STATEMENT:
DESIGN FOR THE PUNCHING MACHINE OF PUNCH OF 10 mm DIAMETER IN ALLUMINIUM SHEET OF THICKNESS 0.05 mm .

DESIGN:
I) COMPONENTS-
1. Punch pin
2. Rod or bar
3. Support casing
4. Spring
5. Pins
6. Lever
7. I-section rod


II) MATERIAL USED-
1. Punch pin: Carbon Steel C-20 (Hot rolled)
2. Rod or bar: Carbon Steel C-20 (Hot rolled)
3. Support casing: cast iron
4. Spring: Oil Hardened steel
5. Pins: : Steel FeE
6. Lever: Steel FeE
7. I-section rod: Carbon Steel C-20 (Hot rolled)

III) CALCULATIONS:
From design data book,
Max. Shear stress for Al-sheet metal is
ts=70 N/mm2

Assuming the efficiency of the mechanism as 40%.
Now the force required to punch the Al-sheet is
F= ts × (Area of shear)
= ts × (2 × pi × r × t)
Where, r=rad. of punch =5 mm and
t= thickness of sheet= 0.05 mm

Hence, F=109.95 N
Since the efficiency is 0.4, hence the force required for punch can be taken as
F=275 N

a. Design for punch pin/rod:
Material used: Carbon steel C-20
syt= 24.6 N/mm2
Assuming factor of safety= 5.
scp= syt/5 =4.92 N/mm2
Now,
Taking dia. of rod as d=10 mm,
And length of the rod = 70 mm
As the rod is undergoes the compressive stress;
sc=F/ (cross-section area of rod),
sc=F/ (p × d2 /4)
Hence, sc=3.5 N/mm2
Which is less than allowable stress (sc < scp)
Hence, the design is safe.
The lower end of the rod is sharpened to get the required punching.
b. Support casing:
Support-casing is made up of cast iron and the clearance is given for the motion of the rod during the punching.
We have used H8d8 fine loose running clearance fit for 10mm shaft & hole pair.
It is seen that 10mm lies in diameter steps of 10-14mm. Thus, value of D (geometrical mean) = (10×14) ^½ =11.83mm.
Standard tolerance unit, i =0.45(D^?) +0.001D
=0.45(11.83^?) + 0.012
= 1.04microns.
For hole quality of 8, standard tolerance, 25i = 25×1.04 = 26 microns = 0.026mm
For “H” hole the FD =0.
Hence, the hole limits are 10 mm & (10+0.026) = 10.026 mm.
For shaft quality of 8, the standard tolerance=25i=25×1.04= 0.026mm.
For “d” shaft the FD = -16(D^0.44) =-16(11.83^0.44) =-48.36microns=-0.048mm.
The shaft limits are (10.000-0.048) =9.952 mm & (10.000-0.048+0.026) =9.978 mm.
c. Design for Spring:
Material: Oil hardened & tempered steel
sut= 1370 N/mm2
ts=0.3(sut) =411 N/mm2
Mod. Of the rigidity= G=81370 N/mm2
Assuming
Di= inner dia. of spring = 12mm,
d= dia. of the spring wire,
Hence,
Do= outer dia. of the spring = (Di +d) =18 mm
Let, Dm =Mean dia. of spring= (Do+Di )/2=15 mm
C= spring index = Dm /d,
K= spring stiffness,
d =deflection of the spring,
Assuming, spring force as 10 N.
But due to losses let spring force=25N
Now, Shear stress in the spring is given by
ts=K× (8×P× Dm /( p × d3))
Assuming K=1,
Putting the values of ts, P and Dm
We get, d=3.03 mm
Hence C=4.95
K=1.314
Thus putting value of new K again in ts
We get d=3.32mm but taking standard value as 3.5 mm
Now; deflection of the spring is given by
d= (8×P×Dm3×N)/ (G × d4)
Where, N= Number of the coils
Also,
Length of spring when the punch is in open =43 mm
Length of spring when the punch applied =38 mm
Hence,
d =43-38=5 mm
Hence, N=7.53=8
Since, the spring has the square and ground end inactive no. of the coils is 2.
Nt=N+2=10
Again,
d’ = (8×P× Dm 3×Nt)/ (G×d4)
Hence, d’ =6.63 mm
Let the total gap between the adjacent coils is
=15% × d’
=0.995 mm
Solid length = Nt × d=35 mm
Free length=solid length +total axial gap+ d’
=42.625 mm or 43mm
Now,
Pitch of the coil = (free length)/ (Nt -1) =4.78 mm
d. Design of the lever (upper):
Material: Steel FeE
syt=250 N/mm2
t = 125 N/mm2
Factor of the safety =5





Here; F= 300 N
Length of the lever as 90 mm.
Taking L1= 75 mm & L2= 15 mm
F × L2 =P × L1

Hence P=60 N
Therefore, R=F-P
R= 300-60=240 N

In case of, two arm levers bending moment is zero at point of application of force (P & F) and maximum at boss of the lever.


Now, maximum bending moment is
Mb= P × L1
=4500 N-mm
Now, for rectangular cross-section
sb= (Mb × y)/I
Taking, b= distance parallel to neutral axis =6 mm
d= distant perpendicular to the neutral axis
Taking, d= 2 × b=12mm
Now,
I= (b × d3)/12=864 kg-mm2, y=d/2
Hence, sb= 31.25 N/mm2 << 50 N/mm2
Thus, the design is safe!

e. Design of the Fulcrum pin:
Material: Steel FeE
syt=250 N/mm2
t = 125 N/mm2
Factor of the safety =5
From the design of the lever
R=240 N
p=permissible bearing pressure =10 N/mm2
d1=dia. of the fulcrum pin and
L1=length of the fulcrum pin

Now, R=p (L1 × d1)
Assuming, d1=6 mm and L1= (1.25 × d1) =7.5 mm
Hence, p=5.333 N/mm2
This is less than 10 N/mm2.
Hence, the design is safe!

f. Miscellaneous:
Now, the lever becomes weak due to the pinhole at the fulcrum pin.
To check the critical c/s which is shown in the fig.
Dia. of the pin=5 mm
Thickness of the bush= 0.5mm
Hence, the moment of the inertia of the pin is given by
I= (1/12) × 10 × (123-63)
I=1260 kg-mm2
sc= (Mb × y)/I
sc=21.42 N/mm2 <50 N/mm2
Hence, the design is safe.
g. Design of the I-section:
Material: Carbon steel
sc=115 N/mm2
Factor of safety= 5
The rod is taken of the section-I
I- section undergo the Buckling.
Let, Wb=Buckling load=60 N
L’=length of the I-section
t=thickness of the hub,
Now,
Assuming t=3 mm and L’=50 mm
Wb= (sc × A)/ (1+ (a × (L’/Kxx) 2))
Now, Kxx= 1.78 × t=radius of gyration
a=1/7500,
A=area of the I-section= (11 × t × t)
From this we get,
sc=0.216 N/mm2 <<<< 23N/mm2
Hence, the design is safe!
h. Design of lever (Lower):
Material: Steel FeE
syt=250 N/mm2
t = 125 N/mm2
Factor of the safety =5

Here; F= 60 N
Assuming, P=60 N (‘P’ should be applied manually)

R=F-P
R=0 N (P=60, F=60N)

In case of, two arm levers bending moment is zero at point of application of force (P & F) and maximum at boss of the lever.

Taking total length of the lever (L) as 80 mm
Hence, L1+L2=L=80
L1=L2=40 mm

Now, maximum bending moment is
Mb= P × L1
=2400 N-mm
Now, for rectangular cross-section
sb= (Mb × y)/I
Taking, b= distance parallel to neutral axis =6 mm
d= distant perpendicular to the neutral axis
Taking, d= 2 × b=12 mm
Now,
I= (b × d3)/12=864 kg-mm2, y=d/2
Hence, sb= 16.67 N/mm2 << 50 N/mm2
Thus, the design is safe!
i. Design of the pins:
Material: Steel FeE
syt=250 N/mm2
t = 125 N/mm2
Factor of the safety =5
Bearing pressure pin= p1=10 N/mm2
Dia. of pin= d1=2.5mm
Length of the pin=L”=20 mm
i. Pin1:
(Connection between lever1 and rod)
F=p1 × d1 × L”
Here, F=300 N,
Hence, p1=6 N/mm2 < 10N/mm2
Therefore, design is safe!

ii. Pin2 and 3:
Pin2 for the connection between lever and I-section and pin3 for the connection between I-section and lever2
F=p1 × d1 × L”
Here, F=60 N,
Hence, p1=1.2 N/mm2 < 10N/mm2
Therefore, design is safe!