Tribology in Machine Design Episode 6 potx

138 Tribology in machine design would have the effect of tightening the band, and the brake would be self- locking. Here, the direction off must be reversed to tighten the band on the drum. From the above conclusions it follows that if e /0 >x/y> 1, downward movement of the force P would tend to slacken the band. Hence for successful action x must be less than y. When the brake is used in the manner indicated above there is no relative sliding between the friction surfaces, so that/is the limiting coefficient of friction for static conditions. The differential tightening effect of the band brake is used in the design of certain types of friction brake dynamometers. 4.11.2. The curved brake block Figure 4.41 represents a brake block A rigidly connected to a lever or hanger LE pivoted at E. The surface of the block is curved to make contact with the rim of the flywheel B, along an arc subtending an angle 2\l/ at the centre, and is pressed against the rim by a force P, at the end L of the lever. In general, the normal pressure intensity between the contact surfaces will vary along the length of the arc in a manner depending upon the conditions of wear and the elasticity of the friction lining material of the brake block surface. Let p=the intensity of normal pressure at position 0, i.e. p is a function of 0 and varies from 0=0 to 0 = 2i//, a = the radius of the contact surfaces, b = the thickness of the brake block, K=the resultant force on the rim due to the normal pressure intensity p, $=the inclination of the line of action of R to the position 0=0. Figure 4.41 Friction, lubrication and wear in lower kinematic pairs 139 Hence, for an element of length a d0 of the arc of contact normal force = pab d0, tangential friction force =fpab d0. The latter elementary force can be replaced by a parallel force of the same magnitude acting at the centre 0 together with a couple of moment dM=fpa 2 b-dQ. Proceeding as for the rim clutch and resolving the forces at 0 in directions parallel and perpendicular to the line of action of /?, we have for the normal force: for the tangential force: If p is given in terms of 0, the vanishing integral determines the angle /?. Further, the resultant force at 0 is and is inclined at an angle 0=tan ^fto the direction of R. Again, the couple M together with force R i at 0 can be replaced by a parallel force R! acting at a perpendicular distance h from 0 given by The circle with centre 0 and radius h is the friction circle for the contact surfaces, and the resultant force on the wheel rim is tangential to this circle. In the case of symmetrical pressure distribution, f$ = ty, and the line of action of R bisects the angle subtended by the arc of contact at the centre 0. The angle 0 is then more conveniently measured from the line of action of R, and the above equations become 140 Tribology in machine design Now, it is appropriate to consider the curved brake block in action. Three cases shall be discussed. (/) Uniform pressure Figure 4.42 represents the ideal case in which the block is pivoted at the point of intersection C of the resultant R^ and the line of symmetry. Since /? = (/> and the pressure intensity p is constant, eqns (4.115) and (4.116) apply, so that Figure 4.42 resisting torque and also so that (ii) Uniform wear Referring to Fig. 4.42, it is assumed that the vertical descent 6 is constant for all values of 0. Hence, measuring 0 from the line of symmetry, normal wear at position 0 = <5cos0 and applying the condition for uniform wear, pa is proportional to <5cos 0 or Applying the integrals as in the preceding case Friction, lubrication and wear in lower kinematic pairs 141 or, resisting torque (Hi) Block pivoted at one extremity Figure 4.43 shows a brake block or shoe pivoted at or near one extremity of the arc of contact. For a new well-fitted surface, the pressure distribution may be approximately uniform. Wear of the friction lining material will, however, occur more readily at the free end of the shoe, since the hinge may be regarded as being at a constant distance from the centre 0. Taking the radius through the pivot centre as representing the position 0=0, let (5 = the angular movement of the shoe corresponding to a given condition of wear. xz = movement at position 0 = 2asin^0<> and normal wear at position 0 = 2asin^0cos^0<!> = c)asin0. Hence, pa is proportional to da sin 0 or Figure 4.43 In this case, eqns (4.109) to (4.112) will apply, and so Expanding the term cos(/? — 0) and integrating, this becomes For the angle ft we have from eqn (4.110) Again, expanding sin(/? — 0) and integrating 142 Tribology in machine design Using this value of /? the equation for R becomes For the retarding couple we have and substituting for k in terms of R this reduces to In all three cases, as the angle \l/ becomes small, the radius of the friction circle approaches the value and the torque This corresponds to the flat block and the wheel rim. In the general case we may write where/' is the virtual coefficient of friction as already applied to friction in journal bearings. Thus Friction, lubrication and wear in lower kinematic pairs 143 and for zero wear at one extremity In every case the retarding couple on the flywheel is Numerical example A brake shoe, placed symmetrically in a drum of 305 mm diameter and pivoted on a fixed fulcrum £, has a lining which makes contact with the drum over an arc as shown in Fig. 4.44. When the shoe is operated by the force F, the normal pressure at position 0 is p =0.53 sin0MPa. If the coefficient of friction between the lining and the drum is 0.2 and the width of the lining is 38 mm, find the braking torque required. If the resultant R of the normal pressure intensity p is inclined at an angle /? to the position 0 = 0, discuss with the aid of diagrams the equilibrium of the shoe when the direction of rotation is (a) clockwise and (b) anticlockwise. Solution Applying eqn (4.127), the braking torque is given by Figure 4.44 Since R is the resultant of the normal pressure intensity, p, the angle /? i given by 144 Tribology in machine design Expanding sin(/? — 0) and integrating, this equation becomes and proceeding as follows Substituting the numerical values For the radius of the friction circle Alternatively, In Fig. 4.44, R^ =Rsec</> is the resultant force opposing the motion of the drum. R\, equal and opposite to R lf is the resultant force on the shoe. The reaction Q at the hinge passes through the point of intersection of the lines of action of R\ and F. As the direction of Q is known, the triangle offerees representing the equilibrium of the shoe can now be drawn. The results are as follows: (a) clockwise rotation, F=1507N; (b) anticlockwise rotation, F = 2710N. 4.11.3. The band and block brake Figure 4.45 shows a type of brake incorporating the features of both the simple band brake and the curved block. Here, the band is lined with a Friction, lubrication and wear in lower kinematic pairs 145 number of wooden blocks or other friction material, each of which is in contact with the rim of the brake wheel. Each block, as seen in the elevation, subtends an angle 2\l/ at the centre of the wheel. When the brake is in action the greatest and least tensions in the brake strap are T t and T 2 , respectively, and the blocks are numbered from the point of least tension, T 2 . Let kT 2 denote the band tension between blocks 1 and 2. The resultant force R\ exerted by the rim on the block must pass through the point of intersection of T 2 and kT 2 . Again, since 2i// is small, the line of action of/?i will cut the resultant normal reaction R at the point C closely adjacent to the rim, so that the angle between R and R\ is 0=tan~ 1 f. Suppose that the angle between R and the line of symmetry OS is /?, then, from the triangle of forces xyz, we have If this process is repeated for each block in turn, the tension between blocks 2 and 3 is k. Hence, if the maximum tension is 7\, and the number of blocks is n, we can write If the blocks are thin the angle (3 may be regarded as small, so that Figure 4.45 or so that 4.12. The role of The maximum possible acceleration or retardation of a vehicle depends friction in the propulsion upon the limiting coefficient of friction between the wheels and the track. and the braking of Tnus if vehicles R=the total normal reaction between the track and the driving wheels, or between the track and the coupled wheels in the case of a locomotive, 146 Tribology in machine design F=the maximum possible tangential resistance to wheel spin or skidding, then Average values of/are 0.18 for a locomotive and 0.35 to 0.4 for rubber tyres on a smooth road surface. Here/is called the coefficient of adhesion and F is the traction effort for forward acceleration, or the braking force during retardation. Both the tractive effort and the braking force are proportional to the total load on the driving or braking wheels. During forward motion, wheel spin will occur when the couple on the driving axle exceeds the couple resisting slipping, neglecting rotational inertia of the wheels. Conversely, during retardation, skidding will occur when the braking torque on a wheel exceeds the couple resisting slipping. The two conditions are treated separately in the following sections. Case A. Tractive effort and driving couple when the rear wheels only are driven Consider a car of total mass M in which a driving couple L is applied to the rear axle. Let /t =the moment of inertia of the rear wheels and axle, 1 2 =the moment of inertia of the front wheels, FI =the limiting force of friction preventing wheel spin due to the couple L, F 2 =the tangential force resisting skidding of the front wheels. Also, if v is the maximum possible acceleration, a the corresponding angular acceleration of the wheels, and a their effective radius of action, then Figure 4.46 Referring to Fig. 4.46, case (a), the following equations can be written Friction, lubrication and wear in lower kinematic pairs 147 Adding eqns (4.142) and (4.143) and eliminating (F l -F 2 ) from eqn (4.144), then Also, from eqns (4.143) and (4.144) and eliminating a This equation gives the least value of F t if wheel spin is to be avoided. For example, suppose M = 1350 kg, 7 1 = 12.3kgm 2 ; / 2 =8.1kgm 2 and a=0.33m, then or so that, if L exceeds this value, wheel spin will occur. The maximum forward acceleration Equation (4.145) gives the forward acceleration in terms of the driving couple L, which in turn depends upon the limiting friction force F t on the rear wheels. The friction force F 2 on the front wheels will be less than the limiting value. Thus, if R t and R 2 are the vertical reactions at the rear and front axles, then To determine /? t and R 2 , suppose that the wheel base is b and that the centre of gravity of the car is x, behind the front axle and y, above ground level. Since the car is under the action of acceleration forces, motion, for the system as a whole, must be referred to the centre of gravity G. Thus the forces F! and F 2 are equivalent to: (i) equal and parallel forces FI and F 2 at G (Fig. 4.46, case (b) (ii) couples of moment F { y and F 2 y which modify the distribution of the weight on the springs. Treating the forces RI and R 2 in a similar manner, and denoting the weight of the car by W, we have [...]... Tribology in machine design can be distinguished; free rolling, braking, accelerating, cornering or any combination of them Figure 4.54 shows the loads acting on the tyre during (a) a free rolling, (b) a braked rolling and (c) a driven rolling In all cases, longitudinal tractive forces are produced in the contact zone, giving rise to net forces Fr, Fb, Fd acting on the tread and the reaction force W acting... equatorial line in a deflected state In the contact zone Figure 4.51 which corresponds to a force Q' = — 2Gcc(y + c) At the rear of the contact zone, there is a discontinuity in dk/dx which gives rise to an infinite traction q"(c) corresponding to a force 154 Tribology in machine design Q" = —2GcKy(y + c) The total cornering force is thus Self-aligning torque can be found by taking moments about O The infinite... force If, through varying conditions of limiting friction at either of the front wheels, or because of uneven wear in the brake linings, the braking torques on the two wheels are not released simultaneously, a couple tending to 150 Tribology in machine design rotate the front axle about a vertical axis will be instantaneously produced, resulting in unsteady steering action This explains the importance... as and, eliminating F^ and F2 from eqn (4.173) maximum retardation = 4.13 Tractive resistance In the foregoing treatment of driving and braking, the effects of friction in the bearings were neglected However, friction in the wheel bearings and in the transmission gearing directly connected to the driving wheels is always present and acts as a braking torque Therefore, for a vehicle running freely on... not continuous but end abruptly within the tread Their main role is to displace bulk water from the tyre footprint They also permit the macro-movement of the tread during the wiping action 4.14.5 The mechanism of rolling and sliding Both rolling and sliding can be experienced by a pneumatic tyre Pure sliding is rather rare except in case of a locked wheel combined with flooding due to heavy rainfall... concentrated in the centre of the contact zone This is mainly due to the tread 152 Tribology in machine design 4.14.1 Creep of an automobile tyre An automobile tyre will tend to creep longitudinally if the circumferential strain in the contact patch is different from that in the unloaded periphery In accordance with the theory of the membrane, there is a shortening in the contact patch of the centre-line of... resisted However, in the absence of this pressure the fluid would continue to be drawn into the cavity with the interface advancing to the right It has been shown experimentally, that when the meniscus reaches the end of the constricted passage it begins to turn itself inside out as indicated in Fig 4 .63 Owing to the contamination of engineering surfaces, the contact angles of oil against synthetic rubber... components Instances of failure of the barrier elements by fatigue are usually due to aeroelastic instability which could be avoided by suitable design There are computer programmes available to design a labyrinth seal 4.15 .6 Wear in mechanical seals The sealing elements (the primary ring and the mating ring), of a nominally contact type seal, usually operate in uni-directional sliding Reciprocating motion... under laminar conditions is given by Figure 4 .66 where fi = (h + c)/c orh/c+l,y= b/(a + b) and L is the effective length of the screwed portion At high Reynolds numbers (Re Js 60 0-1000), turbulent flow conditions lead to a more effective sealing action Typical values for the design 164 Tribology in machine design parameters of helical seals are as follows: a = 10°-20 ; (3 = 4 -6; y =0.5-0.8 Taking mean... vehicles) of the dynamic hydroplaning limit is shown in Fig 4.58, case (a) It is not difficult to show that, according to hydrodynamic theory, twice the speed is required under sliding compared with rolling to attain the dynamic hydroplaning when Ph = W This is because both surfaces defining the converging gap attempt to drag the water into it when rolling, whereas during sliding usually only one of the . cos(/? — 0) and integrating, this becomes For the angle ft we have from eqn (4.110) Again, expanding sin(/? — 0) and integrating 142 Tribology in machine design Using this value . couple tending to 150 Tribology in machine design rotate the front axle about a vertical axis will be instantaneously produced, resulting in unsteady steering action. This explains . rise to an infinite traction q"(c) corresponding to a force 154 Tribology in machine design Q" = —2G c Ky(y + c). The total cornering force is thus Self-aligning torque