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.
Biswajit Roy said:
1 decade ago
Shear stress*Pi*dt = Crushing stress*Pi*d^2/4.
d = t.
If wrong why?
d = t.
If wrong why?
Deepak said:
9 years ago
S.S = ( 1/4) * C.S.
F/Pi dt = (1/4) * (4F/pi * d^2).
Thus, d = t.
F/Pi dt = (1/4) * (4F/pi * d^2).
Thus, d = t.
Mukesh said:
9 years ago
1/4 s.s = c.s
1/4 * F/pi d t = F/pi d d.
d = 4t.
1/4 * F/pi d t = F/pi d d.
d = 4t.
Adarsh said:
1 decade ago
min d = [4t(max ss of plate)/max crushing stress of punch].
Ramkumar said:
8 years ago
According to me, the correct answer is "d=t" only!
Ramya said:
9 years ago
What is shear stress? Can anyone explain easily?
MANPREET SINGH said:
1 decade ago
Shear stress = Crushing stress in rivet joint.
Surya said:
6 years ago
The correct answer is option A) d=t.
Akshay said:
1 decade ago
How is shear stress equal to l*t?
Rajeeev said:
10 years ago
D = T should be correct answer.
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