1. Shafts – Definition
Generally shafts are members which rotate in order to transmit power or motion. They are usually circular in cross section, and that’s the type we will analyze.
Shafts do not always rotate themselves, as in the case of an axle – but axles support rotating members.

2. Common Shaft Types $ $$
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3. Elements Attached to a Shaft
Shoulders provide axial positioning location, & allow for larger center shaft diameter – where bending stress is highest.

4. Common Shaft Materials
Typically shafts are machined or cold-drawn from plain hot-rolled carbon steel. Applications requiring greater strength often specify alloy steels (e.g., 4140).
Some corrosion applications call for brass, stainless, Ti, or others.
Aluminum is not commonly used (low modulus, low surface hardness).

5. Shafts for Steady Torsion
Often the rotating mass & static load on a shaft are neglected, and the shaft is sized simply to accommodate the transmitted power. In such cases, the engineer typically seeks to limit the maximum shear stress max to some value under the yield stress in shear (Sys), or to limit the twist angle  .

6. Shafts in Steady Torsion
Chapter 1 review equations:
kW = FV/1000 = Tn/9600
hp = FV/745.7 = Tn/7121
kW = kilowatts of power
F = tangential force (N)
V = tangential velocity (m/s)
T = torque (N x m)
n = shaft speed (rpm)

7. U.S. Power Units Review equation: hp = FV/33,000 = Tn/63,000 where,
hp = horsepower
F = tangential force (lb.)
V = tangential velocity (ft/min)
T = torque (lb - in.)
n = shaft speed (rpm)

8. Steady State Shaft Design
Because shafts are in torsion, the shear stress is generally the limiting factor. Recall that
max = Tc/J
where c = radius, and, for a circular shaft,
J = d4/32
As always, use a safety factor of n to arrive at
all = max /n

9. Limiting the Twist Angle
In some cases, it is desired to limit the twist angle to a certain value. Recall:
 = TL/GJ
L = length
G = shear modulus
 is always in radians (deg. x /180)

11. Combined Static Loads The axial stress is given by:
x = Mc/I + P/A = 32M/ D3 + 4P/ D2
(M = bending moment, P = axial load, D = diameter)
The torsional stress is given by:
xy = Tc/J = 16T/ D3
(T = Torque, J = polar moment of inertia, c = radius)
(For circular cross sections.)

12. Maximum Shear Stress Theory
Typically the axial load P is small compared to the bending moment M and the torque T, and so it is neglected. (Notice how direct shear is completely omitted.)
Recall the maximum shear stress criterion:
Sy/n = (x2 + 4 xy 2)1/2

13. Maximum Shear Stress Theory
Substitute the previous values for x and xy into MSST to obtain:
This equation, or the related eq. for the maximum energy of distortion theory (MDET), is useful for finding either D or n. Note that this would be for steady loads.

14. Fluctuating Loads
In their support of rotating members, most shafts are subject to fluctuating loads, possibly including a shock component as well. We’ve covered fatigue & impact in previous lectures, and that material is directly applied to the design of shafts.

15. Shock Factors
In shaft design, shock loading is typically accounted for by yet more fudge factors, Ksb (bending shock) and Kst (torsional shock). The values of these factors range from 1.0 to The shock factors are applied to their respective stress components.

16. Shaft Design Formulas
There are a number of shaft design formulas that incorporate failure theories (MSST or MDET) with fatigue theories (Goodman or Soderberg).
In practice, using MDET with the Soderberg criterion is probably the most accurate.

17. Shaft Design Formulas
; MDET with the Goodman criterion and shock factors. For Soderberg, recall that you use Sy instead of Su.

18. Fully-Reversed Bending
In analyzing a rotating shaft for fatigue life, you will need to compute Mm and Ma. The moment might be due to a rotating imbalance or due to the tension from a belt, or radial loading from gears. No matter the case, because the shaft is rotating, it experiences both tension and compression from the bending loads: therefore, typically, Mm = 0, and Ma = Mmax. (A sinusoidal variation about zero….)

19. Example 9.2
Find required dia. of shaft using MDET & Soderberg fatigue relation. Surface is ground. Su = 810 MPa, and Sy = 605 MPa. Torque varies by +/- 10%. The fatigue stress [] factor Kf = Temp = 500 oC, and n = 2. Survival rate = 50%.

20. Critical Speeds of Shafts
All structures exhibit one or more natural, or resonant frequencies. When a shaft rotates at speeds equal or close to the natural frequencies, resonance may occur. This is usually to be avoided, although some designs feature resonance.
Generally the designer tries to keep the speed at least 25% lower than o. But in some cases, the operating speed is higher.

21. The Rayleigh Equation ncr = (1/2)[ (gW)/(W2)]1/2
ncr = critical speed (rev/sec)
g = gravitational acceleration (9.81 m2/s)
W = concentrated weight including load (kg)
 = respective static deflection of the weight.

22. Shaft Attachments
Many different methods, each with pros and cons of both function, ease of use, and cost: the designer must balance between these factors.
Some methods are very weak compared to the shaft (e.g., a set screw), others are stronger than the shaft itself.

23. Shaft Attachments: Keys
Square (w ~ D/4) Flat Round (or tapered)
Shaft Attachments: Keys
Gib head
Woodruff key

24. Shaft Attachments: Pins
Straight Tapered Roll

25. Shaft Attachments: Tapered Clamps
            
          

26. Stresses in Keys
Distribution of force is quite complicated. The common assumption is that the torque T is carried by a tangential force F acting on radius r:
T = Fr

27. Stresses in Keys
From T = Fr, both shear and compressive bearing stresses may be calculated from the width and length of the key.
The safety factor ranges from n = 2 (ordinary service) to n = 4.5 (shock).
The stress concentration factor in the keyway ranges from 2 to 4.

28. Splines
Splines permit axial motion between matching parts, but transmit torque. Common use is automotive driveshafts – check your car.

29. Couplings
In many designs involving shafts, two shafts must be connected co-axially. Couplings are used to make these connections.
Couplings are either rigid or flexible. Rigid couplings require very close alignment of the shafts, generally better than .001” per inch of separation.

30. Rigid Couplings – Sleeves
The simplest type of coupling is the simple sleeve coupling. But this also has the lowest torque capacity.

31. Rigid Couplings - Flanged
Taper locked
Keyed to shaft
Great web resource:

32. Flexible Couplings
There are many types of flexible couplings as well. Generally a flexible element is sandwiched in between, or connected to, rigid flanges attached to each shaft.
Alignment is still important! Reaction forces increase with misalignment, and often bearings are not sized properly for reaction forces. “Mechanics” often assume that because the coupling is flexible, alignment is unimportant.

33. Two-piece “Donut” (or toroidal) flexible coupling

34. Universal Joints
U-joints are considered linkages rather than couplings, but serve the same purpose of transmitting rotation.
Very large angular displacements may be accommodated.
Single joints are not constant-velocity. Almost always, two joints are used. The angles must be equal for uniform velocity.

35. Shafts parallel
but offset
Shafts not parallel but intersecting

36. It’s Not Nanotechnology, But You Could Get Rich!
Despite decades of research and 1000s of Ph.D. theses, highly engineered shafts and components fail all too frequently. Even NASA can’t always get it right.
Often the connections are to blame: keys, splines, couplings, and so on. Fatigue wear failure is the culprit.

37. Bearing Definition
“ A device that supports, guides, and reduces the friction of motion between fixed and moving machine parts.”

38. Bearing Types
Three major types: hydrodynamic or journal bearings, rolling-element bearings, and sleeve bearings.

39. Design of Journal Bearings
Nomenclature:
r = journal radius
c = radial clearance
L = length of bearing
= viscosity
n = speed (rps)
W = radial load
P = load per projected
area (W/2rL)
In this figure, U = tangential velocity and F = frictional force

40. Journal Bearing Design Charts
Procedure: generally, you first calculate the dimensionless Sommerfeld Number, from,
S = (r/c)2(n/P)
This characteristic number is used along with the L/D ratio of the bearing to enter the Design Charts. In some cases, you find the Sommerfeld Number from given data.

41. ugu2155x_1013.jpg

42. ugu2155x_1014.jpg

43. ugu2155x_1015.jpg

44. Journal Design Examples
Problem 10.6:
A 4-in. diameter  2-in. long bearing turns at 1800 rpm; c/r = 0.001; h0 = in. SAE 30 oil is used at 200F. Through the use of the design charts, find the load W.

45. Journal Design Example I
Looking at Figure 10.7, find the viscosity for SAE 30 wt. Oil at 200oF, = 1.2 x 10-6 reyns
In this problem, we don’t have enough data to calculate S, but we can look it up on the charts

46. Oil Viscosity
Fig. 10.7, p. 385

47. Journal Design Example I
We are given r = 2”, and c/r = Therefore, the clearance c = .002.”
We are also given the minimum film thickness, ho = .001.”
This enables us to enter Design Chart with L/D = 0.5, and ho/c = 0.5.
Then, you can find S = 0.5 on the chart.

49. Journal Design Example
With S = 0.5, we can go back to the definition of the Sommerfeld #, from Equation :
S = (r/c)2(n/P)
Rearranging this to solve for P, we have
P = [(r/c)2 n]/S

50. Journal Design Example I
P = [(r/c)2 n]/S
S = 0.5
(c/r) = .001, so (r/c) = 1000
= 1.2 x 10-6 psi-sec
N = 1800 rpm = 30 rps
Therefore P = 72 psi, and,
W = P*L*D = 576 lbs.

51. Journal Design Example II
A 25mm diameter by 25mm long bearing carries a radial load of 1.5 kN at 1000 rpm; c/r = ,  = 50 mPa-sec. Use charts to find:
A) The minimum oil film thickness ho
B) The friction power loss

52. Journal Design Example II
In this case, we have enough information to calculate the Sommerfeld #,
S = (r/c)2(n/P)
P = W/DL = 1500/(.025*.025) = 2.4 MPa
n = 1000 rpm = rps
c/r = .0008, so r/c = 1250
 = 50 mPa-sec
S = 0.543

53. Journal Design Example II
With S = 0.543, and L/D = 1.0, we can once again chart to find
ho/c = 0.75
We are given c/r = .0008, and r = 12.5mm, so c = .01mm.
Therefore, ho = .75*.01 = .008mm (part A)

55. Journal Design Example II
Next, to find the friction power loss, we can use chart. We have S = .543, and L/D = From that we can look up the coefficient of friction variable
(r/c)*f = 11
Since c/r is given as .0008, f = .0088

57. Journal Design Example II
Knowing the coefficient of friction f, we can then use equation to calculate the friction torque, Tf:
Tf = fWr = .0088*1500*.0125 = .165 N-m
Then the friction power loss is found from equation:
Power = Tf*n/159
= (.165*16.67)/159 = .017 kW

58. Rolling Element Bearings
Rolling element, or “anti-friction” bearings, make use of spherical or cylindrical rolling elements captured between inner and outer rings. The rolling elements support the load, and transmit rotation by rolling, rather than sliding.

59. Rolling Element Bearings
A major benefit of rolling versus sliding is that the coefficient of friction is much lower. Recall that for journal bearings operating hydrodynamically,
0.002 < f < 0.010
For rolling element bearings,
0.001 < f < 0.002

60. Rolling Element Benefits
Observe that f is much more uniform. In addition, f is much less a function of rotational speed. This means that friction power loss is more predictable, and remains constant over a range of speeds. Rolling element bearings also experience much less wear at slower speeds than do journal bearings.

61. Ball Bearings & Roller Bearings
There are two types of rolling element bearings, ball bearings and roller bearings.
In general, ball bearings can operate at higher speeds (but with less load), and roller bearings operate at lower speeds but with heavier loads. The difference is due to point contact versus line contact.

62. Ball Bearings
There are many types of ball bearings: deep-groove, double or triple row, angular contact, thrust, cam followers, etc. Each is best suited for a particular application.
For different types, there are “series” numbers, usually in increasing order of cross section (i.e., thicker rings, larger spheres, etc.)

64. Ball Bearing Dimensions

65. Roller Bearings
The same situation exists with roller bearings: there are single and double row, removable inner or outer race, tapered or straight rollers, thrust bearings, and spherical bearings. Again, each is best suited for a particular application.

66. Bearing Examples
Double row spherical bearing from the axis of the earth: high load rating with angular misalignment capability.
“NU” bearing, straight cylindrical rollers, for radial loads only – note translational ability
Light-weight single row ball bearing
Tapered roller bearing: common type of automotive wheel bearing. Car example, 1.79 x 108 revolutions with no maintenance.

67. Bearing Load & Life
There is a “basic load rating” associated with each bearing. It is nominally the radial load that a bearing can support for 106 revolutions. These numbers, however, are for comparison purposes only. In practice, the design load for most bearings is only a few % of the “basic load rating.”

68. Equivalent Radial Load
The “basic load rating” is given for purely radial loads only. However, most bearings need to support both radial and axial loads.
Equations are used to calculate an equivalent radial load given actual radial and axial loads, and the geometry of the design:

69. Equivalent Radial Load, P
P = XVFr + YFa
P = VFr (cyl. rollers, gen.)
Fr = applied radial load
Fa = applied axial load (thrust)
V = rotation factor, 1.0 for inner-ring rotation, 1.2 for outer-ring rotation
X = a radial factor
Y = a thrust factor
NOTE that straight cyl. roller bearings cannot support much thrust.

70. Equivalent Load with Shock
P = Ks(XVFr + YFa)
P = KsVFr
Ks is a shock or service factor, find in table. Ks ranges from 1.0 to 3.0 depending on the type of bearing and the service.

71. The L10 Life
Bearing life is an important consideration in many designs. The desired lifetime could range from a few million to a few billion revolutions. It doesn’t take long for 1000 rpm running 24/7/365 to add up. (0.5x109)
The L10 life refers to the expected life (hours or revs) under a given load at which 90% of the bearings will survive.

72. L10 Life in Revolutions L10 = life rating in 106 revolutions
C = basic load rating from manufacturer or Tables 10.3 and NOTE difference between C and Cs.
P = equivalent radial load
a = 3 for ball bearings or 10/3 for roller bearings.

73. L10 Life in Hours
L10 = rating life in hours
n = rotational speed, rpm

74. L5 and Beyond
The L10 life is based on a 90% survival rate. If the application requires higher reliability, then a “life adjustment [fudge] factor,” Kr, is applied. Kr is found in chart, and ranges from 1.0 (90% reliability) to 0.2 (99% reliability). “L5” is the name given to any reliability > 90%.