Mechanic second year
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Mechanic second year

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الحقيبة التعليمية لمادة ميكانيك السيارات للمرحلة الثانية لقسم المكائن والمعدات - معهد اعداد المدربين ...

الحقيبة التعليمية لمادة ميكانيك السيارات للمرحلة الثانية لقسم المكائن والمعدات - معهد اعداد المدربين التقنيين - العراق- بغداد
من اعداد المهندس : صلاح مهدي خليل

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    Mechanic second year Mechanic second year Presentation Transcript

    • Ministry of higher education and scientific research Foundation of technical education Institute of technical instructor preparing Machine and equipment department Automobile division6/28/2012 mechanic , 2nd year - machine and equipment 1
    • The lesson aims toTo study and learning the forces and stresses affected on the automobile, the varioussystem design and the power transmitted from the different components Week Item 1 Automotive performance , the total resistance affecting car motion 2 Traction effort 3-4 Surplus effort & examples 5-6 Gears , types gearing system , motion between two gears , selecting the best gear ratio , ear axle ratio , overall gear ratio examples 7 Bearing types , calculations and design of sliding bearing 8 Shafts , types , calculation and design of the shafts 9-10-11 Clutch , types , design , power transmitted , calculation 12-13-14-15 Belts . types , system types , calculation of power transmitted from flat and v. type. 16-17-18-19- Brakes , types systems function , calculation of stopping 20 distance , declaration , load transfer during brake , braking force on front and rear wheel , wheel piston diameter , all these calculation based on disc and shoes brake type. 21-22 Suspension system types advantages and disadvantages Calculation of leaf and coil spring 23-24 Steering system , calculations , types 25-26 Overturning and sliding speed 27 Piston , types , calculation of thermal and tensile stress 28 Crankshaft , types , calculation of thermal and tensile stress 29-30 Study of various design car system ( car with front engine mounted and rear wheel drive , car with front engine and rear wheel drive , car with rear engine mounted and wheel drive system6/28/2012 mechanic , 2nd year - machine and equipment 2
    • THE TRAFFIC RESISTANCE6/28/2012 mechanic , 2nd year - machine and equipment 3
    • LEARNING OBJECTIVES  At the end of these lessons you will be able to  Account the different type of traffic resistance  Draw the relation between the speed and the different type of traffic resistance  Solve the different types of problem by using the different type of equations that describe the traffic resistance6/28/2012 mechanic , 2nd year - machine and equipment 4
    • THE TRAFFIC RESITANCEThere are some forces which prevent the car motion like :-•The air resistance (Ra)•The gradient resistance (Rg)•The rolling resistance (Rr)THE AIR RESISTANCE (Ra)This resistance depend on , the car shape , speed and the front areaof the car .RA=K.A.V2Where ;K= the coefficient of air resistance(<1)V= the car speed (m/s)A=the car frontal area (m2)6/28/2012 mechanic , 2nd year - machine and equipment 5
    • THE GRADIENT RESIATANCE (Rg)The resistance (force )which prevent the car from moving up when the car climb ongradient roadRg = W sin ƟRg = W.(H/L)Where sin Ɵ= H/L as shown in figure besideƟ it might given as angle = 45, 40 , 20 ….Or as ratio = 1:20 ,, 1:12…Or as percentage 18%,, 20% ….And for the gradient resistance in relative tocar speed can be presenting as figure beside 6/28/2012 mechanic , 2nd year - machine and equipment 6
    • The rolling resistance (Rr)It is caused by the frictionbetween the wheel and theroad , its depends on , the typeof wheels , the type of roadand the car weight . 6/28/2012 mechanic , 2nd year - machine and equipment 7
    • THE SUM OF RESISTANCERT=Ra +Rg +RrWhereRa only depend on the speed ofcarRg = zero on horizontal roads orways6/28/2012 mechanic , 2nd year - machine and equipment 8
    • TRACTIVE EFFORT AND SURPLUS EFFORTif we draw the relation between the effortand speed the shape will be like as infigure beside , so we can see there isdifference between the tractive effort andresistance effort and that what we call thesurplus effort . SE= TE-RTMax value as shown when there is huge difference Between TE and RTMin value when TE=TR at point BThe advantage from calculate the SE is to define What acceleration we need to makethe car moving That meanIncrease SE »» increase acceleration »» increase the speed When we change thespeed by gear box for example SE: the difference between the TE and TR The great amount of SE lead to high acceleration for the car moving6/28/2012 mechanic , 2nd year - machine and equipment 9
    • THE AMOUNT OF ACCELERATION AT ANY SPEED a = (SE/W). g = (SE/m) where : a=acceleration (m/s2) g=9.81 (m/s2) W= car weight (n) :. F= m.a Then the gradient overall ƞ=(SE/W).1006/28/2012 mechanic , 2nd year - machine and equipment 10
    • Ex:-1------------------------------------- a long its mass (1000 kg ) the car going up a hill which gradient (1:25) and rolling resistance (250 n) the speed increase from (45 km/h ) to ( 75km/h) at 12 sec find the tractive effort . Now if the car moving at gradient (1:15) and the engine stopped , what is the speed after the going down (200 m ) Assume the rolling resistance the same as two cases Sol:- The going up V1=(45*1000)/(60*60)= 12.5 m/s V2= (75*1000)/(60*60)= 20 m/s V2=V1+at6/28/2012 mechanic , 2nd year - machine and equipment 11
    • THE MOVING UP CASEV2-V1= atV22-V12= 2ax:. a = (V2-V1)/t = (20-12.5)/12= 0.625 m/s2( the acceleration required to up)The acceleration force (F):. F= m.a1000* 0.625=625 nRg= W(h/L)= (1000*9.81)*(1/25)= 392.4 n(According to force direction to up )TE=RT+FTE=(Rg+Ra+Rr)+F ( since Ra so smallthen its equal to zero)Then TE= (392.4+0+250)+625= 1267.4 n 6/28/2012 mechanic , 2nd year - machine and equipment 12
    • THE GOING DOWN CASE6/28/2012 mechanic , 2nd year - machine and equipment 13
    • Ex:-2--------------------------------------- Along its mass (700 kg) the car is moving at (36 km/h) at horizontal way , when the gear on bush , what is the distance which car moving it beyond the stopping ?if the rolling resistance (155 n) Sol: RT= Rr= 155 n F=m.a :. a=F/m that’s mean a = 155/700 = 0.221 m/s2 V22-V12=2aX X=( V22-V12)/2a X=(36*(1000/3600))2/(2*0.221)=226.216/28/2012 mechanic , 2nd year - machine and equipment 14
    • Ex:- 3-------------------------------------- A car of mass (2.5 ton) it is moving on horizontal way at tractive effort (1.3 kn) , if the rolling resistance (180 n/ton) , find the car acceleration ? Sol :- Ra= 0 (not emotion) Rg= 0 (horizontal road) M= 2.5 ton ==== 2500 kg TE= 1.3 kn ==== 1300 n Rr=150 n/ton * 2.5 ton = 450 n SE=TE-RT SE= 1300-(450+0+0)= 850 n :. a = SE/m = 850/2500= 0.34 m/s26/28/2012 mechanic , 2nd year - machine and equipment 15
    • Ex:-4---------------------------------- A car of mass (4.5 ton ) it is going up a hill its gradient (1:20) at constant speed (40 km/h) if the rolling resistance (70 n/ton) . calculate the tractive effort ? if the engine stopped find the distance when the car stops beyond it ? Sol:- To find TE Rg=W*(h/L) = (4.5*1000*9.81)*(1/2) = 2205 n Rr=70*4.5= 315 n :. TE = Rr+RG= 2205+315 = 2520 n Now if the engine stopped , the required force to stopping the car F=RT = 2520 n :. F= m.a :. a = F/m = (2520)/(4.5*1000) = 0.56 m/s2 V12-V22=2aX ((40*1000)/60*60))2-0=0*0.56*X Then X = 110.23 m6/28/2012 mechanic , 2nd year - machine and equipment 16
    • GEAR TRAIN6/28/2012 mechanic , 2nd year - machine and equipment 17
    • GEARSGears are toothed cylindrical wheels used for transmittingmechanical power from one rotating shaft to another.Several types of gears are in common use. This LESSONSintroduces various types of gears and details the design,specification and selection of spur gears in particular. 6/28/2012 mechanic , 2nd year - machine and equipment 18
    • LEARNING OBJECTIVES  At the end of these lessons you should be :  • familiar with gear nomenclature;  • able to select a suitable gear type for different applications;  • able to determine gear train ratios;  • able to determine the speed ratios  • able to describe the main gear terminology6/28/2012 mechanic , 2nd year - machine and equipment 19
    • PRINCIPLE TERMINOLOGYVarious definitions used fordescribing gear geometryare illustrated in Figure atside and listed below. For apair of meshing gears, thesmaller gear is called the‘pinion’, the larger is calledthe ‘gear wheel’ or simplythe ‘gear’. 6/28/2012 mechanic , 2nd year - machine and equipment 20
    • PRINCIPLE TERMINOLOGY6/28/2012 mechanic , 2nd year - machine and equipment 21
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    • At side6/28/2012 mechanic , 2nd year - machine and equipment 23
    • Note6/28/2012 mechanic , 2nd year - machine and equipment 24
    • TYPES OF GEAR TRAINS Following are the different types of gear trains, depending upon the arrangement of wheels :  1. Simple gear train,  2. Compound gear train,  3. Reverted gear train, and In the these three types of gear trains, the axes of the shafts over which the gears are mounted are fixed relative to each other.6/28/2012 mechanic , 2nd year - machine and equipment 25
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    • The advantage of a compound train over a simple gear train is that a much larger speed reduction from the first shaft to the last shaft can be obtained with small gears. If a simple gear train is used to give a large speed reduction, the last gear has to be very large. Usually for a speed reduction in excess of 7 to 1, a simple train is not used and a compound train or worm gearing is employed. Note: The gears which mesh must have the same circular pitch or module. Thus gears 1 and 2 must have the same module as they mesh together. Similarly gears 3 and 4, and gears 5 and 6 must have the same module.6/28/2012 mechanic , 2nd year - machine and equipment 34
    • EX-1Consider the gear train shown in Figure atside . Calculate the speed of gear five.sol-:- N2=- (T1/T2 )*N1 N3=-(T2/T3)*N2 N4=N3 …. SAME SHAFT N5=-(T4/T5)*N4 N5=-((T4/T5) (T2 /T3)(T1/T2) )N16/28/2012 mechanic , 2nd year - machine and equipment 35
    • EX-2 T1 For the double reduction gear train shown in Figure at side , if the input speed is 1750 rpm in a clockwise direction what is the output speed? sol-:- T2 T3 N2=- (T1/T2 )*N1 N3=N2 ….. SAME SHAFT N4=-(T3/T4)*N3 T4 N4= ((T3 /T4 )*(T1/ T2))*N1 N4 =(-18/54)*(-20/70)*(-1750) N4= 166.7 RPM6/28/2012 mechanic , 2nd year - machine and equipment 36
    • EX-3 For the double reduction gear train with an idler shown in Figure at side , if the input speed is 1750 rpm in a clockwise direction what is the output speed? sol-:- N5=- (T4/T5 )*N4 N4=-(T3/T4)*N3 N3=N2 ….. SAME SHAFT N2= ((T1 /T2 )* N1 N5=-((T4/T5)(T3/T4)(T1/T2)) *N1 N5 =(-22/54)*(-18/22)*(-20/70)) (-1750) N5= 166.7 RPM6/28/2012 mechanic , 2nd year - machine and equipment 37
    • HW -1 a- Find the speed ratio ( N8/N1) for the gearshown ? b- If the angular velocity N4 = 50 r.p.s and T1=20 teeth T2=45 teeth T3 =30 teeth T4=50 teeth T5 =30 teeth T6= 60 teeth T7 =40 teeth T8=30 teeth Find all the angular velocities (ω rad/sec )? 6/28/2012 mechanic , 2nd year - machine and equipment 38
    • HW-2:- a- Find the speed ratio ( N9/N1 )For thegearbox shown bellow ? b- If the angular velocity N4=40 r.p.s andT1=20 teeth T2=60 teeth T3=30 teeth T4=50teeth T5=60 teeth T6=30 teeth T7=40 teethT8=30 teeth T9=50 teeth T10=20 teethT11=80 teeth T12=30 teeth. Find the speeds of N12 , N7 , N5 and N9 ?6/28/2012 mechanic , 2nd year - machine and equipment 39
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    • BEARINGS 6/28/2012 mechanic , 2nd year - machine and equipment 47
    • BEARINGS The purpose of a bearing is to support a load, typically applied to a shaft, whilst allowing relative motion between two elements of a machine. The aims of these lessons are to describe the range of bearing technology, to outline the identification of which type of bearing to use for a given application, to introduce journal bearing design and to describe the selection of standard rolling element bearings6/28/2012 mechanic , 2nd year - machine and equipment 48
    • LEARNING OBJECTIVES At the end of these lessons you should be able to:  • distinguish what sort of bearing to use for a given application;  • specify when to use a boundary lubricated bearing and select an appropriate bearing material to use for given conditions;  • determine the principal geometry for a full film boundary lubricated bearing;  • determine the life of a rolling element bearing using the life equation;  • select an appropriate rolling element bearing from a manufacturer’s catalogue;  • specify the layout for a rolling bearing sealing and lubrication system.6/28/2012 mechanic , 2nd year - machine and equipment 49
    • INTRODUCTION The term ‘bearing’ typically refers to contacting surfaces through which a load is transmitted. Bearings may roll or slide or do both simultaneously. The range of bearing types available is extensive, although they can be broadly split into two categories: sliding bearings also known as plain surface bearings, where the motion is facilitated by a thin layer or film of lubricant, and rolling element bearings, where the motion is aided by a combination of rolling motion and lubrication. Lubrication is often required in a bearing to reduce friction between surfaces and to remove heat. At side illustrates two of the more commonly known bearings: a deep groove ball bearing and a journal bearing .A general classification scheme for the distinction of bearings is given in Figure at next slide 6/28/2012 mechanic , 2nd year - machine and equipment 50
    • BEARING CLASSIFICATION 6/28/2012 mechanic , 2nd year - machine and equipment 51
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    • SLIDING BEARINGSThe term ‘sliding bearing’ refers to bearings where two surfaces move relative to each other without thebenefit of rolling contact. The two surfaces slide over each other and this motion can be facilitated bymeans of a lubricant which gets squeezed by the motion of the components and can generate sufficientpressure to separate them, thereby reducing frictional contact and wear. A typical application ofsliding bearings is to allow rotation of a load-carrying shaft. The portion of the shaft at the bearing isreferred to as the journal and the stationary part, which supports the load, is called the bearing (seeFigure 4.4). For this reason, sliding bearings are often collectively referred to as journal bearings,although this term ignores the existence of sliding bearings that support linear translation ofcomponents. Another common term is ‘plain surface bearings’. This section is principally concernedwith bearings for rotary motion and the terms ‘journal’ and ‘sliding’ bearing are used interchangeably.There are three regimes of lubrication for sliding bearings:1. boundary lubrication;2. mixed film lubrication;3. full film lubrication. 6/28/2012 mechanic , 2nd year - machine and equipment 54
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    • The performance of a sliding bearing differs markedly depending on which type of lubrication is physically occurring .This is illustrated in Figure in next slide , which shows the variation of the coefficient of friction with a group of variables called the ‘bearing parameter’ which is defined by: where , viscosity of lubricant (Pa s); N, speed (for this definition normally in rpm); P, load capacity (N/m2) given by where W, applied load (N);L, bearing length (m); D, journal diameter (m). The bearing parameter, N/P, groups several of the bearing design variables into one number. Normally, of course, a low coefficient of friction is desirable. In general, boundary lubrication is used for slow speed applications where the surface speed is less than approximately 1.5 m/s. Mixed film lubrication is rarely used because it is difficult to quantify the actual value of the coefficient of friction6/28/2012 mechanic , 2nd year - machine and equipment 56
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    • LUBRICANTSAs can be seen from Figure at side , bearingperformance is dependent on the type oflubrication occurring and the viscosity of thelubricant .The viscosity is a measure of a fluid’sresistance to shear. Lubricants can be solid, liquid or gaseous,although the most commonly known are oilsand greases. The principal classes of liquidlubricants are mineral oils and synthetic oils.Their viscosity is highly dependent ontemperature as illustrated in Figure In nextslide .They are typically solid at 35°C, thin asparaffin at 100°C and burn above 240°C. Manyadditives are used to affect their performance 6/28/2012 mechanic , 2nd year - machine and equipment 58
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    • DESIGN OF BOUNDARY LUBRICATED BEARINGSGeneral considerations in the design of a boundary lubricated bearing are:• the coefficient of friction (both static and dynamic);• the load capacity;• the relative velocity between the stationary and moving components;• the operating temperature;• wear limitations; and• the production capabilityThis approach is set out as a step-by-step procedure below.1. Determine the speed of rotation of the bearing and the load to be supported.2. Set the bearing proportions. Common practice is to set the length to diameter ratio between 0.5 and 1.5. If thediameter D is known as an initial trial, set L equal to D.3. Calculate the load capacity, P W/(LD).4. Determine the maximum tangential speed of the journal.5. Calculate the PV factor.6. Multiply the PV value obtained by a factor of safety of 2.7. Interrogate manufacturer’s data or Table 4.2 to identify an appropriate bearing material with a value for PVfactor greater than that obtained in (6) above. 6/28/2012 mechanic , 2nd year - machine and equipment 60
    • EX1-A bearing is to be designed to carry a radial load of 700 N for a shaft of diameter 25 mm running at a speed of75 rpm (see Figure 4.8). Calculate the PV value and by comparison with the available materials listed in Tablein next slide determine a suitable bearing material.Sol:-The primary data are W= 700N, D = 25mm and N = 75 rpm.Use L/D = 1 as an initial suggestion for the length to diameterratio for the bearing. L =25mm. 6/28/2012 mechanic , 2nd year - machine and equipment 61
    • Sliding bearing Roller6/28/2012 mechanic , 2nd year - machine and equipment 62
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    • SHAFTS 6/28/2012 mechanic , 2nd year - machine and equipment 67
    • SHAFTS Shaft is rotating machine element which is used to transmit power from one place to another . The power is delivered to the shaft by some tangential forces and the resultant torque (twisting moment ) setup with in shaft permits the power to be transferred to various machines linked up to the shaft6/28/2012 mechanic , 2nd year - machine and equipment 68
    • Material used in shaft The material used for shaft should have following properties: 1- its should have high strength. 2- it should have good machinability . 3- it should have notch sensitivity factor . 4- it should have good heat treatment . 5- it should have wear resistant properties. The material used for ordinary shaft is carbon steel of grade 40C8 , 45C8 , 50C4 and 50C12 . The mechanical properties of these grade carbon steel are shown in table bellow6/28/2012 mechanic , 2nd year - machine and equipment 69
    • Material used in shaft NO Indian standard Ultimate Yield (ASME) strength strength (Mpa ) (Mpa ) 1 40 C 8 650- 670 320 2 45 C 8 610-700 350 3 50 C 4 640-760 370 4 50 C 12 MIN 700 390When shaft of high strength required , then any alloysteel such as nickel , nickel – chromium or chrome –vanadium steel is used6/28/2012 mechanic , 2nd year - machine and equipment 70
    • Stress in shafts The following stresses are induced in the shaft 1- shear stress due to the transmission of torque (due to torsional loads ) 2- bending stress (tensile or compression ) due to force acting upon machine element like gears , pulleys , etc. as well as due to the weight of shaft it self 2- stress due to the sum of above two kind that mention before in 1, 2 .6/28/2012 mechanic , 2nd year - machine and equipment 71
    • Max permissible working stress According to ASME (American Society of Mechanical Engineering) this is the reference stresses for shaft design .Design case Compression , tensile Torsional stress ( mpa ) (mpa ) (bending stress) (shear stress)With key way 84 42Without key way 112 5628 June 2012 mechanic , 2nd year - machine and equipment 72
    • Shaft Subjected to Twisting Moment onlyWhen the shaft is subjected to torque only , then the diameter of shaft can obtained fromtorsion equation which is 𝑻 𝝉 = where T = twisting moment (torque) 𝑱 𝒓 J= polar moment of inertia of shaft about axes of rotation 𝜏=torsional shear stress r = radius of shaft = d/2Case 1 : for solid shaft 𝝅 𝟒𝑱= 𝒅 𝟑𝟐So then 𝑻 𝝉 𝝅 = that mean 𝑻= ∗ 𝝉∗ 𝒅𝟑𝝅 𝟒 𝒅/𝟐 𝟏𝟔 𝒅𝟑𝟐 d6/28/2012 mechanic , 2nd year - machine and equipment 73
    • Shaft Subjected to Twisting Moment onlyThe twisting moment (T) can be found byP= (2πNT)/60Where: P= power (watt) N = speed of shaft (rpm) T= twisting momentThen T=(P*60)/(2 π N)In case of belt drive or driven shaft the torque is found by (we will illustratedthis in belt and rope lecturer)T= (T1-T2)* RWhere T1= tight side tension =(Tl) large tension side (n) T2 = slack side tension =(Ts)= small tension side (n) R= radius of bully 6/28/2012 mechanic , 2nd year - machine and equipment 74
    • Example 1A line shaft rotating at 200 rpm to transmit 20 kw . The shaft assumed to bemade of mild steel with allowable shear stress of 42 mpa . Determine thediameter of the shaft neglecting the bending moment of the shaft .Sol :N= 200 rpm P=20 kw = 2000 wτ= 42 mpa = 42 n/mm2T=(P*60)/(2πN)= (20*1000*60)/(2 π*200)=955 n.m= 955*1000 n.mmT= (π/16)*τ*d3955*1000= (π/16)*42*d3 =8.25 d3 =====> d=955*1000/8.25= 48.7 mm ≈ 50 mm 6/28/2012 mechanic , 2nd year - machine and equipment 75
    • Example 2 Find the diameter of a solid steel shaft to transmit 20 kW at 200 r.p.m. The ultimate shear stress for the steel may be taken as 360 MPa and a factor of safety as 8. If a hollow shaft is to be used in place of the solid shaft, find the inside and outside diameterwhen the ratio of inside to outside diameters is 0.5. Solution. Given : P = 20 kW = 20 × 103 W ; N = 200 r.p.m. ; τu = 360 MPa = 360 N/mm2 ; F.S. = 8 ; k = di / do = 0.5 6/28/2012 mechanic , 2nd year - machine and equipment 76
    • Shafts Subjected to Bending Moment Only 6/28/2012 mechanic , 2nd year - machine and equipment 77
    • Examplepair of wheels of a railway wagon carries a load of 50 kN on each axle box,acting at a distance of 100 mm outside the wheel base. The gauge of therails is 1.4 m. Find the diameter of the axle between the wheels, if the stressis not to exceed 100 Mpa 6/28/2012 mechanic , 2nd year - machine and equipment 78
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    • THE CLUTCH6/28/2012 mechanic , 2nd year - machine and equipment 80
    • THE CLUTCH THE CLUTCH FUNCTION -To engagement or disengagement of the gears when the car is moving without damaging the gears teeth . -Transmit the power from the engine to the wheels smoothly and gradually. -By using the clutch the car speed may reduce with the same engine speed . TYPE OF CLUTCH PLATE CLUTCHES It is shown in figure at side Where :- S=spring factor R1= outer diameter R2 = inter diameter n = pairs of surface in contact µ= coefficient of friction N= r.p.m6/28/2012 mechanic , 2nd year - machine and equipment 81
    • F= µ*S T=F*R = µ.S.R F=µ.S Where R = (r1+r2)/2 Thus T = n.µ.S (R1+R2)/2 power = P= T.W W=(2π.N)/606/28/2012 mechanic , 2nd year - machine and equipment 82
    • Ex:1-------------------------------- A clutch has : S= 2.5 kn ===== 2500 n r1=0.1 m n = 2 surfaces r2=0.05 m µ= 0.35 N = 3000 rpm Find .. T = torque and p = power Sol:- R= (r1+r2)/2= (0.1+0.05)/2 = 0.075 m T= n .µ.S.R =2* 0.35*2500*0.075= 131.25 n.m P=T.W =131.25*(2π.N)/60=41250 w6/28/2012 mechanic , 2nd year - machine and equipment 83
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    • Ex: 2----------------------------------------- A centrifugal clutch has : N = 5000 rpm µ= 0.3 n=4 S= 500 n R=20 cm m = 8 kg r = 16 cm find : T = torque and p= power Sol: ω= (2π.N)/60 = (2 π .5000)/60= 52.37 rad FC= 8*(52.37)2*(16/100)=3510.54 n :. T = n .µ (FC-S) .R= 4*0.3 *(3510.54-500)*(20/100)= 722.53 n.m P= T.ω = 722.53*52.37= 37.839 watt6/28/2012 mechanic , 2nd year - machine and equipment 85
    • Belts 6/28/2012 mechanic , 2nd year - machine and equipment 86
    • BELTS The belts or ropes are used to transmit power from one shaft to another by means of pulleys which rotate at the same speed or at different speeds. The amount of power transmitted depends upon the following factors : 1. The velocity of the belt. 2. The tension under which the belt is placed on the pulleys. 3. The arc of contact between the belt and the smaller pulley. 4. The conditions which the belt is used. It may be noted that (a) The shafts should be properly in line to insure uniform tension across the belt section. (b) The pulleys should not be too close together, in order that the arc of contact on the smaller pulley may be as large as possible. (c) The pulleys should not be so far apart as to cause the belt to weigh heavily on the shafts, thus increasing the friction load on the bearings6/28/2012 mechanic , 2nd year - machine and equipment 87
    • Type of belts6/28/2012 mechanic , 2nd year - machine and equipment 88
    • Belt speed consideration Torque 22.5 m/s Speed6/28/2012 mechanic , 2nd year - machine and equipment 89
    • Coefficient of friction 6/28/2012 mechanic , 2nd year - machine and equipment 90
    • TYPES OF FLAT BELT DRIVESThe power from one pulley to another may be transmitted by any of the followingtypes of belt drives: 1. OPEN BELT DRIVE. The open belt drive, as shown in Figure , is used with shafts arranged parallel and rotating in the same direction. In this case, the driver A pulls the belt from one side (i.e. lower side RQ) and delivers it to the other side (i.e. upper side LM). Thus the tension in the lower side belt will be more than that in the upper side belt. The lower side belt (because of more tension) is known as tight side whereas the upper side belt (because of less tension) is known as slack side, as shown in Figure bellow6/28/2012 mechanic , 2nd year - machine and equipment 91
    • 2. CROSSED OR TWIST BELT DRIVE.The crossed or twist belt drive, as shown inFigure side , is used with shafts arranged paralleland rotating in the opposite directions. In thiscase, the driver pulls the belt from one side (i.e.RQ) and delivers it to the other side (i.e. LM).Thus the tension in the belt RQ will be morethan that in the belt LM. The belt RQ (becauseof more tension) is known as tight side, whereasthe belt LM (because of less tension) is knownas slack side, as shown in Figure above , A littleconsideration will show that at a point where thebelt crosses, it rubs against each other and therewill be excessive wear and tear. In order to avoidthis, the shafts should be placed at a maximumdistance of 20 b, where b is the width of belt andthe speed of the belt should be less than 15 m/s. 6/28/2012 mechanic , 2nd year - machine and equipment 92
    • 3. QUARTER TURN BELT DRIVE. The quarter turn belt drive also known as rightangle belt drive, as shown in Fig. (a) bellow, is used with shafts arranged at right anglesand rotating in one definite direction. In order to prevent the belt from leaving thepulley, the width of the face of the pulley should be greater or equal to 1.4 b, where b isthe width of belt. In case the pulleys cannot be arranged, as shown in Fig. 11.5 (a), orwhen the reversible motion is desired, then a quarter turn belt drive with guide pulley,as shown in Figure . (b) bellow, may be used. 6/28/2012 mechanic , 2nd year - machine and equipment 93
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    • Velocity ratio in belt drive 6/28/2012 mechanic , 2nd year - machine and equipment 95
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    • Now the new concept is as shown in figure side Tl= large tension side (n) Ts= small tension side (n) µ= coefficient of friction Ɵ= angle of contact (rad) TL/TS=eµƟ TL/TS=e(µƟ/sin β) Where β= angle of belt shape6/28/2012 mechanic , 2nd year - machine and equipment 97
    • TL/TS=eµƟ TL/TS=e(µƟ/sinβ) Where β= angle of belt shpe THE POWER TRANSMITTED BY BELT T=(TL-TS).r P=T.ω P=(TL-TS).r.ω6/28/2012 mechanic , 2nd year - machine and equipment 98
    • Ex :-1 ------------------------------- Find the powertransmitted by a V- belt where :v- angle = 300diameter of bully = 60 cmangular speed = 200 rpmcoefficient of friction = 0.25angle of contact = 1600the large tension = 250 nsol:TL/TS=e(µƟ/sinβ)Ɵ=1600*(2π/360)=2.793 radµƟ/sin β=(0.25*2.793)/sin 15 = 2.6978TS=TL/ e(µƟ/sinβ) =250/14.84749= 16.837 nPower= (TL-TS).r.ω=(250-16837)*(30/100)*((2π*200)/60)=1465 w6/28/2012 mechanic , 2nd year - machine and equipment 99
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    • Ex . 1 :------------------------- Find the length of the belt where :- Radius of bully 1 = 2.4 m Radius of bully 2 = 0.4 m Distance between them = x= 12 m Sol:--------------------------------------- r 1 = 2.4 r 2= 0.4 sin ɸ= (r1-r2)/x ɸ= sin-1 ((2.4-0.4)/12)=sin-1(0.1666)=9.5940 from right side 2ɸ= 180-Ɵ ===== Ɵ=180-2ɸ = 180-2(9.594)=160.80 Arc fab = r2*Ɵrad = 0.4*(160.8*(2π/360)=1.122 m Arc cde = r1(360-Ɵ)rad = 2.4((360-160.8)*(2π/360))=8.343 m Line bc = ef = x.cos ɸ = 12*cos (9.594)=11.832 :. L= 1.122+11.832+8.343+11.832= 33.129 m6/28/2012 mechanic , 2nd year - machine and equipment 101
    • Ex:2 ---------------------------Find the length of the belt as infigureSol:- X=120 CM = 120/100= 1.2 M r 1 =20 cm = 20/100= 0.2 mr 2= 16 cm = 16/100= 0.16 mnow we need to find ɸsin (ɸ)=(r1-r2)/X= (0.2-0.16)/1.2=0.1:. ɸ=sin-1(0.1)=5.7302ɸ=180-Ɵ that’s lead to Ɵ=180-2ɸ =180-2(5.73)= 168.540Arc length = radius *opposite angle in radian.: arc hkf=r2*Ɵrad=0.16*((168.54*π) /180)=0.4706 mArc EJG=r1*(360-Ɵ)rad=0.2*((360-168.54)*(π/180))=0.66832 mXLine FE =HG=X*cos ɸ = 1.2*cos (5.730)=1.194mThe total lengthL=hkf+fe+ejg+gh= 0.4706+1.194+0.66832+1.194=3.5269 m 6/28/2012 mechanic , 2nd year - machine and equipment 102
    • Ex 2:-------------------------------related with previous ex : Find the power transmitted by a V-belt where; the V- angle = 500 angular velocity of pulley 2 =300 rpm coefficient of friction = 0.3 the large tension = 100 n sol:- V-angle=500 =2β --- β=50/2=250 , µ=0.3 , TL=100 n From the previous example Ɵ=168.540 Ɵ0=168.540*(π/180)=2.9415 rad µƟ/sinβ=(0.3*2.9415)/sin(250)=0.88245/0.4226=2.088 so e(µƟ/sinβ) =e2.088=8.069202 From the previous example Ɵ=168.540 TL/TS=e(µƟ/sinβ) = 100/TS=8.069202 TS=100/8.069202= 12.392799 n :. Power= P=(TL-TS).r2 .ω2 ‫ فً قانون القدرة‬r=r2 ‫ فعلٌه نستخدم‬r2 ‫بما انه اعطً فً السؤال السرعة الدورانٌة للبكرة الثانٌة‬ ‫ فً قانون‬Ɵ=(360-Ɵ) ‫ فً قانون القدرة وكذلك فان‬r1= r ‫ فسوف نستخدم‬r1 ‫ولو اعطً فً سؤال اخر السرعة الدروانٌة للبكرة االولى‬ ‫االحتكاك‬ According to what denoted above P=(100-12.392799)*0.16*((2π*300)/60)= 440.361 watt6/28/2012 mechanic , 2nd year - machine and equipment 103
    • Brake system 6/28/2012 mechanic , 2nd year - machine and equipment 104
    • BACKGROUND ON AUTOMOTIVE BRAKE SYSTEMS AND STATE OF THE ART FRICTION BRAKES The brake is a mechanism, which is used to absorb the kinetic energy of the vehicle with the aim to stop or retard the motion. Brakes transform kinetic energy into heat. Since the acceleration required during an emergency brake maneuver is much higher than the acceleration during normal operation, the brake power must be much higher than the motor power of the vehicle. Even for small vehicles a maximum brake power in the order of several hundred kilowatts is the rule rather than the exception. The energy to be dissipated in braking from speed v on a slope of height h is E = ((m v2)/2) + mg h where m is the vehicle mass. The first term of the equation is the kinetic energy while the second term of the equation evaluates the potential energy when moving downhill. The energies which must be dissipated are enormous and result in strong demands on the materials used in the friction contact which must withstand extremely high temperatures. Brake linings can be classified into organic, metallic and carbon. Modern brake pads are composed of many different ingredients.6/28/2012 mechanic , 2nd year - machine and equipment 105
    • BRAKE SYSTEM WITH IT POWER FLOW6/28/2012 mechanic , 2nd year - machine and equipment 106
    • DISC BRAKE Disc brakes consist of a rotor (disc) and a caliper. The rotor turns with the wheel. Each side of the rotor is a friction surface. The caliper is attached to an anchor plate or mounting bracket on the vehicle suspension. The hydraulic piston or several pistons convert the hydraulic pressure into an applied force that presses the pad against the rotor. This generates the friction forces needed for the braking. Fig bellow shows principal designs of typical disc brakes6/28/2012 mechanic , 2nd year - machine and equipment 107
    • DRUM BRAKESDrum brakes consist of a drum and a brake mechanism with two brake shoes that are curved toconform to the inside diameter of the drum. The drum brake has a steel or iron drum to which thewheel is bolted. The hydraulic pressure pushes the shoe-actuating pins out; hence the brake shoes areforced against the rotating drum. The resulting friction between the brake lining and the drum slowsor stops the car.Drum brakes may be different in appearance and construction, but functionally they are all the same.There are three types of drum brakes: Simplex, duplex and duo-servo drum brakes. Early automotivebrake systems used a drum design at all four wheels. They were called drum brakes because thecomponents were housed in a round drum that rotated along with the wheel. The shoes were madeof a heat-resistant friction material similar to that used on clutch plates. 6/28/2012 mechanic , 2nd year - machine and equipment 108
    • INTERNAL EXPANDING BRAKEAn internal expanding brake consists of two shoes S1 and S2as shown in Figure bellow The outer surface of the shoes arelined with some friction material (usually with Ferodo) toIncrease the coefficient of friction and to prevent wearingaway of the metal. Each shoe is pivoted at one end about afixed fulcrum O1 and O2 and made to contact a cam at theother end. When the cam rotates, the shoes are pushedoutwards against the rim of the drum. The friction betweenthe shoes and the drum produces the braking torque andhence reduces the speed of the drum. The shoes arenormally held in off position by a spring as shown in Figurebellow. The drum encloses the entire mechanism to keep outdust and moisture. This type of brake is commonly used inmotor cars and light trucks.TYPE OF BRAKEHydraulic system = fluidMechanical system = cable as in hand brakePneumatic system = compressed air 6/28/2012 mechanic , 2nd year - machine and equipment 109
    • SHOES TYPE BRAKE THEORY 6/28/2012 mechanic , 2nd year - machine and equipment 110
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    • Heat to be Dissipated during Braking6/28/2012 mechanic , 2nd year - machine and equipment 112
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    • Example A vehicle of mass 1200 kg is moving down the hill at a slope of 1: 5 at 72 km / h. It is to be stopped in a distance of 50 m. If the diameter of the tyre is 600 mm, determine the average braking torque to be applied to stop the vehicle, neglecting all the frictional energy except for the brake. If the friction energy is momentarily stored in a 20 kg cast iron brake drum, What is average temperature rise of the drum? The specific heat for cast iron may be taken as 520 J / kg°C. Determine, also, the minimum coefficient of friction between the tyres and the road in order that the wheels do not skid, assuming that the weight is equally distributed among all the four wheels. 6/28/2012 mechanic , 2nd year - machine and equipment 117
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    • Steering system 6/28/2012 mechanic , 2nd year - machine and equipment 120
    • Lesson objective At the end of lecturer the student will be able to : 1- determine the angle of turning 2- determine the space of turning 3- draw steering mechanism according to Ackerman condition6/28/2012 mechanic , 2nd year - machine and equipment 121
    • STEERING MECHANISMThe condition for correct steering is :-∆ IAPTan Ɵ1=b/x ==== cotƟ1 =x/b∆ IBPTan Ɵ2= b/x+c ==== cotƟ2=(x+c)/bCot Ɵ2 = (x/b )+(c/b):. cotƟ2 = cotƟ1 +c/bcotƟ2-cotƟ1=c/b ‫للحفظ‬ Ɵ2 ‫ اكبر من قٌمة‬Ɵ1 ‫حٌث الشرط المهم هنا ان تكون قٌمة‬And this what we call of Ackerman condition 6/28/2012 mechanic , 2nd year - machine and equipment 122
    • ex1:----------------------------- from the shape above, find Ɵ1 ifƟ2=200 and c=1.2 m and b= 2.7 msol:-(1/tan Ɵ2)-(1/tan Ɵ1)=c/b(1/tan200 )-(1/tanƟ1)=1.2/2.7(1/0.36397)-(1/tanƟ1)=0.44442.747474-0.4444=1/tanƟ11/tanƟ1=2.30303tanƟ1=1/2.30303=0.43421Ɵ1=tan-1(0.43421)= 23.47090Since Ɵ1> Ɵ2 that’s mean the solutionis correct .h.w : by using the same sketch above find Ɵ1 if Ɵ2 = 300 c=1.2 b=2.7 ans = 37.8341590 6/28/2012 mechanic , 2nd year - machine and equipment 123
    • SPACE REQUIREMENT. FOR TURNING 6/28/2012 mechanic , 2nd year - machine and equipment 124
    • SPACE REQUIREMENT.The kinematic steering condition can be used to calculate the space requirement of a vehicleduring a turn. Consider the front wheels of a two-axle vehicle, steered according to theAckerman geometry as shown in Figure above (previous slide) The outer point of the front ofthe vehicle will run on the maximum radius RMax, whereas a point on the inner side of thevehicle at the location of the rear axle will run on the minimum radius Rmin. The front outerpoint has an overhang distance g from the front axle. The maximum radius RMax isTherefore, the required space for turning is a ring with a width 4 R, which is a function of thevehicle’s geometry.6/28/2012 mechanic , 2nd year - machine and equipment 125
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    • Home work  from the shape bellow, find ∆R if δo= 100 and W=1.2 m and L= 2.7 m and g= 0.5 m6/28/2012 mechanic , 2nd year - machine and equipment 127
    • Overturning speed and skidding speed 6/28/2012 mechanic , 2nd year - machine and equipment 128
    • Lesson objective At the end of the lesson the student will be able to :- 1- analyze the force acting during turning 2- determine the overturning speed & skidding speed 3- specify the condition of safe turning6/28/2012 mechanic , 2nd year - machine and equipment 129
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    • FIRST AT REST w= m.g RA= RB= w/2 = m.g/2 SECOND AT MOVING ‫ نحو االعلى قلٌال لذلك سوف ٌكون‬A ‫عند لحظة االستدارة سوف تتولد قوة طرد مركزي تحاول ان ترفع االطار‬ B ‫اسناد لوزن السٌارة على االطار‬ + ∑ MB = 0 FC*h= W*d/2 m*(V2/r )h= m.g (d/2) V2=r/h*(g.d/2) …………. lead to ………V= ……. Overturning speed6/28/2012 mechanic , 2nd year - machine and equipment 131
    • Ex1:----------------------------------------A car travels on circular bath whereRadius of curve = r = 50 mHeight of c.g above ground level = h = 0.7 mDistance between the wheels of the car =d = 1.4 mCoefficient of friction = µ=0.7Calculate the maximum speed of turning : (a)Without overturning(b) Without skidding outwardsSol:(a) Overturning speed = Vo=V= = 22.1 m/s *3.6 =79.74 km/hr(b) Skidding speed = Vs= μ gr ==18.53 m/s *3.6 = 66.71 km/hr 6/28/2012 mechanic , 2nd year - machine and equipment 132
    • CRANK SHAFT6/28/2012 mechanic , 2nd year - machine and equipment 133
    • Lesson objective At the end of the lesson the student will be able to :- 1- specify of the force acting on the crankshaft 2- calculate the force acting on crankshaft 3- specify the special cases to determine the max bending stress and max twisting (shear stress)6/28/2012 mechanic , 2nd year - machine and equipment 134
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    • Ex 1:--------------------------------------From the figure above if you now that:-r= 0.0255 m d=0.024 m L= 0.046 mƟ= 10 0 ɸ=2.488 F=4521.6 n Find the shear stress and bending stress?Sol:-FC=F/cosɸ= 4521.6/cos2.488= 4526n nFt=FC. Sin (Ɵ+ɸ)= 4526*sin(100+2.488)=978.7 nFr= FC.cos(Ɵ+ɸ)=4526*cos(10+2.488)=4418.9 n Rr= Fr/2 =4418.9/2=2209.4 nRt= Ft/2 =978.7/2= 489.35 nMv= Rr*(L/2)= 2209.4*(0.046/2)=50.8 n.mMh=Rt*(L/2)=489.35*(0.046/2)=11.25 n.mM=√Mv2+Mh2M=√50.82+11.252 = 52.03078T=Rt *r= 489.35 *0.0255= 12.5 n.mShear stress (twist) =τMAX=(16 T)/(πd3)= (16*12.5)/(π*(0.024)3)= 46051778.8 n/m2 Bending stress = Ϭb= 32.M/πd3= 32*52.030784/π(0.024)3=3833776.1 n/m2h.w :- same question but take Ɵ=200 and ɸ=4.905 0hw:- same question , but take Ɵ=00 and ɸ=0 06/28/2012 mechanic , 2nd year - machine and equipment 136
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    • ‫مالحظة :-‬ ‫الحظ ان قمٌة ‪ Se>Sg‬بمقدار 05% من قمٌة ‪ Sg‬وهذة الحالة غٌر طبٌعٌة‬ ‫بمعنى اخر سوف تحدث استطالة (هطول ) غٌر منتظم . بمعنى اخر لو كان‬ ‫االجهاد متساوي ‪ Sg =Se‬سوف ٌكون ‪ deflection‬غٌر متساوي ولحل‬ ‫هذة المشكلة ٌتم اعطاء انحناء اكبر للشرائح السفلٌة بدءا من ثانً شرٌحة وهذا‬ ‫ما ٌعرف بالشرائح المجهدة مسبقا وكما تم شرحها سابقا .‬ ‫حٌث ان الطبٌعً فً التصمٌم هو ان تتساوى قٌم االجهاد وتتساوى بنسبة‬ ‫معقولة قٌم االنفعال فً الشرائح الكاملة االطوال مع الشرائح المتدرجة اسفل‬ ‫منها .‬‫2102/82/6‬ ‫‪mechanic , 2nd year - machine and equipment‬‬ ‫051‬
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    • SOMETHING TO SAY Design methods used to make use of monographs, however spreadsheets are now used. The Society of Automotive Engineers (SAE) publish a ‘Spring Design Manual’ that contains information about design and design methodology, reliability and materials. Normally a design will start with some constraints about space available, governing D, required spring rate, limits of motion, availability of wire diameter, material, maximum allowable stress when the spring is ‘solid’. Some iterations will probably be needed to reach the best solution. Fatigue testing is commonly carried out on new designs of springs destined for critical applications.6/28/2012 mechanic , 2nd year - machine and equipment 155
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    • Cam is a rotating machine element which gives reciprocating or oscillating motion to another element known as follower. The cam and the follower have a line contact and constitute a higher pair. The cams are usually rotated at uniform speed by a shaft, but the follower motion is predetermined and will be according to the shape of the cam. The cam and follower is one of the simplest as well as one of the most important mechanisms found in modern machinery today. The cams are widely used for operating the inlet and exhaust valves of internal combustion engines, automatic attachment of machineries, paper cutting machines, spinning and weaving textile machineries, feed mechanism of automatic lathes etc6/28/2012 mechanic , 2nd year - machine and equipment 159
    • Classification of Followers The followers may be classified as discussed below:6/28/2012 mechanic , 2nd year - machine and equipment 160
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    • BALANCING OF ROTATING MASSES high speed of engines and other machines is a common phenomenon now-a-days. It is, therefore, very essential that all the rotating and reciprocating parts should be completely balanced as far as possible. If these parts are not properly balanced, the dynamic forces are set up. These forces not only increase the loads on bearings and stresses in the various members, but also produce unpleasant and even dangerous vibrations. In this view slides we shall discuss the balancing of unbalanced forces caused by rotating masses, in order to minimize pressure on the main bearings when an engine is running6/28/2012 mechanic , 2nd year - machine and equipment 166
    • THE THEORYMost of theory depends on the low of moment ( force and arm) and some ofsketch skills and some attention to what we do first6/28/2012 mechanic , 2nd year - machine and equipment 167
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    • B 4 kg 0.3 m C 3.75 600 kg 0.2 m 900 A ً‫مقٌاس للرسم لك‬ 0.35 m 2.5 900 kg 520 ‫نتمكن من اظهار‬ 0.25 m 670 ‫الرسم بقٌاس معقول‬ 0.25 m ‫ومقبول‬ D 5 kg 2.5 kg 900 From solution 0 60 670 5 cm 900 5206/28/2012 mechanic , 2nd year - machine and equipment 169