上海交通大学：《Design and Manufacturing》课程教学资源（参考文献）Shigley's Mechanical Engineering Design ch8 screws, fasteners and non permenent joints.pdf_P11-P15

Budynas-Nisbett:Shigley's Ill.Design of Mechanical 8.Screws,Fasteners,and T©The McGraw--Hill Mechanical Engineering Elements the Design of Companies,2008 Design,Eighth Edition Nonpermanent Joints Screws,Fasteners,and the Design of Nonpermanent Joints 405 the coordinate system of Fig.8-8,we note 6F 0x= Txy=0 πdrnip 16T 0y=0 ty:= πd 4F 02=- πd Tx=0 then use Eq.(5-14)of Sec.5-5. The screw-thread form is complicated from an analysis viewpoint.Remember the origin of the tensile-stress area A,which comes from experiment.A power screw lift- ing a load is in compression and its thread pitch is shortened by elastic deformation. Its engaging nut is in tension and its thread pitch is lengthened.The engaged threads cannot share the load equally.Some experiments show that the first engaged thread carries 0.38 of the load,the second 0.25,the third 0.18,and the seventh is free of load. In estimating thread stresses by the equations above,substituting 0.38F for F and set- ting n to I will give the largest level of stresses in the thread-nut combination. EXAMPLE 8-1 A square-thread power screw has a major diameter of 32 mm and a pitch of 4 mm with double threads,and it is to be used in an application similar to that in Fig.8-4. The given data include f=fe=0.08,de =40 mm,and F=6.4 kN per screw. (a)Find the thread depth,thread width,pitch diameter,minor diameter,and lead. (b)Find the torque required to raise and lower the load. (c)Find the efficiency during lifting the load. (d)Find the body stresses,torsional and compressive. (e)Find the bearing stress. (f)Find the thread stresses bending at the root,shear at the root,and von Mises stress and maximum shear stress at the same location. Solution (a)From Fig.8-3a the thread depth and width are the same and equal to half the pitch,or 2 mm.Also dm=d-p/2=32-4/2=30mm Answer d,=d-p=32-4=28mm 1=np=2(4)=8mm (b)Using Eqs.(8-1)and (8-6),the torque required to turn the screw against the load is TR= FdmI+πfdm) Ffede 2πdm-fl/ 2 6.4(30)8+π(0.08)(30)1 6.4(0.08)40 2 π(30)-0.08(8) 2 Answer =15.94+10.24=26.18N.m

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints 408 © The McGraw−Hill Companies, 2008 Screws, Fasteners, and the Design of Nonpermanent Joints 405 the coordinate system of Fig. 8–8, we note σx = 6F πdrnt p τxy = 0 σy = 0 τyz = 16T πd3 r σz = − 4F πd2 r τzx = 0 then use Eq. (5–14) of Sec. 5–5. The screw-thread form is complicated from an analysis viewpoint. Remember the origin of the tensile-stress area At , which comes from experiment. A power screw lifting a load is in compression and its thread pitch is shortened by elastic deformation. Its engaging nut is in tension and its thread pitch is lengthened. The engaged threads cannot share the load equally. Some experiments show that the first engaged thread carries 0.38 of the load, the second 0.25, the third 0.18, and the seventh is free of load. In estimating thread stresses by the equations above, substituting 0.38F for F and setting nt to 1 will give the largest level of stresses in the thread-nut combination. EXAMPLE 8–1 A square-thread power screw has a major diameter of 32 mm and a pitch of 4 mm with double threads, and it is to be used in an application similar to that in Fig. 8–4. The given data include f = fc = 0.08, dc = 40 mm, and F = 6.4 kN per screw. (a) Find the thread depth, thread width, pitch diameter, minor diameter, and lead. (b) Find the torque required to raise and lower the load. (c) Find the efficiency during lifting the load. (d) Find the body stresses, torsional and compressive. (e) Find the bearing stress. ( f) Find the thread stresses bending at the root, shear at the root, and von Mises stress and maximum shear stress at the same location. Solution (a) From Fig. 8–3a the thread depth and width are the same and equal to half the pitch, or 2 mm. Also dm = d − p/2 = 32 − 4/2 = 30 mm Answer dr = d − p = 32 − 4 = 28 mm l = np = 2(4) = 8 mm (b) Using Eqs. (8–1) and (8–6), the torque required to turn the screw against the load is TR = Fdm 2 l + π f dm πdm − f l + F fcdc 2 = 6.4(30) 2 8 + π(0.08)(30) π(30) − 0.08(8) + 6.4(0.08)40 2 Answer = 15.94 + 10.24 = 26.18 N · m

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints © The McGraw−Hill 409 Companies, 2008 406 Mechanical Engineering Design Using Eqs. (8–2) and (8–6), we find the load-lowering torque is TL = Fdm 2 π f dm − l πdm + f l + F fcdc 2 = 6.4(30) 2 π(0.08)30 − 8 π(30) + 0.08(8) + 6.4(0.08)(40) 2 Answer = −0.466 + 10.24 = 9.77 N · m The minus sign in the first term indicates that the screw alone is not self-locking and would rotate under the action of the load except for the fact that the collar friction is present and must be overcome, too. Thus the torque required to rotate the screw “with” the load is less than is necessary to overcome collar friction alone. (c) The overall efficiency in raising the load is Answer e = Fl 2πTR = 6.4(8) 2π(26.18) = 0.311 (d) The body shear stress τ due to torsional moment TR at the outside of the screw body is Answer τ = 16TR πd3 r = 16(26.18)(103) π(283) = 6.07 MPa The axial nominal normal stress σ is Answer σ = − 4F πd2 r = −4(6.4)103 π(282) = −10.39 MPa (e) The bearing stress σB is, with one thread carrying 0.38F, Answer σB = −2(0.38F) πdm(1)p = −2(0.38)(6.4)103 π(30)(1)(4) = −12.9 MPa ( f) The thread-root bending stress σb with one thread carrying 0.38F is σb = 6(0.38F) πdr(1)p = 6(0.38)(6.4)103 π(28)(1)4 = 41.5 MPa The transverse shear at the extreme of the root cross section due to bending is zero. However, there is a circumferential shear stress at the extreme of the root cross section of the thread as shown in part (d) of 6.07 MPa. The three-dimensional stresses, after Fig. 8–8, noting the y coordinate is into the page, are σx = 41.5 MPa τxy = 0 σy = 0 τyz = 6.07 MPa σz = −10.39 MPa τzx = 0 Equation (5–14) of Sec. 5–5 can be written as Answer σ = 1 √2 {(41.5 − 0)2 + [0 − (−10.39)]2 + (−10.39 − 41.5)2 + 6(6.07)2 } 1/2 = 48.7 MPa

410 Budynas-Nisbett:Shigley's Ill.Design of Mechanical 8.Screws,Fasteners,and ©The McGraw-Hil Mechanical Engineering Elements the Design of Companies,2008 Design,Eighth Edition Nonpermanent Joints Screws,Fasteners,and the Design of Nonpermanent Joints 407 Alternatively,you can determine the principal stresses and then use Eq.(5-12)to find the von Mises stress.This would prove helpful in evaluating max as well.The prin- cipal stresses can be found from Eq.(3-15);however,sketch the stress element and note that there are no shear stresses on the x face.This means that or is a principal stress.The remaining stresses can be transformed by using the plane stress equation. Eq.(3-13).Thus,the remaining principal stresses are -10.39 -10.39 +6.072=2.79,-13.18MPa 2 Ordering the principal stresses gives o1,o2,o3 =41.5,2.79,-13.18 MPa.Substi- tuting these into Eq.(5-12)yields 0 41.5-2.7+2.79-(-13.18)2+[-13.18-41.51p Answer =48.7MPa The maximum shear stress is given by Eq.(3-16),where Tmax=t1/3,giving Answer mx=1、=41.5-(-13.18) =27.3MPa 2 Table 8-4 Screw Nut Screw Bearing Material Material Safe Por psi Notes Pressure pb Steel Bronze 2500-3500 Low speed Source:H.A.Rothbart, Steel Bronze 1600-2500 10 fpm Mechanical Design and Cast iron 1800-2500 8 fpm Systems Handbook,2nd ed., McGraw-Hill,New York, Steel Bronze 800-1400 20-40fpm 1985. Cast iron 600-1000 20-40fpm Steel Bronze 150-240 50 fpm Ham and Ryan'showed that the coefficient of friction in screw threads is inde- pendent of axial load,practically independent of speed,decreases with heavier lubri- cants,shows little variation with combinations of materials,and is best for steel on bronze.Sliding coefficients of friction in power screws are about 0.10-0.15. Table 8-4 shows safe bearing pressures on threads,to protect the moving sur- faces from abnormal wear.Table 8-5 shows the coefficients of sliding friction for common material pairs.Table 8-6 shows coefficients of starting and running friction for common material pairs. Ham and Ryan,An Experimental Imvestigation of the Friction of Screw-threads.Bulletin 247.University of Illinois Experiment Station,Champaign-Urbana,Ill..June 7,1932

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints 410 © The McGraw−Hill Companies, 2008 Screws, Fasteners, and the Design of Nonpermanent Joints 407 Screw Nut Material Material Safe pb, psi Notes Steel Bronze 2500–3500 Low speed Steel Bronze 1600–2500 10 fpm Cast iron 1800–2500 8 fpm Steel Bronze 800–1400 20–40 fpm Cast iron 600–1000 20–40 fpm Steel Bronze 150–240 50 fpm Table 8–4 Screw Bearing Pressure pb Source: H. A. Rothbart, Mechanical Design and Systems Handbook, 2nd ed., McGraw-Hill, New York, 1985. Alternatively, you can determine the principal stresses and then use Eq. (5–12) to find the von Mises stress. This would prove helpful in evaluating τmax as well. The principal stresses can be found from Eq. (3–15); however, sketch the stress element and note that there are no shear stresses on the x face. This means that σx is a principal stress. The remaining stresses can be transformed by using the plane stress equation, Eq. (3–13). Thus, the remaining principal stresses are −10.39 2 ± −10.39 2 2 + 6.072 = 2.79, −13.18 MPa Ordering the principal stresses gives σ1, σ2, σ3 = 41.5, 2.79, −13.18 MPa. Substituting these into Eq. (5–12) yields Answer σ = [41.5 − 2.79]2 + [2.79 − (−13.18)] 2 + [−13.18 − 41.5]2 2 1/2 = 48.7 MPa The maximum shear stress is given by Eq. (3–16), where τmax = τ1/3, giving Answer τmax = σ1 − σ3 2 = 41.5 − (−13.18) 2 = 27.3 MPa 1 Ham and Ryan, An Experimental Investigation of the Friction of Screw-threads, Bulletin 247, University of Illinois Experiment Station, Champaign-Urbana, Ill., June 7, 1932. Ham and Ryan1 showed that the coefficient of friction in screw threads is independent of axial load, practically independent of speed, decreases with heavier lubricants, shows little variation with combinations of materials, and is best for steel on bronze. Sliding coefficients of friction in power screws are about 0.10–0.15. Table 8–4 shows safe bearing pressures on threads, to protect the moving surfaces from abnormal wear. Table 8–5 shows the coefficients of sliding friction for common material pairs. Table 8–6 shows coefficients of starting and running friction for common material pairs.

Budynas-Nisbett:Shigley's Ill.Design of Mechanical 8.Screws,Fasteners,and ©The McGraw-Hil 411 Mechanical Engineering Elements the Design of Companies,2008 Design,Eighth Edition Nonpermanent Joints 408 Mechanical Engineering Design Table 8-5 Screw Nut Material Coefficients of Friction f Material Steel Bronze Brass Cast Iron for Threaded Pairs Steel,dry 0.15-0.25 0.15-0.23 0.150.19 0.15-0.25 Source:H.A.Rothbart, Steel,machine oil 0.11-0.17 0.10-0.16 0.10-0.15 0.11-0.17 Mechanical Design and Systems Handbook,2nd ed Bronze 0.08-0.12 0.04-0.06 0.06-0.09 McGraw-Hill,New York, 1985. Table 8-6 Combination Running Starting Thrust-Collar Friction Soft steel on cast iron 0.12 0.17 Coefficients Hard steel on cast iron 0.09 0.15 Source:H.A.Rothbart, Soft steel on bronze 0.08 0.10 Mechanical Design and Systems Handbook,2nd ed., Hard steel on bronze 0.06 0.08 McGraw-Hill,New York 1985. 8-3 Threaded Fasteners As you study the sections on threaded fasteners and their use,be alert to the stochastic and deterministic viewpoints.In most cases the threat is from overproof loading of fasteners,and this is best addressed by statistical methods.The threat from fatigue is lower,and deterministic methods can be adequate Figure 8-9 is a drawing of a standard hexagon-head bolt.Points of stress con- centration are at the fillet,at the start of the threads (runout),and at the thread-root fillet in the plane of the nut when it is present.See Table A-29 for dimensions.The diameter of the washer face is the same as the width across the flats of the hexagon. The thread length of inch-series bolts,where d is the nominal diameter,is 2d+inL≤6in LT= (8-13) 2d+号in L6in and for metric bolts is 2d+6 L≤125 d≤48 LT= 2d+12 125<L≤200 (8-141 2d+25 L>200 where the dimensions are in millimeters.The ideal bolt length is one in which only one or two threads project from the nut after it is tightened.Bolt holes may have burrs or sharp edges after drilling.These could bite into the fillet and increase stress con- centration.Therefore,washers must always be used under the bolt head to prevent this.They should be of hardened steel and loaded onto the bolt so that the rounded edge of the stamped hole faces the washer face of the bolt.Sometimes it is necessary to use washers under the nut too. The purpose of a bolt is to clamp two or more parts together.The clamping load stretches or elongates the bolt;the load is obtained by twisting the nut until the bolt has elongated almost to the elastic limit.If the nut does not loosen,this bolt tension

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints © The McGraw−Hill 411 Companies, 2008 408 Mechanical Engineering Design 8–3 Threaded Fasteners As you study the sections on threaded fasteners and their use, be alert to the stochastic and deterministic viewpoints. In most cases the threat is from overproof loading of fasteners, and this is best addressed by statistical methods. The threat from fatigue is lower, and deterministic methods can be adequate. Figure 8–9 is a drawing of a standard hexagon-head bolt. Points of stress concentration are at the fillet, at the start of the threads (runout), and at the thread-root fillet in the plane of the nut when it is present. See Table A–29 for dimensions. The diameter of the washer face is the same as the width across the flats of the hexagon. The thread length of inch-series bolts, where d is the nominal diameter, is LT = 2d + 1 4 in L ≤ 6 in 2d + 1 2 in L > 6 in (8–13) and for metric bolts is LT = ⎧ ⎪⎨ ⎪⎩ 2d + 6 2d + 12 125 < 2d + 25 L ≤ 125 d ≤ 48 L ≤ 200 L > 200 (8–14) where the dimensions are in millimeters. The ideal bolt length is one in which only one or two threads project from the nut after it is tightened. Bolt holes may have burrs or sharp edges after drilling. These could bite into the fillet and increase stress concentration. Therefore, washers must always be used under the bolt head to prevent this. They should be of hardened steel and loaded onto the bolt so that the rounded edge of the stamped hole faces the washer face of the bolt. Sometimes it is necessary to use washers under the nut too. The purpose of a bolt is to clamp two or more parts together. The clamping load stretches or elongates the bolt; the load is obtained by twisting the nut until the bolt has elongated almost to the elastic limit. If the nut does not loosen, this bolt tension Screw Nut Material Material Steel Bronze Brass Cast Iron Steel, dry 0.15–0.25 0.15–0.23 0.15–0.19 0.15–0.25 Steel, machine oil 0.11–0.17 0.10–0.16 0.10–0.15 0.11–0.17 Bronze 0.08–0.12 0.04–0.06 — 0.06–0.09 Table 8–5 Coefficients of Friction f for Threaded Pairs Source: H. A. Rothbart, Mechanical Design and Systems Handbook, 2nd ed., McGraw-Hill, New York, 1985. Combination Running Starting Soft steel on cast iron 0.12 0.17 Hard steel on cast iron 0.09 0.15 Soft steel on bronze 0.08 0.10 Hard steel on bronze 0.06 0.08 Table 8–6 Thrust-Collar Friction Coefficients Source: H. A. Rothbart, Mechanical Design and Systems Handbook, 2nd ed., McGraw-Hill, New York, 1985

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints 412 © The McGraw−Hill Companies, 2008 Screws, Fasteners, and the Design of Nonpermanent Joints 409 Figure 8–9 Hexagon-head bolt; note the washer face, the fillet under the head, the start of threads, and the chamfer on both ends. Bolt lengths are always measured from below the head. l (a) L H A l l L H (b) L H A 80 to 82° (c) A D D D Figure 8–10 Typical cap-screw heads: (a) fillister head; (b) flat head; (c) hexagonal socket head. Cap screws are also manufactured with hexagonal heads similar to the one shown in Fig. 8–9, as well as a variety of other head styles. This illustration uses one of the conventional methods of representing threads. remains as the preload or clamping force. When tightening, the mechanic should, if possible, hold the bolt head stationary and twist the nut; in this way the bolt shank will not feel the thread-friction torque. The head of a hexagon-head cap screw is slightly thinner than that of a hexagon-head bolt. Dimensions of hexagon-head cap screws are listed in Table A–30. Hexagon-head cap screws are used in the same applications as bolts and also in applications in which one of the clamped members is threaded. Three other common capscrew head styles are shown in Fig. 8–10. A variety of machine-screw head styles are shown in Fig. 8–11. Inch-series machine screws are generally available in sizes from No. 0 to about 3 8 in. Several styles of hexagonal nuts are illustrated in Fig. 8–12; their dimensions are given in Table A–31. The material of the nut must be selected carefully to match that of the bolt. During tightening, the first thread of the nut tends to take the entire load; but yielding occurs, with some strengthening due to the cold work that takes place, and the load is eventually divided over about three nut threads. For this reason you should never reuse nuts; in fact, it can be dangerous to do so. 1 64 H Approx. in 30° R W