Mechanical Engineers Data Handbook Episode 4 pdf

APPLIED MECHANICS 65 2.2 Belt drives 2.2. I Flat, vee and timing belt drives Formulae are given for the power transmitted by a belt drive and for the tensions in the belt. The effect of centrifugal force is included. A table of information on timing belt drives is included. Symbols used: F, =belt tension, tight side F, = belt tension, slack side r, =radius of pulley a rb =radius of pulley b N, = speed of pulley a N, = speed of pulley b m=mass of belt per unit length P = power transmitted p =codficient of friction between belt and pulley F, = initial belt tension 6, =arc of belt contact pulley a eb=arc of belt contact pulley b L = distance between pulley centres s = percentage slip u = belt velocity N, rb (100-s) Speed ratio -=- ~ N, r, 100 (when pulley b is the driver) Arc of contact (r, >rb): ea= 180" + 2 sin-' - L (ra -rb) Ob= 180" -2 sin-' - L Tension ratio for belt about to slip: For pulley a >=epe* For pulley 'b' 2 = ereb where: e=base of natural logarithms (=2.718). Power capacity P=o(F, -F,) where: belt velocity u=2nraN,= 2nr,N, (no slip). Pulley torque T,=r,(Fl -F,); T,=r,(F, -F2) (F +F ) Initial tension F, = -L L ' ,F F2 F2 F 2 Effect of centrifugal force: the belt tensions are reduced by mu2 so that F, -mu2 F, -mu2 - d9 Vee belt The 'wedge' action of the vee belt produces a higher effective coefficient of friction p' p'=- P sin a where: a=the 'half angle' of the vee (p'=2.9p for a = 20"). 66 MECHANICAL ENGINEER’S DATA HANDBOOK Timing belts Timing belts have teeth which mate with grooves on the pulleys. They are reinforced with high strength polymer strands to give power capacity up to three times that of conventional belts at three times the speed. There is no slip so a constant ratio is main- tained. A large number of speed ratios is available. Belts are made in several strengths and widths. Timing belt sizes (BS 4548: 1WO) Type Meaning Pitch (mm) Widths (mm) Constant, K XL Extra light 5.08 6.4, 7.9, 9.6 - L Light 9.53 12.7, 19.1, 25.4 1.53 H Heavy 12.70 19.1, 25.4, 38.1, 50.8, 76.2 5.19 XH Extra heavy 22.23 50.8, 76.2, 101.6 12.60 XXH Double extra heavy 31.75 50.8, 76.2, 101.6, 127.0 - ~~ ~ Service factor Hours of service per day < 10 1&16 > 16 % full power 100 72 67 Class Applications % full power 1 Typewriters, radar, light domestic 100 4 Blowers, paper machines, piston pumps, textile machines 58 5 Brickmaking machines, piston compressors, hoists, crushers, mills 54 2 Centrifugal pumps, fans, woodworking machines, light conveyors 69 3 Punching presses, large fans, printing machines, grain conveyors 63 where: K = size constant (see table) N =number of revolutions per minute T= teeth in smaller pulley W= width of belt (mm) Power capacity P=KNTWx kilowatts Example: Type H belt, W=50.8mm, N=1500 revmin-’, T=20, for large fan working 12 hours per day. From tables, K=5.19 service factors 72% and 63 Yo. P=5.19~ 1500~20~ 50.8 x ~0.63 ~0.72=3.59kW Note: at high speeds and with large pulleys the power capacity may be up to 25% less. See manufacturer’s tables. APPLIED MECHANICS 67 2.2.2 Winches and pulleys Winch R Velocity ratio VR=- r W Wr VR -‘R Force to raise load F= where: q =efficiency. F Pulleys Velocity ratio VR-2 W Force to raise load F = q - 2 F 7 Block and tackle Velocity ratio VR = n where: n=number of ropes between the sets of pulleys (= 5 in figure). W Force to raise load F = q- n Differential pulley 2 Velocity ratio VR=- W Force to raise load F = q - VR 68 MECHANICAL ENGINEER’S DATA HANDBOOK U where: a=- R 2.2.3 Hoist Symbols used: m-mass of load I=moment of inertia of drum, etc. R=drum radius T= torque to drive drum T, =friction torque a =acceleration of load d = deceleration of load a = angular acceleration/deceleration Load being raised and accelerating Torque T= T, + Ia + mR(a + g) Load rising and coming to rest, no drive T,- la =mR(d -9) (WR+ T,) (mR+i) Deceleration d = Load being lowered and accelerating, no drive T, + la = mR(g -a) Load falling and being brought to rest T=Ia- T, +mR(g +d) 2.3 Balancing 2.3. I Rotating masses Out of balance due to one mass Balancing of rotating components is of extreme im- portance, especially in the case of high-speed machin- ery. Lack of balance may be due to a single mass in one plane or masses in two planes some distance apart. The method of balancing is given. For mass m at radius r and angular velocity o: Out of balance force F = mrd This may be balanced by a mass mb at ib SO that mbrb= mr APPLIED MECHANICS 69 Several out of balance masses in one plane The forces are: m1r102, m2r2w2, etc. These are re- solved into vertical and horizontal components: Fv=m,r,w2sin8, +m,r202sin8,+. . . F,=m,r,02cos8,+m2r2w2cos8,+. . . Resultant force F, = dm at an angle to horizontal axis 6, =tan - Dynamic unbalance, forces in several planes For a force mrw2 acting at x from bearing A, the moment of the force about the bearing is mrw2x. This has components: mrdx sin 8 vertically mrw2xcos 6 horizontally For several forces: Total vertical moment M,=mlrlw2x, sin 6, -+m,r,w2x, sin 8, Total horizontal moment M,=m,r,w2x, cos8, +m,r,w2x,ws82 . . . Resultant moment ,M, = ,/= acting at ob = tan- * .M The reaction at B is: Rb=L L where: L=span. The process is repeated, by taking moments about end B, and R, found. Method of balancing Complete 'dynamic balance' is achieved by introducing forces equal and opposite to R, and R,. In practice, balancing is carried out at planes a short distance from the bearings. m mr c-? cos0 \ To balance a mass mb at rb such that mbrb=i Fr w is required at an angle e,+ 180". 70 MECHANICAL ENGINEER’S DATA HANDBOOK If this distance is c then the balancing forces are This introduces small errors due to moments which can be corrected for as shown in the figure. A further very small error remains and the process may be repeated until the desired degree of balance is achieved. 23.2 Reciprocating masses For the piston, connecting rod, crank system shown in the figure there exists a piston accelerating force which varies throughout a revolution of the crank. The force can be partially balanced by weights on the crankshaft. Let: m=mass of piston w = angular velocity of crank r = radius of crank L=length of conrod 0 =crank angle Force to accelerate piston F- -mrd (approximately, see Section 2.4.1) Maximum forces F, =mm2 (at crankshaft speed, which can be balanced) r L F, =ma2- (at twice crankshaft speed) Eflect of conrod mass The conrod mass may be divided approximately between the crankpin and the gudgeon pin. If mc is the conrod mass: Effective mass at gudgeon m, =m,- added to piston mass. a L b Effective mass at crankpin m2 = m, 1 2.4 Miscellaneous machine elements 2.4. I Simple engine mechanism Using the same symbols as in the previous section: Piston Piston velocity v = 1 1 K3 Piston acceleration a=m2 cos8+Kcos2B+-((cos2B-cos48)+. . . t 4 APPLlED MECHANICS 71 r where: K = L If K is under about 0.3, it is accurate enough to use only the first two terms containing B in each formula. z4-2 nywfmds Flywheels are used for the storing of enesgy in a rotating machine and to limit speed fluctuations. Formulae are given for the calculation of the moment of inertia of flywheels and for speed and energy fluctuation. Angular velocity o==2nN Angular acceleration a = - Acceleration torque T=la where: I=mk2. 102 Energy stored E = - 2 (02-01) t Example The power of an ongine is 100 kW at a mean speadof250nvmin-'.Theencrgy to beabsorbed by the flywheel between maximum and minimum speeds is 10% of the work done per revolution. Calculate the required moment of inertia for the flywheel if the spad fluctuation is not to cxaed 2%. 2x x 250 60 K, 10.02, KB=O.l, UJ~,,, =- = 26.2 rad S- ' looooo = 24 OOo J Energy per revolution E = 254) Values of1 and k (radius of gyration) Solid disk: Mass m=Hb Radius of gyration k=- r fi mr2 pxr4b Moment of inertia I = mk2 = - = - 22 Calculation of I for given speed jluctuation If P = power, P Energy from engine per revolution=- N Cafficitnt of speed fluctuation Coefficient of energy fluctuation KEE KNOL Required moment of inertia I= Example For flywheel in previous example (J = 175 kg-m*. If the flywheel is a solid disc with thickness of the diameter, and the density is 7000 kgm-3, determine the dimensions. Thus: diameter D= 1088mm, thickneas 6- 181 mm. 72 MECHANICAL ENGINEER'S DATA HANDBOOK Annular ring: m =pn(ri -rf)b I Thin ring: If rm =mean radius, A =cross-sectional area. m = 2nr,Ap k=rm I = mr, 2 Spokes of uniform cross-section: m= p(r, -rl)A I, = mk2 Spoked wheel: The hub and rim are regarded as annular rings. I =l,,u,,+lrim+nl, where: n = number of spokes. 2.4.3 Hooke's joint (cardan joint) This is a type of flexible shaft coupling used extensively for vehicle drives. They are used in pairs when there is parallel misalignment. Symbols used : N = input speed N, = output speed a = angle of input to output shaft 0 =angle of rotation N2 N cos a N, 1-sin2acos2a Speed ratio>= 1 cos a Maximum speed ratio = ~ (at 0 =O" or 180") Minimum speed ratio=cosa (at 0=90" or 270") N +1 2=1, when 0=cos-' Nl JG APPLIED MECHANICS 13 2.4.4 Cams A cam is a mechanism which involves sliding contact and which converts one type of motion into another, e.g. rotary to reciprocating. Most cams are of the radial type, but axial rotary cams are also used. Cams may have linear motion. The motion is transmitted through a ‘follower’ and four types are shown for radial cams. Tangent cam with roller follower On the flank: Lift y- (rl + rJ(sec0- 1) where: 0=angle of rotation. Velocity v=w(rl +ro) sec0 tan0 where: w=- the angular velocity. d0 dt (1 + 2 tanZ e) Acceleration a = w2(rl + r,) cos e On the nose: the system is equivalent to a con- rodlcrank mechanism with crank radius d and conrod length (ro+r2) (see Section 2.4.1). Maximum lift y,,, = d - rl + r2 Circular arc cam with pat follower On flank: Lift y = (R -rl)( 1 -cos e) Velocity o=o(R-r,) sin0 Acceleration a=w2(R-rl) cos0 On nose: Lift y = (r2 - r + d cos(a - e) Velocity u = od sin@ - 0) Acceleration a = - w2d cos(a - 0) Maximum lift y,,=d-r, +r, 74 MECHANICAL ENGINEER'S DATA HANDBOOK Simple harmonic motion cam Lift y = d( 1 -cos 6) where: d =eccentricity. Velocity u = od sin 6 Acceleration a = dd cos 6 Maximum lift y,, = d The shape of the cam is a circle. Velocity v= - (2:) Acceleration a=O during rise and fall but infinite at direction reversal. Constant acceleration and deceleration cam, roller follower The following refers to the motion of the roller centre. Lift y=2ym,, ( - 61)2 (for first half of lift). y = 2ym,, [ f - rey] (for second half of lift). Velocity u=40ymx6 (for first half of lift) 6LX emax u = 4oymar 7 (emax - 6) (for second half of lift) Acceleration and deceleration a = 4w2y,x (constant) Cx I Constant velocity cam, knife-edge follower where: 6,,, = angle for y,,,. Axial cam vace cam) The cam profile is on the end of a rotating cylinder and the follower moves parallel to the cylinder axis. [...]... resistance, N =wheel reaction  84 MECHANICAL ENGINEER'S DATA H A * 2.7 .4 The wedge Wedge angle u =tan - (k) Force Q normal to wedge face F = 2Q(p cos u + sin a) Force Q normal to force (F), F = 2Q tan(u + 4) where: p = tan 4 o=tan-'(z) i (for n starts) Torque to lower load TL=-WD tan(6 - 4) 2 Torque to raise load TR=-tan(6 + 4) WD 2 Efficiency q = tan 6 tan@ 4) + Maximum efficiency qma,= Mechanical advantage... force F 1 J(1 -r2)2+4R2r2 82 MECHANICAL ENGINEER'SDATA HANDBOOK "c where: R = - and r = - w a , W" Phase angle a =tan- 2Rr (1 - r 2 ) 5.0 0 4. 0 b a c - 9 e 3.0 Frequency ratio, r = - & 0, 2.0 Simple harmonic force applied to mass due to rotary unbalance B 1 O 1 0 - 2 Frequency ratio r 3 F =m , w 2 cos ut (due to m s m, rotating at radius a as angular velocity u) 0, Q= r2 J(1 -r2)2+4R2rZ r Simple harmonic... used: D = mean diameter of thread p = pitch of thread 0 =thread angle =tan- ZD 4 =friction angle =tan- l p p =coefficient of friction Mechanical advantage MA= 1 1 RD Velocity ratio VR =- = tan0 p U MA tan0 Efficiency =-= VR tan@+ 4 ) + tan@ 4 ) 76 MECHANICAL ENGINEER’SDATA HANDBOOK Effective coefficientof friction (vee thread) po = p sec fl where: fl- half angle thread Acme thread Used for power transmission... hard sand 0.02 0. 04 0.08 0.1-0.3 78 2.5.3 MECHANICAL ENGINEER’S DATA HANDBOOK Aerodynamic drag Symbols used: C,=drag coefficient A, =frontal area (approx 0.9 bh m’) p =air density (==1.2 kg m - 3, v=velocity (ms-’) aerodynamic drag force: 02 F, = CdAfPT Typical valws of drag coefficient cd cd Sports car, sloping rear Saloon, stepped rear Convertible, open top Bus Truck 0.2-0.3 0 .4- 0.5 0.6-0.7 0.6-0.8... on ice Cermet on metal Dry Dry Dry Slightly lubricated Dry Slightly lubricated Dry Slightly lubricated Dry Dry Dry Dry Dry Dry 0.20 average 0.20 0.21 0.15 0 .49 0.19 0.60 average 0.20 average 0 .4 average 0 .40 0.90 average 0.3-0.5 0.5 0. 04 0.02 0 .4 Dry Coefficient of friction Maximum temperature (0°C) Maximum pressure (bar) 8 8-13 7 13 Materials Wet Dry Cast iron/cast iron Cast iron/steel Hard steel/hard... lubricated Slightly lubricated Very slightly lubricated Dry 0 .47 0.12 0.28 0.38 0.18 - Machine tool s l i i Rubber sliding ~ Pressure (bars) Materials Cast iron/cast iron Cast iron/steel Steel/steel 0.5 1.0 1.5 2.0 ~ Surface 4. 0 0.15 0.20 0.20 0.25 0.30 0.15 0.20 0.25 0.30 0.35 0.15 0.25 0.30 0.35 0 .40 Wet Dry Asphalt Concrete 0.2 54. 75 0 .45 -0.75 0.60-0.85 0.50-0.80 87 APPLIED MECHANICS 2.8 2.8.1 Brakes,... 2.6 2.6 I Vibrations Simple harmonic motion Let : x =displacement X =maximum displacement t =time f=frequency t , =periodic time rn =vibrating mass k =spring stiffness 4 =phase angle 0 =angle of rotation 80 MECHANICAL ENGINEER’SDATA HANDBOOK Definition o simple harmonic motion f where x, =static deflection Referring to the figure, point A rotates with constant angular velocity w at radius AB The projection... P,=required engine power Te=engine torque Ne=engine speed N,=wheel speed r =wheel effective radius F,=wheel force (4 wheels) 0.5-0.6 0.3-0 .4 0.2 0.1 79 APPLIED MECHANICS F,v Engine power P , =- (a-Phh) Rear wheels torque T, =prmg L ?o D 'e Engine torque Te=2nNe Wheel force (for 4 wheels) F w T 4 2 = 3 N r Nw Acceleration power Pa=mvi where: a = acceleration, vi =instantaneous speed Transmission efficiency... steel Hard drawn phosphor bronze/ hard drawn chrome plated steel Powder metal/cast iron or steel 0.05 00 6 0.15-0.2 0.15-0.2 0.05 0.03 - 150 250 250 250 0.03 - 250 10 0.05-0.1 0.1-0 .4 500 10 86 MECHANICAL ENGINEER’SDATA HANDBOOK Clutches and brakes (continued) Coefficient of friction Materials Wet Dry Maximum temperature (O°C) Powder metal/chrome plated hard steel Wood/cast iron or steel Leather/cast... motion is about to commence, and the ‘dynamic’ value, which is smaller, when there is motion Angle of repose 4 = tan-’fi; or when fi= tan 4 If the angle of the plane is greater than the angle of repose, the body will slide down the plane Force horizontal I F = Wtan(0 +4) (up plane) F = Wtan(0 -4) (down plane) t F Coefficient of friction p = N 2.7.2 Friction on an inclined plane 2.7.3 Rolling friction . belt sizes (BS 45 48: 1WO) Type Meaning Pitch (mm) Widths (mm) Constant, K XL Extra light 5.08 6 .4, 7.9, 9.6 - L Light 9.53 12.7, 19.1, 25 .4 1.53 H Heavy 12.70 19.1, 25 .4, 38.1, 50.8,. 1 J(1 -r2)2+4R2r2 and Q= 82 MECHANICAL ENGINEER'S DATA HANDBOOK "c w where: R=- and r=- W" a, 2Rr Phase angle a = tan- - (1 -r2) 5.0 0 4. 0 b a c 3.0. =tan- 4 =friction angle = tan- lp p =coefficient of friction ZD 1 tan@ + 4) Mechanical advantage MA= 1 RD Velocity ratio VR = - = - tan0 p MA tan0 VR tan@ + 4) Efficiency