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 6 of 7.

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.

Shiyas said:   10 years ago
Answer is A.

(pi*(D^2) /4)*Crushing stress = pi*D*t*(Crushing stress/4).

===> D=t;

Rajeeev said:   10 years ago
D = T should be correct answer.

MANPREET SINGH said:   1 decade ago
Shear stress = Crushing stress in rivet joint.

Adarsh said:   1 decade ago
min d = [4t(max ss of plate)/max crushing stress of punch].

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.

Biswajit Roy said:   1 decade ago
Shear stress*Pi*dt = Crushing stress*Pi*d^2/4.

d = t.

If wrong why?

Srikanth said:   1 decade ago
How could it be @Rajesh there is nothing explained in your answer. Can you please explain it briefly?

UPPU HARISH said:   1 decade ago
I couldn't understand this.

If we take a punch punching a hole in a plate then,
Shear stress = P/pi*D*t.
Crushing stress = 4*P/pi*D*D.

When these two are equated with the relation given then we get d=t.
Please tell me where I am wrong?

Satya said:   1 decade ago
Shear stress = Pi/4*d2*load.

Crushing stress = load*d*t.

Given that s.s = 1/4*c.s.

Solving this we get d = t/pi.


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