Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 17)
17.
The maximum diameter of the hole that can be punched from a plate of maximum shear stress 1/4th of its maximum crushing stress of punch, is equal to (where t = Thickness of the plate)
t
2t
4t
8t
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
65 comments Page 4 of 7.

Akahai said:   9 years ago
A steel punch can be worked to a compressive stress of 800N/mm^2.

Find the least diameter of hole which can be punched through a steel plate of 12mm thick, if its ultimate shear strength is 315N/mm^2.

Can anyone answer this?

Siva Kumar G said:   9 years ago
σ : crushing or normal stress, t: thickness, D: diameter, τ: Shear stress.
Given τ = σ /4.
Crushing force = shearing force.
==> σ [(π/4)* D^2] = τ[π D t].

Solving this equation, we get D=t.

Nilesh said:   9 years ago
d = t is the right answer.

Gaurav1995 said:   9 years ago
d = t is the correct answer.

Ramya said:   9 years ago
What is shear stress? Can anyone explain easily?

Saumya singh said:   9 years ago
d = t is the answer.

Himanshu said:   9 years ago
d = t is the answer.

Rabin said:   9 years ago
The question has asked the maximum diameter that can be punched. I think the smallest diameter should be equal to t. But again, I don't know how. Please explain in most basic tone.

Ananth said:   9 years ago
Shear stress = Load(P)/φ * D * t.
φ * D * t = ?

(area= L*T or (φ * D2)/4).

Koribilli hemanth said:   9 years ago
Answer was wrong.

The right answer is T = D.
Shear force = pi * D * T * shear stress.
Crushing force = (pi/4) * D^2 * crushing stress.

The condition is that "crushing stress = 4 * shear stress " at equilibrium condition " shear force = crushing force"

pi * D * T * shear stress = (pi/4) * D^2 * crushing stress.

By canceling the terms finally we will get D = T.


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