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 2 of 7.

Jitesh said:   6 years ago
Yes, the answer will be d=t as the shearing area will be π*d*t and normal area for the plate will be π*d^2.

Pratipalsinh said:   6 years ago
D = t is not correct answer, in this condition D = t is the smallest diameter of the hole that can be punched in a plate, not maximum.

Juhi said:   6 years ago
While punching a hole in a plate.

Shear stress = F/c.s area.
F/(π/4)d^2.

Crushing stress= F/2π*r*h.
= F/π*d*t (since h is the thickness of the plate).

Now,
Shear stress = 4* crushing stress.
4F/(π*d*d)= 4*F/(π*d*t).
d= t.

Hence the dia of the smallest hole that can be punched will be equal to the thickness of the plate.

Surya said:   6 years ago
The correct answer is option A) d=t.

Bhavik said:   6 years ago
The correct answer is d=t.

Shivani said:   6 years ago
The correct answer is d = t.

Mehul Patel said:   6 years ago
Here given that,

Crushing stress = 1/4 * Shear stress
P/A. = 1/4 * F/As
P/πd2/4 = 1/4 * F/πdt.
d = t. (Here Let P=F).

Abhilasha B R Gowda said:   7 years ago
The max. shear stress = (1/4) * max. crushing stress.
(P/A) = (1/4) * (P/dt).
A= area of hole= π*d^2.
substitute A and cancel P & d,
(P/π*d^2) = (1/4) * (P/dt).
4t = π*d.
'd' is directly proportional to 4t.

The answer is d=4t.

Pratik said:   7 years ago
Hence, d=4t.

Pratik said:   7 years ago
Here, it is for a maximum shear stress 1/4th of its maximum crushing stress i.e. (1/4 * maximum shear stress=maximum crushing stress).


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