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Shafts and Shafts Components

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An academic presentation that highlights main shafts applications and conduct stress and fatigue analysis in shafts as shafts being an essential part in the automotive manufacturing

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Shafts and Shafts Components

  1. 1. Shafts and Shafts Components Hussein Basma Lebanese American University Please note that this presentation is totally based on “Shigley’s Mechanical Engineering Design Book” as the only used reference including all the definitions, examples and details.
  2. 2. Introduction about shafts Shaft materials. Shaft layout Shaft design for stress Deflection consideration Critical Speeds for shafts Miscellaneous Shaft Components Limits and fits Outline
  3. 3. Introduction  Shafts are different than axels.  Shafts are rotating members, usually of circular cross section.  Axels are nonrotating members analyzed as static beams.  Shafts are used to transmit power or motion.  They are essential part in many machine designs.
  4. 4.  Material selection.  Geometric layout.  Stress and strength static strength fatigue strength.  Deflection and rigidity  Vibration due to natural frequency What to examine about shafts?
  5. 5. Shaft Materials  Deflection is affected by stiffness not strength.  Stiffness is related to the modulus of elasticity, thus the material.  Modulus of elasticity is constant for all steels, thus material decisions will not control rigidity.  Rigidity will be controlled only by geometric decisions.
  6. 6. Shaft Layout  The design of the shafts does not follow a certain rule.  It depends mainly on the application.  Certain conventions to follow in shaft design: -Avoid long shafts to minimize deflection -Support the loads between bearings -Avoid using more than two bearings.
  7. 7. Shaft Layout  In many shaft applications the aim is to transmit torque from one gear to another gear or pulley.  The shaft must be sized to support the torsional stress and deflection.  To transmit this torque certain torque-transfer elements are used such as:  Torque transmission
  8. 8. Keys
  9. 9. splines
  10. 10. setscrews
  11. 11. pins
  12. 12. Tapered fits
  13. 13. Shrink fits
  14. 14. Shaft Design for Stress  Locate the critical locations along the shaft.  Critical locations are on the outer surface where bending moment is large, torque is present and stress concentrations exist.  Bending moment is determined by shear and bending moment diagrams. (1 or more planes could be needed)  A steady bending moment will produce a completely reversed moment on rotating shaft (σ 𝑚 =0)  Axial stresses can be neglected.  Critical Locations
  15. 15. Shaft Design for Stress  Shaft stresses  We will deal with bending, torsion and axial stresses as midrange and alternating components.
  16. 16.  For analysis it is appropriate to combine the different stresses into alternating and midrange von Mises stresses (again we can neglect axial stresses): Shaft Design for Stress  P.S: For ductile materials the use of the stress-concentration factors is sometimes optional.
  17. 17. Shaft Design for Stress  Now we can take the acquired von Mises stresses and use these values in any of the known fatigue failure criteria to evaluate the factor of safety n or, for design purposes, the diameter d.  Reminder of the fatigue failure criteria:  DE-Goodman  DE-Gerber  DE-ASME Elliptic  DE-Soderberg
  18. 18. Shaft Design for Stress
  19. 19.  DE-Goodman  Plugging in the values of the von Mises stresses will yield the following equations, one solved for n and one for d :
  20. 20.  DE-Gerber
  21. 21.  DE-ASME Elliptic
  22. 22.  DE-Soderberg
  23. 23.  For rotating shafts with constant bending and torsion, the bending stress is completely reversed and the torsion is steady. ( )  We should always consider the possibility of static failure in the first load cycle. ( check for yielding)  The ASME Elliptic criteria takes yield into account but it is not entirely conservative. Shaft Design for Stress
  24. 24.  For this purpose a von Mises maximum stress is calculated: Shaft Design for Stress  To check for yielding, we compare the maximum stress to the yield strength:
  25. 25. Shaft Design for Stress  Example: At a machined shaft shoulder the small diameter d is 28 mm, the large diameter D is 42 mm, and the fillet radius is 2.8 mm. The bending moment is 142.4 N.m and the steady torsion moment is 124.3 N.m. The heat-treated steel shaft has an ultimate strength of 𝑆 𝑢𝑡 =735 MPa and a yield strength of 𝑆 𝑦 = 574 MPa. Our reliability goal is 0.99. (a) Determine the fatigue factor of safety of the design using each of the fatigue failure criteria described in this section. (b) Determine the yielding factor of safety.
  26. 26. Shaft Design for Stress At a machined shaft shoulder the small diameter d is 28 mm, the large diameter D is 42 mm, and the fillet radius is 2.8 mm. The bending moment is 142.4 N.m and the steady torsion moment is 124.3 N.m. The heat-treated steel shaft has an ultimate strength of 𝑆 𝑢𝑡 =735 MPa and a yield strength of 𝑆 𝑦 = 574 MPa. Our reliability goal is 0.99
  27. 27. Shaft Design for Stress (a) D/d = 42/28 = 1.50, r/d 2.8/28= 0.10, 𝐾𝑡 = 1.68 (Fig. A–15–9). 𝐾𝑡𝑠 = 1.42 (Fig. A–15–8), q = 0.85 (Fig. 6–20), 𝑞 𝑠ℎ𝑒𝑎𝑟 = 0.92 (Fig. 6–21). From Eq. (6–32), 𝐾𝑓 = 1 + 0.85(1.68 − 1) = 1.58 𝐾𝑓𝑠 = 1 + 0.92(1.42 − 1) = 1.39 Eq. (6–8) : 𝑆 𝑒’ = 0.5(735) = 367.5 MPa Eq. (6–19): 𝑘 𝑎 =4.51(735)−0.265 Eq. (6–20): 𝑘 𝑏 =( 28 7.62 )−0.107 𝑘 𝑐= 𝑘 𝑑 = 𝑘 𝑓 =1  Solution:
  28. 28. Shaft Design for Stress  Solution: Table 6–6: 𝑘 𝑒 = 0.814 𝑆 𝑒 =0.787(0.870)0.814(367.5) = 205MPa For a rotating shaft, the constant bending moment will create a completely reversed bending stress. 𝑀 𝑎= 142.4 N. m 𝑇 𝑚 = 124.3 N.m 𝑀 𝑚= 𝑇𝑎= 0
  29. 29. Applying Eq. (7–7) for the DE-Goodman criteria gives: 1 𝑛 = 16 𝜋(0.028)3{ [4 1.58× 142.4 2] 1 2 205× 106 + [3 1.39× 124.3 2] 1 2 735× 106 }= 0.615 n=1.62
  30. 30.  Similarly, plugging the values of the mean and amplitude of the torque and bending moment in the equations of the different failure criteria, we get:  P.S: Note that the safety factor calculated using Soderberg relation is the smallest one since this criteria is the most conservative one.
  31. 31. (b) To calculate the yielding safety factor, we use: σ′max= [ 32×1.58×142.4 𝜋 0.028 3 2 + 3 16×1.39×124.3 𝜋 0.028 3 2 ] 1 2 = 125.4 MPa 𝑛 𝑦 = 574 125.4 = 4.85
  32. 32. Shaft Design for Stress  The stress concentration zones are mainly due to bearings, gears fillets and change in shaft geometry or radius.  The stress concentration factor 𝐾𝑡depends on the type of bearing or gear.  Figs. A-15-16 and A-15-17 give values of the concentration factors of flat-bottomed grooves  Table 7-1 gives the typical stress-concentration factors for the first iteration in the design of a shaft.  Estimating Stress Concentration:
  33. 33. Deflection Consideration  As we mentioned earlier, we need to know the complete shaft geometry before we can perform deflection analysis.  Therefore we should design the dimensions at critical locations to handle the stresses and estimate the other dimensions.  Once the complete geometry is known, we can perform the deflection analysis.  Deflection should be checked at gears and bearing supports.  Chapter 4 deals with the methods used to calculate deflection, including singularity functions and numerical integration.
  34. 34. Deflection Consideration  For certain shafts of complex geometries we may need three dimensional analysis and then use superposition to calculate the overall deflection.  Deflection analysis is straight forward, but it incorporates long procedures and tedious steps.  This can be simplified by using certain three dimensional deflection analysis software.
  35. 35. Deflection Consideration  Table 7-2 includes the maximum allowable deflection for different types of bearings and gears.
  36. 36. Deflection Consideration  Where 𝑦 𝑎𝑙𝑙 is the allowable deflection and 𝑛 𝑑 is the design safety factor.  After calculating the deflection and slope in the shaft, we compare it with the allowable values provided by the different tables.  Sometimes, if the calculated deflection is greater than the allowable values, changes in dimensions must be considered and the diameter of the shaft must be changed according to this equation:
  37. 37.  The preceding equation can be applied on the slope as well: Deflection Consideration  Where (𝑠𝑙𝑜𝑝𝑒) 𝑎𝑙𝑙 is the allowable slope.
  38. 38. Deflection Consideration  Shearing deflections are usually 1% of the bending deflections and are usually ignored.  In cases of short shafts, where 𝑙 𝑑 <10 , shearing deflections become important.  The angular deflection is represented by:  For constant torque through a homogeneous material:
  39. 39. Critical Speeds of Shafts  A common problem encountered when dealing with rotating shafts is the problem of critical speeds.  At certain speeds, when the frequency of rotation becomes close to the shaft’s natural frequency, the shaft becomes unstable with increasing deflections that may lead to failure.  For simple geometries, as a simply supported, constant-diameter shaft, the first critical speed can be estimated using the following equation: where m the mass per unit length, γ is the specific weight.
  40. 40. Critical Speeds of Shafts  For an ensemble of attachments, Rayleigh’s method for lumped masses gives: where 𝑤𝑖 is the weight of the ith location and 𝑦𝑖 is the deflection at the ith body
  41. 41.  At this stage, we will define something called the influence coefficients.  An Influence coefficient is the transverse deflection at location i on a shaft due to a unit load at location j on the shaft.  For a simply supported beams with a single unit load, the influence coefficients are given by: Critical Speeds of Shafts
  42. 42. Critical Speeds of Shafts
  43. 43. Critical Speeds of Shafts  Assume three loads, the influence coefficients can be expressed:  From the influence coefficients we can calculate the deflections 𝑦1, 𝑦2 and 𝑦3 as follows:
  44. 44.  The force 𝐹𝑖 arises from weights attached or centrifugal forces 𝑚𝑖 ω2 𝑦𝑖. The equation can be expressed as: Critical Speeds of Shafts  Solving for the roots of this system will yield the following relation:
  45. 45. Critical Speeds of Shafts  If we had one mass 𝑚1, the critical speed denoted by ω11 will be expressed as 1 ω11 2 = 𝑚1 𝛿11  Similarly, 1 ω22 2 = 𝑚2 𝛿22 and 1 ω33 2 = 𝑚3 𝛿33  Then the equation can be expressed as:  1 ω1 2 ≫ 1 ω2 2 + 1 ω3 2 since ω1 is way smaller than ω2 and ω3
  46. 46. Critical Speeds of Shafts  This will simplify to:  The preceding equation can be extended to n-body shafts:  This equation is known as the Dunkerley’s equation.
  47. 47.  Principle of superposition: Critical Speeds of Shafts  This principle includes calculating an equivalent load placed on the center of the shaft. In other words, we transform each load into an equivalent load placed at the center denoted by the subscript c.  This equivalent load can be found from:
  48. 48. Critical Speeds of Shafts  The critical speed ω 1 can be calculated after summing the equivalent load of each load on the shaft:
  49. 49. Critical Speeds of Shafts  Example: Consider a simply supported steel shaft , with 25 mm diameter and a 775 mm span between bearings, carrying two gears weighing 175 and 275 N. (a) Find the influence coefficients. (b) Find 𝑤𝑦 and 𝑤𝑦2 and the first critical speed using Rayleigh’s equation. (c) From the influence coefficients, find ω11 and ω22. (d) Using Dunkerley’s equation, estimate the first critical speed. (e) Use superposition to estimate the first critical speed.
  50. 50. Critical Speeds of Shafts  Solution: a) I = πd4 64 = π(25)4 64 = 19175 𝑚𝑚4 6E Il = 6(207000)(19175)(775)= 18.5× 1012 N.𝑚𝑚3
  51. 51. Critical Speeds of Shafts δ11= 600(175)(7752−6002−1752) 18.5×1012 = 0.00119 𝑚𝑚 𝑁 δ12= δ21= 275(175)(7752−2752−1752) 18.5×1012 = 0.00129 𝑚𝑚 𝑁 δ22= 275(500)(7752−2752−5002) 18.5×1012 = 0.00204 𝑚𝑚 𝑁 b) 𝑦1= 𝑤1δ11+ 𝑤2δ12 = (175)(0.00119) + (275)(0.00129)= 0.56 mm 𝑦2= 𝑤1δ21+ 𝑤2δ22 = (175)(0.00129) + (275)(0.00204)= 0.79 mm
  52. 52. Critical Speeds of Shafts b) 𝑤𝑦 = 175(0.56)+275(0.79) = 315.3 N.mm 𝑤𝑦2= 175 (0.56)2+ 275(0.79)2= 226.5 N.𝑚𝑚2 Using Rayleigh’s equation, ω = 9.81(315.3)×10−3 226.5×10−6 = 117 rad/s c) 1 ω11 2 = 𝑚1 𝛿11, and 𝑚1= 𝑤1 𝑔 , then: 1 ω11 2 = 𝑤1 𝑔 𝛿11= 175×0.00119×10−3 9.81 1 ω22 2 = 𝑚2 𝛿22, and 𝑚2= 𝑤2 𝑔 , then: 1 ω22 2 = 𝑤2 𝑔 𝛿22= 275×0.00204×10−3 9.81 ω11= 217 rad/s ω22= 132 rad/s
  53. 53. d) Using Dunkerley’s equation: Critical Speeds of Shafts 1 ω1 2 = 1 2172 + 1 1322 = 7.863× 10−5 Implies, ω1=113 rad/s e) Using superposition: δ 𝑐𝑐= b𝑐x 𝑐(𝑙2−b𝑐𝑐 2 −x 𝑐𝑐 2 ) 18.5×1012 = 387.5 387.5 7752−387.52−387.52 18.5×1012 = 0.00244 mm/N
  54. 54. 𝑤1𝑐 = 175(0.00119) 0.00244 = 85.3 N 𝑤2𝑐 = 275(0.00204) 0.00244 = 229.9 N Then, ω = 𝑔 δ 𝑐𝑐 𝑤 𝑖𝑐 = 9.81 0.00244(85.3+229.9)10−3 = 112.9 𝑟𝑎𝑑/𝑠 Critical Speeds of Shafts
  55. 55. Critical Speeds of Shafts Method Rayleigh Dunkerly Superposition First Critical Speed 117 rad/sec 113 rad/sec 112.9 rad / sec Since designers seek first critical speed at least twice the operating speed, so the difference has no effect on our design. The critical speed calculated by Dunkerley and Superposition methods are always expected to be less than Rayleigh’s method because they neglected the effect of the other critical speeds

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