Question

Find the blanking force required to punch 10 mm diameter holes in a steel sheet of 3 mm thickness. Given shear strength of material = 400 MPa, penetration = 40% and shear provided on the punch = 2 mm.

1. 22.6 kN
2. 37.7 kN
3. 61.6 kN
4. 94.3 kN

Answer 1

Correct Answer - Option 1 : 22.6 kN

Concept:

The energy or work required in blanking is the same irrespective of shear. Shear just reduces the blanking force.

Work done without shear W = τ × π × d × t × p

where, τ is the shear stress, d is the diameter of the hole, t is thickness of the plate and p is the penetration.

Work done with shear W = F₂ × s

where, F₂ is the blanking force required for shear on punch and s is the shear on punch.

As the work done W is same

∴ F₂ × s = τ × π × d × t × p

[Note: Penetration is always calculated in terms of thickness.]

Calculation:

Given:

d = 10 mm, t = 3 mm, τ = 400 MPa ⇒ 400 N/mm², p = 40 % ⇒ 0.40 × 3 mm = 1.2 mm and s = 2 mm

\({F_2} = \frac{{\tau\; × \;\pi \;× \;d \;× \;t \;×\; p}}{s}\)

\({F_2} = \frac{{400 \;× \;\pi \;× \;10 \;× \;3 \;× \;1.2}}{2} \) = 22.619 kN