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 mechanic , 2nd year - machine and equipment 1

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
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
system

THE TRAFFIC RESISTANCE


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
resistance


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

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


TRACTIVE EFFORT AND SURPLUS EFFORT
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

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

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).100


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


THE MOVING UP CASE
V2-V1= at
V22-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 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
V22-V12=2aX
X=( V22-V12)/2a
X=(36*(1000/3600))2/(2*0.221)=226.21

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

V12-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 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 RPM


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.


Sliding
bearing

Roller


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
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 700 390

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

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
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 42

Without key way 112 56

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

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 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/2
Case 1 : for solid shaft
𝝅 𝟒
𝑱= 𝒅
𝟑𝟐
So then
𝑻 𝝉 𝝅
= that mean 𝑻= ∗ 𝝉∗ 𝒅𝟑
𝝅 𝟒 𝒅/𝟐 𝟏𝟔
𝒅
𝟑𝟐 d

Shaft Subjected to Twisting Moment only
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


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


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


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

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


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

Type of belts


Belt speed consideration

Torque

22.5 m/s

Speed

Coefficient of friction


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

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.

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

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

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 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.


INTERNAL EXPANDING BRAKE
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.
TYPE OF BRAKE
Hydraulic system = fluid
Mechanical system = cable as in hand brake
Pneumatic system = compressed air

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


STEERING MECHANISM
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 قٌمة‬Ɵ1 ‫حٌث الشرط المهم هنا ان تكون قٌمة‬

And this what we call of Ackerman condition


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

SPACE REQUIREMENT. FOR TURNING


SPACE REQUIREMENT.
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.


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 ٌ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 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= μ gr =

=18.53 m/s *3.6 = 66.71 km/hr

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= 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 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


‫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

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


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

520

Mechanic second year

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