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P1-16 The imported band saw shown in Fig

Driven pulley D -400 mm Saw band Frame Drive V-belt pulley Electric motor d 125 mm Motor base Figure 1-11 Band saw. (Problem 1-16) 1-11 is to be equipped with a domestic electric motor. This tool is characterized by having a taut, endless band of steel, with saw teeth on one edge, that passes over two large, flat pulleys or wheels, of which the lower is the driver. The manufacturer recommends an operating speed of 18 to 19 m/s and a minimum power requirement of 3.5 kW (a) Calculate the speed of the drive pulley and diameter D of a driven V-belt pulley for a motor speed of 1200 rpm (b) Select a suitable motor based on intermittent operation and an 8-h working day (c) Calculate the cutting force F which the saw blade exerts on the work piece. Assume 95% effi- ciency

8 PRIMARY CONSIDERATIONS through three machine members with effi- ez, e. By passing through the frst TABLE 1-2 Mechanical Efficier Typical Values member, E loses a small amount of energy and thereby reduces the output to (e)(E). The output of the first member becomes the input of the second member. Again, energy is lost, and the output of the second member (e)(e2)(E) be- comes the input of the third member. Total out- put E, is therefore Efficiency, e Machine Member High Low 1. Ball bearings 2. Silent chains 0.999 0.99 0.99 0.97 0.985 _ 3. Spur and helical gears (including bearings) 4. Roller bearings 0.98 For n machine members with efficiencies e5 e,6. el gears Synchronous belts ez . . . e, total output is E(e)(e2). .. 6. Bevel gears (e)Ei, and total efficiency becomes the product, not the sum, of the individual efficiencies. 7. V-belts 8. Roller chains 9. Worm gearing 0.97 0.95 0.95 e = (e)(ez) . . . (en) 10. Bal bearing screws 0.90 Power screws (multithreads 0.84 0.38 -single threads) Table 1-2 presents typical values of the effi-1 ciency for common machine elements inge 13. Screw fasteners 0 EXAMPLE 1-1 An electric motor drives a machine by means of a gear train with three sets of spur gears and a roller chain drive. Find the efficiency of the drive system 1-7 POWER Power is the rate of doing work or transmitting energy. The SI power unit is the watt (W). Be- cause the watt is such a small unit, the kilowatt (1000 W) is preferred in most technical calcu- lations. With proper units, the power equations SOLUTION 7 Using values from Table 1-2 in Eq. 1-5, we| obtain: ku14 P 1000 Tn 9550 e (0.9850.96) 0.92 kW Efficiency can be measured directly or calcu lated, but it can also be evaluated by means of noise or heat generated. Although not always recognized, noise represents an excellent me dium for evaluating machine performance. En ergy expended in unwanted machine motion or vibration reduces the useful output and results in noise. Generally, the condition of least noise corresponds to minimum loss in efficiency and 1000 33,000 (1-10) 63,000 1-8 where P- power; kW, hp F - force; N, lb minimum wear.

NATURE AND COMPOSITION OF MACHINES 11 rti ng, TABLE 1-3 Standard Capacities and Speeds for Three-Phase, 60-Hz, Alternating-Current Induction or dis- Motors engine hp pm rpm power e tor- result 0.25 1800 1200 3600 1800 1200 100 1800 1200 900 1800 1200 900 600 450 ad of rations0.50 1800 1200 1800 1200 3600 1800 1200 900 30 125 1800 1200 900 720 to the utches well as 600 450 1800 1200 0.75 7.5 1800 1200 3600 800 1200 40 1800 1200 900 150 rs are 450 3600 1800 1200 10 50 200 1800 1200 1800 1200 1200 rs are lifting Is in 600 450 1.5 3600 1800 1200 3600 1800 1200 60 1800 1200 250 1800 1200 n types 600 can be arting ven at ume oad is 450 360 1800 1200 3600 1800 3600 1800 1200 20 300 1200 600 up this stored energy at an equally high rate s said to vibrate when it describes an oscillating motion about a reference point. Such oscillation occurs because of the dynamic effects of manu facturing tolerances, clearances, rolling and rub- bing contact between machine parts, and out-of Mechanical vibration is motion that is usually balanc forces in rotating and reciprocating nintentional and unwanted. A machine member members. Apparently insignificant vibrations func- while slowing down. en 1-9 EFFECTS OF VIBRATION give

NATURE AND COMPOSITION OF MACHINES 13 because it often determines the size of the power source If power and speed are known, torque can be determined by calculation. If several speeds are used, choose the lowest, because that is gen erally where the torque requirement will be greatest Engine Most power sources will produce torque in excess of their nominal values, although not for long periods of time. For instance, the output torque of electric motors at start may momen- tarily reach values of three to five times nominal torque (Fig. 1-3). Although excess torque capa- bilities aid in overcoming inertia during periods of acceleration, they can also raise temperatures unduly and overload couplings, clutches, shafts or gears. The designer must recognize these po- tential hazards and allow for them Vehicle for power sources electric motors. Time, t Figure 1-5 Time-torque curve for common vchi clesclutch engagement delay. s requires a more clutch) so that the refore power prior ows the time delay gines are used in which may be 25 ing torque, can be Tng toque: Nm, ft-lb running torque multiplier (obtained from Table 1-4) running torque, N. m, ft-lb EXAMPLE 1-4 The running torque of a stamping machine ing torque and the motor size SOLUTION torque, can be; running torque is The running torque multiplier is always ating the starting torque is important Characteristics of greater than one and is determined from experi- was measured at 100N m. Estimate the start g torque and the motor s TABLE 1-4 Starting Torque This machine has peak load zones and sub- Various Machines sequent average torque multipliers of 2.5 to 6 Therefore Running Torque Multiplier Types of Applications Minimum starting torque (2.5(100) General machines with ball and roller bearings General machines with sleeve 1.3-1.6 bearings Conveyors and machines with 6-2.5 excess sliding friction Machines that have high load spots in their cycle (e.g some printing and punch presses and machines with cam- or crank-operated mech anisms) 1.2-1.3 -250 N m Maximum starting torque (5)(100) -500 N m Because output of electric motors at start may reach values of three to five times nominal torque, power requirements can be based on a torque value of 100 N m. For a 1200-rpm motor, this corresponds to roughly 15 hp 2.5-6.0 Tn 9550 ( 100 N-m)( 1 200 rpm) kW for various driven 9550 Source. ZERO-MAX, Variable Speed Drives. Zero- and pumps Max Industries, Inc., Minneapolis, 1973 (with permission) (12.57 kW34 hp/kw) - 16.8 hp and hoists

NATURE AND COMPOSITION OF MACHINES 15 anufac- experi- oads on utches In each is mul- an part is must catalog ulated Power source of service 2 (1-12) Engines 2.5 r is ob- One cylinder Two cylinders Four cylinders 2.0 greater are de- ble ca- or 1.5 Steam angine Reciprocating hydraulic or oneumatic motor Water turbine Steam turbine Electric motor Rotary hydraulic or pneumatic motor roxi on 0.6 the en- condi Figure 1-6 Service factor diagram. Service factors help to ensure optimal service life of machine components. (Developed by W. Richter and H. Ohlendorf.) ates and 16 head- indus- As might be expected, the service factor n creases when 1. 4. The duration of services increases from 1 to power nce a type re ac- gen- from g point the 24 h/day. Smoothly running power sources based on rotary motion are replaced by reciprocating engines The starting load becomes heavier and the number of starts increases. The frequency of full-load conditions and shock phenomena increases Only the curves for sensitivities have a nega- tive slope, thus effectively indicating relative smoothness of operation for the seven most com- mon means of power transmission. Spur gears for instance, require a service factor nearly twice that of worm gearing and 50 to 60% higher than 2- ers 3. those of belt and chain drives

Please show step for step. Thank you. The answers in the back of the book say (a) D= 170mm, (b) 7.5 hp motor, (c) Fc = 266 N approx.

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