UNIT-1 VEHICLE FRAME AND SUSPENSION

Transcription

1 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.

2 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.

3 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 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. 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) = = R1 = /180= 197 kgf R2 = = =500.5 kgf. From maximum bending moment, = kgfcm f = 600kgf/cm 2 ( 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 = 64.1cm 3 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

4 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. 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 = W W = ( )/1.5= 10kN For equilibrium total upward forces must be equal to total downward forces Ra + Rb = Total downward forces 8.5+ Rb = 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) = l 1.l 2 E = Young s modulus (pa) W = width of the spring (m) I = moment of inertia = wt 3 /12 (m 4 ) n = number of leaves Deflection, δ = 2.85Pl 3 /Enwt (m) t = thickness of spring (m)

5 Initial radius, R = l 2 /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/wt 2 fb fb = bending stress (pa) rate of spring, r = Enwt 3 l 3 /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 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. P = 4905N, l = 900 mm, w = 75 mm, fb = kpa, E = x 10 6 kpa Solution: Number of leaves, n = 3pl/wt 2 fb n = 3 x 4905 x 0.9/0.075 x x x 10 3 n=8 Deflection, δ = 2.85Pl 3 /Enwt δ = 2.85 x 4905 x x 10 3 /192.2 x 10 9 x x 8 x δ = mm intial radius, R = l 2 /8Z + Z/2 = l 2 /8 δ + δ/2 R= /8 x /2 R = 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

6 fs = allowable stress in the spring (pa) G = modulus of rigidity = x 10 6 pa Deflection, δ = 8nPD 3 /Gd 4 (m) Shear stress, fs = Gd δ/πd 2 n (N/m 2) Energy stored in the spring = P δ/2 (Nm) Volume of steel in the spring = πd 2 Dn/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 N. Calculate the shear stress and the deflection. If the maximum shear stress of kpa is allowable in the material then what is the possible clearance. N = 10, d = 15 mm, fsmax = kpa, D = 125 mm, P = N, G = x 10 3 kpa Solution: δ = 8nPD 3 /Gd 4 = 8 x 10 x x x 10 3 /73575 x 10 6 x δ = 148 mm fs = Gd δ/πd 2 n = x 10 3 x x 0.148/3.14 x 15 2 x 10 fs = kpa δmax = πd 2 fsma x /Gd = π x 1252 x 10 x /73575 x δmax = 284 mm Clearance = = 126 mm.

7 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 = πd 4 /32 (m 4 ) Angular deflection, θmax = 5760TL/π 2 d 4 G (rad) L = length of bar (m) Diameter of bar, d = 3 16Tmax / π fsmax (m) fs = G θd/2l (pa) load rate, r = πd 4 G + 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 N 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 kpa. Calculate the diameter of torsion bar, the effective length and the load rate. L = 250 mm, α = 30 degree, θ = 90 degree, fs = kpa Dynamic load = 2 x static load = 2 x = 6867 N Effective length of the arm = lcos30 = 250 x = mm Solution: Moment acting on the bar, T = Wy = 6867 x = Nm. Diameter of the bar, d = 3 16Tmax / π fsmax d = 3 16 x /π x x 10 3

8 d = 3.12/10 2 x 10 3 d = 31.2 mm length of the bar, L = π x x 10 6 x 90 x 31.2 x 10 3 /360 x x 10 3 L = 2.3 m Deflection above the horizontal, X = lsin30 = 25 x 0.5 = 12.5 cm. r = πd 4 G + 32LW X /32L (l 2 - X 2 ) r = π x x x x 2.3 x 6867 x 0.125/32 x 2.3( ) r = /3.45 = N/m

9 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)

10 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/m 2 ) y=d/2 (m) I = bd 3 -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/I p = 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) I p = (π/32) d 4 for circular section = (π/32) d 3 b 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 R w = the reaction of the wheel on the spindle acting vertically through the center of contact of tyre on ground R t = the load on thrust bearing R u = the load on the upper bearing (knuckle pin) R t = 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.

11 Then M c = 0 gives R w c - R l (d + e) = 0 R l = c/d + e (R w ) M B = 0 gives R w a - R u (d + e) = 0 R u = a/a + e (R w ) M A = 0 gives R t b - R l e - R u d = 0 R t b = R l e + R u d Substitution the expression of R l and R u gives R t = ce + ad/b (d + e) (R w ) 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. 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 = 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 d 3 ) ½(0.45d x (06d) 3 ) =1/12 0.6d 4 ( x ) = 0.6 x 0.838/12 (d 4 ) = d 4 cm / d 4 = 915/d/2

12 d 3 = 12136/ 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 = 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.

13 Problems 1. A motor car has wheel base of m and pivot center of m. The front and rear wheel track is 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. Wheel base b = m Pivot center c = m Wheel track a = m Solution: For correct steering, cot φ cot θ = c/b Cot φ = 1.065/ cot 40 = = φ = 32.4 Turning circle radius of the front wheel (outer), R ot = b/sin φ + a c/2 = 2.743/ sin ( )/2 = 2.743/ /2 = m Turning circle radius of inner wheel (rear), R ir = b/sin θ - a c/2 = 2.743/sin = ( x 2.743) = 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. 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 = /2 x 0.17 sin α = 0.178, α = degree

14 sin (α + θ) + sin (α - φ) = 2 sin α sin ( ) + sin (16.12 φ) = 2 x = = = sin (-9.44) Φ = = For correct steering, cot φ cot θ = c/b cot cot 30 = 1.37/b = 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 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. Pivot center = 1.3 m Length of track arm = 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 = sin α = c d /2r = /2 x = 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) = = sin (-5.8) φ = = 25 degree cot φ cot θ = cot 25 cot 30 = = putting, θ = 35 degree, sin (19.2 φ) = = sin (-8.9)

15 φ = = 28.1 degree cot φ cot θ = cot 28.1 cot 35 = = putting, θ = 37 degree, sin (19.2 φ) = = sin (-10.05) φ = degree cot φ cot θ = cot cot 37 = = putting, θ = 39 degree, sin (19.2 φ) = = sin (-11.2) φ = 30.4 degree putting, θ = 40 degree, sin (19.2 φ) = = sin (-11.7) φ = 30.9 degree cot φ cot θ = cot 30.9 cot 40 = = 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, R if = b/ sin θ a c/2 = 2.85/ sin 37 = /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

16 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

17 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. r 1 and r 2 = 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 (r r 2 1 ) (N/m 2 ) Total frictional torque, T = 2/3µW (r r 3 1 )/ r r 2 1 ) (Nm) Uniform rate of wear: Total axial load, W = 2πC (r 2 - r 1 ) (N) Pr = C, if the rate of wear is assumed to be constant. Frictional torque, T = µwr (Nm) R = mean radius of friction surface = r 1 + r 2 /2 For a single plate clutch having a pair of contact surfaces, T = µw (r 1 + r 2 ) (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 I B ω 2 (Nm) I B = mass moment of inertia attached to driven shaft of the clutch ω = angular speed Energy dissipated due to clutch slip = Tωt Iω 2 /2g (Nm) Problems 1. An automobile power unit gives a maximum toque of Nm. The clutch is a single plate dry disc having effective clutch lining of both sides of plate disc. The coefficient of

18 friction is 0.3 and the maximum axial pressure is 8.29 x 10 4 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. T = Nm p max = 8.29 x 10 4 µ = 0.3 r 2 = 1.25r 1 Solution: T = µπc (r r 2 1 ) C = pr 1 = 8.29 x 10 4 r = 2 x π x 0.3 x 8.29 x r 1 ( r 1 - r 2 1 ) r 3 1 = 13.56/2 x π x 0.3 x 8.29 x 10 4 x r 1 = 53.6 mm r 2 = 1.25 x 53.6 = 67 mm Total axial pressure, W =2πC (r 2 - r 1 ) W = 2 x π x 8.29 x 10 4 r 1 (r 2 r 1 ) = 2 x π x 8.29 x 10 4 x ( ) = 2 x π x 8.29 x 5.36 x 1.34 W = 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 10 4 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 N = 2100 rpm P = 5.9 kw

19 Solution: T = p w 6000/2πN = x 5.9/2π x 2100 = nm T = x 1.35 = Nm T = πµc (r r 2 1 ) = 2 x π x 0.3 x 6.84 x 10 4 x ((r 1 /0.55) 2 - r 2 1 ) 2 = π x x (1/ ) r 1 r 3 1 = x 0.303/ π x x 10 4 x = 49.5 mm r 1 = 49.5 mm r 2 = 90 mm 3. A friction clutch is required to transmit 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 10 4 pa. If the outer diameter of the plate is to be m. Find the required inner diameter of the clutch ring and the total force exerted by the springs. P = kw p w = 6.87 x 10 4 pa N = 2000 rpm d 2 = m Solution: T = p w 60000/2πN T = x 33.12/2 x π x 2000 T = Nm T =2πµC (r r 2 1 ) = 2 x π x 0.3 x 6.87 x 10 4 r 1 ( r 2 1 ) = r 1 ( r 2 1 ) r = r r = By trial we get r 1 = 112 mm or 63.5 mm Inner diameter d 1 = 224 mm or 127 mm When r 1 = m and r 1 = m W = 2πC (r 2 - r 1 ) = 2 x π x 6.87 x 10 4 x x = 1958 N When r 1 = m and r 2 = m

20 W = 2πC (r 2 - r 1 ) = 2 x π x 6.87 x 10 4 x x = 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 N. The outside diameter of friction surface is 250 mm. Both sides of the plate are effective and the µ between the contact surfaces is 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 Nm 2. Calculate the time lapse before the engine attains full speed of 2100 rpm. P = 22.1 kw W = N I = 20.7 Nm 2 N = 2100 rpm d 2 = 250 mm Solution: T = p w /2πN = x 22.1/2 x π x 2100 = Nm T = µw r 2 + r 3 /2 x = x (r 2 + r 1 ) r 1 = / x 0.35 = 202 mm r 1 = = 77 mm T = Iω/gt = 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. 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 3 T = π/16 x fs x d s

21 = π/16 x 420 x d s Diameter of shaft, d s = 25mm Area of friction faces, A = 2πRb W = A x p = 2πRb T = µwrn = π/2µr 3 pn = π/2 x 0.25 x R 3 x 0.07 x 2 = 0.055R 3 R 3 = 79600/0.055 = 1.45 x 10 6 R = 114 mm b = R/4 = 28.5 mm b = r 2 - r 1, r 2 - r 1 = 28.5 mm R = r 1 + r 2 /2 r 1 + r 2 = 2 x 114 = 228 mm r 1 = mm r 2 = mm W = 2πRbp = 2 x π x 114 x 28.5 x 0.07 W = N Total load on springs = 1.25W = 1.25 x = N Maximum load on each spring, Ws = /6 = N Wahls stress factor, K = 4C 1/4C /C = 4 x 6 1 /4 x /6 = Maximum shear stress, 420 = K x 8WsC/πd 2 = x 8 x x 6/ πd 2 d 2 = 5697/420 = = 3.68 mm D = Cd = 6 x = mm δ = 8WsC 3 n/gd = 8 x x 6 3 x 4/84 x 10 3 x = 6.03 mm total no. turns, n = n + 2 = =6 free length of spring, l = n d + δ δ = 6 x x 6.03 = mm Design of multi plate clutch For multi plate clutch having n pair of contact surfaces, T = µwn (r 1 + r 2 )/2 Nm To have n pair of contact surfaces, there must be n + 1 number of discs or plates. If there are n A number of discs on the driving shaft and n B number of discs on the driven shaft, then number of pairs of contact surfaces are n = n A + n B 1

22 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. n A = 3 n = = 4 p w = 25 kw n B = 2 µ = 0.3 N = 1475 rpm r 2 = 120 mm r 1 = 60 mm Solution: ω = 2 x π x 1575/60 = 52.5π rad/s p w = Tω T = p w /ω = 25 x 10 3 /52.5 x π = Nm for uniform pressure condition, T = n2/3µw(r 2 - r 1 / r 2 - r 2 1 ) = 4 x 2/3 x 0.3W ( / ) 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 10 3 N/m Reduction in spring force = total wear x no. of springs x stiffness per spring = 0.01 x 6 x 13 x 10 3 = 780 N New axial load = = 575 N Uniform wear condition, T = nµw (r 2 + r 1 /2) = 4 x 0.3 x 575 ( /2) = 62 Nm P w = Tω = 62.5 x 52.5π = watt = 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

23 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 r 1 = outer radius of the cone r 2 = inner radius of the cone w = breadth of the cone face = r 1 r 2 /sin θ W n = normal load on frictional surface R n = normal reaction µr n = frictional force Area of the ring = 2πrdw Normal load on ring, dp p2πrdr/sin θ 2 Considering uniform pressure, p = W/π (r 2 - r 2 1 ) 3 T = 2/3µW/sin θ (r 2 - r / r 2 - r 2 1 ) Considering uniform rate of wear, pr = C T = 2πµPR 2 W (Nm) Power capacity of the clutch = 2πNT/6000 (kw) Axial force to engage the clutch = W n sin θ + µw n cos θ Axial force to disengage the clutch = µw n cos θ - W n sin θ Force required for effective normal pressure for steady state operation = W n sin θ = R n sin θ Energy supplied = Tωt (Nm) Energy of flywheel = Iω 2 /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 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 Nm 2.

24 θ = 15 degree, µ = 0.3, W = N, N = 1200 rpm, I = 3.4Nm 2 Solution: d m = d 2 + d 1 /2 T = µ (r 2 + r 1 ) 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 = rad/s 2 t = w/ α = 2πN/60 α = 2 x π x 1200/60 x T = 4.11 sec Energy supplied = Tωt =t (2πN/60) T = 106 x 2π x 1200 x 4.11/60 = Nm Energy of flywheel = Iω 2 /2g = 3.4/2 x 9.81 (2 x π x 1200/60) 2 = Nm Energy lost during slip = = 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 1/4 th of mean diameter and the normal pressure between the contact faces is not to exceed 8.27 x 10 4 pa. Allowing the coefficient of the clutch and the axial force required. P = 11.9 kw θ = 12 degree µ = 0.2 N = 750 rpm p = 8.2 x 10 4 pa T= p w 60000/2πN = x 11.9/2 x π x 750 = Nm W = ¼ x d 2 + d 1 /2 = r 2 + r 1 /4 = r 2 r 1 /sin θ = r2 r 1 / sin 12 4 (r2 r1) = sin 12 (r 2 + r 1 ) = (r 2 + r 1 ) 3.792r 2 = 4.208r 1 r2 = (4.208/3.792) r 1

25 r2 = 1.11r1 T = µπc/sin θ (r 2 2 r 2 1 ) = µπpr1/sin θ (r 2 2 r 2 1 ) = (0.2π x 8.27 x /sin 12) r 1 ( ) r 1 = (0.2π x 8.27 x x 0.231/0.208) r 1 3 = r 1 r 3 1 = 142.5/ = r 1 = m r 2 = 1.11 x = 0.14 m w = /4 = m W = 2πC (r 2 r 1 ) = 2πp (r 2 r 1 ) r 1 = 2π x 8.27 x 10 4 x x W = 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

26 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

27 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. T A = number of teeth on annular, internal or ring gear T B = number of teeth on sun or center gear T P = number of teeth on planet gear TC = number of effective teeth on arm or planet carrier T A = T S + 2T P and T C = T S + T A First gear ratio: 1 + T A /T S = T S + T A /T S Second gear ratio: 1 + T S /T A = T S + T A /T A Reverse gear ratio: T A /T S 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: T A /T A + T S 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: = P L (P L + P S + S)/P L (P L + P S + S) + P S S S = number of annulus gear teeth P S = number of small planet gear teeth P L = number of large planet gear teeth Layout of gear boxes Sliding mesh gear box:

28 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. G 1 = N 1 /N 2 x N 5 /N 6 = T 2 /T 1 x T 6 /T 5 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. G 2 = N 1 /N 2 x N 4 /N 3 = T 2 /T 1 x T 3 /T 4 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. G 3 = 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. G r = N 1 /N 2 x N 8 /N 7 x N 7/ N 6 = T 2 /T 1 x T 7 /T 8 x T 6 /T 7

29 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:

30 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.

31 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. p Fm = ( (F i n i t i )/ (n i t i )) 1/p The mean load for continuously fluctuating load is Fm = (1/t o 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

32 is to be 110 mm approximately. Gear teeth of module 3.25 mm. find the number of gear teeth. Center distance = 110 mm Solution: Reverse gear ratio = T B /T A x T J /T I 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. T A + T B = T C + T D = T E + T F =110 x 2/3.25 = 67.6 = 68 In general, the gear ratios are kept in a geometric progression, If G1, G2, G3 are 1 st, 2 nd and 3 rd or top gear ratios respectively, then G = G1.G3 = 1 x 3.3 = First gear ratio, G1 = T B /T A x T D /T C = 3.3 Adopting the relation, T B /T A = T D /T C = 3.3 = Hence T A + T B = T A =60 so T A = 68/2.187 = 24.1 = 24 T B = = 44 adopted T C = 24 and T D = 44 adopted Exact speed reduction, G1 = (44/24) 2 = 3.36:1 Second gear ratio, G2 = T B /T A x T F /T E = T F /T E = 1.817(T A /T B ) = x 24/44 = T E = 68/1.991 = = 34 adopted T F = =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 T I + T J < 68 Speed ratio = T B /T A x T J /T I = 3.3 approx. Adopting T I = 22 and T J = 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

33 the face width of the constantly meshing gear using the modified Lewis formula. The engine torque is 910kgcm and allowable stress is 900 kgf/cm 2 bore diameter = 75 mm Engine torque, Te = 910 kgfcm Stroke length = 100 mm allowable stress = 900kg/cm 2 Solution: G1 = 4 and G3 = 1 Taking gear ratios in geometrical progression, G2 = G1.G3 First gear ratio, G1 = T B /T A x T D /T C = 4 giving T B /T A = T D /T C = 4 = 2 Adopting T A = T C = 16 to avoid interference the T B = T D = 32 T A + T B = T C + T D = T E + T F = 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 = T B /T A x T F /T E = 2 T F /T E = 2 (T A / T B ) = 2 (15/3) = 1 T E = T F = 24 adopted. Pitch diameter of pinion E and gear F = 4.25 x 24 = 102 mm G3 = 1:1 Te = FD/ = F x 75/2000 so that F= x 200/75 = N Modified lewis formula gives, F = (cfb/1000) (m/1000) C = form factor fb = allowable stress (N/m 2 ) b = face width of gear (mm) m = module of gear (mm) = 0.07 x 8829 x 10 4 x b/1000 x 4.25/1000 b = x 2000/0.007 x 8892 x 10 4 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.

34 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. G1 = 3.8:1 G3 = 1.38:1 G2 = 2.24:1 G4 = 1:1 Solution: First gear ratio, G1 = T B /T A x T D /T C = 3.8 Speed of layout is half of engine shaft then N A /N B = T B /T A = 2 T D /T C = 3.8 T A / T B = 3.8/2 = 1.9 T A + T B = T C + T D = T E + T F = T G + T H 3T A = 2.9 T C T A = 29 adopted T C = 30, T B = 29 x 2 = 58 and T D = (T A + T B ) - T C = = 57 adopted G1 = 58/29 x 57/30 = 3.8:1 Second gear ratio, G2 = T B /T A x T F /T E = 2.24 T F /T E = 2.24 T A / T B = 2.24/2 = 1.12 T E + T F = 87 = 2.12 T E T E = 87/2.12 = 41.5 = 41 adopted T F = = 46 adopted G2 = 58/29 x 46/48 = 2.24:1 Third gear ratio, G3 = T B /T A x T H /T G = 1.38 T H /T G = 1.38 T A / T B = 1.38/2 = 0.69 T G + T H = 87 = 1.69T G T G = 87/1.69 = 51.5 = 51 adopted T H = = 36 adopted G3 = 58/29 x 36/51 = 1.41:1 G4 = 1:1 Reverse gear, Gr = T B /T A x T I2 / T C x T D /T I1 = /29 x T I2 /30 x 57/ T I1 = 3.8

35 T I1 / T I2 = 3.82/2 x 30/57 = 1 T I1 = T I2 = 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? 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 = T B /T A x T D /T C = 3.8 T B /T A = T D /T C 3.93 = 1.98 Adopting T A = T C = 15 the lowest value given T B = T D = 1.98 x 15 = 29.7 = 30 adopted G1 = 30/15 x 30/15 = 4:1 T A + T B = T C + T D = T E + T F = T G + T H = 45 Second gear ratio, G2 = T B /T A x T F /T E = 2.28 T F /T E = 2.28 T A / T B = 2.28 x 15/30 = 1.14 T E + T F = 2.14 T E = 45 T E = 45/2.14 = 21 adopted and T F = =24 adopted G2 = 30/15 x 24/21 = 2.28:1 Third gear ratio, G3 = T B /T A x T H /T G = 1.46 T H /T G = 1.46/2 = 0.73 T G + T H = 45 hence T G = 45/1.73 = 26 adopted T H = = 19 adopted G3 = 30/15 x 19/26 = 1.46:1 G4 = 1:1 Centre distance between the shafts = (3.25 x 45)/2 = mm

36 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 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 10 3 kpa for MS. P = 28 kw N = 1500 rpm bottom gear ratio = 3.06 do = 40mm fs = 55 x 10 3 kpa Solution: 2π x 1500 x T/60000 = 28 T = 28 x 60000/ 2π x 1500 = Nm Torque to transmitted by the propeller shaft shaft = torque of engine x gear ratio = x 3.06 = Nm According to torsion equation T/Ip = fs/y Moment of inertia Ip = π/32(do 4 di 4 ) m / π/32(do 4 di 4 ) = 55 x 10 6 /0.02 (do 4 di 4 ) = 32 x 0.02 x / π x 55 x 10 6 = 2 x 10 6 mm 4 di 4 = 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

37 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 = (W R B.6W R R r ) = SI/C M = maximum bending moment in in lbs around the center of the outer bearing. W R = maximum rear end weight of the vehicle in pounds. (.6W R = side skid load.) B = axial distance from the center of the wheel to the center of the outer bearing. R r = 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 = (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: