Mechanic second year

Ministry of higher education and scientific research Foundation of technical education Institute of technical instructor preparing

Machine and equipment department Automobile division

6/28/2012 1 mechanic , 2nd year - machine and equipment

The lesson aims to To study and learning the forces and stresses affected on the automobile, the various system design and the power transmitted from the different components

6/28/2012 mechanic , 2nd year - machine and equipment 2

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.

16-17-18-19-20

Brakes , types systems function , calculation of stopping 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 system

THE TRAFFIC RESISTANCE


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 resistance


LEARNING OBJECTIVES

THE TRAFFIC RESITANCE

There 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 area

of the car .

RA=K.A.V2 Where ; K= the coefficient of air resistance(<1) V= the car speed (m/s) A=the car frontal area (m2)


THE GRADIENT RESIATANCE (Rg) The resistance (force )which prevent the car from moving up when the car climb on gradient road Rg = W sin Ɵ Rg = W.(H/L) Where sin Ɵ= H/L as shown in figure beside Ɵ 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 to car speed can be presenting as figure beside 6/28/2012 6 mechanic , 2nd year - machine and equipment

The rolling resistance (Rr) It is caused by the friction between the wheel and the road , its depends on , the type of wheels , the type of road and the car weight .


THE SUM OF RESISTANCE RT=Ra +Rg +Rr Where Ra only depend on the speed of car Rg = zero on horizontal roads or ways


if we draw the relation between the effort and speed the shape will be like as in figure beside , so we can see there is difference between the tractive effort and resistance effort and that what we call the surplus effort . SE= TE-RT

TRACTIVE EFFORT AND SURPLUS EFFORT

Max value as shown when there is huge difference Between TE and RT Min value when TE=TR at point B The advantage from calculate the SE is to define What acceleration we need to make the car moving That mean Increase SE »» increase acceleration »» increase the speed When we change the speed 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 moving


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).100

THE AMOUNT OF ACCELERATION AT ANY SPEED


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+at


V2-V1= at V2

2-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.a 1000* 0.625=625 n Rg= W(h/L) = (1000*9.81)*(1/25)= 392.4 n (According to force direction to up ) TE=RT+F TE=(Rg+Ra+Rr)+F ( since Ra so small then its equal to zero) Then TE= (392.4+0+250)+625= 1267.4 n

THE MOVING UP CASE


THE GOING DOWN CASE


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 V2

2-V12=2aX

X=( V22-V1

2)/2a X=(36*(1000/3600))2/(2*0.221)=226.21 6/28/2012 14 mechanic , 2nd year - machine and equipment

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/s2


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 V1

2-V22=2aX

((40*1000)/60*60))2-0=0*0.56*X Then X = 110.23 m


GEAR TRAIN


GEARS Gears are toothed cylindrical wheels used for transmitting mechanical power from one rotating shaft to another. Several types of gears are in common use. This LESSONS introduces various types of gears and details the design, specification and selection of spur gears in particular.


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 terminology



PRINCIPLE TERMINOLOGY Various definitions used for describing gear geometry are illustrated in Figure at side and listed below. For a pair of meshing gears, the smaller gear is called the ‘pinion’, the larger is called the ‘gear wheel’ or simply the ‘gear’.

PRINCIPLE TERMINOLOGY




At side


Note

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.










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.


EX-1 Consider the gear train shown in Figure at side . 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) )N1

EX-2 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-:-


N2=- (T1/T2 )*N1 N3=N2 ….. SAME SHAFT N4=-(T3/T4)*N3 N4= ((T3 /T4 )*(T1/ T2))*N1 N4 =(-18/54)*(-20/70)*(-1750) N4= 166.7 RPM

T1

T2 T3

T4

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 RPM

HW -1 a- Find the speed ratio ( N8/N1) for the gear shown ?

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 )?


HW-2


:- a- Find the speed ratio ( N9/N1 )For the gearbox shown bellow ? b- If the angular velocity N4=40 r.p.s and T1=20 teeth T2=60 teeth T3=30 teeth T4=50 teeth T5=60 teeth T6=30 teeth T7=40 teeth T8=30 teeth T9=50 teeth T10=20 teeth T11=80 teeth T12=30 teeth. Find the speeds of N12 , N7 , N5 and N9 ?


BEARINGS


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 bearings


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.


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



BEARING CLASSIFICATION



SLIDING BEARINGS


The term ‘sliding bearing’ refers to bearings where two surfaces move relative to each other without the benefit of rolling contact. The two surfaces slide over each other and this motion can be facilitated by means of a lubricant which gets squeezed by the motion of the components and can generate sufficient pressure to separate them, thereby reducing frictional contact and wear. A typical application of sliding bearings is to allow rotation of a load-carrying shaft. The portion of the shaft at the bearing is referred to as the journal and the stationary part, which supports the load, is called the bearing (see Figure 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 of components. Another common term is ‘plain surface bearings’. This section is principally concerned with 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.



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 friction


LUBRICANTS


As can be seen from Figure at side , bearing performance is dependent on the type of lubrication occurring and the viscosity of the lubricant .The viscosity is a measure of a fluid’s resistance to shear. Lubricants can be solid, liquid or gaseous, although the most commonly known are oils and greases. The principal classes of liquid lubricants are mineral oils and synthetic oils. Their viscosity is highly dependent on temperature as illustrated in Figure In next slide .They are typically solid at 35°C, thin as paraffin at 100°C and burn above 240°C. Many additives are used to affect their performance


DESIGN OF BOUNDARY LUBRICATED BEARINGS


General 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 capability

This 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 the diameter 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 PV factor greater than that obtained in (6) above.

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 of 75 rpm (see Figure 4.8). Calculate the PV value and by comparison with the available materials listed in Table in 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 diameter ratio for the bearing. L =25mm.


Roller

Sliding bearing





SHAFTS


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 shaft


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 bellow


Material used in shaft

When shaft of high strength required , then any alloy steel such as nickel , nickel – chromium or chrome – vanadium steel is used


NO Indian standard (ASME)

Ultimate strength (Mpa )

Yield strength (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 700 390

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 .


Max permissible working stress

Design case Compression , tensile (mpa ) (bending stress)

Torsional stress ( mpa ) (shear stress)

With key way 84 42

Without key way 112 56

28 June 2012 mechanic , 2nd year - machine and equipment 72

According to ASME (American Society of Mechanical Engineering) this is the reference stresses for shaft design .

Shaft Subjected to Twisting Moment only When the shaft is subjected to torque only , then the diameter of shaft can obtained from torsion equation which is

𝑻

𝑱=

𝝉

𝒓 where T = twisting moment (torque)

J= polar moment of inertia of shaft about axes of rotation

𝜏=torsional shear stress

r = radius of shaft = d/2

Case 1 : for solid shaft

𝑱 =𝝅

𝟑𝟐𝒅𝟒

So then

𝑻

𝝅𝟑𝟐

𝒅𝟒

=𝝉

𝒅/𝟐 that mean 𝑻 =

𝝅

𝟏𝟔∗ 𝝉 ∗ 𝒅𝟑


d

The twisting moment (T) can be found by P= (2πNT)/60 Where: P= power (watt) N = speed of shaft (rpm) T= twisting moment Then T=(P*60)/(2 π N) In case of belt drive or driven shaft the torque is found by (we will illustrated this in belt and rope lecturer) T= (T1-T2)* R Where T1= tight side tension =(Tl) large tension side (n) T2 = slack side tension =(Ts)= small tension side (n) R= radius of bully


Shaft Subjected to Twisting Moment only

Example 1 A line shaft rotating at 200 rpm to transmit 20 kw . The shaft assumed to be made of mild steel with allowable shear stress of 42 mpa . Determine the diameter of the shaft neglecting the bending moment of the shaft .

Sol :

N= 200 rpm P=20 kw = 2000 w

τ= 42 mpa = 42 n/mm2

T=(P*60)/(2πN)= (20*1000*60)/(2 π*200)=955 n.m= 955*1000 n.mm

T= (π/16)*τ*d3

955*1000= (π/16)*42*d3=8.25 d3 =====> d=955*1000/8.25= 48.7 mm ≈ 50 mm


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


Example 2

Shafts Subjected to Bending Moment Only


Example pair 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 the rails is 1.4 m. Find the diameter of the axle between the wheels, if the stress is not to exceed 100 Mpa



THE CLUTCH


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.m


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)/60


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 w 6/28/2012 83 mechanic , 2nd year - machine and equipment


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 watt


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 bearings

BELTS


Type of belts


Belt speed consideration


Torque

Speed

22.5 m/s

Coefficient of friction


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 bellow

TYPES OF FLAT BELT DRIVES The power from one pulley to another may be transmitted by any of the following types of belt drives:


2. CROSSED OR TWIST BELT DRIVE. The crossed or twist belt drive, as shown in Figure side , is used with shafts arranged parallel and rotating in the opposite directions. In this case, 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 more than that in the belt LM. The belt RQ (because of more tension) is known as tight side, whereas the belt LM (because of less tension) is known as slack side, as shown in Figure above , A little consideration will show that at a point where the belt crosses, it rubs against each other and there will be excessive wear and tear. In order to avoid this, the shafts should be placed at a maximum distance of 20 b, where b is the width of belt and the speed of the belt should be less than 15 m/s. 6/28/2012 92 mechanic , 2nd year - machine and equipment

3. QUARTER TURN BELT DRIVE. The quarter turn belt drive also known as right angle belt drive, as shown in Fig. (a) bellow, is used with shafts arranged at right angles and rotating in one definite direction. In order to prevent the belt from leaving the pulley, the width of the face of the pulley should be greater or equal to 1.4 b, where b is the width of belt. In case the pulleys cannot be arranged, as shown in Fig. 11.5 (a), or when the reversible motion is desired, then a quarter turn belt drive with guide pulley, as shown in Figure . (b) bellow, may be used.



Velocity ratio in belt drive



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 shape


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.ω


Ex :-1 ------------------------------- Find the power transmitted by a V- belt where : v- angle = 300 diameter of bully = 60 cm angular speed = 200 rpm coefficient of friction = 0.25 angle of contact = 1600 the large tension = 250 n sol: TL/TS=e(µƟ/sinβ) Ɵ=1600*(2π/360)=2.793 rad µƟ/sin β=(0.25*2.793)/sin 15 = 2.6978 TS=TL/ e(µƟ/sinβ) =250/14.84749= 16.837 n Power= (TL-TS).r.ω =(250-16837)*(30/100)*((2π*200)/60)=1465 w 6/28/2012 99 mechanic , 2nd year - machine and equipment


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 m


Ex:2 --------------------------- Find the length of the belt as in figure Sol:- X=120 CM = 120/100= 1.2 M

r 1 =20 cm = 20/100= 0.2 m r 2= 16 cm = 16/100= 0.16 m now we need to find ɸ sin (ɸ)=(r1-r2)/X= (0.2-0.16)/1.2=0.1 :. ɸ=sin-1(0.1)=5.730 2ɸ=180-Ɵ that’s lead to Ɵ=180-2ɸ =180-2(5.73)= 168.540 Arc length = radius *opposite angle in radian .: arc hkf=r2*Ɵrad =0.16*((168.54*π) /180)=0.4706 m Arc EJG=r1*(360-Ɵ)rad =0.2*((360-168.54)*(π/180))=0.66832 m XLine FE =HG=X*cos ɸ = 1.2*cos (5.730)=1.194m The total length L=hkf+fe+ejg+gh = 0.4706+1.194+0.66832+1.194=3.5269 m 6/28/2012 102 mechanic , 2nd year - machine and equipment

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 القدرة r=r2فعلٌه نستخدم r2بما انه اعطً فً السؤال السرعة الدورانٌة للبكرة الثانٌة فً قانون Ɵ=(360-Ɵ)فً قانون القدرة وكذلك فان r1= rفسوف نستخدم r1ولو 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 watt


Brake system


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.


BRAKE SYSTEM WITH IT POWER FLOW


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 brakes


DRUM BRAKES Drum brakes consist of a drum and a brake mechanism with two brake shoes that are curved to conform to the inside diameter of the drum. The drum brake has a steel or iron drum to which the wheel is bolted. The hydraulic pressure pushes the shoe-actuating pins out; hence the brake shoes are forced against the rotating drum. The resulting friction between the brake lining and the drum slows or 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 automotive brake systems used a drum design at all four wheels. They were called drum brakes because the components were housed in a round drum that rotated along with the wheel. The shoes were made of a heat-resistant friction material similar to that used on clutch plates.


An internal expanding brake consists of two shoes S1 and S2 as shown in Figure bellow The outer surface of the shoes are lined with some friction material (usually with Ferodo) to Increase the coefficient of friction and to prevent wearing away of the metal. Each shoe is pivoted at one end about a fixed fulcrum O1 and O2 and made to contact a cam at the other end. When the cam rotates, the shoes are pushed outwards against the rim of the drum. The friction between the shoes and the drum produces the braking torque and hence reduces the speed of the drum. The shoes are normally held in off position by a spring as shown in Figure bellow. The drum encloses the entire mechanism to keep out dust and moisture. This type of brake is commonly used in motor cars and light trucks.

INTERNAL EXPANDING BRAKE

TYPE OF BRAKE Hydraulic system = fluid Mechanical system = cable as in hand brake Pneumatic system = compressed air 6/28/2012 109 mechanic , 2nd year - machine and equipment

SHOES TYPE BRAKE THEORY



Heat to be Dissipated during Braking






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.



Steering system


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 condition


The condition for correct steering is :- ∆ IAP Tan Ɵ1=b/x ==== cotƟ1 =x/b ∆ IBP Tan Ɵ2= b/x+c ==== cotƟ2=(x+c)/b Cot Ɵ2 = (x/b )+(c/b) :. cotƟ2 = cotƟ1 +c/b cotƟ2-cotƟ1=c/b للحفظ

Ɵ2 اكبر من قٌمة Ɵ1 حٌث الشرط المهم هنا ان تكون قٌمة

And this what we call of Ackerman condition

STEERING MECHANISM


ex1:----------------------------- from the shape above, find Ɵ1 if

Ɵ2=200 and c=1.2 m and b= 2.7 m

sol:- (1/tan Ɵ2)-(1/tan Ɵ1)=c/b (1/tan200 )-(1/tanƟ1)=1.2/2.7 (1/0.36397)-(1/tanƟ1)=0.4444 2.747474-0.4444=1/tanƟ1 1/tanƟ1=2.30303 tanƟ1=1/2.30303=0.43421 Ɵ1=tan-1(0.43421)= 23.47090 Since Ɵ1> Ɵ2 that’s mean the solution is 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 123 mechanic , 2nd year - machine and equipment


SPACE REQUIREMENT. FOR TURNING


The kinematic steering condition can be used to calculate the space requirement of a vehicle during a turn. Consider the front wheels of a two-axle vehicle, steered according to the Ackerman geometry as shown in Figure above (previous slide) The outer point of the front of the vehicle will run on the maximum radius RMax, whereas a point on the inner side of the vehicle at the location of the rear axle will run on the minimum radius Rmin. The front outer point has an overhang distance g from the front axle. The maximum radius RMax is Therefore, the required space for turning is a ring with a width 4 R, which is a function of the vehicle’s geometry.

SPACE REQUIREMENT.


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 m



Overturning speed and skidding speed

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 turning



FIRST AT REST w= m.g RA= RB= w/2 = m.g/2

SECOND AT MOVING نحو االعلى قلٌال لذلك سوف ٌكون Aعند لحظة االستدارة سوف تتولد قوة طرد مركزي تحاول ان ترفع االطار

Bاسناد 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 speed


Ex1:---------------------------------------- A car travels on circular bath where Radius of curve = r = 50 m Height of c.g above ground level = h = 0.7 m Distance between the wheels of the car =d = 1.4 m Coefficient of friction = µ=0.7 Calculate the maximum speed of turning : (a)Without overturning (b) Without skidding outwards Sol: (a) Overturning speed = Vo= V= = 22.1 m/s *3.6 =79.74 km/hr (b) Skidding speed = Vs= = =18.53 m/s *3.6 = 66.71 km/hr 6/28/2012 132 mechanic , 2nd year - machine and equipment

μ gr

CRANK SHAFT


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)



Ex 1:-------------------------------------- From the figure above if you now that:- r= 0.0255 m d=0.024 m L= 0.046 m Ɵ= 100 ɸ=2.488 F=4521.6 n Find the shear stress and bending stress? Sol:- FC=F/cosɸ= 4521.6/cos2.488= 4526n n Ft=FC. Sin (Ɵ+ɸ)= 4526*sin(100+2.488)=978.7 n Fr= FC.cos(Ɵ+ɸ)=4526*cos(10+2.488)=4418.9 n Rr= Fr/2 =4418.9/2=2209.4 n Rt= Ft/2 =978.7/2= 489.35 n Mv= Rr*(L/2)= 2209.4*(0.046/2)=50.8 n.m Mh=Rt*(L/2)=489.35*(0.046/2)=11.25 n.m M=√Mv2+Mh2 M=√50.82+11.252 = 52.03078 T=Rt *r= 489.35 *0.0255= 12.5 n.m Shear 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/m2 h.w :- same question but take Ɵ=200 and ɸ=4.905 0 hw:- same question , but take Ɵ=00 and ɸ=0 0 6/28/2012 136 mechanic , 2nd year - machine and equipment















-: مالحظة طبٌعٌة غٌر الحالة وهذة Sg قمٌة من %50 بمقدار Se>Sg قمٌة ان الحظ

كان لو اخر بمعنى . منتظم غٌر ( هطول) استطالة تحدث سوف اخر بمعنى ولحل متساوي غٌر deflection ٌكون سوف Se=Sg متساوي االجهاد

وهذا شرٌحة ثانً من بدءا السفلٌة للشرائح اكبر انحناء اعطاء ٌتم المشكلة هذة . سابقا شرحها تم وكما مسبقا المجهدة بالشرائح ٌعرف ما

بنسبة وتتساوى االجهاد قٌم تتساوى ان هو التصمٌم فً الطبٌعً ان حٌث

اسفل المتدرجة الشرائح مع االطوال الكاملة الشرائح فً االنفعال قٌم معقولة . منها






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.





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 etc


Classification of Followers The followers may be classified as discussed below:






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 running


THE THEORY Most of theory depends on the low of moment ( force and arm) and some of sketch skills and some attention to what we do first




600

900

670

520

900

5 cm

3.75

kg

5 kg

2.5

kg

4 kg

2.5 kg

900

600

900

670

A

B

D

C

0.35 m

0.3 m

0.2 m

0.25 m

0.25 m

From solution

520 مقٌاس للرسم لكً

نتمكن من اظهار

الرسم بقٌاس معقول

ومقبول