UNIT-1 VEHICLE FRAME AND SUSPENSION

Study of loads and moments on frame members

The chassis frame supports the various components and the body and keeps them in

correct positions. The frame must be light but sufficiently strong to withstand the weight and

rated load of the vehicle without having appreciable distortion. The material most commonly

used for frame construction is heat treated alloy steel.

The chassis generally experience four major loading situations or loads acting on the

chassis frame are vertical bending, longitudinal torsion, lateral bending and horizontal lozenging.

a) Vertical bending

Consider a chassis frame is supported at its ends by the wheel axles and a weight

equivalent to the vehicle’s equipment, passengers and luggage is concentrated around the middle

of its wheel base, then the side members are subjected to vertical bending causing them to sag in

the central region.

b) Longitudinal torsion

When diagonally opposite front and rear road wheels roll over bumps simultaneously, the

two ends of the chassis are twisted in opposite direction so that both the side and cross members

are subjected to longitudinal torsion.

c) Lateral bending

The chassis is exposed to side force that may be due to the camber of the road side wind

centrifugal force while turning a corner or collision with some object. The adhesion reaction of

the road wheel tyre opposes these lateral forces.

d) Horizontal lozenging

A chassis frame if driven forward or backward is continuously subjected to wheel impact

with road obstacles such as pot holes, road joints, surface bumps and curbs. Theses conditions

cause the rectangular chassis frame to distort a parallelogram shape known as lozenging.

Study of stresses on frame members

The various chassis member cross section shapes are solid round or rectangular cross

sections, enclosed thin wall hollow round or rectangular box section, open thin wall rectangular

channel such as C,I or top hat sections.

The side members should resist bending stresses and side and cross members should resist

torsional stresses. The square solid bar will resist bending stresses. The round solid bar will resist

bending stresses. The circular tube with longitudinal slit will resist torsional stresses. The

circular closed tube will resist both bending and torsional stresses. C section will resist both

bending and torsional stresses. The top hat and I section will resist bending stresses. Rectangular

box section will resist both bending and torsional stresses.

Bending stiffness:

Square bar= 1.0

Round bar = 0.95

Round hollow tube =4.3

Rectangular C channel= 6.5

Square hollow sections= 7.2

Longitudinal split tube enclosed hollow tube= 62.0

Open c channel= 1.0

Closed rectangular box section=105.0

Types of frame design

Box section frame:

This frame does not have sufficient rigidity against torsion. If the body is not designed to

resist these stresses, the problems like movement between doors and pillars, broken wind screens

and cracking of body panels may occur. Most of the cars have independent suspension, so the

frame must be extremely rigid at the points of joining the main components with the body. TO

achieve this box section members are welded together.

Back bone frame:

In this construction, two longitudinal box section members are welded together at the

centre and separated at front and rear to accommodate the main components.

Energy absorbing frame:

In this by constructing the front and rear end of the frame in a manner during collision and

absorb the impact energy.

Problems:

1. Calculate the maximum bending moment and maximum section modulus:

Wheel base=180cm, overall length=360cm, Equal overhang on either side. 270 kgf acting

on C.G of the load 45cm in front of front a x le. 180 kgf acting on C.G of load 45cm in

front of rear axle. 67.5 kgf acting on C.G of load 45cm behind rear a x le. In addition

there is a uniformly distributed load of 1.75 kgf per cm run over the entire length of

chassis.

Given:

Wheel base = 180cm

Overall length = 360cm

Solution:

∑MR2 = 0

R1 x 180 = (67.5 x 225) + (180 x 135) + (180 x 45) - (270 x 45)

= 4578.5 - 12150=35437.5.

R1 = 35437.5/180= 197 kgf

R2 = 270 + 180 +180 + 67.5 - 197=697.6 - 197=500.5 kgf.

From maximum bending moment,

12150 +7088 = 19238 kgfcm

f = 600kgf/cm2

( allowable stress)

z = section modulus

Bending moment =fz , Maximum bending moment due to dynamic force is twice that due

to static forces.

Z =2 x 19238/600 = 64.1cm3

2. A bus chassis 5.4m long, consists of 2 side members and a number of cross members.

Each side member can be considered as beam, simply supported at two points A and B,

3.6m apart. A being positioned 0.9m from the front end of the frame and subjected to the

following concentrated loads. Engine support(front) 2kN, engine support(rear) 25kN,

gear support 0.5Kn and body W kN. The distances of these loads from the front end of

the frame are respectively 0.6m, 1.8m, 2.4m and 3m. If the reaction at A is 8.5kN.

determine

a) The magnitude of load W due to vehicle body.

b) B) the magnitude of support reaction at B.

Given:

Overall length = 5.4m

Wheel base =3.6m

Solution:

To determine the magnitude of W, take moments about B. For equilibrium the

resultant moment must be zero.

This means clockwise about B= Anticlockwise aboutB

(8.5 x 3.6) = (2 x 3.9) + (2.5 x 3.7) + (0.5 x 2.1) + (W x 1.5)

30.6 = 7.8 + 6.75 + 1.05 + 1.5W

W = (30.6-11.6)/1.5= 10kN

For equilibrium total upward forces must be equal to total downward forces

Ra + Rb = Total downward forces

8.5+ Rb = 2+2.5+0.5+10

Rb = 6.5kN

Design of Leaf spring

P1,P2 = load on each end of the spring (N)

R = radius of curvature of spring (m)

l = length of the spring (m) = l1.l2

E = Young’s modulus (pa)

W = width of the spring (m)

I = moment of inertia = wt3/12 (m

4)

n = number of leaves

Deflection, δ = 2.85Pl3/Enwt (m)

t = thickness of spring (m)

Initial radius, R = l2/8Z + Z/2 = l

2/8 δ + δ/2 (m)

Z = perpendicular distance between the line joining the centers of the eyes of a half

elliptic spring.

Number of leaves, n = 3pl/wt2fb

fb = bending stress (pa)

rate of spring, r = Enwt3l3/1.425(N/m)

Problems

1. A vehicle spring of semi elliptical type has leaves of 75mm width and 10mm thickness

and effective length 900mm. If the stress is not to exceed 220725 kpa when the spring is

loaded to 4905N. Estimate the required number of leaves and the deflection under this

condition. If the spring is just flat under load, what is the initial radius.

Given:

P = 4905N, l = 900 mm, w = 75 mm, fb = 220725 kpa, E = 196.2 x 106

kpa

Solution:

Number of leaves, n = 3pl/wt2fb

n = 3 x 4905 x 0.9/0.075 x 0.012 x 220725 x 10

3

n=8

Deflection, δ = 2.85Pl3/Enwt

δ = 2.85 x 4905 x 0.93 x 10

3/192.2 x 10

9 x 0.075 x 8 x 0.01

3

δ = 86.57 mm

intial radius, R = l2/8Z + Z/2 = l

2/8 δ + δ/2

R= 9002/8 x 86.57 + 86.57/2

R = 1212.9 mm

Design of coil spring suspension

P = axial load on the spring (N)

d = diameter of the spring wire (m)

D = mean diameter of coil (m)

n = number of active coils

fs = allowable stress in the spring (pa)

G = modulus of rigidity = 73575 x 106 pa

Deflection, δ = 8nPD3/Gd

4 (m)

Shear stress, fs = Gd δ/πD2n (N/m

2)

Energy stored in the spring = P δ/2 (Nm)

Volume of steel in the spring = πd2Dn/4 (m

3)

Coil springs are subjected to torsion and failure occurs due to shear. The stresses varies

uniformly form a maximum at the surface to zero at the centre of circular cross section. The

average stress is equal to two third the maximum.

Problems:

1. A typical coil suspension spring has 10 effective coils of mean diameter 125mm and

made out of wires of diameter 15mm. The spring is designed to carry a maximum static

load of 3531.6N. Calculate the shear stress and the deflection. If the maximum shear

stress of 375650kpa is allowable in the material then what is the possible clearance.

Given:

N = 10, d = 15 mm, fsmax = 637650 kpa, D = 125 mm, P = 3531.6N, G = 73575 x

103kpa

Solution:

δ = 8nPD3/Gd

4

= 8 x 10 x 3531.6 x 0.125

3 x 10

3/73575 x 10

6 x 0.015

4

δ = 148 mm

fs = Gd δ/πD2n

= 73575 x 103 x 0.015 x 0.148/3.14 x 15

2 x 10

fs = 332915 kpa

δmax = πD2fsma x /Gd

= π x 1252 x 10 x 637650/73575 x 10315

δmax = 284 mm

Clearance = 284-148 = 126 mm.

Design of torsion bar spring

The torsion bar spring is a bar of spring anchored to the frame at one end while the other

end is freely supported and connected to a lever arm. A rod or tube acting in torsion can work as

torsion bar spring.

W = load acting on the lever arm (N)

L = length of the arm

D = diameter of the bar (m)

θ = angular deflection of bar (rad)

fs = torsional stress (pa)

G = modulus of rigidity (pa)

J = polar moment of inertia = πd4/32 (m

4)

Angular deflection, θmax = 5760TL/π2d

4G (rad)

L = length of bar (m)

Diameter of bar, d = 3√16Tmax / π fsmax (m)

fs = G θd/2L (pa)

load rate, r = πd4G + 32LW X /32L(l

2- X

2) (N/m)

Problems

1. A torsion bar suspension is to be designed to support a ma x imum static load of 3433.5N

at the end of a lever arm 250 mm long. The deflection of the lever above the horizontal is

to be 30 degree with a total angle of deflection of 90 degree. Assuming a safe allowable

stress of 784800 kpa. Calculate the diameter of torsion bar, the effective length and the

load rate.

Given:

L = 250 mm, α = 30 degree, θ = 90 degree , fs = 784800 kpa

Dynamic load = 2 x static load = 2 x 3433.5 = 6867 N

Effective length of the arm = lcos30 = 250 x 0.866 = 216.5 mm

Solution:

Moment acting on the bar, T = Wy = 6867 x 0.2165 = 1486.7 Nm.

Diameter of the bar, d = 3√16Tmax / π fsmax

d = 3√16 x 1486.7/π x 784800 x 10

3

d = 3.12/102 x 10

3

d = 31.2 mm

length of the bar, L = π x 73575 x 106

x 90 x 31.2 x 103/360 x 784800 x 10

3

L = 2.3 m

Deflection above the horizontal, X = lsin30 = 25 x 0.5 = 12.5 cm.

r = πd4G + 32LW X /32L (l

2- X

2)

r = π x 0.03124

x 73575 x 106 + 32 x 2.3 x 6867 x 0.125/32 x 2.3(0.25

2-0.125

2)

r = 218916.4 + 63176.4/3.45 = 81766 N/m

UNIT-2 FRONT AXLE AND STEERING SYSTEM

Study of loads and stresses at different sections of front axle

Front axle carries the weight of the front part of the automobile as well as facilitates

steering and absorbs shocks due to road surface variations. The front axles are generally dead

axles, but are live axles in small cars of compact designs and also in case of four wheel drive.

The front axles generally dead axles which does not transmit power. The dead front axles are

three types. In Elliot type front axles the yoke for king spindle is located on the ends of I beam.

The reverse elliot front axles have hinged spindle yoke on spindle itself instead of on the axle.

This type is commonly used as this facilitates the mounting of brake backing plate on the

forged legs of the steering knuckle. In the lemoine type front axle, instead of yoke type hinge, an

L shaped spindle is used which is attached to the end of the axle by means of pivot. It is normally

used in tractors. To prevent interference due to front engine location and for providing greater

stability of safety at high speeds by lowering the center of gravity of the road vehicles, the entire

center portion of the axle is dropped. The axle beam is of I section or H section and is

manufactured from alloy forged steel for rigidity and strength. As compared to front dead axles,

a totally different type of swiveling mechanism is used on the live front axles.

Front axles are subjected to both bending and shear stresses. In the static condition, the

axles may be considered as beam supported vertically upwards at the ends. The vertical bending

moment thus caused is zero at the point of support and rises linearly to a maximum at the point

of loading and then remains constant.

Design of front axle beam

Maximum bending moment = Wl (Nm)

W = load on the wheel (N)

l = distance between the center of the wheel and spring pad (m)

magnitude of torque = Rδ (nm)

R = resistance of motion (N)

δ = drop from the spindle axis to the center of the section (m)

shear stress in the axle = µWr (Nm)

r = the road wheel radius (m)

µ = coefficient of adhesion road and tyre

For I section, the maximum bending moment is M/I = fb/y

M = maximum bending moment (Nm)

fb = allowable bending stress (N/m2)

y=d/2 (m)

I = bd3-ch

3/12 (m

4)

d = the overall depth of I section (m)

b = flange width (m)

t = flange thickness (m)

w = web thickness (m)

c = b-t h = d-2w

Generally d = 6t, b = 4.25t, w = 2.5t

Max torsion is T/Ip = fs/y

T = maximum torques in plane of section (Nm)

fs = allowable shear stress in the material (pa)

y=d/2 (m)

d = diameter for circular section (m)

Ip = (π/32) d4

for circular section

= (π/32) d3b for oval section with minor axis b.

Load on lower knuckle pin, Rl = c/d + e

Load on upper knuckle pin, Ru = a/a + e

Load on thrust bearing, Rt = (ce + ad/bd + de) Rw

Determination of bearing loads on king pin or steering knuckle

Let Rw = the reaction of the wheel on the spindle acting vertically through the center of contact

of tyre on ground

Rt = the load on thrust bearing

Ru = the load on the upper bearing (knuckle pin)

Rt = the load on lower knuckle pin

B and C represent the centers of lower and upper knuckle pin bearings respectively. A is point on

the spindle axis in the center plane of wheel.

Then ∑ Mc = 0 gives Rwc - Rl (d + e) = 0

Rl = c/d + e (Rw)

∑ MB = 0 gives Rwa - Ru (d + e) = 0

Ru = a/a + e (Rw)

∑ MA = 0 gives Rtb - Rle - Rud = 0

Rtb = Rle + Rud

Substitution the expression of Rl and Ru gives Rt = ce + ad/b (d + e) (Rw)

The other loads acting on knuckle pin bearings are those due to the rolling resisitance and road

shocks. These loads are proportional to the static load and hence can be accounted for.

Problems

1. The load distribution between the front and rear axle of a motor vehicle weighing 1350

kgf is that 48% of total load is taken by front axle. The width of track is 140 cm and the

distance between the centers of the spring pads is 66 cm. Design a suitable I section for

the front axle assuming that the width of flange and its thickness are 0.6 and 0.2 of the

overall depth of the section respectively and thickness of web 0.2 of the width of the

flange.

Given:

Total load of vehicle = 1350 kgf

Load taken bu the front axle = 0.48 x 1350 = 648 kgf

From bending moment diagram, maximum bending moment = 37 x 328 = 12136 kgfcm

Flange width = 0.6d cm

Flange thickness = 0.2d cm

Web thickness = 0.2d cm

Solution:

M/I = fb/y

I = 1/12(0.6d x d3) – ½(0.45d x (06d)

3)

=1/12 0.6d4(1 – 0.45 x 0.6

2)

= 0.6 x 0.838/12 (d4)

= 0.0419 d4

cm4

12136/0.0412 d

4 = 915/d/2

d3 = 12136/0.0419 x 2 x 915

d = 5.42 cm

Dimensions of I section are: flange = 0.6 x 5.42 = 3.25 cm

Flange thickness = 0.2 x 5.42 = 1.08 cm

Web thickness = 0.15 x 5.42 = 0.813 cm

Choice of bearings

The bearing should withstand loads and stresses of weight of the body and components.

The large splaying out effect of the wheel takes place. The wheels are pushed by the force R,

which is opposed by the resistance R. These two forces cause a couple Fx, whose magnitude

becomes very large when the force of front brakes are applied.

Steering becomes heavy because of the distance between king pin and wheel centre. The

wheel moves in an arc of radius x around the pin. Large bending stress occurs in the stub axles

and king pin. To overcome these problems the wheel and king pin should have minimum

possible offset distance x. When the offset is eliminated, the center line of the wheel meets the

centre line of the king pin at the road surface. The condition which can be obtained through

camber, swivel axis inclination. Although the center point steering appears to be ideal, but the

spread effect of the pneumatic tyre cause the wheel to scrub and produce hard steering and tyre

wear. Positive offset is obtained when the center of the line of the wheel meets the swivel at a

point just below the road. The offset distance measure at the road surface between the two center

lines should be equal to balance the inward or outward pull of the wheels.

Castor angle is tilt of the king pin or ball joint center line from the vertical outwards either

the front ( negative castor) or rear ( positive castor) of the vehicle. The weight of automobile

having positive castor tends to turn a wheel inward to allow the body to lower. Negative castor

causes an outward turning effect. Toe in is the amount by which the front wheel rims are set

closed together at the front than at the rear with the wheels in a straight ahead position when the

vehicle is stationary.

Problems

1. A motor car has wheel base of 2.743 m and pivot center of 1.065 m. The front and rear

wheel track is 1.217 m. Calculate the correct angle of outside lock and turning circle

radius of the outer front and inner rear wheels when the angle of inside lock is 40.

Given:

Wheel base b = 2.743 m

Pivot center c = 1.065 m

Wheel track a = 1.217 m

Solution:

For correct steering, cot φ – cot θ = c/b

Cot φ = 1.065/2.743 + cot 40

= 0.388 + 1.19175 = 1.57975

φ = 32.4

Turning circle radius of the front wheel (outer), Rot = b/sin φ + a – c/2

= 2.743/ sin 32.4 + (1.217 – 1.065)/2

= 2.743/0.537 – 0.152/2 = 5.196 m

Turning circle radius of inner wheel (rear), Rir = b/sin θ - a – c/2

= 2.743/sin 40 – 0.076

= (1.19175 x 2.743) – 0.076 = 3.2 m

2. A track has pivot pins 1.37 m apart, the length of each track arm is 0.17 m and the track rod is

behind front axle and 1.17 m long. Determine the wheel base which will give true rolling for all

wheels when the car is turning so that the inner wheel stub axle is 60 degree to the center of the

car.

Given:

Pivot center = 1.37 m

Length of each track arm = 0.17 m

Length of track rod = 1.17 m

θ = 30 degree

Solution:

sin α = c – d/2r = 1.37 – 1.17/2 x 0.17

sin α = 0.178, α = 16.12 degree

sin (α + θ) + sin (α - φ) = 2 sin α

sin (16.12 + 30) + sin (16.12 – φ) = 2 x 0.178 = 0.556

0.55 – 0.720 = -0.164 = sin (-9.44)

Φ = 16.12 + 9.44 = 25.56

For correct steering, cot φ – cot θ = c/b

cot 25.26 – cot 30 = 1.37/b

2.08– 1.732 = 1.37/b

b = 1.37/0.349

Wheel base, b = 1.37/0.349 = 3.92 m

3. The distance between the kingpins of a car is 1.3 m. The track arms are 0.152 m long and

the length of the track rod is 1.2 m. For a track of 1.42 m and a wheel base of 2.85 m.

Find the radius of curvature of the path followed by the near side front wheel at which

correct steering is obtained when the car is turning to the right.

Given:

Pivot center = 1.3 m

Length of track arm = 0.152 m

Length of track rod = 1.2 m

Wheel base = 2.85 m

Wheel track = 1.42 m

Solution:

cot φ – cot θ = c/b = 1.3/2.85 = 0.4562

sin α = c – d /2r = 1.3 – 1.2/2 x 0.152 = 0.328

α = 19.2 degree

The value of θ is to be calculated for correct steering. This can be done conveniently by

drawing a graph between θ and cot φ – cot θ. using the relation,

sin (α + θ) + sin (α - φ) = 2 sin α

putting, θ = 30 degree, sin (19.2 – φ) = 0.656 – 0.757 = sin (-5.8)

φ = 19.2 + 5.8 = 25 degree

cot φ – cot θ = cot 25 – cot 30 = 2.1445 – 1.732 = 0.4124

putting, θ = 35 degree, sin (19.2 – φ) = 0.656 – 0.811 = sin (-8.9)

φ = 19.2 + 8.9 = 28.1 degree

cot φ – cot θ = cot 28.1 – cot 35 = 1.8728 – 1.428 = 0.446

putting, θ = 37 degree, sin (19.2 – φ) = 0.656 – 0.831 = sin (-10.05)

φ = 29.25 degree

cot φ – cot θ = cot 29.25 – cot 37 = 1.7833 – 1.377 = 0.4562

putting, θ = 39 degree, sin (19.2 – φ) = 0.656 – 0.850 = sin (-11.2)

φ = 30.4 degree

putting, θ = 40 degree, sin (19.2 – φ) = 0.656 – 0.859 = sin (-11.7)

φ = 30.9 degree

cot φ – cot θ = cot 30.9 – cot 40 = 1.6709 – 1.1917 = 0.4791

we got the value of θ for correct steering without drawing any graph which is equal to 37

degree.

Radius of curvature of the path followed by the near side front wheel,

Rif = b/ sin θ – a – c/2 = 2.85/ sin 37 = 1.42 – 1.3/2 = 4.68 m

Steering error

The steering geometry errors with up and down wheel movement are generally due to the following

causes.

I) Incorrect relative length of cross steering tubes and linkage arms.

II) Incorrect alignment of steering tube and linkage.

III) Incorrect relative lengths of the instantaneous radii of suspension and steering linkages.

IV) Incorrect relative heights of the ends of the two effective radii.

After knowing the values of φ, the corresponding angles θ and φ are laid off on opposite ends of

line C. A curve is then drawn through the intersection of line describing the angle θ and φ. The

curve drawn is called steering error curve. For correct steering layouts, the effective length of

steering tube has a definite value when looked along the suspension linkage can be either double

wishbone or the strut and link. There will be an inward movement of link and arm end as it rises

from a horizontal position due to misalignment of link and tube.

The double wishbone has horizontal and parallel links of lengths r (upper) and R (upper)

their ends apart by a distance d and the end of the lower link is above the ground by a. The strut

and link type is effectively an inverted version of the double link type, so that the movement

itself provides the straight line motion with its instantaneous center at infinity on a line

perpendicular to the sliding motion of the strut starting from its mounting pivot point.

UNIT-3 CLUTCH

Design of single plate clutch

The clutch enables smooth transmission of the rotary motion of the engine crankshaft to a

gearbox shaft. It enables rapid disengagement and re-engagement of engine from the

transmission.

r1 and r2 = internal and external radii of contact surface (m)

W = axial load exerted by actuating springs (N)

µ = coefficient of friction between the contact surfaces

Uniform intensity of pressure:

Intensity of pressure, p = W/π (r22 - r1

2) (N/m

2)

Total frictional torque, T = 2/3µW (r23 - r1

3)/ r2

2 - r1

2) (Nm)

Uniform rate of wear:

Total axial load, W = 2πC (r2 - r1) (N)

Pr = C, if the rate of wear is assumed to be constant.

Frictional torque, T = µWR (Nm)

R = mean radius of friction surface = r1 + r2/2

For a single plate clutch having a pair of contact surfaces, T = µW (r1 + r2) (Nm)

Note:

i) In case of new clutch, the intensity of pressure is approximately uniform, but in an old

clutch the uniform wear theory is more appropriate.

ii) The uniform pressure theory gives a high friction torque than the uniform wear theory.

Therefore in case of friction clutches, uniform wear theory should be considered

unless otherwise stated.

Energy lost during engagement, E = 1/2g IBω2 (Nm)

IB = mass moment of inertia attached to driven shaft of the clutch

ω = angular speed

Energy dissipated due to clutch slip = Tωt – Iω2/2g (Nm)

Problems

1. An automobile power unit gives a maximum toque of 13.52 Nm. The clutch is a single

plate dry disc having effective clutch lining of both sides of plate disc. The coefficient of

friction is 0.3 and the maximum axial pressure is 8.29 x 104 pa and external radius of

friction surfaces is 1.25 times the internal radius. Calculate the dimensions of clutch plate

and total axial pressure that must ne exerted by clutch springs.

Given:

T = 13.56 Nm pmax = 8.29 x 104

µ = 0.3 r2 = 1.25r1

Solution:

T = µπC (r22 - r1

2)

C = pr1 = 8.29 x 104r1

13.56 = 2 x π x 0.3 x 8.29 x 104r1 (1.5625 r1

2 - r1

2)

r13 = 13.56/2 x π x 0.3 x 8.29 x 10

4 x 0.5625

r1 = 53.6 mm

r2 = 1.25 x 53.6 = 67 mm

Total axial pressure, W =2πC (r2 - r1)

W = 2 x π x 8.29 x 104 r1 (r2 – r1)

= 2 x π x 8.29 x 104 x 0.0536 (0.067 – 0.0536)

= 2 x π x 8.29 x 5.36 x 1.34

W = 373.92 N

2. A motor car develops 5.9 bkw at 2100 rpm. Find the suitable size of clutch plate having

friction linings riveted on both sides to transmit the power under the following

conditions.

a) Intensity of pressure on surface not to exceed 6.87 x 104 pa.

b) Slip torque and losses due to wear is of 35% of engine torque

c) Coefficient of friction is 0.3

d) Inside diameter of the friction plate is 0.55 times the outside diameter

Given:

N = 2100 rpm

P = 5.9 kw

Solution:

T = pw 6000/2πN = 60000 x 5.9/2π x 2100 = 26.84 nm

T = 26.84 x 1.35 = 36.23 Nm

T = πµC (r22 - r1

2) 2

36.23 = 2 x π x 0.3 x 6.84 x 104 x ((r1/0.55)

2 - r1

2)2

= π x 4.122 x 104(1/0.303 – 1) r1

2

r13 = 36.23 x 0.303/ π x 4.122 x 10

4 x 0.697 = 49.5 mm

r1 = 49.5 mm r2 = 90 mm

3. A friction clutch is required to transmit 33.12 kw at 2000 rpm. It is to be of single plate

clutch with both sides of the plate effective the pressure being applied axially by means of

springs and limited to 6.87 x 104 pa. If the outer diameter of the plate is to be 0.305 m. Find

the required inner diameter of the clutch ring and the total force exerted by the springs.

Given:

P = 33.12 kw pw = 6.87 x 104 pa

N = 2000 rpm d2 = 0.305 m

Solution:

T = pw60000/2πN

T = 60000 x 33.12/2 x π x 2000

T = 158.22 Nm

T =2πµC (r22 - r1

2)

158.22 = 2 x π x 0.3 x 6.87 x 104 r1 (0.1525

2 - r1

2)

= 129430.8 r1 (0.02325 - r12)

r13 = 0.02325 r1

+ - 0.00122

r13 + 1222.42 = 232.562

By trial we get r1 = 112 mm or 63.5 mm

Inner diameter d1 = 224 mm or 127 mm

When r1 = 0.112 m and r1 = 0.1525 m

W = 2πC (r2 - r1)

= 2 x π x 6.87 x 104 x 0.112 x 0.0405

= 1958 N

When r1 = 0.0635 m and r2 = 0.1525 m

W = 2πC (r2 - r1) = 2 x π x 6.87 x 104 x 0.0635 x 0.089 = 2439.5 N

4. An automobile is fitted with a single plate clutch to transmit 22.1 kw at 2100 rom. The total

axial load on the clutch plate is 1422.5N. The outside diameter of friction surface is 250 mm.

Both sides of the plate are effective and the µ between the contact surfaces is 0.35. Assuming

uniform rate of wear condition. Calculate inner diameter. The rotating parts attached to the

driven parts of clutch are initially at rest MT of 20.7 Nm2. Calculate the time lapse before the

engine attains full speed of 2100 rpm.

Given:

P = 22.1 kw W = 1422.5 N I = 20.7 Nm2

N = 2100 rpm d2 = 250 mm

Solution:

T = pw/2πN = 60000 x 22.1/2 x π x 2100 = 100.55 Nm

T = µW r2 + r3/2 x 2

100550 = 00.35 x 1422.5 (r2 + r1)

1.25 + r1 = 100550/1422.5 x 0.35 = 202 mm

r1 = 202 – 125 = 77 mm

T = Iω/gt

100.55 = 20.7/9.81 (2 x π x 2100/60)/t

t = 4.62 seconds

Design of clutch components

1. A single dry plate clutch is to be designed to transmit 7.5 kw at 900 rpm. Find the

diameter of shaft, mean radius and face width of friction lining assuming the ratio of

mean radius to the face width as 4, outer and inner radii of clutch plate, dimensions of the

springs, assuming no. of springs are 6 and spring index = 6.

Given:

P = 7.5 kw, N = 900 rpm, R/b = 4, no. of springs = 6, C = D/d = 6, fs = 420 Mpa

Solution:

T= p x 60 /2πN = 7500 x 60/2 x π x 900 = 79.6 Nm

T = π/16 x fs x ds3

79600 = π/16 x 420 x ds3

Diameter of shaft, ds = 25mm

Area of friction faces, A = 2πRb

W = A x p = 2πRb

T = µWRn = π/2µR3pn

79600 = π/2 x 0.25 x R3 x 0.07 x 2 = 0.055R

3

R3

= 79600/0.055 = 1.45 x 106

R = 114 mm b = R/4 = 28.5 mm

b = r2 - r1, r2 - r1 = 28.5 mm

R = r1 + r2/2 r1 + r2 = 2 x 114 = 228 mm

r1 = 128.25 mm r2 = 99.75 mm

W = 2πRbp = 2 x π x 114 x 28.5 x 0.07

W = 1429.2 N

Total load on springs = 1.25W = 1.25 x 1429.2 = 1786.5 N

Maximum load on each spring, Ws = 1786.5/6 = 297.75 N

Wahls stress factor, K = 4C – 1/4C – 4 + 0.615/C

= 4 x 6 – 1 /4 x 6 – 4 + 0.615/6 = 1.252

Maximum shear stress, 420 = K x 8WsC/πd2

= 1.2525 x 8 x 297.75 x 6/ πd2

d2 = 5697/420 = 13.56 = 3.68 mm

D = Cd = 6 x 4.064 = 24.38 mm

δ = 8WsC3n/Gd = 8 x 297.75 x 6

3 x 4/84 x 10

3 x 4.064 = 6.03 mm

total no. turns, n’ = n + 2 = 4 + 2 =6

free length of spring, l = n’d + δ + 0.15 δ = 6 x 4.064 + 6.03 + 0.15 x 6.03 = 31.32 mm

Design of multi plate clutch

For multi plate clutch having n pair of contact surfaces, T = µWn (r1 + r2)/2 Nm

To have n pair of contact surfaces, there must be n + 1 number of discs or plates. If there are nA

number of discs on the driving shaft and nB number of discs on the driven shaft, then number of

pairs of contact surfaces are n = nA + nB – 1

1. A plate clutch has 3 discs on the driving shaft and 2 discs on the driven shaft, providing 4

pairs of contact surfaces. The outside diameter of the contact surfaces is 240 mm and inside

diameter 120mm. Assuming uniform pressure and µ = 0.3. Find the total spring load pressing

the plates together to transmit 23 kw power at 1475 rpm. If there are 6 springs each of

stiffness 13 kN/m and each of the contact surface has worn away by 1.25 mm. Find the

maximum power that can be transmitted.

Given:

nA = 3 n = 3 + 2 - 1 = 4 pw = 25 kw

nB = 2 µ = 0.3 N = 1475 rpm

r2 = 120 mm r1 = 60 mm

Solution:

ω = 2 x π x 1575/60 = 52.5π rad/s

pw = Tω T = pw/ω = 25 x 103/52.5 x π = 151.6 Nm

for uniform pressure condition, T = n2/3µW(r23

- r13 / r2

2 - r1

2)

151.6 = 4 x 2/3 x 0.3W (0.123

- 0.063/0.12

2 - 0.06

2)

W = 1355 N

No. of springs = 6

Contact surfaces of the spring = 8

Wear on each contact surface = 1.25 mm

Total wear = 1.25 x 8 = 10 mm

Stiffness of spring = 13kN/m = 13 x 103 N/m

Reduction in spring force = total wear x no. of springs x stiffness per spring

= 0.01 x 6 x 13 x 103 = 780 N

New axial load = 1355 – 780 = 575 N

Uniform wear condition, T = nµW (r2 + r1/2) = 4 x 0.3 x 575 (0.12 + 0.06/2) = 62 Nm

Pw = Tω = 62.5 x 52.5π = 10230 watt = 10.23 kw

Design of cone clutch

In cone clutch the contact surfaces between the driving and driven shaft forma a part of

the cone. The effectiveness of clutch is increased due to wedging action of the cone so that

the normal force on the lining increases. This increases in normal force results in increased

torque.

W = total axial force required to engage the clutch supplied by spring

θ = semi angle of the cone

r1 = outer radius of the cone

r2 = inner radius of the cone

w = breadth of the cone face = r1 – r2/sin θ

Wn = normal load on frictional surface

Rn = normal reaction

µRn = frictional force

Area of the ring = 2πrdw

Normal load on ring, dp – p2πrdr/sin θ

Considering uniform pressure, p = W/π (r22

- r12)

T = 2/3µW/sin θ (r23

- r13/ r2

2 - r1

2)

Considering uniform rate of wear, pr = C

T = 2πµPR2W (Nm)

Power capacity of the clutch = 2πNT/6000 (kw)

Axial force to engage the clutch = Wnsin θ + µWncos θ

Axial force to disengage the clutch = µWncos θ - Wnsin θ

Force required for effective normal pressure for steady state operation = Wnsin θ = Rnsin θ

Energy supplied = Tωt (Nm)

Energy of flywheel = Iω2/2g (Nm)

Problems

1. In a cone clutch the semi angle of cone is 15 degree, coefficient of friction is 0.35 and the

contact surfaces have an effective mean diameter of 80 mm. If the axial force applied is

196.2 N. Find the torque required to produce the slipping of the clutch. Calculate the time to

attain the ful speed and also the energy lost in the slipping of the clutch, if the clutch is

employed at 1200 rpm, with a flywheel which is stationary and moment of inertia of 3.4

Nm2.

Given:

θ = 15 degree, µ = 0.3, W = 193.2 N, N = 1200 rpm, I = 3.4Nm2

Solution:

dm = d2 + d1/2

T = µ (r2 + r1) W/2sin θ = 0.35 x 0.08 x 196.2/2 sin 15

T = 10.6 N/m

T = (T/g) α 10.6 = (3.4/9.81)α

α = 9.81 x 10.6/3.4 = 30.58 rad/s2

t = w/ α = 2πN/60 α

= 2 x π x 1200/60 x 30.58

T = 4.11 sec

Energy supplied = Tωt =t (2πN/60) T

= 106 x 2π x 1200 x 4.11/60 = 5471.9 Nm

Energy of flywheel = Iω2/2g

= 3.4/2 x 9.81 (2 x π x 1200/60)2

= 2733.8 Nm

Energy lost during slip = 5471.9 – 2733.8 = 2738.1 Nm

2. A cone clutch with cone semi angle of 12 degree is to transmit 11.9 kw at 750 rpm. The

width of the face is 1/4th

of mean diameter and the normal pressure between the contact faces

is not to exceed 8.27 x 104 pa. Allowing the coefficient of the clutch and the axial force

required.

Given:

P = 11.9 kw θ = 12 degree µ = 0.2

N = 750 rpm p = 8.2 x 104 pa

T= pw60000/2πN = 60000 x 11.9/2 x π x 750 = 142.5 Nm

W = ¼ x d2 + d1/2 = r2 + r1/4

= r2 – r1/sin θ = r2 – r1/ sin 12

4 (r2 – r1) = sin 12 (r2 + r1)

= 0.208 (r2 + r1)

3.792r2 = 4.208r1 r2 = (4.208/3.792) r1

r2 = 1.11r1

T = µπC/sin θ (r22 – r1

2) = µπpr1/sin θ (r2

2 – r1

2)

142.5 = (0.2π x 8.27 x 104/sin 12) r1 (1.231 – 1) r1

2

= (0.2π x 8.27 x 104 x 0.231/0.208) r1

3

= 57678.5 r13

r13 = 142.5/57678.5 = 0.0024706

r1 = 0.135 m r2 = 1.11 x 0.135 = 0.14 m

w = 0.135 + 0.14/4 = 0.7125 m

W = 2πC (r2 – r1) = 2πp (r2 – r1) r1

= 2π x 8.27 x 104 x 0.135 x 0.014

W = 1051.7 N

Torque capacity of clutch

The transmitted torque of the clutch can be raised by increasing the number of pairs of

rubbing surface. Theoretically the torque capacity is directly proportional to the number of pairs

for a given clamping load. Since the single driven plate clutch has two pairs of surfaces of given

clamping load, since then a twin or triple driven plate clutch for the same spring thrust should

ideally exhibit twice or three times the torque transmitting capacity respectively in comparison to

that the single driven plate unit. Considering the difficulty in dissipating the extra heat generated

in a clutch unit, a safely factor is introduced due to which the torque capacity is generally of the

order 80% per pair of surfaces relative to single driven plate clutch. The increase in number of

pairs of rubbing surface a;sp improves lining life because wear is directly related to the energy

dissipation per unit area.

Design details of sprag and roller clutch

A sprag type clutch resembles a roller bearing, but instead of allowing the elements to roll

freely in both directions the rolling elements are allowed to roll freely in one direction.

Contained with a sprag clutch bearing there are insert elements consisting of cage, sprags and a

spring to preload the friction contact between those sprags and mating parts. The sprag clutch

design shows dt the contact surfaces a sophisticated geometrical shape, the engagement curve

which results in a certain pitch angle with the round mating parts. The engagement curve has to

meet the operating condition (α< = µ) (µ = coefficient of friction). The size of engagement angle

is determined by the applied torque and the force of reaction of the expansion of the mating

parts. The engagement angle creates an angle of twist between outer and inner ring which will

remain equal with constant operating conditions.

UNIT-4 GEAR BOX

Gear train calculations

Epicyclic gear train (simple): An epicyclic gear train consists of an internally toothed

annular (ring) A with band brake encircling it. In the center of this gear is sun gear S which

forms a part of the input shaft. The sun gear and the annular gear are connected by a number of

planet gears P which are mounted on a carrier C and is integral with the output shaft. For

transmission of torque, either the sun gear the carrier or the annular gear must be held stationary.

TA = number of teeth on annular, internal or ring gear

TB = number of teeth on sun or center gear

TP = number of teeth on planet gear

TC = number of effective teeth on arm or planet carrier

TA = TS + 2TP and TC = TS + TA

First gear ratio: 1 + TA/TS = TS + TA/TS

Second gear ratio: 1 + TS/TA = TS + TA/TA

Reverse gear ratio: TA/TS

In the first gear ratio, the annular gear is held stationary and the planet carrier is driven by the

power supplied to sun gear. In second gear ratio the sun gear is held stationary, the planet carrier

is driven member and the annular gear is driven by the sun gear to which the power is applied.

Overdrive: TA/TA + TS

Epicyclic gear train (compound): A simple epicyclic gear train cannot provide adequate velocity

ratios. Therefore a compound epicyclic gear train is used in a gear box to give higher velocity

ratios and to allow several ratios to be obtained. A compound gear train is obtained by joining

together all the arms of simple gear train

Overdrive: = PL (PL + PS + S)/PL (PL + PS + S) + PSS

S = number of annulus gear teeth

PS = number of small planet gear teeth PL = number of large planet gear teeth

Layout of gear boxes

Sliding mesh gear box:

It comprises of clutch gear fixed to the end of clutch shaft and an main shaft which is

splined to accommodate the first and the reverse gear, second and the top speed gear. The gears

have sliding as well as rotary motion. Lay shaft or the countershaft with the gears comprises one

forging. An idler gear is used to reverse the direction of rotation of the main shaft. The main and

the lay shaft are supported on ball bearing pressed in the gear box casing. SAE 90 grade of oil is

filled in the casing up to the desired level for lubrication purpose. The desired gear is engaged by

moving the gear lever, the end of which fits in a slot made in selector fork. The spring loaded

ball resists the movement of selector fork

First gear: The first gear is slid to the left to engage it with gear. The drive from the clutch shaft

passes to the gear 2 of the lay shaft and then through the first seed gears to the main shaft.

G1 = N1/N2 x N5/N6 = T2/T1 x T6/T5

Second gear: The gear 3 is slid to the right to mesh with gear of the lay shaft. The drive from the

clutch shaft passes to gear 2 and then to the main shaft through gears 4 and 3.

G2 = N1/N2 x N4/N3 = T2/T1 x T3/T4

Top gear: The gear is slid to the left so that it directly meshes with the clutch gear. The dog

clutches on gears 1 and 3 meshes with each other to accomplish direct drive.

G3 = 1:1

Reverse gear: The reverse gear 6 on main shaft is slid to the right such that it meshes with the

idler gear 7.

Gr = N1/N2 x N8/N7 x N7/ N6 = T2/T1 x T7/T8 x T6/T7

Constant mesh gear box:

The sliding mesh gear box gives high mechanical efficiency but has certain disadvantages like

gear noise, rough in operation. The constant mesh gear box uses helical gears for quiet operation.

The main shaft is splined and carries gears mounted on bushes. These gears are in constant mesh

with the lay shaft gears. Thrust washes located between gears and casing resist the axial bushes.

The sliding dog is slid by the selector forks to the left or right to obtain in the required gear ratio.

The forks have internal splines and so can move over the splined main shaft. The sliding dog

clutch is positive locking device whose purpose is to allow the power flow from the primary

shaft to the output shaft when the friction clutch has disengaged the gear box from the engine.

The dog cutch has an inner and outer hub. The inner hub contains both internal and external

splines and is fixed to the output main shaft through internal splines. The outer hub carries a

single groove formed round the outside to position a selector fork and is internally splined to

mesh with the splines of the inner hub.

First gear: The sliding dog B is moved to the left to engage with the first speed gear. The drive

from the clutch gear passes to the main shaft.

Second gear: The sliding dog A is slid to the right to engage with the second speed. This locks

the second gear to main shaft.

Top gear: The dog A is now moved to the left so that it meshes with the clutch gear directly.

Reverse gear: The dog B is slid to the right to engage with the reverse gear on main shaft. The

idler gear makes the main shaft rotate in a direction opposite to the clutch shaft.

Synchromesh gear box:

Synchro means nearly equal i.e. the speed of the main shaft gear and the dog clutch must be

nearly the same when the dog clutch must be nearly the same when the dog clutch is being

engaged with any of the main shaft gear. Double declutch is eliminated. It comprises of a toothed

ring or sleeve with internal splines, an internal cone which fits into the external toothed ring a

spring loaded ball which maintains pressure on the toothed spring and a cone which is tapered to

match the surface of the internal cone. Dogs form an integral part of this cone. The device is

usually placed on the second and the top gear in the case of a 3 speed gear box. The device is slid

towards the desired gear wheel by selection forks having their ends in the recess in the sleeve.

The friction between the two adjusts the speed of gear wheel to a suitable value. A little pressure

on the gear wheel allows the sleeve to override and mesh positively with the dogs on the gear

wheel. The equalizing speeds between the cones depend upon spring pressure, cone angles,

coefficient of friction between the two and depth of groove in the sleeve. Synchronization would

take a longer time if any of these factors are disturbed. Too fast a movement of the gear lever

causes clashing of gears.

Calculation of bearing loads and selection of bearings

To compute bearing loads, the forces which act on the shaft being supported by the bearing

must be determined. Loads which act on the shaft and its related parts include dead load of the

rotator, load produced when the machine performs work, and load produced by transmission of

dynamic force. These can theoretically be mathematically calculated, but calculation is difficult

in many cases. A method of calculating loads that act upon shafts that convey dynamic force,

which is the primary application of bearings.

The mean bearing load, Fm, for stepped loads is calculated from formula F1, F2 ……. Fn are the

loads acting on the bearing; n1, n2…….nn and t1, t2……. tn are the bearing speeds and

operating times respectively.

Fm = (∑ (Fip

ni ti)/ ∑ (ni ti)) 1/p

The mean load for continuously fluctuating load is Fm = (1/to ∫ F (t) p

dt) 1/p

The mean load for linear fluctuating load, Fm, can be approximated by formula

Fm = Fmin + 2Fmax /3

The dynamic equivalent radial load is expressed by formula Pr = XFr + YFa

Pr = Dynamic equivalent radial load, N {kgf}

Fr = Actual radial load, N {kgf}

Fa = Actual axial load, N {kgf}

X = Radial load factor

Y = Axial load factor

The dynamic equivalent axial load for these bearings is given in formula Pa = Fa + 1.2Fr

Pa = Dynamic equivalent axial load, N {kgf}

Fa = Actual axial load, N {kgf}

Fr = Actual radial load, N {kgf}

The bearings mostly used in gear boxes are roller ball bearing, cylindrical roller bearing and

tapered roller bearing, spherical roller bearing.

The factors to be considered while designing a bearing for gear box are:

Load conditions

Speed of rotation in bearing

Shaft arrangements

Friction

Bearing life

Lubrication in bearing

Environmental conditions

Strength of connecting parts

Design of 3 speed gear box

1. An automotive gear box gives 3 forward speeds and one reverse with a top gear of unity and

bottom and reverse gear ratio of approximately 3.3:1. The center distance between the shafts

is to be 110 mm approximately. Gear teeth of module 3.25 mm. find the number of gear

teeth.

Given:

Center distance = 110 mm

Solution:

Reverse gear ratio = TB/TA x TJ/TI

Since the pitch is same for all wheels and the center distance is the same for all pair of

mating wheels, the total number of teeth must be same for each pair.

TA + TB = TC + TD = TE + TF =110 x 2/3.25 = 67.6 = 68

In general, the gear ratios are kept in a geometric progression, If G1, G2, G3 are 1st , 2

nd and

3rd

or top gear ratios respectively, then G =√G1.G3 = √1 x 3.3 = 1.817

First gear ratio, G1 = TB/TA x TD/TC = 3.3

Adopting the relation, TB/TA = TD/TC = √3.3 = 1.817

Hence TA + TB = 2.817 TA =60 so TA = 68/2.187 = 24.1 = 24

TB = 68 – 24 = 44 adopted

TC = 24 and TD = 44 adopted

Exact speed reduction, G1 = (44/24)2 = 3.36:1

Second gear ratio, G2 = TB/TA x TF/TE = 1.817

TF/TE = 1.817(TA/TB) = 1.817 x 24/44 = 0.991

TE = 68/1.991 = 34.05 = 34 adopted

TF = 68 – 34 =34 adopted

G2 = 34/34 x 44/24 = 1.83:1

Top gear ratio G3 = 1:1

Reverse gear ratio:

Presence of an idler gives TI + TJ < 68

Speed ratio = TB/TA x TJ/TI = 3.3 approx.

Adopting TI = 22 and TJ = 40

Exact reduction = 44/24 x 40/22 = 3.33:1

2. Then maximum gear box ratio of an engine 75 mm bore and 100 mm stroke is 4. The

pitch diameter of the constantly meshing gear is 75% of piston stroke. If the module is

4.25 mm. Calculate the size and number of teeth gears for a 3 speed gear box. Calculate

the face width of the constantly meshing gear using the modified Lewis formula. The

engine torque is 910kgcm and allowable stress is 900 kgf/cm2

Given:

bore diameter = 75 mm Engine torque, Te = 910 kgfcm

Stroke length = 100 mm allowable stress = 900kg/cm2

Solution:

G1 = 4 and G3 = 1

Taking gear ratios in geometrical progression, G2 = √G1.G3

First gear ratio, G1 = TB/TA x TD/TC = 4 giving TB/TA = TD/TC = √4 = 2

Adopting TA = TC = 16 to avoid interference the TB = TD = 32

TA + TB = TC + TD = TE + TF = 48

Pitch diameter of constantly meshing gear i.e. gear A = 0.75 x 100= 75 mm

Pitch diameter of pinion C = module x number of teeth = 4.25 x 16 = 68 mm

Gear D and pinion B = 4.25 x 32 = 136 mm

Second gear ratio, G2 = TB/TA x TF/TE = 2

TF/TE = 2 (TA/ TB) = 2 (15/3) = 1

TE = TF = 24 adopted.

Pitch diameter of pinion E and gear F = 4.25 x 24 = 102 mm

G3 = 1:1

Te = FD/2000 89.27 = F x 75/2000 so that

F= 89.27 x 200/75 = 2380.5 N

Modified lewis formula gives, F = (cfb/1000) (m/1000)

C = form factor

fb = allowable stress (N/m2)

b = face width of gear (mm)

m = module of gear (mm)

2380.5 = 0.07 x 8829 x 104 x b/1000 x 4.25/1000

b = 2380.5 x 2000/0.007 x 8892 x 104 x 4.25 Face width, b = 90.6 mm

Design of 4 speed gear box

1. Sketch a section through a sliding gear type box with 4 forward and one reverse speeds and

explain clearly how the different speed ratios will be explained in the following cases.

Gear ratio on top gear = 1:1

Gear ratio on third gear = 1.38:1

Gear ratio on second gear =2.24:1

Gear ratio on first gear = 3.8:1

Gear ratio on reverse gear = 3.8:1

Assume counter shaft or layout shaft speed is half that of the engine speed and smallest gear is

not to have less than 15 teeth.

Given:

G1 = 3.8:1 G3 = 1.38:1 G2 = 2.24:1 G4 = 1:1

Solution:

First gear ratio, G1 = TB/TA x TD/TC = 3.8

Speed of layout is half of engine shaft then NA/NB = TB/TA = 2

TD/TC = 3.8 TA/ TB = 3.8/2 = 1.9

TA + TB = TC + TD = TE + TF = TG + TH

3TA = 2.9 TC TA = 29 adopted

TC = 30, TB = 29 x 2 = 58 and TD = (TA + TB) - TC

= 87 – 30 = 57 adopted

G1 = 58/29 x 57/30 = 3.8:1

Second gear ratio, G2 = TB/TA x TF/TE = 2.24

TF/TE = 2.24 TA/ TB = 2.24/2 = 1.12

TE + TF = 87 = 2.12 TE TE = 87/2.12 = 41.5 = 41 adopted

TF = 87 – 41 = 46 adopted

G2 = 58/29 x 46/48 = 2.24:1

Third gear ratio, G3 = TB/TA x TH/TG = 1.38

TH/TG = 1.38 TA/ TB = 1.38/2 = 0.69

TG + TH = 87 = 1.69TG

TG = 87/1.69 = 51.5 = 51 adopted

TH = 87 – 51 = 36 adopted

G3 = 58/29 x 36/51 = 1.41:1

G4 = 1:1

Reverse gear, Gr = TB/TA x TI2/ TC x TD/TI1 = 3.8

58/29 x TI2/30 x 57/ TI1 = 3.8

TI1/ TI2 = 3.82/2 x 30/57 = 1 TI1 = TI2 = 15 adopted

2. A 4 speed gear box is to be constructed for providing the ratios 1.0, 1.46, 2.28 and 3.93 to

1 as nearly as possible. The diametral pitch of each gear is 3.25 mm and the smallest

pinion is to have at least 15 teeth. Determine the suitable number of teeth of different

gears. What is the distance between the main and layout shaft?

Given:

G1 = 3.93:1 G3 = 1.46:1 Pitch = 3.25

G2 = 2.28:1 G4 = 1:1 module = 15 teeth

Solution:

First gear ratio, G1 = TB/TA x TD/TC = 3.8

TB/TA = TD/TC √3.93 = 1.98

Adopting TA = TC = 15 the lowest value given

TB = TD = 1.98 x 15 = 29.7 = 30 adopted

G1 = 30/15 x 30/15 = 4:1

TA + TB = TC + TD = TE + TF = TG + TH = 45

Second gear ratio, G2 = TB/TA x TF/TE = 2.28

TF/TE = 2.28 TA/ TB = 2.28 x 15/30 = 1.14

TE + TF = 2.14 TE = 45

TE = 45/2.14 = 21 adopted and TF = 45 – 21 =24 adopted

G2 = 30/15 x 24/21 = 2.28:1

Third gear ratio, G3 = TB/TA x TH/TG = 1.46

TH/TG = 1.46/2 = 0.73

TG + TH = 45 hence TG = 45/1.73 = 26 adopted

TH = 45 – 26 = 19 adopted

G3 = 30/15 x 19/26 = 1.46:1 G4 = 1:1

Centre distance between the shafts = (3.25 x 45)/2 = 73.12 mm

UNIT-5 DRIVE LINE AND REAR AXLE

Design of propeller shaft

1. An automobile engine develops 28 kw at 1500 rpm and its bottom gear ratio is 3.06. if a

propeller shaft of 40mm outside diameter is to be used, determine the inside diameter of mild

steel tube to be used, assuming a safe shear stress of 55 x 103 kpa for MS.

Given:

P = 28 kw N = 1500 rpm bottom gear ratio = 3.06 do = 40mm fs = 55 x 103 kpa

Solution:

2π x 1500 x T/60000 = 28

T = 28 x 60000/ 2π x 1500 = 178.34 Nm

Torque to transmitted by the propeller shaft shaft = torque of engine x gear ratio

= 178.34 x 3.06 = 545.72 Nm

According to torsion equation T/Ip = fs/y

Moment of inertia Ip = π/32(do4 – di

4) m

4

545.72/ π/32(do4 – di

4) = 55 x 10

6/0.02

(do4 – di

4) = 32 x 0.02 x 545.72/ π x 55 x 10

6 = 2 x 10

6 mm

4

di4

= 404 - 2 x 10

6 = 56 x 10

4 mm

4

di = 27 mm

Design of final drive

A final drive is that part of a power transmission system between the drive shaft and the

differential. Its function is to change the direction of the power transmitted by the drive

shaft through 90 degrees to the driving axles. At the same time, it provides a fixed reduction

between the speed of the drive shaft and the axle driving the wheels. The reduction or gear

ratio of the final drive is determined by dividing the number of teeth on the ring gear by the

number of teeth on the pinion gear. In passenger vehicles, this speed reduction varies from

about 3:1 to 5:1. In trucks it varies from about 5:1 to 11:1. To calculate rear axle ratio, count the

number of teeth on each gear. Then divide the number of pinion teeth into the number of ring

gear teeth. For example, if the pinion gear has 10 teeth and the ring gear has 30 (30 divided by

10), the rear axle ratio would be 3:1. Manufacturers install a rear axle ratio that provides a

compromise between performance and economy. The average passenger car ratio is 3.50:1. The

higher axle ratio, 4.11:1 for instance, would increase acceleration and pulling power but

would decrease fuel economy. The engine would have to run at a higher rpm to maintain an

equal cruising speed. The lower axle ratio 3:1 would reduce acceleration and pulling power

but would increase fuel mileage. The engine would run at a lower rpm while maintaining the

same speed. The major components of the final drive include the pinion gear, connected

to the drive shaft, and a bevel gear or ring gear that is bolted or riveted to the differential

carrier.

Design of semi floating, fully floating and three quarter floating axle

A) Semi‐Floating Axle: This design employs one wheel support bearing mounted on the

outer end of the axle shaft and inside the axle tube. All the wheel forces including vehicle

weight, wheel side skid, wheel traction, and torsional drive, are supported by the axle shaft. This

design is less complicated, lighter weight, less costly, and is used in passenger cars. The diagram on Figure 7, developed by anti‐friction bearing engineers, has been used to

calculate the axle shaft diameter at the wheel bearing location. The analysis consists of summing

the moments around the center of the outer bearing and equating them to the flexure of length, C,

of the axle shaft:

M = (WRB ‐ .6WRRr) = SI/C

M = maximum bending moment in in‐lbs around the center of the outer bearing. WR = maximum rear end weight of the vehicle in pounds. (.6WR = side skid load.) B = axial distance from the center of the wheel to the center of the outer bearing. Rr = radius of tire in inches. SI/C = allowable axle shaft stress times section modulus.

From the above information, the diameter of the axle shaft is calculated as follows:

D = 2.168 (M/S)1/3

D = the diameter of the axle shaft at the outer bearing center in inches.

B) Three‐Quarter Floating Axle: This design uses one wheel support bearing mounted on

the outer end of the axle tube. The wheel forces due to vehicle weight and tractive effort are

supported by the axle tubes. The vehicle force due to wheel side skid loads and torsional drive

forces are supported by the axle shaft which is then said to be “three‐quarter floating”. This

design has limited application in the automotive industry. Figure 8 has a sketch of a three‐quarter floating axle design. The following equation sums the

moment loading around the outer bearing center which is the result of the side skid load only:

M = .6WRRr

M = moment load in inch‐pounds around outer bearing center. .6WR = maximum side skid load in pounds. Rr = radius of tire in inches. Using the same analysis as was done for the semi‐floating axle above; the diameter of the axle

shaft is as follows:

D = 2.168 (M/S)1/3

D is the diameter of the axle shaft at the outer bearing in inches. M = the moment load in inch pounds calculated above.

S = the maximum allowable shaft stress in inch‐pounds.

In this three‐quarter design configuration, the axial shaft supports the side skid load only. The

radial load is supported by the axle tube.

C) Full‐Floating Axle: In this design, the wheels are supported by two bearings that mount

on the outer end of the axle tubes. Wheel forces due to vehicle weight, vehicle tractive effort, and

side skid load are reacted by the axle tubes. The outer end of the axle shaft is unsupported or

“floats” in the axle tube and transmits torsional drive forces only. This design is used for heavy

duty applications. Figure 9 has a sketch of the full‐floating axle design. The axle shaft supports torsional loads only

while the axle tube supports all other loads. The equation is as follows:

S = Q1/(π/16)D3

Rearranging:

D = 1.721(Q1/SS)1/3

Q1 = One‐half maximum low gear torque in inch‐pounds.

Calculation of axle shaft

The following forces act on a moving wheel:

· The torque due to the traction or braking force (Tw and Tb)

· The traction or braking force (Fw and Fb)

· The lateral force Fy when the vehicle makes a turn or skid

· The normal reaction Rw

Simultaneous appearance of maximum longitudinal and transverse forces at the wheel road

contact is not possible, for joint action is restricted by the adhesion force

The loading conditions of axle shafts and beams reduce to the following three cases:

1- Rectilinear motion

The longitudinal force (Fw or Fb) attain its maximum value equal to Rw φ, Maximum torque is

where:

Tw = wheel torque

Te max = maximum engine torque

ig = gearbox ratio (1st gear)

if = final drive ratio

kd = dynamic factor

kl = the coefficient of differential locking

ma = automobile mass accounted for the driving axle

g = 9.81 m/s2

wt = transferred weight

where

φ = coefficient of adhesion (0.8)

* In this case, Fy = 0

2- Skidding of automobile

In this case a lateral force and normal reaction are acting on the wheel. Assume that the

longitudinal force Fw = 0. The largest lateral force- centrifugal force- whose value is limited by

the wheel-road grip equals

The vertical reactions and lateral forces of the inner and outer wheels are

where

v = vehicle speed (km/h)

R = radius of turn of the road

t = wheel track width

φ = coefficient of road adhesion during sidewise skidding =1.0

+ = plus sign is used for the axle shaft of the wheel which is inner relative to the skidding

direction, and the negative sign, the outer wheel.

3-Driving wheels overcome irregularities

Here, only the vertical force is accounted for

where

kdr = is the dynamic factor of road;

for cars, kdr = 1.75; for trucks, kdr= 2.50

-The axle shaft dimensions are determined for the most dangerous case of loading. For s semi-

floating axle the dangerous cress section lies in the bearing installation zone. For the first

condition, the equivalent stress due to bending and torsion is

where

d = the axle shaft diameter

b = the overhanging length

During skidding the following bending moments and stresses act on the axle shaft

Mi = Rw i b – Fy i rw;

Mo = Rw o b + Fy o rw

where

rw = wheel radius

(the upper sings are used for the inner axle shaft, and the lower sign, for the outer axle shaft

relative to the skidding direction).

-When the driving wheels overcome an irregularity, the bending stress is

- The floating axle is calculated only for torsion at the maximum traction force

The axle shaft is calculated also for the maximum twist angle

where

L = the length of the axle shaft

G = the shear modulus

J = the moment of inertia of the cross section of axle shaft

sb = 55 MPa for shafts without keyway

40 MPa for shafts with keyway

* The permissible twist angle is θ = 8o for 1 m length of the shaft.

# Number of splines of the axle shaft is form 10 (for cars) to 18 (for trucks)

# the shaft factor of safety = 2.0- 2.5