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.
Srikanth said:
1 decade ago
How could it be @Rajesh there is nothing explained in your answer. Can you please explain it briefly?
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?
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.
Adarsh said:
1 decade ago
min d = [4t(max ss of plate)/max crushing stress of punch].
MANPREET SINGH said:
1 decade ago
Shear stress = Crushing stress in rivet joint.
Rajeeev said:
10 years ago
D = T should be correct answer.
Shiyas said:
10 years ago
Answer is A.
(pi*(D^2) /4)*Crushing stress = pi*D*t*(Crushing stress/4).
===> D=t;
(pi*(D^2) /4)*Crushing stress = pi*D*t*(Crushing stress/4).
===> D=t;
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.
AMAL said:
10 years ago
Here a hole is punched then we need to take area of that hole that is (in shear) pi*d^2/4 and in crushing pi*d*t. So the right answer is d=t.
Sharan said:
1 decade ago
Shear stress = force/area.
Area = length*thickness.
So, 1/4 shear stress = l*t.
Stress = 4*t*l.
Area = length*thickness.
So, 1/4 shear stress = l*t.
Stress = 4*t*l.
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