing life is 5000 Hours. 2b: Will the selected bearing be able to survive for more than 10,000 Hours? Justify your answer.

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88

Shafts † Example: A hollow shaft must carry torque of 3400 N.m at shearing stress of 55 MPa. Assume di=0.65 do. Calculate value of outside diameter. † ANS: 72.6 mm

3/17/2015

89

Torsion of circular shaft Stress in a circular shaft of uniform cross section loaded at the ends by Torque T,

τr =

φ G r1 L

1

⇒ τ max or,

T r 16 T = = J π d3

τ max

16 T d o = π d o4 − d i4

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(

) 90

Torsion of circular shaft Example: Design a shaft so that angular deformation should not exceed 1° in a length of 1.8 m. Permissible shear stress = 83 MPa and modulus of rigidity = 77 GPa. ANS: d=222.34 mm NOTE: Design of shaft consists of determining correct shaft diameter from strength and rigidity considerations. 3/17/2015

91

A complete shaft design has much interdependence on the design of the components: Steps size; Threads; Loading; etc.

Stress & Strength Deflection & Stiffness Vibration localized

3/17/2015

In shaft design, usual practice is locate the critical areas, size these to meet The strength requirements, and then Size the rest of the shaft to meet Requirements of shaftsupported element, manufacturing steps, or mass criterion.

92

Maximum static stress 2

† Often shafts carry combined loads of bending and torque. σ max

σ

τ max

2

or τ max or τ max

⎛σ ⎞ = + ⎜ ⎟ +τ 2 2 ⎝2⎠ 2

2

or σ max or σ max

⎛ 32 M ⎞ ⎛ 16 T ⎞ 32 M ⎟ +⎜ ⎟ = + ⎜⎜ 3 3⎟ 3⎟ ⎜ 2π d ⎝ 2π d ⎠ ⎝ π d ⎠ 16 ⎡ 2 2⎤ ( ) ( ) M M T = + + ⎥⎦ π d 3 ⎢⎣

3/17/2015

⎛σ ⎞ = ⎜ ⎟ +τ 2 ⎝2⎠ ⎛ 32 M ⎞ ⎛ 16 T ⎞ ⎟ ⎟ +⎜ = ⎜⎜ 3⎟ 3⎟ ⎜ ⎝ 2π d ⎠ ⎝ π d ⎠ 16 2 2 ( ) ( ) M T = + π d3

2

Maximum shear stress theory Maximum normal stress theory

Stress Concentration ???? 93

2

Design of shafts for fluctuating loads Ta r Tm r τ a = K fs τ m = K fsm J J where K fs and K fsm are torsional fatigue stress concentration factors. If shaft is subjected to fluctuatin g moment

Ma r Mm r σa = K f σ m = K fm I I where K f and K fm are bending fatigue stress concentrat ion factors. If shaft is subjected to fluctuating axial load

σ a _ axial = K t

Fa A

σ m _ axial = K t

Fm A

Theory of failure ???? Distortion energy (von-Mises) theory 3/17/2015

Design for fluctuating bending, fluctuating axial & fluctuating torsion von - Mises stresses

σ a′ = σ m′ =

(σ a + σ a _ axial )2 + 3τ a2 (σ m + σ m _ axial )2 + 3τ m2

von - Mises stresses 2

⎛ K f M a r K t Fa ⎞ ⎛ K fsTa r ⎞ ⎟⎟ + 3 ⎜⎜ ⎟⎟ σ a′ = ⎜⎜ + A ⎠ ⎝ I ⎝ J ⎠ 2

2

⎛ K fsmTm r ⎞ ⎛ K fm M m r K t Fm ⎞ ⎟⎟ ⎟⎟ + 3 ⎜⎜ σ m′ = ⎜⎜ + I A ⎠ ⎝ ⎝ J ⎠ 3/17/2015

2

95

Goodman Line Criterion σ

' a

σ

' m

1 + = S e Sut N

? N ≥1

N ≥2

σ a' + σ m' ≤ S ys

where 2

⎛ K fsTa r ⎞ ⎛ K f M a r K t Fa ⎞ ⎟⎟ ⎟⎟ + 3 ⎜⎜ σ a′ = ⎜⎜ + A ⎠ ⎝ J ⎠ ⎝ I 2

2

⎛ K fsmTm r ⎞ ⎛ K fm M m r K t Fm ⎞ ⎟⎟ ⎟⎟ + 3 ⎜⎜ σ m′ = ⎜⎜ + I A ⎠ ⎠ ⎝ J ⎝

3/17/2015

or

2

Static failure ? Fatigue failure ?

96

Example: Assume D=42mm, d= 28mm, fillet radius =2.8mm, M = ± 142.4 N.m, T= 124.3 N.m, Sut = 735 MPa, Sy=574 MPa, required reliability =99%.

• Find stress concentration factors • Find endurance limit and factors reducing value of endurance limit -Æ Find endurance strength • Use Goodman criterion, Find factor of safety. • Check safety against yielding

3/17/2015

97

r 2.8 = = 0.1 d 28 D 42 = = 1.5 d 28

K f = K fm = 1.68

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98

r 2.8 = = 0.1 d 28 D 42 = = 1.5 d 28

K fs = K fsm = 1.38

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99

For Steel Se′ = 0.5 Sut Se′ = 0.45 Sut

bending

Se′ = 0.29 Sut

Torsion

Axial

Probability of survival, %

Reliability factor, kr

99

0.814

⇒ Se′ = 0.5 Sut = 367.5 MPa K finish = a(Sut in MPa ) = 0.787 b

Finishing method

Constant a Exponent b

Machined or cold drawn

K size = 1.24 d −0.107 K size = 1.24 (28)

− 0.107

3/17/2015

4.51

-0.265

2.79 ≤ d ≤ 51 mm = 0.870

S e = 367.5 * 0.787 * 0.870 * 0.814 S e = 205 MPa

100

Goodman Line Criterion

σ

' a

σ

' m

1 + = S e Sut N

where 2

K f M ar ⎛ K f M ar ⎞ 2 + 0 ⎟⎟ + 3 (0 ) = σ a′ = ⎜⎜ I ⎝ I ⎠ 2

⎛ K fsmTm r ⎞ ⎛ K fsmTm r ⎞ 2 ⎟⎟ = 3 ⎜⎜ ⎟⎟ σ m′ = (0 + 0) + 3 ⎜⎜ ⎝ J ⎠ ⎝ J ⎠

σ a' + σ m' ≤ S ys 3/17/2015

Design (ASME method) for fully reversed bending and steady torsion Introduce a factor of safety, N f 2

⎛ σ ⎞ ⎛ τ ⎞ ⎜⎜ N f a ⎟⎟ + ⎜ N f m ⎟ = 1 S e ⎠ ⎜⎝ S ys ⎟⎠ ⎝ Sy we know S ys = 3 2

2

⎛ ⎛ σa ⎞ τ m ⎞⎟ ⎜ ⎜ ⎟ ⇒⎜Nf ⎟ + 3⎜ N f S ⎟ = 1 S e ⎠ y ⎠ ⎝ ⎝ substitute expressions for σ a & τ m for solid shaft 2

⎛ 32 M a ⎞ ⎟ ⎜Nf Kf 3 ⎜ π d S e ⎟⎠ ⎝

2

2

⎛ 16 Tm ⎞⎟ ⎜ + 3 N f K fsm =1 3 ⎜ ⎟ π d Sy ⎠ ⎝ 1 2⎤6

2 ⎡⎛ ⎛ ⎞ M 32 16 Tm ⎞⎟ ⎥ a ⎟ ⎢ ⎜ 3/17/2015 ⎜ d = ⎜Nf Kf + 3 N f K fsm ⎜ ⎢⎝ π S e ⎟⎠ π S y ⎟⎠ ⎥ ⎝ ⎦ ⎣

102

Factor of safety = N f 2

⎛ σ a ⎞ ⎛⎜ τ m ⎞⎟ ⎜⎜ N f ⎟⎟ + N f =1 S e ⎠ ⎜⎝ S ys ⎟⎠ ⎝ 2

2

⎛ ⎞ ⎛ σa ⎞ τ m ⎟ =1 ⎟⎟ + 3 ⎜ N f ⎜⎜ N f ⎜ ⎟ S S e ⎠ y ⎠ ⎝ ⎝ 2

N f = 1.73

Example: Design a shaft that must transmit 2 hp at 1725 rpm. Shaft is loaded with a spur gear and a sheave. Assume stress concentration for 2.25 for step radii in bending, 1.57 for step radii in torsion, and 2.5 at keyways. Assume corrected endurance strength = 50 MPa and yield strength is 150 MPa. Ref: Machine Design: An Integrated Approach.. R. L. Norton

Fg

Fr

∑ M A = R2 b + Fg p + Fs q = 0 ⇒ R2 = −0.4 Fg − 1.35 Fs F =R1 + Fg + R2 + Fs = 0 ∑3/17/2015

⇒ R1 = −0.6 Fg + 0.35 Fs

104

R2 x = −0.4 (40) − 1.35(162 ) ⇒ R2 x = −234.7 N R2 y = −0.4 (− 110 ) − 1.35(0 ) ⇒ R2 y = 44 N

R1x = −0.6(40 ) + 0.35(162) ⇒ R1x = 32.7 N R1 y = −0.6(− 110 ) + 0.35(0) ⇒ R1 y = 66 N

Fg

Fr

3/17/2015

105

M xzB = 1.2426 N .m M xzC = 7.2329 N .m M xzD = 1.0769 N .m M yzB = 2.508 N .m M yzC = −0.088 N .m M yzD = −0.088 N .m

Moment function in xz plane at various sections M xz = R1x 〈 z − 0〉1 + Fgx 〈 z − 50〉1 + R2 x 〈 z − 127〉1 + Fsx 〈 z − 172〉1 M xz = (32.7 )〈 z − 0〉1 + (40 )〈 z − 50〉1 + (− 234.7 )〈 z − 127〉1 + (162 )〈 z − 172〉1

Moment function in yz plane at various sections M yz = R1 y 〈 z − 0〉1 + Fgy 〈 z − 50〉1 + R2 y 〈 z − 127〉1 + Fsy 〈 z − 172〉1 M yz = (66 )〈 z − 0〉1 + (− 110)〈 z − 50〉1 + (44)〈 z − 127〉1 Calculate moment

M B = 2.7989; M C = 7.2334; M D = 1.0805

Using ASMEmethod 1 2⎤6

2 ⎡⎛ ⎛ ⎞ 16 Tm ⎞⎟ ⎥ M 32 B ⎢ ⎜ ⎟⎟ + 3 N f K fsm d1 = ⎜⎜ N f K f ⎜ ⎢⎝ π Se ⎠ π S y ⎟⎠ ⎥ ⎝ ⎦ ⎣

1 2⎤6

2 ⎡⎛ ⎛ ⎞ M 32 16 Tm ⎞⎟ ⎥ C ⎟ ⎢ ⎜ ⎜ + 3 N f K fsm d2 = ⎜ N f K f ⎜ ⎢⎝ π S e ⎟⎠ π S y ⎟⎠ ⎥ ⎝ ⎦ ⎣

1 2⎤6

2 ⎡⎛ ⎛ ⎞ 16 Tm ⎞⎟ ⎥ M 32 D ⎢ ⎜ ⎟ + 3 N f K fsm d 3 = ⎜⎜ N f K f ⎜ ⎢⎝ π S e ⎟⎠ π S y ⎟⎠ ⎥ ⎝ ⎣ ⎦

ANS: d1=11.7 mm

2 * 746 ⎛ 2 π ⎞(1725) ⎜ 60 ⎟ ⎝ ⎠ Tm = 8.3 N .m Tm =

d2=15.0 mm d3=09.8 mm As per available drawing d1>d2. Therefore select d3=10mm, d2=17mm, and d1=20 mm.

Corrected endurance strength !! 107

Design for fluctuating bending & fluctuating torsion Using factor of safety N f in Goodman line 1 σ a′ σ m′ = + N f Se Sut ⎡ 32 N f d =⎢ ⎢ π ⎣

3/17/2015

⎧ ⎪ ⎨ ⎪⎩

(K

f M a ) + 0.75 (K fsTa ) 2

2

Se

+

(K

2 2 ⎫⎤ ) ( ) + M K T 0 . 75 ⎪⎥ fm m fsm m ⎬ Sut ⎪⎭⎥⎦

1 3

108

Example: Design a shaft to support attachments shown in Figure. Torque & moment on shaft are both varying with time in repeated fashion, i.e., their alternating & mean components are of equal magnitude. Mean & alternating components of torque are both 17 N.m. There is no axial loads. Assume stress concentration for 2.25 for step radii in bending, 1.57 for step radii in torsion, and 2.5 at keyways. Assume corrected endurance strength = 50 MPa and ultimate strength is 250 MPa.

3/17/2015

109

⎡ 32 d =⎢ ⎢⎣ π

(K f M )

2

[(

d = 0.0063 K f M

)

1 ⎫⎤ 3

⎧1 1 + 0.75 K fsT ⎨ + ⎬⎥ S S ut ⎭⎥ ⎩ e ⎦

2

(

)

(

)]

+ 0.75 K fsT

2

2 1/ 6

M B = 2.8 N .m M C = 7.3 N .m M D = 1.1 N .m T = 17 N .m S e = 50 MPa S ut = 250 MPa Nf = 2

ANS: d1=28.6 mm d2=30.2 mm d3=28.3 mm As per available drawing d1>d2. Therefore select d3=29mm, d2=31 mm, and d1=33 mm. 3/17/2015

110

Deflection/Stiffness consideration † Critical speed Æ Unstable shaft. t = t4 t4 > t1 t = t1 t=0

3/17/2015

111

Equilibrium This image cannot currently be display ed.

Shaft rotates slightly off of its true centerline. Such rotor deflection is referred to as mode shape. First/Second mode….

† Speed at which shaft is unstableÆ Critical. „ Continuous increase in deflection without upper bound. 3/17/2015

112

Critical Speed of Shaft † Speed at which shaft is unstable. „ Continuous increase in deflection without upper bound. Tranmissibility as function of ζ 4 3.5

Tramissibility

3 2.5 2 1.5 1 .5 0

0

3/17/2015

1

2

3

4

5 Frequency ratio

6

7

8

9

113

10

Transmissibility † Ratio (T) of output to input. † T is function of operating frequency. † T > 1 Æ Amplification. Max amplification when forcing frequency (ω) and natural frequency of system (ωn) coincides. † T < 1 Æ Isolation (i.e. passenger compartment from automobile chassis). „ Minimize natural frequency of system.

† Often transmissibility is referred as “Q factor”. 3/17/2015

114

Critical Speed of Shaft m &x& + c x& + kx = F0 sin(ω t ) „ Estimation of critical speed helps to decide maximum operating speed. „ Rate of increase in amplitude estimates the allowable time to shoot the speed above critical speed.

Tramissibility

Tranmissibility as function of ζ 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1

Frequency of external force to Natural frequency of system.

ω 0.5 > > 1.42 ωn / d 0

3/17/2015

.2

.4

.6

.8

Frequency ratio

1

1.2

1.4 115

Estimating Natural Frequency k xi = W

m &x& = ∑ F = W − k ( xi + x ) x = X sin (ω t )

assume

− m ωn2 x = − k x k ωn = m 1 fn = 2π 3/17/2015

k m

1 fd = 2π

k ⎛ c ⎞ ⎟⎟ − ⎜⎜ m ⎝ 2m ⎠

2

116

Damped Natural Frequency m &x& + c x& + kx = 0

kx

C dx/dt

assume x = e st s1, 2

s1, 2

c =− ± 2m

2

⎛ c ⎞ k ⎜⎜ ⎟⎟ − ⎝ 2m ⎠ m

c k ⎛ c ⎞ ⎟⎟ =− ±i − ⎜⎜ 2m m ⎝ 2m ⎠

3/17/2015

2

1 fd = 2π

k ⎛ c ⎞ ⎟⎟ − ⎜⎜ m ⎝ 2m ⎠

2

117

harmonic function

How to estimate Transmissibility ?

1 .75 .5 .25 0 -.25 -.5 -.75 -1

Force and displacement

0

.02

.04

.06

.08

.1

m &x& + c x& + kx = F0 sin(ω t )

.12

.14

.16

.18

time

x = X sin (ω t − φ ) X = amplitude φ = phase angle

m (− ω 2 X sin (ω t − φ )) + c (ω X cos(ω t − φ )) + k ( X sin (ω t − φ )) = F0 sin(ω t ) F0 (k − mω )sin(ω t − φ ) + cω cos(ω t − φ ) = X sin(ω t ) 2

F0 = X 3/17/2015

(k − mω ) + (cω ) 2 2

2

T=

(kX )2 + (cωX )2 F0

(k )2 + (cω )2 (k − mω 2 )2 + (cω )2

X

T= X

118

.2

How to estimate Transmissibility ? F0 = X

kx

C dx/dt

(k − mω ) + (cω ) 2 2

2

† Ratio (T) of output to input. T=

(kX )2 + (cωX )2 F0

(k )2 + (cω )2 2 2 (k − mω ) + (cω )2

X

T= X

3/17/2015

T=

⎛ cω ⎞ 1+ ⎜ ⎟ ⎝ k ⎠

2

⎛ m 2 ⎞ ⎛ cω ⎞ ⎜1 − ω ⎟ + ⎜ ⎟ k k ⎠ ⎝ ⎠ ⎝ 2

2

119

Mathematical Model of Transmissibility T=

⎛ cω 2 m k 1 + ⎜⎜ ⎝ cc k ⎛ ω ⎞ ⎜⎜1 − 2 ⎟⎟ ⎝ ωn ⎠ 2

T=

2

⎛ cω 2 m k + ⎜⎜ ⎝ cc k

⎛ cω 1 + ⎜⎜ 2 ⎝ cc

m⎞ ⎟ k ⎟⎠

2

⎛ ω ⎞ ⎛ cω ⎜⎜1 − 2 ⎟⎟ + ⎜⎜ 2 ⎝ ωn ⎠ ⎝ cc 2

⎞ ⎟ ⎟ ⎠

2

⎞ ⎟ ⎟ ⎠

2

2

m⎞ ⎟ k ⎟⎠

2

1 + (2ςλ )

2

T=

λ=

(1 − λ ) + (2ςλ ) 2 2

ω c 3/17/2015 ς= ωn cc

2

120

ζ = 0.5

0

.5

1

1.5

2

2.5

1.5 1.4 1.3 1.2 1.1 1 .9 .8 .7 3.5 .6 .5 .4 .3

Tramissibility

Tramissibility

ζ = 0.1 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 .5 0

3

Frequency ratio

0

.5

1

1.5

2

2.5

3

3.5

2.5

3

3.5

Frequency ratio ζ=1 1.2 ζ = 1.5

1

1.1

.9

1.05

.8

1

.7

.95

.6 .5

0

.5

1

1.5

2

Frequency ratio

2.5

3

Tramissibility

Tramissibility

1.1

.9 .85 3.5 .8 .75 .7

0

.5

1

1.5

2

Frequency ratio

3/17/2015

121

Tranmissibility as function of ζ 4 3.5

Tramissibility

3 2.5 2

Robust design ???

1.5

Vibration free design ????

1 .5 0

0

1

2

3

4

5

6

7

8

9

10

Frequency ratio

3/17/2015

122

Question: Determine the natural frequency of mass M on the end of a cantilever beam (length L) of negligible mass Deflection δ of cantilever beam under a concentrated load P 3

M

PL P = δ= 3E I k

1 fn = 2π

k M

1 fn = 2π

3E I L3 M

3/17/2015

123

Question: Determine the natural frequency of mass M on the end of a simply supported beam (length L) of negligible mass Deflection δ of simply supported beam under a concentrated load P 3

M

PL P = δ= 48 E I k

1 fn = 2π

k M

1 fn = 2π

48 E I 4 ⇒ fn = 3 2π L M

3/17/2015

Four times !!!! 3E I L3 M 124

Dunkerley equation • Estimates fundamental critical speed of shaft carrying a number of components (gears, pulley, coupling, etc.) • Estimate critical speed of each individual subsystem by direct formula. • To obtain critical speed of shaft-system, combine individual frequencies using following equation. 1 1 1 1 ω1 = natural speed if only mass 1 exists

ω

2 cr

=

ω

2 1

3/17/2015

+

ω

2 2

+ ... +

ωn2

ω2 = natural speed if only mass 2 exists ωn = natural speed if only mass n exists 125

200

Disk 3

200

Disk 2

200

Disk 1

Ex: Assume shaft (negligible mass, dia =40mm) shown in Fig. is made of steel with E=207 GPa. Mass of each disk is 13.5 kg. Estimate fundamental frequency of system.

200

• Disk 1 and Disk 3 have same mass and are symmetrically placed, therefore natural frequency of disk 1 and disk 3 will be same. • Find natural frequencies of Disk 1 & Disk 2 3/17/2015

126

Natural frequency of Disk 2 can be estimated using P L3 P = δ= 48 E I k

fn =

1 2π

1 fn = 2π

k M 48 E I 4 ⇒ fn = 3 2π L M

3E I L3 M

Natural frequency of Disk 1 can be estimated using

δ=

P ⎛ 600 ⎞ P ( 0 . 2 ^ 3 ) + 3 * ( 0 . 2 ^ 3 ) − 2 * ( 0 . 2 ^ 2 ) * 0 . 8 − ( 0 . 2 ^ 4 ) /( 0 . 8 ) ⎟= ⎜ 6 E I ⎝ 800 ⎠ k a

fn = 255.1 3/17/2015

P δx = 6E I

b

l ⎡b 3 a3 x ⎤ 3 2 ⎢ x − x − a + 3ax − 2alx − ⎥ l ⎦ ⎣l

Using Dunkerley equation 1

ωcr2

=

1

ω12

+

1

ω22

+

1

ω32

ω1 = natural speed if only Disk 1 ω2 = natural speed if only Disk 2 ω3 = natural speed if only Disk 3

1 2 1 = 2+ 2 2 f cr f1 f2

3/17/2015

128

Question † A simply supported 25-mm diameter (E=207 GPa, Specific weight = 75 kN/m^3) uniform steel shaft is 600 mm long. „ Find the lowest critical speed of the shaft. „ If aim is to double the critical speed, find the new diameter. „ What will be the critical speed if the shaft length is reduced from 600mm to 400 mm. 3/17/2015

129

Shafts are meant to transmit power through sheave (pulley: Flat, V), gear (spur, helical, bevel, etc.), sprocket, etc. Shaft is positioned on supports (bearings, etc.). Key joint (flat key, taper pin), coupling, or interference fit is required to mount required elements on the shaft. 3/17/2015

130

Keys:

Key Joint

ASME defines a key “demountable element which when assembled into keyways, provides a positive means for transmitting torque between shaft and hub.

Keyways

Square

Rectangular

Gibhead taper

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† Key: „ Primary function: Transmit torque from shaft to hub of mating element and viceversa. „ Prevent relative rotation between shaft & joining element.

† Keyway: „ A recess or slot on shaft and/or hub to accommodate key. „ Keyway results in stress concentration (initial value ~ 2.5) in shaft and hub. 3/17/2015

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Flat Key Assembly NOTE: Key is inexpensive and relatively easy to replace if keyway is undamaged. Therefore relatively softer compared to keyway material (having lesser strength) is used for the key.

Ductile failure

Two modes of failures:

1. Shear Failure (AS = w.l) Æ Fatigue failure 2. Crushing/Bearing Failure (Abearing=(h/2.l) Æ static failure

Parallel or taper key Assumptions: Axis, shear force at top/bottom/sides, 3/17/2015 133 turning couple, uniform contact

Failure of KEY 3/17/2015

Rounding of corners 134

Standard keys Shaft diameter (mm)

Key width* Height (mm*mm)

8