S.NoComponentMaterialNumber
1WheelCast Iron2
2ShaftMild Steel
3GearsAluminium2
4Furrow OpenerCast Iron1
5Delivery PipeCast Iron1
6Nut
7Bolt
COMPARISON BETWEEN THE DESIGNED MODEL AND THE EXISTING TYPES OF
CULTIVATING EQUIPMENT
To do the general cost estimation the general cost of many
metals we do use for the fabrication of this design are:1kg cast
iron=6$, 1kg alloy steel=10$1kg aluminium= 4$, 1kg mild
steel=7$
By considering all factors and a general size that can be used
in a field to cultivate the farm. The general size of various
components and the amount of material used for the fabrication is
approximately taken as:Length of shaft=1.5mDiameter of the
shaft=6cmDiameter of the Wheel (diameter) =180cmLength of the
pulling rod=280cmDiameter of the bearings=7cmApproximate amount of
Material required in kilograms:Cast Iron=12kgs, Aluminium=3kgsMild
Steel=3kgs.Alloy steel=1kgTotal Cost as per the calculated and
requirements of the amount of material required:Cast Iron=72$,
Aluminium = 12$Mild Steel = 21$, Alloy Steel = 10$Total Cost of the
Material = 115$(approximately Rs5750/-)Material Handling costs and
also Work Cost = 204$ (Approximately Rs1000/-)The total cost the
equipment is Rs 7000/- max
Main Shaft connecting wheels Material - Mild SteelLength 680
mmDiameter 15 mm
Shaft Connected to Seed Metering DeviceMaterial - Mild
SteelLength 180 mmDiameter 15 mm
Gear on Main shaft Material Aluminium No. of Teeth 18 Gear on
Shaft connected to seed metering deviceMaterial Aluminium No. of
Teeth 12
Driving WheelMaterial Cast IronOuter Dia 36 cmInner Dia 26 cmNo.
Of Projection 12
Furrow openerMaterial Cast ironLength 10 mmHeight 2.5 mm
Teeth on driving sprocket , T1 = 18Diameter of driving sprocket,
d1 = 7.5cm Teeth on driven sprocket , T2 = 12Diameter of driven
sprocket, d2 = 5cm Speed of driving sprocket, N1 = 85 rpmLet Speed
of driving sprocket = N2Now, N2 X T2 = N1 X T1 N2 = (T1/T2) X N1 =
(18/12) X 85 = 127.5Pitch , p = d1 sin(/T1) = 1.3cmVerifying the
calculated value of pitchTerm (r1 + r2) = p/2 X (T1 = T2)PUTTING
VALUES OF r1, r2, T1, T2 ,we get p = 1.3cmCalculating centre
distance between the sprockets:Assuming the centre distance between
two sprockets, x = 15 cmLength of chain, L = p/2 (T1 + T2) + 2x +
p/2 ((cosec( - cosec() / 2x L = 49.58Therefore no. of links, n =
49.58/1.3 = 38.14Assuming n = 40Length becomes, L = 40 X 1.3 =
52cmNow from the formula, L = p/2 (T1 + T2) + 2x + p/2 ((cosec( -
cosec() / 2x x comes out to be 16cm
DESIGN OF SHAFTAssuming the beam to be supported in fixed
supports. Let length of shaft be lWhole weight is supposed to act
at the centre of the beam.For a fixed beam with load at its centre
: M = Wl/8Here,W = 12 kg = 118 N and l = 68 cm = 680 mmBending
Moment, M = 118680/8 = 10N-mTo Find Torque TLet speed, v=4km/hr =
1111mm/s(As avg speed of a walking man is 5 km/hr. With this much
load speed must be reduced. Here we assume it to be 4 km/hr ,
neglecting friction )Dia of Driving wheel, r = 260 mmTherefore, =
V/r = 8.54 rad/sNow, Power required to pull the Machine , P = FVLet
F = 159.81 N = 150 N (approx)(Additional load due to friction and
other losses are considered)P = 166.5 KJ/sEnergy requied to move
machine per unit per unit is 166.5 KJ/sNow we know , P = TTherefore
T = 19.5 N-mNow Considering Combined Bending and torsional effects
on the shaft,Let the allowable shear stress , = 40N/sq.mLet
allowable bending stress , = 80 N/sq.mNow , acc to max shear
stresss theory , Equivalent torsional moment,Te = (M + T) = 22
N-mNow /16 d = TeThis implies d = 14.1 mm
According to Normal stress Theory,Equivalent bending moment , Me
= 1/2(( M + (M + T)) = 16 N-mNow , /32 d= Med = 12.68 mmNow for
safety taking larger diameter d = 14.1 mm for shaft therefore dia
of shaft = 15mm.
