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Design of Shaft
A shaft is a rotating member usually of circular cross-section (solid or hollow), which transmits power and rotational motion.
Machine elements such as gears, pulleys (sheaves), flywheels, clutches, and sprockets are mounted on the shaft and are used to transmit power from the driving device (motor or engine) through a machine.
Press fit, keys, dowel, pins and splines are used to attach these machine elements on the shaft.
The shaft rotates on rolling contact bearings or bush bearings.
Various types of retaining rings, thrust bearings, grooves and steps in the shaft are used to take up axial loads and locate the rotating elements.
Couplings are used to transmit power from drive shaft (e.g., motor) to the driven shaft (e.g. gearbox, wheels).
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2. The connecting shaft is loaded primarily in torsion.

3. Combined bending and torsion loads on shaft: Shaft carrying gears.
From power and rpm find the torque (T), which gives rise to shear stress.
From Torque (T) and diameter (d), find Ft = 2T/d. From Ft and pressure angles of gears you can find Fr and Fa.
Fr and Ft are orthogonal to each other and are both transverse forces to the shaft axis, which will give rise to normal bending stress in the shaft. When shaft rotates, bending stress changes from tensile to compressive and then compressive to tensile, ie, completely reversing state of stress.
Fa will give rise to normal axial stress in the shaft.

4. Loads on shaft due to pulleys
Pulley torque (T) = Difference in belt tensions in the tight (t1) and slack (t2) sides of a pulley times the radius (r), ie
T = (t1-t2)xr
Left pulley torque
T1 = ( )x380=1,710,000 N-mm
Right pulley has exactly equal and opposite torque:
T2 = ( )x380=1,710,000 N-mm
FV2
Bending forces in vertical (Fv) and horizontal (FH) directions:
At the left pulley: FV1=900N; FH1= = 9900N
At the right pulley: FV2= =9900N; FH2=0

5. Torque and Bending moment diagrams for the pulley systemFH
1,710,000 N-mm
2,227,500 MH
Torque diag.
900N
9900N
From Horizontal forces (FH) and vertical forces (Fv), Bending moments MH & MV are drawn separately.
Then the resultant moments at various points on the shaft can be found from FV
911,250
M V
2,227,500
The section of shaft where the left pulley is located has obviously the highest combination of Torque (1,710,000 N-mm) and Bending moment (2,406,685 N-mm)
2,227,500
2,406,685
Resultant bending moment

6. Power, toque & speed For linear motion: Power = F.v (force x velocity)
For rotational motion
Power P = Torque x angular velocity
= T (in-lb).w (rad/sec) in-lb/sec
= T.(2 p n/60) in-lb/sec [n=rpm]
= T.(2 p n/(60*12*550)) HP [HP=550 ft-lb/sec]
= T.n/63,025 HP
or, T= 63,025HP/n (in-lb), where n = rpm
Similarly, T= 9,550,000kW/n (N-mm), where n = rpm

7. Shear (t) and bending (s) stresses on the outer surface of a shaft, for a torque (T) and bending moment (M)
For solid circular section:
For hollow circular section:

8. Principal Normal Stresses and Max Distortion Energy Failure criterion for non-rotating shafts
The stress at a point on the shaft is normal stress (s) in X direction and shear stress (t) in XY plane.
From Mohr Circle:
Max Distortion Energy theory:
Putting values of S1 & S2 and simplifying:
This is the design equation for non rotating shaft

9. Design of rotating shafts and fatigue consideration
The most frequently encountered stress situation for a rotating shaft is to have completely reversed bending and steady torsional stress. In other situations, a shaft may have a reversed torsional stress along with reversed bending stress.
The most generalized situation the rotating shaft may have both steady and cyclic components of bending stress (sav,sr) and torsional stress (tav,tr).
From Soderberg’s fatigue criterion, the equivalent static bending and torsional stresses are:
Using these equivalent static stresses in our static design equation, the equation for rotating shaft is:

10. Bearing mounting considerations and stress concentration

11. Conventional retaining (or snap) rings fit in grooves and take axial load, but groves cause stress concentration in shaft
Retaining rings are standardized items, available in various standard sizes with various axial load capacities.

12. Push type retaining rings – no grooves required, less stress concentration, but less axial support

13. Various types of keys for transmitting torque

14. Other common types of keys

15. Various types of collar pins

16. Integrated splines in hubs and shafts allow axial motion and transmits torque
All keys, pins and splines give rise to stress concentration in the hub and shaft

17. Chapter 11 - Keys, Couplings and Seals
Chapter Objectives:
How attach power transmission components to shaft to prevent rotation and axial motion?
Torque resistance: keys, splines, pins, weld, press fit, etc..
Axial positioning: retaining rings, locking collars, shoulders machined into shaft, etc….
What is the purpose of rigid and flexible couplings in a power transmission system?
Specify seals for shafts and other types of machine elements.

18. Keys
Most common for shafts up to 6.5” is the square and rectangular keys:
Cost effective means of locking the
Can replace damaged component
Ease of installation
Can use key as “fuse” – fails in shear at some predetermined torque to avoid damaging drive train.
Advantages:
Figure 11.1

19. Square and rectangular keys:
Step 1 – Determine key size based on shaft diameter
Step 2 – Calculate required length, L, based on torque (11.4)

20. Step 3 – Specify appropriate shaft and bore dimensions for keyseat:
See Figure 11.2
For 5/16” key
SHAFT
BORE
Note, should also specify fillet radii and key chamfers – see Table 11-2

22. Other types of keys:
Tapered key – can install after hub (gear) is installed over shaft.
Gib head key – ease of extraction
Pin keys – low stress concentration
Woodruff key – light loading offers ease of assembly

23. 11.4 Design of Keys – stress analysis to determine required length:=
Torque being transmitted
No load
T = F/(D/2) or F = T/(D/2) this is the force the key must react!!!

24. Bearing stress
Shear stress
Required Length based on Bearing Stress:
Required Length based on Shear Stress:
Typical parameters for keys:
N = 3, material 1020 CD (Sy = 21,000 psi)

25. Solution (note since key is weakest material, focus analysis on key!):
Example: Specify the complete key geometry and material for an application requiring a gear (AISI OQT 1000) with a 4” hub to be mounted to a 3.6” diameter shaft (AISI 1040 CD). The torque delivered through the system is 21,000 lb-in. Assume the key material is 1020 CD (Sy = 21,000 psi) and N = 3.
Solution (note since key is weakest material, focus analysis on key!):
See handout

26. Splines
Advantages:
Can carry higher torque for given diameter (vs keys) or
Lower stress on attachment (gear)
Better fit, less vibration (spline integral to shaft so no vibrating key)
May allow axial motion while reacting torque
Disadvantage:
Cost
Impractical to use as fuse

27. Splines “Axial keys” machined into a shaft
Transmit torque from shaft to another machine element

28. Advantages Uniform transfer of torque Lower loading on elements
No relative motion between “key” and shaft
Axial motion can be accommodated (can cause fretting and corrosion)
Mating element can be indexed with a spline
Generally hardened to resist wear

29. Spline Types Straight SAE 4, 6, 10 or 16 splines Involute
Pressure angles of 30, 37.5, or 45 deg.
Tend to center shafts for better concentricity

30. SAE Spline Sizes A: Permanent Fit B: Slide without Load
C: Slide under Load
Pg 504

31. Two types of splines:
Straight Sided
Involute:

32. Use this for spline design – SAE formulas based on 1,000 psi bearing stress allowable!!
Use this to get diameter. Then table 11.4 to get W, h, d

33. Torque Capacity
Torque capacity is based on 1000 psi bearing stress on the sides of the splines
T = 1000*N*R*h
N = number of splines
R = mean radius of the splines
h = depth of the splines

34. Torque Capacity Cont’d
Substituting R and h into torque equation:

35. Torque Capacity Cont’d
Further refinement can be done by substituting appropriate values for N and d.
For 16 spline version, with C fit,
N = 16 and d = .810D
Torque in IN-LBS/INCH of spline
Required D for given Torque

36. Torque Capacity for Straight Splines
Pg 505

37. Torque Capacity for Straight Splines

38. T = torque capacity in in-lbs
Example: A chain sprocket delivers 4076 in-lbs of torque to a shaft having a 2.50 inch diameter. The sprocket has a 3.25 inch hub length. Specify a suitable spline having a B fit.
T = kD2L
T = torque capacity in in-lbs
kD2 = torque capacity per inch (from Table 11-5)
L = length of spline in inches

39. Example Continued
From Table 11-5, use 6 splines

40. Torque Capacity for Straight Splines
4076/3.25
2.5

41. Example: Specify straight spline for the previous problem (i. e
Example: Specify straight spline for the previous problem (i.e. Torque = 21,000 lb-in and shaft is 3.6 in diameter.

42. Taper & Screw Expensive – machining Good concentricity
Moderate torque capacity
Can use a key too

43. Couplings
Used to connect two shafts together at their ends to transmit torque from one to the other.
Two kinds of couplings:
RIGID
FLEXIBLE

44. Rigid Couplings NO relative motion between the shafts.
Precise alignment of the shafts
Bolts in carry torque in shear. N = # of bolts.

45. Flexible Couplings Transmit torque smoothly
Permit some axial, radial and angular misalignment

46. Flexible Couplings

47. Flexible Couplings

48. Lord Corp. Products

49. Flexible Coupling

50. Universal Joints Large shaft misalignments permissible
Key factors in selection are Torque, Angular Speed and the Operating Angle
Output not uniform wrt input
Output IS uniform wrt input

51. Axial Constraint Methods
Spacers
Retaining ring
Retaining ring
Shoulders

52. Retaining Rings

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Locknuts
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