INTRODUCTION:
Design of portable mill machine. The drive shaft is rotated by motor input of 4 Hp. The power is transmitted to the driven shaft by the belt and pulleys. The mill-machine works on the principle that the food grains are sheared in between rock wheels having rough surface. (One rock wheel is fixed.)
I) COMPONENTS-
1. Shaft
2. Pulley
3. Wheel
4. Belt
5. Keys
II) MATERIAL USED
1. Pulley - Grey cast iron FG-150
2. Shaft - Plane carbon steel 40 C8
3. Wheel – rock wheel
4. Belt - Leather
5. Keys - Commercial Steel

1. CALCULATIONS:

Let us assume the efficiency of the mill machine to be 60%.
Power required for the application is 2.4 horse power.
Hence Input Power applied by the motor is 4 Hp=2.984 Kw
Design of pulley1:
Material- Grey cast iron FG-150
Sut=150 N/mm2
FOS=5
st=Sut/FOS=30 N/mm2
Power is 4 Hp=2.984 Kw
Assuming number of revolution as 120 r.p.m. for safety purpose.
N=120 r.p.m.
Assuming the diameter of pulley1=200mm
Torque transmitted by pulley is
Mt=power / angular velo.=2984/12.56=237459.18N-mm
Assuming that one half of the arms carry the load at any instance, the tangential force at each arm is given by
F=Mt/(R × N/2) =1187.295N
Taking hub dia. =80 mm
Arm length for bending moment at section X-X near the hub is given by
L=R-r=100-40=60 mm
At section XX
Mb=F× L=71237.7 N-mm
Now, for an elliptical cross section with a & b as minor and major axis respectively.
I=pi × a ×b3 /64 and b= (2×a)
Bending stresses in the arm is given by
sb= (Mb × y)/I and y=a
Putting values of Mb, y, I we get
a=18. 22 or=20 mm
And b=40 mm
Design of pulley2:
Material- Grey cast iron FG-150
Sut=150 N/mm2
FOS=5
Power transmitted is 2.984 Kw
As N2d2=N1d1
N2=96 r.p.m.
Assuming the diameter of pulley2=250mm
Torque transmitted by pulley is
Mt=power / angular velo.=2984/10.05=296823.9689 N-mm
Assuming that one half of the arms carry the load at any instance, the tangential
Force at each arm is given by
F=Mt/(R × N/2) =1187.295N
Taking hub dia. =80 mm
Arm length for bending moment at section X-X near the hub is given by
L=R-r=125-40=85 mm
At section XX
Mb=F× L=100920.08 N-mm
Now, for an elliptical cross section with a & b as minor and major axis respectively.
I=pi × a ×b3 /64 and b= (2×a)
Bending stresses in the arm is given by
sb= (Mb × y)/I and y=a
Putting values of Mb, y, I we get
a=20.46 or=21 mm and b=42 mm

DESIGN OF SHAFT 1:
Material: Plane carbon steel 40 C8
Syt=380 N/mm2
FOS=5
Let d0 be the outer diameter of the shaft and di is inner diameter of shaft
Assuming do=60mm, C=di/do=0.6. Thus di=36 mm
Permissible Shear stress =Ssy/FOS=0.5Syt/FOS=38 N/mm2
Torque transmitted by the shaft, Mt= (T1-T2) × R
Here T1=3753.96 N, T2=1492.33 N and R=100 mm
Therefore, Mt=226163 N-mm
Neglecting weight of the pulley as compared to T1, T2.
Bending moment at the center of the pulley
Mb= (T1+T2) ×R
=524629 N-mm
Now, tmax= (16/ (pi×do3 × (1-C4))) × (Mb2+Mt2)1/2
tmax=15.48 N/mm2 < 38 N/mm2
Hence the design is safe!


DESIGN OF SHAFT 2:
Material: Plane carbon steel 40 C8
Syt=380 N/mm2
FOS=5
Let d0 be the outer diameter of the shaft and di is inner diameter of shaft
Assuming do=60mm, C=di/do=0.6. Thus di=36 mm
Permissible Shear stress =Ssy/FOS=0.5Syt/FOS=38 N/mm2
Torque transmitted by the shaft, Mt= (T1-T2) × R
Here T1=3753.96 N, T2=1492.33 N and R=125 mm
Therefore, Mt=283954 N-mm
Neglecting weight of the pulley as compared to T1, T2.
From resultant bending moment diagram, B.M. is maximum at point C and is equal to 262314.7 N-mm
Mb= 262314.7 N-mm
Now, tmax= (16/ (pi × do3 (1-C4))) (Mb2+Mt2)1/2
tmax =10.47 N/mm2 < 38 N/mm2
Hence the design is safe!





Design of the Key1:
Key: Rectangular sunk key
Material: Commercial Steel
Syt=230N/mm2
FOS=3
For the key material, sc=230/3=76.667 N
According to the shear stress theory of failure,
Ssy=0.5 Syt=115 N/mm2
t=Ssy/FOS=38.33 N/mm2
Mt= (power / ?) =237459.18 N-mm
The length of the key is given by
Taking Standard cross section for key is 16 ×10 mm
L=2 Mt/ (t db) =15.49 mm,
Also, L=4 Mt/ (sc × d × h) =24.68mm
Thus taking length of key =25mm
Hence dimensions of key are 16 ×10 ×25 mm3
Design of the Key2:
Key: Rectangular sunk key
Material: Commercial Steel
Syt=230N/mm2
FOS=3
For the key material, sc=230/3=76.667 N
According to the shear stress theory of failure,
Ssy=0.5 Syt=115 N/mm2
t=Ssy/FOS=38.33 N/mm2
Mt= (power / ?) =296823.97 N-mm
The length of the key is given by
Taking Standard cross section for key is 16 ×10 mm2
L=2 Mt/ (t db) =19.36 mm,
Also, L=4 Mt/ (sc×d × h) =30.98 mm
Thus taking length of key =31 mm
Hence dimensions of key are 16 ×10 ×31 mm3

Design of Belt:
Material: Leather
For the belt material mass density = 0.001 gm/mm3
Permissible stress, ft=2.59 N/mm2
Diameter and speed of smaller pulley are 100 mm and 120 r.p.m. resp.
Assuming belt thickness, t=10 mm
Coefficient of friction is, µ =0.3
Assuming angle of contact=1800
Now, V= ? (R1 +t/2) =1.3194 m/s
Power, P= (T1-T2) ×V
Thus, T1-T2=2261.63 N
Also, T1/T2=eµ? =2.515
Therefore, T1=3754.455N, T2=1492.83 N
Now, Tc=mV2= (width × thickness × length ×density) × V2
= (b×0.010 ×1 × 1000) × 1.31942=17.40 b N
Tmax=T1+Tc=3754.455+ (17.40 b) =ft × (b×t)
3754.455+ (17.40 b) =2.5 ×106 ×b ×0.010
Thus b=150 mm
Thus belt is under the safe conditions!
Auxiliary parts:
The rock wheels used in the milling machine have rough surface-texture to shear the food-grain particles. Out of the two rock wheels one is fixed and other is connected to the driven shaft.
A conical part is provided for the food grain particles to enter in between the rock-wheels so that process of milling is done and the required output (wheat) is obtained.