上海交通大学：《Design and Manufacturing》课程教学资源（参考文献）Shigley's Mechanical Engineering Design ch7 shaft and shaft components

60 Budynas-Nisbett:Shigley's Ill.Design of Mechanical 7.Shafts and Shaft T©The McGraw-Hill Mechanical Engineering Elements Components Companies,2008 Design,Eighth Edition Shafts and Shaft Components 357 DE-Gerber +(月 (7-91 =(++(》 (7-10 where A=4(K Ma)2+3(KisTa)2 B =4(Ky Mm)2+3(KfsTm)2 DE-ASME Elliptic (“)+()+()广+()门 (7-11 -()+()(+(门 7-12) DE-Soderberg 日-{发KMP+3K,P+[(K,M.2+3K,9\ (7-13) d=(g{传[aKM+3K,] kM产+3K,e) + (7-14 For a rotating shaft with constant bending and torsion,the bending stress is com- pletely reversed and the torsion is steady.Equations(7-7)through(7-14)can be sim- plified by setting Mm and Ta equal to 0,which simply drops out some of the terms. Note that in an analysis situation in which the diameter is known and the factor of safety is desired,as an alternative to using the specialized equations above,it is always still valid to calculate the alternating and mid-range stresses using Egs.(7-5) and(7-6),and substitute them into one of the equations for the failure criteria,Eqs. (6-45)through (6-48),and solve directly for n.In a design situation,however,hav- ing the equations pre-solved for diameter is quite helpful. It is always necessary to consider the possibility of static failure in the first load cycle. The Soderberg criteria inherently guards against yielding.as can be seen by noting that its failure curve is conservatively within the yield (Langer)line on Fig.6-27,p.297.The ASME Elliptic also takes yielding into account,but is not entirely conservative

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 7. Shafts and Shaft Components 360 © The McGraw−Hill Companies, 2008 Shafts and Shaft Components 357 DE-Gerber 1 n = 8A πd3 Se ⎧ ⎨ ⎩ 1 + 1 + 2BSe ASut 2 1/2 ⎫ ⎬ ⎭ (7–9) d = ⎛ ⎝ 8n A πSe ⎧ ⎨ ⎩ 1 + 1 + 2BSe ASut 2 1/2 ⎫ ⎬ ⎭ ⎞ ⎠ 1/3 (7–10) where A =  4(Kf Ma)2 + 3(Kf sTa)2 B =  4(Kf Mm)2 + 3(Kf sTm)2 DE-ASME Elliptic 1 n = 16 πd3 4 Kf Ma Se 2 + 3 Kf sTa Se 2 + 4 Kf Mm Sy 2 + 3 Kf sTm Sy 2 1/2 (7–11) d = ⎧ ⎨ ⎩ 16n π 4 Kf Ma Se 2 + 3 Kf sTa Se 2 + 4 Kf Mm Sy 2 + 3 Kf sTm Sy 2 1/2 ⎫ ⎬ ⎭ 1/3 (7–12) DE-Soderberg 1 n = 16 πd3  1 Se  4(Kf Ma) 2 + 3(Kf sTa) 21/2 + 1 Syt  4(Kf Mm) 2 + 3(Kf sTm) 21/2 (7–13) d = 16n π  1 Se  4(Kf Ma) 2 + 3(Kf sTa) 21/2 + 1 Syt  4(Kf Mm) 2 + 3(Kf sTm) 21/2 1/3 (7–14) For a rotating shaft with constant bending and torsion, the bending stress is com￾pletely reversed and the torsion is steady. Equations (7–7) through (7–14) can be sim￾plified by setting Mm and Ta equal to 0, which simply drops out some of the terms. Note that in an analysis situation in which the diameter is known and the factor of safety is desired, as an alternative to using the specialized equations above, it is always still valid to calculate the alternating and mid-range stresses using Eqs. (7–5) and (7–6), and substitute them into one of the equations for the failure criteria, Eqs. (6–45) through (6–48), and solve directly for n. In a design situation, however, hav￾ing the equations pre-solved for diameter is quite helpful. It is always necessary to consider the possibility of static failure in the first load cycle. The Soderberg criteria inherently guards against yielding, as can be seen by noting that its failure curve is conservatively within the yield (Langer) line on Fig. 6–27, p. 297. The ASME Elliptic also takes yielding into account, but is not entirely conservative

Budynas-Nisbett:Shigley's Ill.Design of Mechanical 7.Shafts and Shaft T©The McGraw-Hill 361 Mechanical Engineering Elements Components Companies,2008 Design,Eighth Edition 358 Mechanical Engineering Design throughout its entire range.This is evident by noting that it crosses the yield line in Fig.6-27.The Gerber and modified Goodman criteria do not guard against yielding. requiring a separate check for yielding.A von Mises maximum stress is calculated for this purpose. ax=[【om+a)2+3(m+a)2]p +(门 (7-15) To check for yielding,this von Mises maximum stress is compared to the yield strength,as usual. S (7-16) For a quick,conservative check,an estimate for can be obtained by simply adding o and o(o)will always be greater than or equal to oax and will therefore be conservative. EXAMPLE 7-1 At a machined shaft shoulder the small diameter d is 1.100 in,the large diameter D is 1.65 in,and the fillet radius is 0.11 in.The bending moment is 1260 Ibf.in and the steady torsion moment is 1100 lbf.in.The heat-treated steel shaft has an ultimate strength of Sur=105 kpsi and a yield strength of Sy=82 kpsi.The reliability goal is0.99. (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. Solution (a）D/d=1.65/1.100=1.50,r/d=0.11/1.100=0.10,K,=1.68Fig.A-15-9), Ka=1.42(Fig.A-15-8),q=0.85(Fig.6-20),qshear=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%=0.5(105)=52.5kpsi Eq.(6-19y: ka=2.70(105)-0.265=0.787 1.100 -0.107 Eq.(6-20: kb=0.30) =0.870 ke=ka=k灯=1

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGraw−Hill 361 Companies, 2008 358 Mechanical Engineering Design throughout its entire range. This is evident by noting that it crosses the yield line in Fig. 6–27. The Gerber and modified Goodman criteria do not guard against yielding, requiring a separate check for yielding. A von Mises maximum stress is calculated for this purpose. σ max =  (σm + σa) 2 + 3 (τm + τa) 21/2 = 32Kf (Mm + Ma) πd3 2 + 3 16Kf s (Tm + Ta) πd3 2 1/2 (7–15) To check for yielding, this von Mises maximum stress is compared to the yield strength, as usual. ny = Sy σ max (7–16) For a quick, conservative check, an estimate for σ max can be obtained by simply adding σ a and σ m . (σ a + σ m ) will always be greater than or equal to σ max, and will therefore be conservative. EXAMPLE 7–1 At a machined shaft shoulder the small diameter d is 1.100 in, the large diameter D is 1.65 in, and the fillet radius is 0.11 in. The bending moment is 1260 lbf · in and the steady torsion moment is 1100 lbf · in. The heat-treated steel shaft has an ultimate strength of Sut = 105 kpsi and a yield strength of Sy = 82 kpsi. The 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. Solution (a) D/d = 1.65/1.100 = 1.50, r/d = 0.11/1.100 = 0.10, Kt = 1.68 (Fig. A–15–9), Kts = 1.42 (Fig. A–15–8), q = 0.85 (Fig. 6–20), qshear = 0.92 (Fig. 6–21). From Eq. (6–32), Kf = 1 + 0.85(1.68 − 1) = 1.58 Kf s = 1 + 0.92(1.42 − 1) = 1.39 Eq. (6–8): S e = 0.5(105) = 52.5 kpsi Eq. (6–19): ka = 2.70(105)−0.265 = 0.787 Eq. (6–20): kb = 1.100 0.30 −0.107 = 0.870 kc = kd = kf = 1

Budynas-Nisbett:Shigley's Ill.Design of Mechanical 7.Shafts and Shaft ©The McGraw-Hil Mechanical Engineering Elements Components Companies,2008 Design,Eighth Edition Shafts and Shaft Components 359 Table 6-6: ke=0.814 Se=0.787(0.870)0.814(52.5)=29.3kpsi For a rotating shaft,the constant bending moment will create a completely reversed bending stress. Ma 1260 Ibf.in Tm =1100 Ibf.in Mm=Ta =0 Applying Eq.(7-7)for the DE-Goodman criteria gives 1 16 [4(1.58.12609]2 [3(1.39.11002/2 =0.615 nπ(1.1)序 29300 105000 Answer n=1.62 DE-Goodman Similarly,applying Eqs.(7-9),(7-11),and(7-13)for the other failure criteria, Answer n=1.87 DE-Gerber Answer n=1.88 DE-ASME Elliptic Answer n=1.56 DE-Soderberg For comparison,consider an equivalent approach of calculating the stresses and apply- ing the fatigue failure criteria directly.From Eqs.(7-5)and (7-6), 3> 2 32.1.58.1260 0= =15235psi π(1.1)3 1/2 16.1.39.1100 =10134psi π(1.1)3 Taking,for example,the Goodman failure critera,application of Eq.(6-46) gives 1_4+ 15235 10134 =0.616 n Se Sut 29300+105000 n=1.62 which is identical with the previous result.The same process could be used for the other failure criteria. (b)For the yielding factor of safety,determine an equivalent von Mises maximum stress using Eq.(7-15) 32(1.58)(1260)12 1/2 +3 161.39)(1100) 2 =18300psi π(1.1)3 π(1.1)3 82000 Answer Sy ny= =4.48 18300 For comparison,a quick and very conservative check on yielding can be obtained by replacing with This just saves the extra time of calculatingo if o and o have already been determined.For this example

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 7. Shafts and Shaft Components 362 © The McGraw−Hill Companies, 2008 Shafts and Shaft Components 359 Table 6–6: ke = 0.814 Se = 0.787(0.870)0.814(52.5) = 29.3 kpsi For a rotating shaft, the constant bending moment will create a completely reversed bending stress. Ma = 1260 lbf · in Tm = 1100 lbf · in Mm = Ta = 0 Applying Eq. (7–7) for the DE-Goodman criteria gives 1 n = 16 π(1.1)3  4 (1.58 · 1260) 21/2 29 300 +  3 (1.39 · 1100) 21/2 105 000  = 0.615 Answer n = 1.62 DE-Goodman Similarly, applying Eqs. (7–9), (7–11), and (7–13) for the other failure criteria, Answer n = 1.87 DE-Gerber Answer n = 1.88 DE-ASME Elliptic Answer n = 1.56 DE-Soderberg For comparison, consider an equivalent approach of calculating the stresses and apply￾ing the fatigue failure criteria directly. From Eqs. (7–5) and (7–6), σ a = 32 · 1.58 · 1260 π (1.1) 3 2 1/2 = 15 235 psi σ m = 3 16 · 1.39 · 1100 π (1.1) 3 2 1/2 = 10 134 psi Taking, for example, the Goodman failure critera, application of Eq. (6–46) gives 1 n = σ a Se + σ m Sut = 15 235 29 300 + 10 134 105 000 = 0.616 n = 1.62 which is identical with the previous result. The same process could be used for the other failure criteria. (b) For the yielding factor of safety, determine an equivalent von Mises maximum stress using Eq. (7–15). σ max = 32(1.58) (1260) π (1.1) 3 2 + 3 16(1.39) (1100) π (1.1) 3 2 1/2 = 18 300 psi Answer ny = Sy σ max = 82 000 18 300 = 4.48 For comparison, a quick and very conservative check on yielding can be obtained by replacing σ max with σ a + σ m . This just saves the extra time of calculating σ max if σ a and σ m have already been determined. For this example,

Budynas-Nisbett:Shigley's Ill.Design of Mechanical 7.Shafts and Shaft ©The McGraw-Hil 363 Mechanical Engineering Elements Components Companies,2008 Design,Eighth Edition 360 Mechanical Engineering Design Sy 82000 ny=- +a%=15235+10134=3.23 which is quite conservative compared with n=4.48. Estimating Stress Concentrations The stress analysis process for fatigue is highly dependent on stress concentrations. Stress concentrations for shoulders and keyways are dependent on size specifications that are not known the first time through the process.Fortunately,since these elements are usually of standard proportions,it is possible to estimate the stress concentration factors for initial design of the shaft.These stress concentrations will be fine-tuned in successive iterations,once the details are known. Shoulders for bearing and gear support should match the catalog recommenda- tion for the specific bearing or gear.A look through bearing catalogs shows that a typical bearing calls for the ratio of D/d to be between 1.2 and 1.5.For a first approx- imation,the worst case of 1.5 can be assumed.Similarly,the fillet radius at the shoul- der needs to be sized to avoid interference with the fillet radius of the mating com- ponent.There is a significant variation in typical bearings in the ratio of fillet radius versus bore diameter,with r/d typically ranging from around 0.02 to 0.06.A quick look at the stress concentration charts (Figures A-15-8 and A-15-9)shows that the stress concentrations for bending and torsion increase significantly in this range.For example,with D/d =1.5 for bending.K,=2.7 at r/d =0.02,and reduces to K,=2.1 at r/d =0.05,and further down to K =1.7 at r/d =0.1.This indicates that this is an area where some attention to detail could make a significant difference. Fortunately,in most cases the shear and bending moment diagrams show that bend- ing moments are quite low near the bearings,since the bending moments from the ground reaction forces are small. In cases where the shoulder at the bearing is found to be critical,the designer should plan to select a bearing with generous fillet radius,or consider providing for a larger fillet radius on the shaft by relieving it into the base of the shoulder as shown in Fig.7-9a.This effectively creates a dead zone in the shoulder area that does not Sharp radius Shoulder Large radius undercut relief groov Stress flow Bearing Shaft (c) Figure 7-9 Techniques for reducing stress concentration at a shoulder supporting a bearing with a sharp rodius.(a)Large radius undercut into the shoulder.(b)Large radius relief groove into the bock of the shoulder.(d Large radius relief groove into the small diameter

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGraw−Hill 363 Companies, 2008 360 Mechanical Engineering Design ny = Sy σ a + σ m = 82 000 15 235 + 10 134 = 3.23 which is quite conservative compared with ny 4.48. Estimating Stress Concentrations The stress analysis process for fatigue is highly dependent on stress concentrations. Stress concentrations for shoulders and keyways are dependent on size specifications that are not known the first time through the process. Fortunately, since these elements are usually of standard proportions, it is possible to estimate the stress concentration factors for initial design of the shaft. These stress concentrations will be fine-tuned in successive iterations, once the details are known. Shoulders for bearing and gear support should match the catalog recommenda￾tion for the specific bearing or gear. A look through bearing catalogs shows that a typical bearing calls for the ratio of D/d to be between 1.2 and 1.5. For a first approx￾imation, the worst case of 1.5 can be assumed. Similarly, the fillet radius at the shoul￾der needs to be sized to avoid interference with the fillet radius of the mating com￾ponent. There is a significant variation in typical bearings in the ratio of fillet radius versus bore diameter, with r/d typically ranging from around 0.02 to 0.06. A quick look at the stress concentration charts (Figures A–15–8 and A–15–9) shows that the stress concentrations for bending and torsion increase significantly in this range. For example, with D/d = 1.5 for bending, Kt = 2.7 at r/d = 0.02, and reduces to Kt = 2.1 at r/d = 0.05, and further down to Kt = 1.7 at r/d = 0.1. This indicates that this is an area where some attention to detail could make a significant difference. Fortunately, in most cases the shear and bending moment diagrams show that bend￾ing moments are quite low near the bearings, since the bending moments from the ground reaction forces are small. In cases where the shoulder at the bearing is found to be critical, the designer should plan to select a bearing with generous fillet radius, or consider providing for a larger fillet radius on the shaft by relieving it into the base of the shoulder as shown in Fig. 7–9a. This effectively creates a dead zone in the shoulder area that does not Sharp radius Bearing Shaft Large radius undercut Stress flow (a) Shoulder relief groove (b) Large-radius relief groove (c) Figure 7–9 Techniques for reducing stress concentration at a shoulder supporting a bearing with a sharp radius. (a) Large radius undercut into the shoulder. (b) Large radius relief groove into the back of the shoulder. (c) Large radius relief groove into the small diameter

Budynas-Nisbett:Shigley's Ill.Design of Mechanical 7.Shafts and Shaft ©The McGraw-Hill Mechanical Engineering Elements Components Companies,2008 Design,Eighth Edition Shafts and Shaft Components 361 carry the bending stresses,as shown by the stress flow lines.A shoulder relief groove as shown in Fig.7-9b can accomplish a similar purpose.Another option is to cut a large-radius relief groove into the small diameter of the shaft,as shown in Fig.7-9c. This has the disadvantage of reducing the cross-sectional area,but is often used in cases where it is useful to provide a relief groove before the shoulder to prevent the grinding or turning operation from having to go all the way to the shoulder. For the standard shoulder fillet,for estimating K,values for the first iteration, an r/d ratio should be selected so K,values can be obtained.For the worst end of the spectrum,with r/d =0.02 and D/d =1.5.K,values from the stress con- centration charts for shoulders indicate 2.7 for bending,2.2 for torsion,and 3.0 for axial. A keyway will produce a stress concentration near a critical point where the load- transmitting component is located.The stress concentration in an end-milled keyseat is a function of the ratio of the radius r at the bottom of the groove and the shaft diameter d.For early stages of the design process,it is possible to estimate the stress concentration for keyways regardless of the actual shaft dimensions by assuming a typical ratio of r/d =0.02.This gives K,=2.2 for bending and Kis =3.0 for tor- sion,assuming the key is in place. Figures A-15-16 and A-15-17 give values for stress concentrations for flat- bottomed grooves such as used for retaining rings.By examining typical retaining ring specifications in vendor catalogs,it can be seen that the groove width is typically slightly greater than the groove depth,and the radius at the bottom of the groove is around 1/10 of the groove width.From Figs.A-15-16 and A-15-17,stress concen- tration factors for typical retaining ring dimensions are around 5 for bending and axial, and 3 for torsion.Fortunately,the small radius will often lead to a smaller notch sen- sitivity,reducing Kf. Table 7-1 summarizes some typical stress concentration factors for the first iter- ation in the design of a shaft.Similar estimates can be made for other features.The point is to notice that stress concentrations are essentially normalized so that they are dependent on ratios of geometry features,not on the specific dimensions.Conse- quently,by estimating the appropriate ratios,the first iteration values for stress con- centrations can be obtained.These values can be used for initial design,then actual values inserted once diameters have been determined. Table 7-1 First Iteration Estimates Bending Torsional Axial for Stress Concentration Shoulder fillet-sharp (r/d=0.02) 2.7 2.2 3.0 Factors K Shoulder fillet-well rounded (r/d =0.1) 1.7 1.5 19 Warning:These factors are End-mill keyseat (r/d =0.02) 2.2 3.0 only estimates for use when actual dimensions are not Sled runner keyseat 1.7 yet determined.Do not Retaining ring groove 5.0 3.0 5.0 use these once actual dimensions are available Missing values in the table are not reodily available

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 7. Shafts and Shaft Components 364 © The McGraw−Hill Companies, 2008 Shafts and Shaft Components 361 carry the bending stresses, as shown by the stress flow lines. A shoulder relief groove as shown in Fig. 7–9b can accomplish a similar purpose. Another option is to cut a large-radius relief groove into the small diameter of the shaft, as shown in Fig. 7–9c. This has the disadvantage of reducing the cross-sectional area, but is often used in cases where it is useful to provide a relief groove before the shoulder to prevent the grinding or turning operation from having to go all the way to the shoulder. For the standard shoulder fillet, for estimating Kt values for the first iteration, an r/d ratio should be selected so Kt values can be obtained. For the worst end of the spectrum, with r/d = 0.02 and D/d = 1.5, Kt values from the stress con￾centration charts for shoulders indicate 2.7 for bending, 2.2 for torsion, and 3.0 for axial. A keyway will produce a stress concentration near a critical point where the load￾transmitting component is located. The stress concentration in an end-milled keyseat is a function of the ratio of the radius r at the bottom of the groove and the shaft diameter d. For early stages of the design process, it is possible to estimate the stress concentration for keyways regardless of the actual shaft dimensions by assuming a typical ratio of r/d = 0.02. This gives Kt = 2.2 for bending and Kts = 3.0 for tor￾sion, assuming the key is in place. Figures A–15–16 and A–15–17 give values for stress concentrations for flat￾bottomed grooves such as used for retaining rings. By examining typical retaining ring specifications in vendor catalogs, it can be seen that the groove width is typically slightly greater than the groove depth, and the radius at the bottom of the groove is around 1/10 of the groove width. From Figs. A–15–16 and A–15–17, stress concen￾tration factors for typical retaining ring dimensions are around 5 for bending and axial, and 3 for torsion. Fortunately, the small radius will often lead to a smaller notch sen￾sitivity, reducing Kf . Table 7–1 summarizes some typical stress concentration factors for the first iter￾ation in the design of a shaft. Similar estimates can be made for other features. The point is to notice that stress concentrations are essentially normalized so that they are dependent on ratios of geometry features, not on the specific dimensions. Conse￾quently, by estimating the appropriate ratios, the first iteration values for stress con￾centrations can be obtained. These values can be used for initial design, then actual values inserted once diameters have been determined. Table 7–1 First Iteration Estimates for Stress Concentration Factors Kt . Warning: These factors are only estimates for use when actual dimensions are not yet determined. Do not use these once actual dimensions are available. Bending Torsional Axial Shoulder fillet—sharp (r/d 0.02) 2.7 2.2 3.0 Shoulder fillet—well rounded (r/d 0.1) 1.7 1.5 1.9 End-mill keyseat (r/d 0.02) 2.2 3.0 — Sled runner keyseat 1.7 — — Retaining ring groove 5.0 3.0 5.0 Missing values in the table are not readily available.