Question

calculate the force required for punching a hole of
20mm diameter through a mild steel plate 5mm
thick. The maximum shear stress do mild steel is
250 N/mm. Also find the compressive stress developed
on punch​

Answer 1

Concept:

We need to apply the equation of punching force. It is expressed as- Punching force = (πDt) × S

Given:

Diameter of punching hole, D = 20mm

The thickness of a mild steel plate, t = 5mm

The maximum shear stress of mild steel, S = 250 N/mm

Find:

We need to determine:

the force required for punching a hole

the compressive stress developed on punch

Solution:

Punching force is determined by the equation-

Punching force = (πDt) × S

We have, D as the diameter of the punching hole, t as the thickness and S as the shear stress

The equation of Punching force = (πDt) × S becomes-

Punching force = 22/7 × 20 × 5 × 250

Punching force = 78,571.42 N

Compressive stress = Punching force/area

Compressive stress = 78,571.42/ πr²

We have diameter = 20 mm therefore radius = 20/2 = 10mm

Therefore, the equation of Compressive stress becomes-

Compressive stress = 78,571.42/ 22/7 × (10)²

Compressive stress = 249.99 ≈ 250 N/mm²

Thus, the force required for punching a hole is 78,571.42 N and Compressive stress is 250 N/mm²

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Answer 2

Answer:

The force required for punching a hole is 78,571.42 N and Compressive stress is 250 N/mm².

Explanation:

Concept to be used is Punching force = (πDt) × S

It is given that

Diameter of punching hole, $D = 20mm$

The thickness of a mild steel plate, $t = 5mm$

The maximum shear stress of mild steel, $S = 250 N/mm$

We need to calculate the force required for punching a hole and the compressive stress developed on punch.

Punching force = (πDt) × S

Punching force = 22/7 × 20 × 5 × 250 = 78,571.42 N

Also, Compressive stress = Punching force/area

and

Compressive stress = 78,571.42/ πr²

Hence, Compressive stress = 78,571.42/ 22/7 × (10)² = 249.99 ≈ 250N/mm²

#SPJ3