Presentation on theme: "Chapter 5 Efficiency and Self-lock of Machineries"— Presentation transcript:
1 Chapter 5 Efficiency and Self-lock of Machineries §5－1 Friction in Kinematic Pairs§5－2 Efficiency of Machineries
2 §5－1 Friction in Kinematic Pairs 一、 Friction in the sliding pairsFN21The slider 1 acted on by drive force F moves forward at a constant velocity, considering vertical load G acting on the slider, normal reaction force FN21, thus the friction Ff21 isv1221Ff21FGFf21＝f FN21If the load G is constant , the friction Ff21 is related with the geometry of elements of pairs.G21θ1FN21G212PlanarVee-slotCylinderGFN212FN21= GFf21=f G/sinθFN21= kGFf21=f GFf21= kf GFN21=G/sinθk =1~π/2
3 fv —equivalent coefficient of friction The friction of sliding can be calculated by the following general formulafv —equivalent coefficient of frictionFf21= f FN21= fv GφFR21Planar friction： fv = fv12GFFf2121FN21Vee-slot friction： fv = f / sin θCylinder friction： fv = fDetermination of total reaction force in sliding pairs：Total reaction force FR21—vector sum of FN21 and Ff21Frictional angleφ—is the angle between FR21 and FN21.φ= arctan fThe angle between FR21 and v12 is constant （90＋）.
4 For example （一）Slope friction The slider 1 moves at a constant velocity on slope 2, let us consider the drive force F.nφFNFR21Ffv12GF1 ．Slider moves up12αFAccording to equilibrium conditions of force: F + FR21+G = 0FR21α+φGFrom figure,F = Gtan(α+φ)2 ． Slider moves down12αnv12GF ’φF’R21F’According to equilibrium conditions of force:F’ + F’R21+G = 0F’R21Gα-φFrom figure,F’ = Gtan(α-φ)
5 （二）Friction in the Helical Pairs 1.Friction in the rectangular-thread helical pairsA rectangular-thread helical pair consists of bolt 1 and nut 2 .We can suppose that forces produced between bolt 1 and nut 2 all act on the cylinder surface. If we expand this cylinder surface the thread will form an inclined plane with a slope angle α . Now friction in helical pairs is simplified into friction of slider on the slope.
6 If nut 2 moves up at a constant speed under the acting of the torque M and axial load G, this is similar to the slider moving along the slope at a constant speed. So we obtainG/212d2d3d1FG1If nut 2 moves down at a constant speed under the acting of the torque M and axial load G, this is similar to the slider moving along the slope at a constant speed. So we obtain2πd2lα
7 fv = f / sin(90°-β) = f / cosβ 2. Friction in the triangular-thread helical pairsFriction in the triangular-thread helical pairs is very similar to friction in the rectangular-thread helical pairs. We can introduce equivalent coefficient of friction fv and equivalent frictional angle φv to solve this problem.fv = f / sin(90°-β) = f / cosβφv = arctan fvTorque required in tightening a nutTorque required in unscrew off a nut
8 二、Friction in the Revolute Pairs （一）Friction in shaft neckIn Fig. , radial load G and drive moment of couple Md act on shaft neck 1, which turns at a constant angular speed 12 .ω12MdGr12MfFR21——is vector sum of FN21and Ff21According to equilibrium conditions of force:FR21FN21Ff21ρFR21=- G Md =－Mf = FR21ρ =Gρ
9 Mf= F21 r = fv G r fv = kf = (1~π/2 )f ω12MdGr12Mfρ= fv rMf= fv G r = GρTo a specific shaft neck-bearing pair, because fv and r are constant, is also constant. is the radius of frictional circle.FR21FN21Ff21ρr12Mfω12MdFR21FN21Ff21GNow we know that if shaft neck turns in bearing, the total reaction force FR21 will be always tangent to the frictional circle.
10 （二）Friction at the shaft end Mf21GωMShaft end is the part of shaft that sustains axial load.When shaft end 1 turns on thrust bearing 2, frictional torque Mf is generated between them.2R2r1.For new shaft endTo new shaft end, we may suppose the pressure on the contact surface p is constant, so we have2. For old shaft endTo old shaft end, p = constant. We have
11 §5－2 Efficiency of Machineries When a machine is running: Wd= Wr+WfEfficiency of Machineries is the ratio of output work to input work, usually denoted by η.FvFF0vFη＝Wr / Wd ＝1－Wf /WdIf Nd ,Nr and Nf represent input power, output power and lost power respectively, we haveη＝Nr / Nd ＝1－Nf /NdIdeal machineryη0Actual machineryηGvGη＝ F0 / F ＝M0 / Mη＝Nr / Nd ＝G vG /F vFη0＝Nr / Nd ＝G vG /F0vF =1
12 Example 1. Slope F = Gtan(α+φ) F0 = Gtanα η= F0 / F = tanα / tan(α+φ) η ’ = G / G0 = F0 / F ’ = tan(α-φ) / tanα2. Helical pairsscrewunscrewEfficiency of machineries may also be calculated by experiment method. The efficiencies of some general mechanisms and pairs are listed in Table 5－1.
13 二、Efficiency of a group of mechanisms 1. Serial structure＝123…k2.Parallel structure3. Parallel-serial structure Its total efficiency can be calculated by combining above two equations.
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