Budynas-Nisbett:Shigley's Ill.Design of Mechanical 8.Screws,Fasteners,and T©The McGraw-Hil 403 Mechanical Engineering Elements the Design of Companies,2008 Design,Eighth Edition Nonpermanent Joints 400 Mechanical Engineering Design Table 8-3 d,in 子各是之吾石1 1H1为1是2223 Preferred Pitches for Acme Threads P.in 6日立b日名名古多日子日行五 are sometimes modified to a stub form by making the teeth shorter.This results in a larger minor diameter and a somewhat stronger screw. 8-2 The Mechanics of Power Screws A power screw is a device used in machinery to change angular motion into linear motion,and,usually,to transmit power.Familiar applications include the lead screws of lathes,and the screws for vises,presses,and jacks. An application of power screws to a power-driven jack is shown in Fig.8-4.You should be able to identify the worm,the worm gear,the screw,and the nut.Is the worm gear supported by one bearing or two? Figure 8-4 The Joyce wormgear screw jack.(Courtesy Joyce-Dayton Corp.,Dayton,Ohio.)
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 403 Companies, 2008 400 Mechanical Engineering Design Figure 8–4 The Joyce worm-gear screw jack. (Courtesy Joyce-Dayton Corp., Dayton, Ohio.) are sometimes modified to a stub form by making the teeth shorter. This results in a larger minor diameter and a somewhat stronger screw. 8–2 The Mechanics of Power Screws A power screw is a device used in machinery to change angular motion into linear motion, and, usually, to transmit power. Familiar applications include the lead screws of lathes, and the screws for vises, presses, and jacks. An application of power screws to a power-driven jack is shown in Fig. 8–4. You should be able to identify the worm, the worm gear, the screw, and the nut. Is the worm gear supported by one bearing or two? d, in 1 4 5 16 3 8 1 2 5 8 3 4 7 8 1 11 4 11 2 13 4 2 21 2 3 p, in 1 16 1 14 1 12 1 10 1 8 1 6 1 6 1 5 1 5 1 4 1 4 1 4 1 3 1 2 Table 8–3 Preferred Pitches for Acme Threads
404 Budynas-Nisbett:Shigley's Ill.Design of Mechanical 8.Screws,Fasteners,and T©The McGraw-Hil Mechanical Engineering Elements the Design of Companies,2008 Design,Eighth Edition Nonpermanent Joints Screws,Fasteners,and the Design of Nonpermanent Joints 401 Figure 8-5 Portion of a power screw. F/2 Figure 8-6 Force diagrams:(a)lifting the load;(b)lowering the lood. (B) In Fig.8-5 a square-threaded power screw with single thread having a mean diameter dm,a pitch p,a lead angle and a helix angle is loaded by the axial compressive force F.We wish to find an expression for the torque required to raise this load,and another expression for the torque required to lower the load. First,imagine that a single thread of the screw is unrolled or developed (Fig.8-6) for exactly a single turn.Then one edge of the thread will form the hypotenuse of a right triangle whose base is the circumference of the mean-thread-diameter circle and whose height is the lead.The angle A,in Figs.8-5 and 8-6,is the lead angle of the thread.We represent the summation of all the unit axial forces acting upon the normal thread area by F.To raise the load,a force Pg acts to the right (Fig.8-6a),and to lower the load, PL acts to the left(Fig.8-6b).The friction force is the product of the coefficient of fric- tion fwith the normal force N,and acts to oppose the motion.The system is in equilib- rium under the action of these forces,and hence,for raising the load,we have >Fn PR-N sin).-fN cos=0 (a) >Fv=F+fN sinx-N cos=0 In a similar manner,for lowering the load,we have ∑FH=-PL-Nsin入+fN cosi入=0 6 ∑Fv=F-fNsin-Ncos入=0
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints 404 © The McGraw−Hill Companies, 2008 Screws, Fasteners, and the Design of Nonpermanent Joints 401 F⁄ 2 p F F⁄ 2 Nut dm Figure 8–5 Portion of a power screw. dm l F PR fN N dm (a) (b) l F fN PL N Figure 8–6 Force diagrams: (a) lifting the load; (b) lowering the load. In Fig. 8–5 a square-threaded power screw with single thread having a mean diameter dm , a pitch p, a lead angle λ, and a helix angle ψ is loaded by the axial compressive force F. We wish to find an expression for the torque required to raise this load, and another expression for the torque required to lower the load. First, imagine that a single thread of the screw is unrolled or developed (Fig. 8–6) for exactly a single turn. Then one edge of the thread will form the hypotenuse of a right triangle whose base is the circumference of the mean-thread-diameter circle and whose height is the lead. The angle λ, in Figs. 8–5 and 8–6, is the lead angle of the thread. We represent the summation of all the unit axial forces acting upon the normal thread area by F. To raise the load, a force PR acts to the right (Fig. 8–6a), and to lower the load, PL acts to the left (Fig. 8–6b). The friction force is the product of the coefficient of friction f with the normal force N, and acts to oppose the motion. The system is in equilibrium under the action of these forces, and hence, for raising the load, we have FH = PR − N sin λ − f N cos λ = 0 (a) FV = F + f N sin λ − N cos λ = 0 In a similar manner, for lowering the load, we have FH = −PL − N sin λ + f N cos λ = 0 (b) FV = F − f N sin λ − N cos λ = 0
Budynas-Nisbett:Shigley's Ill.Design of Mechanical 8.Screws,Fasteners,and I©The McGraw-Hil Mechanical Engineering Elements the Design of Companies,2008 Design,Eighth Edition Nonpermanent Joints 402 Mechanical Engineering Design Since we are not interested in the normal force N,we eliminate it from each of these sets of equations and solve the result for P.For raising the load,this gives F(sinλ+fcos入) PR= (d cosλ-fsinλ and for lowering the load, F(fcos入-sin) PL= (d) cosλ+fsinλ Next,divide the numerator and the denominator of these equations by cosA and use the relation tan=l/d (Fig.8-6).We then have,respectively, PR= F[(l/dm)+f] (e) 1-(fl/πdm) FLf-/πdm)】 PL= 1+(fl/πdm) (0 Finally,noting that the torque is the product of the force P and the mean radius dm/2, for raising the load we can write Fdm/l+πfdm TR= πdm-fi 8-1) 2 where TR is the torque required for two purposes:to overcome thread friction and to raise the load. The torque required to lower the load,from Eg.(f),is found to be Fdmπfdm-l\ TL=- (8-2) 2 πdm+fl This is the torque required to overcome a part of the friction in lowering the load.It may turn out,in specific instances where the lead is large or the friction is low,that the load will lower itself by causing the screw to spin without any external effort.In such cases, the torque TL from Eq.(8-2)will be negative or zero.When a positive torque is obtained from this equation,the screw is said to be self-locking.Thus the condition for self-locking is πfdm>l Now divide both sides of this inequality by rdm.Recognizing that l/ndm tanA,we get f>tan入 (8-31 This relation states that self-locking is obtained whenever the coefficient of thread friction is equal to or greater than the tangent of the thread lead angle. An expression for efficiency is also useful in the evaluation of power screws.If we let f =0 in Eq.(8-1),we obtain FI To= 2π (g)
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 405 Companies, 2008 402 Mechanical Engineering Design Since we are not interested in the normal force N, we eliminate it from each of these sets of equations and solve the result for P. For raising the load, this gives PR = F(sin λ + f cos λ) cos λ − f sin λ (c) and for lowering the load, PL = F( f cos λ − sin λ) cos λ + f sin λ (d) Next, divide the numerator and the denominator of these equations by cos λ and use the relation tan λ = l/πdm (Fig. 8–6). We then have, respectively, PR = F[(l/πdm) + f ] 1 − ( f l/πdm) (e) PL = F[ f − (l/πdm)] 1 + ( f l/πdm) (f) Finally, noting that the torque is the product of the force P and the mean radius dm/2, for raising the load we can write TR = Fdm 2 l + π f dm πdm − f l (8–1) where TR is the torque required for two purposes: to overcome thread friction and to raise the load. The torque required to lower the load, from Eq. (f), is found to be TL = Fdm 2 π f dm − l πdm + f l (8–2) This is the torque required to overcome a part of the friction in lowering the load. It may turn out, in specific instances where the lead is large or the friction is low, that the load will lower itself by causing the screw to spin without any external effort. In such cases, the torque TL from Eq. (8–2) will be negative or zero. When a positive torque is obtained from this equation, the screw is said to be self-locking. Thus the condition for self-locking is π f dm > l Now divide both sides of this inequality by πdm . Recognizing that l/πdm = tan λ, we get f > tan λ (8–3) This relation states that self-locking is obtained whenever the coefficient of thread friction is equal to or greater than the tangent of the thread lead angle. An expression for efficiency is also useful in the evaluation of power screws. If we let f = 0 in Eq. (8–1), we obtain T0 = Fl 2π (g)
06 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 403 which,since thread friction has been eliminated,is the torque required only to raise the load.The efficiency is therefore To Fl e= TR 2TR (8-4④ The preceding equations have been developed for square threads where the nor- mal thread loads are parallel to the axis of the screw.In the case of Acme or other threads,the normal thread load is inclined to the axis because of the thread angle 20 and the lead angle A.Since lead angles are small,this inclination can be neglected and only the effect of the thread angle(Fig.8-7a)considered.The effect of the angle o is to increase the frictional force by the wedging action of the threads.Therefore the frictional terms in Eq.(8-1)must be divided by cos a.For raising the load,or for tightening a screw or bolt,this yields Fdm TR=2 l+πf d sec a (8-51 πdm-fl sec In using Eq.(8-5),remember that it is an approximation because the effect of the lead angle has been neglected. For power screws,the Acme thread is not as efficient as the square thread,because of the additional friction due to the wedging action,but it is often preferred because it is easier to machine and permits the use of a split nut,which can be adjusted to take up for wear. Usually a third component of torque must be applied in power-screw applications. When the screw is loaded axially,a thrust or collar bearing must be employed between the rotating and stationary members in order to carry the axial component.Figure 8-7b shows a typical thrust collar in which the load is assumed to be concentrated at the mean collar diameter de.If fe is the coefficient of collar friction,the torque required is Ffede Te= 2 (8-6) For large collars,the torque should probably be computed in a manner similar to that employed for disk clutches. Figure 8-7 (a)Normal thread force is increased because of angle a; b)thrust collar has frictional F/2 2 diameter d -Collar Nut 2a= Thread angle (a) (b)
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints 406 © The McGraw−Hill Companies, 2008 Screws, Fasteners, and the Design of Nonpermanent Joints 403 Thread angle Collar Nut F cos F⁄ 2 (a) (b) F⁄ 2 F 2 = F⁄ 2 F⁄ 2 Figure 8–7 dc (a) Normal thread force is increased because of angle α; (b) thrust collar has frictional diameter dc. which, since thread friction has been eliminated, is the torque required only to raise the load. The efficiency is therefore e = T0 TR = Fl 2πTR (8–4) The preceding equations have been developed for square threads where the normal thread loads are parallel to the axis of the screw. In the case of Acme or other threads, the normal thread load is inclined to the axis because of the thread angle 2α and the lead angle λ. Since lead angles are small, this inclination can be neglected and only the effect of the thread angle (Fig. 8–7a) considered. The effect of the angle α is to increase the frictional force by the wedging action of the threads. Therefore the frictional terms in Eq. (8–1) must be divided by cos α. For raising the load, or for tightening a screw or bolt, this yields TR = Fdm 2 l + π f dm sec α πdm − f l sec α (8–5) In using Eq. (8–5), remember that it is an approximation because the effect of the lead angle has been neglected. For power screws, the Acme thread is not as efficient as the square thread, because of the additional friction due to the wedging action, but it is often preferred because it is easier to machine and permits the use of a split nut, which can be adjusted to take up for wear. Usually a third component of torque must be applied in power-screw applications. When the screw is loaded axially, a thrust or collar bearing must be employed between the rotating and stationary members in order to carry the axial component. Figure 8–7b shows a typical thrust collar in which the load is assumed to be concentrated at the mean collar diameter dc . If fc is the coefficient of collar friction, the torque required is Tc = F fcdc 2 (8–6) For large collars, the torque should probably be computed in a manner similar to that employed for disk clutches.
Budynas-Nisbett:Shigley's Ill.Design of Mechanical 8.Screws,Fasteners,and T©The McGraw-Hill 07 Mechanical Engineering Elements the Design of Companies,2008 Design,Eighth Edition Nonpermanent Joints 404 Mechanical Engineering Design Nominal body stresses in power screws can be related to thread parameters as follows.The maximum nominal shear stress t in torsion of the screw body can be expressed as 16T T= πd (8-7 The axial stress a in the body of the screw due to load F is F 4F =A=元d 8-8) in the absence of column action.For a short column the J.B.Johnson buckling formula is given by Eq.(4-43),which is (份)=-() (8-91 Nominal thread stresses in power screws can be related to thread parameters as follows.The bearing stress in Fig.8-8,og,is 2F B=一 dan,p/2= (8-101 πdmnip where n,is the number of engaged threads.The bending stress at the root of the thread op is found from -42心-4pM= 6 sO M Fp 24 6F %=元=4dn,p=4m,p (8-11) The transverse shear stress r at the center of the root of the thread due to load F is 3V 3 F 3F 【=2A=2rd,np/2-Td,np (8-12) and at the top of the root it is zero.The von Mises stress o'at the top of the root"plane" is found by first identifying the orthogonal normal stresses and the shear stresses.From Figure 8-8 Geometry of square thread useful in finding bending and transverse shear stresses at the thread root
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 407 Companies, 2008 404 Mechanical Engineering Design dm dr F p ⁄ 2 p ⁄ 2 z x Figure 8–8 Geometry of square thread useful in finding bending and transverse shear stresses at the thread root. Nominal body stresses in power screws can be related to thread parameters as follows. The maximum nominal shear stress τ in torsion of the screw body can be expressed as τ = 16T πd3 r (8–7) The axial stress σ in the body of the screw due to load F is σ = F A = 4F πd2 r (8–8) in the absence of column action. For a short column the J. B. Johnson buckling formula is given by Eq. (4–43), which is F A crit = Sy − Sy 2π l k 2 1 C E (8–9) Nominal thread stresses in power screws can be related to thread parameters as follows. The bearing stress in Fig. 8–8, σB, is σB = − F πdmnt p/2 = − 2F πdmnt p (8–10) where nt is the number of engaged threads. The bending stress at the root of the thread σb is found from I c = (πdrnt)(p/2)2 6 = π 24 drnt p2 M = Fp 4 so σb = M I/c = Fp 4 24 πdrnt p2 = 6F πdrnt p (8–11) The transverse shear stress τ at the center of the root of the thread due to load F is τ = 3V 2A = 3 2 F πdrnt p/2 = 3F πdrnt p (8–12) and at the top of the root it is zero. The von Mises stress σ at the top of the root “plane” is found by first identifying the orthogonal normal stresses and the shear stresses. From