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 2 of 7.
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?
Indrajeet said:
9 years ago
Option A is the correct answer.
Crushing stress = 4 * Load(P)/φ * D * D.
Shear stress = Load(P)/φ * D * t.
Since Crushing stress = 4 * Shear stress.
After resolve, We get, D = t.
So option A is right.
Crushing stress = 4 * Load(P)/φ * D * D.
Shear stress = Load(P)/φ * D * t.
Since Crushing stress = 4 * Shear stress.
After resolve, We get, D = t.
So option A is right.
JAYADEB DASH said:
6 years ago
The answer is d = t.
1. Max shear stress in plate = 1/4 max crushing stress in the punch.
2. Shear stress in plate = p/ dt.
3. Crushing stress in punch = 4p/ d^2.
4. Now p/' dt = 4p/4 d^2.
And d = t.
1. Max shear stress in plate = 1/4 max crushing stress in the punch.
2. Shear stress in plate = p/ dt.
3. Crushing stress in punch = 4p/ d^2.
4. Now p/' dt = 4p/4 d^2.
And d = t.
(1)
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.
Shalabh Suradhaniwar said:
1 decade ago
@Uppu Harish perfect answer & absolute correct explanation.
d=t, shall be the correct answer.
Crystal clear depiction of force & area in shear & crush.
d=t, shall be the correct answer.
Crystal clear depiction of force & area in shear & crush.
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].
Kuruu said:
1 decade ago
@Akshay I want to explain you about the question.
Actually @Rajesh had done exactly no need again just check once if at all u didnt get that will see.
Actually @Rajesh had done exactly no need again just check once if at all u didnt get that will see.
Ashok kumar Chauhan said:
10 years ago
Shear stress = F/pi*d*t.
Crushing stress = 4F/pi*d^2.
Here pi = 22/7.
According to question Shear stress = Crushing stress/4.
That is why d = t.
Crushing stress = 4F/pi*d^2.
Here pi = 22/7.
According to question Shear stress = Crushing stress/4.
That is why d = t.
NITESH BORKAR said:
9 years ago
The equation will be:
(Pi/4 * d^2) * Crushing strength,
= Pi * d * t * Shear strength (or crushing strength * 2).
Which comes out to be t only.
(Pi/4 * d^2) * Crushing strength,
= Pi * d * t * Shear strength (or crushing strength * 2).
Which comes out to be t only.
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