Unit-2-me331

ME-331 DESIGN OF MACHINE ELEMENTS UNIT 2 MECHANICAL SYSTEMS AND MANUFACTURING CONSIDERATIONS IN DESIGN

DETERMINATION OF POWER CAPACITY INTRODUCTION: Almost all mechanical systems have power transmitted from one place, Say, an electric motor to another say, a centrifugal compressor. To Power, various elements are used. They are: Belt and Rope Drives Chain Drives Gear Drives “ Power Capacity” means the ability of the transmission element to Transmit a required amount of power, without any form of failure. BELT DRIVES & ROPE DRIVES Flat belt & V-belts are employed to transmit power from one shaft D another. Belt design involves the selection of a proper belt size table to the power requirements. . TYPES OF BELT DRIVES Belt drives are of the following types 1. Flat belt drive a. open belt There are shown in the fig. (2a I)

ANALYSIS OF V —BELT Let 2 be the groove angle of the V-belt section. Then

ADVANTAGES OF FLAT BELT (i) Simple, low cost (ii) Flexible and good shock absorption (iii) Different velocity ratios possible using stepped pulley. (iv) Best suited for long center distances. Up to 15m. LIMITATIONS OF FLAT BELT: (i) Velocity ratio not constant due to slip & Creep. (ii) Requires a tensioning device. (iii) Shorter life for high speed operations. ADVANTAGES OF V —BELT: (i) Due to wedging action. power capacity is more. (ii) Compact constructions (iii) Velocity ratio is constant due to negligible slip. Limitations of V —belt: (i) Less durable. (ii) Complicated design of pulley grooves. (iii) Costly DESIGN STEPS FOR FLAT BELT DRIVE: (i) Find 1 and 2 for smaller and larger pulleys. (ii) Find e 1 and e 2 (iii) Compare the values of e 1 and e 2 whichever is less. that it the governing pulley. So design should be based on that particular pulley. (iv) If both pulleys are of same material ( same) then, smaller pulley is the deciding one. (v) Velocity ratio is given by . N1/N2 =d2/d1 ROPE DRIVES

Rope drives are meant for large amount of power over a considerable distance. The advantages of rope drives are. (i) Smooth, quiet operation (ii) Slight misalignflients of shafts may be accommodated. (iii) Mechanical efficiency is high. Important note: All the formula pertaining to V-belts are valid for rope drives also. GEARS Gears are a means of power transmission as belt and ropes. But, importantly as slip is absent in gear drives, they provide a positive drive compared to ropes or belts. Classification of gear drives: Gear drives may be classified as follows: 1. According to the position of axes of the shaft: a. Parallel b. intersecting c. Non intersecting and non parallel Spur gears are used for case (a) Bevel gears are used for case (b) Spiral gears are used for case 2. According to peripheral velocity of gears: a. Low velocity (V< 3rns) b. Medium velocity (3
.y for gear For which element o .y is smaller that is weaker. If both are of same material, the pinion is weaker and design is based on pinion. (ii) From the grail module and no. of teeth, find PCD (Pitch Circle Diameter). (iii) Using v = DN/60 find peripheral velocity of the gear (or) pinion. (iv )power capacity,P=W T x v(use lewis equation to find W T o

PROBLEMS ON POWER TRANSMITTING ELEMENTS PROBLEM Power is transmitted by means a belt drive. If the angle of lap on the pulley is 210 , tensions on tight and slack slides are 2500N and 1100N respectively and the pulley rotates at speed of 1500rpm and pulley diameter is 750mm, find the power transmitted. Given data:

To find: Power, P © Solution: Peripheral velocity,

Power.

Result Power transmitted = 82.418kW. PROBLEM A fan is driven by a belt from a motor running at 74 rpm. A medium double ply leather belt 8mm thick and 250mm wide is used. The diameters of the motor pulley and driven pulley rerespective 350mm

and 1370mm. The center distance is 1370mm and both pulley’ s are made of CI. for which co-efficient of friction with leather is 0.35. The allowable stress for the belt is 2.4Mpa which allows for the factor of safety and also .t the fact that a double ply belt does not have double the capacity’ of a single ply belt. Belt mass is 970kg/m. What is the power capacity of the belt? Given data:

To find: P Solution: As both pulleys are made of same material, the smaller pulley the design criterion. For smaller pulley.

PREFERRED NUMBERS: Often, a manufacturer has to decide about the range and sizes of his product. He has to decide what sizes will cover a certain range efficiently with minimum number of sizes. Generally, ranges are covered by sizes by two methods. 1. 2.

Arithmetic progression. Geometric progression.

In Arithmetic progression is used for a product size, a certain ‘ amount’ is added 10 the first size to get the second size arid so on. For example, if the first size is 10 and the last is 100, it is done as, 10, (10 + 18 = 28), (28 + 18 = 46), (46 + 18 = 64), (64 + 18 = 82), (82 + 18 = 100) In this case,‘ 18’ is added to the previous size to get the second one. Arithmetic progression is suitable for cases where the range is limited. In general, if a series is designated as, Rn It means, The step ratio is (10)1/n. The last number becomes 10 fold in n’ steps. There are totally n+1 numbers in that series. LIMIT, FITS AND TOLERANCES NTRODUCTION In spite of recent advances in manufacturing methods it is 11 possible to produce a part exactly conforming to the requirements. To means that the parts arc produced within a range of dimensions rays than exact dimension. To specify the permissible range of dimension limit, fits and tolerances are used. Shaft:

This refers not only to the diameter of circular shaft but also to external dimension on a component. Hole: This refers to any internal dimension on a component, not only diameter of circular hole. Male and Female surfaces: When an assembly is made of two parts, the enveloped surface is known as ‘ Male’ and the enveloping surface is known as ‘ female surface. Basic size: Basic (or) nominal size is the standard size for the part arid is same both for hole and shaft. Actual size: This is the dimension as measured actually on a part. Limits of size: These are the maximum and minimum permissible sizes of the part.

Maximum Limit: This is the maximum permissible size of the part. Minimum Limit: This is the minimum permissible size of the part. Tolerance: This is the difference between maximum limit and minimum limit.

-0005 For example. if it is given 250 000 it is unilateral. Because variation s allowed in one direction only (Positive). If tolerance is specified on both the sides, it is bilateral tolerance. Allowance: Allowance is the difference between the basic sizes of the mating parts. This is usually referred to the maximum material condition of the mating parts (hole & shaft). Maximum material condition: This refers to the condition of dimensions when there will be maximum material left in the part. For a shaft, this corresponds to the maximum size and for a hole, this is the minimum hole size. Unilateral & bilateral tolerance: When tolerance is specified on one side of the basic size (either positive or negative) it is called unilateral tolerance. 0 005 For example, 25 0000 is the case of bilateral tolerance because variation is allowed both in positive and negative directions.

tolerance Zone: The region between the maximum and minimum limit size is called tolerance zone. Deviation:

Algebraic difference between a size and corresponding basic size

Upper deviation: Algebraic difference between the maximum limit and the basic size. Lower deviation: Algebraic difference between the minimum limit and the basic size. Actual deviation: Algebraic difference between the actual size and corresponding basic size. Zero line: This is a straight line to which the deviation are referred to. It is a line of zero deviation and represents the basic size. Positive and negative deviations are shown above and below the zero line. Fundamental deviation: It is one of the two deviations which is chosen to define the position of tolerance zone in relation to zero line. SURFACE FINISH The surface finish (or) roughness of a part depends on various Factors like sharpness of tool cutting conditions, tool material etc. Achievement of required surface finish is important because it affects the performance of the part concerned. No part can be made mirror like (i without any irregularities. Yet, it should be tried to achieve the best possible surface finish.

Need for good surface finish i) To maintain a lubricant film between surfaces. ii) To minimize the effect of stress raisers. iii) To achieve required tolerance. Fig. iv) To protect from wear, corrosion etc.

A machined surface will have micro irregularities in the form of peak and valleys (ups and downs) and so, it is not perfectly smooth as shown in fig (2.21) The cost of production of a high quality surface is obviously less .than that of a poor quality surface. The relation between surface finish, tolerance and production cost is Shown in fig.(a)

Fig (A) It is important to know to what extent the surface is irregular. The following are some important terms used in surface finish studies. Terminologies in surface roughness studies 1. Nominal surface:

Any machined surface with irregularities is called nominal surface. 2. Roughness: The micro irregularity on a surface produced by the cutting action of the tool is called roughness Fig. (b ) 3. Waviness:

Fig(b) Surface undulation of a higher value, with large spacing than roughness is called waviness [ curve + Waviness curve]=Profile curve 4. Lay: Lay is the principal direction of tool marks that make a pattern on the machined surface. Direction of lay depends on the production method. 5. Sampling length (or) Cut off length:

This is the length of the machined surface taken for the evaluation of surface finish. Sampling length is selected based on the production method. STANDARDS • There are machine parts which are used in number of machines such as • Fasteners, seals, handles, belts, ropes, chains, couplings, lubricating devices, pumps, motors, etc., • The sizes and ratings of these parts should be standardized because the cost of a standard part is only a fraction of the cost of the same made to order. • Use of standard parts permits interchangeability and help in servicing as well.

• The work of standardization is accomplished by National and International organizations, in India Bureau of Indian Standards(BIS) is entrusted with the task of standardization. • Unless standards are followed the product may not stand in the International competition.

TWO MARK QUESTIONS 1. State the advantages of interchangeable manufacture. 2 What is maximum material condition? 3. How are clearance fits classified? 4. Give the types of interference fits and their applications. 5. What are the types of geometric variations? Explain. 6. With examples. Show how parallism, perpendicular straightness and flatness arc indicated on drawings. 7. Give the indication method for circularity, cylinder city coaxiality and run out. 8. Explain the need for good surface finish. 9. Explain the various methods of indicating Ra values on drawings. 10. How roughness is indicated in case of identical values for majority of surface? 11. Give the step ratios for R5, RIO, R20, R40 series. 12 What are preferred numbers? Why are they used? PROBLEMS 1 V-belt drive is to transmit 18. 5kw from a 250mm pitch diameter sheave operating at 1800rpm to a 900mm diameter flat Pulley, center distance being 1m Groove angle is 40º µ between belt and Sheave is 0.2, µ between belt and flat pulley is 0.2. Belt cross section b2= 38mm; d = 25mm; b1= 19mm; each belt weighs llkn/m and allowable tension per belt is 900N. How many belts are required? 2. Power is transmitted by means of a Flat belt drive. The belt cross sectional area is 1000mm and allowable stress is 2Mpa.Distance between the pulleys is 800mm and pulleys rotate in the same direction. Pulley diameters and speeds are given below: Smaller Larger 1500rpm 700rpm

125rpm — Find the power transmitted coefficient of friction between belt & pulleys is 0.25. 3.201 A belt 100mm wide and 10mm thick transmits power@ l000m/min. The net driving tension is 1.8 times the tension on slack side. If the safe permissible stress on the belt section is 1.6Mpa, calculate the maximum power that can be transmitted at this speed. Assume density of leather as 1000kg/m3. Calculate the absolute maximum power that can be transmitted by that belt and the speed at which this is transmitted. 4. A compressor, 100kw power is to run at 2 The drive is by V-belts from an electric motor running at 7 Diameter of puller on compressor shaft is limited to 1m and center distance between pulleys is limited to 1.75m Belt speed should not exceed l600m/min. Determine the no. Of V-belts required if Each belt has an effective cross sectional area of 375mm2 Density 1000kg/m3and allowable stress of 2.5Mpa. µ=0.25, Groove angle is 35º . 5. A rope drive transmits 600kw .from a pulley of Diameter 4m, running at 90rpm. Angle of lap is 160.Groove angle45º .µ = 0.28; mass of rope = 1.5kg/m and allowable tension in each Rope is 2400N.Find the number of ropes required.