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 7 of 7.
Kuruu said:
1 decade ago
@Akshay I want to explain you about the question.
Actually @Rajesh had done exactly no need again just check once if at all u didnt get that will see.
Actually @Rajesh had done exactly no need again just check once if at all u didnt get that will see.
Akshay said:
1 decade ago
How is shear stress equal to l*t?
Rajesh said:
1 decade ago
Max.shear stress of plate = 1/4 of max crushing stress.
And plate thickness = t.
We know, shear stress = L*t and crushing stress = d*L.
Therefore,L*t = 1/4 of max. crushing stress.
or, 4*t*L = crushing stress=d*L.
or, 4t = dia of hole.
And plate thickness = t.
We know, shear stress = L*t and crushing stress = d*L.
Therefore,L*t = 1/4 of max. crushing stress.
or, 4*t*L = crushing stress=d*L.
or, 4t = dia of hole.
Jammy khan said:
1 decade ago
Shear stress = F/A.
F = 1/4, A = L*thickness.
Put the value in given equation,
Shear stress = (4/1)*(L*thickness).
F = 1/4, A = L*thickness.
Put the value in given equation,
Shear stress = (4/1)*(L*thickness).
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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