Buckling of Punches:-
Whenever Press tool is worked upon within the press, The punches mounted in that tool, are subjected to compression stresses. But if due consideration of stresses are overlooked during designing of the tool, the thin punch within the tool may fail by buckling.
Whenever Press tool is worked upon within the press, The punches mounted in that tool, are subjected to compression stresses. But if due consideration of stresses are overlooked during designing of the tool, the thin punch within the tool may fail by buckling.
Hence by maximum force, which a punch can withstand without buckling can be calculated by using the following formula.
Fb = [Ï€² × E × I ] y Lp²
Fb= Maximum Force beyond which buckling occurs.
E = Modulus of Elasticity ( For steel Modulus of Elasticity varies from 200 to 220 GN /m²)
I= Moment of Inertia in mm4
Lp= Length of punch in mm
The ultimate condition is when,
Buckling Force = Cutting Force required for the operation
= Shear force on the punch.
Fb = [Ï€² × E × I ] y Lp²
Fb= Maximum Force beyond which buckling occurs.
E = Modulus of Elasticity ( For steel Modulus of Elasticity varies from 200 to 220 GN /m²)
I= Moment of Inertia in mm4
Lp= Length of punch in mm
The ultimate condition is when,
Buckling Force = Cutting Force required for the operation
= Shear force on the punch.
Example 1:
Is it possible to punch 1mm brass sheet with a 5mm square punch?
Is it possible to punch 1mm brass sheet with a 5mm square punch?
T max = 320N/mm²
Length of Punch = 60mm
Shear Force required
to pierce the hole = L×S×Tmax
L = Cut length in mm
S = Sheet thickness in mm
Tmax = Shear force in N/mm²
L = 5 × 4 = 20mm
S = 1mm
Tmax= 320N/mm²
Length of Punch = 60mm
Shear Force required
to pierce the hole = L×S×Tmax
L = Cut length in mm
S = Sheet thickness in mm
Tmax = Shear force in N/mm²
L = 5 × 4 = 20mm
S = 1mm
Tmax= 320N/mm²
Shear Force = 20 × 1 × 320
= 6400N ( y1000)
= 6.4KN
Buckling Force, Fb = [Ï€² × E × I ] y Lp²
E = 210GN/mm²
I = a4 y 12
= 54 y 12
= 52.08 mm4
Lp = 60mm
Buckling Force
= [Ï€²×210×52.08×10 -12 ]y0.06²
= 2.99835 × 10-5 GN
= 29.9835 KN
As the punch can withstand a force of 29.35KN and the force coming on the punch is only 6.4KN, it is possible to use the punch.
Example 2:
To find the smallest diameter of the punch to pierce 2mm Mild Steel sheet.
Length of the punch = 60mm
E = 210GN/m²
Assume Fb = 800N
Fb = [Ï€² × E × I ] y Lp²
800 × 10-9 = [Ï€² × 210 × I] y 0.06²
800 × 10-9 × 0.06² = Ï€² × 210 × I
I = [800 × 10-9 × 0.06²] y[Ï€² × 210]
= [2.88 × 10-9 ] y [2070.516]
I = 1.389547 × 10-12 m4
E = 210GN/m²
Assume Fb = 800N
Fb = [Ï€² × E × I ] y Lp²
800 × 10-9 = [Ï€² × 210 × I] y 0.06²
800 × 10-9 × 0.06² = Ï€² × 210 × I
I = [800 × 10-9 × 0.06²] y[Ï€² × 210]
= [2.88 × 10-9 ] y [2070.516]
I = 1.389547 × 10-12 m4
I = 1.389547 × 10-12 m4
I = [Ï€d4 ] y 64
1.389547 × 10-12 = [Ï€d4 ] y 64
d4 = 1.389547 × 10-12 × 64
d4 = 2.8307619 × 10-11
d = 2.3066mm
2.31mm
I = [Ï€d4 ] y 64
1.389547 × 10-12 = [Ï€d4 ] y 64
d4 = 1.389547 × 10-12 × 64
d4 = 2.8307619 × 10-11
d = 2.3066mm
2.31mm
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