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Chapter 15 • Fasteners and Power Screws 15.1

An Acme-threaded power screw with a crest diameter of 1 1/8 in. and single threads is used to raise a load of 25,000 lbf. The collar mean diameter is 1.5 in. The coefficient of friction is 0.1 for both the thread and the collar. Determine the following: a) Pitch diameter of the screw b) Screw torque required to raise the load c) Maximum thread coefficient of friction allowed to prevent the screw from self-locking if collar friction is eliminated.

Notes: Recognizing that this is an Acme screw, some data is obtained from Table 15.2 on page 671. In addition, Equations (15.4), (15.8) and (15.10) are used in the solution of this problem. Solution: This is an Acme screw, so referring to Table 15.2 on page 671 with a crest diameter of 1 1/8 in, n=5 per inch. Therefore, from Equation (15.1), p=1/n=0.2in. The pitch diameter is calculated from Equation (15.4): dp=dc-0.5p-0.01in=1.125in-0.5(0.2in)-0.01=1.015in If this is a single thread then the lead, l, equals the pitch, or l=0.2 in. From Equation (15.5), α is given by  l   0 .2in  α = tan−1  = tan−1   = 3 .589 °   π (1 .015in )   πd p  From Figure 15.5, it can be seen that β=29°, so from Equation (15.8), θ is 29°  β   θn = tan−1  cos α tan  = tan−1 cos 3 .589° tan = 14 .47 °    2 2  The torque to raise the load is given by Equation (15.10):   d p / 2 (cos θn tan α + µ ) Tr = W  + rc µ c  cos θn − µ tan α    

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 (1.015in / 2 )( cos 14.47° tan 3.589° + 0.1)  1 .5in  + = (25kip )  ( 0.1) = 3995in − lb  2  cos14 .47 ° − 0 .1 tan3 .589 °   If collar friction is eliminated, then the load will lower if the numerator of Equation (15.12) is zero, or µ =cosθntanα=cos(14.47°)tan(3.589°)=0.0607

15.2

A car jack consists of a screw and nut. The car is lifted by turning the screw. Calculate the torque needed to lift a load of 1 ton. The lead of the thread l=9mm, its pitch diameter is 22 mm, and its thread angle is 30°. The coefficient of friction is 0.10 in the threads and zero elsewhere.

Page 15-1

Notes: The torque is calculated from Equation (15.10), which required calculation of α and θ from equations (15.5) and (15.8), respectively. Note that a metric ton is 1000 kg, which is the load used here. Some students may interpret the load as an English ton, or 2000 lb. Solution: From Equation (15.5), α is given by

 l  −  9mm  α = tan−1  = tan 1   = 7.42°   π ( 22mm)   πd p  From Equation (15.8), θ is

30°  β   = 14 .9 ° θn = tan−1  cos α tan  = tan−1 cos7 .42 ° tan   2 2  The torque to raise the load is given by Equation (15.10):  d p / 2 (cos θ n tan α + µ )  Tr = W  + rcµ c cos θ n − µ tan α    

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 ( 0. 022 m / 2)( cos 14. 9 °tan 7. 42 ° + 0. 1)  = (1000 kg) 9. 81 m/ s2  + 0  = 25 .56 Nm cos14 .9 ° − 0 .1 tan 7 .42 °  

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15.3

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A power screw gives the axial tool motions in a numerically controlled lathe. To get high accuracy in the motions, the heating and power loss in the screw have to be low. Determine the power efficiency of the screw if the coefficient of friction is 0.10, pitch diameter is 30 mm, lead is 6 mm and thread angle is 25°.

Notes: The efficiency is the output work divided by the input work. If this is evaluated over a given distance, such as one screw revolution, then the torque equation can be used. In this regards, the analysis is similar to Problem 15.2. Note however, that we’re worried about the efficiency of the screw, not the collar, so we’ll ignore the collar losses, that is, we’ll take µ c=0. Solution: The input work is the product of the torque and the rotation. Any reference rotation can be used, so use one revolution or 2π radians as the reference. The torque is given by Equation (15.10), but first, from Equation (15.5), α is given by  l  −  6mm  α = tan−1  = tan 1   = 3.65°   π ( 30mm)   πd p  From Equation (15.8), θ is 25 °  β   = 12 .47° θ n = tan−1  cos α tan  = tan−1  cos 3 .65 ° tan  2 2  The torque to raise the load is given by Equation (15.10):

Page 15-2

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  d p / 2 (cos θ n tan α + µ )  (0.030m / 2) (cos 12.47° tan 3.65° + 0.1)  Tr = W  + rc µc  = W   = (0.00251m )W cos 12.47 ° − 0.1 tan 3.65 ° cos θn − µ tan α       If the lead is 6mm, this is the distance the load will be moved in one revolution. The output work is the product of the load and the distance. Therefore, the ratio of the output work to the input work is: Wl W (0 .006 m) = = 0 .381 = 38 .1 % e= Tθ W(0 .00251 m) (2 π )

15.4

Sketch a shows a stretching device for steel wires used to stabilize the mast of a sailing boat. Both front and side views are shown and all dimensions are in millimeters. A screw with square threads (β=0), a lead and pitch of 4 mm and an outer diameter of 20 mm is used. The screw can move axially but is prevented from rotating by flat guiding pins (side view in sketch a). Derive an expression for the tightening torque as a function of the stretching force P when the coefficient of friction of all surface contacts is 0.20. Also, calculate the torque needed when the tightening force is 1000N.

Notes: This problem requires a derivation similar to that on pages 673-674 for the particular circumstances of this problem. Solution:

Page 15-3

A free-body diagram of the threads is given to the right above. From Equation (15.5), 4 mm l tanα = = = 0.07074 2π rm 2π (9 mm) Equilibrium in the vertical direction of the free body diagram gives: ΣFy=0=-Pax+Pncos α-µ Pnsinα ; Pax=Pncos α-µ Pnsinα and in the horizontal direction, ΣFx=0=Pt-Pnsinα-µ Pncos α; Pt=Pnsinα+µPn cos α Therefore we can write

tan α + µ Pt sinα + µ cosα = = Pax cos α − µ sin α 1 − µ tanα

Vertical equilibrium of the screw gives

∑ Fy = 0 = Pax − P −

µPt rm µP r ; Pax = P + t m L L

Torque equilibrium of the nut gives ΣT=0=T-µPaxrm-Ptrm; T=µPaxr M+Ptrm Eliminating Pax and Pt gives

  rm ( µ + tan α )  1 − µ tanα + µrm   T = P  1 − µ rm ( tanα + µ )    L(1 − µ tan α)  

If P=1000N, then substituting µ=0.2, r m=9mm, rM=15mm, L=20mm and tanα=0.0707, this expression yields T=5.61Nm.

15.5

A flywheel of a motorbike is fastened by a thread manufactured directly in the center of the flywheel as shown in sketch b. The flywheel is mounted by applying a torque T. The cone angle is γ. Calculate the tensile force W in the shaft between the contact line at N and the thread as a function of D1, D2, γ, and T. The lead angle is α at the mean diameter D1. The shaft is assumed not to deform.

Page 15-4

Notes: This problem is solved by performing torque equilibrium on the flywheel and then using Equation (15.10) for the screw torque. Solution: The forces acting on the shaft are shown above right. Note that when the flywheel is mounted, the contact slides in the circumferential and the axial direction. Torque equilibrium on the flywheel gives  D  ∑ T = 0 = T − µ N cos φ  22  − Tt where Tt is the thread torque and φ is the angle as shown defined by D tan α tan φ = 1 D2 cos γ Horizontal force equilibrium gives ΣFx=0=W-Nsinγ -µ Nsinφ cosγ ; W=N(sinγ+µsin φcos γ ) Substituting this into Equation (15.10) and then the resulting expression into the torque equation given above yields WD1( cosθn tanα + µ ) / 2 µ D2 W cos φ / 2 T− =0 − sinγ + µ sin φ cosγ cosθ n − µ tanα This could be solved for W if desired.

15.6

To change its oil, a 10-ton truck is lifted a height of 1.7m by a screw jack. The power screw has Acme threads and a crest diameter of 5 in. with two threads per inch, and the lead equals the pitch. Calculate how much energy has been used to lift and lower the truck if the only friction is in the threads, where the coefficient of friction is 0.08

Notes: A ton is taken here as 2000lb, although it may be interpreted as the weight of 1000 kg. The approach is similar to Problems (15.1) to (15.3), but one must also use Equation (15.12) to calculate the torque needed to lower the load. Solution: For a crest diameter of 5 in, Table 15.2 gives the number of threads per inch as 2, so that the pitch is 0.5in. From Equation (15.4), the pitch diameter is dp=dc-0.5p-0.01in=5in-0.5(0.5in)-0.01=4.74in Since the lead, l, equals the pitch, l=0.5 in. From Equation (15.5), α is given by  l  −1 0 .5in  α = tan−1  = 1. 923°  = tan  π π d  ( 4. 74in )   p From Figure 15.5, it can be seen that β=29°, so from Equation (15.8), θ is 29°  β   = 14. 49° θn = tan −1  cos α tan  = tan −1 cos 1. 923° tan   2 2  The torque to raise the load is given by Equation (15.10) (note that µc=0):

Page 15-5

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  d p / 2 (cos θn tan α + µ) + rc µ c Tr = W    cos θn − µ tan α   ( 4. 74 in / 2)( cos 14. 49° tan 1. 923° + 0. 08)  = ( 20 kip)   = 5523 in − lb cos14 .49° − 0 .08 tan1 .923°   The energy is the product of the torque and the rotation. To travel 1.7m=66.93in, the screw needs to rotate 66.93/l=66.93/0.5=133.9revolutions=841.1rad. The energy needed to raise the load is then Er=(841rad)(5523in-lb)=4645kip-in or 387kip-ft The torque to lower the load is obtained from Equation (15.12):   d p / 2 ( µ − cos θn tan α ) + rc µ c  Tl = −W  cos θn + µ tan α    

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( 4. 74 in / 2)( 0. 08 − cos 14. 49° tan 1. 923° )  = −( 20 kip)   = −2318 in − lb cos14 .49° + 0 .08 tan1 .923°   and the energy needed to lower the load is then El=(841rad)(2318in-lb)=1950kip-in=162kip-ft

15.7

A single-threaded M32x3.5 power screw is used to raise a 12-kN load at a speed of 25 mm/s. The coefficients of friction are 0.08 for the thread and 0.12 for the collar. The collar mean diameter is 55 mm. Determine the power required. Also, determine how much power is needed for lowering the load at 40 mm/s.

Notes: The power is calculated from the torque needed to raise the load or lower the load, obtained from Equations (15.10) and (15.12), respectively. Solution: The bolt designation gives dc=32mm and p=3.5mm. From Equation (15.4) for metric threads, dp=dc-0.5p-0.25=32mm-(0.5)(3.5mm)-0.25mm=30mm Since there is a single thread, the lead is the same as the pitch, or l=3.5mm. The torque is given by Equation (15.10), but first, from Equation (15.5), α is given by  l  −  3 .5mm  α = tan−1  = tan 1   = 2.13°   π ( 30mm)   πd p  Note from Figure 15.5 that β=29°. From Equation (15.8), θ is β 29 °  θ n = tan−1  cos α tan  = tan−1  cos 2 .13 ° tan = 14 .49°   2 2  The torque to raise the load is given by Equation (15.10):

Page 15-6

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  d p / 2 (cos θn tan α + µ) + rc µ c Tr = W  cos θn − µ tan α     ( 0.030 m / 2) ( cos14.49° tan 2.13° + 0.08) (0 .055 m)(0 .12 )  + = ( 12kN)   = 61 .23 Nm cos 14.49° − 0.08 tan 2.13° 2   Since the lead is l=3.5mm, and the load is raised at 25mm/s, the screw must be rotating at a speed of ωr=25/3.5=7.14rev/s=44.88rad/s. Therefore the power is the product of torque and angular velocity, or hpr=Trωr=(61.23Nm)(44.88rad/s)=2748W The torque to lower the load is given by Equation (15.12) as   d p / 2 ( µ − cos θn tan α ) + rc µ c  Tl = −W  cos θn + µ tan α     . / . cos . tan . 0 . 055 14 49° 2 13°) ( ( 0 030 m 2) ( 0 08 − m)( 0.12)  + = −( 12kN)   = −47 .75 Nm cos 14.49° + 0.08 tan 2.13° 2   The lowering speed is ωl=40/3.5=11.43rev/s=71.81rad/s, so the power needed is hpl=Tlωl=(47.75Nm)(71.81rad/s)=3429W

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15.8

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A double-threaded Acme power screw is used to raise a 1350-lb load. The outer diameter

of the screw is 1.25 in and the mean collar diameter is 2.0 in. The coefficients of friction are 0.13 for the thread and 0.16 for the collar. Determine the following a) Required torque for raising and lowering the load b) Geometrical dimensions of the screw c) Efficiency in raising the load. d) Load corresponding to the efficiency if the efficiency in raising the load is 18%. Notes: This problem is similar to the previous problems, especially Problem 15.2 and 15.7. The new concepts introduced are the use of a double thread and more in-depth use of the equation for efficiency (Equation (15.13)). Solution: From Table 15.2 on page 671, for a crest diameter of 1.25 in, there are n=5 threads per inch, so that p=1/n=0.2in, Since this is a double thread, the lead is twice the pitch, or l=0.4in. The pitch diameter is obtained from Equation (15.4) as dp=dc-0.5p-0.01=1.25in-0.5(0.2in)-0.01in=1.14in The torque is given by Equation (15.10), but first, from Equation (15.5), α is given by  l  −  0 .4in  α = tan−1 = tan 1  = 6 .376 °   π (1 .14 in)   π dp Note from Figure 15.5 that β=29°. From Equation (15.8), θ is

Page 15-7

29°  β   = 14 .4 ° θ n = tan−1  cos α tan  = tan−1 cos 6 .376 ° tan    2 2  The torque to raise the load is given by Equation (15.10):   d p / 2 (cos θn tan α + µ ) + rc µ c  Tr = W    cos θn − µ tan α  

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  (1. 14 in / 2 )(cos 14. 4° tan 6.376 ° + 0.13) + (1.0 in) (0.16)  = 408.1 in − lb = ( 1350 lb)  cos 14.4° − 0. 13 tan 6.376°   The torque to lower the load is given by Equation (15.12) as  d p / 2 ( µ − cos θ n tan α )  Tl = −W  + rc µ c  = − 233inlb cosθn + µ tanα     From Equation (15.13), the efficiency is given by 100Wl (100 )(1350 lb)( 0 .4 in) = 21 .06 % = e= 2π ( 408inlb) 2π T

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If the efficiency is 18% at the same torque, the load is obtained from Equation (15.13) as 100Wl 2 πeT 2 π(18 )(408inlb ) = 1154lb e= = ;W = 2πT 100 l 100 (0.4in )

15.9

A 25-kN load is raised by two Acme-threaded power screws with a minimum speed of 35 mm/s and a maximum power of 1750 W per screw. Because of space limitations the screw diameter should not be larger than 45mm. The coefficient of friction for both the thread and the collar is 0.09. The collar mean diameter is 65mm. Assuming that the loads are distributed evenly on both sides, select the size of the screw to be used and calculate its efficiency.

Notes: This problem requires selection of a screw from table 15.2, then analysis of this screw. Solution: For a thread diameter of 45mm=1.77 in, the largest screw which can be used is, from Table 15.2, a 1.75 inch crest diameter screw with 4 threads per inch. Therefore, the pitch is 0.25in=6.35mm. The pitch diameter is calculated from Equation (15.4) as dp=dc-0.5p-0.01in=1.75in-0.5(0.25)-0.01in=1.615in=41.02mm At first, use a single thread so that l=p=6.35mm. The torque is given by Equation (15.10), but first, from Equation (15.5), α is given by  l  −1  6 .35mm  α = tan −1   = tan   = 2.82°  π ( 41.02mm)   πd p  Note from Figure 15.5 that β=29°. From Equation (15.8), θ is β 29 °  θ n = tan−1  cos α tan  = tan−1  cos 2 .81 ° tan = 14 .48°   2 2 

Page 15-8

The torque to raise the load is given by Equation (15.10), using W=12,500N since there are two screws:  d p / 2 (cos θn tan α + µ )  Tr = W  + rc µ c  cos θn − µ tan α    

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  (.04102 m / 2)( cos14.48° tan 2.82° + 0.09) + (0 .0325 m) (0 .09 ) = 73 .2 Nm = (12.5kN ) ° − ° 14 48 0 09 2 82 cos . . tan .   To raise the load at 35mm/s, the angular velocity ω=35/6.35=5.51rev/s=34.63rad/s. Therefore the power is hp=Tω=(73.2Nm)(34.63rad/s)=2540W This horsepower is too high. Using double threads, the lead is l=2p=12.70mm. Using the same equations, one obtains α=5.6°, θn=14.4°, Tr=86.24Nm, ω=17.31rad/s, hp=1490W. Therefore, a double thread screw satisfies the power requirement.

15.10

A screw with Acme thread can have more than one entrance to the thread per screw revolution. A single thread means that the pitch and lead are equal, but for a double and triple thread the lead is larger than the pitch. Determine the relationship between the number of threads per inch n, the pitch p and the lead l.

Notes: If a student has difficulty visualizing this problem, the concept can be illustrated by wrapping a single string around a pencil, a double thread by wrapping two strings (preferably of different colors) and a triple thread by wrapping three strings around a pencil. Solution: If there are m threads, then the lead is related to the pitch and the threads per inch by m l = mp = n

15.11

An M12, coarse pitch, class 5.8 bolt with a hexagonal nut assembly is used to keep two machine parts together as shown in sketch c. Determine the following: a) Bolt stiffness and clamped member stiffness by using Wileman’s method. b) Maximum external load that the assembly can support for a load safety factor of 2.5. c) Safety factor guarding against separation of the members d) Safety factor guarding against fatigue if a repeated external load of 10kN is applied to the assembly.

Page 15-9

Notes: This problem is long only because of the many parts; each part is very straightforward. The equations used are for part (a), (15.21), (15.25) and (15.26). For part (b), (15.17) and (15.31), for part (c), (15.32), and for part (d), (15.40). Solution: I. Bolt and Member Stiffness Note from the inside front cover that for steel, Es=207GPa and for aluminum Eal=72GPa. From the sketch we see that Lt=20mm and Ls=40mm. From Table 15.7 on page 689, for a crest diameter of 12mm and coarse threads, p=1.75mm, At=84.3mm2. From Equation (15.2), ht is 0 .5 p 0 .5(1 .75mm ) = ht = = 1. 516mm tan 30° tan 30° From Figure 15.4, the root diameter is: dr = dc − 2 (0. 625 ht ) = 12 mm − 2 (0. 625 )(1.516 mm) = 10.105 mm Therefore, the bolt stiffness is, from Equation (15.21),  0.04 m + 0.4(0.012 m) 0. 02m + 0.4( 0.010105m )  1 4  Ls + 0.4 d c Lt + 0.4 dr  4 + = +   =  2 2 kb π E  ( 0 .012m)2 dc dr ( 0.010105 m)2  π (207GPa)   or kb=297.5MN/m. For the member, we use Equation (15.26) with the proper constants taken from Table 15.3. Referring to the aluminum as member 1 and the steel as member 2, we can write )( ) / (. / k j1 = Ed cAal e Bal d c L = (72GPa )( 0 .012 m) (0 .7967) e 0 63816 12 35 = 0.857GN /m ( . / ( )/ k = Ed A e Bal dc L = ( 207GPa)( 0. 012 m )(0. 78715 e 0 62873) 12 25 = 2. 644GN/ m j2

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c steel

Therefore, the joint stiffness, from equation (15.25) is 1 1 1 1 1 = + = + ; k = 0.647 GN / m k j k j1 k j2 0 .857 GN / m 2 .644 GN / m j The stiffness parameter Ck is

Ck =

kb 297 .5 MN / m = = 0.315 k b + k j 297 .5MN / m + 647MN / m

II. Maximum Load From Table 15.5, for a 5.8 grade bolt, the proof strength is Sp=380MPa, Sut=520MPa and Sy=415MPa. If we assume this is a reused connection, then from Equation (15.33), 2

Pi=0.75Pp=0.75SpAt=0.75(380MPa)(84.3mm )=24,025N~24kN The maximum load is obtained from Equation (15.31): 84 .3 mm2 ( 380 MPa) − 24 kN At Sp − Pi At Sp − Pi ; Pmax,b = = = 10. 2 kN n sb = 2. 5( 0. 315) Pmax,b Ck nsb Ck

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III. Joint Separation The safety factor against joint separation is given by Equation (15.32) as 24 kN Pi = = 3.45 nsj = Pmax, j ( 1− Ck ) (10 .2 kN ) (1 − 0 .315) IV. Fatigue Analysis Assuming the threads are rolled, then Kf=2.2 from Table 15.8. We don’t know how the load is applied, but if we assume the loading is axial, then from Equation (7.7) the endurance limit is

Page 15-10

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Se=0.45Su=0.45(520MPa)=234MPa. The prestress is σi=Pi/At=(24kN)/(84.3mm )=285MPa. The alternating stress is given by Equation (15.36) as 0 .315(5000 N ) C P σa = k a = = 18. 68MPa At 84. 3mm 2 and the mean stress is calculated from Equation (15.37) as P + Ck Pm 24kN + ( 0 .315) (5000 N) = σm = i = 303.4MPa At 84.3mm 2 Therefore, the safety factor against fatigue failure is given by Equation (15.40) as 520 MPa − 285 MPa Sut − σ i = = 2 .14 ns =  Pa  Sut  Pm  ...