DESIGN OF TRANSMISSION FOR A LOW COST RURAL VEHICLE
AUTHORS
Dr. V. JAI GANESH1
J.AMIRTHARAJA*
S.N.BARATH*
T.DEEPANCHAKRAVARTHY*
1- Professor & Dean, Department of Mechanical Engineering, Sriram Engineering College, Chennai – 602 024.
*- Final Year Students, Department of Mechanical Engineering, Sriram Engineering College, Chennai – 602 024.
ABSTRACT
The main objective of this paper is to replace the “Existing gearbox in which the gear ratios are modified to enhance the performance, in terms of speed and grade ability under fully laden conditions” The Transmission system comprises of clutch, gearbox, propeller shaft and differential unit. We have focused on design of the gearbox in the rural vehicle. Also our work involves the study and analysis of the gear ratios of commercial gearbox and their functioning and relative merits and demerits. This transmission must be able to cater all the need of an existing transmission system in the market and it must be of a low cost.
1. TRANSMISSION SYSTEM
Automobiletransmission systemrefers to the assembly of gears and associated automotive components by means of which the torque of the engine is multiplied or modified and transmitted to the driving axle. It is also the system through which a driver select gears by the action of hand on the gear lever or gearstick. A manual transmission system is known by several different terms such as standardtransmission, stick shift, stick, or straight drive. With the transmissions featuring a driver-operated clutch and a movable gear selector, most of them enable the driver to select any of the forward gears at any time.
Automobile transmission can either be manual or automatic. The manual transmissions feature different gear ratios which can be switched to by engaging pairs of gears inside the transmission.
On the other hand theautomatictransmissionstypically have gears controlled by brake bands or clutch packs for selecting a gear ratio. Another category of transmission is the semi-automatic transmission, which allows the driver to select the current gear manually.
There are a number of variables to consider when selecting a power transmission drive for a given application. The most important of these variables are:
• Horsepower or torque to be transmitted
• Required speeds (revolutions per minute, rpm)
• Duty cycle
2. GEARBOX
Gear box is essential equipment in a gear assembly. Gear Box is also known as Gear head, Gear reducer and Speed reducer. The fundamental principle of a gearbox is to transmit the cause of mechanical rotation between two shafts. In this order, there is a structural support present in between the two shafts. Generally, gearboxes are kept inside the casings. This helps the gearboxes in their structural support, provides protection and ensures in doing safe functioning. Normally, the gearboxes are designed in reducing the speed, but sometimes, a gearbox may be designed for speed enhancing duties. The shafts inside the gearboxes are placed for the purpose of accepting and delivering the machinery rotation. This machinery rotation (torque) is achieved in the form of spines that should be suitable to connector join to another unit. The mechanical rotation which is generated by the engine is consumed through the gearbox. This in turn, is being converted into a force at the road surface.
3. EXISTING GEAR BOX
Presently available gear Box is having 8 Speed and cost of the gear box is also high, so it cannot be used for the low cost rural Vehicle.
INSERT TABLE 1- EXISTING GEAR RATIO
4. MODIFIED GEAR BOX
6 Speed gear box is highly sufficient for our application. By reducing Number of speeds, we can easily reduce the cost of the gearbox.
INSERT TABLE 2- MODIFIED GEAR RATIO
INSERT TABLE 3- VEHICLE SPECIFICATION
5. GEAR REDUCTIONS
In our 6 speed gear box reduction of speed takes place at two stages. The 1st reduction takes place between input shaft and the counter shaft which forms the constant meshing gears.
The second reduction takes place between the counter shaft and the output shaft.
6. ANALYTICAL STUDIES
6.1. CALCULATION of centre distance:
Centre distance (a) = (i+1) 3√ (0.7/σc) 2* (E*Mt)/ (i*ψ)
i = 2.3
σc = CR*HRC*KCl
= 280x55x0.585 = 9009 kgf/cm2
Mt = 156Nm (or) 1560 kgf/cm
E = 2.1x106kgf/mm2
ψ = 0.5
a = (2.3+1) 3√ (0.7/9009)2x (2.1x106x 1560) / (2.3x0.5)
a = 85 mm
INSERT FIG 1- CENTRE DISTANCE BETWEEN TWO GEARS
6.2. DESIGN Calculation Of Forward & Reverse Gear:
6.2.1. FIRST GEAR:
Module m = 2a / (Z1+Z2)
Assume Z1 = 11
Z2 = i x Z1 = 3.9x11 = 43
m= 2 x 85 x / (11+43)
m = 3.14 mm
d1 = mn * Z1
= 3.14 x 11
d1 = 34.54mm
d2 =3.14 x 43 = 135 mm
D = (d1+d2) = 169.5mm
Face width (b) = ψ*a = 0.5 x 85 = 42.5mm
Output speed:
i = N1 / N2
N2 = (3000/ 8.89) = 338 rpm
Addendum = 0.8*m = 0.8x2.95 =2.36mm
Dedendum = 1*m =1x2.95 = 2.95mm
Minimum total depth = 1.80*m =1.80x2.95 =5.31mm
Working depth = 1.60*m =1.60x2.95 =4.72mm
Tooth thickness = 1.5708*m =1.57x2.95 =4.63mm
6.2.2. SECOND GEAR:
Module m = 2a cos β / (Z1+Z2)
Assume Z1 = 13
Z2 = i x Z1 = 2.55x13 = 33,
m = 2 x 85 x cos 20ₒ/ (13+33)
m = 3.47mm
d1 = mn * Z1 / (cos β)
= 3.47 x 13 / (cos 20ₒ )
d1 = 48mm
d2 = 3.47 x 33 / (cos 20ₒ ) = 121.85mm
D = (d1+d2) = 169.85mm
Output speed:
i = N1 / N2
N2 = (3000/ 5.865) =512 rpm
6.2.3. THIRD GEAR:
Module m = 2a cos β / (Z1+Z2)
Assume Z1 = 15
Z2 = i x Z1 = 1.67 x 15 = 25,
m = 2 x 85 x cos 20ₒ/ (15+25)
m = 3.9mm
d1 = mn * Z1 / (cos β)
= 3.9 x 15 / (cos 20ₒ )
d1 = 62.25mm
d2 = 3.9 x 25 / (cos 20ₒ ) = 103.75mm
D = (d1+d2) = 166mm
Output speed:
i = N1 / N2
N2 = (3000/ 3.85) = 781 rpm
6.2.4. FOURTH GEAR:
Module m = 2a cos β / (Z1+Z2)
Assume Z1 = 19
Z2 = i x Z1 = 1.06 x 19 = 21
m = 2 x 85 x cos 20ₒ/ (19+21)
m = 3.99mm
d1 = mn * Z1 / (cos β)
= 3.99 x 19 / (cos 20ₒ )
d1 = 80.67mm
d2 = 3.99 x 21 / (cos 20ₒ ) = 89.16mm
D = (d1+d2) = 169.83mm
Output speed:
i = N1 / N2
N2 = (3000/ 2.45) = 1224 rpm
6.2.5. FIFTH GEAR:
Module m = 2a cos β / (Z1+Z2)
Assume Z1 = 23
Z2 = i x Z1 = 0.72 x 23 = 17
m = 2 x 85 x cos 20ₒ/ (23+17)
m = 3.99mm
d1 = mn * Z1 / (cos β)
= 3.99 x 23/ (cos 20ₒ )
d1 = 97.65mm
d2 = 3.99 x 17 / (cos 20ₒ ) = 72.18mm
D = (d1+d2) = 169.83mm
Output speed:
i = N1 / N2
N2 = (3000/ 1.66) = 1807 rpm
6.2.6. Reverse Gear:
Module m = 2a / (Z1+Z2)
Assume Z1 = 11
Z2 = i x Z1
= 3.1 x 11 = 35
m = 2 x 85 / (11+35)
m = 3.69mm
d1 = m* Z1
= 3.69 x 11
d1 = 40.59mm
d2 = 3.69 x 35 = 129.15mm
D = (d1+d2) = 169.74mm
Output speed:
i = N1 / N2
N2 = (3000/ 7.12) = 421 rpm
Addendum = 0.8*m = 0.8x3.69 =2.95mm
Dedendum = 1*m =1x3.69 = 3.69mm
Minimum total depth = 1.80*m =1.80x3.69 =6.61mm
Working depth = 1.60*m =1.60x3.69 =5.9mm
Tooth thickness = 1.5708*m =1.57x3.69 =5.79mm
7. STRESS CALCULATIONS FOR GEAR MATERIALS:
7.1. Bending stress:
Bending stress (σb) = (i+1) [Mt] / (a*m*b*y)
i = 2.3
Mt = 156Nm (or) 1560 kgf/cm
a = 85mm
m = 4.07mm
b = 42.5 mm
y = 0.452 for Z = 29
(σb) = (2.3+1) x (156x103) / (85 x 4.07 x 42.5 x 0.452)
= 67 N/mm2
Bending stress (σb) = 67 N/mm2
7.2. Contact stress:
Contact stress (σc) = 0.74*(i+1/a) √ (i+1/i*b)*E [Mt]
= 0.74 x (2.3+1 / 85) √ [3.3/ (2.3x42.5)] x 2.1x105 x 156x103
= 955.3 N/mm2
Contact stress (σc) = 955.3 N/mm2
7.3. Calculation Of Allowable Stress:
Bending stress (σb) = [1.4 kbl / n kσ] x σ-1
Kbl = life factor for bending
= 0.7 (PSG design data book)
Kσ = stress concentration factor
= 1.2 (PSG design data book)
N = factor of safety
= 2.0 (PSG design data book)
σ-1 = Endurance limit
= 0.35 σu + 120
= 0.35 x 630 + 120 (σu = ultimate strength)
= 340 N/mm2
(σb) = [1.4 x 0.7 / 2 x 1.2] x 340
= 138N/mm2
Comparing the induced bending stress and allowable bending stress,
Induced bending stress < allowable bending stress
i.e., 67 N/mm2 < 138 N/mm2
Contact stress (σc) = CR*HRC*KCl
= 280 x 60 x 0.585
= 982.8 N/mm2
Comparing the induced contact stress and allowable contact stress,
Induced contact stress < allowable contact stress
i.e., 955.3 N/mm2 < 982.8 N/mm2
Therefore THE DESIGN IS SAFE.
Therefore both the induced bending and contact stress are within the limits of allowable induced bending and contact stress.
Hence THE CHOSEN GEAR MATERIAL IS SATISFACTORY.
INSERT FIG 2: 2D DIAGRAM OF 6 SPEED GEARBOX
INSERT FIG 3: GEARBOX WITH HOUSING
8. DESIGN CALCULATIONS FOR SHAFTS:
8.1. Design of shaft 3:
Torque transmitted by the shaft:
T = P*60/ (2pN)
Power input = 49kw
Module = 2.95
R1 = 26mm
R2 = 58mm
R3 = 17.25mm
R4 =67.45mm
Torque on gear T1 = P*60/ (2pN)
= 49x10^3x60 / (2x3.14x3000)
= 155.9N-m
Torque on gear T2 = P*60/ (2pN)
= 49x103x60 / (2x3.14x1304)
= 358.8 N-m
Torque on gear T3 = T2
Torque on gear T4 = P*60/ (2pN)
= 49x10^3x60 / (2x3.14x338)
= 1384.3 N-m
Tangential force Ft4 = T4 / r4
= 1384.3 / 0.067
= 20538.5N
Radial force Fr4 = Ft4 * tanα
= 20538.5 x tan 20
= 7475.4N
Shaft 3 is subjected to bending in the vertical plane due to Fr4 and in the horizontal plane due to Ft4.
Bending moment in the vertical plane,
BMv = Fr4 * l3
= 7475.4 x 0.056
= 418.62 N-m
Bending moment in the horizontal plane,
BMh = Ft4 * l3
= 20538.5 x 0.056
= 1150.15 N-m
Resultant Bending Moment,
BMR = √ ( BMv 2 + BMh 2)
= √ (418.622 + 1150.152)
= 1223.96N-m
Torque equivalent Te = √ ( BMR 2 + T4 2)
= √ (1223.962 + 1384.3 2)
= 1847.79N-m
Assume case hardened alloy steel as shaft material take permissible stress
t = 95N/mm2
16* Te/ (pd33) = [t]
16 x 1847.79x103/ (pd33) = 95
D3 = 46mm
8.2. Design of Shaft 2:
Bending moment in the horizontal plane BMh = Ft2* l2
= 2598.12 N-m
Bending moment in the vertical plane BMv = Fr2 * l2
= 945.60 N-m
Resultant Bending Moment BMR = √ (BMv 2 + BMh 2)
= 2764.64 N-m
Torque equivalent Te = √( BMR 2 + T22)
= 2787.52 N-m
D2 = 53mm
8.3. Design of Shaft 1:
Bending moment in the horizontal plane BMh = Ft2* l2
= 335.78 N-m
Bending moment in the vertical plane BMv = Fr2 * l2
= 122.21 N-m
Resultant Bending Moment BMR = √ (BMv 2 + BMh 2)
= 357.32 N-m
Torque equivalent Te = √( BMR 2 + T22)
= 389.84N-m
D1 = 28mm
9. CONCLUSION:
In design of transmission for a low cost rural vehicle, existing gear box design has been replaced with a 6 speed gear box design. In which the number of gears have been reduced. In order to reduce the cost and size.The reductions of gears have been done by iterative method. Gear ratios, module, center distance, number of teeth’s on gear and diameter of the gear & shaft and also various stresses that act on material calculated , in order to know the design safe or not. That the proposal model of 6 speed gear box design is found to be feasible. This transmission will be able to cater all the need of an existing transmission system in the market and it will be of a low cost.
10. REFERENCES:
1. Bhandhari.K (1984), “Design Of Machine Elements”, Tata McGraw Hill Publishing Co.Ltd, pp. 306-325
2. Sadhu singh (1989), “Theory of Machines” Pearson publications, pp 408-432
3. “PSG Design Data Book” (1978) , pp. 7.1-7.18, 8.1-8.25
4. Puraja.K.K, Juneja.L & Bhandhari.N.C (1983), “Machine Design”, Dhanapat Rai & Sons, pp. 265-288
5. R.B.Gupta (1977) “Automobile Engineering”, Tech india Publications, pp.378-401.
6. R.S. Khurmi (1992), “Machine Design”, S.Chand publishers, New delhi pp. 115-138.
7. Prabhu .T.J (1988), “Design of transmission system”, S k publishers,PP. 1.7- 1.58, 2.3-2.23
8. www.wikipedia.com
9. www.howstuffworks.com
Table 1. EXISTING GEAR RATIO Table 2. MODIFIED GEAR RATIO
GEAR / RATIO1st / 10
2nd / 5.87
3rd / 4
4th / 3.7
5th / 2.5
6th / 2.35
7th / 1.48
8th / 1
GEAR / RATIO
1st / 8.87
2nd / 5.875
3rd / 3.85
4th / 2.45
5th / 1.48
6th / 1
Table 3: VEHICLE SPECIFICATIONS
Number of cylinders / 2Bore/ stroke / 79 / 92 (mm)
Displacement / 2700 cc
Compression ratio / 18 : 1
Rated speed / 3000 RPM
Mean piston speed / 9.20 m/s
Power rating at rated speed for Automotive engine / 54 HP
Mean effective pressure / 5.99 bar
Max Torque @ Speed 1800~2000RPM / 156 Nm
Min. Idle Speed / 1150 ~ 1250 RPM
Specific fuel consumption @ rated Speed / 195 ~ 200 g/HP hr