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 5 of 7.
Ramkumar said:
8 years ago
According to me, the correct answer is "d=t" only!
Nitesh said:
7 years ago
Here, d= t is the right answer.
Aarif said:
7 years ago
d = t is the right answer.
Ravikumar said:
7 years ago
The Answer is d = t.
Because shear stress = 1/4 crashing stress.
Now,
p/(πdr) = p/(πdd/4)4.
d=t.
Because shear stress = 1/4 crashing stress.
Now,
p/(πdr) = p/(πdd/4)4.
d=t.
Bidhan Das said:
7 years ago
The answer will be d=t.
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).
Pratik said:
7 years ago
Hence, d=4t.
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.
(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.
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).
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).
Shivani said:
6 years ago
The correct answer is d = t.
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