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)
Discussion:
65 comments Page 4 of 7.
Ramya said:
9 years ago
What is shear stress? Can anyone explain easily?
Gaurav1995 said:
9 years ago
d = t is the correct answer.
Nilesh said:
9 years ago
d = t is the right answer.
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.
Given τ = σ /4.
Crushing force = shearing force.
==> σ [(π/4)* D^2] = τ[π D t].
Solving this equation, we get D=t.
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?
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?
Neeraj said:
9 years ago
@Akahai.
7.62 apply same formula and this answer of questions is D=tCrushing force = shearing force.
==> σ [(φ/4)* D^2] = τ[φD t].
7.62 apply same formula and this answer of questions is D=tCrushing force = shearing force.
==> σ [(φ/4)* D^2] = τ[φD t].
Shubham Soni said:
9 years ago
Shear stress = P/π/4 * D^2 = 4P/π.D^2.
Crushing Stress = P/D.t.
So, Shear stress = 1/4 of Crushing Stress.
Therefore, 4P/pi.D^2 = P/D.t.
Hence, 4/pi. D = 1/t.
D = 4t/π.
i.e. D is directly proportional to the 4t.
Hence the answer is 4t.
Crushing Stress = P/D.t.
So, Shear stress = 1/4 of Crushing Stress.
Therefore, 4P/pi.D^2 = P/D.t.
Hence, 4/pi. D = 1/t.
D = 4t/π.
i.e. D is directly proportional to the 4t.
Hence the answer is 4t.
Bharadwaz said:
8 years ago
Almost all are saying using shear in rivets formula,
But in question it is mentioned punching so shear is load(p)/area(pi*d*t).
Shear stress = P/pi*D*t.
Crushing stress = 4*P/pi*D*D,
Shear=crushing/4,
P/pi*D*t.=(4*P/pi*D*D)*(1/4),
P/pi*D*t=P/pi*D*D,
d=t.
But in question it is mentioned punching so shear is load(p)/area(pi*d*t).
Shear stress = P/pi*D*t.
Crushing stress = 4*P/pi*D*D,
Shear=crushing/4,
P/pi*D*t.=(4*P/pi*D*D)*(1/4),
P/pi*D*t=P/pi*D*D,
d=t.
Balu said:
8 years ago
d = t is right answer.
Sachin Sharma said:
8 years ago
I also agree with d=t
Beacuse, shearing stress is given 1/4th of crushing stress hence and shearing force is calculated by shearing stress into the shearing area and hence shearing area is (π/4 d^2) x t Hence shearing force will be,
(πd) x t x (shearing stress)
and crushing stress of punch will be equal to π/4 d^2) x σc
now (πd) x t x (σc/4) =π/4 d^2) x σc
Hence after solving d=t.
Beacuse, shearing stress is given 1/4th of crushing stress hence and shearing force is calculated by shearing stress into the shearing area and hence shearing area is (π/4 d^2) x t Hence shearing force will be,
(πd) x t x (shearing stress)
and crushing stress of punch will be equal to π/4 d^2) x σc
now (πd) x t x (σc/4) =π/4 d^2) x σc
Hence after solving d=t.
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