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Shafts – Definition

Shafts – Definition

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Shafts – Definition

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  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 $ $$ $$$

  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)

  10. 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.)

  11. 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

  12. Maximum Shear Stress Theory Substitute the previous values for x and xyinto 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.

  13. 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.

  14. 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 2.0. The shock factors are applied to their respective stress components.

  15. 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.

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

  17. 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….)

  18. 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 = 1.4. Temp = 500 oC, and n = 2. Survival rate = 50%.

  19. 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.

  20. 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.

  21. 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.

  22. ShaftAttachments: Keys Square (w ~ D/4) Flat Round (or tapered) Gib head Woodruff key

  23. Shaft Attachments: Pins Straight Tapered Roll

  24. Shaft Attachments: Tapered Clamps www.ringfeder.com

  25. 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

  26. 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.

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

  28. 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.

  29. Rigid Couplings – Sleeves The simplest type of coupling is the simple sleeve coupling. But this also has the lowest torque capacity. http://www.grainger.com/Grainger/wwg/start.shtml

  30. Rigid Couplings - Flanged Taper locked Keyed to shaft Great web resource: http://www.powertransmission.com/pages/couplings.htm

  31. 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.

  32. Two-piece “Donut” (or toroidal) flexible coupling http://viva.rexnord.com/content/features.html

  33. 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.

  34. Shafts parallel but offset Shafts not parallel but intersecting

  35. 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.

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

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

  38. 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

  39. 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.

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

  41. 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

  42. Oil Viscosity Fig. 10.7, p. 385

  43. Journal Design Example I • We are given r = 2”, and c/r = .001. 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.

  44. 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

  45. 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.